Difference between revisions of "2016 AIME I Problems/Problem 15"

m (Solution 1)
(Solution)
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We can cancel <math>30\cdot 104\cdot y</math> from both sides, finding <math>37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2</math>. Therefore,  
 
We can cancel <math>30\cdot 104\cdot y</math> from both sides, finding <math>37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2</math>. Therefore,  
 
<cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath>
 
<cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath>
 +
 +
==Solution 3==
 +
<asy>
 +
size(9cm);
 +
import olympiad;
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real R1=45,R2=67*R1/37;
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real m1=sqrt(R1^2-23.5^2);
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real m2=sqrt(R2^2-23.5^2);
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pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);
 +
draw(circle(o1,R1));
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draw(circle(o2,R2));
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pair q=(-R1/(R2-R1)*o2.x,0);
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pair a=tangent(q,o1,R1,2);
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pair b=tangent(q,o2,R2,2);
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pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0];
 +
pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1];
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dot(a^^b^^x^^y^^c^^d);
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draw(x--y);
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draw(a--y^^b--y);
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draw(d--x--c);
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draw(a--b--c--d--cycle);
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draw(x--a^^x--b);
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label("$A$",a,NW,fontsize(9));
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label("$B$",b,NE,fontsize(9));
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label("$C$",c,SE,fontsize(9));
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label("$D$",d,SW,fontsize(9));
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label("$X$",x,2*N,fontsize(9));
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label("$Y$",y,3*S,fontsize(9));
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</asy>
 +
First of all, since quadrilaterals <math>ADYX</math> and <math>XYCB</math> are cyclic, we can let <math>\angle DAX = \angle XYC = \theta</math>, and <math>\angle XYD = \angle CBX = 180 - \theta</math>, due to the properties of cyclic quadrilaterals. In addition, let <math>\angle BAX = x</math> and <math>\angle ABX = y</math>. Then, since quadrilateral <math>ABCD</math> is cyclic as well, we have the following sums:
 +
<cmath>\theta + x +\angle XCY + y = 180^{\circ}</cmath>
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<cmath>180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}</cmath>
 +
Cancelling out <math>180^{\circ}</math> in the second equation isolating <math>\theta</math> yields <math>\theta = y + \angle XDY + x</math>. Substituting <math>\theta</math> back into the first equation, we obtain
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<cmath>2x + 2y + \angle XCY + \angle XDY = 180^{\circ}</cmath>
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Since
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<cmath>x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}</cmath>
 +
<cmath>x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}</cmath>
 +
we can then imply that <math>\angle DAY = x + y</math>. Similarly, <math>\angle YBC = x + y</math>. So then <math>\angle DXY = \angle YXC = x + y</math>, so since we know that <math>XY</math> bisects <math>\angle DXC</math>, we can solve for <math>DY</math> and <math>YC</math> with Stewart’s Theorem. Let <math>DY = 37n</math> and <math>YC = 67n</math>. Then
 +
<cmath>37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n</cmath>
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<cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath>
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<cmath>n^2 = \frac{270}{2479}</cmath>
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Now, since <math>\angle AYX = x</math> and <math>\angle BYX = y</math>, <math>\angle AYB = x + y</math>. From there, let <math>\angle AYD = \alpha</math> and <math>\angle BYC = \beta</math>. From angle chasing we can derive that <math>\angle YDX = \angle YAX = \beta - x</math> and <math>\angle YCX = \angle YBX = \alpha - y</math>. Now it is clear that <math>\triangle DAY \sim \triangle AYB \sim \triangle YBC</math>. Therefore the length of <math>AY</math> is the geometric mean of the lengths of <math>DA</math> and <math>YB</math> (from <math>\triangle DAY \sim \triangle AYB</math>). However, due to the proportion <math>\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}</math>, the length of <math>AB</math> is the geometric mean of the lengths of <math>DY</math> and <math>YC</math>.
 +
We can now simply use arithmetic to calculate <math>AB^2</math>.
 +
<cmath>AB^2 = DY \cdot YC</cmath>
 +
<cmath>AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}</cmath>
 +
<cmath>AB^2 = \boxed{270}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2016|n=I|num-b=14|after=Last Question}}
 
{{AIME box|year=2016|n=I|num-b=14|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:19, 15 October 2016

Problem

Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$. The three points $C$, $Y$, $D$ are collinear, $XC = 67$, $XY = 47$, and $XD = 37$. Find $AB^2$.

Solution

Solution 1

By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$.

