AMC USA(J)MO USAMO geometry parallelogram inequalities algebra

Is there enough interest for me to post Olympiad practice problems in this forum, or should we just leave those to the Pre-Olympiad/Olympiad forums?

Okay let's start with this one:

[b]Problem 1:[/b]

Solve in real numbers the system of equations:

$ x^4 \plus{} 2y^3 \minus{} x \equal{} \minus{} \frac {1}{4} \plus{} 3\sqrt {3}$
$ y^4 \plus{} 2x^3 \minus{} y \equal{} \minus{} \frac {1}{4} \minus{} 3\sqrt {3}$

Proposed by Titu Andreescu, University of Texas at Dallas

There's also a nice geometry problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199290]here[/url].

This seems too simple to cover all of the solutions, but...

[hide="Number 1"]
Add the two equations:

$ x^4\plus{}2x^3\minus{}x \plus{} y^4\plus{}2y^3\minus{}y\equal{}\minus{}\frac{1}{2}$
$ (x^2 \plus{} x \minus{} \frac{1}{2})^2 \plus{} (y^2 \plus{} y \minus{} \frac{1}{2})^2\equal{}0$

The only way to solve this in real numbers is if both the x and y expressions are equal to 0.

So $ (x,y)\equal{} (\frac{\minus{}1\pm\sqrt{3}}{2}, \frac{\minus{}1\pm\sqrt{3}}{2})$

A little experimentation shows that the only pair that works is $ \boxed{(\frac{\minus{}1\minus{}\sqrt{3}}{2}, \frac{\minus{}1\plus{}\sqrt{3}}{2})}$

[/hide]

That's correct. Here are the next two problems.

[b]Problem 2:[/b]
Find the least n such that no matter how we color an nxn lattice point grid with two colors we can always find a parallelogram with all vertices of the same color.

[b]Problem 3:[/b]
Let x, y, and z be positive real numbers such that $ xyz \equal{} 1$. Show that the following inequality holds

$ \frac{1}{(x^2\plus{}1) \plus{} y^2 \plus{} 1} \plus{} \frac{1}{(y^2\plus{}1) \plus{} z^2 \plus{} 1} \plus{} \frac{1}{(z^2\plus{}1) \plus{} x^2 \plus{} 1} \leq \frac{1}{2}$

[hide="attempt at #2"]We can easily find counterexamples for n=1,2,3.  Experimenting, we conjecture that n=4 works, and we try to prove it:

The conditions would be hardest to satisfy when we have an equal number of each color (lets call them red and green).  So if we can prove it for 8 green dots, and 8 red dots, then we have it proved.  (That's intuitive, but I'm not sure how to justify that, or if it even needs justification).  We see that if there is more than 1 row with 3 points with the same color then the conditions are satisfied.  So we look at the situations where there are either 1 or 0 rows with 3 green points... but I'm not sure where to go from there.

[/hide]

When I get a chance to think about it (and actually remember what the solution is), I'll come back with a comment Sean.

OK, I think I have it now, but there's probably something prettier than this:

[hide="number 2"]
First off, we can find a counterexample for n=3.  After some experimentation, we cannot find a counterexample for n=4, so we conjecture that n=4.

WLOG, call the colors blue and red.

We now consider 3 cases:
1) There is a row with exactly 4 blues.
2) There is a row with exactly 3 blues.
3) Every row has exactly 2 blues.

If any row has less than 2 blues, and does not satisfy cases 1 or 2, then we can invert blues and reds, and come up with one of the above cases by symmetry.  Thus, these 3 cases cover all possible scenarios.  If we can prove our conjecture for each one of these cases, then we will be done.

Consider the distance between two points of the same color on the same row.  If a distance found on row x can be found on any other row, then a parallelogram exists. (Condition 1)

If there are 2 rows with 3 or more points of the same color, we can find a parallelogram with all vertices the same color. (Condition 2)

So if either condition is satisfied, we are done.

[b]Case 1: There is a row x with exactly 4 blues.[/b]
If the other rows each have 1 or less blues on them, then we have at least 3 reds on each row.  Condition 2.
If One or more of the other rows has 2 or more blues on it, then the distance between these blues will be either 1, 2, or 3.  These distances all exist on row x.  Condition 1.

[b]Case 2: There is a row x with exactly 3 blues.[/b]
There exist at least 2 out of the 3 possible different distances on this row.  If both of the other rows have the third distance, then they satisfy Condition 1.  Otherwise, one of the other rows will have a distance found on row x, satisfying Condition 1.

[b]Case 3: Every row has exactly 2 blues.[/b]
In this case, each row will have exactly one distance.  There are 3 possible distances, and 4 rows, so at least one distance must be represented twice, and Condition 1 is satisfied.

Since every possible case will result in satisfaction of one of the conditions for existence of a parallelogram with all vertices of the same color, any 4x4 lattice point grid with two colors will contain at least one parallelogram with all vertices of the same color.  $ \text{QED}$
[/hide]

and now i have no idea on #3...

My number 2 with n=4 goes like this:
[hide]Note that if a row has a color on two adjacent squares, then no other row can have that color on two adjacent squares because otherwise those pairs of adjacent squares wold make a parallelogram. At the very most we have 2 rows of such in a 4x4 square (with two colors), so there are at least 2 rows with alternating colors. We can see that two of these alternating rows can always make a paralleogram, so we are done. QED.  :) [/hide]
Someone tell me if it would be a valid proof :)

Also number 3 has a flaw: x=y=z=1 produces $ \frac {3}{4} \le \frac{1}{2}$ which is false.

Both of your solutions for number 2 look correct! I can't see any flaws in them, but I also can't say what a USAMO grader would think. Both solutions are essentially the same, Sean just spelled the cases out in more detail. I don't know if Ravi's solution is explained enough by USAMO standards or not. I'll see if I can get one of the veteran Olympians to take a look and give you feedback.

I'm very sorry on number 3. That was a terrible mistake for me to make. The problem should be:

[b]Problem 3:[/b]
Let x, y, and z be positive real numbers such that $ xyz \equal{} 1$ . Show that the following inequality holds

$ \frac {1}{(x \plus{} 1)^2 \plus{} y^2 \plus{} 1} \plus{} \frac {1}{(y \plus{} 1)^2 \plus{} z^2 \plus{} 1} \plus{} \frac {1}{(z \plus{} 1)^2 \plus{} x^2 \plus{} 1} \leq \frac {1}{2}$

Very sorry for that. :oops:

By the way, the problems I've been posting come from here: http://reflections.awesomemath.org/archives.html

Problems 1, 2, and 3 are J1, J5, and J63 respectively.

J63 is something else...

EDIT: nevermind, its just not respectively  :P
EDIT2: Arggh! #3 is in the only problem set without solutions... I will try to get one tomorrow...

Note that that should have been Problems 1, 2, and 3 are J1, [b]J63[/b], and [b]J5[/b] respectively.

I think you guys' solutions to problem 2 are both good, though Anaxerzia's is surprisingly brief. Sean's solution is almost identical to the solution given on the website, so I bet his would receive a full score by USAMO standards.

My only critique of Anaxerzia's solution would be with the second sentence. It's not completely clear what configurations of colors the "2 rows of such" refers to - although it can be inferred from the conclusion of the sentence. I still think his solution is good enough for at least 6 out of 7 points and I like how it really cuts to the heart of the matter.

Anyway, here's [b]Problem 4:[/b]

Let $ a_1, a_2, a_3, \ldots, a_{2n\plus{}1}$ be distinct positive integers not exceeding $ 3n\plus{}1$. Prove that among them there are two such that $ a_i \minus{} a_j \equal{} m$ for all $ m \in \{1, 2, 3, \ldots n\}$.

Proposed by Ivan Borsenco, University of Texas at Dallas

Okay I failed at those two last ones. Can you post the solution please?  :oops:   :blush:


Wanted:The most Elegant solution gets .....ummmm.....knowledge?Forget it..just find the most elegant solution to the following problems,mostly easy ones....

1)(AIME,i think)Find $x^2+y^2$ if
xy+x+y=71,$x^2y+xy^2=880$

2)Find integer sols-
$n^4+n^3+2n^2+2n+1=m^2$

3)Find triplets of integers satisfying-
$x^3+y^3+z^3=(x+y+z)^3$

The reason i want elgant solutions is that i'm getting all by boring ol' methosds(not short ones)[/b]

You might not consider this as elegant but... after expansion, $x^3 + y^3 + z^3 = (x + y + z)^3$ is seen to be equivalent to $3(x + y)(y + z)(z + x) = 0$.

(1)
Let $S=x+y$ and $P=xy.$  Then we have \begin{eqnarray*}S+P &=& 71\\ SP &=& 880\end{eqnarray*}

So, let $S$ and $P$ be the roots of a quadratic in $w$.  By Viete's, we have this quadratic as $w^2-71w+880$ which has roots 55 and 16.

Gotta go...but the rest is fairly easy.

[quote="shadysaysurspammed"]Wanted:The most Elegant solution gets .....ummmm.....knowledge?Forget it..just find the most elegant solution to the following problems,mostly easy ones....

3)Find triplets of integers satisfying-
$x^3+y^3+z^3=(x+y+z)^3$

The reason i want elgant solutions is that i'm getting all by boring ol' methosds(not short ones)[/b][/quote]

given equation is symmetric in x, y and z. Taking any one term to right hand side and using the formulae for $a^3+b^3$ and $a^3-b^3$, it is easy to see that $x+y$, $y+z$ or $z+x$ have to be zero.  once sum of two variables is zero, the equation satisfies for any value of the remaining variable.

Number 1:

We can see by factoring that:

$xy+x+y=71$ and $xy(x+y)=880$

By letting $x+y=a$ and $xy=b$, we want to find out $a^2-2b$

So we have $a+b=71$ and $ba=880$, now the way we can solve this is by letting it be part of a quadratic equation with $b$ and $a$ as the roots, since we know that the product of the roots (or $ab$)=$\frac{b}{a}$ and the sum of the roots are $\frac{-b}{a}$, we can make the equation: $c^2-71c+880$ which we can factor to be $(c-55)(c-16)$, so we have that $c=55,16$ so $x+y=16$ and $xy=55$, thus $(x+y)^2-2xy=16^2-110=\boxed{146}$.

Or we can find out what $x$ and $y$ are individually $(5,11)$ or $(11,5)$.

Number 2:

$(n+2n)(n+1)=(m+1)(m-1)$

still working it...jk...I have no idea where do go from there :(

I think I have found quite a nice one for the first.

xy + x + y = 71
xy = 71 - (x+y)

yxx + xxy = 880

x[71 - (x+y)] + y[71 - (x+y)] = 880
-(xx + yy + 2xy) + 71(x + y) - 880 = 0

this is a quadratic in (x+y), and:

(x + y - 55)(x + y - 16) = 0
so either x + y = 55 or x + y = 16

put this into the first equation,,

if x + y = 55, then xy = 71 - (x+y) = 16
if x + y = 16, then xy = 71 - (x+y) = 55

(x + y)^2 = xx + yy + 2xy

which equals either 146 or 2993

Just currently wondering where 2993 came from....

[quote="shadysaysurspammed"]2)Find integer sols-
$n^4+n^3+2n^2+2n+1=m^2$[/quote]

$n^4+n^3+n^2+n+1+(n^2+n)=m^2\\ \frac{n^5-1}{n-1}+n(n+1)=m^2$

Nvm, this doesn't look very promising.

2:

(m+1)(m-1)=n(n+1)(n^2+2)

Setting entire factors equal:

m+1=n(n+1) and m-1=n^2+2 gives n=4 m=19
m-1=n(n+1) and m+1=n^2+2 gives n=0 m=1

This doesn't guarantee these are [i]all[/i]  the solutions... :o

[hide="2."]

Check $n= 1, 2, 3$ and we find no solutions.

We have
$\left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2 \le n^4+n^3+2n^2+2n+1 \le \left(n^2+\frac{n}{2}+1\right)^2$ for $n \ge 4$ or $n \le 0$, which can be verified by expanding.

Since there are no squares between $\left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2$ and $\left(n^2+\frac{n}{2}+1\right)^2$ (for $n$ both even and odd), our only solutions are the equality conditions:

$n^4+n^3+2n^2+2n+1 = \left(n^2+\frac{n}{2}+1\right)^2$
$0 = \frac{n^2}{4}-n = n\left(\frac{n}{4}-1\right)$,

which yields $n = 0, m = \pm 1$ and $n = 4, m = \pm 19$.

$\left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2 = n^4+n^3+2n^2+2n+1$
$0 = \frac{3}{4}n^2+\frac{3}{2}n+\frac{3}{4} = \frac{3}{4}(n+1)^2$,

which yields $n = -1, m = \pm 1$.

Hence our solutions are $(n,m) = (-1, \pm 1); (0, \pm 1); (4, \pm 19)$.

