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Difference between revisions of "2018 AMC 12B Problems/Problem 21"
m (→‎Solution 2: The radius of a circle cannot equal to 0. Although this solution gives the right answer, it uses the wrong approach ...)
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<cmath>2.5^2-5x+x^2+6^2-12x+x^2=6.5^2-13x+x^2</cmath>
 
<cmath>2.5^2-5x+x^2+6^2-12x+x^2=6.5^2-13x+x^2</cmath>
 
<cmath>x^2=(5+12-13)x</cmath><cmath>x\neq 0\implies x=4.</cmath>Now the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be <math>\boxed{\textbf{E}.}</math>
 
<cmath>x^2=(5+12-13)x</cmath><cmath>x\neq 0\implies x=4.</cmath>Now the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be <math>\boxed{\textbf{E}.}</math>
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== Solution 2 ==
 
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Notice that we can let <math>M = C</math>. If <math>O = \left(0, 0\right)</math>, then <math>C = \left(6, -\frac{5}{2}\right)</math> and <math>I = \left(4, -\frac{1}{2}\right)</math>. Using shoelace formula, we get <math>\left[COI\right] = \frac{7}{2}</math>. <math>\boxed{\textbf E.}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 08:58, 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 $\triangle{MOI}$?

$\textbf{(A)}\ 5/2\qquad\textbf{(B)}\ 11/4\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 13/4\qquad\textbf{(E)}\ 7/2$

Solution 1

Let the triangle have coordinates $(0,0),(12,0),(0,5).$ 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 $\boxed{\textbf{E}.}$

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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