Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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− | ~ | + | ~ Pi Academy |
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Revision as of 17:38, 5 January 2025
Contents
[hide]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 (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 2 (Basic Trigonometry)
If we reflect triangle over line
, we will get isosceles triangle
. By the Pythagorean Theorem, we are capable of finding out that the
. Hence,
. Therefore, as of triangle
, the radius of its inscribed circle
Solution 3
Like solution 2, we reflect over line
and label the reflection of point
as
. As
by the Pythagorean Theorem, we use the formula
, where
is the inradius (what we're trying to find),
is the semiperimeter (
), and
is the area of the triangle in which the incircle is inscribed in. Substitution gives:
~megaboy6679
Video Solution (Fast and Easy!)
https://youtu.be/qZJNi-WM0XY?si=AL2FANuanWBAQu17
~ Pi Academy
Video Solution
- savannahsolver
Video Solution
https://www.youtube.com/watch?v=sOF1Okc0jMc
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.