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.
, there exist pairwise different real numbers 
satisfying
be a positive integer, and let 
denote all positive divisors of , If the following conditions are satisfied:
.
satisfying the equation 
.
.
and 
but 
because 
. I mean 
and 
are perfect squares. 
. a should be 0 or 1 because of . We have two solution. These are $(a,b)=(0,0),(1,0)
it can draw the line which forms the interior angle bisector of 
.
in the plane . Prove that Euclid can use the splitter several times to draw the centre of circle passing through 
and 
.
, satisfying 
, what is the maximum possible value of 
?
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be a cyclic convex quadrilateral and 
be its circumcircle. Let 
be the intersection of the diagonals of 
and 
. Let 
be the center of the circle tangent to sides 
, 
, and 
, and let 
be the midpoint of the arc of not containing 
and 
. Prove that the excenter of triangle 
opposite lies on the line 
.
that has 8097 distinct points with area of a triangle that has 3 points belong to all 
. Prove that there exists a triangle 
that has the area 
contains at least 2025 points that belong to ( each of that 2025 points can be inside the triangle or lie on the edge of triangle )X
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
be the incentre of acute triangle 
with 
. The incircle 
of is tangent to sides 
, and at 
and 
, respectively. The line through perpendicular to 
meets at 
. Line 
meets again at 
. The circumcircles of triangle 
and 
meet again at 
.
and 
meet on the line through perpendicular to 
.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
, and the distance AB' equals the semiperimeter of triangle APQ. Since the perimeter of triangle APQ equals 2, its semiperimeter must be 1; thus, we have AB' = 1. On the other hand, we know that AB = 1 (in fact, the square ABCD has sidelength 1). Since the points B and B' both lie on the ray AP, we thus can conclude that the points B and B' coincide. Hence, becomes 
; in other words, 
. Hence, the point C' lies on the perpendicular to the line AB at the point B. But since ABCD is a square, the perpendicular to the line AB at the point B is the line BC. Hence, the point C' lies on the line BC. Similarly, the point C' lies on the line DC. As the lines BC and DC have only one common point, namely the point C, it follows that the point C' coincides with the point C. Since we have defined the point C' as the center of the excircle of triangle APQ opposite to the vertex A, it follows that the point C is the center of the excircle of triangle APQ opposite to the vertex A. Thus, this point C lies on the external angle bisector of the angle APQ. Hence, 
. Similarly, 
. Hence, by the sum of the angles in triangle PCQ, we have
.
.
be a quadrilateral for which 
, 
and 
, 
, 
be two points so that 
. Show that 
.
is equal to 
, 
and the point 
belongs to the bisector of the angle 
. Therefore, the point is the 
- exincenter of the triangle . Generally, 
, i.e. 
.
and 
Prove that
,
.
is 
find the angle 
.
then 
?
and 
.
and 
.
is 
we have 
.
.
,
Hence 
.