Question

In $\triangle ABC$ with side lengths $AB=13,$ $AC=12,$ and $BC=5,$ let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle ABC.$ What is the area of $\triangle MOI?$

$\dfrac{5}{2}$

$\dfrac{11}{4}$

$3$

$\dfrac{13}{4}$

$\dfrac{7}{2}$

Answer: E

Concepts:coordinate geometry circumcircle, circumcenter, and circumradius incircle, incenter, and inradius tangent circles

Difficulty rating: 2360

Solution:

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

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

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

Thus, the correct answer is E.

2018 AMC 12B Problem 21

Answer: E

Difficulty rating: 2360

Solution:

Problem 21 in Other Years