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); | ||
label("$5$", (12, -2.5), E);</asy> | label("$5$", (12, -2.5), E);</asy> | ||
− | We can see that Circle <math>O</math> is the incircle of <math> | + | We can see that Circle <math>O</math> is the incircle of <math>ABB'.</math> We can use the formula for finding the radius of the incircle to solve this problem: 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>5+13 = 18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore a<math>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math> |
Asymptote diagram by Mathandski | Asymptote diagram by Mathandski |
Revision as of 18:25, 24 December 2020
Contents
Problem 22
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 the formula for finding the radius of the incircle to solve this problem: The area of a triangle
. The area of
is
The semiperimeter is
Simplifying
Our answer is therefore a
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
Video Solution
https://youtu.be/3VjySNobXLI - Happytwin
~savannahsolver
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.