You could look at the fourth differences (which should be constant since this is a 4th-degree polynomial) and extrapolate from there. This will certainly work, but only because 4th-degree is still a relatively small number.

For a more general solution, look up Lagrange Polynomials. They give you ways to construct polynomials of least degree that go through some points; in this case you are looking for the lowest-degree polynomial that goes through the points $(1,81),(2,106),(3,114),(4,188),(5,127)$, so you could do that also.

But for such a small number as four, I think examining the fourth difference will be best.

[quote="indianamath"]what's rref?[/quote]
Row-reduced echelon form. Most graphing calculators have this command.

[quote="13375P34K43V312"]uh

if this is a homework problem you shouldn't be posting it

but it's pretty trivial by letting it equal ax^4+bx^3+cx^2+dx+e and doing rref on your calculator[/quote]
It's from Intermediate Algebra midterm.

[quote="indianamath"]
I need help with this one. I tried using substitution but there were too many fractions within fractions. help! solutions will be great![/quote]
Uh, dude, we get full solutions after, don't post the problems please (someone may unintentionally see it, ruining the fun)

uh..isnt there a function on a calculator that if will find you the equation of a 4th degree polynomial if you give it values..so you dont hvae to use matrices.


[quote="khashi70"]Hameye adadhaye tabi'ee $a,b$ ra tori peida konid ke $a^{3}+6ab+1$ va $b^{3}+6ab+1$ har do moka'abe kamel bashand .[/quote]
farz mikonim$a\geq{b}$dar natije:
$a^{3}<a^{3}+6ab+1<(a+2)^{3}\Rightarrow{a^{3}+6ab+1=(a+1)^{3}}\Rightarrow{6ab=3a^{2}+3a}\Rightarrow{2b=a+1}$
bayad $b^{3}+6ab+1$niz mokabe kamel bashad:
$b^{3}+6ab+1=b^{3}+6b(2b-1)+1=b^{3}+12b^{2}-6b+1$baraye b>3 darim:
$(b+2)^{3}<b^{3}+12b^{2}-6b+1<(b+4)^{3}$ pas bayad dashte bashim:
$b^{3}+12b^{2}-6b+1=(b+3)^{3}\Rightarrow{12b^{2}-6b+1=9b^{2}+27b+27}$
ke be peimane 3 gheire momkan ast.baraye b<4 ham faghat be ezaye b=1$b^{3}+12b^{2}-6b+1$ mokaab mishavad pas tanha javab a=b=1 ast


[hide]

Using the quadratic equation gives you

This goes to  which means that, since BOTH solutions must fit the inequality, 3>m>-3 and 5>m>-1, which narrows down to 3>m>-1, the answer.[/hide]

[quote="MCrawford"]Find all real values of $m$ such that both roots of the equation
\[ x^2 - 2mx + m^2 - 1 = 0 \]
are greater than -2 but less than +4.[/quote]

Well, (x-m)^2 = 1.

Hence  we have (x-m = -1) U (x-m = 1) such that x E (-2,4)
It follows that m > -1 and that m < 5 from the first, and:
m > -3 and m < 3 from the second.  Hence the union is m --> (-1,3)


prove that k^2 + k+ 1 cannot be equal to a square...
( do not use completing the square method )


Is there any way to change the original full path of the directory where the executables are? How?

Build -> Define Output Profiles -> Wizard button at bottom left


If $ f(x) \equal{} ax^2 \plus{} x \plus{} c$, and $ f(1)\equal{}0$, find $ f(\minus{}1)$.

[hide="solution"]Since $ 1^2\equal{}(\minus{}1)^2,$ the terms $ ax^2$ and $ c$ are the same in both equations. Since $ |\minus{}1\minus{}1|\equal{}2,$ we subtract $ 2$ from $ 0$ to get $ \boxed{\minus{}2}.$[/hide]

$ f(x)\equal{}ax^2\plus{}x\plus{}c$
$ f(1)\equal{}0$

$ f(1)\equal{}a(1)^2\plus{}(1)\plus{}c$
$ 0\equal{}a\plus{}c\plus{}1$
$ \minus{}1\equal{}a\plus{}c$

$ f(\minus{}1)\equal{}a(\minus{}1)^2\plus{}(\minus{}1)\plus{}c$
$ f(\minus{}1)\equal{}a\minus{}1\plus{}c$
$ f(\minus{}1)\equal{}a\plus{}c\minus{}1$
$ f(\minus{}1)\equal{}\minus{}1\minus{}1$
$ f(\minus{}1)\equal{}\minus{}2$

Giving us our answer...by a different approach

Or you could just set $ c\equal{}\minus{}2$, $ a\equal{}1$, and $ b\equal{}1$ and go from there:
$ 0\equal{}1^2\plus{}1\minus{}2$
So then just plug in -1 as x.
$ (\minus{}1)^2\plus{}(\minus{}1)\minus{}2\equal{}\boxed{\minus{}2}$

So you could get the same answer from different values of a, b, and c.


For a set S of nonnegative integers, let r_S(n) denote the number of ordered pairs (s_1,s_2) such that s_1 element of S, s_2 element of S, s_1 != s_2, and s_1+s_2=n. Is it possible to partition the nonnegative integers into two sets A and B in such a way that r_A(n)=r_B(n) for all n?

[hide]Let $ f(x) = \sum_{n=0}^{\infty} a_n x^n$ where $ a_n \in \{ 0, 1 \}$ depending on whether or not $ n \in A$

Therefore $ f(x)^2 - f(x^2)$ is the generating function for the number of ordered pairs in question.

Therefore we must find $ f$ which satisfies the functional equation:

$ \displaystyle f(x)^2 - f(x^2) = \left ( \frac{1}{1-x} - f(x) \right )^2 - \left ( \frac{1}{1-x^2} - f(x^2) \right )$

Rearranging, $ 0 = \frac{1}{1-x} - 2f(x) - \frac{1}{1+x} + 2f(x^2)(1-x)$

Therefore ${ 0 = \sum_{n=0}}^{\infty} x^n - \sum_{n=0}^{\infty} 2a_nx^n - \sum_{n=0}^{\infty} (-1)^n x^n + \sum_{n=0}^{\infty} \left ( 2a_n x^{2n} - 2a_n x^{2n+1}  \right )$

Comparing coefficients, when $ n = 2k$: $ 0 = 1 - 2a_{2k} - 1 + 2a_k \Rightarrow a_{2k} = a_k$
when $ n = 2k+1$: $ 0 = 1 - 2a_{2k+1} +1 - 2a_k \Rightarrow a_{2k+1} = 1-a_k$

Therefore we can find $ A$ and $ B$ explicitly, and we're done[/hide]


geometry
3D geometry
sphere
calculus
integration
real analysis
geometric transformation

I am studying integration theory. The book I am using doesn't really discuss any specific measures and all types of specific properties of the Lebesgue measures are stored away in the exercises. I would like some hints on a couple. So far I have shown in the exercises (they all seem to build on each other in the order they appear):

1) The Borel Sets are Lebesgue Sets
2) The measure of open and closed "intervals" is as expected.
3) Any Lebesgue set is of the form Borel set - Null Set

The next exercises are to show that the measure of a hyperplane in $R^{n}$ is 0 and that the boundary of a ball in $R^{n}$ has measure 0. I can't really figure out how to do this in any reasonable way. I don't know about rotation invariance or anything like that yet. Can I get some hints? Thanks.

A hyperplane has measure zero if any finite rectangular (box-shaped; my mental picture is in $\mathbb{R}^{3}$) slice of it has measure zero by countable additivity and translation invariance. It's easy to fit a rectangle into a box with small volume.

For a sphere: Do you have the scaling property that $\mu(cA)=c^{n}\mu(A)$ for any positive real $c$ and measurable set $A$? If so, consider two concentric balls so that the sphere is contained in one and outside the other. With appropriate scaling, the difference in measure between the two balls can be made arbitrarily small (a ball obviously has finite measure, since it's contained in a finite box).

No, I hadn't thought about that property yet, but it doesn't seem hard to show it. That's a nice way to deal with the second problem.

For the hyperplane I don't understand what you are suggesting. It is easy to fit a axis-parallell rectangle into a box with small volume, but to me it's not that easy to do it for a rectangle that makes an angle to the axes. I suppose if I really wanted to I could bruteforce it, but there should be something better right?

Ah, sorry about that. I wasn't thinking about angled hyperplanes.

Suppose that our hyperplane $H$ is given by $\sum a_{i}x_{i}=b$. For each real $c$, $\sum a_{i}x_{i}=c$ is a translation of this hyperplane by $\frac{c-b}{\sum a_{i}^{2}}(a_{1},a_{2},\dots,a_{n})$, and thus has the same measure. WLOG, let $b=\sum a_{i}^{2}=1$.
Consider the intersection $X$ of $H$ and the ball $B$ of radius 2 centered at zero- this set is nonempty and relatively open in $H$. For each $c\in (-1,1)$, $X_{c}=X+(c-1)(a_{1},a_{2},\dots,a_{n})\subset B$, and these sets are disjoint. They all have the same measure and are contained in a set of finite measure, so each has measure zero.

There are a lot of ways to get at this, but the essence is that we can stack infinitely many disjoint translates of a hyperplane.

Ok, thanks!

I modified the construction some. I suppose the plane $K$ passes through the origin and let $e_{1},...e_{n-1}$ be a basis for it. Then I set $E_{k}= \{\sum t_{j}e_{j}|-k \le t_{j}\le k\}\subset K$ and suppose $(a_{1},...,a_{n})$ is a normal to the hyperplane. Each $E_{k}$ is compact and in particular measurable.

Fix a $k$ and let $A_{j}= \frac{1}{2^{j}}(a_{1},...,a_{n})+E_{k}$. Each $A_{j}$ is measurable and disjoint. So we have $\mu\left( \cup A_{j}\right) = \sum \mu(A_{j}) = \sum \mu(E_{k})$. But $\cup A_{j}$ is bounded so it has finite measure, and thus $\mu (E_{k}) = 0$. Now, $K = \cup E_{k}$ and so by subaddivity $K$ has measure 0.

I based this construction on your last comment. Is this approximately what you meant? To be honest I don't understand your proof. Too much algebra I don't want to think about.

About the sphere, I wrote down carefully a solution in my papers earlier. Do you mean to use the scaling just for spheres with center origin and then use translation invariancy to get any sphere, or do you have some clever construction where just scaling is enough?

That's fine- I didn't really expect you to use the exact same construction.
For the spheres, I do mean to use both scaling and translation invariance, as you said.

Suppose the hyperplane is covered by some cubes or other figures of scale $1/n$.  The number of these figures within distance $R$ of $0$ is asymptotic to $n^{d-1}$ and so the volume within that region is $O(1/n)$. This means that the part of the hyperplane within distance $R$ has measure 0, for all $R$.  Conclude either by countable additivity or by decomposing the hyperplane into annuli at distance between $[R,2R]$, $[2R,4R], \dots$ with covering volumes tending to zero.

[quote="Kalle"]It's easy to hand-wave such a construction a bit but it's not really that easy to write it down properly..[/quote]

It does not require handwaving, because the underlying covering can be chosen from a lattice of cubes of size 1/n.  Having chosen this the other statements are straightforward and rigorous.  It is not promised that you can write the proof in 50 words, but there is no step that presents any new difficulty beyond that of notation (using greatest-integer function and the like).  You can also save some calculations by using cubical rather than spherical annuli.


If $ a$ $ b$ and $ c$ are positive reals such that $ ab\plus{}bc\plus{}ca\geq 3$ then prove or disprove that:

$ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\geq 3$

[quote="delegat"]If $ a$ $ b$ and $ c$ are positive reals such that $ ab \plus{} bc \plus{} ca\geq 3$ then prove or disprove that:

$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\geq 3$[/quote]
Try $ a\equal{}b\rightarrow\sqrt3$ and $ c\rightarrow0.$ :wink:


[size=150][b]Solve for all integers x, y, z

x*y*z=x+y+z[/b][/size]

[size=150]Prove that only one solution exists.[/size]

I could find only the {1,2,3} solution.

There are 6 pairs of $ (x,\ y,\ z)$ after neglecting the restriction $ x\leq y\leq z$.

@Kunny

Please post the 6 pairs excluding the cyclic permutations of {1,2,3}

Anybody, any other answers? :lol:

Please prove that no other solution exists. Integers of course.

they are infinities of solutions if $ (x,y,z) \in\Z$

This was the original problem that I saw.

