and 
Prove that
Let 
Prove that
.
-
.
and 
intersect at 
.
such that 
is a square perfect.
friends in this. However, if two people are not friends at least they have a friend in this party. What is the largest possible number of people in the party?
be real numbers such that 
Prove that
Let 
Prove that
Let 
Prove that
be the altitude and 
be the intersection point of the altitudes of triangle 
is symmetric to 
intersects 
.
.
![$f\in\mathbb{C}[X], \ f=aX^3+bX^2+cX+d, \ a,b,c,d\in\mathbb{R}^*$](http://latex.artofproblemsolving.com/8/b/d/8bdc04d75d0654da65463301f6fd0b0af4160b2b.png)
a polynomial whose roots 
are in geometric progression with ration 
.
.
satisfying:
satisfying 
Prove that 
Proposed by Phan Ngoc Chau
,
and let 
.
is:
be a quadrilateral with 
,
,
and 
.
(in degrees) is:
Y by CyclicISLscelesTrapezoid, thanosaops, r00tsOfUnity, Kagebaka, MathLuis, aansc1729, MihaiT, mathmax12, anantmudgal09, Kingsbane2139, SpecialBeing2017, PHSH, dolphinday, Rounak_iitr, vrondoS, Funcshun840
Let 
be a triangle with incenter 
and excenters 
, 
, and 
opposite 
, 
, and 
, respectively. Let 
be an arbitrary point on the circumcircle of 
that does not lie on any of the lines 
, 
, or 
. Suppose the circumcircles of 
and 
intersect at two distinct points and 
. If 
is the intersection of lines 
and , prove that 
.
Proposed by Zach Chroman
be a triangle with incenter 
and excenters 
, 
, and 
opposite 
, 
, and 
, respectively. Let 
be an arbitrary point on the circumcircle of 
that does not lie on any of the lines 
, 
, or 
. Suppose the circumcircles of 
and 
intersect at two distinct points and 
. If 
is the intersection of lines 
and , prove that 
.Proposed by Zach Chroman
This post has been edited 2 times. Last edited by v_Enhance, Jun 19, 2023, 5:10 PM
Reason: fix latex
Reason: fix latex
![[asy]
size(10cm);
defaultpen(fontsize(10pt));
pen pri=mediumblue;
pen sec=purple;
pen tri=fuchsia;
pen fil=invisible;
pen sfil=invisible;
pen tfil=paleblue;
pair O, Ia, Ib, Ic, A, B, C, I, D, X, W, Y, W1, Y1, M, N, U, V, X1, X3, P, Q;
O=(0, 0);
Ia=dir(105);
Ib=dir(210);
Ic=dir(330);
A=foot(Ia,Ib,Ic);
B=foot(Ib,Ia,Ic);
C=foot(Ic,Ia,Ib);
I=intersectionpoint(A--Ia, B--Ib);
X=C+0.25(C-B);
X1=foot(X,A,Ia);
Y=2*X1-X;
X3=foot(X,Ib,Ic);
W=2*X3-X;
D=intersectionpoint(Y--(Y+100*(Y-A)), circumcircle(A, B, C));
P=intersectionpoint(X--Y, circumcircle(D, I, Ia));
Q=intersectionpoint(X--W, circumcircle(D, Ib, Ic));
M=(I+Ia)/2;
N=(Ib+Ic)/2;
Y1=2M-X;
W1=2N-X;
U=intersectionpoint(X--M, circumcircle(D,I,Ia));
V=intersectionpoint(X--N, circumcircle(D,Ib,Ic));
draw(Ia--Ib--Ic--cycle,pri);
draw(A--Ia, pri);
draw(B--Ib, pri);
draw(C--Ic, pri);
draw(A--B--C--cycle, pri);
draw(circumcircle(A, B, C), pri);
draw(C--X, pri);
draw(circumcircle(D, I, Ia), tri);
draw(circumcircle(D, Ib, Ic), tri);
draw(X--Y, dashed+sec);
draw(X--W, dashed+sec);
draw(X--Y1,dashed+sec);
draw(X--W1,dashed+sec);
draw(D--W,pri);
draw(A--X,pri);
label("$I_A$",Ia,dir(90));
label("$I_B$",Ib,dir(180));
label("$I_C$",Ic,dir(0));
label("$A$",A,dir(270));
label("$B$",B,dir(60));
label("$C$",C,dir(120));
label("$I$",I,dir(240));
label("$D$",D,dir(0));
label("$X$",X,dir(150));
label("$W$",W,dir(240));
label("$Y$",Y,dir(300));
label("$W'$",W1,dir(300));
label("$Y'$",Y1,dir(60));
label("$M$",M,dir(120));
label("$N$",N,dir(270));
label("$U$",U,dir(180));
label("$V$",V,dir(270));
label("$P$",P,dir(60));
label("$Q$",Q,dir(180));
[/asy]](http://latex.artofproblemsolving.com/c/c/1/cc1a5550cab9a32c7e42274e888f5adccf1c04f4.png)
be the point on 
;
,
.
