Difference between revisions of "2002 AMC 12A Problems/Problem 18"

Revision as of 16:52, 17 January 2021

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

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$",(-15,0) + 9*dir(225), SW ); label("$D_1$",(10,0), SE ); label("$D_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.

@@ Line 17: / Line 17: @@
 == Solution 1 ==
-'''(C)''' First examine the formula <math>(x-10)^2+y^2=36</math>, for the circle <math>C_1</math>. Its center, <math>D_1</math>, is located at (10,0) and it has a radius of <math>\sqrt{36}</math> = 6. The next circle, using the same pattern, has its center, <math>D_2</math>, at (-15,0) and has a radius of <math>\sqrt{81}</math> = 9. So we can construct this diagram:
+First examine the formula <math>(x-10)^2+y^2=36</math>, for the circle <math>C_1</math>. Its center, <math>D_1</math>, is located at (10,0) and it has a radius of <math>\sqrt{36}</math> = 6. The next circle, using the same pattern, has its center, <math>D_2</math>, at (-15,0) and has a radius of <math>\sqrt{81}</math> = 9. So we can construct this diagram:
 <asy>
 unitsize(0.3cm);
@@ Line 50: / Line 50: @@
 <math>\sqrt{64} = 8</math>
-Finally, the length of PQ is <math>12+8=\boxed{20}</math>, or '''C'''.
+Finally, the length of PQ is <math>12+8=\boxed{20}</math>, or '''(C)'''.
 == Solution 2 ==

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

Page

Toolbox

Search

Difference between revisions of "2002 AMC 12A Problems/Problem 18"

Revision as of 16:52, 17 January 2021

Contents

Problem

Solution 1

Solution 2

See Also