[quote="thegreatindiankhali"][size=150][b]Solve for all positive integers x, y, z

x*y*z=x+y+z[/b][/size][/quote]

WLOG $ x\leq y\leq z$, we have $ xyz\leq 3x$, yielding $ yz \equal{} 1,\ 2,\ 3$.
After some case work, we get finally $ \{x,\ y,\ z\} \equal{} \{1,\ 2,\ 3\}.$

$ (x,\ y,\ z) \equal{} (1,\ 2,\ 3),\ (1,\ 3,\ 2),\ (2,\ 1,\ 3),\ (2,\ 3,\ 1),\ (3,\ 1,\ 2),\ (3,\ 2,\ 1)$.

[quote="thegreatindiankhali"][size=150]anybody plz prove that no other solution except the permutations of {1,2,3} exists.[/size]

@kunny
the original question was quite easy.[/quote]
What do you mean by proving that  $ (1,2,3)$ and its permutations are the only solutions? Kunny's solution shows us consequently that there are only six solutions. Maybe I am not understanding you, but please clarify your question.
Also, kunny - san I think you have  amistake in your solution :oops:  Because $ xyz \equal{} x \plus{} y \plus{} z\leq 3\color{red}{z}$ :)

[quote="kunny(edited)"]This was the original problem that I saw.
[quote="thegreatindiankhali"]
$ xyz \equal{} x \plus{} y \plus{} z$[/quote]
WLOG $ x\leq y\leq z$, we have $ xyz\leq 3z$, yielding $ \color{red}{xy \equal{} 1,\ 2,\ 3}$.
After some case work, we get finally $ \{x,\ y,\ z\} \equal{} \{1,\ 2,\ 3\}.$
$ (x,\ y,\ z) \equal{} (1,\ 2,\ 3),\ (1,\ 3,\ 2),\ (2,\ 1,\ 3),\ (2,\ 3,\ 1),\ (3,\ 1,\ 2),\ (3,\ 2,\ 1)$.[/quote]
Since $ 3z \minus{} xyz\geq 0 ;$ we must have $ 3\geq xy \implies xy \equal{} 1,2,3$ Now all that is left is casework.... If you want to show this casework, then here it is:
$ \boxed{\text{Case I: } xy \equal{} 1} \implies (x,y) \equal{} (1,1).$ Putting this into the given equation we have $ z \equal{} 2 \plus{} z$ which is impossible.
$ \boxed{\text{Case II: } xy \equal{} 2}\implies (x,y) \equal{} (1,2)$. Putting this into the given equation gives $ z \equal{} 3\implies (x,y,z) \equal{} (1,2,3)$ is a solution.
$ \boxed{\text{Case III: } xy \equal{} 3} \implies (x,y) \equal{} (1,3)$, putting this gives $ z \equal{} 2$ which contradicts $ x\leq y\leq z$.
Therefore $ (x,y,z) \equal{} (1,2,3)$ and its permutations are the only solutions.

[quote="thegreatindiankhali"][size=150]anybody plz prove that no other solution except the permutations of {1,2,3} exists.[/size]

@kunny
the original question was quite easy.[/quote]

First of all, You should appologize it, man. :mad:

Recently we have chance to see person those who change problem like YOU!

They don't have common sense, do you?

[quote="Potla"][quote="Potla"]
@kunny
the original question was quite easy.[/quote]
What do you mean by proving that  $ (1,2,3)$ and its permutations are the only solutions? Kunny's solution shows us consequently that there are only six solutions. Maybe I am not understanding you, but please clarify your question.
Also, kunny - san I think you have  amistake in your solution :oops:  Because $ xyz \equal{} x \plus{} y \plus{} z\leq 3\color{red}{z}$ :)

[quote="kunny(edited)"]This was the original problem that I saw.
[quote="thegreatindiankhali"]
$ xyz \equal{} x \plus{} y \plus{} z$[/quote]
WLOG $ x\leq y\leq z$, we have $ xyz\leq 3z$, yielding $ \color{red}{xy \equal{} 1,\ 2,\ 3}$.
After some case work, we get finally $ \{x,\ y,\ z\} \equal{} \{1,\ 2,\ 3\}.$
$ (x,\ y,\ z) \equal{} (1,\ 2,\ 3),\ (1,\ 3,\ 2),\ (2,\ 1,\ 3),\ (2,\ 3,\ 1),\ (3,\ 1,\ 2),\ (3,\ 2,\ 1)$.[/quote]
Since $ 3z \minus{} xyz\geq 0 ;$ we must have $ 3\geq xy \implies xy \equal{} 1,2,3$ Now all that is left is casework.... If you want to show this casework, then here it is:
$ \boxed{\text{Case I: } xy \equal{} 1} \implies (x,y) \equal{} (1,1).$ Putting this into the given equation we have $ z \equal{} 2 \plus{} z$ which is impossible.
$ \boxed{\text{Case II: } xy \equal{} 2}\implies (x,y) \equal{} (1,2)$. Putting this into the given equation gives $ z \equal{} 3\implies (x,y,z) \equal{} (1,2,3)$ is a solution.
$ \boxed{\text{Case III: } xy \equal{} 3} \implies (x,y) \equal{} (1,3)$, putting this gives $ z \equal{} 2$ which contradicts $ x\leq y\leq z$.
Therefore $ (x,y,z) \equal{} (1,2,3)$ and its permutations are the only solutions.[/quote]
Yes, you are right, Potla. Thank you for pointing out it.

But you have posted the perfect solution, that was in vain, because the original problem has been tampered.

Do neglect the problem. :wink:

@Kunny

I'm sorry, I could'nt get you solution completely. I didnt change the question, (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) were the solutions i could find. Just wanted to check any other solution....  :oops: . Sorry, if i didnt sound polite.

@Potla.
Thanks for your checking too.  :lol:

[quote="thegreatindiankhali"]@Kunny

I'm sorry, I could'nt get you solution completely. I didnt change the question, (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) were the solutions i could find. Just wanted to check any other solution....  :oops: . Sorry, if i didnt sound polite.[/quote]

Nice to meet you,  thegreatindiankhali :)

[quote="thegreatindiankhali"]Solve for all [b]integers[/b] x, y, z[/quote]

Don't forget (0,0,0)  :)  (and of course (-1,-2,-3) permutations).

As a matter of fact, this becomes a more interesting problem. Other solutions include (-x,0,x).

Well it's still not very interesting. :wink:

If any of them is zero then it's $ (m, \minus{} m,0)$ and permutations.

We have solved all three positive or negative with $ \pm(1,2,3)$ and permutations.

If there are two positives $ a,b$ and one negative $ \minus{} c$ we have $ \minus{} abc \equal{} a \plus{} b \minus{} c$ for positive $ a,b,c$, or $ abc \plus{} a \plus{} b \equal{} c$. But $ a,b > 0\implies abc \plus{} a \plus{} b\ge a \plus{} b \plus{} c > c$. And if there's a solution with two negatives $ \minus{} a, \minus{} b$ and one positive $ c$ negating will yield a solution with two positives and one negative.

Good work Mewtwo, R.Maths and Math154!

To make the problem a bit more interesting:
Does it remind you something about the perfect numbers?

6 being the smallest perfect number, has factors 1,2,3 is perfectly made for this question, such that x,y,z={1,2,3)

Cheers! :D


analytic geometry
number theory proposed
number theory

Well quite interesting thing  :) :

[color=blue]Problem[/color]A point like $A=(a,b)$ in cartesian plane is called visible if $a ,b \in Z$ and $A$ can be seen from the origin(it means that they are no points with integer coordinates on the line $AO$.)
Find all of the visible points

Find all visible points?

:huh: Aren't those just the points $(x,y)$ for which $x$ and $y$ are coprime?

yes they are  :)

Actually the problem was :

A point like $A=(x,y)$ is called visible if $gcd(x,y)=1$ ,then prove that for every $T \in N$  there exist a circle with radius  $T$  such that ther are no visible points in the circle .

I find the the first problem(Or i shall say lemma,while solving this one  :) )

This last problem follows easily from :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invisible&t=17055

Pierre.


Okay, there are two teams: the +'s and the -'s.  The +'s add one when they post, the -'s subtract one.
On anybody's first post, they must quote the following list, delete the quote tags, and add themself to the [i]losing team[/i].  That way the team's will be somewhat even.  The first team to achieve 50 or -50 wins.  Also, make sure to always put the number you post in [b]BOLD[/b].

[u]+ Team[/u]
Miyomiyo

[u]- Team[/u]

[b]1[/b]

[quote="miyomiyo"]
+ Team
Miyomiyo

- Team
AstroPhys
[/quote]
Hooray a just game  :) !

[b]0[/b].

Note the "delete the quote tags" part.

[b]1[/b]

+ Team
Miyomiyo

- Team
AstroPhys
perfect628

[b]0[/b]

Let's go -!

+ Team
Miyomiyo
mnmath

- Team
AstroPhys
perfect628

[b]0[/b]

+ Team
Miyomiyo
mnmath

- Team
AstroPhys
perfect628
ProtestanT

[b]0[/b]

+ Team
Miyomiyo
mnmath

- Team
AstroPhys
perfect628
ProtestanT
xpmath

[b]0[/b]

[b]1[/b]

The - team is getting awfully big...

Mods, get rid of this, plz.

it's really unfair, biased, and annoying


inequalities algebra polynomial calculus derivative function geometry

[hide]
This may be completely wrong

From cauchy..
$ (a^2\plus{}b^2\plus{}c^2\plus{}d^2)(a^4\plus{}b^4\plus{}c^4\plus{}d^4) \ge (a^3\plus{}b^3\plus{}c^3\plus{}d^3)^2$
$ \sqrt{(a^4\plus{}b^4\plus{}c^4\plus{}d^4)} \ge (a^3\plus{}b^3\plus{}c^3\plus{}d^3)$

Now..
$ 1\equal{}(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\equal{} a^4\plus{}b^4\plus{}c^4\plus{}d^4 \plus{} 2(a^2b^2 \plus{} a^2c^2 \plus{}a^2d^2 \plus{} b^2c^2\plus{} b^2d^2 \plus{}c^2d^2)$
So we want to maximise $ a^4\plus{}b^4\plus{}c^4\plus{}d^4$

Also..
$ a\plus{}b\equal{}\minus{}(c\plus{}d) \longrightarrow a^2\plus{}b^2\minus{}c^2\minus{}d^2 \equal{} 2cd\minus{}2ab$
Squaring both sides...
$ \Rightarrow a^4\plus{}b^4\plus{}c^4\plus{}d^4 \equal{} 2(a^2b^2 \plus{} a^2c^2...c^2d^2) \minus{} 8abcd$
$ \Rightarrow  a^4\plus{}b^4\plus{}c^4\plus{}d^4 \plus{} 8abcd \equal{} 2(a^2b^2 \plus{} a^2c^2...c^2d^2)$
If we sub this in to the above equation we find
$ 1 \equal{} 2(a^4\plus{}b^4\plus{}c^4\plus{}d^4) \plus{}8abcd$

So we look to minimise $ abcd$,
So  by AM-GM
$ \dfrac{a^2\plus{}b^2\plus{}c^2\plus{}d^2}{4} \ge \sqrt [4] {(abcd)^2}$
$ \Rightarrow \frac{\minus{}1}{16} \le abcd \le \frac{1}{16}$
So
$ a^4\plus{}b^4\plus{}c^4\plus{}d^4 \equal{} \dfrac{1\plus{}\frac{1}{2}}{2}$
And $ a^3\plus{}b^3\plus{}c^3\plus{}d^3$ is a maximum at $ \dfrac{\sqrt{3}}{2}$
[/hide]

[quote="ocha"] This may be completely wrong[/quote]

Yes it's wrong  :)

[quote="ocha"]$ \dfrac{a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4} \ge \sqrt [4] {(abcd)^2}$
$ \Rightarrow \frac { \minus{} 1}{16} \le abcd \le \frac {1}{16}$[/quote]

Here is mistake.

Equality holds, then $ a^2\equal{}b^2\equal{}c^2\equal{}d^2$.

{a,b,c,d} = {2,5;2,5;-2,5;-2,5}

But it may be not correct answer.

consider

(x1 + x2 + x3 + x4)((x1)^2 + (x2)^2 + (x3)^2 + (x4)^2)

= sum (x1^3) + sum (x1)(x2)(x1 + x2)  =  0

we need to minimise

sum (x1)(x2)(x1 + x2)

and I don't know how to do that  :(

I think it's not easy ,,, take your time.
I don't have the answer.

We can cast this as a polynomial problem via Newton Sums.

I use some light calculus below -- namely, the polynomial derivative -- to find the relative extrema.

[hide]Consider a polynomial $ 2x^4 \plus{} ax^2 \plus{} bx \plus{} c$, with roots $ x_1,x_2,x_3,x_4$. I've set the coefficient on $ x^3$ equal to zero because we know the sum of the roots is $ 0$.

