[/quote]
,
be real number such that 
Prove that
Let 
be real number . Prove that
be a triangle. The points 
and 
lie on the segments 
and 
respectively, such that the lines 
and 
intersect in a common point. Prove that it is possible to choose two of the triangles 
and 
whose inradii sum up to at least the inradius of the triangle 
and 
.
divides 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
be a point interior to triangle 
)
,
and 
meet again its circumcircle 
at 
,
,
to 
at 
follows 
.
.
be the midpoint of the arc 
not containing 
similarly. Show that the orthocenter of 
is the incenter 
of 
of positive integers such that 
is an integer.
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
and 
Prove that
![\[y''+by'+cy=g(t)\]](http://latex.artofproblemsolving.com/7/8/8/7882dc954ab6dba2445b68d370d630ce1c66a6db.png)
with 
,
are constants has the form ![\[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]](http://latex.artofproblemsolving.com/6/3/c/63c34df1c0013bf659bc76506a93873a5f879070.png)
and use variation of parameters in each case to find the solution. Is there any short way to calculate solution?
Y by Adventure10, Mango247
For a set 
of nonnegative integers, let 
denote the number of ordered pairs 
such that 
, 
, 
, and 
. Is it possible to partition the nonnegative integers into two sets 
and 
in such a way that 
for all 
?
of nonnegative integers, let 
denote the number of ordered pairs 
such that 
, 
, 
, and 
. Is it possible to partition the nonnegative integers into two sets 
and 
in such a way that 
for all 
?This post has been edited 1 time. Last edited by rrusczyk, May 8, 2018, 12:40 PM
Reason: LaTeX touch-up
Reason: LaTeX touch-up
there are no further choices and the sets 
and 
After that, let 
belong to the same set that 
belong to the other set. We won’t need it in the proof, but we could prove from this description the following characterization: write the number 
’
and if number of 
consider the set 
but delete from this set the number 
if 
where 
if 
if 
in which each pair contains one element of each set. If 
has balanced membership from 
To get the set we want, we must delete 
be the number of such pairs in which both numbers belong to 
be the number of such pairs in which both members belong to 
be the number of such pairs in which one member belongs to 
but we don’t need this equation.
Hence, for all 
and 
where 
iff 
,
,
similarly. It is clear that 
,
.
and the problem statement now desires that 
,
,
,
is well defined, and the reader can verify that 
and 
do exist with coefficients of only 1 and 0, so the sets do exist as desired.
.
and 
is equivalent to the problem statement. Simplify to get 
.
.
(a), 
(b).
times it.
where 
and consider the formal series 
in the ring of formal series ![$R[[X]]$](http://latex.artofproblemsolving.com/d/5/d/d5d26cd52ca7c4e0dbe8eecfe66a0d0cd9cb1888.png)
(the usual operations of addition and multiplication) defined by the relations 
and 
and observe that 
.
and note that 
and 
.
of pairs 
with 
and 
such that 
is simply, the coefficient of 
in the formal series 
.
,
Thus, we have the relation 
and successively replacing 
by 
in this relation and multiplying the equations obtained, we get 
and notice that by assumption, 
and thus, 
are respectively, the set of all non-negative integers with an odd (respectively, even) sum of digits in binary and they clearly satisfy the conditions by reversing the arguments.
by 
and
Then we clearly have 
.
and define ![$\mathbb{Z}[[X]].$](http://latex.artofproblemsolving.com/9/a/b/9abd8ac5b33034db2cb38f459d552aa501855096.png)
Let
and 
By definition of the sequence 
Since 
has the inverse 
,
or
Note, however, that
and so this partition works.
and 
be the generating functions of 
However, 
,
Now, doing a similar trick as in 1996 USAMO 6, notice that 
as we perform arbitrarily many iterations. Thus, 
,
where we exchange the first power of 
.
.
,
.
.
Note that we find that,
as 
.
This is due to the fact that 
represents the number of partitions that 
can be partitioned over all 
and 
to account for 
which is not allowed by the problem condition. Then rearranging the second we have,![\begin{align*}
[F(x) - G(x)] \cdot [F(x) + G(x)] &= F(x^2) - G(x^2)\\
[F(x) - G(x)] \cdot \frac{1}{1-x} &= F(x^2) - G(x^2)\\
\frac{1}{1-x} &= \frac{F(x^2) - G(x^2)}{F(x) - G(x)}
\end{align*}](http://latex.artofproblemsolving.com/b/1/0/b10fcf4b57ab36682c0bc7109bce9cd8ab11f110.png)
Then define 
.
Let 
for constants 
.
Then we find the system,
As such we conclude that 
and 
.
and 
we find that we have,
Then taking the partition of 
and 
we are done. 
and 
.![\[A(x)+B(x)=\sum_{n \ge 0} x^n = \frac{1}{1-x}, \quad A(x)^2-A(x^2) = B(x)^2-B(x^2).\]](http://latex.artofproblemsolving.com/c/6/5/c6582c1148d7bb3c0a78c0256d5d9bc53ad0dd89.png)
,![\[P(x^2) = \frac{P(x)}{1-x} = P(x) \cdot (1+x)(1+x^2)(1+x^4)(1+x^8) \ldots\]](http://latex.artofproblemsolving.com/7/2/b/72b4dfce10953b7170c3cfa3c8918c6ca858843a.png)
![\[\implies P(x) = (1-x)(1-x^2)(1-x^4)(1-x^8)\ldots\]](http://latex.artofproblemsolving.com/5/4/3/543866579182c2f98e6799119bd0a78a79cfbfb5.png)
as desired. Thus our answer is 
,![\[\boxed{\begin{cases} \{A\} =& \text{Integers with an even number of 1's in binary.} \\ \{B\} =& \text{Integers with an odd number of 1's in binary.} \end{cases}}\]](http://latex.artofproblemsolving.com/6/c/9/6c9f41998ff9a1052320487f17d4b8d21698c0bf.png)
be the generating functions for 
so 
but 
so 
satisfies 
which it does since clearly 
are in different sets and clearly the even index terms form the same as the original sequence. thus this genfunc satisfies the equation
We have that 
However, by rearrangement, we see that:
However, since 
we see that
Now, define 
Hence, we have that 
Clearly, we have that 
work since 
.
