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 $AB$ .

$[asy] draw(circle((0,0),4)); draw((-4,0)--(4,0)); draw((-2,-2*sqrt(3))--(-2,2*sqrt(3))); draw((-2.6,0)--(-2.6,0.6)); draw((-2,0.6)--(-2.6,0.6)); dot((0,0)); dot((-2,0)); dot((4,0)); dot((-4,0)); dot((-2,2*sqrt(3))); dot((-2,-2*sqrt(3))); label("A",(-4,0),W); label("B",(4,0),E); label("C",(-2,2*sqrt(3)),NW); label("D",(-2,-2*sqrt(3)),SW); label("H",(-4,0),SE); label("O",(0,0),NE);[/asy]$

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 $OH = \frac{33}{2}$ .)

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
All AIME Problems and Solutions

Page

Toolbox

Search

1983 AIME Problems/Problem 12

Contents

Problem

Solution

See Also