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Difference between revisions of "2017 AMC 8 Problems/Problem 22"

(Created page with "==Problem 22== In the right triangle <math>ABC</math>, <math>AC=12</math>, <math>BC=5</math>, and angle <math>C</math> is a right angle. A semicircle is inscribed in the tria...")
 
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<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}</math>
 
<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}</math>
  
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[[2017 AMC 8 Problems/Problem 22|Solution
 
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==Solution:==
 
==Solution:==
 
We can reflect triangle <math>ABC</math> on line <math>AC.</math> This forms the triangle <math>AB'C</math> and a circle out of the semicircle. Let us call the center of the circle <math>O.</math>
 
We can reflect triangle <math>ABC</math> on line <math>AC.</math> This forms the triangle <math>AB'C</math> and a circle out of the semicircle. Let us call the center of the circle <math>O.</math>
  
 
We can see that Circle <math>O</math> is the incircle of <math>AB'C.</math> We can use the formula for finding the radius of the incircle to solve this problem. The are of <math>AB'C</math> is <math>12\times5 = 60.</math> The semiperimeter is <math>5+13 = 18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore <math>\text{D)}</math> <math>\dfrac{10}{3}.</math>
 
We can see that Circle <math>O</math> is the incircle of <math>AB'C.</math> We can use the formula for finding the radius of the incircle to solve this problem. The are of <math>AB'C</math> is <math>12\times5 = 60.</math> The semiperimeter is <math>5+13 = 18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore <math>\text{D)}</math> <math>\dfrac{10}{3}.</math>

Revision as of 14:40, 22 November 2017

Problem 22

In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle? [asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("$A$", (0,0), W); label("$C$", (12,0), E); label("$B$", (12,5), NE); label("$12$", (6, 0), S); label("$5$", (12, 2.5), E);[/asy]

$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}$

Solution:

We can reflect triangle $ABC$ on line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$

We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The are of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\text{D)}$ $\dfrac{10}{3}.$

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