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>.  
+
In the adjoining figure, two circles with radii <math>8</math> and <math>6</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 such a way that the chords <math>QP</math> and <math>PR</math> have equal length. Find the square of the length of <math>QP</math>.  
  
[[Image:1983_AIME-14.png]]
+
<asy>size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);</asy>
  
 
__TOC__
 
__TOC__
 +
 
== Solution ==
 
== Solution ==
 +
 +
Note that some of these solutions assume that <math>R</math> lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from <math>P</math> passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto <math>QR</math>.
 +
 
=== Solution 1 ===
 
=== 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:
+
Firstly, 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>, and with radius <math>6</math>) that intersects circle <math>A</math> at <math>Q</math>. The rest is just finding lengths, as follows.
  
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
+
Since <math>P</math> is the midpoint of segment <math>BC</math>, <math>AP</math> is a median of <math>\triangle ABC</math>. Because we know <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>. 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
  
 
<center><math>\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.</math></center>
 
<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 = \boxed{130}.</math>
+
Solving this equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math>
=== Solution 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 2 (Easiest)===
 +
<asy>
 +
size(0,5cm);
 +
pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);
 +
draw(b--r--n--b--a--m--n);
 +
draw(a--q--m);
 +
draw(circumcircle(origin,q,p));
 +
draw(circumcircle((14,0),p,r));
 +
draw(rightanglemark(a,m,n,24));
 +
draw(rightanglemark(b,n,r,24));
 +
label("$A$",a,S);
 +
label("$B$",b,S);
 +
label("$M$",m,NE);
 +
label("$N$",n,NE);
 +
label("$P$",p,N);
 +
label("$Q$",q,NW);
 +
label("$R$",r,E);
 +
label("$12$",(14,0),SW);
 +
label("$6$",(23,0),S);
 +
</asy>
 +
'''
 +
 
 +
Draw additional lines as indicated. Note that since triangles <math>AQP</math> and <math>BPR</math> are isosceles, the altitudes are also bisectors, so let <math>QM=MP=PN=NR=x</math>.
 +
 
 +
Since <math>\frac{AR}{MR}=\frac{BR}{NR},</math> triangles <math>BNR</math> and <math>AMR</math> are similar. If we let <math>y=BN</math>, we have <math>AM=3BN=3y</math>.
 +
 
 +
Applying the Pythagorean Theorem on triangle <math>BNR</math>, we have <math>x^2+y^2=36</math>. Similarly, for triangle <math>QMA</math>, we have <math>x^2+9y^2=64</math>.
  
{{incomplete|solution}}
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Subtracting, <math>8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}</math>.
  
 
=== Solution 3 ===
 
=== 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>.
+
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}\left(\frac{{-11}}{24}\right)</math>. Also, angles <math>QPA</math> and <math>BPR</math> equal <math>\cos^{-1}\left(\frac{x}{16}\right)</math> and <math>\cos^{-1}\left(\frac{x}{12}\right)</math>. So we have <center><math>\cos^{-1}\left(\frac{x}{16}\right)+\cos^{-1}\left(\frac{{-11}}{24}\right)=180^{\circ}-\cos^{-1}\left(\frac{x}{12}\right).</math></center> Taking the cosine of both sides, and simplifying using the addition formula for <math>\cos</math> as well as the identity <math>\sin^{2}{x} + \cos^{2}{x} = 1</math>, gives <math>x^2=\boxed{130}</math>.
 +
 
 +
===Solution 4 (quickest) ===
 +
Let <math>QP = PR = x</math>. Extend the line containing the centers of the two circles to meet <math>R</math>, and to meet the other side of the large circle at a point <math>S</math>.
 +
 
 +
The part of this line from <math>R</math> to the point nearest to <math>R</math> where it intersects the larger circle has length <math>6+(12-8)=10</math>. 
 +
The length of the diameter of the larger circle is <math>16</math>.
 +
 
 +
Thus by Power of a Point in the circle passing through <math>Q</math>, <math>R</math>, and <math>S</math>, we have <math>x \cdot 2x = 10 \cdot (10+16) = 260</math>, so <math>x^2 = \boxed{130}</math>.
 +
 
 +
===Full Proof that R, A, B are collinear ===
 +
<asy>
 +
size(0,5cm);
 +
pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);
 +
draw(b--r--n--b--a--m--n);
 +
draw(a--q--m);
 +
draw(circumcircle(origin,q,p));
 +
draw(circumcircle((14,0),p,r));
 +
draw(rightanglemark(a,m,n,24));
 +
draw(rightanglemark(b,n,r,24));
 +
label("$A$",a,S);
 +
label("$B$",b,S);
 +
label("$M$",m,NE);
 +
label("$N$",n,NE);
 +
label("$P$",p,N);
 +
label("$Q$",q,NW);
 +
label("$R$",r,E);
 +
label("$12$",(14,0),SW);
 +
label("$6$",(23,0),S);
 +
label("$T$", t , NW);
 +
</asy>
 +
 
 +
Let <math>M</math> and <math>N</math> be the feet of the perpendicular from <math>A</math> to <math>PQ</math> and <math>B</math> to <math>PR</math> respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so <math>QM = MP = PN = NR</math>, since the problem told us <math>QP = PR</math>.
 +
 
 +
We will show that <math>R</math> lies on <math>AB</math>.
 +
 
 +
Let <math>T</math> be the intersection of circle centered at <math>B</math> with <math>AB</math>. Then <math>BT = TA = 6</math>.  
  
