Referring to shalex proof of PDF file at

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fraction&t=30733

there is a fundamental lemma

For $ a,b,c$ positive reals such that $ \frac{2}{3}\leq a\plus{}b\plus{}c\leq 1$

   $ \sqrt{\frac{a}{1\minus{}a}}\plus{}\sqrt{\frac{b}{1\minus{}b}}\geq 2\sqrt{\frac{a\plus{}b}{2\minus{}a\minus{}b}}$

Have you a nice proof, Shalex?

Write it as
\[ \left(\sqrt{\frac {a}{1\minus{}a}}\minus{}\sqrt{\frac{a\plus{}b}{2\minus{}a\minus{}b}}\right)\plus{}\left(\sqrt{\frac {b}{1\minus{}b}}\minus{}\sqrt{\frac{a\plus{}b}{2\minus{}a\minus{}b}}\right)\geq0.\]
Taking the conjugates, ...

Another inequality of this type is to prove that if $ a,b\geq1$ then
\[ \sqrt{\frac{a}{b\plus{}1}}\plus{}\sqrt{\frac{b}{a\plus{}1}}\geq2\sqrt{\frac{a\plus{}b}{2\plus{}a\plus{}b}}.\]

[quote="pvthuan"]Write it as
\[ \left(\sqrt{\frac{a}{1\minus{}a}}\minus{}\sqrt{\frac{a\plus{}b}{2\minus{}a\minus{}b}}\right)\plus{}\left(\sqrt{\frac{b}{1\minus{}b}}\minus{}\sqrt{\frac{a\plus{}b}{2\minus{}a\minus{}b}}\right)\geq0.\]
Taking the conjugates, ...
[/quote]

Can you explain it better, please?

Thank you very much.

Has anyone understood pvthuan hint?  :blush:

[quote="manlio"]Referring to shalex proof of PDF file at

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fraction&t=30733

there is a fundamental lemma

For $ a,b,c$ positive reals such that $ \frac{2}{3}\leq a\plus{}b\plus{}c\leq 1$

   $ \sqrt{\frac{a}{1\minus{}a}}\plus{}\sqrt{\frac{b}{1\minus{}b}}\geq 2\sqrt{\frac{a\plus{}b}{2\minus{}a\minus{}b}}$
[/quote]
Maybe $ \frac{2}{3}\leq a\plus{}b\leq1$ $ ?$
If so, by Holder's inequality we obtain:
$ \left(\sqrt{\frac{a}{1\minus{}a}}\plus{}\sqrt{\frac{b}{1\minus{}b}}\right)^{2}(a^{2}(1\minus{}a)\plus{}b^{2}(1\minus{}b))\geq(a\plus{}b)^{3}.$
Id est, it remains to prove that $ \frac{(a\plus{}b)^{3}}{a^{2}\plus{}b^{2}\minus{}a^{3}\minus{}b^{3}}\geq\frac{4(a\plus{}b)}{2\minus{}a\minus{}b},$
which equivalent to the following inequality:$ (a\minus{}b)^{2}\left(a\plus{}b\minus{}\frac{2}{3}\right)\geq0.$ :wink:

Thank you very much arqady for your very simple proof. :)

Oh argedy you get it right!
It's also a nice proof  :wink:

I'm sorry for my absence these months for preparing the math competition at school.

[quote="pvthuan"]Another inequality of this type is to prove that if $ a,b\geq1$ then
\[ \sqrt {\frac {a}{b \plus{} 1}} \plus{} \sqrt {\frac {b}{a \plus{} 1}}\geq2\sqrt {\frac {a \plus{} b}{2 \plus{} a \plus{} b}}.
\]
[/quote]
By AM-GM we have:
$ \sqrt{\frac{b\plus{}1}{a}} \leq \frac{\frac{b\plus{}1}{a}\plus{}1}{2}\equal{}\frac{a\plus{}b\plus{}1}{2a}$
=>$ \sqrt {\frac {a}{b \plus{} 1}} \geq \frac{2a}{a\plus{}b\plus{}1}$
Similar we have $ \sqrt {\frac {b}{a \plus{} 1}} \geq \frac{2b}{a\plus{}b\plus{}1}$
Then we need prove that $ \frac{a\plus{}b}{a\plus{}b\plus{}1} \geq \sqrt{\frac{a\plus{}b}{a\plus{}b\plus{}2}}$
or $ (a\plus{}b)(a\plus{}b\plus{}2) \geq (a\plus{}b\plus{}1)^2$
<=>$ ab \geq 1$
This inequality is true because $ a;b \geq 1$
Done!

[quote="chien than"][quote="pvthuan"]Another inequality of this type is to prove that if $ a,b\geq1$ then
\[ \sqrt {\frac {a}{b \plus{} 1}} \plus{} \sqrt {\frac {b}{a \plus{} 1}}\geq2\sqrt {\frac {a \plus{} b}{2 \plus{} a \plus{} b}}.
\]
[/quote]
By AM-GM we have:
$ \sqrt {\frac {b \plus{} 1}{a}} \leq \frac {\frac {b \plus{} 1}{a} \plus{} 1}{2} \equal{} \frac {a \plus{} b \plus{} 1}{2a}$
=>$ \sqrt {\frac {a}{b \plus{} 1}} \geq \frac {2a}{a \plus{} b \plus{} 1}$
Similar we have $ \sqrt {\frac {b}{a \plus{} 1}} \geq \frac {2b}{a \plus{} b \plus{} 1}$
Then we need prove that $ \frac {a \plus{} b}{a \plus{} b \plus{} 1} \geq \sqrt {\frac {a \plus{} b}{a \plus{} b \plus{} 2}}$
or $ (a \plus{} b)(a \plus{} b \plus{} 2) \geq (a \plus{} b \plus{} 1)^2$
<=>$ ab \geq 1$
This inequality is true because $ a;b \geq 1$
Done![/quote]

I think it is wrong. :wink:


Let $ a_n \equal{} 1 \plus{} 2 \plus{} ... \plus{} n$ for every $ n \ge 1$; the numbers $ a_n$ are called triangular. Prove that if $ 2a_m \equal{} a_n$ then $ a_{2m \minus{} n}$ is a perfect square.

[quote="STARS"]Let $ a_n \equal{} 1 \plus{} 2 \plus{} ... \plus{} n$ for every $ n \ge 1$; the numbers an are called triangular. Prove that if $ 2a_m \equal{} a_n$ then $ a_{2m \minus{} n}$ is a perfect square.[/quote]

$ a_{n}\equal{}\frac{n(n\plus{}1)}{2}$-----> $ 2a_m \equal{} a_n$-----> $ m(m\plus{}1)\equal{}\frac{n(n\plus{}1)}{2}$---> $ 2m\minus{}n\equal{}n^{2}\minus{}2m^{2}$---->

$ a_{2m\minus{}n}\equal{}\frac{(2m\minus{}n)(2m\minus{}n\plus{}1)}{2}\equal{}\frac{(2m\minus{}n)^{2}\plus{}2m\minus{}n}{2}\equal{}\frac{(2m\minus{}n)^{2}\plus{}n^{2}\minus{}2m^{2}}{2}\equal{}\frac{2(m\minus{}n)^{2}}{2}\equal{}(m\minus{}n)^{2}$


Find all pairs $(m,n)$ of integers that satisfy the equation \[(m-n)^{2}=\frac{4mn}{m+n-1}.\]

[quote="Peter"]Find all pairs $ (m,n)$ of integers that satisfy the equation
\[ (m \minus{} n)^{2} \equal{} \frac {4mn}{m \plus{} n \minus{} 1}.
\]
[/quote]
[u]Case 1[/u] $ m\equal{}n$ then $ m\equal{}n\equal{}0$
$ m,n$ distint.
Suppose $ m>n$
$ (m\minus{}n)^2|4mn$
Let $ d\equal{}gcd(m,n)$ then
$ m\equal{}da,n\equal{}db$
$ \Rightarrow (a\minus{}b)^2|4ab$
But $ \gcd(a\minus{}b,ab)\equal{}1$ so $ a\minus{}b|2$
So $ a\minus{}b\equal{}1$ or $ a\minus{}b\equal{}2$
[b]1.[/b] $ a\minus{}b\equal{}1$ so $ a\equal{}b\plus{}1$
$ d^2(d(a\plus{}b)\minus{}1)\equal{}4d^2ab$
$ \Rightarrow d(a\plus{}b)\minus{}1\equal{}4ab$
$ d(2b\plus{}1)\minus{}1\equal{}4b(b\plus{}1)$
$ d\equal{}2b\plus{}1$
So $ m\equal{}(2b\plus{}1)(b\plus{}1),n\equal{}(2b\plus{}1)b$
[u]Case 2[/u] $ a\equal{}b\plus{}2$
$ 4d^2(d(a\plus{}b)\minus{}1)\equal{}4d^2ab$
$ d(2b\plus{}2)\minus{}1\equal{}(b\plus{}2)b$
$ \Leftrightarrow d\equal{}\frac{b\plus{}1}{2}$
So $ m\equal{}\frac{(b\plus{}2)(b\plus{}1)}{2},n\equal{}\frac{b(b\plus{}2)}{2}$

$ (m \minus{} n)^{2} \equal{} \frac {4mn}{m \plus{} n \minus{} 1}$

$ (m \minus{} n)^2 \plus{} 4mn \equal{} 4mn(\frac {1}{m \plus{} n \minus{}1}) \plus{} 4mn$

$ (m \plus{} n)^2 \equal{} 4mn (\frac {1}{m \plus{} n \minus{} 1} \plus{} 1)$

$ (m \plus{} n)^2 \equal{} 4mn (\frac {m \plus{} n}{m \plus{} n \minus{}1})$

$ (m \plus{} n) \equal{} \frac {4mn}{m \plus{} n \minus{} 1}$

$ (m \plus{} n) \equal{} (m \minus{} n)^2$

check it.


Find the original 3-digit number.  The hundreds digit is the sum of the tens and units digit.  The units digit is twice the tens digit.  The difference between the number and the new number with reversed digits is 297.

[hide] ans: [b]936[/b] This problem simply involves setting up eqations in the "100x + 10y + z" format.[/hide]


An unfair coin is tossed 10 times. The probability of heads on each toss is 0.4. Let a_n be the number of heads in the first n tosses. Find the probability that a_n/n ≤ 0.4 for n = 1, 2, ... , 9 and a_10/10 = 0.4.

[hide]

The maximum possible number of heads after x number of flips is:

0 0 1 1 2 2 2 3 3 4 (ex. after the fifth flip there can be at most 2 heads, since 2/5= 0.4). The first two flips must be tails and the last flip must be heads. We can divide the other 7 flips into categories:

1) Flips 3 and 4 (max 1 head)

2) Flips 5-7(max 2 heads)

3) Flips 8,9 (max 3 heads)

There are four ways heads can be distributed among these groups:

(1,1,1) (1 head among flips 3 and 4, 1 among 5-7, 1 among 8,9)

(1,0,2)

(0,1,2)

(0,2,1)

It's not too difficult to calculate these 4 individual probabilities, although it is somewhat time-consuming. There is probably a more elegant solution.

[/hide]

On a side note, a probability is not an integer and cannot be an answer to an AIME problem...

Most AIME problems have rational answers, and then they say ``m and n are relatively prime positive integers. Find m+n.''

Another way of doing this:  consider this a "walking across the grid" problem: draw a 4-by-6 grid.  You start at the lower left and are trying to get to the upper right.  You move according to the rule that a head moves you to the right and a tail moves you up.  You can figure out the limitations on your path based on the facts that Jennifer expressed (that is, you can't move to the right on the first 2 moves, and you have to move right on the last one, and so on.)  Then it's easy using standard methods to figure out the total number of such paths, and thus the total probability of getting any of tehm.

