Question

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 $\frac{m}{n},$ where $m$ and $n$ are relatively prime integers. Find $m+n.$

Answer

Answer: 997

Concepts:altitude incircle, incenter, and inradius law of cosines

Difficulty rating: 2560

Small Hint:

Express the distance from $H$ to each tangency point using the semiperimeter of its right triangle

Big Hint:

Find $AH-BH$ from the side lengths without first computing the altitude

Solution:

Let $h=CH.$ In right triangle $ACH,$ the tangent length from $H$ to its incircle is $\frac{AH+h-AC}{2};$ in triangle $BCH,$ it is $\frac{BH+h-BC}{2}.$ Hence $\begin{aligned}RS&=\frac12\left|AH-BH\right.\\&\qquad\left.-AC+BC\right|.\end{aligned}$ The projection formula gives $\begin{aligned}AH-BH&=\frac{AC^2-BC^2}{AB}\\&=\frac{1994^2-1993^2}{1995}\\&=\frac{3987}{1995}.\end{aligned}$ Since $AC-BC=1,$ $RS=\frac12\left(\frac{3987}{1995}-1\right)=\frac{332}{665}.$ Thus $m+n=332+665=997.$

1993 AIME Problem 15

Answer: 997

Difficulty rating: 2560

Small Hint:

Big Hint:

Solution:

Problem 15 in Other Years