Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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<math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math> | <math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math> | ||
+ | <asy> | ||
+ | pair A,B,C,D,E,P; | ||
+ | A=(0,0); | ||
+ | B=(80,0); | ||
+ | C=(20,40); | ||
+ | D=(50,30); | ||
+ | E=(40,0); | ||
+ | P=(33.3,13.3); | ||
+ | draw(A--B); | ||
+ | draw(B--C); | ||
+ | draw(A--C); | ||
+ | draw(C--E); | ||
+ | draw(A--D); | ||
+ | draw(D--E); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | dot(P); | ||
+ | label("A",A,NNW); | ||
+ | label("B",B,NNE); | ||
+ | label("C",C,ENE); | ||
+ | label("D",D,ESE); | ||
+ | label("E",E,SSE); | ||
+ | label("P",P,SSE); | ||
+ | </asy> | ||
==Solution== | ==Solution== |
Revision as of 19:31, 22 February 2013
Problem
In triangle , medians and intersect at , , , and . What is the area of ?
Solution
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