Jennifer, did you get [hide]23*3^6/5^10 = 16767/9765625[/hide]


inequalities
function
inequalities proposed

Let $a,b,c \ge 0$ such that $a+b+c=3$. Prove that
$\sum\frac{(3a-b-c)^{2}}{a^{2}+(b+c)^{2}}\ge \frac{3}{2}$

Notice that this inequality can be easily prove using Jensen's inequality, but there is a nicer way to prove.

For  function which you define in [0,3] you can prove that it is Convex
then use jensen inequality.
for notifying convexity or cncavity of this function you can use second
derivate test or Sketching the graph of the function easily the same problem of this idea are purposed in:Japan-97 or United state 2002 .


inequalities
logarithms
function
inequalities solved

It is a wrong inequality!

please be carefull when posting ...

Sorry about that.
it's :
x^(x 2 +2yz) * y^(y 2 +2zx) * z^(z 2 +2xy)

a quick look at it tells me that you should apply:

apply lg to it and then use Cebasev's inequality.

hope it works. got to go now, bye bye! :D :D

[quote="Lagrangia"]apply lg to it and then use Cebasev's inequality.[/quote]

I don't think Cebasev works for this one.

Taking $\ln$ on both sides, and move the RHS over to the LHS, combining and factorizing, it's equivalent to:

For $x,y,z \geq1$,
\[
(\ln{x})(x-y)(x-z) + (\ln{y})(y-z)(y-x) + (\ln{z})(z-x)(z-y) \geq 0
\]
This is the generalized Schur's inequality since $\ln{x}$ is an increasing function and it's non-negative when $x\geq1$.


function
limit
real analysis
real analysis unsolved

Show by means of an example that $ lim_{x \rightarrow a}[f(x)g(x)]$ may exist even though neither $ lim_{x \rightarrow a}f(x)$ nor $ lim_{x \rightarrow a}g(x)$ exists.

For example, let $ f(x) \equal{} \text{sgn } x \plus{} 2$ (see [url=http://en.wikipedia.org/wiki/Sign_function]sign function[/url]) and let $ g(x) \equal{} \frac{1}{f(x)}$.  Then $ \lim_{x \to 0} f(x)$ does not exist, nor does $ \lim_{x \to 0} g(x)$, but $ f(x) g(x) \equal{} 1$ everywhere.

In general, the product of two discontinuous functions may be continuous.

There should be a theorem that says: a combination of two good things is good, a combination of a good and a bad thing is bad, and a combination of two bad things can be either. But in such generality it would be a theorem in philosophy.  :lol:

Even then, there are exceptions. Multiplication by zero cures almost all.


[quote="luimichael"]Let $x, y, z$ be real numbers such that
$x^2+z^2=1$ and $y^2+2y(x+z)=6$.
Prove that $y(z-x)\leq 4$, and determine when does the equality hold.[/quote]
$y(z-x)\leq4\Leftrightarrow4xy-4yz\geq-16\Leftrightarrow y^2+2xy+2yz+4xy-4yz\geq-10\Leftrightarrow$
$\Leftrightarrow10x^2+y^2+10z^2+6xy-2yz\geq0\Leftrightarrow(3x+y-z)^2+(x+3z)^2\geq0.$
Equality holds when $x+3z=3x+y-z=x^2+z^2-1=y(z-x)-4=0,$
id est when $x=-\frac{3}{\sqrt{10}},$ $y=\sqrt{10},$ $z=\frac{1}{\sqrt{10}}$ or $x=\frac{3}{\sqrt{10}},$ $y=-\sqrt{10},$ $z=-\frac{1}{\sqrt{10}}$ :)

[quote="luimichael"]Aha!
Just can't imagine how you figure out the given inequality is equivalent to the last inequality which involves a sum of two squares. :)[/quote]
$y(z-x)\leq4\Leftrightarrow4xy-4yz\geq-16\Leftrightarrow6+4xy-4yz\geq-10\Leftrightarrow$
$\Leftrightarrow y^2+2xy+2yz+4xy-4yz\geq-10.$ And $-10=-10x^2-10z^2.$ OK?

Also in the Mathscope, you can find a generalization of this inequality. Two Vietnamese IMO gold medalists (Vu Ngoc Minh, Pham Gia Vinh Anh) gave that generalized form.


geometry
geometric transformation
reflection
geometry solved

On the sides of an acute-angled triangle ABC,similar isosceles triangles DBC,ECA,FAB are constructed externally,such that
 <DBC=<DCB=<EAC=<ECA=<FAB=<FBA=<BAC

Let M be the mid-point of BC,and let P and Q be the intersections of DE with AC and of DF with AB,respectively.
Prove that  MP:MQ=AB:AC

Let's just use the fact (which I'll prove later) that $P$ is the midpoint of $DE$.

Let $T$ be the reflection of $C$ in $P$. We have $DT\|CE\Rightarrow \angle DTC=\angle A$, which means that $TBDC$ is cyclic, so $\angle BTC=2A$. Since $MP\|BT$, we get $\angle MPC=2A$, so after we do the same thing for the other side, we get $\angle MPC=\angle MQB$, and the conclusion follows because if $X,Y$ are the projections of $M$ on $CA,AB$, then $\frac{MX}{MY}=\frac{AB}{AC}$, and we have similar right triangles $MXP,MYQ$.

Now to show that $P$ is the midpoint of $DE$ (I'm sure there are nicer ways). Keep everything fixed except for $B$, which we move on the fixed line $AB$. Then $D$ also moves on a fixed line, so the midpoint of $DE$ moves on a fixed line, so in order to determine that line we need only two positions of $B$. Take, for example, $B\to A$ and $B\to\infty$. We find the locus of the midpoint of $DE$ to be precisely $AC$.

Nice proof, Grobber!

A better way to show that the point P is the midpoint of the segment DE is as follows:

As the triangles DBC and ECA are similar, we have CE / CA = BD / BC. Since the triangle DBC is isosceles, we have BD = CD, and thus this becomes CE / CA = CD / BC = CD / CB.

Let S be the reflection of the point E in the line CA. Then, < SAC = < EAC, < SCA = < ECA and CS = CE. Since < SAC = < EAC and < EAC = < BAC, we have < SAC = < BAC, so that the point S lies on the line AB. Since CS = CE, we can rewrite CE / CA = CD / CB as CS / CA = CD / CB. Finally, from < SCA = < ECA and < DCB = < ECA, we get

  < DCS = < DCB + < BCA - < SCA = < ECA + < BCA - < ECA = < BCA.

Together with CS / CA = CD / CB, this implies that the triangles CSD and CAB are similar. Hence, < DSC = < BAC. But < BAC = < ECA; thus, < DSC = < ECA. Since < SCA = < ECA, we get < DSC = < SCA. Hence, DS || CA.

Since the point S is the reflection of the point E in the line CA, the midpoint U of the segment ES lies on the line CA. Then, we can rewrite DS || CA as DS || PU. Hence, the line PU passes through the midpoint U of the side ES of the triangle DES and is parallel to its side DS. Hence, this line is a midparallel in the triangle DES, and it must also pass through the midpoint of its side DE. In other words, the point P is the midpoint of the side DE of the triangle DES. Proof complete.

  Darij


geometry circumcircle calculus integration function inequalities probability

[quote="zhaoli"]I have translated the longlisted problems of IMO 1990 from a Chinese book. The SL problems are following the number listed in the book "The IMO Compendium".[/quote]

Great!

Interestingly, problem (LongList'1990/43) equals problem (IMO'1992/5).

zhaoli says: [i]"I have translated these problems from a Chinese book named "数学奥林匹克 ----- 第31届国际数学竞赛预选题"; when translated into English, the name of the book is "mathematical olympiad ----- the problems proposed to 31st IMO". The book was published by Peking University Press in July, 1991. ISBN7-01517-8/G87. The authors of the book were Hu Datong and Tao Xiaoyong."[/i]

Thank you Zhaoli for giving us the problems of ILL 1990 and answering my questions :)

I posted problems in forums and when I'm trying to list them here, it says :
[color=#FF0000]The submitted form was invalid. Try submitting again.
Your message contains 44636 characters. The maximum number of allowed characters is 40000.[/color]

So I'm posting them in two parts.

[size=150][b]IMO 1990 LongListed Problems: (Posted in Forums) - Part One[/b][/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022341#p2022341]1. (AUS 1)[/url] In triangle $ABC, O$ is the circumcenter, $H$ is the orthocenter. Construct the circumcircles of triangles $CHB, CHA$ and $AHB$, and let their centers be $A_1, B_1, C_1$, respectively. Prove that triangles $ABC$ and $A_1B_1C_1$ are congruent, and their nine-point circles coincide.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=216919#p216919]2. (AUS 2)[/url]Prove that $ \sum_{k = 0}^{995} \frac {( - 1)^k}{1991 - k} {1991 - k \choose k} = \frac {1}{1991}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=360451&#p360451]3. (AUS 3)[/url] [b](SL1990-1)[/b] The integer $ 9$ can be written as a sum of two consecutive integers: $ 9 = 4+5.$ Moreover, it can be written as a sum of (more than one) consecutive positive integers in exactly two ways: $ 9 = 4+5 = 2+3+4.$ Is there an integer that can be written as a sum of $ 1990$ consecutive integers and that can be written as a sum of (more than one) consecutive positive integers in exactly $ 1990$ ways?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225210&#p1225210]4. (CAN 1)[/url] [b](SL1990-2)[/b] Given $ n$ countries with three representatives each, $ m$ committees $ A(1),A(2), \ldots, A(m)$ are called a cycle if

[i](i)[/i] each committee has $ n$ members, one from each country;
[i](ii)[/i] no two committees have the same membership;
[i](iii)[/i] for $ i = 1, 2, \ldots,m$, committee $ A(i)$ and committee $ A(i + 1)$ have no member in common, where $ A(m + 1)$ denotes $ A(1);$
[i](iv)[/i] if $ 1 < |i - j| < m - 1,$ then committees $ A(i)$ and $ A(j)$ have at least one member in common.

