Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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Revision as of 18:03, 1 February 2015
Problem
In triangle , medians and intersect at , , , and . What is the area of ?
Solution
Solution 1
Let us use mass points: Assign mass . Thus, because is the midpoint of , also has a mass of . Similarly, has a mass of . and each have a mass of because they are between and and and respectively. Note that the mass of is twice the mass of , so AP must be twice as long as . PD has length , so has length and has length . Similarly, is twice and , so and . Now note that triangle is a right triangle with the right angle . This means that the quadrilateral is a kite. The area of a kite is half the product of the diagonals, and . Recall that they are and respectively, so the area of is
Solution 2
Note that triangle is a right triangle, and that the four angles that have point are all right angles. Using the fact that the centroid () divides each median in a ratio, and . Quadrilateral is now just four right triangles. The area is
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AMC 10 Problems and Solutions |
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