2018 AMC 12B Problems/Problem 21

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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 $\triangle{MOI}$?

$\textbf{(A)}\ 5/2\qquad\textbf{(B)}\ 11/4\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 13/4\qquad\textbf{(E)}\ 7/2$

Solution

Let the triangle have coordinates $(0,0),(5,0),(0,12).$ Then the coordinates of the incenter and circumcenter are $(2,2)$ and $(2.5,6),$ respectively. If we let $M=(x,x),$ then $x$ satisfies \[\sqrt{(2.5-x)^2+(6-x)^2}+x=6.5\] \[2.5^2-5x+x^2+6^2-12x+x^2=6.5^2-13x+x^2\] \[x^2=(5+12-13)x\]\[x\neq 0\implies x=4.\]Now the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be $\boxed{\textbf{E}.}$


See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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