We can invoke Newton's Sums here. We are given that:

$ S_1 \equal{} 0$ and $ S_2 \equal{} 1$

But $ 2S_2 \plus{} 2a \equal{} 0 \implies a \equal{} \minus{} 1$

Then, $ 2S_3 \plus{} 3b \equal{} 0 \implies S_3 \equal{} \minus{} \frac {3}{2} b$.

which correlates negatively with $ b$. So, essentially, we want to find the minimum value of $ b$ in the equation such that $ f \equal{} 2x^4 \minus{} x^2 \plus{} bx \plus{} c$ still has 4 real roots. Note that so long as this quartic has the right "shape", meaning it has two relative minima and one relative maximum, we can adjust $ c$ to make it have four real roots. So, let's guarantee this.

Take the derivative: $ f' \equal{} 8 x^3 \minus{} 2x \plus{} b$. We need $ b$ such that this has three real roots. Again, take the derivative to find that the relative extrema of this occur when $ 24 x^2 \minus{} 2 \equal{} 0 \implies x \equal{} \pm \sqrt {\frac {1}{12}}$, with the negative producing the relative maximum and the positive producing the relative minimum.

To minimize $ b$, we need $ f' \equal{} 0$ at its relative maximum (the $ \plus{} b$ shifted down the graph as much as possible). So, $ \frac {2}{9} \sqrt {3} \plus{} b \equal{} 0 \implies b_{\textbf{min}} \equal{} \minus{} \frac {2}{9} \sqrt {3}$

Recall that $ S_3 \equal{} \minus{} \frac {3}{2} b$ so $ \boxed{(S_3)_{\textbf{max}} \equal{} \frac {\sqrt {3}}{3}}$[/hide]

Why $ 2S_3\plus{}3b\equal{}0$ ?  :maybe:

See [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums]here[/url].

We have $ a_4 S_3 \plus{} a_3 S_2 \plus{} a_2 S_1 \plus{} 3a_1\equal{}0$ in that link above.

The polynomial has $ (a_4, a_3, a_2, a_1) \equal{} (2,0,a,b)$ from the polynomial, so $ 2S_3 \plus{} a S_1 \plus{} 3b \equal{} 0$

We also know that $ S_1 \equal{} 0$, since it was given, so $ 2 S_3 \plus{}3b \equal{} 0$

We don't have to go too much further for that.

[hide]We made a double-root in $ f'$ at $ x \equal{} \minus{} \frac {\sqrt {3}}{6}$, so $ f$ should have a triple-root here. Thus, $ 2x^4 \minus{} x^2 \minus{} \frac {2}{9} \sqrt {3}x \plus{} c$ should be zero'ed at $ x \equal{} \minus{} \frac {\sqrt {3}}{6}$. Thus, we solve:

$ 2\left( \minus{} \frac {\sqrt {3}}{6} \right) ^4 \minus{} \left( \minus{} \frac {\sqrt {3}}{6} \right)^2 \minus{} \frac {2}{9} \sqrt {3} \left( \minus{} \frac {\sqrt {3}}{6} \right) \plus{} c \equal{} 0 \implies c \equal{} \minus{} \frac {1}{24}$

Thus, the quartic with roots $ x_1, x_2, x_3, x_4$ is $ \boxed{2x^4 \minus{} x^2 \minus{} \frac {2\sqrt {3}}{9} x \minus{} \frac {1}{24}}$.

Fortunately, we know that we have a triple root at $ x \equal{} \minus{} \frac {\sqrt {3}}{6}$, so we can use this to find the other root, which ends up being $ \frac {\sqrt {3}}{2}$

Therefore, $ \boxed{(x_1, x_2, x_3, x_4) \equal{} \left(\frac {\sqrt {3}}{2}, \; \minus{} \frac {\sqrt {3}}{6}, \; \minus{} \frac {\sqrt {3}}{6}, \; \minus{} \frac {\sqrt {3}}{6}, \right)}$ and permutations.[/hide]

[b]@ OP:[/b] Where did you get this problem from?

[quote="TZF"]
[b]@ OP:[/b] Where did you get this problem from?[/quote]
 :maybe: ... from my friend  :D

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=232926]General Version[/url]

I would like to see a more simple solution for this.

[hide]Here's my solution after spending 2 and half hours on this one:
First, let I replace x_1,x_2,x_3 and x_4 by a,b,c and d since it matches my variables.
We have: a^3+b^3+c^3+d^3=[2(c^2+d^2)-1/2].3(c+d)(*)(from (a+b+c+d)^3=0 we easily infer this one)
W.L.O.G, assume that a>=b>=c>=d. We see that since 2(c^2+d^2)-1=<0, we can assume that c+d<0(since we want to find the maximum value and we know that there exists a positive value for this maximum, which means at least one of these variables is=<0). Now we will prove that: f(a,b,c,d)=<f(a,b,(c+d)/2,(c+d)/2), which is equivalent to: 3(c+d)((c+d)^2-2(c^2+d^2)>=0, which is true since -(c-d)^2.(c+d)>=0. Therefore, we now just need to find the maximum value in the case c=d. Rewriting(*), we get: 6d(2d^2-1/2)=a^3+b^3+c^3+d^3 with the given restriction: a+b+2d=0(1) and a^2+b^2+2d^2=1(2) with a>=b>=d.
 From (1) we get: 0>=a+3d, which means -3d>=a. So we have: 9d^2>=a^2(since -3d>=0 and a must be>=0 since a=max(a,b,c,d)).
On the other hand, since a+d=-(b+d)>=0(since if a+d<0, then b+d<0 as well since a>b, which is unreasonable) we have: -d>=b, which means d^2>=b^2.
Therefore, from (2) we have: 1=<12d^2, which means d=<-1/sqrt(12)(since d<0)
Using derivative(yeah we must use it in this situation), we have: f '(d)=36d^2-3. We see that the critical number here must be -1/sqrt(12), since d<0. In the interval (negative infinity, -1/sqrt(12)), f(d) is a decreasing function, so f(-1/sqrt(12)) is a relative maximum of f(d).Since -1/sqrt(12) is also the maximum value of d, so the maximum value of f(d) is: sqrt(3)/3.
The equality occurs when d=b=c= -1/sqrt(12) and a=sqrt(3)/2(Q.E.D) :)
@Mr.Kunny: Unfortunately, my solution doesn't work for your general case since I can't derive the sum of the cube of each variable into the form above :( .[/hide]

Nobody wants to check my solution for this hard problem to see whether it's correct or not? :(

That's because it takes a lot of effort to read. Here are some comments and concerns I have on your solution.

$ \text{[1]} \blacktriangleright$ It would help if you wrote it using $ \text{\LaTeX}$. To see how to implement the elements you use, [hide="click here"]- Put dollar signs around your equations to tell the system you want to typeset them
- Use [color=blue]\frac{a}{b}[/color] to write $ \frac {a}{b}$
- Use [color=blue]\leq[/color] and [color=blue]\geq[/color] to make $ \leq$ and $ \geq$, respectively
- Use [color=blue]\implies[/color] to create the "implies" arrow, $ \implies$, and [color=blue]\iff[/color] to create the iff. arrow, $ \iff$
- Use [color=blue]\infty[/color] to create $ \infty$
- Use the underscore, [color=blue]_[/color], to create a subscript, and the symbol [color=blue]^[/color] to create a superscript. You have done this anyway in your plain-text notation.
- Use [color=blue]\sqrt{a}[/color] to create $ \sqrt {a}$
[/hide]

$ \text{[2]} \blacktriangleright$ Add more white space -- ie, line breaks. It helps to have spaces between equations, too. This makes writing much more readable. Try to group your solution into "paragraphs" too, just like in writing.

$ \text{[3]} \blacktriangleright$ A function maps from one set to another. This means the format of your arguments must be consistent. You did not do this.
- Take a look at where you write "Now we will prove that: f(a,b,c,d) ..." Here, the function f [i](which you did not define) [/i]accepts four arguments.
- Later, you take the derivative [i](with respect to what?)[/i], and then we suddenly see f'(d) and f(d), where f accepts only one argument.

$ \text{[4]} \blacktriangleright$ If an equation is worth labeling, its worth giving its own line. Try to label in roman numerals, or in bold, or in some other format that can't be confused with the equation itself. In LaTeX, you can use the command [color=blue]\qquad[/color] to create a large gap to set your label apart from your equation.

$ \text{[5]} \blacktriangleright$ Some of your equations don't seem to check.
- Take a look at [color=blue]a^3+b^3+c^3+d^3=[2(c^2+d^2)-1/2].3(c+d)[/color] -- it doesn't seem to check. For example, for $ (a,b,c,d) = (1/2, \; 1/2, \; - 1/2, \; - 1/2)$, we have $ a^3 + b^3 + c^3 + d^3 = 0$, but the RHS is not equal to 1.

$ \text{[6]} \blacktriangleright$ Your argument for $ c = d$ doesn't work. Let's look at the statement $ f(a,b,c,d) \leq f \left( a,b,\frac {c + d}{2},\frac {c + d}{2} \right)$. Note that $ a^2 + b^2 + 2 \left( \frac {c + d}{2} \right)^2 \neq a^2 + b^2 + c^2 + d^2 = 1$ unless $ c = d$ to begin with. So, it does not meet the requirement $ x_1 ^2 + x_2 ^2 + x_3 ^2 + x_4 ^2 = 1$, and there is no use computing $ f$ at this point.

Note that $ a + b + c + d = 0$ and $ a^2 + b^2 + c^2 + d^2 = 1$ together define a sphere embedded in four-space. Just like a sphere embedded in 3-space, if you fix two of the coordinates, there are at most two possible points you can be referring to. Here, if you fix $ x_1 = a$ and $ x_2 = b$, the only two possible points you can be referring to are $ (a,b,c,d)$ and $ (a,b,d,c)$.

$ \text{[7]}\blacktriangleright$ I don't understand what you are taking the derivative of when you find $ f'(d)$, but make sure you are not taking a partial derivative if there are $ a,b,c$ terms left, because $ a,b,c$ all vary with $ d$.

Hello TZF
Before I star rephrasing my solution and explaining something more in detail to answer your comment that my argument doesn't work, I want to say:"Thank you very much for your hard work!!". It's very useful to me except the fact that I don't know how to use LATEX.
Now, it's time to rephrase it!!!
[b]1st step[/b]: the identity (a^3+b^3+c^3+d^3)=[(2(c^2+d^2)-1)/2].3(c+d) is not wrong. I think my typo makes you misunderstand this one, right?

[b]2nd step[/b]:Now we see that 2(c^2+d^2)-1=(c^2+d^2-a^2-b^2).3(c+d)/2=f(a,b,c,d). I shouldn't leave it in its "disguised form" though

[b]3rd step[/b]:W.L.O.G, assume that a>=b>=c>=d. We see that since 2(c^2+d^2)-1=<0, we can assume that c+d<0(since we want to find the maximum value and we know that there exists a positive value for this maximum, which means at least one of these variables is=<0)(I copy from my post above)

[b]4th step[/b]: We will prove that: [b]f(a,b,c,d)=<f(a,b,(c+d)/2,(c+d)/2)[/b], which is equivalent to: 3(c+d)((c+d)^2-2(c^2+d^2)>=0(true since -(c-d)^2.(c+d)>=0).
Therefore, we now just need to find the maximum value in the case c=d according to [b]MIXING VARIABLE THEOREM[/b]. If you haven't heard of it, here's the link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113248

However, I don't use this method as a part of my proof although it completely works in this case and even in the more general case. I temporarily name my method here is:[b]" PARTIAL MIXING VARIABLE" [/b]because I don't deduce at most number of equal variables as SMV does.

[b]5th step[/b]: Once you need to find the maximum of the original polynomial, you now can deduce it into the case of 3 variables: a,b and c. In other word, all you have now is: a+b+2d=0 and a^2+b^2+2d^2=0.

This step is used to simplify the problem and you can see how efficient it is when you just need to find the maximum of: a^3+b^3+2d^3, or rewrite it in my form is: [b]6d(2d^2-1/2)[/b](now it becomes 1-variable polynomial) :wink:

[b]6th step[/b](In my opinion, it's an excellent step):
From (1) we get: 0>=a+3d, which means -3d>=a. So we have: 9d^2>=a^2(since -3d>=0 and a must be>=0 since a=max(a,b,c,d)).

On the other hand, since a+d=-(b+d)>=0(since if a+d<0, then b+d<0 as well since a>b, which is unreasonable) we have: -d>=b, which means d^2>=b^2.
Therefore, from (2) we have: 1=<12d^2, which means d=<-1/sqrt(12)(since d<0)

Using derivative(yeah we must use it in this situation), we have: f '(d)=36d^2-3. We see that the critical number here must be -1/sqrt(12), since d<0. In the interval (negative infinity, -1/sqrt(12)), f(d) is a decreasing function, so f(-1/sqrt(12)) is a relative maximum of f(d).