Is it possible to have a cycle of 1990 committees with 11 countries?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022368#p2022368]5. (COL 1)[/url] Given the condition that there exist exactly $1990$ triangles $ABC$ with integral side-lengths satisfying the following conditions:
(i) $\angle ABC =\frac 12 \angle BAC;$
(ii) $AC = b.$
Find the minimal value of $b.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022374#p2022374]6. (COL 2)[/url] Let function $f : \mathbb Z_{\geq 0}^0 \to \mathbb N$ satisfy the following conditions:
(i) $ f(0, 0, 0) = 1;$
(ii)  $f(x, y, z) = f(x - 1, y, z) + f(x, y - 1, z) + f(x, y, z - 1);$
(iii) when applying above relation iteratively, if any of $x', y', z$' is negative, then $f(x', y', z') = 0.$
Prove that if $x, y, z$ are the side lengths of a triangle, then  $\frac{(f(x,y,z) ) ^k}{ f(mx my mz)}$ is not an integer for any integers $k, m > 1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022379#p2022379]7. (CUB 1)[/url] $A$ and $B$ are two points in the plane $\alpha$, and line $r$ passes through points $A, B$. There are $n$ distinct points $P_1, P_2, \ldots, P_n$ in one of the half-plane divided by line $r$. Prove that there are at least $\sqrt n$ distinct values among the distances $AP_1, AP_2, \ldots, AP_n, BP_1, BP_2, \ldots, BP_n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366461&#p366461]8. (CZS 1)[/url] [b](SL1990-3)[/b]. [i]IMO 2[/i] Let $ n \geq 3$ and consider a set $ E$ of $ 2n - 1$ distinct points on a circle. Suppose that exactly $ k$ of these points are to be colored black.  Such a coloring is [b]good[/b] if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $ n$ points from $ E$.  Find the smallest value of $ k$ so that every such coloring of $ k$ points of $ E$ is good.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=404543&#p404543]9. (CZS 2)[/url] [b](SL1990-4)[/b] Assume that the set of all positive integers is decomposed into $ r$ (disjoint) subsets $ A_1 \cup A_2 \cup \ldots \cup A_r = \mathbb{N}.$ Prove that one of them, say $ A_i,$ has the following property: There exists a positive $ m$ such that for any $ k$ one can find numbers $ a_1, a_2, \ldots, a_k$ in $ A_i$ with $ 0 < a_{j + 1} - a_j \leq m,$ $ (1 \leq j \leq k - 1)$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022385#p2022385]10. (CZS 3)[/url] Let $p, k$ and $x$ be positive integers such that $p \geq k$ and $x < [ \frac{p(p-k+1)}{2(k-1)} ]$, where $[q]$ is the largest integer no larger than $q$. Prove that when $x$ balls are put into $p$ boxes arbitrarily, there exist $k$ boxes with the same number of balls.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022387#p2022387]11. (CZS 4)[/url] In a group of mathematicians, every mathematician has some friends (the relation of friend is reciprocal). Prove that there exists a mathematician, such that the average of the number of friends of all his friends is no less than the average of the number of friends of all these mathematicians.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023323#p2023323]12. (CZS 5)[/url] For any permutation $p$ of set $\{1, 2, \ldots, n\}$, define $d(p) = |p(1) - 1| + |p(2) - 2| + \ldots + |p(n) - n|$. Denoted by $i(p)$ the number of integer pairs $(i, j)$ in permutation $p$ such that $1 \leqq < j \leq n$ and $p(i) > p(j)$. Find all the real numbers $c$, such that the inequality $i(p) \leq c \cdot d(p)$ holds for any positive integer $n$ and any permutation $p.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022389#p2022389]13. (FIN 1)[/url] Six cities $A, B, C, D, E$, and $F$ are located on the vertices of a regular hexagon in that order. $G$ is the center of the hexagon. The sides of the hexagon are the roads connecting these cities. Further more, there are roads connecting cities $B, C, E, F$ and $G$, respectively. Because of raining, one or more roads maybe destroyed. The probability of the road keeping undestroyed between two consecutive cities is $p$. Determine the probability of the road between cities $A$ and $D$ is undestroyed.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022392#p2022392]14. (FIN 2)[/url] We call a set $S$ on the real line $R$ "superinvariant", if for any stretching $A$ of the set $S$ by the transformation taking $x$ to $A(x) = x_0 + a(x - x_0)$, where $a > 0$, there exists a transformation $B, B(x) = x + b$, such that the images of $S$ under $A$ and $B$ agree; i.e., for any $x \in S$, there is $y \in S$ such that $A(x) = B(y)$, and for any $t \in S$, there is a $u  \in S$ such that $B(t) = A(u).$ Determine all superinvariant sets.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=246214&#p246214]15. (FRA 1)[/url] [b](SL1990-5)[/b] Given a triangle $ ABC$. Let $ G$, $ I$, $ H$ be the centroid, the incenter and the orthocenter of triangle $ ABC$, respectively. Prove that $ \angle GIH > 90^{\circ}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022396#p2022396]16. (FRA 2)[/url] We call an integer $k \geq 1$ having property $P$, if there exists at least one integer $m \geq 1$ which cannot be expressed in the form $m = \varepsilon_1 z_1^k  + \varepsilon_2 z_2^k + \cdots + \varepsilon_{2k} z_{2k}^k $ , where $z_i$  are nonnegative integer and $\varepsilon _i = 1$ or $-1$, $i = 1, 2, \ldots, 2k$. Prove that there are infinitely many integers $k$ having the property $P.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022399#p2022399]17. (FRA 3)[/url] 1990 mathematicians attend a meeting, every mathematician has at least 1327 friends (the relation of friend is reciprocal). Prove that there exist four mathematicians among them such that any two of them are friends.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022400#p2022400]18. (FRG 1)[/url] Find, with proof, the least positive integer $n$ having the following property: in the binary representation of $\frac 1n$, all the binary representations of $1, 2, \ldots, 1990$ (each consist of consecutive digits) are appeared after the decimal point.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366470&#p366470]19. (FRG 2)[/url] [b](SL1990-6)[/b] [i]IMO 5[/i] Given an initial integer $ n_0 > 1$, two players, $ {\mathcal A}$ and $ {\mathcal B}$, choose integers $ n_1$, $ n_2$, $ n_3$, $ \ldots$ alternately according to the following rules :

[b]I.)[/b] Knowing $ n_{2k}$, $ {\mathcal A}$ chooses any integer $ n_{2k + 1}$ such that
\[ n_{2k} \leq n_{2k + 1} \leq n_{2k}^2.
\]
[b]II.)[/b] Knowing $ n_{2k + 1}$, $ {\mathcal B}$ chooses any integer $ n_{2k + 2}$ such that
\[ \frac {n_{2k + 1}}{n_{2k + 2}}
\]
is a prime raised to a positive integer power.

Player $ {\mathcal A}$ wins the game by choosing the number 1990; player $ {\mathcal B}$ wins by choosing the number 1.  For which $ n_0$ does :

[b]a.)[/b] $ {\mathcal A}$ have a winning strategy?
[b]b.)[/b] $ {\mathcal B}$ have a winning strategy?
[b]c.)[/b] Neither player have a winning strategy?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022402#p2022402]20. (FRG 3)[/url] Could the three-dimensional space be expressed as the union of disjoint circumferences?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022404#p2022404]21. (GRE 1)[/url] Point $O$ is interior to triangle $ABC$. Through $O$, draw three lines $DE \parallel BC, FG \parallel CA$, and $HI \parallel AB$, where $D, G$ are on $AB$, $I, F$ are on $BC$ and $E, H$ are on $CA$. Denote by $S_1$ the area of hexagon $DGHEFI$, and $S_2$ the area of triangle $ABC$. Prove that $S_1 \geq \frac 23 S_2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225213&#p1225213]22. (GRE 2)[/url] [b](SL1990-7)[/b] Let $ f(0) = f(1) = 0$ and

\[ f(n+2) = 4^{n+2} \cdot  f(n+1) - 16^{n+1} \cdot f(n) + n \cdot 2^{n^2}, \quad n = 0, 1, 2, \ldots\]

Show that the numbers $ f(1989), f(1990), f(1991)$ are divisible by $ 13.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225217&#p1225217]23. (HUN 1)[/url] [b](SL1990-8)[/b] For a given positive integer $ k$ denote the square of the sum of its digits by $ f_1(k)$ and let $ f_{n+1}(k) = f_1(f_n(k)).$ Determine the value of $ f_{1991}(2^{1990}).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022409#p2022409]24. (HUN 2)[/url] Find the real number $t$, such that the following system of equations has a unique real solution $(x, y, z, v)$:

\[ \{\begin{array}{cc}x+y+z+v=0\\ (xy + yz +zv)+t(xz+xv+yv)=0\end{array} \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225224&#p1225224]25. (HUN 3)[/url] [b](SL1990-9)[/b] The incenter of the triangle $ ABC$ is $ K.$ The midpoint of $ AB$ is $ C_1$ and that of $ AC$ is $ B_1.$ The lines $ C_1K$ and $ AC$ meet at $ B_2,$ the lines $ B_1K$ and $ AB$ at $ C_2.$ If the areas of the triangles $ AB_2C_2$ and $ ABC$ are equal, what is the measure of angle $ \angle CAB?$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022413#p2022413]26. (ICE 1)[/url] Prove that there exist infinitely many positive integer $n$ such that the number $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers having this property.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=360420&#p360420]27. (ICE 2)[/url] [b](SL1990-10)[/b] A plane cuts a right circular cone of volume $ V$ into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the volume of the smaller part.

[i]Original formulation:[/i]

A plane cuts a right circular cone into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the ratio of the volume of the smaller part to the volume of the whole cone.

[size=150][b]IMO 1990 LongListed Problems: (Posted in Forums) - Part Two[/b][/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022416#p2022416]28. (IND 1)[/url] Let $ABC$ be an arbitrary acute triangle. Circle $\Gamma$ satisfies the following conditions:
(i) Circle $\Gamma$ intersects all three sides of triangle $ABC.$
(ii) In the convex hexagon formed by above six intersections, the three pairs of opposite sides are parallel respectively. (The hexagon maybe degenerate, that is, two or more vertices are coincide. In this case, "opposite sides are parallel" is defined through limit opinion.)
Find the locus of the center of circle $\Gamma$, and explain how to construct the locus.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022420#p2022420]29. (IND 2)[/url] Function $f(n), n \in \mathbb N$, is defined as follows:
Let $\frac{(2n)!}{n!(n+1000)!} = \frac{A(n)}{B(n)}$ , where $A(n), B(n)$ are coprime positive integers; if $B(n) = 1$, then $f(n) = 1$; if $B(n) \neq 1$, then $f(n)$ is the largest prime factor of $B(n)$. Prove that the values of $f(n)$ are finite, and find the maximum value of $f(n).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366460#p366460]30. (IND 3)[/url] [b](SL1990-11)[/b] [i]IMO 1[/i] Chords $ AB$ and $ CD$ of a circle intersect at a point $ E$ inside the circle.  Let $ M$ be an interior point of the segment $ EB$.  The tangent line at $ E$ to the circle through $ D$, $ E$, and $ M$ intersects the lines $ BC$ and $ AC$ at $ F$ and $ G$, respectively.  If
\[ \frac {AM}{AB} = t,
\]
find $\frac {EG}{EF}$ in terms of $ t$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022426#p2022426]31. (IND 4)[/url] Let $S = \{1, 2, ..., 1990\}$. A $31$-element subset of $S$ is called "good" if the sum of its elements is divisible by $5$. Find the number of good subsets of $S.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022436#p2022436]32. (IRN 1)[/url] Using following five figures, can a parallelepiped be constructed, whose side lengths are all integers larger than $1$ and has volume $1990$ ? (In the figure, every square represents a unit cube.)

[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28560&mode=view[/img][/list]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=393236#p393236]33. (IRN 2)[/url] Let $S$ be a set with $1990$ elements. $P$ is the set such that its elements are the ordered sequences of $100$ elements of $S$. Knowing that any ordered element pair of $S$ appears at most in one element of $P$. (If $x = (\cdots a \cdots b \cdots)$, then we call ordered pair $(a, b)$ appeared in $x$.) Prove that $P$ has at most $800$ elements.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022439#p2022439]34. (IRN 3)[/url]There are $n$ non-coplanar points in space. Prove that there exists a circle exactly passes through three points of them.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022443#p2022443]35. (IRN 4)[/url] Prove that if $|x| < 1$, then
\[ \frac{x}{(1-x)^2}+\frac{x^2}{(1+x^2)^2} + \frac{x^3}{(1-x^3)^2}+\cdots=\frac{x}{1-x}+\frac{2x^2}{1+x^2}+\frac{3x^3}{1-x^3}+\cdots\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=360423&#p360423]36. (IRE 1)[/url] [b](SL1990-12)[/b] Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG = DE$. Prove that $ CA = CB$.

