AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, we have:
such that there exists a sequence of positive integers 
, 
,
, 
satisfying: ![\[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\]](http://latex.artofproblemsolving.com/7/b/4/7b4ab849cdc409906514e2ebb688802f17df41ea.png)
for every 
with 
.
, there exists the positive integers 
, such that for 
, With 
may be formed as a triangle side length , and the area of the triangle is a positive integer.
(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.
.
students bought some books in a bookstore. It is known that every student bought exactly three kinds of books, and any two of them shared at least one kind of book. Determine, with proof, how many students bought the most popular book at least? (Note: the most popular book means most students bought this kind of book)
in it, and every other time, Henrik picks a triangle where he writes a 
. If one of the players gets four in a row in some direction (see figure), he wins the game.
on the diagonal 
of a given rectangle 
, such that 
. The point 
is the intersection point between 
and the parallel line to 
that contains 
. Prove that the triangle 
is equilateral if and only if is a square.
satisfies 
. Find the value of 
.
be a convex quadrilateral, such that
and 
. is a kite.
be a triangle with orthocentre 
. The tangent lines from 
to the circle with diameter 
touch this circle at 
and 
. Prove that 
and are collinear.
be a sequence that consists of an initial block of 
positive distinct integers that then repeat periodically. This means that 
are distinct positive integers and 
for every positive integer . The terms of the sequence are not known and the goal is to find the period . To do this, at each move it possible to reveal the value of a term of the sequence at your choice. is one of the first prime numbers, find for which values of there exist a strategy that allows to find revealing at most 
terms of the sequence.
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 
. Prove that no very convex function exists.
exists, then ![\[\displaystyle \frac{f(x)+f(y)}{2} - f\biggl( \frac{x+y}{2} \biggr) \ge 2^n|x-y|\]](http://latex.artofproblemsolving.com/d/e/1/de1963fa03c475ad0e722990a312aa5ba91ca693.png)
for all non negative integers . We are given the base case 
; now, if the inequality holds for a given , then for 
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.
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 is twice-differentiable, but I wonder whether there is some way to rigorize this idea?
prove that 
Prove that
replaced by 
,
must be constant. Try searching, there's a solution there.
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.
,
,![\[ \frac{g(x)+g(y)}{2}\geq g\left(\frac{x+y}{2}\right)+|x-y|,\]](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.
.
and 
,
,
.
and the given condition, we obtain that
hence 
.
incredibly large in 
,
denote the assertion 
We claim that there is no such function. WLOG 
by setting 
for positive integer 
define 
to be the statement: 
for all 
Note that since 
is a real number, there exists a maximum value 
for which 
we get that 
,
to be exact. Since 
,
implies that 
Then, let 
.
for nonnegative integers 
and 
which completes the induction. Similarly, if we let 
,
Now, if we plug in 
,
Thus, from the two above claims, we have 
which is just 
which is clearly a contradiction for sufficiently large 
for some positive constant 
.![\[\frac{f(x+n) + f(x)}{2} \geq f \left( \frac{2x+n}{2} \right) + n\]](http://latex.artofproblemsolving.com/c/4/6/c46ec4098212388a58517aacf5aa3b1327e02797.png)
. Thus we can see that over each infinitessimaly small segment of length 
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 
,
is very convex, and achieves its minimum value at 
.
so 
,
is 
,
,
is also very convex. Thus, define 
,
and also very convex. We now prove the key, contradictory claim.
.
.
is very convex and bounded below by 
as desired. Now, we perform the inductive step. Assume the claim holds when 
,
,
.
This completes the inductive step and the proof.
must be arbitrarily large, and hence not equal to any real value. Thus, our initial assumption that a very convex function 
.
.
for a positive real ![\[f(x) + f(-x) \geq 2f(0)+ 4x = 4x,\]](http://latex.artofproblemsolving.com/6/7/c/67cf7dbe20a75a15417ac809bd027a9c481b040c.png)
as claimed.
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 
works, then so does 
where 
for any 
and 
.
and you'll get 
.
.
Add both the inequalities, to get
which completes the proof by induction. 
for all natural numbers 
we have 
yields 
so the base case holds.
Then, 
yields 
so 
Also, 
yields 
so 
so by the principle of mathematical induction, 
yields 
for all nonnegative integers 
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.