Since -1/sqrt(12) is also the maximum value of d, so the maximum value of f(d) is: sqrt(3)/3.
The equality occurs when d=b=c= -1/sqrt(12) and a=sqrt(3)/2(Q.E.D)
Here's my comment on this problem:
Now after working through it completely, if it is given in a contest requiring only answer(like AIME), but in the general form as Mr. Kunny mentioned above, we really really need to use SMV immediately, or LARANGE MULTIPLIER to know its answer.If it's a "USAMO" problem, then we need to show an elementary proof. In that case, my method here seems to work.

The last one is difficult to read  :(
But please do not use calculus, that's why I wrote "Inequality"  :D
ok my friend said:
you need to prove that $ \minus{}\frac{\sqrt 3}{2}\le x_1,x_2,x_3,x_4\le \frac{\sqrt 3}{2}$.
good luck. :)

[quote="mathson"]The last one is difficult to read  :(
But please do not use calculus, that's why I wrote "Inequality"  :D
ok my friend said:
you need to prove that $ \minus{} \frac {\sqrt 3}{2}\le x_1,x_2,x_3,x_4\le \frac {\sqrt 3}{2}$.
good luck. :)[/quote]
Your comment is very useful.To be honest, I already think of it before but I can't find a way to deal with it.
We can easily have: (a+b+c)^2=d^2=<3(a^2+b^2+c^2)=3(1-d^2)
Therefore, -sqrt(3)/2=<d=<sqrt(3)/2.
using the same argument, we can infer that every variables lie in the interval [-sqrt(3)/2, sqrt(3)/2].
Now, just applying it to my function: f(d)=6d(2d^2-1/2) (d belongs to [-sqrt(3)/2,sqrt(3)/2]. Now calculus seems to be the fastest way(I'm not sure though! I'll do it later when coming home). :)

EDIT: Finally, I'm done with my school's stuffs. Let me add something to the above: I think very carefully about my last step to see whether I can do in a better way(i.e, a shorter way), but I regretfully say "NO" now. Why? In fact, your bound above is "just" useful for the largest variable in your assumption( in my case, this is a). How do I know this? The answer is: Looking at the equality, what is the most significant thing you have to consider? That is, the largest variable uniquely reaches its largest bound: sqrt(3)/2. In other words, if I'm considering my function as a function of 1-variable: a, then it's useful to use this bound. However, in my case, f(d) is the function of the lowest variable(depending on my assumption at the beginning). Therefore, two bounds of d:sqrt(3)/2 and -sqrt(3)/2 will be meaningless unless I use derivative to find the maximum value of f(d).Hope you understand what I meant though!


un problema de combinatoria :

sea $P_{n}(k)$ el numero de permutaciones del conjunto ${1,2,3,.....,n}$ el cual se tiene k puntos fijos . Probar que :

     ( $sumatoria$ $desde$ $k=0$ $hasta$ $n$ )$kP_{n}(k) = n!$

disculpen por no saber latex (el simbolo de sumatoria)


calculus
topology
videos
function
geometry
geometric transformation
reflection

How intelligent do you have to be to participate in the IMO? Do you have to be gifted? Having done Calculus at the age of 5? Or,

Can you work to be at a level of the IMO?

I am in Australia and i would say getting to the IMO is most likely the hardest! The test isn't hard, its the way they select! Unlike the USA and alot of other countries, we dont do one test for which they make another round and continue till they have 6 people. We have about 3-8 tests that dont even tell you if your going to IMO, we have tests that just get looked at when the people select people for the IMO.

But this is not the intended thesis of this post, it is of course to get answers on the name of the post, How intelligent do you have to be, and, do you have to be a genius?

I wouldn't classify myself as genius! I have a passion and crave for mathematics, i truly appreciate the beauty and elegance of mathematics. I skipped 1 year of advanced mathematics and toped that year in all the tests to date. I could have skipped 2 year of adv. math but there are stupid board of studies laws that i wont go into that stopped me from going a head 2 years. I really dislike rural schools as they have held me back! I cannot express my true... intelligence...

I have 3 years until i finish high school, is it possible for me to work harder than before and possibly get into the IMO?

I'll probably get blasted for this but whatever:

Do you need to have some natural abilities?  Of course.  Do you need to have won a Fields Medal already?  Definitely not.  Practice is really really important since by doing old problems, you'll get a sense of what they look like, and how they are solved.  You might also pick up on some tricks.  But no, you don't have to be one of the people who finished calculus at 5.

I agree. Being at a level where you can compete at the IMO for a major country is going to require some raw talent as well as a considerable amount of work.

There are people in the UK training system who have improved for some time and then been in the system for a long time and not seemed to get much better, never getting to an IMO despite having been in training (and having worked pretty darn hard) for two or three more years than other people who do qualify for the IMO.

I suspect there probably is a limit to a person's natural ability to magic solutions to hard problems out of their intuition, or at least a point at which a radically different training approach is required to improve their problem solving standards. I can't say I like this idea. It certainly isn't terribly égalitarian or fair, but then neither is life.

That said, this is no reason not to try to better oneself and to push one's own limits. Just because such a limit might exist, you don't know where it is, and you will only be able to find out if you put in the work and assume you can be good. If you are a genius, you are going to have to do some work for such attributes to actually make a difference.

Btw, I started working towards the IMO with only two years of 'high school' left, and qualified last year and this year for the UK, so the idea that you are 'too late starting' shouldn't be an issue, especially given the IMO is at least 70% about cleverness and there is very little knowledge to be learnt before you are able to answer most IMO questions.

Also, don't knock rural schools. If you're good, just bargain with them until you have your way ;). If you have the passion you seem to then you can motivate your own learning, especially given you have the internet and the manifold of resources there. Rural schooling seemed to be in some ways beneficial to Ildar (from the Australian team last year), I think from Queensland and the only team member not from Sydney, who ended up negotiating his own office with his school where he could happily do lots of independent work.

Finally, of course it's hard. There are 6 places. In a country as large and economically strong as Australia, many more than 6 people will have ambitions to make the team. Don't be discouraged if you don't.

[quote="Ilthigore"]
Btw, I started working towards the IMO with only two years of 'high school' left, and qualified last year and this year for the UK, so [u]the idea that you are 'too late starting' shouldn't be an issue[/u], especially given the IMO is at least 70% about cleverness and there is very little knowledge to be learnt before you are able to answer most IMO questions.
[/quote]

Ok, you're right there but all the good performers at IMO start soonely because their country organizes competitions for young students wich pushes them to work, whereas in other countries which does not, the students generally don't do well on IMO  :huh: apart some exceptions of course :D .
A 2 years experience guy can never be like a 4 years one, and some problems require only knowing the technique or the theorem.
I'm I right?

In morco for example olympiad test starts until 16 years old students, for me I've never known mathematical olympiad until that age...

if someone selected feels like he can't do it, should he quit?  :blush:   :(  :)

[quote="Ilthigore"]I suspect there probably is a limit to a person's natural ability to magic solutions to hard problems out of their intuition, or at least a point at which a radically different training approach is required to improve their problem solving standards. I can't say I like this idea. It certainly isn't terribly égalitarian or fair, but then neither is life.[/quote]

[quote="Ilthigore"]Finally, of course it's hard. There are 6 places. In a country as large and economically strong as Australia, many more than 6 people will have ambitions to make the team.[/quote]

Yes and most of those who make the Australian team will probably be Asian and that fact has to do with culture not genetics.

Your attitude is mistaken, ilthigore. When you talk about response to training, you do not take account of:

1) [i]a young person's level of will-power[/i],

2) [i]their assumptions as to their ability and performance[/i].

both of which might be improved by being among those who foster a better environment (who obviously must have a clue about how to do so). You can call it better "training" if you like. I don't mean things like showing videos about Paul Erdos, as if he's some kind of model. Will-power and assumptions as to ability are not wholly determined by hard work. Two people can do the same amount of very hard work and have very different amounts of will-power and different assumptions regarding their ability and performance, and be related to by trainers in very different ways too... Which brings me to...

3) [i]the assumptions of the "trainers"[/i].

This is an extremely important factor. Take a look at what psychologists term the "Pygmalion Effect". You cannot expect those who believe, deep down, that intelligence is a function of individual genetics to be very bright on these matters.

The experiences you've had that you relate here don't undermine the above three points at all.

Look beyond simply "hard work" plus "genetics" (or "talent" to use a current euphemism) as determiners of success! Even among guys and gals who've spent a lot of time in the same rooms and scenes, other aspects of their environments and backgrounds, and how other people relate to them, will be very very different.

Look too at the statistically significant drop in the average IMO rankings of Britain under the present leader compared with its performance under the last-but-one. (The intervening one didn't do the job long enough for a robust comparison to be made). That's explained...how? Would you exclude the attitudes of the trainers from being a factor?

The present leader is the kind of "leader" who publicly declared 6 months before IMO2008 that Britain had 7 or 8 people who on a very good day might get silver medals 6 months later! Any football manager who showed such appallingly bad leadership would be given the order of the boot. Can you imagine a manager of a top-30 team saying around Christmas time, "The team members are strong enough to get to the quarter-finals of the FA Cup if they're very lucky?" The British guy also said in the same month that four of the six British places for 2009 (that's 18 months in the future) were "taken already". What's interesting is that no-one commented publicly about the dreadful leadership skills he was demonstrating.

Do you think these statements indicated good or bad leadership? And if bad, were they just aberrations which it would be churlish and nitpicky to comment on, or were they deeply significant of his approach?

But it's not just his fault. Look at the general culture.

As for your comment about "life" not being egalitarian, well sure, cultural conditions are anything but. So you're right in that sense. But be aware that the idea that babies are born stupid or clever (or with an "IQ" equal to such-and-such) is so dominant in Britain. If it weren't so dismally sad, it would be risible in a country that has been sinking for decades. But the idea continues to play well at Oxbridge and most of those who run the BMO pat themselves on the back and say "it's the way of the world" when the topic arises of why such a small number of mostly private schools account for the vast majority of places on the BMO team.

But you can't expect those in such positions to study the weaknesses of their own attitudes and culture very often.

Yes, of course there are exceptions to the "private school" rule. For all I know, you may be one of them - and good luck to you if you are. But what about the rule? That isn't explained by genetics, and it isn't explained by references to "life". My granddaughter is a member of the clergy and that sort of reference to "life" is typical of most of the wealthy Christians I've met. Most of them haven't "lived" in the sense of having met a wide range of people from a wide range of backgrounds, and haven't got the basis on which to comment usefully on "life" at all. Mostly of course they don't subject it to reflection, or support it with intellectual arguments. I'm sure you get what I mean about assumptions. But it's stupid regardless, however well they'd perform on IQ tests.

Look at how certain admissions types (whether educated at private schools themselves or having spent the last 20 years sucking up to those who have been educated at private schools and copying and internalising their culture) push the idea that they want students who have "talent" rather than just being "well-taught". Ha ha. Obviously the "well-taught" bit is a typical example of hypocritical sarcasm, which has the hidden meaning of "just crammed for exams". That's the kind of talk that its users think is sophisticated in England, world capital of mindless put-down sarcasm. Pathetically I've seen many academics at Oxford who come from other countries who try to copy it (saying one thing on the surface, a very different thing underneath), as if it really exemplifies the type of sophisticated person they want to be. Yuck! As for the "talent" bit, it's cover for what's basically sociobiology whether it's authors are bright enough to know it or not.

Another comment I'd make about your comment about "life" is how woolly it is, with reference to the thesis that you're attempting to support. You're a bright lad - I think you'll accept that point.

So be aware that there are those who have thought about these matters a very great deal - I first debated them at Oxford in the mid-1950s - who profoundly disagree with you. Throw the Wittgenstein in the bin, my friend! There are things that can and should be spoken of, which don't often get spoken of - for certain reasons which themselves don't often get spoken of. Old Ludwig didn't even touch on that!

D Stone

I would expect the abolishion of Grammer schools in the UK to have an effect and I think you may be able to see it. The some of most capable children will no longer get a the best education.
YOu would expect to see a change in the 1980's and beyond but I am not sure of all the dates and numbers.


Find the sum of all possible values for $ C$ which will make $ \underline{49C345}$ divisible by 15.

It's easy to check that it works for $ C\equal{}2$.  Now 15 doesn't divide 1000 or 2000 but it does divide 3000, so $ C\equal{}5$ and $ C\equal{}8$ will also work.  So the sum is $ 2\plus{}5\plus{}8\equal{}15$.


Given a positive integer $ m$, define the sequence $ \{ a_{n}\}$ as:

[list]$ a_{1} \equal{} \frac{m}{2} ,  \ \ a_{n \plus{} 1} \equal{} a_{n}\lceil a_{n}\rceil,$ for $ n \geq 1$.[/list]
Determine all the values of $ m$, for which $ a_{2007}$ is the first integer in the sequence.

