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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 1

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

Solution 2

Let $O$ be the centroid. Knowing that there are two medians, we need to find the length of the third one. Therefore, we draw the median $AP$ such that $O$ is on $AP$. Then, it follows that $BP = CP = OP = \frac{s}{2}$ by Thales's Theorem, and that $AO = s$. So, $AP = \frac{3}{2}s$, which gives us the idea that $CA = s\sqrt{2}$.

Since $N$ is the median that cuts $CA$, we find out that $CN = AN = \frac{s\sqrt{2}}{2}$. Finally, using Pythagorean again gives $BN=\boxed{{\textbf{(E)}\ \frac{s\sqrt6}{2}}}$.

~elpianista227

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AHSME Problems and Solutions

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