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

Edit by GameMaster402:

It can be shown that in any triangle with side lengths $n-1, n, n+1$, if you draw an altitude from the vertex to the side of $n+1$, and draw the incircles of the two right triangles, the distance between the two tangency points is simply $\frac{n-2}{2n+2)}=\frac{n-2}{2(n+1)}$.

Plugging in $n=1994$ yields that the answer is $\frac{1992}{2(1995)}$, which simplifies to $\frac{332}{665}$

Edit by phoenixfire:

It can further be shown for any triangle with sides $a=BC, b=CA, c=AB$ that $$RS=\dfrac{|b-a|}{2c}|a+b-c|$$ Over here $a=1993, b=1994, c=1995$.