1993 AIME Problems/Problem 15

Problem

Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ , $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relatively prime integers. Find $m + n\,$ .

Solution

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From the Pythagorean Theorem, $AH^2+CH^2=1994^2$ , and $(1995-AH)^2+CH^2=1993^2$ . Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$ . After simplification, we see that $2*1995AH-1995^2=3987$ , or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$ . Note that $AH+BH=1995$ . Therefore we have that $BH=\frac{1995}{2}-\frac{3987}{2*1995}$ . Therefore $AH-BH=\frac{3987}{1995}$ .

Now note that $RS=|HR-HS|$ , $RH=\frac{BH+CH-BC}{2}$ , and $HS=\frac{CH+AH-AC}{2}$ . Therefore we have

$RS=|\frac{BH+CH-BC-CH-AH+AC}{2}|=\frac{|BH-AH-1993+1994|}{2}$ .

Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665}$ .

1993 AIME (Problems • Answer Key • Resources)
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1993 AIME Problems/Problem 15

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