function
arithmetic sequence
IMO
IMO 2009
IMO Shortlist

Call $ d_1,d_2$ the differences in the arithmetic progressions $ s_{s_i}$ and $ s_{s_i\plus{}1}$.  Then, since $ s_i$ is increasing,
\[ s_{s_1\plus{}1}\plus{}(n\minus{}1)d_1>s_{s_1}\plus{}(n\minus{}1)d_1\equal{}s_{s_n}\geq s_{s_{n\minus{}1}\plus{}1}\equal{}s_{s_1\plus{}1}\plus{}(n\minus{}2)d_2,\]
or for all $ n$, $ d_1>(n\minus{}2)(d_2\minus{}d_1)$.  If $ d_2>d_1$, we reach an easy contradiction when $ n$ grows without bounds.  Similarly,
\[ s_{s_1\plus{}1}\minus{}s_{s_1}\equal{}s_{s_n\plus{}1}\minus{}s_{s_n}\plus{}(n\minus{}1)(d_1\minus{}d_2)>(n\minus{}1)(d_1\minus{}d_2),\]
again reaching an easy contradiction if $ d_1>d_2$.  Hence $ d_1\equal{}d_2\equal{}d$ is the common difference between both arithmetic progressions, and $ s_{s_n\plus{}1}\minus{}s_{s_n}\equal{}s_{s_1\plus{}1}\minus{}s_{s_1}\equal{}D$ is constant for all $ n$.

[quote="daniel73"]Hence $ d_1 \equal{} d_2 \equal{} d$ is the common difference between both arithmetic progressions, and $ s_{s_n \plus{} 1} \minus{} s_{s_n} \equal{} s_{s_1 \plus{} 1} \minus{} s_{s_1} \equal{} D$ is constant for all $ n$.[/quote]This doesn't solve the problem though it might be an intermediary step.

[quote="Valentin Vornicu"]This doesn't solve the problem though it might be an intermediary step.[/quote]

That's the idea, I am still working on finishing up the problem using this intermediate step, maybe someone beats me to the answer using this first step. After all, this is not the contest, and I think collaborative efforts are OK...   :D

It's true that what I wrote may seem fairly obvious, but I'll be glad if it helps someone out there working on this problem...  :)

Let $ d$ be a common difference of $ \{ s_{s_{i}} \}$.
Since $ \{ s_{i} \}$ is a strictly increasing sequence, it is obvious that $ |k \minus{} l|\leq|s_{k} \minus{} s_{l}|$ for all natural numbers $ k,l$.
Therefore, $ |s_{i \plus{} 1} \minus{} s_{i}| \leq |s_{s_{i \plus{} 1}} \minus{} s_{s_{i}}| \equal{} d$ $ \forall i \in N$.
Thus, $ |s_{s_{i} \plus{} 1} \minus{} s_{s_{i}}|\leq d$.
Therefore, a common difference of $ \{ s_{s_{i} \plus{} 1} \}$ is $ d$ as well.
It indicates that for all positive integer $ i$, there exists a constant $ c$ such that $ s_{s_{i} \plus{} 1} \equal{} s_{s_{i}} \plus{} c$.
Also, we know that a set of possible values of difference between two consecutive terms is both upper and lower bounded.
Let $ m$ and $ M$ be the maximum and the minimum value of the differences.
Then, there exist positive integers $ m'$ and $ M'$ satisfying $ s_{M' \plus{} 1} \minus{} s_{M'} \equal{} M$ and $ s_{m' \plus{} 1} \minus{} s_{m'} \equal{} m$.
Since $ s_{s_{M' \plus{} 1}} \minus{} s_{s_{M'}} \equal{} d$ and $ s_{s_{M'} \plus{} M} \equal{} s_{s_{M' \plus{} 1}}$, $ s_{s_{M'} \plus{} M} \minus{} s_{s_{M'}} \equal{} d$.
By the definition, $ mM \leq s_{s_{M'} \plus{} M} \minus{} s_{s_{M'}}\leq M^{2}$.
Therefore, $ mM \leq d \leq M^{2}$.
Using the same logic for $ m'$, we also can derive that $ m^{2} \leq d \leq mM$.
Combining two inequalities, $ d \equal{} mM$.
Therefore, all consecutive differences between $ s_{s_{M'} \plus{} M}$ and $ s_{s_{M'}}$ are $ m$, and those between $ s_{s_{m'} \plus{} m}$ and $ s_{s_{m'}}$ are $ M$.
Therefore, $ s_{s_{M'} \plus{} 1} \minus{} s_{s_{M'}} \equal{} m \equal{} c$. In the same way, $ M \equal{} c$.
Therefore, $ m \equal{} M$, which implies that $ \{ s_{i} \}$ is an arithmetic progression.

Note that $ \displaystyle s_{s_{k}\plus{}1}\minus{}s_{s_k}\equal{}s_{s_1\plus{}1}\minus{}s_{s_1}\geq s_{s_{k}\plus{}1}\minus{}s_{s_{k\minus{}1}\plus{}1}\equal{}r$.Also we have that $ s_{s_{1}\plus{}1}\minus{}s_{s_{1}}\leq s_{s_2}\minus{}s_{s_1}\equal{}r$.Now the conclusion is $ s_{s_1\plus{}1}\minus{}s_{s_1}\equal{}r$ where $ r$ the ratio of the arithmeticprogression.The conclusion is that $ s_{k\plus{}1}\equal{}s_{k}\plus{}1$.

Maybe then consider ${ s_{s_{s_{n} + 1}}} - s_{s_{s_{n}}} = d$ and also $ s_{s_{s_{n}} + 1} - s_{s_{s_{n}}} = d$ or maybe I am doing something wrong :blush: ; but first equation is useful I think...

Edit: yeah, I was wrong, it's ${ s_{s_{s_{n} + 1}}} - s_{s_{s_{n}}} = d$; $ s_{s_{s_{n}} + 1} - s_{s_{s_{n}}} = D$; then $ s_{s_{s_{n}} + d} - s_{s_{s_{n}}} = D$, maybe this can lead to something...

I'll define ${ s_{s_{i+1}}-s_{s_i}}=A$, $ s_{s_{i+1}+1}-s_{{s_i}+1}=B$, $ s_{s_{i+1}+1}-s_{s_{i+1}}=d$.
First, we can show $ A=B$. If not, $ s_{s_{i+1}+1}-s_{s_{i+1}}$ is a strictly decreasing or increasing sequence, both makes contradiction.
(It's trivial to prove nondecreasing and if  $ s_{s_{i+1}+1}-s_{s_{i+1}}$ is increasing sequence, there is a number j such that $ s_{s_{j+1}+1}-s_{s_{j+1}} > B$, but this is a contradiction.
Then, $ d$ is a constant.
Suppose that $ s_{i+1}-s_{i} = m$. Then, $ s_{s_{i+1}}-s_{{s_i}}= s_{s_{i}+m}-s_{{s_{i}}}=B$.
This means if $ \min (s_{i+1}-s_{i}) =(s_{z+1}-s_{z}) J$, then $ \max (s_{i+1}-s_{i}) \leq B/J$
If $ B/J$ is not an integer, for pigeonhole principle, there is an integer $ s_{z} \leq Y < s_{z+1}$ such that $ s_{Y+1} -s_{Y} > B/J$.
So $ B/J$ is an integer, and there is a real example for this maximum bound.
Then, all number for $ s_{z} \leq Y < s_{z+1}$ ,  $ s_{Y+1} -s_{Y} = B/J$.
And, all number for $ s_{s_{z}} \leq Z < s_{s_{z+1}}$, $ s_{Z+1} -s_{Z} = J$.
Repeating similar process, for all $ K$, we can find $ K$ consecutive numbers such that $ s_{i+1} -s_{i} = J$.
Also, we can find $ K$ consecutive numbers such that $ s_{i+1} -s_{i} = B/J$
It's easy to prove $ d \leq B$, and we can let $ K>B$, then we can find $ s_{N}$ in K consecutive numbers.
Then, $ s_{s_{N}+1} -s_{s_{N}} = J$. So, $ d=J$.
Similar way, we can prove $ d=B/J$.
So, $ J=B/J$, which shows that the sequence is itself an arithmetic progression.
Sorry for my bad English.

I think the idea to solve this problem is too easy to be an idea of 3rd problem. Is there any trap or something deceptive in the problem? :(

My idea is to see the sequence as if it were a function.
The sequence is f(1), f(2),…

I prove that f(n) is an arithmetic sequence using only that f(f(n)) is an arithmetic sequence. I looked for a mistake, but I didn’t found it. Please someone correct me.

Let $ f(f(n) \equal{} a \plus{} dn$
Let $ g_{m}$=f(f(…f(n)…))  (m iterations of f)  for a fixed n
Then $ g_{m \plus{} 2} \equal{} dg_{m} \plus{} a$
Then $ g_{m} \equal{} w(\sqrt {d})^m \plus{} u( \minus{} \sqrt {d})^m \minus{} \frac {a}{d \minus{} 1}$
w and u are constants that satisfy w>u≥0  , that's because f is increasing
Then $ f(n) \equal{} g_{1} \equal{} (w \minus{} u)\sqrt {d} \minus{} \frac {a}{d \minus{} 1}$  and $ n \equal{} (w \plus{} u) \minus{} \frac {a}{d \minus{} 1}$
Then $ f(n) \plus{} \frac {a}{d \minus{} 1}\leq(n \plus{} \frac {a}{d \minus{} 1})\sqrt {d}$

But this is true for every value of n

Then $ f(f(n)) \plus{} \frac {a}{d \minus{} 1}\leq(f(n) \plus{} \frac {a}{d \minus{} 1})\sqrt {d}\leq(n \plus{} \frac {a}{d \minus{} 1})d$
Then $ f(f(n))\leq(a \plus{} dn) \equal{} f(f(n))$
Then the inequalities are equalities  and then $ f(n) \plus{} \frac {a}{d \minus{} 1} \equal{} (n \plus{} \frac {a}{d \minus{} 1})\sqrt {d}$
Then $ f(n) \equal{} n\sqrt {d} \plus{} c$   (c is a constant)
Then f(n) is an arithmetic sequence

Why can't $ u$ be negative?  The sequence $ f(n)\equal{}4 * 5^n \minus{} (\minus{}5)^n$ is still increasing...

[quote="m.candales"]Then $ g_{m \plus{} 2} \equal{} dg_{m} \plus{} a$
Then $ g_{m} \equal{} w(\sqrt {d})^m \plus{} u( \minus{} \sqrt {d})^m \minus{} \frac {a}{d \minus{} 1}$
w and u are constants that satisfy w>u≥0  , that's because f is increasing
[/quote]I don't think this is correct. In fact it seems that $ g_{2m} \equal{} d^{2m} \left( \frac a{d \minus{} 1} \plus{} n \right) \minus{} \dfrac a{d \minus{} 1}$, if we consider $ g_0 \equal{}n$ ...

u doesn't have to be non-negative; I was wrong asuming that

This was an amusing little problem, but it seems like day 1 was overall quite easy  :maybe:

I've tried to rewrite this problem into:

Suppose there exists a function $ f: Z^ \plus{} \rightarrow Z^ \plus{}$ is strictly increasing and
$ f(a(x \minus{} 1) \plus{} b) \equal{} a(f(x) \minus{} 1) \plus{} b$ for all integers x, where a and b are positive integers.

Show that f is linear.

But now it's very late, and I haven't solved it yet  :mad: . Could anyone help me?

Let $X_i=s_{s_i}$ be an arith. prog. and let $Y=s_{s_i+1}$ be also an arith. prog. Notice that, if the difference in $X$ were greater than the difference in $Y$, then eventually $X_i > Y_{i+C}$ for any constant $C$. But since $X_i < Y_i$, this is impossible. Similarly, if the difference in $Y$ were greater than the difference in $X$, then eventually
$Y_i>X_{i+C}$ for any constant $C$. But since $Y_i \le X_{i+1} < X_{i+2}$, this is impossible. So these two arith. progs. have the same difference, and so

$s_{s_i+1}-s_{s_i}$

is constant. Let $d$ be the common difference of $X$ and $Y$.

Let $a_i=s_{i+1}-s_i$ for all $i$. Now notice that if, for some $i$, $a_i>d$ then

$d=s_{s_{i+1}}-s_{s_i}=a_{s_i}+a_{s_i+1}+...+a_{s_{i+1}-1} > d$

since we have more than $d$ positive numbers. Contradiction, so $a_i \le d$ for all $i$.

Now take $k$ and $l$ such that $a_k=m$ is minimal and $a_l=M$ is maximal.

We have

$d=s_{s_{l+1}}-s_{s_l}=a_{s_l}+...+a_{s_{l+1}-1} \ge mM$

since we have $M$ numbers that are at least $m$. But also

$d=s_{s_{m+1}}-s_{s_m}=a_{s_m}+...+a_{s_{m+1}-1} \le mM$

since we have $m$ numbers that are at most $M$.

So $d=mM$ and those two equations will guarantee

$a_{s_l}=...=a_{s_{l+1}-1}=m$

and

$a_{s_m}=...=a_{s_{m+1}-1}=M$

Finally, recall that $a_{s_p}$ was constant for all $p$. And so $M=m$, therefore $a_p$ is constant for all $p$, and so $s$ is an arith. prog.!

