This kind of combinatorial proof is a bit too hard for the Getting Started forum.

The middle term should be $2\binom{n}{r}$, and then..

[hide="hint"]it follows from the recursive definition of binomial coefficients:

For all $n$, $\binom{n}0=\binom{n}n=1$,

and for $0<k<n$, $\binom{n}k=\binom{n-1}{k-1}+\binom{n-1}k$.[/hide]

well if u know the identity
[hide]${n \choose r+1}+{n \choose r} ={n+1 \choose r+1}$[/hide]
it is quite simply as well. this identity isnt too dificult to prove from the definition.

[quote="BoomHead"]Prove that
$\binom{n}{r+1}+2\binom{n}{r+1} + \binom{n}{r-1} = \binom{n+2}{r+1}$

[color=green][size=75]mod edit: $\text{\LaTeX} \text {ed... did I interpret it correctly? If not, edit this post...you should understand how.}$[/size][/color][/quote]

Collect the coefficient of $x^{r+1}$ in the expansion of $(1+x)^{n+2}$ for the right hand side and
the same in the expansion of $(1+x)^n(1+2x+x^2)$


Daca sunt date numerele naturale $\ a,b,c$ astfel incat $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ si
$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ sunt naturale, demonstrati ca $\ |a|=|b|=|c|$.

Hey..notand $\frac ab=x,\frac bc=y,\frac ac=z$ vom avea ca $x+y+z \in \mathbb{N}$ si $xy+xz+yz \in \mathbb{N}$ de unde am scos urmatoarele relatzii(poate folositoare):
$(*) x^2+y^2+z^2 \in \mathbb{N}$

$(**) x^3+y^3+z^3 \in \mathbb{N}$

$(***)x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2) \in \mathbb{N}$

$(****)xy(x+y)+xz(x+z)+yz(y+z) \in \mathbb{N}$

Uite ideea mea(chiar daca nu e prea buna).Daca avem $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ si
$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ intregi, atunci si diferenta lor va fi un numar intreg, deci obtinem ca
$\frac{(a-b)(b-c)(c-a)}{abc}$ este un numar intreg.De aici putem avea doar posibilitatile $\ abc/(a-b)$ si analoagele.Daca numerele $\ a,b,c$ sunt mai mari ca $\ 1$, atunci $\ abc>a+b+c$, si obtinem usor ca
$\ a+b+c<a-b$, fals, deoarece $\ b,c>0$.Daca, in schimb, numerele sunt mai mici ca $\ 1$, mai avem de discutat
cazurile $\ a+b+c<a-b$ si $\ a+b+c>a-b$.Primul este deja discutat, iar in al doilea obtinem $\ 2b+c>0$.De unde avem ca unul din numere este pozitiv, in caz contrar avand $\ 2b+c<0$, contradictie.Sa consideram ca $\ c>0$,Dar
$\ c<1$.Deci rezulta ca nu avem solutii daca numerele $\ |a|,|b|,|c|$ sunt diferite doua cate doua.
    Poate nu am explicat eu cum trebuie, ca sunt si eu in graba, trebuie sa ma duc la o nunta...

Nu.Dar s-a casatorit ieri verisorul mamei mele.Oricum, sunt prea mic pentru a ma casatori. ;)

Numerele raţionale $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ sunt rădăcinile polinomului $X^3 - \alpha X^2 + \beta X - \gamma$, unde $\alpha = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \in \mathbb Z$, $\beta = \frac{a}{b} \cdot \frac{b}{c} + \frac{b}{c} \cdot \frac{c}{a} + \frac{c}{a} \cdot \frac{a}{b} = \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \in \mathbb Z$ şi $\gamma = \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1$.

Dacă $\frac{p}{q}$ ($\gcd(p,q)=1$) e o rădăcină raţională a polinomului $X^3 - \alpha X^2 + \beta X - 1$, atunci $\frac{p^3 - \alpha p^2 q + \beta p q^2 - q^3}{q^3} = 0$, deci $p|1, \, q|1$. Rezultă că $\frac{a}{b},\frac{b}{c},\frac{c}{a} \in \left\{ \pm 1 \right\}$, iar concluzia cred că este evidentă acum.

Sper că n-am greşit... din nou...

frumoasa solutie perfect_radio. cred ca este cea mai scurta si cea mai eleganta :)

Exista vreo solutie de nivelul clasei a VII-a?Eu am dat una acum doua posturi, dar nu stiu cat de buna e.

daca $abc$ divide $(a-b)(b-c)(c-a)$ atunci nu inseamna neaparat ca $abc$ divide unul din numerele $a-b$, $b-c$ sau $c-a$...

Da,..asa e.Dar atunci avem mai multe cazuri si problema e astfel lunga.
Dar macar cazul discutat de mine e cat de cat bunicel??N-am exolicat asa de bine ca eram in graba.

va trebui sa cauti o solutie care sa acopere toate cazurile problemei... sincer nu cred ca asta este calea cea buna


Let $ n$ be a positive integer and let $ a_i$ be real numbers, $ i \equal{} 1,2,\ldots,n$ such that  $ |a_i|\leq 1$ and $ \sum_{i\equal{}1}^n a_i \equal{} 0$.
Show that $ \sum_{i\equal{}1}^n |x \minus{} a_i|\leq n$, for every $ x\in \mathbb{R}$ with $ |x|\le 1$.

Consider
$ A \equal{} \{i\in N|a_i \geq x\}$
$ B \equal{} \{i\in N|a_i < x\}$
We can resume that $ |B|\geq |A|$ .(if not we consider $ \{ \minus{} x, \minus{} a_i\}$
Easy to check that
$ \sum_{i \in A} a_i \plus{} \sum_{i\in B}a_i \equal{} 0$
$ F \equal{} \sum_{i \equal{} 1}^n |x \minus{} a_i| \equal{} \sum_{i\in A }a_i \minus{} \sum_{i\in B}a_i \plus{} x(|B| \minus{} |A|)$
$ \Leftrightarrow F \equal{} x(|B| \minus{} |A|) \plus{} 2\sum_{i\in A}a_i$
[u] Case 1 [/u] $ x\geq 0$
$ F\leq |B| \minus{} |A| \plus{} 2\max_{i\in A}a_i|A| \leq |A| \plus{} |B| \equal{} n$
[u] Case 2 [/u] $ \minus{} 1\leq x < 0$
So $ x(|B| \minus{} |A|)\leq 0$
$ F\leq 2|A| \leq |A| \plus{} |B| \equal{} n$
So problem claim .

$ F(x)\equal{}\sum |x\minus{}a_i|$ it's convex and that implies that the maximum is touched in the limits of the interval. Hence $ F(1)\equal{}F(\minus{}1)\equal{}n$ the problem is proved.


$0<a_1<a_2<...$ real numbers such that $\sum a_i = +\infty$
Show that $Vect(t^{a_i})$ is dense in $C^0([0;1], \mathbb{R})$

I think the condition is $\sum\frac 1{a_i}=\infty$, and it's called Muntz's Theorem, if I am not mistaken. There's a series of closely-related results just like this one due to Muntz and Sasz (sorry if I misspelled the names).

It was the subject of my TIPE last year.And I know that it was the subject of some MINES/PONTS from 2002 or 2003.If you want I can post the whole TIPE. :)

of course .. i would be very interested to see a proof  :)

See also Rudin  real and complex analysis chapter 15

Check the attached file(And excuse les erreurs de riguer du francais  :lol: )

thanx a lot Killer. I will read your TIPE with great pleasure !


[quote="ali666"][quote="eagerkill"]a,b,c>=0 and a+b+c=0; [/quote]
it give a=b=c=0  :rotfl:[/quote]

Haha.. but if you drop the second condition, and instead have $a+b,b+c,c+a>0$, it should be just a simple application of Rearrangement inequality.

[quote="eagerkill"]:blush:  :blush: oh ! sorry! I get mistake!
it must be a,b,c>=0 and a+b+c=3[/quote]

As I said before, you don't need the condition on $a+b+c$ at all. And you still need the conditions I gave in the previous post for the question to make sense. Then once you have that then it is just Rearrangement.


algebra
polynomial
search
advanced fields
advanced fields unsolved

Let $ f$ and $ g$ be polynomials with rational coefficients, and let $ F$ and $ G$ denote the sets of values of $ f$ and $ g$ at rational numbers. Prove that $ F \equal{} G$ holds if and only if $ f(x) \equal{} g(ax \plus{} b)$ for some suitable rational numbers $ a\not \equal{} 0$ and
$ b$.

[i]E. Fried[/i]


Determine the number of permutations of $1, 2, 3, 4,\ldots , 32$ such that if $m$ divides $n$, the $m$th number divides the $n$th number.

[quote="cosinator"][hide="is it..."] 1??[/hide][/quote]

I think it is because you can work your way up from the first number to the last and the sequence will be forced.

More formally, you could say that there are floor $\frac{32}{m}$ numbers divisible by $m$.

Thus, no divisors of $32$ can be interchanged.

Also, no pairs of adjacent numbers can be interchanged, because of divisibility by $2$.

And then some more stuff of the sort.

[quote="cosinator"][hide="is it..."] 1??[/hide][/quote]
No, the answer isn't 1.

I found [hide="more than 1 :)"]240[/hide] different permutations, if it's right,I can explain.

