calculus
function
derivative
algebra
domain
limit
analytic geometry

That is a correct calculation of a different problem.

You know how to find critical points; you also need to be able to test to find out whether they are maxima, minima, or neither.

Most of the early optimization problems you see will also be accessible to noncalculus methods.

For the specific problem in question: you have $xy=100$ and want to minimize $x+y=x+\frac{100}x$.

[quote="Silverfalcon"]Hi,

I just read the pages about optimization and I'm not sure how to solve them.

Here is a problem that I need help on.

Find two positive numbers whose product is 100 and whose sum is minimum.

[hide="Using non-calculus"]

AM-GM worked out nice here. Set $x = a$ and $y = \frac{100}{x} = b$ and apply AM-GM to $a$ and $b$, I get $(x-10)^2 \geq 0$ which implies $x = y = 10$. [/hide]

But I'm not sure how to use calculus to find the minimum. I know how to do the maximum but I can't do minimum. (Assuming I did the maximum right). So, to be sure I know I can do maximum (if I don't, working on minimum won't help much I guess), here is a problem and solution I did for maximum.

[hide="Max"]

Find two numbers whose sum is 100 and whose product is maximum.

I set $y = 100-x$ and get equation $(100-x)x = 0$. Let this function be $f(x)$. Taking derivative (one), we get

$f'(x) = -2x+100 = 0$ and solving for $x$ gives 50. [/hide]

I hope I'm right.. :|  Thanks for great help![/quote]

If you don't know the calculus approach (the method of Lagrange multipliers) you can always try a brute force attack.

The prime factors of 100 are 2, 2, 5, and 5.

choose (1,100) : sum = 101.

choose (2, 50): sum = 52.

choose (4, 25): sum = 29.

choose (5, 20): sum = 25.

choose (10, 10): sum = 20.

You have exhausted the list of integer factors. The pair with the minimum sum is (10,10), which is the same answer you would get by using the calculus.

[quote="Kent Merryfield"]Dr. No, there is a difference between a discrete optimization problem and a continuous optimization problem (with the discrete ones often being more difficult). The original language of this problem said "positive number," not "positive integer." I think you know that, but (as someone who teaches calculus) I would not want to be in the business of encouraging students to think that the integers are all that matter.

Just to make the point, what two positive numbers whose product is 167 have the smallest sum? The answer, which is $\sqrt{167},$ is not something that you're going to find using the method you just described.[/quote]

You are evidently not a Number Theorist  ;)

Seriously, your point is well taken but not knowing how much calculus Silverfalcon has seen yet I didn't want to blow his mind by giving a full presentation of multivariate optimization techniques.  Whether my approach was pedagogically sound is debatable; that my solution to the problem is the solution is not!

Clearly, Silverfalcon is dealing with the "Calculus I" part of a textbook; the time for discussing Lagrange multipliers has not come yet.

(Now I'm speaking to Silverfalcon.)

Your most important task is to express the thing to me maximized or minimized as a function of one or more variables. At this (single-variable) stage, you need to reduce that to a function of one variable.

But note that it's not enough simply to write a formula for that function. It is vitally important that you also state the domain of that function. In many "practical" problems, that domain is restricted by some physical considerations not apparent from the formula.

You then need to find all the points that need investigating. That always includes all of the points where the derivative is zero, but it includes other things. For piecewise-defined functions, it may not be apparent at first glance whether the derivative exists at the places where one piece changes to another; so include such points as "needing to be investigated." And the endpoints of the domain are always points that need to be investigated. Even if the domain doesn't include endpoints, you want to know how the function behaves in the limit as you approach that end.

In this case, we wish to minimize $f(x)=x+\frac{100}x$ over the domain $x\in(0,\infty).$

There are no endpoints in the domain, but we explore the extreme cases anyway:

$\lim_{x\to0^+}f(x)=\infty$ and $\lim_{x\to\infty}f(x)=\infty.$

We now differentiate the function (it is differentiable everywhere on the domain) and find all $x$ for which $f'(x)=0.$ There is exactly one such point in the domain (you can find it yourself). Since the function tends to infinity at both ends of the domain, this has to be the location of a minimum.

[quote="Kent Merryfield"]

(Now I'm speaking to Silverfalcon.)

Your most important task is to express the thing to me maximized or minimized as a function of one or more variables. At this (single-variable) stage, you need to reduce that to a function of one variable.

But note that it's not enough simply to write a formula for that function. It is vitally important that you also state the domain of that function. In many "practical" problems, that domain is restricted by some physical considerations not apparent from the formula.

You then need to find all the points that need investigating. That always includes all of the points where the derivative is zero, but it includes other things. For piecewise-defined functions, it may not be apparent at first glance whether the derivative exists at the places where one piece changes to another; so include such points as "needing to be investigated." And the endpoints of the domain are always points that need to be investigated. Even if the domain doesn't include endpoints, you want to know how the function behaves in the limit as you approach that end.

In this case, we wish to minimize $f(x)=x+\frac{100}x$ over the domain $x\in(0,\infty).$

There are no endpoints in the domain, but we explore the extreme cases anyway:

$\lim_{x\to0^+}f(x)=\infty$ and $\lim_{x\to\infty}f(x)=\infty.$

We now differentiate the function (it is differentiable everywhere on the domain) and find all $x$ for which $f'(x)=0.$ There is exactly one such point in the domain (you can find it yourself). Since the function tends to infinity at both ends of the domain, this has to be the location of a minimum.[/quote]

Silverfalcon,

The only thing I can add to Mr. Merryfield's explanation is the reason why one examines the points where $f'(x) = 0$.  For any differentiable function $f(x)$, its derivative, $f'(x)$, gives the slope of the line tangent (i.e. parallel to the curve and just touching it) to $f(x)$ at the point $x$.  If one graphs $f(x)$, one can easily see that at its extreme points, maxima or minima, the slopes of the tangent lines at these points is zero; the tangent lines themselves are horizontal.

The reason one examines the end points of the domain is because even though $f'(x) = 0$ indicates an extreme it is still possible for $f$ at the end points to have more extreme values than those given by $f(x)$ where $f'(x) = 0$. Where this is the case, it is those end point values that give the actual extrema.

Thank you for awesome explanations!

Yes, I'm covering Calculus 1 part of the text. For the problem I posted, is this how you mean to do?

$f(x) = x^1 + 100 x^{-1}$
$f'(x) = 1 -100 x^{-2}$

Setting this to zero:

$0 = 1 - \frac{100}{x^2}\\
-1 = - \frac{100}{x^2}\\
1 = \frac{100}{x^2}\\
x^2 = 100\\
x = \pm 10$

But since the problem specified for positive numbers, we get $x = 10$ as the answer.

Is this correct way? I'm still bit unsure why I set the first derivative to 0 though. (I read the above explanations several times but still bit confused)

[quote]I'm still bit unsure why I set the first derivative to 0 though.[/quote]
What's behind this is a theorem usually attributed to and named for Fermat (who lived a generation before there was calculus, but it's still appropriate):

If $f$ has a local maximum or local minimum at $a$ and if $f'(a)$ exists, then $f'(a)=0.$

The proof is fairly simple: suppose $f$ has a local maximum at $a.$ (The case of a local minimum is similar, with the signs reversed.)

Assume $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h$ exists.

For all small enough $h$, $f(a+h)\le f(a).$ (That's what local maximum means.)

So if $h>0,$ $\frac{f(a+h)-f(a)}h\le0$ and $\lim_{h\to0^+}\frac{f(a+h)-f(a)}h\le 0.$

While if $h<0,$ $\frac{f(a+h)-f(a)}h\ge0$ and $\lim_{h\to0^-}\frac{f(a+h)-f(a)}h\ge0.$

Since we were assuming the limit exists, the right and left hand limits must be equal, and equal to $f'(a)$. We have both $f'(a)\le 0$ and $f'(a)\ge0$ so $f'(a)=0.$

This theorem then leads to the definition of a critical point as a point in the domain of the function for which either $f'(a)=0$ or $f'(a)$ doesn't exist. We must also add to the list of points to be investigated endpoints of the domain. (The proof above does not work for endpoints.) Above, I also advocated adding points for which the existence of the derivative is not obvious to the list of "points to be investigated."

Suppose the function were $f(x)=x+\frac{100}x$ for all real numbers for which that makes sense. If asked to name the critical points, a typical calculus beginner would say there are three: $\pm10$ and $0,$ since the derivative does not exist at $0.$

In fact, $0$ is not a critical point; it lies outside the domain of the function. But if we take the attititude that we are assembling a list of "points to be investigated," we realize that we cannot understand the appearance of the graph of this function without understanding how it behaves in a neigborhood of $0.$

[quote]$x=\pm10$

But since the problem specified for positive numbers, we get $x = 10$ as the answer. [/quote]

Domain, domain, domain! In other words, yes, that's right.


Prove: for every positive integer there exists a positive integer having n digits, all of them being  $1$'s and $2$'s only, such that this number is divisible by $2^{n}$.
Is this still true in base $4$ or $6$¿

[hide="Part 1"]For $n=1$ we have $N=2$.

Assume there exists such number $N$ for some given number of digits $n$. Then we have to prove that either $10^{n}+N$ or $2\cdot 10^{n}+N$ is divisible by $2^{n+1}$. But if $N=k\cdot 2^{n}$ where $k$ is integer, then

${10^{n}+N\over 2^{n+1}}={5^{n}+k\over 2}$

${2\cdot 10^{n}+N\over 2^{n+1}}={2\cdot 5^{n}+k\over 2}$

and exactly one of the two numerators is even since their difference is odd. Thus the inductive proof is completed, with the enhancement that such number is [u]unique[/u] for any given $n$.[/hide]

[hide="base 4"]
If we take $n\geq2$, then the number is divisible by $2^{2}= 10_{4}$, and therefore has a $0$ at the end of the number, and does not consist of only 1s and 2s.
[/hide]

The result is true for base $6$ as well, because we can simply replicate Farenhajt's proof with $10$ and $5$ replaced by $6$ and $3$.


geometry
geometric transformation
homothety
trapezoid
perpendicular bisector

Let $ O_1$ and $ O_2$ be the midpoints of $ AC$ and $ AB$, respectively.  Let $ a$ and $ b$ be the radii of $ (O_1)$ and $ (O_2)$, respectively.  It is straight-forward to calculate that $ O_1 D \equal{} O_2 M \equal{} a$ and $ O_1 M \equal{} O_2 E \equal{} b$.  Also, $ \angle O_1 DM \equal{} \angle O_2 ME \equal{} 90^\circ$.  Hence, triangles $ O_1 DM$ and $ O_2 ME$ are congruent.

Then $ \angle DO_1 M \equal{} \angle EO_2 M$, so $ O_1 D \parallel O_2 E$.  Hence, the homothety centred at $ A$ that takes $ (O_1)$ to $ (O_2)$ also takes $ D$ to $ E$.  Therefore, $ A$, $ D$, and $ E$ are collinear.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/7de7654cc183c75cdf913e320f22c64f.png[/img]

[quote="stergiu"]Please , see the attachment.I 'm trying to find a simple solution to this problem, but I can not.I have only an indirect proof.[/quote]
[hide="Solution"]
Let $ AE$ meet the smaller circle at $ D'$.  Let the foot of the altitude from $ M$ to $ AE$ be $ P$.  Furthermore $ CD'\perp AE$ and $ BE\perp AE$ since they are diameters of given circles.  Then, $ D'CBE$ is a trapezoid with midline $ MP$, so $ PD' \equal{} PE$, which means that $ M$ lies on the perpendicular bisector of $ D'E$, so $ \angle MD'E \equal{} \angle AEM \equal{} 90 \minus{} \angle EAM$.  Thus, $ \angle AD'M \equal{} 90 \plus{} \angle D'AM$, so $ \angle CD'M \equal{} \angle AD'M \minus{} \angle AD'C \equal{}  \angle D'AC$, so $ MD'$ is tangent to the smaller circle, which means that $ D \equal{} D'$, so $ ADE$ is a line.  [/hide]


quadratics
superior algebra
superior algebra unsolved

I have showed that $ 1\plus{}\sqrt{\minus{}3}$ is irreducible but not prime in $ Z[\sqrt{\minus{}3}]$.
Is it true in general this?
 Let $ d\equal{}\minus{}p$ where $ p$ is prime and $ d\equal{}1(mod 4)$
for example: -3, -7, -11, -19 and so on
Is it true that $ 1\plus{}\sqrt{d}$ is irreducible but not prime in $ Z[\sqrt{d}]$?

$ (1\plus{}\sqrt{d})(1\minus{}\sqrt{d}) \equal{} 1\minus{}d \equal{} 2 \cdot \frac{1\minus{}d}{2}$. Show that these factorisations are different (note that $ d<\minus{}1$ odd is everything needed, so also the cases $ d \equiv 3 \mod 4$ are handled that way).

