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

m (Solution 1)
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Solving for <math>r</math> gives  
Solving for <math>r</math> gives  
<cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}</cmath>
<cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}</cmath>
==Video Solution==
https://youtu.be/3VjySNobXLI - Happytwin
==See Also==
==See Also==

Revision as of 18:09, 10 May 2020

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 $O.$ [asy] draw((0,0)--(12,0)--(12,5)--cycle); draw((0,0)--(12,0)--(12,-5)--cycle); draw(circle((8.665,0),3.3333)); label("$A$", (0,0), W); label("$C$", (12,0), E); label("$B$", (12,5), NE); label("$B'$", (12,-5), NE); label("$12$", (7, 0), S); label("$5$", (12, 2.5), E); label("$5$", (12, -2.5), E);[/asy] 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 $= \text{Semiperimeter} \cdot \text{inradius}$ . 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 $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

Asymptote diagram by Mathandski

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 \[r = \boxed{\textbf{(D) }\frac{10}{3}}\]

Video Solution

https://youtu.be/3VjySNobXLI - Happytwin

See Also

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

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