Difference between revisions of "2021 Fall AMC 12B Problems/Problem 22"
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− | ==Solution 1 (Analytic Geometry) == | + | ==Solution 1== |
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+ | Let <math>M</math> be the midpoint of <math>AB</math>; so <math>BM=AM=5</math>. Let <math>D</math> be the point such that <math>ABCD</math> is a rectangle. Then <math>MO\perp AB</math> and <math>MP\perp AB</math>. Let <math>\theta = \angle BAC</math>; so <math>\tan\theta = \tfrac 68 = \tfrac 34</math>. Then | ||
+ | <cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \tfrac{35}{12}.</cmath> | ||
+ | |||
+ | ==Solution 2 (Analytic Geometry) == | ||
In a Cartesian plane, let <math>C, B,</math> and <math>A</math> be <math>(0,0),(0,6),(8,0)</math> respectively. | In a Cartesian plane, let <math>C, B,</math> and <math>A</math> be <math>(0,0),(0,6),(8,0)</math> respectively. |
Revision as of 18:28, 6 June 2022
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
![[asy] defaultpen(fontsize(11)+0.8); size(200); pair A,B,C,M,Ic,Ib,O,P; C=MP("C",origin,down+left); A=MP("A",8*right,down+right); B=MP("B",6*up,2*up); draw(A--B--C--A); draw(B--(B+A), gray+0.25); M=MP("M",(A+B)/2,down+left); O=MP("O",extension(B,B+A,M,M+(B-M)*dir(-90)),down); P=MP("P",extension(A,B+A,M,M+(B-M)*dir(-90)),up); draw(M--P^^A--P, gray+0.25); label("$\theta$", A, 7*dir(162)); label("$\theta$", B, 7*dir(-20)); label("$\theta$", P, 7*dir(-110)); label("$6$", B--C, left); label("$8$", A--C, down); label("$D$", A+B, right); [/asy]](http://latex.artofproblemsolving.com/c/8/3/c830ad9adab3b4322991576ff56f376a06675397.png)
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 2
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 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 |
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