1984 AIME Problems/Problem 6
Problem
Three circles, each of radius , are drawn with centers at , , and . A line passing through is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?
Contents
Solution 1
The line passes through the center of the second circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle.
Draw the midpoint of (the centers of the other two circles), and call it . If we draw the feet of the perpendiculars from to the line (call ), we see that by HA congruency; hence lies on the line. The coordinates of are .
Thus, the slope of the line is , and the answer is .
Remark: Notice the fact that the radius is 3 is not used in this problem; in fact changing the radius does not affect the answer.
Solution 2
First of all, we can translate everything downwards by and to the left by . Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem:
Two circles, each of radius , are drawn with centers at , and . A line passing through is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?
Note that this is equivalent to finding a line such that the distance from to the line is the same as the distance from to the line. Let the line be . Then, we have that: We can split this into two cases.
Case 1:
In this case, the absolute value of the slope of the line won’t be an integer, and since this is an AIME problem, we know it’s not possible.
Case 2:
But we also know that it passes through the point , so . Plugging this in, we see that . .
Solution 3 (non-rigorous)
Consider the region of the plane between and . The parts of the circles centered at and in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since is units below the center of the lower circle, we will have the line exit the region units above the center of the upper circle, at . We then find that the slope of the line is and our answer is .
(Note: this solution does not feel rigorous when working through it, but it can be checked easily. In the above diagram, the point is marked. Rotate the aforementioned region of the plane about point , and the fact that certain areas are equal is evident.)
Solution 4
We can redefine the coordinate system so that the center of the center circle is the origin, for easier calculations, as the slope of the line and the congruence of the circles do not depend on it. , and . A line bisects a circle iff it passes through the center. Therefore, we can ignore the bottom circle because it contributes an equal area with any line. A line passing through the centroid of any plane system with two perpendicular lines of reflectional symmetry bisects it. We have defined two points of the line, which are and . We use the slope formula to calculate the slope, which is , leading to an answer of .
Solution by a1b2
Solution 5 (Easy to understand)
Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose is the desired line. Draw lines and both parallel to such that passes through and passes through . Clearly, must be the "average" of and . Suppose . Then . So we have that which yields for an answer of .
~yofro
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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