Difference between revisions of "1982 AHSME Problems/Problem 21"
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\textbf{(B)}\ \frac 32s\sqrt2 \qquad | \textbf{(B)}\ \frac 32s\sqrt2 \qquad | ||
\textbf{(C)}\ 2s\sqrt2 \qquad | \textbf{(C)}\ 2s\sqrt2 \qquad | ||
| − | \textbf{(D)}\ \frac{ | + | \textbf{(D)}\ \frac{s\sqrt5}{2}\qquad |
| − | \textbf{(E)}\ \frac{ | + | \textbf{(E)}\ \frac{s\sqrt6}{2}</math> |
== Solution == | == Solution == | ||
| Line 30: | Line 30: | ||
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> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
== See Also == | == See Also == | ||
{{AHSME box|year=1982|num-b=20|num-a=22}} | {{AHSME box|year=1982|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:14, 14 September 2021
Problem
In the adjoining figure, the triangle 
is a right triangle with 
. Median 
is perpendicular to median 
,
and side 
. The length of 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]](http://latex.artofproblemsolving.com/9/8/f/98f891d054bcb9cbfc2bc384c989c5b6725d2e09.png)
Solution
Let 
be the intersection of 
and 
By the properties of centroids, we have 
and 
Note that 
and 
are both complementary to 
so 
By AA, we conclude that 
with the ratio of similitude 
from which ![\[CP^2=BP\cdot NP=\frac29 BN^2.\]](http://latex.artofproblemsolving.com/f/1/c/f1c1f303db9f1b21ff139ba5ea3a41ffdd37894f.png)
Applying the Pythagorean Theorem to right 
we get 
from which ![\[\frac23 BN^2=s^2.\]](http://latex.artofproblemsolving.com/7/e/3/7e308226122a900de04253cb919e22f4a5f0f921.png)
Solving for gives 
~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. 