[quote="rnwang2"][quote="gunes"]I found [hide="more than 1"]240[/hide] different permutations, if it's right,I can explain.[/quote]
That's right.  :)[/quote]
Good!  :lol:
Firstly, let's identify prime numbers(I will include 1,too):1,2,3,5,7,11,13,17,19,23,29,31
Then lets identify how many multiples they have upto 32:
32,16,10,6,4,2,2,1,1,1,1,1
We can interchange the places of primes that has same number of multiples upto 32. For example, 17,19,23,29,31 can change their places and it's 5! different permutations.
Then, 13 and 11 can change places but with them 22 and 26 have to change their places. (11,22) with (13,26). Because they depend on each other, there are 2!=2 different permutations.
Finally, we have 5!*2!=240 different permutations because there are no more primes that have same number of multiples upto 32.

[quote="JBL"][quote="gunes"]Firstly, let's identify prime numbers(I will include 1,too):1,2,3,5,7,11,13,17,21,23,29,31[/quote]

You got lucky: you missed 19, but you included 21, so it worked out anyhow.[/quote]
 :o

[quote="JBL"][quote="gunes"]Firstly, let's identify prime numbers(I will include 1,too):1,2,3,5,7,11,13,17,21,23,29,31[/quote]

You got lucky: you missed 19, but you included 21, so it worked out anyhow.[/quote]
:mad:
 no it was just a typing error, I solved the question on paper. I don't know how I typed 21, maybe because I was thinking another thing. The logic is true anyway, :wink:
Note= I changed 21 with 19


linear algebra
matrix
function
linear algebra unsolved

What is the inverse matrix of the matrix $A[k]_{m,n}=e^{(2\pi imn)/k}$ as a function of k?

That's not just a Vandermonde matrix- it's the matrix of the Fourier transform. $A^{*}A=AA^{*}=kI$, so $A^{-1}=\frac1kA^{*}=\frac1k\overline{A}$.

Thank you, but I don't know what over line A means, could you explain?

I can't believe it is that simple, it came up in an analysis problem I've been working on for a while. Thank you very much.


I have trouble on looking at these HMMT problems and here's another one.

Find all the roots of $(x^2 + 3x+ 2)(x^2 - 7x + 12)(x^2 - 2x - 1) + 24 = 0$.

I looked at the explanation and I got all of them till here:

If $y = (x-1)^2$ (I get this part), then we can rewrite all expression as $(y-4)(y-9)(y-2) + 24 = (y-8)(y-6)(y-1)$.

What did they do with $+24$?

hey,
basically they took that previous expression, expanded it, and upon adding 24 to it, you could factor it into the new expression

Oh. That's what I thought but I wasn't sure on "why" would they do that way..

I guess to take off 24 I guess.

Then another question. Before we did that step, they made $y = (x-1)^2$. Any idea why? (I thought they were going to set $y = x^2 - 2x$ because the product was written as $(x^2 - 2x - 3)(x^2 - 2x - 8)(x^2 - 2x - 1)$)

note that (x-1)^2 = x^2 - 2x + 1.

as long as your substitution has an x^2 - 2x term in it, your answer should ultimately end up nicely. it's just a matter a preference. i would have made the same substitution as you as well, but the solution preferred something else. no biggie  :)


Give a mechanistic explanation for the reaction :---

$Me CH == CH CH_{2} OH \longrightarrow MeCHClCH == CH_{2}$ 100% , where the reagent is $SOCl_{2}$

Very nice question.

[hide="Hint"]In the first step occurs a reaction similar to the one between an alcohol and an acid chloride.[/hide]


What was the failure in Rutherford's atomic model? And what was needed to be explained through another model of the atom?

it cudn't explain the stability in light of classical concept of electromagnetism

another model was needed to explain how bonding and excitation of electrons took place

Rutherford's model also failed to explain spectral lines of elements.


During jury selection, jurors are chosen from a pool of candidates. For a celebrity trial involving a famous pop singer, the judge wants to be certain there are no jurors who are fans. It is estimated that 67% of the population have heard this singer's music.

Construct a 99% confidence interval for the percent of jurors who will know the singer

I think this problem may be flawed because it doesn't give us the sample size for the juror selection.
If I know the sample size n, then I would use the following equation to determine the interval:

p+2.575*( 2 /frac 7

I guess it is a typical American textbook problem which makes two implicit assumptions:
1) the jury consists of 12 people
2) whoever has heard the singer likes his songs
Anyway, if you can do it for every size $ n$ and the answer really depends on $ n$, nobody can ask you for more (plugging in the value $ n\equal{}12$ or whatever into the final formula is a trivial work).


What disc games do you play? I love Ultimate and I'm just starting to get into Freestyle a little. Personally, I've never really been that big on disc golf, though.

What about you?

[quote="A+MATH"]Ultimate is soooo nerdy you just gotta love it.  :D[/quote]
Dude, ultimate is so not nerdy! ...but ya do gotta love it. :lol:

my favorite thing about any frisbee games is that i only play them in places with high concentration of mathy people (HCSSiM, ARML, MIT) then when i tell someone that's my favorite sport they usually wince at me!

I love Ultimate frisbee, too :lol:
(Unfortunately, it's not very common in Japan, so I hope more people learn it...)


calculus
integration
real analysis
function
geometry
algebra
domain

Is it possible to integrate $ \int_{B(z_0,r)} (z\minus{}z_0)^n d\mu (z)$, where $ B(z_0,r)\equal{}\{z\in\mathbb{C}: |z\minus{}z_0|\leq r\},~r>0$, $ \mu$ Lebesgue- measure, $ n\in\mathbb{N}_0$. For $ n\equal{}0$ it's obvious $ r^2\pi$. But I wonder if there's result for $ n\geq 1$. What bothers me here, is that we are integrating a complex function, over an area, and that's not what we have done in Complex analysis class, we integrated a function over a closed contour, and then used residue theorem and so on.

Of course, it is possible: it is a bounded continuous function, so there is no problem in integrating it with respect to any finite Borel measure. The value of the integral for $ n\ge 1$ is not very interesting, though: you cannot even divide by it ;).

As to whether there is any general theorem about integrals of analytic functions over domains like the Cauchy theorem for contour integration, the only thing that comes to my mind is that the average value of an analytic (or just harmonic) function over a disk equals the value at the center of the disk. For all other domains, you have to integrate in honest.

The area integration that should accompany the contour integration in a decent complex variable class is the integral of $ \bar\partial f$, so, say, the full Cauchy formula that is true for all reasonable functions, not only for analytic ones, is
\[ f(a)\equal{}\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)\,dz}{z\minus{}a}\plus{}\frac 1{\pi}\int_{\Omega}\frac {\bar\partial f(z)\,d\mu(z)}{a\minus{}z}\]
(I hope I have the right sign in the last term...). Needless to say, it is almost never done :(.

Thanks Fedja!

[quote="fedja"]The value of the integral for $ n\ge 1$ is not very interesting, though: you cannot even divide by it ;). [/quote]

Do you mean by that is equals zero, cause that's exactly what's I need. How can I show that? It doesn't seem that obvious.

Well, just do the direct computation in polar coordinates ($ z\equal{}z_0\plus{}\rho e^{i\theta}$, $ 0\le \rho\le r$, $ 0\le\theta\le 2\pi$) and integrate with respect to $ \theta$ first.


geometry
3D geometry
tetrahedron
geometry unsolved

Let $ ABCD$ is a tetrahedron. Denote by $ A'$, $ B'$ the feet of the perpendiculars from $ A$ and $ B$, respectively to the opposite faces. Show that $ AA'$ and $ BB'$ intersect if and only if $ AB$ is perpendicular to $ CD$. Do they intersect if $ AC \equal{} AD \equal{} BC \equal{} BD$?

I saw the solution of this problem in the book titled [i]How to solve the world's mathematical olympiad problems[/i] authored by Steve Dinh.

If you know the solution of problems, you can post to it?
I'm also curious about alternatives to the problem of this tetrahedron. In box geometry of AoPS, I have searched and found very little of all the geometry of space $E^3$   :o , mostly $E^2$ geometry. I  really was surprised.
This problem is interesting and i hope, you will post the solution up, because in the words you say, as you know its answer. :wink:

It's copyright material and I cannot post it here.

Anyway, his books have solutions to most of the three dimensional geometric problems posted in here that have not been solved in decades.

[quote="MathJerk"]It's copyright material ......[/quote]
I do not think so, because the solution is not related to accounting issues of copyright.
 Nothing serious here at all, U shoud share a solution. I think it would be better to advertise the book. Maybe. :)


real analysis
function
real analysis unsolved

Let f:R->R be a Lebesgue measurable function.
Suppose f(t1+t2)=f(t1)+f(t2) for all t1,t2 in R.
Is there any possiblity that f(t) is not of the form a*t(a is in R)?

Seems a easy problem,but I just can't find a counter example.

Use the theorem: if $ A$ is a set of positive Lebesgue measure, then the difference set $ A \minus{} A$ contains an interval about $ 0$.

I Know this Thm.
But I don't konw how to use it here  :(
Could you explain more?
Thanks.


algebra
polynomial
complex numbers
real analysis
real analysis theorems

can we solve equations like this:
$ax^3+bx^2+cx+d=0$ ?
(if this question is not good for here,plz excuse me because i really don't know where should i post it?thanks.) :D

Yes, but the answer is ugly.  You can also do so for 4th degree polynomials, but not generally for 5th degree polynomials or higher.  Of course, you could have just as easily looked this up on mathworld, or wikipedia, or whatever.

