Back in September I posted a geometry problem mentioned by Andy Liu in Math Horizons in November 1997. Several readers recognized it and wrote in with the pretty solution — here it is:

As before, we’re given that ∠DCA = 20°, ∠ACB = 60°, ∠CBD = 50°, and ∠DBA = 30°, and we’re asked to find ∠CAD. Start by extending CD and BA to intersect at O, and draw a circle with O as the center and OB as the radius. Now, because ∠OCB and ∠OBC both measure 80°, BC is one side of an 18-gon inscribed in this circle.

Let E be the fifth vertex of this 18-gon to the left of C and F be the fifth vertex to the right of C. Also let G be the first vertex to the left of C and H be the first vertex to the right of B. Then, by symmetry, EB, GF, and OC meet. And by the central angle theorem ∠EBC is half the measure of ∠EOC, or 50°, so EB, GF, and OC meet at D.

Now, OFH is an equilateral triangle (by symmetry and the fact that ∠FOH is 60°), and ∠GFH is half the measure of ∠GOH, or 30° (again by the central angle theorem). So GF bisects OH.

Finally, by symmetry, AC = AH. But ∠ACD = 20° = ∠AOD, so triangle AOC is equilateral and AC = AO. Then AO = AH, and by symmetry AF bisects OH. And that means that GF passes through A.

Therefore, ∠BAD = ∠OAF, which is half of ∠OAH, or 70°. And from the information given at the start we can infer that ∠CAB = 40°. So ∠CAD = 30°.

I’m told that there are more problems like this in I.F. Sharygin’s 1988 book Problems in Plane Geometry. Thanks to the folks who wrote in about this.

11/06/2015 UPDATE: Another reader pointed out an alternate solution, discovered by Edward Mann Langley in 1922. (Thanks, January.)

In 1726, the Swiss naturalist Johann Jakob Scheuchzer mistook the skull and vertebral column of a large salamander from the Miocene epoch for the “betrübten Beingerüst eines alten Sünders” (sad bony remains of an old human sinner) and dubbed it Homo diluvii testis, “the man who witnessed the Deluge.” The fossil lacked a tail or hind legs, so he thought it was the remains of a trampled human child:

It is certain that this [rock] contains the half, or nearly so, of the skeleton of a man; that the substance even of the bones, and, what is more, of the flesh and of parts still softer than the flesh, are there incorporated in the stone; in a word it is one of the rarest relics which we have of that accursed race which was buried under the waters. The figure shows us the contour of the frontal bone, the orbits with the openings which give passage to the great nerves of the fifth pair. We see there the remains of the brain, of the sphenoidal bone, of the roots of the nose, a notable fragment of the maxillary bone, and some vestiges of the liver.

The fossil made its way to Teylers Museum in the Netherlands, where in 1811 Georges Cuvier recognized it as a giant salamander. Ironically, Scheuchzer’s original belief is reflected in the fossil’s modern name, Andrias scheuchzeri — Andrias means “image of man.”

During a solar eclipse, the splashes of light that appear among the shadows of leaves take on the crescent shape of the sun.

In a pinch you can fashion your hands into a pinhole camera in order to observe an eclipse: Just make a loose fist of one hand and use it to focus the sun’s image onto the palm of your other hand. “The 0.25 cm (0.098 in) aperture f/200 optical system yields a reasonable image of the progress of the eclipse,” writes Peter L. Manly in Unusual Telescopes. “This telescope is easy to use, inexpensive and portable. The tracking system, however, leaves something to be desired.”

Or how about this amazing result. Each of a million men puts his hat into a very large box. Every hat has its owner’s name on it. The box is given a good shaking, and then each man, one after the other, randomly draws a hat out of the box. What’s the probability that at least one of the men gets his own hat back? Most people would answer with ‘pretty slim,’ but in fact the answer is the astonishingly large 0.632! Who would have guessed that?

— Paul J. Nahin, Will You Be Alive 10 Years From Now?, 2014

Hercules is battling a hydra. He manages to sever a head, but finds that a new head is generated according to the following rule: We move down one segment from the point where the head was severed and make a copy of the entire subtree above that point. Worse, the rate is increasing — the nth stroke of Hercules’ sword produces n new subtrees.

What will happen? The situation certainly looks dire, but, amazingly, Hercules cannot lose. No matter how large the hydra is, and no matter the order in which he severs the heads, he will always kill the hydra in finitely many turns.

Why? With each step, the complexity in the network is migrating toward the root — and that can’t continue forever. “We basically say that whenever something goes on K steps away from the root, it’s infinitely times worse than anything that is going on K-1 steps away from the root,” Comenius University computer scientist Michal Forišek explains in Slate.

“Now, whenever you kill a head, you very slightly simplified something that is K steps away. And even though you get a lot of new stuff in return, all that stuff is only K-1 steps away, and hence the entire result is still simpler than it was before.”

(Laurie Kirby and Jeff Paris, “Accessible Independence Results for Peano Arithmetic,” Bulletin of the London Mathematical Society 14 [1982], 285-293.)