Difference between revisions of "1983 AIME Problems/Problem 14"
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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>. | 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>. | ||
− | [[Image: | + | [[Image:1983_AIME-14.png]] |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
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: | 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 | + | 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 [[Stewart's Theorem]] or similar approaches. 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> | + | <center><math>\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.</math></center> |
− | 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> | + | Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> |
− | == Solution | + | === Solution 2 === |
− | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoints of QP be <math>M_1</math>, and the midpoint of PR be <math>M_2</math>. Let x be the length of AM_1, and y that of BM_2 | + | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoints of <math>QP</math> be <math>M_1</math>, and the midpoint of <math>PR</math> be <math>M_2</math>. Let <math>x</math> be the length of <math>AM_1</math>, and <math>y</math> that of <math>BM_2</math>. |
− | == Solution | + | {{incomplete|solution}} |
− | + | ||
+ | === Solution 3 === | ||
+ | Let <math>QP=PR=x</math>. Angles <math>QPA</math>, <math>APB</math>, and <math>BPR</math> must add up to <math>180^{\circ}</math>. By the [[Law of Cosines]], <math>\angle APB=\cos^{-1}(-11/24)</math>. Also, angles <math>QPA</math> and <math>BPR</math> equal <math>\cos^{-1}(x/16)</math> and <math>\cos^{-1}(x/12)</math>. So we have <center><math>\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).</math></center> Taking the <math>\cos</math> of both sides and simplifying using the cosine addition identity gives <math>x^2=130</math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | Let the circles of radius <math>8</math> and <math>6</math> be centered at <math>A</math> and <math>B,</math> respectively. Let the midpoints of <math>QP</math> and <math>PR</math> be <math>N</math> and <math>O.</math> Dropping a perpendicular from <math>B</math> to <math>AN</math> (let the point be <math>K?</math>) gives a rectangle. | ||
+ | |||
+ | Now note that triangle <math>ABK</math> is right. Let the midpoint of <math>AB</math> (segment of length <math>12</math>) be <math>M.</math> Hence, <math>KM = 6 = BM = BP.</math> | ||
+ | |||
+ | By now obvious [[similar triangles]], <math>3BO = 3KN = AN,</math> so it's a quick system of two linear equations to solve for the desired length. | ||
== See also == | == See also == | ||
{{AIME box|year=1983|num-b=13|num-a=15}} | {{AIME box|year=1983|num-b=13|num-a=15}} | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 18:13, 25 April 2008
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 .
Contents
[hide]Solution
Solution 1
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 Stewart's Theorem or similar approaches. 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
Solution 2
This is a classic side chase - just set up equations involving key lengths in the diagram. Let the midpoints of be , and the midpoint of be . Let be the length of , and that of .
Solution 3
Let . Angles , , and must add up to . By the Law of Cosines, . Also, angles and equal and . So we have
Taking the of both sides and simplifying using the cosine addition identity gives .
Solution 4
Let the circles of radius and be centered at and respectively. Let the midpoints of and be and Dropping a perpendicular from to (let the point be ) gives a rectangle.
Now note that triangle is right. Let the midpoint of (segment of length ) be Hence,
By now obvious similar triangles, so it's a quick system of two linear equations to solve for the desired length.
See also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |