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.
be a positive integer. Find, in terms of , all polynomials 
with integer coefficients such that for every integer 
, there exists an integer 
such that 
.
.
, angle 
. Point 
lies on 
such that 
. Line parallel to 
passing trough intersects 
at 
. Prove that 
.
such that
, prove/disprove the following claim: :maybe:
be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at 
. Denote by 
and 
the midpoints of 
and 
. Rays 
and 
meet 
and 
at 
and 
, respectively. Prove that there exists a point 
, lying outside quadrilateral , such that
bisects both angles 
, 
, and
.
be the circumcircle of triangle 
, and let 
be the tangent line of 
passing 
. Let 
be the points each on side 
such that 
. Line 
meets at points 
. The line parallel to passing 
meets at , the line parallel to 
passing meets at 
. Prove that there exists a circle passing four points 
and tangent to line 
.
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of on division by 
are pairwise different; and
denotes the integer part of real number 
, that is, the largest integer that does not exceed .)
be a trapezoid with 
. A point 
is chosen inside the trapezoid, and a point is chosen inside the triangle 
such that 
, 
. Prove that triangles 
and 
have equal areas.
, 
, 
, 
and 
, 
(we are assuming is parallel to -axis), then we should have ![\[ \frac{m_2-y_1}{m_1-a} = \frac{y_2-n_2}{c-n_1} \quad \text{and} \quad \frac{n_2-y_1}{n_1-d} = \frac{y_2-m_2}{b-m_1} \quad \quad (1) \]](http://latex.artofproblemsolving.com/7/a/1/7a1775e52fe0926ceea74b805ee37841cd53c2fd.png)
as and .![\[ \frac{1}{2} \begin{vmatrix}
1 & 1 & 1\\
a & b & n_1 \\
y_1 & y_2 & n_2
\end{vmatrix} = \frac{1}{2} \begin{vmatrix}
1 & 1 & 1\\
c & d & m_1 \\
y_2 & y_1 & m_2
\end{vmatrix} \quad \quad (2) \]](http://latex.artofproblemsolving.com/7/6/3/763b1763b0ab067d37f099290a7f892e77f235ca.png)
as they are areas of triangles (left one) and (right one).
implies 
using some algebraic manipulations.
are outside or inside the specified polygon.
as ![$[X]$](http://latex.artofproblemsolving.com/f/6/3/f631f7a53d834de52b718ed6b496689e9658eadf.png)
be a trapezoid s.t. 
; 
be a point then![$[AECD]=[BECD]$](http://latex.artofproblemsolving.com/1/a/b/1abc88febaeab84e9dd2d39d33e02d2bf0654df5.png)
![$[BAND]=[CAND ]=[CMND ]$](http://latex.artofproblemsolving.com/e/7/1/e71b0dcf4d59f6355d2aa2b23cf1651018a7e9dd.png)
but ![$[BAND]=[BAN ]+[BND ]$](http://latex.artofproblemsolving.com/9/1/b/91b2fe0229d73e2d916d02ff4b6446ea2e5af798.png)
and ![$[CMND]=[DCM ]+[DMN ] $](http://latex.artofproblemsolving.com/b/0/d/b0dfff6d5364c7d9367d43178a81f10a1a791b7a.png)
besides![$[BND ]=[DMN ](\because DN\parallel BM)$](http://latex.artofproblemsolving.com/3/4/f/34f3412f388b0beafd7eee69afbf7c4718db2250.png)
we deduces then ![$[BAN ]=[DCM ]$](http://latex.artofproblemsolving.com/f/7/c/f7c80c95eae5455ed3e2eaed13620a8ae089b562.png)
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
and 
are parallelograms,easy to know that 
and 
.
at 
at 
.
,
,
.
.
and 
.
and let 
.![$$[ABN]=[AYN]+[NYB]+[BYA]=\lambda(YM\cdot YN+YB\cdot YN+YB\cdot AM)=\lambda(YM\cdot YN+YB\cdot AX)$$](http://latex.artofproblemsolving.com/0/8/5/085fa07e11f01fc630d9d6a94278b21837503509.png)
Similarly ![$[CDM]=\lambda(XM\cdot XN+XD\cdot CY)$](http://latex.artofproblemsolving.com/6/f/2/6f227eca071ac3c94c8d33bb7082fda45b956a86.png)
.
and 
(because 
)
to a unit square through an affine transformation.
for some 
.
.
and 
,
for some 
(because 
(because 
.
we get 
.![$$\frac{[ABM]}{[ABC]}=\begin{vmatrix}1&0&0\\0&1&0\\\frac{(y+z)((t+1)x+z)}{z}&-\frac{y((t+1)x+z)}{z}&y-tx\end{vmatrix}=y-tx$$](http://latex.artofproblemsolving.com/b/2/e/b2e9a6fcaf633cb9b20c69943830d43874f12ef8.png)
and![$$\frac{[CDM]}{[ABC]}=\begin{vmatrix}0&0&1\\1&t&-t\\x&y&z\end{vmatrix}=y-tx.$$](http://latex.artofproblemsolving.com/7/f/0/7f0020c1f0c535c4766ebb4844f9786a5d76e7f0.png)
Therefore, ![$[ABM]=[CDM]$](http://latex.artofproblemsolving.com/a/5/f/a5f31256fafabff83c5b84fc662ce8058a454211.png)
,