EDIT: Missed negative $m$'s.
EDIT2: Missed equality condition on the left inequality.[/hide]

I imagine he just got these out of Zeitz, but copied down $2$ incorrectly.


please solve !!!!!!!!!!!!!

solve inequalities below:

     A) sqrt(cos2x)>=0

     B) sinx+1>0

*********************
sqrt(x) = "square root of x"

[hide="A"]
$ \sqrt{\cos 2\alpha} \ge 0 \implies \cos2\alpha \ge 0$

First solve \[ \cos 2\alpha \equal{}0 \implies x_1\equal{}\frac{\pi}{4}\plus{}k\pi, x_2\equal{}\minus{}\frac{\pi}{4}\plus{}k\pi\]
Therefore \[ x \in [\minus{}\frac{\pi}{4}\plus{}k\pi,\frac{\pi}{4}\plus{}k\pi]\][/hide]
[hide="B"] $ \sin x > \minus{}1$

First solve \[ \sin x \equal{} \minus{}1 \implies x_1\equal{}\minus{} \frac{\pi}{2} \plus{} 2k\pi\]  ,and \[ x_2\equal{}\pi\minus{}(\minus{}\frac{\pi}{2})\equal{}\frac{3\pi}{2}\plus{} 2k \pi\]
Therefore $ \sin x > \minus{}1$ has solution in interval \[ x \in (\minus{}\frac{\pi}{2}\plus{} 2k \pi,\frac{3\pi}{2}\plus{} 2k \pi )\] [/hide]


[quote="chess64"]Find all positive integers $x, y$ such that $7^x - 3^y = 4$.[/quote]

This was posted at
http://www.mathlinks.ro/Forum/viewtopic.php?t=66637
http://www.mathlinks.ro/Forum/viewtopic.php?t=1220
http://www.mathlinks.ro/Forum/viewtopic.php?t=41187
http://www.mathlinks.ro/Forum/viewtopic.php?t=48431
http://www.mathlinks.ro/Forum/viewtopic.php?t=3864

I haven't checked which of these threads gives a correct solution, but it looks like not all of them do.

  darij


Find all polynomials $f(x)$ with real coefficients satisfying
\[f(x^{2})=f(x)f(x-1) \quad \forall x \in \mathbb R.\]

[quote="M4RI0"]Find all real polynomials $f(x)$ satisfying

$f(x^{2})=f(x)f(x-1)$ for all $x$[/quote]

Nice question.
Let first consider that f is a complex polynomial.
If $z$ is a complex root, then $z^{2}$ is also a root, and so is $z^{2^{n}}$ $\forall n$. So, it must exist $p$ and $q$ such that  $z^{2^{p}}=z^{2^{q}}$  and $z$ is either $0$, either $e^{\frac{2\pi}{n}i}$.

If $z$ is a complex root, $(z+1)^{2}$ is also a root, so $0$ can't be a root (since $1$ would be a root, and then $4, 25, ...$), and $(z+1)^{2}$ must also be some $e^{\frac{2\pi}{k}i}$,

But the only complex on the unit circle such that $(z+1)$ be also on the unit circle are $j=e^{\frac{2\pi}{3}i}$ and $j^{2}=e^{-\frac{2\pi}{3}i}$.

So the only possible roots are $j$ and $j^{2}$ and we must have $f(x)=a(x-j)^{p}(x-j^{2})^{q}$

Since we are looking for real polynomials, $p=q$ and we must have $f(x)=(x^{2}+x+1)^{p}$ or $f(x)=0$

And it is easy to check that this necessary condition works.

--
Patrick

This is another idea http://www.mathlinks.ro/Forum/viewtopic.php?t=132186  :wink:


function
integration
topology
real analysis
real analysis unsolved

Let $C_{0}(0,1)$ be a set of compactly supported continuous functions. Let the norm in $C_{0}(0,1)$ defined by
\[\parallel f\parallel =\text{max}_{x\in(0,1)}|f(x)|. \]
Define $L: C_{0}(0,1)\to\mathbb{R}$ by
\[L(f)=\int_{0}^{1}\frac{f(x)}{x^{2}}\,dx. \]
Is $L(f)$ defined? If defined, is $L(f)$ always continuous? Show that $L(f)$ is not always bounded.

Yes, no, and no. Let $f_{n}(x)=1/\sqrt{n}$ on the interval $[2/n,3/n]$, let $f_{n}(x)=0$ outside of $[1/n,4/n]$, and let $f$ be affine in between. Then $f_{n}\to 0$ uniformly, but $L(f_{n})\ge \sqrt{n}/9$.

What is an example of such non-continuos function?

I don't understand your question. You wanted the functions to be continuous on $[0,1]$, and now you ask for a non-continuous example?  :huh:

Please forgive me. I asked if $L(f)$ always continuous and you said no. I was looking a (continuous) function $f\in C_{0}(0,1)$ such that $L(f)$ is non-continous.

The value of $L(f)$ isn't a function; it's a scalar. $L$ is a linear functional on $C_{0}(0,1),$ meaning that it is a linear mapping from $C_{0}(0,1)$ into $\mathbb{R}.$

Asking whether $L(f)$ is continuous is a nonsense question; that's simply an element of $\mathbb{R}.$ The interesting question about continuity is to ask whether $L$ is a continuous operator. The answer to that depends on what topology you put on the domain. Use the uniform norm, and mlok has answered the question: no, it isn't. To be exact, it fails to be bounded linear operator, which means that it is not uniformly continuous. (This means its definition cannot be extended to all of $C[0,1].$)

Ok. Thanks for an explanation. Is there some good book or lecture notes where I can study the basics of linear operators?

Try Dunford and Schwartz, "Linear operators I: General theory".


[quote="Drunken_Math"][quote="pascal_1623"]The weight of the tea in a box of 100 identical tea bags is 8 ounces. What is the weight in ounces of the tea in 3 tea bags?[/quote]

[hide]If 100 tea bags weigh 8 ounces, then 1 tea bag weighs 0.08 ounces. So 3 tea bags should weight 0.24 ounces.[/hide]

Correct?[/quote]
Yes.  :)


function
logarithms
limit
inequalities unsolved
inequalities

very easy (did I miss something?)

 for $k=1$, we have to show $n^n\ge(n+1)^{n-1}$ <=>$n^{\frac1{n-1}}\ge(n+1)^\frac1n$.
Now it is easy to show that the function $f(x)=\frac{\ln x}{x-1}$ is decreasing for $x\ge0$ (note $\lim_{x\to 1} f(x)=1$, no singularity), and so is $e^{f(x)}=x^{\frac1{x-1}}$ as well.

For natural $k>1$ we have by this assumption $n{}^k{}^n\ge((n+1){}^k){}^{n-1}=\left(n^k+{k\choose 1}n^{k-1}+...+1\right)^{n-1} \ge (n^k+n^{k-1}+...+1)^{n-1}$. :)


find all function $f:\mathbb{N}\rightarrow \mathbb{R}$ such that
1. $f(x+22)=f(x)$ for all $x \in \mathbb N$, and
2. $f(x^2y)=f(f(x))^2 \cdot f(y)$ for all $x,y \in \mathbb N$.

[quote="Mathx"]find all function $ f: \mathbb{N}\rightarrow \mathbb{R}$ such that:

1.$ f(x \plus{} 22) \equal{} f(x)$
2.$ \forall x,y \in \mathbb{N}$   $ f(x^2y) \equal{} f(f(x))^2.f(y)$[/quote]

First notice that, although $ f(x)$ is stated $ f: \mathbb{N}\rightarrow \mathbb{R}$, we need $ f: \mathbb{N}\rightarrow \mathbb{N}$ in order to have $ f(f(x))$ defined in the equation in point 2.

Then, notice that $ f(x \plus{} 22) \equal{} f(x)$ implies that $ f(n)$ take at most $ 22$ different values when $ n\in\mathbb{N}$

Now, since $ f(x)\in\mathbb{N}$, $ f(f(x))^2\in\mathbb{N}$. Assume that $ \exists x_0\in\mathbb{N}$ such that $ f(f(x_0))>1$. If $ x_0\equal{}1$, take $ x_0\equal{}23$ (since $ f(23)\equal{}f(1)$). So we can consider $ x_0>1$. Then $ f(x_0^2y)>f(y)$, but also $ f(x_0^4y)>f(x_0^2y)>f(y)$ and we can easily find more that $ 22$ different values for $ f(x)$, which is impossible.

So $ f(f(x))\equal{}1$ $ \forall x\in\mathbb{N}$ and $ f(x^2y)\equal{}f(y)$ $ \forall x,y\in\mathbb{N}$

Let then $ x>y\in\mathbb{N}$ :
$ f(22^2x)\equal{}f(x)$
$ f(22^2y)\equal{}f(y)$
$ f(22^2x)\equal{}f(22^2y\plus{}22k)$ with $ k\equal{}22(x\minus{}y)$ and so $ f(22^x)\equal{}f(22^2y)$ and so $ f(x)\equal{}f(y)$ and so $ f(x)\equal{}c$ and so $ f(x)\equal{}1$ (since $ f(f(x))\equal{}1$)

And the only solution to these equations is $ f(x)\equal{}1$ $ \forall x\in\mathbb{N}$

This isn't a problem from APMO 2002, in APMO 2002 equational problem is http://www.mathlinks.ro/Forum/viewtopic.php?p=474887#474887 .

Actually it is similar to APMO 2003 - Problem 7
The conditions are the same but $ f(x^2y)\equal{}(f(x))^2f(y)$


function
algebra
floor function
number theory unsolved
number theory

In a sequence $ (x_n)_{n \equal{} 1}^{\infty}$, the first term $ x_1$ is a rational number greater than $ 1$. For all $ n\geq 1$, $ x_{n \plus{} 1} \equal{} x_n \plus{} \frac {1}{x'_n}$, where $ x'_n$ is the floor function of $ x_n$. Prove that for some $ i$, $ x_i$ is an integer.

let $ (a_{n})$ be the the increasing sequence of all naturals $ k$ for which $ x_{k}\leq [x_{k \plus{} 1}]$

Of course $ [x_{a_{t \plus{} 1} \plus{} 1}] \equal{} [x_{a_{t} \plus{} 1}] \plus{} 1$

Then let $ [x_{a_{t} \plus{} 1}] \minus{} x_{a_{t}} \equal{} \frac {p_{t}}{q_{t}}$
Every rational number can be written as follows:
$ \frac {p_{1}}{q_{1}} \equal{} \Sigma _{0 < j < n \plus{} 1} \frac {1}{b_{j}}$ for some $ n$ and $ b_{1} \leq ...\leq b_{n}$
and $ \frac {p_{1}}{q_{1}} < \frac {1}{[x_{a_{1}}]}$

there are 2 cases in progress of sequence $ \frac {p_{t}}{q_{t}}$ :
1.
 There don't exist such $ j$ that $ b_{j} \equal{} [x_{a_{t}}]$
 then $ \frac {p_{t}}{q_{t}} \equal{} \frac {p_{t \minus{} 1}}{q_{t \minus{} 1}}$

2. $ b_{j} \equal{} [x_{a_{t}}]$ for some $ j$
    then $ \frac {p_{t}}{q_{t}} \equal{} \frac {p_{t \minus{} 1}}{q_{t \minus{} 1}} \minus{} \frac {1}{b_{j}}$

All in all $ (\frac {p_{n}}{q_{n}})$ converges to $ 0$, so for some natural $ n$   ,$ x_{n}$ is integer.

[quote="Imp"]

Of course $ [x_{a_{t \plus{} 1} \plus{} 1}] \equal{} [x_{a_{t} \plus{} 1}] \plus{} 1$

[/quote]

Why is this?

suppose that difference is bigger than $ 1$. Then exist such $ k$ that $ [x_{a_{t} \plus{} 1}] < [x_{k \plus{} 1}] < [x_{a_{t \plus{} 1} \plus{} 1}]$
But it means that exist that $ h$ which holds condition $ x_{h}\leq [x_{h \plus{} 1}]$ , so should be in the $ (a_{n})$ sequence, contradiction


Flash must carry 18 pounds across a bridge.  Flash takes t(w) = w^3-4w^2+10w seconds to cross the bridge while carrying weight w.
(So when he crosses the bridge to get his next load, he crosses instantly.) Flash can carry any positive, real amount of weight.  What is the minimum amount of time it takes him to move the 18 pounds across the bridge?

