2015 AMC 12B Problems/Problem 24
Contents
Problem
Four circles, no two of which are congruent, have centers at ,
,
, and
, and points
and
lie on all four circles. The radius of circle
is
times the radius of circle
, and the radius of circle
is
times the radius of circle
. Furthermore,
and
. Let
be the midpoint of
. What is
?
Solution
First, note that lies on the radical axis of any of the pairs of circles. Suppose that
and
are the centers of two circles
and
that intersect exactly at
and
, with
and
lying on the same side of
, and
. Let
,
, and suppose that the radius of circle
is
and the radius of circle
is
.
Then the power of point with respect to
is
and the power of point with respect to
is
Also, note that .
Subtract the above two equations to find that or
. As
, we find that
. Plug this into an earlier equation to find that
. This is a quadratic equation with solutions
, and as
is a length, it is positive, hence
, and
. This is the only possibility if the two centers lie on the same same of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is
.
Number-Intensive Solution
Start by drawing first, because trying to get all four circles down will take you a few years. Next, because all circles have
and
on them, and since all points on a circle are equidistant from the center, all circle centers lie on the perpendicular bisector of
, and point
is on this bisector.
In order for all the circle radii to be different (because the circles can't be congruent), two circle centers are on the same side of , and two are straddling it. For the latter two circles- just call them
and
- clearly
is 39.
Now, let's take the next case. Then and
lie on the same side. Construct the triangles from your picture, and use the Pythagorean Theorem (maybe divide all lengths by 3 to lessen big numbers) and then you get that the distance from
to the closest circle center is
. Therefore, the answer is
.
Pythagorean Theorem Solution
Since the radical axis is perpendicular to the line connecting the center of the circles, we have hat
and
are collinear. WLOG, assume that
and
are on the same side of
. Let
and let
so that
. Then, we have
Subtracting the second from the first gives
and substituting this in the second equation gives
Since
and
while
Since none of the circles are congruent,
and
must be on the opposite side of
so
Hence,
which is
.
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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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