# 1985 AHSME Problems/Problem 3

## Problem

In right $\triangle ABC$ with legs $5$ and $12$, arcs of circles are drawn, one with center $A$ and radius $12$, the other with center $B$ and radius $5$. They intersect the hypotenuse in $M$ and $N$. Then $MN$ has length

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("M", M, dir(point--M)); label("N", N, dir(point--N)); label("12", (6,0), S); label("5", (12,3.5), E);[/asy]$

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }\frac{24}{5}$

## Solution

Firstly, the Pythagorean theorem gives \begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$, and so $MN = BN-BM = 5-1 = \boxed{\text{(D)} \ 4}$.