2020 AMC 10A Problem 12

Below is the video solution and professionally curated solution for Problem 12 of the 2020 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

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Concepts:median (geometry)centroidcoordinate geometry

Difficulty rating: 1660

12.

Triangle AMCAMC is isosceles with AM=AC.AM = AC. Medians MV\overline{MV} and CU\overline{CU} are perpendicular to each other, and MV=CU=12.MV=CU=12. What is the area of AMC?\triangle AMC?

4848

7272

9696

144144

192192

Video solution:
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Written solution:

Let the centroid be the origin. Since a centroid divides each median in a 2:12:1 ratio, we may place median MVMV horizontally with M=(8,0)M=(8,0) and V=(4,0)V=(-4,0), and median CUCU vertically with C=(0,8)C=(0,8) and U=(0,4)U=(0,-4).

Because UU is the midpoint of AMAM, we get A=2UM=(8,8)A=2U-M=(-8,-8). The area of AMC\triangle AMC is 12(16,8)×(8,16)=12(25664)=96\dfrac12 |(16,8)\times(8,16)|=\dfrac12(256-64)=96. Thus, C is the correct answer.

Problem 12 in Other Years