Difference between revisions of "2021 Fall AMC 12B Problems/Problem 22"
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~Wilhelm Z | ~Wilhelm Z | ||
+ | |||
+ | == Solution 2 == | ||
+ | This solution is based on this figure: [[:Image:2021_AMC_12B_(Nov)_Problem_22,_sol.png]]. | ||
+ | |||
+ | Because the circle with center <math>O</math> passes through points <math>A</math> and <math>B</math> and is tangent to line <math>BC</math> at point <math>B</math>, <math>O</math> is on the perpendicular bisector of segment <math>AB</math> and <math>OB \perp BC</math>. | ||
+ | |||
+ | Because the circle with center <math>P</math> passes through points <math>A</math> and <math>B</math> and is tangent to line <math>AC</math> at point <math>A</math>, <math>P</math> is on the perpendicular bisector of segment <math>AB</math> and <math>OA \perp AC</math>. | ||
+ | |||
+ | Let lines <math>OB</math> and <math>AP</math> intersect at point <math>D</math>. | ||
+ | Hence, <math>ACBD</math> is a rectangle. | ||
+ | |||
+ | Denote by <math>M</math> the midpoint of segment <math>AB</math>. Hence, <math>BM = \frac{AB}{2} = 5</math>. | ||
+ | Because <math>O</math> and <math>P</math> are on the perpendicular bisector of segment <math>AB</math>, points <math>M</math>, <math>O</math>, <math>P</math> are collinear with <math>OM \perp AB</math>. | ||
+ | |||
+ | We have <math>\triangle MOB \sim \triangle CBA</math>. | ||
+ | Hence, <math>\frac{BO}{AB} = \frac{BM}{AC}</math>. | ||
+ | Hence, <math>BO = \frac{25}{4}</math>. | ||
+ | Hence, <math>OD = BD - BO = \frac{7}{4}</math>. | ||
+ | |||
+ | We have <math>\triangle DOP \sim \triangle CBA</math>. | ||
+ | Hence, <math>\frac{OP}{BA} = \frac{DO}{CB}</math>. | ||
+ | Therefore, <math>OP = \frac{35}{12}</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(C) }\frac{35}{12}}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=23|num-b=21}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=23|num-b=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:21, 26 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 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 2
This solution is based on this figure: Image:2021_AMC_12B_(Nov)_Problem_22,_sol.png.
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)
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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