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is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
be a triangle with circumcircle and incentre . A line 
intersects the lines , 
, and 
at points 
, , and , respectively, distinct from the points , 
, 
, and . The perpendicular bisectors 
, 
, and 
of the segments 
, 
, and 
, respectively determine a triangle 
. Show that the circumcircle of the triangle is tangent to .
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
denote the set of single-variable polynomials in with integer coefficients. Find all functions ![$\theta : \mathbb{Z}[x] \to \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/3/2/032a637d4ae21da2c0ddd8dc127723807f11cc5f.png)
(i.e. functions taking polynomials to polynomials)![$p, q \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/7/e/7/7e7eb925258a33e456ef4ff722c1f943dc155d10.png)
, 
;![$p \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/2/6/0267b98c16b400dda1f081877e9b84fe672ce0e7.png)
, 
has an integer root if and only if 
does.
. Then, given consecutive vertices 
(the indices are taken modulo 
), we have 
because 
is a trapezoid. If we denote 
by 
, we thus have 
, but 
, so 
, and we're done: the two diagonals considered above cut each other in their common midpoint , so 
.
has opposite sides parallel 
.
are concurrent if and only if opposite sides are equal.
are concurrent 
.
) (grobber's sol)
.
(the indices are taken modulo 
)
because 
is a trapezoid. If we denote 
by 
,
,
,
,
.
)
,
becomes a parallelogram. Then, if we denote X as an intersection of 
and 
,
.
is a midpoint of