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2018 AMC 12B Problem 21

Attempt Problem 21 of the 2018 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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21.

In △ABC\triangle ABC with side lengths AB=13,AB=13, AC=12,AC=12, and BC=5,BC=5, let OO and II denote the circumcenter and incenter, respectively. A circle with center MM is tangent to the legs ACAC and BCBC and to the circumcircle of △ABC.\triangle ABC. What is the area of △MOI?\triangle MOI?

52\dfrac{5}{2}

114\dfrac{11}{4}

33

134\dfrac{13}{4}

72\dfrac{7}{2}

Answer: E
Concepts:coordinate geometrycircumcircle, circumcenter, and circumradiusincircle, incenter, and inradiustangent circles
Difficulty rating: 2360
Small Hint:

The triangle is right-angled at C;C; put CC at the origin with the legs on the axes

Big Hint:

OO is the midpoint of the hypotenuse and I=(2,2);I=(2,2); the circle at M=(ρ,ρ)M=(\rho,\rho) is tangent to the circumcircle

Solution:

Since 52+122=132,5^2+12^2=13^2, the triangle is right-angled at C.C. Set C=(0,0),C=(0,0), A=(12,0),A=(12,0), and B=(0,5).B=(0,5). Then OO is the midpoint of AB‾,\overline{AB}, namely O=(6,52),O=\left(6,\tfrac52\right), with circumradius 132.\tfrac{13}{2}. The inradius is areas=3015=2,\tfrac{\text{area}}{s}=\tfrac{30}{15}=2, so I=(2,2).I=(2,2).

Because MM’s circle is tangent to both legs, M=(ρ,ρ).M=(\rho,\rho). Internal tangency to the circumcircle gives MO=132−ρ.MO=\tfrac{13}{2}-\rho. Setting this equal to (ρ−6)2+(ρ−52)2\sqrt{(\rho-6)^2+\left(\rho-\tfrac52\right)^2} and solving gives ρ=4,\rho=4, so M=(4,4).M=(4,4).

The shoelace formula on M=(4,4),M=(4,4), O=(6,52),O=\left(6,\tfrac52\right), I=(2,2)I=(2,2) gives area 72.\tfrac72.

Thus, the correct answer is E.

Problem 20#20
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Problem 22#22

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