[/quote]
,
and 
with 
,
and 
,
,
and 
.
be an acute scalene triangle. A point 
varies on its side 
.
and 
are the midpoints of the arcs 
and 
(not containing 
and 
,
passes through a fixed point, independent of the choice of 
are positive integer.
touches the sides 
,
and 
,
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
with 
has at least two cycles in it.
that satisfy the following conditions:
is an interval.
.
intersects 
and 
.
is the midpoint of 
.
be the altitude in 
,
intersects 
be the midpoint of 
be the reflection of 
and 
are similar.
is a pair of real numbers such that
.
such that 
divides 
for any positive integers 
and 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
is orthocenter. Let 
and 
and 
and 
cuts the circumcircle of 
.
and 
concur at a point on 
Y by HWenslawski, jhu08, megarnie, ImSh95, Adventure10, Mango247, DroneChaudhary, Lhaj3
Let 
be an integer. Prove that there exists a set 
of 
positive integers satisfying the following property: For every 
the set can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality 
.
be an integer. Prove that there exists a set 
of 
positive integers satisfying the following property: For every 
the set can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality 
.This post has been edited 1 time. Last edited by psi241, Jul 22, 2019, 6:00 AM
use the set 
.
construct 
and partition
where 
denote 
Fibonacci number.
then a straightforward calculation shows that this partition works.
which generalizes readily: we write 
where 
which gives the desired equalities for 
.
and partition
for each 
.
and 
.
and 
are equal for all 
and partition
A straightforward calculation shows that this works.
by giving examples:
and the relevant sets are:
We now go by induction on 
be the working set for 
be the total of all the elements in 
to be the subset of size 
with total 
.
elements as 
by considering 
and 
which both have total of elements 
is 
.
as follows:
and all the 
have total 
.
by induction we're done for all 
and partition
elements, it can easily be corrected by replacing 
by 
.
(because we can cancel out the denominator to turn rationals into integers) then we only need the following system of equations to hold:
From this we can obtain 
.
:
to get 
.
.
.
to be considerably small, so that 
as follows:
,
,
.
For example, our sequence starts 
,
.
It is clear that 
.
.
Also, note that
Hence,
Hence we have provided a construction for 
It is clear that
Using a similar argument to the odd case, the set 
be the subset of cardinality 
of positive integers such that 
for all 
.
.
Now, the only condition to be checked of is that 
which is a pretty easy task!
and 
for all 
.
works.
for all 
,
by the induction hypothesis, as desired.
take the set 
.
,
where 
is the 
number in the recursion;
This is just the Fibonacci Sequence starting from 1 and 2. Now, notice that we can have
Thus, for all 
,
the set 
with equal sums.
(just scale everything up by LCM of denominators to recover the desired set of positive integers). Consider the set![\[ S = \left\{ \frac{9}{10}, \frac{9}{10^2}, \cdots, \frac{9}{10^{n-1}}, \frac{1}{10}, \frac{1}{10^2}, \cdots, \frac{1}{10^{n-1}}, x, y \right\} \]](http://latex.artofproblemsolving.com/f/9/5/f954bc280283d0ac84ca35d711f925988c7f8fc4.png)
where 
and 
are arbitrary rationals such that the sum of all elements in 
.
.
we pick one of the subsets to be![\[ A = \left\{ \frac{9}{10}, \frac{9}{10^2}, ..., \frac{9}{10^{m-1}}, \frac{1}{10^{m-1}} \right\} \]](http://latex.artofproblemsolving.com/7/f/a/7faf4820da45b08f16205d6c59f896a3fd874d90.png)
It is easy to see that the sum of the elements in 
is 
,
too.
where 
,
for all ![\[S=\left\{f_1, f_2,\ldots, f_{2n-1}, c\right\} \text{with } c=\sum_{i=1}^{2n-4}f_i\]](http://latex.artofproblemsolving.com/9/0/4/904abdc06bf2b364e4ee3237d9249169835421e9.png)
satisfies our desired property. We start with the sets 
and 
.![\[\sum_{i=1}^{2n-2}f_i = f_{2n-2}+f_{2n-3}+\sum_{i=1}^{2n-4}f_i=f_{2n-1}+c\]](http://latex.artofproblemsolving.com/5/b/4/5b47430ef0747c0d48abef68d13b74db70458106.png)
so 
works. Now, we'll show that we can keep on replacing an element of 
with two elements of 
which have the same sum, to reach all desired values of 
,
and 
and 
.
and 
.