[i]Original formulation:[/i]

Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF = DE,$ prove that $ AC = BC.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225228&#p1225228]37. (IRE 2)[/url] [b](SL1990-13)[/b] An eccentric mathematician has a ladder with $ n$ rungs that he always ascends and descends in the following way: When he ascends, each step he takes covers $ a$ rungs of the ladder, and when he descends, each step he takes covers $ b$ rungs of the ladder, where $ a$ and $ b$ are fixed positive integers. By a sequence of ascending and descending steps he can climb from ground level to the top rung of the ladder and come back down to ground level again. Find, with proof, the minimum value of $ n,$ expressed in terms of $ a$ and $ b.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022451#p2022451]38. (IRE 3)[/url] Let $\alpha$ be the positive root of the quadratic equation $x^2 = 1990x + 1$. For any $m, n \in \mathbb N$, define the operation $m*n = mn + [\alpha m][ \alpha n]$, where $[x]$ is the largest integer no larger than $x$. Prove that $(p*q)*r = p*(q*r)$ holds for all $p, q, r \in \mathbb N.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022459#p2022459]39. (IRE 4)[/url] Let $a, b, c$ be integers. Prove that there exist integers $p_1, q_1, r_1, p_2, q_2$ and $r_2$, satisfying $a = q_1r_2 - q_2r_1, b = r_1p_2 - r_2p_1$ and $c = p_1q_2 - p_2q_1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022463#p2022463]40. (ISR 1)[/url]Given three letters $X, Y, Z$, we can construct letter sequences arbitrarily, such as $XZ, ZZYXYY, XXYZX$, etc. For any given sequence, we can perform following operations:

$T_1$: If the right-most letter is $Y$, then we can add $YZ$ after it, for example, $T_1(XYZXXY) =
(XYZXXYYZ).$

$T_2$: If The sequence contains $YYY$, we can replace them by $Z$, for example, $T_2(XXYYZYYYX) =
(XXYYZZX).$

$T_3$: We can replace $Xp$ ($p$ is any sub-sequence) by $XpX$, for example, $T_3(XXYZ) = (XXYZX).$

$T_4$: In a sequence containing one or more $Z$, we can replace the first $Z$ by $XY$, for example,
$T_4(XXYYZZX) = (XXYYXYZX).$

$T_5$: We can replace any of $XX, YY, ZZ$ by $X$, for example, $T_5(ZZYXYY) = (XYXX)$ or $(XYXYY)$ or $(ZZYXX).$

Using above operations, can we get $XYZZ$ from $XYZ \ ?$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022515#p2022515]41. (ISR 2)[/url] Let $n$ be an arbitrary positive integer. Calculate $S_n = \sum_{r=0}^n 2^{r-2n} \binom{2n-r}{n}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022518#p2022518]42. (ITA 1)[/url] Find $n$ points $p_1, p_2, \ldots, p_n$ on the circumference of a unit circle, such that $\sum_{1\leq i< j \leq n} p_i p_j$ is maximal.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022519#p2022519]43. (ITA 2)[/url] Let $V$ be a finite set of points in three-dimensional space. Let $S_1, S_2, S_3$ be the sets consisting of the orthogonal projections of the points of $V$ onto the $yz$-plane, $zx$-plane, $xy$-plane, respectively. Prove that $| V|^2 \leq | S1|\cdot|S2||\cdot |S3|$, where $| A|$ denotes the number of elements in the finite set $A.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022520#p2022520]44. (ITA 3)[/url] Prove that for any positive integer n, the number of odd integers among the binomial coefficients $\binom nh \ ( 0 \leq h \leq n)$ is a power of 2.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022522#p2022522]45. (ITA 4)[/url] The tourist on an island can play the "getting treasure" game. He has to open a series of doors, each door is colored with one of n colors, according to the following rules:

(i) The tourist has n keys, each key with a different color.

(ii) Once a key is used, it is not permitted to change until it is destroyed.

(iii) Each key can open any door, and keeps intact when it opens the door having different color with it, but is destroyed when it opens the door having the same color with it.

Find the least number of doors to ensure that no tourist, no matter how he choose the order of the keys to use, can get the treasure.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017846#p2017846]46. (JAP 1)[/url] Let $P$ be an interior point of triangle $ABC$. Let $Q, R, S$ be the intersections of $AP, BP, CP$ with sides $BC, CA, AB$, respectively. Prove that $S_{QRS} \leq  \frac 14 S_{ABC}$ where $S$ denote the area function.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225231&#p1225231]47. (JAP 2)[/url] [b](SL1990-14)[/b] In the coordinate plane a rectangle with vertices $ (0, 0),$ $ (m, 0),$ $ (0, n),$ $ (m, n)$ is given where both $ m$ and $ n$ are odd integers. The rectangle is partitioned into triangles in such a way that

[i](i)[/i] each triangle in the partition has at least one side (to be called a “good” side) that lies on a line of the form $ x = j$ or $ y = k,$ where $ j$ and $ k$ are integers, and the altitude on this side has length 1;

[i](ii)[/i] each “bad” side (i.e., a side of any triangle in the partition that is not a “good” one) is a common side of two triangles in the partition.

Prove that there exist at least two triangles in the partition each of which has two good sides.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022529#p2022529]48. (JAP 3)[/url] Prove that $\sqrt 2 +\sqrt 3 +\sqrt{1990}$ is irrational.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023330#p2023330]49. (LUX 1)[/url] $AB, AC$ are two chords of the circle centered at $O$. The diameter, which is perpendicular to $BC$, intersects $AB, AC$ at $F, G$ respectively ($F$ is in the circle). The tangent from $G$ tangents the circle at $T$. Prove that $F$ is the projection of $T$ on $OG. $

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022530#p2022530]50. (MEX 1)[/url] During the class interval, $n$ children sit in a circle and play the game described below. The teacher goes around the children clockwisely and hands out candies to them according to the following regulations: Select a child, give him a candy; and give the child next to the first child a candy too; then skip over one child and give next child a candy; then skip over two children; give the next child a candy; then skip over three children; give the next child a candy;...

Find the value of $n$ for which the teacher can ensure that every child get at least one candy eventually (maybe after many circles).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225238&#p1225238]51. (MEX 2)[/url] [b](SL1990-15)[/b] Determine for which positive integers $ k$ the set \[ X = \{1990, 1990 + 1, 1990 + 2, \ldots, 1990 + k\}\] can be partitioned into two disjoint subsets $ A$ and $ B$ such that the sum of the elements of $ A$ is equal to the sum of the elements of $ B.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022535#p2022535]52. (MON 1)[/url] Let real numbers $a_1, a_2, \ldots, a_n$ satisfy $0 < a_i \leq a, \  i = 1, 2, \ldots, n$. Prove that

(i) If $n = 4$, then
\[\frac 1a \sum_{i=1}^4 a_i - \frac{a_1a_2 + a_2a_3 + a_3 a_4 + a_4 a_1}{a^2}  \leq 2.\]

(ii) If $n = 6$, then
\[\frac 1a \sum_{i=1}^6 a_i - \frac{a_1a_2 + a_2a_3 + \cdots + a_5 a_6 + a_6 a_1}{a^2}  \leq 3.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022536#p2022536]53. (MON 2)[/url] Find the real solution(s) for the system of equations
\[\begin{cases}x^3+y^3 &=1\\x^5+y^5 &=1\end{cases}\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022547#p2022547]54. (MON 3)[/url] Let $M = \{1, 2, \ldots, n\}$ and $\phi : M \to M$ be a bijection.

(i) Prove that there exist bijections $\phi_1, \phi_2 : M \to M$ such that $\phi_1 \cdot \phi_2 = \phi , \phi_1^2 =\phi_2^2=E$, where $E$ is the identity mapping.

(ii) Prove that the conclusion in (i) is also true if $M$ is the set of all positive integers.

[size=150][b]IMO 1990 LongListed Problems: (Posted in Forums) - Part Three[/b][/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022541#p2022541]55. (MON 4)[/url] Given points $A, M, M_1$ and rational number $\lambda \neq -1$. Construct the triangle $ABC$, such that $M$ lies on $BC$ and $M_1$ lies on $B{}_1C{}_1$ ($B_1, C_1$ are the projections of $B, C$ on $AC, AB$ respectively), and $\frac{BM}{MC}=\frac{B{}_1M{}_1}{M{}_1C{}_1}=\lambda.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022549#p2022549]56. (MOR 1)[/url] For positive integers $n, p$ with $n \geq p$, define real number $K_{n, p}$ as follows:
$K_{n, 0} = \frac{1}{n+1}$ and $K_{n, p} = K_{n-1, p-1} -K_{n, p-1}$ for $1 \leq p \leq n.$

(i) Define $S_n = \sum_{p=0}^n K_{n,p} , \ n = 0, 1, 2, \ldots$ . Find $\lim_{n \to \infty} S_n.$

(ii) Find $T_n = \sum_{p=0}^n (-1)^p K_{n,p} , \ n = 0, 1, 2, \ldots$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022556#p2022556]57. (MOR 2)[/url] The sequence $\{un\}$ is defined by $u_1 = 1, u_2 = 1, u_n = u_{n-1} + 2u_{n-2} for n \geq 3$. Prove that for any positive integers $n, p \  (p > 1), u_{n+p} = u_{n+1}u_{p} + 2u{}_nu{}_{p-1}$. Also find the greatest common divisor of $u_n$ and $u_{n+3}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366472&#p366472]58. (NET 1)[/url] [b](SL1990-16) [/b][i]IMO 6[/i] Prove that there exists a convex 1990-gon with the following two properties :

[b]a.)[/b] All angles are equal.
[b]b.)[/b] The lengths of the 1990 sides are the numbers $ 1^2$, $ 2^2$, $ 3^2$, $ \cdots$, $ 1990^2$ in some order.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022572#p2022572]59. (NET 2)[/url] Given eight real numbers $a_1 \leq a_2 \leq  \cdots \leq a_7 \leq a_8$. Let $x = \frac{ a_1 + a_2 +  \cdots + a_7 + a_8}{8}$, $y = \frac{ a_1^2 + a_2^2 +  \cdots + a_7^2 + a_8^2}{8}$. Prove that
\[2 \sqrt{y-x^2} \leq a_8  - a_1 \leq 4 \sqrt{y-x^2}.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225240&#p1225240]60. (NET 3)[/url] [b](SL1990-17)[/b] Unit cubes are made into beads by drilling a hole through them along a diagonal. The beads are put on a string in such a way that they can move freely in space under the restriction that the vertices of two neighboring cubes are touching. Let $ A$ be the beginning vertex and $ B$ be the end vertex. Let there be $ p \times q \times r$ cubes on the string $ (p, q, r \geq 1).$

