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

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<u><b>Remark</b></u>
 
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Alternatively, to find <math>[MOI],</math> we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height:
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Alternatively, we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height to find <math>[MOI]:</math>
  
 
* By the Distance Formula, we have <math>MI=2\sqrt2.</math>
 
* By the Distance Formula, we have <math>MI=2\sqrt2.</math>

Revision as of 13:15, 21 October 2021

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 find $[MOI]:$ \[[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.\] Remark

Alternatively, we can use $\overline{MI}$ as the base and the distance from $O$ to $\overleftrightarrow{MI}$ as the height to find $[MOI]:$

  • 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

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
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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