A Quickie

by t0rajir0u, May 2, 2007, 2:10 AM

A short fun problem.

Problem 1: (Engel, Chapter 4, #67) Prove that, among any seven real numbers $y_{1}, ... y_{7}$, there exist two such that

$0 \le \frac{ y_{i}-y_{j}}{1+y_{i}y_{j}}\le \frac{1}{ \sqrt{3}}$.

Solution: The middle should instantly remind you of tangent subtraction, while the RHS should remind you of $\tan \frac{\pi}{6}$. The interpretation is almost automatic:

Let $\tan \theta_{i}= y_{i}$. Then WLOG $-\frac{\pi}{2}< \theta_{i}< \frac{\pi}{2}$, an interval of length $\pi$. Among seven such $\theta_{i}$ we can choose two that are strictly less than $\frac{\pi}{6}$ apart by Pigeonhole. Let these two be $\theta_{i}\ge \theta_{j}$. Then

$0 \le \tan (\theta_{i}-\theta_{j}) = \frac{y_{i}-y_{j}}{1+y_{i}y_{j}}< \tan \frac{\pi}{6}$

As desired. The right inequality turns out to be strict. QED.


One of many unexpected applications of Pigeonhole. The wording of the problem made the pigeons more or less obvious, but the holes much less so. Nearly all of the difficulty of these types of problems is in discerning the identity of the pigeons and the holes.


Practice Problem 1: (Engel, Chapter 4, #61) There are $650$ points inside a circle of radius $16$. Prove that there exists a ring with inner radius $2$ and outer radius $3$ covering ten of these points.

Practice Problem 2: (Engel, Chapter 4, #76) Any of the $n$ points $P_{1}, ... P_{n}$ in space has a smaller distance from point $P$ than from all the other points $P_{i}$. Prove that $n < 15$.

(Can you generalize to lower and/or higher dimensions?)

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6 Comments

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Hold on... is the wording for the first practice problem correct?

by Ubemaya, May 2, 2007, 2:18 AM

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I think so... or 2 was meant to be the distance "between" the inner and outer circles of the ring?

by kohjhsd, May 2, 2007, 2:41 AM

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Hmm, for the second one I'm thinking about projecting all of the points toward $P$ onto a sphere of radius $1$, somehow showing that the $\angle P_{i}PP_{j}$ must be at least some number of radians, and then calculating the area of a circle on a sphere... that doesn't sound particularly easy. I bet there's a better way.

by paladin8, May 2, 2007, 3:07 AM

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Ubemaya wrote:
Hold on... is the wording for the first practice problem correct?

Whoops. Sorry about that.

by t0rajir0u, May 2, 2007, 3:49 PM

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i feel like we are going to need some regular polyhedra here...the analogue to two dimensions is a hexagon...

by Altheman, May 4, 2007, 5:55 AM

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Yeah, that's sort of what I was thinking. But in 2-D it is easy to partition the interval $[0, 2\pi)$ and then apply Pigeonhole so one angle is $< \frac{\pi}{3}$ and get a contradiction. In 3-D, I can't see a good way to apply a similar argument.

by paladin8, May 6, 2007, 3:24 AM

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