AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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 to 
).
board.
candies. A starts first. On a move, if in the box there are 
candies, the player chooses positive integer 
so that 
and 
, and eats candies from the box. The player who eats the last candy wins. Who has winning strategy, in terms of 
.
be a trapezoid with parallel sides 
. Points 
and 
lie on the line segments 
and 
, respectively, so that 
. Suppose that there are points 
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points , , 
and 
are concyclic.
, with 
, and the circle 
with center 
and radius 
. Denote by 
the midpoint of side . The line 
intersects the circle for the second time in 
. Let 
be a point on the circle such that 
and 
. Show that 
.
. Let and be the feet of the altitudes from to 
and from 
to respectively.
and 
as the reflections of across lines 
and , respectively. Let 
be the circumcircle of 
. Denote by 
the second intersection of line 
with , and by 
the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
![\[\sqrt{a+h_B}+\sqrt{b+h_C}+\sqrt{c+h_A}=\sqrt{a+h_C}+\sqrt{b+h_A}+\sqrt{c+h_B}\]](http://latex.artofproblemsolving.com/5/d/6/5d6881be29327bcdc3e7bd76c221968578c7f50b.png)
is isosceles.
be all the divisors of . It is known, that 
are all divisors for some 
(except 
). Find all such 
. and 
be the points where the angle bisectors of the angles and 
intersect the sides 
and , respectively. be the point where the ray 
intersects the circumcircle of triangle .
.
. and be the points where the angle bisectors of the angles and intersect the sides and , respectively. be the point where the ray intersects the circumcircle of triangle ..
DX=x,DY=y,DZ=zfrom
Dto
CB,BA,AC.
x+y=z...(1)
BYD,CZD: \frac{y}{BD}=\frac{z}{CD}..(2)
BXD,AZD: \frac{x}{BD}=\frac{z}{AD}..(3)
\frac{1}{BD}=\frac{1}{AD}+\frac{1}{CD}QED.
, and let 
respectively. We use three things: 
: triangles 
are similar, so 
; 
: triangles 
are similar, so 
; 
from the transversal theorem we get 
. we deivde everything by 
and we get 
.
we get 
, and from 
we get 
, and by dividing this last equality with 
we get exactly what we want. we would have to show that 
, where, in general, 
is the image of 
under the inversion, but this seems much harder and I could make no progress whatsoever.
Prove that
such that 
and 
.
