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 1== | + | ==Solution 1 (Pythagorean Theorem)== |
− | We can | + | 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> | ||
+ | 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 <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>. | ||
− | + | ~MrThinker | |
− | ==Solution 2== | + | ==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== | ||
− | + | 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== | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | https://youtu.be/KtmLUlCpj-I | ||
− | + | - savannahsolver | |
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− | |||
==See Also== | ==See Also== |
Latest revision as of 19:40, 2 November 2024
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 (Pythagorean Theorem)
We can draw another radius from the center to the point of tangency. This angle, , is . Label the center , the point of tangency , and the radius .
Since is a kite, then . Also, . By the Pythagorean Theorem, . Solving, .
~MrThinker
Solution 2 (Basic Trigonometry)
If we reflect triangle over line , we will get isosceles triangle . By the Pythagorean Theorem, we are capable of finding out that the . Hence, . Therefore, as of triangle , the radius of its inscribed circle
Solution 3
Like solution 2, we reflect over line and label the reflection of point as . As by the Pythagorean Theorem, we use the formula , where is the inradius (what we're trying to find), is the semiperimeter (), and is the area of the triangle in which the incircle is inscribed in. Substitution gives:
~megaboy6679
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solutions
- 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.