Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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− | Since <math>ODBC</math> is a kite, then <math>DB=CB=5</math>. Also, <math>AD=13-5=8</math>. Hence, <math> \tan \angle BAC = \frac{5}{12}=\frac{r}{8} </math>. Therefore, 12r=40 with cross multiplication, hence <math>r= | + | Since <math>ODBC</math> is a kite, then <math>DB=CB=5</math>. Also, <math>AD=13-5=8</math>. Hence, <math> \tan \angle BAC = \frac{5}{12}=\frac{r}{8} </math>. Therefore, 12r=40 with cross multiplication, hence <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}</math>. |
~[[User:PowerQualimit|PowerQualimit]] | ~[[User:PowerQualimit|PowerQualimit]] |
Revision as of 20:00, 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, 12r=40 with cross multiplication, hence
.
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.