Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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label("$5$", (12, 2.5), E); | label("$5$", (12, 2.5), E); | ||
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− | We can see that Circle <math>O</math> is the incircle of <math>ABB'.</math> We can use | + | We can see that Circle <math>O</math> 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 | Asymptote diagram by Mathandski |
Revision as of 14:41, 19 July 2021
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. Let us call the center of the circle
We can see that 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
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.