Difference between revisions of "2015 AMC 12B Problems/Problem 24"
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− | Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are all on the perpendicular bisector of <math>PQ</math>. Let us say the distance from <math>A</math> to the line <math>PQ</math> is some <math>a</math>. Therefore, the distance from <math>B</math> to the line <math>PQ</math> is <math>\sqrt{(\frac{8}{5}\sqrt{ | + | Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are all on the perpendicular bisector of <math>PQ</math>. Let us say the distance from <math>A</math> to the line <math>PQ</math> is some <math>a</math>. Therefore, the distance from <math>B</math> to the line <math>PQ</math> is <math>\sqrt{(\frac{8}{5}\sqrt{a^2 + 24^2})^2 - 24^2}</math>, which comes out to be <math>\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576}</math>. Since <math>AB = 39</math>, we have one of <math>\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576} \pm a</math> to be equal to <math>39</math>. We can solve both equations to get that out of the four possible solutions, and only two are positive: <math>7</math> and <math>57</math>. Note that since no two circles can be congruent, we need the radius of one of <math>A</math> or <math>C</math> to be <math>7</math> and the other to be <math>57</math>. Plugging in to find the corresponding radii of <math>B</math> and <math>D</math> gives <math>32</math> and <math>96</math>, and adding everything up gives <math>\boxed{D) 192}</math>. |
~Solution by hyxue | ~Solution by hyxue | ||
Revision as of 19:18, 9 January 2021
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 side 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
.
Solution 2
Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that ,
,
, and
are all on the perpendicular bisector of
. Let us say the distance from
to the line
is some
. Therefore, the distance from
to the line
is
, which comes out to be
. Since
, we have one of
to be equal to
. We can solve both equations to get that out of the four possible solutions, and only two are positive:
and
. Note that since no two circles can be congruent, we need the radius of one of
or
to be
and the other to be
. Plugging in to find the corresponding radii of
and
gives
and
, and adding everything up gives
.
~Solution by hyxue
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 that
and
are collinear. WLOG, assume that
and
are on the same side of
. Let
and let
so that
. Then, we have in right triangles
and
,
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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