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(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be the incircle of triangle 
. Line 
is parallel to side 
and tangent to . Line 
is parallel to side 
and tangent to . It turned out that the intersection point of and lies on circumcircle of 
satisfy all conditions below:
for all 
for all 
for all
.
and 
, then 
, then we also conclude that 
.
and the given condition, we obtain that
hence 
.
incredibly large in . Fixing 
, we obtain contradiction.
denote the assertion 
We claim that there is no such function. WLOG by shifting.
by setting 
for positive integer 
and making indefinitely large.
define 
to be the statement: 
for all 
Note that since 
is a real number, there exists a maximum value 
for which is true. Now, summing these four expressions up:
we get that 
, or 
to be exact. Since 
, 
implies that is true for infinitely large , contradiction.
Then, let 
.
for nonnegative integers . We will use induction. Clearly this is true for 
. We will use induction. Letting 
and we have 
which completes the induction. Similarly, if we let 
, 
Now, if we plug in and 
, we get 
Thus, from the two above claims, we have 
which is just 
which is clearly a contradiction for sufficiently large , QED.
. We claim that 
for some positive constant 
. This proves that the function is impossible because we would have it define two values.![\[\frac{f(x+n) + f(x)}{2} \geq f \left( \frac{2x+n}{2} \right) + n\]](http://latex.artofproblemsolving.com/c/4/6/c46ec4098212388a58517aacf5aa3b1327e02797.png)
. Thus we can see that over each infinitessimaly small segment of length (we choose so that is effectively linear). Notice that we must pass through the point that is units down from the midpoint of this segment (it's also perpendicular). This means that the second derivative is always greater than that constant.
, the very convex condition is equivalent to 
, which will be used extensively below.. Proceeding under this assumption, we now prove two claims.
is very convex, and achieves its minimum value at 
.
so 
is very convex. Also, 
, so the minimum possible value of 
is 
, achieved at . now; observe that for any real 
, the function 
is also very convex. Thus, define 
, which is bounded from below by 
and also very convex. We now prove the key, contradictory claim.
, we have 
.
. Since 
is very convex and bounded below by , we have 
as desired. Now, we perform the inductive step. Assume the claim holds when 
, for some integer 
, and we will show it also holds when 
. We have 
This completes the inductive step and the proof., the value of 
must be arbitrarily large, and hence not equal to any real value. Thus, our initial assumption that a very convex function exists must be false. The proof is complete. 
denote the assertion in the problem. Note that since shifting by a constant amount doesn't affect the inequality, we may assume WLOG that 
. and positive reals, 
.
for a positive real gives us ![\[f(x) + f(-x) \geq 2f(0)+ 4x = 4x,\]](http://latex.artofproblemsolving.com/6/7/c/67cf7dbe20a75a15417ac809bd027a9c481b040c.png)
as claimed. for some . Then, summing 
and 
for a positive real gives us![\[ f(x) + f(-x) \geq 2f( \tfrac{x}{2} ) + 2 f(- \tfrac{x}{2} ) + 4x \geq (4(n+1)) \cdot x, \]](http://latex.artofproblemsolving.com/1/3/7/137bb7eadf476986bf65157c9bb4ab7f726e5975.png)
which finishes our inductive step.
we get ![\[f(a) + \frac{f(a-2d)+f(a+2d)}2 \geq 4d+f(a+d)+f(a-d) \geq 8d+2f(a).\]](http://latex.artofproblemsolving.com/f/e/5/fe540c8893c68e0bc17b0553ab479214593b56ed.png)
In other words, it is true that ![\[\frac{f(a-2d)+f(a+2d)}2 \geq f(a) + 8d.\]](http://latex.artofproblemsolving.com/5/4/f/54f20f59512e2aa77bfe1a97b997e734826aac89.png)
Applying this same method inductively yields a bound of 
for all positive integers , which is clearly impossible.
. Let denote the assertion to the statement. If 
works, then so does 
where is some constant. Therefore, without loss of generality, assume that .
for any 
and 
.. For the base case, just take 
and you'll get 
. For the inductive step, assume 
. Consider and to get
Add both the inequalities, to get
which completes the proof by induction. 
for all natural numbers . Taking to be large enough, we get the desired contradiction finishing the solution. 
be the assertion that 
be an arbitrary nonnegative integer. We claim that, for all integers 
we have 
yields 
so the base case holds. be an integer such that 
Then, 
yields 
so 
Also, 
yields 
so 
so by the principle of mathematical induction, for all integers 
yields 
for all nonnegative integers If 
then setting 
yields 
a contradiction. Thus, there exists no very convex function, as desired. 
This is clearly allowed.
Simple induction implies that 
which is impossible.
very convex if![\[ \frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|
\]](http://latex.artofproblemsolving.com/3/b/a/3ba7733caf51cc70a090920ca21f8b47aad9b1ec.png)
and 
.
replaced by 
,![\[\displaystyle \frac{f(x)+f(y)}{2} - f\biggl( \frac{x+y}{2} \biggr) \ge 2^n|x-y|\]](http://latex.artofproblemsolving.com/d/e/1/de1963fa03c475ad0e722990a312aa5ba91ca693.png)
for all non negative integers 
real, ![\[\displaystyle \frac{f(a)+f(a+2b)}{2} \ge f(a+b)+2^{n+1}|b|,\]](http://latex.artofproblemsolving.com/e/c/a/eca563b9be7b0a4aba090b638dea06c4d01d5e3c.png)
![\[f(a+b)+f(a+3b) \ge 2(f(a+2b)+2^{n+1}|b|),\]](http://latex.artofproblemsolving.com/3/b/1/3b155ce9cd1c1c9f6774b9556864eb38ce666dc6.png)
and ![\[\displaystyle \frac{f(a+2b)+f(a+4b)}{2} \ge f(a+2b)+2^{n+1}|b|,\]](http://latex.artofproblemsolving.com/4/3/0/4304adbe0c667d36e1e0787a355f447658e4d427.png)
Summing the three enqualities and setting 
completes the induction. Now we only have to notice that the inequality we have proved cannot hold for all arbitrarily large n.
