AoPS Academy
differents subjects, each consist of 
students. Given that for any 
different subjects, there exists a student compete in both subjects. Prove that there exists a student who compete in at least 
different subjects.
, define 
and 
, where 
and 
denote the product and sum of the digits of , respectively. Find all solutions to 
for which the following statement holds: there exists at least one triple 
of integers such that
and all triples of real numbers, satisfying the equations, are such that 
are integers.
+1 w
students and a teacher with 
marbles. The teacher then play a Marble distribution according to the following rules. At the start, the teacher distributed all her marbles to students, so that each student receives at least 
marbles from the teacher. Then, the teacher chooses a student , who has never been chosen before, such that the number of marbles that he owns in a multiple of 
. That chosen student then equally distribute half of his marbles to 
other students. The same goes on until the teacher is not able to choose anymore student., such that for some initial numbers of marbles that the students receive, the teacher can choose all the student(according to the rule above), so that each student receiving equal amount of marbles at the end.
be a cyclic quadrilateral with 
. Let be the midpoint of the arc 
not containing 
. Suppose there is a point 
inside such that 
and 
.
, and 
are concurrent.
intersect 
and 
and 
and 
, and let 
intersect 
and 
at 
and 
. Also, let 
and meet at 
. By the second angle condition, 
is cyclic. The arc midpoint also gives 
is cyclic.
and 
are cyclic.
and![\[\angle MAD+\angle ADP=\angle MBC+\angle BCP\]](http://latex.artofproblemsolving.com/1/b/0/1b08f238f9233e2d47482cfba266938c88e30efb.png)
so 
and 
. Thus![\[\measuredangle XAY=\measuredangle CAM=\measuredangle ACP=\measuredangle XCW=\measuredangle XDW=\measuredangle XDY
\]](http://latex.artofproblemsolving.com/a/3/b/a3b0d13c9b6ab4b8bb0c09811b0f269aded3a603.png)
as desired. 
and 
. Radical axis on 
, and taking turns between 
, 
, and 
finishes since all of 
lie on the radical axis of and .
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
real xmin = -23.88782587247942, xmax = 28.11307823832208, ymin = -9.242139859667088, ymax = 6.863695719095153;
draw(circle((0.5809291498702329,-0.01981345647294291), 5.292133555959725), linewidth(1));
draw((0.5542578591817454,5.272252890084698)--(12.770695651820194,-1.2314621937659902), linewidth(0.6));
draw((-1.172584231738878,-5.012995647684546)--(12.770695651820194,-1.2314621937659902), linewidth(0.6));
draw((-2.6235032789327573,4.1918715250220755)--(5.864154960092957,-0.3267378971930568), linewidth(0.6));
draw((-3.8232675432202528,-2.954047438739143)--(5.864154960092957,-0.3267378971930568), linewidth(0.6));
draw((-2.6235032789327573,4.1918715250220755)--(4.616982119434293,-3.4428171083827763), linewidth(0.6));
draw((-3.8232675432202528,-2.954047438739143)--(4.986206114864625,2.9127984230305017), linewidth(0.6));
draw((2.672898183950598,1.372204559170503)--(2.52153455609322,-1.2332843473134645), linewidth(0.6));
draw((4.986206114864625,2.9127984230305017)--(4.616982119434293,-3.4428171083827763), linewidth(0.6));
draw((1.0656729475516833,0.30184064265604144)--(12.770695651820194,-1.2314621937659902), linewidth(0.6) + linetype("4 4"));
dot((-1.172584231738878,-5.012995647684546),dotstyle);
label("$A$", (-1.6429946695254456,-5.883748135844466), NE * labelscalefactor);
dot((0.5542578591817454,5.272252890084698),dotstyle);
label("$B$", (0.4153744515271139,5.672008333222611), NE * labelscalefactor);
dot((4.986206114864625,2.9127984230305017),dotstyle);
label("$C$", (5.146012256051417,3.2886335614775266), NE * labelscalefactor);
dot((4.616982119434293,-3.4428171083827763),dotstyle);
label("$D$", (4.568224432598067,-4.294831621347743), NE * labelscalefactor);
dot((12.770695651820194,-1.2314621937659902),linewidth(4pt) + dotstyle);
label("$Z$", (13.01837135060331,-1.4420042430468083), NE * labelscalefactor);
dot((5.864154960092957,-0.3267378971930568),linewidth(4pt) + dotstyle);
label("$M$", (6.084917469163111,-0.06975816234509281), NE * labelscalefactor);
dot((-2.6235032789327573,4.1918715250220755),linewidth(4pt) + dotstyle);
label("$X$", (-3.3041346619538268,4.4080974694184), NE * labelscalefactor);
dot((-3.8232675432202528,-2.954047438739143),linewidth(4pt) + dotstyle);
label("$Y$", (-4.604157264723864,-3.4642616251335467), NE * labelscalefactor);
dot((1.0656729475516833,0.30184064265604144),linewidth(4pt) + dotstyle);
label("$P$", (0.7764918411854577,-0.5392107689009429), NE * labelscalefactor);
dot((2.672898183950598,1.372204559170503),linewidth(4pt) + dotstyle);
label("$E$", (2.5098553115455076,1.7719405249124727), NE * labelscalefactor);
dot((2.52153455609322,-1.2332843473134645),linewidth(4pt) + dotstyle);
label("$F$", (2.257073138784667,-2.019792066500162), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/5/d/f/5df172c71e6744ae96fc1a147480aa4d8f300304.png)
Let 
meet 
again at 
respectively.
