[/quote]
,
be a positive integer and 
be a family of at most 
distinct subsets of the set 
with the following property: we can consider 
and draw segments connecting these points such that points 
and 
are connected if and only if 
belongs to 
Determine the maximal value that the sum of the cardinalities of the subsets in 
can take.
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
such that the equation 
has an integer solution 
.
such tha
(Adit Aggarwal, India.)
such that 
and 
for all reals 
.
and some point 
on 
.
at 
intersects the circumcircle of 
at 
.
does not depend on the choice of 
of parallelogram 
a point 
is chosen. The perpendicular from 
to 
to 
intersect at 
is the midpoint of 
.
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the 
cars from the 
be a convex hexagon, such that the triangles 
are equilateral and the diagonals 
and 
are concurrent. Prove that 
or 
is prime.
such that 
and 
is squarefree. Prove that for any almost squarefree positive integer 
is an integer.
?
such that![\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]](http://latex.artofproblemsolving.com/d/b/1/db14b64aa71cb3c8dbf53707f719f5d9a43bd7e4.png)
for any positive real numbers 
and for any non-empty subset 
she computes the sum ![\[s_A=\sum_{k
\in A}a_k.\]](http://latex.artofproblemsolving.com/7/8/a/78afe6d1036a006b2252023d8f9619285fe0c6d5.png)
She orders these sums 
Prove that there exists a subset 
with 
elements such that, regardless of the integers 
with 
Y by Davi-8191, anantmudgal09, mijail, Adventure10, Mango247
The incircle of triangle 
touches sides 
and 
at points 
and 
, respectively. Let 
be the intersection of lines 
and 
. The reflections of with respect to 
and 
are 
and 
, respectively. Prove that lines 
and 
are concurrent at a point on line 
, where 
and 
are the incenter and circumcenter of triangle .
touches sides 
and 
at points 
and 
, respectively. Let 
be the intersection of lines 
and 
. The reflections of with respect to 
and 
are 
and 
, respectively. Prove that lines 
and 
are concurrent at a point on line 
, where 
and 
are the incenter and circumcenter of triangle .
are the altitudes of 
.
on 
on 
.
meet the circumcircle again at 
are collinear. Let 
,
and 
is the angle bisector of 
,
lies on 
.
and 
concur is equivalent to show that 
concur. This can be done by trig Ceva. Easy to see that:
and so on …
,
indeed concur at a point, say 
are collinear. Notice that 
,
,
and 
,
and so 
,
,
goes through 
.
are collinear, once again by simple Trig-Ceva on triangle 
,
are collinear, where 
,
,
concur on 
bisects 
hence 
is a line 
similarly, then 
and 
is the ninepoint centre of 
then 
are collinear, and since 
are corresponding we are done.
and 
then the lines 
and 
![\[ux+vy+wz = 0\]](http://latex.artofproblemsolving.com/1/3/d/13d72685675c022d430b098e72dfca13bdee0ebf.png)
will be denoted as ![\[(u \, ; \, v \, ; \, w ).\]](http://latex.artofproblemsolving.com/4/5/a/45acf17235acf70c6ee672359d2cb1216e126b9f.png)
concur at the Gergonne point 
of 
.![\[(\frac{b-c}{a}(-a+b+c) \, ; \, \frac{c-a}{b}(a-b+c) \, ; \, \frac{a-b}{c} (a+b-c))\]](http://latex.artofproblemsolving.com/7/9/2/792112793d0f638561e1cf0c56aa9670e69a0cd4.png)
.![\[\frac{\sin FAX}{\sin XAE} = \frac{FX \cdot \frac{\sin AFX}{AX}}{EX \cdot \frac{\sin AEX}{AX}} = \frac{FG}{EG} \cdot \frac{\sin AFX}{\sin AEX}.\]](http://latex.artofproblemsolving.com/7/e/c/7ecdd62a6a058ccf7c40460b34baaeb86e7f9d4d.png)
A bit of angle chasing yields ![\[\measuredangle AFX = \measuredangle AFE + \measuredangle EFX = \measuredangle FEA + \measuredangle CFE = \measuredangle FEC + \measuredangle CFE = \measuredangle FCE \]](http://latex.artofproblemsolving.com/8/3/0/830687413e6f79491e120388a9093a0245b9c521.png)
(in directed angles) and thus ![\[\frac{\sin FAX}{\sin XAE} = \frac{FG}{EG} \cdot \frac{\sin GCE}{\sin FBG} = \frac{(\frac{\sin GCE}{EG})}{(\frac{\sin FBG}{FG})} = \frac{(\frac{\sin EGC}{CE})}{(\frac{\sin BGF}{BF})} = \frac{BF}{CE} = \frac{a-b+c}{a+b-c}.\]](http://latex.artofproblemsolving.com/5/d/9/5d9bf2adea8a91760404895f587b2dce03aedb86.png)
It follows that ![\[(x : \frac{b}{a-b+c} : \frac{c}{a+b-c}).\]](http://latex.artofproblemsolving.com/0/f/f/0ffeea6e376b3fe167409df1024bc3c80e3caf25.png)
Thus 
concur at the point ![\[Q = (\frac{a}{-a+b+c}:\frac{b}{a-b+c} : \frac{c}{a+b-c}) \]](http://latex.artofproblemsolving.com/d/6/d/d6d2df5beba30d5b02e515cb869d4b405dd6ebf8.png)
which is on line 
.
analogously. First we prove the
concur at a point 
Now 
so we compute 
and similarly we have 
.
