Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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− | ==Problem | + | ==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> | + | 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> |
+ | <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), W); | ||
+ | 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 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> | ||
+ | |||
+ | Asymptote diagram by Mathandski | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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 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>\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 4== | ||
+ | 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> | ||
+ | Solving for <math>r</math> gives | ||
+ | <cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}</cmath> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/3VjySNobXLI - Happytwin | ||
+ | |||
+ | https://youtu.be/KtmLUlCpj-I - savannahsolver | ||
− | + | https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14 | |
==See Also== | ==See Also== |
Latest revision as of 01:19, 16 February 2021
Problem
In the right triangle , , , and angle is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
Solution 1
We can reflect triangle over line This forms the triangle and a circle out of the semicircle. Let us call the center of the circle We can see that Circle is the incircle of We can use the formula for finding the radius of the incircle to solve this problem: The area of a triangle . The area of is The semiperimeter is Simplifying Our answer is therefore
Asymptote diagram by Mathandski
Solution 2
We immediately see that , and we label the center of the semicircle and the point where the circle is tangent to the triangle . Drawing radius with length such that is perpendicular to , we immediately see that because of congruence, so and . By similar triangles and , we see that .
Solution 3
Let the center of the semicircle be . Let the point of tangency between line and the semicircle be . Angle is common to triangles and . By tangent properties, angle must be degrees. Since both triangles and are right and share an angle, is similar to . The hypotenuse of is , where is the radius of the circle. (See for yourself) The short leg of is . Because ~ , we have and solving gives
Solution 4
Let the tangency point on be . Note By Power of a Point, Solving for gives
Video Solution
https://youtu.be/3VjySNobXLI - Happytwin
https://youtu.be/KtmLUlCpj-I - savannahsolver
https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.