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

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==Solution 2==
 
==Solution 2==
 
Let the center of the semicircle be <math>O</math>. Let the point of tangency between line <math>AB</math> and the semicircle be <math>F</math>. Angle <math>BAC</math> is common to triangles <math>ABC</math> and <math>AFO</math>. By tangent properties, angle <math>AFO</math> must be <math>90</math> degrees. Since both triangles <math>ABC</math> and <math>AFO</math> are right and share an angle, <math>AFO</math> is similar to <math>ABC</math>. The hypotenuse of <math>AFO</math> is <math>12 - r</math>, where <math>r</math> is the radius of the circle. (See for yourself) The short leg of <math>AFO</math> is <math>r</math>. Because <math>\triangle AFO</math> ~ <math>\triangle ABC</math>, we have <math>r/(12 - r) = 5/13</math> and solving gives <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}.</math>
 
Let the center of the semicircle be <math>O</math>. Let the point of tangency between line <math>AB</math> and the semicircle be <math>F</math>. Angle <math>BAC</math> is common to triangles <math>ABC</math> and <math>AFO</math>. By tangent properties, angle <math>AFO</math> must be <math>90</math> degrees. Since both triangles <math>ABC</math> and <math>AFO</math> are right and share an angle, <math>AFO</math> is similar to <math>ABC</math>. The hypotenuse of <math>AFO</math> is <math>12 - r</math>, where <math>r</math> is the radius of the circle. (See for yourself) The short leg of <math>AFO</math> is <math>r</math>. Because <math>\triangle AFO</math> ~ <math>\triangle ABC</math>, we have <math>r/(12 - r) = 5/13</math> and solving gives <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}.</math>
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==Solution 3==
 
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Let the tangency point on <math>AB</math> be <math>D</math>. Note <cmath>AD = AB-BD = AB-BC = 8.</cmath> By Power of a Point, <cmath>12(12-2r) = 8^2.</cmath>
 
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Solving for <math>r</math> gives
 
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<cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}.</cmath>
 
  
 
==Solution 4==
 
==Solution 4==

Revision as of 10:47, 24 July 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 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 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)

We can draw another radius from the center to the point of tangency. 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, $DB=CB=5$, and $AB=13$ due to the Pythagorean Theorem. Angle $\angle{ODA}$ is $90^\circ$, so we can use the famous mnemonic SOH CAH TOA. $AD=AB-DB=13-5=8 \Rightarrow \tan \angle BAC = \frac{5}{12}=\frac{r}{8} \Rightarrow 12r=40 \Rightarrow r= \frac{40}{12}= \boxed{\textbf{(D)}\ \frac{10}{3}}$

~PowerQualimit

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/ZOHjUebMNpk

~Education, the Study of Everything

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

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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

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