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

Revision as of 19:24, 30 October 2022

Problem

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 $ABB'.$ We can use a formula for finding the radius of the incircle. The area of a triangle $= \text{Semiperimeter} \cdot \text{inradius}$ . The area of $ABB'$ is $12\times5 = 60.$ The semiperimeter is $\dfrac{10+13+13}{2}=18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

Asymptote diagram by Mathandski

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 $\triangle AFO$ ~ $\triangle ABC$ , 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 $r = \boxed{\textbf{(D) }\frac{10}{3}}$

Solution 5

Let us label the center of the semicircle $O$ and the point where the circle is tangent to the triangle $D$ . The area of $\triangle ABC$ = the areas of $\triangle ABO$ + $\triangle ACO$ , which means $(12 \cdot 5)/2 = (13\cdot r)/2 +(5\cdot r)/2$ . So it gives us $r = \boxed{\textbf{(D)}\ \frac{10}{3}}$ .----LarryFlora

Solution 6 (Pythagorean Theorem)

We can draw another radius from the center to the point of tangency. This angle, $\angle{ODB}$ , is $90^\circ$ . Label the center $O$ , the point of tangency $D$ , and the radius $r$ . $[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("$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); draw((8.665,0)--(7.4,3.07)); label("$O$", (8.665, 0), S); label("$D$", (7.4, 3.1), NW); label("$r$", (11, 0), S); label("$r$", (7.6, 1), W); [/asy]$

Since $ODBC$ is a diamond, then $DB=CB=5$ . Also, $AD=13-5=8$ . By the Pythagorean Theorem, $r^2 + 8^2=(12-r)^2$ . Solving, $r^2+64=144-24r+r^2 \Rightarrow 24r=80 \Rightarrow \boxed{\textbf{(D) }\frac{10}{3}}$ .

~MrThinker

Video Solution

https://youtu.be/Y0JBJgHsdGk

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

https://youtu.be/KtmLUlCpj-I - savannahsolver

https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14

2017 AMC 8 (Problems • Answer Key • Resources)
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