Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1982 AHSME Problems/Problem 21"

(→‎Solution: Made the explanation longer and clearer.)
(Shortened the solution significantly.)
Line 27: Line 27:
  
 
== Solution ==
 
== Solution ==
−
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>
+
Suppose that <math>P</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{BN}.</math> Let <math>BN=x.</math> By the properties of centroids, we have <math>BP=\frac23 x.</math>
−
 
 
−
Note that <math>\angle BCP</math> and <math>\angle CNP</math> are both complementary to <math>\angle NCP,</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 <cmath>CP^2=BC^2-BP^2=s^2-\frac49 BN^2.</cmath>
 
−
Equating the expressions for <math>CP^2</math> gives <cmath>\frac29 BN^2=s^2-\frac49 BN^2,</cmath> from which <math>BN=\boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}.</math>
 
  
 +
Note that <math>\triangle BPC\sim\triangle BCN</math> by AA, with the ratio of similitude <math>\frac{BP}{BC}=\frac{BC}{BN},</math> from which <cmath>\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*}</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  

Revision as of 19:06, 15 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

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, with the ratio of similitude $\frac{BP}{BC}=\frac{BC}{BN},$ from which \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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_21&oldid=162307"