[i](a)[/i] Determine for which values of $ p, q,$ and $ r$ it is possible to build a block with dimensions $ p, q,$ and $ r.$ Give reasons for your answers.
[i](b)[/i] The same question as (a) with the extra condition that $ A = B.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022577#p2022577]61. (NET 4)[/url] Prove that we can fill in the three dimensional space with regular tetrahedrons and regular octahedrons, all of which have the same edge-lengths. Also find the ratio of the number of the regular tetrahedrons used and the number of the regular octahedrons used.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225242&#p1225242]62. (NOR 1)[/url] [b](SL1990-18)[/b] Let $ a, b \in \mathbb{N}$  with $ 1 \leq a \leq b,$ and $ M = \left[\frac {a + b}{2} \right].$ Define a function $ f: \mathbb{Z} \mapsto \mathbb{Z}$ by
\[ f(n) = \begin{cases} n + a, & \text{if } n \leq M, \\
n - b, & \text{if } n \geq M. \end{cases}
\]
Let $ f^1(n) = f(n),$ $ f_{i + 1}(n) = f(f^i(n)),$ $ i = 1, 2, \ldots$ Find the smallest natural number $ k$ such that $ f^k(0) = 0.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225246&#p1225246]63. (POL 1)[/url] [b](SL1990-19)[/b] Let $ P$ be a point inside a regular tetrahedron $ T$ of unit volume. The four planes passing through $ P$ and parallel to the faces of $ T$ partition $ T$ into 14 pieces. Let $ f(P)$ be the joint volume of those pieces that are neither a tetrahedron nor a parallelepiped (i.e., pieces adjacent to an edge but not to a vertex). Find the exact bounds for $ f(P)$ as $ P$ varies over $ T.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022588#p2022588]64. (POL 2)[/url] Given an $m$-element set $M$ and a $k$-element subset $K \subset M$. We call a function $f: K \to M$ has "path", if there exists an element $x_0  \in K$ such that $f(x_0) = x_0$, or there exists a chain $x_0, x_1, \ldots, x_j = x_0 \in K$ such that $_xi = f(x_{i-1})$ for $i = 1, 2, \ldots, j$. Find the number of functions $f: K \to M$ which have path.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225251&#p1225251]65. (POL 3)[/url] [b](SL1990-20)[/b] Prove that every integer $ k$ greater than 1 has a multiple that is less than $ k^4$ and can be written in the decimal system with at most four different digits.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022592#p2022592]66. (POL 4)[/url] Find all the continuous bounded functions $f: \mathbb R \to \mathbb R$ such that
\[f^ 2(x) -f^ 2(y) = f(x + y)f(x - y) \text{  for all } x, y \in \mathbb R.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022635#p2022635]67. (PRK 1)[/url] Let $a + bi$ and $c + di$ be two roots of the equation $x^n = 1990$, where $n \geq 3$ is an integer and $a,b,c,d \in \mathbb R$. Under the linear transformation $f =\left(\begin{array}{cc}a&c\\b &d\end{array}\right)$, we have $(2, 1) \to  (1, 2)$. Denote $r$ to be the distance from the image of $(2, 2)$ to the origin. Find the range of $r.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022636#p2022636]68. (PRK 2)[/url] In coordinate plane, a variable point $M$, starting from the origin $O(0, 0)$, moves on the line $l$ with slope $k$, where $k$ is an irrational number.

(i) Prove that point $O(0, 0)$ is the only rational point (namely, the coordinates of which are both rationals) on the line $l.$

(ii) Prove that for any number $\varepsilon > 0$, there exist integers $m, n$ such that the distance between $l$ and the point $(m, n)$ is less than $\varepsilon.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022721#p2022721]69. (PRK 3)[/url] Consider the set of cuboids: the three edges $a, b, c$ from a common vertex satisfy the condition
\[\frac ab = \frac{a^2}{c^5}\]
(i) Prove that there are $100$ pairs of cuboids in this set with equal volumes in each pair.

(ii) For each pair of the above cuboids, find the ratio of the sum of their edges.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022639#p2022639]70. (PRK 4)[/url] $BC$ is a segment, $M$ is point on $BC$, $A$ is a point such that $A, B, C$ are non-collinear.

(i) Prove that if $M$ is the midpoint of $BC$, then $AB^2 + AC^2 = 2(AM^2 + BM^2).$

(ii) If there exists another point (except $M$) on segment $BC$ satisfying (i), find the region of point $A$ might occupy.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022734#p2022734]71. (PRK 5)[/url] Given a  point $P = (p_1, p_2, \ldots, p_n)$ in $n$-dimensional space . Find point $X = (x_1, x_2, \ldots, x_n)$, such that $x_1 \leq x_2 \leq\cdots \leq x_n$ and $\sqrt{(x_1-p_1)^2 + (x_2-p_2)^2+\cdots+(x_n-p_n)^2}$ is minimal.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022739#p2022739]72. (KOR 1)[/url] Let $n \geq 5$ be a positive integer. $a_1, b_1, a_2, b_2, \ldots, a_n, b_n$ are integers. $( a_i, b_i)$ are pairwisely distinct for $i = 1, 2, \ldots, n$, and $|a{}_1b{}_2 - a{}_2b{}_1| = |a{}_2b{}_3 -a{}_3b{}_2| = \cdots = |a_{n-1}b_n -a{}_nb{}_{n-1}| = 1$. Prove that there exists a pair of indexes $i, j$ satisfying $2 \leq |i - j| \leq n - 2$ and $|a{}_ib{}_j -a{}_jb{}_i| = 1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022744#p2022744]73. (KOR 2)[/url] Let $\mathbb Q$ be the set of all rational numbers and $\mathbb R$ be the set of real numbers. Function $f: \mathbb Q \to \mathbb R$ satisfies the following conditions:

(i) $f(0) = 0$, and for any nonzero $a \in Q, f(a) > 0.$
(ii) $f(x + y) = f(x)f(y) \qquad \forall x,y \in \mathbb Q.$
(iii) $f(x + y) \leq \max\{f(x), f(y)\} \qquad \forall x,y \in \mathbb Q.$

Let $x$ be an integer and $f(x) \neq 1$. Prove that $f(1 + x + x^2+ \cdots + x^n) = 1$ for any positive integer $n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023250#p2023250]74. (KOR 3)[/url] Let $L$ be a subset in the coordinate plane defined by $L = \{(41x + 2y, 59x + 15y) | x, y \in \mathbb Z \}$, where $\mathbb Z$ is set of integers. Prove that for any parallelogram with center in the origin of coordinate and area $1990$, there exist at least two points of $L$ located in it.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225256&#p1225256]75. (ROM 1)[/url] [b](SL1990-21)[/b] Let $ n$ be a composite natural number and $ p$ a proper divisor of $ n.$ Find the binary representation of the smallest natural number $ N$ such that

\[ \frac{(1 + 2^p + 2^{n-p})N - 1}{2^n}\]

is an integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023251#p2023251]76. (ROM 2)[/url] Prove that there exist at least two non-congruence quadrilaterals, each of which having a circumcircle, such that they have equal perimeters and areas.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022761#p2022761]77. (ROM 3)[/url] Let $a, b, c \in \mathbb R$. Prove that
\[(a^2 + ab + b^2)(b^2 + bc + c^2)(c^2 + ca + a^2) \geq (ab + bc + ca)^3.\]
 When does the equality hold?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225273&#p1225273]78. (ROM 4)[/url] [b](SL1990-22)[/b] Ten localities are served by two international airlines such that there exists a direct service (without stops) between any two of these localities and all airline schedules offer round-trip service between the cities they serve. Prove that at least one of the airlines can offer two disjoint round trips each containing an odd number of landings.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366466&#p366466]79. (ROM 5)[/url] [b](SL1990-23)[/b] [i]IMO 3[/i] Determine all integers $ n > 1$ such that
\[ \frac {2^n + 1}{n^2}
\]
is an integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023271#p2023271]80. (SPA 1)[/url] Function $f(x, y): \mathbb N \times \mathbb N \to \mathbb Q$ satisfies the conditions:

(i) $f(1, 1) =1;$

(ii) $f(p + 1, q) + f(p, q + 1) = f(p, q)$ for all $p, q \in  \mathbb N;$

(iii) $qf(p + 1, q) = pf(p, q + 1)$ for all $p, q \in  \mathbb N$

Find $f(1990, 31).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005759&#p2005759]81. (SPA 2)[/url] [b](LL1992 -66)[/b] A circle of radius $\rho$ is tangent to the sides $AB$ and $AC$ of the triangle $ABC$, and its center $K$ is at a distance $p$ from $BC$.

[i](a)[/i] Prove that $a(p - \rho) = 2s(r - \rho)$, where $r$ is the inradius and $2s$ the perimeter of $ABC$.
[i]
(b)[/i] Prove that if the circle intersect $BC$ at $D$ and $E$, then
\[DE=\frac{4\sqrt{rr_1(\rho-r)(r_1-\rho)}}{r_1-r}\]

where $r_1$ is the exradius corresponding to the vertex $A.$

[size=150][b]IMO 1990 LongListed Problems: (Posted in Forums) - Part Four[/b][/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005765&#p2005765]82. (SPA 3)[/url] [b](LL1992 -67)[/b] In a triangle, a symmedian is a line through a vertex that is symmetric to the median with the respect to the internal bisector (all relative to the same vertex). In the triangle $ABC$, the median $m_a$ meets $BC$ at $A'$ and the circumcircle again at $A_1$. The symmedian $s_a$ meets $BC$ at $M$ and the circumcircle again at $A_2$. Given that the line $A{}_1A{}_2$ contains the circumcenter $O$ of the triangle, prove that:

[i](a) [/i]$\frac{AA'}{AM} = \frac{b^2+c^2}{2bc} ;$

[i](b) [/i]$1+4b{}^2c{}^2 = a^2(b^2+c^2)$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023273#p2023273]83. (SWE 1)[/url] Point $D$ is on the hypotenuse $BC$ of right-angled triangle $ABC$. The inradii of triangles $ADB$ and $ADC$ are equal. Prove that $S_{ABC} = AD^2$, where $S$ is the area function.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023274#p2023274]84. (SWE 2)[/url] Let $n \geq 4$ be an integer. $a_1, a_2, \ldots, a_n \in (0, 2n)$ are $n$ distinct integers. Prove that there exists a subset of the set $\{a_1, a_2, \ldots, a_n \}$ such that the sum of its elements is divisible by $2n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023275#p2023275]85. (SWE 3)[/url] Let $A_1, A_2, \ldots, A_n (n \geq 4)$ be $n$ convex sets in plane. Knowing that every three convex sets have a common point. Prove that there exists a point belonging to all the sets.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023276#p2023276]86. (SWE 4)[/url] Given function $f(x) = \sin x + \sin \pi x$ and positive number $d$. Prove that there exists real number $p$ such that $|f(x + p) - f(x)| < d$ holds for all real numbers $x$, and the value of $p$ can be arbitrarily large.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023277#p2023277]87. (THA 1)[/url] Let $m$ be an positive odd integer not divisible by $3$. Prove that $\left[4^m - (2+\sqrt 2)^m\right]$ is divisible by $112.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=360441#p360441]88. (THA 2)[/url] [b](SL1990-24)[/b] Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023280#p2023280]89. (THA 3)[/url] Let $n$ be a positive integer. $S_1, S_2, \ldots, S_n$ are pairwise non-intersecting sets, and $S_k $ has exactly $k$ elements $(k = 1, 2, \ldots, n)$. Define $S = S_1\cup S_2\cup\cdots \cup S_n$. The function $f: S \to S $ maps all elements in $S_k$ to a fixed element of $S_k$, $k = 1, 2, \ldots, n$. Find the number of functions $g: S \to S$ satisfying $f(g(f(x))) = f(x).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023281#p2023281]90. (TUR 1)[/url] Let $P$ be a variable point on the circumference of a quarter-circle with radii $OA, OB$ and $\angle AOB = 90^\circ$. H is the projection of $P$ on $OA$. Find the locus of the incenter of the right-angled triangle $HPO.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023282#p2023282]91. (TUR 2)[/url] Quadrilateral $ABCD$ has an inscribed circle with center $O$. Knowing that $AB = CD$, and $M, K$ are the midpoints of $BC, AD$ respectively. Prove that $OM = OK.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023284#p2023284]92. (TUR 3)[/url] Let $n$ be a positive integer and $m = \frac{(n+1)(n+2)}{2}$. In coordinate plane, there are $n$ distinct lines $L_1, L_2, \ldots, L_n$ and $m$ distinct points $A_1, A_2, \ldots, A_m$, satisfying the following conditions:

i) Any two lines are non-parallel.

ii) Any three lines are non-concurrent.

iii) Only $A_1$ does not lies on any line $L_k$, and there are exactly $k + 1$ points $A_j$'s that lie on line $L_k$ $(k = 1, 2, \ldots, n).$