,
,
,
(latter two are classic properties of the orthic config) so 
.
are the reflections of 
respectively, 
.
reflections of 
are cyclic isosceles trapezoids so if 
then 
are diameters of the two circles. Furthermore, 
.
,
is on 
;
is on 
.
,
and 
be the circumcenters of 
and 
.
.
,
,
,
,
.
,
,
,
.
then 
,
Now, compute 
and 
is
is
The product of the two slopes is 
,
.
,
.
,
,
.
,
,
.
.
,
and 
,
and 
.
as well, so we can solve the system and we have 
and 
.
,
,
and 
.
,
.
and 
.
and we need to find this intersected with 
.
.
,
.
.
is a triangle with orthic triangle 
and orthocenter 
.
lies on the nine-point circle, and point 
lies on 
such that 
and 
are isogonal wrt 
.
and 
.
,
.
be the radical axis of 
and 
be the radical axis of ![\[ \operatorname{Pow}_{(PAH)} (R) = RA \cdot RH = RE \cdot RF = \operatorname{Pow}_{(DEF)} (R),\]](http://latex.artofproblemsolving.com/8/6/3/863c45c1dfabc5b4ee4759561c1705be098cfa2b.png)
where lengths are directed and the second equality comes from Power of a Point on 
.
.
.
.
is cyclic. To see this, consider the circle 
such that![\[ \operatorname{Pow}_\omega(K) = \operatorname{Pow}_{(PBC)} (K) + \operatorname{Pow}_{(PAH)} (K) - \operatorname{Pow}_{(DEF)} (K),\]](http://latex.artofproblemsolving.com/0/2/a/02a92b2088e995113df2f9e31fa76b4fb16bc63e.png)
for all points 
,![\[ \operatorname{Pow}_{(PAH)} (D) + \operatorname{Pow}_{(PBC)} (D) = DA \cdot DH + DB \cdot DC = 0.\]](http://latex.artofproblemsolving.com/5/5/4/554c0419f1063702aeec5e0e0efb6885c0cd8bd5.png)
In addition, 
,
,
,
.
and fixing both 
and 
.
.
.
be the intersection of the reflection of 
with 
.
of 
respectively, i.e. that 
.
be 
.
since 
.
Thus, we get that 
.
gives us
Finally, we have that
Similarly, we can compute
Thus, since these are equal, 
and 
.
.![\[
\measuredangle BAD=\measuredangle BAI+\measuredangle IAD=\measuredangle IAC+\measuredangle E'AI=\measuredangle E'AC=\measuredangle EAC
\]](http://latex.artofproblemsolving.com/2/5/5/255ce855adff8b4d6aeb99d21ebb29eb390de265.png)
so 
are zero.
.
,
is similar to 
.
Thus, since 
are isogonal, reflecting 
and 
to 
give points on 
and 
are cyclic.
and 
are cyclic.
and 
.
However, since 
is the circle with diameter 
.
which for the same reason, gives 
,
,
,
and make 
inversion and let 
the centers of 
and 
respectivily. Now note that 
,
(becuase it moves in the straight bisector of 
(becuase in this case even if it lies in the bisector of 
is tangent to 
so if we want to prove 
colinear we need 
cases!
which is enough since 
(use McLaurin to verify)
(wlog 
)
but also 
so we are done.
but 
or 
with 
the inverse of 
inversion with center 
is cyclic and also 
colinear by the inversion so by 
so we are done.
.
as reference. Then we must have 
and similarly 
,
and 
.
satisfying, ![\[a^2y_1z_1 + b^2x_1z_1 + c^2x_1y_1 = 0.\]](http://latex.artofproblemsolving.com/0/b/8/0b81ac985d1b88565320a0ccd58cf16fac040082.png)
Also take 
,![\[Pow_{\omega_1}(E') = Pow_{\omega_2}(E').\]](http://latex.artofproblemsolving.com/8/7/1/8717b7cc45273ba5821a9779fd4f2ea03e7c47b4.png)
Now we will find the equation of 
These relatively nicely simplify down to,
Now subtracting the first two equations we find 
.
The first equation implies we have 
.
Then clearly we find 
so our circle has equation,
Yay. Now we have to do the same thing for 
This neatly simplifies to,
Adding the first two equations give 
.
