# 2018 AMC 12B Problems/Problem 21

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

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 $\triangle{MOI}$?

$\textbf{(A)}\ \frac52\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{13}{4}\qquad\textbf{(E)}\ \frac72$

## Diagram

$[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, O, I, M; C = origin; A = (12,0); B = (0,5); C = origin; O = circumcenter(A,B,C); I = incenter(A,B,C); M = (4,4); fill(M--O--I--cycle,yellow); draw(A--B--C--cycle^^circumcircle(A,B,C)^^incircle(A,B,C)^^circle(M,4)^^M--O--I--cycle); dot("A",A,1.5*SE,linewidth(4)); dot("B",B,1.5*NW,linewidth(4)); dot("C",C,1.5*SW,linewidth(4)); dot("O",O,1.5*dir((5,12)),linewidth(4)); dot("I",I,1.5*S,linewidth(4)); dot("M",M,1.5*N,linewidth(4)); [/asy]$ ~MRENTHUSIASM

## Solution

In this solution, let the brackets denote areas.

We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$

Since $\triangle ABC$ is a right triangle with $\angle ACB=90^\circ,$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\frac52\right).$ Note that the circumradius of $\triangle ABC$ is $\frac{13}{2}.$

Let $s$ denote the semiperimeter of $\triangle ABC.$ The inradius of $\triangle ABC$ is $\frac{[ABC]}{s}=\frac{30}{15}=2,$ from which $I=(2,2).$

Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or $$\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.$$ Solving this equation, we have $a=4,$ from which $M=(4,4).$

Finally, we apply the Shoelace Theorem to $\triangle MOI:$ $$[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.$$ Remark

Alternatively, we can use $\overline{MI}$ as the base and the distance from $O$ to $\overleftrightarrow{MI}$ as the height for $\triangle MOI:$

• By the Distance Formula, we have $MI=2\sqrt2.$
• The equation of $\overleftrightarrow{MI}$ is $x-y+0=0,$ so the distance from $O$ to $\overleftrightarrow{MI}$ is $h_O=\frac{\left|1\cdot6+(-1)\cdot\frac52+0\right|}{\sqrt{1^2+(-1)^2}}=\frac74\sqrt2.$

Therefore, we get $$[MOI]=\frac12\cdot MI\cdot h_O=\frac72.$$ ~pieater314159 ~MRENTHUSIASM