and 
This follows from angle chasing, since the first condition gives that arcs 
and 
sum to arc 
and the second gives that arcs and are equal. Thus arc equals arc 
and arc equals arc 
and 
Pascal's on 
gives 
since the tangent to at is parallel to 
so 
and 
are homothetic. The center of homothety must be 
so 
collinear as desired.
![[asy]
size(7.5cm);
defaultpen(fontsize(10pt));
import olympiad;
import geometry;
pair A = dir(-20);
pair B = dir(-130);
pair C = dir(110);
pair D = dir(70);
pair M = dir(90);
pair X = 2*foot(-M,C,B) - C;
pair Y = 2*foot(-M,A,D) - D;
pair P = extension(C,Y,D,X);
pair T = 2*foot((0,0),M,P) - M;
draw(circle(A,B,C), linewidth(0.7));
draw(A--B--C--D--cycle, linewidth(1));
draw(C--Y--A, black);
draw(D--X--B, black);
draw(arc(circumcenter(C,X,P),circumradius(C,X,P),70,-80,CW), gray+0.7);
draw(arc(circumcenter(D,Y,P),circumradius(D,Y,P),130,-90,CCW), gray+0.7);
dot("$A$", A, dir(5));
dot("$B$", B, dir(-163));
dot("$C$", C, dir(75));
dot("$D$", D, dir(111));
dot("$M$", M, dir(89));
dot("$X$", X, dir(-66));
dot("$Y$", Y, dir(-100));
dot("$P$", P, dir(7));
dot("$T$", T, 1.5*dir(-19));
[/asy]](http://latex.artofproblemsolving.com/b/8/2/b82a81b6d4c6c01753fe8d94ad930ffb0456783b.png)
and 
, so the angle condition implies 
is cyclic. Consider circles 
and 
. Clearly, 
lies on the radical axis, so it suffices to show that does.
be the intersection of those circles. Observe that
so 
. Moreover, 
, so 
. This implies the conclusion.
and 
, also 
and 
be a point such that 
is a parallelogram.
.
, by isogonality lemma, we have 
.
and analogously, 
, and are isogonal conjugates and this implies 
. 
and meet and at 
. Note that 
so 
is cyclic. Note that 
so if circles 
and 
meet at 
we have that lies on . Using Radical axis we have that 
are concurrent to we need to prove lies on 
which is true since 
.