The first equation gives 
.
Then we easily get 
.![\[-a^2yz - b^2xz - c^2xy + (x+y+z)\left(-bcx + \frac{bc^2x_1}{cy_1 + bz_1}y + \frac{b^2cx_1}{cy_1 + bz_1} \right) = 0\]](http://latex.artofproblemsolving.com/9/3/c/93ce021ff9bd6ad803fb5ec6b489717e2375093e.png)
Now we're on the homestretch. Plugging in 
into the circle equations we have,
Thus it suffices to show,
so we are done. 
denote the reflections of 
, 
,
and 
are cyclic.
.
.
.
as 
is the reflection of 
and 
.
as 
intersect the circumcircle of 
.
,
and 
,
,
.
is an isoceles trapezoid via 2019 IMO P6 and get that 
and 
are congruent, and 
.
.
as a sum of linear functions 
,
,
defined by ![\[ f_1(\bullet) = \text{Pow}(\bullet, (DI_BI_C))-\text{Pow}(\bullet, (II_A)) \]](http://latex.artofproblemsolving.com/8/1/a/81a18cf89fe04b44c75d921e0a736666c5b3a660.png)
![\[ f_2(\bullet) = \text{Pow}(\bullet, (II_A))-\text{Pow}(\bullet, (I_BI_C)) \]](http://latex.artofproblemsolving.com/5/c/a/5caa6561c033342fddd6614c718340ce63edfa0f.png)
![\[ f_3(\bullet) = \text{Pow}(\bullet, (I_BI_C)) - \text{Pow}(\bullet, (DII_A)). \]](http://latex.artofproblemsolving.com/f/8/b/f8bdc2a0b1e1da51cc72d4f0a30dc588490f48c6.png)
Note that 
vanishes at 
and 
since the midpoints of 
and 
lie on the kernels of 
and 
.![\[ \text{Pow}(E, (I_BI_C)) - \text{Pow}(E, (II_A)) = 0, \]](http://latex.artofproblemsolving.com/1/3/3/13394cefaae79f72261bd9df85d1a708198357fd.png)
but this is clear as the radical axis of 
and 
is 
and 
.
and 
.
and 
.
.
on ![\[EU=1\]](http://latex.artofproblemsolving.com/4/5/c/45cd999f9c9e4e1255959f36d326825c95749ac0.png)
![\[ER=\frac{2}{1}\]](http://latex.artofproblemsolving.com/9/0/0/900e3ccd1d5ea6e419f71220a47a4fb09441f2d8.png)
![\[EV=1\]](http://latex.artofproblemsolving.com/8/f/e/8fee1737af99ff13548744d151793be71cb5061c.png)
![\[ES=\frac{2}{1}\]](http://latex.artofproblemsolving.com/6/b/5/6b5d8c7d62f78190c44a8e76a9026713b5643e25.png)
where we found degree for 
(analogously, 
)
as midpoint of 
,
.![\[\frac{3}{1}=\frac{3}{1}\]](http://latex.artofproblemsolving.com/f/6/a/f6a228d36ccdffcd005bbc5199ddcc2bc9c682a7.png)
which essentially means we need to prove a degree four polynomial is equal to zero.
are easy.
is easy. (Because 
is easy. (Because 
,
,
have another intersection point.
.
is totally gaming.
orthocentric system with 
and 
and 
.
form orthocentric system. (this suffices)
![[asy]
size(11cm);
defaultpen(fontsize(10pt));
pair A = dir(115), B = dir(205), C = dir(335), D = dir(40), M1 = dir(270), M2 = dir(90), I = incenter(A,B,C), Ia = 2*M1 - I, Ib = extension(A,M2,C,Ia), Ic = extension(A,M2,B,Ia), E = extension(A,2*foot(D,M1,M2)-D,B,C), E1 = 2M1-E, E2 = 2M2-E, A1 = extension(A,I,B,C), D1 = extension(E,E1,D,A1), D2 = extension(D,extension(A,Ib,B,C),E,M2), F = E + dir(E--D)*abs(E-D1)*abs(E-E1)/abs(E-D), D0 = 2*foot(D,M1,M2)-D;
draw(circumcircle(A,B,C));
draw(arc(circumcenter(D,Ib,Ic), abs(D-circumcenter(D,Ib,Ic)), -140, 5), purple);
draw(A--B--C--A);
draw(Ia--Ib--Ic--Ia);
draw(E1--E2^^M1--M2, red);
draw(D--E, red+dashed);
draw(B--E);
draw(circumcircle(D,I,Ia), purple);
draw(E1--E--E2, red);
draw(A--Ia);
draw(circumcircle(B,C,Ib)^^circumcircle(B,I,C), blue+dotted);
draw(circumcircle(D1,E1,D2), purple+dashed);
draw(A--E, dotted);
dot("$A$", A, dir(115));
dot("$B$", B, dir(210));
dot("$C$", C, dir(330));
dot("$I$", I, dir(180));
dot("$I_a$", Ia, dir(270));
dot("$I_b$", Ib, dir(0));
dot("$I_c$", Ic, dir(225));
dot("$E_1$", E1, dir(315));
dot("$E_2$", E2, dir(0));
dot("$D_1$", D1, dir(225));
dot("$D_2$", D2, dir(225));
dot("$M_1$", M1, dir(270));
dot("$M_2$", M2, dir(90));
dot("$D$", D, dir(90));
dot("$E$", E, dir(180));
dot("$F$", F, dir(90));
dot("$D'$", D0, dir(135));
[/asy]](http://latex.artofproblemsolving.com/8/f/4/8f4afda9c72fe8e369f58cc2d49904f3f22398ee.png)
meet 
,
be the midpoint of arc 
,
.
