Modern Art

modern art puzzle

Which part of this square has the greater area, the black part or the gray part?

	Select Click for Answer
Color the indicated regions dark gray: Now if we add the dark gray regions to the black region, we get triangle BCF, whose area is half that of the whole square (a triangle’s area is one-half base times height). If we add the dark gray areas to the light gray areas, we again produce a region that’s half the area of the square (because it’s the full square minus triangle ABE). So the black part and the light gray part are equal. By V. Proizvolov, from the Russian science magazine Kvant.

May 13, 2015May 24, 2015 | Puzzles

Black and White

shinkman chess problem

By William Anthony Shinkman. White to mate in two moves.

	Select Click for Answer
1. Rc8! (ha!) and the rook will mate. From the Detroit Free Press, 1905.

May 10, 2015May 17, 2015 | Puzzles

Roll Call

A problem from the 2002 Moscow Mathematical Olympiad:

A group of recruits stand in a line facing their corporal. They are, unfortunately, rather poorly trained: At the command “Left turn!”, some of them turn left, some turn right, and some turn to face away from the corporal. Is it always possible for the corporal to insert himself in the line so that an equal number of recruits are facing him on his left and on his right?

	Select Click for Answer
Yes. When the corporal has taken a place in the line, let l denote the number of recruits to his left and facing him, and let r denote the number of recruits to his right and facing him. If the corporal takes the leftmost place in line, then there’s no one to his left (l = 0). If no one to his right happens to be facing him either, then the problem is solved. Otherwise r > 0. Now suppose the corporal moves gradually rightward, one position at a time. If he passes a recruit who is facing right (that is, whose back had been to him), then l increases by 1 and r doesn’t change. If he passes a recruit who’s facing left, l doesn’t change and r decreases by 1. And if he passes a recruit who’s facing the back of the parade ground, then neither l nor r changes. Now consider what happens to the difference l – r as the corporal moves along the line. At first it’s negative (because r is positive and l is 0). As the corporal moves down the line, l – r changes by at most 1 as he passes each soldier. When he reaches the right end of the line, r will be 0 and so l – r will be nonnegative. So we’ve started with a negative number, set it increasing and decreasing by 1 several times, and arrived at a nonnegative number. At some point, then, the difference must have been 0. At that moment l = r, and the corporal has an equal number of soldiers facing him on both sides.

Click for Answer

May 7, 2015May 17, 2015 | Puzzles

Podcast Episode 56: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits! Solve along with us as we explore some strange scenarios using only yes-or-no questions. Many were submitted by listeners, and most are based on real events.

See full show notes …

May 4, 2015October 25, 2017 | Podcast · Puzzles

Black and White

gold chess problem

By Sámuel Gold. White to mate in two moves.

	Select Click for Answer
1. Ra1 Kxg6 2. Qb1# From Benjamin Glover Laws, The Two-move Chess Problem, 1890.

April 27, 2015May 3, 2015 | Puzzles

Star Power

A puzzle by A. Korshkov, from the Russian science magazine Kvant:

It’s easy to show that the five acute angles in the points of a regular star, like the one at left, total 180°.

Can you show that the sum of these angles in an irregular star, like the one at right, is also 180°?

	Select Click for Answer
Join two of the star’s adjacent vertices, such as A and E. Triangles BMD and AME have an angle in common (the congruent angles at M), so the sum of angles B and D in triangle BMD equals the sum of angles A and E in triangle AME. This means that the sum of the star’s five angles equals the sum of the angles of triangle ACE — which is 180°.

April 25, 2015May 3, 2015 | Puzzles · Science & Math

Hoop Dreams

A memorably phrased puzzle from The Graham Dial: “Consider a vertical girl whose waist is circular, not smooth, and temporarily at rest. Around the waist rotates a hula hoop of twice its diameter. Show that after one revolution of the hoop, the point originally in contact with the girl has traveled a distance equal to the perimeter of a square circumscribing the girl’s waist.”

	Select Click for Answer
Hold the hoop steady and let the girl roll around inside it: Image: Wikimedia Commons Since the ratio of the diameters is 2:1, so is the ratio of the circumferences. This means that a point on the girl oscillates back and forth between two opposite points on the hoop, passing through the hoop’s center on the way and producing a straight line (the “Tusi couple”). “The perimeter of circumscribing square equals four girl diameters or two hula hoop diameters which is the total displacement of initial point of contact between hula hoop and the aforementioned vertical girl.” From L.A. Graham, The Surprise Attack in Mathematical Problems, 1968.

Click for Answer

April 22, 2015May 3, 2015 | Puzzles

Quickie

University of Strathclyde mathematician Adam McBride recalls that in his student days a particular teacher used to present a weekly puzzle. One of these baffled him:

Find positive integers a, b, and c, all different, such that a³ + b³ = c⁴.