Note: $ \lceil x \rceil$ is the [url=http://en.wikipedia.org/wiki/Ceiling_function]ceiling function[/url].

Posted before... Before you post a problem you should do a search first (it was on resources  :P ) :wink:
http://www.mathlinks.ro/viewtopic.php?p=925863#925863


linear algebra
matrix
complex numbers
linear algebra unsolved

Let $ x_1,x_2,\cdots,x_n,y_1,y_2,\cdots,y_n$ be $ 2n$ complex numbers,
Define:
$ a_{kj} \equal{} \sum_{i_1 \plus{} i_2 \plus{} \cdots \plus{} i_k \equal{} j}x_{i_1}x_{i_2}\cdots x_{i_k}$
$ b_{kj} \equal{} \sum_{i_1 \plus{} i_2 \plus{} \cdots \plus{} i_k \equal{} j}y_{i_1}y_{i_2}\cdots y_{i_k}$
Define the matrix
$ A \equal{} \left(a_{kj} \right)_{n\times n},B \equal{} \left(b_{kj} \right)_{n\times n}$
Prove that:
$ AB \equal{} E$ if and only if:
$ \sum_{k \equal{} 1}^nx_kb_{kj}\equal{}0,j\equal{}2,\cdots,n$ and $ x_1y_1\equal{}1$

Note that both $ A$ and $ B$ is upper-triangular.

What is E and I don't think you finished the last sentence

Oh,I am sorry $ E$ means the identity matrix.
In fact the later one is the special case of the former one. :)

what's the last sentence? the iff condition.

I am sorry,It's my fault.
Now I have edited it.
Can somebody help me?


algebra
function
domain
Functional Analysis
real analysis
real analysis unsolved

$T$ is a closed operator from a Banach space $X$ to another $Y$.
The null space of T is $\{0\}$. The range of $T$ is closed in $Y$.
Prove that there exists a constant a such that $\|x\| \leq a\|Tx\|$ for all $x \in D(T)$.

Denote the range of $T$ by $\widetilde Y$. It is a closed subspace of a Banach space and, thereby, also a Banach space. Since the null space of $T$ is trivial, we can define $T^{-1} : \widetilde Y\to X$. It is a closed linear operator whose domain is a Banach space. By the Closed Graph Theorem, it must be bounded, which is equivalent to the statement to prove.


limit
induction
real analysis
real analysis unsolved

$ (x_n)_n$ with $ x_0 \equal{} a\le1, x_1 \equal{} b\ge1$ and $ x_{n \plus{} 2} \equal{} \frac23x_{n \plus{} 1} \plus{} \frac13min\{1,x_n\}$.Prove that if $ x_2 \geq1$ then  $ (x_n)_n$ is convergent si and find $ \displaystyle\lim_{n\to\infty}x_n$.

Prove that  $ x_{3} \geq 1$. Then, by induction ($ x_{k} \geq 1 \vee x_{k \plus{} 1} \geq 1 \rightarrow x_{k \plus{} 2} \geq 1$) we have $ x_{n} \geq 1\forall n\geq 2 \Rightarrow x_{n} \equal{} \frac {2}{3}x_{n \minus{} 1} \plus{} \frac {1}{3}$, and now it's easy.

[quote]
That only works if $ a \plus{} 2b \geq 3$. If $ a \plus{} 2b < 3$ then the sequence approaches smaller values.[/quote]
But this is implied in the hypothesis since $ x_{2} \geq 1$.  The other case is a little longer since the recurrence relation is more complicated, but not too hard.

Sorry, you're absolutely right, I missed that condition when reading  :oops:


Given an array {a_1,a_2...a_n}.

you have to find out such sub array containing: {a_i,a_i+1,a_i+2,...,a_j} so that min(of this sub array)*(j-i+1) is maximum.

Actually this problem is like this:

there are lots of tall buildings on X axis. you are given the height of the builds at each position. find out the maximum area in which there are floors of the buildings.

for example, if the bulildings are like: 2,1,0,3,5 then the answer is: 6 (3*2)

What sort of bounds on n are we talking about?

Why is it times 6?  :?

Could you explain a little further.

[code]
    *
    *
   **
*  **
** **
[/code]

So

[code]
    *
    *
   ++
*  ++
** ++
[/code]

"+" denotes the maximum area.

well i don't know the boundary. but i can say it might be: H,W <= 10^6

Here is my solution:

h_i is height of the building at position i.

First you need a Stack of structures that hold two values : height and position
[code]
Loop through all buildings i, every time you loop,
   Pop from the top of the stack, until top(S).height >= h_i and check if top(S).height*(i-top(S).pos+1) gives you maximum area.

   Push (h_i, i) on the stack.
[/code]
That solves the problem in O(n). I think that's the optimal solution. Don't forget to pop the stack after you loop through all buildings.

Solution is based on the simple principle to extend rectange as much as possible.

yes my solution is same! But i don't know whether that would get AC.

(it is a local contest problem. so i can't check it now)

It would, it's simple to prove. You try to extend every height as much as possible -> you can't get more. Since you are trying to extend every possible height, you checked all rectangles, hence maximum is returned.

Also, I saw similar problem in USACO Training Site and Polish Olympiad in Informatics. It's certainly correct  :)

hmm... good news then! but i have not seen such problem at USACO training site. where is that?

Rectangular Barn in Chapter 6. I solved it in O(n^2) by applying that algo to every row.

Hmm how about this:

[code]
*
* *
* * *
* * *
* * *
[/code]

I don't understand how your approach (popping from the stack until we run into a height not less than $h_i$) works above :/

Care to explain?

you need to keep 2 information. 1. height. 2. position.

first you push: (5,1)
then when youare going to push (4,2) you can see that it is useless to keep the first data. so you pop (5,1) and push (4,1) [not (4,2)]

and so on...


Prove that the plan separated in parts by $ n$ circles can be filled with black and white ink, in such a way that every two neighbor parts have different colors.

Ok so if n = 0, I hope you can figure out how to do it.

Ok now assume that for any k circles you can color the plane with 2 colors.  Now consider k+1 circles in the plane, remove one and now you have k circles, so by our assumption you can color it with 2 colors.  Now we add back in the circle we removed, by the Jordan Curve Theorem the circle divides the plane into two disconnected peices, an inside and an outside.  Now take all the regions on the inside and invert their colors, since they were originally colored fine they cannot have any internal problems by doing this, and since the inside only touches the outside on the circle this also ensures no problems between the inside and the outside.

Interestingly, you don't really need circles you just need n simple closed curves that intersect eachother at points rather than along edges.

The difference between circles and arbitrary Jordan curves is that Jordan's theorem (which is quite advanced) isn't needed for the circles ;)

One can just as easily state it noninductively: label each region with the number of circles (curves) that contain it.  Then color all the regions with odd labels black and all the regions with even labels white.


ARML search geometry MATHCOUNTS AMC AIME

Hi,
I'm interested to be part of a ARML team; however, I have no idea what teams are there and how to contact them, so could anyone help me here?

There is a Texas ARML team (one organization with multiple teams, actually), and I'm pretty sure it's based near Austin. A forum search finds no other relevant information, except for the thread you already found pointing to the private Texas forum- does anyone have information for that team to add [url=http://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team]here[/url]?
Given the distances, a separate team based near Houston might make sense. This would require some discussion to satisfy the territory rules.

A search of the ARML forum for the work "Texas" does turn up a few other things, including the name of the person who was organizing the the Texas team in 2005 (Sam Baethge), the information that someone in Texas is mailing out invitations to join ARML, and the fact that "Klebian", an AoPS poster with many posts, is from the Houston area and has been on the ARML team.

sorcererofDM:

yes, there is a Texas ARML team(s)... we send 3 (Gold, Silver, Bronze). Sam usually invites highschool students who do quite well in AMC, Rice, A&M.
8th graders are invited based upon Mathcounts performance, AMC8/10 scores. I've heard TXML performance thrown around a few times.

Since invitation to the team is standing (i.e. once invited, you're always invited until you graduate), there generally are usually less than 10 new-comers each year. I hate to be bearer of bad news, but if you're past 10th grade, there is probably not too much chance of being invited (unless liek you got perfect AMC and AIME, destroyed A&M and Rice, and somehow managed not to be noticed at all before that).

Also, while it says the team is based in Austin, the team is pulled from all over Texas (though mostly from Houston, Austin, Dallas areas)


I didn't really know where to post this so here:
Why doesn't AoPs write books for physics ?
I mean they do have mr.markan (i don't know how to spell his name :D ) who has been to IPhO :o

Just my personal opinion: I think it is a lot harder to write a physics "problem solving" book.  The AoPS books seem to develop a lot of formulas, equations, tricks, and intuition so when you see a problem, you know how to handle.  Just from personal experience, I find this harder with physics.  You can teach the basic formulas and stuff but actually solving the problems might requirea really good physics insight, which may not come naturally.  IPhO is like this, I guess, but it just seems different than the IMO.  Also, to really get into physics, you need calculus and I haven't seen AoPS put out a calculus-based product yet.

the only thing i new before doing intro to algebra was that $ (3)(4) \equal{} 12$ and multiplication facts and division. My experience is that insight by doing problems. And yes i would agree that calculus is required to do physics but since they haven't released a calculus based product yet, it might be start. Because i hear many people asking for good physics books (including me  :D ) and it would be really really [b]really[/b] good if they came out with a book like says Physics vol 1 and Physics vol 2

If you don't know calculus, learn it  :wink: .  Otherwise, if you want some titles:

Ginacoli wrote a good book.  Very in-depth and insightful.  Just make sure you find the one without calculus.

Halliday and Resnick is considered one of the standards, although it does use calculus.  Very, very strong, and the problems are very nice.  It actually won an award as the best physics text of the 20th century (or something like that).

Feynman Lectures.  The classic.  The one and only.  But it uses calculus.

Also consider David Morin's book on classical mechanics and Griffith's Introduction to Electrodynamics.  They have some advanced material, though.

I actually have neither.  I do have the Feynman Lectures and Irodov's physics problem book.  You can use the Feynman Lectures (which have no problems) to learn the material and Irodov (which doesn't cover the material, but assumes you know it already) for the practice problems.

I don't think it's necessary for AoPS to come out with physics textbooks - there are already plenty of books with great problems.

Griffiths?  That is typically used at the advanced undergraduate level (junior year).  Way too difficult for basics, since it requires at least knowledge of calc 3.

If you want REALLY hard, of course you want Jackson. :D

I never did say i didn't know calculus! :!:

Then go with Feynman for the material, Halliday and Resnick for the problems.


You are given a pair of triangles for which two sides of one triangle are equal in length to two sides of the second triangle, and the triangles are similar, but not necessarily congruent. Prove that the ratio of the sides that correspond under the similarity is a number between $ \frac {1}{2}(\sqrt {5} \minus{} 1)$ and $ \frac {1}{2}(\sqrt {5} \plus{} 1)$.

Let $ a,b,c$ are the sides of the first triangle with $ a \ge b \ge c$ and $ b\plus{}c \ge a$
and $ d,e,f$ are the sides of the second triangle with $ d \ge e \ge f$ and $ e\plus{}f \ge d$
and $ r\equal{}\frac{a}{d}\equal{}\frac{b}{e}\equal{}\frac{c}{f}$

there are nine cases for two equal sides:
$ a\equal{}d$, $ b\equal{}e$ then $ r\equal{}1$
$ a\equal{}d$,$ b\equal{}f$ then $ r\equal{}1$
(*) $ a\equal{}e$,$ b\equal{}f$
(**) $ b\equal{}d$,$ c\equal{}e$
$ b\equal{}e$,$ c\equal{}f$ then $ r\equal{}1$
$ b\equal{}d$,$ c\equal{}f$ then $ r\equal{}1$
$ a\equal{}d$, $ c\equal{}e$ then $ r\equal{}1$
$ a\equal{}d$,$ c\equal{}f$ then $ r\equal{}1$
$ a\equal{}e$,$ c\equal{}f$ then $ r\equal{}1$

(*) $ a\equal{}e$,$ b\equal{}f$
$ r\equal{}\frac{a}{d}\equal{}\frac{b}{a}\equal{}\frac{c}{b}$
$ a\equal{}rd$, $ b\equal{}ra$, $ c\equal{}rb$

$ b\plus{}c \ge a$
$ ra\plus{}c \ge a$
$ ra\plus{}rb \ge a$
$ ra\plus{}r^2a \ge a$
$ r^2 \plus{}r\minus{}1 \ge 0$

idem for the other case.


geometry
rhombus
number theory
prime factorization

ok...here is where you discuss the answers to AFIN

Jli, i love the termology of this post.  ;)

Number 1 is pretty obvious. Divide into triangles if you don't realize it at first.

I have to refer to the AFIN chain to do the others i guess.

