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 == | ||
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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
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
Applying the Pythagorean Theorem to right
we get
from which
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.