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

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Let <math>E = DA \cap CB</math>. By Radicaly Axes, <math>E</math> lies on <math>XY</math>. Note that <math>EAXB</math> is cyclic as <math>X</math> is the Miquel point of <math>\triangle EDC</math> in this configuration.
 
Let <math>E = DA \cap CB</math>. By Radicaly Axes, <math>E</math> lies on <math>XY</math>. Note that <math>EAXB</math> is cyclic as <math>X</math> is the Miquel point of <math>\triangle EDC</math> in this configuration.
  
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[b]Claim.[/b] <math>\triangle DXE \sim \triangle EXC</math>
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Claim. <math>\triangle DXE \sim \triangle EXC</math>
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[i]Proof.[/i] We angle chase. <cmath>\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE</cmath>and<cmath>\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square</cmath>
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Proof. We angle chase. <cmath>\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE</cmath>and<cmath>\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square</cmath>
  
 
Let <math>F = EX \cap AB</math>. Note <cmath>FA^2 = FX \cdot FY = FB^2</cmath>and<cmath>EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY</cmath>By our claim, <cmath>\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}</cmath>and<cmath>FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}</cmath>Finally, <cmath>AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47}{4} = \boxed{270}. \blacksquare</cmath>~Mathscienceclass
 
Let <math>F = EX \cap AB</math>. Note <cmath>FA^2 = FX \cdot FY = FB^2</cmath>and<cmath>EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY</cmath>By our claim, <cmath>\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}</cmath>and<cmath>FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}</cmath>Finally, <cmath>AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47}{4} = \boxed{270}. \blacksquare</cmath>~Mathscienceclass

Revision as of 20:13, 21 May 2020

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 $AB^2$.

Solutions

Solution 1

Let $Z = XY \cap AB$. By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$. Furthermore, by simple angle chasing, $\triangle DXE \sim \triangle EXC$. Let $y = EX, x = XZ$. Then $\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67$. Now, by Power of a Point, $AZ^2 = \frac{AB^2}{4}$, $(y-x)x = \frac{AB^2}{4}$, and $x(47+x) = \frac{AB^2}{4}$. Solving, we get $\dfrac{AB ^ 2}{4} = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) \implies AB ^ 2 = 37\cdot67 - 47^2 = \boxed{270}$

Solution 2

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 3

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 4

[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 $\angle ABX = y$. Thus, $\angle ADX = \angle AYX = x$ and $\angle XYB = \angle XCB = y$. 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 and 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 $\angle YCX = \angle YBX = \alpha - y$. From there, since $\angle ADX = x$, it is quite clear that $\angle ADY = \beta$, and $\angle YAB = \beta$ can be found similarly. From there, since $\angle ADY = \angle YAB = \angle BYC = \beta$ and $\angle DAY = \angle AYB = \angle YBC = x + y$, we have $AA$ similarity between $\triangle DAY$, $\triangle AYB$, and $\triangle YBC$. Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\triangle DAY \sim \triangle AYB$). However, $\triangle DAY \sim \triangle AYB \sim \triangle YBC$ yields the proportion $\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}$; hence, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$. We can now simply use arithmetic to calculate $AB^2$. \[AB^2 = DY \cdot YC\] \[AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}\] \[AB^2 = \boxed{270}\]

-Solution by TheBoomBox77

Solution 5 (not too different)

Let $E = DA \cap CB$. By Radicaly Axes, $E$ lies on $XY$. Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\triangle EDC$ in this configuration.

Claim. $\triangle DXE \sim \triangle EXC$ Proof. We angle chase. \[\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE\]and\[\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square\]

Let $F = EX \cap AB$. Note \[FA^2 = FX \cdot FY = FB^2\]and\[EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY\]By our claim, \[\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}\]and\[FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}\]Finally, \[AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47}{4} = \boxed{270}. \blacksquare\]~Mathscienceclass

See Also

2016 AIME I (Problems • Answer Key • Resources)
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