[/quote]
,
be a function satisfying
for all integers 
and 
.
and 
such that 
for all integers 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
,
,
is a multiple of 
.
is the number of integers not more than 
a point in the space and 
be a 
on it, line 
makes a fixed angle with the tangent to 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
be a monic polynomial so that there exists another real coefficients 
that satisfy![\[P(x^2-2)=P(x)Q(x)\]](http://latex.artofproblemsolving.com/6/9/7/697faa929e4fda7a6e9b1cd97849bd42ffc14306.png)
Determine all complex roots that are possible from 
with exactly 9 elements that is subset of 
.
and 
that satisfy the following:
is an empty subset.
is 
and 
.
to 
is 
and 
is 
is 
is tangent to the 
and 
.
.
be the midpoints of sides 
,
be points on segment 
and 
,
is the angle bisector of 
and 
is the angle bisector of 
.
is parallel to 
if and only if 
divides 
for which, for all 
,![\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]](http://latex.artofproblemsolving.com/9/0/c/90c180110402e1a32b70edb2b0a03a28727457d1.png)
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
Y by Adventure10, Adventure10, Mango247
Let 
be a cyclic quadrilateral, and 
a point in its interior such that 
. Let 
, 
and 
be the orthogonal projections of on the sides 
, 
and 
, respectively. Show that triangles 
and 
are similar.
Alternative formulation. A point is in the interior of the cyclic quadrilateral and has the property 
. The feet of the perpendiculars from to , 
and 
are respectively denoted , and . Show that 
and 
are similar.
be a cyclic quadrilateral, and 
a point in its interior such that 
. Let 
, 
and 
be the orthogonal projections of on the sides 
, 
and 
, respectively. Show that triangles 
and 
are similar.Alternative formulation. A point
is in the interior of the cyclic quadrilateral and has the property 
. The feet of the perpendiculars from to , 
and 
are respectively denoted , and . Show that 
and 
are similar.
and 
.
.
seems to be not so easy.
,
.
and the circle (L) at points 
,
and the circle (L) at points 
,
respectively. Since X is the internal homothety center of the circles (K), (L), 
Combining, 
,
are both cyclic and generally not similar. Since (X) is the common P-Apollonius circle of the triangles 
and the common Q-Apollonius circle of the triangles 
,
and 
,
(inversion preserves angles). The quadrilateral vertices A, B, C, D are inverted into the intersections A', B', C', D' of the lines 
are similar. Hence, 
,
,
are parallel. Thus A'B'C'D' is a cyclic trapezoid, isosceles, and B'C' = A'D'. Consequently, and the line x bisecting the angle formed by its diagonals 
is also the perpendicular bisector of the sides A'B', C'D', so that the triangles 
are congruent. As a result, the triangles 
are similar. But the pedal triangle 
of the triangle 
with respect to the point P is similar to the image 
of this triangle in the inversion in the circle (P), so that the triangles 
are similar as well.
)
is a trapezoid (using the condition given in the problem). It must also be cyclic since 
), so it must be an isosceles trapezoid and 
.
and 
,
and 
So we're done. 
.
and by the converse of the Alternate Segment Theorem, 
is the common tangent of the circles through 
and 
.
,
.
is cyclic so 
.
is cyclic so 
.
(This follows from 
)
by the Alternate Segment Theorem, and 
,
.
) but I didn't manage to complete the proof. In fact, I tried considering the second intersection of the circumcircles of 
and 
as well, but that didn't really work - probably because there is no second intersection 
Anyways, the idea of that construction is not thát unnatural...
and the line 
,
of the quadrilateral 
,
for the triangles 
are cyclically and 
,
for which 
and 
.
is the common tangent in the point 
and 
,
.
.
Therefore, 
and 
and two fixed points ![$[AB]\cap d =\emptyset$](http://latex.artofproblemsolving.com/e/4/9/e49820d949d4a6c8d442d622b9074165f8bd3c44.png)
and 
.
,
for which the circumcircles of the triangles 
,
are tangent to the line 
is maximum belongs to the set 
.