[quote="Iversonfan2005"]Who won the top 3 in this round?[/quote]

actually, only 3 answered this round but it was a day early so it doesn't count

[quote="jli"]here are the questions& answers to AFIN3

1.  a rhombus is inscribed in a circle with diagonal of 64.  what is the length of the other diagonal?

2.  what is the sum of the first AFIN1 natural numbers?

3. add the digits of AFIN2.

4.  $\sqrt{53+(AFIN3)\sqrt{125}}$ can be made into the form of $a+b\sqrt{c}$ when simplified fully. find c

5.  if there are AFIN4 digit telephone number how many numbers are there so that there are at least 2 same consecutive digits.

6.  how many numbers are in the prime factorization of AFIN5

1.  64
2. 2080
3. 10
4.  5
5.  34390
6. 4[/quote]

1)
obviously its going to be $\boxed{64}$

2)
$1+2+3+...+63+64=\frac{1+64}{2}\cdot 64=\boxed{2080}$

3)
$2+0+8+0=\boxed{10}$

4)
you dont really need AFIN3 to answer this question.
fist you simplify: $\sqrt{53+5(AFIN)\sqrt{5}}$

the answer is automaticcally 5

5)the total # of cases is 10^5

the number of cases where there are no same consecutive cases is 10*9*9*9*9

so 34390

6)
34390=2*5*19*181 so 4

I might join in the next round if there is one and I'm not on vacation yet.

[quote="Iversonfan2005"]Yeah, I screwed up on number $5$, I wish I got it because I could've made a perfect score  :([/quote]

Me too. I forgot that telephone #s could start with a 0.


Let $ x > 1$ be a real number, and consider the sequence $ \lbrace s_k \rbrace_{k\in \mathbb{N}}$ defined by
\[ \begin{cases} s_0 = 0 \\
s_k = s_{k - 1} + p_k^{ - x}(1 - s_{k - 1}) \end{cases}
\]
where $ p_k$ denotes the $ k$th prime number, with $ p_1 = 2$.  What is ${ \lim_{n\to\infty}s_n}$?

we take $ u_k\equal{}s_k\minus{}1$
$ u_0\equal{}\minus{}1$ and $ u_k\equal{}u_{k\minus{}1}(1\minus{}p_k^{\minus{}x})$ gives $ u_k\equal{}\minus{}\prod_{h\equal{}1}^{k}(1\minus{}p_h^{\minus{}x})$
then $ \lim_{n\to\plus{}\infty}s_n\equal{}1\minus{}\prod_{h\equal{}1}^{\plus{}\infty}(1\minus{}\frac{1}{p_h^{x}})\equal{}......zeta....$

Exactly
$ \lim_{n\to \plus{}\infty}s_n\equal{}1\minus{}\frac{1}{\zeta (x)}$


AMC
USA(J)MO
USAMO
geometry
geometric transformation
homothety
dilation

oops i put yes cause i thought you meant, is the problem possible.

but no, its not possible.

Eh WLOG two circles are tangent, otherwise perform a homothety about the center of one until it becomes tangent. Now consider the circle of radius $\frac{\sqrt{2}}{2}$ tangent to these two circles. We claim that any circle going through this small circle, with radius at least five, must go through one of the other two circles. Assume this is false, then take a homothety about the center of the contradicting circle until it is tangent to one of the two larger circles, and take a homothety about the point of tangency until the other large circle is hit as well. If it cannot be hit, then dilate with infinite ratio, so that the circle becomes a straight line, and rotate the line until it is tangent to the other circle. In each case, if the small circle intersected the new circle before, it also intersects the transformed circle. Now create a new circle that is tangent to all three large circles. Since the circle of radius $\frac{\sqrt{2}}{2}$ is tangent to two of the large circles and intersects the third, it has a radius larger than this new circle. But by Descarte's circle theorem, the new circle has radius greater than $\frac{\sqrt{2}}{2}$, a contradiction. Therefore no circle goes through the points within the circle of radius $\frac{\sqrt{2}}{2}$. But this circle contains a unit square with sides parallel to the coordinate axes, and therefore a lattice point, so we are done.

[quote="JSteinhardt"]But by Descarte's circle theorem, the new circle has radius greater than $\frac{\sqrt{2}}{2}$, a contradiction. [/quote]

 :rotfl:

but yea.. I concur

The only idea i had was that if you take 3 circles and make them tangent and increase the radius of the circles they will get farther away from the center all i needed to prove was that there was a point in the center...btw i put "No QED"

I believe that this will rigorously show that using tangent circles of radius 5 will be optimal.

Consider three circles in the plane. If they are not all tangent, then we may move one of the circles such that they are. It is quite clear that the gap from before the move contains the gap from after the move. This can be made rigorous by noting that the point of tangency is moved out of the original circle, and so the arc that limits the new gap lies completely outside of the old circle. Therefore, we only wish to have gaps between 3 tangent circles.

Now consider one of these such gaps. Suppose you decrease the radius of one of the circles, while keeping it tangent to the other two. Then, the points of tangency both lie outside of the circle, but there is a point on the other side of the circle inside the old circle, unless they never intersect. In both cases, the new circle's arc that limits the gap lies completely outside of the old circle, and thus the gap only gets smaller, and no new gap is introduced.

Wow I did this a completely different way (which is probably wrong).  First I proved (using a not very rigorous argument) that there must be three integer pts A, B, C that form a triangle with sides length 1, 1, and $\sqrt{2}$ with A, B, C each being in a distinct disc $D_{A},D_{B}, D_{C}$ respectively.  Then I considered a disc with radius $5+\sqrt{2}/2$ centered at the midpoint of the side of ABC that has length $\sqrt{2}$.  This disc contains the centers of $D_{A},D_{B}, D_{C}$.  But then at least one of of the side lengths of the triangle formed by the centers of $D_{A},D_{B}, D_{C}$ has length less than 10, which is a contradiction, since $D_{A},D_{B}, D_{C}$ all have radius greater than 5, and they are all nonoverlaping.

[quote="K81o7"][quote="JSteinhardt"]Eh WLOG two circles are tangent, otherwise perform a homothety about the center of one until it becomes tangent. Now consider the circle of radius $\frac{\sqrt{2}}{2}$ tangent to these two circles. We claim that any circle going through this small circle, with radius at least five, must go through one of the other two circles.[/quote]

Aha!  Now here's the way I was looking for to solidify that argument!  Okay, so this is the kind of rigor which would receive a 7...[/quote]

I just said, suppose we have a working configuration. Dilate all the circles until every circle is tangent to some other circle. But this is a pretty simple step.

I knew I had to use Descartes' theorem on this. The problem is that I couldn't remember it off the top of my head. So I had to come up with something else, and what I ended up putting had some major mistakes and I might not get more than a 1.

The basic idea is this: consider three disks of radius at least five that are mutually tangent. Descartes' theorem tells us that the disc that is mutually externally tangent to all three of these discs has radius

$r_{4}= \frac{1}{\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}+2\sqrt{\frac{1}{r_{1}r_{2}}+\frac{1}{r_{1}r_{3}}+\frac{1}{r_{2}r_{3}}}}$

Clearly, to minimize $r_{4}$, the other $r$s must be as small as possible; that gives us $r_{1}=r_{2}=r_{3}=5$. Substituting into the equation, we obtain

$r_{4}=\frac{10\sqrt{3}}{3}-5$

Since this is greater than $\frac{\sqrt{2}}{2}$, at least one grid point must be in the interior of this disc, and therefore there is at least one grid point between every three mutually tangent discs.

I actually had never heard of Descartes' Theorem before these post-USAMO discussions, but I believe that it's unnecessary if you use my solution. Once you establish that the gap between three tangent circles is contained in any gap in the layout, then you can use an equilateral triangle formed by tangents to the circles with side length $20-10\sqrt3$, but the longest a side length can be with such an equilateral triangle is $1+\frac{2}{\sqrt3}$, and so you end up having to prove that $\sqrt{3}>\frac{32}{19}$ but $\sqrt{3}> 1.7 > \frac{32}{19}$ and you're done.

[quote="Hamster1800"]I believe that this will rigorously show that using tangent circles of radius 5 will be optimal.

Consider three circles in the plane. If they are not all tangent, then we may move one of the circles such that they are. It is quite clear that the gap from before the move contains the gap from after the move. This can be made rigorous by noting that the point of tangency is moved out of the original circle, and so the arc that limits the new gap lies completely outside of the old circle. Therefore, we only wish to have gaps between 3 tangent circles.
[/quote]

Consider three circles whose centers are collinear, for example.

Hmm... I took a completely different approach to trying to solve this problem.  I tried to show that if any four points (k,k), (k,k+1), (k+1,k), and (k+1,k+1) are contained in non-overlapping circles of radius >=5, then they are contained in at most two non-overlapping circles of radius >=5.  This leads to a pretty clear contradiction, if all points are assumed to be contained by non-overlapping circles.  However, I thought of this argument with like 30 minutes left, and ran out of time trying to justify everything, so I'm don't really think I'll get many points for it.

[quote="TheAmazingOne"]Your argument may be correct and it expresses what to us seems "obvious," but I was wary of approaching the problem through any argument about "moving circles around" and "narrowing gaps" because there are a lot of cases to consider and it's just sort of iffy to begin with. The nice thing about using Descartes' theorem is that it makes for a brief argument with very little room for error.[/quote]

but what if you didnt know descartes.
then the diff between a 6 and a 1 is whether you know it?

Well, I didn't know it, but that didn't stop me from citing it. You might be able to get away with calling it a well-known theorem.

[quote="kstan013"]What if you had 3 mutually tangent circles and kept inscribing circles into the "gaps" between them infinitely so that the whole grid is filled?[/quote]

Then the inscribed circles will eventually have radius less than 5.

hmm...and where, exactly, did you encounter said theorem of Descartes, JSteinhardt?  i found it on MathWorld just now but have never seen it anywhere else.

but yeah it kills the problem.  completely.

hmm...i didn't even make such a nice argument as hamster's involving points of tangency, although i did use a lot of extremal principle and contradiction to show that you could find three disks in a 'triangle' shape with no other disks 'interfering' (not my terms), which nobody seems to be discussing...yet it's a nontrivial part of the proof.

I encountered Descartes' theorem at a previous math tournament I went to earlier this year. Wikipedia has a pretty good article on it.

I got the circle of radius \frac{\sqrt{2}}{2}, but not Descartes's circle theorem.  Hopefully, I'll get a few points despite the handwaving.

-Ravi

Okay so JSteinhardt made me rigorize my solution more during school today...

Suppose we have some configuration of circles. Consider two circles. Now suppose we place a circle with radius 5 tangent to both of them. This causes no space to be uncovered, and you can show (with points of tangency, etc) that the important (bounding) arc will not intersect a circle unless the disk that we just placed was already there. Then we can shrink the other two down to show that the gap between these three circles becomes even smaller, and no new gap between them is introduced. Then we have the hexagonal tiling of radius 5 circles, which we all know how to tackle.

I still like using the equilateral triangle. It's much stronger than the circle.

[quote="MellowMelon"][hide="stuff"]I had no idea how to first approach this problem. Didn't know Descartes, didn't even think of an enclosed circle of radius 2^(-1/2). So I spun out a totally wacky solution. Unfortunately since I wasted so much time on problem 3, I had to handwave a lemma that for a working solution there had to exist a point covered by a disk that was distance 1 from each of 2 points in` orthogonal directions each covered by a different disk. But with the existence of those 3 points, I proved that a vector of magnitude at least 5 (center of disk covering first point) and two other vectors of magnitude at least 6 (center of disks covering other two points, and I know 6 is a bit big) where each vector was a certain distance from each other provided a contradiction. Obviously a bit more rigorous on my solution paper, but I don't think enough for a 6-7.

The contradiction I came up with would have still worked out if they had asked for disks of at least radius 4 though... I guess I must have done something wrong, because the official solutions said no one knows if 4 is possible or not! :ninja:

(edit: typo)[/hide][/quote]

Woahs it is Palmer Mebane! Did you like disappeared or something because I haven't seen you at a chess tournament in a while. Hope we meet at MOP or something.

Anyways as for how I knew about Descarte's Circle Theorem, I remembered it being mentioned some time a while ago at an ARML practice, then later someone asked me if it was worth knowing and I was like NO! IT IS POINTLESS TO MEMORIZE THIS!

[quote="MellowMelon"]I had no idea how to first approach this problem. Didn't know Descartes, didn't even think of an enclosed circle of radius 2^(-1/2). So I spun out a totally wacky solution. Unfortunately since I wasted so much time on problem 3, I had to handwave a lemma that for a working solution there had to exist a point covered by a disk that was distance 1 from each of 2 points in` orthogonal directions each covered by a different disk. But with the existence of those 3 points, I proved that a vector of magnitude at least 5 (center of disk covering first point) and two other vectors of magnitude at least 6 (center of disks covering other two points, and I know 6 is a bit big) where each vector was a certain distance from each other provided a contradiction. Obviously a bit more rigorous on my solution paper, but I don't think enough for a 6-7.

