2002 AMC 12A Problems/Problem 18

Problem

Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$ ?

$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$

Solution

Circle $C_1$ has center at $S_1=(10,0)$ and radius $r_1=\sqrt{36}=6$ , circle $C_2$ has center at $S_2=(-15,0)$ and radius $r_2=9$ .

$[asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0)); label("$C_1$",(10,0) + 6*dir(-45), SE ); label("$C_2$",(-15,0) + 9*dir(225), SW ); label("$S_1$",(10,0), SE ); label("$S_2$",(-15,0), SW ); label("$Q$", p2[0], NE ); label("$P$", p1[1], SW ); label("$O$", (0,0), SW ); [/asy]$

Let $PQ$ be the inner tangent of the two circles, as shown in the picture above. Then the triangles $S_1PO$ and $S_2QO$ are similar right triangles. As $PS_1:QS_2 = 6:9 = 2:3$ , we also have $OS_1 : OS_2 = 2:3$ , hence $OS_1 = 10$ , $OS_2=15$ , and thus $O=(0,0)$ .

We can now use the Pythagorean theorem to compute $PO = \sqrt{ OS_1^2 - PS_1^2 } = \sqrt{ 10^2 - 6^2 } = 8$ and $QO = \sqrt{ OS_2^2 - QS_2^2 } = \sqrt{15^2 - 9^2} = 12$ , and thus $PQ = 8+12 = 20$ .

The only other option for $PQ$ is the outer tangent of the two circles. We will now show that the outer tangent is always longer than the inner one.

$[asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0)); pair[] p3 = intersectionpoints(C1, circle((35,0),25) ); pair[] p4 = intersectionpoints(C2, circle((22.5,0),37.5) ); dot(p3[1],red); dot(p4[1],red); draw( p3[1]--p4[1], red ); pair X = intersectionpoint( p3[1]--p4[1], p1[1]--(p1[1] + (p1[1]-p2[0])) ); draw( p1[1]--(p1[1] + 0.2*(p1[1]-p2[0])) , blue ); dot(X, blue ); label("$C_1$",(10,0) + 6*dir(-45), SE ); label("$C_2$",(-15,0) + 9*dir(225), SW ); label("$S_1$",(10,0), SE ); label("$S_2$",(-15,0), SW ); label("$Q$", p2[0], NE ); label("$P$", p1[1], SW ); label("$O$", (0,0), SW ); label("$P'$", p3[1], SSE, red ); label("$Q'$", p4[1], SSE, red ); label("$X$", X, SSW, blue ); [/asy]$

Consider the outer tangent $P'Q'$ shown in red in the picture above. Extend $PQ$ to intersect $P'Q'$ in the point $X$ shown in blue. Clearly $XQ > PQ$ .

Now the segments $XQ$ and $XQ'$ are the two tangents from the point $X$ to the circle $C_2$ , hence $XQ=XQ'$ . And as obviously $XQ' < P'Q'$ , we get $PQ < P'Q'$ .

Therefore our answer is the length of the inner common tangent, i.e., $\boxed{20}$ .

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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
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2002 AMC 12A Problems/Problem 18

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