=== Solution 4 ===
+
Let <math>P</math>' be the foot of the perpendicular from <math>T</math> to <math>MN</math>. Then <math>TP'</math> is a midline (or midsegment) in trapezoid <math>AMNB</math>, so <math>P'</math> coincides with <math>P</math> (they are both supposed to be the midpoint of <math>MN</math>). In other words, since <math>\angle TP'N = 90^\circ</math>, then <math> \angle TPN = 90^\circ</math>.
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>
+
Thus, <math>\angle TPR</math> subtends a <math>90^\circ \times 2 = 180^\circ</math> degree arc. So arc <math>TR</math> in circle <math>B</math> is <math>180^\circ</math>, so <math>TR</math> is a diameter, as desired. Thus <math>A</math>, <math>B</math>, <math>R</math> are collinear.
  
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.
+
NOTE: Note this collinearity only follows from the fact that <math>6</math> is half of <math>12</math> in the problem statement. The collinearity is untrue in general.
  
== See also ==
+
== See Also ==
 
{{AIME box|year=1983|num-b=13|num-a=15}}
 
{{AIME box|year=1983|num-b=13|num-a=15}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 15:39, 9 April 2020

Problem

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

[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

Solution

Note that some of these solutions assume that $R$ lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from $P$ passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto $QR$.

Solution 1

Firstly, 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 $C$, and with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths, as follows.

Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \sqrt{56}$. 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

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

Solving this equation, we find that $x^2=\frac{65}{2}$, so $PQ^2 = 4x^2 = \boxed{130}.$

Solution 2 (Easiest)

[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("$A$",a,S); label("$B$",b,S); label("$M$",m,NE); label("$N$",n,NE); label("$P$",p,N); label("$Q$",q,NW); label("$R$",r,E); label("$12$",(14,0),SW); label("$6$",(23,0),S); [/asy]

Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$.

Since $\frac{AR}{MR}=\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$.

Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$.

Subtracting, $8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}$.

Solution 3

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

$\cos^{-1}\left(\frac{x}{16}\right)+\cos^{-1}\left(\frac{{-11}}{24}\right)=180^{\circ}-\cos^{-1}\left(\frac{x}{12}\right).$

Taking the cosine of both sides, and simplifying using the addition formula for $\cos$ as well as the identity $\sin^{2}{x} + \cos^{2}{x} = 1$, gives $x^2=\boxed{130}$.

Solution 4 (quickest)

Let $QP = PR = x$. Extend the line containing the centers of the two circles to meet $R$, and to meet the other side of the large circle at a point $S$.

The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$. The length of the diameter of the larger circle is $16$.

Thus by Power of a Point in the circle passing through $Q$, $R$, and $S$, we have $x \cdot 2x = 10 \cdot (10+16) = 260$, so $x^2 = \boxed{130}$.

Full Proof that R, A, B are collinear

[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("$A$",a,S); label("$B$",b,S); label("$M$",m,NE); label("$N$",n,NE); label("$P$",p,N); label("$Q$",q,NW); label("$R$",r,E); label("$12$",(14,0),SW); label("$6$",(23,0),S); label("$T$", t , NW); [/asy]

Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so $QM = MP = PN = NR$, since the problem told us $QP = PR$.

We will show that $R$ lies on $AB$.

Let $T$ be the intersection of circle centered at $B$ with $AB$. Then $BT = TA = 6$.

Let $P$' be the foot of the perpendicular from $T$ to $MN$. Then $TP'$ is a midline (or midsegment) in trapezoid $AMNB$, so $P'$ coincides with $P$ (they are both supposed to be the midpoint of $MN$). In other words, since $\angle TP'N = 90^\circ$, then $\angle TPN = 90^\circ$.

Thus, $\angle TPR$ subtends a $90^\circ \times 2 = 180^\circ$ degree arc. So arc $TR$ in circle $B$ is $180^\circ$, so $TR$ is a diameter, as desired. Thus $A$, $B$, $R$ are collinear.

NOTE: Note this collinearity only follows from the fact that $6$ is half of $12$ in the problem statement. The collinearity is untrue in general.

See Also

1983 AIME (ProblemsAnswer KeyResources)
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