This works if you assume he makes every trip with the same amount of weight ( I think this will be the min regardless but I'm not sure how to prove it)
[hide]
We are seeking to minimize the function $T(w)$
$T(w) = \frac{18}{w}(w^{3}-4w^{2}+10) = 18(w^{2}-4w+10)$
Using calculus:
$\frac{d}{dw}= 18(2w-4) = 36(w-2)$.
The minimum occurs at  $w = 2$ and will take $108$ seconds. [/hide]

[quote="t0rajir0u"]If you don't make the assumption that each trip uses the same amount of weight, then the problem seeks to minimize the expression

$(w_{1}^{3}+w_{2}^{3}+...+w_{n}^{3})-4(w_{1}^{2}+w_{2}^{2}+...+w_{n}^{2})+10(w_{1}+w_{2}+...+w_{n})$

Given that $w_{1}+w_{2}+...+w_{n}= 18$.  How can this be done?[/quote]
Well equiv to minimize \[(w_{1}^{3}+w_{2}^{3}+...+w_{n}^{3})-4(w_{1}^{2}+w_{2}^{2}+...+w_{n}^{2})\]
Now consider $f(x)=x^{3}-4x^{2}$ and $f''(x)=6x-8$ has ip at $4/3$. Now Jensen's etc. or just boundary conditions.
All problems of this form can be solved with Vasc's method (LCF/RCF theorems) (even those with several ip's in the interval)

[quote="knexpert"][hide="Hint"]
Perhaps the Power Mean Inequality?
[/hide][/quote]

Problematic because the coefficient of the squared terms is negative.  Also problematic because $n$ varies.  Jensen's seems like it could work, but, [b]me@home[/b], could you elaborate on how Jensen's would consider the inflection?

I had this in mind:

[hide]Consider $T(w) = \frac{t(w)}{w}$  This represents time per pound, so the smaller this function is, the better for Flash.
We want the minimum of $T(w) = w^{2}-4w+10 = (w-2)^{2}+6$, so the minimum is when $w=2$
2 divides 18, so the best strategy is to take 2 pounds 9 times, for a total of 108 seconds.[/hide]

It's not enough.  You're still assuming that he takes the same amount of weight each trip.  And what would happen if the minimal value of $w$ did not divide $18$?

I'm not assuming that.

[hide]

$T(w) = \frac{t(w)}{w}$ represents time per pound. Regardless of how Flash carries the weight, he wants T(w) to be as small as possible. So the best case scenario is if the value of w that minimizes T(w) is taken every time.
 I'm not assuming that he takes the same weight every time here, I'm just finding the minimum value of T(w), and hoping that the w that gives that minimum value divides 18.

Since w=2 does divide 18, and this minimizes his time per pound, Flash takes 2 pounds each time.

(I made the problem starting from the minimum value, and then picked 18 because it is divisible by 2)

If there were 19 pounds, the problem would be a lot harder.[/hide]

Fair enough.  But who wants to examine the case where $2 \not | \sum w_{i}$?


I'm having more troubles...  :(

1) Ex Pert calculated that if three times as many people worked twice as fast as the twelve people doing the work now, the job would be finished in eight days. Instead, Ex hired only four new people and all sixteen worked at the same rate as the original twelve. How many days did it take to finish the job?
[hide]My answer (Which I know is incorrect because it was a decimal at first): 10[/hide]

2) The sum of $ n$ consecutive positive integers is 100. What is the greatest possible value of $ n$?

3) The sum of three different positive unit fractions is $ \frac{6}{7}$. What is the least number that could be the sum of the denominators of these fractions?
[hide]My answer: 21[/hide]

[hide="1."]$ \frac{1}{2\cdot3\cdot12\cdot8}\equal{}\text{How much work one worker gets done in one day}$
$ \text{You get: }\frac{1}{576}$

Now, because Ex Pert has a total of 16 people working for him now, you take 576 and divide it by 16.

Hence, the answer is $ {36}$[/hide]

For number 2 I would say... is right....

2. The sum of a sequence is median x number of terms.

Let's minimize the median. If it's 1 we have 100 terms which cnat work. We want the number of terms to be less than the median. Factors of 100: 1,2,4,5,10,20,25,50,100

2-50 terms
4-25
5-20
10-10 Works, but if there is an even number of terms, then the numbers will not be integers.
20-5 YAY!

THus our answer is $ \boxed{5}$ (this is for the sequence 18,19,20,21,22)

Oh yeah, in #3 I sort of guessed. Does anyone have an explanation?

How in the world do you get 21? In any case math154 posted a very elegant and rigorous solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242776]here.[/url]

He probably got $ 21$ through $ \frac17 \plus{} \frac27 \plus{} \frac37$.  However, if you read carefully (which I should learn myself :D), you see the words unit fractions, or any fractions in the form of $ \frac1{x}$.

3. 1/2+1/3+1/42=36/42=6/7. So 47.

2. Let the first term be $ a$. Thus, if there are $ n$ terms, the sequence is $ a,a\plus{}1,a\plus{}2,a\plus{}3,\hdots,a\plus{}n\minus{}1$.

Thus, the sum is $ na\plus{}\frac{n(n\minus{}1)}{2}\equal{}100$.

Therefore, $ 2na\plus{}n(n\minus{}1)\equal{}200$.

Thus, $ 2na\plus{}n^2\minus{}n\equal{}200 \implies a\equal{}\frac{\frac{200}{n}\minus{}n\plus{}1}{2}$.

So, $ n$ is a factor of $ 200$, or $ 1,2,4,5,8,10,20,25,40,50,100,200$.

Since $ a$ is positive, we can throw out $ 200,100,50,40,25,20$.

If $ n\equal{}10$ then $ a$ is not an integer.

If $ n\equal{}5$, $ a\equal{}18$, and $ 18\plus{}19\plus{}20\plus{}21\plus{}22\equal{}100$.

Thus, 5.

Sorry to revive, but the answer to number $ 2$ is actually $ \boxed8$.

[hide="Solution"]
Note that the sum is the average of the first term $ a\in\mathbb Z^\plus{}$ and the last term $ a\plus{}(n\minus{}1),\quad n\in\mathbb Z^\plus{}$ times the number of terms, or $ \frac n2(2a\plus{}n\minus{}1)\equal{}100\implies n(2a\plus{}n\minus{}1)\equal{}200$. Note that $ (2a\plus{}n\minus{}1)\minus{}n\equal{}2a\minus{}1$ is odd, which means that $ n$ and $ 2a\plus{}n\minus{}1$ must have the same parity (i.e. even or odd). Since $ a\ge1$, $ 2a\plus{}n\minus{}1\ge n\plus{}1>n$, so we must have $ 2a\plus{}n\minus{}1$ larger than $ n$. But we are maximizing $ n$, and testing yields $ (a,2a\plus{}n\minus{}1)\equal{}(\boxed8,25)$ which gives the corresponding sequence $ 9,10,\ldots,15,16$.
[/hide]


Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the ﬁgure. She plants one flower per square foot in each region. Asters cost $ \$$1 each, begonias $ \$$1.50 each, cannas $ \$$2 each, dahlias $ \$$2.50 each, and Easter lilies $ \$$3 each. What is the least possible cost, in dollars, for her garden?
[asy]unitsize(5mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

draw((6,0)--(0,0)--(0,1)--(6,1));
draw((0,1)--(0,6)--(4,6)--(4,1));
draw((4,6)--(11,6)--(11,3)--(4,3));
draw((11,3)--(11,0)--(6,0)--(6,3));

label("1",(0,0.5),W);
label("5",(0,3.5),W);
label("3",(11,1.5),E);
label("3",(11,4.5),E);
label("4",(2,6),N);
label("7",(7.5,6),N);
label("6",(3,0),S);
label("5",(8.5,0),S);[/asy]$ \textbf{(A)}\ 108 \qquad
\textbf{(B)}\ 115 \qquad
\textbf{(C)}\ 132 \qquad
\textbf{(D)}\ 144 \qquad
\textbf{(E)}\ 156$

[hide="Click for solution"]
To minimize the cost, Rose should place the most expensive flowers in the smallest region, the next most expensive in the second smallest, etc.  The areas of the regions are, starting from the top-left region and going counter-clockwise, $ 20$, $ 6$, $ 15$, $ 21$, and $ 4$.  Thus, the minimal total cost, in dollars, is $ 3(4)\plus{}2.5(6)\plus{}2(15)\plus{}1.5(20)\plus{}1(21)\equal{}108$, which corresponds to answer choice $ \textbf{(A)}$.
[/hide]

Hope I helped!

The answer is A. 108

Since u want to minimize the cost u want to put the costliest flower in the least area possible.

So put Easter lilies in the middle 2*2 square

Dahlias in the 6*1 rectangle

Cannas in the 5*3 rectangle

Begonias in the 5*4 rectangle

and Asters in the 7*3 rectangle.

Dang just beaten to it.


The $AA_1$,$BB_1$,$CC_1$ bisectors of $\bigtriangleup ABC$ form $\bigtriangleup A_1B_1C_1$.Prove that
                               $P_{A_1B_1C_1}\le\frac{P_{ABC}}{2}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=29083

  darij


Let d be positive integer,$ t_1,t_2,...,t_d$ be integer,$\lambda_1,\lambda_2,...,\lambda_d$ be real numbers,prove that$\sum_{k=1}^{d}\lambda_k.t_k^j$ is an integer for $1\leq j< d$,then also$\sum _{k=1}^{d}\lambda_k.t_k^d$ is an integer.

A lovely problem! But I think there is a problem with the statement, since in this form it is not true for $ d=2 $. If we assume that the sum corresponding to $ j=0 $ is also an integer (which I think its the case), then I have a solution. Let the sum corresponding to $j $ be denoted $ S_j $. Let $ P_1=t_1+t_2+...+t_d,..., P_n=t_1\cdot...\cdot t_d $ be the symmetric sums in $ t_1,...,t_d $. Then the identity $ S_d=P_1\cdot S_{d-1}-P_2\cdot S_{d-2}+...+(-1)^{d-1}\cdot P_d\cdot S_0 $ is easy to show and this implies the conclusion. The identity can be proved by using the fact that $ t_1,...,t_d $ are the zeroes of $ (x-t_1)...(x-t_d) $. Hope Im not wrong again.


geometry
parallelogram
circumcircle
trigonometry
trig identities
Law of Sines
Law of Cosines

In $\triangle ABC$ $AB=3$, $BC=4$, and $\angle ABC=120$. Draw a line perpendicular to $AB$ through $A$ and a line perpendicular to $BC$ through $C$. Denote their intersection by $D$. Compute the distance $CD$.

[hide]
Draw a line perpendicular to $AB$ through $B$. This line will intersect $CD$ in point $M$. Now consider $\triangle BMC$. Clearly it is right, cause $\angle BCM=90 ^{\circ}$, and we see that $\angle MBC=120 ^{\circ} - 90 ^{\circ}=30 ^{\circ}$. $BC=4$, so $CM=\frac{4}{\sqrt{3}}$ and $BM=\frac{8}{\sqrt{3}}$.
Now draw a line perpendicular to $BC$ through $B$. This line will intersect $AD$ in point $N$. For the same reasoning we did before, $\triangle BAN$ is right and $BN=\frac{6}{\sqrt{3}}$. Now we have that $BMDC$ is a parallelogram, because $\angle BND=\angle BMD=120 ^{\circ}$ and $\angle NBM=\angle MDN= 60^{\circ}$, so $BN=MD=\frac{6}{\sqrt{3}}$, $CM=\frac{4}{\sqrt{3}}$ and
$CD=DM+MC=\frac{10}{\sqrt{3}}$[/hide]

nice solution.  :lol:  i'll show you mine now.

Consider quadrilateral $ABCD$. Since $\angle DAB$ and $\angle DCB$ are supplementary, the quadrilateral is cyclic; i.e., its vertices all lie on a unique circle. Draw this circle and denote it as circle $C$. Now draw segment $BD$, which happens to be a diameter of $C$ since it is the hypotenuse to a right triangle inscribed in $C$, $\triangle BDC$.

From here, if we can find the circumradius $R$ of the $\triangle ABC$, which is also the radius of $C$, then we can apply the pythagorean thm. to $\triangle BDC$ to find $CD$.

Let's find $R$. If $A$ denotes the are of $\triangle ABC$, then $A=\frac{1}{2}ac\sin\angle B=\frac{1}{2}(3)(4)(\sin120)=3\sqrt{3}$. Furthermore, from the the extended law of sines we know that $\frac{b}{\sin\angle B}=2R$ so that $\sin\angle B=\frac{b}{2R}$. Thus $A=\frac{1}{2}ab\sin\angle B=\frac{abc}{4R}$. Length $c$ is given by the law of cosines: $c=\sqrt{9+16-(2)(3)(4)(-\frac{1}{2})}=\sqrt{37}$. This implies that $R=\frac{abc}{4A}=\frac{(3)(4)(\sqrt{37})}{4(3)\sqrt{3}}=\sqrt{\frac{37}{3}}$. So $BD=2R=2 \sqrt{\frac{37}{3}}$. Now we can apply the pythagorean thm. to $\triangle BDC$. $DC=\sqrt{4(\frac{37}{3})-16}=\frac{10}{\sqrt{3}}$.