The difference in the two cases mod $ 4$ is that when $ d\equiv 1\mod 4$, the natural ring of "integers" is actually $ \mathbb{Z}\left[\frac{\sqrt{d}\plus{}1}{2}\right]$

In those rings, the first instance of non-unique factorization you encounter is $ \frac{1\plus{}\sqrt{\minus{}15}}{2}\cdot \frac{1\minus{}\sqrt{\minus{}15}}{2}\equal{}2\cdot 2$. For $ \minus{}d$ prime, the first instance is $ \frac{1\plus{}\sqrt{\minus{}23}}{2}\cdot \frac{1\minus{}\sqrt{\minus{}23}}{2}\equal{}2\cdot 3$.


Supposethat to make pizza, Terry uses three ingredients: Tomato sauce, dough, and cheese. Terry initially uses 100 gallons of tomato sauce per day, and the cost of the other ingredients(dough,souce) is 500$ per day.
1)Draw a curve to show the relationship between the price of tomato sauce and the daily cost of production pizza (for prices between $1 and $5)


Sorry if this topic is in wrong place, but do you know which country is the best at each kind of mathematical olympiad problems, such as Algebra, Combinatorics... Can we have an opinion poll ?  :)

Yeah, and i think with Algebra, that is Chinese, and number theory, geometry, inequality..?


limit
integration
real analysis
real analysis unsolved

Let $f:\mathbb{R}\to\mathbb{R}$ a continuous and periodical function. Prove that if there exists
$\lim_{x\to\infty} \int_{0}^{x} f(t)dt$ finite ,  then $f=0$.

Suppose $f$ has a period of $T$. It is clear to see that $\int_x^{x+T} f(x)dx = 0$ for all $x$.

From Cauchy's Criterion, for any $\epsilon > 0$, there exists $X$ such that, $\left|\int_{x_1}^{x_2} f(x)dx\right| < \epsilon$ for all $x_1,x_2>X$.

It is not difficult to see that:

For any $\epsilon > 0$, for any $x_1,x_2\in [0,T]$ we have $\left|\int_{x_1}^{x_2} f(x)dx\right| < \epsilon$. Hence $\int_{x_1}^{x_2} f(x)dx = 0$ for all $x_1,x_2\in [0,T]$. We are going to show that $f(x)\equiv 0$ on $[0,T]$.

Without loss of generality, suppose $f(x_0)>0, x_0\in [0,T)$. Then from the continuity of $f$, we can find $\delta>0$ such that $f$ is positive on $[x_0,x_0+\delta]\subset [0,T)$, thus $\int_{x_0}^{x_0+\delta} f(x)dx > 0$. A contradiction.


function
induction
algebra unsolved
algebra

Determine all functions $ f: \mathbb{Q}\rightarrow\mathbb{R}$ such that

$ f(xy)\equal{}f(x)f(y)$ and $ f(x\plus{}y)\leq \max (f(x),f(y))$ for all $ x,y\in \mathbb{Q}$

Let $ n$ be any positive integer and $ x$ any rational, we can prove by induction $ f(nx)\leq f(x)$. Particularly, $ f(n)\leq f(1)$. Hence,
$ f(nx)\minus{}f(x)\equal{}f(x)(f(n)\minus{}f(1))\leq0$------------------------(1)
We notice that
$ f(0)\equal{}f(0)*f(0)$-----------------(2)
$ f(1)\equal{}f(1)*f(1)$-----------------(3)
$ f(1)\equal{}f(\minus{}1)*f(\minus{}1)$---------------(4)
From now on, all the variables are assumed to be rationals.
Case 1) $ f(0)\equal{}1$
$ f(x)\equal{}f(x)f(0)\equal{}f(0)\equal{}1$

Case 2) $ f(y)\equal{}0$ for some $ y\neq0$
$ f(x)\equal{}f(\frac xy)f(y)\equal{}0$.

Case 3) $ f(0)\equal{}0$, $ f(1)\equal{}1$, $ f(\minus{}1)\equal{}\minus{}1$.
$ f(\minus{}x)\equal{}\minus{}f(x)$. Looking at (1), we have both
$ f(x)*(f(n)\minus{}1)\leq0$ and $ f(\minus{}x)*(f(n)\minus{}1)\equal{}\minus{}f(x)*(f(n)\minus{}1)\leq0$.
Combine them together, $ |f(x)*(f(n)\minus{}1)|\leq0$.
$ f(n)$ has to be $ 1$ for any positive integer $ n$.
Using $ f(xy)\equal{}f(x)f(y)$, we have another solution
$ f(x)\equal{}sign(x)$, where $ sign(x)$ is 1 for positive, 0 for 0, -1 for negative.

Case 4) $ f(0)\equal{}0$, $ f(1)\equal{}1$, $ f(\minus{}1)\equal{}1$. $ f(n)\equal{}1$ for all positive integer $ n$.
Using $ f(xy)\equal{}f(x)f(y)$, we have another solution
$ f(x)\equal{}1$ when $ x\neq0$, $ f(0)\equal{}0$.

Case 5) $ f(0)\equal{}0$, $ f(1)\equal{}1$, $ f(\minus{}1)\equal{}1$. $ f(n)<1$ for some positive integer $ n$.
Obviously $ f(x)\geq0$ for all $ x$, otherwise (1) will not hold.
Moreover, $ f(x)>0$. Otherwise we reduce to case 2).
We know $ f(n)\leq1$ for all positive integer $ n$. Let $ p$ be the smallest positive integer such that $ 0<f(p)\equal{}b<1$. So $ f(m)\equal{}1$ for all $ m<n$. Obviously $ p$ is a prime, otherwise $ p\equal{}p_1*p_2$ implies $ f(p)\equal{}f(p_1)f(p_2)\equal{}1$.
If $ p|n$, $ f(n)\equal{}f(p)f(\frac np)\leq b<1$.
If $ n$ is positive  and has nonzero remainder $ r>0$ when divided by $ p$, $ f(r)\equal{}1\leq\max(f(r\minus{}n),f(n))$. Since $ p|r\minus{}n$, $ f(r\minus{}n)\equal{}f(n\minus{}r)<1$. So $ f(n)\geq1$. But $ f(n)\leq1$ for all integer $ n$. So $ f(n)\equal{}1$.
Now we can create another solution
$ f(0)\equal{}0$ and $ f(x)\equal{}b^t$ if $ x\neq0$ and $ |x|\equal{}p^t\frac mn$ where $ gcd(m,n)\equal{}gcd(m,p)\equal{}gcd(n,p)\equal{}1$ and $ t$ is integer.

Conclusion: we have five possible different solutions.


Stronger inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=25778 .

  Darij

From Cauchy we have  \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \Big( a(b+c)+b(c+a)+c(a+b) \Big) \geq \left( \frac 1a +\frac 1b+ \frac 1c \right)^2 \] and as $\frac 1a + \frac 1b + \frac 1c = ab+bc+ca$ we obtain that \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \geq \frac{ ab+bc+ca} 2 \geq \frac{\sqrt [3]{a^2b^2c^2}}{2} = \frac 32 \] which is what we wanted to prove.

We have  abc=1

   Therefore  1/(a^3)(b+c)   =   bc/(a^2)(b+c)

  From Cauchy we have :

      bc/(a^2)(b+c) +  1/4b + 1/4c = bc/(a^2)(b+c)+(b+c)/4bc >= 1/a

    Similarly  we have  P>= 1/2a + 1/2b +1/2c>=3/2

                                              which is what we wanted to prove

hope this has not been forgotten!  :D  i came to a weird solution, but cool and nice!

we want to prove there exists and $r$ such that
$\frac{1}{a^3(b+c)}\geq \frac{3}{2}\frac{a^r}{a^r+b^r+c^r}$.

take $b=c$ and use that $a=b^{-2}$, then we get to

$\frac{b^6}{2b}\geq \frac{3}{2}\frac{b^{-2r}}{b^{-2r}+2b^r}$ or $b^5\geq \frac{3}{1+2b^{3r}}$ or $f(b)=2b^{3r+5}+b^5-3\geq 0$.

Obviously, $f(1)=0$, so we want $f'(1)=0$, this is
$f'(b)=2(3r+5)b^{3r+2}+5b^4$, so we want $2(3r+5)+5=0$, or $r=\frac{-5}{2}$.

we want to prove that
$\frac{1}{a^3(b+c)}\geq \frac{3}{2}\frac{a^\frac{-5}{2}}{a^\frac{-5}{2}+b^\frac{-5}{2}+c^\frac{-5}{2}}$.

use that $a=\frac{1}{bc}$ and let $x=\sqrt{b}$ and $y=\sqrt{c}$.

we want to prove that
$\frac{x^6y^6}{x^2+y^2}\geq \frac{3}{2}\frac{x^5y^5}{x^5y^5+\frac{1}{x^5}+\frac{1}{y^5}}$ or (after a bit of simplifications)

$2(x^{10}y^{10}+x^5+y^5)\geq 3x^4y^4(x^2+y^2)$

from am-gm and rearrangement we have that

$2(x^{10}y^{10}+x^5+y^5)\geq (x^{10}y^{10}+2x^4y)+(x^{10}y^{10}+2xy^4)\geq 3x^4y^4(x^2+y^2)$ and we're done!

[hide="Sorry for bringing this up again but I just recently did this problem from the inequalities packet"]
Notice that $\sum bc \geq  3\sqrt[3]{(abc)^{2}}=3$ by $AM-GM$. Also,
\[\sum \frac{1}{a^{3}(b+c)}=\sum \frac{b^{2}c^{2}}{a^{3}b^{2}c^{2}(b+c)}=\sum\frac{(bc)^{2}}{ab+ac}\geq \frac{\left(\sum bc \right)^{2}}{2\sum bc}=\frac{\sum bc}{2}\geq \frac32\]
which proves the problem.
[/hide]

the proof is long and i'm too tired to rewrite it so i'll just note:

use well known substition $a = \frac{x}{y}$ and analogically for $b$ and $c$, so the assumption that $abc = 1$ is satisfied, ok

now we multiply everything, we have something like this
$\sum \frac{(y^{3}x)(y^{3}z)}{(x^{3}z)(y^{3}z)+(x^{3}y)(y^{3}x)}$ then we prove it pretty same as Nesbitt

soz lads... i thought this way worths a mention...

[quote="x^n+y^n=z^n"][quote="orl"]Let $ a,b,c$ be positive real numbers such that $ abc=1$.
[/quote]

[quote="Valentin Vornicu"]
\[ \frac{\sqrt [3]{a^{2}b^{2}c^{2}}}{2}= \frac{3}{2}\]
[/quote]

 :huh:?[/quote]

Valentin seemed to have made a typo, it should be ${ \frac{3}{2}\sqrt [3]{a^{2}b^{2}c^{2}}}= \frac{3}{2}$.

Hello. I am really sorry for bringing up such an old topic. I am really new to olympiad inequalities and was just curious as to how you guys would have come up with that solution with Cauchy.

Did you instinctively know that you were looking for $ ab \plus{} bc \plus{} ac$ in some form and know that the expression $ a(b\plus{}c) \plus{} b(a\plus{}c) \plus{} c(a\plus{}b)$ would similarly give you this expression?

Or did you narrow down the possible approaches (from C-S, AM-GM, finding some expression always $ \ge 3/2$ but $ \le$ the left hand side, etc.) and try applying them until something worked?

I've noticed that the most common solution to this problem has not been posted so I will post it for completeness:

Replace $a,b,c$ with $\frac{1}{x},\frac{1}{y},\frac{1}{z}$. We still have $xyz=1$. We aim to prove $\sum\frac{x^3yz}{y+z}\ge \frac{3}{2}$ or equivalently $\sum\frac{x^2}{y+z}\ge\frac{3}{2}$.

From here just about any method will work. First by Cauchy-Schwarz: $\left(\sum y+z\right) \left(\sum\frac{x^2}{y+z}\right)\ge (x+y+z)^2$.
This rearranges to $\sum\frac{x^2}{y+z}\ge \frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}$.

Another way is using Chebyshev and Nesbitt: $3\left(\sum\frac{x^2}{y+z}\right)\ge\left(\sum\frac{x}{y+z}\right)(x+y+z)\ge\frac{9}{2}$.

[quote="WakeUp"]I've noticed that the most common solution to this problem has not been posted so I will post it for completeness:

Replace $a,b,c$ with $\frac{1}{x},\frac{1}{y},\frac{1}{z}$. We still have $xyz=1$. We aim to prove $\sum\frac{x^3yz}{y+z}\ge \frac{3}{2}$ or equivalently $\sum\frac{x^2}{y+z}\ge\frac{3}{2}$.
.[/quote]
From Cauchy: $\frac{x^2}{y+z}+\frac{y+z}{4}\ge x$

[quote="orl"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {1}{a^{3}\left(b \plus{} c\right)} \plus{} \frac {1}{b^{3}\left(c \plus{} a\right)} \plus{} \frac {1}{c^{3}\left(a \plus{} b\right)}\geq \frac {3}{2}.
\][/quote]
let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$. Inequality becomes: $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$$\geq \frac{3}{2}$.
Now, by $AM-GM$ inequality we have: $\frac{x^2}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x$
Also quite similar and we get results: $LHS\geq \frac{z+y+x}{2}\geq \frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}=RHS$. So we are done:)[hide]I don't know English[/hide]

Begin by noticing $ab+bc+ca \ge 3\sqrt[3]{(abc)^2} =3$. Then, we have

\begin{align*}
\sum_{\text{cyc}} \frac{1}{a^3(b+c)} &= \sum_{\text{cyc}} \frac{b^2c^2}{a^3b^2c^2(b+c)} \\
&= \sum_{\text{cyc}} \frac{(bc)^2}{ab+ac} \\
& \ge \frac{(ab+bc+ca)^2}{2(ab+bc+ca)} = \frac{3}{2}. \ \square
\end{align*}

[quote=florr.io]

$Can \ EV \ do \ it \ ?$[/quote]

I didn't solve like that but maybe it can by mixed values.

