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

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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>
 
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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==Similar Triangles Solution==
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==Solution 2==
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We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</math>. Drawing radius <math>OD</math> with length <math>x</math> such that <math>OD</math> is tangent to <math>AB</math>, we immediately see that <math>ODB\cong OCB</math> because of HL 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}}</math>.
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We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</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 HL 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}}</math>.
  
  

Revision as of 16:35, 22 November 2017

Problem 22

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

We can reflect triangle $ABC$ on line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$

We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The area of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

Solution 2

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




Similar Triangles 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) One leg of $AFO$ is $r$. Because $AFO$ ~ $ABC$, we have $r/(12 - r) = 5/13$ and solving gives $r = 10/3 (D)$

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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