and 
be polynomials with integer coefficients. Suppose that the integers 
and 
are roots of 
.
has no integer solutions.
.
.
be three positive real numbers with 
.
,
satisfying![\[
2a^n+3b^n=4c^n.
\]](http://latex.artofproblemsolving.com/0/2/9/0295405b01494edc05ce9867b2ec48a4e069bf26.png)
be a real function such that![\[ \lim_{x \rightarrow +\infty} \frac{f(x)}{e^x}=1\]](http://latex.artofproblemsolving.com/7/f/e/7fe200614381665ef11aed463fbbee51f8461bb3.png)
for all sufficiently large 
.![\[ \lim_{x \rightarrow +\infty} \frac{f'(x)}{e^x}=1.\]](http://latex.artofproblemsolving.com/0/9/6/096b5b71111efdd7565d60424240e20ee3251382.png)
and 
with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true for the polynomials
Prove that 
be a linear transformation on linear space 
satisfying:
but 
, and 
is the eigensubspace of eigenvalue 
.
)
be then 
my question is dose,
My reservations about this is that 
and 
so 
.
is a set and 
that 
we can have, for 
we can have,
for some 
such that 
for all 
order, and if the constituents of its first, second, ..., 
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the 
)
Y by loveGeometrytothemoon
Let 
be a triangle. Point 
moves on the opposite ray of 
such that 
. Let 
be the midpoint of 
. 
cuts 
at 
a) Suppost that
is the circumcenter of 
and 
is the circumcenter of 
, Prove that 
b) Let
be the midpoint of 
and 
be the orthocenter of triangle 
. Prove that when 
varies on the opposite ray of 
, 
go through a fixed point
be a triangle. Point 
moves on the opposite ray of 
such that 
. Let 
be the midpoint of 
. 
cuts 
at 
a) Suppost that
is the circumcenter of 
and 
is the circumcenter of 
, Prove that 
b) Let
be the midpoint of 
and 
be the orthocenter of triangle 
. Prove that when 
varies on the opposite ray of 
, 
go through a fixed pointThis post has been edited 1 time. Last edited by 799786, Mar 4, 2022, 10:31 AM
Reason: typo
Reason: typo
intersects 
at 
and 
,
and 
.
.
and 
are corresponding points. So 
and therefore 
.
,
be the midpoints of 
,
.
and 
have the same radius, so 
are orthocenter of 
are collinear.
.
and 
.
we have 
.
and 
.
are collinear and 
are collinear. We also have 
.
.
or 
.
are midpoints of 
.
is parallelogram. So 
.
then 
and 
.
.
which is a fixed point.
