Difference between revisions of "2021 Fall AMC 12B Problems/Problem 22"
(→Solution 1 (Analytic Geometry)) |
(→Solution 1 (Analytic Geometry)) |
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Through using the distance formula, | Through using the distance formula, | ||
− | <math>OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}=\boxed{\textbf{(C)}}</math>. | + | <math>OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}=\boxed{\textbf{(C) \frac{35}{12}}}</math>. |
~Wilhelm Z | ~Wilhelm Z |
Revision as of 04:41, 24 November 2021
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
?
Solution 1 (Analytic Geometry)
In a Cartesian plane, let and
be
respectively.
By analyzing the behaviors of the two circles, we set 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,
$OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}=\boxed{\textbf{(C) \frac{35}{12}}}$ (Error compiling LaTeX. Unknown error_msg).
~Wilhelm Z
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 |
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