(Via Robert A. Carman, “Mathematical Misteaks,” Mathematics Teacher 64:2 [February 1971], 109-115.)

This worrying result was first published by German mathematician Oskar Schlömilch in 1868. (The discrepancy is explained by minute gaps in the diagonals, as explained here.)

Charles Dodgson (Lewis Carroll) seems to have been taken with the paradox — his papers show that between 1890 and 1893 he was working to determine all the squares that might similarly be converted into rectangles with a “gain” of one unit of area, apparently unaware that V. Schlegel had carried out the same task much earlier.

(Warren Weaver, “Lewis Carroll and a Geometrical Paradox,” American Mathematical Monthly 45:4 [April 1938], 234-236.)

Draw a triangle ABC and pick a point V that’s not on one of its sides. Draw a segment from each of the triangle’s vertices through V. Now draw a new triangle whose sides are parallel to these three segments. Segments drawn from each of this new triangle’s vertices and parallel to the first triangle’s sides, as shown, will meet in a common point.

Proven by James Clerk Maxwell!

Each side of the yellow square is 2 feet long. In the Meno, Socrates asks a slave boy how long would be the side of a square that had twice the yellow square’s area. The boy guesses first 4 feet, then 3, and finds himself at a loss.

Socrates builds a square four times the size of the yellow one, then divides each of its constituent squares in half with a diagonal. The area of the blue square is thus twice that of the yellow one, and its side has the length we’d sought.

“Some things I have said of which I am not altogether confident,” Socrates tells Meno. “But that we shall be better and braver and less helpless if we think that we ought to enquire, than we should have been if we indulged in the idle fancy that there was no knowing and no use in seeking to know what we do not know; that is a theme upon which I am ready to fight, in word and deed, to the utmost of my power.”

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.