Y by gyluo, megarnie, Jc426, Lionking212, sleepypuppy, mathleticguyyy, ImSh95, Adventure10, Mango247
Circles 
and 
intersect at points 
and 
. Line 
is tangent to and at 
and 
, respectively, with line 
closer to point than to . Circle 
passes through and intersecting again at 
and intersecting again at 
. The three points 
, , 
are collinear, 
, 
, and 
. Find 
.
and 
intersect at points 
and 
. Line 
is tangent to and at 
and 
, respectively, with line 
closer to point than to . Circle 
passes through and intersecting again at 
and intersecting again at 
. The three points 
, , 
are collinear, 
, 
, and 
. Find 
.
![[asy]
import olympiad;
size(300);
defaultpen(linewidth(0.8));
pair P = dir(100), D = dir(220), C = dir(320);
pair M = (D+C)/2, pt = bisectorpoint(D,P,C), Z = reflect(P,pt) * M, Y = extension(P,Z,D,C);
draw(P--D--C--P--Y);
pair T1 = P + Y - D, T2 = P + Y - C, A = intersectionpoint(P--D, T2--Y), B = intersectionpoint(T1--Y, P--C);
draw(A--B);
pair[] in = intersectionpoints(circumcircle(A,D,Y), circumcircle(B,C,Y));
pair X = in[0];
pair Q = 2 * A - P, R = 2 * B - P;
draw(A--X--B^^C--X--D,linetype("2 2"));
draw(D--Q--R,linetype("4 4"));
draw(circumcircle(A,D,Y)^^circumcircle(B,C,Y));
label("$A$", A, dir(120));
label("$B$", B, dir(60));
label("$C$", C, dir(X--C));
label("$D$", D, dir(200));
label("$X$", X, 2.7 * dir(195));
label("$Y$", Y, 1.5*S);
label("$M$", (A+B)/2, dir(140));
label("$P$", P, N);
label("$Q$", Q, dir(Y--Q));
label("$R$", R, dir(Y--R));
[/asy]](http://latex.artofproblemsolving.com/c/4/d/c4d7b0e1e0ffa3e2990a8568f6ff9aab277bfc65.png)
.
also passes through 
.
to be cyclic, we need 
.
.
.
is a parallelogram, so letting 
we have 
.
and 
at 
and 
respectively. Since 
,
.
,
is cyclic. This means ![\[DY\cdot YC = QY\cdot YR = AB^2.\]](http://latex.artofproblemsolving.com/3/8/b/38bc4c0bfda28a02df658113e9289a498ca67bd1.png)
To compute 
,![\[\angle YXC=\angle YBC=\angle DPC=\angle DAY=\angle DXY,\]](http://latex.artofproblemsolving.com/e/0/9/e09b2b45222d1889fa4d5fb7f841f1ff09f17ec5.png)
so 
.
and 
for some 
.![\[37^2\cdot 67x+37\cdot 67^2x = 47^2\cdot 104x + 37x\cdot 67x\cdot 104x.\]](http://latex.artofproblemsolving.com/f/3/a/f3a14d18ec12f0015f9ed50de758db12caf7297e.png)
Note that the LHS can be rewritten as 
,
from both sides gives ![\[37\cdot 67 = 47^2 + (37x)(67x) = 47^2+DY\cdot YC.\]](http://latex.artofproblemsolving.com/8/c/2/8c26aed984416b8218e6b0eccf9ce109fb57df26.png)
Hence ![\[AB^2=DY\cdot YC = 37\cdot 67 - 47^2 = \boxed{270}.\]](http://latex.artofproblemsolving.com/7/3/9/73990f77e907f8f23f3c8281500f53a3d07daeba.png)
passes through the external center of homothety of the two circles.
concur. We also know that 
.
collinear in mind, we have ![\[\angle BAP=\angle CDB=\angle YDB=\angle YBA\]](http://latex.artofproblemsolving.com/8/3/5/835ce14df02b2948f17396bfea8f3cc222d5f69b.png)
This means 
and similarly 
,
is a parallelogram.
from more angle chasing. Thus we can obtain 
and 
.![\[AB^2=4AM^2=4MX\cdot MY=4(\frac {\sqrt {37\cdot 67}+47}{2}\cdot \frac {\sqrt {37\cdot 67}-47}{2})=\boxed {270}\]](http://latex.artofproblemsolving.com/1/1/d/11d92da347da5e66a35a4270984c4ab8cf631a02.png)
.
,
and 
(again resulting from the homothety).
