Dara O Briain: School of Hard Sums : Dave

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maurits_meijboom | 17 May 13
Really enjoy the show. But, whilst working out the 'word sum', I noticed that the solution given on the programme is in fact non-unique: the sum 8542 + 0915 = 09457 is also a solution. Just thought I'd share...
AmyB6989 | 15 May 13
I have just been watching the programme, and saw the 'word sum' and it was really funny :) I'm in S3 at high school now, but in first year, we had these sorts of challenges set at the start of the lesson, only we had 3 minutes to complete them. Nobody failed. It was just quite funny watching three adults and a group of students not being able to figure it out after 10 minutes :D
David R - UKTV | 14 May 13
Well thank you everyone for marking our homework. Heck, we're not perfect, we know it. But don't you love Dave all the more for the intriguing and alluring imperfections? Following amends have been made: Dara's breakfast brainteaser – wrong percentage (410%) has been corrected and solution explanation has been changed. But we'll kindly ask you to allow us a little artistic license for not including bacon. From Dara's puzzling quizzes… Dara's everday maths whizz quiz – answer to Question 11 was incorrect and has been amended. Dara's sporting sums quiz – answer to Question 10 was incorrect and has been amended. Dara's code cracker quiz – one of the answers to Question 14 had a typo which has been corrected. Dara's bewildering brain tease quiz – answer to Question 14 was incorrect and has been amended.
David R - UKTV | 10 May 13
We've certainly lit YOUR fuse...
kellyc7c | 10 May 13
martinS31300 | 8 May 13 If you cut the 1 MINUTE FUSE in half lit both ends of half then as soon as it had burnt -15 seconds you lit one end of the other 30 second fuse and waited till it burnt out it would be EXACTLY 45 seconds!! and you could do again with the other 1 minute fuse in this time of austerity!!! That wouldn't work as the rate the fuse burns at is not consistent. For instance one half could burn for 40 seconds and the other for 20 seconds. If it was consistent you could just cut three quarters of the fuse and light that for 45 seconds.

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David R - UKTV | 10 May 13
One of our quiz writers is indeed a muppet. He/she/it (hard to tell) was thrown out of the Jim Henson Workshop for lewd behaviour and we felt sorry for them so we took them on. However, if we find out their work has been shoddy they WILL BE FIRED. We'll investigate and let you know.
David R - UKTV | 10 May 13
PhilB32426 | 10 May 13
Enjoyed having a go at the 'bewildering brain teasers quiz'. However,whoever came up with the answers for questions 14 and 17 is a muppet.
AlisonW84378 | 9 May 13
At least a couple of the quiz answers are wrong (e.g. Average height of basket ball players and room with a rug and a 2m border all round). this is very annoying. This is very annoying!
martinS31300 | 9 May 13
what about the 5 miles fuel left in the broken down car? unless it broke down due to not enough fuel!

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nevY78246 | 9 May 13
The fuse problem in last night's episode is a nonsense. If the fuse burns at an irregular rate then it is impossible to make fuses that burn for 60 seconds as the only way to get the correct length is to burn it; but then it's gone.
Brian Barker | 9 May 13
The solution given to the Homework Puzzle (spoiler alert) in episode 2 ("Lunch Money") is only one of an infinite set of correct answers - and it is hardly ideal. Any starting sum, however large or small, is whittled away by the bullies' thefts and becomes insignificant, being replaced in significance by the results of the bullies' remorseful donations, albeit as diminished by subsequent bullies. The amount of lunch money remaining approaches £2 whatever the starting value. Since, in the long run, the bullies are acting to our advantage, we can benefit by exposing ourselves to more of them. And we can get as close as we like to £2 by choosing a sufficiently large number of bullies: the more the better! The ideal starting amount is thus zero. The first bully takes half of our nothing and donates £1. The second bully now takes half of that and donates another £1, leaving us with £1.50. After the third bully we have £1.75. Now by repeatedly diverting around the back streets and voluntarily revisiting the same bullies, we can increase the sum in our possession arbitrarily closely to £2. There are only three problems with this technique: o the discrete nature of coinage, o the possibility that the bullies may not be stupid enough to remain oblivious of the fact that they are being taken advantage of, and o the necessity (for any finite number of encounters with bullies) to persuade the school dinner lady - one of a known fearsome breed - to let us off the vestigial shortage in payment, however small.
martinS31300 | 8 May 13
If you cut the 1 MINUTE FUSE in half lit both ends of half then as soon as it had burnt -15 seconds you lit one end of the other 30 second fuse and waited till it burnt out it would be EXACTLY 45 seconds!! and you could do again with the other 1 minute fuse in this time of austerity!!!
j_connolly2003 | 6 May 13
60 seconds in an hour?????
norharwil | 4 May 13
Daras Puzzles: EVERY DAY MATHS Surely Dara got this wrong! He puts a 96 sq m carpet into a 14mx10m room and claims this gives a 2m border (all round?). The carpet size must be 60 sq m ie 10mx6m as the border is on all sdes and the official answer 96 sq m must be wrong! Do any "Aids" check the maths?

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TheFont | 2 May 13
Of course, the prisoners’ hats problem is a dumbed-down version of the five-hat problem, in which the rear prisoner would know, if he could see two blue hats, that he was wearing one of the three red hats... But he says nothing. So the other two know they are not both wearing blue hats... So if the middle prisoner could see a blue hat he would know he was wearing a red hat... But he says nothing... So the prisoner at the front knows he is wearing a red hat. This can be extended indefinitely – to n prisoners, with n red hats and n-1 blue hats. In theory, the rearmost prisoner of those who see no red hats should eventually be able to deduce that he is wearing a red hat... But, in practice, how long does the chain(gang) have to get before it becomes impossible for that prisoner to be sure he’s allowed enough time for the mental processes of those behind him, while avoiding the possibility of being beaten to it by one of the blue-hats in front of him – erroneously announcing that he’s wearing a red hat?
robinnt | 2 May 13
The program specialises in the accurate meaning of words, to the point of pedantry. So the solution to the cameras in the gallery was wrong. Marcus clearly stated that the cameras must go on a wall, and that there was a pillar in the middle of the space. Yet Dara placed his cameras on the pillar, which is not a wall. The solution was still 3, only the cameras needed to be on the walls adjacent to the positions selected on the pillar.
johnB15155 | 30 Apr 13
Yes bacon is essential therefore no one had a full English breakfast! why was bacon omitted?
TheFont | 24 Apr 13
Indeed. Seems their working is as shoddy as ever. Even had they said 310 % (which is what they meant), it hardly qualifies as an explanation. But it does turn out that your method is equivalent to subtracting 300 % from the total. Personally, I got the answer by combining ingredients: at least 45 % had both beans and eggs; at least 65 % had both sausage and toast; so at least 10 % had all four... But hang on a minute... Isn’t bacon a necessary ingredient of a full English?
RobinW58213 | 24 Apr 13
Not convinced of the answer for Brainteaser 4 as it doesn't add up to 410%; add up the percentages of those who definitely didn't have the each of the ingredients to get a total of those who possibly didn't have a full breakfast and subtract from 100 to give how many definitely did!

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gollomS10628 | 12 Nov 12
richardD - You ask either guard "which door would the other guard indicate as being the door to paradise". Whichever guard you ask, the answer will indicate the real door to hell (either the lying guard tells a lie about the truth teller or the truth teller tells the truth about the liar). Don't forget to take the other door though!
richardD47509 | 19 Sep 12
can anyone give me the answer to dara's puzzle about being in a room with 2 doors one leads to hell the other to paradise and there is a guard infront of each door. one always tells the truth and one always lies and you can only ask one question. please help guys. its doing my noodle in trying to remember the answer.ta byeeeeeeeee.
LisaJ98019 | 19 Aug 12
Well Dara, fair play to you entirely. A little more explanation regarding the problem of the drinking friends would up the correct response rate(I didn't know where he was sitting and that he wouldn't buy a friend more than one pint). How about the four card problem: each card has a letter on one side and a number on the other. If you wanted to check the rule that every D had a 3 on the other side which cards would you need to turn over if you were looking at the cards f, 3, d, 7?
GaryL56115 | 19 Aug 12
Tell Dara that the neutrinos ARE mutating :-) http://www.rdmag.com/News/2012/08/General-Science-Astronomy-Test-Measurement-New-system-could-predict-solar-flares-give-advance-warning/?et_cid=2796267&et_rid=54736868&linkid=http%3a%2f%2fwww.rdmag.com%2fNews%2f2012%2f08%2fGeneral-Science-Astronomy-Test-Measurement-New-system-could-predict-solar-flares-give-advance-warning%2f
norharwil | 25 Jul 12
Thats a fair comment on an approximation method, a progressive rather than a successive method as Newton would have imagined in the language of his day, but perhaps he did not play cards.

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gollomS10628 | 24 Jul 12
OK then. Successive approximation method (sort of). Take 3 playing cards, one ace and two jacks. Deal them randomly face down. Pick one and put it to one side. Turn theother two face up. Remove the Jack (or a Jack depending on what you have face up). count how many times the remaining card is the ace. The more times you do it the more it approximates to 2/3. Hopefully the analogy is obvious or I'd run off the page.
TheFont | 23 Jul 12
But it's not a matter of... Oh, never mind...
norharwil | 22 Jul 12
TheFont: Yes, you may be right, people are always doing things at my expense, but on the essentials of the subject problem we seem to agree that the Punter should switch, even if the incentive to do so may be a matter of opinion it is always positive.
TheFont | 22 Jul 12
My reasoning is, of course, predicated on the assumption (restated by norharwil) that the host “has to open to a toy”. But, as Monty Hall himself said (for a 1991 New York Times article on the history of the problem), “If the host is required to open a door all the time and offer you a switch, then you should take the switch. But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood.” (This caveat also applies to the doctor in the twins problem – but in that case rational assumptions must be set aside in order to come to the ‘correct’ answer!)
TheFont | 22 Jul 12
Oh, it's even simpler than that, norharwil. I suspect gollomS is having a little fun at your expense. The probability of the first door you guessed being right is one third. And that doesn't change when the second door is opened to reveal a teddy. Ergo the probability that the car is behind the third door is precisely two thirds. If you still don't get it, I'm afraid there's little more we can do about it. Perhaps you could ask a friend to act as the host, and repeat the experiment a few hundred times...

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norharwil | 21 Jul 12
If only it was that simple GollomS, there is no successive or iteration about chance but plenty of approximation! In the 3 door case I think we have to eliminate cheating by moving the car as it is a game show with a large audience and thus too risky to try. Looking at it from the HOST'S point of view, he has to open to a toy or the game would end ignomineously and he always has a choice, in one case in 3 he has 2 choices for opening the door when the Punter has a car behind his own choice of door (one case in 3 again) so the punter should switch as he cannot improve his chances any other way. He may then get a chance (APPROXIMATELY) between 50/50 and 66/33.
TerryB2876 | 19 Jul 12
gollomS10628 | 19 Jul 12
norharwill - I think you could settle this once and for all with one of your itterative/successive approximation solutions...
norharwil | 19 Jul 12
Well what do you think GollomS? If you study the explanation in which you have a 2/3 chance on switching you can see clearly that this is because the HOST has kindly switched the teddy for a car for you behind the door (3) he previously opened for you to show a teddy. This has changed the chances but is clearly cheeting!
gollomS10628 | 18 Jul 12
No arguments about teddy bears?

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norharwil | 15 Jul 12
Thanks TheFont:For some reason I did not spot your method , which is excellent as well as not too hard, as you say: but I tried a harder method, a kind of successive approximation starting at exactly 1.0 o'clock with the hour hand at exactly division 5 (minutes).The minute hand was then moved through the 5 divisions but did not co-incide with the hour hand which had moved on by 5/12 of a division. The minute hand was moved through 5/12 and hour hand through 5/144 etc etc. The infinite series geometrical progression produced was summed and, of course, produced 5/11 or 0.45454545...etc. which leads to the same answer you would get of 1 hour, 5 min and 27.272727...etc sec for the intervals between co-incidences of clock hands, an interesting excercise nevertheless.
PaulB2385 | 15 Jul 12
15-Jul Problem One: Houses and Utilities You neglected that the houses are not dimensionless points. The Electricity of house 2 can pass ***through*** the building of house 3 or house 1.
TheFont | 13 Jul 12
Well, it happens at regular intervals, 11 times every 12 hours, so I don't see how you can really make it harder than 12 divided by 11.
norharwil | 13 Jul 12
The clock problem interests me but I cannot access the Dara/Kit Yates Maths 10 Multiplechoice problems any more, this website has many problems of its own regarding organisation and accessibility. Does anyone know the "hard" way to find when the hour hand and the minute hand of a 12 hour clock co-incide?
norharwil | 11 Jul 12
Hi GollomS again et al: Your problem looks to me the same as the twins problem since one of the items is already defined (ie one ball is white, one twin was a boy) so, again, I would say the chance is 50/50 of the second ball being white (ie both white) or black. If the balls could be either black or white but this remains undefined the chance is one in three of white-white:black-black:white-black.

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gollomS10628 | 10 Jul 12
Ooops sorry all - it wasn't THAT rude, but it got covered with Xs. Just a word for a big thing that holds back water with an "n" on the end.
gollomS10628 | 10 Jul 12
xxxxxx! It's a repeat. Oh well thanks for the series Dave. Please make another one soon. I see we are arguing over the twins again... can't wait for the teddy bears! Norharwill - good point about the multiple choices. However, consider this - someone picks, at random, 2 snooker balls from a very large pile of equal numbers of black ones and white ones, and hides them. He tells you that one of them is white. What is the chance of them both being white?
norharwil | 9 Jul 12
In these 10 multiple choice questions the answers are described in commendable detail. However due to availability of possible answers there is no need to follow hard sum detail and some answers can be found almost at a glance by following the cunning of multiple choice. For example the elephant/monkey problem can be "solved" by substitution from the given answers in a few seconds instead of having to write out and solve simultaneous equations. The same applies in the clock problem. Multiple choice teaches a quick skill, a kind of "overview", but not much maths.
norharwil | 9 Jul 12
I refer to the 10 multiple choice sums set by Dara/Kit Yates Maths. Number 10 is about expected twins where one is known to be a boy. The question is: "what is the chance that both babies will be boys" As one is known to be a boy I would say there is a 50/50 chance of the second also being a boy. Not so say the problem setters. They conclude there is one chance in three; an unbelievable result! What they do is consider 3 so-called combinations: boy+girl : girl+boy : boy+boy : correctly eliminating girl+girl. I object that boy+girl and girl+boy are permutations of one combination and therefore inadmissible. Hence Dara's etc result is clearly wrong.
jefferyB70308 | 8 Jul 12
They got the number of kisses answer ignores the shape of the human in question - if you were lying down on the floor you could be nearest to more than 6 people provided we are not defiining distance from centre of mass rather than distance from the nearest point

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lesley.whitfield1 | 1 Jul 12
Dara's homework puzzle 8 is incorrectly phrased - at least on the written version - or has the wrong answer. If each team really plays twice there will be 10 matches, (2 plays each * 10 teams / 2 teams in each match). But if (as would be more usual) they place EACH (other) team twice there will be 90 matches ( 2 plays each * 10 teams * 9 other teams / 2 teams in each match). This is rather sloppy...
TheFont | 25 Jun 12
That's certainly a possibility.
norharwil | 24 Jun 12
GollomS: The Font: The origin of the swimming pool rescue problem: Is this where Marcus found the problem? Wikipaedia Snell's Law -Derivations and Formulae "In a classical analogy, the area of lower refractive index is replaced by a beach. The area of higher refractive index by the sea, and the fastest way for a rescuer on the beach to get to a drowning person in the sea is to run along a path that follows Snell's law".
norharwil | 23 Jun 12
TheFont: Yes you are right about the triangles changing shape together since they are joined at B whose position is always our result. As far as I know your general solution would be based on pure maths and calculus. My methods are based more on applied maths (applied mechanics as part of physics it used to be called) I will come back to all this of course!
TheFont | 21 Jun 12
But surely you can see that as we move point B along the pool edge seeking the optimal entry point, both triangles are changing shape?

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norharwil | 20 Jun 12
TheFont: OK! so we don't agree, then all we can do is each put forward our own point of view. The way I work out all of the problems is to find the location of point B on in-the-pool triangle ABJ. This fixes x (AB) and BD = 75-x = X. The shape of ABJ, ie angle ABJ depends on R/S. This in turn is affected by any existing upstream triangle down which the rescuer runs at an angle to the pool other than 0 deg. Such upstream triangle is not changed in shape and is unaffected by what happens downstream. There is an upstream and a downstream in this problem. In conclusion, all I can say is that these ideas work for me and the proof of the pudding surely is in the eating. When I get wrong results I will certainly change my thinking.
TheFont | 19 Jun 12
norharwil: No, it will not be agreed that the pool problem is centered on any one “main” triangle. Any general solution “centred on” the triangle with the swum hypotenuse (however we define “centred on”) will work equally well if “centred on” the triangle with the run hypotenuse. That is what I meant by mathematical symmetry.
norharwil | 19 Jun 12
GollomS: TheFont: The swimming pool rescue problem The "two triangles" again. I think it will be agreed that this problem is centered on the main triangle ABJ in the pool where J (Jeremy) is awaiting rescue. But this is a catch 22 situation. Triangle, or angle ABJ, IS the solution all along once it is recognised that sides ratio BJ/AB = run speed/swim speed ratio. This is usually recognisable only after solving the problem the "hard way" hence catch 22. Every time you change the R/S ratio the triangle and angle ABJ must change accordingly, which is the simplest indication of how to solve the problem, if only one new in the beginning. Dara did not go into the interesting aspect arising when "Alex" starts his run far from the poolside. This means he starts from C (say) and runs towards B at an angle to the poolside DBA thus forming triangle CBD as he runs down its hypotenuse CB. This has a profound effect on ABJ, the position of B and thus on the result. In the poolside direction DBA, triangle ABJ sees Alex's motion as increased giving an enhanced (pseudo) R/S. The triangle adjusts itself accordingly keeping to its sides ratio rule. The overall result is found by simply calculating the length of BD, or BC if preferred. Detailed calculations are shown in the previous post 16 June.
TheFont | 18 Jun 12
KenB – Differentiation method: We wish to minimize sqrt (x^2 + 4) + 2sqrt ((75 - x)^2 + 625). I shall use h to represent a half. We apply the chain rule to each term, differentiating each square root function with respect to its operand, then multiplying each result by the derivative of the corresponding operand. This gives us: h(x^2 + 4)^-h * 2x + 2h(6250 - 150x + x^2)^-h * (-150 + 2x). We want this to be zero, so we change the sign of one term, move it to the other side of the equation, and square both sides. We then multiply by (x^2 + 4)(6250 - 150x + x^2), expand, move terms back to the left side of the equation, divide by 3, and we’re left with a quartic equation with coefficients 1, -150, 5422, -800, 30000.
norharwil | 17 Jun 12
KenB91 Integration would not be needed, nor is differentiation if you follow my methods as in many recent posts. However if you insist, look at other previous posts or website wolframalpha will illustrate it all for you, good luck.