[quote="hydro"] $BMDC$ is a parallelogram[/quote]

It's BMD[b]N[/b] of course ! Amazing and very elementary solution ! :roll:
Thanks !

My solution:

Since $\angle DAB=\angle DCB= 90^{\circ}$, the quadrilateral $ABCD$ is inscribed in a circle of diameter DB.

Let $R$ be the radius of this circle and $O$ his center. Since $\angle ADC=60^{\circ}$, it follows that $\angle AOC=120^{\circ}$. In the triangle $AOC$ we find, by the cosine rule, that $AC=R\sqrt3$.
By the same cosine rule in triangle $ABC$, we have $AC^2=3^2+4^2-2\cdot 3\cdot 4\cdot \cos120^{\circ}$
So $AC=\sqrt{37}$
And then $R=\frac{\sqrt{37}}{\sqrt3}$.
In triangle $BCD$, by Pythagoras, we have that $DC^2=(2R)^2-4^2=\frac{100}{3}$ then
finally $DC=\frac{10}{\sqrt3}$

See the attached diagram.

Quadrilateral $ABCD$ is cyclic (with $BD$ being the diameter), hence $\angle ADC=60^\circ$.

Construct the intersection $E$ of $AB$ and $CD$. Also find the point $F$ on $BE$ such that $BF=4$.

Since $\angle CBF=60^\circ$, we have that $\triangle CBF$ is equilateral, hence $CF=4$ and $\angle FCE=30^\circ$. Thus $\triangle FCE$ is isosceles, implying $FE=4$

Hence $AE=11\implies DE={AE\over{\sqrt{3}/2}}={22\sqrt{3}\over 3}$, and from $\triangle CFE$ we have $CE=2\cdot{4\sqrt{3}\over 2}=4\sqrt{3}$, so finally $CD={22\sqrt{3}\over 3}-4\sqrt{3}={10\sqrt{3}\over 3}$.


Source: Alabama Statewide Contest

In the diagram below, Circle $O$ and $P$ are tangent. Line $AB$ is tangent to Circle $O$ and $P$ at $R$ and $S$, respectively. Ray $PA$ is tangent to Circle $O$ at $M$, and Ray $OB$ is tangent to Circle $P$ at $N$. Find the area of $\triangle ABC$, given that the radius of each circle is $3$.

[hide]$12\sqrt{3}-18$.  You can first say that $ONP$ is a 30-60-90 triangle because one leg is half the hypotenuse.  Then so is $OBR$.  So we have $OR=3$.  Thus, $RB=3\sqrt{3}$.  Thus, $AB=2RB-6=6(\sqrt{3}-1)$.  Then we find that the height of $ABC$ is $\sqrt{3}(\sqrt{3}-1)$ by constructing the altitude and finding a 30-60-90 triangle.  Thus, by multiplication, we find that the area of the triangle is $12\sqrt{3}-18$.[/hide]

Wait... why is $OBR$ a right triangle just because $ONP$ is?

Well OBR is a right triangle because BR is tangent to circle O, meaning it is perpendicular to OR

Actually, I was thinking "Why is it a 30-60-90 triangle?"

The picture doesn't make it very evident that it's a right triangle, either, but I know it is. :P

Well...
You can see that the hypotenuse is twice the radius.  And you can see that one leg is the radius.  Thus, 30-60-90.

[quote="K81o7"][quote="JesusFreak197"]I can't see that $OB$ is twice the radius... how do you know that?[/quote]
...not $OB$.  $OP$.  I think $OB$ is actually twice the radius, but then $OBP$ is isosceles.[/quote]

How do you know that OB is twice the radius. Ofcourse it very much seems like it, but how do you know for sure? Is there a way to prove that ON=RB, or that OB=OP? I guess they do look like it, but...

[quote="K81o7"][quote="Iversonfan2005"]
How do you know that OB is twice the radius. Ofcourse it very much seems like it, but how do you know for sure? Is there a way to prove that ON=RB, or that OB=OP? I guess they do look like it, but...[/quote]

Draw $ONP$.  Then you know that that is 30-60-90, and thus the length of the altitude is $\frac{3\sqrt{3}}{2}$.  Then, use similar triangles (put a point $X$ in the circle with center $P$ such that $OSX$ is 30-60-90), using the fact that the perpendicular from $S$ to $OP$ has length 3.  This yields $OB=\frac{3\sqrt{3}}{2}*\frac{2}{\sqrt{3}}=6$[/quote]

Okay thank you, I just needed proof :)


Consider an $n$ by $m$ column grid $G$ such that each element of the grid is a non-negative integer, the sum of the elements in the grid is $n,$ each element in the grid is no less than the element below it or to its right, and no two elements in a single row differ by more than $1.$ Prove that the number of distinct grids $G$ is independent of $m$ and that it is equal to the number of ways of writing $n$ as a sum of non-decreasing positive integers.

[hide="Comments"]
The solution that follows may not be entirely clear, as I wasn't able to use diagrams (and I'm too lazy to draw one and scan it over  :P ).
[/hide]

[hide="Solution"]

Let us denote a row sum of $G$ as a sum of the entries of a row of $G.$ We have two cases: $m \ge n$ and $m < n.$

[b]Case 1: $m \ge n$[/b]

[b]Lemma: All entries are either $1$ or $0.$[/b]

We will prove the lemma by contradiction. Assume that there exists a row in $G$ such that an entry, $a,$ is greater than or equal to $2.$ Since no two elements in the row differ by more than $1,$ let the other entries be $a-1.$ In this particular row, let there be $k$ $a-1$'s. If we take the sum of the elements of the row, we get:

$k(a-1)+a(m-k)=am-k \ge 2m-k \ge 2n-k \ge n$

since $n \ge k.$ Contradiction! So, all entries in the grid must either equal $1$ or $0.$ $\blacksquare$

We shall now proceed with the rest of the proof.

We find that a bijection exists between the number of distinct grids and the number of ways of writing $n$ as a sum of non-decreasing positive integers; for any grid, if we take each row sum and write these sums down in increasing order, we get the bijection. We know that a distinct grid exists per method of expressing $n$ as a sum of non-decreasing positive integers, because the rows and columns of the grid are monotonically decreasing from left to right and from top to bottom, respectively. $\blacksquare$

[b]Case 2: m < n[/b]
[b]
Lemma: Given a fixed $m$ and two fixed integers $k$ and $k-1,$ there is only one way to express a sum $P$ in terms of $m$ values of $k$ and $k-1.$[/b]

Let there be $a$ $k-1$'s. We thus have $a(n-k)+k(a-1) = P \Rightarrow P = an-k.$ Since $P$ and $k$ are fixed and $0 \le a \le m,$ there is only one distinct value of $a.$ Thus, there is only one way to express $P$ in terms of $m$ values of $k$ and $k-1.$ $\blacksquare$

If we look at each row of the $m$ by $n$ grid, we can look at each row sum as each individual component of  a way of writing $n$ as a sum of non-decreasing positive integers, we see that a bijection exists. We know that each row is distinct and each column is distinct since the rows and columns of the grid are monotonically decreasing from left to right and from top to bottom, respectively. So, we are done. $\blacksquare$
[/hide]

EDIT: Yeah, I changed the title... originally, I thought there was a glitch in the problem, because I thought that you could only use 0's and 1's and I completely forgot about the fact that 2's, 3's, etc. could be present. And I started freaking out, but I eventually figured it out  :lol:

Hey I remember doing this problem! We had it during a TJUSAMO contest. I don't understand your title though. The problem is fine. A lot of the elements in the grid are 0.


Prove that every equation:
[b](1)[/b] $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 1$
[b](2)[/b] $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 2$
has infinitely many solutions for $ x,y,z\in\textbf Z$ and $ |x|,|y|,|z| > 1$.

[hide]For example: $ 9^3\minus{}8^3\minus{}6^3\equal{}1$ and $ 7^3\minus{}6^3\minus{}5^3\equal{}2$.[/hide]

Sorry, i wanted to say that $ |x|,|y|,|z| > 1$. I'll edit problem...


geometric inequality
construction
geometry
median
Triangle
IMO Shortlist
IMO Longlist

Suppose medians $m_a$ and $m_b$ of a triangle are orthogonal. Prove that:

a.) Using medians of that triangle it is possible to construct a rectangular triangle.

b.) The following inequality: \[5(a^2+b^2-c^2) \geq 8ab,\] is valid, where $a,b$ and $c$ are side length of the given triangle.

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]  :)

Please don't understand me in the way that you think I have so much time now to spend it on trivial problems like this, but I fear that the below is the only thing I can contribute right now to Orl's nice ISL/ILL project. So here is my solution of this problem.

If ABC is our triangle, and X, Y, Z are the midpoints of its sides BC, CA, AB, then we have the vectorial identities $\overrightarrow{AX}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$, $\overrightarrow{BY}=\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}$, $\overrightarrow{CZ}=\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}$, so that

$\overrightarrow{AX}+\overrightarrow{BY}+\overrightarrow{CZ}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}+\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}+\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}$
$=\frac{\overrightarrow{BC}+\overrightarrow{CB}}{2}+\frac{\overrightarrow{CA}+\overrightarrow{AC}}{2}+\frac{\overrightarrow{AB}+\overrightarrow{BA}}{2}=\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}$.

Thus, the vectors $\overrightarrow{AX}$, $\overrightarrow{BY}$, $\overrightarrow{CZ}$ can be positioned in such a way that they form a triangle. Let DEF be this triangle, where $\overrightarrow{EF}=\overrightarrow{AX}$, $\overrightarrow{FD}=\overrightarrow{BY}$, $\overrightarrow{DE}=\overrightarrow{CZ}$. Then, we have EF || AX, FD || BY, DE || CZ, and $EF=AX=m_a$, $FD=BY=m_b$, $DE=CZ=m_c$. In other words, the sides of the triangle DEF are parallel and equal in length to the medians of triangle ABC.

Since EF || AX and FD || BY, the orthogonality of the medians $m_a=AX$ and $m_b=BY$ implies the orthogonality of the sidelines EF and FD of triangle DEF. Thus, this triangle DEF is right-angled at F. This proves part [b]a)[/b] of the problem.

Since the triangle DEF is right-angled at F, the Pythagoras theorem yields $DE^2=EF^2+FD^2$. But since $EF=m_a$, $FD=m_b$, $DE=m_c$, this rewrites as $m_c^2=m_a^2+m_b^2$, and hence $4m_c^2=4m_a^2+4m_b^2$. Using the well-known formulas $4m_a^2=2b^2+2c^2-a^2$, $4m_b^2=2c^2+2a^2-b^2$, $4m_c^2=2a^2+2b^2-c^2$, this takes the form $2a^2+2b^2-c^2=\left(2b^2+2c^2-a^2\right)+\left(2c^2+2a^2-b^2\right)$, what simplifies to $5c^2=a^2+b^2$. Hence,

$5\left(a^2+b^2-c^2\right) = 5\left(a^2+b^2\right)-5c^2 = 5\left(a^2+b^2\right)-\left(a^2+b^2\right) = 4\left(a^2+b^2\right) \geq 8ab$

from the well-known inequality $a^2+b^2 \geq 2ab$. And hence part [b]b)[/b] is also proven. Note that equality holds if and only if $a=b=\sqrt{\frac52c}$.

  Darij


Does anyone have functional equations or other functional problems about functions with two or more variables?
If you have, please post here.

hello, see here
http://math.stanford.edu/~vakil/putnam05/05putnam4.pdf
Sonnhard.

NO, that is not what I wanted. I am looking for the problems involving the functions of the form $ f(x,y), f(x,y,z),\ldots$.


integration
logarithms
calculus
calculus computations

Hi can anyone help out with this problem?

If $ \displaystyle F(s)\equal{}\mathcal{L}\{f(t)\}$, prove that $ \displaystyle\mathcal{L}\left\{\int_0^{\infty}\frac{t^u \cdot f(u)}{\Gamma (u\plus{}1)}du\right\}\equal{}\frac{F(\ln s)}{s\ln s}$

[quote="Mitochondria"]Hi can anyone help out with this problem?

