AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
. Let 
and 
be the feet of the altitudes from 
to 
and from 
to 
respectively.
and 
as the reflections of across lines 
and , respectively. Let 
be the circumcircle of 
. Denote by 
the second intersection of line 
with , and by 
the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
+1 w
have,where 
and 
are natural numbers.
be the set of the positive real numbers. Find all functions 
such that 
for all positive real numbers 
and 
.
be defined by![\[
a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}
\]](http://latex.artofproblemsolving.com/8/8/f/88f423d87c08b20dc552703fcf30f2a8e9585902.png)
Prove that![\[
a_n^{2025} >n^{2024}
\]](http://latex.artofproblemsolving.com/a/6/8/a6806c817edb7ce29a399981137c1a3e93f63c63.png)
for all positive integers 
.
with 
and the circle 
of radius centered at 
Let 
be the midpoint of 
The line 
intersects a second time at 
Let be a point on such that 
Let 
be the intersection of 
and Prove that 
such that for any set of consecutive integers taken from the given set of integers, there will always be a colour colouring exactly one of them? That is, for all integers 
such that 
, there will be a colour coloring exactly one integer from the set 
.
such that: 
. Find the maximum: 
such that if![\[
1 = d_1 < d_2 < \cdots < d_{k-1} < d_k = n
\]](http://latex.artofproblemsolving.com/a/d/2/ad2cbfe9e2110de2fdb6308d93f58880bd79afbc.png)
are the divisors of , then the sequence![\[
d_2 - d_1,\, d_3 - d_2,\, \ldots,\, d_k - d_{k-1}
\]](http://latex.artofproblemsolving.com/d/2/3/d239210af0c2aa005ee1e5b820678d4aadc366c0.png)
forms a permutation of an arithmetic progression.(Kritesh Dhakal,Nepal)
be a finite set of integers. Prove that there exists a number 
depending on such that for each non-constant polynomial 
with integer coefficients the number of integers 
satisfying 
does not exceed 
.
such that 
and 
for all reals 
.
, 
be positive integers such that 
is a multiple of 
for all positive integers . Prove that 
.
be triangle with altitude 
are concurrent at 
. 
cuts perpendicular bisector of 
at 
. Prove that 
when is circumcenter of .
be midpoint of 
then 
is perpendicular bisector of . Easily seen 
is radical axis of 
and 
. This means 
when 
are midpoints of 
.
with altitude 
are concurrent at 
with note that, 
line of triangle is Euler line of triangle . We are done.
![[asy] unitsize(6cm);
pathpen=black; pointpen=black;
pair Ia = Drawing("I_a", dir(-85));
pair Ib = Drawing("I_b", dir(50), dir(0));
pair Ic= Drawing("I_c", dir(160), dir(180));
draw(Ia--Ib--Ic--cycle);
draw(unitcircle);
pair V = Drawing("V", circumcenter(Ia,Ib,Ic), dir(-110));
pair I = Drawing("I", orthocenter(Ia,Ib,Ic), dir(-100));
pair A = Drawing("A", foot(Ia, Ib, Ic), dir(90));
pair B = Drawing("B", foot(Ib, Ia, Ic), dir(180));
pair C = Drawing("C", foot(Ic, Ib, Ia), dir(45));
draw(A--B--C--cycle);
draw(Ia--A);
draw(Ib--B);
draw(Ib--C);
pair O = Drawing("O", circumcenter(A, B, C));
pair f = foot(Ia, O, I);
draw(I--f);
pair x1 = reflect(I,f)*Ia;
pair c1 = rotate(180,C)*B;
pair X = Drawing("X", intersectionpoint(B--c1,Ia--x1), dir(0));
draw(Ia--X);
draw(B--X);
draw(A--Ia);
pair m1 = foot(X,A,Ia);
draw(X--m1);
pair M = Drawing("M", midpoint(Ia--Ic), dir(-120));
pair N = Drawing("N", midpoint(Ia--Ib), dir(-70));
draw(X--M);
draw(circumcircle(Ia,B,C), dotted);
draw(circumcircle(Ia,M,N), dotted);
draw(Ia--V);
pair P = Drawing("P", midpoint(Ia--I), dir(180));
pair Q = Drawing("Q", midpoint(Ia--V), dir(0));
draw(P--Q);
[/asy]](http://latex.artofproblemsolving.com/9/0/5/905229cc1015a7a1750c538e76fd1fd0dc7478ef.png)
be the Bevan point of 
, or the circumcenter of 
. The Bevan point lies on , which we prove later. Let 
be the midpoints of 
, so that is the perpendicular bisector of 
. Let 
. Note that is the orthic triangle of , and is the orthocenter.
is the radical axis of circles 
and 
, so 
, where 
are the midpoints of 
. Hence 
, so that is the line 
, and 
concur, as required.
are concyclic, maththuz, I have a general problem,
be two isogonal conjugate point on bisector from of triangle . 
cut 
again at 
. Prove that pependicular bisector of 
and the line passing through and perpendicular to 
intersect on with is circumcenter of .
over the bisector to obtain 
, hence 
. Reflect over , hence lies on 
(circumcircle of ). Let the perpendicular bisector of meets the perpendicular at , hence it's clear that 
. Apply inversion center at with radius 
, it takes 
, let 
, hence 
and 
lies on 
since 
are concyclic. Now reflect the line 
over the bisector, as a result 
and lies on and we are done.
are concyclic, maththuz, I have a general problem, be two isogonal conjugate point on bisector from of triangle . cut again at . Prove that pependicular bisector of and the line passing through and perpendicular to intersect on with is circumcenter of . intersects at 
be reflection of in .
is the center of 
, then 
and 
Prove that
are circumcenter, incenter, A-excenter of the triangle 
passes through 
and perpendicular to 
such that 
,
are midpoints od arcs 
that do not contain 
respectively.
such that its projection on chord 
is 
and its projection on chord 
is 
.
and 
are "perpendicular". It is an easy calculation.
are collinear on the Euler line of 
.
,
is the pole of 
,
are collinear, which is obvious by Pascal.