2020 AMC 10A Problem 12

Question

Triangle

AMC

is isosceles with

AM = AC.

Medians

\overline{MV}

and

\overline{CU}

are perpendicular to each other, and

MV=CU=12.

What is the area of

\triangle AMC?

48

72

96

144

192

Put the centroid at the origin and align the perpendicular medians with the axes

The centroid divides each median in a

2:1

ratio

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Let the centroid be the origin. Since a centroid divides each median in a

2:1

ratio, we may place median

MV

horizontally with

M=(8,0)

and

V=(-4,0)

, and median

CU

vertically with

C=(0,8)

and

U=(0,-4)

.

Because

U

is the midpoint of

AM

, we get

A=2U-M=(-8,-8)

. The area of

\triangle AMC

is

\dfrac12 |(16,8)\times(8,16)|

=\dfrac12(256-64)=96

. Thus, C is the correct answer.

Problem 12 in other years