The contradiction I came up with would have still worked out if they had asked for disks of at least radius 4 though... I guess I must have done something wrong, because the official solutions said no one knows if 4 is possible or not! :ninja:

(edit: typo)[/quote]

I think I did something pretty similar.  I also proved that there has to be a triangle with vertices at integer pts with side lengths 1,1, and $\sqrt{2}$ with each point in a distinct disc (say $D_{1}$, $D_{2}$, $D_{3}$), which looks like just a different way of saying what you proved with your lemma (I also used a handwaving argument, but I think the result is still probably true).  From there, though, I proceeded in a different way than you.  First you can assume that $D_{1}$, $D_{2}$, $D_{3}$ all have radius 5 - if they don't then nest a disc of radius 5 within each of the offending ones that still contains the required pt (and clearly still doesn't overlap with the other discs).

I then considered a disc $B$ of radius $5+\sqrt{2}/2$ centered at the midpoint of the side of length $\sqrt{2}$ of the afore mentioned triangle.  This disc contains all the centers of $D_{1}$, $D_{2}$, $D_{3}$.  You can then use the Law of Cosines to show that any triangle with in $B$ has a side of length less than 10 (on the test I just guessed the arithmetic would work out, which it does).  Obviously this is a contradiction since $D_{1}$, $D_{2}$, $D_{3}$ don't overlap and all have radius of 5 and form a triangle in $B$.

Incidentally the official solutions don't actually say whether it is known if 4 is a lower bound on the minimum radius required.  It only says that the minimum is greater than 1.8something and less than 4.5something and that it is not known exactly what the value is.

My ideas on number 2 are as follows (this is an outline of my proof):

Choose the first lattice point P outside of some circle A. Say you have a second circle B which covers it.  After establishing several bounds, by the Law of Cosines, we find the angle APB to be huge.

So take the second "first" lattice point Q not in A or in B.  The circle C that goes through this point must have radius less than a circle that goes through Q and is tangent to A and B.  We can show the impossibility of this by considering triangle ABC and with the knowledge that angles APB, BPC, CPA must sum to 360.

I also never mentioned the Descartes' Theorem.

But, somehow I managed to grab the circle part with rational thinking after 2 hours of agony.  :ninja:

Oh well, I am even thinking about going to MOSP anyway. I am happy if I get more than 9 or 10. I am only an 8th grader... :roll:


since they ask for the difference between the two averages, it'll be.......

85+86+87=258 258/3=86

86+87+88=261  261/3=87

87-86=1

i hope i did well on my AMC-8!!


trigonometry
inequalities unsolved
inequalities

Let triangle $ ABC$. Prove that
$ 3a^2\plus{}2b^2\geq c^2\plus{}4S$

$ \Leftrightarrow 3a^2\plus{}2b^2\geq a^2\plus{}b^2\minus{}2ab\cos C\plus{}2ab\sin C\Leftrightarrow 2a^2\plus{}b^2\geq 2ab(\sin C\minus{}\cos C)$
It is OK because $ 2a^2\plus{}b^2\geq 2\sqrt{2}ab\geq 2ab(\sin C\minus{}\cos C)$


Find all functions from §§\{R^{+}}§§  to §§\{R^{+}}§§
Such that for all x and y ,§§\0<y<x§§ ; §§\f(\frac{x-y}{f(y)})=\frac{f(x)}{f(y)}§§.
 :ninja:

I guess it should look like this:

[quote="arkhammedos"]Find all functions from $R^{+}$  to $R^{+}$
Such that for all x and y ,$0<y<x$ ; $f(\frac{x-y}{f(y)})=\frac{f(x)}{f(y)}$.
 :ninja:[/quote]


geometry
search
analytic geometry
MATHCOUNTS
similar triangles

Can someone post the solution for $ \#10$ of the $ 2008$ National Team Round? No matter how hard I try, I can't seem to get it.

$ \text{In the figure, what is the area of triangle ABD? Express your answer as a common fraction.}$ [asy]unitsize(5mm); defaultpen(linewidth(.8pt)); draw((0,0)--(0,4)--(7,0)--cycle); label("$A$",(0,4),NW); label("$B$",(7,0),E); label ("$C$",(0,0),SW); draw((0,2)--(7,0)); draw((3,0)--(0,4)); label("$E$",(0,2),W); label("$F$",(3,0),S);label("$D$",(intersectionpoint((0,2)--(7,0),(3,0)--(0,4))),NE); label("$3$",(0,0)--(3,0),S);[/asy]

Triangle $ ABC$ is a right triangle. $ AE=2, EC=2, CF=3, FB=4$.

Nikeballa, I don't UNDERSTAND those solutions.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1167436233&t=268977]Search.[/url] There are 3 solutions there.

Could you say what you don't understand about those three?  (Focus on the first two, 1st-similar triangles, 2nd-coordinate geometry, the third uses stuff to advanced for mathcounts)


i need some problems of this part  :|
about Abel :|

problem like this :

[b]Let a,b,c are positive real numbers satify a <= b <= c ; (1/a)+(2/b)+(3/c) >= 3 ; (2/b)+(3/c) >= 2 ; 3/c >= 1 ! Prove that (a^3)+(b^3)+(c^3) <= 36 [/b]

Thanks a lot   :welcomeani: !!!!!!!!

[hide]We have $ \frac{2}{b}\plus{}\frac{3}{c}\ge2$. Because $ \frac{3}{c}\ge1$, we have $ \frac{2}{b}\plus{}1\ge2 \implies \frac{2}{b}\ge1$. In the same way, we have $ \frac{1}{a}\plus{}\frac{2}{b}\plus{}\frac{3}{c}\ge3 \implies \frac{1}{a}\plus{}1\plus{}1\ge3 \implies \frac{1}{a}\ge1$. Thus, we have the following inequalities:

$ \frac{1}{a}\ge1$

$ \frac{2}{b}\ge1$

$ \frac{3}{c}\ge1$

Since $ a$, $ b$, and $ c$ are positive, we may rewrite them

$ a\le1$

$ b\le2$

$ c\le3$

Now, $ a^3\plus{}b^3\plus{}c^3$ is maximized when $ a$, $ b$, and $ c$ are maximized, and their maximums are $ 1$, $ 2$, and $ 3$, respectively. Thus, the maximum value of $ a^3\plus{}b^3\plus{}c^3$ is $ 1^3\plus{}2^3\plus{}3^3\equal{}36$. Therefore,

$ a^3\plus{}b^3\plus{}c^3\le36$.[/hide]


How many strings of length k can be written with the following rules?

1) Each element is a letter (A, B, C, ..., Z)
2) Each element is greater than or equal to the element immediately before it (B > A, C > B > A, etc.)

[hide]For k<26

(25+k)Ck is the answer[/hide]

I have to leave now. I will explain after I come bacik.

I'll also say [hide]C(25+k,25).  Imagine k stars, where a letter can be placed on each star.  Then imagine 25 dividers.  Every star to the left of the first divider is an A, in between the first two is a B, etc.  We want to place the 25 dividers in between (or on the outside of) the stars.  So, we're looking for all of the arrangements of stars and dividers (bars).  Since there are 25+k total spaces in which either a star or a bar can be placed, and there are k stars and 25 bars, there are C(25+k,25) or C(25+k,k) ways to do this.[/hide]

I realise that some confusion could result from the phrasing of the question. I've attempted to make it clearer... sorry. ^_^"

[quote="Osiris"]Caroline, a question - how do you differentiate AAB and AAC, for example?[/quote]

AAB would start with **\*\, whereas AAC would start with **\\*.  These are different choices of placement for the stars and bars.

My method:

[hide]Let A=1, B=2, C=3.....Z=26. The problem is equivalent to find all possible sequence [tex]a_1, a_2,...a_k[/tex] that satisfies 1 :le: [tex]a_1[/tex] :le: [tex]a_2[/tex]... :le: [tex]a_k [/tex]:le: 26.

Here is the creative part(I learned from kalva).

Imagine 26 black guns and k white guns arranged in a sequence that it must start with a black gun. Then value of a_n corresponds to number of black guns to the left of the nth white gun. It always works.  The total number of such possible sequence is (25+k)Ck[/hide]

Now I'm not sure I understand this (after the edit and then looking at solutions...), but the way I see the problem:

[hide]Have k each of A, B, C, etc.  That gives 26k to start with.  We need to kill off 25k in order to get a sequence of length k.  Given k items, there is always exactly one way to arrange the letters to meet the second criterion.  Therefore, I get C(26k, 25k).  I probably don't understand the question, got meddled by the other solutions, or -- ok, no more excuses -- can't do this problem.[/hide]

Probability -- imagine, in your solution, that we eliminated the first A and no other A's.  Then imagine that we did it again and eliminated the second A and no other A's.  We would end up with the same resulting solution.  You overcount heavily.


A drunkard is unaware of the path he takes. He has equal probability of making a step in any of the eight directions. He starts moving one step away from the well and takes 9 steps. What is the probability that he will fall into the well at the 9th step?? :D  :D  :D

edited:
NEW APPROACH.
Assume that the initial point is (0,0,0,0,0,0,0,0) and the final point be ((+ or -)1,0,0,0,0,0,0,0) or a permutation of it (i.e. one or -1 can be in any place)
Now since we have to travel to the point in 9th step = prob of him travelling to the pt in the 9th step - // prob he falls before that (commenting the subtraction dunno if necessary)
let the no. of steps he take in the ith direcn be Ai .
Implies A1 = even ; A2 =odd ; ..............  ; A8 = odd or a permutation of it.
no. of ways w1= coeff $ x^9$ in the expansion of
$ 16\cdot(x^1 \plus{} x^3 \plus{} x^5 \plus{} x^7 \plus{} x^9)\cdot (1 \plus{} x^2 \plus{} x^4 \plus{} x^6 \plus{} x^8)^7$
no of places he can go in 9 steps is w2= coeff x^9 in the expansion of
$ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6 \plus{} x^7 \plus{} x^8 \plus{} x^9)^8$
Therefore probability = w1/w2
( Dunno if the 16 is also necessary )
:)

btw why did ya post it thrice?
 :spam:  :spam:

[color=blue]MOD EDIT:I have deleted the other two posts by srinitrs[/color]

Dei he can move anywhere in the 8 coord system so hence assumed that 8 coords.
Also in multinomial done for 8 diff expressions.
And i think wat i've done is right
bcos i've not counted the expressions where he reaches in fewer steps bcos then the coeffs would change  :maybe:
And by the way i'll post a link in pre olympiad section so we'll see what's the real answer  :D

[quote="srinitrs"]semma seen approach but where are considering the possibity that in making a step he can move in any of the 8 directions ??? is it the same if he had ony 2 directions to move in or 4??? :lol:   :D  and one more thing shouldnt we consider the possibilty that he must not fall into the well before he goes upto the 9th step no???  :spider:  this is how the drunkard now is in???so your multinomial theorem fails here as it considers all the posibbilties for w1 your w2 is perfect of course??? :huh:  :ninja:[/quote]

Well, if you want for considering he may fall at any step, just extend rohits logic.

Prob (total) = Prob(he falls in one step) + Prob(2 steps)+.....+... Prob(9).

A set of multinomials..


Hi Folks, I know that Metro Atlanta's Chapter will be on Sat, Feb 6, but does anyone know when Cobb's event will be? Someone told me it might be held on Fridays - true?

Thanks!

Let me ask an easier question: have Cobb Chapter MC's been on Fridays in the past?

My memories are pretty stale and dusty by now, but I think it was on a Saturday back in the days when I was in middle school.  (This is Miles, by the way.)  There are some dates here :  http://www.gspe.org/mathcounts.htm , and Cobb's is listed as TBA.

I also remember MathCounts chapter being on Saturdays as well, but again, that was like 2005 or earlier, and my brain is currently hammered from the Putnam.

So how'd everyone do?

Once again, I could not come because of piano stuffs. :*(

MATHCOUNTS in Cobb has always been on Saturday recently. Usually the weekend of Valentine's Day (or the state swim meet depending on what's more important! :-)

Whoops, posted the Lassiter thing as a reply instead of a topic.  Ignore it please!

I think it's supposed to be on Saturdays, but I don't know.  I've never been to the Cobb MC thing.


function
calculus
derivative
analytic geometry
graphing lines
slope
limit

The task im having trouble with:

Explain how Laplace transformations can be used to solve differentional equations with constant coefficients where you have a pulse starting at t = 0 , the pulse lasts for 0,100s.
During this time it rises lineaer in 0,070s to the value 1, then it falls linear to 0.

