A Mathless Math Puzzle

$hess bug puzzle$

Richard Hess posed this problem in the Spring 1980 issue of Pi Mu Epsilon Journal. At noon on Monday, a bug departs the upper left corner, X, of a p × q rectangle and crawls within the rectangle to the diagonally opposite corner, Y, arriving there at 6 p.m. He sleeps there until noon on Tuesday, when he sets out again for X, crawling along another path within the rectangle and reaching X at 6 p.m. Prove that at some time on Tuesday the bug was no farther than p from his location at the same time on Monday.

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Well, suppose there are two bugs making their journeys simultaneously — one departs X as the other departs Y, each headed for the opposite corner, and they travel different paths within the rectangle, arriving at their destinations simultaneously. At some time during their journeys they’ll both be the same horizontal distance from the left side of the rectangle. Since both paths remain within the rectangle, at that moment the distance between them can’t be greater than p. The editors received this answer from six readers, including Hess. “Their solutions were characterized by the complete absence of mathematical symbols and mathematical jargon,” they noted. “While there is certainly no objection to mathematical solutions to mathematical problems, a simple word-solution intelligible to any layman is to be preferred. Some of the other submitted solutions were profuse with subscripts, coordinates, inequalities, vinculi, functional relations, intermediate value theorems, Greek symbols, graphs, continuous functions and derivatives — all reminiscent of the sledgehammer method of swatting a fly.” The original problem is here (PDF, page 135).

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March 19, 2017July 17, 2017 | Puzzles

Pentalpha

During a visit to Crete in 1938, Miss L.S. Sutherland described a game she saw played on a pentagram:

You have nine pebbles, and the aim is to get each on one of the ten spots. You put your pebble on any unoccupied spot, saying ‘one’, and then move it through another, ‘two’, whether this spot is occupied or not, to a third, ‘three’, which must be unoccupied when you reach it; these three spots must be in a straight line. If you know the trick, you can do this one-two-three trick, for each of your nine pebbles and find it a berth, and then you win your money. If you don’t know the trick, it’s extremely hard to do it.

To make this a bit clearer: The figure has 10 “spots,” the five points of the star and the five corners of the pentagon in the middle. A move consists of putting a pebble on any unoccupied spot, moving it through an adjacent spot (which may be occupied) and continuing in a straight line to the next adjacent spot, which must be unoccupied. You then leave the pebble there and start again with a new pebble, choosing any unoccupied spot to begin this next move. If you can fill 9 of the 10 spots in this way then you’ve won.

Can you find a solution?

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Image: Wikimedia Commons One solution that’s easy to remember is to start your first move at a corner of the pentagon, arriving at a point of the star, and then contrive each subsequent move to end at the start of the preceding move (as above). Another solution:

March 17, 2017March 26, 2017 | Puzzles

Black and White

By T.R. Dawson. “The problem is for each White man to capture the corresponding Black man in such a way that the routes traversed by each man never come in contact with one another.”

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From Strand Magazine, October 1911.

March 14, 2017March 26, 2017 | Puzzles

Ships That Pass

For four months in 1840 Edgar Allan Poe conducted a puzzle column for the Philadelphia newspaper Alexander’s Daily Messenger. In that time he defied his readers to send him a cryptogram that he could not solve, and at the end of his tenure he declared himself undefeated. One of the later challenges came from 17-year-old Schuyler Colfax of New Carlisle, Iowa, who would grow up to become vice president of the United States:

Dear Sir — As you have in your Weekly Messenger defied the world to puzzle you by substituting arbitrary signs, figures, etc. for the different letters of the alphabet, I have resolved to try my utmost to corner you and your system together, and have manufactured the two odd looking subjects which accompany this as avant couriers. … If you succeed in solving the accompanying, I will, of course, as you request, acknowledge it publicly to my friends.

Poe responded: “We have only time, this week, to look at the first and longest cypher — the unriddling of which, however, will no doubt fully satisfy Mr. Colfax that we have not been playing possum with our readers.” Here’s Colfax’s cryptogram:

8n()58†d w!0 b† !x6n†z k65 !nz k65,8l†n b)x 8nd)Pxd !zw8x 6k n6 36w-†nd!x86n;

x=†0 z†,5!z† x=† w8nz 8n 8xd 62n †dx††w !nz k653† 8x x6 5†36l†5 8xd P†l†P b0 5†l†n,†.

()n8)d

What’s the solution?

