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

(Video Solution)
(Video Solution by OmegaLearn)
(7 intermediate revisions by 4 users not shown)
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==Solution 1==
 
==Solution 1==
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 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.  
 
<asy>
 
<asy>
 
draw((0,0)--(12,0)--(12,5)--cycle);
 
draw((0,0)--(12,0)--(12,5)--cycle);
 
draw((0,0)--(12,0)--(12,-5)--cycle);
 
draw((0,0)--(12,0)--(12,-5)--cycle);
 
draw(circle((8.665,0),3.3333));
 
draw(circle((8.665,0),3.3333));
label("$A$", (0,0), W);
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label("$A$", (0,0), hhhhhhhhhhhhhhhhhhhhhhhh);
 
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("$5$", (12, 2.5), E);
 
label("$5$", (12, 2.5), E);
 
label("$5$", (12, -2.5), E);</asy>
 
label("$5$", (12, -2.5), E);</asy>
We can see that Circle <math>O</math> 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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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>
  
 
Asymptote diagram by Mathandski
 
Asymptote diagram by Mathandski
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==Solution 4==
 
==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 ACO</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
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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>.
 +
 
 +
--LarryFlora
  
 
==Solution 5 (Pythagorean Theorem)==
 
==Solution 5 (Pythagorean Theorem)==
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~MrThinker
 
~MrThinker
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==Solution 6 (Basic Trigonometry)==
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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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 +
~[[User:Bloggish|Bloggish]]
 +
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/ZOHjUebMNpk
 +
 +
~Education, the Study of Everything
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
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https://youtu.be/Y0JBJgHsdGk
 
https://youtu.be/Y0JBJgHsdGk
  
https://youtu.be/3VjySNobXLI - Happytwin
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https://youtu.be/3VjySNobXLI  
 +
 
 +
- Happytwin
 +
 
 +
https://youtu.be/KtmLUlCpj-I
  
https://youtu.be/KtmLUlCpj-I - savannahsolver
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- savannahsolver
  
 
Vertical videos for mobile phones:
 
Vertical videos for mobile phones:

Revision as of 18:58, 1 April 2023

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

We can reflect triangle $ABC$ over line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. [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 $12\times5 = 60.$ The semiperimeter is $\dfrac{10+13+13}{2}=18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

Asymptote diagram by Mathandski

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)

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

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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