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 2)
(→‎Solution 2)
Line 40: Line 40:
 
==Solution 2==
 
==Solution 2==
  
−
Let <math>O</math> be the centroid. Knowing that there are two medians, we need to find the length of the third one. Therefore, we draw the median <math>AP</math> such that <math>O</math> is on <math>AP</math>. Then, it follows that <math>BP = CP = OP = \frac{s}{2}</math> by Thales's Theorem, and that <math>AO = s</math>. So, <math>AP = \frac{3}{2}s</math>, which gives us the idea that <math>CA = s\sqrt{2}</math>. \\\\
+
Let <math>O</math> be the centroid. Knowing that there are two medians, we need to find the length of the third one. Therefore, we draw the median <math>AP</math> such that <math>O</math> is on <math>AP</math>. Then, it follows that <math>BP = CP = OP = \frac{s}{2}</math> by Thales's Theorem, and that <math>AO = s</math>. So, <math>AP = \frac{3}{2}s</math>, which gives us the idea that <math>CA = s\sqrt{2}</math>. \\
 
Since <math>N</math> is the median that cuts <math>CA</math>, we find out that <math>CN = AN = \frac{s\sqrt{2}}{2}</math>. Finally, using Pythagorean again gives <math>\boxed{BN = \frac{s\sqrt{6}}{2} \textbf{(E)}}</math>.
 
Since <math>N</math> is the median that cuts <math>CA</math>, we find out that <math>CN = AN = \frac{s\sqrt{2}}{2}</math>. Finally, using Pythagorean again gives <math>\boxed{BN = \frac{s\sqrt{6}}{2} \textbf{(E)}}</math>.
  

Revision as of 21:14, 8 December 2024

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

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 $\boxed{BN = \frac{s\sqrt{6}}{2} \textbf{(E)}}$.

~elpianista227

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=236544"