I know the transform linearity is: L(af + bg) = aL(f) + bL(g)

I would be greatful if I someone could give me some hints how to solve this task, some pointers for me to work on. A short desc about my goal would be great..
Im stuck, I dont "understand" all the relationships with linearity and the laplace transfer.

Offtopic: my first post, searched alot in this forum.. great resource

Let's see if I have this right: you have a function $f(t)$ which is a piecewise continuous linear function with vertices at $(0,0),\,(.07,1),$ and $(.1,0)$ and zero after that. We'd like to compute the Laplace transform of that.

It would be convenient to compute the derivative of the function first, using the relationship $\mathcal{L}(f')(s)=s\mathcal{L}(f)(s)-f(0).$ In fact, it's even better with two derivatives, provided, of course, that we interpret these as "weak" derivatives:

$f'(t)$ is a step function which takes on the value $\frac1{.07}$ on $(0,.07)$ and the value $-\frac1{.03}$ on $(.07,1).$

$f''(t)=\frac1{.07}\delta(t)-\left(\frac1{.07}+\frac1{.03}\right)\delta(t-.07)+\frac1{.03} \delta(t-.1).$

$\mathcal{L}(f'')(s)=\frac1{.07}-\left(\frac1{.07}+\frac1{.03}\right)e^{-.07s}+\frac1{.03} e^{-.1s}.$

$\mathcal{L}(f)(s)=\frac{\frac1{.07}-\left(\frac1{.07}+\frac1{.03}\right)e^{-.07s}+ \frac1{.03} e^{-.1s}}{s^2}.$

To use this to solve a differential equation or functional equation, we'd have to have that equation so we could apply it.

[b]Thank you so much, that did clear alot of things up![/b]

There are no equation to apply it to in the task, so we can choose what diff equation to apply it to.

Are there any good or bad choices regarding which equation to apply it to?

I dont know all the tweaks/tricks with all equations so I can trap myself by choosing the wrong one maybe?

Thanks again for all the help Kent Merryfield..

Just as a trial run, how about this?

$y''+4y'+3y=f(t);\,y(0)=0,\,y'(0)=0.$ where $f$ is the function above.

What is the solution $y(t)$?

Thanks for the math task you gave me :P perfect.. im getting to that later :)

..but im not that far yet. I've attached a image of the step function, I had to add the +3.33 at the end to make it correct.. I didnt figure any other way to make it correct, but I must have mist something, or is it an okey solution?

On the interval $(.07,1),\,f(t)$ is a linear function with slope $-\frac1{.03}.$ Thus its equation in that interval is $f(t)=-\frac1{.03}t+b.$ Applying the condition $f(.1)=0$ gives us that $b=\frac{.1}{.03}=\frac{10}3.$ (Of course, your "3.33" was only an approximation.)

I notice you told your software to differentiate it, and it came back with the classical derivative, telling you that the function is not differentiable at the vertices. For the weak derivative, we don't care about those single points.

Thank you, that cleared it up for me why 3.33 worked :) Replaced it with your more accurate value +(0.1/0.03), and now i understand why it's there too :)

And nice that you explained why we dont care about the weak derivative :)

I'm trying to solve the y(t) task you gave me, but as you can see the output isn't really nice, so im expecting something is wrong, or is the solution correct?

Except that problem, what else do you think should be added to this solution ? Some text explaining what's going on, do you recommend more math like you wrote in your first post to me ?

Btw, math application the screenshot is from is Maple 9.5

The solution won't be that nice. To start with, let's write the numerator of my expression in #2 as $a-(a+b)e^{-\alpha s}+be^{-\beta s},$ where $a=\frac1{.07},\, b=\frac1{.03},\, \alpha=.07,\,\beta=.1.$ (Note: I had to fix a couple of typos up there, so check it again.)

Let $y(t)$ be the solution of our differential equation, and let $Y(s)$ be its Laplace transform.

$(s^2+4s+3)Y(s)=\frac{a-(a+b)e^{-\alpha s}+be^{-\beta s}}{s^2}.$

$Y(s)=\frac1{s^2(s^2+4s+3)}\left(a-(a+b)e^{-\alpha s}+be^{-\beta s}\right).$

Now we have a partial fractions problem:

$\frac1{s^2(s^2+4s+3)}=-\frac49\cdot\frac1s+\frac13\cdot\frac1{s^2}+\frac12\cdot \frac1{s+1}-\frac1{18}\cdot\frac1{s+3}$

That means that $\frac1{s^2(s^2+4s+3)}$ is the Laplace transform of the function

$g(t)=-\frac49+\frac13\,t+\frac12\,e^{-t}-\frac1{18}\,e^{-3t}.$

Now, multiplcation by $e^{-\alpha s}$ on the Laplace transform side is the same as translating by $\alpha$ (and simultaneously multiplying by $u(t-\alpha)$ where $u$ is the Heaviside unit step function.)

Thus $y(t)=ag(t)-(a+b)g(t-\alpha)u(t-\alpha)+bg(t-\beta)u(t-\beta).$

Putting in the definitions of $a,\,b,$ and $g$ and writing all of that out in detail would result in a rather complicated looking expression. It would probably be worth your while to find a way to draw the graph in Maple of that expression.

Thank you so much for your help, you have no idea how much i appreciate it :)

Maybe I'll step into some problems, or maybe not, your explanation seemed very good and great deatils explaining it easy.

Been in a meeting all day, so I'm gonna work with this and your reply in a few hours..

PS! If you need some computer help, just ask! :P hehe..

I've put in the values and did plot the solution to have a graphical view of it.

I've only taken values for t from 0 to 1.1, which is the value that's interesting for me (correct?).
What seems abit strange is that it goes as far as 1.2 before it crosses the t axes, and it starts at t=0, the value is aprox. 9,5. I have a hard time realising what it graphically means, it's the "system" (the diff.) that is exposed for the pulse , right? What "happends" when this happen is that output i've plotted.. have i understood it correctly ?

The graph is clearly wrong. That's not how it behaves.

I see one mistake in the definition of $g(t)$: that should be $-\frac49$, not $\frac49.$

But that's clearly not all. That mistake alone would give you the value of $\frac89$ at zero, not 9.5.

I see a bigger mistake: you twice wrote u*(t-alpha) or u*(t-beta). That should be u(t-alpha) and u(t-beta). And make sure that $u$ has been properly defined somewhere in the system.

If you're going to include values to the left of zero, then $y(t)=0$ for $t<0.$ It would probably be best to write

$y(t)=ag(t)u(t)-(a+b)g(t-\alpha)u(t-\alpha)+bg(t-\beta)u(t-\beta).$

The correct solution of this differential equation should satisfy these properties:

$y(0^+)=0,$

$y(t)>0$ for all $t>0.$

$\lim_{t\to\infty}y(t)=0.$


function
algebra
polynomial
complex analysis
complex analysis unsolved

I imagine that Fermat's "last theorem" for polynomials (there are no nonconstant polynomials $f,g,h$ satisfying $f^{n}+g^{n}=h^{n}$ for $n >2$) has appeared on this forum, more than once. Let's consider Fermat's equation $f^{n}+g^{n}= h^{n}$ with $f,g,h \in \mathcal{O}(\mathbb{C})$ entire functions.

1. Suppose $n \geq 3$, and assume that one of $f,g,h$ are constant. Show that *all* functions $f,g,h$ are constant. Equivalently, show that for $n \geq 3$ the equations $f^{n}\pm g^{n}= 1$ have no solutions in nonconstant entire functions $f,g$.

2. However, show that for $n =3$ there DO exist nonconstant entire solutions. Find an explicit such triples $f,g,h$ of entire functions satisfying $f^{3}+g^{3}=h^{3}$.

3. For $n \geq 4$, show that $f^{n}+g^{n}=h^{n}$ has no solutions in entire nonconstant functions $f,g,h$.

Enjoy!
--Vesselin

A stronger version of (1) follows from Picard's theorem (which states that a non-constant entire function can miss at most one value in the complex plane), namely:

If $f^{n}+g^{n}=h^{n},\ n\ge 3$ for non-constant $f,g,h$, then each of them has zeros.

Assume, for example, that $g$ has no zeros. This means that $f^{n}-h^{n}=(f-h)(f-\zeta h)\ldots(f-\zeta^{n-1}h)$ has no zeros, where $\zeta$ is a primitive $n$'th root of unity. In particular, $u=f-h,u-(\zeta-1)h,u-(\zeta^{2}-1)h$ have no zeros. Dividing the last two by $u$, we find that the (entire, because $\frac{1}{u}$ is entire) function $\frac{h}{u}$ misses $\frac 1{\zeta-1},\frac 1{\zeta^{2}-1}$, so by Picard's theorem it must be constant. The conclusion follows easily from this.

I'll think about the other two.

Indeed, this is the solution I knew. But note that this is no more general than what I have formulated: if, say, $g$ has no zeros, then $f_{1}= f/g, h_{1}= h/g$ are entire solutions to $h_{1}^{n}-f_{1}^{n}= 1$.


Draw a triangle with these information:
1. $(t_a,t_b,t_c)$
2. $(t_a,m_b,b:c)$
3. $(t_a,m_b,h_c)$

$t_a,t_b,t_c$ are bisectors of angels A,B,C

That's what I was afraid of :).

Check [url=http://www.cut-the-knot.org/triangle/TriangleFromBisectors.shtml]this[/url] out. You can't (in general) construct a triangle with the ruler and compass (I assume that's what you mean by construction), given just the lengths of its three bisectors.

[color=red][Moderator edit: This shows that problem [b]1.[/b] cannot be solved by compass and ruler. Problem [b]3.[/b] cannot be solved by compass and ruler as well. I am not sure about problem [b]2.[/b] yet. Darij][/color]


[quote="Peter"]Let $ x, y$, and $ z$ be integers with $ z > 1$. Show that
\[ (x \plus{} 1)^{2} \plus{} (x \plus{} 2)^{2} \plus{} \cdots \plus{} (x \plus{} 99)^{2}\neq y^{z}.
\]
[/quote]
$ 99x^2\plus{}2x(1\plus{}...\plus{}99)\plus{}(1^2\plus{}..\plus{}99^2)\equal{}y^z$
$ \Rightarrow 3|y$ imply that $ 9|y^z$
But $ 9\not |99x^2\plus{}2x(1\plus{}...\plus{}99)\plus{}(1^2\plus{}..\plus{}99^2)$
So it hasn't solution.

[hide=Solution]Let $k = x + 45$. We wish to show $$(k - 44)^2 + (k-43)^2 + \cdots + (k+44)^2 = y^z$$ has no solutions. Once the terms on the LHS cancel, we are left with $$99k^2 + 2\cdot \sum_{i=1}^{44} i^2 = 99k^2 + 44\cdot 15\cdot 89 = y^z.$$ Clearly by taking $\pmod 3$ on the equation, we get that $y\equiv 0\pmod 3$.  However, taking the equation under $\pmod{9}$, we get that $y^z\equiv 0\pmod 9$ since $z>1$ but $99k^2+44\cdot 15\cdot 89\not\equiv 0\pmod 9$, a contradiction.[/hide]


For any positive integer n, we have $\sum_{d|n}\phi(d) = n$.

hint: think of reducing the fractions 1/n, 2/n, 3/n, ..., n/n, you might be able to come up with a semi-rigorous proof.


construction
geometry
IMO
altitudes
IMO 1960

Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

I found two triangles. Is this correct?

Well, suppose we have constructed the triangle $ABC$ with the altitudes $AD=h_a, BE=h_b$ and the median $AM=m_a$

The perpendicular from $M$ to $AC$ intersects the line $AC$ at a point $K$.

From the triangle $BCE$ we can see that $MK = \dfrac{h_b}{2}$.

So, let's start with the construction.
We construct first the triangle $ADM$.
It's easy to construct it, since we know that it is right with hypotenuse $AM$ and one perpendicular side $AD$.

(so it should be $m_a > h_a$)

Then, we construct another right triangle $AMK$ with the same hypotenuse $AM$ and one perpendicular side $MK$ equal to $\dfrac{h_b}{2}$.

So the other restriction is $m_a > \dfrac{h_b}{2}$

There is not just one point $K$ with the above properties. There are two points $K_1, K_2$ on opposite sides of $AM$.

For each $K$ ($K_1$ or $K_2$) , the ray $AK$ intersects the line $DM$ at the point $C$. Now it's easy to find the point $B$, which is the symmetric of $C$ w.r.t. $M$.

The triangle $ABC$ has the altitude $h_a=AD$, and the median $m_a=AM$.
If we bring the altitude $BE$ from $B$, then $BE=2 \cdot MK = h_b$

In my scetch there are two triangles, one for each place of $K$

I have discussed this problem on my YouTube channel:
Video link: [url=https://youtu.be/M0_UdvxH890]Video Solution[/url]