.
,
with 
and 
to obtain the case where 
,
and 
and 
are still in 
,
.
.
.
works. Basically the first set that works is 
.
is already a Fibonacci number and in our set. For this we can use the construction 
.
,
is the 
t
work since 
.
whereupon we can choose the subsets ![\[\{5, 7\}, \{1, 2, 3, 6\}\]](http://latex.artofproblemsolving.com/5/2/3/52370ef4afd428164fbae56c328f201666f7e3ca.png)
and ![\[\{1, 5, 6\}, \{2, 3, 7\}\]](http://latex.artofproblemsolving.com/d/6/d/d6df580b25a2334e81ae5eedac3bc75ca6f108ea.png)
for 
and 
respectively.![\[\left\{3^{n - 1}, \frac{3^{n - 1} - 3}{2}\right\} \cup \bigcup_{0 \leq i \leq n - 2} \{3^i, 2 \cdot 3^i\}\text{.}\]](http://latex.artofproblemsolving.com/0/8/2/0822eaa26eb147b5f79929b58b3c8c7245ba0fce.png)
This is a well-defined set since
and since 
,
:![\[\left\{3^4, \frac{3^4 - 3}{2}\right\}, \{1, 2, 3, 2 \cdot 3, 3^2, 2 \cdot 3^2, 3^3, 2 \cdot 3^3\}\]](http://latex.artofproblemsolving.com/e/e/3/ee3aa3e8ddfd21a831cc3b6ccf33d4277451a9c8.png)
which evidently work. Now we swap the 
on the left-hand side with the 
and 
on the right-hand side:![\[\left\{3^3, 2 \cdot 3^3, \frac{3^4 - 3}{2}\right\}, \{1, 2, 3, 2 \cdot 3, 3^2, 2 \cdot 3^2, 3^4\}\text{.}\]](http://latex.artofproblemsolving.com/3/5/4/3549258058d145c3bb9688f2aa697855379c47ba.png)
This still works since 
.
and 
on the right-hand side again. This argument can continue until we reach the sets![\[\left\{3, 2 \cdot 3, 2 \cdot 3^2, 2 \cdot 3^3, \frac{3^4 - 3}{2}\right\}, \{1, 2, 3^2, 3^3, 3^4\}\]](http://latex.artofproblemsolving.com/9/c/3/9c3a2139eaed1ced20c797e9f01dffc25a7024d5.png)
at which point we're done. (We can actually apply the argument again, but we don't need to.)
,
,
t
,
,
are integers such that 
and none of 
(just make 
be the sum of all elements in 
which finishes.
where we define 
.
Clearly each of these partitions work so we're done. 
I'm here
with the partitions 
and 
for 
and 
for 
.
.
where 
.![$S[0]$](http://latex.artofproblemsolving.com/0/9/2/092537da06a23754e13353afc4213eb300c733a0.png)
and ![$S[1]$](http://latex.artofproblemsolving.com/4/d/a/4dae8683b68a3027467d5e116f19954247d81e06.png)
imply the first and the second element of the set ![$S_i[0] + S_i[1] = S_{i+1}[0]$](http://latex.artofproblemsolving.com/a/f/8/af862d9b60677c4de505ff6f2583b0e2200982f4.png)
.
.![\[ S_i[0] = S_i[1] \iff 2^{2n} = 2^{2i} + \sum_{k=i}^{n-1} 2^{2k} \iff \nu_{2}(2^{2n}) = \nu_{2}\left(2^{2i+1} + \sum_{k=i+1}^{n-1} 2^{2k}\right) = \nu_{2}(2^{2i+1}) \iff 2n = 2i+1 \]](http://latex.artofproblemsolving.com/7/a/6/7a6348d0407ae0c77bec0e001bd82a88fffb8ec4.png)
which is clearly false. So all the all the elements of set 
are distinct.
