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Difference between revisions of "1984 AIME Problems/Problem 6"

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=== Solution 1 ===
 
=== Solution 1 ===
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Draw the [[midpoint]] of <math>\overline{AC}</math> (the centers of the other two circles), and call it <math>M</math>. If we draw the feet of the [[perpendicular]]s from <math>A,C</math> to the line (call <math>E,F</math>), we see that <math>\triangle AEC\cong \triangle CFM</math> by [[HA congruency]]; hence <math>M</math> lies on the line. The coordinates of <math>M</math> are <math>\left(\frac{19+14}{2},\frac{84+92}{2}\right) = \left(\frac{33}{2},88\right)</math>.  
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Draw the [[midpoint]] of <math>\overline{AC}</math> (the centers of the other two circles), and call it <math>M</math>. If we draw the feet of the [[perpendicular]]s from <math>A,C</math> to the line (call <math>E,F</math>), we see that <math>\triangle AEM\cong \triangle CFM</math> by [[HA congruency]]; hence <math>M</math> lies on the line. The coordinates of <math>M</math> are <math>\left(\frac{19+14}{2},\frac{84+92}{2}\right) = \left(\frac{33}{2},88\right)</math>.  
  
 
Thus, the slope of the line is <math>\frac{88 - 76}{\frac{33}{2} - 17} = -24</math>, and the answer is <math>\boxed{024}</math>.
 
Thus, the slope of the line is <math>\frac{88 - 76}{\frac{33}{2} - 17} = -24</math>, and the answer is <math>\boxed{024}</math>.

Revision as of 14:15, 20 July 2012

Problem

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

1984 AIME-6.png

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 1

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 2

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 $y = mx + (76 - 17m)$. $\displaystyle AE = CF \displaystyle$ by the HA argument above and CPCTC, so we can use the distance of a point to a line formula and equate.

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

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

$m = -24$

And $|-24| = 24$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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