lies on the radical axis of the circumcircles of 
and 
. This is sufficient as lies on both circles and 
lies on the radical axis by power of a point. with respect to the two circumcircles, let lines 
and 
intersect the circumcircles of and at points 
and 
, respectively. We want to show 
, which is equivalent to 
as 
by definition. We go about computing 
and 
by using the trigonometric variant of Ptolemy's theorem for the cyclic 
and 
:
Luckily, 
, so now we just want to show![\[DA \sin \angle PDM - DP \sin\angle ABM = CB \sin\angle PCM - PC\sin\angle BAM.\]](http://latex.artofproblemsolving.com/7/3/d/73d67c662ebce3a998f130c074a64dd747f987e2.png)
At this point, it's only logical that 
. Indeed, assume it's not and let 
be such that 
and lies in the same halfplane as and 
with respect to line 
. Then clearly 
, and furthermore:![\[\angle ADP' = \angle ADC - \angle P'DC = \angle ADC - \angle MAB\]](http://latex.artofproblemsolving.com/3/c/b/3cbe50a82d6715e114101364efa97e18630ec6bf.png)
![\[\angle BCP' = \angle BCD - \angle P'CD = \angle BCD - \angle MBA\]](http://latex.artofproblemsolving.com/2/5/8/25851f9b7fad91f3caa218a461a1dc98d3f6735d.png)
As 
, we get 
. It's now obvious that 
as![\[\angle P'DA = \frac{1}{2}(\angle DP'C+\angle DXC) = \frac{1}{2}(\angle DPC + \angle DXC) = \angle PDA.\]](http://latex.artofproblemsolving.com/a/9/a/a9aa46875c9b2a5f79cf70cb751095c43780362a.png)
We are now ready to prove the line from above as 
, 
, 
, and 
from 
. Denote 
, 
, 
, and 
. Then
where 
is the radius of the circumcircle of . Analogously,
With this, the solution is complete.
and 
.
. We have 
This means that 
However, by the Incenter-Excenter Lemma, is the angle bisector of 
, so 
, which gives 
, implying that . Analogously, we obtain 
, as desired. 
meet again at , 
meet again at . It follows that 
and 
are isoceles trapezoids and 
.
and 
are isoceles trapezoids.
and 
are half the lengths of equal arcs, they must equal. Thus, 
, but is cyclic, so is an isoceles trapezoid. Similarly, 
, so is also an isoceles trapezoid and 
. 
and 
meet at , and meet at ., , are collinear.
, 
, and 
, 
and 
are homothetic. Thus, , , 
concur at the center of homothety , giving the claim. , , are collinear.
and 
are homothetic triangles, so 
, , 
concur at the center of homothety similar to above. 
are collinear, which is exactly the desired condition, so we are done. 
. We'll prove 
is the radical axis 
of 
and 
. The identity 
means 
. A simple angle chase gives that 
which means 
is the tangent to at . Therefore, 
; similarly 
. Since 
, 
. Notice that ![\[\angle YXC = \angle BXC = 180^{\circ}-\angle CBA-\angle PCB=\angle ADC - \angle ADP = \angle PDC = \angle YDC,\]](http://latex.artofproblemsolving.com/6/c/0/6c0cb1aa4129337f284451e2200998bdf6631c81.png)
meaning 
is cyclic. Then 
, which finishes the solution.
is unique, since must vary until we get that the angle between the two lines is 
Let the line through parallel to and the line through 
parallel to intersect at point 
We claim that 
Second, we see that 
so satisfies the problem conditions and thus is equal to ![$$\frac{[BPC]}{[APD]} = \frac{[BMC]}{[AMD]}.$$](http://latex.artofproblemsolving.com/7/5/f/75f61feac9a7526987d40df8d7023ba8c95bf65c.png)
This finishes, since that will readily imply that 
passes through 
at which point we are done.
while the area on bottom is equal to 
Dividing the two and noting that 
we get ![$$\frac{[BPC]}{[APD]} = \frac{BC \cdot CP}{AD \cdot DP}.$$](http://latex.artofproblemsolving.com/b/7/d/b7dfcd6c32f7805fd8731c574d0af04db0bdea04.png)
Now we compute the second ratio. The area on top is equal to 
while the area on bottom is equal to 
Dividing the two and noting that 
we get ![$$\frac{[BMC]}{[AMD]} = \frac{BC \cdot BM}{AD \cdot AM}.$$](http://latex.artofproblemsolving.com/8/5/8/8585b2df61a5c1abd6fd0470d142eab09e672740.png)
It thus suffices to show that 
In fact, 
, and if hits again at point 
then 
so 
and so 
Thus 
giving the desired length ratio, and we are done.
be a cyclic quadrilateral with . Let be the midpoint of the arc not containing . Suppose there is a point inside such that and ., and are concurrent.
be the circumcircle of and let 
and 
. We have 
. So 
and lie on a circle 
. Let 
be the center of . By angle chasing we get 
and thus 
. So 
and analogously 
and lies on . and are two points on and the perpendicular bisector of . We have 
since B and lie on the perpendicular bisector of 
. We have 
since 
. So 
and 
is a diameter of .
