Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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<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> | ||
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==Solution 2== | ==Solution 2== |
Revision as of 19:04, 27 May 2022
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 2
We immediately see that , and we label the center of the semicircle
and the point where the circle is tangent to the triangle
. Drawing radius
with length
such that
is perpendicular to
, we immediately see that
because of
congruence, so
and
. By similar triangles
and
, we see that
.
Solution 3
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 4
Let the tangency point on be
. Note
By Power of a Point,
Solving for
gives
Solution 5
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
Video Solution
https://youtu.be/3VjySNobXLI - Happytwin
https://youtu.be/KtmLUlCpj-I - savannahsolver
https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14
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.