If $ \displaystyle F(s) \equal{} \mathcal{L}\{f(t)\}$, prove that $ \displaystyle\mathcal{L}\left\{\int_0^{\infty}\frac {t^u \cdot f(u)}{\Gamma (u \plus{} 1)}du\right\} \equal{} \frac {F(\ln s)}{s\ln s}$[/quote]

\[ \displaystyle\mathcal{L}\left\{\int_0^{\infty}\frac {t^u \cdot f(u)}{\Gamma (u \plus{} 1)}\, du\right\} \equal{}\int_0^{\infty}e^{\minus{}st}\int_0^{\infty}\frac {t^u \cdot f(u)}{\Gamma (u \plus{} 1)}\, du dt\]
\[ \equal{}\int_0^{\infty}\frac {f(u)}{\Gamma (u \plus{} 1)}\int_0^{\infty}e^{\minus{}st}t^u \, dt du\equal{}\int_0^{\infty}\frac {f(u)}{\Gamma (u \plus{} 1)}\displaystyle\mathcal{L}\left\{t^u\right\}\, du\]
\[ \equal{} \int_0^{\infty}\frac { f(u)}{\Gamma (u \plus{} 1)}\cdot\frac { \Gamma (u \plus{} 1)}{s^{u\plus{}1}}\, du\equal{} \int_0^{\infty}\frac { f(u)}{s^{u\plus{}1}}\, du\]

\[ \equal{} \frac{1}{s}\int_0^{\infty} e^{\minus{}u\ln s}f(u)\, du \equal{} \frac{F(\ln s)}{s}\]

Are you sure about the result? I verified my result for $ f(u)\equal{}u$ and $ u^2$ with maple.


HI...

I have a few questions about Discrete Mathematics...I was thinking over nites, but i still couldn't get it solved...
I'll have a test the day after tomorrow~
Thanks for the help >"<
It's so hardddddd..........

@ Prove that for every positive n, there are n consecutive composite integers.
@ Convert 1100001100112 from its decimal expansion to its hexadicimal expansion.

Thank you guys

1.  $ (n\plus{}1)! \plus{} 2, (n\plus{}1)! \plus{} 3, (n\plus{}1)! \plus{} 4, ... (n\plus{}1)! \plus{} (n\plus{}1)$ are all composite.
2.  You mean from its binary expansion to its hexadecimal expansion?  Try thinking about how the base-$ 2$ and base-$ 2^4$ expansions of a number are related.


complex analysis
complex analysis unsolved

let $ G_{1}, G_{2}$ be simply connect regions neither of which is the whole complex plane. Let $ f$ be analytic, one-to-one, onto mapping from $ G_{1}$ to $ G_{2}$, let $ a\in G_{1}$  and $ \alpha \equal{} f(a)$. Prove that for any one-to-one analytic map $ h$ from $ G_{1}$ into $ G_{2}$ with $ h(a) \equal{} \alpha$ , prove that $ |h'(a)|\leq |f'(a)|$?

thanks

WLOG we can assume that $ G_1=G_2=\mathbb E$ and $ a=\alpha=0$, by Riemann mapping theorem. In this situation one of the crucial definition  in a classical proof of the Riemann mapping theorem is \[ \beta=\sup\{ .\mid h'(0)\mid\mid h: \mathbb E\to\mathbb E \mbox{ holomorphic and injective} \}\]Then the supremum $ \beta <\infty$ is a maximum and the proof shows that, for every injective holomorphic mapping $ h: \mathbb E\to\mathbb E$ it holds $ h$ is surjective iff $ \mid h'(0)\mid =\beta$.


find the erea of the triangle bounded by the lines

$x+y =2005$

$\frac{x}{2005}+\frac{y}{2006}=1$

$\frac{x}{2006}+\frac{y}{2005}=1$


n teams play in a football competition(win - 3 points, draw - 1 point, lose - 0 points). Each two teams plays two times between them. If P is the minimal sum of all points of the teams after all the matches and Q is the maximal sum, find P and Q.

Are P and Q sum of all points of all teams after the tournament? If yes, answer is obvious. (P-case of all draws, Q- case of all non-draws).

It's STUPIDLY easy.

If there is a tie, you add two to the addition
If else, you add three.

The minimum would be all tied, and the maximum, no game tied. so:

P=2n(n+1), Q=3n(n+1), because there are n(n+1)/2 forms for choose pairs of teams, and you multiply for two, because there are two matches between any pair of teams. so, there are n(n+1) matches, and using the argument above, you find the solution.

Lets make some variations on the theme... Here's one problem related to this, just a little bit harder. What is the least possible number of points in a league consisted of n teams (each playing two matches with all other opponents) won by the winner of the league?

BTW, this topic is definitely not Olympiad section material... more of high school math...


calculus
geometry
3D geometry
sphere
symmetry
integration
calculus computations

How do you solve this problem?

A tank is a sphere of radius 50 feet.  Calculate the depths of the water when the tank is filled to one-fourth and three-fourths of its total capacity.

(Answer:  One-fourth: 32.64 feet;  three-fourth: 67.36 feet)

I tried using area by cross-section, but I could not get the answer...

Notice by symmetry that the depth of $\frac{3}{4}$ capacity is $100$ ft - depth of $\frac{1}{4}$ capacity.

$\text{Volume of a sphere of radius, r} = \int_{-r}^{r} \pi \sqrt{r^2-x^2}^2 \, dx = \frac{4}{3} \pi r^3$

$\frac1{3} \pi r^3 = \int_{-r}^{X_p} \pi \sqrt{r^2-x^2}^2 \, dx = \pi (r^2 \cdot X_p - \frac{X_p^3}{3}+r^3-\frac{r^3}{3})$

$\frac1{3}r^3 = \frac{2}{3} r^3 + X_p r^2 - \frac{X_p^3}{3} \rightarrow X_p r^2 - \frac{X_p^3}{3} + \frac1{3}r^3 = 0$

For $r = 50$ ft.

$2500 X_p - X_p^3 + \frac{125000}{3} = 0$

Solving this should yield the correct result, for $-r < X_p < 0$

Then the particular point, $X_p$, can be used to find the depth.


trigonometry
algebra unsolved
algebra
trig identities

Calculator says it's wrong, be it degrees or radians. Please review the statement.

Excuse my typo. The problem is $tg^{2}10^{\circ}+tg^{2}50^{\circ}+tg^{2}70^{\circ}=9$

[b]General[/b]
We have: $\tan^{2}{\alpha}+\tan^{2}({60^{\circ}-\alpha})+\tan^{2}({60^{\circ}+\alpha})=9\tan^{2}{3\alpha}+6$
But my proof is not nice and who can help me?

For $\alpha : =10^{\circ}$ doesn't verify ! I wish see your solution, even if it is ... very, very ugly.

I'm sorry. You can see last edited of mine.


A derangement is a permutation in which none of the objects appear in their "natural" place.
An example:
After 4 students take a quiz, they exchange their quizzes with each other such that no person has his/her own test. The number of possible arrangements is called a derangement. In this case, there are $ 9$ such arrangements.

Prove that the number of derangements of $ n$ objects is given by
\[ n!\sum_{k \equal{} 0}^n\frac {\left( \minus{} 1\right)^k}{k!}
\]
[hide="Hint"]Use PIE - Principle of Inclusion-Exclusion[/hide]

[hide="Solution."]
=====
The total number of permutations of $ n$ objects is $ n!$. We seek to subtract from this the number of permutations such that at least one object appears in its "natural place".

Let $ P(k)$ be the number of ways to permute the $ n$ objects such that at least $ k$ objects are in their natural place. We first place some $ k$ objects and then permute the remaining $ n - k$ objects in $ (n - k)!$ ways. Thus $ P(k) = (n - k)!$.

We now use the symmetric form of PIE (AoPS 2), summing $ k$ from $ 1$ to $ n$; the total number of derangements is therefore:
\begin{eqnarray*} n! - \sum_{k = 1}^{n}( - 1)^{k + 1}\binom{n}{k}P(k) & = & n! + \sum_{k = 1}^{n}( - 1)^{k}\frac {n!}{k!(n - k)!}(n - k)! \\
& = & n! + n!\sum_{k = 1}^{n}\frac {( - 1)^{k}}{k!} \\
& = & n!\left(1 + \sum_{k = 1}^{n}\frac {( - 1)^k}{k!}\right) \\
& = & n!\sum_{k = 0}^n\frac {( - 1)^k}{k!},\end{eqnarray*}
as desired.
=====
[/hide]

I'm surprised this isn't in the Counting chapter in AoPS 2... it would've fit right into the symmetric PIE section.

And can someone explain why there's a stray $ (2)$ in my solution above? It's annoying me...

[quote="jvenezuela716"]And can someone explain why there's a stray $ (2)$ in my solution above? It's annoying me...[/quote] You have \begin{eqnarray*} but \end{eqnarray}, and so LaTeX is confused.

Ah, it's the closing \right) that has no accompanying \left.


I was wondering if AP Chemistry is difficult.  Does anyone have any opinions?  How do you get a $ 5$?  Thanks!

[quote="JRav"]You get a 5 by getting most of the questions right. :wink:

I got a 5 on it last year.  I didn't find it overwhelmingly easy but I didn't find it that hard.  If you put time into it, there's no reason you can't get a 4 or a 5.[/quote]

How did you prepare for the exam?  Thanks for your advice.

AP Chem is offered in May, and last I checked, it is June so you are preparing for next year.  Then you have plenty of time.  A good starting point would be to actually take the course.  After that, stay on top of the workload, do the homework, study for the exams and then, come April, read through a review book and do practice exams (personally, I like Princeton Review).

I'm planning on taking AP chemistry next year (as a sophmore). I heard you had twice the number of labs (compared to AP biology) and the write-ins were very strict.

I heard that Barron's was also decent for chemistry, but I didn't verify it. Opinions?

I don't know if AP Chem has "required" labs but since you are very likely to be tested on labs and lab techniques, there are a lot that you should do.  What do you mean the "write ins" were very strict?

I also have Barron's.  It is a lot bigger and has a lot more problems.  Some of the problems are nice, some are hard.  But there are 6 tests to do.  Barron's has a lot of extra stuff so if you want to be overprepared, then choose Barron's, but Princeton Review got me a 5.

Wouldn't being overprepared also get you a 5? Or would that just kill your nerves during the test?

The write-ins (for the labs) are graded to the letter by my [future] teacher. If a grammar error is spotted, significant points are deducted. If this number isn't in sig figs, many more points are deducted (and I hate sig figs). Just think what will happen when you get something [i]incorrect[/i].  :|

dude I strongly doubt they take points off for grammar

stop stressing about APs

sheesh

[quote="MysticTerminator"]dude I strongly doubt they take points off for grammar[/quote]

[i]They?[/i] You mean him, right? Since you know my AP chemistry teacher so well? You know how he grades and what he likes on labs?

yeah dude your chem teacher and I go way back

AP Chem = An easy 5 if science and math is your thing ;)

Barrons is probably the way to go for a review book in my opinion.  If you want to truly master predicting equations, then I highly recommend http://www.amazon.com/Ultimate-Chemical-Equations-Handbook/dp/1877991627

Bonne Chance!

If I go into AP Chemistry as a freshman with no background at all, how difficult do you think the course will be?

Thanks!

[quote="elcric"]If I go into AP Chemistry as a freshman with no background at all, how difficult do you think the course will be?[/quote]

Pretty difficult. I attempted AP Biology with no background and it was hard competing with the seniors that already had a standard level. But my teacher was nice and was patient with the other freshman with me when we didn't get something that we should have.

I'm doing AP chemistry next year as a sophmore, and our school requires us to take a CTY course in basic chem before we go to AP. No exceptions. I'd advise you to pick up a decent textbook (preferably from a lower level chem course) and complete ALL of it.

It's possible, but frequently more difficult since intro level courses tend to ground you in the basics, so that when you go on to more advanced stuff, you can just quickly pick up without having to worry about understanding the easier stuff.

Does anyone have any links that I can use to practice for AP Chemistry?  Thanks!

I just received my AP Scores today, and got a 5 on AP Chem :)

It's not difficult if you can pick things up fast, and if you have a really strong background in math and science.  There's simply a lot of things to juggle and you have to know how to sort these things.  If you just understand what's going on, you should be fine.  Obviously a strong background of Chemistry is needed along with some algebraic skills, and possibly Physics.

And lastly, if you are interested, you will get a good grade... I was really interested when i took the class


Both $ x$ and $ y$ are positive real numbers, and the point $ (x,y)$ lies on or above both of the lines having equations $ 2x\plus{}5y \equal{} 10$ and $ 3x\plus{}4y \equal{} 12$. What is the least possible value of $ 5x\plus{}9y$?

No matter what x and y are, to be minimized, they satisfy those two equations. $ 9x \plus{} 5y$ is their sum, or $ 10 \plus{} 12 \equal{} 22$.


[quote]Did you mean $3^{3452}$ or did you mean something else...? The question looks quite unclear.[/quote]
if it is.....
it's $1$

I'm almost sure it is $3^{3452}$, because it is highly probable that he forgot the curly brakets.
Now,
$\\3\\ 3^2=9\\ 3^3=27\\ 3^4=81\\ 3^5=243\\ \vdots$
The units digit have period 4, thus
$\\3452\equiv 0\pmod{4}\\ \implies$ The units digit of $3^{3452}$ is $\boxed{1}$.

