1983 AIME Problems/Problem 12

Problem

The length of diameter $AB$ is a two digit integer. Reversing the digits gives the length of a perpendicular chord $CD$ . The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$ .

Solution

Let $AB=10x+y$ and $CD=10y+x$ . It follows that $CO=\frac{AB}{2}=\frac{10x+y}{2}$ and $CH=\frac{CD}{2}=\frac{10y+x}{2}$ . Applying the Pythagorean Theorem on $CO$ and $CH$ , $OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}=\frac{3}{2}\sqrt{11(x+y)(x-y)}$ .

Because $OH$ is a positive rational number, the quantity $\sqrt{11(x+y)(x-y)}$ cannot contain any square roots. Therefore, $x+y$ must equal eleven and $x-y$ must be a perfect square (since $x+y>x-y$ ). The only pair $(x,y)$ that satisfies this condition is $(6,5)$ , so our answer is $65$ .

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1983 AIME Problems/Problem 12

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