2016 AIME I Problems/Problem 15

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

Using the radical axis theorem, the lines $\overline{AD}, \overline{BC}, \overline{XY}$ are all concurrent at one point, call it $F$ . Now recall by Miquel's theorem in $\triangle FDC$ the fact that quadrilaterals $DAXY$ and $CBXY$ are cyclic implies $FAXB$ is cyclic as well. Denote $\omega_{3}\equiv(FAXB)$ and $Z\equiv\ell\cap\overline{FXY}$ .

Since point $Z$ lies on the radical axis of $\omega_{1},\omega_{2}$ , it has equal power with respect to both circles, thus $AZ^{2}=\text{Pow}_{\omega_{1}}(Z)=ZX\cdot ZY=\text{Pow}_{\omega_{2}}(Z)=ZB^{2}\implies AZ=ZB.$ Also, notice that $AZ\cdot ZB=\text{Pow}_{\omega_{3}}(Z)=ZX\cdot ZF\implies ZY=ZF.$ The diagonals of quadrilateral $FAYB$ bisect each other at $Z$ , so we conclude that $FAYB$ is a parallelogram. Let $u:=ZX$ , so that $ZY=ZF=u+47$ .

Because $FAYB$ is a parallelogram and quadrilaterals $DAXY, CBXY$ are cyclic, $\angle DFX=\angle AFX=\angle BYX=\angle BCX=\angle FCX~~\text{and}~~\angle XDF=\angle XDA=\angle XYA=\angle XFB=\angle XFC$ so we have the pair of similar triangles $\triangle DFX~\sim~\triangle FCX$ . Thus $\dfrac{37}{2u+47}=\dfrac{2u+47}{67}\implies 2u+47=\sqrt{37\cdot 67}\implies u=\dfrac{1}{2}\left(\sqrt{37\cdot 67}-47\right).$ Now compute $AB^{2}=4AZ^{2}=4\cdot ZX\cdot ZY=4u(u+47)=37\cdot 67-47^{2}=\textbf{270}.$

Solution 1

Let $Z = XY \cap AB$ . By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$ . Moreover, $\triangle DXP \sim \triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$ . Then $\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.$ Now, $AZ^2 = \tfrac 14 AB^2$ , and by power of a point, $\begin{align*} x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ x(47+x) &= \tfrac 14 AB^2 \end{align*}$ Solving, we get $\tfrac 14 AB^2 = \tfrac 12 (y-47)\cdot \tfrac 12 (y+47) \qquad \implies$ $\qquad AB ^ 2 = 37\cdot 67 - 47^2 = \boxed{270}$

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 Radical 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, $AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare$ ~Mathscienceclass

Solution 6 (No words)

$AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 7 (Linearity of Power of a Point)

Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $P$ . Let $M$ be the midpoint of segment $AB$ . Then by radical axis on $(ADY)$ , $(BCY)$ and $(ABCD)$ , $P$ lies on $XY$ . By the bisector lemma, $M$ lies on $XY$ . It is well-known that $P$ , $A$ , $X$ , and $B$ are concyclic. By Power of a point on $M$ with respect to $(PAXB)$ and $(ADY)$ , $|\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY,$ so $MP=MY$ . Thus $AB$ and $PY$ bisect each other, so $PAYB$ is a parallelogram. This implies that $\angle DAY = \angle YBC,$ so by the inscribed angle theorem $\overline{XY}$ bisects $\angle DXC$ .

Claim: $AB^2 = DY \cdot YC$ .

Proof. Define the linear function $f(\bullet) := \text{Pow}(\bullet, (ADY)) - \text{Pow}(\bullet, (ABCD))$ . Since $\overline{BY}$ is parallel to the radical axis $\overline{AD}$ of $(ADY)$ and $(ABCD)$ by our previous parallelism, $f(B)=f(Y)$ . Note that $f(B)=AB^2$ while $f(Y)=DY \cdot YC$ , so we conclude. $\square$

By Stewart's theorem on $\triangle DXC$ , $DY \cdot YC=37 \cdot 67 - 47^2 = 270$ , so $AB^2=\boxed{270}$ .

~ Leo.Euler

Video Solution by MOP 2024

https://youtu.be/qFfgB15fYS8

~r00tsOfUnity

Video Solution

https://youtu.be/QoVIorvv_I8

~MathProblemSolvingSkills.com

Video Solution by The Power of Logic

https://youtu.be/lTZx6tp2Fvg

2016 AIME I (Problems • Answer Key • Resources)
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