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Difference between revisions of "1983 AIME Problems/Problem 12"
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== Problem ==
 
== Problem ==
 
The length of diameter <math>AB</math> is a two digit integer. Reversing the digits gives the length of a perpendicular chord <math>CD</math>. The distance from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</math>.
 
The length of diameter <math>AB</math> is a two digit integer. Reversing the digits gives the length of a perpendicular chord <math>CD</math>. The distance from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</math>.
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== Solution ==
 
== Solution ==

Revision as of 02:42, 21 January 2007

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 $AB$. 1983number12.JPG

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 $65$.

See also

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