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 $\triangle{MOI}$? $\textbf{(A)}\ \frac52\qquad$ $\textbf{(B)}\ \frac{11}{4}\qquad$ $\textbf{(C)}\ 3\qquad$ $\textbf{(D)}\ \frac{13}{4}\qquad$ $\textbf{(E)}\ \frac72$
Solution
We know the incircle has radius $2$, and the circumcenter is the midpoint of the hypotenuse $AB$ (circumradius $\frac{13}2$).
Build the coordinate system with $C(0,0)$, $A(12,0)$, $B(0,5)$. Then $I(2,2)$, $O(6,\frac52)$, and $M(t,t)$ with radius $t$.
The circle $M$ is internally tangent to the circumcircle, so $OM=\frac{13}2-t$, which gives $(t-6)^2+(t-\frac52)^2=(\frac{13}2-t)^2$.
Solving, $t=4$, so $M$ has coordinate $(4,4)$.
To use shoelace's theorem, we move $(2,2)$ to the origin, then $O$ will be $(4, \frac12), M$ will be $(2,2)$, so the area is $\frac12(8-1)=\frac72$.
In $\triangle ABC$ with $AB=20,AC=16,BC=12$ (a right triangle), let $O,I$ be the circumcenter, incenter; a circle centered at $M$ is tangent to legs $AC,BC$ and to the circumcircle. What is the area of $\triangle MOI$?
