[/quote]
,
.
and 
lie on 
and 
.
meet 
meet 
and 
have an intersection lie on line 
.
be a positive integer. Alex plays on a row of 9 squares as follows. Initially, all squares are empty. In each turn, Alex must perform exactly one of the following moves:
Choose a number of the form 
,
a non-negative integer, and place it in an empty square.
Choose two (not necessarily consecutive) squares containing the same number, say 
and erase the number in the other square.
,
and![\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/c/8/9/c895ae7fdf8d7f284ac9fc94cc077d6edad6cbf0.png)
be a convex hexagon with 
and 
.
which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
,
.
be a point in the plane, distinct from the vertices of 
.
the reflections of 
and 
,
are collinear if and only if 
coincide, prove that 
given that there exist positive integers 
satisfying
and 
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be an odd prime number. How many 
of 
are there, the sum of whose elements is divisible by 
be real numbers satisfying
be a fixed square and 
.
is constructed such that 
and 
.
be the intersection point of lines 
and 
.
board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place 
.
.
denote the number of positive integers less than 
has at least 
solutions in 
that can be written in the form
where 
are pairwise coprime positive integers such that 
,
,
.
Y by KevinYang2.71, EthanWYX2009, ehuseyinyigit, GeoKing, Rounak_iitr, AlexCenteno2007, MathLuis, ohiorizzler1434, ItsBesi, buratinogigle, menpo, Jt.-., KhaiMathAddict, GreenTea2593, cubres, Osim_09
Let 
be a triangle with 
. Let the incenter and incircle of triangle be 
and 
, respectively. Let 
be the point on line 
different from 
such that the line through parallel to 
is tangent to . Similarly, let 
be the point on line different from 
such that the line through parallel to 
is tangent to . Let 
intersect the circumcircle of triangle at 
. Let 
and 
be the midpoints of and , respectively.
Prove that
.
Proposed by Dominik Burek, Poland
be a triangle with 
. Let the incenter and incircle of triangle be 
and 
, respectively. Let 
be the point on line 
different from 
such that the line through parallel to 
is tangent to . Similarly, let 
be the point on line different from 
such that the line through parallel to 
is tangent to . Let 
intersect the circumcircle of triangle at 
. Let 
and 
be the midpoints of and , respectively.Prove that
.Proposed by Dominik Burek, Poland
This post has been edited 4 times. Last edited by sarjinius, Jul 17, 2024, 4:20 PM
,
and 
,
.![\[ \measuredangle ZXB = \measuredangle ZXC = \measuredangle ACX = \measuredangle ACB = \measuredangle APB = \measuredangle ZPB, \]](http://latex.artofproblemsolving.com/2/3/9/2399de8c84c8ca94ce44bd633b239a88b17e0586.png)
so 
cyclic, similarly 
cyclic. Now note that 
by homothety (note the parallelogram) so![\[ \measuredangle BZC + \measuredangle XPY = \measuredangle BZP + \measuredangle PZC + \measuredangle XPY = \measuredangle BXP + \measuredangle PYC + \measuredangle XPY = \measuredangle YXP + \measuredangle PYX + \measuredangle XPY = 0. \]](http://latex.artofproblemsolving.com/d/a/d/dada29dd5233512d53650a503b3d28e94ec2eb13.png)
be the tangency point of the incircle to 
and 
be the tangency point of 
to the incircle (different from 
)
,
are collinear and this 
.
.
be the reflection of 
across 
gives 
.
and hence 
and 
are collinear.
,
is cyclic. Similarly, quadrilateral 
is cyclic as well.
,
cyclic. If we draw in the tangent line 
to the circumcircle at 
.
(and similarly, 
)
.
meet 
.
,
is cyclic. Thus, the claim is equivalent to 
are collinear. To do this, we first note that
so we have
and by projection from 
are collinear, as desired.
,
and 
are tangent to the incircle.
is the image of 
under a homothety of ratio 
![[asy]
size(11cm); pair A = dir(105); pair B = dir(200); pair C = dir(340); pair P = dir(270); filldraw(unitcircle, invisible, blue); pair I = incenter(A, B, C); filldraw(incircle(A, B, C), invisible, blue); draw(A--B--C--cycle, blue); pair E = foot(I, C, A); pair F = foot(I, A, B); pair T = 2*I-A; pair U = 2*I-E; pair V = 2*I-F; draw(U--T--V, deepgreen); draw(B--T--C, red); pair K = midpoint(A--B); pair L = midpoint(A--C); draw(K--I--L, red+dashed); pair X = extension(U, T, B, C); pair Y = extension(V, T, B, C); draw(A--P, grey); draw(circumcircle(B, X, P), grey+dashed); draw(circumcircle(C, Y, P), grey+dashed);
dot("$A$", A, dir(A)); dot("$B$", B, dir(160)); dot("$C$", C, dir(20)); dot("$P$", P, dir(45)); dot("$I$", I, dir(250)); dot("$T$", T, dir(225)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(310));
/* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(11cm); A = dir 105 B 160 = dir 200 C 20 = dir 340 P 45 = dir 270 unitcircle / 0.1 lightcyan / blue I 250 = incenter A B C incircle A B C / 0.1 lightcyan / blue A--B--C--cycle / blue E := foot I C A F := foot I A B T 225 = 2*I-A U := 2*I-E V := 2*I-F U--T--V / deepgreen B--T--C / red K = midpoint A--B L = midpoint A--C K--I--L / red dashed X = extension U T B C Y 310 = extension V T B C A--P / grey circumcircle B X P / grey dashed circumcircle C Y P / grey dashed */
[/asy]](http://latex.artofproblemsolving.com/5/5/b/55b83e43c6c4150178f22b5a434180bc131228d4.png)
and 
are cyclic.
and similarly. (Some people call this Reim's theorem.) ![\[ \measuredangle CTB = \measuredangle CTP + \measuredangle PTB = \measuredangle CYP + \measuredangle PXB = \measuredangle XYP + \measuredangle XYP = \measuredangle XPY \]](http://latex.artofproblemsolving.com/4/9/d/49de0c710fdebebd153a37b6dd4e55781dac14b4.png)
as desired. (The length conditions 
ensure that 
as 
translates to the undirected 
.
be the intouch triangle and use complex numbers with 
etc., so 
and 
.
