AoPS Academy
are given. Prove that one can only use straightedge (which draws a straight line passing two drawn points) and compass (which draws a circle with a chosen radius equal to the distance of two drawn points and centered at a drawn points) to construct a line and two points on it with length 
in only 
steps (Namely, the total number of circles and straight lines drawn is at most .)
inscribed in a circle with center 
. 
is any point not on (O). 
intersect 
at 
. Let 
be the Lemoine points of triangle 
respectively. Prove that 
are collinear.
be a prime number and 
. Prove that![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
has property 
if :![\[
\begin{cases}
\text{the set } A \text{ has at least } 4 \text{ elements} \\
\text{the element } 1+1 \text{ is invertible}\\
x+x^4=x^2+x^3 \text{ holds for all } x \in A
\end{cases}
\]](http://latex.artofproblemsolving.com/b/6/9/b6993ee76eb3792e08d441f5d5c463247c392c4a.png)
a) Prove that if a ring has property , and 
are distinct elements, such that 
and 
are units, then 
is also a unit, but 
is not a unit..
be a function differentiable at 0 with 
. Find
be a smooth function on 
where 
for all 
and satisfies
. Prove that 
for all 
.
Where 
denotes the n-th Pell Number given by 
and 
invertible, for an infinite number of 
:
such that 
is a scalar
![\[
I = \int_0^1 \frac{(2^x + 2)\, dx}{1 + x^4}
\]](http://latex.artofproblemsolving.com/e/3/9/e391d4208a008146dc3922ff31ac1f544172b5b4.png)
using the trapezoidal rule with accuracy 
.
be the object circle and 
the perpendicular line to 
through 
is the inverse of under the inversion with pole 
and power 
and 
are the inverses of 
Thus 
But from 
we deduce that
intersects second time at and circle 
intersects second time at 
, it's not big deal to prove 
, wherefrom 
and applying p.o.p. we get 
and 
. Adding 
and 
side by side and taking into account 
, we get the desired relation.
Determine 
such that 
Determine 
.
and 
at inner points 
,![\[|AB|\cdot |AM | + |AD|\cdot |AN|=|AK|\cdot |AC|\]](http://latex.artofproblemsolving.com/7/f/2/7f274fdf1cc8ad23cc1419c116667674eb5c6d83.png)
(1)
and 
so 
is similar to 
and therefore:
, 
and 
. Substituting this into (1) we get the desired result.
is linear. Consequently,