Masoud Zargar

[hide="How to prove that the units digit cycles every 4"]
Euler proved that $a^{\phi(n)} \equiv 1 \pmod {n} \forall a \perp n$. Thus $3^{4} \equiv 1 \pmod {10}$, and raising both sides to the $k$th power, we find that $3^{4k} \equiv 1 \pmod {10}$. Similarly we find $3^{4k+r} \equiv 3^r \pmod {10}$. $\boxed {}$
[/hide]

[hide="Without using mods"]
$3^{3452}=(3^4)^{863}=81^{863}=(80+1)^{863}$
Now think about rewriting this using binomial expansion.
Apparently all the terms are multiples of $10$ except for the last term, which is $1$.
Therefore, the answer is $1$.[/hide]

[hide]3452 mod 4 = 0, so the units digit is [b]1.[/b][/hide]


Did you mean "What is [i]April's[/i] number"?  May's number is already given...  If so...
[hide="Solution?"]For April to be in the middle of the line, there must be an odd number of participants.  Since schools are represented in groups of $ 3$, the number of participants must be a multiple of $ 3$.  Let the number of people be $ n$.  The person in the middle (April in this case) has the number $ \frac {n \plus{} 1}{2}$.  We know that $ \frac {n \plus{} 1}{2} < 19\implies n\le 37$, because the middle number must be less than May's number, 19, and $ n\ge 28$, because the number of participants must be at least $ 28$, June's number.  The only whole number $ n$ that fits the criteria and is an odd multiple of $ 3$ is $ 33$.  Thus, there must be a total of $ 33$ participants, meaning $ \frac{33}{3}\equal{}\boxed{11}$ schools, and April's number is $ \frac {33 \plus{} 1}{2} \equal{} \boxed{17}$[/hide]

[quote="AndrewTom"]April, May and June are three girls representing their school at the cross-country race made up of teams of $ 3$ from different schools. When lining up to collect their numbers, April finds she is in the middle of the line. May is behind her and is $ 19$th in line; June is $ 28$th in line. What is May's number and how many schools competed?[/quote]

April is 20th, and 39 schools competed.

I'm probably misunderstanding the question though.


calculus
integration
logarithms
trigonometry
algebra proposed
algebra

hi, can someone please find the solution of ∫ ln4x + cos4x + 3 dx between and including x= 4 to 6  and
∫ ln3x+cos2x+2 dx between and including 2 to 4 so I can verify the answers I've got. Thank You , I really need someone to help me!

Thank You again :)

yes I know this, but i need someone to find the solution in numbers so i can check with what i got, thank you


pigeonhole principle
combinatorics unsolved
combinatorics

Prove that in a company of 17 people, in which each person knows exactly four others, there exist two people not  knowing each other and having no common acquaintances.

Suppose the contrary. Consider a person Alan and each of the 12 people he does not know. They each share at least one common acquaintance, and each of Alan's acquaintances (call them Alaan, Alaaan, Alaaaan, and Aln) must know exactly three of them by the problem conditions. But consider each of the 12. They must share acquaintances with every other member of the 12, but by Pigeonhole they would have no common acquaintances with at least of the other 11 if they did not know at least one of Alaan, Alaaan, Alaaaan, Aln, who in turn would know 5 or more people - contradiction.


A rectangular band formation is a formation with $ m$ band members in each of $ r$ rows, where $ m$ and $ r$ are integers. A particular band has less than 100 band members. The director arranges them in a rectangular formation and finds that he has two members left over. If he increases the number of members in each row by 1 and reduces the number of rows by 2, there are exactly enough places in the new formation for each band member. What is the largest number of members the band could have?

Setting an equation we get $2m=r.$ The product is $2mr$ meaning the largest number of members is the largest even integer under 100, which is 98.

Did you get this: [hide]$rm+2=(r-1)(m+2)$[/hide]?


farz konid N adadi tabie bashad .dar har marhale jame adad zoj on adad ra ba khode on jame mikonim sobet konin bad az chand marhale donbale sabet mishavad .masalan:
$ 12 \rightarrow 14 \rightarrow 18 \rightarrow 26 \rightarrow 34 \rightarrow 38 \rightarrow 46 \rightarrow 56 \rightarrow 62 \rightarrow 70$


If $ a,b$ are positive real numbers such that $ ab\geq1$, then

$ \frac {1}{1 \plus{} \sqrt {1 \plus{} ka^2}} \plus{} \frac {1}{1 \plus{} \sqrt {1 \plus{} kb^2}}\geq\frac {2}{1 \plus{} \sqrt {1 \plus{} kab}}$

holds if and only if $ k\geq8.$

[quote="Ji Chen"]If $ a,b$ are positive real numbers such that $ ab\geq1$, then

$ \frac {1}{1 \plus{} \sqrt {1 \plus{} ka^2}} \plus{} \frac {1}{1 \plus{} \sqrt {1 \plus{} kb^2}}\geq\frac {1}{1 \plus{} \sqrt {1 \plus{} kab}}$

holds if and only if $ k\geq8.$[/quote]
Maybe $ \frac {1}{1 \plus{} \sqrt {1 \plus{} ka^2}} \plus{} \frac {1}{1 \plus{} \sqrt {1 \plus{} kb^2}}\geq\frac {2}{1 \plus{} \sqrt {1 \plus{} kab}}$ $ ?$


prove there are no integers $x$,$y$ suh that :$x^{5}+31=y^{2}$

see that $(x+2)|y^{2}+1$

now, is easy to see that $x\equiv1(mod.4)$ (if no, we have that $y^{2}\equiv2,3(mod.4)$)

so, (x+2) have a prime factor p of form 4k+3, and this is impossible, because p can't divide y²+1


linear algebra
matrix
number theory
greatest common divisor
linear algebra unsolved

Related to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289976 I received a question about the case where $ n$ is odd. Here is one result.

Let $ n > 2$ be an odd integer, and $ A\in \mathcal{M}_n(\mathbb{Z})$.
Show that there exist $ B, C, D \in \mathrm{SL}_n(\mathbb{Z})$, s.t. $ A \equal{} B \plus{} C \plus{} D$

A partial result:
According to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289976
it is sufficient to prove the result for a diagonal $ 3\times{3}$ matrix $ A\equal{}diag(u,v,w)$ where $ u|v|w$ if they are not $ 0$ (smith normal form).
$ diag(\alpha,\beta,\gamma)\equal{}\begin{pmatrix}\alpha&\minus{}1&a\\1&0&b\\1&1&c\end{pmatrix}\plus{}\begin{pmatrix}0&1&\minus{}a\\\minus{}1&\beta&\minus{}b\\\minus{}1&\minus{}1&\gamma\minus{}c\end{pmatrix}\equal{}p\plus{}q$. $ det(p)\equal{}det(q)\equal{}1$ give $ c\minus{}b\equal{}\alpha{b}\minus{}a\plus{}1\equal{}\minus{}a\plus{}\gamma\minus{}\beta{a}\minus{}1$. Thus $ \alpha{b}\plus{}\beta{a}\equal{}\gamma\minus{}2$ must be satisfied. We can always write  $ diag(\alpha,\beta,\gamma)\equal{}p\plus{}q$ iff $ gcd(\alpha,\beta)|\gamma\minus{}2$. In particular $ gcd(\alpha,\beta)\equal{}1$ or $ \gamma\equal{}2$ is a sufficient condition.
Now $ A\equal{}diag(u\minus{}\epsilon_1,v\minus{}\epsilon_2,w\minus{}\epsilon_1\epsilon_2)\plus{}diag(\epsilon_1,\epsilon_2,\epsilon_1\epsilon_2)$ where $ \epsilon_i\equal{}\pm{1}$.
Conclusion: if at least one among the following $ 4$ relations: $ gcd(u\minus{}\epsilon_1,v\minus{}\epsilon_2)\equal{}1$ is satisfied, then we are done.


What n-somes of real numbers (a_1, a_2, ..., a_n) is the following inequality held for, if a_1*a_2*...*a_n = 1?

 _____1_____ + _____1_____ +[color=white].....[/color]+ _____1_____  \geq _n_
 a_1(a_2 + 1)[color=white].....[/color]a_2(a_3 + 1)[color=white]....[/color]......[color=white]....[/color]a_n(a_1 + 1)[color=white]......[/color]2

Does anybody have any suggestion? Please, answer.
[color=white]....[/color]

Isn't it some Cauchy-Schwarz ?

\sum 1/ai(a(i+1) + 1) *  \sum ai(a(i+1) + 1) >= n

now  \sum ai >= n and \sum ai a(i+1) >= n by AM-GM inequality.


ratio
geometry
3D geometry
sphere
rectangle

The ratio of the volume of a sphere of radius $ r$ to the volume of the largest cylinder that fits inside the sphere can be written as $ a\sqrt{b}: c$, where $ a$ and $ c$ share no common factors, and $ b$ is not divisible by the square of any prime. Find $ a^2\plus{}b^2\plus{}c^2$.

largest in terms of volume?

like if i told u to inscribe the largest possible rectangle into a circle, should u assume that i mean largest area, and thus incsribing a square.

maybe a cylinder can be inscribed? i really have no clue :blush:

[quote="mathking123"]largest in terms of volume?[/quote]
Yes, I meant largest by volume. Sorry for not making that clear.


videos
email
abstract algebra
IMO
IMO 2007

There will be many available as soon as I get back from Vietnam (uploading here is very slow).

Yeah, and they take random videos every day, do you know where we can find them after IMO ie maybe download, etc.? :maybe:

http://www.imo2007.edu.vn/images/Category/Category_1629DSCN0765.jpg

Pretty, the girl who doesn't speak  :D Does someone has her yahoo email ? Haha  :roll:

 :arrow:

Here is a link to a collection of video clips and pictures on IMO'2007:
http://ultrametric.googlepages.com/imo2007

Can anybody upload the contents of the DVD which was given for each team at the end of IMO?
In fact can anybody tell what was there on that DVD?

Thanks.

Interesting... The Luxembourg team was not given any DVD at the end of the imo. Should the DVD have been provided by the guide (for this would account  for an explanation)?

There is a small collection of "IMO 2007 pictures" here: http://www.mathlinks.ro/weblog.php?w=872  :D

I have created an album in our galleries for this: http://www.mathlinks.ro/album.php?t=161568
Of course it's a good idea to upload only relevant images.

Here :

http://www.imo2007.edu.vn/index.php?module=ViewPhoto&NewId=60


geometry
geometric transformation
rotation

Suppose in each day on a certain planet, there are only 10 hours and
every hour has 100 minutes. What is the measure, in degrees, of the
acute angle formed by the hour hand and minute hand at 6 o’clock 75
minutes?

I gotta go..(I'm posting this in school..)
But I'll give an info. before I go.

[hide]Hour hand creates = $ {\frac12}^{\circ}$per minute.
Minute hand creates = $ {6}^{\circ}$per minute.[/hide]

[quote="abacadaea"]unlike a normal clock, one rotation is 1000 minutes. so we can express its as a percent.

[hide]the hour hand is %67.5 of the way around a clock.
the minute hand is %75. 75-67.5=7.5.
%7.5 of 360 is [b]27[/b][/hide][/quote]

I haven't thought about that way.. Nice.


invariant
linear algebra
linear algebra unsolved

Let $ A$ be a family of $ n\times n$ complex matrices and assume that $ {0}$ and $ \mathbb{C}^n$ are the only $ A$-invariant subspaces of $ \mathbb{C}^n$. Then the algebra generated by these matrices is $ M_n\left(\mathbb{C}\right)$.

as far as I know this theorem is due to Burnside and for infinite dimensional version you can find in "Invariant subspaces"Springer-Verlag.for this version the idea is to prove that there is transformation with rank 1 and then prove that all of them exists in algebra .If you wish I will send complete solution.

