be an acute-angled triangle with circumference 
.
and 
intersect 
.
be the intersection point of these angle bisectors. Let 
and 
be the respective reflections of 
and 
,
cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
and 
,
for all 
,
from the deck, 
then 
and 
for 
.
is increasing.
,
,
of 
such that ![\[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]](http://latex.artofproblemsolving.com/9/8/0/98097fb721722b9103ef65d72322956877e262cb.png)
for every 
.
.
.
?
since the coordinates correspond to the 6th roots of unity scaled by the radius. Since the sum of all n-th roots of unity is zero, the coordinates add up to zero.
,
,
are respectively the midpoints of the major and minor arcs 
.
be 
,
is the intersection point of the line through 
intersects 
at 
,
on 
so that 
.
intersects 
,
intersects 
at 
.
is perpendicular to 
.
into what she calls the 
numbers, 
,
.
except that 
,
)
,
such that 
and both 
and 
.
For this to be equal to 
,
must be equal to 
,
,
,
,
and 
.![\begin{align*}
4&=[(t+ui)j]^4 \\
4&=(t+ui)^4 \\
4&=r^4(\cos4\theta+i\sin4\theta)
\end{align*}](http://latex.artofproblemsolving.com/9/3/e/93e59e5b32eea971180f54a729046aeaf0dbd599.png)
.
![\begin{align*}
4&=(\pm iB+ Bj)^4 \\
4&=[B(\pm i+j)]^4 \\
4&=B^4(-4)\\
-1&=B^4
\end{align*}](http://latex.artofproblemsolving.com/f/e/7/fe772c742c5f665ba9d8d42bc7a1066ebf4b6c9e.png)
and![\begin{align*}
4&=(A \pm Aij)^4 \\
4&=[A(1\pm ij)]^4 \\
4&=A^4(-4)\\
-1&=A^4
\end{align*}](http://latex.artofproblemsolving.com/9/f/a/9fa539c17f93e97ad28ba51a9785c9ae940cb800.png)
are the fourth roots of unity of 
t
.![\[
f(x) = \frac{1}{\sqrt[3]{x^2+2x+1}+\sqrt[3]{x^2-1}+\sqrt[3]{x^2-2x+1}}
\]](http://latex.artofproblemsolving.com/2/0/8/20806cb4860ace2b69aa76ea0a429a283ec76637.png)
for all positive integers 
,![\[
f(1)+f(3)+f(5)+\dots +f(7999)
\]](http://latex.artofproblemsolving.com/9/0/b/90bd38f627e03a496e81078a5944067084963afa.png)
be a triangle such that 
.
lie on segments 
,
and 
are the angle bisectors of triangle 
.
.
,![\[
S=\left\{ x \in \mathbb{Z} | \frac{x^2 - 2x +7}{2x-1} \in \mathbb{Z} \right\}
\]](http://latex.artofproblemsolving.com/d/0/6/d0681591c7ca805f605caafb20be9ff10e14ca06.png)
Hence find the number of elements in 
.
rolls have been made. If Alice makes the first roll, the probability that Bob makes the last roll can be written as 
,
and 
are relatively prime positive integers. What is 
?
Y by Adventure10, Mango247
Let 
be a point in the exterior of a circle of center 
, and let 
be the points where the tangents from touch the circle. Let 
be an arbitrary point on the small arc 
. The tangent in at the circle intersects the line 
in 
, and the lines 
and 
intersect in 
. We denote by 
the intersection of the lines and 
.
Prove that the lines
and 
are perpendicular.
be a point in the exterior of a circle of center 
, and let 
be the points where the tangents from touch the circle. Let 
be an arbitrary point on the small arc 
. The tangent in at the circle intersects the line 
in 
, and the lines 
and 
intersect in 
. We denote by 
the intersection of the lines and 
.Prove that the lines
and 
are perpendicular.
.
is the pole of 
,
.
and we're done.
, but it's a solution, right?
,
.
and let 
.
,
.
