make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
denote the set of all polynomials in three variables 
with integer coefficients. Let 
denote the subset of 
with 
.
for all non-negative integers 
satisfying 
.
of integer numbers, we define 
as the sum of the elements of 
and 
whose intersection is empty, whose union is the set 
and such that the product 
is a perfect square.
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
.
.
for every 
or 
for every 
The altitudes 
, intersect
.
o
.
b
.
intersects 
,
intersects 
at 
lie on the same circle.
intersects the circle 
The line 
.
are collinear.
such that 
is an exact 5th power, 
is an exact 6th power, and 
is an exact 7th power.
denote the set of natural numbers and let 
denote 
ordered pairs in 
.
distinct elements in the set 
for 
whose product is a perfect cube.
. Prove that
have real roots and real coefficients with 
.
?
,
.
have roots 
.
.
(I won't be proving this here). This new information allows us to utilize the mean inequality chain as follows:
Similarly, applying AM-HM yields 
.
.
such that 
is a perfect cube.
digits are distinct?
Solve for 
,
,
,
.
be the point on 
,
be the midpoint of 
.
is a point such that 
is a rectangle, compute the area of 
.
has a sidelength of 
,
,
,
,
,
,
,
are length 
.
.
satisfies for all integers 
If 
and 
,
.
children have black hair, all but 
have red hair, and all but 
have blond hair. How many children are there in the class?
can be partitioned into n Akash subsets.
:
.
Let 
.
Y by
What is the smallest value of 
such that for any polynomial 
of degree 
with real coefficients, there exists a polynomial 
of degree at most with real coefficients such that the graphs of 
and 
intersect at exactly points?
such that for any polynomial 
of degree 
with real coefficients, there exists a polynomial 
of degree at most with real coefficients such that the graphs of 
and 
intersect at exactly points?This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
real roots counting multiplicity. So if 
the considering the 
-
gives that for any three given reals 
always has two real root if 
and 
lol).
works, first clean all the polynomial of degree 
using 
and from there many a suitable pick, in fact 
are not in our control, so that the polynomial has all real roots with no multiplicity and none of them are zero.
we can obviously shift the parabola by a constant acording to what is needed due to discriminant criteria so we should be fine here, due to continuity in fact you can just avoid a root being zero or the resulting parabola being tangent to the 
you have something unfree of the form 
and so you pick some 
on the free part where 
have freedom acording to needs like if 
and could just be suficiently small given we put 
to have two roots and that in order to not have repeated roots or the cubic being tangent to the 
you basically have control over a depressed cubic upon doing a similar setup as seen above just shift everything by a small constant at the end to satisfy all the given conditions needed and now we will show this is the kind of work you have to do.
then we prove it for 
,
so we just use the free term to focusing on a polynomial of the kind 
for 
,
is a root twice then we just shift by a very small constant as needed whether is above or below, one of them should work by continuity and IVT, since have 
as a root is only a one thing thing on a small interval we can also use continuity to avoid this and still satisfy the no double roots or more condition, hence by using the inductive hypothesis this is also completed and thus our claim is proven.
.
.
with degree less than 
has 100 roots for 
.
as the degree of 
such that 
.
as the real roots of 
and 
is equal to 
and 
which is a contradiction. It follows that 
and 
with 
being the coefficient of 
being the leading coefficient. Obviously this is possible. From here we just alter the other coefficients so that each one of the coefficients becomes equal to the coefficients of the polynomial that has all these 
and so we are done. 
.
