[/quote]
,
and 
Prove that
such that 
and 
for all reals 
.
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 
.
,
be positive integers such that 
is a multiple of 
for all positive integers 
.
.
be a positive integer and 
be a family of at most 
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.)
and some point 
on 
intersects the circumcircle of 
at 
and the perpendicular from 
intersects the circumcircle of 
at 
.
does not depend on the choice of 
of parallelogram 
a point 
is chosen. The perpendicular from 
to 
and the perpendicular from 
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.
?
Y by canhhoang30011999, harshmishra, Adventure10, Mango247, GeoKing
The incircle of a non-isosceles triangle 
with the center 
touches the sides 
at 
respectively. The line 
meets the circumcircle of at 
. The line 
meets the line 
at 
and the line 
meets the circumcircle of at 
. Define 
similarly. Prove that the lines 
are concurrent.
with the center 
touches the sides 
at 
respectively. The line 
meets the circumcircle of at 
. The line 
meets the line 
at 
and the line 
meets the circumcircle of at 
. Define 
similarly. Prove that the lines 
are concurrent.
is not harmonic in general. The concurrency point is not the Lemoine point of 
i.e. the perspector of 
be the projections of 
on 
and the arc 
of the circumcircle 
we get 
is cyclic 
Furthermore, 
implies that 
bisects 
externally or 
Since 
then 
therefore isosceles 
are similar with corresponding cevians 
and 
Hence since 
is cyclic on account of the right angles at 
we deduce that 
are collinear and analogously 
and 
By Cevian Nest Theorem the conclusion follows.
is circumcircle of triangle 
.
is center of spiral similar goes 
to 
.
is on the ray 
.
.
is tangent to the circle 
and 
is tangent to the circle 
.
and 
.
we have 
and 
.
.
to get that,
.
and using them in (4), we get that ,
.
are concurrent.
To jayme, yes indeed, my solution shows this directly. Let the perpendicular to 
,
.
are collinear and perpendicular to 
,
are concurrent at 
of 
,
are collinear . ... 
is the external bisector of 
,
we get 
.
are concurrent
) :
be the projection of 
on 
and 
be the midpoint of arc 
.
,
,
is the bisector of 
and 
.
,
is the bisector of 
,
.
,
,
we get 
.
of 
and 
,
from my post, we can do this:
be the radical axis of 
.
be 
and note that 
by harmonic conjugates (fact 1). So, note 
for obvious reasons, and because of the fact 1 we conclude 
meets 
contains the foot of the 
altitude of 
;
concurrent.
.
,
we get that 
is harmonic. This yields 
,
is the bisector of 
,
.
which is equivalent to the fact that 
are concurrent.
be a circle tangent to 
and tangent to 
at some point, say 
.
similarly. We claim that 
etc.
denote the midpoint of arc 
denotes the homothety centered at 
are collinear. Since 
is a diameter of 
.
is a harmonic pencil. But 
and hence 
is the bisector of 
,
.
gives 
are collinear, so 
.
etc. are concurrent at the 
lie on same line 
.
images of points 
wrt this projective transformation. So we have that 
,
,
.
,
,
concurrent at point 
,
,
concurrent at point which is isogonal to 
.
and hence 
so we're done.
denote lengths 
,
be the semiperimeter of 
denote the circumcenter and circumradius of 
,
,
is the polar of 
,
or
so 
,
or
Now, by the Incenter-Excenter Lemma, we know that 
.
(tangents), we must have that 
.
be the point at infinity along 
or
and symmetry implies that 
and 
are also satisfied.
Which implies the concurrence by the trigonometric form of Ceva's Theorem, AWD.
are the midpoints of 
respectively and let 
be the midpoints of 
respectively. Then 
.
.
concur at 
from this problem 
is the orthic triangle of 
,
.
and 
.
intersects the circle again at 
,
intersects at 
Thus, 
is an angle bisector, and 
.
