is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
,
,
,
,
satisfy 
and 
.
of 
such that 
.
passes through 
and 
again at points 
)
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
are similar.
such that 
, where 
be the second intersection of the circumcircle of triangle 
with the internal bisector of the angle 
.
be the circumcenter of triangle 
and 
the orthocenter of triangle 
.
is equal to the circumradius of triangle 
such that the following property holds: if we list the divisors of 
in increasing order as 
,![\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]](http://latex.artofproblemsolving.com/a/d/c/adc94af88b3579164aabba51bd021442133e364a.png)
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
be the incircle of a fixed equilateral triangle 
be a variable line that is tangent to 
at points 
,
is chosen such that 
and 
.
be positive real numbers such that ![\(a+b+c=4\sqrt[3]{abc}\)](http://latex.artofproblemsolving.com/2/8/2/28214a1814762b71858db1543fae1a50212eaafa.png)
.![\[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]](http://latex.artofproblemsolving.com/7/0/4/7043a170ac5535465a5a1b4daccb1f1de3043280.png)
Y by Davi-8191, A-Thought-Of-God, HamstPan38825, Hasin_Ahmad, Adventure10, Mango247
Let 
be an acute triangle and let 
be the foot of the 
-bisector. Moreover, let 
be the midpoint of 
. The circle 
with diameter 
meets 
at 
, while the circle 
with diameter 
meets 
at 
. Assume that and lie inside . Prove that 
, , , 
are concyclic.
be an acute triangle and let 
be the foot of the 
-bisector. Moreover, let 
be the midpoint of 
. The circle 
with diameter 
meets 
at 
, while the circle 
with diameter 
meets 
at 
. Assume that and lie inside . Prove that 
, , , 
are concyclic.
be the intersection of 
,
be the second intersection of the two circles, which lies on 
is cyclic.
and 
similar.
.
which is Humpty-Point WRT 
in 
.
be the orthocenter of 
be the perpendicular foot from 
.
.
.
WRT 
.
which gives 
.
.
which finishes the problem. 
,
at 
intersect 
.
,
at 
.
.
.
.
. 
, similarily, 
,
.
.
is cyclic.
![[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.
*/
pair A = (-57.30375,53.83617);
pair B = (-83.34998,-39.17961);
pair C = (43.19882,-39.61374);
pair H = (-57.62315,-39.26787);
pair D = (-31.07334,-39.35895);
pair L = (-70.32686,7.32827);
pair K = (-7.05246,7.11121);
pair P = (-20.20788,-77.96332);
pair X = (-22.03030,7.16259);
pair Y = (61.56501,6.87581);
pair M = (-44.18855,7.23860);
pair I = (-40.28433,-6.63285);
pair E = (-69.06976,-22.25320);
pair F = (-23.36015,-3.92844);
import graph;
size(12cm);
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
real xmin = -130, xmax = 85, ymin = -100, ymax = 85;
draw(A--B, linewidth(0.5));
draw(B--C, linewidth(0.5));
draw(C--A, linewidth(0.5));
draw(circle((-19.96366,-6.77439), 71.18934), linewidth(0.5));
draw(circle(L, 48.29685), linewidth(0.5) + blue);
draw(circle(K, 68.61788), linewidth(0.5) + blue);
draw(circle(P, 74.10196), linewidth(0.5) + linetype("4 4") + red);
draw(B--M, linewidth(0.5) + red);
draw(M--C, linewidth(0.5) + red);
draw(A--P, linewidth(0.5));
draw(H--A, linewidth(0.5));
draw(L--Y, linewidth(0.5));
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$H$", H, dir(270));
dot("$D$", D, SE);
dot("$L$", L, NW);
dot("$K$", K, NE);
dot("$P$", P, dir(270));
dot("$X$", X, NW);
dot("$Y$", Y, NE);
dot("$M$", M, NW);
dot("$I$", I, NW);
dot("$E$", E, NW);
dot("$F$", F, NE);
clip((xmin,ymin)--(xmax,ymin)--(xmax,ymax)--(xmin,ymax)--cycle);
[/asy]](http://latex.artofproblemsolving.com/7/2/a/72a31842d51aa7cfb341cb67a8f51875a6b3aa9d.png)
and 
be the incenter of 
.
and 
respectively such that 
be the midpoints of 
The inversion 
swaps 
.
Firstly, apply a suitable scale such that 
.
and 
.
and 
Now we multiply everything by a scale factor of 
.
,
,
and 
.
.
and 
.
.
,
and we are done. 
.
swaps 
.
are orthogonal. So the inversion 
fixes 
.
and 
.
.
and similarly 
.
and 
are collinear and we are done.
invert at 
.
be 
intersects 
at 
intersects 
at 
is cyclic.
and likewise.
as 
is isoceles trapezoid 
bisector of 
and 
)
is a line, see that 
And done. 
,
is isoceles trapezoid and hence cyclic.