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KenB91869 | 17 Jun 12
Please, please can you show us how to do the differentiation or integration for the swimming pool puzzle? I'm sure I could have done it 40 years ago!
KenB91869 | 17 Jun 12
Please, please can you show us how to do the differentiation or integration for the swimming pool puzzle? I'm sure I could have done it 40 years ago!
norharwil | 16 Jun 12
TheFont: Sorry but I don't agree with you about the origin of the 1/2/sqrt3 triangle. I have quoted one source: Smartmathguy and I doubt if this had anything to do with any form of Snell's Law. Anyone who calculated the Hard Sum correctly would know of the 1/2/sqrt3 triangle. Does any of it matter. All I did was put it together and make it work. Who else has done this? My interest is in making things gel and work if possible rather than semantics. And what about symmetry? Perhaps I am ignorant and only understand simple symmetry by how it looks to the naked eye. So be it, whats in a word.
norharwil | 16 Jun 12
GollomS: TheFont:"The case of the two triangles" continued: Part 2 Impact of the second triangle ( on the first). Dara's program included a 2 mtr offset from the pool for Alex to start from. This forms such a small, narrow triangle that it has little impact. It has complicated calculus procedures but is easy to deal with by my method. To study real effects of such a triangle TheFont proposed looking at larger offsets (40 mtrs or so) and this we did. I mistakenly assumed there would be little impact on the first triangle but this proved erroneous. After correcting this I posted a note that my rule applied to both triangles and was able to prove this. The effect can be explained by assuming (eg) the second triangle CDB, rt angled at D, has a 60 deg angle CBD with the poolside DB extending to A where DA is 75 mtrs. As Alex runs down BC his speed in the direction DBA must be doubled as "equivalent speed" to find its true impact on triangle ABJ. Thus the ratio BJ/AB now is = 2*2 = 4 and ABJ is changed in shape. It may now be evaluatd by simple calculation without any iterations to give AB = x = 6.455, DB = AD-x = 68.545mtrs BJ = y = 25.820mtrs. Triangle BDC remains unchanged in shape so as DB is known the offset DC and running side CB are easily found. The true R/S ratio is still 2/1 for running down CB. When the offset length is given and angle CBD is unknown, recalcs ie iterations are needed. Any other details must await another post, I fear.
TheFont | 16 Jun 12
I didn't get to what we’re calling Snell’s law by analogy with optics: I got to it by thinking about the problem at the level of infinitesimal changes. That's where the 1:2:sqrt(3) triangle comes from: so-called Snell’s law in the special case where the angle of arrival is zero. We already knew that in this case the cosine of the angle of pool entry equals the swim/run speed ratio. (And I would have to take issue with you on my mathematical use of the word symmetry: I feel it was the mot juste.)

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norharwil | 15 Jun 12
Hi GollomS: TheFont: "The strange case of the two triangles" this could be a novel about the swimming pool rescue. Part (1) The big triangle starting "The Penny has dropped" posted 25 May. After a complex calculation involving calculus Smartmathguy 23 May revealed a simple exact result of X = 75 - 25/sqrt3. TheFont 24 May stated maximum required swim distance was 50/sqrt3 = 28.868 To me this was a revelation and I put the two equations together to define the big triangle as a 2/1/sqrt3 triangle. As run/swim speed was also 2/1 the connection was obvious, I didn't need the analogous Snell's optical law to tell me the same thing. Therefore on 26 May I proposed my General law on this. General means one could change the R/S ratio to any reasonable value and apply it to the triangle eg 3/1 would mean the triangle would be a 3/1/sqrt8 triangle. Thus the triangle would immediately give the result for a new pool rescue problem with R/S = 3. This is strange behaviour, but these are strange triangles.
norharwil | 15 Jun 12
OK TheFont: You asked me to look into your "speedo" problem and I have spelt out the answer, now you chide me that it was not necessary because of your symmetry proposal which I should have swallowed but "freely admit.....fail to see. Yes I think "symmetry" is the wrong word and is misleading.
TheFont | 15 Jun 12
The point is: we don't have to go through the calculations again (apart from subtracting from 75 m) because we've already done them... thanks to the mathematical symmetry you freely admit you fail to see.
norharwil | 15 Jun 12
TheFont: gollomS HI guys: The "speedo" problem we can wrap up as the counterpart to the original pool problem with 2 mtrs offset. J becomes C, A bec D, D to A C to J only B the same and AD stays at 75 mtrs AJ is 2 mtrs CD is 25 mtrs. Triangle BCD remains on land and ABJ in the water. Set BC/BD at 2/1 CBD initially then 60 deg. Alex hobbles from C to B at half his swim speed, then swims from B to rescue Jeremy at J. x = 14.423 AB = 75 - x = 60.5767 BJ = 60.610 BC = y = 28.86227 S/R stays at 2/1 but BC/BD increases to 2.001 and 60 deg becomes 60.018 while angle ABJ stays at 1.891 deg.
norharwil | 14 Jun 12
gollomS Thanks very much for your note of appreciation, I appreciate it in turn! I agree with your sentiments about the next show, having little else to go on we have rather done this pool problem to death, but it has so many fascinating aspects - even still.

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gollomS10628 | 14 Jun 12
norharwill - Yes I now understand your easycalc. Thanks for your considerable efforts to explain. We've got it working on a very thin, very tall and medium shaped triangle so I think it's pretty much a wrap. Very well done indeed. Let's hope there is something meaty in the next show (if it's a new one) and not just pairs of socks!
norharwil | 14 Jun 12
Can anyone find any use for Snell's Law in solving the pool rescue problem? If so can we hear of it please? If not can we dicard it as a "red herring"
norharwil | 14 Jun 12
There is loads of stuff about Snell's Law, I heard it was known around 980 AD as TheFont says. Descartes is said to have first published it.
| 14 Jun 12
Got the first one...
| 14 Jun 12
Does anyone know how to do either of the following: Switch off the email notifications about postinga? Delete one's profile?

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TheFont | 14 Jun 12
Although named after Dutch astronomer Willebrord Snellius (1580–1626), Snell’s law was first accurately described by the Arab scientist Ibn Sahl in 984, when he used the law to derive lens shapes that focus light without geometric aberration.
norharwil | 14 Jun 12
TheFont: re Fermat "still centuries away" I don't think so. It was about 1620 and Snell and Fermat were contemporaries and both in France. By the way you should read "Fermat's Last Theorum" by Simon Singh. Great stuff- there was a Horizon program on this about 10 years ago.
norharwil | 14 Jun 12
GollomS the hobbling question: If we are in any way following the original program script Alex has to run/hobble to the point where he drops into the water. It is difficult to do it the other way round, I think?
norharwil | 14 Jun 12
Hi Further to these results (Previous post) we need to fill in the swimming distance y = 8.262*2*1.5157 = 25.0454 mtrs and angles ABJ = 86.531 deg and CBD = 83.048 deg. I made some silly errors early on eg AB1 should be 1.252 not 2.5126 BD1 then is 73.748 also AB2 should be 1.521 not 1.252 BD2 then is 73.479 Since successive convergence is so rapid these errors hardly affect the result. Need I say more?
norharwil | 13 Jun 12
Hi GollomS, TheFont: Poolrescue, easy calc example: Symbols used are as in recent posts. AJ = 25 mtrs: AD = 75 mtrs CD = 602.6231 R/S = 2/1. The preliminary calc is done on triangle ABJ with angle ABJ = 60 deg equiv to R/S = BJ/AB =2/1. Given AJ is 25, AB = x = 14.434 (by Pythagoras) so BD = 75-x =60.566 Then in triangle BDC DC = 602.6261 BC = 605.662(Pythagoras). Correction factor F1 = CB/DB = 10.0 Now we start the first iteration, or corrective recalc, AB1 = x1 =25/sqrt((2*10)^2-1) =2.5126 BD1 =75-2.5126 =72.4874 then from BD1 and CD get CB1 = 607.12(Pythag) then CB1/DB1 = F2 =8.232 2nd recalc: AB2 = x2 =25/sqrt((2*8.232)^2-1) =1.252 BD2 = 75-1.252 = 73.748 and then to CB2 = 607.12 and CB2/DB2 = F3 =8.262 Then 3rd: AB3 = x3 = 25/sqrt((2*8.262)^2-1) = 1.5157 and BD3 = 75-1.5157 = 73.484 then to CB3 = 607.1 and CB3/BD3 = F4 = 8.261 Then 4th: AB4 = x4 = 25/sqrt((2*8.261)^2-1) = 1.5158 OK this is enough, just tidy up what we've got BD4 = 75-1.5158 = 73.484 mtrs and BC4 = 607.09 F5 = BC4/BD4 = 8.262 if you want it. In the next post I will review these results.

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norharwil | 13 Jun 12
TheFont GollomS: Your first post 12 June: you query if easy calc is only for very thin triangles. This is a fair question so I feel I should offer some clarification: My method IS an iterative (ie repeating) form of calculation. It can be programmed but does not need to be, there is no requirement. This is due to the extremely rapid convergence of the "iterations", usually only three to five are needed. It is feasible to use an ordinary non-programmable calculator and to do these iterations manually as simple repeated calculations to achieve an accurate converged result. If I may GollomS, I will go through your latest example calculation ( 600+ mtrs offset) to illustrate the details of my method, in another post.
gollomS10628 | 13 Jun 12
Yes all thinking along the same lines. I was going to say just get Jeremy to rescue Alex and it is the original problem. Since time will be the same in either direction that's it done. However, norharwil, I am intigued by your suggestion that he can't swim before he hobbles - why not?
norharwil | 13 Jun 12
Hello TheFont: This "speedos" problem is a bit of a malteser. I am still pondering it. A solution might be by reversing the original problem so that Alex starts where Jeremy was (J) and progresses to point (C) where Alex previously started but Jeremy now would be if only there was some water. As this means swimming before he hobbles it cannot be allowed. Therefor the pool must be placed where the land was and vice-versa. Swimming will now become running and vice-versa. The triangles won't notice this because they can't see very well.
TheFont | 13 Jun 12
The way to get from the solution of the original problem to the solution of the Speedos variant is to subtract it from 75 m. The Speedos problem is the original problem... just rotated through 180 deg. That's because the two triangles do possess mathematical and algebraic (if not geometric) symmetry within the parameters of the problem.
gollomS10628 | 12 Jun 12
norharwil - ok thanks for tweaking my misunderstanding. I'll have another go when I stop giggling at the image created by TheFont's "speedos" post. Show next on 8th July. Anyone know if this is a repeat or does the show continue? re your last post I use the Windows 7 calculator in scientific view and an old fashioned pen and paper for extra memories!

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norharwil | 12 Jun 12
GollomS, Incidentally how do you do your calcs? Do you have reasonable functions on your calculator besides sqrt,squaring (or x+), memory MR MC M+ preferably 2 memories: inverse(reciprocal) eg 1/entry? Do you use a spreadsheet preferably Excel an old version is good enough even Lotus 123 1984
norharwil | 12 Jun 12
GollomS: TheFont: My error calcs are simple. First you calc the correct final result then you go back to each prelim result and divide by the correct result then make as %.
norharwil | 12 Jun 12
Hello GollomS, The Font: Now have a load of posts in to answer: Your very large CD =602.6261 no you have not made a mistake, only an omission, but you are expecting a miracle-remember the prototype calc you are working on is basd on CD = 2 mtrs! (5June) still even this we can break down. Your omission was in not multiplying R/S by the updated factor BC/BD = 8.232 in fact you did not even include the vital R/S =2. Instead of 10 put in 2*8.232 and get x = 1.252. Your Snells Law application then gives 2.429 not right but keep on calculating. These are my methods NOT Snells Law which, unfortunately, The Font mistakenly imagines is my method. Finally you will get (without programming) 86.531 degrees and 83.048 deg x=1.51582 BC = 607.089 the rest you can easily deduce, good luck!
TheFont | 12 Jun 12
As for the rope and trolley, norharwil... it's not mine: it's PaulH’s. He put it forward as a practical (i.e. ‘brawn’) alternative to repeated runnings and swimmings. I gave it another airing partly because I felt it hadn't received the credit it was due at the time, and partly because it's a convenient framework to illustrate the method I’d used when I originally derived the swimming pool analogue of Snell’s law. (And don't forget: when the original, light version of the law was first expounded they didn't really know about the speed of light – refractive index was just some constant, and Fermat’s principle of least time was still centuries away.) The physical analogy might fall apart (literally) for run to swim ratios requiring more complicated pulley arrangements, but the theory works for all ratios and all combinations of distances. Without drawing you a diagram (literally and metaphorically) it’s tricky to explain the analysis involved, but try it yourself and hopefully you'll get the principle.
gollomS10628 | 12 Jun 12
I know how I'd do it, but it won't help us to discover anything new so I'll keep my nose out for now.

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TheFont | 12 Jun 12
Before I address the matter of symmetry, norharwil, perhaps you could let me know how, having solved the original problem, you might approach a new problem, where Jeremy is now only 2 m from the edge, but Alex starts 25 m from the pool and, crucially, has his Speedos tangled round his ankles – so he can only shuffle along on land at half his swim speed. (The distance between the perpendiculars is still 75 m.)
gollomS10628 | 12 Jun 12
norharwill - I think we were both posting at the same time! Yes you seem to have an iterrative method that works. Can you enlighten me as to how you are working out the errors at each stage?
gollomS10628 | 12 Jun 12
norharwill - Sorry but I think that your easy calc method only works for very thin triangles (if I have understood it correctly). I tried working it through with a large distance CD, namely 602.6261 metres (you'll see why in a bit). Ok so the first part, running along the edge is fine and gives the same results as before. Now we work out the hypotenuse CB and it happens to be 605.662 giving a ratio of 10 for CB to DB. Now if I understand it correctly you now make the new AB = 25/(10^2-1)^0.5. This gives a new AB of 2.5126 and hence a new DB of 72.4874. Now if we check this out using Snells law, cos(CBD) should be 2*cos(ABJ). So using Atan to get the angles I get Angle CBD = 83.1411 deg and Angle ABJ =84.2608 deg. Then cos(CBD)= 0.1194 and cos (ABJ) = 0.1 so the ratio of 2:1 does not hold. Assuming I haven't made a mistake with the number crunching! Interested to see your comments (and TheFont's.)
norharwil | 12 Jun 12
Hi GollonS: What you can expect if you start Alex off 40 mtrs from the edge, using an unprogrammable calculator for hand calcs: (A) set up triangle ABJ at 60 deg for R/S = 2/1 giving DB =75-x = 60.5662 mtrs: calc(1) gives BC(hypotenuse)= 72.58 mtrs: error from true = -3.11%: calc(2) 75.07 error +0.21%: (3) 74.90 error 0.013% (4) 74.912 error 0.001% (5) 74.9111 error 0.00005% BD = 75-x now =63.3378 x = 11.6622 y = 27.5864 mtrs swim distance angle ABJ =65.0 deg CBD = 32.27 deg_
norharwil | 11 Jun 12
TheFont Your rope and trolley example is all very well as an analogy but hardly the real maths thing. Does it all start 2 mtrs from the edge as Marcus seems to require? I have already dealt with these proofs in the real cases in posts: 1st post 26 May, 28 May, and 2nd post 31 May. You start at 60 degree because this is the given case 2/1 run/swim speed, not "could be anywhere" Yes I "consistently failed to see the symmetry between the triangles" but this is because they are extremely different shapes ie far from symmetrical.

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norharwil | 11 Jun 12
I should comment in the matter of "originality". I have worked on this "Hard Sum" on the pool rescue problem in good faith without being aware that I may have unwittingly re-discovered Snell's Law, like so many others. If so I claim priority, in its application to this problem, in so far as it co-incides with my discoveries. I claim that my discoveries are original unless and until it is proved that they are not due to prior anticipation.
norharwil | 11 Jun 12
Hi TheFont: GollomS You'r getting the right idea! Previous posts show how TheFont and I both calculated with Alex starting a good distance away from the poolside. I should say that on 30 May I posted that my general rule applies to BOTH triangles, until then I thought it only applied to that on the poolside with runs in line with the poolside. I claim that my easycalc methods solve both versions of the problem (ie with & without the 2 mtr offset) without calculus,spreadsheet or programming, I do not accept that any one repeat of a calculation is programming. All that is needed is a pocket calculator with square root facility. Regarding Snells Law, What use is this when it has only produced quartic equations. Please show me posts where procedures and results match mine and my methods! Incidentally I only claim my "discoveries" in relation to this problem. They are not likely to be original any more than Snell's were, many of his contemporaries (around 1620) discovered his "Law" and according to Google it was known in India about 981 AD
TrevorB67314 | 10 Jun 12
Regarding the 3 bottles problem (8 litre, 5 litre and 3 litre), I found a nice visual method to illustrate how to reach the answer:- If (x, y, z) represents the amount of liquid in the 3 bottles, largest to smallest, then we begin with the value (8, 0, 0) and want to reach the value (4, 4, 0). There are actually only 16 possible values for (x, y, z) and the values and possible moves between them can be best illustrated thus:- In the middle of a piece of paper write (8, 0, 0), (5, 0,3), (3, 5, 0) and (0, 5,3) as the four corners of a square, then around the outside of the paper, going clockwise write (5, 3, 0), (2, 3, 3), (2, 5, 1), (7, 0, 1), (7, 1, 0), (4, 1, 3), (4, 4, 0), (1, 4, 3), (1, 5, 2), (6, 0, 2) (6, 2, 0), (3, 2, 3). From each of the values on the outer ring, the only possible next moves are to go one step clockwise, one step anti-clockwise, or into one or more of the four central values. For each of the four central values there are multiple moves between each other, but the only move out of these four is either from (5, 0, 3) to (5, 3, 0) or from (3, 5, 0) to (3, 2, 3). Hence to get from (8, 0, 0) to (4, 4, 0) you either go:- (8, 0, 0) -> (5, 0, 3) -> (5, 3, 0) -> clockwise -> (4, 4, 0) = 8 moves or (8, 0, 0) -> (3, 5, 0) -> (3, 2, 3) -> anti'wise -> (4, 4, 0) = 7 moves
gollomS10628 | 10 Jun 12
Ok guys I think I am up to speed on the swimming rescue now. The Font - I like your proof of the Cosine ratio since it essentially proves Snells law without formal calculus so we are not just repeating the original method. norharwil - Hi yes I see your idea now. The fact that the ratio of run speeds equal the ratio of the cosines follows from Snell as well. (Sin of the angle to the normal is same as cos of the angle between the running line and the pool and = 1 when you run parallel to it). I want to investigate what happens if Alex starts much further from the pool so the first estimate (running parallel to the pool) is not so close to the answer. So when I've got time I'll have a go at starting Alex 40m from the edge. Hope the swimmer doesn't drown waiting! PS sorry if that's repeating anything but I haven't read back through all the posts on this.
norharwil | 10 Jun 12
gollomS Hi ref BTW please see my method involving two very simple calcs, my post 5 June. I will answer any queries.

