Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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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> | ||
− | ==Solution 1 | + | ==Solution 1 (Pythagorean Theorem)== |
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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>. | 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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~MrThinker | ~MrThinker | ||
− | ==Solution | + | ==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> | |
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− | + | ~[[User:Bloggish|Bloggish]] | |
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− | + | ==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!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
− | ==Video | + | ==Video Solutions== |
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https://youtu.be/KtmLUlCpj-I | https://youtu.be/KtmLUlCpj-I | ||
- savannahsolver | - 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.