2015 AMC 12B Problems/Problem 24
Contents
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 side 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 .
Solution 2
Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that , , , and are all on the perpendicular bisector of . Let us say the distance from to the line is some . Therefore, the distance from to the line is , which comes out to be . Since , we have one of to be equal to . We can solve both equations to get that out of the four possible solutions, and only two are positive: and . Note that since no two circles can be congruent, we need the radius of one of or to be and the other to be . Plugging in to find the corresponding radii of and gives and , and adding everything up gives .
~hyxue
Solution 3
Let's start by drawing . Because all circles contain and , all the centers lie on the perpendicular bisector of , and point is on this bisector.
For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of , and two are on the opposite side of . For the latter two circles--call them and --.
Let's consider the next case, where and lie on the same side. Construct right triangles from the picture, and use the Pythagorean Theorem (divide by 3 to negate big numbers). You will get that the distance from to the closest circle center is . Therefore, the answer is .
Solution 4 (Pythagorean Theorem bash)
Since the radical axis is perpendicular to the line connecting the center of the circles, we have that , , , , and are collinear. WLOG, assume that and are on the same side of and let and let so that .
Then, using the Pythagorean Theorem on right triangles and , Subtracting the from gives Substituting into gives Taking the positive solution (), and .
Since none of the circles are congruent, and must be on the opposite side of so . Hence, .
Solution 5 (Fast)
Note that the four circles are coaxial, meaning are all collinear. Let . By Pythagorean Theorem, the radius of circle squared would be and the radius of circle squared would be Since , Solving this will give , or ; . The same equation will apply to ; would be the other root: (the negative signals that is the negative, or opposite, direction of , about ). Thus, , , , , implying .
~sml1809
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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