Difference between revisions of "1983 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
| + | In the adjoining figure, two circles with radii <math>6</math> and <math>8</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in sich a way that the chords <math>QP</math> and <math>PR</math> have equal length. (<math>P</math> is the midpoint of <math>QR</math>) Find the square of the length of <math>QP</math>. | ||
| + | [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=793[/img] | ||
== Solution == | == Solution == | ||
| + | First, notice that if we reflect <math>R</math> over <math>P</math> we get <math>Q</math>. Since we know that <math>R</math> is on [[circle]] <math>B</math> and <math>Q</math> is on circle <math>A</math>, we can reflect circle <math>B</math> over <math>P</math> to get another circle (centered at a new point <math>C</math> with radius <math>6</math>) that intersects circle <math>A</math> at <math>Q</math>. The rest is just finding lengths: | ||
| + | |||
| + | Since <math>P</math> is the midpoint of segment <math>BC</math>, <math>AP</math> is a median of triangle <math>ABC</math>. Because we know that <math>AB=12</math>, <math>BP=PC=6</math>, and <math>AP=8</math>, we can find the third side of the triangle using stewarts or whatever else you like. We get <math>AC = \sqrt{56}</math>. So now we have a kite <math>AQCP</math> with <math>AQ=AP=8</math>, <math>CQ=CP=6</math>, and <math>AC=\sqrt{56}</math>, and all we need is the length of the other diagonal <math>PQ</math>. The easiest way it can be found is with the [[Pythagorean Theorem]]. Let <math>2x</math> be the length of <math>PQ</math>. Then | ||
| + | |||
| + | <math>\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}</math>. | ||
| + | |||
| + | Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = 130.</math> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | * [[1983 AIME Problems/Problem 13|Previous Problem]] | ||
| + | * [[1983 AIME Problems/Problem 15|Next Problem]] | ||
| + | * [[1983 AIME Problems|Back to Exam]] | ||
== See also == | == See also == | ||
| − | * [[ | + | * [[AIME Problems and Solutions]] |
| + | * [[American Invitational Mathematics Examination]] | ||
| + | * [[Mathematics competition resources]] | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 23:22, 23 July 2006
Problem
In the adjoining figure, two circles with radii 
and 
are drawn with their centers 
units apart. At 
, one of the points of intersection, a line is drawn in sich a way that the chords 
and 
have equal length. ( is the midpoint of 
) Find the square of the length of .
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=793[/img]
Solution
First, notice that if we reflect 
over we get 
. Since we know that is on circle 
and is on circle 
, we can reflect circle over to get another circle (centered at a new point 
with radius ) that intersects circle at . The rest is just finding lengths:
Since is the midpoint of segment 
, 
is a median of triangle 
. Because we know that 
, 
, and 
, we can find the third side of the triangle using stewarts or whatever else you like. We get 
. So now we have a kite 
with 
, 
, and 
, and all we need is the length of the other diagonal 
. The easiest way it can be found is with the Pythagorean Theorem. Let 
be the length of . Then
.
Doing routine algebra on the above equation, we find that 
, so 
