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

m (→‎Solution)
(→‎Solution)
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==Solution==
 
==Solution==
−
By radical axis theorem <math>AD, XY, BC</math> concur at point <math>E</math>.
 
  
−
Let <math>AB</math> and <math>EY</math> intersect at <math>S</math>. Note that because <math>AXDY</math> and <math>CYXB</math> are cyclic, by Miquel theorem <math>AXBE</math> are cyclic as well. Thus
+
===Solution 1===
 +
By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>.
 +
 
 +
Let <math>AB</math> and <math>EY</math> intersect at <math>S</math>. Note that because <math>AXDY</math> and <math>CYXB</math> are cyclic, by Miquel's Theorem <math>AXBE</math> is cyclic as well. Thus
 
<cmath>\angle AEX = \angle ABX = \angle XCB = \angle XYB</cmath>and
 
<cmath>\angle AEX = \angle ABX = \angle XCB = \angle XYB</cmath>and
−
<cmath>\angle XEB = \angle XAB = \angle XDA = \angle XYA.</cmath>Thus <math>AY // EB</math> and <math>YB // EA</math> so <math>AEBY</math> is a parallelogram. Hence <math>AS = SB</math> and <math>SE = SY</math>. But notice that <math>DXE</math> and <math>EXC</math> are similar by <math>AA</math> Similarity, so <math>XE^2 = XD \cdot XC = 37 \cdot 67</math>. But
+
<cmath>\angle XEB = \angle XAB = \angle XDA = \angle XYA.</cmath>Thus <math>AY \parallel EB</math> and <math>YB \parallel EA</math>, so <math>AEBY</math> is a parallelogram. Hence <math>AS = SB</math> and <math>SE = SY</math>. But notice that <math>DXE</math> and <math>EXC</math> are similar by <math>AA</math> Similarity, so <math>XE^2 = XD \cdot XC = 37 \cdot 67</math>. But
 
<cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 37 \cdot 67 - 47^2 = 270.</math>
 
<cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 37 \cdot 67 - 47^2 = 270.</math>
 +
 +
 +
===Solution 2===
 +
First, we note that as <math>\triangle XDY</math> and <math>\triangle XYC</math> have bases along the same line, <math>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}</math>. We can also find the ratio of their areas using the circumradius area formula. If <math>R_1</math> is the radius of <math>\omega_1</math> and if <math>R_2</math> is the radius of <math>\omega_2</math>, then
 +
<cmath>\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}.</cmath>
 +
Since we showed this to be <math>\frac{DY}{YC}</math>, we see that <math>\frac{R_2}{R_1}=\frac{67}{37}</math>.
 +
 +
We extend <math>AD</math> and <math>BC</math> to meet at point <math>P</math>, and we extend <math>AB</math> and <math>CD</math> to meet at point <math>Q</math> 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 <math>ABCD</math> is cyclic, we know that <math>\angle BCD=180-\angle DAB=\angle BAP</math>. But then as <math>AB</math> is tangent to <math>\omega_2</math> at <math>B</math>, we see that <math>\angle BCD=\angle ABY</math>. Therefore, <math>\angle ABY=\angle BAP</math>, and <math>BY\parallel PD</math>. A similar argument shows <math>AY\parallel PC</math>. These parallel lines show <math>\triangle PDC\sim\triangle ADY\sim\triangle BYC</math>. Also, we showed that <math>\frac{R_2}{R_1}=\frac{67}{37}</math>, so the ratio of similarity between <math>\triangle ADY</math> and <math>\triangle BYC</math> is <math>\frac{37}{67}</math>, or rather
 +
<cmath>\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.</cmath>
 +
We can now use the parallel lines to find more similar triangles. As <math>\triangle AQD\sim \triangle BQY</math>, we know that
 +
<cmath>\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.</cmath>
 +
Setting <math>QA=37x</math>, we see that <math>QB=67x</math>, hence <math>AB=30x</math>, and the problem simplifies to finding <math>30^2x^2</math>. Setting <math>QD=37^2y</math>, we also see that <math>QY=37\cdot 67y</math>, hence <math>DY=37\cdot 30y</math>. Also, as <math>\triangle AQY\sim \triangle BQC</math>, we find that
 +
<cmath>\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.</cmath>
 +
As <math>QY=37\cdot 67y</math>, we see that <math>QC=67^2y</math>, hence <math>YC=67\cdot30y</math>.
 +
 +
Applying Power of a Point to point <math>Q</math> with respect to <math>\omega_2</math>, we find
 +
<cmath>67^2x^2=37\cdot 67^3 y^2,</cmath>
 +
or <math>x^2=37\cdot 67 y^2</math>. We wish to find <math>AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2</math>.
 +
 +
Applying Stewart's Theorem to <math>\triangle XDC</math>, we find
 +
<cmath>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).</cmath>
 +
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>
  
 
==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 17:02, 6 March 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 $AB^2 = 37 \cdot 67 - 47^2 = 270.$


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, \[AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.\]

See Also

2016 AIME I (Problems • Answer Key • Resources)
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