Difference between revisions of "2018 AMC 12B Problems/Problem 21"

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 $\boxed{\textbf{E}.}$

Solution

Notice that we can let $M = C$. If $M = \left(0, 0\right)$, then $C = \left(6, -\frac{5}{2}\right)$ and $I = \left(4, -\frac{1}{2}\right)$. Using shoelace formula, we get $\left[COI\right] = \frac{7}{2}$. $\boxed{\textbf E.}$