Difference between revisions of "2021 AIME I Problems/Problem 13"
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− | Since <math>X</math> is on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, it has equal power, so | + | Since <math>X</math> is on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, it has equal power with respect to both circles, so |
<cmath> O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d} </cmath>since <math>O_1X+O_2X=d</math>. Now we can solve for <math>O_1X</math> and <math>O_2X</math>, and in particular, | <cmath> O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d} </cmath>since <math>O_1X+O_2X=d</math>. Now we can solve for <math>O_1X</math> and <math>O_2X</math>, and in particular, | ||
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 23:56, 13 March 2021
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
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,
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.