[b]Generalization 2[/b]
Let $a_{1},a_{2},\cdots,a_{n},n,p$ be positive reals ($n\geq 2$ ,$p\geq 1$) such that $\prod{a_{1}}=1$. Then prove that

$$\sum_{cyc}{\left(\dfrac{1}{a_{1}^{2p+1}\left(x_{1}.\dfrac{\prod{a_{1}}}{a_{1}a_{n}}+x_{2}.\dfrac{\prod{a_{1}}}{a_{1}a_{2}}\right)^{p}}\right)}\geq \dfrac{n}{\left(x_{1}+x_{2}\right)^{p}}$$

By Titu,
\begin{align*}
\sum_{\text{cyc}} \left(\frac{\left(\frac1a\right)^2}{a(b+c)}\right) &\ge \frac{\left(\frac1a+\frac1b+\frac1c\right)^2}{\sum_{\text{cyc}} a(b+c)}\\
&=\frac{(ab+bc+ca)^2}{2(ab+bc+ca)}\\
&=\frac12 (ab+bc+ca),
\end{align*}
so the problem reduces to showing that $ab+bc+ca \ge 3$, which is clear.

[b]Generalization 4[/b]
Let $a_{1},a_{2},\cdots,a_{n},n,p,k,s$ be positive reals ($n\geq 2$, $p\geq 1$, $2s\leq n-1$) such that $\prod{a_{1}}=k$. Then prove that

$$\sum_{cyc-j}{\left(\dfrac{k^{p+1}}{\left(a_{j}a_{j+1}\cdots a_{j+s}\right)^{2p+1}\left(x_{1}\left(a_{j+s+1}a_{j+s+2}\cdots a_{j+n-s-2}\right)+x_{2}\left(a_{j+2s+2}a_{j+2s+3}\cdots a_{j-1}\right)\right)^{p}}\right)}\geq \dfrac{n\sqrt[n]{k^{n-s-1}}}{\left(x_{1}+x_{2}\right)^{p}}$$

Here
\begin{align*}
    \sum \frac{1}{a^3 (b + c)} &= \sum \frac{\frac{1}{a^2}}{ab + ac} \\
    &\geq \frac{(\sum \frac{1}{a})^2}{2\sum ab}\text{ (by Titu)} \\
    &= \frac{(\sum ab)^2}{2a^2 b^2 c^2 \sum ab} \\
    &= \frac{\sum ab}{2a^2 b^2 c^2} \\
    &= \frac{1}{2} \sum ab \\
    &\geq \frac{3}{2}\text{ (by AM-GM)}
\end{align*}
as desired.

$=\sum_{cyc}\frac{b^2c^2}{a(b+c)} \geq \frac{ab+bc+ca}{2} \geq \frac{3}{2}$ by Titu's lemma and AM-GM respectively.
We are thus done.

By Holder's inequality we get \[\left(\sum_{\text{cyc}} a(b+c)\right)^{\frac{1}{2}} \left(\sum_{\text{cyc}} \frac{1}{a^3(b+c)} \right)^{\frac{1}{2}} \geq \left(\frac 1a + \frac 1b + \frac 1c \right) = (ab+ac+bc)\] Thus it remains to prove that $\frac{ab+ac+bc}{2} \geq \frac{3}{2}$. This is obvious from Muirhead with $\left(\frac 43, \frac 43, \frac 13 \right) \succ (1, 1, 1)$

$a=b=c$ works
Multiplying by the denominators  and using $abc = 1$ gives
$$\sum{a^3b^2 a^2b^3+ bc + b^4c^4} \ge 3 + \frac{3}{2}\sum{a^2b + b^2a}$$
From Muirhead's Inequality we have
$$\sum{a^2b + b^2a} = \sum{a^{\frac{8}{3}} b^{\frac{5}{3}} c^{\frac{2}{3}}} \le \sum{a^3b^2 + a^2b^3} $$
$$\sum{a^2b + b^2a} = \sum{a^{\frac{11}{3}} b^{\frac{8}{3}} c^{\frac{5}{3}}} \le \sum{a^4b^4} $$
$$ab+ac+bc \ge 3$$
Adding them all together gives the desired inequality.

$$\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}$$
As $abc=1, $Multiplying $(abc)^2$ on the $LHS$, we get:
$$\frac{(bc)^2}{a(b+c)} + \frac{(ca)^2}{b(c+a)}+\frac{(ab^2)}{c(a+b)} \geq \frac{(ab+bc+ca)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(ab+bc+ca)}{2}$$ [By Titu's Lemma]

Again, $\frac{ab+bc+ca}{2} \geq \frac{3 \sqrt[3]{a^2b^2c^2}}{2} = \frac{3}{2}$ by AM-GM. Hence, we're done!

[quote=anduran]a variation
If $a+b+c=3,$ then
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$$[/quote]

Write$b+c = 3-a$ etc, and use Jensen on $f(x) = \frac{1}{x^3(3-x)}$.


hi.could you put the answers of USA AMC 8 2009 in the site?
thanks...

Clicking on the blue-boxed question numbers [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=42&year=2009]here[/url] will satisfy your inquiry.

For, the solutions, yeah. Just in case answers is what you want: http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2009-8a/2009-amc8answers.shtml


function
algebra
difference of squares
special factorizations

For each positive integer m>1, let P(m) denote the greatest prime factor of m.  For how many positive integers n is it true that both P(n)=$ \sqrt[2]{n}$ and P(n+48)=$ \sqrt[2]{n\plus{}48}$?

using difference of squares a^2-b^2=48
(a+b)(a-b)=48
So we have solutions (a,b)=(7,1) (fail)
(a,b)=(8,4) (fail)
(a,b)=(9.5,6.5) (fail)
(a,b)=(13, 11) (pass)
(a,b)=(24.5,23.5) (fail)
So the only solution with both prime integers is (a,b)=(13,11) and n=121. the answer is 1 solution

well what you have to do is find a number that is a square and then test it

so it will have to be a square $ 1\minus{}13$ because the difference the the difference of n and n+48 is even so it will be around 1/4 of 48 as the highest amount

now lets test the squares $ 1^2,2^2,3^2...$

so $ 1\plus{}48\equal{}49$ and that works because $ 49\equal{}7^2$

so 1 works

now since we know the next square is $ 8^2$we know 2 and 3 will not work

but 4 will because $ 64\minus{}48\equal{}4^2$

if you keep on doing it you get $ 3$

because the next one is $ 11^2$

so the answer is 3

[quote="calculatorwiz"]well what you have to do is find a number that is a square and then test it

so it will have to be a square $ 1 \minus{} 13$ because the difference the the difference of n and n+48 is even so it will be around 1/4 of 48 as the highest amount

now lets test the squares $ 1^2,2^2,3^2...$

so $ 1 \plus{} 48 \equal{} 49$ and that works because $ 49 \equal{} 7^2$

so 1 works

now since we know the next square is $ 8^2$we know 2 and 3 will not work

but 4 will because $ 64 \minus{} 48 \equal{} 4^2$

if you keep on doing it you get $ 3$

because the next one is $ 11^2$

so the answer is 3[/quote]1) "1/4 of 48" has no logical sense whatsover.
2) 1 does not work, 1 is not considered a prime, at least in AMC-sense.
3) 4 is not a prime! Consider re-reading how the function P(m) is defined.
4) Multiple posts of the same exact content are not allowed, but it looks like you were lagging.

Note to nohomo332: the two solutions (9.5,6.5) and (24.5,23.5) were not necessary if you were aware of the fact that the factors $ (a\minus{}b),(a\plus{}b)$ should both be even when the difference of squares are even for $ a,b\in\mathbb{Z}$.

@calculatorwiz, next time don't rush to press the submit button three times; be patient and wait  :wink:

You do realize that was very mean saying that that person has no logical sense.

[quote="calculatorwiz"]1/4 of 48 does have logical sense if you are smart enough to understand it unlike...

the difference of square you wanta get two different squares.because 48 is an even numberit cant be the difference of two consecutive integer squares

so you to find numbers that are twoo apart and add up to 24

that would be eleven  and thirteen and clearly 1/4 of 48=12 is between 11,13

if you do not know how to count the numbers go 1,2,3,4,5,6,7,8,9,10,11,12,13,14

you do not have logical sense[/quote]n as 1 and 16 don't work because it says it has to be a prime factor, 1 and 4 aren't prime.

[quote="calculatorwiz"]1/4 of 48 does have logical sense [b]if you are smart enough[/b] to understand it unlike...

the difference of square you wanta get two different squares.because 48 is an even numberit cant be the difference of two consecutive integer squares

so you to find numbers that are twoo apart and add up to 24

that would be eleven  and thirteen and clearly 1/4 of 48=12 is between 11,13

[b]if you do not know how to count the numbers[/b] go 1,2,3,4,5,6,7,8,9,10,11,12,13,14

[b]you do not have logical sense[/b][/quote]Even if the "1/4 of 48" make some sense, your intuitive explanation is poor. The bolded words that I have highlighted in the quote are more than enough proof that you have just embarrassed yourself. BTW, you still got the wrong answer.

To show his "1/4 of 48" in a mathematical way, here is a more comprehensive way to look at it:
Consider $ a^2 \minus{} b^2 \equal{} 4k$. Taking the pair $ a \minus{} b \equal{} 2$, $ a \plus{} b \equal{} 2k$ yields $ a \equal{} k \plus{} 1,b \equal{} k \minus{} 1$.


calculus
integration
Gauss
function
trigonometry
complex analysis

find the mean value of $ x^2 \minus{} y^2 \plus{} 2y$ over the circle $ |z \minus{} 5 \plus{} 2i| \equal{} 3$
i have solved like
any point on the circle is $ z \minus{} 5 \plus{} 2i \equal{} 3(cos\theta \plus{} i sin\theta )$
so $ x \plus{} iy \minus{} 5 \plus{} 2i \equal{} 3(cos\theta \plus{} i sin\theta)$
$ x \minus{} 5 \equal{} 3cos\theta, y \plus{} 2 \equal{} 3sin\theta$
$ x \equal{} 5 \plus{} 3cos\theta, y \equal{} \minus{} 2 \plus{} 3sin\theta$
$ dx \equal{} \minus{} 3sin\theta, dy \equal{} 3 cos\theta$
after putting in the integral $ \int (x^2 \minus{} y^2 \plus{} 2y)(dx \plus{} idy)$ we get 17 but ans is 5
if we have to use gauss mean value theorum then will we have to convert $ x^2\minus{}y^2\plus{}2y$ into complex form if yes then how?
help me

please help me . i am in trouble.
thankyou.

I don't see how you're getting 5.  This is how I'd approach it:

It's a little awkward the way you wrote it when you say "over the circle $ |z \minus{} (5 \minus{} 2i)| \equal{} 3$ which implies the mean of a complex-valued function over this contour in the complex plane.  The function $ f(x,y) \equal{} x^2 \minus{} y^2 \plus{} 2y$ is harmonic but the complex-valued function $ f(z) \equal{} f(x,y) \plus{} 0 i$ is not analytic so Gauss' Mean Value Theorem does not apply in the complex sense.  However,since $ f(x,y)$ is harmonic, it has the (real) mean-value property $ f(x_0,y_0) \equal{} \frac {1}{2\pi R}\int_C f(z)ds \equal{} \frac {1}{2\pi}\int_0^{2\pi}f(R,\theta)d\theta$ where $ z(s) \equal{} z(s(\theta)) \equal{} (5 \plus{} 3\cos(\theta), \minus{} 2 \plus{} 3\sin(\theta))$ and $ f(5, \minus{} 2) \equal{} 17$.


Who can give me generalization of this problem?

