Difference between revisions of "2015 AMC 12B Problems/Problem 24"
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On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that <math>O_1 R + O_2 R = O_1 O_2 = 39</math>. Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is <math>57+96+39 = \boxed{\textbf{(D)}\; 192}</math>. | On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that <math>O_1 R + O_2 R = O_1 O_2 = 39</math>. Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is <math>57+96+39 = \boxed{\textbf{(D)}\; 192}</math>. | ||
+ | ==Number-Intensive Solution== | ||
+ | Start by drawing <math>PQ</math> first, because trying to get all four circles down will take you a few years. Next, because all circles have <math>P</math> and <math>Q</math> on them, and since all points on a circle are equidistant from the center, all circle centers lie on the perpendicular bisector of <math>PQ</math>, and point <math>R</math> is on this bisector. | ||
+ | |||
+ | In order for all the circle radii to be different (because the circles can't be congruent), two circle centers are on the same side of <math>PQ</math>, and two are straddling it. For the latter two circles- just call them <math>A</math> and <math>B</math>- clearly <math>AR+BR</math> is 39. | ||
+ | |||
+ | Now, let's take the next case. Then <math>C</math> and <math>D</math> lie on the same side. Construct the triangles from your picture, and use the Pythagorean Theorem (maybe divide all lengths by 3 to lessen big numbers) and then you get that the distance from <math>R</math> to the closest circle center is <math>57</math>. Therefore, the answer is <math>39+2*57+39=192 \boxed{(D)}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=25|num-b=23}} | {{AMC12 box|year=2015|ab=B|num-a=25|num-b=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:56, 4 February 2018
Problem
Four circles, no two of which are congruent, have centers at , , , and , and points and lie on all four circles. The radius of circle is times the radius of circle , and the radius of circle is times the radius of circle . Furthermore, and . Let be the midpoint of . What is ?
Solution
First, note that lies on the radical axis of any of the pairs of circles. Suppose that and are the centers of two circles and that intersect exactly at and , with and lying on the same side of , and . Let , , and suppose that the radius of circle is and the radius of circle is .
Then the power of point with respect to is
and the power of point with respect to is
Also, note that .
Subtract the above two equations to find that or . As , we find that . Plug this into an earlier equation to find that . This is a quadratic equation with solutions , and as is a length, it is positive, hence , and . This is the only possibility if the two centers lie on the same same of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is .
Number-Intensive Solution
Start by drawing first, because trying to get all four circles down will take you a few years. Next, because all circles have and on them, and since all points on a circle are equidistant from the center, all circle centers lie on the perpendicular bisector of , and point is on this bisector.
In order for all the circle radii to be different (because the circles can't be congruent), two circle centers are on the same side of , and two are straddling it. For the latter two circles- just call them and - clearly is 39.
Now, let's take the next case. Then and lie on the same side. Construct the triangles from your picture, and use the Pythagorean Theorem (maybe divide all lengths by 3 to lessen big numbers) and then you get that the distance from to the closest circle center is . Therefore, the answer is .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.