[/quote]
,
be positive reals. Prove that![\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]](http://latex.artofproblemsolving.com/6/9/3/6937791380a8642edca3e773dc72dbf3f0e0f206.png)
be a perfect square then, is it true that it must be of the form
be points on the sides 
respectively of a triangle 
.
and 
meet at 
and 
,
and 
,
,
.
to the area of 
.
and 
intersect at different points 
and 
.
.
be a point on 
line intersects 
again, the 
line intersects again to 
and the line 
intersects again to the circumference 
.
line passes through the midpoint of the 
segment.
,![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
,such that 
P
,
are such that 
and 
and 
,
and 
,
is a right angle. Prove that the points 
are collinear.
.![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.
and 
intersects at point 
.
be a point on ray 
.
intersects line 
at 
![\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]](http://latex.artofproblemsolving.com/4/3/1/431b0ed375e037ee3d19e48a2bb4099f12f87784.png)
Proposed by David Stoner
such that![\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad \forall x,y \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/b/0/f/b0f1a1d3b9de5d23f1b5a2e9be3b88d8bd575883.png)
is not divisible by 
for any positive integer 
.
,
formed by using the digits 
once each such that the number 
is divisible by 
for 
.
Y by Amir Hossein, Davi-8191, rustam, Kayak, Tenee, Purple_Planet, centslordm, jhu08, megarnie, Adventure10, Mango247, lian_the_noob12, Sedro
A rectangle 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.
Proposed by Jeck Lim, Singapore
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.Proposed by Jeck Lim, Singapore
This post has been edited 2 times. Last edited by MarkBcc168, Jul 15, 2018, 12:57 PM
.
board. Color 
and 
with 
,
,
.
.
.
,
.
is even and so 
is odd. Repeat in the other direction (up and down instead of left and right).
whose sides are of odd length.
,
tile it with dominoes and for 
mark the leftmost corner and tile the remaining rectangle with dominoes. Observe that each tiling 
maps to a tiling 
consisting only of dominoes and boxes such that 
white if 
,
squares have at most 2 corners.
,
be the side of one other small rectangle as well?
are both even or odd; we also know for sure that the distance from the horizontal sides of this rectangle to 
![$[0,a]\times [0,b]$](http://latex.artofproblemsolving.com/6/0/1/601076a1c0a357d6bd2c71085669c8c3074b0129.png)
and we are considering ![$[x,x+c]\times [y,y+d]$](http://latex.artofproblemsolving.com/3/6/b/36b9f1541a35e6b750fd01d3ff4dbb3765c805c2.png)
,
,
are odd and 
is possible. Let us just focus on the first half. We can note that by weighting each cell by 
,
must be odd, so there must be some rectangle with 
odd.
.
.
,
black corners. Iff the rectangle's distances from the four sides are all odd or all even, then it has 
white and 
,
in the dissection has the following property: the distance from 
,
to be the number of black squares contained in it, and 
to be the number of white squares in it. Then the condition on our required rectangle 
.
for every rectangle. Then 
where 
refers to the rectangle covering the entire of 
refers to the partition. But this is impossible due to 
,
Such a small rectangle would have odd area and thus odd side lengths, meaning the distances from each pair of parallel sides have the same parity (i.e. the distance from the top and bottom would both be odd or both be even)
,
,
and 
rectangles. Suppose such a rectangle doesn't exists. Then, for any rectangle 
which is contained in some division of 
If a smaller rectangle 
and 
or in a 
and 
will never generate rectangles with that.
By a 
rotation, it suffices to consider the first case. Note that the piece 
squares. Color the cells chessboard-style with the corners all white and number the white cells 
with even area
with even area
with even area
with odd area
with odd area
with odd area
![\[
f(\mathcal{D}) = \int\int_\mathcal{D} e^{\pi i(x+y)}
\]](http://latex.artofproblemsolving.com/7/2/e/72e842bd369b5836cf555d4b33a018fc33ed45fd.png)
is nonzero iff it has all odd side lengths. Since 
is nonzero, the 
of one of the rectangles in the partition must also be nonzero.