[hide=Solution]
Denote $s_{s_n}=a_1n+b$ and $s_{s_n+1}=a_2n+c$.
[b]Claim 1:[/b] $a_1=a_2$.
[b]Proof:[/b] $s_{s_n+1}>s_{s_n}$ gives $a_2n+c>a_1n+b$, so letting $n\to\infty$, we see that $a_2\geq a_1$. $s_{s_{n+1}}\geq s_{s_n+1}$ gives $a_2n+a_2+c\geq a_1n+b$, so letting $n\to\infty$, we see that $a_1\geq a_2$. Thus $a_1=a_2$.

Denoting $a_1=a_2=a$ and $c-b=\Delta$, $s_{s_n}=an+b$ and $s_{s_n+1}=an+b+\Delta$. Thus $\Delta>0$.

Define $u=\min(s_{n+1}-s_n)_{n\geq1}$ and $v=\max(s_{n+1}-s_n)_{n\geq1}$.
[b]Claim 2:[/b] $u,v$ are finite.
[b]Proof:[/b] This is obvious for $u$. For $v$, it suffices to prove that $\max(s_{n+1}-s_n)_{n\geq s_1}$ is finite, since $s_1$ is finite. Suppose $n\geq s_1$, and let $k$ be the maximum $k$ such that $s_k\leq n$ (This exists since $k=1$ works, and is finite since $\{s_n\}$ is obviously unbounded). Note that this implies $s_{k+1}\geq n+1$, so
\[s_{n+1}-s_{n}\leq s_{s_{k+1}}-s_{s_k}=a.\]
Thus, $\max(s_{n+1}-s_n)_{n\geq s_1}$ is bounded, so $\max(s_{n+1}-s_n)_{n\geq1}$ is too, and thus $v$ is finite.

Let $x,y$ be such that $s_{x+1}-s_x=u$ and $s_{y+1}-s_y=v$.

[b]Claim 3:[/b] $a=uv$.
[b]Proof:[/b] Note that
\[a=s_{s_{x+1}}-s_{s_x}=s_{s_x+u}+s_{s_x}=\sum_{i=1}^{u}s_{s_x+i}-s_{s_x+i-1}\leq\sum_{i=1}^{u}v=uv\]
and
\[a=s_{s_{y+1}}-s_{s_y}=s_{s_y+v}+s_{s_y}=\sum_{j=1}^{v}s_{s_y+j}-s_{s_y+j-1}\geq\sum_{j=1}^{v}u=uv\]
imply that $a=uv$.

[b]Claim 4:[/b] $u=v=\Delta$.
[b]Proof:[/b] Since $a=uv$, equality must hold in the above inequalities. Thus $s_{s_x+i}-s_{s_x+i-1}=v$ for $i=1,2,\ldots,u$ and $s_{s_y+i}-s_{s_y+i-1}=u$ for $j=1,2,\ldots,v$. Letting $i=j=1$,
\[u=s_{s_y+1}-s_{s_y}=\Delta=s_{s_x+1}-s_{s_x}=v\]
as desired.

[b]Claim 5:[/b] $s_1,s_2,s_3,\ldots$ is an arithmetic sequence.
[b]Proof:[/b] From the definitions of $u$ and $v$, we see that
\[\min(s_{n+1}-s_n)_{n\geq1}=\max(s_{n+1}-s_n)_{n\geq1},\]
so $s_{n+1}-s_n$ is constant for all $n\geq1$. Thus $s_1,s_2,s_3,\ldots$ is an arithmetic sequence.
[/hide]

The first claim is that the sequences $\{s_{s_i}\}$ and $\{s_{s_i+1}\}$ have the same common difference. Note that the sequence $\{s_{s_i+1}-s_{s_i}\}$ is an arithmetic sequence as it is the difference of two arithmetic sequences. It must also have a nonnegative common difference, as otherwise we would eventually have $s_{s_i}>s_{s_i+1}$, contradicting that $\{s_i\}$ is increasing. But we have that $s_{s_{s_k}+1}-s_{s_{s_k}}\le s_{s_{s_k+1}}-s_{s_{s_k}}$ is the common difference of $\{s_{s_i}\}$, so $\{s_{s_{s_i}+1}-s_{s_{s_i}}\}$ is bounded above, implying that $\{s_{s_i+1}-s_{s_i}\}$ is also bounded above. This means that it is constant, so $\{s_{s_i}\}$ and $\{s_{s_i+1}\}$ have the same common difference, as desired. Let this common difference be $d$, and let $c=s_{s_i+1}-s_{s_i}$, which is the fixed difference between $\{s_{s_i}\}$ and $\{s_{s_i+1}\}$.

Now, we claim that $\{s_{i+1}-s_i\}$ is bounded above. Indeed, if $k$ is the unique integer such that $i\in[s_k,s_{k+1})$, then we have that $s_{i+1}-s_i\le s_{s_{k+1}}-s_{s_k}=d$, as desired. Now, let $M$ and $m$ be the maximum and minimum value of $\{s_{i+1}-s_i\}$ over all positive integers $i$. Suppose that $k$ is a positive integer such that $s_{k+1}-s_k=M$. Then, we have that $c+(M-1)m\le s_{s_k+1}-s_{s_k}+\sum_{i=s_k+1}^{s_k+M-1}(s_{i+1}-s_i)\le c+(M-1)M$, so $d-c\in[Mm-m,M^2-M]$ as the middle sum is simply $s_{s_{k+1}}-s_{s_k}$. Similarly, if $k$ is a positive integers such that $s_{k+1}-s_k=m$, then we have that $c+(m-1)m\le s_{s_k+1}-s_{s_k}+\sum_{i=s_k+1}^{s_k+m-1}(s_{i+1}-s_i)\le c+(m-1)M$. This implies that $d-c\in[m^2-m,Mm-m]$. As $m\le M\implies m^2-m\le Mm-M\le Mm-m\le M^2-M$, we must have that $m=M$ in order for the intervals $[Mm-m,M^2-M]$ and $[m^2-m,Mm-M]$ to intersect. But this then implies that the sequence $\{s_{k+1}-s_k\}$ is constant, so $\{s_k\}$ is an arithmetic sequence, as desired.

Wait, does this actually work? If so, then it's a very nice problem! :)

[quote=IMO 2009/3]Suppose that $ s_1,s_2,s_3, \ldots$ is a strictly increasing sequence of positive integers such that the sub-sequences \[s_{s_1},\, s_{s_2},\, s_{s_3},\, \ldots\qquad\text{and}\qquad s_{s_1+1},\, s_{s_2+1},\, s_{s_3+1},\, \ldots\] are both arithmetic progressions. Prove that the sequence $ s_1, s_2, s_3, \ldots$ is itself an arithmetic progression.

[i]Proposed by Gabriel Carroll, USA[/i][/quote]

[hide=solution] Let $d_1, d_2$ denote the common difference of the sequences $(s_{s_n})_{n \ge 1}$ and $(s_{s_n+1})_{n \ge 1}$ respectively. Write $s_{s_n}=a+nd_1$ and $s_{s_n+1}=b+nd_2$; we claim that $d_1=d_2$. Indeed, note that $s_{k+1} \ge s_k+1>s_k$ so $$a+(k+1)d_1=s_{s_{k+1}} \ge s_{s_k+1}=b+kd_2 >s_{s_k}=a+kd_1,$$ so letting $k \rightarrow \infty$ gives the conclusion. Let $d$ be this common difference for both the sequences.

Let $x_k=(s_{k+1}-s_k)$ for all $k \ge 1$. Notice the very useful identity $$s_m-s_n=\sum^{m-1}_{k=n} x_k$$ for all $m>n$. We claim that the sequence $(x_k)_{k \ge 1}$ is bounded from above and below. Indeed, the lower bound property holds due to well-ordering of $\mathbb{N}$ and upper bound occurs because $$d=(s_{s_{k+1}}-s_{s_k})=\sum^{s_{k+1}-1}_{j=s_k} x_j \ge (s_{k+1}-s_k-1).$$ Call the greatest lower bound of this sequence $m$ and the least upper bound as $M$. It suffices to show $m=M$.

Again, we use the very useful identity to obtain $$(s_{k+1}-s_k-1)m \le d=(s_{s_{k+1}}-s_{s_k}) \le (s_{k+1}-s_k-1)M,$$ Therefore, $d \le \min_{k \ge 1} {(x_k-1)M}=(m-1)M$ and $d \ge \operatorname{max}_{k \ge 1} {(x_k-1)m}=(M-1)m$; hence, $$(M-1)m \le d \le (m-1)M \implies M \le m \implies \boxed{M=m}$$ as desired. $\blacksquare$

[/hide]

Nice problem :D. Rephrase the problem with the flavour of functional equations:
[quote]
Suppose $f : \mathbb{N} \to \mathbb{N}$ is a strictly increasing function such that
\[ f(f(1)), f(f(2)), f(f(3)), \dots \ \ \text{and} \ \ f(f(1) + 1), f(f(2) + 1), f(f(3) + 1) \]
are both arithmetic progressions.
Prove that $f(1), f(2), f(3), \dots$ is itself an arithmetic progression.
[/quote]

Now, we have for any integer $a,n,m \in \mathbb{N}$. Let
\[ f(f(n)) = f(f(m)) + a(n - m) \]
\[ f(f(n) + 1) = f(f(m) + 1) + b(n - m) \]

[b]Claim 01.[/b] $a  = b$
[i]Proof.[/i] Since $f$ is strictly increasing, we have
\[ (a - b)(n - m) < f(f(m) + 1) - f(f(m)) \]
And since $(a - b)(n - m)$ is bounded from above and $n - m$ could grow sufficiently large. We must have $a \le b$.

Now, we know that $f(n + 1) \ge f(n) + 1$, since $f$ is strictly increasing. Therefore,
\[ f(f(n + 1)) \ge f(f(n) + 1) \]
\[ f(f(1)) + an \ge f(f(1) + 1) + b(n - 1) \]
\[ an - bn + b \ge f(f(1) + 1) - f(f(1)) > 0 \]
\[ (a - b)n + b > 0 \]
\[ b > (b - a)n \]
Notice that we have $b \ge a$, and $n$ could grow sufficiently large. Therefore, it is clearly a contradiction for $b > a$. Hence, we must have $b = a$.

[b]Claim 02.[/b] $f(n) - f(m) \le (n - m)a$ for any $n,m \in \mathbb{N}$
[i]Proof.[/i] Notice that for $n > m$, we have
\[ f(f(n)) = f( f(m) + f(n) - f(m) ) \ge f(f(m)) + f(n) - f(m) \]
Thus,
\[ (n - m)a = f(f(n)) - f(f(m)) \ge f(n) - f(m) \]

Define $d_n = f(n + 1) - f(n)$, now
[b]Claim 03.[/b] $d_{f(f(n))} = d_n$ for all $n \in \mathbb{N}$.
[i]Proof.[/i] From [b]Claim 02[/b], $d_n$ is bounded above by $a$.
For simplicity, let $f(f(n)) = Dn + x$ and $f(f(n) + 1) = Dn + y$.
Take $k$ that maximizes the value of $d_k$. Now, consider counting $f(f(f(n + 1))) - f(f(f(n)))$ in two ways.
Then, we have
\[ f(Dn + D + x) - f(Dn + x) = D( f(n + 1) - f(n) ) = D \cdot d_n \]
\[ \sum_{i = 0}^{D - 1} f(Dn + x + i + 1) - f(Dn + x + i) = D \cdot d_n \]
\[ \sum_{i = 0}^{D - 1} d_{Dn + x + i} = D \cdot d_n \]
Since we choose that $d_n$ is maximum over all possible $d_i$s, then we must have
\[ d_{Dn + x} = d_{Dn + x + 1} = \dots = d_{Dn + x + D - 1} = d_n \]
and therefore, we have $d_n = d_{Dn + x} = d_{f(f(n))} = f( f(f(n)) + 1) - f(f( f(n))) = y - x $ for all $n \in \mathbb{N}$.
Therefore, $f(n)$ is an arithmetic progression as well. Al fine ~

For ease of notation, we replace $s_n$ with $f(n)$. Then $f$ is a strictly increasing function from $\mathbb{Z}_+$ to $\mathbb{Z}_+$ such that $f(f(1)),f(f(2)),f(f(3)),\dots$ and $f(f(1)+1),f(f(2)+1),f(f(3)+1),\dots$ are arithmetic progressions.

Let $f(f(2))-f(f(1)) = d_1$ and $f(f(2)+1)-f(f(1)+1) = d_2$. We claim that $d_1 = d_2$.

Proof: Suppose $d_1 \ne d_2$. We split this into two cases: $d_2 < d_1$ and $d_1 < d_2$.

Suppose that $d_2 < d_1$. We note that for sufficiently large $N$,
\[f(f(N)+1)-f(f(1)+1) = (N-1)d_2 \le (N-2)d_1 = f(f(N))-f(f(2)).\]
But this is absurd because $f(f(N)) < f(f(N)+1)$ and $f(f(2)) > f(f(1))$.

Now, suppose that $d_1 < d_2$. For sufficiently large $N$, we have that
\[f(f(N))-f(f(1)) = (N-1)d_1 \ge (N-2)d_2 = f(f(N-1)+1) - f(f(1)+1).\]
Again, this absurd because $f(f(N)) \ge f(f(N-1)+1)$ and $f(f(1)) < f(f(1)+1)$.

Thus, neither $d_1 < d_2$ nor $d_2 < d_1$ holds, meaning that $d_1 = d_2$. $\blacksquare$

We let $d_1 = d_2 = d$. Thus, for all $n$, some $r_1$, and some $r_2$, we have that $f(f(n)) = dn+r_1$ and $f(f(n)+1)=dn+r_2$. For all $n$, we then have that $f(f(n)+1)-f(f(n)) = r_2-r_1$. Consider the sequence
\[f(2)-f(1),f(3)-f(2),f(4)-f(3),\dots \qquad (1).\]
By the well-ordering principle, this sequence has a minimal element $m$.

