[/quote]
,
and 
,
such that![\[
f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1,
\quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/7/a/5/7a51a7be2306d18f4e80db1f5aac6e9bbf25c25c.png)
be an acute-angled triangle with 
.
and 
intersect at 
be the midpoint of 
.
is chosen on the extension of 
beyond 
is chosen on the segment 
such that 
.
and 
onto 
.
bisects 
.
so that
such that 
and 
is a perfect square.
L
such that 
again at 
and 
cuts 
again at 
and 
is 
and 
be the intersection of 
with 
and 
respectively such that 
,
,
are coaxial circles
,
,
,
be 
integers. If 
is an integer, prove that the only solution to
is is 
.
be a cyclic quadrilateral an let 
The diagonals 
at 
The line through 
mmets the extension of the side 
beyond 
at 
The line through 
meets the extension of the side 
Prove that the circumcircles of the triangles 
and 
are tangent .
such that 
,
such that 
such that 
is a perfect square but none of the integers 
are perfect squares.
and ![$P_1(x),P_2(x), \ldots P_n(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/c/6/4/c64176e52d94032e571b4579e35f9e6adfde675a.png)
such that 
.
such that every equation: 
has no real roots
Y by
Pick a point 
on a semicircle with diameter 
. Let 
and 
be two points on segment such that 
and 
. The point 
is the center of the circumscribed circle of triangle 
and point 
is the orthocenter of triangle . Prove that for all posible locations of point , the line 
is passing through a fixed point.
(Mykhailo Sydorenko)
on a semicircle with diameter 
. Let 
and 
be two points on segment such that 
and 
. The point 
is the center of the circumscribed circle of triangle 
and point 
is the orthocenter of triangle . Prove that for all posible locations of point , the line 
is passing through a fixed point.(Mykhailo Sydorenko)
This post has been edited 1 time. Last edited by parmenides51, Dec 13, 2023, 3:35 AM
has angles of 
.
bisect 
is the incenter of 
.
is the midpoint of 
not containing 
centered at 
and 
.
,
of 
,
of the way from 
and the perpendicular bisector of 
is constant. By the Ratio Lemma on 
,
which by LoS in 
equals 
.