,
,
,
concur. By radical axis on 
,
,
,
concur. Thus![\[-1 = (B,C;\overline{AI} \cap \overline{BC},\overline{AM_2} \cap \overline{BC})\overset D= (B,C;D_1,D_2).\]](http://latex.artofproblemsolving.com/3/4/d/34d7feceb7277fddc14f4708ec14ff9d0fe2b865.png)
Since 
are arc midpoints, we then have 
,
,
and 
be the reflection of 
.
depending on if ![\[f(E) = f(D_1) = \frac{f(\overline{AI} \cap \overline{BC})}{f(D)} = f(A)f(D'),\]](http://latex.artofproblemsolving.com/8/b/b/8bb3eb504776dc2a99938e30166f8dcdb19e1e2a.png)
so 
be the reflection of 
for 
.![\[M_1D_1 \cdot M_1E_1 = M_1D_1 \cdot M_1E = M_1B^2 = M_1I \cdot M_1I_a,\]](http://latex.artofproblemsolving.com/1/d/6/1d64e39ba9742eba4e58091ea06f961567fc67fb.png)
so 
and similarly 
.
we have 
is cyclic by Reim's, so we are done by radical axis on 
.
.
,
then homothety centered at 
implies 
.
.
.
and 
are images under an inversion at 
composed with a reflection over ![\[AD \cdot AE'=AB \cdot AC=AI \cdot AI_a.\]](http://latex.artofproblemsolving.com/0/0/b/00b0a7418934b38dd67040b76fa0a8b053d44c15.png)
Let 
and 
.
.
.
,
By Stewart on 
,
Plugging this into the expression for 
,![\[XI \cdot XI_a=2XA \cdot XM-XD \cdot XE'=XD\left(\frac{2XA \cdot XM}{XD}-XE'\right)=XD(2XP-XE'),\]](http://latex.artofproblemsolving.com/4/c/6/4c66827657c633ec6eb84ff9207a786a3a827a72.png)
as desired. 
,
,
?
is the circumcircle of 
is cyclic, therefore 
,
are collinear
are collinear
and 
,
are all cyclic
the intersection of 
and 
,
with 
is 
,
at 
pass through the Miquel point of 
,
are collinear, and by 
we have 
are collinear, 
is the midpoint of arc 
are collinear, which is equivalent to show that 
are collinear, which means 
and 
.
are collinear by radical axis
are collinear, then 
,
to be the center of 
,
,
,
are collinear because they all lie on the perpendicular bisector of 
,
,
is a parallelogram, this implies 
,
and so if 
intersects 
.
and similarly the power of 
.
or 
which has to be the equation of a line in 
.
