2017 AMC 8 Problems/Problem 22: Difference between revisions

Revision as of 06:58, 27 August 2024

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 (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 2 (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, $\tan \frac{\angle BAD}{2} = \tan \angle BAC = \frac{5}{12}$ . Therefore, as of triangle $ABD$ , the radius of its inscribed circle $r = \frac{tan \frac{\angle BAD}{2}\cdot (AB + AD - BD)}{2} = \frac{\frac{5}{12} \cdot 16}{2} = \boxed{\textbf{(D) }\frac{10}{3}}$

~Bloggish

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/ZOHjUebMNpk

~Education, the Study of Everything

Video Solutions

https://youtu.be/KtmLUlCpj-I

- savannahsolver

@@ Line 34: / Line 34: @@
 ~MrThinker
+==Solution 2 (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 BAD}{2} = \tan \angle BAC = \frac{5}{12} </math>. Therefore, as of triangle <math> ABD </math>, the radius of its inscribed circle <math> r = \frac{tan \frac{\angle BAD}{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 (CREATIVE THINKING + ANALYSIS!!!)==

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