Hidden among his characteristic satire of average American life, *The Simpson* It is full of mathematical Easter eggs. The show’s writers have boasted an impressive pedigree of Ivy League mathematicians who couldn’t resist infusing America’s Oldest Sitcom with inside jokes, sprinkled like sprinkles on Homer’s donuts.

Already in the first shot of the show’s second episode, Maggie, the always one-year-old baby, stacks her alphabet blocks to read. EMCSQU. Without a doubt, a tribute to Einstein’s famous equation *mi = mc ^{2}*.

There’s an episode where Homer tries to become an inventor and designs some crazy ideas, including a shotgun that shoots makeup in your face and a recliner with a built-in toilet. During a frantic brainstorm, Homer scribbles some equations on a whiteboard, including:

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

This refers to Fermat’s Last Theorem, one of the most infamous equations in the history of mathematics. The encapsulated version, if you haven’t found it yet: 17th century mathematician Pierre de Fermat wrote that the equation *to ^{north} +b^{north} =c^{north}* has no integer solutions when n is greater than 2. In other words, you cannot find three integers (not decimal numbers like 1, 2, 3)

*to*,

*b*and

*c*such that

*a*either

^{3}+b^{3}=c^{3 }*a*, and so on. Fermat wrote that he had discovered a truly wonderful proof of this but he could not include it in the margin of his text. Later mathematicians found this message and, despite the simple appearance of the statement, failed to prove it. For more than four centuries until Andrew Wiles finally achieved it in 1994. The Wiles test is based on techniques much more advanced than those available. in Fermat’s time, which leaves open the tantalizing possibility that Fermat knew a more elementary proof that we have yet to discover (or his supposed proof was flawed).

^{4}+b^{4}=c^{4}Enter Homer’s equation into your calculator. It is verified! *The Simpson *Find a counterexample to Fermat’s Last Theorem? It turns out that Homer’s trio of numbers constitutes a near error. Most calculators do not display enough precision to detect the slight discrepancy between the two sides of the equation. Writer David

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

The ultimate problem of tiebreaker geometry is more difficult than it seems. I hope it doesn’t make you scream Doh!

*Did you miss last week’s puzzle? check it **here**and find your solution at the end of today’s article. Be careful not to read too far ahead if you haven’t solved Last Week yet!*

### Puzzle #20: The Simpsons M

Add three straight lines to the diagram to create nine nonoverlapping triangles.

Triangles can share sides, but they 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 along with a new puzzle. Do you know of an interesting puzzle that you think should appear here? ? Send me a message on Twitter @JackPMurtagh or email me at [email protected]

### Solution to puzzle #19: Mental illusions

How was last week? issues? I compared them to optical illusions because both puzzles seem at first glance to require some complicated calculations. But once you perceive the hidden trick, the solution comes into focus as neck cubes abruptly reversing. Both puzzles are actually tricks, with the right perspective. Shout out to reader McKay, who emailed two correct answers.

1. It will take at most one minute for all the ants to fall off the end of the meter stick. It seems complicated to follow the oscillating behavior of each ant. Couldn’t they swing back and forth forever? When you squint, let’s see that the condition in which two colliding ants immediately change direction is no different from the case in which the ants move through each other. other! In both cases, there will be ants at exactly the same points along the stick walking in the same direction.

Imagine that each ant was wearing a small top hat and every time two collide they instantly swap hats before continuing in the opposite direction. Make a single top-hat stroke and notice that it simply goes in a straight line toward one end of the club at a constant pace throughout the entire stroke. time. Since ants move at one meter per minute and the longest any ant could travel is the full length of the one-meter rod, all the ants will reach the end of the rod in one minute.

2. How about the geometry problem?

What is the length of AC?

Looks ready for the SAT. Maybe the Pythagorean theorem is in order. Maybe a trigonometric identity or two. Blink twice and the illusion of complexity appears and 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 the circle! The diagram tells us the radius of the circle along the x axis: 6+5 = 11, our answer.

This content has been automatically translated from the original material. Due to the nuances of machine translation, there may be slight differences. For the original version, click here.