In 1980, Japanese astrophysicist Kōryō Miura worked out a pattern of parallelograms that permit a map to be folded much more compactly than conventional right-angle creases. So efficient is the pattern that a map can be opened or refolded with a single motion by pulling on opposite ends, rather like an accordion. Today it’s used to fold surgical devices, furniture, and solar panel arrays on spacecraft.

During lunch one day at Los Alamos, Richard Feynman told his colleagues, “I can work out in sixty seconds the answer to any problem that anybody can state in ten seconds, to 10 percent!”

He had completed several challenges when mathematician Paul Olum walked past.

‘Hey, Paul!’ they call out. ‘Feynman’s terrific! We give him a problem that can be stated in ten seconds, and in a minute he gets the answer to 10 percent. Why don’t you give him one?’

Without hardly stopping, he says, ‘The tangent of 10 to the 100th.’

“I was sunk: you have to divide by pi to 100 decimal places! It was hopeless. … He was a very smart fellow.”

(From Surely You’re Joking, Mr. Feynman!, 1985.)

Above: A valid maze can be generated recursively by dividing an open chamber with walls and creating an opening at random within each wall, ensuring that a route can be found through the chamber. The secondary chambers themselves can then be divided with further walls, following the same principle, to any level of complexity.

Below: Valid mazes can even be generated fractally — here a solution becomes available in the third panel, but an unlucky solver might wander forever in the depths of self-similarity at the center of the image.

This is Euclid’s proof of the Pythagorean theorem — Schopenhauer called it a “brilliant piece of perversity” for its needless complexity:

1. Erect a square on each leg of a right triangle. From the triangle’s right angle, A, draw a line parallel to BD and CE. This will intersect BC and DE perpendicularly at K and L.
2. Draw segments CF and AD, forming triangles BCF and BDA.
3. Because angles CAB and BAG are both right angles, C, A, and G are collinear.
4. Because angles CBD and FBA are both right angles, angle ABD equals angle FBC, since each is the sum of a right angle and angle ABC.
5. Since AB is equal to FB, BD is equal to BC, and angle ABD equals angle FBC, triangle ABD is congruent to triangle FBC.
6. Since A-K-L is a straight line that’s parallel to BD, rectangle BDLK has twice the area of triangle ABD, because they share base BD and have the same altitude, BK, a line perpendicular to their common base and connecting parallel lines BD and AL.
7. By similar reasoning, since C is collinear with A and G, and this line is parallel to FB, square BAGF must be twice the area of triangle FBC.
8. Therefore, rectangle BDLK has the same area as square BAGF, AB².
9. By applying the same reasoning to the other side of the figure, it can be shown that rectangle CKLE has the same area as square ACIH, AC².
10. Adding these two results, we get AB² + AC² = BD × BK + KL × KC.
11. Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC.
12. Therefore, since CBDE is a square, AB² + AC² = BC².

The diagram became known as the bride’s chair due to a confusion in translation between Greek and Arabic.

By Wikimedia user Cmglee, a visual proof that a³ – b³ = (a – b)(a² + ab + b²):

Each of the first 18 multiples of 526315789473684210 contains all 10 digits:

526315789473684210 x  1 =  526315789473684210
526315789473684210 x  2 = 1052631578947368420
526315789473684210 x  3 = 1578947368421052630
526315789473684210 x  4 = 2105263157894736840
526315789473684210 x  5 = 2631578947368421050
526315789473684210 x  6 = 3157894736842105260
526315789473684210 x  7 = 3684210526315789470
526315789473684210 x  8 = 4210526315789473680
526315789473684210 x  9 = 4736842105263157890
526315789473684210 x 10 = 5263157894736842100
526315789473684210 x 11 = 5789473684210526310
526315789473684210 x 12 = 6315789473684210520
526315789473684210 x 13 = 6842105263157894730
526315789473684210 x 14 = 7368421052631578940
526315789473684210 x 15 = 7894736842105263150
526315789473684210 x 16 = 8421052631578947360
526315789473684210 x 17 = 8947368421052631570
526315789473684210 x 18 = 9473684210526315780

The 19th multiple, alas, is 9999999999999999990.

Is every integer the sum of four perfect cubes?

Interestingly, no one knows. It’s conjectured that the answer is yes, but no one yet has been able to prove this.

“Good tests kill flawed theories; we remain alive to guess again.” — Karl Popper

“There are two possible outcomes: if the result confirms the hypothesis, then you’ve made a measurement. If the result is contrary to the hypothesis, then you’ve made a discovery.” — Enrico Fermi

“This particular thesis was addressed to me a quarter of a century ago by John Campbell, who … told me that all theories are proven wrong in time. … My answer to him was, ‘John, when people thought the Earth was flat, they were wrong. When people thought the Earth was [perfectly] spherical, they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together.’ The basic trouble, you see, is that people think that ‘right’ and ‘wrong’ are absolute; that everything that isn’t perfectly and completely right is totally and equally wrong. However, I don’t think that’s so. It seems to me that right and wrong are fuzzy concepts.” — Isaac Asimov, The Relativity of Wrong, 1988