I imagine Burnside found that as a refinement of Molien's theorem, since the last one is quite old. But anyway, the only proof for Burnside's theorem I saw in the same written exam at X, but it was absolutely abstract and very long and tedious. So, if you have some other nicer proof, please post it, sam-n. Thanks.

according to alekk's private message I enliven this old thread :)
assume that $\cal{A}$ is the algebra generated by $A$.note that $x\neq 0$ implies that $\{Cx: C\in \cal{A}\}\= \mathbb{C}^n $.we begin by showing that $\cal{A}$contains some $T_0$ of rank $1$.To see this ,choose $T_0\in \cal{A}$with minimal nonzero rank,say $d$.if $d>1$,choose $x_1$ and $x_2$ so that $\{ T_0x_1,T_0x_2\}\ $ is linearly independent ,and then choose $C\in \cal{A}$ so that CT_0x_1=x_2$ then T_0CT_0x_1$ and $T_0x_1$ are linearly independent.and $T_0CT_0-\alpha T_0$ is different from $0$ for all scalars $\alpha$.since the field of scalars is algebraically closed,there is a $\alpha_0$ such tha restriction of $T_0C-\alpha_0$ to $T_0(\mathbb{C}^n)$ is not invertible .then $(T_0C-\alpha_0)T_0$ has rank less than $d$.cotradiction implise $d=1$.every linear transformation on $\mathbb{C}^n$ is sum of linear transformation of rank $1$;so to finish we should show the algebra contains all transformation of rank $1$.we know there exists nonzero transformation $T_0$ such that $T_0x=\phi_0(x)y_0$for $y\in \mathbb{C}^n$ $\phi(x)$ is linear functional.now the se set of all linear functinal $\phi$ such that $Tx=\phi(x)y_0$ such that $T\in cal{A}$ includes $\phi(Cx)$ for each $C\in \mathbb{C}^n$.
thus every $T$ of the form $Tx=\phi(x)y_0$ for some $\phi$ is in $cal{A}$.for any $y$ and $\phi$ if $Tx=\phi(x)y_0$ and if $C\in\cal{A}$ si chosen so that $Cy_0=y$.then $CTx=\phi(x)y$ is chosen to be in $\cal{A}$.we are done .for infinite dimentional version of Burnside theorem refer to ,Invariant Sunbspaces,Heydar Radjavi and Peter Rosenthal.

thanx Sam-n, I will carefully read your proof  :)


At the conclusion of a party, a total of 28 handshakes was exchanged was exchanged. Assuming that each guest was equally polite toward all the others, that is, each guest shook hands with each of the others, find the number of guests,n, at the party

[hide]$ \binom82=28$, meaning that $ n=8$

>_>[/hide]


See picture. Square ABCD is a unit square, and E,F,G,and H are midpoints of the 4 sides. The product of the lengths of the three highlighted segments can be expressed as $\sqrt{\frac ab}$, where a and b are relatively prime positive integers. Find $a+b$.

The length of the green segment is:
\[ \sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{\sqrt{2}}{2} \]
The length of the red segment plus the half of the length of green segment equal to the length of the diagonal, therefore:
\[ l_{green}= \sqrt{2} - \frac{\sqrt{2}}{4}=\frac{3}{4} \sqrt{2} \]
Let the length of the blue segment be $x$, then we have:
\[ x^2+4x^2=1 \]
\[ x=\frac{\sqrt{5}}{5} \]
So the product of the three segments are:
\[ \frac{3\sqrt{5}}{20} \]
Therefore $a+b=65$.

But it's actually all of $\sqrt{\frac{a}{b}}$ so you would get $\frac{3\sqrt{5}}{20} = \sqrt{\frac{45}{400}} \Rightarrow 45+400 = 445$ but $45$ and $400$ are not relatively prime like the problem asks for.

I think that  shobber computed the length of the blue segment incorrectly.
The blue segment is really composed of 2 parts, part to the left and part to the right of the intersection with the blue and red line. Using some basic geometry, we see that the blueline is actually just half BH. So, the length of the blue line is: $\frac{1}{2}\sqrt{\frac{1}{4}+1} = \frac{\sqrt{5}}{4}$

So, the product will be $\sqrt{\frac{180}{1024}} = \sqrt{\frac{45}{256}}$. So, $a + b = 45 + 256 = 301$

[quote="Hashbrownie"]the blueline is actually just half BH.[/quote]
 :(

For the blue line, perhaps it would help to draw FD?  :)

AE=EB, AG||CG.
Now, do you notice your mistake?


not induction! I am not allowed to use that! forgot to add that!
 :(  sorry!

I think that this is the cure for the itch! :D

Let a1,a2,a3,.. be an arithmetical progression of numbers! you can show that:
a1/a2+a1*a3/(a2*a4)+..+a1*a3*..*a(2n+1)/(a2*a4*..*a(2n))=1/r(a1*a3*..*a(2n+1)/(a2*a4*..*a(2n))-a1)

then apply it to your problem! :D
tschuess! bye!

okay.. here is a proof of the identity I gave:

we know that a(2k+1)-a(2k)=r then the general term of the sum can be written like so:
a1*a3*..*a(2k-1)/(a2*a4*..*a(2k))=1/r* a1*a3*..*a(2k-1)(a(2k+1)-a(2k))/(a2*a4*..*a(2k))=1/r*
a1*a3*..*a(2k+1)/(a2*a4*..*a(2k))-1/r* a1*a3*..*a(2k-1)/(a2*a4*..*a(2k-2)).

now we give values to k from 2 to n and we add the identities we get with the following thing a1/a2=1/r (a1a3/a2 - a1) and we get what we want!

hope it's true!
bye! :D

that is wonderful ... where did u get that?
10x a lot!

well here are other identities that may help you some day in solving problems like these!

if (an)n>=1 is an arithmetical progression of natural numbers <>0 and the ratio r<>0 we have:
1) a1a3/a2+a1a3a5/(a2a4)+..+a1a3..a(2n+1)/(a2a4..a(2n))=1/3r *(a1a3..a(2n+3)/(a2a4..a(2n))-a1a3)

2) a1/a2a4+a1a3/(a2a4a6)+..+a1a3..a(2n-1)/(a2a4..a(2n+2))=1/r*
(a1/a2-a1a3..a(2n+1)/(a2a4..a(2n+2)).

okay! the proof of these two is simillar to the other one!
if you have other's please post them here!
bye!


geometry
circumcircle
projective geometry

Let $\triangle{ABC}$ be an acute angled triangle. The circle with diameter AB intersects the sides AC and BC at points E and F respectively. The tangents drawn to the circle through E and F intersect at P.
Show that P lies on the altitude through the vertex C.

Connect $OE$ and $OF$. So $\angle{OEA}=\angle{OAE}$, $\angle{OFB}=\angle{OBF}$. And $\angle{OEP}=\angle{OFP}=90^{o}$.
Since the sum of the angles of a pentagon is $540^{o}$, so:
 \[\angle{A}+\angle{B}+\angle{BFO}+\angle{OFP}+\angle{FPE}+\angle{PEO}+\angle{OEA}=540^o\]
Therefore: $2\angle{A}+2\angle{B}+\angle{FPE}=360^{o}$
Since $2\angle{A}+2\angle{B}+2\angle{C}=360^o$, so $\angle{FPE}=2\angle{C}$.

Now since $PE$ and $PF$ are tangents, so $PE=PF$. Therefore $P$ is the orthocentre of $\triangle{CEF}$.
Hence: $\angle{A}+\angle{ACP}=\angle{OEA}+\angle{CEP}=90^{o}$, which means $CP \perp AB$.

[quote="marko avila"]P  is the circumcenter not the orthocenter  ;)[/quote]
It was a typo... Thanks for pointing it out!  :)

After two years...  :)

Let $H$ be the foot of altitude from $C$.
Since $PE = PF$, $\measuredangle PEF = \measuredangle PFE$. $\measuredangle CEP = 90^{\circ}-\measuredangle CAH = \measuredangle HCE$. Also $\measuredangle CFP = \measuredangle HCF$.
Consider the cevians $CH, FP, EP$ of $\Delta CEF$. By the trigonometric Ceva they concur. Hence $P$ lies on $CH$.

Or very short with projective geometry:

Call $EF \cap AB = {G}$. Now, the polar of $G$ goes through $C$ and is perpendicular to $AB$, so it is the altitude through $C$.
Also, we know that $EF$ is the polar of $P$, and $G$ lies on it. So $P$ lies on the polar of $G$, $\mathbb{QED}$  :D

Denote $H=AF \cap BE$, of course, $H$ is the ortochenter of $\triangle{ABC}$. Using Pascal's Theorem for the cyclic hexagon $AEEBFF$ we get $C-P-H$ collinear, qed.

[quote][b][u]PP[/u].[/b] Let $\triangle{ABC}$ be an acute triangle with the orthocenter $H$ . The circle $w$ with the diameter $[BC]$  intersects $AC$ and $AB$ at points $E$
and $F$ respectively, i.e. $H\in BE\cap CF$ . The tangents drawn to the circle $w$ through $E$ and $F$ intersect at $P$ . Show that $P\in AH$ .[/quote]
[b][u]Proof[/u].[/b] Denote the midpoints $M$ , $N$ of the segments $[BC]$ , $[AH]$ respectively. $\{\begin{array}{ccc}
NF=NH & \implies & m(\widehat{NFC})=B\\\
NE=NH & \implies & m(\widehat{NEB})=C\end{array}$ .

Since  $\{\begin{array}{c}
m(\widehat{PFC})=B\\\
m(\widehat{PEB})=C\end{array}$ obtain that  $P\in FN\cap EN\implies$ $P\equiv N\in AH$ .

$ABFE$ is cyclic, $\widehat{FEC}=\widehat{ABC}$. Easy angle chasing gives $\widehat{EOF}=180^\circ- 2\cdot m(\angle BCA)$, i.e. $EO, FO$ are tangent to circle $\odot (CEF)$.
$PEOF$ cyclic gives $\widehat{PEF}=\widehat{PFE}=\frac{\widehat{FEO}}{2}=90^\circ-m(\angle ACB)$, hence $P$ is the circumcenter of $\odot (CEF)$, or $\widehat{FCP}=90^\circ-\widehat{CEF}=90^\circ-\widehat{ABC}$, $\equiv CP\bot AB$.

Best regards,
sunken rock

Clearly $BE$ and $CF$ are the altitudes of triangle $ABC$.suppose the common point of altitudes is $H$.Again $CEHF$ is cyclic with diameter $CH$,this is enough to prove that $P$ is the circumcenter of $CEHF$,now supposing $X$ is the midpoint of $CH$, and extended $XE$ to $G$.now $XE=XC$.......(1)
${\angle}ABE={\angle}AFE$,${\angle}HFE={\angle}HCE$,now using (1) we get ${\angle}AEG={\angle}ABE$,so $XG$ is tangent to circumcircle of $EABF$,hence $X$ and $P$ are not distinct,hence we are done.


Be ABC triangle with incircle (I,r) and circumcircle (O,R). A external tangent circle opposite A, touch circumcircle and AB,AC at A1,D,E, similarly the e.t.c.o. B, touch circumcircle and AB,BC at B1,F,G also the e.t.c.o. C, touch circumcircle
and AC,BC at C1,M,N. The lines DE,FG,MN perform the triangle XYZ.
Now:
1) I is the circumcenter of XYZ.
2) [XYZ]/[ABC]=8R/r (where: [XYZ]=area XYZ)

Just the first part:

It can be shown, by applying Casey's Theorem, that, for example, $I_a$, the $a$-excenter of $ABC$, is the midpoint of $DE$ (this result is analogous the the one stating that if the mixtilinear circle corresponding to vertex $A$ touches $AB,AC$ in $U,V$ respectively, then $I$ is the midpoint of $UV$), and I'll use this without proving it. Certainly, the mixtilinear circles have been discussed lots of times on ML :).

Let $X=FG\cap MN$, and let $I^a$ be the projection of $I$ on $BC$. $ABC$ is the orthic triangle of $I_aI_bI_c$, so $BCI_bI_c$ is cyclic, meaning that $\angle IBC=\angle II_cI_b=\angle IXI_b\ (1)$, the last equality following from the fact that the previous paragraph implies $I_bX\perp I_bB,\ I_cX\perp I_cC$, so $II_bI_cX$ is cyclic. At the same time, we have $\angle(II^a,I_bX)=\angle IBC\ (2)$, because $II^a\perp BC,\ I_bX\perp IB$. $(1)$ and $(2)$ now imply that $II^a=IX$.

If $I^b,I^c$ are the projections of $I$ on $CA,AB$ respectively, then analogous results to the one above show that $I$ is the intersection between $XI^a,YI^b,ZI^c$, and since $XYZ,I^aI^bI^c$ are homothetic (their sides are parallel) and $I$ is the circumcenter of $I^aI^bI^c$, it must also be the circumcenter of $XYZ$, as desired.

Dear Juan Carlos
You could ad to 1) that the circumcircle of XYZ is centered at I and radius 4R.
In fact, XYZ is homothetic of the contact triangle of the incircle in (I, -4R/r).

Dear friends:
Thanks for your remarks. You can see the another solution:
http://www.personal.us.es/rbarroso/trianguloscabri/sol/sol239sal.htm
Greetings.

[quote="jpe"]Dear Juan Carlos
You could ad to 1) that the circumcircle of XYZ is centered at I and radius 4R.
[/quote]

This follows from the fact that the circumradius of $XYZ$ is $IX$, which is also the diameter of $(II_bI_c)$, which is, in turn, the diameter of the circumcircle of $I_aI_bI_c$, since $I$ is the orthocenter of $I_aI_bI_c$. This diameter must be twice the diameter of the circumcircle of $ABC$ (and hence $4R$), since $ABC$ is the orthic triangle of $I_aI_bI_c$, so it's circumcircle is the nine-point circle of $I_aI_bI_c$.


