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

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 (vibe check solution 😎😎😎)

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 keeping the idea that cos(θ) = $\frac{5}{13}$ , we find that tan(θ/2) = sqrt((1-cosθ)/(1+cosθ)) = $\sqrt{\frac{4}{9}}$ = $\frac{2}{3}$ . That's rockin and all, but the real deal starts when we acknowledge that we can find the length of BC in terms of the radius of the incircle, which I'll be calling r. Using the knowledge that $\triangle RIB$ is right, and some trigonometric tomfoolery, 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. Absolutely radical. 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 collectively determine that $MQ$ = $BM$ - $BQ$ = 13/2 - 3 = 7/2. Fantabular. We also know that $IQ$ has a measure of 2 since it's the radius of circle I (me thinks you would find this relatable (cause u have an iq of 2 lol haha)).

Now we get vibey. We know very little about the placement of $M$ , but that's all about to change. First, we can conclude that $\triangle IOQ$ is part of $\triangle MOI$ . This much is true. We also feel out that $M$ has to be higher up than $O$ . I mean, it just makes sense. Now for the vibey part. We're about to feel out this solution. We find the area of $\triangle IOQ$ . It's 7/2. We glance at the answer choices. 7/2 is the greatest option. erm whart the sigma??? I sincerely doubt 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 could only mean one thing...*GASP*! This whole time, $\overline{QO}$ was on $\overline{AB}$ !? THEREFORE WE CAN CONCLUDE $\triangle IOQ=\triangle MOI$ . That means the answer must be $\boxed{\textbf{(E)}\ \frac72}$ , and therefore, this question passes the vibe check. Thanks for following fam and I'll catch you in a jiffy! ~me

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 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 $\overline{AC}$ . 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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