[hide]I got the solution from farenhajt :)  in one of the marathons[/hide]
[hide="hint"]If p and q are positive integers such that
\[ 1 \plus{} \frac{1}{2} \minus{} \frac{2}{3} \plus{} \frac{1}{4} \plus{} \frac{1}{5} \minus{} \frac{2}{6} \plus{} \ldots \plus{} \frac{1}{478} \plus{} \frac{1}{479} \minus{} \frac{2}{480}
\equal{} 2 \cdot \sum^{159}_{k\equal{}0} \frac{641}{(161\plus{}k) \cdot (480\minus{}k)}.\]
The sum can be written as

$S=\left(1+{1\over 2}+{1\over 3}+\cdots+{1\over 480}\right)-3\left({1\over 3}+{1\over 6}+\cdots+{1\over 480}\right)$

$S=\left(1+{1\over 2}+{1\over 3}+\cdots+{1\over 480}\right)-\left(1+{1\over 2}+{1\over 3}+\cdots+{1\over 160}\right)$

$S={1\over 161}+{1\over 162}+\cdots+{1\over 480}$

Hence $S=\sum_{k=1}^{320}{1\over k+160}$. Pairing up k-th and (321-k)-th terms, we get ${1\over k+160}+{1\over 481-k}={641\over (k+160)(481-k)}$ and the conclusion follows.[/hide]


Hello Everyone,

Here is a online calculus textbook that I found...if you want, you can download it, free of charge:

http://faculty.etsu.edu/knisleyj/calculus/final.pdf

Enjoy :)

[quote="splintercell990"]Here is a online calculus textbook that I found...if you want, you can download it, free of charge:

http://faculty.etsu.edu/knisleyj/calculus/final.pdf
[/quote]

I would like to know what the experienced calculus learners here think of that publication. I found it in an earlier Web search, and wondered if what is new and improved about that book is sound and helpful.

my personal preference being that I like the theoretical stuff a lot better.  But this will probably work better than say Spivak for a physics major


hi i need help with my ti-86
if i want to find the zeros for x squared plus 1 = 0
how can my calculator solve this for me other than graphing.
does it have to do with the 2nd "solve" button?

please help me
thank you

There's probably a "SOLVER" feature somewhere in the math menu. At least that's how it is for 83s.

[quote="mathnerd314"]You wouldn't do that, you'd just solve it by hand.

The roots are imaginary, and I don't know if the TI-86 can find imaginary roots.[/quote]

Yes, it can.  2nd --> POLY.  It's the button labeled PRGM.  It leads you to a general polynomial solver (well, approximator to enough digits for general use :P ).


geometry
perimeter
ratio
area of a triangle
Heron\u0027s formula

hi
i am new to this forum

i wanted to know whether there was any relation between area of a triangle and its perimeter...

Maybe Heron's formula is something you want: If $a,b,c$ are the sides of a triangle $ABC$ and $a+b+c=2p$, then the area of $ABC$ satisfies
\[|ABC|^{2}=p(p-a)(p-b)(p-b).\]
In general, if you have a triangle with fixed perimeter, its area is not fixed.

A related issue:
If a triangle has perimeter, p, what is the maximum area?

[hide]
The triangle with maximum area will be equilateral.
The area of an equilateral with side length, s is
$s^{2}\frac{\sqrt{3}}{4}$

Let $s = \frac{p}{3}$ and we have the answer:
$A = \frac{p^{2}\sqrt{3}}{36}$
[/hide]

well
this is the question that i am trying to solve...

ABC has sides a, b, c. The triangle A'B'C' has sides a + b/2, b + c/2, c + a/2. Show that its area is at least (9/4) area ABC.

i am not sure how to approach this problem
the ratio of the perimeters of these two triangles is 3/2 ....

can someone help me solve this please...

Yep.  Heron's is what you want there.  That and some familiarity with inequalities.

Can you post the solution? I'm having difficulty with the inequality.


IMO Shortlist
game
combinatorics
invariant
Hi

A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a [i]final configuration[/i]. For each $n$, show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119189]IMO ShortList 2001, combinatorics problem 7, alternative[/url]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

1) First of all, we need to show that we do eventually have to stop. This is clear because each time we move a pebble goes to the right, and we can't go arbitrarily far to the right, because each pebble (except for those on the first column) has another pebble immediately to its left, so we would need infinitely many pebbles. This shows that there always is a final configuration.

2) Another thing we notice is that a final configuration consists of columns of decreasing length (from left to right).

This is because before and after each move, this property is clearly conserved. Moreover, the difference between a column and its right neighbour is $0$ or $1$, because otherwise we would be able to continue moving the pebbles.

3) In a final configuration we can't have three consecutive columns of equal lengths.

This is true because if we look at the last move performed on one of the three columns, then we see that no matter how we make that move, before we make it we need two consecutive columns s.t. the one on the left has fewer pebbles than the one on the right, and this is impossible. In fact, this shows that we never have three equal consecutive columns (not just in final configurations).

4) In a final configuration we can't have two different pairs of equal consecutive columns.

Consider two such pairs $c_1,c_2$ and $c_1',c_2'$, s.t. $c_1'$ is to the right of $c_2$, which don't have any other such pairs between them. Since this is a final config, the columns between $c_2$ and $c_1'$ inclusive form a decreasing arithmetic progression with common difference $-1$. We can now retrace the steps until we get two pairs of equal columns which are adjacent and which differ in length by $1$, and this is clearly impossible (consider, for example, $3,3,2,2$; no matter how we try to obtain this, we can't avoid either two consecutive solumns s.t. the one on the left is shorter, or three equal consecutive columns, and we have examined these situations before).

These conditions show that the final configuration is unique, because each $n$ has only one decomposition as a sum of different numbers except, maybe, for a pair, and this decomposition is obtained as follows: assume we have $T_{k-1}=\frac{(k-1)k}2<n\le \frac{k(k+1)}2=T_k$. Then we write $n$ as $1+2+\ldots+(k-1)+n-T_{k-1}$. These are also the lengths of the columns.

For any $n,$ let $F_n$ denote [b]a[/b] final configuration. Also, let $T_n$ denote the $n$-th triangular number.

Firstly, for any pair $(x,y)$ with $x$ on the left, we will always have $x \ge y.$ The proof is easy. Note that $x$ can decrease if and only if it gives one pebble to $y.$ Hence if $x<y,$ then we must have had the pair $(x+1,y-1).$ But here, $(x+1)-(y-1)=x-y+2<2,$ a contradiction.

[b]Key observation:[/b] $F_n$ has exactly one pair of consecutive elements equal except when $n$ is triangular, in which case there is no equal pair.
[i]Proof:[/i] Firstly, we show we cannot have the consecutive triple $(x,x,x).$ As argued above, the triple $(a,b)$ can only be obtained from $(a+1, b-1).$ Note that all three cannot turn equal in one move if they are all distinct. Hence either the first two or the last two must be equal beforehand. But then we get a contradiction in both the cases by the line above.

If $(x, x, \cdots y, y)$ occurs, with $(x, x), (y, y)$ being the closest (closest by the number of elements in between). Then we must have that all the elements between $(x, x), (y, y)$ are distinct with a distance $=1.$ So the terms that occur are $(x, x, x-1,x-2 \cdots y+1, y, y).$ Again, $(a, b)$ can only be obtained from $(a+1, b-1)$ so we can backtrack and find that a consecutive equal triple must exist between $(x,x),  (y,y)$ some moves before. But this case has already been ruled out. $\square$

Let $T_k \le n < T_{k+1}.$ We will show $k \le |F_n|<k+2,$ and $k=|F_n|$ if and only if $n=T_k.$
Assume $|F_n| \ge k+2,$ then $n=S(F_n) \ge 1+1+2+\cdots+k+1=T_{k+1}+1>n,$ a contradiction.
Similarly, if $|F_n| \le k,$ then $n=S(F_n) \le 1+2+\cdots+k-2+k-1+k=T_{k} \le n,$ which is only possible if $n=T_k.$

By these two observations, we find that no two elements of $F_n$ are at a distance $\ge 2,$ as well as at most one pair is at a distance $0.$ So the last element must be $1,$ and we know that the elements of $F_n$ are in decreasing order, hence the elements are fixed, i.e. from $k$ to $1$ in the same order. Also, we know the sum, so we can determine which element repeats twice. So $F$ is unqiue and has the form:

[list]
[*] If $n=T_k:$ $F_n=(k, k-1, \cdots 1).$

[*] If $n=T_{k}+l, l>0,$ then for any $F_n=(k,k-1, \cdots,l+1,l , l, l-1 \cdots, 1).$
[/list]

With Aatman Supkar, Alan Chen, Alexandru Girban, Anushka Aggarwal, Daniel Sheremeta, Derek Liu, Dhrubajyoti Ghosh, Eric Shen, Gene Yang, Jeff Lin, Paul Hamrick, Raymond Feng, Robin Son, Sean Li:

First, note that the process does eventually terminate (the configuration, when viewed as a tuple, decreases lexicographically). So we only have to show there is at most one possible final configuration.

The proof proceeds in three claims about showing what statements must hold at any step.

[b][color=red]Claim:[/color][/b]  	The sizes of piles (from left to right) are non-increasing.
[i]Proof.[/i]  	Obvious. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	It is impossible to achieve $(\dots,a,a,a,\dots)$ 	for any positive integer $a$.
[i]Proof.[/i]  	Clear by looking one step immediately before 	such a configuration is achieved. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	It is impossible to achieve 	$(\dots,a,a,a-1,a-2,\dots,b+1,b,b,\dots)$ 	for any integers $a > b > 0$.
[i]Proof.[/i]  	Again, look one step immediately before 	such a configuration is achieved, 	and proceed by induction on $a-b$. $\blacksquare$

Finally, note that in a final configuration, any two adjacent piles differ by at most $1$. For $2001$ pebbles, one may check that \[ T = (62,61,\dots,49,48,48,47,46,\dots,2,1) \] is the unique configuration with $2001$ pebbles satisfying all four of the conditions. In the general case, since there is only one decomposition of any integer $n$ in the form \[ n = (1+2+\dots+k)+\ell \qquad 	\text{where} \; 1 \le \ell \le k \] this gives the unique possible final configuration.

[b][color=red]Remark:[/color][/b]  	The second claim is really better interpreted 	as a special case of the third, when $a=b$.

[b][color=red]Remark:[/color][/b]  	For concreteness, the final 	configurations for $1 \le n \le 10$ are shown below. 	\begin{align*} 		1 &= \mathbf{\color{red} 1} \\ 		2 &= 1+\mathbf{\color{red} 1} \\ 		3 &= \mathbf{\color{red}2}+1 \\ 		4 &= 2+1+\mathbf{\color{red}1} \\ 		5 &= 2+\mathbf{\color{red}2}+1 \\ 		6 &= \mathbf{\color{red}3}+2+1 \\ 		7 &= 3+2+1+\mathbf{\color{red}1} \\ 		8 &= 3+2+\mathbf{\color{red}2}+1 \\ 		9 &= 3+\mathbf{\color{red}3}+2+1 \\ 		10 &= \mathbf{\color{red}4}+3+1+1. 	\end{align*}

Number the columns $1, \dots, k$ with $C_1, \dots C_k >0$ stones in them (we let $k$ and $C_1, \dots, C_k$ be dynamic as the procedure unfolds). Note that the given precondition for a move ensures that we have $C_1 \ge C_2 \ge \dots \ge C_k$ at all times. We write $i \to i+1$ to indicate moving a stone from column $i$ to column $i+1$.

[b]Lemma: [/b]Suppose $C_i = C_{i+1}$ at some point in time. Then the most recent move which affected $(C_i, C_{i+1})$ was $i \to i+1$.
[b]Proof: [/b]Use the fact that $C_1 \ge \dots \ge C_k$ at all times. $\Box$

[b]Corollary: [/b]We cannot have $C_i = C_{i+1} = C_{i+2}$ at any point in time.
[b]Proof: [/b]Assume the contrary and consider the most recent move that affected $C_{i+1}$. By the lemma we have a contradiction. $\Box$

We say the configuration of stones is [i]trianglish[/i] if it has the following characterization: $C_k=1$, and $C_{i+1}=  C_i - 1$ for all but at most one pair of [i]bad[/i] adjacent columns, which instead satisfy $C_i = C_{i+1}$. We say a configuration is [i]semi-trianglish[/i] if it is trianglish in all regards, except it has at least two bad pairs of columns (keep in mind that the corollary forbids three consecutive columns with the same number of stones, so the bad pairs of adjacent columns all have different numbers of stones).
[asy]
size(6 cm);
void token(pair P) { fill(circle(P, 0.2), black); }
void col(int x, int height){
for (int j=0; j<height; ++j) {token((x, j));}
}

col(1, 4);
col(2, 3);
col(3, 3);
col(4, 2);
col(5, 1);
label("Trianglish", (3, -1.5), fontsize(10pt));
col(8, 5);
col(9, 4);
col(10, 4);
col(11, 3);
col(12, 2);
col(13, 2);
col(14, 1);
label("Semi-trianglish", (11, -1.5), fontsize(10pt));
[/asy]Check with some easy computations that there is a unique trianglish configuration on $n$ stones. We claim that the final configuration of our process is the trianglish one.

[b]Lemma 2: [/b]At any point in time and any $m \le k$, the configuration of stones formed by the rightmost $m$ columns is not semi-trianglish.
[b]Proof: [/b]In any semi-trianglish configuration, let $(i, i+1)$ and $(j, j+1)$ be the rightmost two bad pairs of columns, where $i<j$. We will induct on $j-i$. For the base case, suppose we arrive at semi-trianglish configuration $\mathcal{B}$ with $j-i=2$. By Lemma 1, the most recent move acting on $(i, i+1)$ was $i \to i+1$; let us examine the configuration $\mathcal{A}$ just before that move. Since $i\to i+1$ is the only move to act on column $i+1$ between $\mathcal{A}$ and $\mathcal{B}$, the moves to the left of column $i+1$ do not "interfere" with the moves to the right of column $i+1$ (e.g. all of the stones to the right of column $i+1$ in $\mathcal{B}$ must have already been to the right of column $i+1$ in $\mathcal{A}$).

