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Source: 2-nd Taiwanese Mathematical Olympiad 1993

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#1 h Jan 14, 2007, 10:10 AM • 2 Y

Y by Adventure10, Mango247

In the Cartesian plane, let $C$ be a unit circle with center at origin $O$ . For any point $Q$ in the plane distinct from $O$ , define $Q'$ to be the intersection of the ray $OQ$ and the circle $C$ . Prove that for any $P\in C$ and any $k\in\mathbb{N}$ there exists a lattice point $Q(x,y)$ with $|x|=k$ or $|y|=k$ such that $PQ'<\frac{1}{2k}$ .

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#2 h Jan 16, 2007, 2:38 AM • 2 Y

Y by Adventure10, Mango247

Consider the case when $OP$ makes an angle $\theta$ such that $0\leq\theta\leq\frac{\pi}{4}$ . The argument then extends to all of $C$ by symmetry. Let $k$ be given. Consider all the lattice points of the form $Q(k,y)$ with $0\leq y\leq k$ . Then by drawing all the rays $OQ$ , we separate the eighth-circle into $k$ arcs. The length of the $j^{th}$ arc is $\tan^{-1}\left(\frac{j}{k}\right)-\tan^{-1}\left(\frac{j-1}{k}\right)$ by the formula $s=r\theta$ , with $r=1$ in this case. Since $\tan x$ is concave down, i.e. $\frac{d^{2}}{dx^{2}}\tan x \leq 0$ , when $x\geq 0$ , the longest arc will be the first one, between the first and second rays, formed by the origin and $(k,0)$ and $(k,1)$ , respectively. This arc has length $\tan^{-1}\left(\frac{1}{k}\right)-\tan^{-1}\left(\frac{1-1}{k}\right)=\tan^{-1}\left(\frac{1}{k}\right)$ . Since $\tan^{-1}x<x$ for all $x>0$ , $\tan^{-1}\left(\frac{1}{k}\right)<\frac{1}{k}$ . So all the arcs have length less than $\frac{1}{k}$ . This means that no matter what arc $P$ lies in, the length of the arc between $P$ and the closest $Q'$ must be less than $\frac{1}{2k}$ , which implies $PQ'<\frac{1}{2k}$ , as required.

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