[url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Tartaglia.html]Tartaglia.[/url]

[url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cardan.html]Cardano.[/url]

Another opportunity to plug the excellent St. Andrews website on the history of mathematics.

Warning: the Tartaglia-Cardano formulas can result in some mighty strange repesentations of answers.

For instance, $x^3+6x-20=0$ has a single real root, $x=2.$ (It also has the complex roots $-1\pm 3i$.) The Tartaglia-Cardano formulas render that as

$x=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}$

Believe it or not, that's 2.

Worse: the equation $x^3-15x-4=0$ has three real roots: $x=4$ and $x=2\pm\sqrt{3}.$

The Tartaglia-Cardano formulas give its roots as:

$x=\sqrt[3]{2\pm\sqrt{-121}}-\sqrt[3]{-2\pm\sqrt{-121}}.$

What a way to write a real answer! What a way to write 4.

One more historical reference, to the beginnings of how complex numbers came into mathematics: [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Bombelli.html]Bombelli.[/url]

[quote="mathmanman"]There was also Ferrari involved if I'm not mistaken ;)[/quote]
And, of course, he's mentioned on the Cardano page in the St. Andrews website, and there's a link to his biography from the Cardano page. It's difficult to untangle what portion of the [i]Ars Magna[/i] (with Cardano as the listed author) is due to Ferrari's influence, as Ferrari was Cardano's servant. The primary reason I didn't mention him above is that I associate Ferrari with the quartic equation, rather than the cubic.

The ugly dispute between Cardano and Tartaglia over Cardano's publication of the cubic method in the [i]Ars Magna[/i] was "settled" (or not, as the case may be) by a public competition between Tartaglia and Ferrari in 1548.


[hide]Add the first and last terms of the sequence and see what happens.

Also, use your knowledge of complementary angles.[/hide]

[hide]

notice that the first and last angles (as well as the second and second-to-last angles, etc.) are complementary. Also remember that the tangent of an angle and the tangent of its complement are reciprocals.

$\frac{1}{1+\tan^{4}(x)}+\frac{1}{1+\frac{1}{\tan^{4}(x)}}=\frac{1}{1+\tan^{4}(x)}+\frac{\tan^{4}(x)}{1+\tan^{4}(x)}=\frac{1+\tan^{4}(x)}{1+\tan^{4}(x)}=1$
[/hide]

[quote="mathgeniuse^ln(x)"][quote="lingomaniac88"]I once posed a similar problem that uses a similar technique:

$\text{Compute }\prod_{i=1}^{89}\tan i^\circ\text{.}$[/quote]

[hide]This is just 1, since you can pair terms n and 90-n, and their product is 1 so the answer is 1 overall.[/hide][/quote]

just a note: tan 45 has no pair, but it is equal to 1 anyway

[quote="mathgeniuse^ln(x)"][quote="lingomaniac88"]I once posed a similar problem that uses a similar technique:

$\text{Compute }\prod_{i=1}^{89}\tan i^\circ\text{.}$[/quote]

[hide]This is just 1, since you can pair terms n and 90-n, and their product is 1 so the answer is 1 overall.[/hide][/quote]

Yeah, that's much simpler.  I asked it to some friends in school.  One person managed to get it right away.


algebra
polynomial
number theory proposed
number theory

Determine a polynomial $ P(x)$ with integer coefficients such that $ P(10) \equal{} 400$, $ P(14) \equal{} 440$, and $ P(18) \equal{} 520$.

[quote="Victory.US"]Determine the polynomial $ P(x)$ with integer coefficient such that $ P(10) \equal{} 400$ ; $ P(14) \equal{} 440$; $ P(18) \equal{} 520$[/quote]

If such a polynomial exists, so infinitely many exist (just add $ (x\minus{}10)(x\minus{}14)(x\minus{}18)Q(x)$ for any $ Q\in\mathbb Z[X]$ and so the question  "Determine [u][b]the [/b][/u]polynomial ..." need to be clarified.

IMHO.

No such polynomial exists

Let $ Q(x) \equal{} P(x) \minus{} 10x \minus{} 300 \implies Q(10) \equal{} Q(14) \equal{} 0$ and $ Q(18) \equal{} 40$

$ \therefore Q(x) \equal{} (x\minus{}10)(x\minus{}14)R(x)$ where $ R(x)$ has integer coefficients and $ 32 R(18) \equal{} 40 \implies R(18) \equal{} \frac{5}{4}$
but if $ R(x)$ has integer coefficients it cannot take on fractional values for integer arguments, contradiction.

if $ P(x)$ exist

WLOG , we can write $ P(x)\equal{}(x\minus{}10)(x\minus{}14)(x\minus{}18)Q(x)\plus{}R(x)$ with $ R(x) \equal{}ax^2\plus{}bx\plus{}c$

rewrite $ R(x)\equal{}R(x\minus{}4)\plus{}R(x\plus{}4)\minus{}2R(x)\minus{}32a$ with $ x\equal{}14$ . it follow $ 32a\equal{}40$ which is contradiction


analytic geometry
calculus
calculus computations

if $ r \equal{} e^{\theta}$ find points on curve where tangent is horizontal and vertical.

I got $ (e^{n\pi\minus{}\frac{\pi}{4}},\pi n\minus{}\frac{\pi}{4})$  for horizontal tangents.

Is this correct?


Let $ n$ be a positive integer. Prove that at least two numbers of $ \sqrt{n},\ \sqrt{n\plus{}1},\ \sqrt{n\plus{}2}$ are irrational numbers.

[hide="Solution"]we will prove this by contradiction.
Let us assume that two of them are rational. Hence, they can be represented as $ k$ and $ (k \plus{} 1)$ where $ k$ is a positive integer.

Then
 $ 1\le (k \plus{} 1)^2 \minus{} k^2\le 2 \\
1\le 2k \plus{} 1\le 2$
which is impossible as $ k$ is positive integer.
Hence, assumption is wrong.

So not more than one can be a rational.[/hide]

That's correct.

Let me say one thing: How did you prove the ''difference 2''?

[quote="kunny"]That's correct.

Let me say on thing: How did you prove the ''difference 2''?[/quote]
Sorry, didn't get you.

[quote="kunny"]I'm sorry, the diffenrece means the difference of maximum value and minimum value among $ \sqrt {n},\ \sqrt {n \plus{} 1},\ \sqrt {n \plus{} 2}$.[/quote]
ok
The minimum value will be 3 for 2k+1, hence difference 2 and 1 both gets rejected. I don't think there is a need to prove any further.

Whats your opinion?


Two fair 6-sided dice are rolled.  What is the probability the sum rolled is 9?

There are $ 4$ ways to get a $ 9$, so the probability is $ \frac4{6^2}\equal{}\boxed{\frac19}$.

What if unordered number combination? The probability should be 2/21.


In each of the 1 X 1 squares of a 10 X 10 board, an integer number is written, in such a way that the difference between two numbers written in two adjacent squares is not greater than 1 (two squares are adjacent if they have a side in common).
a) Show a board filled with the maximum possible quantity of different numbers.
b) Prove that there exists a number that is written in at least six different squares.

both qns are answered in one shot:

 a) we show that the maximum number of distnct entries in an m*n rid is<= m+n-1 and equality can be achieved.thus it would follow that any such square has atmost 19 distinct entries in it.(to get one such,simply create 'peels' in the following manner:
         1  2   3   4   5   6   7   8  9  10
         2  3   4   5   6   7   8   9 10 11
          .  .             .   .   .   .   .   .  .
          .  .             .   .   .   .   .   .  .

        10 11 12 13 14 15 16 17 18 19

and that is one such grid).this also solves the second part since if every entry occurs atmost 5 times then there are atleast  100/5=20 distinct entries.-><-.

so it suffices to prove the claim i made above.that follows in the following manner:
  consider the grid graph with the cells being vertices and adjacency by adjacentcy of cells(cells sharing an edge).in this graph,the distance between 2 cells A and B is an upper bound on the number of distinct entries on any 2 disjoint paths from A to B.this is so since on any path,the number of distinct entries is atmost the length of the path.thus if we start a walk from a smallest cell(least entry) to a largest cell,then the distance between 2 such cells is certainly an upper bound on the total number of distinct entries in the grid.but in the grid graph,one easily sees that the maximum distance between a pair of vertices is m+n-1.hence the claim.

i think i can simplify all that i was saying into something much simpler:
suppose we have 2 cells with entries a and b.then notice that |a-b| <= the shortest path between the corresponding cells.hence if we consider the maximum and the minimum entries in the grid as a and b,a-b+1 = # of distinct entries in the grid.but then a-b<=m+n-2(since  the path from one end of a diagonal to anotehr is (m-1)+(n-1))hence a-b+1<=m+n-1.


there are 10 points in the plane . if we choose 5 points arbitrarily there are
4 points of them such that lie on a circle . find the minimum number of points should lie on same circle?

This is a problem of the 1997 Iranian olympiad.

More generally, I prove (and this solution appears in Crux vol.29, October 2003, p.378-381) that :

Let n  \geq  5 be an integer and S be a set of n points in the plane such that among any five of them, four are concyclic. Denote by f(n) the largest integer k for which we can correctly say ' At least k of the n points lie on a circle', then :
f(6) = 4, and
f(n) = n-1 for each n  \geq  5 and n =/= 6.

