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be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
id defined by![\[
f(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}
\]](http://latex.artofproblemsolving.com/0/2/c/02c7b4af3cac77da8e3cc102c79a7db2e95f437f.png)
now what is CDF or cumulative distribution function ![$F_X(x)=P[X\leq x]$](http://latex.artofproblemsolving.com/7/9/1/79150c7527581feec873c9c790f8aeddc8f8efb8.png)
such that 
for all 
.
![$f(x)\in C[0,3]$](http://latex.artofproblemsolving.com/8/e/2/8e2d8e1cf446975c3e3aa894610ac8b6371737b4.png)
satisfy 
for all 
and
Show that
be the set of all real monic polynomial functions of degree 
. Prove that for any 
, 
be an integral domain and ![$A[X]$](http://latex.artofproblemsolving.com/8/f/c/8fc58a509a695c9718230e43a63ed145a6ce2835.png)
be its associated ring of polynomials. For every integer 
we define the map ![$\varphi_n : A[X] \to A[X],$](http://latex.artofproblemsolving.com/4/6/6/466be576a494f1b7fc51e8a0dacdc953fcbac5a3.png)
and we assume that the set ![$$M= \Big\{ n \in \mathbb{Z}_{\ge 2} : \varphi_n \mathrm{~is~an~endomorphism~of~the~ring~} A[X] \Big\}$$](http://latex.artofproblemsolving.com/b/b/8/bb8acb92661d9f087b34ef4abca876d46317b463.png)
is nonempty.
such that 
be a finite group and 
a fixed element. Define the set 
Show that:
, then 
;
is divisible by 
.
is a triangle with incentre 
. The feet of the altitudes from to 
are 
respectively, and the line through 
parallel to 
intersects 
and 
at 
and 
respectively. Prove that the circles with diameters 
and 
have a common point on the circumcircle of 
is a triangle with incentre . The feet of the altitudes from to are respectively, and the line through parallel to intersects and at and respectively. Prove that the circles with diameters and have a common point on the circumcircle of 
Define 
and 
See that 
and 
swap under incircle inversion. Also, 
swaps with the nine-point circle of 
. Hence, 
(sharky-devil)
be the 
-Antipode in 
. Applying Pascal's Theorem on 
we get 
, since 
. Thus, we have 
. Similarly, 
so we are done.
where 
and 
