Difference between revisions of "2002 AMC 12A Problems/Problem 18"
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</math> | </math> | ||
− | == Solution == | + | == Solution 1 == |
− | + | 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 34: | Line 34: | ||
label("$C_1$",(10,0) + 6*dir(-45), SE ); | label("$C_1$",(10,0) + 6*dir(-45), SE ); | ||
label("$C_2$",(-15,0) + 9*dir(225), SW ); | label("$C_2$",(-15,0) + 9*dir(225), SW ); | ||
− | label("$ | + | label("$D_1$",(10,0), SE ); |
− | label("$ | + | label("$D_2$",(-15,0), SW ); |
label("$Q$", p2[0], NE ); | label("$Q$", p2[0], NE ); | ||
label("$P$", p1[1], SW ); | label("$P$", p1[1], SW ); | ||
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 | + | 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 <math>D_2QO</math> and <math>D_1PO</math> similar by AA, with a scale factor of <math>6:9</math>, or <math>2:3</math>. Next, we must subdivide the line <math>D_2D_1</math> in a 2:3 ratio to get the length of the segments <math>D_2O</math> and <math>D_1O</math>. The total length is <math>10 - (-15)</math>, or <math>25</math>, so applying the ratio, <math>D_2O</math> = '''15''' and <math>D_1O</math> = '''10'''. These are the hypotenuses of the triangles. We already know the length of <math>D_2Q</math> and <math>D_1P</math>, '''9''' and '''6''' (they're radii). So in order to find <math>PQ</math>, we must find the length of the longer legs of the two triangles and add them. |
<math>15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144</math> | <math>15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144</math> | ||
Line 50: | Line 50: | ||
<math>\sqrt{64} = 8</math> | <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 == | ||
+ | Using the above diagram, imagine that segment <math>\overline{QD_2}</math> is shifted to the right to match up with <math>\overline{PD_1}</math>. Then shift <math>\overline{QP}</math> downwards to make a right triangle. We know <math>\overline{D_2D_1} = 25</math> from the given information and the newly created leg has length <math> \overline{QD_2} + \overline{PD_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. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/wVadnulGLok | ||
== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:13, 1 September 2023
Problem
Let and be circles defined by and respectively. What is the length of the shortest line segment that is tangent to at and to at ?
Solution 1
First examine the formula , for the circle . Its center, , is located at (10,0) and it has a radius of = 6. The next circle, using the same pattern, has its center, , at (-15,0) and has a radius of = 9. So we can construct this diagram: 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 and similar by AA, with a scale factor of , or . Next, we must subdivide the line in a 2:3 ratio to get the length of the segments and . The total length is , or , so applying the ratio, = 15 and = 10. These are the hypotenuses of the triangles. We already know the length of and , 9 and 6 (they're radii). So in order to find , we must find the length of the longer legs of the two triangles and add them.
Finally, the length of PQ is , or (C).
Solution 2
Using the above diagram, imagine that segment is shifted to the right to match up with . Then shift downwards to make a right triangle. We know from the given information and the newly created leg has length . Hence by Pythagorean theorem .
, or C.
Video Solution
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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