As mentioned above, the solution uses 4 pages, and it is convenient to draw some figures so I cannot give it here....

Pierre.

A similar problem (with 13 birds, such that four birds out of any five are concyclic) was proposed in the Flanders Mathematics Olympiad 1997. The solution is about one page long.

:D  well... I always do longer...

Pierre.

there is probably a solution shorter, there is always a solution shorter.


Hi Valentin,

Can we have a new section called

Math  In Your Country

in wich the participant of the forum Mathlinks  will write the contains of what he study in mathematics scientifics classes grade 12th, 11th, 10th.
I think it will be interesting for  teachers, students, solvers problem.

Thank you very much,

oubinol


i want solve this problem
determine greatest reality K such as for all realities A and B and C checking : A+B+C<=K we have to: ABC<=k??

take k->infinity , a->infinity, b->infinty and c->0

so there is not a maximum k!! :)

[quote="akhrif"]it's not clear[/quote]
so is your original problem


[hide]$ \frac {1}{\frac {2}{3}} \equal{} \frac {1}{6}$ turkey is in the each slice. Thus there are $ \frac {\frac {1}{3}}{\frac {1}{6}} \equal{} \frac {6}{3} \equal{} \boxed{2}$ slices that Martin can eat.[/hide]

[quote="besmart"]please help me with these problem(can you show me the work also)

1. Martin was on a diet and bought 1/2 lb of turkey. The turkey cam in 3 slices each weighing the same amount. When he got home and read his diet book, he discovered that he is allowed to eat only 1/3 lb of the turkey. How can Martin get 1/3 lb of turkey out of 3 slices.[/quote]

Well, we start by finding the amount in each slice. We get

\[ \frac{(\frac{1}{2})}{3}\equal{}\frac{1}{6}\] Since he wants $ \frac{1}{3}$, we see that he can eat $ \frac{\frac{1}{3}}{\frac{1}{6}}\equal{}2$ slices.


uhm, i just downloaded Latex recently and i had some build errors just now.
\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}

\pdfpagewidth 8.5in
\pdfpageheight 11in

\begin{document}

Fourscore and seven years ago our fathers brought forth on this continent a new nation, conceived in liberty and dedicated to the proposition that all men are created equal.

\indent\indent Now \indent we \indent are engaged in a great civil war, testing whether that nation or any nation so conceived and so dedicated can long endure. We are met on a great battlefield of that war. We have come to dedicate a portion of it as a final resting place for those who died here that the nation might live. This we may, in all propriety do. But in a larger sense, we cannot dedicate, we cannot consecrate, we cannot hallow this ground. The brave men, living and dead who struggled here have hallowed it far above our poor power to add or detract. The world will little note nor long remember what we say here, but it can never forget what they did here.

\noindent It is rather for us the living, we here be dedicated to the great task remaining before us--that from these honored dead we take increased devotion to that cause for which they here gave the last full measure of devotion--that we here highly resolve that these dead shall not have died in vain, that this nation shall have a new birth of freedom, and that government of the people, by the people, for the people shall not perish from the earth.

Now, if we can prove
\[ \Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m
\ge 2^{m+1} \]
then we will be done. Dividing both sides of this inequality by $2^m$ yields
\[ \Big(\frac{a}{b}\Big)^m + \Big(\frac{b}{a}\Big)^m \ge 2, \]
which we can verify easily by AM-GM on the reciprocals that make up the LHS,
thus we have our desired
\[ \Big(1 + \frac{a}{b}\Big)^m + \Big(1 + \frac{b}{a}\Big)^m \ge
\Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m \ge 2^{m+1}. \]

Now, if we can prove
\[ \Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m
\ge 2^{m+1} \]

\noindent then we will be done. Dividing both sides of this inequality by $2^m$ yields
\[ \Big(\frac{a}{b}\Big)^m + \Big(\frac{b}{a}\Big)^m \ge 2, \]

\noindent which we can verify easily by AM-GM on the reciprocals that make up the LHS,
thus we have our desired
\[ \Big(1 + \frac{a}{b}\Big)^m + \Big(1 + \frac{b}{a}\Big)^m \ge
\Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m \ge 2^{m+1}. \]

\end{document}

i was just copying and pasting the stuff in the tutorial and i got 3 errors
can any1 explain why?

You can use the \documentclass or any other header commands on the forum. Just the text between the \begin{document} and \end{document} tags ;)

Hey sheepwarrior,

I copy-pasted what you posted into a new .tex file. The only things I changed were the additions of dollar signs where it looks like your code should be in math mode, and since the expressions properly registered on this forum, I assume you already have them anyways. When you mentioned the 3 errors, were you referring to "(There were 3 error messages)" or "LaTeX-Result: 3 Error(s), 0 Warning(s), 0 Bad Box(es), 1 Page(s)"? Although your statement of build errors usually implies the latter, I actually got the same first message, "(There were 3 error messages)", but then again, it's always there  from my experience, for some strange reason. Your code compiles perfectly as there are no errors, and the output is rather handsome too. You can always double click on the error message to see which line it is in. If you can't get it to work, you can post your error messages, and I'll see what I can do to help.

thx samson, but when i clicked view output, it would show all the \usepackages and \begin{documents} etc
(this is jonathan btw)
and waht do those dollar signs do?


Find two lines through the point(2,8) that are tangent  to the curve y= x^3

Since the point is on the curve, one obvious choice is $ y\equal{}12x\minus{}16$, the tangent to the curve at the given point.  Assume otherwise that at a point $ (x_0,x_0^3)$ there is a tangent line $ y\equal{}3x_0^2x\minus{}2x_0^3$ that passes through $ (2,8)$.  Hence $ 0\equal{}x_0^3\minus{}3x_0^2\minus{}4\equal{}(x_0\minus{}2)^2(x_0\plus{}1)$, and the only other tangent line is $ y\equal{}3x\plus{}2$, tangent to $ y\equal{}x^3$ at $ (\minus{}1,\minus{}1)$.


Lately, all my problems have been not on the topics I originally had, like probability questions. I have counting numbers in lists, venn diagrams, and counting with multiplication. Even though I get some correct, these subjects still don't appear in the report. Am I supposed to answer like 20 of these questions correct? It didn't happen before when I got new subjects.

Currently when you run out of questions in the topics you are ready for you are presented with a random question. This may change in the future.


algebra
difference of squares
special factorizations

Evaluate $ (x\plus{} y)(x\minus{}y)$ when $ x\equal{} 15$ and $ y\equal{} 5$.

very discriptive, jongao

i'll provide a more detailed explanation

here is our equation
$ (x\plus{}y)(x\minus{}y)$

we have $ x\equal{} 15$ and $ y\equal{} 5$

we can plug them into our equation and solve

$ (x\plus{}y)(x\minus{}y) \equal{} (15 \plus{} 5)(15 \minus{} 5) \equal{} (20)(10) \equal{} 200$

I doubt anyone will do it this way, but we can also use the difference of squares property $ (a^2 \minus{} b^2)$, which is 225-25, or 200.

[quote="ernie"][quote="dragon96"]I doubt anyone will do it this way, but we can also use the difference of squares property $ (a^2 \minus{} b^2)$, which is 225-25, or 25.[/quote]

Yeah, you can use differences of squares, but the answer would be $ \boxed{100}$.

Also, you could find the average of the two numbers and square that. (almost same method anyways)[/quote]
It'd be 200.

[quote="ernie"][quote="dragon96"]I doubt anyone will do it this way, but we can also use the difference of squares property $ (a^2 \minus{} b^2)$, which is 225-25, or 25.[/quote]

Yeah, you can use differences of squares, but the answer would be $ \boxed{100}$.

Also, you could find the average of the two numbers and square that. (almost same method anyways)[/quote]

But that... doesn't... work...

Averaging the numbers and then squaring is a lot different than the difference of squares property. That's why it doesn't work.

Averaging is: $ \frac{a\plus{}b}{2}$

Difference of squares is: $ a^2 \minus{} b^2$

$ \frac{a\plus{}b}{2} \neq a^2 \minus{} b^2$


[quote="GameBot"]Calculate $ 0.375\times1608$ and express your answer as a decimal.[/quote]
Keeping in mind that .375 is equivalent to the fraction 3/8, this problem is made slightly easier:

$ \dfrac{3}{8}\times 1608\equal{} 3\times 201\equal{} \boxed{603}$


Is there a place where I can discuss really really basic high school topics? My algebra book talks a lot about logrithm properties, but how do you change the base of these logrithms? there are a lot of problems in competitions like these.

I'm only in 7th grade, so I'm not too worried.

Log base changes?

Do you mean this?

$\displaystyle \log_a{b}=\frac{\log_c{b}}{\log_c{a}}$ .

Or something else?

I'd recommend the section Classroom Math in the Junior High area of the forums.  They could probably help you out there.

