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

$[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]$

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 $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$ .

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

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