1983 AIME Problems/Problem 12

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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 $AB$.

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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, $1+0=1 \leq x+y \leq 9+9=18$, 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 $\boxed{065}$.

See Also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions