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

(Solution)
(Solution 1)
(31 intermediate revisions by 12 users not shown)
Line 3: Line 3:
 
Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>.
 
Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>.
  
==Solution==
 
  
===Solution 1===
+
==Solution 1==
 +
 
 +
Let <math>Z = XY \cap AB</math>. By the radical axis theorem <math>AD, XY, BC</math> are concurrent, say at <math>P</math>. Moreover, <math>\triangle DXP \sim \triangle PXC</math> by simple angle chasing. Let <math>y = PX, x = XZ</math>. Then <cmath>\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.</cmath> Now, <math>AZ^2 = \tfrac 14 AB^2</math>, and by power of a point, <cmath>\begin{align*}
 +
        x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\
 +
        x(47+x) &= \tfrac 14 AB^2
 +
    \end{align*}</cmath> Solving, we get <cmath>\tfrac 14 AB^2 = \tfrac 12 (y-47)\cdot \tfrac 12 (y+47) \qquad \implies</cmath>
 +
<cmath> \qquad AB ^ 2 = 37\cdot 67 - 47^2 = \boxed{270}</cmath>
 +
 
 +
==Solution 2==
 
By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>.
 
By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>.
  
Line 11: Line 18:
 
<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 \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>\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===
+
==Solution 3==
 
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  
 
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>
 
<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>
Line 65: Line 71:
 
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 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 <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>. Thus, <math>\angle ADX = \angle AYX = x</math> and <math>\angle XYB = \angle XCB = 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>
 +
<cmath>180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}</cmath>
 +
Cancelling out <math>180^{\circ}</math> in the second equation and isolating <math>\theta</math> yields <math>\theta = y + \angle XDY + x</math>. Substituting <math>\theta</math> back into the first equation, we obtain
 +
<cmath>2x + 2y + \angle XCY + \angle XDY = 180^{\circ}</cmath>
 +
Since
 +
<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>
 +
<cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath>
 +
<cmath>n^2 = \frac{270}{2479}</cmath>
 +
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>. From there, since <math>\angle ADX = x</math>, it is quite clear that <math>\angle ADY = \beta</math>, and <math>\angle YAB = \beta</math> can be found similarly. From there, since <math>\angle ADY = \angle YAB = \angle BYC = \beta</math> and <math>\angle DAY = \angle AYB = \angle YBC = x + y</math>, we have <math>AA</math> similarity between <math>\triangle DAY</math>, <math>\triangle AYB</math>, and <math>\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, <math>\triangle DAY \sim \triangle AYB \sim \triangle YBC</math> yields the proportion <math>\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}</math>; hence, 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>
 +
 +
'''-Solution by TheBoomBox77'''
 +
==Solution 5 (not too different)==
 +
Let <math>E = DA \cap CB</math>. By Radical 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.
 +
 +
Claim. <math>\triangle DXE \sim \triangle EXC</math>
 +
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) - 47^2}{4} = \boxed{270}. \blacksquare</cmath>~Mathscienceclass
 +
 +
 +
 +
==Solution 6 (No words)==
 +
[[File:2016 AIME I 15.png|450px|right]]
 +
[[File:2016 AIME I 15b.png|400px|left]]
 +
<math> - \\ - </math>
 +
 +
'''Shelomovskii,  vvsss,  www.deoma-cmd.ru'''
 +
 +
==Video Solution==
 +
https://youtu.be/QoVIorvv_I8
  
 
==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 03:37, 23 June 2022

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 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)

2016 AIME I 15.png
2016 AIME I 15b.png

$- \\ -$

Shelomovskii, vvsss, www.deoma-cmd.ru

Video Solution

https://youtu.be/QoVIorvv_I8

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png