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Difference between revisions of "1982 AHSME Problems/Problem 21"
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== Solution ==
 
== Solution ==
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Let <math>P</math> be the intersection of <math>\overline{CM}</math> and <math>\overline{BN}.</math> By the properties of centroids, we have <math>BP=\frac23 BN</math> and <math>PN=\frac13 BN.</math>
+
Let <math>P</math> be the intersection of <math>\overline{CM}</math> and <math>\overline{BN}.</math> By the properties of centroids, we have <math>BP=\frac23 BN</math> and <math>NP=\frac13 BN.</math>
  
 
Note that <math>\angle BCP</math> and <math>\angle CNP</math> are both complementary to <math>\angle PCN,</math> so <math>\angle BCP=\angle CNP.</math> By AA, we conclude that <math>\triangle BCP\sim\triangle CNP,</math> with the ratio of similitude <math>\frac{BP}{CP}=\frac{CP}{NP},</math> from which <cmath>CP^2=BP\cdot NP=\frac29 BN^2.</cmath>
 
Note that <math>\angle BCP</math> and <math>\angle CNP</math> are both complementary to <math>\angle PCN,</math> so <math>\angle BCP=\angle CNP.</math> By AA, we conclude that <math>\triangle BCP\sim\triangle CNP,</math> with the ratio of similitude <math>\frac{BP}{CP}=\frac{CP}{NP},</math> from which <cmath>CP^2=BP\cdot NP=\frac29 BN^2.</cmath>
−
Applying the Pythagorean Theorem to right <math>\triangle BCP,</math> we get <math>BP^2+PC^2=BC^2,</math> from which <cmath>\frac23 BN^2=s^2.</cmath> Solving for <math>BN</math> gives <math>BN=\boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}.</math>
+
Applying the Pythagorean Theorem to right <math>\triangle BCP,</math> we get <math>BP^2+CP^2=BC^2,</math> from which <cmath>\frac23 BN^2=s^2.</cmath> Solving for <math>BN</math> gives <math>BN=\boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 00:24, 14 September 2021

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 $NP=\frac13 BN.$

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 $BP^2+CP^2=BC^2,$ from which \[\frac23 BN^2=s^2.\] Solving for $BN$ gives $BN=\boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}.$

~MRENTHUSIASM

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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