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| | <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> |
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| − | ==Solution 1== | + | ==Solution 1 (Pythagorean Theorem)== |
| − | 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.
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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), hhhhhhhhhhhhhhhhhhhhhhhh);
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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 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> 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>
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| − | Asymptote diagram by Mathandski
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| − | ==Solution 2==
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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>
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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>
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| − | ==Solution 4==
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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 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>.
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| − | --LarryFlora
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| − | ==Solution 5 (Pythagorean Theorem)==
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| | 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>. | | 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> | | <asy> |
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| | ~MrThinker | | ~MrThinker |
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| − | ==Solution 6 (Basic Trigonometry)==
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| − | 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>.
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| − | <asy>
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| − | draw((0,0)--(12,0)--(12,5)--(0,0));
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| − | 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>
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| − | Since <math>ODBC</math> is a kite, <math>DB=CB=5</math>, and <math>AB=13</math> due to the [[Pythagorean Theorem]]. So, <math>AD=13-5=8</math>. Hence, <math> \tan \angle BAC = \frac{5}{12}=\frac{r}{8} </math>. Therefore, <math>12r=40</math> with cross multiplication, hence <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}</math>.
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| − |
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| − | ~[[User:PowerQualimit|PowerQualimit]]
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| | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
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| | ~Education, the Study of Everything | | ~Education, the Study of Everything |
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| − | ==Video Solution by OmegaLearn== | + | ==Video Solutions== |
| − | https://youtu.be/FDgcLW4frg8?t=3837
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| − | - pi_is_3.14
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| − | ==Video Solutions==
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| − | https://youtu.be/Y0JBJgHsdGk
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| − | https://youtu.be/3VjySNobXLI
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| − | - Happytwin
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| | https://youtu.be/KtmLUlCpj-I | | https://youtu.be/KtmLUlCpj-I |
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| | - savannahsolver | | - savannahsolver |
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| − | Vertical videos for mobile phones:
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| − | * https://youtu.be/tKmJlyspyAI
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| − | * https://tiktok.com/@problemsolvingchannel/video/7162571579854032130
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| | ==See Also== | | ==See Also== |
, 
, 
, and angle 
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]](http://latex.artofproblemsolving.com/1/7/f/17f66d4cef8b689eb626ec8c226380e1e72d8719.png)
, is 
. Label the center 
, the point of tangency 
, and the radius 
.
![[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]](http://latex.artofproblemsolving.com/3/8/f/38fe1daab146801175ebc91b54d9343afa54e6b3.png)
is a kite, then 
. Also, 
. By the Pythagorean Theorem, 
. Solving, 
.
