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Hidden amid its trademark satire of average American life, *The Simpsons* is riddled with mathematical Easter eggs. The show’s writing staff has boasted an impressive pedigree of Ivy League mathheads who couldn’t resist infusing America’s longest-running sitcom with inside jokes, scattered about like sprinkles on Homer’s doughnuts.

As early as the opening shot of the show’s second episode, the perpetually one-year old baby, Maggie, stacks her alphabet blocks to read EMCSQU. No doubt an homage to Einstein’s famous equation *E = mc ^{2}*.

There’s an episode where Homer tries to become an inventor and he engineers a few harebrained ideas, including a shotgun that blasts make-up on your face and a recliner with a built-in toilet. During a brainstorming frenzy, Homer scribbles some equations on a chalkboard including:

1987^{12} + 4365^{12 }= 4472^{12 }

This references Fermat’s Last Theorem, one of the most infamous equations in math history. The potted version, if you haven’t come across it: 17th century mathematician Pierre de Fermat wrote that the equation *a ^{n} + b^{n} = c^{n}* has no whole number solutions when n is greater than 2. In other words, you can’t find three whole numbers (non-decimal numbers like 1, 2, 3…)

*a*,

*b*, and

*c*such that

*a*or

^{3}+ b^{3}= c^{3 }*a*, and so on. Fermat wrote that he had “discovered a truly marvelous proof of this” but couldn’t fit it in the margin of his text. Later mathematicians found this message and, despite the simple appearance of the claim, failed to prove it. It went unproven for over four centuries until Andrew Wiles finally cracked it in 1994. Wiles’ proof relies on techniques far more advanced than what was available in Fermat’s day, which leaves open the tantalizing possibility that Fermat knew of a more elementary proof that we have yet to discover (or his supposed proof had a bug).

^{4}+ b^{4}= c^{4}Plug Homer’s equation into your calculator. It checks out! Did *The Simpsons *find a counterexample to Fermat’s Last Theorem? It turns out that Homer’s trio of numbers constitute a near-miss. Most calculators don’t display enough precision to detect the slight discrepancy between the two sides of the equation. Writer David X. Cohen wrote his own computer program to search for near-miss solutions to Fermat’s notorious equation all for this split-second gag.

This week’s puzzle comes from the season 26 finale, in which the denizens of Springfield participate in a mathlete competition. The episode is packed with mathematical goodies, including the little joke below posted outside of the competition. Can you decipher it?

The climactic tie-breaking geometry problem is tougher than it looks. I hope it doesn’t make you shout, “D’oh!”

*Did you miss last week’s puzzle? Check it out **here**, and find its solution at the bottom of today’s article. Be careful not to read too far ahead if you haven’t solved last week’s yet!*

### Puzzle #20: The Simpsons M

Add three straight lines to the diagram to create nine non-overlapping triangles.

The triangles may share sides, but shouldn’t share interior space. For example, the left-hand figure below depicts two triangles, whereas the right-hand figure only counts as one triangle, because the larger triangle overlaps with the smaller one.

I’ll post the answer next Monday along with a new puzzle. Do you know a cool puzzle that you think should be featured here? Message me on Twitter @JackPMurtagh or email me at gizmodopuzzle@gmail.com

### Solution to Puzzle #19: Mental Illusions

How did you fare on last week’s problems? I compared them to optical illusions because both puzzles appear at first blush to require some involved calculation. But once you perceive the hidden trick, the solution snaps into focus like Necker cubes abruptly inverting. Both puzzles are actually gimmes, with the right perspective. Shout-out to reader McKay, who submitted two correct answers over email.

1. It will take at most one minute for all of the ants to fall off an end of the meter stick. It seems complicated to track the oscillating behavior of each ant. Couldn’t they bobble back and forth forever? When you squint your eyes, you’ll see that the condition where two colliding ants immediately switch their directions is no different from the case where the ants move right through each other! In both cases, there will be ants at exactly the same points along the stick walking in the same direction.

Imagine each ant was wearing a little top hat and whenever two collide they instantly swap hats before carrying on in the opposite direction. Track a single top hat’s path and you’ll notice that it just beelines for one end of the stick at a constant pace the whole time. Since ants move at one meter per minute and the longest any ant could have to travel is the full length of the meter stick, all of the ants will reach an end of the stick within one minute.

2. How about the geometry problem?

What is the length of AC?

It appears SAT-ready. Maybe the Pythagorean theorem is in order. Perhaps a trigonometric identity or two. Blink twice and the illusion of complexity vanishes. The line connecting points O and B is also a diagonal of the rectangle and will have the same length as AC. Only OB is more useful because it’s a radius of the circle! The diagram tells us the circle’s radius along the x-axis: 6+5 = 11, our answer.

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