[/quote]
,
that satisfy the equation 
for all real numbers 
and 
.
be an interior point of triangle 
.
and 
intersect sides 
at points 
and 
,
and 
are parallelograms. Suppose lines 
and 
intersect at a point 
.
and 
of circle 
meets 
and let 
be the reflection of 
.
passes through the midpoint of 
.
be the orthogonal projection of 
is an altitude. Let 
and 
be the orthogonal projections of 
is convex and has the same area as triangle 
grid, each cell contains a card placed face down, with a real number written on the front. Let the number on the card in the 
-
-
,
,
.
,
,
,
.
such that there exists a positive real constant 
satisfying the following: for any positive integers 
and any grid of numbers adhering to the above property, the player can guarantee that by flipping no more than 
cards, they can identify the smallest number in the grid.
of positive integers such that 
for all positive integers 
such that 
.
A sequence of positive integers is 
if there exists a constant 
for all 
.
if 
for all 
with reminder 
or 
modulo 
prove that the number triples 
is odd.
be non-negative real numbers. Setting![\[x = \frac{3(a^2 + b^2 + c^2 + d^2)}{2(ab + bc + ca + da + db + dc)}.\]](http://latex.artofproblemsolving.com/3/c/d/3cd85503ab5de8b5af352ec1c30fcf2cb79f09bd.png)
Prove that![\[\frac{a}{b + c + d} + \frac{b}{c + d+a} + \frac{c}{d+a + b } + \frac{d}{a + b + c} \geqslant \frac43 + \frac15\left(x - \frac1x\right).\]](http://latex.artofproblemsolving.com/4/f/d/4fd1f976f9d50bb35869533cdce8461810064992.png)
(2) Let 
be non-negative real numbers. Prove that![\[\sum_{i=1}^{n} \frac{a_i}{\displaystyle \sum_{j=1}^{n} a_j - a_i} \geqslant \frac{n}{n - 1} + \frac{1}{2n-3}\left[\frac{\displaystyle (n - 1)\sum_{i=1}^{n} a_i^2}{\displaystyle \left(\sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2} - \frac{\displaystyle \left(\sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2}{\displaystyle (n - 1)\sum_{i=1}^{n} a_i^2}\right].\]](http://latex.artofproblemsolving.com/9/7/5/975e5faa294dc6e4debc130b1edf31bb29e27016.png)
.
.
and the length of 
is 
intersects 
.
such that no one is contained in another and no two cover whole 
?
,
,
be positive integers. A lemonade stand owns 
![\[(n - 1) \displaystyle\sum_{i=1}^{n} \frac{a_i^2}{\displaystyle\sum_{j=1}^{n} a_j - a_i}
\geq \sqrt{n \displaystyle\sum_{i=1}^{n} a_i^2}.\]](http://latex.artofproblemsolving.com/0/7/5/07579ac517d06493e5e30509ccc0f7213a2ecdf9.png)
be a 
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
Y by
Let 
be a circle with center 
. Let 
and 
be two fixed points on the circle and let 
be a variable point on . We denote by 
the intersection point of lines 
and 
, assuming 
. Let 
be the circumcircle of triangle 
. Let 
be the second intersection point of with . The tangent to at intersects at 
. The line 
intersects at 
. The perpendicular bisector of segment 
intersects line 
at 
, and line 
intersects at 
. We denote by 
the second intersection point of the circumcircle of triangle 
with . Prove that, as point varies, points and remain fixed.
be a circle with center 
. Let 
and 
be two fixed points on the circle and let 
be a variable point on . We denote by 
the intersection point of lines 
and 
, assuming 
. Let 
be the circumcircle of triangle 
. Let 
be the second intersection point of with . The tangent to at intersects at 
. The line 
intersects at 
. The perpendicular bisector of segment 
intersects line 
at 
, and line 
intersects at 
. We denote by 
the second intersection point of the circumcircle of triangle 
with . Prove that, as point varies, points and remain fixed.
is diameter of 
.
and 
so 
. Hence 
, 
are isosceles triangle with 
then 
.
then 
intersect 
are concyclic
are concyclic. Hence 
are collinear
so the power from 
equal to the power from 
are collinear. Hence 
Let 
be the center of 
Then
Note that the coordinate of 
.
denote the coordinate of 
Now we find the coordinate of 
Since 
Intersecting these gives
Let 
Then if 
,![\begin{align*}
u' &= \frac{p't'(\overline{p'} - \overline{t'})}{\overline{p'}t' - p'\overline{t'}} \\
&= \frac{-\frac{b^8}{ac^3(ac-b^2)}\left[-\frac{ac^2}{b^4} + \frac{ac^2}{b^2(ac-b^2)}\right]}{-\frac{ac}{ac-b^2} - \frac{b^2}{ac-b^2}} \\
&= \frac{\frac{b^6}{c(ac-b^2)^2} - \frac{b^4}{c(ac-b^2)}}{\frac{ac}{ac-b^2} + \frac{b^2}{ac-b^2}} \\
&= \frac{b^6 - b^4(ac-b^2)}{c(ac-b^2)(ac+b^2)} \\
&= \frac{b^4(2b^2-ac)}{c(a^2c^2-b^4)}
\end{align*}](http://latex.artofproblemsolving.com/5/3/9/539c16882c11c15654951d16f56b9f1744563bd0.png)
and then
So the coordinate of 