Consider an alternate process beginning at $\mathcal{A}$ in which we first skip the move $i \to i+1$ and instead perform all moves which act on columns to the right of column $i+1$, until that rightmost section matches its appearance in $\mathcal{B}$: trianglish with a bad pair $(j, j+1)$ just to the right of $i+1$.
[asy]
size(4cm);
void token(pair P) { fill(circle(P, 0.2), black); }
void col(int x, int height){
for (int j=0; j<height; ++j) {token((x, j));}
}
col(0, 5);
col(1, 3);
col(2, 3);
col(3, 3);
col(4, 2);
col(5, 1);

draw((1.5, -0.5)--(1.5, 5), dotted);
label("$i$", (0,-1), fontsize(8pt));

label("$j$", (2,-1), fontsize(8pt));

label("The altered procedure produces a contradiction.", (3, -2), fontsize(10pt));
[/asy]Since the original procedure achieves this without interference from the left, our altered procedure is perfectly valid; however, it contradicts the corollary, since we find a moment in time when consecutive columns satisfy $C_{i+1} = C_j = C_{j+1}$. This establishes the base case.

The inductive step proceeds identically: if we arrive at a semi-trianglish configuration $\mathcal{B}$ with $j-i > 2$, we consider the configuration $\mathcal{A}$ just before the move $i \to i+1$, and then we (validly) alter the sequence of moves to skip $i \to i+1$ and act so that the columns to the right of $i+1$ match their appearance in $\mathcal{B}$.
[asy]
size(4cm);
void token(pair P) { fill(circle(P, 0.2), black); }
void col(int x, int height){
for (int j=0; j<height; ++j) {token((x, j));}
}
col(0, 6);
col(1, 4);
col(2, 4);
col(3, 3);
col(4, 2);
col(5, 2);
col(6, 1);

draw((1.5, -0.5)--(1.5, 5), dotted);
label("$i$", (0,-1), fontsize(8pt));

label("$j$", (4,-1), fontsize(8pt));

label("The altered procedure produces a smaller semi-trianglish configuration.", (3, -2), fontsize(10pt));
[/asy]This yields a procedure and a point in time when the rightmost columns, starting at $i+1$, form a semi-trianglish configuration with smaller $j-i$. This is impossible, so the induction is complete. $\Box$.

To finish, consider a stable, final configuration. No more moves are possible, so number of stones in each pair of adjacent columns differs by at most $1$. Lemma 2 implies that there is at most one pair of bad columns with $C_i = C_{i+1}$, i.e. the configuration is trianglish, and we are done.

Interestingly, no one posted a solution along these lines.

Solution by [b][url=https://artofproblemsolving.com/community/user/347186]p_square[/url][/b]:

Denote by $n_i$ the number of pebbles in the $i$th column from the left. Make a graph with $n$ vertices $v_i$, where for each suitable $i$, the vertex $v_i$ has $n_{i+1} - n_{i}$ chips. Chip fire!

Very hard to explain.

Label the columns $1,2,\dots$ from left to right and let $a_i$ denote the number of pebbles in column $i$ at any given time. Rephrase the problem as follows.\\\\ We are given several columns containing a total of $n$ pebbles. An $\textit{operation}$ consists of moving a pebble from column $i$ to column $i-1,$ where column $i$ contains at most as many pebbles as column $i-1.$ Given that it is possible to apply some series of operations to reach the configuration with all pebbles in column $1,$ find all possible maximal initial configurations.\\\\ Say a column $i$ is "friendly" if $a_{i-1}=a_i\ne 0.$ \\\\ $\textbf{Claim: }$ Initially, we have $a_{i-1}-a_{i}\in\{0,1\}$ for all $i.$\\\\ $\textbf{Proof: }$ If $a_{i-1}-a_i<0$ for some $i$ initially, then column $i$ will always contain more pebbles than column $i+1,$ so the final configuration cannot be the one with all pebbles in column $1.$\\\\ On the other hand, if $a_{i-1}-a_{i}\ge 2$ for some $i$ initially, then there is a configuration that could have occurred before the initial configuration, contradicting its maximality.  $\blacksquare$\\\\ $\textbf{Corollary: }$ At $\textit{any}$ point in time, the number of pebbles is non-increasing from left to right.\\\\\\ $\textbf{Claim: }$ The initial configuration contains at most one friendly column.\\\\ $\textbf{Proof: }$ We will show the more general statement that there are no friendly columns $i$ and $i+d,$ for some positive integer $d,$ at $\textit{any}$ point in time. To prove this, we will induct on the minimum value of $d.$\\\\ $\textit{Base Case: }$ Assume, for the sake of contradiction, that there is a point in time when $a_{k-1}=a_{k}=a_{k+1}\ne 0$ for some $k.$ Since the final configuration contains no such columns, at least one operation affecting columns $k-1,k,$ or $k+2$ must have been performed. \\\\But it is easy to see that performing any such operation does not keep the number of pebbles non-increasing from left to right, so we are done.\\\\ $\textit{Inductive Step: }$ Assume the claim is true for $n=1,2,\dots,m-1.$ Suppose, for the sake of contradiction, that $a_{k}=a_{k-1}\ne 0$ and $a_{k+m}=a_{k+m-1}\ne 0$ for some $k$ at some point in time.\\\\ Then, by our first claim, we must have $$a_{k-1}=a_{k}=a_{k+1}+1=a_{k+2}+2=\dots=a_{k+m-1}+m-1=a_{k+m}+m-1.$$ Note that the final configuration does not contain pebbles in this pattern, so we must have performed at least one move affecting columns $k-1,k,\dots,k+m.$\\\\ But it is easy to see that the first operation we perform that affects columns $k-1,k,\dots,k+m$ must result in a configuration with two friendly columns at a smaller distance, which is impossible by the inductive hypothesis. $\blacksquare$\\\\ There is only once initial configuration satisfying the conditions above, so we are done.

There's a general strategy which applies to these sorts of problems. Replace everywhere the height of the \(i^{\text{th}}\) column with the height of the \(\left(i+1\right)^{\text{st}}\) column subtracted from it.

[b]Definition(s) 0 [Generic] :[/b] Define a [i]state[/i] to be an element \(\xi\) of the \(\mathbb{Z}\)-module \(\mathcal{S}:=\prod_{i\colon\omega}\mathbb{Z}\), a [i]finite state[/i] to be a \(\xi\) such that \[\sum_{i\colon\omega}\left|\xi\left(i\right)\right|<\infty,\] and a [i]real state[/i] to be a \(\xi\) such that \[\forall i\colon\omega,\ \xi\left(i\right)\geq 0.\] Equip \(\mathcal{S}\) with the left \(\mathcal{G}:=\bigoplus_{i\colon\omega}\mathbb{Z}\) action induced by the collection of maps \[\mathcal{G}\left(j\right)+\xi\left(i\right):=\begin{cases}\xi\left(i\right)-2\text{ if }i=j\\ \xi\left(i\right)+1\text{ if }i=j+1\\ \xi\left(i\right)\text{ otherwise }\end{cases}.\] Say that a state \(\xi\) is [i]terminal[/i] iff it is real and there exists no \(j\colon\omega\) such that \(\mathcal{G}\left(j\right)+\xi\) is real and [i]quasiterminal[/i] iff it is the difference of two terminal states, and define a [i]game[/i] \(J\) of [i]length[/i] \(n\colon\omega\) with [i]seed[/i] state \(\xi\) to consist of a map \(J\colon n\to\omega\) satisfying that for each \(n'\colon n\), \[\left(\sum_{k\colon n'}\mathcal{G}\left(J\left(k\right)\right)\right)+\xi\] is real. \(J\) is said to itself be [i]terminal[/i] iff \[\left(\sum_{k\colon n}\mathcal{G}\left(J\left(k\right)\right)\right)+\xi\] is terminal.

[b]Proposition 1 [Generic] :[/b] State \(\xi\) is terminal iff \[\forall i\colon\omega,\ 0\leq \xi\left(i\right)< 2\] and quasiterminal iff \[\forall i\colon\omega,\ -2<\xi\left(i\right)< 2.\]

[b]Proof:[/b] Clear from the definitions. \(\Box\)

[b]Definition 2 [Specific]:[/b] Given a finite state \(\xi\), define the [i]zeroth moment[/i] of \(\xi\) to be \[m\left(0,\xi\right):=\sum_{i\colon\omega}\xi\left(i\right).\]

[b]Proposition 3 [Specific] :[/b] For any finite state \(\xi\) and \(j\colon\omega\), \[m\left(0,\mathcal{G}\left(j\right)+\xi\right)=m\left(0,\xi\right)-1.\]

[b]Proof:[/b] Immediate from the definitions. \(\Box\)

[b]Proposition 4 [Specific] :[/b] For any real finite state \(\xi\), \[m\left(0,\xi\right)\geq 0.\]

[b]Proof:[/b] Immediate from the definitions. \(\Box\)

[b]Proposition 5 [Generic] :[/b] For every finite state \(\xi\), there exists a unique minimal \(\ell\left(\xi\right)\colon\omega\) such that every game with seed \(\xi\) is of length less than \(\ell\left(\xi\right)\). (And in particular, \(\ell\left(\xi\right)>0\) if and only if \(\xi\) is real.)

[b]Proof:[/b] Immediate from the definitions, [b]Proposition 3[/b], and [b]Proposition 4[/b]. \(\Box\)

[b]Proposition 6 [Generic] :[/b] The only element \(g\) of \(\mathcal{G}\) such that \(g+0\) is quasiterminal is \(0\).

[b]Proof:[/b] Clear from the definitions and [b]Proposition 1[/b]. \(\Box\)

[b]Proposition 7 [Generic] :[/b] Given real finite state \(\xi\) and terminal games \(J_{0}\) and \(J_{1}\) of lengths \(n_{0}\) and \(n_{1}\) respectively on \(\xi\), \[\left(\sum_{k\colon n_{0}}\mathcal{G}\left(J_{0}\left(k\right)\right)\right)+\xi=\left(\sum_{k\colon n_{1}}\mathcal{G}\left(J_{1}\left(k\right)\right)\right)+\xi.\]

[b]Proof:[/b] Apply [b]Proposition 6[/b] to \[g=\sum_{k\colon n_{0}}\mathcal{G}\left(J_{0}\left(k\right)\right)-\sum_{k\colon n_{1}}\mathcal{G}\left(J_{1}\left(k\right)\right).\ \Box\]

Porting through the evident analogies, we win. \(\blacksquare\)

[quote=jj_ca888][hide = "Solution"][b][color=#00f]Remark:[/color][/b] [b]Bulgarian Solitaire[/b] is very similar.

Move everything to the positive lattice points of the coordinate plane. Initially, let all $n$ pebbles are on the points $(1, 1) \to (1, n)$. Also, note that it is clear the column sizes remain nonincreasing from left to right. Now consider the following:

[b][color=#f00]Claim:[/color][/b] The sum of all $x$ and $y$ coordinates is nonincreasing. In fact, it is unchanged if the move consists of moving a pebble between two columns with size diffence exactly $2$, and is strictly decreasing otherwise.
[b][color=#f00]Proof:[/color][/b] Assume a move is done on columns $i$ and $i+1$ which go from $(i, 1) \to (i, k + \ell)$ and $(i+1, 1) \to (i+1, k)$, respectively, where $\ell \geq 2$. Then, the quantity changes by\[\delta = ((i + 1) + (k + 1)) - ((i) + (k + \ell)) = 2 - \ell\]which is indeed equal to $0$ if $\ell = 2$ and $< 0$ otherwise.

Note that we can keep making moves until the difference in size between any adjacent stacks is $\in \{0, 1\}$ by definition. At this point, we will have $n$ points oriented in the positive coordinate lattice for which the quantity\[\sum_{i : 1\to n} (x_i + y_i)\]for all $n$ pebbles is minimized. This is true because if we suppose not, and that if the quantity is not minimized, then there must be two columns with difference $\ell > 2$ and hence we can make another move and this is not the end state. If we write $n = (1 + 2 + \ldots + m) + m'$ where $m' \in [0, m - 1]$, then a graphical interpretation tells us that the sum is minimized if pebbles cover all points $x + y \leq m + 1$ that have positive integer coordinates and the remaining $m'$ are sprinkled along $x + y = m + 2$. However, they must be consecutively placed from the bottom up, so in fact we will have piles from left to right of sizes\[\{m, \ldots , m', m', \ldots, 1\}\]which will always be our end state.[/hide][/quote]

This is incorrect. Here is a [b]hopefully[/b] correct solution.

