Difference between revisions of "2017 AMC 8 Problems/Problem 22"
(→Similar Triangles Solution) |
Redjack-512 (talk | contribs) m (→Solution 1) |
||
(38 intermediate revisions by 26 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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 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 | |
==Solution 2== | ==Solution 2== | ||
− | We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</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 HL 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}}</math>. | + | 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> | ||
+ | ==Solution 5== | ||
+ | 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*5)/2 = (13*r)/2 +(5*r)/2</math>. So it gives us <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}</math>.----LarryFlora | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/Y0JBJgHsdGk | ||
+ | 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== |
Revision as of 14:41, 19 July 2021
Contents
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 a formula for finding the radius of the incircle. 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
Solution 5
Let us label the center of the semicircle and the point where the circle is tangent to the triangle . The area of = the areas of + , which means . So it gives us .----LarryFlora
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.