It is now clear that the claim must hold. As an immediate corollary, we see 
.
be the isogonal conjugate of 
is a line. Note that 
.
Now we undertake the arduous task of plugging in 
Compute the RHS as 
On the other hand, compute LHS as 
Consequently, it suffices to check 
which is clearly true. 
bisects 
?
from 
you can see.
denote the foot of altitude from 
similarly. We will prove that 
,
,
,
,
bisects 
.
.
,
concur at the homothetic centre of the two triangles, say 
.
are concurrent on the Isogonal Mittenpunkt Point 
of 
which lies on 
.
is the foot of the altitude from 
.
meets 
,
intersect at a point 
and moreover 
,
and 
are reflections about 
is collinear. Now we know that 
to the nine-point circle of 
at 
respectively.
Let 
be the reflection of 
across 
and let 
be the foot of the altitude from 
are collinear.
and let 
so 
collinear. Furthermore, 
collinear and 
to 
It is easy to see that this homothety sends 
so we are done. 
are the feet of the altitudes from 
to 
respectively. Now the concurrency follows from the Cevian Nest Theorem, so let 
so 
Similarily, 
and 
so triangles 
and 
the circumcenter of 
is the nine-point center of 
as desired.
its circumcircle and 
the altitude from 
to 
its reflection WRT 
are collinear.
.
are collinear, AKA the 
since 
bissects 
,
,
cuts 
such that 
.
similarly. An easy angle chasing gives us that 
are homothetic 
are concurrent (as well as 
)
is the ortocenter of 
are collinear. This implies that 
be the orthic triangle of 
,
.
,
,
.
,
,![[asy]
size(12cm);
pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair G = extension(B, E, C, F); pair X = -G+2*foot(G, E, F); pair Y = -G+2*foot(G, F, D); pair Z = -G+2*foot(G, D, E); pair J = foot(D, E, F); pair K = foot(E, F, D); pair L = foot(F, D, E);
filldraw(A--B--C--cycle, invisible, blue); filldraw(incircle(A, B, C), invisible, blue); filldraw(D--E--F--cycle, invisible, red); draw(D--J, red); draw(E--K, red); draw(F--L, red); draw(A--D, blue); draw(B--E, blue); draw(C--F, blue); draw(A--J, deepgreen); draw(B--K, deepgreen); draw(C--L, deepgreen); draw(G--X, dashed+orange); draw(G--Y, dashed+orange); draw(G--Z, dashed+orange);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(280)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$J$", J, dir(120)); dot("$K$", K, dir(K)); dot("$L$", L, dir(315));
/* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 130 B = dir 210 C = dir 330 I = incenter A B C D = foot I B C E = foot I C A F = foot I A B G 280 = extension B E C F X = -G+2*foot G E F Y = -G+2*foot G F D Z = -G+2*foot G D E J 120 = foot D E F K = foot E F D L 315 = foot F D E A--B--C--cycle / 0.1 lightcyan / blue incircle A B C / 0.1 lightcyan / blue D--E--F--cycle / 0.1 yellow / red D--J / red E--K / red F--L / red A--D / blue B--E / blue C--F / blue A--J / deepgreen B--K / deepgreen C--L / deepgreen G--X / dashed orange G--Y / dashed orange G--Z / dashed orange */
[/asy]](http://latex.artofproblemsolving.com/5/2/0/520d0adf0a8d8ce018f9cbb4065037c73a3728bb.png)
![\[ \measuredangle JKF = \measuredangle JKD = \measuredangle JED = \measuredangle FED = \measuredangle BFD = \measuredangle BFK \implies \overline{BA} \parallel \overline{JK}. \]](http://latex.artofproblemsolving.com/0/7/d/07de3eafd6c9a7c251c2eac88c026538e55a100f.png)
Now the incenters of 
and 
are the reflections of 
and 
be the tangential triangle. Then 
,
and 
concur on the Euler line.
a
and 
the reflection of the symmedian point of triangle 
be the reflection of 
the intersection of line 
with the circumcircle of triangle 
with line ![\[-1=(AC;BK_b) \overset{B}{=}(AL;TK)\overset{D}{=}(\infty,M;T,DK \cap MT)\]](http://latex.artofproblemsolving.com/f/5/c/f5c39904bb9125bec7152598d8859356c21cde44.png)
But this implies that 
and 
which proves the claim.
and 
