Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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==Solution 6 (Basic Trigonometry)== | ==Solution 6 (Basic Trigonometry)== | ||
− | + | 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>. 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]] | ~[[User:Bloggish|Bloggish]] |
Revision as of 19:54, 17 September 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)
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,
. 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.