Lemma: The greatest element $k$ of sequence $1$ exists and satisfies $d \ge km$.

Proof: Consider any element $k$ of sequence $1$. Since $k$ is an element of this sequence, it can be expressed as $f(a+1)-f(a)$ for some $a$. We have that $f(f(a+1))-f(f(a)) = d$. We rewrite this as
\[d = \sum_{i=0}^{k-1} \left(f(f(a)+i+1)-f(f(a)+i)\right) \ge \sum_{i=0}^{k-1} m = km,\]
and we are done. $\blacksquare$

Since $m$ is an element of sequence $1$, there exists some $b$ such that $f(b+1)-f(b) = m$. Now, we observe that
\[d=f(f(b+1))-f(f(b)) = \sum_{i=0}^{m-1} \left(f(f(b)+i+1)-f(f(b)+i)\right) \le km.\]
But we already have that $d \ge km$, so we see that $d = km$.

Now, we return to the $a$ defined in our lemma to see that because equality holds, we must have that $f(f(a)+i+1)-f(f(a)+i) = m$ for all $0 \le i < k$. In particular, we have that $r_2-r_1 = f(f(a)+1)-f(f(a)) = m$.

Now, since $d = km$, we also have that $f(f(b)+i+1)-f(f(b)+i) = k$ for all $0 \le i < m$. In particular, this shows that $r_2 - r_1 = f(f(b)+1)-f(f(b)) = k$. But this means that $k = m$. Since $k$ is the greatest element of sequence $1$ and $m$ is the least element of sequence $1$, we see that sequence $1$ is constant. So, we see that for all $n$, we have
\[f(n)-f(1) = \sum_{i=1}^{n-1} \left(f(i+1)-f(i)\right) = (n-1)m,\]
implying that $f$ is the arithmetic progression $f(n) = (n-1)m + f(1) = nm + f(1)-m$. We are done. $\blacksquare$

Rephrase this as a functional equation $f:\mathbb{N}\to\mathbb{N}$, where $f(x)=s_x$. Let $f(f(x))=ax+b$ where $a\ge 1$ and $b\ge 0$. Since $f$ is increasing, we have \[ax+b=f(f(x))<f(f(x)+1)\le f(f(x+1))=ax+b+a,\] so in fact, \[f(f(x)+1) = ax+b+c\] where $1\le c\le a$. Note that \[f(ax+b)=f(f(f(x))) = af(x)+b\] and \[f(ax+b+c) = f(f(f(x)+1))=a[f(x)+1]+b.\] Call $L$ [i]good[/i] if $f(x+1)\ge f(x)+L$ for all $x$. We see that $1$ is good.

[b]Lemma:[/b]  If $L$ is good, then so is $1+\lceil L(1-c/a)\rceil$.

[i]Proof:[/i]  Indeed, we have \[af(x+1)+b=f(a(x+1)+b)\ge L(a-c)+f(ax+b+c) = L(a-c)+a[f(x)+1]+b,\] so \[f(x+1)\ge L\frac{a-c}{a}+1+f(x),\] so the desired result follows. $\blacksquare$

Clearly, all large enough numbers are not good, so let $L$ be the largest good number. Now, if $L<a/c$, then \[L(1-c/a)=L-(c/a)L>L-1,\] so by the lemma, there is a larger good number. Thus, $L\ge a/c$. If $L>a/c$, then \[a+af(x)+b=f(ax+b+c)\ge cL+f(ax+b)>a+af(x)+b,\] which is a contradiction, so $L=a/c$. In fact, the above implies that $f(ax+b+1)=f(ax+b)+a/c$, else we get a similar contradiction.

Thus, \[af(x)+b+a/c = f(ax+b+1)=f(f(f(x))+1)=af(x)+b+c,\] so $a/c=c$, so $c^2=a$. So to summarize, we have \[f(f(x))=c^2x+b\] and $f(x+1)\ge f(x)+c$. This is actually enough to finish.

Call $x$ [i]normal[/i] if $f(x+1)=f(x)+c$, and [i]weird[/i] otherwise.

[b]Claim:[/b]  There are only finitely many weird numbers.

[i]Proof:[/i]  Suppose to the contrary. Then, for any $N$, we have \[f(x)\ge f(1)+c(x-1)+N\] for large enough $x$. Thus, for large enough $x$, we have \[c^2x+b=f(f(x))\ge f(1)+c(f(1)+c(x-1)+N-1)+N = c^2x+g(N),\] where $g$ is an increasing function of $N$. This is clearly a contradiction, so the claim is proved. $\blacksquare$

Thus, all large enough numbers are normal, so $f(x)=cx+d$ for large enough $x$. We see that $c(cx+d)+d=c^2x+b$ for large enough $x$, so $b=d(c+1)$. Redefine normal $x$ such that $x$ is normal if and only if $f(x)=cx+d$. Note that if $f(x)$ is normal, then \[c^2x+d(c+1) = f(f(x)) = cf(x)+d,\] so $f(x)=cx+d$, so $x$ is normal. Thus, every $x$ is normal, since eventually $f^n(x)$ is normal for large enough $x$. Thus, $f(x)=cx+d$ for all $x$ as desired.

[b]Remark:[/b]  Once I got $f(ax+b)=af(x)+b$ and $f(ax+b+c)=af(x)+b+a$, I felt I should be able to use these and increasing to get quite strong conditions. Indeed, I let $a=5$, $b=0$, $c=2$, and tried to construct a single sequence satisfying just these two conditions. Everything I tried eventually contradicted increasing, and when I dug deeper as to why, the lemma fell out naturally. The rest of the solution given the lemma was quite easy.

Let \(a_i=s_i-s_{i-1}\) for \(i \ge 2\) and \(a_1=s_1\). Note that \(a_i \ge 1\) for all \(i\), and \(s_i=a_1+\ldots+a_i\). We are given
\[a_{s_i+1}+a_{s_i+2}+\ldots+a_{s_{i+1}}=c_1\]
for some constant \(c_1\), and
\[a_{s_i+2}+a_{s_i+2}+\ldots+a_{s_{i+1}+1}=c_2\]
for some constant \(c_2\).

[b]Lemma:[/b] \(c_1=c_2,\) or \(a_{s_i+1}\) is constant.

[i]Proof:[/i] Assume by contradiction \(c_2 \neq c_1\). Note \(a_{s_{i+1}+1}-a_{s_i+1}=c_2-c_1\), so by induction on \(i\), \(a_{s_i+1}\) is unbounded, which is impossible since the value must always be between \(1\) and \(c_1\). Thus \(c_1=c_2\). \(\blacksquare\)

Now, define the constant \(c_3=a_{s_i+1}\). Note that for every \(i\), we have
\[1 \le a_i \le a_{s_i+1}+a_{s_i+2}+\ldots+a_{s_{i+1}} = c_i,\]
wwith the second inequality following from each term on the RHS being at least \(1\). Thus, \(a_i\) is bounded. Let \(k,l\) be integers so that \(a_k \ge a_i \ge a_l\) for all integers \(i\).

[b]Main claim[/b](which finishes the problem): \(a_k=c_3\) and \(a_l=c_3\).

[i]Proof:[/i] Let \(S_i\) denote the sum \(a_{s_i+1}+a_{s_i+2}+\ldots+a_{s_{i+1}}\). Note \(S_i\) has value \(c_1\) and has a total of \(a_i\) terms. Now, the sum
\[S_{s_k+1}+S_{s_k+2}+\ldots+S_{s_{k+1}}\]
has a total of \((a_{s_k+1}+\ldots+a_{s_{k+1}})=c_1\) terms, and contains \(s_{k+1}-s_k=a_k\) sub-sums, each with value \(c_1\). Thus, the value and number of terms of the sum are \(a_kc_1\) and \(c_1\) terms, respectively. But since each term is at most \(a_k\) , all terms are in fact equal to \(a_k\). In particular,
\[a_k=a_{s_{s_k+1}+1}=c_3,\]
so \(a_k=c_3\) as desired. \(a_l=c_3\) follows analogously. \(\blacksquare\)

Since the maximum and minimum of \(a_i\) are the same, then \(a_i\) is constant, and \(s_i\) forms an arithmetic sequence as desired.

Translating into functional equation language, we have an increasing function $f:\mathbb{N}\to\mathbb{N}$ such that both
\[f(f(1)),f(f(2)),f(f(3)), \dots \text{ and } f(f(1)+1),f(f(2)+1),f(f(3)+1),\dots\]
are arithmetic progressions. Observe that
\[f(1) < f(1)+1\le f(2) < f(2)+1 \le f(3) < f(3)+1 \le \dots\]
so
\[f(f(1)) < f(f(1)+1) \le f(f(2)) < f(f(2)+1) \le f(f(3)) < f(f(3)+1) \le \dots\]
This means that the two arithmetic progressions have the same common difference, say $N$.

Define $g(x) = f(x+1)-f(x)$ for each $x\in\mathbb{N}$. Let $n = g(f(1)) = g(f(2)) = \dots$ and furthermore define $m = \min_{x\in\mathbb{N}} g(x)$, $M=\sup_{x\in\mathbb{N}} g(x)$. The goal is to prove that $m=M$.

First, we know that $g(x) = f(x+1)-f(x) \le f(f(x+1)) - f(f(x)) = N$, so $M$ is a positive integer. Now, take $x$ such that $g(x) = m$. Then
\[x\to f(x) \to f(f(x)),\ x+1\to f(x)+m \to f(f(x))+N\]
This means that $\frac{N}{m} \le M$ by pigeonhole. On the other hand, take $x$ such that $g(x) = M$, then
\[x\to f(x) \to f(f(x)),\ x+1\to f(x)+M \to f(f(x))+N\]
This means that $\frac{N}{M}\ge m$ by pigeonhole. This implies that $mM \ge N \ge mM$, so equality is achieved. This means that $f(f(x)+1) = f(x) + n = f(x) + m$ in the first case, and $f(f(x)+1) = f(x) + n = f(x) + M$ in the second case, so $m = n = M$ as desired.

First, note that both the given subsequences must have the same common difference. To see this, let $s_{s_i} = a + xi$ and $s_{s_{i}+1} = b + yi$ with $x,y$ as the common differences.

If $x > y$, then pick $i$ sufficiently large so that $s_{s_i} > s_{s_i+1}$. If $y > x$, then again take $i$ sufficiently large so that we have $s_{s_{i}+1} > s_{s_{i+1}}$, which are both contradictions.

Let $d$ be the common common difference of the $2$ subsequences.

Now, suppose $S$ is the smallest difference between consecutive $s_i$'s and let $L$ be the largest. [hide=Why is there a largest L?]Because $s_i$ is strictly increasing, we have $d = s_{s_{i+1}} - s_{s_i} \ge s_{i+1} - s_i$, and so this sequence has an upper bound.[/hide]

Notice that we have $d = s_{s_{i+1}}-s_{s_i} \ge (s_{i+1}-s_{i})S$ since there are at least those many terms in between and each difference is at least $S$. Similarly, we have that $d = s_{s_{j+1}}-s_{s_j} \le (s_{j+1}-s_{j})L$

So, we have $(s_{i+1}-s_i)S \le d \le (s_{j+1}-s_{j})L$. But, since we have this true for any $i,j$, pick them such that maximum is achieved on LHS and minimum on RHS. This forces $LS \le d \le LS$ and so we have $d = LS$

Assume now, ftsoc that $L > S$

We shall now use the following scam idea. Consider the first $i$ such that $s_{i+1} - s_i$ is [i]either[/i] $L$ or $S$. WLOG assume its $S$ since the other thing is similar. We have that all terms in between $s_{s_{i+1}}-s_{s_{i}}$ differ by $L$. But since we had $s_{s_{i+1}+1} - s_{s_{i}+1} = d = s_{s_{i+1}} - s_{s_{i}}$, which since $s_{s_{i}+1} = s_{s_{i}} + L$, means that $s_{s_{i+1}+1} = s_{s_{i+1}} + L$.

Similarly, we obtain that $s_{s_{i+k}+1} = s_{s_{i+k}} + L$ for all $k \ge 1$. Now, consider the first $j$ such that $s_{j+1}-s_j = L$, obviously $j > i$ since that's how we defined it. But now, all terms between $s_{s_{j+1}}$ and $s_{s_{j}}$ must differ by $S$. But this is impossible! Because we had that $s_{s_{i+k}+1} = s_{s_{i+k}} + L$ and so there is at least one pair of consecutive terms that differs by $L$ and so u can never have all of them differing by $S$.

Therefore, we have that $L = S$ and so $s_{i+1} - s_i$ is constant and so it is indeed an arithmetic progression, as desired. $\blacksquare$

Let $S$ be the sequence $s_1, s_2, \ldots$, and let $S_1$ and $S_2$ be the subsequences $s_{s_1}, s_{s_2}, \ldots$ and $s_{s_1+1}, s_{s_2+1}, \ldots$ respectively. Define $x = s_{s_2}-s_{s_1} = s_{s_{i+1}}-s_{s_i}$ and $y = s_{s_1+1}-s_{s_1} = s_{s_i+1}-s_{s_i}$ for all positive integers $i$.

[color=#3D85C6][b]Claim: [/b][/color]Let $m = \min(s_{i+1}-s_i), i \in \mathbb Z ^ +$. If $m < y$, then there does not exist a sequence $S$.

