Difference between revisions of "2002 AMC 12A Problems/Problem 18"
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== Solution 1 == | == 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); | ||
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<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 == | == Solution 2 == |
Latest revision as of 16:52, 17 January 2021
Contents
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 SQO and SPO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line DD in a 2:3 ratio to get the length of the segments DO and DO. The total length is 10 - (-15), or 25, so applying the ratio, DO = 15 and DO = 10. These are the hypotenuses of the triangles. We already know the length of DQ and DP, 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.
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.
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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