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 $m + n\,$.
From the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/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 $\frac{332}{665}$
~minor edit by [Yiyj1](https://artofproblemsolving.com/wiki/index.php/User:Yiyj1)


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 $a=1993, b=1994, c=1995$, so using the formula gives \[RS = \dfrac{|1994 - 1993|}{2 \cdot 1995}|1993 + 1994 - 1995| = \dfrac{1 \cdot 1992}{2(1995)} = \frac{332}{665}.\]
~minor edit by [Yiyj1](https://artofproblemsolving.com/wiki/index.php/User:Yiyj1)
Note: We can also just right it as $RS=\frac{|b-a|(a+b-c)}{2c}$ since $a+b-c \geq 0$ by the triangle inequality. ~[Yiyj1](https://artofproblemsolving.com/wiki/index.php/User:Yiyj1)