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 1

Let the triangle have coordinates $(0,0),(5,0),(0,12).$ Then the coordinates of the incenter and circumcenter are $(2,2)$ and $(6,2.5),$ 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 2

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

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2018 AMC 12B Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

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