-
and 
.
since 
.
.
is real. We compute these expressions separately.
Hence the desired expression equals 
,
be the intersection of the tangent with 
.
we have 
,
.
unique hence 
the symmetric of 
.
(the south pole) belong to the bisector of 
are concyclic (as well as 
)
and 
.
.
,
it is well known in config that 
concur at 
meet at 
are homothetic with ratio 
at 
so 
are cyclic, and now by homothety 
,
and 
.
and 
are similar.
bisects the angle between both tangents from 
.
,
.
and 
.
,
.
,
,
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/b/0/6/b06cd9ec213142dc35196c237776ad3ffdb8b3f8.png)
are cyclic.
We have 
and 
Also, since 
are the midpoints of 
Therefore we get, 
we get, 
also 
Also we get, 
Therefore we have to prove that 
be 
and the line parallel to 
be 
You can easily prove that point 
lies on 
by converse of midpoint theorem we get, 
to a point on the line 
Now, since 
where 
and 
are the inradius. From this we easily say that 
Therefore 
are concyclic therefore 
are concyclic. By similar logic we say that 
are cyclic.
be 
.
,
.
,
,
,
,
,
.
by the angle bisector theorem and a 
homothety at 
.
,
,
.
lies on 
,
,
.
,
is the inradius of 
is the length of the altitude from 
and 
are both twice the area of 
.
,
,
,
is just image of 
are cyclic quadrilaterals.
and 
are both subtended by the same arc 
.![\[
\Rightarrow \angle CAP = \angle CBP
\]](http://latex.artofproblemsolving.com/b/1/6/b1630dde6b33cece7076a1dd2b0d038e386b51e9.png)
![\[
\angle CAI = \angle XA'A \quad \text{[Alternate angle]}
\]](http://latex.artofproblemsolving.com/2/9/6/29646e4a410496e50abf0f1fc0e833221852d618.png)
![\[
\angle PA'X + \angle XA'A = 180^\circ
\]](http://latex.artofproblemsolving.com/0/4/9/049c8cc40790385900d18970365f20b7dbedc800.png)
![\[
\angle PA'X + \angle PBX = 180^\circ
\]](http://latex.artofproblemsolving.com/7/4/7/74792bc8bd147f63d3379e881a9b6d867fd991bd.png)
:![\[
\Rightarrow BXA'P \text{ is a cyclic quadrilateral.}
\]](http://latex.artofproblemsolving.com/9/a/a/9aa994f554d33581c780bf5f8c8dce85eddd765c.png)
![\[
\angle A'BX = \angle A'PX \quad \text{[angle subtended by same arc]}
\]](http://latex.artofproblemsolving.com/1/8/a/18a3e161ca4b7287b7f432dbad4fae01409708db.png)
Similarly,![\[
\angle YCA' = \angle YPA'
\]](http://latex.artofproblemsolving.com/a/a/8/aa87fbd8d23670fb87d51fb92fbd5878954e1ba9.png)
![\[
\Rightarrow \angle YCA' + \angle XBA' = \angle XA'P + \angle YPA' = \angle XPY
\]](http://latex.artofproblemsolving.com/c/c/6/cc65af44783f66b19927ca7b19465443b521e396.png)
![\[
\angle BA'C + \angle XCA + \angle XBA' = 180^\circ
\]](http://latex.artofproblemsolving.com/9/2/e/92e04ead6085da858c1e467eec640d4b5fe0e474.png)
![\[
\boxed{\Rightarrow \angle KIL + \angle XPY = 180^\circ}
\]](http://latex.artofproblemsolving.com/b/f/2/bf221f613fc3c203d24320a688b54153a41c5218.png)
and 
is both tangent quadrilateral and paralellogram, 
.
and 
are cyclic quadrilaterals.
are antiparallel and 
and 
.
as desired.
.
with points 
and 
at 
.
is tangent to 
and 
form circumcircles, which means that 
and 
.
Finally, notice that 
and 
and 
respectively. In other words, 
.
:
be 
.
be the semi perimeter, 
be the area, 
be the 
altitudes o
meet 
concur.
then 
is a parallelogram. But 
is an inscribed circle of a parallelogram, hence 
bisects 
and consequently, 
.
and that 
)
and 
.
and we are done. 
and 
and so on.
and 
notice that by rotating 
)
and similarly 
by evaluating the determinat gives us that 
which is trivial..
and 
.
are tangent to 
.
is a rhombus since it is a parallelogram and has an inscribed circle, so ![\[ \angle BPT = \angle BCA = \angle BXM = 180^\circ - \angle BXT \]](http://latex.artofproblemsolving.com/c/2/4/c24fbc5de9064e9d942dd063ae49bb19faa45bd8.png)
so 
is cyclic. Similarly, 
is cyclic too so![\[ \angle XPY = \angle XPT + \angle YPT = \angle BXT + \angle YCT = 180^\circ - \angle BTC \]](http://latex.artofproblemsolving.com/6/9/7/697b96aa3ff2d33c9c69274a4570e3e3f6613d1d.png)
as needed.