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TheFont | 10 Jun 12
Of course we've applied Snell’s law, norharwil. I stated on 24 May that it gives you the same quartic equation as differentiation (although I wasn't calling it Snell’s law at the time), and gollomS confirmed my finding on 26 May. And it's also the basis of your own iterative method!
TheFont | 10 Jun 12
They were on the right track... but they were getting lapped! Anyway... Quick recap of the pool problem... Well, you did ask! norharwil seems to have stumbled on an iterative approach, but it fails to live up to what was promised – on two counts. First, being iterative, it does require a form of programming. Second, it depends on what we shall, for the sake of convenience, refer to as Snell’s law. Since it would be begging the question somewhat to derive this ‘law’ by reverse-engineering empirical results from the field of optics, we resort to analysis. Note that formal differentiation is not required. It is easier to consider the position of the entry point using first principles: consider PaulH's trolley, constrained to move on a frictionless rail, with a rope fixed to its centre, looped round Jeremy (resisting all temptation to get dragged to safety), passed back through an aperture on the trolley, and so to Alex, who pulls the whole shooting match taut. Since the paid-out rope is at a minimum, we can in principle move the trolley an infinitesimal distance delta x to the right (or left) such that viewed from the trolley the ropes remain parallel to their original positions – without Alex having to let out any more rope. Consideration of the (also infinitesimal) amount of rope passed through the trolley aperture reveals that the cosines of the angles of arrival and pool entry (with respect to the pool edge) must be in the same ratio as the speeds in the respective [for want of a better word] media (in this case 2:1). if you don't like this approach, tough: the rules of differentiation rely on precisely this kind of consideration. Any road up, by consistently failing to see the symmetry between the two right-angled triangles, and starting at the limiting case of 60 deg for the watery one (though, in all honesty, you could probably start anywhere), norharwil has seemingly hit upon an approach whereby (if I have it aright) you repeatedly scale your estimate of x and feed it back into the equation. It seemed to make sense when I was sober, so I might try verifying it and scoping it in the morning...
norharwil | 9 Jun 12
Hello again TheFont I note that Snells Law has been mensioned to solve the pool rescue problem but no-one has applied it. I find it fits in well with my "discoveries" when 2 triangles are involved but with the single triangle ABJ total internal refraction must be invoked. It does not help except as a scientific curiosity. I have set up the triangles ABJ and BCD as mirror images of each other so CD = AJ = 25 mtrs AB = BD = 37.5 mtrs: straight line CBJ = 90 mtrs. Solving this problem for R/S = 3/1 needs only 2 calculations/iterations and gives angles ABJ & CBD of 71.81 & 20.52 By Snells law the ratio of cosines of these angles is 3/1 the R/S ratio used.
gollomS10628 | 9 Jun 12
Well, put like that it is a generalised solution. Interesting that the students on the show took the LCM of the frequencies, so they were on the right track but mixed up the two methods.
TheFont | 9 Jun 12
Well, I took the lowest common multiple of the intervals between lappings to get the interval between simultaneous lappings; he took the highest common factor of the frequencies of lappings, and then flipped that. Not much difference... but I would contend that mine is the more intuitive approach.

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gollomS10628 | 9 Jun 12
TheFont - well sort of, but there is no mention of the general solution of finding common factors of the differences, which DavidW was referring to. BTW have you rescued that swimmer yet? Can't find a method other than differentiate and get a quartic, even by iterrating. I think we may be just walking around Konigsberg.
norharwil | 9 Jun 12
Hi TheFont My post 30 May: I said I had the whole solution with BOTH triangles. Reverting to my 5 June calculation under review: I surmised that the thin triangle (base = 2 mtrs) behaved in the same way as the main one ABJ in that the speed along the hypotenuse BC would be less than along BD as is swim speed along BJ less than run speed along the poolside. Thus the run speed along CB should be increased by factor CB/DB to give an equivalent run speed along poolside DB as only this direction can be considered in respect of triangle ABJ. ABJ was then adjusted (ie recalculated) accordingly. The results are all in the subject calculation. The result is accurate and not requiring iteration only because angle CBD is tiny (1.9 degrees or so). I have further interesting comments to make.
TheFont | 9 Jun 12
gollomS ... and as I said two days before that!
norharwil | 8 Jun 12
I note, TheFont, what you say and you seem to agree with me without actually saying so. As far as iteration, or simple repeated calculation, goes for the more complex examples convergence is extremely fast. On a spreadsheet it is a fraction of a second for 5 decimal places compared with 18 minutes or so when I used the spreadsheet manually/interactively on 21 May.
gollomS10628 | 8 Jun 12
DavidW ....as I said two posts below you.

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TheFont | 8 Jun 12
No, seriously, norharwil, when considering the question as set, we really do have to ignore results from those who misheard Marcus and are consequently answering a different question. That said, the magic rule you discovered is not without merit, in that it is the first step of an iterative approach. My numerical analysis is a little rusty, so I can't say offhand under what conditions it is convergent. I shall look into it...
norharwil | 8 Jun 12
OK TheFont I accept that, from evidence gathered on re-seeing the program etc the 2 mtrs offset was part of the problem. But half of the comments are from people who obtained valid solutions of 60.5662 mtrs by assuming running was along the poolside only. Surely these results should not be ignored. My posts 26 to 28 May describe my general rule applied when the run is only along the poolside. TheFont 26&28 May appears to agree with me. Can this be confirmed? If so we can move on to considering the more controversial aspects.
DavidW18092 | 8 Jun 12
It's dead simple! All three can complete a whole number of laps in an hour, so they will all meet after one hour, but can they meet sooner? They can if we can divide the hour by a number that leaves the DIFFERENCES between their distances as whole numbers of laps! To do this, note the differences (6-2=4 and 14-6=8), find the highest common factor (4), and divide the hour by this number. Therefore, after a quarter of an hour, they will have travelled 2/4=½, 6/4=1½ and 14/4=3½ laps. Q.E.D.
gollomS10628 | 8 Jun 12
DermotW - Arghh typos! Sorry. For l2 read L2 and for 2(16-7) read 2/(16-7).
gollomS10628 | 8 Jun 12
DermotW - it's easier than that. If the Number of laps per unit time are L1 and l2, and the time unit is T (eg T= 1hr if L1 and L2 are in laps /Hr), then the time to line up is T/(L1-L2) in the same units as T. eg say 9 and 17 laps per hour then 1/(17-9)=1/8 Hrs. Check back - they will have done 1+1/8 and 2+1/8 laps. Three runners is a bit more difficult, you need to take L1-L2 and L2-L3 then find the highest common factor of the two differences and divide it into T. If you have fractional laps per hour you just multiply up eg say 8 and 3.5 laps per hour just make it 16 and 7 laps per 2 Hrs so T now equals 2, so time to line up is 2(16-7) Hrs.

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norharwil | 7 Jun 12
Hi TheFont Thanks for your offer, which I accept,but it needs a bit of time. The answers to your queries and much more are contained in previous posts from 21 May onwards. I will repeat some to give you an immediate reply. Of course I don't claim the R/S ratio but I claim the simplified calculation method which results when the R/S=2/1 is applied directly to the sides of the triangle. ie 2/1 on the triangle becomes 60 degrees and defines the optimum run length = 75-25/sqrt(2^2-1). We must deal separately with the way to handle the 2 mtr side triangle to prevent confusion. I will come back to this and Snells rule etc
AndrewM12421 | 7 Jun 12
Homework Puzzles 6 and 8 are rather ambiguous. 6 states that there are 19 friends and yourself, making 20 people, yet the image clearly shows 20 seats and Dara, 21. Assuming Dara is representing the generous friend, which stool is he meant to be sitting on? 8 states that each "each team plays twice", when the solution uses the fact that each team plays each other team twice. They are completely different things, and the second should not be assumed from stating the first. If someone had a tray of different biscuits and said you could have two, they'd probably be a bit annoyed if you took two of each kind
| 7 Jun 12
I know this will seem long-winded but I wanted to figure out the general solution to the lap problem so here's my tuppence hapworth: If two runners are running around a track at different speeds we can calculate when they lap as follows. Let the first runner be R1 and the second R2, let the length of the track be L, and the time each takes to run one lap be T1 and T2 respectively. Then we have the speeds of each runner as follows: S1 = L/T1 S2 = L/T2 Now, if one runner is faster than the other, then at some point the faster runner will overtake the slower. This may occur after the faster runner has completed a lap of the track. When this occurs, both runners will have been running for the same time (t) but have covered different distances (D1 and D2). Since distance = speed X time we have for each runner: D1 = S1t (1) D2 = S2t (2) Now, assume R2 is the faster runner. This means that his distance may include one or more additional lengths of the track so: D2 = D1 + nL, where n is an integer. (3) Imagine the second runner runs 600 metres in the time it takes the first runner to run 200 metres. On a 400 metre track, 600m = 400m + 200m so we can see that both runners will be at the same place on the tack after this time. Combining equations (1), (2) and (3) we get: S2t = S1t + nL (4) Rearranging, we get: nL = t(S2 – S1) (5) Now, to find the time (t) at which the lapping takes place, we solve using integer values of n, i.e. n = 1, 2, 3 etc. If R1 runs 2 laps in the time it takes R2 to run 6 laps then we have: S1 = 800 and S2 = 2400 Substituting in (5), with L = 400 we find: 400n = t(2400 – 800) = 1600t (6) n = 1600t/400 = 4t (7) For n = 1, t = ¼ = 15 mins (8) So the faster runner laps the slower runner every 15 minutes. If there is a third runner, R3, with speed S3 = 5600 then: 400n = t(5600 – 800) = 4800t (9) n = 4800t/400 = 12t (10) For n = 1, t = 1/12 = 5 mins (11) For n = 2, t = 1/6 = 12 mins (12) For n = 3, t = ¼ = 15 mins (13) Since (8) and (13) both have solutions where t = 15 minutes this means that all three runners overtake at 15 minute intervals.
TheFont | 7 Jun 12
Well, norharwil, I can critique your analysis if you really want me to. But before I do so could you clarify a couple of points in your post of 5 June that don't appear to make sense? First, what exactly is it that you are claiming as your “discovery”? (It seems to be the fact that Alex can run twice as fast as he can swim – but surely that is in the question.) Second, why is Alex’s 2 m distance from the pool merely a suggestion? (That too is part of the original question – even if the graphics department conveniently ignored it.)
norharwil | 7 Jun 12
Hi TheFont Hello Again: You will have seen my effort of 2 June. Do you agree that this works now you have seen it (remember "I'll believe it when I see it")?

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gollomS10628 | 6 Jun 12
TheFont Good point well made. So much so that the probability of me trying that one is a (very) round number.
TheFont | 6 Jun 12
Nice try, gollomS, but I'm way ahead of you. I was careful to say “probably”. You've assumed that the “tiny amount” is a rational fraction. But (unless you know something about the probability density function that I don't) that (while not impossible) would appear to be infinitely improbable! [For a hard sum, assume that the pensioner manages precisely (e – 0.7183) laps per hour.]
gollomS10628 | 6 Jun 12
TheFont - No they will line up eventually. For instance say the pensioner does 1.9998 laps/hr. After 15 min A and B line up and the pensioner is .0002 laps behind. After 5000 15 minute intervals the pensioner will be 1 lap behind and they will line up. So that's 1250 Hrs or 52 days and 2 hrs. Now we can wait with anticipation for all the posts about pee breaks, food breaks etc! Thanks for the only hard (ish) sum this week!
TheFont | 6 Jun 12
Well, Hell-O, HelenO. Just got in from failing to see the sun, let alone the transit of Venus (or Lucifer, as it used to be known). I fear you have misheard the question, Helen. The jogger only manages 14 laps in an hour, so he laps the pensioner 12 times per hour (i.e. every 5 minutes), and he laps the walker 8 times per hour (i.e. every 7.5 minutes). Clearly, then, he passes both simultaneously every 15 minutes. A lot simpler than Dara made it look! And indeed, we can see that the walker laps the pensioner 4 times per hour – which is consistent with our result. Of course, this all depends on the quoted number of laps being accurate to an arbitrary number of decimal places. Were the pensioner, say, to slow by a tiny amount from 2.0000 laps per hour, the 3 athletes would probably never meet exactly.
HelenO69568 | 5 Jun 12
I Need some Help/ Clarification. The question was When do all 3 runners meet if Runner A runs 16 laps an hour, B 6 laps an hour and c 2 laps an hour. SO; A: 1Lap = 60/16 = 3.75 mins/ Lap B: 1Lap = 60/6 = 10 mins/ Lap C: 1Lap = 60/2 = 15 mins/ Lap Therefore, at 15 Mins; A = At the start Line having ran 4 laps (15/3.75 = 4) B = Halfway around on Lap 2 (15/ 10 = 1.5) C = Halfway around on Lap 1 (15/30 = 0.5) So whilst B & C meet A doesn't? How is this incorrect?!?

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GuyK20625 | 5 Jun 12
90 matches...
norharwil | 5 Jun 12
Pretty slim pickings now, so I will revert to the pool rescue problem which most people, including Marcus, believe is a "Hard" sum. Here is my easycalc solution: Jeremy is drowning at point J, a perpendicular distance of 25 mtrs from point A at the poolside. Lifesaver Alex starts at point D on the poolside 75 mtrs from A and runs along the poolside to B where he decides he can save most time by swimming from B to J although his swim to run speed is only 1/2. The solution is to set triangle ABJ so that angle ABJ has cosine 1/2, following my discovery that this is also the swim to run speed. Then poolside length AB is half of swim length BJ and is 25/(2^2-1)^0.5 = 14.4338 Subtracting this from 75 gives running distance DB of 60.5662 mtrs. It is suggested, to make this a "Hard" sum, that Alex really starts running from a point C 2 mtrs from D on the poolside where CD is at right angles to the poolside. So Alex runs along the hypotenuse CB of triangle CDB to B. The longer distance CB is sqrt(DB^2+2^2) = 60.5993 The ratio CB/DB is calculated to be : 1.000545 Triangle ABJ is then altered to take account of this (my rule) so the new ratio becomes 2*1.000545 = 2.001090 Then the new AB is 25/sqrt(2.00109^2-1) ie 14.4233 mtrs. Thus DB is now increased to 75-14.4233 = 60.5767 This the number found by a difficult application of the calculus but no calculus is required.
TheFont | 4 Jun 12
You're right, lordforbes. Nearly all their questions could do with being specified more clearly.
norharwil | 4 Jun 12
Congrats The Font,you did it then! The most interesting bit is when 4 vs 4 don't balance and you have only two weighings left. You say you have some choice here. Yes you don't need to use a good'un. I remember now how I did it many years ago and it wan't entirely my own work either. Essentially you weigh 3 vs 3 and then 1 vs 1 as you did.I named left hand side (LHS) as potential heavies ABCD and RHS as potential lights EFGH. Then put, say, ABE vs CDF. if LHS down then either AB heavy or F is light one. If LHS up then either CD heavy or E light one. Finally put A vs B or C vs D (if this applies). All is then known. Interesting logic.
lordforbes | 4 Jun 12
Thanks The Font. I see what you mean now, although, in fairness that isn't strictly what was asked on the show. It's all in how the question is asked isn't it? And it's a big weakness of a show which aims to demonstrate how maths helps in real life situations that the answers to many of its problems differ from the theoretical ones. Similarly the three doors problem is sloppily constructed in that it neglects to mention (until the solution) that the game show host is committed IN ADVANCE to offering the opportunity to change doors. Without this piece of knowledge the answer is most definitely STICK!

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TheFont | 4 Jun 12
lordforbes: I suspect you'll need a 7th move to combine the 1 litre with the 3 in the same jar – which I think is the consensus view of what is required. But if you really can get 2 x 4 litres in 6 moves, please enlighten us.
lordforbes | 4 Jun 12
You can do the truffle oil problem in 6.
TheFont | 4 Jun 12
Note that had there been 13 coins instead of 12 there would still be fewer than 27 possible cases to test for. And indeed we would still be able to find the wrong’un – and determine its modality of wrongness – in 3 weighings... provided we have a 14th coin, which we know to be good.
TheFont | 4 Jun 12
Hallo again, norharwil. Not entirely my own work, but we've got 12 coins, any one of which is a potential light or a potential heavy. That gives us 24 possible cases to test for. Each weighing has 3 possible outcomes, so 3 weighings gives us 3 trits of information – meaning we can plausibly distinguish up to 27 possible cases. Game on! The trick is going to be ensuring that each weighing eliminates roughly two thirds of the (remaining) possible coin/(light/heavy) combinations. Our first move is thus constrained to be weighing 4 against 4. This guarantees an initial reduction from 24 to 8 possibilities: either 4 coins, but we don't know whether the wrong’un is light or heavy; or 8 coins comprising 4 potential lights and 4 potential heavies. If the former, we simply weigh 3 of the 4 unknowns against 3 of the 8 known good’uns. If they balance, we've identified the wrong’un, and we've 1 weighing left to establish whether it's light or heavy; otherwise we've narrowed it down to 3, but we do know whether the wrong’un is light or heavy, so it can be identified by weighing 1 of the 3 against any other 1 of the 3. Now for the ‘latter’ case: 4 potential lights, 4 potential heavies. We have a certain amount of choice here. I decided to take 3 potential heavies out of the equation (so if our second weighing balances then we can use the third weighing, as above, to tell which of them really is the heavy). Then I split the 4 potential lights, putting 2 of them with the 1 remaining potential heavy, and I put 1 of the 4 known good’uns in the other pan so we have 3 coins in each pan. We've already covered the case where these balance. The other possible outcomes leave us with either: just 2 potential lights (which can then be weighed against each other – or 1 against a good’un); or: 2 potential lights plus (in the other pan) 1 potential heavy. This is the final case to consider: we weigh the potential lights against each other. If they don't balance, we know which is light; if they do, the 1 potential heavy gets promoted to actual heavy. QED.
steveM92309 | 3 Jun 12
I was watching the episode where the prof, sorry I can't remember his name, mentioned a way of multiplying on the fingers. This method also work in bases other than 10. For example base 8 where you would have only 4 finger on each hand. If you were calculation in base 8 the answers would be in base 8 too. Steve

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norharwil | 2 Jun 12
Sorry: on the TB/door problem,if one of the 2 doors left is opened I omitted to say that, equally, a TB may be won so the car is then lost to the person betting on that door.
norharwil | 1 Jun 12
Can anyone remember the solution to the problem of weighing 12 balls against each other to find the odd one, which may be heavier or lighter than the other 11. Which, heavier or lighter, has also to be fixed: the whole in no more than 3 weighings.
norharwil | 1 Jun 12
Hi guys: On this TB/car puzzle it all depends how the thing is operated. If the bet is placed when there are only 3 doors hiding locked compartments and one door is opened to reveal a TB the bet stays valid on the chosen door, now one of two, whatever happens. So the chances remain updated to one in two, unless one of the two doors is opened when, either the car is won, or the chance becomes updated to 1 in 1 ie 100%. Adding 98 more doors complicates procedures which must then be redefined, meanwhile the chances remain as before. The punter must be very patient while these 98 teddies are revealed before his bet can be properly excercised (excorcised? What if the car is moved around by the Host? All bets are off then I think.
norharwil | 1 Jun 12
Hi The Font What doesn,t is it the caculator? I don't see this, it is perfectly possible to use a calculator just 4 times on this problem and get an accurate result even when the odds seem stacked against you when Alex starts 55.6012 mtrs from the pool. I will agree that the data from a previous calculation must be fed to the calculator in a certain (not random) way. Would you call this programming?
rdmontgomerie | 31 May 12
Ah - got it - if you assume you cannot change then - the chances are that you are wrong !!!!!!!!!!!!!!!!!!!!!! Thanks for clarification- and my apologies (if spelt right)

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gollomS10628 | 31 May 12
rdmontgomerie. Afraid TheFont is right (he usually is). You pick a TB. Probability that the host opens 98 TBs is 100% (because he knows where the car is and will not open it). So probability of your original choice being a car is 1/100. Probability of other unopened door being a car is 99/100. And as a check the sum of all possible outcomes must always be equal to 1. The key thing (which was not very clear in the question) is that the host knows where the car is and always opens a TB (or 98TBs in the 100 door case).
rdmontgomerie | 31 May 12
Sorry TheFont - dont think you get it - if you have chosen the car the odds are 1/100 - the odds he shows you a TB is 100% ie the odds you picked the car are 1/100 - this we agree on. The odds you picked a TB are 99/100 the odds he showed you a TB not the one you are on is 98/99 . 99/100*98/99= 98/100 ratio is now 1/100 to 98/100 i.e. 1:98 or odds of 1/99
TheFont | 31 May 12
rdmontgomerie: They have, and those who agree with you are wrong, I'm afraid. Where you've gone wrong is in assuming that the magician opens doors at random. He doesn't. He deliberately avoids the one with the car behind it... assuming you didn't guess right first time. You know he’s going to reveal 98 teddy bears, whatever happens. So when he does just that it tells you nothing you didn't already know about your original choice... but almost everything about the door he's chosen to leave closed. If you were wrong with your first guess (99 %), the one remaining closed door will definitely be hiding the car.
rdmontgomerie | 31 May 12
I dont know if others have commented on problem 02 (Homework- car and teddy bears) - the answer is incorrect - there is an equal chance! There is NO advantage in changing. Consider 100 doors - and magician reveals 98 TB's - do you really think the probability of changing now increases your chance of winning to 99% ? I agree the chances of choosing the car in first instance do not change i.e. 1/100 but, the cannces if having chosen the TB given one (2 ,3 ,..) is/are revealed are 98/99, 97/98 , ....... changing the odds of the TB choice to 1/100 i.e. 50:50 same odds.
TheFont | 31 May 12
No, norharwil, it doesn't!