Clearly, from left to right, the size of the piles must be non-increasing, else we would have had to make a move on two adjacent piles with size difference $< 2$, clearly not possible. Since this process clearly ends (for example, consider the monovariant $\sum ia_i$ where piles are numbered $i: 1 \to \infty$ from the left and there are $a_i$ pebbles in pile $i$), we need to show that there is a unique final configuration (which exists).

Furthermore, in the end state, the difference between adjacent piles must $\in \{0, 1\}$ or else it is still possible to make a move. First, notice that it is impossible to arrive at three piles $i-1, i, i+1$ with $a_{i-1} = a_i = a_{i+1}$, else if we rewind one step, the piles are not in nonincreasing order. Furthermore, I claim that it is impossible for two different adjacent pairs of piles $i, i+1$ and $j, j+1$ to have $a_i = a_{i+1}$ and $a_j = a_{j+1}$. Suppose we have some two $i < j$ with $i - j > 2$ minimal such that $a_i = a_{i+1}$ and $a_j = a_{j+1}$. Then, there are no two adjacent piles between $i$ and $j$ that have the same size. Hence, if we let $a_i = a_{i+1} = x$ and $a_j = a_{j+1} = x - (j - i -  1)$ then\[\{a_i, a_{i+1}, \ldots , a_{j+1}\} = \{x, x, x-1,\ldots , y+1, y, y\}\]where everything between $x - 1 \to y + 1$ is consecutive. Backtracking far enough, we see that there must have been some $i < x, y, z < j$ with $a_x = z_y = a_z$, clearly impossible.

Hence, the final state must be piles decreasing by $1$ every adjacent pair with the exception of one adjacent pair. So, we must write\[n = (1 + \ldots m) + m'\]for some $m$ such that $m' \in [0, m - 1]$. Clealy each $n$ has a unique such representation, and since $m'$ is fixed, the pair of adjacent piles with the same size is also known, so the final configuration must be of the form\[\{m, m-1, \ldots , m', m', \ldots ,1\}\]from left to right, as desired. $\blacksquare$

Let $T_{k}$ be the kth triangular number, such that $T_{k}\leq n < T_{k+1}$. Then, I claim the final configuration is where we have columns of size $1$ to $k$ from right to left, then there is are two adjacent columns of size $n - T_{k}$ (unless $n = T_{k}$).

Observe that if column $i$ was to the left of column $j$, the number of pebbles in column $i$ is at least the number of pebbles in column $j$. It's not hard to see that this process must terminate then (we can only make finite number of moves from the first column due to that constraint, finite from the second column, etc.). Let a pair of adjacent columns be "sparkle" pairs. I claim that we can not have two sparkle pairs in the final configuration (which proves that we only have one such configuration, as described above).

Assume there existed two sparkle pairs, let them be in columns $i, i+1, j, j+1$ where $i + 1 < j$ such that no sparkle pair is between them. Then, since this is the final configuration, and there are no sparkle pairs between them, each column $i + 1 < k < j$ must have exactly one more pebble than the one to its right (or else we can make another move). Now, consider the last operation made on a column between $i-1$ and $j + 1$. It could not have been on $i-1$, since that means column $i+1$ would've had more pebbles than column $i$ before that operation. Similarly, it can not be on column $j + 1$. If it was on column $d$, then before that operation was made, column $d$ had the same number of pebbles as column $d-1$, so now we have two new sparkle pairs that are closer to each other. We can keep on considering the last moves until the two sparkle pairs pairs coincide - that is we have $3$ columns of the same size. Let these three columns be $k, k + 1, k+2$. If the last operation was on $k+1$ or $k+2$, before that operation it would've had more pebbles than it's left column. If it was on $k-1$ or $k$, then before that operation that column would've had less pebbles than it's right column. Both cases result in a contradiction, so we can conclude that it is impossible to have two pairs of sparkle columns. Thus, the configuration we described is the only possible final configuration.

Let $T=\tfrac{k(k+1)}{2}$ be the largest trianglular number less than or equal to $n$; we show the final configuration is the multiset $\{1,2, \dots , k, n-T \}$ arranged in nonincreasing order.

Note that in the final configuration, adjacent columns differ by $0$ or $1$. Then to solve the problem it clearly suffices to prove the following two claims:

[b][color=#0000ff]Claim:[/color][/b] We never have the sequence $(\dots, a, a, a, \dots)$ for $a>0$.

Proof: Assume otherwise and consider the first time when this happens; there's no possible previous move.

[b][color=#0000ff]Clima:[/color][/b] We never have the sequence $(\dots , a, a, a-1, a-2, \dots , b+1, b, b, \dots)$ for $a>b>0$.

Proof: Assume otherwise and consider the first time when this happens; it's easy to see that one move ago, there must also have been a subsequence of this form, contradicting that this is the first time.

Very hard to writeup, but easy for a C7??

[hide=sol]
The move clearly forces the columns to be non-decreasing. This forces the process to terminate rather quickly. Now the key is to consider how many columns can be equal in a final configuration. I claim that at most one pair of columns can be equal. Indeed, assume otherwise, that columns $a, a+1; b, b+1$ have the same number of stones, where $b$ is possible equal to $a+1$. Then consider the possible previous moves. One of the stones in column $b+1$ must have come from the previous columns, and since they differ by one this would make $a, a+1, a+2$ have the same number of stones. But moving a stone from $a+2$ to $a+1$ would make the column $a+1$ larger than $a$, which is not possible. Thus, there is at most one pair of equal columns and the rest are differing by one. Now in fact this is equivalent to a staircase with a triangular number of dots, plus possibly adding one extra dot per column starting from the zero column. For $2001$ stones we find that column $x$ has
$$\begin{cases}
63-x$ stones$ & x\leq 15 \\
64-x$ stones$ & 16\leq x\leq 63
\end{cases}$$
[/hide]

The unique configuration is $62,61,60\ldots 48,48,47,46\ldots 1$. We will prove the following two claims, which together imply the desired result.

Claim 1: there can be no three consecutive stacks of the same height at any point.
Proof: Suppose FTSOC that $(n,n,n)$ exists. must have had one of $(n-1,n,n),(n+1,n-1,n),(n,n+1,n-1),(n,n,n+1)$ beforehand, none of which are legal positions as trivially we can't have a lower stack behind a higher stack.

Claim 2: there is one pair of two consecutive stacks of the same height in the final position.
Proof: Call such pairs [i]pearly.[/i] Let $i,i,i-1\ldots j+2,j+1,j,j$ contain two consecutive pearly pairs. Then, the only possible final move in this section would be $(j+1,j-1)\rightarrow (j,j)$, from which we can ignore $j-1$ and similarly backtrack repeatedly to arrive at that $(i,i,i)$ must have happened.

We will show by induction that if $n=\frac{a(a+1)}2+k$ with $k\leq a$, then the final configuration will have piles of sizes $a$, $a-1$, $a-2$, $\ldots$, $k+1$, $k$, $k$, $k-1$, $\ldots$, $2$, $1$, where the pile of size $a$ is at the location of the original pile.

Base Case: $n=1$
Then, no moves are possible, so the final configuration will have one pile of size $1$.

Inductive Step:
Suppose that a total of $m=\frac{x(x+1)}2+y$ moves are done on the position of the original column. Then, we must have $m<n$ since it is impossible to remove a pebble from a pile with $1$ pebble. Any final configuration must be able to be reached from a pile with $n-m$ pebbles and a pile with $m$ pebbles. Therefore, by the inductive hypothesis, a final configuration must have pile sizes of the form $n-m$, $x$, $x-1$, $x-2$, $\ldots$, $y+1$, $y$, $y$, $y-1$, $\ldots$, $2$, $1$. However, the pile sizes must be nonincreasing and differ by at most $1$. Therefore, we must have $|n-m-x|\leq1$, so either $n-m=x$ or $n-m=x+1$. This means that $n=\frac{(x+1)(x+2)}2+y-1$ or $n=\frac{(x+1)(x+2)}2+y$. In the first case, either $k=y-1$ and $a=x+1$, which means that the piles have sizes $a-1$, $a-1$, $a-2$, $\ldots$, $k+2$, $k+1$, $k+1$, $k$, $\ldots$, $2$, $1$, or $y=0$, $x=a$, and $k=x$, which means that the piles have sizes $a$, $a$, $a-1$, $a-2$, $\ldots$, $2$, $1$. In the second case, we must have $a=x+1$ and $y=k$, so the piles have sizes $a$, $a-1$, $a-2$, $\ldots$, $k+1$, $k$, $k$, $k-1$, $\ldots$, $2$, $1$. Now, suppose that the final configuration has piles of sizes $a-1$, $a-1$, $a-2$, $\ldots$, $k+2$, $k+1$, $k+1$, $k$, $\ldots$, $2$, $1$. Every move, the difference between consecutive piles changes by increasing one difference by $1$, decreasing the next difference by $2$, and increasing the last difference by $1$. The differences in the final configuration are $-a+1$, $0$, $1$, $1$, $\ldots$, $1$, $0$, $1$, $\ldots$, $1$. If this is a possible configuration, then it is possible to add $2$ to a number and subtract $1$ from the adjacent numbers and eventually get the sequence $-n$, $n$, $0$, $0$, $\ldots$, $0$. The number $2$ must be added to the second $0$ at least once. After that, the number to the left becomes $0$, so then eventually the second and the third number must be $0$. However, this means that it is impossible to make the second number equal to $n$. Therefore, the final configuration must have piles of sizes $a$, $a-1$, $a-2$, $\ldots$, $k+1$, $k$, $k$, $k-1$, $\ldots$, $2$, $1$.

Therefore, the final configuration must have piles of sizes $a$, $a-1$, $a-2$, $\ldots$, $k+1$, $k$, $k$, $k-1$, $\ldots$, $2$, $1$. If $n=2001$, then since $2001=\frac{62\cdot63}2+48$, this means that the piles must have sizes $62$, $61$, $60$, $\ldots$, $49$, $48$, $48$, $47$, $\ldots$, $2$, $1$.

Represent a configuration as a sequence. Obviously, the terms in the sequence are nonstrictly decreasing, and in any final configuration the difference between consecutive terms is either $0$ or $1$. We will prove the two following claims by reversing the process, starting from the finishing position and moving stones to the left so long as we preserve the nonstrictly decreasing nature of the terms.\\

[color=#00f][b]Claim 1:[/b][/color] We cannot ever have three consecutive equal terms.\\
[color=#00f]Proof:[/color] Suppose the terms are $a,a,a$. We can't ever move onto the last $a$, nor move from any of the $a$'s, without forming a nondecreasing subsequence within these three terms. $\blacksquare$\\

[color=#00f][b]Claim 1:[/b][/color] We can't ever have terms $a,a,a-1,\ldots,b+1,b,b$ where $b>0$.\\
[color=#00f]Proof:[/color] Note that the degenerate case of $a=b$ is actually claim 1. Now, we use downwards induction: to go from such a configuration to the starting configuration (2001 pebbles in one column), we have to either move from one of the terms listed, since we cannot move to the last $b$ term. It's easy to check that the first time we move, the listed terms are preserved, and making any of the specified moves creates either $a,a,a-1,\ldots,c+1,c,c$ or $d,d,d-1,\ldots,b+1,b,b$ for $c>b$ or $a>d$. $\blacksquare$\\

These two claims, along with the nature of the final configuration, are enough to imply that the difference between any two consecutive terms is $1$, except for at most one pair of terms where the difference is $0$. Hence the final configuration looks something like $n,n-1,\ldots,m+1,m,m,m-1,\ldots,1$, where $0 \leq m \leq n$ (where $m=0$ corresponds to no pair of terms being equal), which has sum $\tfrac{n(n+1)}{2}+m, 0 \leq m \leq n$. Simple algebra yields that for $2001$, the corresponding unique configuration has $n=62,m=48$, yielding $62,61,\ldots,49,48,48,47,\ldots,1$. $\blacksquare$

Suppose two adjacent columns are $a$ and $b$, then if you can move a pebble from the $a$ column to the $b$ column then $a\ge b+2$, so after the move, $a-1\ge b+1$ which means that at each stage, the column heights are non-increasing.

Let $a_1,a_2,\dots ,a_n$ be the column heights. Note that we can consider "unmoves" as moves that put a pebble from a column to the column directly to its left while keeping the sequence of column heights non-increasing. Note that having $a_i=a_{i+1}=a_{i+2}$ is impossible because any "unmove" will bring make the sequences $a_i$s non-increasing.

Suppose we have $a_i=a_{i+1}>a_{i+2}>\dots a_{j}=a_{j+1}$. As before, any "unmove" will retain the fact that there are two instances of consecutive columns being same height, so we go back to the previous case. Thus, in the final configuration we must have either zero or one adjacent piles that are the same.

There is a unique configuration from these conditions: find the least number $k$ such that $\tfrac{k(k+1)}{2}$ and then make the sequence $k,k-1,\dots ,1$ then add one to some of them, suitably.a

I'm fairly sure that this solution is valid (I may be wrong); just need someone to check that the main claim outlined is actually necessary for the proof.

