Difference between revisions of "2002 AMC 12A Problems/Problem 18"
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− | 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> |
Revision as of 03:12, 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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