Hidden amidst its characteristic satire of average American life, The simpsons is riddled with mathematical Easter eggs. The show’s writing team boasted an impressive pedigree of Ivy League mathematicians who couldn’t help but infuse America’s longest-running sitcom with inside jokes, scattered like sprinkles on Homer’s donuts.
From the first shot of the series’ second episode, Maggie, a perpetually one-year-old baby, stacks her alphabet blocks to read EMCSQU. No doubt a tribute to Einstein’s famous equation E = mc2.
There’s an episode where Homer tries to become an inventor and he designs some wacky ideas, including a shotgun that puts makeup on your face and a recliner with a built-in toilet. During a brainstorming frenzy, Homer scribbles some equations on a board, including:
198712 +436512 = 447212
This refers to Fermat’s Last Theorem, one of the most infamous equations in the history of mathematics. The potted version, if you haven’t already encountered it: 17th-century mathematician Pierre de Fermat wrote that the equation Anot +bnot =cnot has no integer solutions when n is greater than 2. In other words, you cannot find three integers (non-decimal numbers like 1, 2, 3…) A, bAnd vs such as A3 +b3 =c3 Or A4 +b4 =c4, And so on. Fermat writes that he “discovered a truly marvelous proof” but that he cannot put it in the margins of his text. Later mathematicians discovered this message and, despite the simple appearance of this statement, failed to prove it. This went unproven for over four centuries until Andrew Wiles finally discovered it in 1994. Wiles’s proof relied on techniques far more advanced than those available in Fermat’s time, which which leaves open the tantalizing possibility that Fermat knew a more basic proof that we have yet to discover (or his supposed proof had a bug).
Plug Homer’s equation into your calculator. It’s verified! Did The simpsons find a counterexample to Fermat’s last theorem? It turns out that Homer’s trio of ciphers constitutes a near miss. Most calculators are not accurate enough to detect the slight difference 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 famous equation, all for this split-second gag.
This week’s puzzle comes from the season 26 finale, in which the residents of Springfield participate in a math competition. The episode is full of math goodies, including the little joke below posted outside of the competition. Can you decipher it?
The decisive geometry problem is more difficult than it seems. I hope this doesn’t make you scream, “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 if you haven’t yet solved last week’s!
Riddle #20: The Simpsons M
Add three straight lines to the diagram to create nine non-overlapping triangles.
Triangles can share sides, but must not share interior space. For example, the left figure below represents two triangles, while the right figure only counts as one triangle, because the larger triangle overlaps the smaller one.
I will post the answer next Monday with a new puzzle. Do you know of a cool puzzle that you think should be featured here? Message me on Twitter @JackPMurtagh or email me at firstname.lastname@example.org
Solution to riddle #19: Mental illusions
How did you get on last week? problems? I likened them to optical illusions because both puzzles appear at first glance to require complex calculations. But once you perceive the hidden trick, the solution appears as Neck cubes reversing abruptly. Both puzzles are actually gifts, with the right perspective. Thanks to reader McKay, who submitted two correct answers via email.
1. It will take at most one minute for all the ants to fall from one end of the meter. It seems complicated to follow the oscillating behavior of each ant. Couldn’t they oscillate indefinitely? When you squint, you will see that the situation where two colliding ants immediately change direction is no different from the case where the ants are moving through each other! In both cases, there will be ants in exactly the same places along the stick, walking in the same direction.
Imagine that each ant wears a small top hat and that every time two collide, they instantly swap hats before continuing in the opposite direction. Follow the path of a single top hat and you’ll notice that it’s just heading towards one end of the stick at a constant pace the entire time. Since ants move at one meter per minute and the longest distance an ant could have to travel is the entire length of the meter, all ants will reach one end of the stick in one minute.
2. What about the geometry problem?
How long is AC?
He looks ready for SATs. Maybe the Pythagorean theorem is in order. Maybe one or two trigonometric identities. Blink twice and the illusion of complexity disappears. 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 is a radius of a circle! The diagram tells us the radius of the circle along the x-axis: 6+5 = 11, our answer.