Replace $2001$ with $s$ and note that the process must terminate eventually. It is important to note that we can interchange any operations in the sequence as long as the resulting sequence remains valid.
[b]Claim:[/b] We can interchange operations in any such sequence so that the leftmost available operation is always the next to be moved.
[b]Proof:[/b] Consider the state of the pebbles at any point $t_1$ in time. Denote the position of the leftmost available operation by $p$. Evidently, without making any operations at position $p$, the process has to terminate eventually; therefore, at some point $t_2$, operation $p$ must be performed. If we consider the sequence of moves from $t_1$ to $t_2$ (ending with $p$), then if we simply move $p$ to the start of this sequence, we find that we have created an entirely valid new sequence. Now repeating this algorithm for all such points in time proves the claim. $\blacksquare$

With this claim, we can assume that one final configuration must exist. Now we can induct to show that if $s=1+2+\dots+n+r$ with $0\le r\le n$, the final positions of the piles must be
\[n,n-1,\dots,r,r,r-1,\dots,1,\]
with this definition being trivially altered on the edge cases. The procedure is simple: clearly the base case is trivial, and otherwise we can WLOG that one pebble stays at the first column, and note that after all other pebbles are moved to completion, the lone pebble travels down to the first value equal to $r$ and increments that value. The answer for $s=2001$ is then
\[62,61,\dots,48,48,47,\dots,1\]
and we are done. $\blacksquare$

We write the piles as a sequence of their lengths. Define a \textbf{reverse-operation} as reversing the operation.

[b][color=#00f]Claim 1:[/color][/b]
The sequence must be non-increasing.
[i]Proof[/i]
    This is obvious due to the rule that an operation can't occur if there aren't at least two more pebbles to the right of it.

This proves that there will eventually be no operations possible. Now, we prove that the final configuration is unique.

[color=#00f][b]Claim 2[/b][/color]
The final configuration can only differ by $0$ or $1$ for every consecutive pile.
[i]Proof[/i]
    Or it won't be final.

[b][color=#00f]Claim 3[/color][/b]
The final configuration cannot have at least $3$ of the same numbers in a row.
[i]Proof[/i]
    For the sake of contradiction, we assume that it is possible to have at least $3$ of the same numbers in a row. Then, we must have
    \[\cdots, a, a, a, \cdots\]
    There's no possible way to achieve this because any reverse-operation we apply to this creates an increasing sequence, contradiction.

[b][color=#00f]Claim 4[/color][/b]
The final configuration cannot have more than $1$ pair of consecutive equal integers. For example, this configuration is impossible
\[\cdots, n+k, n+k, n+k-1, n+k-2, \cdots, n+1, n, n, \cdots\]
[i]Proof[/i]
    For the sake of contradiction, we assume that it is possible. We wish to perform reverse-operations until the configuration becomes one pile. We must perform a reverse-operation on the pairs, but eventually they will become next to each other, like so
    \[\cdots, n+k+1, n+k, n+k, n+k-1, n+k-1, n+k-2, \cdots \]
    We can't perform any more reverse operations without creating a row of $3$ or an increasing sequence, contradiction.

Note that these claims create a unique final sequence. To find this unique configuration, let $k = \lfloor\frac{-1+\sqrt{1+8n}}{2}\rfloor$. $k$ is the greatest integer such that $n> \frac{(k)(k+1)}{2}$. Then, we form the sequence, $k, k-1, k-2, \cdots, 2, 1$ and add on another $n-\frac{(k)(k+1)}{2}$ to the sequence.

For $n = 2001$, we have.
\[\boxed{62, 61, \cdots, 49, 48, 48, 47, \cdots, 1}\]

First, we will show that the process must terminate as long as its initial position is finite. We do this by strong induction. The base case is trivial. Now for the inductive step, suppose we have $n$ pebbles. Notice that there must be at least one pebble that never leaves the first pile, since if the first pile reaches a height of $1$ then its pebble cannot be moved. Now consider the configuration after all pebbles that leave the first pile have left. There must be at most $n-1$ pebbles in the piles starting from the second pile, so starting from this point, the process terminates. Therefore the whole process terminates, completing the induction.

Now, let the difference between the heights of pile $i$ and pile $i+1$ be $a_i.$ Notice that for any $i,$ we have that pile $i+1$ cannot be taller than pile $i.$ Any operation is now equivalent to subtracting $2$ to $a_i$ and adding $1$ to $a_{i-1}$ and $a_{i+1}$ if they exist. In our initial position, we have $a_1=n$ and $a_i=0$ for $i>1.$ Clearly in our final position, we must have $a_i \in \{0,1\}$ for all $i.$ We will show that there does not exist $i,j$ with $i<j$ such that in our final position we have $a_i=a_j=0$ and $a_{j+k}=1$ for some $k \ge 1.$ Suppose we do. We may assume without loss of generality that there is no $k$ with $i<k<j$ and $a_k=0.$ Now consider the last step that changed an $a_k$ with $i \le k \le j.$ If this changed $a_i$ then we know that it was an operation on pile $i$ since operating on pile $i+1$ or $i-1$ increases $a_i.$ Thus, we know that prior to this operation, we had $a_{i+1}=0,$ since in the final step we have $a_{i+1}=1.$ Similarly, we find that if the last step changed $a_j$ we would have had $a_{j-1}=0$ prior to it. Finally, notice that if this final step operated on some pile $k$ with $i+1<k<j-1$ then we would have had $a_{k-1}=a_{k+1}=0$ prior.

Notice that each of these steps decreases the distance between a pair of nearest zeros, since $j-(i+1)=(j-1)-i<j-i$ and $(k-1)-i<j-i.$ Additionally, we can notice that the $a_x$ between each of these pairs are all still $1,$ and we can also notice that if at any point during this working backwards we reach a pair $x,y$ with $x<y$ and $a_x,a_y$ a pair of nearest zeros, then we have $a_{y+1}$ is nonzero. Therefore, working backwards from here, we eventually find a $k$ such that $a_k=a_{k+1}=0,$ and $a_{k+2} \ne 0.$ This final observation must mean that pile $k+1$ has been operated before, and we clearly see that this implies that pile $k$ has also been operated before. However, consider the first operation changing one of either $a_k$ or $a_{k+1}.$ This cannot be an operation on piles $k-1$ or $k+1$ since this would have had to start with $a_k$ being negative. However, this also cannot be an operation on piles $k$ or $k+2,$ since these would have had to start with $a_{k+1}$ being negative. However, these are the only four piles that can affect $a_k$ or $a_{k+1},$ so this is a contradiction, and accordingly we find that in the final state such a pair $i,j$ does not exist.

Next, we easily see that we have $n=a_1+2a_2+3a_3+\dots+ia_i$ so let $a_k$ be the index that is zero in the final state with $a_{k+1}=1$ if it exists. Then we have $n+k=\frac{i(i+1)}{2},$ with $1 \le k < i.$ From here we can easily see that given any positive integer $n$ there is a unique choice of $k$ and $i$ unless $n$ is a triangular number, in which case we find $a_k$ does not exist and $\frac{i(i+1)}{2}=n.$ Therefore, we can define $f(n)$ as the smallest triangular number greater than or equal to $n,$ and we may let it be the $g(n)$th triangular number. Then, converting this back into pile heights we find that the $i$th pile will end up having height $g(n)-i$ if $i \le f(n)-n$ and $g(n)-i+1$ otherwise.

In particular we find that for $n=2001$ we get the configuration $62,61,\dots,49,48,48,47,\dots,2,1.$

\textbf{Claim: } Let use have $m$ total pebbles. All final configurations are of the form \[(n, n-1, n-2, \ldots n-x+2, n-x+1, n-x +1, n-x, \ldots 3, 2, 1)\] where $n$ is the smallest integer such that $\binom{n+1}{2} \geq m$ and $x = \binom{n+1}{2} - m$. Notice that if $x = $ then we never have two numbers the same.

Our base case is obvious. Now for the inductive step. We now have that our pile must go to the piles \[(n+1, n-1, n-2, \ldots n-x+2, n-x+1, n-x +1, n-x, \ldots 3, 2, 1)\] Notice that we can (and must) operate an all piles until we get to the point where we have the two consecutive $n-x+1$. Thus we get \[n, n-1, n-2, \ldots n-x+3, n-x+2, n-x+2, n-x+1, \ldots 3, 2, 1\] and our induction is complete. $\blacksquare$

Can someone please verify this solution :maybe: . This problem felt very very rigid in that sense.

Let $m_i$ denote the move on the $i$th pile.

Note that if after swapping $m_i$ and $m_{i+1}$ in the sequence of the moves, the final sequence of moves can be still executed fully, then swapping them does not change the final configuration after all the moves. This is kind of obvious.

We use induction to show that the final configuration is unique. Base case for $1$ is clearly true. Now we assume the inductive hypothesis for $k$ stones and then prove for $k+1$ stones.

If there is any move that uses the extra stone added to the first pile, we push the move forward in the sequence. If the move gets restricted by some other move, we push both of them forward and keep doing so until we cannot anymore. Now note that all the moves that are currently at the end of the sequences that we pushed forward are the moves related to that extra stones that we add. But then the set of moves preceding the ones we pushed forward are the ones that are used in the case with $k$ stones. So this is more like performing all the moves on the case with $k$ stones and then after we reach final configuration, then we add the extra stone into the first pile. After adding it, note that our set of moves is totally fixed which ensures that there is only one final configuration until we can stop.

-----
$\textbf{CLAIM: }$ If $n = \dfrac{t(t+1)}{2}$, then the configuration is like $\left( t, t-1, \ldots , 1, 0 \right)$. And for any $\dfrac{t(t+1)}{2} < n < \dfrac{(t+1)(t+2)}{2}$, then we get the configuration by adding $1$ to each of the $n - \dfrac{t(t+1)}{2}$ from the end.

[b]Proof:[/b] We first proceed by using induction on $t$. This is clearly true for $t=1$. So now we assume the base case for some $t=k$ and we then prove it for $t=k+1$. Thus from the inductive hypothesis, we know that for $n=\dfrac{(t+1)(t+2)}{2}-1$, our final configuration is somewhat like $(t, t, t-1, \ldots , 2,1)$. Now if we add $1$ to the first pile, then the final configuration just simply becomes $(t+1,t, \ldots , 1, 0)$.

Now we use induction inside inside induction lmao. We now induct on $n$. We already proved the base case for $n = \dfrac{(k+1)(k+2)}{2}$. Now assume the hypothesis for some $n=\dfrac{(k+1)(k+2)}{2} + l$ where $0\le l\le k$. So for our base case, the sequence is something like $(t+1 , t, \ldots, l,l,l-1,\ldots, 1)$. Now adding one stone to the first pile, note that we keep passing the stone on until we reach the first $l$. Then that $l$ becomes $l+1$ and thus our induction is complete. Thus our claim is proved. :yoda:
-----

Now note that $\dfrac{62\times 63}{2} < 2001 < \dfrac{63\times 64}{2}$. Now $2001 - \dfrac{62 \times 63}{2} = 48$. Note that the configuration for $n=\dfrac{62\times 63}{2}$ is just $(62,61,\ldots, 1,0)$. Now by our lemma, we get the configuration for $2001$ is $(62,61,\ldots, 48,48,47,\ldots, 1)$.

Nice problem!
Let the number of pebbles in $i^{th}$ stack be $a_i$
Call a subsequence $(s_i,s_{i+1})$ weird if $s_i=s_{i+1}$.
We claim that there are at most $2$ weird subsequences in the final configuration
FTSOC assume that there exist two $k,l$ such that $(s_k,s_{k+1});(s_l,s_{l+1})$ are weird, then there must exist a [b]first[/b] configuration $\mathbb{K}$ in which there are two wierd subsequences. Notice the move made before $\mathbb{K}$ must be played in $k^{th}$ or $m^{th}$ place both of which imply the configuration before it must also have $2$ weird subsequences, which contradicts the minimality of $\mathbb{K}$
Now this determines the final configuration, because for any non-weird subsequence $s_{i+1}=s_i-1$.

We can easily prove the following observations by induction on the number of moves that happened to reach a configuration(the inductive hypothesis should be all of these at the same time).
\begin{itemize}
    \item If $a<b$, then $a$ $b$ is not a substring of the configuration.
    \item $a$ $a$ $a$ is not a substring of the configuration.
    \item For any $k\in\mathbb N$, $a+k$ $a+k$ $a+k-1$ $a+k-2$ $\dots$ $a+1$ $a$ $a$ is not a substring of the configuration.
\end{itemize}

Since the ending configuration also cannot have $a$ $b$ where $a\ge b+2$, the configuration must be just $k$ $k-1$ $k-2$ $\dots$ $1$ along with at most one repeat for some $k$.  Finding such a configuration with a given sum is indeed unique. Take the largest triangular number that is at most $n$, then make the repeat as necessary.

Now we will prove that this process indeed terminates. But since the numbers must be nonincreasing, the $n+1$-th position(1-indexed) will never have any pebbles, so only finitely many moves may be performed. $\blacksquare$


Έστω $ M,$ σταθερό σημείο στο εσωτερικό δοσμένου ευθύγραμμου τμήματος $ AB.$ Προσδιορίστε σημείο $ N$ μεταξύ των $ M$ και $ B,$ ώστε να είναι $ (AM)^{2} \plus{} (BN)^{2} \equal{} (MN)^{2}.$

Κώστας Βήττας.

