1993 AIME Problem 15

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

Express the distance from

H

to each tangency point using the semiperimeter of its right triangle

Find

AH-BH

from the side lengths without first computing the altitude

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.

Problem 15 in other years