In triangle $ABC$, angle $C$ is obtuse.  The bisectors of the exterior angles at $A$ and $B$ meet $BC$ and $AC$ extended at $D$ and $E$ respectively. If $AB=AD=BE$, prove that $\angle ACB =108$

This question is really weird, I keep on getting $\angle ACB = 36$

Are you sure its $AB=AD=BE$?  This would make $\angle C$ acute, like you found, since $AB<AC \wedge AB<BC$.

Yeah, thats what the book says...thats why I think there is a typo in the book  :maybe:


What are some good math topics to look into for ISEF?

Truly independent math research is difficult at the high school level.  Rather than ask us for help finding a topic, you're better off contacting a professor at a local university and asking him for general guidance, which includes finding a topic.


inequalities
limit
inequalities proposed

This is Carleman's inequality. It's also problem 6 from second day of 1996 IMC

[hide="Not easy!"]This is a simply case of Theorem 1.2 in

http://www.emis.de/journals/JIPAM/images/165_06_JIPAM/165_06_www.pdf[/hide]


How many positive integers less than or equal to 100 have a prime factor that is greater than 4?

[hide="Solution"]Integers that have only $ 2$'s and $ 3$'s in their prime factorizations (i.e. don't have any prime factors greater than four) are of the form $ 2^a3^b$.

There are $ 20$ of these numbers (I won't list them out), so the answer is $ 100\minus{}20\equal{}\boxed{80}$.[/hide]

hello, No... Anyway, I used this.

[hide="Solution"]Numbers with 5 as a factor: 20
7:14
11: 9
13: 7
...
97:1

Total=88

5,7 link: -2
5, 11 link: -1
...
5:19 link: -1
7, 11 lknk: -1
7, 13 link: -1
Total=-8

88-8=80, Angry_Bird_010112[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Hide your solutions[/url].[/color]

[quote=math154](I won't list them out)[/quote]

Aight
[hide=Solution]
We approach this with complementary counting. The primes less than 4 are 2 and 3, so every number must be of the form $2^a3^b$, and since there are less powers of 3 that are less than 100 than powers of 2, we will use powers of 3 as cases.

Case 1 ($3^0$): We have 7 possible integers, the powers of 2 from 0 to 6.
Case 2 ($3^1$): We have 6 possible integers, the powers of 2 from 0 to 5, multiplied by 3.
Case 3 ($3^2$): We have 4 possible integers, the powers of 2 from 0 to 3, multiplied by 9.
Case 4 ($3^3$): We have 2 possible integers, the powers of 2 from 0 to 1, multiplied by 27.
Case 5 ($3^4$): We have 1 possible integer, $2^0\cdot 81$.

Thus, we have $7+6+4+2+1=20$ integers that do not fit the restriction listed by the problem, so our answer is $100-20=\boxed{80}$.

i was 4 days old when this question was posted (also just list them LOL)

*Rages because she forgot about 1* :wallbash_red:

Is there any better way then to find cases?

We use complementary counting. These numbers must be in the form $2^{a}\cdot3^{b}$. So we have $\{1, 2, 4, 8, 16, 32, 64, 3, 9, 27, 81, 6, 12, 24, 48, 96, 18, 36, 72, 54\}$. These consist of $20$ numbers. So our answer is $100 - 20 = \boxed{80}$.

[quote=ilovepizza2020]Oops accidentally entered 20 on my first try[/quote]

LOL YES SAME!!!!

This is more like a casework problem than complementary counting


Billy Bob thinks that since it's summer, he should invite his best buddies like Rilly Rob and Killy Kob, for barbecue parties. He wants to use the same gas tank that's in the grill. It has 10 hours of gas left on it. Billy Bob takes 20 minutes to cook food for one party. Now, they have agreed to have two parties a week. How many weeks will the tank last for? Show an explanation along with the answer.

Billy Bob- :gleam:

me-  :juggle:

[hide]Time needed to cook food--20 minutes

there are 30 "20 minutes" in 10 hours. Take the first party off and there are 29 cooking times. There are 2 parties every week, so the tank will last for $\frac{29}{2}=14\frac{1}{2}$ weeks.[/hide]

[hide="solution"]Let's say that the tank will run for x weeks. So it will last 20x minutes, or 10 hours, if they have 1 party per week. There are 600 minutes in 10 hours. So 20x=600, 10x=300, x=30 If there are two parties, the usage will be twice as fast, so the answer is $15$[/hide]


Let n be a positive integer, and let p be a prime number.Prove that p^p|n! imlies that p^(p+1)|n!

If $n < p^2$ we deduce that the only multiples of $p$ which may be not greater than $n$ are $p,2p,3p, \cdots, (p-1)p$ so that the exponent of $p$ in $n!$ is at most $p-1.$
Thus, if $p^p$ divides $n!$ it follows that $n \geq p^2.$ Thus $p^{p+1}$ divides $n!$.

Pierre.


Suppose your calculator broke the only keys that work the 4 key the square root the + the - and * and divide key assuming the calculator does not round then what number from one to twenty is no posible with only pressing 4 four times.

[quote="seungsoo"]Pressing only 4 times - including number keys and enter? And how can that be if the only keys that work are the square root, +, -, *, and divide keys?[/quote]

Yeah i was wondering about that too

[quote="seungsoo"]One more question: does the calculator follow the order of operations? Some calculators that I've seen don't, especially ones that calculate the square root of the current value when the sq. rt. button is pressed.[/quote]

All graphing calculators and 2-line display calculators (at least those made by TI) follow order of operations. For all other calculators: they might and they might not.


nice problems but you should post each of them individually..


inequalities
trigonometry
geometry
circumcircle
complex numbers
angle bisector
geometry solved

triangle ABC has circumradius 1
any point M on the circumcircle, MA*MB*MC<=2
show it is equilateral

I guess that trigonometry or complex numbers could come in hand for this problem... but there may be a lot amount of calculation..

cheers! :D

i don't have a full solution but i think the result is true for a n-sided polygon,coz for a regular polygon inscribed in a circle of radius 1 and a point P on it,the product of all the distances is
 \sqrt [2 - 2cos(n \emptyset )]and u can make it equal to 2 by choosing an appropriate \emptyset
(where  \emptyset  denotes the angle between  P and the  vertex of the polygon).mayb it's easier to prove this using complex numbers,summing the affixes of the vertices over roots of unity and stuff,but trigo gets me nowhere
best regards

Ok, it was a misunderstanding. Thank you!

[quote="bugzpodder"][quote="siuhochung"][quote="Eckard"]Could it be that the correct inequality must be MA*MB*MC <= 8?
Consider a unit circle with points A, B, C close together and point M diametral opposite to them. Then each of MA, MB, MC is only a little bit less than 2, so their product must be greater than 2.

If this is right, a proof follows directly from an application of the AGM inequality.

Best regards[/quote]

ABC is not an equilateral triangle in that case[/quote]

Soarer, you must recognize this as one of meowth's questions  :D[/quote]

yes, i know, and i wonder if there's a pure geometry solution

Solution by Michael Lipnowski

Let ABC be such at triangle where MA*MB*MC<=2
Let A be the largest angle, and let the angle bisector of A intersect the circle at M.  Suppose A >60, then we can find B' C' on the minor arc BC such that <MB'A=<MC'A=30, ie B'AC' is equilateral => MB'*MC'*MA=2
But MB'<MB, MC'<MC which implies MB*MC*MA >2, a contradiction
therefore the largest angle is at most 60 degrees => ABC is equilateral


[quote="Potato Theory"]I saw this on a math competition, and I don't really know how to solve it:

${\lfloor \log_{2}(\log_{10}(2006!)}\rfloor$

Choices are 9, 10, 11, 12, and 13.

I know $\log_{10}$ of a factorial is simply $\sum_{k}^{n}\log_{10}k$ for $k!$, but I don't really know how that helps.[/quote]

[hide]Well, the numbers that contribute to 3 from the $\log_{10}{2006!}$ are the numbers from 1000 to 2006, which 1007

The numbers that contribute to 2 from the $\log_{10}{2006!}$ are the numbers from 100 to 999, which is 900

The numbers that contribute to 1 from the $\log_{10}{2006!}$ are the numbers from 10 to 99, which is 90

And then we realize that the $5<\log_{10}{9!}<6$.

So, the total number from each log that is contributed is at least

$1007*3+900*2+5=4826$

And also, take note that it cannot be more than 4*2006=8024, since the most contributed is $\log_{10}{2006}<4$.

So the $4824<\log_{10}{2006!}<8024$

So, now we know that the $\log_{2}{4824}<\log_{2}{\log_{10}{2006!}}<\log_{2}{8024}$

This implies that

$12<\log_{2}{\log_{10}{2006!}}<13$

So the answer is $12$

If someone has a better way, please explain it.

PS Your Statement helps[/hide]


Show that in any simple graph with m vertices and n edges, the chromatic number is at least m^2/(m^2-2n), where the chromatic number is the minimum number of colors you need to paint the vertices such that any two adjacent vertices have distinct colors.

Suppose the chromatic number is k.And suppose the vertices painted i-th color are $a_{i}$ for i=1,2...k. Then we have :
$n\leq \sum_{i\neq j}a_{i}a_{j}$ But since $\sum a_{i}= m$  the maximum of $\sum_{i\neq j}a_{i}a_{j}$ occurs when all of $a_{i}$ are as equal as possible we have:
$n \leq\frac{m^{2}(k-1)}{2k}$
(this is fot $a_{i}= \frac{m}{k}$)

and this after simplifying we get what we want.

(NOTE: i'll switch values n and m)

We have graph $G = (V,E)$ such that $|G| = n$ and $\parallel G\parallel = m$.

we know that $\chi(G) \geq \omega(G)$. and $\frac{n^{2}}{n^{2}-2m}\leq \frac{n^{2}}{n^{2}-2{n \choose 2}}= n$

but we know that $\omega(G) \leq n$

$QED$

ps. lol i hope i didn't screw it up
ps2. but i have strange feeling that actually, i did :D

[b]gollywog[/b], i think your prove is wrong.Because you have
$\frac{n^{2}}{n^{2}-2m}\leq n$
and you have
$\chi(G) \geq \omega(G)$ and if you want to finish the prove from here you need to have $\omega(G)\geq n$ but you have the opposite.

well i tried to make it with graph theory facts, i failed  :blush:


calculus
integration
vector
complex analysis
complex analysis unsolved

I know complex integration sufficiently well but I was interested why define:
$ \int_{\gamma}f(z)dz = \int_{a}^{b}f(\gamma(t))\gamma'(t) dt$.

Im asking because Im teaching somebody complex analysis and thought he would like to know.

Note: This looks stricking similar to the [i]line integral[/i] from Vector Analysis.

We need something that can be calculated by parametrizing the curve, but is independent of the exact parametrization. We'd also like something that works with antiderivatives and the Fundamental Theorem. This is the way to do it.


geometry parallelogram AMC USA(J)MO USAMO

[quote="Scrambled"]Can a convex 17-agon be divided into parallelagrams?[/quote]

I think that only if it is a regular 17-agon, because with every regular n-gon, you can form equilateral or iscoceles triangles so I guess...

I don't see how a nonregular 17agon can be divided into parallelograms.

[quote="Scrambled"][quote="Palytoxin"]interesting, i always thought only those $n$-agon that can be divided into parallelograms are those in which $n$ is divisible by 6 except $n=4$
also, do you mean regular 17-agon or all 17-agon in general?[/quote]
an octogon can be tiled[/quote]
 :?: tiles? i wasn't talking about tiles... and it is only some thoughts, my statement is not true because i can divide an octagon into parallelograms... :?

[quote="Iversonfan2005"]So is it or isn't it Scrambled?[/quote]
why? i wouldn't want Scrambled to tell us the answer right now, let's solve it  :flex:

[quote="Scrambled"][quote="chess64"][hide]
I say no, because of the odd number of sides.
[/hide][/quote]
you'll need a better explanation than that[/quote]

If it can, then it has to have an even number of sides. Think about it. For some reason, I can't explain it. :sleeping:

lol, I would love to see that written in a USAMO competition

"Can't you see it?! It doesn't have an even number of sides, duh!"

Heh, I'm just playing.

But I agree with chess's intuition about the odd numbered shape not being able to be split up into parallelograms, but alas I can't explain it either (I just stopped by to do a little psuedo-mocking :))

Well, a parallelogram has $4$ sides which is an even number and since a 17agon doesn't have an even number of sides, it can't be split up into parallelograms because it needs an extra side.....okay, I can't explain it either :blush:

[quote="WindSlicer"]lol, I would love to see that written in a USAMO competition

"Can't you see it?! It doesn't have an even number of sides, duh!"[/quote]

 :rotfl:  :rotfl: Lol! No, better yet: "The answer is obviously no, because it doesn't have an even number of sides--DUH! 17 is ODD, not EVEN!"

Well, it is pretty hard to explain. Scrambled, can you tell us the answer now?