To change the base of a logarithm:
Say you have
$\log_2(x) = y$
$2^{y} = x$ <- know this relationship
Then you can change it to log base anything by taking that log of both sides.  For example 10:
$\log_{10}(2^{y}) = \log_{10}(x)$

If it's more help here are the properties of logarithms:
1) $\log_{a}b^{n} = n\log_{a}b$
2) $\log_{a}(b) + \log_{a}(c) = \log_{a}bc$
3) $\log_{a}(b) - \log_{a}(c) = \log_{a}b/c$
4) $(\log_{a}b)(\log_{c}d) = (\log_{a}d)(\log_{c}b)$
5) $\frac{\log_{a}b}{\log_{a}c} = \log_{c}b$
6) $log_{a^n}b^n = log_{a}b$


I'm new at math competitions, and I've got some questions to ask those of you that are more experienced. First of all, my school doesn't have AMC. If I want to take it in February, do I have to register early? And also, how much math do you guys do everyday, and how much of it corresponds to how well you do on competitions and stuff? Any suggestions as to how to prep for AMC? Are there any advices that you can give me to help me succeed? Thanks guys.

The best thing you could do would be to find a local school that does offer the AMC and make arrangements to take it there. You just have to scout around and ask people if they wouldn't mind if you take it too. Some schools allow students from other schools to take the AMC, and some don't, so you may have to ask quite a few places before you get one that allows you to take it. If that doesn't work, you could either convince your school to offer it (which would probably be quite difficult) or find a group of people independently who would like to take it and a teacher who would be willing to proctor and pay for the tests and do them outside the auspices of any school. (I think you can do that, but someone should correct me if I'm wrong.) You'd have to register for it by December or so, so you should do your research fairly soon.

In terms of preparation, you should get the Art of Problem Solving books and go through them. By going through them I mean you have to solve the problems and not just read the theory and the solutions. You might try purchasing some past exams from AMC, whose website is [url]http://www.unl.edu/amc[/url].

Well, I was in the same shoes you are in right now. I didn't feel like taking the AMC at another school because I was afraid I'd miss school or something important so our local math club coordinated it.

First, I would suggest asking your favorite math teacher, or a teacher you know that likes you and is willing to support you in this matter. Ask them (nicely!) if they would do so. Fill out the appropriate forms yourself so they don't have to (i.e. be courteous/considerate) and in terms of money I believe it's cheaper if you register early but you can verify that on the registration form. If you are truly destitute write the AMC for a waiver otherwise you might have to put up the money yourself or have the test takers split it. Or, you could do like I did and have it taken out of the club budget  :twisted:

Then just follow the instructions (read the manuals that come so you know what to do) I have found it to be more worthwile to just tell the advisor/proctor what to do because mine was really busy...

In terms of preparing, I would seriously consider buying AoPS I & II that is if you have the cash. Furthermore buy old tests, get your hands on as many problems as you can. Keep a notebook of problems you've solved and especially the tricks you used along the way. That will help you a lot. Another thing is to utilize this forum by asking questions, helping others and providing suggestions. Doing these things with a sincere effort will get you to the big leagues. The real secret is solving problems as much as you can. But don't go psycho, remember breathe and that sleep is a good thing.

I wish you luck and feel free to ask some more questions!
IW.OAV

yeah. get lotsssss of sleep. though i dont and never will do that. but i would like to

I thought it would be fun to reply to the oldest post ever.

Get a good night's rest.
Forget everything you ever learned.


This year, Math is Cool changed their division set-up from grade level to class level, the same divisions as MATheta spring.  What do you guys think about the division set-up at competitions like MIC this year and MATheta spring?
Personally, it kinda bugs me that they separate students by class level.  For example, at MIC Masters two sophomores in calculus from my school got 27s on their individual test. If you have a good enough memory, 6th-10th place (calc division) was a giant 5-way tie for 28 (with possibly more 28s not in the top ten, i dont remember). These sophs were forced to compete against the best juniors and seniors in the state, and ultimately got shafted from some hardware.
Another place you see problems is the MATheta sweepstakes. Schools like Garfield and Bellevue, with arguably the strongest programs in the state this year, didn't come close to winning because they couldnt fill out the Theta and Alpha divisions.
Ok, this is kind of a rant, but let me know what you guys think.

I preferred Math is Cool the way it used to be; the new system particularly  hurts Bellevue because we've always been hard-pressed to find non-Mu's in our entirely AP-driven courseload. It really makes it difficult for deserving freshmen and sophomores to win stuff.

On the other hand, it is good that we have at least one (MAO) competition that does allow people to compete against those in their own class level. It encourages participation, and everyone who wins still deserves their trophy, but it's not quite fair to those who are challenging themselves by taking harder math classes.

As for Sweepstakes, we are satisfied with the 3rd place (quite far from both 2nd and 4th) we got this year. This was really the first time we have tried to fill the non-Mu spots with people, and it paid off to an extent. It does sort of bug me that we won 4 of the 7 main Mu events (Ciphering, Individual*, NT*, Probability, S&S*, Calc AB, Calc BC* - we won the starred ones) and ended up 3rd, but that's life. MAO was still a blast.

I was kinda angry about that :mad:
Especially since my grade (sophmore) really lacks ppl good at math :roll:
Oh well, life goes on


A and  B are odd positive integers and A<B.

The sum of all the integers greater that A and less than B is 1000.

Find A and B.

The problem clearly implies $1000 = \sum_{i=1}^{B-1} i  - \sum_{j=1}^{A-1} j$.

Set $A = 2x+1$, $B = 2y+1$ where x,y are natural. (because A,B odd this can be done)

Then we have

$1000 = \sum_{i=1}^{2Y} i  - \sum_{j=1}^{2X} j$

implying

$1000 = Y(2Y+1) - (X)(2X+1) = (Y-X)(2Y + 2X + 1)$

The parity of the second factor is odd.  This implies 8 divides $y - x$.  So the second factor can only be $125, 25, 5, 1$.  We rule out the last 3 because the second factor is bigger than the first.  So $Y-X = 8$, $2Y + 2X + 1 = 125$, from whence $Y = 35$, $X = 27$.

edit:  this is VERY straight foward.. i really dont think this is preolympiad.


Prove that in triangle ABC  two circles touching side BC, circumcircle and
   Gergonne cevian from A have same radius.

   Maj.  Pestich

 P.S.  This problem was inspired by Da Vinci's  statue 'David' in Florence.

[quote="pestich"]Prove that in triangle ABC  two circles touching side BC, circumcircle and
   Gergonne cevian from A have same radius.[/quote]

This proposizione seems false (see the attached figure) or, perhaps I have erroneously interpreted the text of the problem !

Leon

Hello Leon,

  You are not from  Firenze,  are you? Otherwise you would have corrected the mistake
  made.

  As for the problem, the tangency to the circumcircle is external.

  Lt.  Pestich

If $(I_{a},r_{a})$ is the excircle of $ABC$ opposite $A$, and $AD$ is the Gergonne cevian. Let $(O_{1},r_{1}),(O_{2},r_{2})$ be external circles touch $BC,AD$ and circumcircle $(O,R)$ of $ABC$. Then: $O_{1},O_{2},I_{a}$ are collinear by Thebault's theorem hence it's enough prove that $O_{1}I_{a}$ //$BC$ or $r_{1}=r_{a}$, similarly $O_{2}I_{a}$//$BC$ or $r_{2}=r_{a}$.

Perhaps the same Thebault's solution?:
We known $I_{a}M$ //$AD$ ($M$ is midpoint of $BC$). The diameter of the excircle $(I_{a})$ orthogonal through $I_{a}$ to $AD$ is $EF$, the two parallels through $E$,$F$ cut $BC$ at $P$ and $Q$, so $P$,$Q$ are the touch points of the excircles $(O_{1})$ and $(O_{2})$.Then$PB=CQ$,also the tangents from $P$ and $Q$ to the circumcircle $(O)$ are equals. If $S$ and $T$ are the touch points of $(O_{1}),(O_{2})$ with $(O)$ resp. now $PS=QT$ so $ST$//$BC$, hence $r_{1}=r_{2}=r_{a}$.
Analogous result we have with the incircle and the mixtilinear incircles with the circumcircle and the Nagel's cevian from A vertex.


Let a,b,c,d positive real numbers such that a + b + c + d = 1.
Find the maximum of b a^2 + c b^2  + d c^2  +  a d^2 .

I have solved with my teacher! :)
Bye  :wink:

[quote="Goxer"]Let $a,b,c,d$ positive real numbers such that $a+b+c+d = 1.$
Find the maximum of $b a^{2}+c b^{2}+d c^{2}+a d^{2}.$[/quote]
I think, $\sup_{a+b+c+d=1,\{a,b,c,d\}\subset\mathbb R}(b a^{2}+c b^{2}+d c^{2}+a d^{2})=\frac{4}{27}.$
What has turned out at you with your teacher? Thank you.

[quote="arqady"]I think, $ \sup_{a,b,c,d\in R,a \plus{} b \plus{} c \plus{} d \equal{} 1}{a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a} \equal{} \frac {4}{27}$.[/quote]$ \sup_{a,b,c,d\in R,a \plus{} b \plus{} c \plus{} d \equal{} 1}{a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a}$

$ \geq\sup_{d\in R,a \equal{} 0, b \equal{} \minus{} d, c \equal{} 1}{a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a} \equal{} \sup_{d\in R}{d^{2} \plus{} d} \equal{} \plus{} \infty$.

[quote="Goxer"]Let $ a,b,c,d$ nonnegative real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 1$.
The maximum of $ a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a$ is $ \frac {4}{27}$.[/quote]Actually, the following stronger inequality holds:

$ 27(a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a)\leq4(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 37(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} c \equal{} d$.

