Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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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. | + | 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), | + | 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); | ||
Line 27: | Line 27: | ||
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 | + | 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 | + | 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 | ||
− | ==Video Solution== | + | ==Solution 6 (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]] | ||
+ | |||
+ | ==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/Y0JBJgHsdGk | ||
− | https://youtu.be/3VjySNobXLI - Happytwin | + | https://youtu.be/3VjySNobXLI |
+ | |||
+ | - Happytwin | ||
− | https://youtu.be/KtmLUlCpj-I | + | https://youtu.be/KtmLUlCpj-I |
− | + | - savannahsolver | |
Vertical videos for mobile phones: | Vertical videos for mobile phones: |
Revision as of 18:58, 1 April 2023
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. We can see that our 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
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 3
Let the tangency point on be . Note By Power of a Point, Solving for gives
Solution 4
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
Solution 5 (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 6 (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
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=3837
- pi_is_3.14
Video Solutions
- Happytwin
- savannahsolver
Vertical videos for mobile phones:
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.