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

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==Problem 22==
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==Problem==
  
 
In the right triangle <math>ABC</math>, <math>AC=12</math>, <math>BC=5</math>, and angle <math>C</math> is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
 
In the right triangle <math>ABC</math>, <math>AC=12</math>, <math>BC=5</math>, and angle <math>C</math> is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
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<asy>
 
<asy>
 
draw((0,0)--(12,0)--(12,5)--(0,0));
 
draw((0,0)--(12,0)--(12,5)--(0,0));
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==Solution 1==
 
==Solution 1==
We can reflect triangle <math>ABC</math> over line <math>AC.</math> This forms the triangle <math>AB'C</math> and a circle out of the semicircle. Let us call the center of the circle <math>O.</math>
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We can reflect triangle <math>ABC</math> over line <math>AC.</math> This forms the triangle <math>AB'C</math> and a circle out of the semicircle. Let us call the center of the circle <math>O.</math>
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<asy>
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draw((0,0)--(12,0)--(12,5)--cycle);
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draw((0,0)--(12,0)--(12,-5)--cycle);
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draw(circle((8.665,0),3.3333));
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label("$A$", (0,0), W);
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label("$C$", (12,0), E);
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label("$B$", (12,5), NE);
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label("$B'$", (12,-5), NE);
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label("$12$", (7, 0), S);
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label("$5$", (12, 2.5), E);
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label("$5$", (12, -2.5), E);</asy>
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We can see that Circle <math>O</math> is the incircle of <math>ABB'.</math> We can use a formula for finding the radius of the incircle. The area of a triangle <math>= \text{Semiperimeter} \cdot \text{inradius}</math> .  The area of <math>ABB'</math> is <math>12\times5 = 60.</math> The semiperimeter is <math>\dfrac{10+13+13}{2}=18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore <math>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math>
  
We can see that Circle <math>O</math> is the incircle of <math>AB'C.</math> We can use the formula for finding the radius of the incircle to solve this problem. The area of <math>AB'C</math> is <math>12\times5 = 60.</math> The semiperimeter is <math>5+13 = 18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore <math>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math>
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Asymptote diagram by Mathandski
  
 
==Solution 2==
 
==Solution 2==
We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</math> And the point where the circle is tangent to the triangle <math>D</math>. Drawing radius <math>OD</math> with length <math>x</math> such that <math>OD</math> is perpendicular to <math>AB</math>, we immediately see that <math>ODB\cong OCB</math> because of <math>\operatorname{HL}</math> congruence, so <math>BD=5</math> and <math>DA=8</math>. By similar triangles <math>ODA</math> and <math>BCA</math>, we see that <math>\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{(D)}\ \frac{10}{3}}</math>.
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We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</math> and the point where the circle is tangent to the triangle <math>D</math>. Drawing radius <math>OD</math> with length <math>x</math> such that <math>OD</math> is perpendicular to <math>AB</math>, we immediately see that <math>ODB\cong OCB</math> because of <math>\operatorname{HL}</math> congruence, so <math>BD=5</math> and <math>DA=8</math>. By similar triangles <math>ODA</math> and <math>BCA</math>, we see that <math>\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{(D)}\ \frac{10}{3}}</math>.
  
 
==Solution 3==
 
==Solution 3==
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>AFO</math> ~ <math>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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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>
  
 
==Solution 4==
 
==Solution 4==
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Solving for <math>r</math> gives  
 
Solving for <math>r</math> gives  
 
<cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}</cmath>
 
<cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}</cmath>
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==Solution 5==
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Let us label the center of the semicircle <math>O</math> and the point where the circle is tangent to the triangle <math>D</math>. The area of <math>\triangle ABC</math> = the areas of <math>\triangle ABO</math> + <math> \triangle ACO</math>, which means <math>(12*5)/2 = (13*r)/2 +(5*r)/2</math>. So it gives us <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}</math>.----LarryFlora
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==Video Solution==
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https://youtu.be/Y0JBJgHsdGk
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https://youtu.be/3VjySNobXLI - Happytwin
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https://youtu.be/KtmLUlCpj-I - savannahsolver
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https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 14:41, 19 July 2021

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 2

We immediately see that $AB=13$, and we label the center of the semicircle $O$ and the point where the circle is tangent to the triangle $D$. Drawing radius $OD$ with length $x$ such that $OD$ is perpendicular to $AB$, we immediately see that $ODB\cong OCB$ because of $\operatorname{HL}$ congruence, so $BD=5$ and $DA=8$. By similar triangles $ODA$ and $BCA$, we see that $\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{(D)}\ \frac{10}{3}}$.

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 $(12*5)/2 = (13*r)/2 +(5*r)/2$. So it gives us $r = \boxed{\textbf{(D)}\ \frac{10}{3}}$.----LarryFlora

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

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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