2002 AMC 12A Problems/Problem 18

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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 $r_2=9$ .

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.

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 • 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

From AoPSWiki

Problem

Solution

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