[/quote]
,
is an integer. Prove that 
are integers.
(
well). i want to study so if you guys have some easy or normal problems please send me
and 
.
.
of degree 
.![$[a;b]$](http://latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
and Maxim will give in response the maximum value of 
![\[
v_p(n!) \leq \frac{n}{p - 1}
\]](http://latex.artofproblemsolving.com/5/7/c/57ca4d0a6532fb4b97687a97b9c1fb35ad02b9f2.png)
and 
Prove that
Let 
and 
Prove that
not isosceles, the incircle 
.
is the intersection of 
.
onto 
.
cut 
.
is the projection of 
onto 
.
cut 
respectively. Prove 
are collinear
be the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
,
for all integers 
.
.
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
.
in natural numbers m and n.
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
.
and 
intersect at 
of the two circles touches 
at 
respectively. Let 
be the symmetric points of 
intersects circles 
respectively which are distinct from 
and 
are equal.
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 
,
.
,
of 
,
of the way from 
and the perpendicular bisector of 
is constant. By the Ratio Lemma on 
,
which by LoS in 
equals 
.