Κώστα αν δεν έχει γίνει κανένα λάθος πρέπει να είναι κάπως έτσι ...

Θεωρώντας ΑΒ = λ σταθερό και  ΑΜ = α, επίσης σταθερό ( λόγω υπόθεσης )

Βγαίνει ότι ΑΝ = $ \frac{\lambda^2}{2(\lambda\minus{}a)}$

 :)

[quote="vangelismak"]Κώστα αν δεν έχει γίνει κανένα λάθος πρέπει να είναι κάπως έτσι ...

Θεωρώντας ΑΒ = λ σταθερό και  ΑΜ = α, επίσης σταθερό ( λόγω υπόθεσης )

Βγαίνει ότι ΑΝ = $ \frac {\lambda^2}{2(\lambda \minus{} a)}$ $ ,(1)$  :)[/quote]
Βαγγέλη σ’ ευχαριστώ και δεν νομίζω ότι υπάρχει κάποιο λάθος στη σχέση $ (1)$ που ανέφερες.

Εγώ βέβαια ξέχασα να αναφέρω έναν πρόσθετο περιορισμό που υπάρχει στην εκφώνηση. Το να γίνει ο προσδιορισμός του σημείου $ N,$ με γραφικό τρόπο αν είναι εφικτό, χωρίς υπολογισμό της απόστασής του από κάποιο σταθερό σημείο.

Επειδή καθυστέρησα να απαντήσω, βάζω στο ίδιο σχήμα τη λύση όπως προκύπτει από τη σχέση $ (1)$ και τη γραφική λύση, χωρίς τον υπολογισμό του $ AN.$

Στο ορθογώνιο τρίγωνο $ ACD,$ με κάθετες πλευρές $ AC \equal{} 2(\lambda \minus{} \alpha)$ και $ AD \equal{} AB \equal{} \lambda,$ φέρνουμε δια της κορυφής του $ D,$ ευθεία κάθετη επί την υποτείνουσα $ CD,$ η οποία τέμνει την ευθεία $ AB,$ στο ζητούμενο σημείο $ N,$ όπως αποδεικνύεται εύκολα από τα όμοια ορθογώνια τρίγωνα $ \bigtriangleup ACD$ και $ \bigtriangleup AND.$

Στη γραφική λύση τώρα, ξεκινάμε από το ορθογώνιο ισοσκελές τρίγωνο $ \bigtriangleup OAB,$ με υποτείνουσα το δοσμένο ευθύγραμμο τμήμα $ AB$ και γράφουμε τον κύκλο $ (O),$ με κέντρο την κορυφή του $ O$ και ακτίνα $ OA \equal{} OB.$

Δια του $ A,$ φέρνουμε κάθετη ευθεία επί την ευθεία $ OM,$ η οποία τέμνει τον κύκλο $ (O),$ στο σημείο έστω $ P.$

Η κάθετη ευθεία από το $ O,$ επί την ευθεία $ PB,$ τέμνει επίσης την ευθεία $ AB$ στο ζητούμενο σημείο $ N,$ όπως αποδεικνύεται εύκολα από τις γωνιακές σχέσεις, που δίνουν ως ορθογώνιο, το τρίγωνο $ \bigtriangleup PMN,$ όπου $ \angle MPN \equal{} 90^{o}.$

Κώστας Βήττας.

Ναι Κώστα έπρεπε να το φανταστώ οτι ήθελε με γεωμετρικό τρόπο γιατί διαφορετικά είναι αρκετά απλή η άσκηση ...  :)

Καλησπέρα! Έχω δύο λύσεις. Αν και λάτρης των κύκλων, η δεύτερη μου άρεσε περισσότερο..

[hide="1η λύση - με κύκλους"]
$ \boxed{(AM)^{2} \plus{} (BN)^{2} \equal{} (MN)^{2}} \ \ (1)$

$ \Longrightarrow (MN)^{2} \minus{} (AM)^{2} \equal{} (BN)^{2}\ \ (2)$

Έστω ο κύκλος $ k \equal{} (M,MA)$

Από τη $ (2)$ προκύπτει $ D_k^N \equal{} BN^2 \ \ (3)$

Έστω ο κύκλος $ f \equal{} (N,NB)$ και $ B'$ αντιδιαμετρικό του $ B$

[img]12167[/img]

Η σχέση $ (3)$ μας λέει ότι οι κύκλοι $ k,f$ τέμνονται, και μάλιστα κάθετα

Άρα το $ B'$ είναι αντίστροφο του $ B$ ως προς τον κύκλο $ f$

Το $ B'$ κατασκευάζεται αν ξέρουμε τα $ A,M,B$ και το $ N$ είναι μέσον του $ BB'$
$ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$[/hide]
[hide="2η λύση - με ορθογώνιο τρίγωνο"]
Έστω ότι έχουμε κατασκευάσει το σημείο $ N$.
Από το $ M$ φέρουμε κάθετη ημιευθεία $ Mx$ στην $ AB$ και παίρνουμε τα διαδοχικά σημεία $ E,K\in Mx$ έτσι ώστε $ ME \equal{} BN$ και $ EK \equal{} MN$. Τότε είναι $ MB \equal{} MK$

Είναι $ (AM)^{2} \plus{} (ME)^{2} \equal{} (AE)^{2}$ και από την $ (1)$ παίρνουμε $ MN \equal{} AE$

Επομένως το τρίγωνο $ EAK$ είναι ισοσκελές

[img]12169[/img]
[i][b]Αντίστροφα[/b][/i]:
Ορίζω το σημείο $ K$ έτσι ώστε $ MK \perp MB$ και $ MK \equal{} MB$

Αν η μεσοκάθετος του $ AK$ τέμνει το [u]ευθύγραμμο τμήμα[/u] $ MK$ σε σημείο $ E$, τότε:

$ EK^2 \equal{} AM^2 \plus{} ME^2 \ \ (4)$

και επειδή $ ME \plus{} EK \equal{} MB$ μπορώ να βρώ $ N$ μεταξύ των $ M,B$ έτσι ώστε $ MN \equal{} EK$ και $ NB \equal{} ME$

και τελικά η $ (4)$ γράφεται $ MN^2 \equal{} AM^2 \plus{} NB^2,$ που είναι η ζητούμενη[/hide]

[color=olive]Λουκάς[/color]

[color=indigo][i][u]Παρατήρηση[/u]
Αν ΜΑ>ΜΒ τότε δεν έχουμε λύση, όπως φαίνεται και από το  $ \frac {\lambda^2}{2(\lambda \minus{} a)}$[/i][/color]

Λουκά σε ευχαριστώ για τις όμορφες λύσεις σου.

Αν δεν κάνω λάθος, από την πρώτη λύση, προκύπτει ότι το ζητούμενο σημείο $ N,$ είναι το μέσον του ευθύγραμμου τμήματος $ B'B,$ όπου $ B',$ είναι το αρμονικό συζυγές του $ B,$ ως προς τα $ A,$ $ C.$

Νομίζω ότι το πρόβλημα έχει λύση και όταν $ MA > MB,$ όπως φαίνεται στο σχήμα και αποδεικνύεται εύκολα.

Εύκολα επίσης αποδεικνύεται ότι $ \angle MON \equal{} 45^{o}$ και αξίζει εδώ να σημειωθεί ότι αν θεωρήσουμε την περιστροφή της γωνίας αυτής περί το σημείο $ O,$ τότε οι πλευρές της τέμνουν την ευθεία $ AB,$ σε δύο σημεία $ M$ και $ N,$ τέτοια ώστε $ (AM)^{2} \plus{} (BN)^{2} \equal{} (MN)^{2}.$

Αυτή η όμορφη ιδέα, δόθηκε σαν λύση από τον [b][size=100]Νίκο Δεργιαδέ[/size][/b], σε παρόμοιο πρόβλημα που εμφανίστηκε παλιότερα στο φόρουμ ΥΑΚΙΝΘΟΣ.

$ ($ Τελικά τίποτα δεν έρχεται από τον ουρανό. Όλα έχουν κάποια σχέση με κάτι που έχουμε ξαναδεί, έστω κι αν έχει ξεχαστεί $ ).$

Κώστας Βήττας.

[quote="pontios"][quote="vittasko"]Νομίζω ότι το πρόβλημα έχει λύση και όταν $ MA > MB,$ όπως φαίνεται στο σχήμα και αποδεικνύεται εύκολα.[/quote]
Ναι αλλά το $ N$ δεν είναι μεταξύ των $ M$ και $ B$[/quote]
Λουκά έχεις δίκιο. Δεν κατάλαβα ότι αναφέρθηκες στην εκφώνηση όπου ζητείται το $ N$ να είναι μεταξύ των $ M,$ $ B$ και νόμισα ότι μιλούσες γενικότερα. :blush:

Θέλοντας να απλοποιήσω την άσκηση, καθόρισα το που να βρίσκεται το $ N$ και δεν έγραψα για τον προσδιορισμό του με γραφική λύση, που έπρεπε να αναφερθεί. :(

Σε ευχαριστώ, Κώστας Βήττας.

Α εντάξει τότε. Όλα ok!  :)

Απλώς μπορούμε να αλλάξουμε την εκφώνηση.

Δεν το έχω ελέγξει σ αυτήν την περίπτωση, αλλά νομίζω ότι ακόμη και στο ορθογώνιο τρίγωνο δουλεύει το ίδιο αν δεχτούμε το $ N$ να κινείται και εκτός του $ MB$


Find all triples $(x,y,z)$ or real numbers for which:
$x+y+z=3$
$x^3+y^3+z^3=27$

Hi im new...

[hide]By setting z=3 and x=-y we get infinate solutions. However is there a more elegant way of doing it?[/hide]

Yes you are right, [b]seamusoboyle[/b], but we must show that all the solutions are $x=3, y=-z$.
[hide="hint"]Show that $yz=-3x$.[/hide]

Now you can even try to solve this one:
$x+y+z=t$
$x^{3}+y^{3}+z^{3}=t^{3}.$ :)

Yes Kunny, you gave solution like original was :)
Note that for positive x,y,z you can use inequalities...


Suppose $ x\plus{}\frac{1}{y}\equal{}y\plus{}\frac{1}{z}\equal{}z\plus{}\frac{1}{x}$

Find $ xyz$.

[hide="answer"]

We have
$ x\minus{}y\equal{}\frac{z\minus{}y}{yz}$

$ x\minus{}z\equal{}\frac{x\minus{}y}{xy}$

$ z\minus{}y\equal{}\frac{z\minus{}x}{xz}$

Multiply the three equations above
$ (x\minus{}y)(x\minus{}z)(z\minus{}y)\equal{}\frac{(z\minus{}y)(x\minus{}y)(z\minus{}x)}{(xyz)^2}$
$ xyz\equal{}\pm1$

[/hide]


How many perfect squares are divisors of the product $ 1!\cdot 2!\cdot 3!\cdots 9!$?

$ \textbf{(A)}\ 504 \qquad \textbf{(B)}\ 672 \qquad \textbf{(C)}\ 864 \qquad \textbf{(D)}\ 936 \qquad \textbf{(E)}\ 1008$

I think if you re-write it as $ (1)^{9}(2)^{8}(3)^{7}\cdots(9)^{1}$ it would help you see it better.

[hide="Solution"]Factoring the product $ 1! \cdot 2! \cdot 3! \cdots 9!$, we have $ 2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$.

Since a perfect square must have at least one pair of prime divisors, the number must be in the form:
$ 2^{2a}\cdot 3^{2b} \cdot 5^{2c} \cdot 7^{2d}$, where $ a, b, c, d$ are positive integers...
$ 2^{30}$ contributes $ 16$ perfect squares.
$ 3^{13}$ contributes $ 7$ perfect squares.
$ 5^5$ contributes $ 3$ perfect squares.
$ 7^3$ contributes $ 2$ perfect squares.

Multiplying, we have: $ 16 \cdot 7 \cdot 3 \cdot 2 \equal{} 672 \Rightarrow \boxed{\text{none of the answer choices}}$

You posted the wrong answer choices. Choice $ B$ should be $ 672$ instead of $ 682$.[/hide]

Hmm. Does anyone else think this is pretty easy for a #23 on the 12? Especially compared to this year's #23?

Yea, but I think this year's number 23 was easy too. (For a number 23)

Just factor and count the # of possibilities where all the exponents are even. The answer is 672.

If we just multiply 16*7*3*2, wont we overcount the square $1$ several times?

No, because $1=2^0*3^0*5^0*7^0$, so it is only satisfied if all of the exponents are 0.

[b]Solution:[/b] I think the key to doing it fast is realizing you can just write the prime factorization as $(2)^8 \cdot (3)^7 \cdot (4)^8 \cdot \ldots \cdot (8)^2 \cdot (9)^1$ and then simplify from there. The resulting factorization should be $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$. Since squares have even prime factors, our answer is $16 \cdot 7 \cdot 3 \cdot 2 = \boxed{672}$.