[/quote]
,
and some point 
on 
intersects the circumcircle of 
at 
and the perpendicular from 
at 
.
does not depend on the choice 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 
coins, all of which show 
on one side and 
on the opposite side. The coins are laid out in a row on the table. Peer and Solveig alternate taking turns. On his turn, Peer may turn over one or more coins, so long as he does not turn over two adjacent coins. On her turn, Solveig picks precisely two adjacent coins and turns them over. When the game begins, all the coins are showing 
for which Solveig can keep Peer from winning.
be an angle in the interval 
.
is irrational, and that 
and ![$ \cos[(k + 1)\theta ]$](http://latex.artofproblemsolving.com/b/3/a/b3a14c57a1a8b5a6585195c394181c5124f8c471.png)
are both rational for some positive integer 
,
.
be three positive reals such that 
.![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
.
-
at 
and 
on lines 
and 
such that quadrilateral 
is a rhombus.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
:
,![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
such that the following inequality holds for all 
:![\[
ab + bc + ca \leq \frac{(a + b + c)^2}{3} + k \cdot \max \{ (a - b)^2, (b - c)^2, (c - a)^2 \} \leq a^2 + b^2 + c^2
\]](http://latex.artofproblemsolving.com/1/c/e/1ce66163068b499383adb4a4ecec86c5f4edaec3.png)
be positive real numbers such that : 
. Prove that : 
where 
are lengths of parallel sides and 
are other two sides.
,
,
,
,
in some order.
board that is entirely covered by dominoes, with no overlap. He jumps along the vertical lines of the board, always going from the midpoint of the vertical segment of a 
square to another midpoint of the vertical segment, according to the rules:
When the domino is horizontal, the cricket jumps to the opposite vertical segment (such as from 
to 
)
When the domino is vertical downwards in relation to its position, the cricket jumps diagonally downwards (such as from 
to 
When the domino is vertically upwards relative to its position, the cricket jumps diagonally upwards (such as from 
)
board.
Prove that
candies. A starts first. On a move, if in the box there are 
so that 
and 
,
Y by FaThEr-SqUiRrEl, Adventure10
In a triangle 
, the external bisector of 
intersects the ray 
at 
. The feet of the perpendiculars from 
and 
to line 
are
and 
, respectively and the foot of the perpendicular from to 
is 
. Show that 
.
, the external bisector of 
intersects the ray 
at 
. The feet of the perpendiculars from 
and 
to line 
are
and 
, respectively and the foot of the perpendicular from to 
is 
. Show that 
.This post has been edited 1 time. Last edited by Amir Hossein, Jun 18, 2018, 8:59 AM
entonces 
(*)
entonces 
es ciclico, luego 
(1)
,
y 
,
.
es ciclico,
entonces 
.
(2)
.
intersect at 
.
.
.
,
.
we have that 
are concyclic. So,
.
are cyclic as well.
means that 
are concyclic. So, 
.
,
meets 
then the ratio 
is harmonic and because 
are the perpendicular projections from 
respectively, then the ratio 
is also harmonic.
and 
then 
bisects 
.
,
is cyclic, hence if 
meets 
at 
,
is cyclic hence 
.
is cyclic, so 
.
is similar to 
then 
and 
.
a
)
intersects 
,
) and properties of bisectors
We can easily see that (from the properties of bisectors, again) 
.
.
.
.
.
.
.
