Difference between revisions of "1982 AHSME Problems/Problem 21"
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\textbf{(E)}\ \frac{s\sqrt6}{2}</math> | \textbf{(E)}\ \frac{s\sqrt6}{2}</math> | ||
− | == Solution == | + | == Solution 1 == |
Suppose that <math>P</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{BN}.</math> Let <math>BN=x.</math> By the properties of centroids, we have <math>BP=\frac23 x.</math> | Suppose that <math>P</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{BN}.</math> Let <math>BN=x.</math> By the properties of centroids, we have <math>BP=\frac23 x.</math> | ||
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==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>BN=\boxed{{\textbf{(E)}\ \frac{s\sqrt6}{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>BN=\boxed{{\textbf{(E)}\ \frac{s\sqrt6}{2}}}</math>. | ||
Latest revision as of 01:30, 9 December 2024
Contents
Problem
In the adjoining figure, the triangle is a right triangle with . Median is perpendicular to median , and side . The length of is
Solution 1
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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