[color=#3D85C6][b]Proof: [/b][/color]Assume otherwise. Let $i$ be an index such that $s_{i+1}-s_i = m$. Notice that $s_{s_{i+1}}-s_{s_i+1} = x-y$ so for $j \in [s_i+1, s_{i+1}-1]$, the average value of $s_{j+1}-s_j$ is $\frac{x-y}{s_{i+1}-s_i-1} = \frac{x-y}{m-1}$. This means there exists some $k$ such that $s_{k+1}-s_k = \left\lceil\frac{x-y}{m-1}\right\rceil$.

Now, we again have $s_{s_{k+1}}-s_{s_k+1} = x-y$, which means for $j \in [s_k+1, s_{k+1}-1]$, the average value of $s_{j+1}-s_j$ is $\frac{x-y}{s_{k+1}-s_k-1} \le \frac{x-y}{\left\lceil\frac{x-y}{m-1}\right\rceil-1} = M$. It suffices to show $M \le m$ with equality when $m = \frac{x-y}{m-1}$. Let $\left\lceil\frac{x-y}{m-1}\right\rceil = \frac{x-y}{m-1}+k$ for some nonnegative real $k$. Notice that $$\frac{x-y}{\left\lceil\frac{x-y}{m-1}\right\rceil-1} \le m \implies m\le m\left(\frac{x-y}{m-1}+k\right)-(x-y) \implies m \le \frac{x-y}{m-1}+mk.$$ Notice that $x = s_{s_2}-s_{s_1+1}+s_{s_1+1}-s_1 \ge m(m-1)+y$, so substituting gives $$m \le \frac{m(m-1)+y-y}{m-1}+mk = m+mk$$ with equality when $m = \frac{x-y}{m-1} \implies k = 0$. If equality doesn't exist, then we have that $M \le m-1$ which means there exists an index $p$ such that $s_{p+1}-s_p = m-1 < m$, a contradiction.

We now tackle the equality case separately. Because $y > m = \frac{x-y}{m-1}$, we define $a = s_{s_1}$ and $b = s_{s_1+1}$, which gives us $s_b-s_{a+1} = x-y$, and that for $j \in [a+1, b-1]$, the average value of $s_{j+1}-s_j$ is $\frac{x-y}{b-a-1} = \frac{x-y}{y-1} = \frac{m(m-1)+y-y}{y-1} < m$. Therefore, there exists an index $k$ such that $s_{k+1}-s_k = m-1 < m$, a contradiction. $\square$

Notice that our claim is symmetric with the claim that if $y < n = \max(s_{i+1}-s_i), i \in \mathbb Z ^ +$, then there does not exist a sequence $S$. Therefore, all $s_{i+1}-s_i = y$ and thus $S$ is an arithmetic sequence. $\blacksquare$

[b][color=#f00]Remarks: [/color][/b]Cool bounding problem. I was kinda lazy at the end and saying it's "symmetrical" is not rigorous at all but the computation for the following is basically identical.

[hide = Solution]
Let $s_n = f(n)$, and suppose $f(f(n)) = a_1 + d_1n$ and $f(f(n) + 1) = a_2 + d_2n$, for all $n$.

[color=#0B5394][b]Claim:[/b][/color] $d_1 = d_2$.

[i]Proof.[/i] Assume not. If $d_1 > d_2$, then for sufficiently large $n$ we have \[a_1 + d_1n > a_2 + d_2n \implies f(f(n)) > f(f(n) + 1),\] contradiction because $f$ is increasing. Meanwhile, if $d_1 < d_2$, for sufficiently large $n$ we have \[a_1 + d_1n < a_2 + d_2(n - 1) \implies f(f(n)) < f(f(n - 1) + 1).\] Since $f$ is increasing this means $f(n) < f(n - 1) + 1$, again a contradiction because $f$ is increasing. $\blacksquare$

Now let $d_1 = d_2 = d$.

[color=#0B5394][b]Claim:[/b][/color] The consecutive differences $f(n + 1) - f(n)$ are bounded.

[i]Proof.[/i] Since $f$ is increasing, $f(m + 1) \geq f(m) + 1$. So then \[d = f(f(n + 1)) - f(f(n)) \geq f(n + 1) - f(n),\] which means $f(n + 1) - f(n)$ is bounded. $\blacksquare$

[color=#0B5394][b]Claim:[/b][/color] All consecutive differences $f(n + 1) - f(n)$ are equal.

[i]Proof.[/i] Let one consecutive difference be $k$, so $f(n) = m$ and $f(n + 1) = m + k$ for some positive integers $m$ and $n$. Then we have $f(m) = f(f(n)) = a_1 + dn$, so \[f(a_1 + dn + 1) = f(f(m) + 1) = a_2 + dm.\] Meanwhile, $f(m + k) = f(f(n + 1)) = a_1 + d(n + 1)$, so \[f(a_1 + dn + d) = f(f(m + k)) = a_1 + d(m + k).\] If $d = 1$, then $a_2 + dm = a_1 + d(m + k)$, so $k$ is fixed. Otherwise, consider the average of the consecutive differences between $a_1 + dn + 1$ and $a_1 + dn + d$, which is \[\frac{f(a_1 + dn + d) - f(a_1 + dn + 1)}{d - 1} = \frac{a_1 + dk - a_2}{d - 1} = k + \frac{a_1 - a_2 + k}{d - 1}.\] Now let $k_{\text{min}}$ and $k_{\text{max}}$ be the minimum and maximum consecutive difference (which exist by the second claim), and assume for contradiction that $k_{\text{min}} < k_{\text{max}}$. Then \[a_1 - a_2 + k_{\text{min}} < a_1 - a_2 + k_{\text{max}},\] so either $a_1 - a_2 + k_{\text{min}} < 0$ or $a_1 - a_2 + k_{\text{max}} > 0$. In the first case, take $k = k_{\text{min}}$ (and the corresponding value of $n$). Then the average consecutive difference between $a_1 + dn + 1$ and $a_1 + dn + d$ is strictly less than $k$, so one of these differences must be less than $k$, contradiction as $k$ is the minimum. Similarly, in the second case take $k = k_{\text{max}}$. Then the average consecutive difference is strictly greater than $k$, so one of these differences must be greater than $k$, contradiction as $k$ is the maximum. $\blacksquare$

So $f(n + 1) - f(n)$ is constant, which means $f(n)$ is an arithmetic progression.
[/hide]

Redefine $(s_i)$ as $f(i)$. The problem condition is equivalent to $f(f(1)), f(f(2)), \ldots, $ and $f(f(1) + 1), f(f(2) + 1), \ldots, $ are arithmetic progressions and $f(1), f(2), \ldots, $ is strictly increasing.

Claim: Both arithmetic sequences have the same common difference.
Proof: Suppose not and their differences were equal to $d_1$ and $d_2$, respectively. So $f(f(n)) = d_1 n + e_1$ and $f(f(n) + 1) = d_2 n + e_2$ for some constants $e_1, e_2$.  If $d_1 > d_2$, then there exists large $i$ such that $f(f(i)) > f(f(i) + 1),$ which is absurd since $f$ is strictly increasing. If $d_1 < d_2$, there exists large $i$ such that $f(f(i+1))  <  f(f(i) + 1)$ (since $d_1 i + (d_1 + e_1) <d_2 i + e_2$ for large $i$). This implies that $f(i+1) < f(i) + 1$, so $f(i+1) \le f(i)$, contradiction. $\square$

Let the common differences be equal to $d$ and $g(n) = f(n+1) - f(n)$. We see that \[ d = f(f(n+1)) - f(f(n))  = g(f(n)) + g(f(n) + 1) + \cdots + g(f(n+1) -1)
 \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\]

Since every positive integer is in the interval $[f(n), f(n+1) - 1]$ for one positive integer $n$, so $g(x) \le d\forall x\in \mathbb{N}$. Since $g$ is upper (and obviously lower) bounded, we may let $c$ be the minimum value of $g$ and $C$ be the maximum.  Let $g(m) = c$ and $g(M) = C$ for some positive integers $m, M$. Denote $g[a,b]$ for $a<b$ as $g(a) + g(a+1) + \cdots  + g(b)$.

Notice that $d = g[f(m), f(m+1) - 1]$, which is the sum of $g(m) = c$ values of $g$, and thus at most equal to $C\cdot c$. We see that $d$ is also equal to $g[f(M), f(M+1) - 1]$, which is the sum of $g(M) = C$ values of $g$, and thus at least equal to $c\cdot C$. Hence $c \cdot C \le d \le C\cdot c$, so $d = c\cdot C$.

Claim: $g(f(i))$ is constant over all positive integers $i$.
Proof: Note that \[ d = f(f(n+1) + 1) - f(f(n) + 1) = g(f(n) + 1) + g(f(n) + 2) + \cdots + g(f(n+1)) \] Comparing this with $(1)$ gives that $g(f(n)) = g(f(n+1))$, so $g(f(i))$ must be constant. $\square$

Now we have that the $c$ numbers $g(f(m)), g(f(m)+1), \ldots, g(f(m+1) - 1)$ are all at most $C$ and add up to $d = c\cdot C$, so they are all equal to $C$. Also, the $C$ numbers from $g(f(M)), g(f(M) + 1), \ldots, g(f(M+1) - 1)$ are all at least $C$ and add up to $d = C\cdot c$, so they are all equal to $c$. Therefore, $c = g(f(m)) = g(f(M)) = C$, which means $g$ is constant, so $f$ is an arithmetic sequence.

Let $D$ and $D'$ be the respective differences of the given arithmetic progressions, and for all $i$ let $d_i=s_{i+1}-s_i$. Because $s_i$ is strictly increasing, $D=s_{s_{i+1}}-s_{s_i} \geq s_{i+1}-s_i$, hence $d_i$ is bounded. Therefore, let $m$ and $M$ be its minimum and maximum respectively.

By considering some $i$ with $d_i=M$, it follows that $\tfrac{D}{M} \geq m \implies D \geq mM$, else by expectation we can find some $d_j<m$. By considering some $i$ with $d_i=m$, it follows that $\tfrac{D}{m} \leq M \implies D \leq mM$. This means that $D=mM$, and similarly $D'=mM$, so $d_{s_i}$ is constant.

On the other hand, due to equality considerations, if $d_i=M$, then $d_{s_i},d_{s_i+1},\ldots,d_{s_{i+1}-1}$ must all equal $m$, and if $d_j=m$, then $d_{s_j},d_{s_j+1},\ldots,d_{s_{j+1}-1}$ must all equal $M$. But since $d_{s_i}=d_{s_j}$, it follows that $M=m$, so $d_i$ is constant and hence $s_i$ is arithmetic. $\blacksquare$

Let $d_1=s_{s_{i+1}}-s_{s_i}$ and $d_2=s_{s_{i+1}+1}-s_{s_i+1}$. We present two claims.

[b][color=red]Claim:[/color][/b] $d_1\geq d_2$.
[i]Proof.[/i] Notice that \[s_{s_{i+d_2}}-s_{s_i}=d_1d_2=s_{s_{i+d_1}+1}-s_{s_i+1}.\] Therefore, we have \[s_{s_{i+d_1}+1}-s_{s_{i+d_2}}=s_{s_i+1}-s_{s_i}>0 \iff s_{s_{i+d_1}+1}>s_{s_{i+d_2}}\] as the sequence is strictly increasing. Using this again, we get \[s_{i+d_2}<s_{i+d_1}+1\leq s_{i+d_1+1}\] and again to get $i+d_2<i+d_1+1$ or $d_1+1>d_2$ implying the claim. $\blacksquare$

[b][color=red]Claim:[/color][/b] $d_1=d_2$
[i]Proof.[/i] Suppose $d_1-d_2\geq 1$. Then, subtracting $s_{s_{i+k}}-s_{s_i}=kd_1$ and $s_{s_{i+k}+1}-s_{s_i}=kd_2$ we obtain \[\left(s_{s_{i+k}}-s_{s_{i+k}+1}\right)+\left(s_{s_i+1}-s_{s_i}\right)=k\left(d_1-d_2\right)\geq k\] But $s_{s_{i+k}}-s_{s_{i+k}+1}$ is negative and $s_{s_i+1}-s_{s_i}$ is a constant, hence we get a contradiction with sufficiently large $k$. $\blacksquare$

Now, let $a_i=s_{i+1}-s_i$; the claim above gives $a_{s_i}$ is constant. Also, let $B\leq a_i\leq A$ so that $A$ and $B$ are both hit by $a_i$, which exist since clearly $a$ is bounded. Therefore, we have $d_1=\sum_{k=s_i+1}^{s_{i+1}}a_k$ so $a_i\cdot A\geq d_1 \geq a_i\cdot B$ implying $AB=d_1$ since there is some $a_i=A,B$. Then, for $a_i=A$, we have \[AB=d_1=\sum_{k=s_i+1}^{s_{i+1}}a_k\geq A\cdot B\] and hence all summands are equal to $B$, which forces $B=a_{s_{i+1}}=a_{s_i}=A$ as desired.


i have a situation in which i have three line segments having perpendicular bisectors concurrent.

can call the 6 points concyclic?

what other condition should they satisfy so that they can be called concyclic?

We can't tell exactly as it merely depends on the problem conditions. But for instance, if we have six points A,B,C,D,E,F such that perpendicular bisectors of AB,CD,EF concur at some point O, then A,B,C,D,E,F are concyclic if and only if the perpendicular bisectors of another pair of distinct segments pass through O.

