[/quote]
,
,
and 
are both integers.
and 
Prove that
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
,
denote the greatest prime factor of 
.
and 
such that there is no integer 
.
?
Prove that
be a 
-
.
,
in 
such that the sets ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
be a triangle. Let 
be the feet of the altitudes from 
respectively. Let 
be the projections of 
.
.
and 
Prove that
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
and let 
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
Y by A-Thought-Of-God, samrocksnature, Ya_pank, ohhh
Let 
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
Proposed by Mahavir Gandhi
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.Proposed by Mahavir Gandhi
This post has been edited 4 times. Last edited by Rg230403, May 13, 2021, 11:41 AM
is the incircle of 
are the intouch points. Let 
be the foot of perpendicular from D onto EF. 
are defined analogously.
.
and define 
analogously. Let the tangent to 
meet AB, AC at 
and 
respectively. Now by Newton's Theorem on quadrilateral 
,
= 
and 
and 
= 
.
at 
and 
, 
and let the point of tangency between the A-Mixtilinear incircle and 
.
passes through U.
= 
athrough 
,
)
)
.
.
, 
are
.
=
.
= 
= 
.
is cyclic as required.
= 
.
and 
,
and let 
= 
.
and 
are tangent to 
and 
by Miquel's Theorem, 
.
lies on 
= 
.
.
.
is cyclic. So YZ is antiparallel to BC wrt 
.
.
.
.
.
and 
.
is similar to 
.
.
is similar to 
.
concur at T.
denote the incircle, 
the circle centered at 
,
the circumcircle; 
the respective intouch points opposite to 
respectively, and 
the midpoint of arc 
.![\[\measuredangle DIM_A = \measuredangle AM_AO = \measuredangle OAM_A = \measuredangle AIA'.\]](http://latex.artofproblemsolving.com/7/0/6/7069ff7ae7abe5dee411c0940a45e696deef1636.png)
As both 
are perpendicular to 
are perpendicular to 
.
and 
.
is tangent to 
,
and 
(as 
,
.
(as 
)
to 
with 
meets 
.
denote the 
be the intouch points of 
respectively.
,
intersect 
,
is tangent to both 
,
is the radical axis of 
)
and we get the desired
,
-
and 
to be the respective mixtilinear incircles.
,
.
and 
.
concur at 
are isogonals, where 
is the 
,
.
to 
,
.
goes through the exsimillicenter of the incircle and circumcircle.
be the incircle.
.
are equal. For this, observe that 
,
which is equivalent to the two inradii being equal.
mixtilinear incircle and the claim follows.
.
,
.
and 
be the center of 
since both lie on its perpendicular bisector. Thus, we also have 
.
on 
but this point is 
to 
and 
to ![[asy]
size(13cm);
defaultpen(fontsize(11pt));
import olympiad;
pair A = dir(116), B = dir(195), C = dir(-15), O = (0,0);
pair I = incenter(A,B,C);
pair D = foot(I,C,B), E = foot(I, A, C), F = foot(I, A, B);
pair Ma = 2*foot(O,A,I) - A;
path w = circumcircle(I, 2*foot(I,A,B)-I, 2*foot(I,A,C)-I);
pair[] P12 = intersectionpoints(w, circumcircle(A,B,C));
pair A1 = P12[0], A2 = P12[1];
w = circumcircle(I, 2*foot(I,C,B)-I, 2*foot(I,A,C)-I);
P12 = intersectionpoints(w, circumcircle(A,B,C));
pair C1 = P12[0], C2 = P12[1];
w = circumcircle(I, 2*foot(I,A,B)-I, 2*foot(I,B,C)-I);
P12 = intersectionpoints(w, circumcircle(A,B,C));
pair B1 = P12[0], B2 = P12[1];
w = incircle(A,B,C);
draw(A--B--C--cycle, royalblue);
pair A0 = foot(I, A1, A2), B0 = foot(I, B1, B2), C0 = foot(I, C1, C2);
pair X = extension(A, A0, B, B0);
pair Mb = 2*foot(O, I, B) - B, Mc = 2*foot(O, I, C) - C;
draw(circumcircle(A,B,C), royalblue);
draw(w, fuchsia);
draw(A1--Ma--A2, fuchsia);
draw(A1--A2, orange);
draw(B1--B2, orange + dashdotted);
draw(C1--C2, orange + dashdotted);
draw(A0--X--B0, brown);
draw(X--C0, brown);
draw(A--A0, brown + longdashed);
draw(C--C0, brown + longdashed);
draw(B--B0, brown + longdashed);
draw(O--I, black+dashed);
draw(X--I, black);
dot("$A$", A, dir(A));
dot("$A_1$", A1, dir(A1));
dot("$A_2$", A2, dir(A2));
dot("$B_1$", B1, dir(B1));
dot("$B_2$", B2, dir(B2));
dot("$C_1$", C1, dir(C1));
dot("$C_2$", C2, dir(C2));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M_A$", Ma, dir(Ma));
dot("$O$", O, dir(Ma));
dot("$I$", I, dir(Ma));
dot("$A'$", A0, dir(A0));
dot("$B'$", B0, dir(B0));
dot("$C'$", C0, dir(C0));
dot("$X$", X, dir(X));
[/asy]](http://latex.artofproblemsolving.com/3/e/b/3eb986c295c2f63c7edf08e04da8e1fc8a6695d2.png)
) .
,
similarly. The key claim of the problem is that 
.
are concyclic, we deduce that 
and 
and similarly.
.
and similarly 
.
so we have 
.
is isoceles and 
,
.
,
and 
are isoceles with two sides parallel, so 
,
not containing 
be points on 
such that 
and 
are tangent to incircle of 
is the angle bisector of 
.
is tangent to incircle of 
.
.
and the result follows from fact 5 
be the tangency points of incircle with 
respectively.
.
not containing 
be the 
but this would clearly imply 
are isogonal w.r.t 
and hence done 
similiarly for others.
and also 
so 
at 
.
.
,
is tangent to 
and so 
.
,
and so 
and so 
and now its easy enough to prove by angel chasing that 
.
are already tangent to the incircle, it must be the excircle. So, 
is perpendicular to 
,
.
is tangent to the incircle of 
,
is some permutation of 
,
and 
be the intouch triangle and circumcevian triangle of 
,
,
and 
at 
,![\[\measuredangle AYX=90^\circ-\measuredangle OAC=\measuredangle CBA\]](http://latex.artofproblemsolving.com/6/4/a/64a0305b41bc722fd9c7977660e83c0ee98322b6.png)
and thus quadrilateral 
is bicentric. It is then well-known that 
.
,
.
must also be homothetic as well. Therefore, 
and 
are concurrent at the homothetic center of two triangles, which lies on 
.
are constant and 
s.t. 
is the mixti touch point then 
concurr
be the arc midpoint of 
has both the same circumcircle and incircle as 
is the bisector of 
,
is 
is symmetric in 
.
is parallel to 
,
,
M
takes 
as 
,
and we get that the desired concurrence point is the exsimillicenter of 
become collinear since 
are concurrent and we are done 
concurrency and then you notice your configs and win.
