Difference between revisions of "1982 AHSME Problems/Problem 21"
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− | 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
Contents
[hide]Problem
In the adjoining figure, the triangle is a right triangle with
. Median
is perpendicular to median
,
and side
. The length of
is
Solution
Suppose that is the intersection of
and
Let
By the properties of centroids, we have
Note that by AA. From the ratio of similitude
we get
~MRENTHUSIASM
Solution 2
Let be the centroid. Knowing that there are two medians, we need to find the length of the third one. Therefore, we draw the median
such that
is on
. Then, it follows that
by Thales's Theorem, and that
. So,
, which gives us the idea that
. \
Since
is the median that cuts
, we find out that
. Finally, using Pythagorean again gives
.
~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 |
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