Problem: (2000 Putnam - A5) Three distinct points with integer coordinates lie in the plane on a circle of radius $ r > 0 $. Show that two of these points are separated by a distance of at least $ r^{1/3} $.
Solution: I will admit that this is not my solution. But it's just so amazingly awesome it deserves to be here.
Let the triangle formed by the segments between the three points have side lengths $ a, b, c $. By a well-known formula, the area of the this triangle is $ \frac{abc}{4r} $.
On the other hand, by Pick's Formula, we know the area is $ I+\frac{B}{2}-1 \ge \frac{3}{2}-1 = \frac{1}{2} $. So
$ \frac{abc}{4r} \ge \frac{1}{2} \Rightarrow abc \ge 2r $.
But by AM-GM, we have $ \max\{a, b, c\} \ge (abc)^{1/3} \ge (2r)^{1/3} > r^{1/3} $. QED.
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Comment: This is a really neat solution because it uses some well-known formulas to prove a very interesting result. Finding it would probably be considerably more difficult than reading it (as with many proofs), but the conciseness of the proof is almost too good to be true.
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Practice Problem: (2000 Putnam - A4) Show that the improper integral $ \displaystyle \lim_{B \rightarrow \infty} \int_0^B \sin{(x)} \sin{(x^2)} dx $ converges.
That's not AM-GM, that just follows from the definition of a maximum and from the fact that x^{1/3} is an increasing function.
ReplyDeletePIck's is definitely one of the coolest things ever, but then that bound is pretty weak...
Oh, well AM-GM does it, too... lol. Yeah, anyway...
ReplyDeleteyou were sorta right on the practice problem
ReplyDeleteYeah, the square is just on the inside.
ReplyDelete