“The previous puzzles had been relatively easy but this one had me stumped,” he wrote later. He created three columns headed a³, b³, and c⁴ and spent hours looking for a sum that would work. On the night before the deadline, he found one: 70³ + 105³ = 35⁴.

“This shows how sad a person I was! However, I then realised also how stupid I had been. I had totally missed the necessary insight.” What was it?

	Select Click for Answer
“All you need to do is start with 2³ + 3³ = 35 and multiply by 35³.” (From Adam McBride, “Mathematics: The Greatest Subject in the World,” The Mathematical Gazette, vol. 89, no. 516 [November 2005].)

April 19, 2015April 26, 2015 | Puzzles

All Relative

1. A puzzle from J.A.H. Hunter’s Fun With Figures, 1956:

Tom and Tim are brothers; their combined ages make up seventeen years. When Tom was as old as Tim was when Tim was twice as old as Tom was when Tom was fifteen years younger than Tim will be when Tim is twice his present age, Tom was two years younger than Tim was when Tim was three years older than Tom was when Tom was a third as old as Tim was when Tim was a year older than Tom was seven years ago. So how old is Tim?

2. Another, by Sam Loyd:

“How fast those children grow!” remarked Grandpa. “Tommy is now twice as old as Maggie was when Tommy was six years older than Maggie is now, and when Maggie is six years older than Tommy is now their combined ages will equal their mother’s age then, although she is now but forty-six.” How old is Maggie?

3. According to Wirt Howe’s New York at the Turn of the Century, 1899-1916, this question inspired an ongoing national debate when it appeared in the New York Press in 1903:

Brooklyn, October 12

Dear Tip:

Mary is 24 years old. She is twice as old as Anne was when she was as old as Anne is now. How old is Anne now? A says the answer is 16; B says 12. Which is correct?

John Mahon

	Select Click for Answer
1. Let x be Tom’s age, and let y be Tim’s age. Now start with the statement of fact (“Tom was two years younger,” etc.) and work backward: When Tim is twice his present age, Tim will be 2y. When Tom was 15 years younger than that, Tom was (2y – 15). When Tim was twice as old as that, Tim was (4y – 30). When Tom was as old as that, Tom was (4y – 30). Working backward from the end, we get: Seven years ago, Tom was (x – 7). When Tim was 1 year older than that, Tim was (x – 6). When Tom was a third of that, Tom was (x – 6) / 3. When Tim was 3 years older than that, Tim was (x + 3) / 3. When Tom was 2 years younger than that, Tom was (x – 3) / 3. So that gives us 4y – 30 = (x – 3) / 3, and also x + y = 17. Tom is 9 years old, and Tim is 8. 2. I don’t have Loyd’s solution, but here’s my stab: T = Tommy’s age M = Maggie’s age “Tommy is now twice as old as Maggie was when Tommy was six years older than Maggie is now”: T = 2(M – (T – (M + 6))) “When Maggie is six years older than Tommy is now their combined ages will equal their mother’s age then, although she is now but forty-six”: (T + 6) + (T + (T – M) + 6) = 46 + (T – M) + 6 If we combine these equations, we find that Tommy is 20 and Maggie is 12. 3. I get a third answer. Maybe it’s time for another debate.

Click for Answer

April 18, 2015April 26, 2015 | Puzzles

Three Chess Puzzles

rockwell chess problem

1. By A.F. Rockwell. White to mate in two moves. (Solutions are below.)

2. In 1866 Sam Loyd asked: Suppose you’re playing White against an opponent who’s required to mirror every move you make — if you play 1. Nf3 he must play 1. … Nf6, and so on. Can you design a game in which your eighth move forces your opponent to checkmate you with a nonmirror move?

miles chess problem

3. An endgame study by J.A. Miles. “White to play and draw the game.”

	Select Click for Answer
1. Qc5! and the queen or bishop will mate. (From A Collection of Chess Problems, 1890.) 2. Loyd’s solution is 1. e4 e5 2. Ke2 Ke7 3. Ke3 Ke6 4. Qf3 Qf6 5. Ne2 Ne7 6. b3 b6 7. Ba3 Ba6: Now 8. Nd4+ forces 8. … exd4, mating White. (From Alain C. White, Sam Loyd and His Chess Problems, 1962.) 3. This is the frontispiece to the collection Chess Stars: A Galaxy of Self-Mates, from 1888. Unfortunately no solution is given! White is already nearly stalemated — if he can get rid of his queen he’ll have his draw. But the only line I can find is the ridiculous and ungainly 1. Qf1+ Nc4 2. Qxc4+ Kb6 3. Qa6+ Kc7 4. Qc8+ Kb6 5. Qb7+, where Black must take the queen. All Black’s moves are forced if he wants to avoid a draw, but this is such an unbeautiful line that I feel sure it can’t be the intended solution. Any ideas?

Click for Answer

April 11, 2015April 19, 2015 | Puzzles