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Difference between revisions of "2002 AMC 12A Problems/Problem 18"
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== Solution 2 ==
 
== Solution 2 ==
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Using the above diagram, imagine that segment <math>\overline{QS_2}</math> is shifted to the right to match up with <math>\overline{S_1P}</math>. Then shift <math>\overline{QP}</math> downwards to make a right triangle. We know <math>\overline{S_2S_1} = 25</math> from the given information and the newly created leg has length <math> \overline{QS_2} + \overline{S_1P} = 9 + 6 = 15</math>. Hence by Pythagorean theorem <math>15^2 + {\overline{QP}}^2 = 25^2</math>.
+
Using the above diagram, imagine that segment <math>\overline{QS_2}</math> is shifted to the right to match up with <math>\overline{PS_1}</math>. Then shift <math>\overline{QP}</math> downwards to make a right triangle. We know <math>\overline{S_2S_1} = 25</math> from the given information and the newly created leg has length <math> \overline{QS_2} + \overline{PS_1} = 9 + 6 = 15</math>. Hence by Pythagorean theorem <math>15^2 + {\overline{QP}}^2 = 25^2</math>.
  
 
<math> \overline{QP} = \boxed{20}</math>, or C.
 
<math> \overline{QP} = \boxed{20}</math>, or C.

Revision as of 00:19, 20 January 2014

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 1

(C) First examine the formula $(x-10)^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: [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] Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles S$_2$QO and S$_1$PO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line D$_2$D$_1$ in a 2:3 ratio to get the length of the segments D$_2$O and D$_1$O. The total length is 10 - (-15), or 25, so applying the ratio, D$_2$O = 15 and D$_1$O = 10. These are the hypotenuses of the triangles. We already know the length of D$_2$Q and D$_1$P, 9 and 6 (they're radii). So in order to find PQ, we must find the length of the longer legs of the two triangles and add them.

$15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$

$\sqrt{144} = 12$

$10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$

$\sqrt{64} = 8$

Finally, the length of PQ is $12+8=\boxed{20}$, or C.

Solution 2

Using the above diagram, imagine that segment $\overline{QS_2}$ is shifted to the right to match up with $\overline{PS_1}$. Then shift $\overline{QP}$ downwards to make a right triangle. We know $\overline{S_2S_1} = 25$ from the given information and the newly created leg has length $\overline{QS_2} + \overline{PS_1} = 9 + 6 = 15$. Hence by Pythagorean theorem $15^2 + {\overline{QP}}^2 = 25^2$.

$\overline{QP} = \boxed{20}$, or C.

See Also

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
All AMC 12 Problems and Solutions

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