. So 
and are tangent to . So is the polar of w.r.t. . By pole-polar duality lies on the polar of w.r.t. . By Brocard 
is the polar of w.r.t. . Thus 
and concur at .
and meet 
at and . Then since , 
is an isosceles trapezoid with 
. Now let 
be the point such that 
and 
. Then since 
, lies on , and we also have 
, so 
. Since 
and 
are homothetic, we see that 
, 
and 
are concurrent at a point . Now Pascal's theorem on the hexagon 
shows that , and 
are collinear so we are done.
and 
(this has already been proved in previous solutions, so I won't write it). Now let 
, 
and 
. Then by Thales/homothety we get that
so 
is on the radical axis of 
and 
. Finally, the inversion with centre and radius maps to and to , which means that also lies on the radical axis of and . Done.
and meet the circumcircle of 
again at 
, respectively. Denote, the intersections of 
and 
, respectively.
, 
and 
. Hence, 
and 
are homothetic, implying 
are collinear.
, we obtain that 
are also collinear, so we are done. 
and 
. Define 
, 
and 
. By construction 
, 
and 
are all cyclic (!) so that radical axis finishes.
be a cyclic quadrilateral with . Let be the midpoint of the arc not containing . Suppose there is a point inside such that and ., and are concurrent.
, 
, 
and 
. Let 
and 
. Because 
and 
. Additionally, 
. This shows 
, 
and 
due to the fact that 
.
is the homothety center of 
and 
. This shows that 
. By Pascal Theorem on 
, 
which is sufficient.
and 
. Then 
gives 
cyclic. Let 
and 
. Then 
gives 
cyclic. Since ![\[ \angle DYB = \angle QYC = 180^\circ - \angle YQC - \angle QCY = 180^\circ - \angle YQC - \angle ADB = 180^\circ - \angle YQC - \angle CPD = \angle QCR = \angle QDR = \angle YDB, \]](http://latex.artofproblemsolving.com/0/f/5/0f525db411217bd06a2f41b91597e4d38c4ac94f.png)
we have 
, and similarly 
. Then 
, so 
and similarly 
. be the second intersection of with . Then 
gives 
cyclic, and similarly 
is cyclic. The desired claim follows from radical axis on , , and .
in triangle 
where 
.
be the isogonal conjugate. Then
As 
, we find 
. Now, let the perpendicular bisector of intersect at 
and 
at 
and 
with in the minor arc . Then
Thus we find 
. I.e. 
bisects 
. Appolonius circle gives that 
bisects 
, and therefore 
and 
are isogonal. Hence pass through by construction of .
intersect again at point , and let intersect at 
, and similarly define 
, we have that 
are concyclic, so we can just angle chase for 
and 
and 
, so 
and 
are both cyclic is a radical axis of these two circles and since is also cyclic, we have that , , and are concurrent on 
be the intersection of the line passing through and parallel to with . Since 
, then 
. 
. It is also equal to 
. So 
. Since point is the midpoint, 
. It can be seen that 
Since 
it is 
. Now we have to prove that 
for the point , which is the intersection of the line passing through and parallel to 
with the line passing through 
and . Let 
. Let 
and 
. Since 
, then 
. 
is cyclic. Thus 
. 
. Also with 
, by the similarity in the triangle 
, 
. Since as we just found, 
becomes. Thus, 
in the triangle 
. The proof ends.
intersect and at and respectively and let intersect at 
. Let the arcs 
, 
, and be equal to 
, 
, 
and 
respectively. We have that 
is parallel to and is the midpoint of arc . Quick angle chasing in 
gives 
and 
. Also note that 
. We apply trig ceva in 
with point 
, 
and 
. So from trig ceva in 
with point we get: 
we get that 
and 
so , and are collinear. Done!
and 
meet 
and 
at and .
and 
are homothetic.
. Let the second intersections of and with be and , respectively. We are going to show 
.
.
, 
and 
. So we conclude that 
and 
are homothetic.
, 
and are concurrent. Let be the intersection point of these lines. So we know that 
.
gives us 
. and implies .
