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Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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~megaboy6679
 
~megaboy6679
  
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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==Video Solution by Pi Academy (Fast and Easy!)==
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https://youtu.be/ZOHjUebMNpk
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https://youtu.be/qZJNi-WM0XY?si=AL2FANuanWBAQu17
  
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~Education, the Study of Everything
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~ Pi Academy
  
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==Video Solutions==
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==Video Solution==
  
 
https://youtu.be/KtmLUlCpj-I  
 
https://youtu.be/KtmLUlCpj-I  
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- savannahsolver
 
- savannahsolver
  
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== Video Solution only problem 22's by SpreadTheMathLove==
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== Video Solution==
 
https://www.youtube.com/watch?v=sOF1Okc0jMc
 
https://www.youtube.com/watch?v=sOF1Okc0jMc
  

Latest revision as of 20:06, 19 January 2025

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

Solution 3

Like solution 2, we reflect $\triangle ABC$ over line $\overline{AC}$ and label the reflection of point $B$ as $D$. As $AB = AD = 13$ by the Pythagorean Theorem, we use the formula $rs=A$, where $r$ is the inradius (what we're trying to find), $s$ is the semiperimeter ($\frac{\overline{AB}+\overline{AD}+\overline{BD}}{2}$), and $A$ is the area of the triangle in which the incircle is inscribed in. Substitution gives: \[r=\frac{\frac{10\cdot12}{2}}{\frac{13+13+10}{2}}\] \[r=\frac{60}{18}\] \[r=\boxed{\textbf{(D) }\frac{10}{3}}\]

~megaboy6679

Video Solution by Pi Academy (Fast and Easy!)

https://youtu.be/qZJNi-WM0XY?si=AL2FANuanWBAQu17

~ Pi Academy

Video Solution

https://youtu.be/KtmLUlCpj-I

- savannahsolver

Video Solution

https://www.youtube.com/watch?v=sOF1Okc0jMc

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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