numbers. Now consider the quantity ![$\lambda = \sum_{i=1}^{n-1} S_i[0] + S_i[1]$](http://latex.artofproblemsolving.com/a/4/b/a4b0e6660c7617273e05b6fd4102806d4ce6bc6d.png)
.![\[ \nu_2(\lambda) = \nu_{2}\left(\sum_{i=1}^{n-1} S_i[0] + S_i[1]\right)= \nu_{2}\left(\left(\sum_{i=2}^{n-1} S_i[0]\right) + 2^{2n}\right) = \nu_2{2^{2\cdot 2}} = 4. \]](http://latex.artofproblemsolving.com/9/d/6/9d6cced0b9a9b8913c87bdbf5b626e0fe459d4c6.png)
.
argument (the condition 
), it is easy to note that ![$\lambda -2^{2n} \not= S_2[0]$](http://latex.artofproblemsolving.com/6/5/e/65e508a4aa230333c2f7ab888ece2ddd1292f8af.png)
or ![$S_2[1]$](http://latex.artofproblemsolving.com/1/9/a/19ab99086b9b5ab9dda6a4f136d712eaa199a8a6.png)
.
are all unique.
is our desired set.
and 
.![$S_{n}[0]$](http://latex.artofproblemsolving.com/3/9/9/3996bbf8e14e46253bd2613ca24100b47ef171da.png)
with ![$\left\{S_{n}[0], S_{n}[1]\right\}$](http://latex.artofproblemsolving.com/c/d/f/cdf24dfb8b60840ce7e939e5de77b561be0696a3.png)
as they both have the same sum. This increases ![$S_{i+1}[0]$](http://latex.artofproblemsolving.com/2/d/d/2dd139c6e9fa4f9bdf5811cf3859657e0de28045.png)
with ![$\left\{S_i[0], S_i[1]\right\}$](http://latex.artofproblemsolving.com/1/a/3/1a3b8a0e6df8df30d91f65e3bdb6187c5272a6e5.png)
till we reach 
.
It literally took me 
hours to rigorize the idea and then type it. 
,
,
,
and 
.
for 
and some 
for 
.![\[\{a_0,a_1,\dots, a_{n-2}, s_1, s_2, \dots, s_{n-2}, X_1, X_2, C\}\]](http://latex.artofproblemsolving.com/1/9/1/1918382360a99f1bcdaff4c290078d6df0f2c23b.png)
for sufficiently large 
and 
such that 
is half the sum of the whole set works. Indeed, the subsets that are half the sum of the whole thing are![\[ \{C, s_{n-2}\}, \{C, s_{n-3}, a_{n-2}\}, \{C, s_{n-4}, a_{n-2}, a_{n-3}\}, \dots, \{C, s_{1}, a_{n-2}, a_{n-3}, \dots, a_2\}, \{C, a_{n-2}, a_{n-1}, \dots, a_{0}\} \]](http://latex.artofproblemsolving.com/d/b/b/dbb287f9a69a3a162c089c859d93dcdbfac8a7f9.png)
.
for 
,
for 
.
and be a scaled version of an already working set with 
elements be 
.
with sum 
.
,
elements with sum 
from the first 
.
,
elements with sum 
,
.
.
,
,![\[U=\{F_{2n-2m+4},F_{2n-2m+6},\dots,F_{2n-2}\}\cup\{F_{2n-2m+3},F_{2n-2}-2\}\]](http://latex.artofproblemsolving.com/1/3/d/13d1d0841347dfced618548e751c29d623003910.png)
and 
.
.
,
)
element from right.
which clearly works.
)
from 
and 
.
,![\[\sum \text{$1$ thing} = \sum \text{$2$ things} = \cdots = \sum \text{$n-1$ things} = x\]](http://latex.artofproblemsolving.com/0/3/7/037a9530cab25ed20840d86b44c60a54a4967071.png)
To show that this is true, assume that the sum of the 
,
different sets above. Note that this requires 
.![\[ F_1, F_2, F_3, F_4, \dots, F_{2n-1} .\]](http://latex.artofproblemsolving.com/0/5/9/059a4a96b9f88052c03672a3664b5f6448e4f60a.png)
Here, 
and 
.
.