![[asy]
/* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (7.77952,50.31114); pair B = (-7.41116,-7.83103); pair C = (41.85666,-8.11885); pair D = (51.58016,23.65726); pair M = (50.74270,6.53777); pair P = (18.65367,15.52170); pair F = (104.39862,-8.48421); pair X = (37.38301,20.14939); pair Y = (32.01336,1.91008);
import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5;
pen ffxfqq = rgb(1.,0.49803,0.); draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle((17.36719,16.75068), 34.90312), linewidth(0.6)); draw(A--D, linewidth(0.6)); draw(D--M, linewidth(0.6)); draw(M--C, linewidth(0.6)); draw(B--M, linewidth(0.6) + ffxfqq); draw((24.59535,0.07721)--(23.56486,-2.04245), linewidth(0.6) + ffxfqq); draw((24.59535,0.07721)--(22.69624,1.47304), linewidth(0.6) + ffxfqq); draw((21.66576,-0.64662)--(20.63528,-2.76630), linewidth(0.6) + ffxfqq); draw((21.66576,-0.64662)--(19.76666,0.74919), linewidth(0.6) + ffxfqq); draw(D--F, linewidth(0.6) + linetype("4 4") + red); draw(F--C, linewidth(0.6) + linetype("4 4") + red); draw(P--F, linewidth(0.6) + linetype("4 4") + red); draw(P--D, linewidth(0.6) + ffxfqq); draw((38.04650,20.31333)--(37.01602,18.19365), linewidth(0.6) + ffxfqq); draw((38.04650,20.31333)--(36.14740,21.70915), linewidth(0.6) + ffxfqq); draw((35.11691,19.58948)--(34.08643,17.46980), linewidth(0.6) + ffxfqq); draw((35.11691,19.58948)--(33.21781,20.98531), linewidth(0.6) + ffxfqq); draw(P--C, linewidth(0.6) + blue); draw((32.36897,1.54776)--(30.01987,1.35631), linewidth(0.6) + blue); draw((32.36897,1.54776)--(32.60426,3.89287), linewidth(0.6) + blue); draw((30.25517,3.70142)--(27.90607,3.50997), linewidth(0.6) + blue); draw((30.25517,3.70142)--(30.49046,6.04653), linewidth(0.6) + blue); draw(A--M, linewidth(0.6) + blue); draw((31.37491,26.27079)--(29.02581,26.07934), linewidth(0.6) + blue); draw((31.37491,26.27079)--(31.61021,28.61591), linewidth(0.6) + blue); draw((29.26111,28.42446)--(26.91201,28.23301), linewidth(0.6) + blue); draw((29.26111,28.42446)--(29.49641,30.76957), linewidth(0.6) + blue);
dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, NE); dot("$M$", M, NE); dot("$P$", P, NW); dot("$F$", F, NE); dot("$X$", X, N); dot("$Y$", Y, dir(270)); [/asy]](http://latex.artofproblemsolving.com/3/2/c/32c7fb60ae8f8f7a8f0b851c14d19e2fd8fa9a03.png)
and 
. and 
.
and 
. Firstly, note that, ![\[ \measuredangle YPX=\measuredangle CPD=\measuredangle ADB =\measuredangle AMB = \measuredangle XMY .\]](http://latex.artofproblemsolving.com/a/7/0/a7070400ecfb77e49f2ddfcac57a171c2a0e5b7c.png)
Now note that, 
Now since the opposite angles of the quadrilateral 
are equal, we must have that is a parallelogram. This gives us our desired result. as the intersection of line through parallel to with the line through parallel to 
.
be a triangle and let denote the midpoint of arc 
not containing . Let be any arbitrary point on 
. Let the line through parallel to be 
. Similarly, let the line through parallel to be 
. Define 
. Prove that , and are concurrent., and . We animate 
projectively on 
. Firstly note that has degree 
(as it is also fixed). So, the line is fixed and has degree . Thus also has degree . Furthermore note that the line at infinity is also fixed as 
and 
are fixed. is fixed, is also fixed.![\[ B\mapsto MB\mapsto MB\cap \ell_{\infty}\mapsto D\infty_{MB}\mapsto D\infty_{MB}\cap CP \text{ via }\odot(ACD)\mapsto \mathcal{P}(M) \mapsto \ell_{\infty}\mapsto \mathcal{P}(D) \mapsto CP \]](http://latex.artofproblemsolving.com/c/d/3/cd3a99662abb4df4702fa929d01707c7ae09f584.png)
is projective. This means that has degree . So the line has degree 
![\[ B\mapsto CB\text{ via }\odot(ACD)\mapsto \mathcal{P}(C) \]](http://latex.artofproblemsolving.com/4/7/a/47aedfc17dd81e6e7f0a53bf80b3db18feb20d59.png)
is projective. This means that the line 
has degree . Thus finally, the condition that , and has degree 
. Firstly note that when 
, we can reduce the degree by using Zack's Lemma.
