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Difference between revisions of "2017 AMC 8 Problems/Problem 22"
(Undo revision 231272 by Megaboy6679 (talk))
(Tag: Undo)
 
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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. Let us call the center of the circle <math>O.</math>  
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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>.
 
<asy>
 
<asy>
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draw((0,0)--(12,0)--(12,5)--cycle);
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draw((0,0)--(12,0)--(12,5)--(0,0));
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draw((0,0)--(12,0)--(12,-5)--cycle);
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draw(arc((8.67,0),(12,0),(5.33,0)));
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draw(circle((8.665,0),3.3333));
 
 
label("$A$", (0,0), W);
 
label("$A$", (0,0), W);
 
label("$C$", (12,0), E);
 
label("$C$", (12,0), E);
 
label("$B$", (12,5), NE);
 
label("$B$", (12,5), NE);
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label("$B'$", (12,-5), NE);
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label("$12$", (6, 0), S);
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label("$12$", (7, 0), S);
 
 
label("$5$", (12, 2.5), E);
 
label("$5$", (12, 2.5), E);
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label("$5$", (12, -2.5), E);</asy>
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draw((8.665,0)--(7.4,3.07));
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We can see that Circle <math>O</math> is the incircle of <math>ABB'.</math> We can use the formula for finding the radius of the incircle to solve this problem: 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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label("$O$", (8.665, 0), S);
 +
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);
 +
</asy>
  
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Asymptote diagram by Mathandski
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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>.
  
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==Solution 2==
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~MrThinker
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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 2 (Basic Trigonometry)==
 +
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>
 +
 
 +
~[[User:Bloggish|Bloggish]]
  
 
==Solution 3==
 
==Solution 3==
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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>
+
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>
 +
<cmath>r=\frac{60}{18}</cmath>
 +
<cmath>r=\boxed{\textbf{(D) }\frac{10}{3}}</cmath>
 +
 
 +
~megaboy6679
 +
 
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/ZOHjUebMNpk
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solutions==
 +
 
  
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==Solution 4==
 
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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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==Video Solution==
 
  
−
https://youtu.be/3VjySNobXLI - Happytwin
 
  
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https://youtu.be/KtmLUlCpj-I - savannahsolver
+
https://youtu.be/KtmLUlCpj-I  
  
−
https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14
+
- savannahsolver
  
 
==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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