See: http://www.mathlinks.ro//viewtopic.php?t=148360
[quote="BoesFX"]Can you post your solution?  :)[/quote]$ 4(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 27(a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a) \minus{} 37(bcd \plus{} cda \plus{} dab \plus{} abc)$

$ \equal{} a\left[16(a \minus{} b \plus{} c \minus{} d)^2 \plus{} 5(a \minus{} c)^2 \plus{} 5(b \minus{} d)^2\right] \plus{} \frac {7}{4}(d \minus{} a)(a \minus{} b)^2$

$ \plus{} \frac {b \minus{} a}{34}(a \plus{} 11b \minus{} 22c \plus{} 10d)^2 \plus{} \frac {30b \plus{} 68c \plus{} 17d \minus{} 115a}{68}(3a \minus{} b \plus{} 2c \minus{} 4d)^2\geq 0$,

which is clearly true for $ a \equal{} \min\{a,b,c,d\}$. :blush:

$ f(a,b,c,d) \equal{} a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a$.
Wlog, suppose
$ a \ge b$,
$ a \ge c$,
$ a \ge d$.
It can be shown that
$ f(a \plus{} c,b \plus{} d,0,0) \equal{} (a \plus{} c)^2(b \plus{} d) \ge f(a,b,c,d)$.
$ x \equal{} a \plus{} c$ and $ 1 \minus{} x \equal{} b \plus{} d$, we have $ Max[x^2(1 \minus{} x)] \equal{} 4/27$.
Therefore, the maximum of $ f(a,b,c,d)$ is 4/27.


induction
floor function
modular arithmetic
algebra unsolved
algebra

If roots of $x^{2}-6x+1=0$ are $x_{1}$ and $x_{2}$  then prove that $x_{1}^{n}+x_{2}^{n}$ is not an  integer divisible by $5$.

[hide]\[x_{1}^{n}+x_{2}^{n}=\left( 3+2\sqrt{2}\right)^{n}+\left( 3-2\sqrt{2}\right)^{n}\]
\[=2\sum_{k=0}^{\lfloor n/2\rfloor}3^{n-2k}2^{2k}2^{k}\binom{n}{2k}\]
\[\equiv 2\cdot 3^{n}\sum_{k=0}^{\lfloor n/2\rfloor}2^{k}\binom{n}{2k}\pmod{5}\]
(We use $2\equiv-3$.) This integer is $3^{n}$ times the integer $\left( 1+\sqrt{2}\right)^{n}+\left( 1-\sqrt{2}\right)^{n}$. Squaring this integer gives $\left( 3+2\sqrt{2}\right)^{n}+\left( 3-2\sqrt{2}\right)^{n}+2(-1)^{n}$.

Now if our integer $\left( 3+2\sqrt{2}\right)^{n}+\left( 3-2\sqrt{2}\right)^{n}$ were divisible by 5, so would $\left( 1+\sqrt{2}\right)^{n}+\left( 1-\sqrt{2}\right)^{n}$, then so would its square, then so would $2(-1)^{n}$. But clearly this last integer is not divisible by 5, so we get a contradiction. [/hide]


Find (a,b) such that (1+i) is a root of the equation x^3 - 5x^2 + ax + b = 0

If a and b are real numbers then 1 - i is also a root

 so cubic is evenly divisible by

 (x - 1 - i)(x - 1 + i)

 = x^2 - 2x + 2

  Divide the cubic by this quadratic to get
 a quotient of x - 3 and a remainder of

  (a - 8)x + b + 6

  This remainder is zero when a = 8 and b = -6

   (a,b) = (8,-6)


[quote="Shakthi S"]Select numbers a and b between 0 and 1 independently and at random, and let [i][][/i] be their sum. Let  A,B  , and C  be the results when a , b , and  c, respectively, are rounded to the nearest integer. What is the probability that A+B=C  ?[/quote]

I typed "I think you are missing something here" where I think you are missing something.

Select numbers a and b between 0 and 1 independently and at random, and let c be their sum. Let A, B, and C be the results when a, b, and c, respectively, are rounded to the nearest integer. What is the probability that A+B=C?

there we go. :D

[quote="Shakthi S"]Select numbers a and b between 0 and 1 independently and at random, and let  be their sum. Let  A,B  , and C  be the results when a , b , and  c, respectively, are rounded to the nearest integer. What is the probability that A+B=C  ?[/quote]

Hey IHatePIe answer=.........?

[quote="Shakthi S"]Select numbers a and b between 0 and 1 independently and at random, and let  be their sum. Let  A,B  , and C  be the results when a , b , and  c, respectively, are rounded to the nearest integer. What is the probability that A+B=C  ?[/quote]

The answer is 3/4

You figure out why :D

Yeah, :D

Just graph a bunch of points on an a-b axis and see if it satisfies the conditions. This was certainly one of the hardest alcumus problems that I encountered. (notice "was", im sad since now im out)


superior algebra
superior algebra unsolved

(A,+,*) is a finite ring, 0=/=1, A isn't a field. If a,b\in A s.t. ab=0 then a=0 or b\in {ka | k\in Z}. Show that |A|=p2 for some prime p.

char(A) cannot be the product of two coprime numbers u and v, because we would have that v is a multiple of u or 0 (contradiction with the fact that gcd(u,v)=1 or contradiction with the fact that char(A) is the smallest number with its property) or that u=0 (so u<char(A)..), it follows that it's a power of a prime. If char(A)=p^k with k>=3, p^(k-1) p = 0 , so p^(k-1) is 0 (p^(k-1)<char(A)) or p is a multiple of p^(k-1) (it's not divisible by p^2). The only possible cases are char(A) = p or char(A) = p^2.

If char(A)=p, then there are a,b in A such that ab=0 and a,b!=0, because it's not a field. b=ka with k in Z and gcd(k,p)=1 (or otherwise b=0, contradiction). ka^2=0, so a^2=0. We prove that M(a) = {u + va | 0<=u,v<=p-1} has p^2 elements. If u1 + v1 a = u2 + v2 a, we have that (u1-u2) = (v2-v1)a. If v1!=v2, a is in {0,1,..p-1}, because we can multiply with the inverse of v1-v2 mod p, so a must be 0, contradiction. Thus, v1=v2, so u1=u2. Consequently, M(a) has p^2 elements. Let x be an element in A\M(a). a * ax = 0, so ax=ka or ax=0. If ax=ka=ak , then a(x-k)=0, so x-k=la => x=k+la in M(a), contradiction. If ax=0, x = ka in M(a), contradiction. So in fact A=M(a) => |A|=|M(a)|=p^2.

If char(A)=p^2, we have that {0,1,..,p^2-1} is a subset of A. We prove that it contains no other element. Let x be another element, then p * px = 0,  therefore since p is not 0, px=0 or px = kp. If px=0, x = pk, with k in Z so x is in {0,1,...p^2-1}. If px=kp=pk => p(x-k) = 0 => x-k = lp => x=k+lp, which is in {0,1,2...,p^2-1}.


$ 17)$ If $ f(x) \equal{}\frac{(x\plus{}1)(x\plus{}3)(x\plus{}5)}{(x\minus{}2)(x\minus{}3)(x\minus{}4)}$, find $ f^{\minus{}1}(\minus{}8)$

$ 18)$ For how many values of $ n$ does $ 4$ choose n = $ (n\plus{}1)!$

[hide="solution to 18"]

$ {4\choose n}\equal{} (n\plus{}1)!$

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

$ 4! \equal{} n!(n\plus{}1)!(4\minus{}n)!$

$ 4! \equal{} (n\plus{}1)(4\minus{}n)!(n!)^{2}$

Fortunately, the case-work is easy. The domain of our function is obviously $ n\in\; [0,4]$ so just test $ n \equal{} 1...4$.

By doing this, we get our solutions as $ \boxed{n \equal{}\{0,2\}}$

[/hide]

[quote="hunter34"][hide="17"]
First take the inverse of the function which is $ \frac{(x\minus{}2)(x\minus{}3)(x\minus{}4)}{(x\plus{}1)(x\plus{}3)(x\plus{}5)}$. Now just substitute $ x$ with $ \minus{}8$ to get $ \boxed{\frac{88}{7}}$
[/hide][/quote]

That's not the inverse.  The inverse function is the function $ f^{\minus{}1}(x)$ such that $ f( f^{\minus{}1}(x)) \equal{} f^{\minus{}1}( f(x) ) \equal{} x$.

[hide="17"] Note that $ f(1) \equal{}\frac{2\cdot 4\cdot 6}{(\minus{}1)\cdot (\minus{}2)\cdot (\minus{}3)}\equal{}\minus{}8$.  Hence $ f^{\minus{}1}(\minus{}8) \equal{}\boxed{1}$.

I'd rather not do this problem algebraically :P [/hide]

To torajirou

I'm pretty sure that $ f(f^{\minus{}1}(x)) \equal{} f^{\minus{}1}(f(x)) \equal{} x$ not $ 1$

Typo.   :oops:  An identity is an identity in my mind sometimes.