,
![[asy]
import graph; size(15cm);
real a=105;
real b=205;
real c=335;
real r=10;
path omega=circle(origin, r);
pair A=r*dir(a);
pair B=r*dir(b);
pair C=r*dir(c);
pair I=incenter(A,B,C);
pair A1=foot(I,B,C);
pair B1=foot(I,C,A);
pair C1=foot(I,B,A);
dot(A1);
dot(B1);
dot(C1);
draw(A--B--C--cycle);
draw(omega,blue);
pair X=10*C1-9*B1;
pair P=10*B-9*C;
pair A3=intersectionpoints(B1--X,B--P)[0];
pair A2=r*dir((b+c)/2);
pair B2=r*dir((a+c)/2+180);
pair C2=r*dir((a+b)/2);
dot(A2);
dot(B2);
dot(C2);
pair A4=intersectionpoints(A2--A3,omega)[1];
draw(A2--A3);
dot(A4);
draw(B1--A3);
draw(A3--A2, magenta);
pair D=(B+C)/2;
dot(D);
draw(B--A3);
draw(A1--A4--A2--D--cycle,red+ linewidth(1));
draw(A--A4,green);
label("$A$",A,dir(90));
label("$B$",B,dir(-135));
label("$C$",C,dir(-45));
label("$D$",D,dir(-45));
label("$A_1$",A1,dir(90));
label("$A_2$",A2,dir(-90));
label("$A_3$",A3,dir(-90));
label("$A_4$",A4,dir(-90));
label("$B_1$",B1,dir(45));
label("$C_1$",C1,dir(-45));
draw(B--A4--C,dotted);
label("$B_2$",B2,dir(45));
label("$C_2$",C2,dir(135));
[/asy]](http://latex.artofproblemsolving.com/b/b/8/bb8a97e3f97d77c934ebfc5768d27d52bb75409d.png)
,
(
,
is cyclic. Because 
,
.
bisects 
.
(
by Ceva's theorem.
, 
and 
.
are collinear.
.
we get that 
is tangent to both 
and 
.
.
.
.
are concurrent by the converse of Ceva's Theorem. 
.
and 
so 
are collinear.
be the feet of the 
-
-
.
denote the orthocenter of 
are concurrent, which finishes. 
.
.
.
,![\[ -1=(BC;A_3A_1)\stackrel{A_2}{=}(BC;A_4X)\implies \frac{BX}{CX}=\frac{BA_4}{CA_4} \]](http://latex.artofproblemsolving.com/5/6/b/56b860053dc17eaece66d7e99a8c402033d04ab9.png)
Now we have ![\[ \frac{\sin \angle BAA_4}{\sin \angle CAA_4}=\frac{\sin \angle BCA_4}{\sin \angle CBA_4}=\frac{BA_4}{CA_4}=\frac{BX}{CX}=\frac{BA_1}{CA_1} \]](http://latex.artofproblemsolving.com/1/a/1/1a1336e3306e21f7c50b024e18ec02d1e8c07255.png)
Now by sine ceva we are done.
.
are concyclic which means 
.
are collinear where 
is a midpoint of arc 
and 
are collinear where 
is a midpoint of arc 
and 
is a midpoint of arc 
and 
are concurrent, as desired.
,
similarly. Now ![\[ (A_3, A_1; B, C) \overset{A_2}{=} (A_4, A_5; B, C). \]](http://latex.artofproblemsolving.com/d/e/4/de4b0d3bbe49d49fcdf737631c1348156375aae7.png)
Thus, if 
is the foot of the altitude from 
,
collinear (this is well-known), and Cevian Nest finishes.
.
are collinear. By spiral similarity, we know that 
maps to 
,
and 
are collinear. Since 
are concurrenct by Ceva, harmonic lemmas give 
harmonic, projecting over 
harmonic, projecting through 
harmonic, projecting through 
harmonic, so 
are collinear with 
.
,
