Difference between revisions of "1982 AHSME Problems/Problem 21"
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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> | 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 < | + | 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> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 17:48, 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
Equating the expressions for
gives
from which
~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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