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

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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>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math>
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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 area 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>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math>
  
 
==See Also==
 
==See Also==

Revision as of 14:42, 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 area of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 AJHSME/AMC 8 Problems and Solutions

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