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norharwil | 31 May 12
Pool Jeremy rescue: I suggested a scale drawing (28 May) to help to explain my procedure which can avoid/minimise using maths. It may help to sketch in a 45 degree triangle and one with BJ/AB ratio of 3. If "Alex" can swim at 1/2^0.5 of the speed he can run then his optimum point to enter the water will be at X = 50 mtrs at angle B of the 45 degree triangle. Similarly for S/R speed ratio of 0.33 his entry point should be at X= 75-8.84 = 66.16 mtrs. To explain this consider the triangle with side AB=x AJ=25 mtrs and BJ = y = sqrt(x^2+25^2) (Pythagoras). Then dy/dx =x/sqrt(x^2+25^2) = x/y . This 1st derivative diminishes in Alex's running direction (to 0 at angle A). As Alex approaches B of the 45 degree triangle he should enter the water only if his S/R ratio is 1/2^0.5 ie 0.707 since at this point dy/dx = 0.707 also. If his S/R ratio is 0.5 as in the given problem, he should continue to run until he reaches triangle with angle B = 60 degrees where dy/dx = 0.5 also. Similarly for the S/R ratio of 0.33 where the triangle has angle B of 70.53 degrees for coincident dy/dx = 0.33 This may be difficult to follow but I hope it explains how and hopefully, why the rule works.
norharwil | 31 May 12
Hi TheFont : Not so. As I said I always use the spreadsheet because it is to hand and my calculator is lost or without a battery. A calculator can easily do two to four recalcs without much effort. What accuracy do you need anyway?
TheFont | 31 May 12
norharwil: Ah... so not “without... programming or spreadsheet” after all, then.
norharwil | 30 May 12
TheFont: I set up a spreadsheet (because I always do this). This was not the usual miscalled "trial and error method, which is really based on the calculus of finite differences" but an iterative non-interactive method. By feeding in a couple of changes, from land perpendicular 30 mtrs to your 55.6012 mtrs and from 75 to 55 mtrs edge I instantly received correct accurate results as in our recent posts. 3 iterations gave 2 decimal place accuracy and 5 gave 3/4 place accuracy. I will describe the simple calculation in a future post.
TerryB2876 | 30 May 12
The point I'm trying to make is, the solution is the same: make two piles, 90 and 10, turn over the ten. But the words used to describe that can make it seem different. What the piles are actually called ("taken out", "left behind") makes no difference to the answer, it's the same result, just from a different viewpoint.

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TerryB2876 | 30 May 12
Heads and Tails: just to say, there is a second solution to this little problem. On the programme, the good practical answer was, to take ten coins from the heap, and then turn all those ten over. Result, as many tails in the ten as are left in the 90 remaining. But instead, another (impractical) (but symmetrical) solution would be, to take 90 coins from the heap, leaving just ten; and then turn those remaining ten over, instead of the 90 you took. Impractical though, I'd lose count before I got to 90, so taking ten is better, - as they did. Of course, Simon or Dara would probably say, that's the other side of the coin ......
TheFont | 30 May 12
James: The question explicitly excluded movement away from the exit. norharwil: I'll believe that when I see it!
norharwil | 30 May 12
Thanks TheFont: I now believe I have the whole solution to the pool problem, having pondered over it for the last 8 days. My new general rule applies to Both triangles! So everything can be solved in a simple way without calculus, programming or spreadsheet. It is so simple but so difficult to explain so I will come back to it later.
james.barnaby1 | 30 May 12
Precisely my point ... the question was "how many possible exit routes are there?" but the program neglected to take into account routes that involved going backwards i.e. back to his desk ...
TheFont | 30 May 12
The maths wasn't that smart, James. Basically, from the stationery cupboard to the exit is 6 moves, from which we can pick any 3 to be transverse – giving us 6C3 (= 6! / (3!*3!) = 20), which we then double. (x+y)Cy works for any x by y grid, and cuts down on the “totting up” when the grid gets bigger.

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james.barnaby1 | 29 May 12
you recently had a problem that involved escaping an office, after visiting the loo and the stationary cupboard. Using a bit of smart maths you came up with the answer of 40 possible escape routes... but what if the wily security guard had forced you back to your desk ... aha!
TheFont | 29 May 12
You're quite right, norharwil. I now get 72.5019 m.
norharwil | 29 May 12
TheFont:Your result is close enough to mine that we agree on essentials. I note you used 56.6012 (not specified by me) but your original was 55.6012 mtrs perpendicular on land. This would explain my slightly dfferent results 72.501 and 26.395 swimming. I didn't do the angles. Out of curiosity I tried altering the initial offset of 2 mtrs between 0, 1, 2, 5 and 10. The results were as we would expect,(except for the 5 mtr case) on the main triangle's sides: 14.434/28.8676 : 14.4310/28.8661 : 14.4230/28.8621 : 14.37/28.8357 : 14.2/28.7513 . I guess the programme could have been even more machiavelian and specified a 25 mtre initial offset. This would have scuppered my simple solution using my (general) rule. I don't think I was the first discoverer, it could have been known to Newton! Ha Ha.
TheFont | 28 May 12
norharwil: With 56.6012 m, 55 m and 25 m respectively of land, edge and water (as you specified) the optimal diagonal running distance would be 75.8708 m (at an angle of 50.5241 deg). The swimming distance would then be 26.3676 m (at an angle of 71.4656 deg). Unless one of the given lengths is zero, or the speed ratio is 1:1, the problem really is as complicated as we say it is. Both triangles have to be taken into account, and the swim angle really isn't going to be 60 deg unless you start on the edge. It's unfortunate that they glossed over the 2 m on the programme, but it really does add that extra centimetre – when you use calculus and/or the right quartic equation.
norharwil | 28 May 12
This pool rescue problem is extraordinarily fascinating. It is about two triangles sitting on each side of the edge of a swimming pool. The thinner triangle complicates calculations on the main one and smacks of a "red herring". The calculation is really very simple with hindsight but has been built up with calculus and quartic equations. I have looked at the main triangle which IS the calculation. It is best to take a sheet of A4 paper in landscape positon and draw the triangle at the bottom left hand corner "A" and to mark the apex "J" for Jeremy (to be rescued) 12.5cm up on the left (to scale for 25 mtrs). Along the bottom is the "adjacent" triangle side along the edge of the pool reaching to "B" the 60 degree angle at which the runner enters the pool from the right. B is joined to J forming the hypotenuse of the triangle. The runners speed is twice the swimmers speed. This defines the ratio of hypotenuse BJ to adjacent side AB length and the angle "B" of 60 degrees. The runner starts 75 mtrs away from the left side of the triangle but at the right hand side running along the bottom pool edge from the right. The result is X mtrs from the start to angle "B". If to scale the distance AB must be doubled to represent cms as mtrs just as AJ (12.5cms) is doubled to indicate 25 mtrs. For accuracy the result is calculated as 75-25/3^0.5 =60.5662

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norharwil | 28 May 12
Hi The Font Thanks for your comments. Now can you do a calc. for me please? If the poolside difference between the two perpendiculars, in your problem, is increased from 25 mtrs to 55 mtrs what then is the running distance in the minimum total equivalent distance case?
TheFont | 27 May 12
Also... Alex apparently made six attempts, of which five were shown. But two of them (“bit less sprinty” and “a bit curvaceous”) were labelled attempt 3. So Dara and the students are not the only ones on the programme to have trouble with numbers.
TheFont | 27 May 12
I've now watched the repeat, and can report that Marcus does mention the 2 m once when he sets the problem, along with the 75 m and the 25 m. But unlike the 75 m and the 25 m, the 2 m does not get an arrow superimposed on the screen. The graphical recap similarly omits mention or representation of the 2 m, although the Alex figure does not appear to be at the edge of the pool. In Alex’s attempts (and their graphical representations), only the “very swimmy” one shows non-parallel running (i.e. straight to the pool). When Dara analyses the problem using on-screen graphics, he does use two right-angled triangles... except in the case of maximal swimming, where the 2 m is ignored for some reason. Dara and the students each come up with quartic equations, but unfortunately Dara’s has no real root, and the students’ would lead to running less than 60 cm before diving in.
TheFont | 27 May 12
Yes, norharwil. I think what you're calling z is what I'm calling x, but your figures agree with mine (to the nearest centimetre). Notice that we’re now swimming at exactly 80 degrees to the edge of the pool. (And running at 69.678 degrees.) The equations are the same – just with different numerical values.
norharwil | 27 May 12
Hello The Font The revised answer is X = run distance =59.29 with z =20.594 from your perpendicular to Jeremy's ditto. For a straight line z would be 17.25 mtrs. This calc. also is not straightforward. These results seem reasonable to me but I have not checked them. OK for now.

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norharwil | 27 May 12
Again Hello TheFont: Your point about the straight line is well made. Back to the spreadsheet, I think calculus would be even harder here and it would be tricky to formulate equations to feed to website wolframalpha. Intinctively I feel my new "general" rule can only apply to running along the poolside in line with the poolside side of the 2/1 triangle so my calculation to you was conceptually wrong! Consider your problem: If you reduce your poolside length from 25 mtrs to zero so that you start inline with Jeremy's perpendicular you must achieve the absolute minimum running AND swimming distance.
TheFont | 27 May 12
gollomS: I told you that would happen!
TheFont | 27 May 12
PaulH: Brilliant. If Alex had gone with that approach rather than all that running and jumping malarkey, I think Marcus and Dara would have been really impressed.
TheFont | 27 May 12
Hello again, norharwil: You have correctly understood my variant of the problem. But if you draw the diagram, you'll see that your route involves more swimming than the straight-line route. Ergo, your 60 degree theory does not hold when starting away from the edge. (As it happens, the optimal swim angle in my variant is a round number... just not that one.)
PaulH84040 | 26 May 12
Just wanted to add a different perspective to the solution of this problem. It doesn't make the maths any easier, but would allow a solution without equations if you used pulleys. Imagine a trolley constrained to travel frictionlessly along the pool's edge. There is a string from you to a pulley on the trolley which then goes to a pulley on the drowning swimmer and then back to the trolley. Pull the string in tight and the trolley will come to rest at the position of the correct answer. Not maths and not trial and error but a kind of analogue computer. For different ratios of water speeds to land speeds one would have to alter the ratio of loops between the swimmer and the life guard... all idealised and frictionless of course. Interestingly the Sagrada Familiar used the same idea to design its arches to balance the loads in its structure, they contsructed an upside down model of the church out of string, with little bags of sand to represent the weights of each of the bits of the church. The structure is very complicated and impossible to solve without a computer, or else letting the strings find the solution for them. A diiferent kind of string theory! Ha.

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norharwil | 26 May 12
Hi TheFont OK! I will check out your problem but I need to be sure of the facts. I assume that your distance 55.6012 mtrs is perpendicular to the poolside and at this poolside point is 25 mtrs along the poolside away from the 25 mtrs perpendicular reaching out to Jeremy. My general rule is not really relevant since the calc. has already been done,for example by smartmathguy as 25/SQRT3 = 14.43376 mtrs. This poolside triangle side length from perpendicular to Jeremy STILL HOLDS for your case and subtracting it from your 25 mtrs gives 10.56624 mtrs poolside distance from your 55.6012 perpendicular. Squaring these triangle sides, adding and taking the SQRT gives hypothenuse length X = 56.60 mtrs. Your answer.
TheFont | 26 May 12
Hi, norharwil. Your result only applies to the special case where the lifeguard starts at the edge of the pool. It's the distance from the edge that makes the problem tricky to solve. Admittedly, the first 2 m adds only a centimetre to the answer, but greater distances have more effect.
gollomS10628 | 26 May 12
Just tried the Snell's law approach. Can't really write it all out here, but guess what? You the same quartic as with differentiation. Aaaargh! Oh well I guess that proves Snell's law!
norharwil | 26 May 12
Hi TheFont: I did say "this kind of problem" meaning running alongside the pool which was the problem given to us. I avoided starting away from the pool, by even 2 mtrs. Adding 1 or 2 cms is to show how tiny this effect is, folks can then make up their own minds. I agree it is a nonsense! I show high accuracy only to demonstrate what can be achieved. Most people have no idea what can be done even with 15 year old spreadsheets and think "trial and error" is only rough and ready and inaccurate.
norharwil | 26 May 12
Back to my earlier post: It could be argued that my result is a fluke or special case. To counter this I set up other triangles with different hypotenuse to adjacent side ratios,such as a 45 degree triangle with sqrt2 hypotenuse ie 1.4142...value and this ratio to other unitary sides. This corresponds to a run to swim speed ratio also of 1.4142...Additionally I examined a triangle with Hyp/Adj side ratio of 3 for run/swim speed ratio of 3. These triangles were set up using the same perpendicular 3rd side length of 25 mtrs reaching to Jeremy. The calculated results were excellent and it was easy to verify their correctness by spreadsheet as they were already at optimal values. Results: For 45 degree triangle X = 50.000 mtrs (swim distance = 35.3553) mtrs. For 3/1 ratio triangle having about 70/90 degree angles X = 66.1612 mtrs (swim distance = 26.5165 mtrs). I also did checks that triangles had optimum side ratios by adding/subtracting increments to poolside lengths and observing corresponding changes to hypotenuse lengths*their multipliers. This showed in each case that optimum lengths had been corrct.

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TheFont | 26 May 12
norharwil, I'm disappointed. The person who gave us the answer to a tenth of a millimetre now resorting to retrospective guesswork involving indeterminate numbers of centimetres. If yours really is an accurate and general rule, what is x if the lifeguard starts 55.6012 m from the edge of the pool, but only 25 m along from the person to be rescued?
norharwil | 26 May 12
Further to my yesterdays comment Re run/swim pool problem: I have now discovered a simple (and I believe general) rule to easily and accurately solve this kind of problem. It does not require calculus, programming or even a spreadsheet. The solution is to take the ratio of run speed to swim speed ie 2/1 and apply it to the main triangle. Thus the hypotenuse to poolsideSide length ratio is to also be 2/1. Knowing the remaining side as 25 metrs perpendicular to poolside reaching to Jeremy the triangle is fully known. The simple calculation is then: X = 75-25/(3^0.5) = 60.56624 mtrs. To allow for initial 2 mtr offset add 1 or 2 centimetrs. I must go now but I will come back to this and proofs of generality.
TheFont | 25 May 12
To be honest, PeterB, I didn't hear Marcus mention the 2 m, but it definitely appears on Dara’s diagram – if not in Alex’s experiments. I only caught the end of the programme, so I'll be watching the repeat on Sunday.
TheFont | 25 May 12
PeterB: Instead of adding x to 2s in your equation, you add sqrt (x^2 + 4). Unfortunately, differentiating that gives you a much harder equation to solve – a quartic (x^4 etc.) instead of a quadratic. You can get there without differentiating: considering an infinitesimal change in x reveals that x/r has to be twice (75 - x)/s, where r is the (diagonal) distance run. (i.e. Snell’s law does apply here.) When x = r, we’re back to the simpler problem, and the 60-degree angle comes out. Unfortunately, the 2 m offset gives us that pesky quartic equation again. I might look up the formula tomorrow...
JanL4405 | 25 May 12
Dog matmatics is very complicated. After many years of observing dogs I can postulate that they have only a few numbers: 1, SOME, MANY, LOTS, MORE THAN EXPECTED and NOT ENOUGH. Where this can get very confusing is that SOME, MORE THAN EXPECTED and NOT ENOUGH can all be the same number, especially when applied to dog treats. I appreciate that Dara, as a cat owner, will not have had the chance to observe this strange and wonderful canine maths.

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PeterB584 | 25 May 12
how would you take that 2m into account with my solution?
PeterB584 | 25 May 12
by the way, did they mention about starting 2 m from the edge of the pool in the program? I thought the only 2 things they gave us was 75m along and 25m in.
PeterB584 | 25 May 12
ah, thank you, i didn't understand how they got 60.58m. not like that matters any way. we make it a simple problem by assuming the speeds are related and constant where realistically they would vary. Also you ignore the resistance forces due to air and water so that extra centimeter doesn't matter too much. also you would jump into the swimming pool which would reduce the swimming distance. But thank you, i kept thinking it was rounding mistakes.
TheFont | 25 May 12
Sorry, PeterB. To two decimal places the answer is 60.58 m. The extra centimetre is the result of having to start 2 m from the edge of the pool – a condition they pretty much sneaked under the radar on the programme.
PeterB584 | 25 May 12
This is how I solved the problem: Start by drawing the right angled triangle. Left corner is the starting point, top right is the max running distance and the bottom right is where Jeremy is. So the the triangle has horizontal length R=75m and vertical length y=25m. Make two smaller triangles within the original by starting at the point in the pool and going to some point between x=0 and x=R=75m. The right hand side triangle should be a right angled triangle with vertical length y=25m, horizontal length d=R-x and a hypotenuse which i denote as s for swimming distance. Using pythogoras we get s=sqrt((R-x)^2 + y^2). The time taken for running distance x at velocity v is t1=x/v. The time taken for swimming distance s at velocity 0.5v is t2=2s/v. The total time is then t=(x + 2s)/v. Now you differentiate t w.r.t. x and set to 0: (missing out a few algebra rearranging) this gives you: 3x^2 - 6Rx + 3R^2 - y^2 = 0 R=75, y=25: 3x^2 - 450x + 16250 = 0 now use the quadratic formula: x = (-b+-sqrt(b^2 - 4ac))/2a which gives two solutions. x=89.43m and x=60.57m. However, 0<=x<=75 so the answer is x=60.57m. Done.

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TheFont | 25 May 12
Further apologies, norharwil. By correcting the value out of context, I fear I may have misled you. 28.8623 m is the optimal swim in the problem as stated. 50 m over sqrt 3 (28.8675 m) is simply the limit as the pool length tends to infinity, the run becomes more parallel, and the problem resolves into the simpler one we all thought we’d heard until we spotted the 2 m in Dara’s working. (Alex was certainly running parallel to the side of the pool. I await the next repeat with interest.) Without the 2 m the swim angle is 60 deg, because its cosine is half cos 0. But I’m afraid that if we want to solve the full problem without using Excel, we probably can't avoid solving the quartic (sic) equation I specified earlier.
norharwil | 25 May 12
THE PENNY HAS DROPPED; Re the Jeremy pool rescue problem. Firstly I smelt a rat with Smartmathguys (23 May) elegant solution. It seemed to me remarkable to square and shuffle these complex equations and end up with the simplistic solution X = 75-25/sqrt3 . Also TheFont (24 May) stated that the optimum swim distance is always 50/sqrt3 regardless of pool length. Now put these simple equations together and you get a one-two right angled triangle with other side sqrt3 (25 mtrs acually). NICE ONE MARCUS! you really lead us up the garden path. Forgetting about the 2mtrs starting offset (giving 1.3 inches effect) X = 75-25/sqrt3 = 60.56624..etc as answer. As TheFont has said the swum distance is constant (at around 28.8674 mtrs) regardless of pool length. The important item is that (75-X) is 28.8674/2 = 14.4337 and 75-14.4337=60.5663mtrs and about right! I have checked this out, using 200 instead of 75 (pool length), by reducing the run distance by 0.5 mtr and 1 mtr, getting 0.506 and 1.026 mtr increases in equivalent swim distance. This suggests that the simple triangle represents the optimum situation, rather surprisingly.
nclarke | 25 May 12
Saw the show.....The optimum distance to run surely is given by Mach's principle. There is an optical analogy and Mach's principle in refractive optics is Snell's law. Snell's law (sini/sinr) = (velocity of light in rare medium)/(velocity of light in dense medium)= n. We know n (walking speed in air)/(walking speed in water) and we know the geometry (start position of rescuer and position of victim) so it's possible to calculate the least action.
PeterB584 | 25 May 12
i take it i cant upload attachments here?
norharwil | 24 May 12
Hi SwissTony1969 Thanks for your details.I think we can make progress here. In your D column you put C1+B1. This means the calculation gives result X = 0 (approx) and minimum swum distance in straight line to Jeremy of about 79 mtrs. Run and swim speeds are equal! More seriously C1+B1 should be entered as 2*C1+B1 to allow for the half speed swimming. I note that you got the correct result,60.58 mtrs, but you are uneasy about the method. You shouldn't be. In column D you used about 5 decimal places (I imagine). If you want more accuracy than X = 60.58 you need to increase decimals to 10 or 12 in Col D and continue to develop Col A with more decimals in regular ascending values while scanning column D by eye, keeping minimum values in view (by readjusting values in Col A). It takes a bit of practice but I am confident you can do it. You should get to these pairs: 57670(35):57671(27):57672(24):57673(26):57674(33). The 5 figures are decimals in Col A and in brackets are 11th/12th decimals in ColD. Result is X = 60.57672 Do forget about drawing graphs, they are useless at this level of accuacy. Good luck!

