AIME 1983 · 第 12 题

Solution

Let $AB=10x+y$ and $CD=10y+x$. It follows that $CO=\frac{AB}{2}=\frac{10x+y}{2}$ and $CH=\frac{CD}{2}=\frac{10y+x}{2}$. Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on $2CO$, $2OH$ and $2CH$, we deduce

$$(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)$$ Because $OH$ is a positive rational number and $x$ and $y$ are integral, the quantity $99(x+y)(x-y)$ must be a perfect square. Hence either $x-y$ or $x+y$ must be a multiple of $11$, but as $x$ and $y$ are different 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 $\boxed{065}$. (Therefore $CD = 56$ and $OH = \frac{33}{2}$.)

Alternate start to solution 1

Since H is the midpoint of $CD$, by Power of a Point, $CH^2=(AH)(BH)$. Because $AH=r-OH$ and $BH=r+OH$, where $r$ is the radius of the circle, we deduce the relation $CH^2=r^2-OH^2$. Thus $OH^2=\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2$. Continue as above.

~anduran