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

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(Undo revision 231272 by Megaboy6679 (talk))
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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==
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==Solution 1 (Pythagorean Theorem)==
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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.
 
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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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==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)==
 
 
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 Trignometry)==
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==Solution 2 (Basic Trigonometry)==
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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>.  
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If we reflect triangle <math> ABC </math> over line <math> AC </math>, we will get isosceles triangle <math> ABD </math>. By the [[Pythagorean Theorem]], we are capable of finding out that the <math> AB = AD = 13 </math>. Hence, <math> \tan \frac{\angle BAD}{2} = \tan \angle BAC = \frac{5}{12} </math>. Therefore, as of triangle <math> ABD </math>, the radius of its inscribed circle <math> r = \frac{tan \frac{\angle BAD}{2}\cdot (AB + AD - BD)}{2} = \frac{\frac{5}{12} \cdot 16}{2} = \boxed{\textbf{(D) }\frac{10}{3}}</math>
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<asy>
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draw((0,0)--(12,0)--(12,5)--(0,0));
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~[[User:Bloggish|Bloggish]]
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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]]. 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>
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==Solution 3==
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Like solution 2, we reflect <math>\triangle ABC</math> over line <math>\overline{AC}</math> and label the reflection of point <math>B</math> as <math>D</math>. As <math>AB = AD = 13</math> by the Pythagorean Theorem, we use the formula <math>rs=A</math>, where <math>r</math> is the inradius (what we're trying to find), <math>s</math> is the semiperimeter (<math>\frac{\overline{AB}+\overline{AD}+\overline{BD}}{2}</math>), and <math>A</math> is the area of the triangle in which the incircle is inscribed in. Substitution gives: <cmath>r=\frac{\frac{10\cdot12}{2}}{\frac{13+13+10}{2}}</cmath>
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<cmath>r=\frac{60}{18}</cmath>
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<cmath>r=\boxed{\textbf{(D) }\frac{10}{3}}</cmath>
  
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~PowerQualimit
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~megaboy6679
  
 
==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==
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==Video Solutions==
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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
 
  
 
https://youtu.be/KtmLUlCpj-I  
 
https://youtu.be/KtmLUlCpj-I  
  
 
- 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
 
  
 
==See Also==
 
==See Also==

Latest revision as of 19:40, 2 November 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 (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 2 (Basic Trigonometry)

If we reflect triangle $ABC$ over line $AC$, we will get isosceles triangle $ABD$. By the Pythagorean Theorem, we are capable of finding out that the $AB = AD = 13$. Hence, $\tan \frac{\angle BAD}{2} = \tan \angle BAC = \frac{5}{12}$. Therefore, as of triangle $ABD$, the radius of its inscribed circle $r = \frac{tan \frac{\angle BAD}{2}\cdot (AB + AD - BD)}{2} = \frac{\frac{5}{12} \cdot 16}{2} = \boxed{\textbf{(D) }\frac{10}{3}}$

~Bloggish

Solution 3

Like solution 2, we reflect $\triangle ABC$ over line $\overline{AC}$ and label the reflection of point $B$ as $D$. As $AB = AD = 13$ by the Pythagorean Theorem, we use the formula $rs=A$, where $r$ is the inradius (what we're trying to find), $s$ is the semiperimeter ($\frac{\overline{AB}+\overline{AD}+\overline{BD}}{2}$), and $A$ is the area of the triangle in which the incircle is inscribed in. Substitution gives: \[r=\frac{\frac{10\cdot12}{2}}{\frac{13+13+10}{2}}\] \[r=\frac{60}{18}\] \[r=\boxed{\textbf{(D) }\frac{10}{3}}\]

~megaboy6679

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/ZOHjUebMNpk

~Education, the Study of Everything

Video Solutions

https://youtu.be/KtmLUlCpj-I

- savannahsolver

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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