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

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(Solution 1)
 
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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?
 +
 
<asy>
 
<asy>
 
draw((0,0)--(12,0)--(12,5)--(0,0));
 
draw((0,0)--(12,0)--(12,5)--(0,0));
Line 13: Line 14:
 
<math>\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}</math>
 
<math>\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}</math>
  
==Solution==
+
==Solution 1==
We can reflect triangle <math>ABC</math> on 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>
+
 +
<asy>
 +
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));
 +
label("$A$", (0,0), hhhhhhhhhhhhhhhhhhhhhhhh);
 +
label("$C$", (12,0), E);
 +
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>
 +
We can see that our circle 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> We use the pythagorean triples to find out that AB is 13. 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>
+
~CHECKMATE2021
  
 
==Solution 2==
 
==Solution 2==
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)}\ \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>
  
 
==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 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==
 +
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 BCO</math>, which means <math>(12 \cdot 5)/2 = (13\cdot r)/2 +(5\cdot r)/2</math>. So, it gives us <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}</math>.
 +
 
 +
--LarryFlora
 +
 
 +
==Solution 5 (Pythagorean Theorem)==
 +
We can draw another radius from the center to the point of tangency. This angle, <math>\angle{ODB}</math>, is <math>90^\circ</math>. Label the center <math>O</math>, the point of tangency <math>D</math>, and the radius <math>r</math>.
 +
<asy>
 +
draw((0,0)--(12,0)--(12,5)--(0,0));
 +
draw(arc((8.67,0),(12,0),(5.33,0)));
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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("$12$", (6, 0), S);
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label("$5$", (12, 2.5), E);
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draw((8.665,0)--(7.4,3.07));
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label("$O$", (8.665, 0), S);
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label("$D$", (7.4, 3.1), NW);
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label("$r$", (11, 0), S);
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label("$r$", (7.6, 1), W);
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</asy>
 +
 
 +
Since <math>ODBC</math> is a kite, then <math>DB=CB=5</math>. Also, <math>AD=13-5=8</math>. By the [[Pythagorean Theorem]], <math>r^2 + 8^2=(12-r)^2</math>. Solving, <math>r^2+64=144-24r+r^2 \Rightarrow 24r=80 \Rightarrow \boxed{\textbf{(D) }\frac{10}{3}}</math>.
 +
 
 +
~MrThinker
 +
 
 +
==Solution 6 (Basic Trigonometry)==
 +
We can draw another radius from the center to the point of tangency. Label the center <math>O</math>, the point of tangency <math>D</math>, and the radius <math>r</math>.  
 +
<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 <math>ODBC</math> is a kite, <math>DB=CB=5</math>, and <math>AB=13</math> due to the [[Pythagorean Theorem]]. Angle <math>\angle{ODA}</math> is <math>90^\circ</math>, so we can use the famous mnemonic SOH CAH TOA. <math>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}}</math>
 +
 
 +
~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
 +
 
 +
Vertical videos for mobile phones:
 +
* https://youtu.be/tKmJlyspyAI
 +
* https://tiktok.com/@problemsolvingchannel/video/7162571579854032130
  
 
==See Also==
 
==See Also==

Latest revision as of 21:20, 20 January 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

[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), hhhhhhhhhhhhhhhhhhhhhhhh); 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 our circle 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 $12\times5 = 60.$ We use the pythagorean triples to find out that AB is 13. The semiperimeter is $\dfrac{10+13+13}{2}=18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

~CHECKMATE2021

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 3

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 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

Vertical videos for mobile phones:

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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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