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1993 AIME Problems/Problem 15

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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

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

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

Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$ .

See also

1993 AIME (Problems • Resources)
Preceded by Problem 14	Followed by Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Category: Intermediate Geometry Problems

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

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