# Swatting flies with a sledgehammer

Prompted by a recent exchange with Gerry Myerson, I was wondering if anyone has a favorite example of a relatively simple problem with a rather elementary (though perhaps complicated) answer for which there's another answer that relies on an elegant use of a powerful result that's almost certainly beyond the background of the poser of the question.

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I don't see any reason for the downvote. That's just mean. Downvoter should explain himself. –  Patrick Da Silva Jul 18 '12 at 17:52
How about Pascal's mystic hexagon? It can be proved by elementary planar geometry, but also be attacked in many ways, including Bézout's theorem in algebraic geometry. –  sos440 Jul 18 '12 at 18:00
There are a number of such proofs for the existence of infinitely many prime numbers (and probably stackexchange and/or mathoverflow have threads devoted to this), so I suggest that proofs of infinitely many primes be excluded in order to leave room for examples that are not so well known. –  Dave L. Renfro Jul 18 '12 at 18:01
(continued) In re-reading the question (often a useful thing to do!), I see that the wording of the question (i.e. though perhaps complicated followed by elegant use) seems to eliminate the prime number examples I was thinking of. –  Dave L. Renfro Jul 18 '12 at 18:04
We have mathoverflow.net/questions/42512/… a while ago on MO –  Mariano Suárez-Alvarez Jul 18 '12 at 18:17

If $2^{1/n}$ were a rational $a/b$, with $n>2$, then $a^n=b^n+b^n$, which would contradict Wiles proof of Fermats Last Theorem.

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I don't think it gets any more OP than this ^^ –  Olivier Bégassat Jul 18 '12 at 17:59
I remembered that this example had been mentioned in this old MathOverflow question and in a comment to the answer, BCnrd explains that the argument is basically circular. –  Adrián Barquero Jul 18 '12 at 18:29

OK, here's my proof that $49 < 50$.

There's one point in the first quadrant where the circle $x^2+y^2=1$ intersects the line $x=y$, and the tangent line to the circle at that point has slope $-1$ and $x$-intercept $\sqrt{2}$. Every line with the same slope and a slightly smaller $x$-intercept intersects the circle twice, and those with a larger intercept do not intersect the circle.

Now suppose we use $7/5$ as an approximation to $\sqrt{2}$, and draw the line with slope $-1$ and that $x$-intercept. Lo and behold, it passes through the points $(4/5,3/5)$ and $(3/5,4/5)$, which are on the circle. So it's a secant line, not a tangent line. Therefore we conclude that $$\frac75<\sqrt{2}.$$ Therefore $$49 = 7^2 < 5^2\cdot2=50.$$

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Last semester our grad Analysis II class proved the Isoperimetric Inequality using Fourier analysis.

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This is pretty standard... –  Olivier Bégassat Jul 18 '12 at 18:01
Sure, but it was new to me! –  Andrew Jul 18 '12 at 18:02
What would be the elementary way to prove the isoperimetric inequality in this case? I've always been under the impression that this already is the elementary way (the use of Fourier analysis certainly is elegant and I like it, too, but this is not what's being asked here). –  Sam Jul 19 '12 at 9:23
I was under the impression that this was "known to the Greeks" whatever that may mean. Doing a quick search it definitely seems that they may have discovered a rigorous proof, e.g.: www.cs.nyu.edu/faculty/siegel/SCIAM.pdf –  Andrew Jul 19 '12 at 15:10