-- Then note that 
. So, 
.
-- Then note that 
which implies that 
. Then clearly 
.
, and intersect again at , and respectively. Let 
. Let 
. Let 
, and 
be and respectively. implies and arcs 
, 
, 
, and 
have equal length. Therefore 
is an isosceles trapezoid along with 
. Observe by pascals on 
, , , and must be collinear. By pascals on 
. Therefore 
is cyclic. Observe that 
therefore 
is tangent to 
. Now Pascals on 
gives , , collinear. This combined with , , collinear gives us the desired result.
intersect 
at and 
intersect 
at . The key claim is the following:
and 
.
and 
on and satisfying the length conditions, then show that 
satisfies both angle conditions; obviously it satisfies the latter one as 
is cyclic (see below).
, so triangles 
and 
are similar. So ![\[\measuredangle CPD = \measuredangle(\overline{E'C}, \overline{F'D}) = \measuredangle(\overline{AE'}, \overline{DB}) = \measuredangle ADB\]](http://latex.artofproblemsolving.com/0/d/a/0da0d240664dcd8b12c91420ca039ad07e102fce.png)
as needed. and to 
meet at as 
. Thus by Pascal on 
we have the desired concurrence.
and 
are homothetic, which just like solves the problem. Geometry is too hard.
denote @rclength. The 
angle condition suggests projecting 
through to the circle at 
, so that 
and 
is a CyclicISLscelesTrapezoid. Now, rewrite the angles in the 
condition:
Equating, we see that the @rclengths 
, 
, 
, and 
are all equal. Besides the aforementioned and the degenerate 
, these produce 
or 
more CyclicISLscelesTrapezoids:
, which gives 
;
, which gives 
;
, which gives 
; and
, which gives 
.
is homothetic to 
, where 
. 
is concurrent with rays and at a point , since is the 
vertex of the nested similar simplices 
and 
(more precisely, the projection onto the plane at hand). Hey, it isn't my fault the diagram looks hyperdimensional 
are isogonal with respect to 
Also note that 
from the first condition. Also note that 
from arc midpoint definition.
We also know that 
so 
Thus, 
which finishes.
so 
Let 
Observe that 
since 
is cyclic. Thus, 
are homothetic, which finishes.
be the intersection of and 
, and let be the intersection of and .
, and by the midpoint of arc condition, 
, so triangles 
and 
are similar. The condition that and then yields that 
is a parallelogram. and intersect again at and , respectively. Then let 
and 
meet at . Note that 
is a trapezoid, so 
, and thus triangles 
and 
are homothetic with center . Thus 
are collinear.
it follows that 
and the intersection of and are collinear. Thus 
and are concurrent.
, 
and 
,
, and are concyclic
Points ,, and are concyclic 
and 
are concyclic and both points 
and 
are collinear 
and 
and 
and 
we get that the quadrilateral -is a parallelogram 
and 
and , and are concurrent., and 
we get that triangles 
and 
are homothetic 
, and 
meet at one point 
Lines , and are concurrent 
then easy to see that 
concyclic
and 
concyclic and done with radical axis.
is cyclic and 
,
and 
is homothetic, and their homotheric center is 
,
is cyclic. We claim that 
Similarly, 
.
and 
.
and 
Analogously: 
We know: 
and 
and 
Since 
By 
By 
and 
,
,
and 
.
is cyclic, so 
,
is the angle bisector of 
.
are collinear by radical axis. We have that 
,
.
and 
are collinear (due to the angle bisector), which finishes the problem.
.
homothetic which finishes.
and 
then 
,
,
are colleniar. Done!
and 
.
.
concur at 
and 
.
.
and 
we have that 
are colinear. Let 
We have that 
but since 
we have 
colinear. Claim: 
are collinear. By applying Pascal's on 
we get our result.
~ 
.
we have that 
is a isosceles trapezoid , 
are colinear. Let 
so by ratios we have 
.
.
colinear and the lines