Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 38: Line 38:
 
Let <math>s</math> denote the semiperimeter of <math>\triangle ABC.</math> The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</math>
 
Let <math>s</math> denote the semiperimeter of <math>\triangle ABC.</math> The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</math>
  
−
Since <math>\odot M</math> is tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath>
+
Since <math>\odot M</math> is also tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath>
 
Solving this equation, we have <math>a=4.</math>
 
Solving this equation, we have <math>a=4.</math>
  

Revision as of 23:10, 20 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 $a=4.$

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

  • 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&oldid=163836"