Difference between revisions of "1984 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
− | Three circles, each of radius <math>\displaystyle 3</math>, are drawn with centers at <math>\displaystyle (14, 92)</math>, <math>\displaystyle (17, 76)</math>, and <math>\displaystyle (19, 84)</math>. A line passing through <math>\displaystyle (17,76)</math> 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? | + | Three circles, each of [[radius]] <math>\displaystyle 3</math>, are drawn with centers at <math>\displaystyle (14, 92)</math>, <math>\displaystyle (17, 76)</math>, and <math>\displaystyle (19, 84)</math>. A [[line]] passing through <math>\displaystyle (17,76)</math> 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? |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | {{ | + | [[Image: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 <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 \displaystyle \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>\displaystyle \frac{88 - 76}{\frac{33}{2} - 17} = -24</math>, and the answer is <math>\displaystyle 024</math>. | ||
+ | |||
+ | ''Remark'': 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 <math>E,F</math> to be the feet of the perpendiculars from <math>A,C</math> to the line (same as above). The equation of the line is <math>y = mx + b</math>; substituting <math>y=76,x=17</math> gives us that <math>b = 76 - 17m</math>, so the line is <math>y = mx + (76 - 17m)</math>. <math>\displaystyle AE = CF \displaystyle</math> by the HA argument above and [[CPCTC]], so we can use the distance of a point to a line formula and equate. | ||
+ | |||
+ | <div style="text-align:center;"><math>\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|</math><br /> | ||
+ | <math>-3m - 16 = -2m + 8</math><br /> | ||
+ | <math>m = -24</math></div> | ||
+ | |||
+ | And <math>|-24| = 24</math>. | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=1984|num-b=5|num-a=7}} | {{AIME box|year=1984|num-b=5|num-a=7}} | ||
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* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 17:09, 10 September 2007
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?
Solution
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 (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: 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 to be the feet of the perpendiculars from to the line (same as above). The equation of the line is ; substituting gives us that , so the line is . by the HA argument above and CPCTC, so we can use the distance of a point to a line formula and equate.
And .
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 |