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

(New page: == Problem == Let <math>C_1</math> and <math>C_2</math> be circles defined by <math>(x-10)^2 + y^2 = 36</math> and <math>(x+15)^2 + y^2 = 81</math> respectively. What is the length of the ...)
 
(→‎Solution)
Line 17: Line 17:
 
== Solution ==
 
== Solution ==
  
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Circle <math>C_1</math> has center at <math>S_1=(10,0)</math> and radius <math>r_1=\sqrt{36}=6</math>, circle <math>C_2</math> has center at <math>S_2=(-15,0)</math> and radius <math>r_2=9</math>.  
+
'''(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:
−
 
 
 
<asy>
 
<asy>
 
unitsize(0.3cm);
 
unitsize(0.3cm);
Line 41: Line 40:
 
label("$O$", (0,0), SW );
 
label("$O$", (0,0), SW );
 
</asy>
 
</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<math>_2</math>QO and S<math>_1</math>PO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line D<math>_2</math>D<math>_1</math> in a 2:3 ratio to get the length of the segments D<math>_2</math>O and D<math>_1</math>O. The total length is 10 - (-15), or 25, so applying the ratio, D<math>_2</math>O = '''15''' and D<math>_1</math>O = '''10'''. These are the hypotenuses of the triangles. We already know the length of D<math>_2</math>Q and D<math>_1</math>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.
  
−
Let <math>PQ</math> be the inner tangent of the two circles, as shown in the picture above. Then the triangles <math>S_1PO</math> and <math>S_2QO</math> are similar right triangles. As <math>PS_1:QS_2 = 6:9 = 2:3</math>, we also have <math>OS_1 : OS_2 = 2:3</math>, hence <math>OS_1 = 10</math>, <math>OS_2=15</math>, and thus <math>O=(0,0)</math>.
+
<math>15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144</math>
−
 
 
−
We can now use the [[Pythagorean theorem]] to compute <math>PO = \sqrt{ OS_1^2 - PS_1^2 } = \sqrt{ 10^2 - 6^2 } = 8</math> and <math>QO = \sqrt{ OS_2^2 - QS_2^2 } = \sqrt{15^2 - 9^2} = 12</math>, and thus <math>PQ = 8+12 = 20</math>.
 
−
 
 
−
The only other option for <math>PQ</math> 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);
 
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path C1=circle((10,0),6);
 
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path C2=circle((-15,0),9);
 
−
 
 
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draw(C1); draw(C2);
 
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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 <math>P'Q'</math> shown in red in the picture above. Extend <math>PQ</math> to intersect <math>P'Q'</math> in the point <math>X</math> shown in blue. Clearly <math>XQ > PQ</math>.
+
<math>\sqrt{144} = 12</math>
  
−
Now the segments <math>XQ</math> and <math>XQ'</math> are the two tangents from the point <math>X</math> to the circle <math>C_2</math>, hence <math>XQ=XQ'</math>. And as obviously <math>XQ' < P'Q'</math>, we get <math>PQ < P'Q'</math>.
+
<math>10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64</math>
  
−
Therefore our answer is the length of the inner common tangent, i.e., <math>\boxed{20}</math>.
+
<math>\sqrt{64} = 8</math>
  
 +
Finally, the length of PQ is <math>12+8=\boxed{20}</math>, or '''C'''.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2002|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2002|ab=A|num-b=17|num-a=19}}

Revision as of 08:50, 27 January 2011

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

(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.

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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