For more results see also [url=http://mathworld.wolfram.com/CyclicHexagon.html]Cyclic Hexagon[/url],  [url=http://mathworld.wolfram.com/FuhrmannsTheorem.html]Fuhrmann's Theorem[/url] and elsewhere.


probability
floor function
limit
logarithms
Euler

Hello,

  Let $ n \ge 2$  be a positive integer. A real number $ x$ is chosen uniformly at random in the interval $ [0,n]$. Let $ P(n)$ be the probability that $ \lfloor x^2 \rfloor \equal{} {\lfloor x \rfloor}^2$.

Prove that :  $ \lim_{n\rightarrow \infty} \frac{nP(n)}{\ln(n)} \equal{} \frac{1}{2}$ .

[hide]By [url=http://www.mathlinks.ro/viewtopic.php?p=1473884#1473884]this result[/url], we have that

$ x\in\bigcup_{k\equal{}0}^{n\minus{}1}\left[k,\sqrt{k^2\plus{}1}\right)$

By geometric probability principles, $ nP(n)$ is the sum of the lengths of all the above intervals:

$ S(n): \equal{}nP(n)\equal{}\sum_{k\equal{}0}^{n\minus{}1}\left(\sqrt{k^2\plus{}1}\minus{}k\right)\equal{}\sum_{k\equal{}0}^{n\minus{}1}{1\over\sqrt{k^2\plus{}1}\plus{}k}$

But for sufficiently large $ n$

$ 1\plus{}{1\over 1\plus{}\sqrt{2}}\plus{}\sum_{k\equal{}2}^{n\minus{}1}{1\over 2(k\minus{}1)}<\sum_{k\equal{}0}^{n\minus{}1}{1\over\sqrt{k^2\plus{}1}\plus{}k}<1\plus{}{1\over 1\plus{}\sqrt{2}}\plus{}\sum_{k\equal{}2}^{n\minus{}1}{1\over 2k}$

and the result follows by The Two Policemen.[/hide]

Hello Farenhajt ,

I dont see how to get $ \lim_{n\rightarrow\infty}\frac{nP(n)}{\ln(n)}\equal{}\frac{1}{2}$ from your result, we have $ \lim_{n\rightarrow\infty}\sum_{k\equal{}2}^{n\minus{}1}{1\over 2k}\equal{} \lim_{n\rightarrow\infty}\sum_{k\equal{}2}^{n\minus{}1}{1\over 2(k\minus{}1)}\equal{}\plus{} \infty$  and $ \lim_{n\rightarrow \plus{}\infty} \ln(n)\equal{} \plus{}\infty$

[quote="linda2005"]Hello Farenhajt ,

I dont see how to get $ \lim_{n\rightarrow\infty}\frac {nP(n)}{\ln(n)} \equal{} \frac {1}{2}$ from your result, we have $ \lim_{n\rightarrow\infty}\sum_{k \equal{} 2}^{n \minus{} 1}{1\over 2k} \equal{} \lim_{n\rightarrow\infty}\sum_{k \equal{} 2}^{n \minus{} 1}{1\over 2(k \minus{} 1)} \equal{} \plus{} \infty$  and $ \lim_{n\rightarrow \plus{} \infty} \ln(n) \equal{} \plus{} \infty$[/quote]

It is known that $ \lim_{n\to\infty}\left(\left(\sum_{k \equal{} 1}^n{1\over k}\right) \minus{} \ln n\right)$ is finite, and is called Gamma constant or Euler-Mascheroni constant and is denoted by $ \gamma$.

Using that result, $ {1\over{\ln n}}\sum_{k \equal{} 1}^n{1\over k}$ tends to $ 1$ when $ n$ tends to infinity. Now just pull the $ {1\over 2}$ factor out of my sums and you get the result.


inequalities
number theory unsolved
number theory

Hi!

I don't know if it is the right place for that... I tried to solve it using inequalities, but cannot go on... Anyway I thought it was the only way for doing it
Here is the problem:

Find all triples of positive integers $ (a, b, c)$ such that
\[ \left(1 \plus{} \frac {1}{a}\right)\left(1 \plus{} \frac {1}{b}\right)\left(1 \plus{} \frac {1}{c}\right) \equal{} 2\]
Well, I used HM-GM, and substituing $ \alpha \equal{} \sqrt [3]{abc}$, I got

$ \alpha^3 \minus{} 3\alpha^2 \minus{} 3\alpha \minus{} 1\le 0$

Thank you in advance! :D

These problems are not hard by brute force. [hide]You know one of $ a,b,c$ must be less than $ 4$ or the left side would be less than $ 2$. So plug in $ 1,2,3$ for $ a$. If $ a \equal{} 1$, no solution. If $ a \equal{} 2$, $ (1 \plus{} \frac {1}{b})(1 \plus{} \frac {1}{c}) \equal{} \frac {4}{3}$. One of $ b,c$ must again be less than $ 5$, so letting $ b \equal{} 1,2,3,4$, we have $ (2,4,15)$ as one solution. If $ a \equal{} 3$, $ (1 \plus{} \frac {1}{b})(1 \plus{} \frac {1}{c}) \equal{} \frac {3}{2}$. One of $ b,c$ must again be less than $ 5$, so letting $ b \equal{} 1,2,3,4$, we find $ (3,3,8)$ and $ (3,4,5)$ are solutions.
So in summary, all solutions are permutations of $ (2,4,15)$,$ (3,4,5)$, and $ (3,3,8)$[/hide]

[quote="lambruscokid"]http://www.mathlinks.ro/viewtopic.php?t=110648[/quote]
Thank you lambruscokid, I searched but didn't find... :D

[quote="dgreenb801"]These problems are not hard by brute force.  Click to reveal hidden content[/quote]
Thanks dgreenb801! So, this kind of problems isn't very instructive... Do you know a good book for learning basic inequalities... I really have to learn how to solve them! ;)


calculus
integration
calculus computations

Assume that $ f(x)\geq0$ such that $ \int_{\minus{}\frac{1}{a}}^{a}{xf(x)\mathrm{d}x}\equal{}0$

Prove that: $ \int_{\minus{}\frac{1}{a}}^{a}{f(x)\mathrm{d}x}>\int_{\minus{}\frac{1}{a}}^{a}{x^2f(x)\mathrm{d}x}$ :blush:

Dear mathlinkers, can you show me some ideas about this problem?  :blush:

[hide]We want to show $ \int^a_{\minus{}\frac{1}{a}}\left(1\minus{}x^2\right)f(x)dx>0$. We also have $ 0\equal{}k\int^a_{\minus{}\frac{1}{a}}xf(x)dx$. Then it's sufficient to find $ k$ so that $ 1\minus{}x^2>kx$ for $ x\in\left(\minus{}\frac{1}{a},a\right)$. $ k\equal{}\frac{1}{a}\minus{}a$ works.[/hide]

[quote="Bictor717"][hide]We want to show $ \int^a_{ \minus{} \frac {1}{a}}\left(1 \minus{} x^2\right)f(x)dx > 0$. We also have $ 0 \equal{} k\int^a_{ \minus{} \frac {1}{a}}xf(x)dx$. Then it's sufficient to find $ k$ so that $ 1 \minus{} x^2 > kx$ for $ x\in\left( \minus{} \frac {1}{a},a\right)$. $ k \equal{} \frac {1}{a} \minus{} a$ works.[/hide][/quote]

Thank you for you key to the problem, I will have a try :lol:


Give me an example of numbers $x,y,z$ different from each other, for which that condition holds. :)

[quote="Megus"]$x,y,z$ are nonnegative reals for which $x+y \geq 2z$, $y+z\geq 2x$, $z+x \geq 2y$ holds. Prove that:

$x^{3}(y+z)+y^{3}(z+x)+z^{3}(y+x)+2xyz(x+y+z) \geq 4((xy)^{2}+(yz)^{2}+(zx)^{2})$

P.s. hope it's not trivial  :)[/quote]

if you add up your conditions you will obtain an equality :wink:

[quote="Megus"]$x,y,z$ are nonnegative reals for which $x+y \geq 2z$, $y+z\geq 2x$, $z+x \geq 2y$ holds. Prove that:

$x^{3}(y+z)+y^{3}(z+x)+z^{3}(y+x)+2xyz(x+y+z) \geq 4((xy)^{2}+(yz)^{2}+(zx)^{2})$

P.s. hope it's not trivial  :)[/quote]

strange condition, Megus!  :D

if $x\geq y \geq z$, we have that $2x \geq y+z \geq 2x$, so, x = y = z!

[quote="Megus"]oh sh....  :lol:  I must have mistaken somethin. Does above inequality hold without any additional condition?[/quote]

no, it doesn`t work. see x=y=1/2 z=1/4. it gives 23/64 is greater or equal than 24/64. so close :P


Find  the  positive  integer solutions  of the  equation $3^x+29=2^y.$

I've made a stupid mistake, sorry. Previous solution was wrong

[quote="dondigo"]$3^{2k}\equiv{1}\pmod{8}$
$3^{2k+1}\equiv{2}\pmod{8}$[/quote]
Sure ?

As has been said, $x = 1, y = 5$ gives the first solution.  There are no solutions for $x > 1, y > 5$.

(Sorry if this proof is long.  There is probably a better way.)

Take the equation $\bmod 9$.  Since $x > 1$, this gives $2 \equiv 2^y \bmod 9 \Rightarrow 1 \equiv 2^{y-1} \bmod 9$.  Euler's Theorem gives $2^{\phi(9)} = 2^{6} \equiv 1 \bmod 9$, so $y = 6n+1, n \ge 1$.

Take the equation $\bmod 32$.  Since $y > 5$, this gives $3^x - 3 \equiv 0 \bmod 32 \Rightarrow 3^x \equiv 3 \bmod 32 \Rightarrow 3^{x-1} \equiv 1 \bmod 32$.  Euler's Theorem gives $3^{\phi(32)} = 3^{16} \equiv 1 \bmod 32$, so $x = 16m+1, m \ge 1$.

Finally, take the equation $\bmod 7$.  This gives $3^{16m+1} + 1 \equiv 2^{6n+1} \bmod 7$.  By Euler's Theorem, $2^6 \equiv 1 \bmod 7$, so $2^{6n+1} \equiv 2 \bmod 7$.  This implies $3^{16m+1} \equiv 1 \bmod 7$, which cannot be true since, by Euler's Theorem, $3^6 \equiv 1 \bmod 7$, and $16m + 1$ cannot be a multiple of 6.

Therefore, there are no solutions $x > 1, y > 5$.

[quote="agolsme"][quote="Andreas"]$(1, 5)$.
But how to prove $3^{2k + 1} + 29 = 2^{2h + 1}$ hasn't other solutions?[/quote]

$3^{x} + 29$ is a strictly increasing function and $2^{y}$ is a strictly decreasing function they can intersect at one place only.[/quote]

But $x \neq y$, we're looking for if $3^x + 29$ and $2^x$ take on the same integer y-values, not if they intersect  :)

Edit:  Well, if you mean the functions $y = 3^x + 29$ and $x = 2^y$, then you would be implying that $y = x$, so you're really looking at the intersection of those two functions and the line $y = x$ (and there aren't any intersections).  You should instead be looking at the function $f(x, y) = 3^x + 29 - 2^y$, which is too ugly to solve  :P

And in any case, $2^y$ isn't a strictly decreasing function...


[quote="ckck"]A silo is created by attaching a hemisphere to the top of a right-circular cylinder with an equal radius. If h dentoed the height and r the radius of the cylinder, then $h-\frac{1}{3}r=5$. The volume of the silo is $80\pi$. Find its surface area(including the bottom).[/quote]
[hide]The vol. of the silo is $r^{2}h\pi+\frac{2r^{3}\pi}{3}=80\pi$
$r^{2}(3h+2r)=240$
$r=3h-15$
$(9h^{2}-90h+225)(6h-15)=240$
$(h^{2}-10h+25)(2h-5)=\frac{80}{9}$
Have fun solving[/hide]


[quote="nayel"]Find, in terms of $ a,b,c$, the number of unit cubes that the diagonal of an $ a\times b\times c$ cuboid intersects.[/quote]
[hide]
$ a\plus{}b\plus{}c\minus{}(a,b)\minus{}(b,c)\minus{}(a,c)\plus{}(a,b,c)$[/hide]

How do you deduce that? It is correct but I'm interested in the solution rather than the answer.

It is not difficult to show that a diagonal of an $ a\times b$ rectangle intersects $ a\plus{}b\minus{}(a,b)$ unit squares. How can we use this to solve our problem?


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Describe the Galois group of $x^{p}-2$, with $p>0$ prime.

