Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
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<asy> | <asy> | ||
draw((0,0)--(12,0)--(12,5)--cycle); | draw((0,0)--(12,0)--(12,5)--cycle); | ||
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> We use the pythagorean triples to find out that AB is 13. 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> |
− | + | ~CHECKMATE2021 | |
==Solution 2== | ==Solution 2== | ||
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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 | ||
+ | |||
+ | ==Solution 6 (Basic Trigonometry)== | ||
+ | We can draw another radius from the center to the point of tangency. 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, <math>DB=CB=5</math>, and <math>AB=13</math> due to the [[Pythagorean Theorem]]. Angle <math>\angle{ODA}</math> is <math>90^\circ</math>, so we can use the famous mnemonic SOH CAH TOA. <math>AD=AB-DB=13-5=8 \Rightarrow \tan \angle BAC = \frac{5}{12}=\frac{r}{8} \Rightarrow 12r=40 \Rightarrow r= \frac{40}{12}= \boxed{\textbf{(D)}\ \frac{10}{3}}</math> | ||
+ | |||
+ | ~PowerQualimit | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
+ | https://youtu.be/ZOHjUebMNpk | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== | ||
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https://youtu.be/Y0JBJgHsdGk | https://youtu.be/Y0JBJgHsdGk | ||
− | https://youtu.be/3VjySNobXLI - Happytwin | + | https://youtu.be/3VjySNobXLI |
+ | |||
+ | - Happytwin | ||
+ | |||
+ | https://youtu.be/KtmLUlCpj-I | ||
− | + | - savannahsolver | |
Vertical videos for mobile phones: | Vertical videos for mobile phones: |
Latest revision as of 03:46, 22 April 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
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 We use the pythagorean triples to find out that AB is 13. The semiperimeter is Simplifying Our answer is therefore
~CHECKMATE2021
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)
We can draw another radius from the center to the point of tangency. Label the center , the point of tangency , and the radius .
Since is a kite, , and due to the Pythagorean Theorem. Angle is , so we can use the famous mnemonic SOH CAH TOA.
~PowerQualimit
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.