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PeterB584 | 24 May 12
by using calculus i have got an answer of 60.57m again which agrees with my java program. once again this could be due to rounding errors
PeterB584 | 24 May 12
i wrote a java program to try and solve that swimming pool problem when you had to work out the optimal distance to run before going into the swimming pool to save a person who is 25m from the side of the pool. 0 <= x <= 75 . the given answer was 60.58m, when i finished my program i calculated it to be x = 60.56999999999989 m which would be 60.57m to 2 d.p. granted there could be some error in my code but i thought i'd let you know.
TheFont | 24 May 12
Apologies, norharwil. It was your value (which is the same as Marcus’s, but accurate to an extra couple of decimal places) that fits my quartic equation so nicely.
TheFont | 24 May 12
Corrigendum: In my earlier response to FrankB (and AdrianR) I should have given the maximum required swimming distance for any length of pool as 28.868 (50 over sqrt 3).
norharwil | 24 May 12
Hi smartmathguy: I congratulate you for your elegant solution with the same result as my first spreadsheet, but you have ignored the start 2 mtrs from the pool edge. This comment was made against me (RE RichardS43 21 May and my reply 22 May where I got 60.5767 since confirmed by others including web site wolframalpha.com The difference is about 1.5 inches. I still think my spreadsheet method is quickest and most practical!

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TheFont | 24 May 12
I get the same nasty-looking quartic (not quadratic) equation (with coefficients 1, -150, +5422, -800, +30000) by differentiation as I do by eschewing formal calculus and simply considering an infinitesimal change in x from the optimum (such that the change in angles is negligible). The latter method shows the cosines of the angles with the pool edge to be proportional to the speeds involved. Which agrees with my recollections of optics. Which is nice. And Mucous Saturday’s value for x is a root of my quartic. Which is also nice.
SwissTony1969 | 24 May 12
Sorry guys. Same post 5 times. Something wrong with my browser
SwissTony1969 | 24 May 12
Hi Norwhail, In Excel I put in column A the integers 0-75. Column B =(A1^2+4)^0.5 (you could use SQRT as well). Column C = ((75-A1)^2+25^2)^0.5. Column D =C1+B1. Did a graph and then found the minimum. I then did the same again for values 60 to 61 with 0.01 intervals. The minimum was, indeed, 60.58 but the difference in time taken was so small (less than 1 * 10^-5) for about half a dozen values each side. When I did my management accounting exams several years ago I had a similar problem. Maximum profitability for a company was where marginal cost equals marginal income. The marginal cost involved a learning curve. Again, I managed to get the formulae correctly and drew a graph and had to approximate the answer as I was mucking about with a -.618 root of a number. I passed, by the way, but only needed 50%.
Damien Dodd | 24 May 12
Still loving this show.
TheFont | 24 May 12
FrankB (and AdrianR), your belief that you should always swim one third of the route is a red herring, I'm afraid. What if the pool was a mile long? Would you swim a third of a mile? In fact, the optimal route never involves swimming more than 28.8623 m. FrankB’s route may be 0.683 m shorter overall than the calculated one, but that is more than made up for by the extra 0.734 m in the water.

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JulesBlue | 24 May 12
Sorry, forgot to square D: x + S * sqrt((W - x)^2 + D^2)
JulesBlue | 24 May 12
I have just discovered this program, and am please and surprised to find it. The last time I used any maths was 2 decades ago at A-Level; and watching last night, the swimming pool rescue problem inspired me to have a go myself (as the show did not have time to provide the mathematical solution), and re-learn how to do derivatives. I expressed the problem in a general form. W = 75 metres, D = 25 metres, S = 2 (runs 2 x swim speed). Therefore it is the smallest number x which satisfies: x + S * sqrt((W - x)^2 + D). This is when the derivative of this equation dy/dx = 0. I then had to re-discover that calculating the derivative requires the "chain" rule, because of the sqrt of a power. Once the derivative is calculated, it was a matter of soliving the resulting quadratic equation. Interestingly, the trial and error approach of doing the actual running and swimming could have led into a discussion about approximation methods for finding the roots of the above equation, e.g. such as Newton's iterative method. I would suggest that they add an additional person on the show whose job is to sove the problem using a computer algorithm.
FrankB11007 | 24 May 12
The swimming pool problem. I solved this a different way without calculus. By deriving two quadratic equations for the two triangles. y^2 = 2^2 + x^2 because we know that the distance in water has to be half of the distance on land the hypotenuse is 0.5y so 0.5y^2 = (75-x)^2 + 25^2 multiplying by 4 y^2 = 25000 - 600x + 4x^2 substitute y^2 from the formula for the first triangle and we get 0 = 3x^2 - 600x + 24996 this is a quadratic equation and we can use the formula to solve this giving x = 59.159 or (140.841) checking the answer y (on land) = 59.1928m and 0.5y (in water) = 29.5962m (exactly half) the total distance traveled = 88.789m Using the value given by others of 60.5767 the distance on land is not double the distance in water and the sum of the distance is 89.472m making it 0.683m longer! And therefore incorrect?
TheFont | 24 May 12
I notice that the earlier debate on the three doors problem (always change your door, by the way) has overshadowed the problems with the original quiz question 10. Only one of the necessary assumptions is stated. Nothing is said about the possibility of identical twins, for instance. But, more importantly, nothing is said about how much the doctor actually knows, or how he knows it. In real life, we might assume from what he says either that he only knows the sex of one twin (in which case the probability of the other one also being a boy is 1 in 2) or that he knows that there is one of each. Here we are expected to assume that he does know the sex of each twin, but that he will only ever tell us one is a boy – even if he knows they both are!
norharwil | 23 May 12
Re SwissTony1969 Re my previous comment: I should have said speed is not involved except in a relative way to define the equivalent swum distance as double the actual distance. If the run and swim speeds were the same there would be no problem as the minimum distance would be the straight line to the victim (about 79 mtrs). You say that you obtained an "inelegant solution" using Excel. Please can you say if this was a correct solution. As I see it the only inelegant solutions are the tedious and the incorrect. Among the tedious are some involving the calculus with horrific equations easily leading to wrong results.

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AllanR16452 | 23 May 12
RE:adrianR91771 Once I had got the sqrt(2^2+x^2)+2 sqrt((75-x)^2+25^2) I used the site www.wolframalpha.com and typed in "minimum t= sqrt(2^2+x^2)+2 sqrt((75-x)^2+25^2)". It then plotted the graph & calculated the minimum value. RE:SwissTony1969 Your memory of maths is better than mine - I'd forgotten about derivatives but if you go on www.wolframalpha.com and type "derivative SQRT(X^2+2^2)+2*SQRT((75-X)^2+25^2)" (you should then be able to click on show steps) hopefully it will make more sense to you. I just thought that once I'd got the equation it was easier to type it into a website than write a program or set up a spreadsheet.
smartmathguy | 23 May 12
3JUGS:800->350->323->620->602->152->143->440 7 MOVES [DO A TREE DIAGRAM WITH ALL POSSIBILIES AND ELIMINATE DEAD ENDS]
smartmathguy | 23 May 12
MAIDS N VAMPS:__________________________________________________________ (3V3M,0)->(V3M,2V),(2V2M,MV)->(2V3M,V)->(3M,3V)->(3M1V,2V)->(MV,2M2V)-> (2M2V,MV)->(2V,3M1V)->(3V,3M)->->(2V,3M1V)->(0,3M3V)=11 MOVES..........
smartmathguy | 23 May 12
BEACH HUT EXTENSION:................................................... A=MAN'S VERT DIST FROM H20 B=HUT'S VERT DIST FROM H20................... X=DIST ALONG BEACH WHERE REBOUND HAPPENS,L=TOTAL HORIZ DIST............. I.E.X=L1 AND L=L1+L2 SO L2=L-X......................................... TOTAL DIST D=SQRT(X^2+A^2)+SQRT((L-X)^2+B^2)............................ AT MIN D:dD/dX=0=2X/[2SQRT(X^2+A^2)]+2*(L-X)*(-1)/[2*(SQRT((L-X)^2+B^2)] =>(XB/A)^2=(L-X)^2=>[1-(B/A)^2]X^2-2LX+L^2=0............................ SOLVE FOR X.
norharwil | 23 May 12
Hi SwissTony19 Worried about run speed? but this problem has nothing to do with run speed! I thought this at first but found speed in itself does not matter. I note that your equation (as those in many other comments) is calculating distance not time and speed is not in it. We are really trying to find the minimum equivalent distance that can be run and swum and the corresponding distance (x say) along the pool side where the rescuer must enter the water. Equivalent denotes that the swimming distance must be doubled due to half speed before adding to x. Total distance to be a minimum = x+2y where y = swum distance.

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norharwil | 23 May 12
further to
smartmathguy | 23 May 12
POISON PROBLEM:ANOTHER WAY OF LOOKING AT IT. GLASSES A B C D TUBES E F A->E & F ,C->F , D->E {A,B}=THE SET OF ELEMENTS A B P(E)={1,0} =>P(AUD)={1,0}<=>P(BUC)={0,1} P(F)={1,0} =>P(AUC)={1,0}<=>P(BUD)={0,1} LOGIC TREE: P(F)=1 AND P(E)=1 <=>P(A)=1 P(F)=1 AND P(E)=0 <=>P(C)=1 P(F)=0 AND P(E)=1 <=>P(D)=1 P(F)=0 AND P(E)=0 <=>P(B)=1 KEY: POISON P:A->P(A) P(AUB):AUB->P(A) OR P(B) I.E. P MAPS A ONTO P(A) -A IS POISONED AUB MEANS "A OR B" SYMBOLS "=>" MEANS "IMPLIES" , "<=>" MEANS "IF & ONLY IF" OR "IMPLIES & IS IMPLICATE"
smartmathguy | 23 May 12
EASY:X=DISTANCE TRAVELLED ON BEACH,V=SWIM SPEED ,SO, 2V=RUN SPEED TOTAL TIME T=TIME ON BEACH+TIME IN WATER=X/2V+SQRT((75-X)^2+25^2)/V min T when 0=dT/dX=[1/2+0.5*2*(75-X)*(-1)/SQRT((75-X)^2+25^2)]/V SO AS V<>0: SQUARING AND SHUFFLING, 4*(75-X)^2 =(75-X)^2+25^2 OR: 3*(75-X)^2 =25^2 ,X=75-25/SQRT(3)~60.566243270259355887271280487451 TADA!!!!!!!!!!!!!!!
adrianR91771 | 23 May 12
I arrived at a different answer myself. Since walking speed is twice swimming, I walk twice swim distance to achieve quickest time. The max walking distance covers X and 2m. x + 2 = 2(25 + (75-x)). This works out as X=66m, giving a swim distance of 26.57 (the hypotenuse of the triangle worked out with 25m and 9m for sides)
adrianR91771 | 23 May 12
For instance, take a right angle with sides 3 and 4 mtrs. the hypotenuse is SQRT 3^2 + 4^2. This is not the same as 3 + 4 even though 3 is the SQRT of 3^2 and 4 of 4^2.

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adrianR91771 | 23 May 12
I am baffled as to how people derive 't' and 'x' esp AlanR16452 from the equation t=SQRT2^2+x^2 + 2SQRT((75-x)^2)+ 25^2). As long as x is unknown, I can't see how the value of the squares and square roots be ascertained. The first part of the above equation for instance refers to the sum of SQRT of the sum of 2^2 and x^2. We need to get this sum first before finding the SQRT. And for this we need 'x'. The same applies to '75-x' which isn't two separate numbers but has one value.
SwissTony1969 | 23 May 12
Further to my previous post, I think I can swim at 1m/s. I would be jogging at 2m/s with a note pad and pen, tongue stuck out and a tape measure working out the optimal distance to run before I jumped in and then precisely measuring the distance so that I can get to Jeremy in the shortest time possible. And Dara, I have bought "Tickling the English" as well, so don't feel left out. Cheers
SwissTony1969 | 23 May 12
Hi, A long time since I did A Level maths. Can somebody take me through the calculus part of the swimming pool problem. I got the time as SQRT(X^2+2^2)+2*SQRT((75-X)^2+25^2) but couldn't get this to x= and certainly couldn't do the first derivative. I know we have to solve for first derivative = 0. My inelegant solution used excel. If you can show me the step by step algebra to get to x= and then the first derivative from there I would be most grateful. BTW Dara and Marcus, excellent show. I bought "The Number Mysteries" off the back of it. Thanks guys
SwissTony1969 | 23 May 12
Hi guys, The fundamental problem with the swimming pool puzzle is the running v swimming issue. If I run the marathon at world record pace, I'm doing approximately 5.6 m/s. If I then swim 25m at 2.8 m/s, my time would be 8.92 seconds. The world record for the 50 free short course is 20.30. So I would jog to the edge of the pool and then murder the world record of 10.15. I wouldn't be there to save the guy because I would be doing a drugs test after my latest win at the Olympics
DazM3764 | 23 May 12
On the beach hut fire problem, there is an even more fundamental issue. The fact that the fire is in a hut on the beach. You would use the sand not the water to put the fire out.

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DavidM64550 | 22 May 12
I got the time t equation as (Dr = distance run): t= [Dr+2*SQRT(Dr^2-150Dr+6250)]/(run speed)so differentiating to get the minimum gives: 1+(2Dr-150)/SQRT(Dr^2-150Dr+6250)=0(independent of run speed) giving Dr=60.56624 (squaring and solving quadratic)phew!
AllanR16452 | 22 May 12
Swimming pool problem: It's basically 2 right angled triangles. The first triangle has a side of 2m and a side 'x' somewhere between 0 & 75. The other triangle has a side of 25m and a side of 75-x. So there is 2 hypotenuse to calculate and the 2nd hypotenuse will take twice as long (time wise (t)). therefore: minimum t= sqrt(2^2+x^2)+2 sqrt((75-x)^2+25^2) typing minimum t= sqrt(2^2+x^2)+2 sqrt((75-x)^2+25^2)into www.wolframalpha.com gives:t=118.334 and x as 60.5767. Though this doesn't take into account if it is feasible to go into the pool while changing direction and not incur a sideways motion.
jefferyB70308 | 22 May 12
I would like to add that even if the beach was stoney the direct route is still best. You would dig a small pit in the beach with the bucket. The pit will now fill with water which you can fill the bucket with. If a bucket load of water is not enough the pit will top itself up whereas the other method requires you to run back to the sea.
norharwil | 22 May 12
Hi RichardS,Thanks for your correction. I hadn't noticed the 2 mtr distance from pool at start,the poor fellow might have fallen in too soon! I copied and revamped my spreadsheet, adding a new column Z for formulae to calculate the distance run parallel to the poolside which is 0.033 mtrs less than actually run on X,now the new hypothenuse. Findings were: X = 60.6097 Z = 60.5767 (Dara got 60.58 and you confirmed this). Swimming distance Y = 28.8623 mtrs. Minimum time was 19.7224 seconds for 6 mtr/sec run speed and double this,of course, for 3 mtr/sec
jefferyB70308 | 22 May 12
Personally I wouldn't even bother with the water - go there direct and put out the fire with sand :)

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RichardS43226 | 21 May 12
Hi Norharwil, You're ignoring that he starts 2m from the edge of the pool before he starts running, so the distance covered running is also given by Pythagoras, sqrt(x^2 + 2^2). Using this and your spreadsheet method I get close to 60.58. Which is answer they gave on the show. But your answer showed me where I was going wrong, so thanks!
RichardS43226 | 21 May 12
Ignore my comment - made a fundamental error : t = d/v not v/d ...oops
norharwil | 21 May 12
Tonights episode was great, some real math. I found it was easiest not to use calculus (I'd be very rusty) on the swimming pool problem but to put on a spreadsheet. Set up a column for X the distance along the side and one for Y the hypothenuse used to swim along and calculated from Pythagoras using 75-X and 25 for triangle sides. A 3rd column is used to calculate the time taken (X+2Y)/6. (6 is runners speed M/sec but other speeds 4 and 10 don't alter the result. By copying (mouse dragging) selected values in column for X and the formulas in the other columns the result is quickly found in column X. This corresponds with the minimum time in the 3rd column. My results were X = 60.5662 metres Y = 28.8675 metres (swimming)
RichardS43226 | 21 May 12
Did I write down the correct distances for the swimming pool rescue problem: 2m, 75m and 25m? If so, I get a different answer. Using a computer to calculate following equation for various values of x, and refining to close in on answer, the answer lies between 56.943 and 56.945. f(x) = (2/sqrt(x^2 + 2^2)) + (1/sqrt((75-x)^2 + (25^2))) x = 56.943, f(x) = 0.0675274780044 x = 56.944, f(x) = 0.0675274780026 x = 56.945, f(x) = 0.0675274780234 Would mean Dara was closest after all, with his guess of 55m! Dara wins!!
JaiminT38439 | 21 May 12
I disagree with the solution to the burning beach hut problem. You haven't factored in the need to carry the water to the burning beach hut. Carrying the water would slow you down due to the extra weight and trying not to spill it. Therefore you need the shortest distance possible between the water and the beach hut to maximise the time spent travelling without carrying the water.

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lindaS52608 | 21 May 12
love the show/dara/maths. i do wish we got more than a glympse of dara's and the students' workings out. i thought i could at least find them on the website.
MalcolmB7878 | 21 May 12
The swimming pool rescue problem needs a physicist. Think optics and critical angle. sin(c) = v1/v2 where c is the critical angle and v1 and v2 the speeds in the two media. Then it is a bit of trigonometry or Pythagoras. There is a fundamental physics principle in play here. The path taken by light through an optical system is the minimum time so the equations used in optics also solve many least journey time problems.
jefferyB70308 | 21 May 12
In tonights episode the solution for putting out the fire could well be wrong if you are nearer the fire than the sea. It is possible in this circumstance that the fastest solution is to run directly to the fire, put the fire in the bucket, then take it directly to the sea and drop it in :)
DavidP21078 | 21 May 12
I was watching the episode with the oil and found that the 4L of oil could be done in 6 and not 7 moves. Start with 8L in 8L jug Fill 5L jug from this Fill 3L from 5L jug Pour 3L from 3L jug into 8L jug Transfer 2L from 5L jug to 3L jug Pour 5L from 8L jug into 5L jug Fill up 3L jug from this leaves 4L in 5L jug Is this correct or have I missed a 'golden rule'?
paulgreenwood59 | 21 May 12
I am a teacher of a Year 5 class (ages 9-10) and had a fantastic lesson this morning tacking the maths investigations in one of your shows. It would be fantastic for these to be used in school, but in the episode used (vampires and maidens challenge) there were a couple of occasions where the language was unsuitable (talking about sex etc). Love your programme but please make sure the conversations are suitable for young ears.

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PatrickH62447 | 21 May 12
jefferyB you answer is ingenious, but not correct. You are confusing volume with capacity. The capacity of the 3-jar is 3 units, but its volume with a lid on must be more than that because of the volume of the material it is made from.
jefferyB70308 | 21 May 12
also the pouring of oil problem can be done in 3 moves ... 1. Put the lid on 3 jar, place it inside the 5 jar then fill the 5 jar (this will require 2 litres). 2. pour the 5 jar contents into the 3 jar, put the lid back on, then again place the 3 jar into the 5 jar. 3. once again fill the 5 jar from the 8 - this will leave 4 litres in the 8 jar ... job done
jefferyB70308 | 21 May 12
of course if you are not forced to use straight lines you can make as many pieces as you like ... start with a spiral cut, then cut into quarters on top of the spiral
jefferyB70308 | 21 May 12
I'm surprised they didn't see the 3 cuts in a pizza puzzle solution of 8 pieces. First two slices cuts it evenly into 4 pieces. The final cut is horizontal cutting all four slices into 8.
PedA58760 | 20 May 12
Can whoever is responsible for the graphics for homework 5 please remove the "p" from the answer box. An amount of money can either be given in pounds (£) OR in pence (p) but not both! I expect better from a tv programme about maths... ;-)

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Damien Dodd | 18 May 12
This show just gets better and better. Love it.
gollomS10628 | 17 May 12
Aposrop'mise - We likes it we does. Have a Gollom award for best post on this board so far. FrancisH - Pretty good too. You wait days for a really good post then 2 come at once!
v.strom | 17 May 12
FrancisH27413 | 17 May 12
I liked Dara's solution to the doughball problem using modulus 6 but it made me think of a different approach using 3s that gives a complete list of all the numbers that can't be ordered using 6s, 9s and 20s. Arrange all the intergers in rows of 3 with 1,2,3 on the first row, 4,5,6 on the second and so forth. So column 3 has all the multiples of 3. Using 6s and 9s you can make all the multiples of 3 (all of column 3) except 3 itself. This is easy to prove. But you can't make any numbers in the other columns. If you put in a 20, that is in column 2 and then using 6s and 9s you can make all the other numnbers in column 2 except 23. Using two 20s, 40 is in column 1 and now you can fill in all of the rest of column 1 except 43. So 43 is the highest number you can't make plus you have all of column 1 below 40, all of column 2 below 20 - and 3!
Apostrop'mise | 17 May 12
A way to avoid trial and error: Because 199=2*98+3, you can rearrange the equation 98x-199y=50 to 98(x-2y)-3y=50, or x-2y=(50+3y)/98. This helps because as x-2y is an integer, (50+3y)/98 must also be an integer. So 50+3y must be a multiple of 98. This gives y=16 (or 114 or other numbers >100).

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LlinosH30499 | 16 May 12
Ooh homework issue's been fixed, happy days!
TamaraW44770 | 16 May 12
Truffle oil problem. I think I managed to do it in 6 moves.. Move 1 : Pour 5 litres into the 5 litre jar. Which leaves 3l in one and 5l in another. Move 2 : From the 5l jar, pour 3l into the 3l jar. Leaving 3l 2l and 3l. Move 3 : Pour the 3l jar back into the 8l jar. Leaving 6l, 2l and 0l. Move 4 : From the 5l jar, pour the 2l into the 3l jar. Leaving 6l, 0l, 2l. Move 5 : From the 8l jar, pour 5l into the 5l jar. Leaving 1l, 5l, 2l. Move 6 : Fill the 3l jar up using the oil in the 5l jar. Which leaves 4 litres in the 5 litre jar.
CarlW13653 | 16 May 12
I can solve the truffle oil problem very easy. All you do is take the 8 litre jar and pour a 3rd into the 3 litre jar, then pour that 3rd into the 5 litre jar. Next you fill the 3 litre jar to the top and pour that into the 5 litre jar giving you 4 litres. So 8 litre jar has 4 litres in and the 5 litre jar has 4 litres in. That was done in 4 MOVES.
tom72 | 15 May 12
Assuming the jars are round and symmetrical, you could : Pour 3 litres into the 3litre jar, then pour HALF of it into the 5 litre. you know youve poured half when the level of the liquid is from the rim of the top of the jar to the rim of the bottom of the jar (IE the diagonal!) Then do the same to the 5 LitreJar, Then pour the 1.5L into the 5 Litre jar to give 4 Litre: 5 Pours! 8,0,0 5,0,3 5,1.5,1.5 1.5,5,1.5 4,2.5,1.5 4,4,0
gollomS10628 | 15 May 12
Anyone got a better way of solving the homework? The solutions given involve either trial and error with a calculator (must confess that's what I did but it's a bit hit and miss) or plotting a graph with a slope of 199/98 and a y axis intercept of 50/98 but with enough resolution to see whole numbers at 16. Not very practical! any better ideas?

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v.strom | 15 May 12
Truffle oil: if I remember correctly, the solution shown started with filling the 3-liter bottle: 8,0,0 - 5,0,3 - 5,3,0 - 2,3,3 - 2,5,1 - 7,0,1 - 7,1,0 - 4,1,3 That's seven steps. 4,4,0 would be an 8th step. There is no shorter way starting with the 3-liter bottle (I wrote a program doing a full search). Stephen Templer's solution (also found by the program) 8,0,0 - 3,5,0 - 3,2,3 - 6,2,0 - 6,0,2 - 1,5,2 - 1,4,3 starts with filling the 5 liter bottle and takes only six steps, or seven if we count in 4,4,0. One step less than claimed in the show anyway.
gollomS10628 | 15 May 12
Thanks all for the tip about the homework. I kept typing it in and getting told it was wrong. Just looked at the cheat and it was right all allong! Mind you I had to log in 4 times before it accepted my password so maybe the web site is poorly. Ref the oil pouring, I think the 6/7 issue is just a question of whether you pour the 3 into the 1 as a 7th move, so not really a maths issue.
tettehturkson | 15 May 12
I've watched the truffle oil bit again. I'm afraid we "6ers" are the victims of poor editing. Whilst it is never clearly stated to be a requirement to have 2 containers containing 4 litres, the solution of 7 moves on the show seems to go that further step.
Starjammer | 15 May 12
Just watched the episode with the Truffle Oil and your overall answer is wrong as I can get it in 6. Start 8=8 5=0 3=0 Move 1 8=3 5=5 3=0 Move 2 8=3 5=2 3=3 Move 3 8=6 5=2 3=0 Move 4 8=6 5=0 3=2 Move 5 8=1 5=5 3=2 MOVE 6 <---- 8=1 5=4 <--- 3=3
LlinosH30499 | 14 May 12
StephenH42090: LEGEND!!!

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StephenH42090 | 14 May 12
Looks like Homework Question number 5 is wrong - you have to type in the change you have after buying the chewing gum, not the change you should have been given as it asks for.
LlinosH30499 | 14 May 12
tettehturkson, good point!
talviruusu | 14 May 12
So frustrating when I can only do 1 out of 3 problems - I used to have a maths brain !! :-(
LlinosH30499 | 14 May 12
No worries :-) have you tried the 5th one? THat's the one that doesn't seem to be working for me. What sign are you putting between the pounds and pence? I tried '.' and ':' but still doesn't work, it's weird :S
tettehturkson | 14 May 12
LlinosH30499, they didn't say that you need to have 2 containers with 4l in them - why would that be necessary when 1 person could simply take 2 jugs as someone must - and my recollection is that their solutions didn't leave 4 in each of 2 either.

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LlinosH30499 | 14 May 12
The puzzle is to get 2 containers with 4l in them, so 2 people can take 4l away with them. So your 6th move ends up with 1.4.3, you then need to move the 3l into the 8l to get the 7th move. Hope that clears it up :-) has anyone tried the homework puzzle? I put in the correct answer but it keeps saying it's wrong. I then check their answer, and it's exactly the same as mine :S
PaulN72349 | 14 May 12
Thought I was going mad as I made the 4l in 6 moves.
JoannaD16321 | 14 May 12
Tonight's episode was brilliant. Maths of Italian restaurants - so Dougls Adams - and also I liked that water pouring puzzle. It was also in Die Hard With a Vengeance.
tettehturkson | 14 May 12
UsmanJ29100, that was my solution to the oil as well. Glad I'm not going mad.
tettehturkson | 14 May 12
elliG6952, that is to assume that both the 5 litre jug and the 8 litre are the same diameter, which did not seem to be the case.

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elliG6952 | 14 May 12
The 4 litre oil puzzle can be completed in one move by filling the 5 litre jug till the level of the oil is the same as the amount of oil left in the 8 litre jug (ie half the oil, 4 litres is in the 5 litre jug and half is in the 8 litre jug.)
UsmanJ29100 | 14 May 12
800 350 first 323 620 602 152 143 That's right n 6 gos I think
UsmanJ29100 | 14 May 12
And thozmeister the choices were 20,9 n 6
MonkeyBot598898 | 14 May 12
D'oh! No line breaks. (1)a=3 b=5 c=3 >>>> (2)a=3 b=2 c=3 >>>> (3)a=6 b=2 c=0 >>>> (4)a=6 b=0 c=2 >>>> (5)a=1 b=5 c=2 >>>> (6)a=1 b=4 c=3
UsmanJ29100 | 14 May 12
True monkey bot I wanted to post it but the xxxxxx activation of the account took ages BRAVO

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MonkeyBot598898 | 14 May 12
The jugs of oil puzzle can be completed in 6 moves. You have 8, 5 and 3 litre jugs I'll call A, B and C. You start of with all 8 litres in the same jug so you have A=8, B=0 & C=0. You make the following moves. A B C 1. 3 5 0 2. 3 2 3 3. 6 2 0 4. 6 0 2 5. 1 5 2 6. 1 4 3 The trick is to create a SPACE of 1 litre in the 3-litre jug instead of trying to get a jug with 1 litre of liquid.
thozmeister | 14 May 12
Please correct me if i'm wrong on the dough ball problem, but their answer was 43 and i'm pretty sure that can be made with servings of 10, 9 and 6? 10+9+9+9+6 equals 43 unless i'm very much mistaken.... surely the answer is 33?
DavidD46181 | 14 May 12
Unless I have missed something, the 3 maidens & 3 vampires puzzle can be done in 9 moves. The only stipulations were vampires <= maidens on either level, max 2 people in the lift and 1 person must be in the lift to move it up or down.
norharwil | 14 May 12
norharwil The 9 coin puzzle is too easy since it is known that the odd coin is lighter.Thus it can be solved in two weighings. If there were 27 coins it would take 3 and if 81 coins only 4 weighings! Why not try the puzzle when the "odd" coin is either heavier or lighter than the rest but not known at the outset. It is also required to find which it is - heavier or lighter? It can be done in 3 weighings using either 9 coins or 12 coins. The logic is brilliant but it is not mine!
plubrown | 13 May 12
yeah you can definitely do the 3 vampires 3 maidens question in less thn 11 moves, 9 moves seems to be the lowest amount

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etchi | 13 May 12
That's not how Vampires work, some victims are cattle, some become thralls, only a select few are actually converted. If we are going to stipulate the existence of vampires, we should at least stick to accepted internal logic.
PeterV2985 | 13 May 12
I'm confused by homework question 4. The answer given is fine if you only have three pairs of socks, but that isn't specified in the question. My answer would be to take half the number of socks as pairs of socks you own, plus one.
AdamD51852 | 11 May 12
I HAVE IT WITH 5 MOVES!!!!!!!!!!!!!!!!!!!!!!!
megmasters | 10 May 12
haha sorry, i meant to bring M down on the second :P this brainteaser gets confusing at times, so thanks for the tips guys:)
michael1980 | 10 May 12
the whole arguement is stupid. if you leave 3 vampires with 3 maidens the maidens are going to get eaten anyway. one vampire could eat three maidenss. the 9 move solution works and even the 11 move solution has vampires out numbering maidens at loads of places

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JonB16963 | 10 May 12
gollomS10628 | 10 May 12
Megamasters - Sorry should have said M down on the second move.
gollomS10628 | 10 May 12
Phew! Well here we go. Megamasters - after MV up, V down, MM up you have VVM at the bottom. Stephanie K - Each lift movement up or down is a move; after your 6th move you have MMV at the top and MV at the bottom with V coming down. When V gets to the bottom it's nom nom nom. Tara L after MV up M down MV up you have VVM at the top. Josie T - Oh yes he does, he's a vampire not a volounteer lift operator! I think this one has got more debate than the 3 doors now.
JosieT71810 | 10 May 12
He doesn't get out the lift :)
PatrickH62447 | 10 May 12
JosieT - After your third move, VM up, there will two vampires and a maiden at the top.

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JosieT71810 | 10 May 12
I did it VV up, V down. VM up, V down (MV up, VVMM down), MM up, V down (MMM up, VVV down), VV up V down, (MMMV up, VV down) VV up. (MMM VVV) Easy 9 :)
KevinT11402 | 10 May 12
Just stopped the stupid error in my working out, 1M 2V Downstairs after the 4th move, forgot about the Vamps downstairs *sigh*
KevinT11402 | 10 May 12
I just want to argue the point for 9 being the answer.. If you start off with 1M & 1V first rather than 2V then you can cut down the moves by 2 and the maidens are never outnumbered. 1M 1V LIFT UP (1 MOVE) 1M 1V UPSTAIRS 1M LIFT DOWN (2 MOVES) 2M LIFT UP (3 MOVES) 2M 1V UPSTAIRS 1V LIFT DOWN (4 MOVES) 1V 1M LIFT UP (5 MOVES) 1V 3M UPSTAIRS 1V LIFT DOWN (6 MOVES) 2V LIFT UP (7 MOVES) 3M 2V UPSTAIRS 1V LIFT DOWN (8 MOVES) 2V LIFT UP (9 MOVES) 3M 3V UPSTAIRS
taraL12116 | 9 May 12
its definately 9! take up m+v take down m up m+v down m up m+m down v up m+v down v up v+v!
StephanieK63157 | 9 May 12
Hi it can be done in 5move: 1v+1m go up, m comes back down ( 1v up, 1m lift, 2v+2m down) 2v go up, v back down ( 2v up, 1v lift, 3m down) 2m go up, v back down ( 1v+2m up, 1v lift, 1v+1m down) 1v+1m go up, v back down ( 1v+ 3m up, 1v lift, 1v down) 2v go up ( 3v+3m up)

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megmasters | 9 May 12
I think it can be done in 9 moves too because: You take a maiden and vampire up and drop off the vampire, then only bring the maiden down, take another maiden up with her, bring only one of the maidens back down, take a vampire up with the maiden, then drop off the maiden and only bring the vampire down, he takes up a maiden with him and then comes back down along to pick up the last vampire. I didn't use formulae just a lot of arrows and letters haha:P please tell me if 9 isn't possible :)
JosieT71810 | 9 May 12
It's 9 moves. I've checked a few times. Key is taking up 2M up on the 5th move & bringing the V back down - so 3 M up the top, then the V's bring themselves up. Bam - 9.
gollomS10628 | 9 May 12
Well after a bit of messing about, I reckon it's still 11 even with a player unless a) player is a third person in the lift (not allowed - only 2 in a lift) or b) player can pop up at top or bottom(about as realistic as getting Scotty to beam 'em all up in one). In the case of a) it could be done in 5 moves as below. So I reckon it's 11 either way. Anyone got any other answers?
gollomS10628 | 9 May 12
If the player exists in the vampire game you can do it in 5. 1 VM up 2. Player comes down 3. VM up 4. Player comes down 5. VM up Done. That aint the real puzzle though. The player doesn't exist, and even if he/she did, they wouldn't be able to get to the top to do any protecting in those 9 move solutions. Might be worth trying a variation where the player does exist but you can still only have 2 in the lift and see how that works out. Off to get a pen & paper.
GuyK20625 | 9 May 12
Oh ok so im not a genius... lol

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MattB77289 | 9 May 12
PatrickH62447 | 9 May 12
TimS and JonandBertie are right. In the fox, chicken, grain question the farmer is needed to row them across the river, and when he is there he stops the fox eating the chicken and the chicken eating the grain. But the vampires and maidens have to sort out the problem on their own.
TimS11223 | 9 May 12
But the player doesn't exist in the puzzle - only the vampires and maidens.
GuyK20625 | 9 May 12
But isnt the player there to protect her? The vampires dont eat the maidens when they are being protected by the player...
jonandbertie | 9 May 12
i havent checked the wording used however i seem to remember the lift has 2 person capacity and so there isnt room for the player to accompany them....i do understand the confusion as you referenced with regard the fox, chicken, grain problem but i think this one is set slightly differently. if you are however allowed to protect the maidens during the changeovers then yes the solution would be 9 moves and your way is one of several possible solutions. if as i suspect, protection during changeovers isnt allowed then i believe there are only 2 possible solutions: 1. VM up 2. M down 3. VV up 4. V down 5. MM up 6. VM down 7. MM up 8. V down 9. VV up 10. V down 11. VV up.... or the nearly identical... 1. VV up 2. V down 3. - 11. as above

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IvanD20076 | 9 May 12
If you are correct that it is when opening the door that no maidens can be on the same floor, then it is 11, but it is not clear that that is the case. the reference is to the old farmer quiz of fox, chicken and grain - in which the farmer does indeed at some points have both the chicken and the grain on the same side for a second. it seems that alot of people have taken it the other way - in which the controlling player prevents the vampires eating the maidens, and it does say that they may not be left on the same floor. (unless im mistaken). here is my 9 move solution, please let me know if i'm wrong: 1: VV go up 2: V comes down 3: VV goes Up 4: V comes down 5: MM goes up 6: V comes down (which is when if my understanding fails - i lose) 7: MV goes up 8: V comes down 9: VV goes up
jonandbertie | 9 May 12
people dont seem to fully grasp the lift puzzle.....at no point including as the lift arrives at a floor can more vampires be present than maidens so: Luke: on your second trip up the lift opens and 2 vampires nom nom the maiden. Guy: On your 3rd trip up door opens and 3 vampires nom nom the maiden before she can go back down..... Lawrence: 2nd trip up doors open and 2 vampires nom nom the maiden
TimS11223 | 9 May 12
After move 5 you will have 3V and 1M up - she's a goner!
LukeN46227 | 8 May 12
Hi I missed the answer tv cut out. I'm sure It's possible in 7 or have I miss understood the question ok . 3V 3M . 2V up = but bring 1V down so 2V 3M bottom 1V top. 1M 1V up and 1V back down = 2V 2M bottom 1V 1 M top. 1V 1M up non down = 1V 1M bottom 2V 2 M top then brig 1V 1M up do 3M 3V up. Is this possible or have I got it wrong as I read the question as the vampires could only attack if left alone with the maidens while your at the opposite side.
AlunC33149 | 8 May 12
With the 3 door problem (HW2), don't you need to know if the game show host knows rather than randomly reveals a cuddly toy? If he knows, swap. If it's random, swap or not, same chance. I take it back you don't need to know, you should always swap and it doesn't matter whether the host knows or not.