[hide="what I have so far"]
* Certainly the splitting field of $x^{p}-2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(2^{\frac1{p}},\zeta_{p})$.
* Denote $E=\mathbb{Q}(2^{\frac1{p}},\zeta_{p})$, then is $[E: \mathbb{Q}]=p(p-1)$. This can be seen easily because $(p,p-1)=1$.
* Hence $|E|=p(p-1)$ too.
* I noticed that $\mathbb{Q}(\zeta_{p})$ is a Galois extension of degree $\varphi(p)=p-1$.
* If I would find an element of order $p$, then I have two generators for $E$ and am I done.
[/hide]

$\sigma: \sqrt[p]2 \mapsto \zeta_{p}\sqrt[p]2$ induces an element of order $p$.
The Galois group should be the semi direct product of $\mathbb{Z}/p\mathbb{Z}$ and $(\mathbb{Z}/p\mathbb{Z})^{*}$, the latter acting canonically on the first.

i) Let $G$ be the galois group of $x^{p}-2$. Let $u=\sqrt[p]{2}$. $\sigma: \zeta^{k}{u}\rightarrow\zeta^{k+1}u$ is of order $p$ in $G$ (cf Zetax). Let $\alpha$ be a generator of $\mathbb{F}_{p}^{*}$. $\tau_{\alpha}: \zeta^{k}{u}\rightarrow\zeta^{\alpha{k}}u$ is of order $p-1$ in $G$. {$\tau_{\alpha},\sigma$} generates $G$ or $G$ is a transitive subgroup of $S_{p}$ which admits as a generator: "($1,2,\cdots,p$),($1,\alpha,\alpha^{2},\cdots,\alpha^{p-2}$)". $\sigma$ has no fixed point and $\tau_{\alpha}$ has one fixed point :$u$.
   ii) Definition and proposition: $F_{m}(n)$ is THE (up to isomorphism) transitive subgroup of $S_{n}$ with $m$ elements s.t. each non trivial element fixes 0 or one point and some non trivial element fixes one point. $F_{m}(n)$ is called a Frobenius subgroup. The complement of $F_{m}(n)$ is the subgroup $H$ fixing $n$  (for example). The kernel of $F_{m}(n)$ is formed by the identity element together with all elements not in any conjugate of $H$; $F_{m}(n)$ is the semi direct product of $K$ and $H$.
    iii) Here $G=F_{p(p-1)}(p)$ where $H$ is of order $p-1$ and admits as generator {($1,\alpha,\alpha^{2},\cdots,\alpha^{p-2})|\alpha$ is a generator of $\mathbb{F}_{p}^{*}$} and $K=[(1,2,\cdots,p)]$ is of order $p$.
Conclusion: $G$ is the semi direct product of $K$ and $H$.

Two remarks:
i) My semi direct product is isomorphous with that of Zetax.
ii) There exists an isomorphism between $G=F_{p(p-1)}(p)$ and $Aff(\mathbb{F}_{p})=\{x\rightarrow{ax+b}|a\in\mathbb{F}_{p}^{*},b\in\mathbb{F}_{p}\}$.


I've seen some pdf files that the option to copy the text is unavailable.

How can I put those kind of "protections" in my $ \text{\LaTeX}$ file?

Thanks

You will probably need commercial software to alter the pdf file; as far as I'm aware, you can't do it in LaTeX.


geometry
geometric transformation
geometry solved

i know its a well-none problem but i want to see u solutions.
i think i solved it very very nice .and im waititng for u nice solutions .
we have three lines (AB&CD&EF)that meet each other at point p, in a circle.and we have:
angleAPC=angleCPE=60.prove that:
PA+PD+PE=PB+PC+PF

i my self use this:
let ABC be a triangle that has equal sides.from each point like p on circum circle if PB>PA>PC
we have:PB=PA+PC

I also used this fact. Here is my complete proof:

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/8067?expand=1]Hyacinthos message #8067[/url]"][b]From:[/b] Darij Grinberg
[b]Subject:[/b] Re: Fermat point problem

Dear Barry Wolk,

In Hyacinthos message #8064, you wrote:

>> An advanced high school student came to me with the
>> following problem. I know I can get some help here.
>>
>> The first Fermat point F is the intersection of the
>> lines AA', BB', CC', where the triangles A'BC, B'CA,
>> and C'AB are equilateral. Let d be the circle
>> through A'B'C'. Line AFA' meets d at A' and A*, and
>> B* and C* on d are defined cyclically. Show that
>>    |AA'| = |AA*| + |BB*| + |CC*|
>>
>> A synthetic solution is preferred. And of course
>> he knows the classical result that
>> |AA'| = |BB'| = |CC'|.

Another high school student (me) is giving a solution
he has found right off. I guess it can be simplified
a lot.

I will use the POMPEIU THOEOREM [I am not sure of the
correct spelling of Pompeiu, btw]:

If P is a point on the circumcircle of an
equilateral triangle ABC, namely on the arc BC, then
PA = PB + PC.

This is a corollary from the Ptolemaios theorem
applied to the cyclic quadrilateral PCAB.

Now we come to the solution of the problem. Call M
the center of the circle through A', B', C'. The
segments A'A*, B'B* and C'C* are chords in this
circle, and hence the projections X, Y, Z of M on
these segments bisect the segments.

Since the angles MXF, MYF and MZF are right angles,
the points X, Y, Z lie on the circle with diameter
MF. Hence, all five points M, F, X, Y, Z lie on one
circle.

Now, we know that angle BFC = angle CFA = angle AFB
= 120°. Hence, (on my figure - on yours it will be
similar)

   angle XYZ = angle XFZ = 180° - angle CFA
             = 180° - 120° = 60°;
   angle YZX = 180° - angle YFX = 180° - angle AFB
             = 180° - 120° = 60°;
   angle ZXY = angle ZFY = 180° - angle BFC
             = 180° - 120° = 60°.

Hence, triangle XYZ is equilateral. (In fact, two
of the equations above would be sufficient.)

From the Pompeiu theorem, FZ = FX + FY or a similar
equation if the points on your sketch are arranged
differently than the ones on mine.

Then we have (again depending on the arrangement of
points)

      A*F - A'F + B*F - B'F + C*F - C'F
 =   (A*X - FX) - (A'X + FX)
   + (B*Y - FY) - (B'Y + FY)
   + (C*Z + FZ) - (C'Z - FZ)
 = - 2 FX - 2 FY + 2 FZ
             (remember that X, Y, Z bisect the
             chords A'A*, B'B*, C'C*)
 = 0         (since FZ = FX + FY),

and thus

   A*F + B*F + C*F = A'F + B'F + C'F.

Now,

   A*A + B*B + C*C
 = (A*F + B*F + C*F) - (AF + BF + CF)
 = (A'F + B'F + C'F) - (AF + BF + CF).

Finally, apply the Pompeiu theorem to the point F on
the circumcircle of the equilateral triangle BCA':

  A'F = BF + CF.

Similarly, B'F = CF + AF and C'F = AF + BF. Hence,
A'F + B'F + C'F = 2 (AF + BF + CF), and

   A*A + B*B + C*C
 = 2 (AF + BF + CF) - (AF + BF + CF) = AF + BF + CF
 = AA',

qed..

Thank you for the nice problem.

  Best regards,
  Sincerely,
  Darij Grinberg[/quote]

By the way, if you have a solution of your problem, then you should post it into "Proposed&Own Problems", not into "Unsolved Problems".

  darij

darij,u solved it exactly the same as me.
i want to know if there is any different solution or not.

no no no no no........ :o  :o  :!:  :!:  :mad:  :mad:
dont say that there are two diferent problems.
the problem that u say,is the special form of it.
then we cant salve it by what u say.
its clear that we cant use the method that u said.but if u want the reason ,i can tell u ;)

[quote="stergiu"]Please read  the comment in the attachment. Can this be a geometrical proof. Or it is no proof ? Grobber uses this method very often.

   Babis[/quote]

I am not sure that solution like this is perfectly correct, though I saw once one
theorem I can't remember name that is, in original, about some circles (i think five
or four of them)and some concurrency and I know that theorem stands also for
degenerate circles-points so that case is similar like this...

But there is very nice solution only if you expend a little that your solution. What
I want to say is (I'm using notation from the article): Let's say that parallel to
$FE$ passing through $K$ is $F'E'$, so when we have what you proved and if we show
that $F'K-FM=KE'-ME$ (which is not hard at all) we will have the wanted equality, so I think that combination of this is the easeist solution.

I just want to add that this is not the first time that I saw a problem where you
can use some special case where it is easier to prove the wanted equality or
anything else, and show that some things don't change beside transformation that you
make (in this case translation a chord). I, myself, solved about 4-5 problems in
that way.

m@re

Well , as I already have said in my post , I don't persist  that this is a solution but I ask the opinion of  all you. It will be an open question to find under which conditions such operations are allowed.For example , the area of a cyrcle is a limit and only in this way we can prove the formula.Second  , I had asked Grobber to give me a proof for a problem(British olympiad) and he used Newtoms Theorem , when the quadriteral is degenerated to a triagle.

  If you want my opinion , I do not see it(what I wrote in the comment) as a pure geometrical solution , even though the method could be right.Probably a counter example would destroy this method  and I ' ll try to find one>>.

  Thanks for your efforts and comunication.

 Babis
   ( to ashegh : it is dangerous  in mathematics  to through away a thought , becouse it does not like us , or to adopt it , since we like it.This means that we must have objections but we must examine them carefully.[b]Keep sending so beautiful geometry problems!)   [/b]

[quote="stergiu"]I had asked Grobber to give me a proof for a problem(British olympiad) and he used Newtoms Theorem , when the quadriteral is degenerated to a triagle.
[/quote]

I would relly like to see how Newton's theorem can be used on degenerated quadraliteral
so if it isn't so hard can you post it...

It was an old comunication with Grobber - out of the forum. I' ll send it , just when I find it.This moment I' m very busy.Sorry!  But don't warry. I' ll find it.

Babis

[i] In a greek book ( Kontogiannis , Geometry - Olympiads  , which I recomend warmly to all greek students or to those who can read Greek) I saw today the same solution as Darij Grinberg ' s.I think that this must be the best solution.In this book I saw and a second solution with vectors. I' ll study it carefully and I' ll send it , if it is not dificult.
Babis[/i]

(that  (which is not hard at all) we will have the wanted equality, so I think that ....).

i want usolution about what u said isnt not hard at all. :lol: .

[u]To silouan![/u]

 It is solved in the coresponding section. This moment I can not see it.If you can not find it , plase write.You ...know.
Babis

I don't have the problem in front of me right now but this is the steps which you should folow. When you have two parallel chords in a circle ($AB$ and $CD$) the trapizium $ABCD$ has it's rays ($BC$ and $AD$) equal, so you construct some paralels (to obtain segment with lenght of diferenc of  two segments) and show that some two traingles are congruent and there you go if there is need to write the whole solution I will do it just ask

[b]Generalization[/b]

  The initial statement( analogously formulated) is true , if we take  $2n + 1 $ chords through P and all angles are equal to :

    $\frac{\pi}{2n + 1}$

Babis

Is that a question about generalization or you are saying that's true? What I'm trying to aks: is that a problem for solving or a statement?[/quote]

No , it is not posted  to be solved but only as a remark ! Solution is done with vectors(this hint is given in the book).

    Babis

dear stergiu.sorry , i made a mistake.
the method that u said .is completly true.
did u know that?
the writer of this problem,first saw the method that u said ,and then wanted to use limit and changed triangle to
a point.
but i want to say that if grobber didnt see that problem,it wasnt too easy to solve this problem.


Prove that $\forall a,b\in\mathbb R ; \exists x,y\in[0,1]:$
\[ |xy-ax-by|\geq\frac13  \]
If the right hand side $\frac13$ change to $0.33334$, is the inequality also true?

Intriguing problem ...  :)

If $|a| \geq 1/3$ then we may take $x=1$ and $y=0$

So we may consider only $|a|,|b| < 1/3$

if $a,b \geq 0$ take $x=y=1$ : $1 - a - b \geq 1/3$
if $a < 0$ or $b < 0$ take $x=y=1$ also.

$1/3$ is sharp by looking at case $a=b=1/3$


function
limit
Putnam
IMC
college contests

Let $ C$ be a nonempty closed bounded subset of the real line and $ f: C\to C$ be a nondecreasing continuous function. Show that there exists a point $ p\in C$ such that $ f(p) \equal{} p$.
(A set is closed if its complement is a union of open intervals. A function $ g$ is nondecreasing if $ g(x)\le g(y)$ for all $ x\le y$.)

[quote="perfect_radio"]Doesn't the [b]OLD[/b] trick of considering $ m = \sup_{x \in C}\left\{ x : f(x) \geq x \right\}$ work?[/quote]Old for Romanians used with previous National MO questions maybe ;)

Yeah but it requires a bit more caution than "usual" (?) because the set may not be connected... (I lost points by not being careful enough :P)

There's also a much quicker and nicer solution found by one of my friends, James Tener from the Budapest Semesters in Math team. See if you can get it! It's like 2 lines. ;)

is it necessary to confirm that $ f$ is continuous?

Assume that $ f(x) \ne x$ for all $ x \in C$. We also assume that $ C = \bigcup_{k = 1}^{n}{\left[{a_{k},b_{k}}\right]}$, where $ b_{k}\leq a_{k+1}$ for all $ k$. Since $ f : C \to C$ then $ f(a_{1}) > a_{1}$, therefore $ f(x) > x$ for all $ x \in \left[{a_{n},b_{n}}\right]$. This is a contradiction since $ f(b_{n}) \leq b_{n}$.

$ f$ doesn't have to be continuous.

Xevarion is right with his objections... I would probably have overlooked this (I had in mind the argument for a compact interval).

The best would be to say that if $ t = f(m) > m$, then $ f(t) \geq f(m) = t$, contradiction.

When $ C$ is unbounded, how about the question?

[quote="bookworm_vn"]Let $ a =\min C$. Suppose that $ f(a) > a$ then consider $ u_{n}= f\left({u_{n-1}}\right)$, $ u_{0}= a$. It is easy to see that $ \left\{{u_{n}}\right\}$ is increasing and bounded from above then there exists $ c \in C$ such that $ \mathop{\lim }_{n \to \infty }u_{n}= c$ which gives $ f(c) =c$.[/quote]

Very nice. This is the short solution I was referring to. This doesn't work in case $ C$ is unbounded above, though. In fact the theorem no longer holds; $ f(x) > x$ can be all of $ C$.

Also regarding your earlier suggestion: you can't write $ C$ as a finite union of closed intervals. Consider the Cantor set. (I nearly thought the statement was false because of the Cantor set but then I realized it doesn't matter.)

