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 $\triangle{MOI}$ ?

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

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 1 (Coordinate Geometry)

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

Solution 2

Let points Q and R be the points of tangency between the incircle and lines $AB$ and $BC$ . Notice that $\angle RIB$ is half of $\angle ABC$ . Let $m\angle ABC$ = θ. Using the half angle tangent formula and remembering that cos(θ) = $\frac{5}{13}$ , we find that tan(θ/2) = sqrt((1-cosθ)/(1+cosθ)) = $\sqrt{\frac{4}{9}}$ = $\frac{2}{3}$ . Now we can find the length of BC in terms of the radius of the incircle, which will be r. Using the knowledge that $\triangle RIB$ is right with some trigonometry, we find that $BR$ is 3r/2. We also find that $CR$ is the r, and so we can create the equation $BR$ + $CR$ = $BC$ -> 3r/2 + r = 5 -> 5r/2. We conclude r = 2. We now accept the fact that quadrilateral $BQIR$ is a kite, so $BR$ = $BQ$ = 3. We also know that O lies on $BA$ and divides $BA$ in half. Next we determine that $MQ$ = $BM$ - $BQ$ = 13/2 - 3 = 7/2. We also know that $IQ$ has a measure of 2 since it's the radius of circle I

We know very little about the placement of $M$ , so we work on that. First, we can conclude that $\triangle IOQ$ is part of $\triangle MOI$ . We guess that $M$ has to be higher up than $O$ . We find the area of $\triangle IOQ$ to be 7/2, which is the highest answer choice. I don't think that $M$ is below $\overline{AB}$ , but if it were to go past $\overline{AB}$ then the answer would be GREATER than 7/2. This means $\overline{QO}$ is on $\overline{AB}$ . Therefore $\triangle IOQ=\triangle MOI$ . The answer must be $\boxed{\textbf{(E)}\ \frac72}$ .

~me

~Judokid (revisions)

Solution 3

The circle with center $M$ is the $C$ -mixtilinear incircle $\omega$ of $\triangle ABC$ . Let $T$ be the intersection between $\omega$ and the circumcircle $\Omega$ . Then, there is a homothety centered at $T$ sending $\omega$ to $\Omega$ . As such, the tangent $\overline{AC}$ gets sent to a parallel tangent to $\Omega$ , which thus must be the arc midpoint $M_B$ of arc ${CA}$ . Thus, by inscribed angle theorem $CM_C$ and $BM_B$ intersect at $I$ , so $EF$ passes through $I$ where $E$ , $F$ are the tangency points of $\omega$ with $\overline{BC}$ and $\overline{AC}$ by Pascal's Theorem. Thus, we see since $\angle BCA = 90 ^{\circ}$ , $r_{\omega} = 2 r_{\text{incircle}} = 4$ . Set up a coordinate plane and apply Shoelace to obtain $[MOI] = \boxed{ \textbf{(E)} \frac{7}{2}}$ .

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