be a triangle with circumcircle 
,
-
,
.
at 
and meets 
.
parallel to 
,
respectively, so that 
meets 
.
is constant.
.
be the feet of the altitudes from 
to 
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
is the circumcenter of 
can be inscribed within a circle.
and 
. Prove that
Where 
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 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 
.
and 
Prove that
Where 
Find the range of values that can be taken by :
are positive reals.
![\[ a = b\cos C + c\cos B. \]](http://latex.artofproblemsolving.com/7/e/d/7ed699f18ede191e5b89bc79d7f0592294347818.png)
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
,
such that
for all reals 
satisfying 
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.
Y by khan.academy, cubres, Mathdreams
Consider a triangle 
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .
(Shining Sun, USA)
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .(Shining Sun, USA)
This post has been edited 6 times. Last edited by Bata325, Apr 13, 2025, 2:54 AM
Reason: italics
Reason: italics
.
.
denote the foot of the altitude from ![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - 2\angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/e/2/d/e2d9375826c14c5e7197102e8c1d14f2dad41068.png)
Hence, the points 
are collinear. Similarly 
are all collinear.
hence 
are the centres of the circles 
respectively
and 
such that 
and 
and 
are the perpendicular bisectors of the radial axes 
and 
hence 
and 
so 
and 
and the required claim is proved
![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - \angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/7/d/9/7d972f3079ea48f9b57a8f171777f9dabada44e0.png)
Hence, the points 
hence 
where ![\[\measuredangle BAQ = \measuredangle CXQ = 90 + \measuredangle BCA\]](http://latex.artofproblemsolving.com/9/4/2/9429375d4f0c28c6a6939e228bc971645d77e8ec.png)
which implies that ![\[\measuredangle CAQ = \measuredangle CXP = 90 + \measuredangle CBA\]](http://latex.artofproblemsolving.com/0/0/a/00ab53c14582f60a8519bd3daef06f3048629879.png)
which implies that 