There is a mistake in my previous solution (can you see it?). Actually, my approach is as following.

Let $ A = \left\{{x \in C: f\left( x \right) > x}\right\}$ and $ c = \sup A$. We will show you that $ f(c) = c$. Indeed, if $ c \in A$ then $ f(c) > c$, this leads us to $ f(f(c)) > f(c)$ which is a contradiction (case $ f(f(c)) =f(c)$ is done) since $ A \ni f(c) > c = \sup A$. Therefore $ c \not \in A$. Suppose that there are $ A \supset \left\{{x_{n}}\right\}_{n}\to c$. It is easy to verify that (
\[ x_{n}< f\left({x_{n}}\right) < f\left( c \right) < c \]
Letting $ n \to+\infty$ we get $ c \leq f(c) < c$ (if $ f(c) = c$ then done). Done.

Since a compact subset of the real line can be organized in an obvious way as a complete lattice under the usual ordering, this is a particular case of the [url=http://planetmath.org/encyclopedia/KnasterTarskiTheorem.html]Knaster-Tarski Theorem[/url]: an order-preserving map $ f: L\to L$ on a complete lattice $ L$ has a fixed point.

The proof of the theorem boils down to what perfect_radio suggested.

My solution on the exam was the same that [b]bookworm_vn[/b] posted  :)

A little observation!!!
We don't know if the set $ A\equal{}(x\in C : f(x)>x)$is empty or not!!

I think the first solution of bookworm, with a =min C, be right but he use continuity of f.
 :wink:
Maybe is an important hypotesis.

[quote="gpzes"]A little observation!!!
We don't know if the set $ A \equal{} (x\in C : f(x) > x)$is empty or not!![/quote]

Very easy, since $ C$ is closed bouded in $ \mathbb R$ then there exists $ C\ni c\equal{}\inf C$. From that we obtain $ f(x)\geq c$ for all $ x\in C$. If $ f(c)\equal{}c$ then nothing to do, otherwise, $ f(c) > c$ this yields $ A$ is nonempty set. OK?

[quote="gpzes"]I think the first solution of bookworm, with a =min C, be right but he use continuity of f.
 :wink:
Maybe is an important hypotesis[/quote]

Exactlly, that's what I mean.

$ \ \begin{math}\text{Ok bookworm, but I don't know if }f(x_{n}) < f(c)\text{ in your second solution.}\\ \text{Anyway, I considered }a_{1}= f(b)\text{ and }a_{n+1}= f(a_{n})\text{ where }b\in C\text{ is the point existing by hypothesis.}\\ \text{Then }\{a_{n}\}_{n\in N}}\text{ is bounded and being }f\text{ monotone}\\ \text{we have that }\exists\text{ }\mathop{lim}_{n\to\infty }a_{n}=\lambda\in C(closed set).\text{ (If }f(b) = b\text{ we are done.)}\\ \text{So }\lambda =\mathop{lim}_{n\to\infty}a_{n+1}=\mathop{lim}_{n\to\infty}f(a_{n}) = f(\mathop{lim}_{n\to\infty}a_{n}) = f(\lambda )\\ \text{using the }continuity\text{ of }f.\\ \end{math}$\

[quote="bookworm_vn"]We also assume that $ C \equal{}\bigcup_{k \equal{} 1}^{n}{\left[{a_{k},b_{k}}\right]}$, where $ b_{k}\leq a_{k\plus{}1}$ for all $ k$.[/quote]No we can't... Ever heard of the Cantor set, for example?

Dear friend, there are 3 solutions in my posts but 2 of then are false. The only right solution is the last one. In my first solution, we cannot assume $ C \equal{}\bigcup_{k \equal{} 1}^{n}{\left[{a_{k},b_{k}}\right]}$ and in my second solution, we are using continuty of $ f$. Are you satisfying?  :oops:

It seems that we can bypass the  continuty  of f how bookworm showed.  :wink:
It remains that using continuity we arrive to a shorter solution and maybe  to a
more "natural" one (like I proposed) :blush:
Anyway we are here to profundize ,true?! :lol:
Thank you all!

In a book entitled [b]Putnam and Beyound[/b] by [i]R. Gelca [/i] and [i]T. Andreescu[/i], Problem 348 states as following: let $ : [a,b]\to [,b]$ be an increasing function. Show that there exists $ \xi\in [a,b]$ such that $ f(\xi)\equal{}\xi$.

Here is the solution.

[img]http://img487.imageshack.us/img487/8881/clipboard04gf8.jpg[/img]

Hm, can anybody tell me if I am missing something in the following:

(the motivation for this is the proof of the Contraction Mapping Principle):

Define the sequence $x_{n} = f(x_{n-1}), x_0=a,$ for some $a \in C.$ But $C$ is closed and bounded in $\mathbb{R}$ hence compact, by the Heine-Borel Theorem, and thus $f$ is bounded (image of a compact set under a continuous function is compact, which in $\mathbb{R}$  means closed and bounded). I.e., $x_n$ is non-decreasing and bounded, and thus has a limit, $L,$ and $L$ is in $C$ as $x_n \in C$ which is closed.
Putting everything together: $$ L = \lim_{n \to \infty} x_n = \lim_{n \to \infty} f(x_n) = f(L),$$ where we used the continuity of $f$ in the last step.


Numerator gets bigger than the denominator. Sum goes to infinity.

Sorry,i read the posts just today,yup,i do require the answer in terms of 'n',pls could someone give the method with the answer

[hide]You're looking for $\sum_{k=1}^{n}\sum_{m=1}^{k}\frac{1+m!}{2^{k}}$, but I don't know a closed form for it because of the $m!$ term.[/hide]

I don't think so, otherwise, summing factorials would be very "easy".

A look at
http://mathworld.wolfram.com/FactorialSums.html
would, I bet, not be very encourageing for any further attempts.

[hide="My First Thoughts on Factorial Sums"]At first I thought to use the Gamma Function, but clearly this doesn't get anywhere very quickly: \[\sum_{k=1}^{n}k!=\sum_{k=1}^{n}\int_{0}^{\infty}e^{-x}x^{k}\,dx=\int_{0}^{\infty}\sum_{k=1}^{n}e^{-x}x^{k}\,dx=\int_{0}^{\infty}e^{-x}\left(\frac{1-x^{k+1}}{1-x}-1\right)\,dx\] [/hide]

[quote="amcavoy"]Is there a closed form for $\sum_{k=1}^{n}k!$?[/quote]
I haven't read the solutions properly, but there is a closed form for $\sum k\cdot k!$.

$\sum k\cdot k! = \sum (k+1-1)k! = \sum (k+1)!-k!$.

I am not sure if this helps in the solution in anyway.


calculus
integration
trigonometry
calculus computations

$ \int \frac{x \plus{}\sin x}{1\plus{} \cos x} \ dx$

This solution isnt mine, its from giannis1990
$ \begin{array}{l}
 I \equal{} \int {\frac{{x \plus{} \sin x}}{{1 \plus{} \cos x}}dx\mathop  \equal{} \limits_{dx \equal{} 2du}^{x \equal{} 2u} } 2\int {\frac{{2u \plus{} \sin 2u}}{{1 \plus{} \cos 2u}}du \equal{} } \int {\frac{{2u \plus{} 2\sin u\cos u}}{{{{\cos }^2}u}}du \equal{} }  \\
  \equal{} 2\int {u{{\left( {\tan u} \right)}^\prime }du}  \plus{} 2\int {\tan udu}  \equal{} 2u\tan u \plus{} c \equal{} x\tan \frac{x}{2} \plus{} c \\
 \end{array}$

$ \int\frac {x \plus{} \sin x}{1 \plus{} \cos x}\ dx \equal{} \int \frac {(x \plus{} \sin x)(1 \minus{} \cos x)}{\sin^2x}\,dx$

$ \equal{} \int x \csc^2x\,dx \minus{} \int x \cos x \sin^{ \minus{} 2}x\,dx \plus{} \int \csc x\,dx \minus{} \int \cot x\,dx$

$ \equal{} \minus{} x \cot x \plus{} \int \cot x \,dx \plus{} x \csc x \minus{} \int \csc x\,dx \plus{} \int \csc x\,dx \minus{} \int \cot x \,dx$

$ \equal{} x (\csc x \minus{} \cot x) \plus{} C$

$ \equal{} x \tan \frac{x}{2} \plus{} C$.


Mcdonalds dollar menu have five items: apple pie, chicken sandwich, hamburger, fries, and milkshake.  John is going to buy three items.  How many different combinations can he have?

My solution is 5x5x5/3!.

My reasoning is that for the first item, I have 5 choices, same for the second and the third item.

To account for the overcount, I divide the product by 3! (hamberger, pie, and fries is the same as pie, fries, and hamberger).

What's wrong with my approach?

Thanks in advance for your help.

Tech.

Hi techin. First of all, $ \frac{5 \times 5 \times 5}{3!}\equal{} \frac{125}{6}\equal{} 20\frac{5}{6}$ is not a whole number.

I'm not sure if I understand the question but this is how I see it - I may well be wrong:

There are $ 5$ posibilities of having $ 3$ selections the same.

There are $ 5 \times 3\equal{}15$ ways of getting $ 2$ the same and $ 1$ different.

There are $ \binom{5}{3} \equal{} 10$ combinations of $ 3$ different items.

Hope this helps.

Thanks, Andrew.

The answer is 56.  I guess my question is: will there be a possibility that there are other cases that are not accounted for?  For example, to get the number of 2 the same and 1  different, should that be 5x4? The first two will have 5 choices. For each double, you would have 4 other choices for the third item.

How about if there are n choices (instead of 5 choices) and I want to buy k items(instead of 3 items), what will be the general formula?

Again your response is greatly appreciated.

Tech

We can work in cases.
Case 1: All items are different; That is 5C3=10
Case 2: 2 items are similar; There are 5 choices for which item to be similar and then 4 choices for the last item (there arent 5 choices for the last item becuase then all 3 would be the same, which is not what we are examining in our case)
Case 3: 3 items are similar; There are 5 choices. Thats a total of 35???
Darnit what am I doing wrong?

As for your original solution, the 5x5x5 accounts for,say, hamburger-hamburger-hamburger, which is not overcounted and doesnt need to be divided by 6. But it also includes ,say, hamburger-frenchfry-milkshake and all of its permutations, which needs to be divided by 6.

EDIT: Oh, so its correct? Techin, where did you get this problem from?

I think 35 is correct ...

The following approach is called sticks and stones in my book ... I have seen it called balls and walls ... [size=67][color=red]AIME15: The mod has seen it as stars and bars :P[/color][/size]it is a variation of balls and urns with indistinguishable balls in distinguishable urns ... the thought is as follows:

You have three items, they will be represented by O's below.

Arrange four dividers (represented as |'s below) to separate the items into five categories:  pie, chicken sandwich, burger, fries, milkshake.

So we are looking for the number of arrangements of four |'s and three O's.

For example, O|||OO| would represent a pie and two orders of fries.  O|O|O|| represents a pie, a chicken sandwich, and a burger.

There are 7C3 = 35 ways to arrange the sticks and stones/balls and walls, etc.  56 would be the answer if there were six dollar menu items (you would need 5 divider 'sticks' and and 3 'stones': 8C3 = 56).

I hate writing general formulas ... use the logic above.  More info, google: distinguishability, balls and urns, etc.

I have asked the teacher.  The answer was wrong.  It should be 35.

Thanks again for all the help, guys.

Tech.

Hi techin.

I made (yet another) silly mistake. My $ 5\times 3$ should have been $ 5\times4$.

For your $ n, k$ question, we have:

All the same; $ k\minus{}1$ the same, and $ 1$ different; $ k\minus{}2$ the same and $ 2$ different; ...; $ 2$ the same and  $ k\minus{}2$ different; all different.

$ \binom{n}{1}\binom{n\minus{}1}{0} \plus{} \binom{n}{1}\binom{n\minus{}1}{1} \plus{} \binom{n}{1}\binom{n\minus{}1}{2} \plus{} ... \plus{} \binom{n}{1}\binom{n\minus{}1}{k\minus{}2} \plus{} \binom{n}{k}$.

However, following the approach suggested by jbatterson, we get the equivalent but much easier expression $ \binom{n\plus{}k\minus{}1}{k}$, which for $ n\equal{}5$ and $ k\equal{}3$ is $ \binom{7}{3} \equal{} 35$.

Thank you, Andrew, for the detail answer.

Tech.


AMC
AIME
algebra
polynomial
trigonometry
symmetry
number theory

hey, can you regrade my paper?  i calculate a 117 instead of a 113.5

thanks

Can we dispute questions?  Good job on the most part, but I think the answer to #12 is 33...

Missed first place by 1 point  :( ...... ohh well.... this mock AMC seems easy

12. What is the number of distinct solutions (x,y,z) to the equation x^2 + yz = y^2 + xz if x, y
and z are integers between 1 and 5 inclusive?
(A) 31 (B) 8 (C) 6 (D) 25 (E) 37

This was the problem. Would anyone care to share why they got 33? I got this one wrong, so it doesn't really affect me, since I didn't get the right answer or 33.

Okay, my answer for #12:

First, we re-arrange the given equation:

\[x^2-y^2=xz-yz\]

or

\[(x+y)(x-y)=(x-y)z\]

which is true iff one (or both) of the following cases holds:
(1) $x-y=0$, so $x=y$, or,
(2) $x+y=z$.
For the first case, there are clearly 25 (5 values for $x$ and $y$ times 5 values for $z$) solutions.  For the second case, there are 8 solutions that don't fall under the first case: (1,2,3), (1,3,4), (1,4,5), (2,1,3), (2,3,5), (3,1,4), (3,2,5), and (4,1,5).  Try all you might, you won't find more.  Thus, there are 25+8=33 solutions.

