AoPS Academy
, 
.
.
are positive integers.
are positive real numbers.
be the set of all functions 
with the following properties: is defined for all real numbers and takes only real values.
the following equality holds: 
such that
,
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
be a parallelogram with 
. A circle tangent to sides 
, 
, and 
intersects diagonal 
at points and 
with 
, as shown. Suppose that 
, 
, and 
. Then the area of can be expressed in the form 
, where 
and 
are positive integers, and is not divisible by the square of any prime. Find 
.
and 
intersect at points 
and 
. Line 
is tangent to and at 
and 
, respectively, with line 
closer to point than to . Circle 
passes through and intersecting again at 
and intersecting again at 
. The three points 
, , 
are collinear, 
, 
, and 
. Find 
.
, we know that lines 
and intersect at 
, lines and 
intersect at 
, and lines 
and 
intersect at 
. Suppose that the circumcircle of 
intersects line 
at and , and the circumcircle of 
intersects line at and , where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at , then 
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
equations, and the last one is
, 
, 
, 
... and consider the systems (for all 
)
, for all ... so we have that
...
, since 
for all 
... taking two distinct indexes, say 
, it follows that 
...
, and we're done 
. Therefore, 
. But if 
or 
are greater than 
, we immediately arrive at a contradiction. Thus, there are two cases: If 
, it follows that 
and 
(subtract the second equation from the first). If 
, it follows that 
, or 
, 
. Both cases clearly lead to a contradiction. Thus, 
, 
, 
, , 
, , , . 
. Therefore, . But if or are greater than , we immediately arrive at a contradiction. Thus, there are two cases: If , it follows that and (subtract the second equation from the first). If , it follows that , or , . Both cases clearly lead to a contradiction. Thus, , , , , , , , . 
-s are integers, which need not be the case?
such that:
,
,
)![\[ \begin{cases} \displaystyle x_1+ \phantom{2^2} x_2+ \phantom{3^2} x_3 + \cdots + \phantom{n^2} x_n &= n+2, \\ x_1 + 2\phantom{^2}x_2 + 3\phantom{^2}x_3 + \cdots + n\phantom{^2}x_n &= 2n+2, \\ x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n &= n^2 + n +4, \\ x_1+ 2^3x_2 + 3^3x_3+ \cdots + n^3x_n &= n^3 + n + 8. \end{cases} \]](http://latex.artofproblemsolving.com/0/8/f/08fc55e0cc886c4c44f678e75eb07e4d814c4a77.png)
and 
.
and 
.
.
.
,
.
for every 
.
and 
(or for the initial 
)