Ooh...this is a toughie ;)  I've never done a problem like this where you have to found out the length but I found the area of the two regions: 8pi/3-(5^2)(3^1/2) but I really don't know what to do from there  :ewpu:  :read:

Let $P, Q$ be the intersections of the circles $(O_1), (O_2)$. Then $PQ \perp O_1O_2$ and by symmetry, $M \equiv PQ \cap O_1O_2$ is the midpoint of both the segment $O_1O_2$ and $AB$. The minor arc $PQ$ of the circle $(O_2)$ cuts the area of the circle $(O_1)$ in half, hence

$S = \frac{\pi r^2}{2}$

On the other hand, the area bounded by the minor arc $PQ$ of the circle $(O_1)$ and the segment $PQ$ is equal to $\frac S 2$ by symmetry. This area is equal to the area of the circlular sector $PO_1Q$ take away the area of the isosceles triangle $\triangle PO_1Q$ with $O_1P = O_1Q = r$. Denote $\phi = \angle PO_1Q$ (in radians).

$\frac S 2 = \frac{\phi}{2 \pi} \cdot \pi r^2 - \frac{r^2 \sin \phi}{2} = \frac{r^2}{2} (\phi - \sin \phi)$

Combining the 2 equations gives an equation for the angle $\phi$:

$\frac{\pi r^2}{2} = r^2 (\phi - \sin \phi)$

$\sin \phi - \phi + \frac \pi 2 = 0$

If the angle $\phi$ is known, then

$r = O_1B = O_1M + MB = r \cos{\frac \phi 2} + \frac{AB}{2}$

$\frac{AB}{r} = 2\left(1 - \cos{\frac \phi 2}\right)$

The equation for the angle $\phi$ has to be solved numerically:

$\phi \doteq 2.310\ \text{rad} \doteq 132.35^o$

$\frac{AB}{r} \doteq 1.192$


[quote="amir2"]Prove that:
$\sin\beta.\sin\gamma.\sin(\beta+\gamma)+\sin\alpha.\sin\gamma.\sin(\alpha+\gamma)+ \sin\alpha.\sin\beta .\sin(\alpha+\beta)=\sin(\alpha+\beta).\sin(\alpha+\gamma).\sin(\beta+\gamma)$[/quote]

This does not appear to be true in general.  For example, if $\alpha$, $\beta$, and $\gamma$ are the sides of a triangle, then the LHS is equal to $3 \sin \alpha \sin \beta \sin \gamma$, and the RHS is equal to $\sin \alpha \sin \beta \sin \gamma$.  Not sure how to correct it.


inequalities
induction
inequalities proposed

hmm ...where did you take this problem (ku e more ) ?

Albanian national mathematical olympiad (Olimpiada kombetare e matematikes  )

(This solution can be rewritten to ignore the substitution but the motivation is more clear here).

Let $ p_k\equal{}x_1x_2...x_k$ with $ p_0\equal{}1$. Then $ x_k\equal{}\frac{p_n}{p_{n\minus{}1}}$.

The inequality can be rewritten as

\[ \frac{p_0^2}{p_1}\plus{}\frac{p_1^2}{p_2}\plus{}\frac{p_2^2}{p_3}\plus{}...\plus{}\frac{p_n^2}{p_{n\plus{}1}}\plus{}4p_{n\plus{}1}\ge 4\]

By AM-GM, $ \frac{p_n^2}{p_{n\plus{}1}}\plus{}4p_{n\plus{}1}\ge 4p_n$. (*)

With this in mind, let us proceed by induction, $ n\equal{}0$ gives

$ \frac{p_0^2}{p_1}\plus{}p_1\ge 4$ which is obvious by AM-GM since $ p_0\equal{}1$.

Now suppose this result holds for $ n\minus{}1$, consider the inequality for $ n$...

\[ \frac{p_0^2}{p_1}\plus{}\frac{p_1^2}{p_2}\plus{}\frac{p_2^2}{p_3}\plus{}...\plus{}\frac{p_n^2}{p_{n\plus{}1}}\plus{}4p_{n\plus{}1}\ge \frac{p_0^2}{p_1}\plus{}\frac{p_1^2}{p_2}\plus{}\frac{p_2^2}{p_3}\plus{}...\plus{}\frac{p_{n\minus{}1}^2}{p_{n}}\plus{}p_n\ge 4\]
by (*) and inductive hypothesis. So we are done.

Comment: we can rewrite this as:

\[ \left(\frac{p_0}{\sqrt{p_1}}\minus{}2\sqrt{p_1}\right)^2\plus{}\left(\frac{p_1}{\sqrt{p_2}}\minus{}2\sqrt{p_2}\right)^2\plus{}\left(\frac{p_2}{\sqrt{p_3}}\minus{}2\sqrt{p_3}\right)^2\plus{}...\plus{}\left(\frac{p_n}{\sqrt{p_{n\plus{}1}}}\minus{}2\sqrt{p_{n\plus{}1}}\right)^2\ge 0\]

Nice (telescoping) inequality!!

this problem was really difficult, or ... I found it difficuld, I resolved it in the way it was resolved on Russia 2007 ... not the way it is resolved here ...
Enio, you competed this year? (me too)


calculus
Putnam
geometry
Princeton
college
function
summer program

Grade: 10

[b]Classes (1st semester): [/b]
-Phsyics
-Biology
-Statistics AP
-Adv. English 10
-French 2
-Cultures
-Calculus III (at college)
[b]
Classes (2nd semester)[/b]
-Phsyics
-Biology
-Statistics AP
-Adv. English 10
-French 2
-Psychology AP
-Differential Equations (at colege)

Nice schedule.. :)

Supposedly you want other people to list their classes, too..
so here I go.

Grade: officially 10...

[b]1st semester[/b]
Physics
French II
AP Chem
Graphic Design
AP Econ (Micro and Macro sort of merged)
Honors English II
World History II
AP Calc BC

[b]2nd semester[/b]
Physics (self-studying for AP B)
French II
AP Chem
PE (Team Sports)... aaargh
AP Econ
Honors English II
World History III
AP Calc BC

Grade 9:

[b][u]Semester 1[/u][/b]
P.E.
English 9 Honors
Chamber Choir
Biology I Honors

[b][u]Semester 2[/u][/b]
Spanish I
Chamber Choir
Health
Algebra II Honors

**my school runs on a block schedule, so we have each class for an hour and a half each day, which explains the only 4 classes per semester**

[quote="MathFiend"]**my school runs on a block schedule, so we have each class for an hour and a half each day, which explains the only 4 classes per semester**[/quote]
4 classes! Wow!
My school also has a block schedule, with 4 85-minute classes each day,
but we still get 8 classes per semester because we take classes every other day.

Grade: 7
1st Semester
English 7
Social Studies 7
Algebra I
Industrial Technology 1A
Art 1A
Science 7
PE 7
2nd Semester
All the same except that I have Computer Programming instead of Ind Tech, and Web Design 1A instead of art.

Grade: 10

Classes:
-Spanish 3
-Elective: WWW publishing, Photography
-AP World History
-Elective: Health, Economics
-AP Physics B
-AP Calculus BC
-Honors Soph English

I like my roster next year though =P (we just started registration):
-AP American Lit
-Advanced (Multivariable) Calculus
-AP US History
-AP Physics C
-AP Chemistry
-AP Macro Economics

sounds like fun =)

Grade: 14

Fall:
-- Complex Analysis
-- Modern Algebra (Group Theory)
-- Functional Analysis
-- Putnam Seminar
-- Prime Number Theorem
-- Tuning and Temperament
-- 18th century counterpoint

Winter:
-- Complex Analysis
-- Set Theory
-- Modern Algebra (Ring Theory)
-- Canon and Fugue
-- Hungarians!
-- F. Scott Fitzgerald
-- Yiddish

Spring:
(I'm not sure yet.)

Grade: 11

Semester 1:

AP Calculus BC
AP Physics C
AP Biology
English 11
Marketing
French 4

Semester 2:

AP Calculus BC
AP Physics C
AP Biology
AP English Composition
Swimming :)
French 4

Grade: 10

Semester 1:
AP Calc AB
Honors French III
Religions of North America (req'd  :( )
Honors American History
Honors Orchestra
Honors American Literature
Honors Biology

Semester 2:
The same, but with Computer Life Skills instead of Religions.

There is [i]very[/i] little flexibility at my school schedule-wise.

Grade 9

Semester 1:
-Honors World Geography
-Elective
-Honors Literature
-Elective
-Adv Geometry
-Honors Biology

Semester 2:
same xept one diff elective

D:

[b]Grade: 9th[/b]
[b]1st Semester[/b]
Pre-AP Algebra II
Pre-AP English I
Pre-AP World Geography
Pre-AP Biology
Spanish II Enhanced
Foundations of Personal Fitness
Intro to Information Technology
Art I

[b]2nd Semester[/b]
Pre-AP Algebra II
Pre-AP English I
Pre-AP World Geography
Pre-AP Biology
Spanish II Enhanced
Health
Intro to Information Technology
Art I

grade 10

1st semester
IB Spanish 5
IB Math Methods Pre Calculus (although i have pointed out that i could well succeed in calculus)
IB World History Semester 1
IB English
IB chemistry
PE (fill-in.... required... :( )

2nd semester
IB spanish6
IB MM Pre-Cal
IB World History
IB English
IB CHemistry
AP PHYsics B2 (1st semester i did physics AP B 1 independent study)

[quote="billzhao"]Where do you guys take your Multivariate Calculus / Linear Algebra? Does your high school offer them, for do you go to a local university for it?[/quote]

It's usually taught at one school in my district. Next year its at my school (yay!). I'm in a big district with a middle school gifted program so they need Multivariate Calculus or people won't have enough math classes. I guess I just got lucky.

