1961 AHSME Problems/Problem 21
Problem
Medians and of intersect in . The midpoint of is . Let the area of be times the area of . Then equals:
Solution
Let and . The altitudes and bases of and are the same, so . Since the base of is twice as long as and the altitudes of both triangles are the same, .
Since the altitudes and bases of and are the same, . Additionally, since and have the same area, and also have the same area, so .
Also, the altitudes and bases of and are the same, so the area of the two triangles are the same. Thus, Thus, the area of is , so the area of is the area of . The answer is .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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