|
|
(54 intermediate revisions by 6 users not shown) |
Line 14: |
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 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>
| |
− | 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), hhhhhhhhhhhhhhhhhhhhhhhh);
| |
− | 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 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
| |
− | | |
− | ==Solution 2==
| |
− | 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 3==
| |
− | 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 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 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)==
| |
| 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> |
Line 64: |
Line 34: |
| | | |
| ~MrThinker | | ~MrThinker |
− |
| |
− | ==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!!!)== | | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
Line 75: |
Line 40: |
| ~Education, the Study of Everything | | ~Education, the Study of Everything |
| | | |
− | ==Video Solution by OmegaLearn== | + | ==Video Solutions== |
− | https://youtu.be/FDgcLW4frg8?t=3837
| |
− | | |
− | - pi_is_3.14
| |
| | | |
− | ==Video Solutions==
| |
| | | |
− | https://youtu.be/Y0JBJgHsdGk
| |
| | | |
− | https://youtu.be/3VjySNobXLI
| |
| | | |
− | - Happytwin
| |
| | | |
| https://youtu.be/KtmLUlCpj-I | | https://youtu.be/KtmLUlCpj-I |
| | | |
| - savannahsolver | | - savannahsolver |
− |
| |
− | Vertical videos for mobile phones:
| |
− | * https://youtu.be/tKmJlyspyAI
| |
− | * https://tiktok.com/@problemsolvingchannel/video/7162571579854032130
| |
| | | |
| ==See Also== | | ==See Also== |
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
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
https://youtu.be/ZOHjUebMNpk
~Education, the Study of Everything
Video Solutions
https://youtu.be/KtmLUlCpj-I
- savannahsolver
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.