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

Let $P$ be the intersection of $\overline{CM}$ and $\overline{BN}.$ By the properties of centroids, we have $BP=\frac23 BN$ and $PN=\frac13 BN.$

Note that $\angle BCP$ and $\angle CNP$ are both complementary to $\angle PCN,$ so $\angle BCP=\angle CNP.$ By AA, we conclude that $\triangle BCP\sim\triangle CNP,$ with the ratio of similitude $\frac{BP}{CP}=\frac{CP}{NP},$ from which $CP^2=BP\cdot NP=\frac29 BN^2.$ Applying the Pythagorean Theorem to right $\triangle BCP,$ we get $BP^2+PC^2=BC^2,$ from which $\frac23 BN^2=s^2.$ Solving for $BN$ gives $BN=\boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}.$

~MRENTHUSIASM

1982 AHSME (Problems • Answer Key • Resources)
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1982 AHSME Problems/Problem 21

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