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

Revision as of 22:04, 28 January 2023

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. $[asy] draw((0,0)--(12,0)--(12,5)--cycle); draw((0,0)--(12,0)--(12,-5)--cycle); draw(circle((8.665,0),3.3333)); label("$A$", (0,0), hhhhhhhhhhhhhhhhhhhhhhhh); label("$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 our circle 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 2

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 3

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 4

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 $\triangle ABO$ + $\triangle BCO$ , 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 5 (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 kite, 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

Solution 6 (Basic Trigonometry)

If we reflect triangle $ABC$ over line $AC$ , we will get isosceles triangle $ABD$ . By the Pythagorean Theorem, we are capable of finding out that the $AB = AD = 13$ . Hence, $\tan \frac{\angle A}{2} = \tan \angle BAD = \frac{5}{12}$ . Therefore, as of triangle $ABD$ , the radius of its inscribed circle $r = \frac{tan \frac{\angle A}{2}\cdot (AB + AD - BD)}{2} = \frac{\frac{5}{12} \cdot 16}{2} = \boxed{\textbf{(D) }\frac{10}{3}}$

~Bloggish

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=3837

- pi_is_3.14

Video Solutions

https://youtu.be/Y0JBJgHsdGk

https://youtu.be/3VjySNobXLI

- Happytwin

https://youtu.be/KtmLUlCpj-I

- savannahsolver

Vertical videos for mobile phones:

@@ Line 64: / Line 64: @@
 ~MrThinker
+==Solution 6 (Basic Trigonometry)==
+If we reflect triangle <math> ABC </math> over line <math> AC </math>, we will get isosceles triangle <math> ABD </math>. By the [[Pythagorean Theorem]], we are capable of finding out that the <math> AB = AD = 13 </math>. Hence, <math> \tan \frac{\angle A}{2} = \tan \angle BAD = \frac{5}{12} </math>. Therefore, as of triangle <math> ABD </math>, the radius of its inscribed circle <math> r = \frac{tan \frac{\angle A}{2}\cdot (AB + AD - BD)}{2} = \frac{\frac{5}{12} \cdot 16}{2} = \boxed{\textbf{(D) }\frac{10}{3}}</math>
+~[[User:Bloggish|Bloggish]]
 ==Video Solution by OmegaLearn==

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