2021 Fall AMC 12B Problems/Problem 22
Contents
Problem
Right triangle has side lengths , , and .
A circle centered at is tangent to line at and passes through . A circle centered at is tangent to line at and passes through . What is ?
Diagram
Solution 1
Let be the midpoint of ; so . Let be the point such that is a rectangle. Then and . Let ; so . Then
Solution 2 (Analytic Geometry)
In a Cartesian plane, let and be respectively.
By analyzing the behaviors of the two circles, we set to be and be .
Hence derive the two equations:
Considering the coordinates of and for the two equations respectively, we get:
Solve to get and
Through using the distance formula,
.
~Wilhelm Z
Solution 3
Because the circle with center passes through points and and is tangent to line at point , is on the perpendicular bisector of segment and .
Because the circle with center passes through points and and is tangent to line at point , is on the perpendicular bisector of segment and .
Let lines and intersect at point . Hence, is a rectangle.
Denote by the midpoint of segment . Hence, . Because and are on the perpendicular bisector of segment , points , , are collinear with .
We have . Hence, . Hence, . Hence, .
We have . Hence, . Therefore, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution
~MathProblemSolvingSkills.com
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=ctx67nltpE0
2021 Fall AMC 12B (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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.