2015 AMC 12B Problems/Problem 24

Revision as of 18:15, 22 December 2018 by Jzhang21 (talk | contribs) (Number-Intensive Solution)


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 $AR+BR+CR+DR$ ?

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


First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ 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 $C_1$ is $r$ and the radius of circle $C_2$ is $\tfrac{5}{8}r$.

Then the power of point $R$ with respect to $C_1$ is

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

and the power of point $R$ with respect to $C_2$ 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 $57+96+39 = \boxed{\textbf{(D)}\; 192}$.

Number-Intensive Solution

Start by drawing $PQ$ first, because trying to get all four circles down will take you a few years. Next, because all circles have $P$ and $Q$ on them, and since all points on a circle are equidistant from the center, all circle centers lie on the perpendicular bisector of $PQ$, and point $R$ 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 $PQ$, and two are straddling it. For the latter two circles- just call them $A$ and $B$- clearly $AR+BR$ is 39.

Now, let's take the next case. Then $C$ and $D$ 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 $R$ to the closest circle center is $57$. Therefore, the answer is $39+2*57+39=192 \boxed{(D)}$.

=Pythagorean Theorem Solution

Since the radical axis $PQ$ is perpendicular to the line connecting the center of the circles, we have hat $A,B,C,D,$ and $R$ are collinear. WLOG, assume that $A$ and $B$ are on the same side of $R$. Let $AR=y$ and let $BP=x$ so that $AP=\frac{5}{8}x$. Then, we have \[(39+y)^2+24^2=x^2\] \[y^2+24^2=\frac{25}{64}x^2\] Subtracting the second from the first gives $x^2=64(2y+39)$ and substituting this in the second equation gives $y^2-50y-399=0=(y-57)(y+7).$ Since $y>0, y=57$ and $AR=57$ while $BR=AR+39=96.$ Since none of the circles are congruent, $C$ and $D$ must be on the opposite side of $R$ so $CR+DR=CD=39.$ Hence, $AR+BR+CR+DR=57+96+39=192,$ which is $\boxed{D}$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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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