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

Revision as of 12:33, 30 August 2019

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 1

We can reflect triangle $ABC$ over 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: Area of a triangle = Semi-perimeter $\times$ inradius . 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 $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

Solution 2

We immediately see that $AB=13$ , and we label the center of the semicircle $O$ And the point where the circle is tangent to the triangle $D$ . Drawing radius $OD$ with length $x$ such that $OD$ is perpendicular to $AB$ , we immediately see that $ODB\cong OCB$ because of $\operatorname{HL}$ congruence, so $BD=5$ and $DA=8$ . By similar triangles $ODA$ and $BCA$ , we see that $\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{(D)}\ \frac{10}{3}}$ .

Solution 3

Let the center of the semicircle be $O$ . Let the point of tangency between line $AB$ and the semicircle be $F$ . Angle $BAC$ is common to triangles $ABC$ and $AFO$ . By tangent properties, angle $AFO$ must be $90$ degrees. Since both triangles $ABC$ and $AFO$ are right and share an angle, $AFO$ is similar to $ABC$ . The hypotenuse of $AFO$ is $12 - r$ , where $r$ is the radius of the circle. (See for yourself) The short leg of $AFO$ is $r$ . Because $AFO$ ~ $ABC$ , we have $r/(12 - r) = 5/13$ and solving gives $r = \boxed{\textbf{(D)}\ \frac{10}{3}}$

Solution 4

Let the tangency point on $AB$ be $D$ . Note $AD = AB-BD = AB-BC = 8$ By Power of a Point, $12(12-2r) = 8^2$ Solving for $r$ gives $r = \boxed{\textbf{(D) }\frac{10}{3}}$

@@ Line 16: / Line 16: @@
 We can reflect triangle <math>ABC</math> over 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: Area of a triangle = Sem-perimeter <math> \times </math> inradius .  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>
+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: Area of a triangle = Semi-perimeter <math> \times </math> inradius .  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>
 ==Solution 2==

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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