Prove that there exist a unique polynomial $p(x, y)$ with degree $n$ satisfying $p(A_1) = 1$ and $p(A_j) = 0$ for $j = 2, 3, \ldots, m.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366471#p366471]93. (TUR 4)[/url] [b](SL1990-25)[/b] [i]IMO 4[/i] Let $ {\mathbb Q}^ +$ be the set of positive rational numbers. Construct a function $ f : {\mathbb Q}^ + \rightarrow {\mathbb Q}^ +$ such that
\[ f(xf(y)) = \frac {f(x)}{y}
\]
for all $ x$, $ y$ in $ {\mathbb Q}^ +$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023286#p2023286]94. (USA 1)[/url] Given integer $n > 1$ and real number $t \geq 1$. $P$ is a parallelogram with four vertices $(0, 0), (0, t), (tF_{2n+1}, tF_{2n}), (tF_{2n+1}, tF_{2n} + t)$. Here, ${F_n}$ is the $n$-th term of Fibonacci sequence defined by $F_0 = 0, F_1 = 1$ and $F_{m+1} = F_m + F_{m-1}$. Let $L$ be the number of integral points (whose coordinates are integers) interior to $P$, and $M$ be the area of $P$, which is $t^2F_{2n+1}.$

[b][i]i)[/i][/b] Prove that for any integral point $(a, b)$, there exists a unique pair of integers $(j, k)$ such that$ j(F_{n+1}, F_n) + k(F_n, F_{n-1}) = (a, b)$, that is,$ jF_{n+1} + kF_n = a$ and $jF_n + kF_{n-1} = b.$

[i][b]ii)[/b][/i] Using [i][b]i)[/b][/i] or not, prove that $|\sqrt L-\sqrt M| \leq \sqrt 2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=205354&#p205354]95. (USA 2)[/url] [b](SL1990-26)[/b] Let $ p(x)$ be a cubic polynomial with rational coefficients. $ q_1$, $ q_2$, $ q_3$, ... is a sequence of rationals such that $ q_n = p(q_{n + 1})$ for all positive $ n$. Show that for some $ k$, we have $ q_{n + k} = q_n$ for all positive $ n$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005928]96. (USA 3)[/url] Triangles $XYZ$ and $ABC$ are similar, and the vertices are corresponded in that order. Points $X, Y, Z$ are on $BC, CA, AB$, respectively. Prove that the orthocenter of triangle $XZY$ and the circumcenter of triangle $ABC$ coincide.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023334#p2023334]97. (USA 4)[/url] In convex hexagon $ABCDEF$, we know that $\angle BCA = \angle DEC = \angle AFB = \angle CBD = \angle EDF.$ Prove that $AB = CD = EF.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225284&#p1225284]98. (USS 1)[/url] [b](SL1990-27)[/b] Find all natural numbers $ n$ for which every natural number whose decimal representation has $ n - 1$ digits $ 1$ and one digit $ 7$ is prime.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023294#p2023294]99. (USS 2)[/url] Given a $10 \times 10$ chessboard colored as black-and-white alternately. Prove that for any $46$ unit squares without common edges, there are at least $30$ unit squares with the same color.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225290&#p1225290]100. (USS 3)[/url] [b](SL1990-28)[/b] Prove that on the coordinate plane it is impossible to draw a closed broken line such that

[i](i)[/i] the coordinates of each vertex are rational;
[i](ii)[/i] the length each of its edges is 1;
[i](iii)[/i] the line has an odd number of vertices.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023302#p2023302]101. (USS 4)[/url] The side-lengths of two equilaterals $ABC$ and $KLM$ are $1$ and $1/4$, respectively. And triangle $KLM$ located inside triangle $ABC$. Denote by $\Sigma$ the sum of the distances from $A$ to lines $KL, LM$ and $MK$. Find the location of triangle $KLM$ when $\Sigma$  is maximal.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023303#p2023303]102. (USS 5)[/url] In coordinate plane, we call a point $(x, y)$ "lattice point" if both $x$ and $y$ are integers. Knowing that the vertices of triangle $ABC$ are all lattice points, and there exists exactly one lattice point interior to triangle $ABC$ (there might exist lattice points on the sides of $ABC$). Prove that the area of triangle $ABC$ is no larger than $\frac 92.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023304#p2023304]103. (VIE 1)[/url] Find the minimal value of the function
\[\begin{array}{c}\ f(x) =\sqrt{15 - 12 \cos x} + \sqrt{4 -2 \sqrt 3 \sin x}+\sqrt{7-4\sqrt 3 \sin x} +\sqrt{10-4 \sqrt 3 \sin x - 6 \cos x}\end{array}\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023307#p2023307]104. (VIE 2)[/url] Let $x,y,z$ be positive reals and $x \geq y \geq z$. Prove that
\[\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023310#p2023310]105. (YUG 1)[/url] Let $S, T$ be the circumcenter and centroid of triangle $ABC$, respectively. $M$ is a point in the plane of triangle $ABC$ such that $90^\circ  \leq  \angle SMT < 180^\circ$. $A_1, B_1, C_1$ are the intersections of $AM, BM, CM$ with the circumcircle of triangle $ABC$ respectively. Prove that $MA_1 + MB_1 + MC_1 \geq MA + MB + MC.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023313#p2023313]106. (YUG 2)[/url] Let $S$ be the incenter of triangle $ABC$. $A_1, B_1, C_1$ are the intersections of $AS, BS, CS$ with the circumcircle of triangle $ABC$ respectively. Prove that $SA_1 + SB_1 + SC_1 \geq SA + SB + SC.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023314#p2023314]107. (YUG 3)[/url] Let $a, b, c$ be the side lengths and $P$ be area of a triangle, respectively. Prove that
\[(a^2+b^2+c^2-4\sqrt 3 P) (a^2+b^2+c^2) \geq 2 \left(a^2(b - c)^2 + b^2(c - a)^2 + c^2(a - b)^2\right).\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2023315#p2023315]108. (YUG 4)[/url] Let $\{ a_1, a_2, \ldots, a_n\} = \{1, 2, \ldots, n\}$. Prove that
\[\frac 12 +\frac 23 +\cdots+\frac{n-1}{n} \leq \frac{a_1}{a_2} + \frac{a_2}{a_3} +\cdots+\frac{a_{n-1}}{a_n}.\]

[url=https://artofproblemsolving.com/community/c4021_1990_imo_longlists]This is a very hard job.[/url]
Thank you very much.


[quote="xxxxx"]Find the sum of the solutions to the equation $|x-2|+|x-5|=4$[/quote]
Normally in the equation $|x|+|y|=z$ you split it up into 4 cases: $x+y, x-y,-x+y,-x-y$. [hide]Since there is an x in both terms, "$x-y,-x+y$" won't work. Therefore you only have 2 cases:
1.$x-2+x-5=4$
2.$-x+2-x+5=4$
In case 1, $x=\frac{11}{2}=5.5$ In case 2, $x=\frac{-3}{2}=-1.5$
$\frac{11}{2}+\frac{-3}{2}=\frac{8}{2}=\boxed{4}$[/hide]

[quote="Geómetra"][quote="Quevvy"]In case 2, $x=\frac{-3}{2}=-1.5$
$\frac{11}{2}+\frac{-3}{2}=\frac{8}{2}=\boxed{4}$[/quote]

Little detail there.

It must be $x=\frac{3}{2}$, and the sum would be $\frac{11}{2}+\frac{3}{2}=\frac{14}{2}=7 \ \blacksquare$[/quote]
You're right, I'll edit it


I have just recently been invited to participate in this competition at my school.  What are the problems like, and, in your opinion, what was the hardest and easyest AMC problem so far?

Description of the Exams:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693

You can purchase old tests on their website:
http://www.unl.edu/amc

You can find some old tests here:
http://goodlymaths.0catch.com/contests/amc/amc_archive.htm


Hmm, due to the quick decline of the last game, I am going to add a few more rules.

1. There will be one task per week. Anyone who fails to complete the task will be substituted/eliminated.

2. Ok, 20 ppl can join this time.

3. Each person starts off with 15 points.

4. We need 25 people to sign up, just in case we need substitutes (see rule 1) :D

5. So like, only join if u absolutely can do this.

The rules stay the same.

SIGN UP QUICKLEEEEEE

or quickly. whichever, quicklee sounds cooler.

Yeah, probably. But like, whats ur name?

his first name is my first name and he's on my nats team

Current list

1. chenhsi
2. budi713
3. pacman2812
4. ilord

Actually, 20's gonna take WAY too long. keep it to ten ppl, but 20 points.


Ring Theory
algebra
function
domain
superior algebra
superior algebra solved

Hi,

first of all , a UFD means a unique factorization domain, this is a ring, with a 1 element, commutative, and with no zero divisors. the UF stands for the fact that every element (no unit) can be written as a unique product of primes (except for changes in order of the primes and switching units) (so 6=2*3=(-3)*(-2) are to be considered the same factorization)

it is a well known results that all finite domains, are division rings, and by wedderburn, they are commutative and thus fields, galois fields thus.  So no interesting finite unique factorization domain exists.

However, are there infinite unique factorization domains with only finitely many primes?  I tried to prove the converse, but failed.  We all know the euclidean style proof
I wanted to take the product of all primes if they would be finitely many, and wanted to add up a unit
p1*.....*pm +u1
and wanted to prove existence of some prime dividing this number, but can it happen that for every unit u1, the resulting number is another unit u2, so my approach fails?

When proving it in Z, this difficulty doesn't show up because we know the number is to big to be -1 or 1.

Can anyone help me with this problem?

Thanks,

frederic

Lemma1: Every prime ideal of an UFD contains a prime element.

Proof: Let $P$ be a non-zero prime ideal of an UFD $A$. Then $P$ contains a non-zero, non-invertible element $a$. Since $A$ is UFD, there exist primes $p_1,p_2\ldots p_n$ not necessarily distinct such as $a=p_1\cdot p_2\ldots p_n$. Since $P$ is a prime ideal one of the $p_k$'s must be in P, so P contains a prime element.

Lemma2: The intersection of all prime ideals of a ring $A$ is the set of all nilpotent elements of $A$.

Proof: You can find this in most commutative algebra books.

Now to the problem. Assume there exists an UFD with finitely many non-zero prime elements, but at least one. Let $p_1,p_2\ldots p_n$ be those primes. Let $p=p_1\cdot p_2\cdot\ldots\cdot p_n$. Let $P$ be a prime ideal of $A$. Then by lemma1, $P$ contains a prime element of $A$. WLOG assume $p_1\in P$. Then $p\in (p_1)\subset P$. So $p$ belongs to all prime ideals of $A$, so by lemma2, $p$ is nilpotent.

But $A$ is UFD, so the only nilpotent element of $A$ is $0$. So $p=0$ (contradiction).

So the only UFD's that have finitely many prime elements are fields.

I think there might be something wrong with that argument...
Consider the localization of $\mathbb{Z}$ at (2), $R = \{ a/b \in \mathbb{Q} : 2\nmid b\}$.
In this case, $R$ has only one prime ideal, $P=(2)$, but 2 is not nilpotent.
Of course, $R$ still has infinitely many primes (but only one up to associates).

I guess $0$ counts as a prime ideal in lemma 2 making it useless.
Lemma 1 should also be changed to any non-zero prime ideal contains a prime element.

so back to unsolved

not really unsolved, since the previous post answers your question:

all local rings are UFD, since every ideal is a power of the maximal ideal. Of course, there are two prime ideals, the maximal one and $(0)$.

Andrei


When Rachel divides her favorite number by 7, she gets a remainder of 5. What will the remainder be if she multiplies her favorite number by 5 and then divides by 7?

Let Rachel's favorite number be $ x$. Then, we have that $ x\equiv5\pmod7$. So, $ 5x\equiv25\equiv\boxed{4}\pmod7$.

Or, just letting Rachel's favorite number be 12 (as $12 \equiv 5$ [mod 7]) would be also fast.

