be a triangle with incircle 
,
and 
at points 
and 
,
and 
on the line 
such that 
and 
.
be the intersection of lines 
and 
.
and 
as the intersections of 
and 
with lines 
and 
,
is the intersection of 
and 
.
.
such that 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
,
and orthocenter 
.
is constant.
such that![\[
f(f(x)+y) = x+f(f(y))
\]](http://latex.artofproblemsolving.com/9/d/7/9d780faa5901646f7035e88400a9e3d8023f04ff.png)
for all 
.
Y by MS_asdfgzxcvb
Let 
be a set of 2025 positive real numbers. For a subset 
, define 
as the median of 
when all elements of are arranged in increasing order, with the convention that 
. Define
![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](//latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set of 2025 positive real numbers, the following inequality holds:
![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](//latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in .
be a set of 2025 positive real numbers. For a subset 
, define 
as the median of 
when all elements of are arranged in increasing order, with the convention that 
. Define![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set of 2025 positive real numbers, the following inequality holds:![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in .
,
be the elements of 
.
and 
as linear combinations of the 
.
unless 
.
by Vandermonde, so
,
are the middle two elements of a subset 
of 
by Vandermonde again, and so
We can collapse this double sum to a single sum using the hockey stick identity. We write![\begin{align*}
Q(A) &= \frac12\sum_{j=1}^{N-1}x_j\sum_{k=j+1}^N\binom{N-k+j-1}{j-1} + \frac12\sum_{k=2}^Nx_k\sum_{j=1}^{k-1}\binom{N-k+j-1}{j-1} \\
&= \frac12\sum_{j=1}^{N-1}\binom{N-1}jx_j + \frac12\sum_{k=2}^N\binom{N-1}{k-2}x_k \\
&= \frac12\sum_{k=1}^N\left[\binom{N-1}k + \binom{N-1}{k-2}\right]x_k,
\end{align*}](http://latex.artofproblemsolving.com/3/3/d/33d57acdff6dcb6f5a34899638b32e02a010928c.png)
and then![$$P(A) - Q(A) = \sum_{k=1}^N\left[-\frac12\binom{N-1}k + \binom{N-1}{k-1} - \frac12\binom{N-1}{k-2}\right]x_k.$$](http://latex.artofproblemsolving.com/2/9/5/295dc6f03843509249132cebba7b2ed6e48b33d1.png)
Now to account for the fact that 
,
denote their consecutive differences; that is, 
and 
for 
.
for each 
,
Then we rearrange and telescope the double sum to get![\begin{align*}
P(A) - Q(A) &= \sum_{k=1}^N\sum_{j=1}^k\left[-\frac12\binom{N-1}k + \binom{N-1}{k-1} - \frac12\binom{N-1}{k-2}\right]a_j \\
&= \sum_{j=1}^Na_j\sum_{k=j}^N\left[-\frac12\binom{N-1}k + \binom{N-1}{k-1} - \frac12\binom{N-1}{k-2}\right] \\
&= \frac12\sum_{j=1}^N\left[\binom{N-1}{j-1} - \binom{N-1}{j-2} + 1\right]a_j.
\end{align*}](http://latex.artofproblemsolving.com/7/7/a/77a55eb014b012ab67a7b8f266fb0c9ad4574b4d.png)
Meanwhile,
Now for each 
,![$$c_j = \frac12\left[\binom{N-1}{j-1} - \binom{N-1}{j-2} + 1\right]$$](http://latex.artofproblemsolving.com/b/3/2/b32540eb826f5dfab3a8af48551ba97b4e6cd058.png)
be the coefficient of 
in 
.
,
,
by taking 
to be arbitrarily close to zero. However, if 
for all 
term-by-term. So the answer we seek is![$$C = \max_{1\le j\le N}c_j = \frac12 + \frac12\max_{1\le j\le N}\left[\binom{N-1}{j-1} - \binom{N-1}{j-2}\right].$$](http://latex.artofproblemsolving.com/0/c/4/0c4276e1c6226b976a5a0341087267988660e184.png)
and
Subtracting these gives![$$\left[\binom{N-1}j - \binom{N-1}{j-1}\right] - \left[\binom{N-1}{j-1} - \binom{N-1}{j-2}\right] = \left[\frac{N-2j}j - \frac{N-2j+2}{N-j+1}\right]\binom{N-1}{j-1}.$$](http://latex.artofproblemsolving.com/6/5/f/65fda319d018414ea15d4f22ee737b61cc4ef7ce.png)
If this is positive, then 
gives a larger difference than 
This quadratic has roots
So as long as 
and so the least possible value of 
is
for this 
,
and so
.
and 
so
The expression 
is positive when 
.
.
and 
.
