1984 AIME Problems/Problem 6

Problem

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ , $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ 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?

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.

Solution 2

Draw the midpoint of $\overline{AC}$ (the centers of the other two circles), and call it $M$ . If we draw the feet of the perpendiculars from $A,C$ to the line (call $E,F$ ), we see that $\triangle AEM\cong \triangle CFM$ by HA congruency; hence $M$ lies on the line. The coordinates of $M$ are $\left(\frac{19+14}{2},\frac{84+92}{2}\right) = \left(\frac{33}{2},88\right)$ .

Thus, the slope of the line is $\frac{88 - 76}{\frac{33}{2} - 17} = -24$ , and the answer is $\boxed{024}$ .

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 3

Define $E,F$ to be the feet of the perpendiculars from $A,C$ to the line (same as above). The equation of the line is $y = mx + b$ ; substituting $y=76,x=17$ gives us that $b = 76 - 17m$ , so the line is $y = mx + (76 - 17m)$ . $AE = CF$ by the HA argument above and CPCTC, so we can use the distance of a point to a line formula and equate.

$\left|\frac{m(14) - 92 + (76 - 17m)}{\sqrt{m^2 + 1}}\right| = \left|\frac{m(19) - 84 + (76 - 17m)}{\sqrt{m^2 + 1}}\right|$

$-3m - 16 = -2m + 8$

$m = -24$

And $|-24| = 24$ .

Solution 4 (non-rigorous)

Consider the region of the plane between $x=16$ and $x=17$ . The parts of the circles centered at $(14,92)$ and $(19,84)$ 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 $(17,76)$ is $8$ units below the center of the lower circle, we will have the line exit the region $8$ units above the center of the upper circle, at $(16,100)$ . We then find that the slope of the line is $-24$ and our answer is $\boxed{024}$ .

(Note: this solution does not feel rigorous when working through it, but it can be checked easily. In the above diagram, the point $M$ is marked. Rotate the aforementioned region of the plane $180^\circ$ about point $M$ , and the fact that certain areas are equal is evident.)

Solution 5

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. $O_1=(-3, 16)$ $O_2=(0,0)$ , and $O_3=(2,8)$ . 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 $(0,0)$ and $(-\frac{1}{2},12)$ . We use the slope formula to calculate the slope, which is $-24$ , leading to an answer of $\boxed{024}$ . $QED \blacksquare$

Solution by a1b2

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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