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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>.  
  
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[[Image:1983problem14.JPG]]
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[[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:
  
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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
+
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
  
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<math>\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}</math>.
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<center><math>\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.</math></center>
  
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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>
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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>
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== Solution II==
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=== Solution 2 ===
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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>.
  
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== Solution III==
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{{incomplete|solution}}
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LetQP=PR=x. Angles QPA, APB, and BPR must add up to 180. By the Law of Cosines, angle APB=arccos(-11/24). Also, angles QPA and BPR equal arccos(x/16) and arccos(x/12). So we have arccos(x/16)+arccos(-11/24)=180-arccos(x/12). Taking the cos of both sides and simplifying gives <math>x^2=130</math>.
+
 
 +
=== 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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* [[AIME Problems and Solutions]]
 
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* [[American Invitational Mathematics Examination]]
 
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* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 19:13, 25 April 2008

Problem

In the adjoining figure, two circles with radii $6$ and $8$ are drawn with their centers $12$ units apart. At $P$, one of the points of intersection, a line is drawn in sich a way that the chords $QP$ and $PR$ have equal length. ($P$ is the midpoint of $QR$) Find the square of the length of $QP$.

1983 AIME-14.png

Solution

Solution 1

First, notice that if we reflect $R$ over $P$ we get $Q$. Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$, we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths:

Since $P$ is the midpoint of segment $BC$, $AP$ is a median of triangle $ABC$. Because we know that $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \sqrt{56}$. So now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then

$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$

Doing routine algebra on the above equation, we find that $x^2=\frac{65}{2}$, so $PQ^2 = 4x^2 = \boxed{130}.$

Solution 2

This is a classic side chase - just set up equations involving key lengths in the diagram. Let the midpoints of $QP$ be $M_1$, and the midpoint of $PR$ be $M_2$. Let $x$ be the length of $AM_1$, and $y$ that of $BM_2$.

Template:Incomplete

Solution 3

Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\circ}$. By the Law of Cosines, $\angle APB=\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\cos^{-1}(x/16)$ and $\cos^{-1}(x/12)$. So we have

$\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).$

Taking the $\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$.

Solution 4

Let the circles of radius $8$ and $6$ be centered at $A$ and $B,$ respectively. Let the midpoints of $QP$ and $PR$ be $N$ and $O.$ Dropping a perpendicular from $B$ to $AN$ (let the point be $K?$) gives a rectangle.

Now note that triangle $ABK$ is right. Let the midpoint of $AB$ (segment of length $12$) be $M.$ Hence, $KM = 6 = BM = BP.$

By now obvious similar triangles, $3BO = 3KN = AN,$ 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
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