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GuyK20625 | 8 May 12
U can do the vampire thing in 9, no? 2V up, 1V down, 2V up, 1V down, 1V 1M up, 1M down, 2M up, 1V down, 1V 1M up Done, 9 moves...
lawrenceP1947 | 8 May 12
yes it is posible to do it in 9 1:VM up 2:M down 3:VM up 4:M down 5:VM up 6:M down 7:MM up 8:V down 9:VM up. as shown the maidens are not left alone over powering the maidens.
DeepBlueC | 8 May 12
Ant..you are right, it was only a proof that 120 degrees gave the optimum solution for that particular approach. As for my comments about not being hard enough, I guess what I really meant was that I have seen most of these problems in some guise before. I was hoping for more original problems rather than more difficult ones.
gollomS10628 | 8 May 12
DavidV3322. Afraid not. 1 VM up = VM at top VVMM at bottom (OK) but 2. V down gives VVVMM at bottom (nom nom nom). also after 3 you have VVV at the bottom and 4 gives VVVM at the bottom and it's nom nom nom again.
gollomS10628 | 8 May 12
DeepBlue C - yes cracked it by a similar method (I actually used the angle top left as theta and it comes out as 30 deg so same thing). I can see why they didn't put it on the TV now!

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chris.jones | 8 May 12
Would just like to say thanks for a noble effort of a show, I'm home-tutoring a highly intelligent school refusenik and it really works for us! I'm still (un)convinced by both sides of the 3-door problem though...
DavidV3322 | 8 May 12
It is possible to do it in 9 moves 1.VM up, 2.V down, 3.MM up, 4.M down, 5.VM up, 6.V down, 7.VV up, 8.V down, 9.VV up I`ve not even started my GCSE`s... just sayin`
DavidV3322 | 8 May 12
It is possible to do it in 9 moves 1.VM up, 2.V down, 3.MM up, 4.M down, 5.VM up, 6.V down, 7.VV up, 8.V down, 9.VV up I`ve not even started my GCSE`s... just sayin`
PatrickH62447 | 8 May 12
In the solution at no time are there more vampires than maidens in any place - in RobertC's example when there are VVMM at the top and VM at the bottom, the next move should be VM down (not M down), followed by MM up. That leaves two vampires at the bottom and the three maidens and one vampire at the top. This vampire twice goes down and brings back up one of the vampires. 1. VM up, 2. M down, 3. VV up, 4. V down, 5. MM up, 6. VM down, 7. MM up, 8. V down, 9. VV up, 10. V down, 11. VV up.
RobertC54996 | 8 May 12
I didn't like the vampire one. Strictly speaking it's impossible (as the tame mathematicians first thought). The stated soln. had 3 vamps at the top and maidens at the bottom at one point. one vamp down then two maidens up leaves Top(2v+2m) Bottom(v+m) with lift at the top If a vamp takes the lift down, there are now two vamps vs one maiden at the bottom. If a maiden goes down, there are two vamps vs one maiden at the top. Only way it works is if the vamps can only attack once the lift is in transit and not the instant the doors open

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PatrickH62447 | 8 May 12
Michael1980 - You cannot have more vampires than maidens at any point. When the vampire leaves a maiden at the top and returns there are then 3 vampires and 2 maidens at the bottom.
michael1980 | 8 May 12
i think the vampire works in 9 goes, first a vampire + maiden go up then bring down the vampire then the other two maidens then bring one back down then the last maiden and vampire the bring the vampire down then all vamps. easy!!!
Captain Paralytic | 7 May 12
Anyone else notice that the sock puzzle was phrased in such a way that the answer is 2? The question stated that one only has "black and white" socks. Assuming the all the socks had the same black and white pattern, then one would only need to take 2 socks. To avoid the ambiguity, he could have said "black socks and white socks". Full marks for maths but none for English!
BrendaS74973 | 7 May 12
I can do the vampire maiden lift puzzle in 9 turns too. cheers.
BenF77062 | 7 May 12
I believe I can do the vampire maiden lift puzzle in only 9 turns.

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StephenD27225 | 6 May 12
I think thats a lot simpler to implement than mine.. Another nice one!
KevinU62200 | 6 May 12
I like your four-lock solution Stephen - here's another one! The master key for the safe is stored in a small metal box attached to the top of the safe. The box has a flat lid which is held in place by 4 padlocks, one in the centre of each edge. If any 3 of the padlocks are removed, the lid can open using the 4th lock as a hinge, revealing the master key.
StephenD27225 | 5 May 12
I've got a solution where I use two rings, four locks and four keys. The outer ring has four chains connected to the outer ring and all chains are connected in the middle like a cross in the ring. These chains are connected to the outer ring with one padlock each. The inner ring has four wire loops linked to it through which you thread each branch of chain. If any three locks are unlocked, the chain cross can slip through the wire loops and the two rings may be seperated. This works for 2 or more bankers, but there must always be one less than the total bankers to open it. The other puzzles in the series have been really hard, but enjoy the solutions anyway.. Great series!
KevinU62200 | 4 May 12
I have a less efficient but possibly interesting solution to the 4 bankers problem! This involves duplicating not the keys but the locks! Each banker has a single key, and each one opens a different lock: A, B, C or D. These are cycle-type locks, basically a loop with a padlock. The door to the safe is restrained by a chain made from these locks. The chain consists of 4 links and each link consists of 3 locks in "parallel". The first link is consists of locks A, B and C. The second link from locks A, B and D, the third from A, C, and D, and the fourth from B, C, and D. The safe is opened by removing one link of the chain - and no matter which 3 bankers are present, there is a link they can remove!
AntA63261 | 4 May 12
DeepBlueC I'm not sure you've completely proved it. You've shown the length of a particular arrangement of paths is minimised if they join at 120 degrees - that's correct, I've no problem with that. But how do you know that there is not another arrangement of paths with an even shorter distance? (There isn't but you haven't proved it).

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v.cooper | 4 May 12
In relation to DeepBlueC's comment about it not being a "hard sum" I would first of all like to say thank you for the link to nrich (a brilliant website which we have used loads). But also my class and I disagree with it not being hard enough! Our approach to mathematics is to do with developing mathematical thinking and reasoning therefore if you have to explain something to someone else using mathematical language that is the difficulty. A problem is only a problem if you cant do it straight away. Also, the word 'sum' is being used instead of equation/calculation/operation/algorithm to get everyone's attention and make mathematics accessible to as many people as possible and demystify it. Has to be a good thing right?
DeepBlueC | 3 May 12
http://i796.photobucket.com/albums/yy250/Deep_Blue_C/hut%20problem/4hutproblem.jpg
TerryB2876 | 3 May 12
Just wanted to say, I'm enjoying these programmes very much. I shall be sorry when the series ends! I guess it takes some "bottle" to try a show like this on TV? And for those appearing in it, it certainly takes courage, to be so exposed. Very entertaining. Thankyou!
DeepBlueC | 3 May 12
quote...gollomS10628 | 30 Apr 12 Anyone got a solution (not just an answer) for the 4 cow sheds problem? On the show we saw the resulting answer, and if you were very attentive you could see that it was when the lines met at 120 degrees. But is there a proof of this? I can't see how to do it or find a solution on the web.. Yes I got a proof. it involves some simple calculus...do you want to see it?..it is hard to post algebra here..
DeepBlueC | 3 May 12
An equivalent version of The chilli roulette can be found on the NRICH website under the name GOT IT. http://nrich.maths.org/1272 you can play to take the last or leave the last. Raises the question that if it is a problem for primary school...is it a "Hard Sum"?

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AntA63261 | 3 May 12
gollomS: I didn't see the programme but from what I gather it's a particular case from a class of problems known as the 'Steiner tree problem'. The general case is finding the shortest tree connecting N points. That belongs to a special class of computationally difficult problems, classed as 'NP-complete'. If you google 'Steiner tree problem' or 'Euclidean Steiner tree problem' you should get some clues.
AntA63261 | 3 May 12
PeterD Bankers A and C can open the safe. As can B and D. The whole point was that it needed 3 to open the safe and any 2 could not.
v.cooper | 3 May 12
I am a primary school teacher and teach children who are 9 and 10 years old. I am also studying on the Maths as a specialist teacher postgraduate module and loved the chilli v chocolate game! We played it in class today substituting sour skittles for the chilli and my class have some fantastic conjectures (plus photos of the headteacher playing).Can we send our work in?!
bowen.simon | 2 May 12
What you on about christophe? If he goes first, takes one, then you take one. He takes another one, then you take two... he can carry on taking one for as long as he wants, always loses. Pay attention.
christophe53618 | 2 May 12
assuming you follow the 4 rule!!!!

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christophe53618 | 2 May 12
regarding your chocolates and chilli pub trick. if you have 13 chocs and the mark goes first and takes 1 choc and continues to take 1 on his turn then you will not win and will have to eat the chilli. deeming this not quite the trick it should have been. shame on you for teaching bad maths. ha ha.x.
AllanR16452 | 2 May 12
1 lock & 4 keys Think of the lock as a cog system with 3 cogs missing and the keys as cogs. Each banker has a key (cog) each. As long as 3 bankers are there - then the system can be completed. Or think of it in computer terms with a login screen saying "Bankers line up from A - D and type in your key" and 1 input box. In this case there would be 4 possible unlock codes.
RobertC54996 | 1 May 12
You can do it with 4 keys and 4 locks. Based on boolean algebra the solution is: (A+B)CD+AB(C+D) where ABCD are locks with the keys given to the bankers (one each). By interlinking the locks this can be done physically (I have the diagram to prove this), or you use logic in the safe with 4 locks on the outside.
gollomS10628 | 1 May 12
PeterD - See my answer to Vincent.
PeterD15217 | 1 May 12
For the Banker/locks question you stated ypu need 6 locks. Surely 4 locks & 2 keys! Banker A has key 1&2, Banker B has keys 2&3, C has keys 3&4 and banker D has keys 1&4. Any three bankers have keys 1,2,3 &4.

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gollomS10628 | 1 May 12
vincentB2199. Yes you might have misunderstood it. It says in the question that no group of 2 bankers should be able to open all the locks. In your solution banker A has keys 1 and 4 and banker C has 2 and 3 so they can open all 4 (and similarly with bankers B and D).
darinM15460 | 1 May 12
This weeks homework states that you must not go back over your lines yet the solution Requires you do just that.I think the instructions need to be clearer/ more accurate. I have done similar ones easly with clear instructions. Not to go over your lines is misleading. If you want to be clever q.I level show then get the fundamentals correct
vincentB2199 | 30 Apr 12
Hello, I don't understand why the answer to the 4 bankers is 6 locks. Can't it be 4 locks as follows. Lock 1 with keys given to banker A and B Lock 2 with keys given to Bankers B and C Lock 3 with keys given to Bankers C and D Lock 4 with keys given to Banker D and A Each bank can only open 2 locks but any 3 of them can open all locks. Have I misunderstood the question? Thanks
gollomS10628 | 30 Apr 12
Anyone got a solution (not just an answer) for the 4 cow sheds problem? On the show we saw the resulting answer, and if you were very attentive you could see that it was when the lines met at 120 degrees. But is there a proof of this? I can't see how to do it or find a solution on the web.
DaveA93076 | 29 Apr 12

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AntA63261 | 29 Apr 12
PatrickH - The host opening a door which he knows not be one in front of the car gives us no information about the door chosen by the contestant, so it cannot change the probability of the contestant being right. Correct - the probability of the initial choice remains 1/3. What it tells you about is the remaining door NOT chosen by the contestant. It's probability of being correct doubles.
DaveA93076 | 29 Apr 12
Oops! My bad! I miss-heard what Dara said in the trailer. I thought he was mis-pronouncing the word Academia, when he's actually saying "a comedian". Erm...Sorry...
PatrickH62447 | 29 Apr 12
gollom - The host opening a door which he knows not be one in front of the car gives us no information about the door chosen by the contestant, so it cannot change the probability of the contestant being right.
DeepBlueC | 28 Apr 12
I cant believe we are still discussing this old chestnut! What chance we have the prisoners Dilemma or similar next week?...Come on Dara, find something original! (OK, I accept it probably isn't Dara's choice - perhaps they just open an envelope to see this week's problem...from a selection of 3..Hmmm...
gollomS10628 | 27 Apr 12
2 ways to prove the door problem. First the simple way. (C=car, T=toy throughout). There are three cases:- a) CTT b) TCT c) TTC You chose first door. Host opens one of the other two (KNOWING that it isn't the car - that's the important bit). You have a 1/3 chance if you stick (ie case a). But if you change you are guarenteed a win in either case b) or case c) because the host will always eliminate the toy leaving you with the car. The other way of proving it - the mathematical way:- Probability of first choice being car = 1/3 Probability of hosts choice being car = 0 (see important bit above) Sum of all probabilities = 1 (always) so the probability of the third door (ie the one neither of you opened) being a car is 1 - 1/3 - 0 = 2/3. Some comments say that the opening of the second door gives you no information since you knew the probabilty of a car was 0 for that door anyway. Not true - opening the door tells you WHICH door has a probability of 0 and that's the key info. Ahh my precious

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KevinD39500 | 27 Apr 12
MartinS85608 | 27 Apr 12
ChrisG97224 | 27 Apr 12
Ref to Littlewood: From lions and Christians in circular arenas to an an infinty of balls in a box where there are none: Littlewood was something special.
ChrisG97224 | 27 Apr 12
http://en.wikipedia.org/wiki/A_Mathematician%27s_Miscellany Is a better link.
ChrisG97224 | 27 Apr 12
Dara (and Marcus) you really MUST read Littlewood's Miscelany!! http://en.wikipedia.org/wiki/Littlewood%27s_law

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JonB16963 | 26 Apr 12
Hahaha, I do so love this topic, both sides have perfectly plausible arguments, I have to admit the only side I'm on is the debate, this time it had to be the 1/2 side contrary to the puzzle in order to start the ball rolling, put in the situation for real I don't know what I'd do, but I tell you this the decision to stick or switch has a 1/2 chance of being the right choice :-P lol
DaveA93076 | 26 Apr 12
Johnny Ball's "Think Of A Number" was much better!
AntA63261 | 26 Apr 12
Here's the problem I posed earlier and the solution. If you have a rectangle and you cut it along a diagonal you end up with 2 identical pieces. To specify this mathematically assume that the operation of cutting removes the points on the diagonal and for simplicity we'll assume the rectangle is open in the usual topology of R^2 so after the cut we are left with 2 open sets. Now the problem, you have an open set which looks like a 3 x 3 square with a triangular bite taken out of one of the corners with sides 1,2,sqrt(5) and out of the opposite corner a quadrant shaped bite is removed radius 1. Now cut the resultant open set into 2 identical pieces. (Hint this problem is only solvable by mathematicians).

Solution: Hopefully the lack of symmetry of the triangular cut removed from the corner should convince you that nothing straightforward will solve this. The clue ought to tell you to use something a bit more high powered.

First notice the shape is homeomorphic to the open square. Construct a space filling curve in an open square and map it into the shape using the homeomorphism. Cut along the curve and the shape falls into 2 identical empty sets.
AntA63261 | 26 Apr 12
Perhaps the problem some people have is due making wrong assumptions about the game. Besides knowing where the car is, the implicit assumptions are that the host always has to open a door (not the one you've chosen) and offer the contestant a chance to change their choice. So the host is not being malicious and offering the chance to change only when the contestant guesses correctly.

Also more subtly if the host has two choices of which door to open they choose at random. If it was known that they always chose the leftmost door when the contestant guessed correctly that would affect the probability, since if the rightmost door is opened you know with certainty that you haven't chosen correctly and you should swap.

The easiest way to convince yourself you should change is to consider the million door problem. you make a choice and the host opens 999,998 doors. The host can always open 999,998 doors showing just a cuddly toy. Are you thinking to yourself 'Wow I'm in with a chance because it's 50-50 between the door I chose and that one over there'? Or are you thinking 'I bet its behind that one over there that he mysteriously didn't open'? The chances of your initial choice being correct is unchanged, it is still 1 in 1,000,000 because the host will always be able to open 999,998 doors without showing a car.

By the same token, in the 3 door problem, after the host opens a door, the chances of your initial choice being correct is still 1 in 3. The car must be behind one of the remaining doors so the chances of it being behind other door is 2 in 3.
GeraldW46047 | 26 Apr 12
Hi re door choosing. If you're really desperate to "prove" this one way or the other (I was when I first saw the puzzle) set up your own model with three doors (and an indulgent partner to be the host)and one prize and run a series of tests. Over time you win far more often if you change your choice than if you don't.

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PatrickH62447 | 26 Apr 12
I think that MontyG agreed that when making the choice of doors the probability that the contestant chooses the correct door is one third. I do not understand why anyone should think that this probability changes when the host opens a door which he knows not to have a car behind it. The probability of the host opening a door with a car behind it was zero, so you have been no more information about the likelihood of your choice being right, and the probability that you original choice was wrong remains at two thirds.
MontyG3466 | 26 Apr 12
As I said.. neither side will be convinced by the other.. I now withdraw and leave the field open for more lucid philosophers to debate the issue.
MontyG3466 | 26 Apr 12
@AntA. Agreed. You have exerted 'some' control over which door he can open, but you have not removed his option of revealing a cuddly toy. He will always be able to do that, therefore the amount of control you have is irrelevant.
You have not changed the conditions of the second choice. It is still this door or that door. It always would have been and it always will be. (I'm sure Confucius might have phrased that better).
AntA63261 | 26 Apr 12
MontyG3466 first selection is irrelevant and therefore pointless.

It's relevant because in most cases your initial choice affects which door the host can open to reveal a cuddly toy. You have exerted some control over which door they can open. That is a very different scenario to a game where the host opens one of the doors to reveal a cuddly toy then you choose one of the others. That would be 50-50, but then you've had no control over which door is opened.

When you make your first choice, it is more likely that the car is behind one of the 2 doors which you didn't choose than behind the door that you did choose. The host then shows you that choosing one of the 2 doors would be a bad idea.
MontyG3466 | 26 Apr 12
OK.. before the math nerds correct me.. yes, you can put an unknown number into an equation (we do it all the time using x's and y's) but my point is that there is no way in which you can substitue out that unknown number and calculate a result.

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MontyG3466 | 26 Apr 12
@Patrick. As I said, the probability is irrelevant since the door won't be opened. It doesn't matter if you pick it or not. If you want then I will cede that the odds are 1-in-3 but, my argument is that that can have no bearing on the odds of your next choice. You still will not know which door the car is behind and you have only 2 doors to choose from. Your choice may have been correct or it may have been wrong but you do not know. You cannot put an unknown number into an equation. The solution presented tries to argue that you have made 2 attempts to pick the location of the car if you switch but that is only valid if each selection is validated. @Matt. Very true. This argument will run and neither side will be convinced by the other. But then that's probably what it is designed to do. The programme is, after all, supposed to be light entertainment.
MattB77289 | 26 Apr 12
I put a few posts on earlier this week as I too was a firm believer in the 1 in 2 answer. This problem is based on the Monty Hall one, which you'll be pleased to know was very devisive, in fact i'm led to believe that Nobel Prize winning physicists refused to accept the answer! So don't worry if like me you just don't get it or refuse to accept it. You're in very good company!
PatrickH62447 | 26 Apr 12
MontyG - what is the probability of your choice of door being right before the host opens one of the other two doors?
DaveA93076 | 26 Apr 12
I too am surprised that Dara would have anything to do with a programme this shoddy! I wasn't holding my breath that it would be any good since in the trailer for the show Dara talks about acamedia. Shouldn't it be Academia? Very poor!
MontyG3466 | 25 Apr 12
Well.. I am more than a little miffed. For Dara to allow his name to be associated with this is, frankly, amazing. He has always struck me as a highly intelligent man of both wit and common sense. I think he's being ill-advised (assuming he has anything to do with the questions being posed here). As to this weeks homework. What a load of rubbish! The answer has nothing to do with 1-in-3 or 2-in-3 or anything-else-in-3. The location of the car is not revealed until the end, AFTER 1 DOOR HAS BEEN OPENED. As has already been mentioned the host knows the location of the car and thus will always reveal a cuddly toy. Since the car is not revealed then the first selection is irrelevant and therefore pointless. It doesn't matter if you pick the car, a toy or make no choice whatsoever. You will always be left with a choice of 2 doors. The only selection that matters is AFTER the first door has been opened. The odds therefore become 1-in-2. A pure 50-50 gamble. Assuming they can't move the car after you pick! It makes no difference whether you stick or switch since both doors have the same odds.

