This page contains the best puzzles that I have found over many years of searching plus a few original ideas of my own. I offer them in the hope of being offered similar puzzles in return. The puzzles are not all original although I have extended some of them and some of the solutions are mine. My address is:
"dick" followed by "@" then "mod2" and finally ".org "This is an original puzzle.
Lord Emsworth has a large circular flower bed in his park, whose centre is marked by a single tall rose stem. It is divided into two parts by a straight line through the centre. He now wishes to divide it into four equal quadrants with a second diameter at right angles to the first. The gardeners can mark straight lines by sighting along lines of pegs and hammering in a fresh peg where two lines cross, but cannot measure distances or draw any kind of circular arc. How can they find the second diameter?
I have just been reading a biography of the Indian mathematical genius, S. Ramanujan, The Man Who Knew Infinity: A Life of the Genius Ramanujan, and the book has given me several puzzles. This is a puzzle he posed in an Indian journal and received no solutions. The book does not give his solution so I have provided one.
Let sq(x) imply the positive square root of a positive number.
Consider a sequence of functions as follows:-
f1(x) = sq(1+sq(x))
f2(x) = sq(1+sq(1+2*sq(x)))
f3(x) = sq(1+sq(1+2*sq(1+3*sq(x))))
.
.
fn(x) = sq(1+sq(1+2*sq(1+3*sq(... (1+n*sq(x))..)))
Evaluate this function as n tends to infinity.
While Ramanujan was at Cambridge, a friend posed a problem to him which was taken from the "Strand magazine". Ramanujan answered immediately, giving a continued fraction, whose convergents were the general solution to the problem. His friend was astounded. How, he asked Ramanujan, had he done it? He replied: "Immediately I heard the problem, it was obvious that the answer was a continued fraction; I then thought, 'Which continued fraction?', and the answer came to my mind!"
Slightly shortened, the problem was as follows.
A friend of mine lived in a long street with houses numbered one, two, three, and so on. The numbers of the houses below him added up to exactly the same as the numbers above him. There were between fifty and five hundred houses in the street. What was the number of my friend's house?
This is a very old puzzle which I was shown when I first started work just after the war. It was rumoured to have been created by the Germans during the war and circulated in British defence establishments to waste engineers' time. It is very frequently posted in various user groups. I include it because of my extra requirement, which is that you mark the coins in weightless chalk and declare the weighings in advance.
You are given 12 identical looking coins, and a balance type weighing machine. 11 of the coins are identical. The twelfth differs in weight and may be heavy or light. You may make 3 weighings, putting coins in either scale pan.
Find which coin is different and whether it is heavy or light.
Find the area of the largest ellipse that can be inscribed in a triangle with sides 5cm, 12cm and 13cm.
Find the area of the smallest ellipse that can be drawn through the vertices of a triangle with sides 5cm, 12cm and 13cm.
This puzzle dates from 1850 and is quoted in W.W. Rouse Ball's book, Mathematical recreations and essays. He devotes a chapter to finding solutions without a computer and I shan't repeat his methods. However, I invite programmers to find and count all the solutions, excluding permutations.
A School Mistress was in the habit of taking her pupils on a daily walk. There were 15 of them and they walked in five groups of three. The problem is to arrange them so that, for seven days, no girl will walk with any of her school-fellows in any triplet, more than once.
This is based on a section in R.K. Guy's book Unsolved Problems in Number Theory (Problem Books in Mathematics / Unsolved Problems in Intuitive Mathematics). The solver is asked to find 9 points, no more than 4 on any straight line or any circle, such that the distance between any pair of points is rational.
This book offers a solution by J. Leech. I have extended this to require a family of solutions (excluding re-scalings and rotations).
An explorer is at the edge of a desert 800 miles wide. He has a truck, but, when full of petrol, it only has a range of 500 miles . He may however, drive out into the desert, dump some petrol for later use and drive back. Assuming no losses from evaporation, spillage, etc., what is the smallest number of 500 mile loads he must use to cross the desert?
This puzzle has a long history. Lewis Carroll's diaries have a reference:
"Up all night with a new puzzle, just in from New York. The problem is to find 3 right-angled triangles with integer sides and equal areas. I could two but not three!"
