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of positive integers such that 
is an integer.
be an acute triangle and 
its orthocenter. Let 
be a point on the parallel through 
to 
such that 
. Define 
and 
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
be an acute triangle with 
, its orthocenter and 
it's circumcenter. Let be a point on the parallel through to such that 
and and are on different sides of . Denote by 
the intersection of the circumcircle of and 
, where 
is the reflection of over , 
the midpoint of 
, the intersection of 
and the parallel through to 
, the intersection of 
and the perpendicular through to 
and 
a point on such that 
, where 
is the midpoint of 
. Prove that 
lie on a line.
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
of the integers with the following property: for any integer 
there is exactly one solution of 
with 
.
; the given condition is equivalent to 
, which immediately gives it away: recall that unique binary expansion implies the identity
. In other words, 
consists of the set of positive integers whose binary expansions in base 
contain only 
s and 
s. (The bijective proof is straightforward: consider the base- expansion of 
and isolate its digits of 
and 
, etc.) Agrippina's problem is similar: the identity is 
and we can take 
.
by studying the generating functions of 
and 
because 
gives all the possible sums at once. For example, the following can be solved by similar means.
of non-negative integers let 
denote the number of ordered pairs 
such that 
. Is it possible to partition the non-negative integers into disjoint sets and such that 
for all ?
, then the set of all sums of distinct elements of is given by 
(why?). The rest is computation, and once you've figured out what the answer should be it is not hard to give a direct proof.
such that every 
is distinct. Suppose this set misses the value . Then add to the set 
, so now we have included as a sum. It is clear that if you take 
arbitrarily large it wont overlap with any previous sums. So such a set exists.
where 
. Uniqueness is evident: if we have 
then suppose 
is the first natural number such that 
. Then
and 
. Then that means that, dividing by 
, we have 
, so 
is divisible by 4, which is impossible because 
and 
are not equal and between 0 and 3. So the uniqueness is proved.
numbers 
where . The minimal number possible is 
and the maximal number is 
, which cover a range of
numbers in the possible range and all numbers are expressed at most once, all numbers must be expressed exactly once. Thus base -4 exists and we can work with it like any normal base., take its base -4 expansion 
. If 
, let 
; if 
, let 
and 
; if 
, let 
and 
; and if 
, let 
, so that in any case, 
. Then let 
and 
also. Then clearly 
by construction.
, we have
.
is either 
or 
. Thus 
where 
is an integer. Hence 
is divisible by 
and not 
, so the highest power of two that divides 
is also a power of 4.
for 
, then 
. If 
then let us look at the highest power of 2 that divides this common difference. It must be a power of 4 that divides 
, but also is a power of 4 that divides 
and so is twice a power of 4 that divides 
. No number is both a power of 4 and twice a power of 4, so this contradiction shows uniqueness and we're done.
(the same base -4 thing)
, where 
is a pretty huge number, tends to infinity,
which solves the problem.
, which this one does not. The generating function approach does not prove the answer, it just points you in the direction. You need another approach to show that it actually works. In the limit, what is ? How do you know that the generating function doesn't just become 
and stop there?
. I know it makes absolutely no sense if you considered as an actual number, but this wasn't the point here. approaches infinity.
by setting 
; he had no rigorous justification for saying two power series were equal just because they had exactly the same roots. And this is a bit like what you're doing; you're saying that the power series 
, whatever means in the limiting case.
as or goes to infinity and saying that it equals the product of 
and 
is unjustified. It would be justified if it converged for some number , but it doesn't.
, of 
; by limit, I just meant that the larger integers (in absolute value) could be represented uniquely once we increased the size of our sequence. I'm not ever using as any evaluation for a numerical answer, I'm just using conveniently because it just simplified the concept of "looking at all sums".
, where 
is bounded by some growing bound dependent on the number of factors. This is the case here. of Laurents sums whose exponent's modulus is bounded by . This is just a less comprehensible way to state what I said in the previous paragraph, though.
. The error here is that to apply geometric series, you would need it to be a module over power series' (i.e. multiplying any Laurent series with a power series would need to make sense; try to multiply the above one by 
to see the problem). But the problem is not that we lack a common are of convergence for those sums.
and 
) and as an identity of meromorphic functions, this is completely correct!
. The necessary and sufficient condition for to satisfy the condition is for 
for all 
. One can notice that we have 
so the infinite product 
so the function 
can be checked to work. to be the set of numbers that can be represented as the sum of some distinct nonnegative powers of 4.
such that 
. 
. 
. 
. We could keep going like this forever, extending in both directions. Therefore, is 
works.
, our condition requires![\[A(x) A(x^2) = \ldots + x^{-2} + x^{-1} + x^0 + x^1 + x^2 + \ldots.\]](http://latex.artofproblemsolving.com/9/b/9/9b9e4ccede78394b579be5db68b07e86779cce44.png)
is simply![\[\boxed{\{\mathcal{X}\} = \text{Integers with only 0 and 1 as digits in base -4}}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/9/c/4/9c4fca57f81d66cbc0f36b57a1e02647241b16f8.png)
then we require 
However observe that 
works since every number has a unique representation in base 
such that 
Note that 
So, we let 
and 
meaning for all 
there is exactly one solution to 
with 
To prove this, just use the generating function.