1983 AIME Problems/Problem 12

Problem

Diameter $AB$ of a circle has length a $2$ -digit integer (base ten). Reversing the digits gives the length of the 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 $CH$ , we deduce $OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}=\frac{3}{2}\sqrt{11(x+y)(x-y)}$

Because $OH$ is a positive rational number, the quantity $\sqrt{11(x+y)(x-y)}$ must be rational, so $11(x+y)(x-y)$ must be a perfect square. Hence either $x-y$ or $x+y$ must be a multiple of $11$ , but as $x$ and $y$ are digits, $1+0\leqx+y\leq9+9=18$ (Error compiling LaTeX. Unknown error_msg), so the only possible multiple of $11$ is $11$ itself. However, $x-y$ cannot be 11, because both must be digits. Therefore, $x+y$ must equal $11$ and $x-y$ must be a perfect square. The only pair $(x,y)$ that satisfies this condition is $(6,5)$ , so our answer is $\boxed{065}$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1983 AIME Problems/Problem 12

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