,
Hence, 
Similarly, 
Then since 
,
Similarly, 
Because 
by Power of a Point, multiplying these relations gives 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = 25.561273455869426, xmax = 251.43455114359386, ymin = 45.70873344820468, ymax = 159.632334179233; /* image dimensions */
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */
/* draw figures */
draw(circle((123.90174878101861,80.37090307120728), 23.786991826712317));
draw((xmin, 0.030438670316427052*xmin + 52.80148983853563)--(xmax, 0.030438670316427052*xmax + 52.80148983853563)); /* line */
draw(circle((165.45131811180127,94.34666730069631), 36.49215802761939));
draw(circle((137.59059581752737,108.42508624472585), 42.923890684144), linetype("2 2") + green);
draw((46.11140346430864,54.20505964641348)--(174.93682540909933,129.5844701383543), red);
draw((142.70941966835346,65.80750731479884)--(115.91289192359766,145.4728600692278), red);
draw(circle((144.3889574163432,96.80650504615912), 44.80588392709177), dotted + blue);
/* dots and labels */
dot((124.62545799779946,56.59492306756442),linewidth(3.pt) + dotstyle);
label("$A$", (124.1164676529477,54.02741221445552), S * labelscalefactor);
dot((142.70941966835346,65.80750731479884),linewidth(3.pt) + dotstyle);
label("$X$", (142.44575984982922,63.051063757507265), S * labelscalefactor);
dot((130.08476718136293,103.3402579516797),linewidth(3.pt) + dotstyle);
label("$Y$", (127.21834787088149,105.06744125484197), NW * labelscalefactor);
dot((166.56157666392102,57.87140275799297),linewidth(3.pt) + dotstyle);
label("$B$", (165.9918505950539,55.296363212697166), S * labelscalefactor);
dot((46.11140346430864,54.20505964641348),linewidth(3.pt) + dotstyle);
label("$Z$", (45.58250031707846,51.34851566261202), S * labelscalefactor);
dot((174.93682540909933,129.5844701383543),linewidth(3.pt) + dotstyle);
label("$C$", (175.43848580421593,130.446461219675), NE * labelscalefactor);
dot((100.84847094295199,86.2332712859296),linewidth(3.pt) + dotstyle);
label("$D$", (98.31446402195296,86.17417083657737), NW * labelscalefactor);
dot((115.91289192359766,145.4728600692278),linewidth(3.pt) + dotstyle);
label("$M$", (113.6828705562613,146.37884597537575), NW * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/5/b/f5b767157857f084590bee09b7951399e5dacbd3.png)
and 
,
By cyclic quadrilaterals, we infer 
Hence, if 
for a second time at 
and 
,
Then by Power of a Point,
![[asy]
import olympiad;
size(300);
defaultpen(linewidth(0.8));
pair P = dir(100), D = dir(220), C = dir(320);
pair M = (D+C)/2, pt = bisectorpoint(D,P,C), Z = reflect(P,pt) * M, Y = extension(P,Z,D,C);
pair T1 = P + Y - D, T2 = P + Y - C, A = intersectionpoint(P--D, T2--Y), B = intersectionpoint(T1--Y, P--C);
draw(A--B--C--D--cycle);
pair[] in = intersectionpoints(circumcircle(A,D,Y), circumcircle(B,C,Y));
pair X = in[0];
pair Q = 2 * A - P, R = 2 * B - P;
draw(circumcircle(A,D,Y)^^circumcircle(B,C,Y));
label("$A$", A, dir(120));
label("$B$", B, dir(60));
label("$C$", C, dir(X--C));
label("$D$", D, dir(200));
label("$X$", X, 2.7 * dir(195));
label("$Y$", Y, 1.5*S);
draw(C--X--Y);
draw(D--X);
draw(A--Y--B);
draw(circumcircle(X,C,D));
pair K = 0.5*C + 0.5*D;
pair L = K+dir(90)*(K-D);
pair N = extension(X,Y,K,L);
draw(Y--N);
draw(C--N--D);
label("$M$",N,S);
[/asy]](http://latex.artofproblemsolving.com/d/c/0/dc0f43fe730375fb4ee25cc402464eb1006ddcd6.png)
and similarly 
.
and 
.
,
,
are all similar to each other. Furthermore the ratio of similarity of 
and the ratio of similarity of 
.
.
.
and 
,
.
be the incenter of 
and 
and let its radius be 
again at 
then 
by symmetry. Also, the circumcircle of 
is tangent to 
because 
.
,
and 
to intersect at a point 
.
is a cyclic quad because of Miquel's Theorem. Now, let 
and 
.
.
,
.
.
is a parallelogram. Now from this, we have that 
(last equality from 
cyclic). Thus, 
.
.
is similar to 
,
,
.
,
.
,
.
.
and 
?
?
,
concur, say at 
,
are swapped. Then 
are fixed. Since 
is also tangent to these circles. Then radical axis on 
implies 
is the symmedian of 
.
,
,
,
is similar to 
so 
.
,
,
.
,
.
,
and 
,
and so we get that 
and 
.
.
.
and 
are cyclic. Now, let the circumcircle of 
.
. By power of a point (about 
.
Thus, we get that 
and 
hence, 
and 
.
and similarly, 
Multiplying both expressions, we get that 
which gives us 
.
.
and 
from the tangency condition and that 
from the fact that 
,
.![\[\frac{AB}{DY} \stackrel{(*)}= \frac{BY}{YA} \stackrel{(\dagger)}= \frac{CY}{AB},\]](http://latex.artofproblemsolving.com/a/4/5/a458d170434065ed252e4d4d2fcd511c8ea851d9.png)
so 
.
because 
.