$12 \cdot 5 = 60.
60 \equiv \boxed {4}$ (mod 7).


Why do we define that $ \text{span}\,(\emptyset) \equal{}\{\mathbf 0\}$? Also, can we define the dimension of an "empty space"?

Because $ \{0\}$ is linearly dependent. Generally we define it as above , becuase we want our proofs be correct for all sets. Maybe we can define $ \text{dim}\{\}\equal{}\minus{}\infty$  :maybe:

The dimension of the zero space is $ 0$ by definition, and if we want to define the dimension of vector spaces as the cardinality of any basis for them without excluding the zero space, then what is the best we can do? Yeah, we consider the empty set as the basis of the zero space, by convention.

For the second part of your question, I don't know what you mean exactly by "empty space", but it does not matter. Since at any rate that would not be a "vector space", so why bother defining dimension for it?

[quote="The Captain"]For the second part of your question, I don't know what you mean exactly by "empty space", but it does not matter. Since at any rate that would not be a "vector space", so why bother defining dimension for it?[/quote]
But today I read a book which said an empty set has dimension -1 by convention.

That's a terrible convention -- is there any nice rule that it is consistent with?  (For example, it violates the rule that the dimension of a product is the sum of the dimensions.)

Another potential explanation is that this is just an instance of an empty sum equal to zero.

I agree with JBL: that's a useless convention.

The only convention that makes sense to me: the word dimension is only meaningful in the context of a vector space (or, in the case of a manifold, for something where there is a vector space in the definition somewhere). The axioms of a vector space require it to have a zero element. Therefore, the empty set is not a vector space and we need not try to define its dimension.

By the same token, $ \{0\}\equal{}\text{span}(\emptyset)$ as a space of dimension $ 0$ makes perfect sense and works in JBL's formula and in all other similar places as well - for instance, the rank-nullity theorem.


Suppose you have an open-topped box that is 5cm x 5cm x 5cm.
The boxes sides are .01 meters thick. Inside the box is a steel perfectly spherical ball that is 25 cubic centimeters in size. Next to the box is a 2 liter pail filled with mercury. How many cubic centimeters of mercury would you have to pour into to box to completely submerge the ball? Round to the nearest 100th of a cubic centimeter if needed.

[quote="cincodemayo5590"][hide]
First, we find the diameter of the ball:
\[25 = \frac{4}{3}\pi r^{3}\implies r =\sqrt[3]{ \frac{75}{4 \pi}}\]
For convenience, we will keep this as $r$. Remember that $d=2r$.

I don't see how the sides can be $10 \text{ cm}$ thick, since the box is only $5 \text{x}5 \text{x}5$. Were the situation possible, we would just figure out the new dimensions of the available space and continue the problem with those numbers.

I'm going to assume that the walls are so thin that their effect on the volume is negligible. The amount of mercury that we need is just a rectangular solid with dimensions $5, 5, 2r$ and from this subtracted the volume of the sphere. Thus our volume of mercury is $50r-25 \approx 9.30 \text{ cm}^{3}$. [/hide][/quote]

isn't .01 meters 1 centimeter? anyways wen i said .01 meters i meant 1 centimeter
just try it that way

btw even if u don't count the walls, ur wrong

[hide]Steel floats in mercury, so good luck getting the ball to sink![/hide]

[quote="cincodemayo5590"][quote="Ihatepie"][quote="cincodemayo5590"][hide]
First, we find the diameter of the ball:
\[25 = \frac{4}{3}\pi r^{3}\implies r =\sqrt[3]{ \frac{75}{4 \pi}}\]
For convenience, we will keep this as $r$. Remember that $d=2r$.

I don't see how the sides can be $10 \text{ cm}$ thick, since the box is only $5 \text{x}5 \text{x}5$. Were the situation possible, we would just figure out the new dimensions of the available space and continue the problem with those numbers.

I'm going to assume that the walls are so thin that their effect on the volume is negligible. The amount of mercury that we need is just a rectangular solid with dimensions $5, 5, 2r$ and from this subtracted the volume of the sphere. Thus our volume of mercury is $50r-25 \approx 9.30 \text{ cm}^{3}$. [/hide][/quote]

btw even if u don't count the walls, ur wrong[/quote]

Umm...explain?[/quote]
Check the density of mercury, and then check the density of steel.

[quote="randomdragoon"][/quote][quote="cincodemayo5590"][quote="Ihatepie"][quote="cincodemayo5590"][hide]
First, we find the diameter of the ball:
\[25 = \frac{4}{3}\pi r^{3}\implies r =\sqrt[3]{ \frac{75}{4 \pi}}\]
For convenience, we will keep this as $r$. Remember that $d=2r$.

I don't see how the sides can be $10 \text{ cm}$ thick, since the box is only $5 \text{x}5 \text{x}5$. Were the situation possible, we would just figure out the new dimensions of the available space and continue the problem with those numbers.

I'm going to assume that the walls are so thin that their effect on the volume is negligible. The amount of mercury that we need is just a rectangular solid with dimensions $5, 5, 2r$ and from this subtracted the volume of the sphere. Thus our volume of mercury is $50r-25 \approx 9.30 \text{ cm}^{3}$. [/hide][/quote]

btw even if u don't count the walls, ur wrong[/quote]

Umm...explain?[/quote][quote="randomdragoon"]
Check the density of mercury, and then check the density of steel.[/quote]

[b]randomdragoon[/b], when he said this, I don't think he was referring to the problem that the ball will float.

Anyways, if the ball will float in mercury, why even try to submerge the thing? It'll just float up and out of the box.

[quote="cincodemayo5590"][/quote][quote="randomdragoon"][/quote][quote="cincodemayo5590"][quote="Ihatepie"][quote="cincodemayo5590"][hide]
First, we find the diameter of the ball:
\[25 = \frac{4}{3}\pi r^{3}\implies r =\sqrt[3]{ \frac{75}{4 \pi}}\]
For convenience, we will keep this as $r$. Remember that $d=2r$.

I don't see how the sides can be $10 \text{ cm}$ thick, since the box is only $5 \text{x}5 \text{x}5$. Were the situation possible, we would just figure out the new dimensions of the available space and continue the problem with those numbers.

I'm going to assume that the walls are so thin that their effect on the volume is negligible. The amount of mercury that we need is just a rectangular solid with dimensions $5, 5, 2r$ and from this subtracted the volume of the sphere. Thus our volume of mercury is $50r-25 \approx 9.30 \text{ cm}^{3}$. [/hide][/quote]

btw even if u don't count the walls, ur wrong[/quote]

Umm...explain?[/quote][quote="randomdragoon"]
Check the density of mercury, and then check the density of steel.[/quote][quote="cincodemayo5590"]

[b]randomdragoon[/b], when he said this, I don't think he was referring to the problem that the ball will float.

Anyways, if the ball will float in mercury, why even try to submerge the thing? It'll just float up and out of the box.[/quote]

That's why this problem was posted in Puzzles & Brainteasers.

Oh, wow. Lol, sorry. I thought I was in the HSB forum, for some reason...   :P


invariant
linear algebra
matrix
algebra
polynomial

Let $ A$ be an $ n\times n$ matrix with characteristic polynomial
$ f(t)\equal{}(\minus{}1)^{n}t^{n}\plus{}a_{n\minus{}1}t^{n\minus{}1}\plus{}\cdots\plus{}a_{1}t\plus{}a_{0}$

(a) Prove that $ A$ is invertible if and only if $ a_{0}\neq 0$.

(b)Prove that if $ A$ is invertible, then
$ A^{\minus{}1}\equal{}(\minus{}1/a_{0})[(\minus{}1)^{n}A^{n\minus{}1}\plus{}a_{n\minus{}1}A^{n\minus{}2}\plus{}\cdots\plus{}a_{1}I_{n}]$

I can figure out half of part a but that's it.  Thank you.

1.
$ a_0$ (up to a sign) is just the determinant of the matrix.

2.
Apply Cayley Hamilton.

What do you mean it is the determinant of the matrix?  Which matrix is this?  I know that if $ a_{0}\equal{}0$, then the polynomial in question has a root of $ t\equal{}0$ hence it is not invertible.  I can't figure out the other way though.

a_0 is just the product of eigenvalues (up to a sign), so it's the determinant of A.


Five points $ A,B,C,D$ and $ E$ are on a circle.Let $ M,N,P$ and $ Q$ be the orthogonal projections of $ E$ on the lines
$ AB,BC,CA$ and $ DA$.Prove that the orthogonal projections of point $ E$ on the lines $ MN,NP,PQ$ and $ QM$ are concyclic

I thinks your problem is prove that $ 4$ this point are collinear!
Let $ I,J,K,L$ be the orthogonal projections of $ E$ on the lines $ MN,NP,PQ,QM$ respectively.
We have:
$ (\widehat{IL,LE}) \equal{} (\widehat{IM,ME}) \equal{}(\widehat{NM,ME}) \equal{}(\widehat{NB,BE}) \equal{} (\widehat{CB,BE}) \equal{} (\widehat{CD,DE}) \equal{} (\widehat{PD,DE}) \equal{} (\widehat{PQ,QE}) \equal{} (\widehat{KQ,QE}) \equal{} (\widehat{KL,LE})(mod\pi)$
Then $ I,K,L$ are collinear!

1). $ E =$ $ {Miquel}$ $ {point}$ of lines $ MN$, $ NP$, $ PQ$ and $ MQ$. Let $ \{S\} = MQ\cap PN$ and $ \{R\} = MN\cap PQ$.
2). points $ I$, $ J$, $ K$ and $ L$ lies on the $ {Simson}$ ${ line}$ of point $ E$ for the four triangles $ MRQ$, $ MNS$, $ PQS$ and $ PRN$.


A5/174 In cate moduri se pot aseza 6 persoane la o masa circulara? In carte se da 5! raspunsul. Enuntul este ambiguu. Ce se intelege prin "mod"? Daca se tine seama de pe ce scaun sta fiecare persoana sunt 6! moduri. Daca se tine seama numai de pozitia relativa a persoanelor intre ele sunt 5! moduri. ?

A8/pag. 169 In cate moduri se pot aseza 3 baieti si 3 fete la o masa circulara astfel incat sa nu stea alaturi 2 persoane de acelasi sex? In carte (de fapt in culegerea atasata cartii, tot de Burtea) se da 72 de moduri la rezultat. Eu spun ca sunt numai 12 daca se tine seama numai de pozitia relativa a persoanelor una fata de alta. Daca se ia in considerare si scaunul pe care sta fiecare, sunt 72 de moduri. ???

A10/ pag 169 Sa se gaseasca numarul de numere naturale de 5 cifre in baza 10 formate din: a) cifre pare b) cifre pare distincte . Eu spun la a) 5 la puterea a 5-a minus 5 la puterea a 4-a. b) 5! - 4!. In carte la rezultat a) este 5 la a 4-a b) la fel ca mine ????

E3/pag 174 Un student are de dat 5 examene in 15 zile. In cate moduri pot fi programate aceste examene? Dar daca primul examen se da in prima zi? Din nou enuntul este putin ambiguu. Sunt 5 discipline la care se pot sustine examen in orice ordine? Sau nu? Eu am spus la prima intrebare: aranjamente din 15 luate cate 5. La a doua: 5 x aranjamente din 14 luate cate 4. In carte la prima e la fel. La a doua insa spune: aranjamente din 14 luate cate 4. ???

Gresesc eu sau cartea? Este o chestiune de conventii sau greseala de rationament? Cine este acest Burtea si cine i-a dat voie sa scrie manuale cu atatea greseli? (Manuale dupa care trebuie sa invete copiii nostri).

