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

Revision as of 16:36, 15 June 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, $AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.$

@@ Line 11: / Line 11: @@
 <cmath>\angle AEX = \angle ABX = \angle XCB = \angle XYB</cmath>and
 <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 = \boxed{270}.</math>
 ===Solution 2===

2016 AIME I (Problems • Answer Key • Resources)
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