Difference between revisions of "2021 AIME I Problems/Problem 13"
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==Solution 5 (Official MAA)== | ==Solution 5 (Official MAA)== | ||
− | Like in other solutions, let <math>O</math> be the center of <math>\omega</math> with <math>r</math> its radius; also, let <math>O_{1}</math> and <math>O_{2}</math> be the centers of <math>\omega_{1}</math> and <math>\omega_{2}</math> with <math> | + | Like in other solutions, let <math>O</math> be the center of <math>\omega</math> with <math>r</math> its radius; also, let <math>O_{1}</math> and <math>O_{2}</math> be the centers of <math>\omega_{1}</math> and <math>\omega_{2}</math> with <math>R_{1}</math> and <math>R_{2}</math> their radii, respectively. Let line <math>OP</math> intersect line <math>O_{1}O_{2}</math> at <math>T</math>, and let <math>u=TO_{2}</math>, <math>v=TO_{1}</math>, <math>x=PT</math>, where the length <math>O_{1}O_{2}</math> splits as <math>u+v</math>. Because the lines <math>PQ</math> and <math>O_{1}O_{2}</math> are perpendicular, lines <math>OT</math> and <math>O_{1}O_{2}</math> meet at a <math>60^{\circ}</math> angle. |
+ | |||
+ | Applying the Law of Cosines to <math>\triangle O_{2}PT</math>, <math>\triangle O_{1}PT</math>, <math>\triangle O_{2}OT</math>, and <math>\triangle O_{1}OT</math> gives <cmath>\begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}</cmath> | ||
Adding the first and fourth equations, then subtracting the second and third equations gives us <cmath>\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)</cmath> | Adding the first and fourth equations, then subtracting the second and third equations gives us <cmath>\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)</cmath> |
Revision as of 10:25, 16 July 2022
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 (Properties of Radical Axis)
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 (Linearity)
Let ,
, and
be the centers of
,
, and
with
,
, and
their radii, respectively. Then, the distance from
to the radical axis
of
is equal to
. Let
and
the orthogonal projection of
onto line
. Define the function
by
Then
By Linearity of Power of a Point,
Notice that
and
, thus
Dividing both sides by
(which is obviously nonzero as
is nonzero) gives us
so
. Since
and
, the answer is
.
Solution 3
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 4 (Quick)
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
Solution 5 (Official MAA)
Like in other solutions, let be the center of
with
its radius; also, let
and
be the centers of
and
with
and
their radii, respectively. Let line
intersect line
at
, and let
,
,
, where the length
splits as
. Because the lines
and
are perpendicular, lines
and
meet at a
angle.
Applying the Law of Cosines to ,
,
, and
gives
Adding the first and fourth equations, then subtracting the second and third equations gives us
Since lies on the radical axis of
and
, the power of point
with respect to either circle is
Hence which simplifies to
The requested distance is therefore equal to
.
Solution 6 (Geometry)
Let circle tangent circles
and
respectively at distinct points
and
. Let
be the centers (the radii) of
and
respectively. WLOG
Let
be the point of
such, that
Let
be the midpoint
be the radical axes of
and
Then is radical center of
and
In is bisector of
is median
and
are colinear.
is inscribed in
is diameter
as they intercept the same arc
in
Since and
, the answer is
.
Shelomovskii, vvsss, www.deoma-cmd.ru
Video Solution
https://youtu.be/gN7Ocu3D62M ~Math Problem Solving Skills
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.