# 1982 AHSME Problems/Problem 21

## Problem

In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$, and side $BC=s$. The length of $BN$ is

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10));real r=54.72; pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M); draw(M--C--A--B--C^^B--N); pair point=P; markscalefactor=0.01; draw(rightanglemark(B,C,N)); draw(rightanglemark(C,P,B)); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("M", M, S); label("N", N, dir(C--A)*dir(90)); label("s", B--C, NW); [/asy]$

$\textbf{(A)}\ s\sqrt 2 \qquad \textbf{(B)}\ \frac 32s\sqrt2 \qquad \textbf{(C)}\ 2s\sqrt2 \qquad \textbf{(D)}\ \frac{s\sqrt5}{2}\qquad \textbf{(E)}\ \frac{s\sqrt6}{2}$

## Solution

Suppose that $P$ is the intersection of $\overline{CM}$ and $\overline{BN}.$ Let $BN=x.$ By the properties of centroids, we have $BP=\frac23 x.$

Note that $\triangle BPC\sim\triangle BCN$ by AA. From the ratio of similitude $\frac{BP}{BC}=\frac{BC}{BN},$ we get \begin{align*} BP\cdot BN &= BC^2 \\ \frac23 x\cdot x &= s^2 \\ x^2 &= \frac32 s^2 \\ x &= \boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}. \end{align*} ~MRENTHUSIASM