.
past 
so 
,
,
.
are collinear
Let 
.
.
.
so 
.
.
.
so multiplying the two ratios gives 
,
.
is cyclic, since 
.
,
and using the three similar triangles we found earlier gives 
So, 
,
by radical axis. Furthermore, let 
.
hence the result. 
by the parallelogram in the previous claim. 
,
at 
.
is cyclic because 
so 
by power of a point. Set 
via the angle bisector theorem. Then Stewart's theorem implies 
directly. Observe the similarity 
,
so we can compute 
without the construction: note by angle chasing that 
and 
.
by 
.
,
,
so 
and then radical axis on 
by definition.
.
Let 
and 
.
lies on 
.
is cyclic, so 
(where we consider the absolute value of all quantities).
,
is a parallelogram.
which yields 
.
finishes.
which implies 
which is easy to find via currently known lengths.
is the 
.
so 
.
by Radical Axis. Notice 
is cyclic since 
Hence, 
and similarly 
Therefore, 
and 
and 
Also, 
so 
and similarly 
Thus, 
is the midpoint of 
and 
Then, 
so 
,
meets at a same point, call it 
,
are concyclic. We do little angle chasing.
,
,
,
is a parallelogram. Then, 
bisect each, call their intersection 
.
.
,
in, we attain 
are all concurrent at one point, call it 
the fact that quadrilaterals 
and 
are cyclic implies 
is cyclic as well. Denote 
and 
.
lies on the radical axis of 
,![\[AZ^{2}=\text{Pow}_{\omega_{1}}(Z)=ZX\cdot ZY=\text{Pow}_{\omega_{2}}(Z)=ZB^{2}\implies AZ=ZB.\]](http://latex.artofproblemsolving.com/8/3/5/835825f8818eacf30b964169ab446b9f657923e6.png)
Also, notice that ![\[AZ\cdot ZB=\text{Pow}_{\omega_{3}}(Z)=ZX\cdot ZF\implies ZY=ZF.\]](http://latex.artofproblemsolving.com/a/7/2/a72c48864a787c2bde011b7cdea7299714fe996a.png)
The diagonals of quadrilateral 
bisect each other at 
,
.
are cyclic, ![\[\angle DFX=\angle AFX=\angle BYX=\angle BCX=\angle FCX~~\text{and}~~\angle XDF=\angle XDA=\angle XYA=\angle XFB=\angle XFC\]](http://latex.artofproblemsolving.com/5/d/4/5d4832eb3d8228af3052b127d813bca879896964.png)
so we have the pair of similar triangles 
.![\[\dfrac{37}{2u+47}=\dfrac{2u+47}{67}\implies 2u+47=\sqrt{37\cdot 67}\implies u=\dfrac{1}{2}\left(\sqrt{37\cdot 67}-47\right).\]](http://latex.artofproblemsolving.com/5/f/c/5fc4b9c6fdb7a8233ec31ab27ef34f52e49fe4b7.png)
Now compute ![\[AB^{2}=4AZ^{2}=4\cdot ZX\cdot ZY=4u(u+47)=37\cdot 67-47^{2}=\textbf{270}.\]](http://latex.artofproblemsolving.com/3/3/2/33215ce12dd16279badb093615343767b84abae9.png)
by the 
tangency and 
by 
or 
.![\[\angle BYC=180^\circ-\angle ABC=\angle ADC\iff BY\parallel AD.\]](http://latex.artofproblemsolving.com/e/0/6/e060676d558344d1f8b4ebf921e41c07b8a9dd42.png)
These parallelities and tangencies imply that 
and 
.
by AA Similarity, and additionally because 
,
.
i.e. 
for some 
,
But 
,
.
and 
to meet at point 
,
and 
,
and ![\[ |\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY, \]](http://latex.artofproblemsolving.com/e/5/9/e59c2730e33c252a8a4c327b75fd9312eb2312bc.png)
so 
.![\[ \angle DAY = \angle YBC, \]](http://latex.artofproblemsolving.com/d/a/2/da2d764a579592449e81d32a90cdf8f4413b9777.png)
so by the inscribed angle theorem 
.
.
is parallel to the radical axis 
.
while 
,
,
,
.
and 
.
is in fact a parallelogram.![\[ \angle DXY = \angle DAY = \angle DZC = \angle YBC = \angle YXC,\]](http://latex.artofproblemsolving.com/f/9/4/f94381b4eff9dcf7cc5764f8acbce482772750b7.png)
so 
.![\[ XD \cdot XC - XY^2 = YD \cdot YC, \]](http://latex.artofproblemsolving.com/c/e/3/ce34fc7cfac1480b7d8c1ea3e42f13d8078d2ff3.png)
so 
.
,
,
form a geometric sequence in that order, hence![\[ AB^2 = 67 \cdot 37 - 47^2 = \boxed{270}.\]](http://latex.artofproblemsolving.com/1/c/c/1cc2091f601f0ffa776499124f8cdf517077a067.png)
and 
.
which implies 
.
to extract 
.
Let 
.
is cyclic since 
is a parallelogram because 
we have 
or 
but 