be the number which is "right next to 
on the number line." Now, let 
,
,
.![\[ \frac {f(m - \epsilon) + f(m + \epsilon)}{2}\ge f(\frac {m + \epsilon + m - \epsilon}{2}) + |m + \epsilon - (m - \epsilon)| = f(m) + 2\epsilon
\]](http://latex.artofproblemsolving.com/3/d/f/3df4bed6b0aa6421c4af4d359be4464419cb92aa.png)
![\[ \frac {a + c\epsilon + a - b\epsilon}{2}\ge a + 2\epsilon\implies \epsilon(c - b)\ge 4\epsilon\implies c - b\ge 4
\]](http://latex.artofproblemsolving.com/9/e/e/9eec7ac50e19ac194d2888277d0098325f3e6a28.png)
and 
is at least 
less than the slope between 
.
.
less than the slope between 
.
less than the slope between 
and 
.
,
.
has infinitely many values, namely those on the vertical line. Hence, 
![\[ f(x) \ge |2nx| - (2^n - 1)f(0) + 2^n f(\frac {x}{2^n})
\]](http://latex.artofproblemsolving.com/c/b/4/cb469aed313b367ca4ec19efa234d26cc239f6aa.png)
is above. Assume it's true for 
.
![\[ f(\frac {x}{2^n}) \ge |\frac {x}{2^{n - 1}}| - f(0) + 2f(\frac {x}{2^{n + 1}})
\]](http://latex.artofproblemsolving.com/6/d/0/6d05a85e1a4fedbeb554904eb38441e922acede1.png)
![\[ 2^n f(\frac {x}{2^n}) \ge |2x| - 2^n f(0) + 2^{n + 1} f(\frac {x}{2^{n + 1}})
\]](http://latex.artofproblemsolving.com/1/3/d/13dd33522ef5cc16e2789f0985db7aa859f8a22d.png)
![\[ f(x) \ge |2nx| - (2^n - 1)f(0) + 2^n f(\frac {x}{2^n}) \ge |2nx| - (2^n - 1)f(0) + |2x| - 2^n f(0) + 2^{n + 1} f(\frac {x}{2^{n + 1}}) = |2(n + 1)x| - (2^{n + 1} - 1) f(0) + 2^{n + 1} f(\frac {x}{2^{n + 1}})
\]](http://latex.artofproblemsolving.com/0/9/c/09cf9b99d95b903e549750120e56f3478f2e8c42.png)
and 
and solving for 
we have![\[ f(\frac {1}{2^n}) \le \frac {(2^n - 1)f(0) - |2n| + f(1) } {2^n}
\]](http://latex.artofproblemsolving.com/1/7/8/17864dfc3e7769df504a1c25ff615ed24963ab40.png)
![\[ f(\frac { - 1}{2^n}) \le \frac {(2^n - 1)f(0) - |2n| + f( - 1) } {2^n}
\]](http://latex.artofproblemsolving.com/f/5/b/f5bca2833cbb701f9521bf6bbaf9bb6dac9abad7.png)
sufficiently large that 
.
.
and 
shows that it is not very convex and we are done.
as![\[ \lim_{h \to 0} \frac{\lim_{h \to 0} \frac{f(x+2h)-f(x+h)}{h} - \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}}{h} = \lim_{h \to 0} \frac{f(x+2h)+f(x) - 2f(x+h)}{h^2} \ge \lim_{h \to 0} \frac{|h|}{h^2} = \infty\]](http://latex.artofproblemsolving.com/4/6/8/4688f8ccea6c0cbaa8aa835ef049a1389ff513d1.png)
where the operations (i.e. taking the two h's to be the same) can be justified by uniform convergence. This would only work if 
,
,![\[ \frac{g(x)+g(y)}{2}\geq g\left(\frac{x+y}{2}\right)+|x-y|,\]](http://latex.artofproblemsolving.com/4/2/d/42d73975d11996b1aa6f74a4b310b5205a6c7f06.png)
so we can consider the very convex function 
instead, with 
.
,![\[ \frac{g(x)}{2} \geq g\left(\frac{x}{2}\right)+|x|.\quad \left(*\right)\]](http://latex.artofproblemsolving.com/d/a/4/da475a3f76131be7fe82b732f45670a167f94ee8.png)
.
,
.
,![\[ \frac{g(x)}{2}\geq 2^{k-1} g\left(\frac{x}{2^{k}}\right)+|kx|. \quad \left(**\right)\]](http://latex.artofproblemsolving.com/6/c/2/6c233a433a8202e8a898399303405c184e11c414.png)
Plugging in 
into 
.
gives us 
,
gives us ![\[ \frac{g(x)}{2}\geq 2^{k} g\left(\frac{x}{2^{k+1}}\right)+|(k+1)x|,\]](http://latex.artofproblemsolving.com/2/d/7/2d72d3bdc14e4b57b48783b2f2af0067fb1a3a6d.png)
completing the inductive process.
denote the assertion. Note that for any 
,
and 
Therefore, we have that 
Now, note that 
so 
so 
i.e. 
Then, it follows that 
so 
is very convex. However, by the above, this would imply that 
However, then we can deduce that 
is convex by dividing everything by 
,
is convex for every nonnegative integer 
,
Then, 
Plugging in 
,
,
for all nonnegative integers 
