Pursuit

A problem from Sam Loyd’s Cyclopedia of Puzzles, 1914:

Here is the puzzle of Tom the Piper’s Son, who, as told by ‘Mother Goose,’ stole the pig and away he run. It is known that Tom entered the far gate shown at the top on the right hand. The pig was rooting at the base of the tree 250 yards distant, and Tom captured it by always running directly towards it, while the pig made a bee-line towards the lower corner as shown. Now, assuming that Tom ran one-third faster than the pig, how far did the pig run before he was caught?

Intriguingly, Loyd adds, “The puzzle is a remarkable one on account of its apparent simplicity and yet the ordinary manner of handling problems of this character is so complicated that solvers are asked merely to submit approximately correct answers, based upon judgment and common sense, just to see who can make the best guess. The simple rule for solving it, however, which will doubtless be quite new to our puzzlists, is based upon elementary arithmetic.” What’s the answer?

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Loyd’s answer: In the puzzle of Tom the Piper’s Son, who stole the pig, Mother Goose reports that Tom had to run 571 3/7 yards to catch the pig, while the porker ran three-quarters of that distance, which would be 428 4/7 yards. The simple rule for solving problems of this class is to halve the distance the man would have to travel to catch the pig in a straight line added to the distance he would go if they advanced toward each other. Tom is 250 yards from the pig, and as he goes one-third faster, in a straight line, he would catch him in 1,000 yards. If they run toward each other, Tom goes four-sevenths of the distance, viz.: 142 6/7 of the 250 yards, which, added to 1,000, equals 1,142 6/7 yards. Half of this equals 571 3/7 yards, which is the correct answer. Loyd doesn’t really explain his “simple rule,” but in the Pi Mu Epsilon Journal (Volume 6, Number 9, Fall 1978, page 545), Tom Apostol of Caltech shows that Loyd’s method is generally valid. The editors add, “The charm of this problem lies in the insidious way the stated result defies our intuitive notions. At first we suspect that the result, if true, might accidentally apply only to the special case involving the given distances and the relative speeds but then the denouement affirms the generality of Loyd’s method of solution.” In his edited collection of Loyd’s puzzles, Martin Gardner added, “The man’s path forms one of the simplest of ‘pursuit curves,’ the study of which forms an interesting branch of what one might call ‘recreational calculus.'”

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September 26, 2020September 25, 2020 | Puzzles · Science & Math

Black and White

Smyth-Helms 1915

In 1915 James Ferguson Smyth invited Hermann Helms to play a practice game at the Manhattan Chess Club. They had reached the position above when Helms, as Black, found a brilliant two-move mate. What is it?

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22. … Qg2+! 23 Kxg2 Rxg3# The whole game is here. Helms’s brilliancy was so inspiring that Anthony Santasiere produced an oil painting of the position, which for many years hung in the Marshall Chess Club.

September 22, 2020September 21, 2020 | Puzzles

Red Square

A problem from the 1999 Russian Mathematical Olympiad:

Each cell of a 50×50 square is colored in one of four colors. Show that there exists a square which has cells of the same color as it directly above, directly below, directly to the left, and directly to the right of it (though not necessarily adjacent to it).

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A 50×50 square contains 2500 cells. At least a quarter of these, 625 cells, must be of the same color, say red. Of these 625 red squares, at most 50 can be the uppermost red squares in their columns; at most 50 can be the lowermost red squares in their columns; at most 50 can be the leftmost red squares in their rows; and at most 50 can be the rightmost red squares in their rows. Together these make up at most 200 squares. The remaining 425 (or more) red squares have none of these distinctions, so each must have a red square directly above, directly below, directly to the left, and directly to the right of it. From Titu Andreescu and Zuming Feng, Mathematical Olympiads 1999-2000: Problems and Solutions from Around the World, 2002.

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September 20, 2020September 19, 2020 | Puzzles

A Perfect Square

An old problem from the Soviet Mathematical Olympiad:

Find the 4-digit number aabb that is a perfect square.

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The smallest 4-digit perfect square is 1024 (32²). The largest is 9801 (99²). And aabb = 1000a + 100a + 10b + b = 1100a + 11b = 11(100a + b). So 11 is one of the factors of aabb, and aabb is of the form (11k)². That leaves only six candidates to check (33², 44², 55², 66², 77², 88²), and it turns out the answer is 88² = 7744. (Thanks, Danesh.)

September 15, 2020September 14, 2020 | Puzzles

Competing Squares

competing squares

A square has been inscribed in each of two congruent isosceles right triangles. Which square is larger?

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Divide each figure into a grid of triangles. This shows that the square on the left occupies 2/4 of its triangle, while the square on the right occupies 4/9 of its triangle. So the square on the left is about 12.5 percent larger. (Duane Detemple and Sonia Harold, “A Round-Up of Square Problems,” Mathematics Magazine 69:1 [February 1996], 15-27.)

September 10, 2020October 11, 2020 | Puzzles

Black and White

tchepizhni chess problem

This problem, by V. Tchepizhni, won fifth prize in the Bohemian Centenary Tourney of 1962. It’s four puzzles in one:

(a) The diagrammed position.
(b) Turn the board 90 degrees clockwise.
(c) Turn the board 180 degrees.
(d) Turn the board 270 degrees clockwise.

Each of the resulting positions presents a helpmate in two: Black moves first, then White, then Black, then White, the two sides cooperating to reach a position in which Black is checkmated. What are the solutions?

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(a) 1. c1=R Rxg5 2. Rc3 Bc2# (b) 1. b5 Bc3+ 2. Kc5 Ba5# (c) 1. b3 Rb4 2. f6 Bf7# (d) 1. g2 Bf4+ 2. Kf2 Bh2# (From John M. Rice, An A.B.C. of Chess Problems, 1970.)

September 5, 2020September 4, 2020 | Puzzles

A Stressful Game

A puzzle by Ezra Brown and James Tanton:

Three gnomes sit in a circle. An evil villain puts a hat on each gnome’s head. Each hat is either rouge or puce, the color chosen by the toss of a coin. Each gnome can see the color of his companions’ hats but not of his own.

At the villain’s signal, all three gnomes must speak at once, each either guessing the color of his own hat or saying “Pass.” If at least one of them guesses correctly and none guesses incorrectly, all three gnomes will live. But if any of them guesses incorrectly, or if all three pass, they’ll all die.

They may not communicate in any way during the game, but they can confer beforehand. How can they arrange a 75 percent chance that they’ll survive?

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Each gnome will pass unless he sees that both of his companions are wearing hats of the same color. In that case he will guess the opposite color. This produces a win in six of the eight possible arrangements of hats. (Ezra Brown and James Tanton, “A Dozen Hat Problems,” Math Horizons 16:4 [April 2009], 22-25.)

August 27, 2020August 26, 2020 | Puzzles

Black and White

glass chess problem

By F.G. Glass, 1903. White to mate in four moves. (Easier than it sounds.)

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If the white king just circles the pawn, all Black’s moves are forced: 1. Kf1 Kd2 2. Kf2 Kd1 3. Ke3 Ke1 4. Rc1#

August 15, 2020August 14, 2020 | Puzzles

Inside Business

A puzzle from MIT Technology Review, September-October 1999: A popular pastime at Bell Labs in the 1970s was to identify a word that contains a given string of letters — for example, OOKKEE is found in BOOKKEEPER. Michael Foster found English words that contain HIPE, HCR, and UFA. What are they?