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“Injuries may be atoned for and forgiven, but insults admit of no compensation; they degrade the mind in its own esteem, and force it to recover its level by revenge.” — Junius

March 9, 2017March 19, 2017 | History · Literature · Puzzles

Black and White

By J. Berger. White to mate in two moves.

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1. Qc8! and the queen will mate. From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

February 28, 2017March 12, 2017 | Puzzles

The Candy Thief

A problem by Wayne M. Delia and Bernadette D. Barnes:

Five children — Ivan, Sylvia, Ernie, Dennis, and Linda — entered a candy store, and one of them stole a box of candy from the shelf. Afterward each child made three statements:

Ivan:

1. I didn’t take the box of candy.
2. I have never stolen anything.
3. Dennis did it.

Sylvia:

4. I didn’t take the box of candy.
5. I’m rich and I can buy my own candy.
6. Linda knows who the crook is.

Ernie:

7. I didn’t take the box of candy.
8. I didn’t know Linda until this year.
9. Dennis did it.

Dennis:

10. I didn’t take the box of candy.
11. Linda did it.
12. Ivan is lying when he says I stole the candy.

Linda:

13. I didn’t take the box of candy.
14. Sylvia is guilty.
15. Ernie can vouch for me, because he has known me since I was a baby eight years ago.

If each child made two true and one false statement, who stole the candy?

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Not as hard as it sounds! If Ivan was the thief, then he told more than one lie. In the same way we can exclude Ernie, Dennis, and Linda. So Sylvia is the thief, and the false statements are #3, #4, #9, #11, and #15. From Pi Mu Epsilon Journal, Fall 1980 (PDF) (solution by John Wesley Emert of Knoxville, Tenn.).

February 25, 2017March 5, 2017 | Puzzles

The Windmill Algorithm

Suppose we have a finite set of points in the plane, no three of which are collinear. A line drawn through one of them pivots around that point until it encounters another point, when it adopts that point as the new pivot. Call this line a “windmill”; it continues indefinitely, always rotating in the same direction. Show that we can choose an initial point and line so that the resulting windmill uses each point as a pivot infinitely many times.

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Draw the line vertically through the centermost of the points, and color all the points to the left of it blue and all the points to the right red. After the line has turned 180 degrees, all the points to the left of it will now be red, and those to the right will be blue. Hence every point has changed color. Each time the pivot changes, the old pivot point returns to the same side of the line that the new one comes from. That means that the total number of blue points remains unchanged; so does the number of red points. Since a point can change color only by serving as a pivot, this means that every point must act as a pivot at some stage. And if the line turns forever, then each point will change color infinitely many times. This is laid out more rigorously here. By Geoffrey Smith of the United Kingdom for the 52st International Mathematical Olympiad in Amsterdam, 2011. The animation is from the excellent Symmetry tumblr.

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February 9, 2017February 19, 2017 | Puzzles

Black and White

weiß chess problem

By Berthold Weiß. White to mate in two moves.

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1. Qh8!, threatening 2. Qa8#. If 1. … Qb8 then 2. Qa1#! From The Good Companion, 1917.

February 3, 2017February 12, 2017 | Puzzles

The Mengenlehreuhr

Further to Saturday’s triangular clock post, reader Folkard Wohlgemuth points out that a “set theory clock” has been operating publicly in Berlin for more than 40 years. Since 1995 it has stood in Budapester Straße in front of Europa-Center.

The circular light at the top blinks on or off once per second. Each cell in the top row represents five hours; each in the second row represents one hour; each in the third row represents five minutes (for ease of reading, the cells denoting 15, 30, and 45 minutes past the hour are red); and each cell in the bottom row represents one minute. So the photo above was taken at (5 × 2) + (0 × 1) hours and (6 × 5) + (1 × 1) minutes past midnight, or 10:31 a.m.

Online simulators display the current time in the clock’s format in Flash and Javascript.

If that’s not interesting enough, apparently the clock is a key to the solution of Kryptos, the enigmatic sculpture that stands on the grounds of the CIA in Langley, Va. In 2010 and 2014 sculptor Jim Sanborn revealed to the New York Times that two adjacent words in the unsolved fourth section of the cipher there read BERLIN CLOCK.

When asked whether this was a reference to the Mengenlehreuhr, he said, “You’d better delve into that particular clock.”

January 17, 2017January 16, 2017 | Puzzles · Science & Math