H.E. Dudeney, in his book The Canterbury Puzzles (Illustrated Edition) (Dodo Press), uses the same idea in a puzzle called 'The Four Princes' which asks for four such triangles, and, in his solution, gives a formula which allows the finding of as many such triangles as are required. He claims to have discovered the formula himself. However, T.L. Heath, in his translation of the works of the Greek Mathematician Diophantus, attributes the same formula to the 16th century French mathematician Pierre Fermat.
None of these authors could find all the solutions, or prove that their solution is the smallest. I invite programmers to list all such triangles with area less than 1 million square units and thus find the smallest set of 4 integer sided right angled triangles with equal area.
(Sir Arthur Eddington was a famous mathematician who wrote many papers on Relativity.
The animal names are taken from Lewis Carroll's delightful nonsense poem, "Jabberwocky").
I took some nieces and nephews to the zoo, and we halted at a cage marked
Tovus Slithius, male and female
Borogovous Mimsius, male and female
Rathus Momus, male and female
Jabberwockius Vulgaris, male and female
The eight animals were asleep in a row and the children began to guess which was which. "That one at the end is Mr. Tove." "No, no! It's Mrs. Jabberwock", and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right.
As the four species were easily distinguishable, no mistake would arise in pairing the animals; naturally, a child who identified one animal as Mr. Tove identified the other animal of the same species as Mrs. Tove.
The keeper, who consented to judge the lists, scrutinised them carefully. "Here's a queer thing. I take two of the lists, say John's and Mary's. The animal which John supposes to be the animal which Mary supposes to be Mr. Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs. Tove. It is just the same for every pair of lists, and all four species.
Curiouser and curiouser! Each boy supposes Mr. Tove to be the animal which he supposes to be Mr. Tove; but each girl supposes Mr. Tove to be the animal which she supposes to be Mrs. Tove. And similarly for the other animals. I mean, for instance, that the animal Mary calls Mr. Tove is really Mrs. Rath, but the animal she calls Mrs. Rath is really Mrs. Tove."
"It seems a little involved." I said, "but I suppose it is a remarkable coincidence."
"Very remarkable," replied Mr Dodgson (whom I had supposed to be the keeper) "and it could not have happened if you had brought any more children."
How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right?
This puzzle is based on one by Hugh ApSimon. I don't have his book to hand so I must improvise.
The villages of Amwell, Boxford and Crofton lie on a plain. Each has a church with a tower and the views are extensive. A feature of the plain are four old trees, an ash, a beech, a cypress and an oak.
It is well known that, from the church tower at Amwell, the oak partly obscures the tower at Boxford, the ash partly obscures the tower at Crofton and the beech is in line with the cypress.
From the tower at Boxford, the cypress partly obscures the tower at Crofton and the beech is in line with the ash.
From the tower at Crofton the beech is in line with the oak.
Three surveyors were told to survey the area, but were not quite sure what to measure.
Their first thought was that they should measure the area of the triangle formed by the Oak - Beech - Cypress; so they did. It measured 6 acres.
Their second thought was that they should measure the area of the triangle formed by the Ash - Beech - Cypress; so they did. It measured 8 acres.
Their third thought was that they should measure the area of the triangle formed by the Oak - Beech - Ash; so they did. It measured 10 acres.
When they got back, their boss didn't want to know anything about trees at all. He wanted to know the area of the triangle formed by the three villages.
No problem said the surveyors. It is...???
The next two puzzles are original ones that I hope readers will enjoy.
ABCD is a quadrilateral whose diagonals cross at right angles at O. Angle OAB = 40°. Angle OBC = 30°. Angle OCD = 20°. Find angle OAD and prove that it is an exact number of degrees.
Triangles XAB and YAB have the same area. Angle XAB = 30°. Angle XBA = 40°. Angle BAY = 80°. Find angle ABY and prove that it is an exact number of degrees.
There are a surprising number of puzzle pages on the web. I very much enjoy the 'Problem of the week', posed by Stan Wagon at Macalester College, St Paul, Minnesota. These are available at Macalester College Problem of the Week
Another excellent page is Ken Duisenberg's Puzzle of the week, which also has a very comprehensive list of other references.