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IanI50416 | 25 Apr 12
Am a little irked by this week's effort. Dara claimed to have attempted a solution to the problem over 8 pages, but his solution wasn't given any air-time. Is it on the web-site somewhere?
MartinS85608 | 24 Apr 12
The answer to the kissing problem is only 5 if you approximate the people as points. Dara is a bit bigger than that so he can have 6 women around the circle and still be closer to them all than they are to each other.
AdamH14582 | 24 Apr 12
View the problem like this: The door you choose has 1/3 probability of having the car and both of the other doors have a 2/3 combined probability. Now the host opens one of the combinesd two doors without the car behind it, meaning one of the two doors you haven't picked has probability zero and the other door 2/3. Now your original door has 1/3 chance of the car but the other door that's left has a 2/3 chance to have the car behind it so you are better off swapping. Google Game Theory for more information.
PatrickH62447 | 24 Apr 12
AntA is right. The question for MattB and JonB is "Why should the probability of your first pick being right change just because the host has opened a door which he knew not to be in front of the car?" He has not given you any more information about your choice, because you already knew that one of the other doors was wrong.
AntA63261 | 24 Apr 12
MattB77289 I think the problem is you're trying to change the probability of your initial choice. You can't do that - the car doesn't change doors to move itself among the remaining doors. It was made when there was 3 (or a million) doors to choose from. Opening up every other door apart from one makes that remaining door far more likely to have it - because in almost every single case it will be. With the million doors example, almost every single initial choice you make wont have the car. So after host shows 999,998 of the remaining doors to be empty (carefully avoiding the one that they know has the car) what can we say? Whenever the car isn't behind the initial door choice (and that is virtually every time), it will be behind the remaining door. So the remaining door is overwhelming likely to have the car.

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MattB77289 | 24 Apr 12
I agree with JonB16963. When the gameshow host opens the door (or doors depending on how many you wish to use as an example), he effectively resets the game and all bets are off. From that point onwards we have a choice of two doors, not 3 or even a million. So we have a 1 in 2 change of finding the car. I can't pick from a choice of 3 doors any more because one of my choices has been removed by the host. So the solution can't be (1 in 3) or (2 in 3), because the 3 options no longer exist. Anyway what would I know, I'm not a maths professor at Oxford.
AntA63261 | 24 Apr 12
JonB16963. The solution is correct. Consider the case where there a million doors and only 1 with a car. Your first choice has a probability of 0.000001. Then let the host open 999,998 doors. It doesn't make your first guess more likely to be right it still has a probability of 0.000001. There a 2 doors left and it must be behind one door or the other, so the sum of the two probabilities must be 1. So the probability that it is behind the other door is 0.999999. Your horse racing analogy doesn't work because it's like the race has already happened. The car has already been hidden behind one of the doors (and it doesn't move to redistribute itself among the remaining closed doors).
JonB16963 | 23 Apr 12
OHH MY GOD !!! I've just been through the 10 questions in the School Of Hard Sums. Got question 7 wrong, factorising quadratic equations, working out right, but on my small scrap of paper I multiplied by a 36 instead of 38 :-( Question 9 wasn't even a maths question, and Question 10 tries to use my first comment here about looking at ALL possible outcomes for a probability, but fails completely in that in this situation girl+boy is the same thing as boy+girl, once you've been told 1 is a boy, there's only 2 outcomes, mixed gender twins or same gender twins (both boy), maybe if you look into the sequence of delivery you could get 3 outcomes, but there's only 2 for the final gender state of the twins as the question asks ! Not impressed !!
JonB16963 | 23 Apr 12
lol, fascinating thing is it's the same length if your sphere is the earth or a golf ball !
KevinD39500 | 23 Apr 12
Teacher pet Jon :) 2 pie was the answer. It was rope around the earth but I changed it to a sphere to stop the moans about the moutains etc lol

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JonB16963 | 23 Apr 12
sorry 2 x pie so 6.28 you've added 1m at both ends of the diameter
JonB16963 | 23 Apr 12
and in answer to kevins problem 'pie' and the problem I heard was a rope around the earth, it makes you think much bigger when the added length is only 3.14 meters
JonB16963 | 23 Apr 12
I want to point out the solution to todays homework puzzle is wrong !! and it baffles me that so many intelligent mathematicians seeem to think it's true. You can't maintain probability after the criteria that created it have changed, it would be like continuing to offer 20/1 odds on a horse after all the other runners have been shot ! I can prove it mathematically for you ... Using your game show analogy, once the host opens a door to reveal a toy, there are only 2 doors, 1 car and 1 toy. Where your proof fails is in the first line in which you pick the door with the car, the host opens door no.2 to reveal a toy but what your proof doesn't show is the 4th permutation where you pick door 1 with the car and the host opens door 3 to reveal the other toy. Factor the 4 possibilities instead of your 3 and you get the true probability of 1/1 or 50/50 or 6/'half a dozen' call it what you like but it's evens.
KevinD39500 | 23 Apr 12
Great show. Here a problem a work colleage set me today and the answer was surprising. Imagen you tie a rope around the a sphere. now put 1m polls around the same sphere, How long will you need to increase the length of the rope to loop the sphere and polls?
KatyD72440 | 23 Apr 12
Is there anywhere that we can get worksheets for the episodes? It would have been so handy to have the grid worksheet with the people on it for today's show!!

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TonyH824 | 19 Apr 12
The second puzzle in Radio Times (page 21 of 14-20 April), "Move the Matchsticks", has another solution: 6 - 25 = -19. You have to move them around but the number of matchsticks is the same.
IanN9239 | 19 Apr 12
SteveB73332 | 19 Apr 12
AntA63261 | 19 Apr 12
Here's a problem for serious mathematicians. If you have a rectangle and you cut it along a diagonal you end up with 2 identical pieces. To specify this mathematically assume that the operation of cutting removes the points on the diagonal and for simplicity we'll assume the rectangle is open in the usual topology of R^2 so after the cut we are left with 2 open sets. Now the problem, you have an open set which looks like a 3 x 3 square with a triangular bite taken out of one of the corners with sides 1,2,sqrt(5) and out of the opposite corner a quadrant shaped bite is removed radius 1. Now cut the resultant open set into 2 identical pieces. (Hint this problem is only solvable by mathematicians).
AntA63261 | 19 Apr 12
Sometimes even simple problems in probability have no satisfactory answer. For example - this is a simple game where you toss a coin until you get heads, whereupon the game ends. If heads comes up on the first throw you get £2, if the first head appears on the second toss you get £4, if the first head appears on the third you get £8, if the first head appears on the forth £16 and so on. So the amount doubles every time a tail appears. Your job is to work out what would be a fair price to charge for playing this game. Obviously you always win at least £2 so you would have to charge more than that. In fact the probability of winning £2 is 1/2 and of winning £4 is 1/4, and of winning £8 is 1/8 etc. Adding all this up the fair value would seem to be infinity! Yet no one in their right mind would pay £1000 to play the game. This is known as the St Petersburg paradox.

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David R - UKTV | 18 Apr 12
GrahamB95253 - yes, they're going to be changing the episode to Ep 1 on Anytime.
IanN9239 | 18 Apr 12
Recorded the first(?) show a few nights ago, and just got round to watching it. I have one comment on the "four stars" problem, which I believe was wrong as phrased. It was framed in terms of building "a constellation" of four stars, equidistant from each other. That was clearly meant to be mere, colourful description, but it changes the problem. A key thing about constellations is that they're not about where stars actually ARE (where they were when the light left, if you really want to get picky), but by where they APPEAR to be - i.e a constellation is a projection from 3 dimensions onto the surface of a sphere with the observer at its centre. Simply putting four stars in any old tetrahedron doesn't work (with one exception); the resulting projections can't meet the conditions of the problem, any more than four points in a plane can. The exception occurs when the projected points are uniformly distributed around the sphere, forming a tetrahedron with the observer at the centre. The four stars can then be at any arbitrary, large distance in their respective directions. Mind you, it takes a real stretch of the terminology to call such an arrangement a "constellation" (latin, loosely, for "grouped stars"); a more honest answer is that there are no solutions to the problem as posed.
GrahamB95253 | 18 Apr 12
Thanks David R - UKTV - I have a solution that works but wanted to check it. shame the wrong episode is on anytine as I will now have missed episode 1 along with any other watchers on anytime - do you think we could get Episode 1 added to anytime / anytime+ ? Thanks
David R - UKTV | 18 Apr 12
Hi GrahamB95253 - it appears that the wrong episode was shown on Sky Anytime. The episode that went out was Ep 5 rather than Ep 1. This has been corrected. The 'wrong change' puzzle will be on the website on 14th May when Ep 5 goes out so you'll be able to find out the solution then.
StuartM87209 | 18 Apr 12
I have just caught up with Dara's new show on anytime so not sure if anything has been updated to say that the answer of 7 moves for the truffle oil puzzle should be 6? Either I have made a very big mistake or the show has now made me feel extremely bright......I am going to go with bright.

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PhilD6226 | 18 Apr 12
In fact it can be many more if the circle of the centre person touches each of the other people the number nearer to them is only limited by how many really thin people you can get around the middle one before they touch each other
PhilD6226 | 18 Apr 12
In the real world the answer to the people at a dance problem can be 6. If the person in the middle is of a large diameter and the six people around the edge are all thin and stand at 90 degrees to the centre then their nearest point will be neareer to the middle person than to each other. The answer of five assumes all personsd are the same size or have no dimensions at all!
GrahamB95253 | 18 Apr 12
It was shown at the end of the show - Watched on Sky Anytime / Anytime+ A waiter gives you change but he has swapped the pounds for pence and the pence for pounds - having spent 50p you now have exactly twice the amount of change you should have been given by the waiter - What was the actual ammount of change you should have received? Twins again - if the orger of the twins is important then surely the options remaining should be B-G, G-B, B1-B2, B2-B1 4 options remaining and 2 each for a boy or a girl and again we are at 50 50 for the other child being a boy.
GrahamB95253 | 18 Apr 12
It was shown at the end of the show - Watched on Sky Anytime / Anytime+ A waiter gives you change but he has swapped the pounds for pence and the pence for pounds - having spent 50p you now have exactly twice the amount of change you should have been given by the waiter - What was the actual ammount of change you should have received? Twins again - if the orger of the twins is important then surely the options remaining should be B-G, G-B, B1-B2, B2-B1 4 options remaining and 2 each for a boy or a girl and again we are at 50 50 for the other child being a boy.
David R - UKTV | 18 Apr 12
GrahamB95253 - where exactly did you see the "waiters change mistake"? The homework puzzle shown in the first show (just before the credits) was the draw 4 lines through 9 dots that we have up on the website.

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RichardJ61915 | 18 Apr 12
Patrick: Thanks for the reply; my point was that they are similar in that the probability should change when the information is given, because both present what appears to be two possibilities where in fact there is only one. With the boxes, if the prize is behind door A, and the player chooses door A, the host can open door B or door C (two possibilities) whereas in practice it makes no difference (as only one non-prize door is opened). With the twins, though there are four possibilities (B:B, B:G, G:B, G:G) if you are then told one is a boy, that discounts option 4 (G:G), as stated; but it also discounts option 3(G:B) if twin one is the boy, and option 2(B:G) if twin two is the boy, leaving just two options in each case. You can’t count the Boy+Girl option twice in this scenario. But (confusingly!) the given answer (the probability remains one in three) is different. I had thought the idea behind this was to illustrate the MH paradox in a fresh way.
CarolineR55277 | 18 Apr 12
@SteveB73332 Yes, I agree with the answer of five, just think he made a slip-up in his drawing and talking about it... good to know we can all make mistakes!!
GrahamB95253 | 18 Apr 12
Hi Liked the show but I do have a couple of questions. 1. Where can I check the answer to the question on the waiters change mistake that was shown on screen at the end of the program? 2. The twins question - you say that the comment you have a boy rules out the girl girl combination leaving three combinations but it also invalidates one of the other possibilities b g or g b although we do not know if is b-g or g-b that has been invalidated so we are actually down to 2 possibilities not 3 so the possibility of the second child being a boy would be 50 50.
SteveB73332 | 18 Apr 12
Sorry my bad, just realised that he pointed out the 6th would be too close to the first
SteveB73332 | 18 Apr 12
@Caroliner55277 I think you may be right Caroline, you should be able to fit 5 triangles of 61' into a circle (5x61=305) leaving a gap of 55' and after drawing this out I get the answer to be 6 not 5?????

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AntA63261 | 18 Apr 12
@OlwenR44550 That wont give you more than 5. The people on the outer circle will be closer to someone halfway in than the person in the middle.
ChrisP73178 | 18 Apr 12
I have a bone to pick with my homework! The solution doesn't meet the terms of the puzzle. 'Can you connect all 9 dots, with only 4 straight lines, without taking your pen off the paper AND WITHOUT GOING BACK OVER YOURSELF' is impossible. technically, you need to go back over one line that you have already drawn. The question should read '...without visiting any dot more than once?' It's a language thing, but if you're going to be pedantic about mathematics, allow us language geeks to be pedantic about the wording of your questions!
OlwenR44550 | 18 Apr 12
No, forget that, eventually copped that the others on the same line would be nearer to each other than to me in the middle. Off to hit the head against the wall a few times . .
OlwenR44550 | 18 Apr 12
The kiss the girls puzzle seemed so wrong it was weird. If u have a radius of say 30 ft (big hall) then just draw another smaller circle at less than half that radius, use the same angles and put someone on each line at that point, another circle at say 6 ft and you'll fit 5 more people to kiss u, maybe even another 5 thin people at 2.5 ft away. Why not?!
DavidF61394 | 17 Apr 12
I think GrahamT25697 is right. I came up with the same solution and was about to add a comment when I saw his. (There were 4 mathematicians not 3 though: Dara, Marcus & the 2 students.)

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CarolineR55277 | 17 Apr 12
KISSING PROBLEM I might be wrong, but when Marcus du Sautoy said you could only draw four triangles surely he meant five?!
GrahamT25697 | 17 Apr 12
I could be wrong buut the puzzle with the 3 jugs of truffle oil can be solved in 6 moves... starts out 8-0-0 move 1 3-5-0 move 2 3-2-3 move 3 6-2-0 move 4 6-0-2 move 5 1-5-2 move 6 1-4-3 i assume i must be wrong or there was another element to the question that i may have not heard because i would expect at least one of the 3 mathematicians to have come up with that if I did!
PatrickH62447 | 17 Apr 12
Richard, the twins (and coins) problems are not versions of the Monty Hall problem. That problem (3 doors, a car behind one, contestant chooses a door, host opens another door, should contestant swap?) needs someone who knows where the car is to open the other door after a choice is made; the probability of the choice being right first time was and has to still be one third, so the probability of the car being behind the remaining door must be two thirds. For the twins problem it would be better stated if the doctor had said "You do not have two girls". That would have removed one of the four equally likely possibilities so there would now be three equally likely possibilities and the probability of twin boys would now be one third.
RichardJ61915 | 17 Apr 12
The twins problem (and the coins problem) are versions of what's called the Monty Hall problem, which Ant just outlined (and which Marcus du Sautoy illustrated in "The Code") and which causes the same kind of arguments. What's interesting is the solution given here is the opposite of the solution to the MHP/ boxes problem. With the boxes, the probability drops from 1 in 3 to 1 in 2; with the twins the answer stated is still 1 in 3. So, are you sure about this?
EdGa.Jr | 17 Apr 12
Dara's homework says you can't go back over your lines, yet the answer involves lines crossing...

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wyki123 | 16 Apr 12
you can complete dara's homework in three lines, it dosen't say there mathematical points so you can go through at an angle
AntA63261 | 16 Apr 12
Probability is full of subtle problems. Here's a similar one. There are 3 closed boxes A,B and C, one of which has £1 million the other two are empty. If you guess correctly you get to keep the money. You make a guess (say B) and one of the boxes you didn't guess is opened (say A) showing it to be empty. You are then given the chance to change your choice of box (which in this example would be from B to C). Is it worth changing your choice? Which box is more likely to contain the money B or C?
[Answer is C is more likely to contain the money than B so you should change your choice.]
GeraldW46047 | 16 Apr 12
@DavidR-UKTV and AntA3261. Thank you both, I now understand. I look forward to being further baffled, however, when I actually watch the show tonight.
AntA63261 | 15 Apr 12
@GeraldW46047 In the case of 2 coins, each of the following are equally likely. coin1=H coin2=H, coin1=H coin2=T, coin1=T coin2=H and coin1=T coin2=T. However we are given the information that at least one of the coins is H, therefore the last option above can be ruled out and we are left with 3 possibilities each of which is equally likely. These 3 possibilities are: coin1=H coin2=H, coin1=H coin2=T and coin1=T coin2=H The chance of the other coin being heads is therefore 1 in 3 and the chance of it being tails is 2 in 3. This is quite different from the case where you turn over one of the coins and discover its heads - in that case the probability of the other being head is 1 in 2. The fact that you haven't specified which coin is heads in Q8 is what makes the problem subtly different.
RobM97558 | 15 Apr 12
Really looking forward to Dara's new vehicle, big Dara fan me, Will Frankie Boyle guest?...Whats happened to Frankie Boyle,anyone? has he been assinated by Susan or those geezers from The Proclaimers?, come back Frankie, the voices in my head miss you > Peace <

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Damien Dodd | 14 Apr 12
Can't wait to see this show. Love Dara and love puzzles - a perfect match!
GeraldW46047 | 13 Apr 12
OK but the question does rather introduce a medical slant. So let's take it out of the medical realm. Same question using inanimate objects with a 50/50 chance. I have in front of me two coins. One is heads. What is the probability the other is also a head? Please don't tell me it's one in three because the chance of it being tails would therefore be two in three which surely can't be the case.
David R - UKTV | 13 Apr 12
No - it's a mathematical puzzle. We assume the gender of each twin is 50/50 within the puzzle - this isn't based on any medical data.
GeraldW46047 | 13 Apr 12
Thanks for your patience. So the probability that the second child will be a girl is in fact two in three? Is this borne out by data on actual twin births?
David R - UKTV | 13 Apr 12
It does indeed relate to the combination. The doctor hasn't told you which twin is the boy. He’s only told you that one of the twins is, in fact, a boy. So you have to consider the three scenarios of boy + boy, boy + girl, girl + boy. The only scenario that doesn’t apply is girl + girl as the doctor has told you that one of the twins is a boy.

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GeraldW46047 | 13 Apr 12
But "girl+boy" and "boy+girl" are not different outcomes and anyway the question relates to the sex of the unknown child not the actual combination. The second twin can only be a boy or a girl there is no third option.
David R - UKTV | 13 Apr 12
We know it's a prickly question but the sex of the first twin actually is relevant. Even though one of the twins has been identified as a boy there are still three probabilities. It's a counterintuitive question and uses what's called Bayesian mathematics. Here's the answer solutions in more detail. http://uktv.co.uk/dave/article/aid/651809
GeraldW46047 | 12 Apr 12
Answer to Q8 of Dara's brainteasers does not make sense. Sex of first twin irrelevant. Chance of second twin being boy 50/50 or 1 in 2 - same as tossing a coin. tell me where I'm going wrong please.
JoannaD16321 | 21 Mar 12
Sounds weird but brilliant. ! I know from following Dara on Twitter that he loves his maths so can't wait to see this.