Difference between revisions of "2015 AMC 12B Problems/Problem 24"
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==Solution 2== | ==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 <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>. | + | 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{\textbf{(D) }192}</math>. |
− | |||
− | + | ~hyxue | |
− | |||
− | + | ==Solution 3== | |
+ | Let's start by drawing <math>PQ</math>. Because all circles contain <math>P</math> and <math>Q</math>, all the centers lie on the perpendicular bisector of <math>PQ</math>, and point <math>R</math> is on this bisector. | ||
− | + | For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of <math>PQ</math>, and two are on the opposite side of <math>PQ</math>. For the latter two circles--call them <math>A</math> and <math>B</math>--<math>AR+BR=39</math>. | |
− | ==Pythagorean Theorem | + | Let's consider the next case, where <math>C</math> and <math>D</math> lie on the same side. Construct right triangles from the picture, and use the [[Pythagorean Theorem]] (divide by 3 to negate big numbers). You will get that the distance from <math>R</math> to the closest circle center is <math>57</math>. Therefore, the answer is <math>39+2\cdot57+39=\boxed{\textbf{(D) }192}</math>. |
− | Since the radical axis <math>PQ</math> is perpendicular to the line connecting the center of the circles, we have that <math>A,B,C,D | + | |
− | <cmath>(39+y)^2+24^2=x^2</cmath> | + | ==Solution 4 (Pythagorean Theorem bash)== |
− | <cmath>y^2+24^2=\frac{25}{64}x^2</cmath> Subtracting the | + | Since the radical axis <math>PQ</math> is perpendicular to the line connecting the center of the circles, we have that <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>R</math> are collinear. WLOG, assume that <math>A</math> and <math>B</math> are on the same side of <math>R</math> and let <math>AR=y</math> and let <math>BP=x</math> so that <math>AP=\frac{5}{8}x</math>. |
+ | |||
+ | Then, using the [[Pythagorean Theorem]] on right triangles <math>PBR</math> and <math>PAR</math>, | ||
+ | <cmath>(39+y)^2+24^2=x^2\qquad(1)</cmath> | ||
+ | <cmath>y^2+24^2=\frac{25}{64}x^2\qquad(2)</cmath> Subtracting the <math>(2)</math> from <math>(1)</math> gives <cmath>x^2=64(2y+39)\qquad(3)</cmath> | ||
+ | Substituting <math>(3)</math> into <math>(2)</math> gives <math>y^2-50y-399=0=(y-57)(y+7).</math> Taking the positive solution (<math>y>0</math>), <math>y=AR=57</math> and <math>BR=(57)+39=96</math>. | ||
+ | |||
+ | Since none of the circles are congruent, <math>C</math> and <math>D</math> must be on the opposite side of <math>R</math> so <math>CR+DR=CD=39</math>. Hence, <math>AR+BR+CR+DR=57+96+39=\boxed{\textbf{(D) }192}</math>. | ||
+ | |||
+ | ==Solution 5 (Fast)== | ||
+ | Note that the four circles are coaxial, meaning <math>A,B,C,D,R</math> are all collinear. Let <math>AR=x</math>. By [[Pythagorean Theorem]], the radius of circle <math>A</math> squared would be | ||
+ | <math>r_A^2 = x^2+24^2</math> | ||
+ | and the radius of circle <math>B</math> squared would be | ||
+ | <math>r_B^2 = (x+39)^2+24^2.</math> | ||
+ | Since <math>r_A^2 = \dfrac{25}{64}r_B^2</math>, | ||
+ | <cmath>x^2+24^2 = \dfrac{25}{64}((x+39)^2+24^2)</cmath> | ||
+ | Solving this will give <math>x^2-50x-399 = 0</math>, or <math>x=57, -7</math>; <math>AR = 57</math>. The same equation will apply to <math>CR</math>; <math>CR</math> would be the other root: <math>CR=7</math> (the negative <math>7</math> signals that <math>C</math> is the negative, or opposite, direction of <math>D</math>, about <math>R</math>). Thus, <math>AR = 57</math>, <math>BR = 57+39 = 96</math>, <math>CR = 7</math>, <math>DR = 39-7 = 32</math>, implying | ||
+ | <math>AR+BR+CR+DR = \boxed{\textbf{(D) }192}</math>. | ||
+ | |||
+ | ~sml1809 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=25|num-b=23}} | {{AMC12 box|year=2015|ab=B|num-a=25|num-b=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:18, 19 October 2023
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
.
~hyxue
Solution 3
Let's start by drawing . Because all circles contain
and
, all the centers lie on the perpendicular bisector of
, and point
is on this bisector.
For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of , and two are on the opposite side of
. For the latter two circles--call them
and
--
.
Let's consider the next case, where and
lie on the same side. Construct right triangles from the picture, and use the Pythagorean Theorem (divide by 3 to negate big numbers). You will get that the distance from
to the closest circle center is
. Therefore, the answer is
.
Solution 4 (Pythagorean Theorem bash)
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
and let
and let
so that
.
Then, using the Pythagorean Theorem on right triangles and
,
Subtracting the
from
gives
Substituting
into
gives
Taking the positive solution (
),
and
.
Since none of the circles are congruent, and
must be on the opposite side of
so
. Hence,
.
Solution 5 (Fast)
Note that the four circles are coaxial, meaning are all collinear. Let
. By Pythagorean Theorem, the radius of circle
squared would be
and the radius of circle
squared would be
Since
,
Solving this will give
, or
;
. The same equation will apply to
;
would be the other root:
(the negative
signals that
is the negative, or opposite, direction of
, about
). Thus,
,
,
,
, implying
.
~sml1809
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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