Let $AB$ and $EY$ intersect at $S$. Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel's Theorem $AXBE$ is cyclic as well. Thus \[\angle AEX = \angle ABX = \angle XCB = \angle XYB\]and \[\angle XEB = \angle XAB = \angle XDA = \angle XYA.\]Thus $AY \parallel EB$ and $YB \parallel EA$, so $AEBY$ is a parallelogram. Hence $AS = SB$ and $SE = SY$. But notice that $DXE$ and $EXC$ are similar by $AA$ Similarity, so $XE^2 = XD \cdot XC = 37 \cdot 67$. But \[XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.\]Hence $AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.$

Solution 2

First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$. We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$, then \[\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.\] Since we showed this to be $\frac{DY}{YC}$, we see that $\frac{R_2}{R_1}=\frac{67}{37}$.

We extend $AD$ and $BC$ to meet at point $P$, and we extend $AB$ and $CD$ to meet at point $Q$ as shown below. [asy] size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label("$A$",a,NW,fontsize(8)); label("$B$",b,NE,fontsize(8)); label("$C$",c,SE,fontsize(8)); label("$D$",d,SW,fontsize(8)); label("$X$",x,2*WNW,fontsize(8)); label("$Y$",y,3*S,fontsize(8)); label("$P$",p,N,fontsize(8)); label("$Q$",q,W,fontsize(8)); [/asy] As $ABCD$ is cyclic, we know that $\angle BCD=180-\angle DAB=\angle BAP$. But then as $AB$ is tangent to $\omega_2$ at $B$, we see that $\angle BCD=\angle ABY$. Therefore, $\angle ABY=\angle BAP$, and $BY\parallel PD$. A similar argument shows $AY\parallel PC$. These parallel lines show $\triangle PDC\sim\triangle ADY\sim\triangle BYC$. Also, we showed that $\frac{R_2}{R_1}=\frac{67}{37}$, so the ratio of similarity between $\triangle ADY$ and $\triangle BYC$ is $\frac{37}{67}$, or rather \[\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.\] We can now use the parallel lines to find more similar triangles. As $\triangle AQD\sim \triangle BQY$, we know that \[\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.\] Setting $QA=37x$, we see that $QB=67x$, hence $AB=30x$, and the problem simplifies to finding $30^2x^2$. Setting $QD=37^2y$, we also see that $QY=37\cdot 67y$, hence $DY=37\cdot 30y$. Also, as $\triangle AQY\sim \triangle BQC$, we find that \[\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.\] As $QY=37\cdot 67y$, we see that $QC=67^2y$, hence $YC=67\cdot30y$.

Applying Power of a Point to point $Q$ with respect to $\omega_2$, we find \[67^2x^2=37\cdot 67^3 y^2,\] or $x^2=37\cdot 67 y^2$. We wish to find $AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2$.

Applying Stewart's Theorem to $\triangle XDC$, we find \[37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).\] We can cancel $30\cdot 104\cdot y$ from both sides, finding $37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2$. Therefore, \[AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.\]

Solution 3

[asy] size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label("$A$",a,NW,fontsize(9)); label("$B$",b,NE,fontsize(9)); label("$C$",c,SE,fontsize(9)); label("$D$",d,SW,fontsize(9)); label("$X$",x,2*N,fontsize(9)); label("$Y$",y,3*S,fontsize(9)); [/asy] First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\angle DAX = \angle XYC = \theta$, and $\angle XYD = \angle CBX = 180 - \theta$, due to the properties of cyclic quadrilaterals. In addition, let $\angle BAX = x$ and $\angle ABX = y$. Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums: \[\theta + x +\angle XCY + y = 180^{\circ}\] \[180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}\] Cancelling out $180^{\circ}$ in the second equation isolating $\theta$ yields $\theta = y + \angle XDY + x$. Substituting $\theta$ back into the first equation, we obtain \[2x + 2y + \angle XCY + \angle XDY = 180^{\circ}\] Since \[x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}\] \[x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}\] we can then imply that $\angle DAY = x + y$. Similarly, $\angle YBC = x + y$. So then $\angle DXY = \angle YXC = x + y$, so since we know that $XY$ bisects $\angle DXC$, we can solve for $DY$ and $YC$ with Stewart’s Theorem. Let $DY = 37n$ and $YC = 67n$. Then \[37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n\] \[37n \cdot 67n + 47^2 = 37 \cdot 67\] \[n^2 = \frac{270}{2479}\] Now, since $\angle AYX = x$ and $\angle BYX = y$, $\angle AYB = x + y$. From there, let $\angle AYD = \alpha$ and $\angle BYC = \beta$. From angle chasing we can derive that $\angle YDX = \angle YAX = \beta - x$ and $\angle YCX = \angle YBX = \alpha - y$. Now it is clear that $\triangle DAY \sim \triangle AYB \sim \triangle YBC$. Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\triangle DAY \sim \triangle AYB$). However, due to the proportion $\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}$, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$. We can now simply use arithmetic to calculate $AB^2$. \[AB^2 = DY \cdot YC\] \[AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}\] \[AB^2 = \boxed{270}\]

See Also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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