[quote="JGeneson"]you are wrong on #12, i can list > 31 answers[/quote]

i got 31 during the test, but I got 33 now..........

[quote="Nosepuppy"]yea, JGeneson and I also get 33 for 12[/quote]

I got 33 as well.  Rearranging, we get
$(x-y)(x+y-z) = 0$, so
$x = y$ or $x+y = z$.

In the first case, there are 25 solutions: (1,1,1-5), (2,2,1-5), ... (5,5,1-5).  In the latter case, there are 8 additional solutions: (1,2,3), (1,3,4), (1,4,5), (2,3,5), and the four with x and y switched.  In all, 25+8=33.

[quote="white_horse_king88"]This was the problem. Would anyone care to share why they got 33? I got this one wrong, so it doesn't really affect me, since I didn't get the right answer or 33.[/quote]

If x=y, then any z will work, because then we have $x^2+xz=x^2+xz$. That gives us 25 equations. If x≠y, then x+y=z. For z=1,2,3,4,5, (x,y) has 0,0,2,2,4 possible values. So we get 33.

Ok... I think I screwed up number 12.  What I am going to do is go back and give everyone 6 pts. for that one.  Sorry about that.  The results are going to be a little different so just bear with me...

HAHA. I was  just thinking about that. Here's how. sinx = 3cosx, and tanx = 3. Thus x = tan^(-1)3, and x = something, and you substitute. HEHE. Cheap as crazy.

[quote="beta"]How do you do #11 with a calculator?? I don't think calculator will make #11 easier than it already is.[/quote]

If you did #11 the hard way, then a calculator will help a little bit.

Yeah... I noticed that some of the problems were trivial with a calculator.  But I realized that after I had sent the test to some ppl early so I didn't feel like I should change it.  But, hey, this test is for practice.  Using a calculator is legal but wont help you very much when the real thing rolls around.  So I hope that those of you who just used a calculator on some of them (especially #23) try to do them by hand because it is a good problem if you can't use a calculator.

Whoa.  We can use a calculator.

...definite dumb moment there.  Right along with forgetting until 7:45.

I say we give a round of applause to joml88 for writing the problems first  :D

[Applause]

How do you solve #20? And is there a simpler way to solve #21 than reducing it to "find the remainder when 12! * 10! is divided by 13"?

For #17, why is  m + n always 1 greater than the exponent?

For #6, is there a property that says the sum of the coefficients of the expanded form is the sum of the coeffients in the unexpanded form raised to that power? (was that understandable?)

thanks!

For any polynomial of any number of variables and any degree, the sum of the coefficients is f(1), so just say f(x,y) is whatever we had, and by the same principle, f(1,1) is the sum of the coefficients.

[quote="Yrael"]
How do you solve #20? And is there a simpler way to solve #21 than reducing it to "find the remainder when 12! * 10! is divided by 13"?

For #17, why is  m + n always 1 greater than the exponent?

For #6, is there a property that says the sum of the coefficients of the expanded form is the sum of the coeffients in the unexpanded form raised to that power? (was that understandable?)
thanks![/quote]
Here is how I did #20

Let f(x)+x^2=g(x)

Note that roots to g(x) are (-1, 2, -3, 4)

so g(x)=k(x+1)(x-2)(x+3)(x-4), where k is a constant
but since leading coefficient of f(x) is 1, so k must be 1 also.
so g(1)=24=f(1)+1
f(1)=23

#21) if you know Wilson's Theorem, (p-1)!=-1(mod p) if p is odd prime

[quote="beta"]sin^2x=9cos^2x, 9cos^2x+cos^2x=1, cos^2x=1/10,
cos^2x *(tanx)=sinxcosx=3/10.....

doesn't seem to hard to do anyway.......  calculator wouldn't really help much.......

I have to agree that #23 can be easily done with a calculator, but unfortunately I was TOO stupid to realize that :?  that I actually solved it..........[/quote]

There's another simple non-calculator solution for this one.  Clearly, there are two solutions in x from 0 to 2pi; one for which sinx and cosx are both positive, and one for which both are negative, and both have the same value for sinx*cosx, so we can assume both are positive (and x is a first quadrant angle).  From the given, we know that tanx=1/3, so we can draw a right triangle with one leg 1y and another leg 3y, and from the Pythagorean Theorem, we know that the hypotenuse is sqrt(10)*y.  Thus, sinx*cosx=(1/sqrt(10))(3/sqrt(10))=3/10

[quote="Yrael"]
For #17, why is  m + n always 1 greater than the exponent?
[/quote]

Let's see how we got the answer:

Log 2 =.3..., So log 2^2005=603.5..., Then 2^2005 has 604 digits. By the same method, 5^2005 has 1402 digits. 1402+604=2006.

But that's assuming we don't know that m+n is one greater than the exponent. Let the exponent equal x. Then the answer is x*log 2+ x*log 5 + 1. Then we have x(log 2 + log 5)+1=x+1.

[quote="Yrael"]For #6, is there a property that says the sum of the coefficients of the expanded form is the sum of the coeffients in the unexpanded form raised to that power? (was that understandable?)[/quote]

In any polynominal, if you set all the variables to one, then we get the sum of the coefficients. So we just set x and y to one to get (2)^4

[quote="beta"][quote="h_s_potter2002"]This Mock AMC was really easy compared to the others (I had a 16.5 increase), but a lot of the problems could be easily solved with a calculator, like 11 and 23.  I guess 15 could also have been done with a calculator.[/quote]

How do you do #11 with a calculator?? I don't think calculator will make #11 easier than it already is.[/quote]

I just plugged in values for x that worked.  You have to remember that I don't know any trignometry, so this is the easiest way I can do it.

Arg, I can't believe I forgot.  :(

Oh well, at least I can try taking it on my own time. :?

[quote="interesting_move"]Whoa, what exactly happened?  How did I get a 139.5?  Did you give credit for 12, since the answer is supposed to be 33?  What about #17 which you said I missed?  Is it B or E?  Thanks!

-interesting_move[/quote]

#17 is E

arghhhhhhhh, before I was 1 point off the first place, now I am 7 points off the first place and only tied for 3rd place :cry:

Actually I reviewed interesting_move s score and together we decided that he got a 133.5  So that means that JGeneson wins and beta is only 4.5 pts off the leader  :D

Hopefully that is the last of my mistakes  :blush:

Arg I had to film a movie... couldn't make it and didn't find out till too late.

I'll take this contest on my own time and see how I do.

And to joml, I'll get the tourney questions into you this week

I think the reason why this Mock AMC was easier is because there wasn't any tricky question....(I solved 20 problems and didn't get anything wrong)...usually on AMC there were 2-4 tricky questions

Question 15 was also easily solvable with a graphing calculator. Just graph the equation (y instead of 0) and then find the 0.  When done, just find cosine.

oh btw for #23),
what I did was
let  $a=\sqrt[3]{2+\sqrt{5}},  b=\sqrt[3]{2-\sqrt{5}}$

$a^3+b^3=(a+b)^3-3a^2b-3b^2a$
$=(a+b)^3-3ab(a+b)$

Let $a+b=x$
so
$4=x^3+3x$
$x^3+3x-4=0$,
and very lucky that if factors....... what happens if it doesn't factor out nicely?  Is there another approach(without using a calculator :) ) to solve the problem?

A .pdf version can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19527]here[/url].

Number of questions:

1. How do you do #9?

2. I still don't get how you can use Wilson's theorem to get 21.

3. Does anyone have a rigorous solution for 22? I feel pretty sure my reasoning was solid, but never once got one of the right answers  :( . It would be nice to see a right way.

4. How the heck do you do 24 or 25? I am totally gone on those.

Thanks,
Sran

Sorry but does anybody can show me the proof for question 22 and 25??
Here's my proof for question 22(I got it wrong :( )
I consider following cases:
1st case: one of 2 numbers has the unit digits is 0. We see that now every ways to choose y is satisfactory so we will have: 9*99*2-(2+4+6+8+10+12+14+16)(x,y)(we see that there are 8+7+6+5+4+3+2+1 ways to choose x,y having the units digits is 0)
2nd case: one of 2 numbers has the unit digits is 1. We see that there are 9 numbers having units digits is 0 that's already used with this units digits 1. And we see that every numbers have units digits is 9 is omitted. Therefore, the total ways  is: 10*80*2-(2+4+6+8+10+12+14+16+18) pairs (x,y)
3rd case: one of 2 numbers has the units digits is 2. We see again that there are 19 numbers having units digits is 0 and 1 that's already used in 2 cases above. And every numbers have the units digits is 8 or 9 must be omitted. Therefore, we wil have: 10*60*2-(2+4+6+8+10+12+14+16+18) pairs(x,y)
Continue this process until we find the total ways in the case having the units digits is 4(mean 2 more cases). After that, since the symmetry with x to y, we see that we counted all of 9 neccessary cases(i.e: (45,54)=(54,45)). And we also have the ways to choose 2 elements from 99 elements is: P(2,99)(since (x,y) is ordered pair).
Now can anybody help me check it since I can't find my mistake :blush:


If the numbers $\ x_1,x_2,...,x_n$ are strict positive, prove
$\ 4\left(\sum\frac{x_1^3}{x_1+x_2}\right)\geq\ 3\left(\sum x_1^2\right)-\left(\sum x_1x_2\right)$

[quote="stancioiu sorin"]If the numbers $\ x_1,x_2,...,x_n$ are strict positive...
[/quote]
What is this "strict positive"? Thank you.

Original and nice inequality.

\[ 4\Bigl(\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\Bigl)\geq3(a^2+b^2+c^2)-(ab+bc+ca). \]

By Cauchy Schwarz, $4\sum\frac{a^3}{a+b}\ge2\sum a^2\frac{2\sum a^2}{\sum a^2+\sum ab}=2\sum a^2\frac{\sum a^2-\sum ab}{\sum a^2+\sum ab}+2\sum a^2$.
The remaining part should be trivial enough.

Denote $M=\sum x_1^2, N=\sum x_1x_2$

By Cauchy Schwarz, we obtain,

$LHS\geq \frac{4M^2}{M+N}\geq 3M-N=RHS$

The last inequality is equivalent to $(M-N)^2\geq 0$ which is trival.


maybe very easy!!

Let a triangle $ ABC$
Proove that $ tanA+tanB+tanC\geq 3\sqrt{3}$
When the equality holds?

Jensen's.

$ \frac13\cdot(\tan{A}+\tan{B}+\tan{C}) \geq \tan(\frac{A}{3}+\frac{B}{3}+\frac{C}{3})$

$ \tan{A}+\tan{B}+\tan{C}\geq 3\cdot\tan\frac{\pi}{3}$

$ \tan{A}+\tan{B}+\tan{C}\geq 3\sqrt3$

The inequality is true only for an acute triangle.
We have $ \tan A+\tan B+\tan C=\tan A\cdot\tan B\cdot\tan C$
If $  A>\frac{\pi}{2}$ then $ \tan A<0, \tan B>0,\tan C>0\Rightarrow \tan A\tan B\tan C<0$


[hide]Just divide both sides by 2^1995.
Thus you get
2^3 - 2^2-2^1+2^0
8-4-2+1=3  so the answer is C.[/hide]
[size=75][color=blue]Moderator kyyuanmathcount's note:No size changes please, and also, please use hide tags so other students can do the problem without having the answer there, thanks![/color][/size]

[hide]Well, we can factor out $2^{1995}$ from this expression: $2^{1995}(2^3-2^2-2^1+2^0)$.The expression inside the parentheses equals 3, so the answer is C.[/hide]


Given n lines in plane, such that no three lines are concurrent and no two lines are paralel. Prove that we can number each region a number m (-n $\leq$ m $\leq$ n) such that the sum of numbers of such regions that locate in a semiplane with side which is a given line is equal to 0. (every line) (the number of the different region may be numbered a same number)

Sorry. My English is not very good.

Oh! I try to solve it for a long time.
Can you post your solution, please?


can anyone give the proof for problem no 28???

Dei i posted it only today  :)  Think Pardesi or Sanjith or Soumya or Chappli or Vihang or Sai Mali ............... may get it easily. Wait and watch da  :)

seri da therija solutiona post pannu!!!!!!!!!!!!!!!!! :D  :D  :D

[quote="Cyber-Shreyas"]Or Shreyas. :D[/quote]

My dear boy that was elementary so i didnt include ur or my name  :P  :P   :D

@pardesi: Dont nv da. Let others try da so plz delete it

[quote="pardesi"]yes that's actually the Wholestone's theorem i suppose[/quote]

Theorem NV!

[quote="pardesi"]that's not nv but respecting the author of the theorem... :)[/quote]

Google does't know what is Wholestone theorem. :) Have you got the spelling wrong
And Did you know that Charles Wheatstone is the cousin brother of Charles Wholestone.

[quote="pardesi"]yes that's actually the Wholestone's theorem i suppose[/quote]

And wiki says no such theorem  :huh:

ok wrong spelling it's actualy
Wolstenholme's

try another method for the sum or try proving the theorem itself but no cheating  :D  :D   :P

[url=http://en.wikipedia.org/wiki/Wolstenholme's_theorem]here[/url] is the wiki link...
i have already done this many times.. :)

i already found the link but where is the proof of that theorem?  :huh:

[quote="madness"]
And wiki says no such theorem  :huh:[/quote]
i gave the link  because of this  :huh:  :huh:  :ninja: