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Difference between revisions of "2015 AMC 12B Problems/Problem 24"

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==Solution==
 
==Solution==
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First, note that <math>PQ</math> lies the radical axis of any of the pairs of circles. Suppose that <math>O_1</math> and <math>O_2</math> are the centers of two circles that intersect exactly at <math>P</math> and <math>Q</math>, with <math>O_1</math> and <math>O_2</math> lying on the same side of <math>PQ</math>, and <math>O_1 O_2=39</math>. Let <math>x=O_1 R</math>, <math>y=O_2 R</math>, and suppose that the radius of circle <math>O_1</math> is <math>r</math> and the radius of circle <math>O_2</math> is <math>\tfrac{5}{8}r</math>.
  
 +
Then the power of point <math>R</math> in <math>O_1</math> is
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<cmath>(r+x)(r-x) = r^2 - x^2 = 24^2</cmath>
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and the power of point <math>R</math> in <math>O_2</math> is
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<cmath>\left(\frac{5}{8}r + y\right) \left(\frac{5}{8}r - y\right) = \frac{25}{64}r^2 - y^2 = 24^2.</cmath>
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Also, note that <math>x-y=39</math>.
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Subtract the above two equations to find that <math>\tfrac{39}{64}r^2 - x^2 + y^2 = 0</math> or <math>39 r^2 = 64(x^2-y^2)</math>. As <math>x-y=39</math>, we find that <math>r^2=64(x+y) = 64(2y+39)</math>. Plug this into an earlier equation to find that <math>25(2y+39)-y^2=24^2</math>. This is a quadratic equation with solutions <math>y=\tfrac{50 \pm 64}{2}</math>, and as <math>y</math> is a length, it is positive, hence <math>y=57</math>, and <math>x=y+39=96</math>. This is the only possibility if the two centers lie on the same same of their radical axis.
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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 <math>O_1 R + O_2 R = O_1 O_2 = 39</math>. Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is <math>57+96+39 = \boxed{\textbf{(D)}\; 192}</math>.
  
 
==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}}

Revision as of 12:42, 5 March 2015

Problem

Four circles, no two of which are congruent, have centers at $A$, $B$, $C$, and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ = 48$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$ ?

$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$

Solution

First, note that $PQ$ lies the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles that intersect exactly at $P$ and $Q$, with $O_1$ and $O_2$ lying on the same side of $PQ$, and $O_1 O_2=39$. Let $x=O_1 R$, $y=O_2 R$, and suppose that the radius of circle $O_1$ is $r$ and the radius of circle $O_2$ is $\tfrac{5}{8}r$.

Then the power of point $R$ in $O_1$ is

\[(r+x)(r-x) = r^2 - x^2 = 24^2\]

and the power of point $R$ in $O_2$ is

\[\left(\frac{5}{8}r + y\right) \left(\frac{5}{8}r - y\right) = \frac{25}{64}r^2 - y^2 = 24^2.\]

Also, note that $x-y=39$.

Subtract the above two equations to find that $\tfrac{39}{64}r^2 - x^2 + y^2 = 0$ or $39 r^2 = 64(x^2-y^2)$. As $x-y=39$, we find that $r^2=64(x+y) = 64(2y+39)$. Plug this into an earlier equation to find that $25(2y+39)-y^2=24^2$. This is a quadratic equation with solutions $y=\tfrac{50 \pm 64}{2}$, and as $y$ is a length, it is positive, hence $y=57$, and $x=y+39=96$. 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 $O_1 R + O_2 R = O_1 O_2 = 39$. Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is $57+96+39 = \boxed{\textbf{(D)}\; 192}$.

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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