;
![[asy]
import geometry;
unitsize(3cm);
pair A = dir(110); pair B = dir(210); pair C = dir(330); pair L = dir(270); pair D = dir(240); triangle t = triangle(A,B,C); pair I = incenter(t); pair Ia = excenter(t.BC); pair Ib = excenter(t.CA); pair Ic = excenter(t.AB); pair E = intersectionpoint(line(B,C), reflect(line(A,I)) * line(A,D)); pair X = reflect(line(A,I)) * E; pair Y = reflect(line(Ib, Ic)) * E;
draw(circumcircle(A,B,C), red); draw(circumcircle(D, I, Ia), green); draw(circumcircle(D, Ib, Ic), green);
draw(circumcircle(B,I,C), blue); draw(circumcircle(B,C,Ib), blue);
draw(A--B--C--cycle); draw(A--Ia); draw(E--X, dashed+orange); draw(E--Y, dashed+orange); draw(Ia--Ib--Ic--cycle); draw(A--D); draw(A--E); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$L$", L, dir(L)); dot("$I$", I, dir(I)); dot("$I_a$", Ia, dir(Ia)); dot("$I_b$", Ib, dir(Ib)); dot("$I_c$", Ic, dir(Ic)); dot("$E'$", E, dir(E)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); [/asy]](http://latex.artofproblemsolving.com/f/b/4/fb4ddd5575245d193273b3afe405b365627f1bec.png)
.![\[ AD \cdot AE' = AB \cdot AC = AI \cdot AI_a = AI_b \cdot AI_c \]](http://latex.artofproblemsolving.com/8/4/1/8415c868149c5486e8ce28bc69c4bea693f4ca9a.png)
Proof. The first equality follows from 
,
,
and 
.
and 
are cyclic.![\[ AI \cdot AI_a = AD \cdot AE' = AD \cdot AX \]](http://latex.artofproblemsolving.com/3/4/b/34ba9ff4411ca9e9d40b21f0b4a1a67986904e90.png)
so by Power of a Point, ![\[ AI_b \cdot AI_c = AD \cdot AE' = AD \cdot AY \]](http://latex.artofproblemsolving.com/b/5/0/b506fcfd0bc091162059bd8617838aed0032853d.png)
so by Power of a Point, 
and 
for brevity. Let the circumcircles of 
and 
be 
be the power of the point 
.![\[ f(\Gamma_1, E') - f(\omega_1, E') + f(\Gamma_1, X) - f(\omega_1, X) = 0 \]](http://latex.artofproblemsolving.com/1/0/7/107554d5970677ed7472f057b1dda2383c552983.png)
![\[ f(\Gamma_2, E') - f(\omega_2, E') + f(\Gamma_2, Y) - f(\omega_2, Y) = 0 \]](http://latex.artofproblemsolving.com/2/d/d/2ddf12ff954c2010792f92469dc1845ac3b8440f.png)
and additionally ![\[ f(\Gamma_1, X) = 0 \]](http://latex.artofproblemsolving.com/e/f/a/efa6b35094a367c7334abb5467d5067ee6925b80.png)
![\[ f(\Gamma_2, Y) = 0 \]](http://latex.artofproblemsolving.com/4/c/3/4c38abc5eaa0d7c6b5435541a11dca60628d98c9.png)
![\[ f(\omega_1, X) = f(\omega_1, E') \]](http://latex.artofproblemsolving.com/a/b/c/abc7660db880b9f14ebe085bcd924f987e7619e7.png)
![\[ f(\omega_2, Y) = f(\omega_2, E') \]](http://latex.artofproblemsolving.com/0/5/d/05d219fafedb44339e88e900df64c05f531d6f2e.png)
![\[ f(\omega_1, E') = f(\omega_2, E') \]](http://latex.artofproblemsolving.com/b/1/d/b1de4ee39163d38b708c318a511771ccc75bf6e2.png)
Proof. The first two follow by Linearity of Power of a Point; in particular, 
for fixed circles 
,
is a linear function with respect to 
be 
and ![\[ \frac{f(\Gamma_1, E') - f(\omega_1, E') + f(\Gamma_1, X) - f(\omega_1, X)}{2} = f(\Gamma_1, M) - f(\omega_1, M) = 0 \]](http://latex.artofproblemsolving.com/8/a/4/8a427c4eddcf9b4881a57f681fbf72ed8d960f83.png)
which implies the first equation. Let the midpoint of 
be 
and ![\[ \frac{f(\Gamma_2, E') - f(\omega_2, E') + f(\Gamma_2, Y) - f(\omega_2, Y)}{2} = f(\Gamma_2, N) - f(\omega_2, N) = 0 \]](http://latex.artofproblemsolving.com/b/d/e/bdef5f9786a17483303b6b2a38b56597a6dcd95b.png)
which implies the second equation.
and 
.
which implies that 
of 
and 
lie on 
.
,
.
.
,
,
,
lies on 
.
,
is cyclic. Inverting back, this implies that 
is the image of 
,
.
being collinear is equivalent to 
being cyclic. Indeed, 
,
Let 
be the arc midpoint of 
are concurrent by Radax on 
Thus, 
is cyclic.
are collinear. We will prove 
Observe that 
so 
and 
are tangent. Thus, 
implying 
lies on 
Finally, observe that 
or equivalently, the antipode of 
are collinear.
and we obviously have 
Thus, by AA similarity, we have 
which finishes.