Multumesc,
Jean

La prima intebare raspunsul este $5!$ caci masa fiind circulara $123456$ e aceeasi cu $612345$.

A10. Sunt 5 cifre pare care trebuiesc puse pe 5 pozitii, pentru a forma un grup de cinci cifre pare. Ca acest grup sa fie valabil trebuie sa nu aiba 0 pe prima pozitie. Numarul de astfel de grupuri de 5 cifre pare cu 0 pe prima pozitie este de $4^4$ (deoarece sunt 4 cifre de asezat pe 4 pozitii). Deci numarul de numere de 5 cifre, formate numai din cifre pare este de $5^5-4^4$.

E3. Daca primul examen se da in prima zi inseamna ca in cele 14 zile care au ramas se vor da 4 examene. Raspunsul este $A_{14}^4$.

[quote="Daniel Stanica"]
E3. Daca primul examen se da in prima zi inseamna ca in cele 14 zile care au ramas se vor da 4 examene. Raspunsul este $A_{14}^4$.[/quote]
In nici un caz. Dacă ordinea examenelor nu este prestabilită, atunci răspunsul este $5\cdot A_{14}^4$, deoarece examenul din prima zi poate fi ales în $5$ moduri. Dacă ordinea examenelor este prestabilită, atunci răspunsul este $C_{14}^4$, deoarece alegem doar cele $4$ zile în care se vor da examenele.

A5 - sunt 5! posibilitati de aranjare si pentru fiecare din acestea exista 6 posibilitati de rotire in jurul mesei
deci raspunsul e $6!/6$ deci $5!$


[hide="Proof"]$ n^5 \minus{} n \equal{} n\left(n^4 \minus{} 1\right) \equal{} n\left(n^2 \minus{} 1\right)\left(n^2 \plus{} 1\right) \equal{} n(n \minus{} 1)(n \plus{} 1)\left(n^2 \plus{} 1\right)$

Either $ n$, $ n \plus{} 1$, or $ n \minus{} 1$ is divisible by $ 3$, and at least one of those is divisible by $ 2$.

If $ 5$ does not divide any of those, then either $ n\equiv2\mod5$ or $ n\equiv3\mod5$.

[b]Case 1:[/b] $ n\equiv2\mod5$

$ n^2 \plus{} 1\equiv4 \plus{} 1\equiv0\mod5$

[b]Case 2:[/b] $ n\equiv3\mod5$

$ n^2 \plus{} 1\equiv9 \plus{} 1\equiv0\mod5$

In both cases, $ 5\mid n^2 \plus{} 1$, so $ 30\mid n(n \minus{} 1)(n \plus{} 1)\left(n^2 \plus{} 1\right)$, and the proof is complete.[/hide]

[quote="jimi"]Can somebody please help me with this problem?:
'Prove by induction that $ n^5 \minus{} n$ is divisible by 30.'[/quote]
Nice solution, i_like_pie!

I have an alternate solution through induction.
[hide] This obviously works for the base case, when n=1, since $ n^5 \minus{} n \equal{} 0$ when n=1 and 0 is divisible by 30.  So, we assume that $ 30|(n^5 \minus{} n)$ and we must show that
\[ 30|((n \plus{} 1)^5 \minus{} (n \plus{} 1))
\]

\[ (n \plus{} 1)^5 \minus{} (n \plus{} 1) \equal{} n^5 \plus{} 5n^4 \plus{} 10n^3 \plus{} 10n^2 \plus{} 5n \plus{} 1 \minus{} n \minus{} 1
\]

\[ (n \plus{} 1)^5 \minus{} (n \plus{} 1) \equal{} n^5 \plus{} 5n^4 \plus{} 10n^3 \plus{} 10n^2 \plus{} 5n \minus{} n
\]
Now, since $ 30|(n^5 \minus{} n)$ and we wish to show that $ 30|((n \plus{} 1)^5 \minus{} (n \plus{} 1))$, then it suffices to show that $ 30|(n \plus{} 5)^5 \minus{} (n \plus{} 1) \minus{} n^5 \plus{} n)$.
\[ (n \plus{} 1)^5 \minus{} (n \plus{} 1) \minus{} n^5 \plus{} n \equal{} n^5 \plus{} 5n^4 \plus{} 10n^3 \plus{} 10n^2 \plus{} 5n \minus{} n \minus{} n^5
\]

\[ (n \plus{} 1)^5 \minus{} (n \plus{} 1) \minus{} n^5 \plus{} n \equal{} 5n^4 \plus{} 10n^3 \plus{} 10n^2 \plus{} 5n
\]

\[ (n \plus{} 5)^5 \minus{} (n \plus{} 1) \minus{} n^5 \plus{} n \equal{} 5n(n^3 \plus{} 2n^2 \plus{} 2n \plus{} 1)
\]

\[ n^3 \plus{} 2n^2 \plus{} 2n \plus{} 1 \equal{} (n \plus{} 1)^3 \minus{} (n^2 \minus{} n) \equal{} (n^2 \plus{} n \plus{} 1)(n \plus{} 1)
\]

\[ (n \plus{} 5)^5 \minus{} (n \plus{} 1) \minus{} n^5 \plus{} n \equal{} 5n(n^2 \plus{} n \plus{} 1)(n \plus{} 1)
\]
Now, either n or n+1 is even.  So, 5(n)(n+1) is divisible by 5*2=10.  Now, if n or n+1 is divisible by 3, then we have that $ 5n(n^2 \plus{} n \plus{} 1)(n \plus{} 1)$ is divisble by 10*3=30.  If they are not divisible by three, then $ n^2 \plus{} n\equiv (1)(2) \equiv 2 \equiv \minus{} 1\bmod 3$, so $ n^2 \plus{} n \plus{} 1\equiv 0\bmod 3$.  Thus, 3 is a factor of $ n^2 \plus{} n \plus{} 1$ and thus $ 5n(n^2 \plus{} n \plus{} 1)(n \plus{} 1)$ is divisble by 30.  So, it is proven. [/hide]

[hide="Just a little touch to what i_like_pie did"]to avoid casework:

\begin{eqnarray*}n(n - 1)(n + 1)(n^2 + 1) = n(n - 1)(n + 1)(n^2 - 4 + 5) = (n - 2)(n - 1)n(n + 1)(n + 2) + 5(n - 1)n(n + 1)\end{eqnarray*}[/hide]

Thanks for your help guys.
i_like_pie
that is pretty neat.


$ \sum$

How do I make it so that numbers can be on the top and bottom?

Thanks Dr. Sonnhard.

hello, $ \sum_{i\equal{}1}^n i$.
Sonnhard.

How do you make the capital pi for products?

hello, see here
$ \prod_{i\equal{}1}^n i$
Sonnhard.

$ \prod_{\text{rofl}}^{\text{roflcopter}}$

Ah, ok thanks Sonnhard!

Messing around

$ \prod_{|\stackrel{\minus{}\minus{}\minus{}\minus{}}{\minus{}\minus{}\minus{}\minus{}}|}^{\stackrel{|\text{>}}{|\text{ }}}$

If you are planning to use this on a LaTeX document, you must use \displaystyle before the \sum symbol. Otherwise, it would look like $ \Sigma_{i\equal{}1}^n i$, with the bounds superscript rather than on top and on bottom.

[quote="herefishyfishy1"]If you are planning to use this on a LaTeX document, you must use \displaystyle before the \sum symbol. Otherwise, it would look like $ \Sigma_{i \equal{} 1}^n i$, with the bounds superscript rather than on top and on bottom.[/quote]

$ \displaystyle{\prod_{|\stackrel{\minus{}\minus{}\minus{}\minus{}}{\minus{}\minus{}\minus{}\minus{}}|}^{\stackrel{|\text{>}}{|\text{ }}}}$

Ahh....must go revise homework!


Hello everyone,

I'm curious to know:

[b]A different set of natural numbers has median $ 20$ and medium $ 17$.

What will be the largest number in this series?[/b]

What I have done:

Because they are natural numbers then are positive;

also for the calculation of the median must be sorted in increasing.

Assuming that the number of terms is odd, then the formula for calculating the median is: $ \displaystyle\frac {n \plus{} 1 }{2} \equal{} 20$, where $ n \equal{} 39.$

And the formula for calculating the average we have:

$ \displaystyle\frac {\displaystyle\sum_{i \equal{} 1}^n{X_i}}{39} \equal{} 17\longrightarrow{\displaystyle\sum_{i \equal{} 1}^n{X_i \equal{} 663}}$

I guess as I now know how many terms and their sum, I calculate the nth term,

for if we assume that the numbers are ordered increasingly, the latter being the largest.

Any ideas?

Thank you very much.

Greetings

You seem to be thoroughly confused about what the median is ;).

whats the biggest POSSIBLE number in this set i think is your question..

$ Y_n \equal{} X_n \minus{} 20$  are integers bounded strictly from below by -20, and median 0, with average -3

$ \sum_{n \equal{} 1}^{N} Y_n \equal{} \minus{} 3N$

if we are allowed to repeat numbers:
so we need half of the numbers to be negative or zero, and the other half to be positive or zero.
thus, if we make the sets to have N zeros, N values of -7, and one value of x=N-3, we get the desired properties..
since you have no restriction on N, this means we can pick as big of a value as we like..

if we cant repeat numbers:
we should never use a zero, since we get less space to raise up the highest value
if we use $ N\le 19$ negatives we should use $ \minus{} 19, \minus{} 18, ..., N \minus{} 20$
and then we can allow N numbers above zero such that we average to -3 overall,
since $ \sum_{k \equal{} 20 \minus{} N}^{19} \minus{} k \equal{} \minus{} \frac {(39 \minus{} N)N}{2}$, we need to pick N positive numbers which add up to $ \minus{} 3(2N) \plus{} \frac {(39 \minus{} N)N}{2}$
in order to maximize the highest value, we make the rest as low as possible, giving us $ 1,2,..., N \minus{} 1$
which add up to $ \frac {N(N \minus{} 1)}{2}$, and thus the last number must be $ \minus{} 3(2N) \plus{} \frac {(39 \minus{} N)N}{2} \minus{} \frac {N(N \minus{} 1)}{2}$
$ \equal{} \minus{} N^2 \plus{} 14N$
this comes to a max at $ N \equal{} 7$, so the maximum top value we can pick is 49
thats of course.. for the Y values, so shifting back to Xs we can pick a high value of 69


a,b,c are non-negative numbers such that : a+b+c=2, proove that:

$ \frac{{bc}}{{a^2  \plus{} 1}} \plus{} \frac{{ac}}{{b^2  \plus{} 1}} \plus{} \frac{{ab}}{{c^2  \plus{} 1}} \le 1$

$ \sum\frac{bc}{a^2\plus{}1}<\equal{}1$   is   equivalent  to

 $ \sum\frac{bc}{a^2\plus{}(a\plus{}b\plus{}c)^2/4}<\equal{}1$,

<=>

 $ 5cg1\plus{}7cg2\plus{}18cg3\plus{}23cg5\plus{}169cg7>\equal{}0$,

in which

$ cg1\equal{} \sum{a^4(a\minus{}b)(a\minus{}c)}>\equal{}0,$
$ cg2\equal{}\sum{ a^3(b\plus{}c)(a\minus{}b)(a\minus{}c)}>\equal{}0,$
$ cg3\equal{} \sum{a^2(b\minus{}c)^2(a\minus{}b)(a\minus{}c)}>\equal{}0,$
$ cg5 \equal{} \sum{a^2bc(a\minus{}b)(a\minus{}c)}>\equal{}0,$
$ cg7 \equal{} a^2b^2c^2>\equal{}0$.

discussed at http://www.mathlinks.ro/viewtopic.php?t=57558 . no halfways nice solution known (yet).

  darij