Difference between revisions of "2021 AIME I Problems/Problem 13"
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Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>. | Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>O_i</math> and <math>r_i</math> be the center and radius of <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>. | Let <math>O_i</math> and <math>r_i</math> be the center and radius of <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>. | ||
Revision as of 06:28, 23 April 2021
Contents
Problem
Circles and
with radii
and
, respectively, intersect at distinct points
and
. A third circle
is externally tangent to both
and
. Suppose line
intersects
at two points
and
such that the measure of minor arc
is
. Find the distance between the centers of
and
.
Solution 1
Let and
be the center and radius of
, and let
and
be the center and radius of
.
Since extends to an arc with arc
, the distance from
to
is
. Let
. Consider
. The line
is perpendicular to
and passes through
. Let
be the foot from
to
; so
. We have by tangency
and
. Let
.
Since
is on the radical axis of
and
, it has equal power with respect to both circles, so
since
. Now we can solve for
and
, and in particular,
We want to solve for
. By the Pythagorean Theorem (twice):
Therefore,
.
Solution 2 (Official MAA, Unedited)
Denote by ,
, and
the centers of
,
, and
, respectively. Let
and
denote the radii of
and
respectively,
be the radius of
, and
the distance from
to the line
. We claim that
where
. This solves the problem, for then the
condition implies
, and then we can solve to get
.
Denote by and
the centers of
and
respectively. Set
as the projection of
onto
, and denote by
the intersection of
with
. Note that
. Now recall that
Furthermore, note that
Substituting the first equality into the second one and subtracting yields
which rearranges to the desired.
Solution 3 (Inversion)
WLOG assume is a line. Note the angle condition is equivalent to the angle between
and
being
. We claim the angle between
and
is fixed as
varies.
Proof: Perform an inversion at , sending
and
to two lines
and
intersecting at
. Then
is sent to a circle tangent to lines
and
, which clearly intersects
at a fixed angle. Therefore the angle between
and
is fixed as
varies.
Now simply take to be a line. If
intersects
and
and
, respectively, and the circles' centers are
and
, then the projection of
to
at
gives that
is a
triangle. Therefore,
~spartacle
Solution 4 (Radical Axis, Harmonic Quadrilaterals, and Similar Triangles)
Suppose we label the points as shown here. By radical axis, the tangents to at
and
intersect on
. Thus
is harmonic, so the tangents to
at
and
intersect at
. Moreover,
because both
and
are perpendicular to
, and
because
. Thus
by similar triangles.
~mathman3880
Video Solution
Who wanted to see animated video solutions can see this. I found this really helpful.
P.S: This video is not made by me. And solution is same like below solutions.
≈@rounak138
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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