Umm yes... I don't take insanely advanced math classes  :blush:

Grade : 10

AP Computer Science AB
Chemistry
Japanese 2
AP World History
Honors Pre-Calc
Health/Weight Training
Honors English

Nex year though... I am going to take Calc BC. And AP Chem. And AP Physics.

I go to a school where there isn't that much choice in what you take.  Mainly, because it's a non-magnet middle school.

Grade 8:
1st semester

Test Prep
Geometry Enr.
Library/Study Hall
Enr. Social Studies 8
Lunch
Orchestra 8
Spanish I
P.E.
Enr. Science 8
Enr. Language Arts 8 (But it says on the transcript that its actually Language Arts 8/9)

2nd Semester
The same, just replace P.E. with Art

I just added the Enr. (which means Enriched btw) and the numbers to make it look better.  The reason it looks like I have a lot of classes is because I go to the high school in the morning, so I really have 9 periods a day, and for the same reason, my P.E. and art classes are half as long as the other classes.

I'm pretty lucky in terms of math possibilities: Advanced students take Multivariable Calculus/Linear Algebra. More advanced students are able to take courses at Princeton University (I am at public Princeton High)

Me schedule: (Grade 11)
Honors American Literature
Advanced Spanish 5
AP Chemistry

First Semester Math: MAT 214 Numbers, Equations, and Proofs (Intro to NT: phi functions, primality tests, infinity spaces, moduli spaces, primitive roots, quad. reciprocity, you get the idea.)

Second Semester Math: MAT 217 Honors Linear Algebra. Taught by Prof. Conway. Those who went to MathCamp might know him, he's a cool guy.

AP World History
Human Behavior

Wow, looking at all the posts, there are some really alternate forms of scheduling. I never heard of block scheduling until I saw this thread. Honestly, I do not think I would like it. Spending hour and half in spanish? :(  :(

Block scheduling isn't all that bad once you get used to it...
especially for PE (you only have to change half as often) and art (you get to actually concentrate).
Block language classes ARE very tedious, though. (My French teacher makes us sing French songs to keep us awake, and by the time the bell rings I'm speaking Frenglish(?)... lol)

We have kind of a block sceduling too.  It's just that our classes are every other day so it's a lot less boring for me.

I guesss so, It might be fun to try block scheduling for a little bit....

block scheduling is bad in the sense that some teachers teach new stuff BEFORE a test so you can forget whats on the test. Otherwise, having 4 classes a day is pretty good cause of hw  :D . You'd think that twice the class period = twice the amount of hw but that is really not the case. At least in my personal experience. Then again, i'm only a freshman so the hw load is quite up there...

I had block schedule for Bio and Chem last year (2 periods a day for half a year each).  I really liked the program.  I think a few reasons for why I enjoyed it are:

1.  I love science.
2.  I absorb information quicker than most students (especially science or math) and thus as the year went on it became easier for me.
3.  By devoting two periods to one class I felt like I dedicated my mind to that class more.  In other words, I was ability to focus my ability on fewer subjects.
4.  I got 100's in both class  :D ...I'm still not sure how I pulled off a 100 in Chem...that was a hard class and I had like a 95 at the halfway point.

I don't really have any other block classes.  Well, the AP sciences and maths at my school are given two periods for the whole year so I have both physics and calculus for two periods.  I don't really consider that as block schedule though.

First semester:

Calculus III (they wouldn't give me credit :( )
Linear Algebra
Analysis I
Algebraic Systems
Introduction to Computer Programming

Second semester:

Vector Calculus (worst.course.evar)
Discrete Math II
Analysis II
Complex Analysis
Introduction to Macroeconomics

Summer:

NSERC (research on functional analysis)
Differential Equations (ugh)
Elementary Number Theory (yawn)

12
Econ Self-study
AP Gov.
AP European Hist.
AP Statistics
AP English Lit. and Composition
AP Physics
French III

Grade 9

AlgebraII H
Spanish II
Biology H
English I
Physical Education
World History H

I hate school, it takes up my time for mathing :(

Grade:  11

First Semester:
Honors American History
Honors American Literature
AP European History
AP Biology
Instrumental Music
Latin IV
Cross Country (I hate muddy fields)

Second Semester:
Same as first semester, EXCEPT
Track (way better than cross country)
AND (oh man I love bragging way too much). . .
Real Analysis through the Graduate/Advanced Undergrad education program at Johns Hopkins University Applied Physics Lab (I work as an intern at APL)

All right, I live in a terrible school district in a small state. Thankfully I am moving after this school year ends.

[u]9th grade[/u]
1st semester [i][b]and[/b][/i] 2nd semester:
  Freshman civics
  English I
  AP Calculus AB (BC Calc isn't offered here, grrr... :mad:)
  Symphonic Band
  Freshman Physical Science
  Spanish II
  Sophomore Biology (I still don't understand why this isn't a freshman course)

However, next year seems a little better:
[u]10th grade[/u]
1st semester:
  AP Chemistry
  AP US History
  AP English: Lang/Comp
  Spanish III Honors
  Healthy Choices
  Band
  Adv. Computer Studies Honors
  Calculus 2 (If I can figure out how to schedule it...)

2nd semester:
  AP Chemistry
  AP US History
  AP English: Lang/Comp
  Spanish III Honors
  Ultimate Challenge
  Band
  Web Design
  Linear Algebra (see above, next to "Calculus 2")

grade 9

1st Semester
English I
Biology (weird teacher...)
Art I
Geometry
Chemistry
Economics
AP Spanish Lang (I live in Mexico... :) )
Health

2nd Semester
English I
Biology
Ceramics
Geometry
Chemistry (better teacher this time)
World History I
AP Spanish Lang
Public Speaking
[hide]
hehe.. first post :blush: [/hide]

Grade:9

First semester:
Study Hall
History 9
Orchestra II
English 9
Chinese 1
Pe/health 1
Alg II/Trig
Chemistry/biology (Lab Science 9)

Second semester
same except biology/physics (science is divided into trimesters)

everythingwith nine after it and pe are required...so yea not much freedom in ninth grade, but next year im doing AP US, AP stats and pre-calc!

Grade: 14 (That is, college sophomore)
Fall 2004:
Modern Algebra 1
Differential Geometry
Women Writers in Imperial China
Introduction to the theory of computing

Spring 2005:
Modern Algebra 2
Complex Analysis
Nature, Science and America
The English Revolution

Fall 2005 (projected):
Lots of math -- probably combinatorics, graph theory, and something else.  Also, beginning Hungarian.  (I'll be studying abroad in Budapest, with any luck.)

Grade, 7..geeze...

follow normal schedules...

French(exploring)
English(7)
math (9th grade math called integrated 1)
biology(7th s***s alot!!!)
social studies(7)
Reading(7)

you guys might laugh at me for listing schedules even though I'm only in 7th grade

Grade: 12

Fall Semester:
Theoretical Probability
Complex Analysis
Computer Systems Research
Philosophy
Geosystems
AP U.S. Government and Politics
AP English Language

Spring Semester:
Numerical Analysis
Differential Equations
Computer Systems Research
Film Study
Geosystems
AP U.S. Government and Politics
AP English Language

It's interesting how all of the technical classes crowd the begining of each day (the first four in each list are one block day, the last three are the other day; all seven classes in order are on Mondays) and the fine arts crowd the end of each day.

[quote="JBL"]
Women Writers in Imperial China
[/quote]

Wise Choice!

(Actually, I don't have any clue but it's about China!)

9th Grade:
Pre-AP Eng 9 (10 next year)
Oklahoma History (Is there a smiley face for "sick"?)/2nd Sem: Pre-AP Gov (APUSH next year)
Pre-AP Bio (AP next year)
Latin II (AP III next year)
Adv Pre-Calc/Trig (Our school has a stupid policy... we have to take Calc AB before BC... )
Tennis (same)
Orchestra. (same)

[quote="plokoon51"]shinwoo, you might be embarassed about listing your schedule while you're in 7th grade but i'm taking geo in 9th grade. that means i'm average!!! And our school isn't lettting 9th,10th graders double up on math and science!!! :o[/quote]

Eh you don't have to be levels ahead of everyone to be good at problem solving.

[quote="beta"]Eh you don't have to be levels ahead of everyone to be good at problem solving.[/quote]

True, but it doesn't feel too good to know that you're beaten by a kid younger than you and with more knowledge of certain math concepts just cause he/she goes to a magnet school.

Ex: at state mathcounts, my friend said he'd slap himself for every 7th grader or lower and every girl (or both) who beat him in the individual round. that was about 30. that doesn't feel too good

Yeah, well I wish I could go to a school like Mildorf does (he goes to Thomas Jefferson, the best highschool in the country) where they offer complex analysis and "theoretical probability" as highschool courses.  I wish I could actually take classes in that stuff instead of teaching it to myself, because since I haven't taken classes in it, I'll have to take the courses anyway in college even if I already know the stuff.

[b]Grade:[/b] not sure.

[b]Semester 1:[/b]
pretty much whatever I want.

[b]Semester 2:[/b]
See above.

Hee hee. I'm homeschooled! :D

grade - 11
First and Second Semester
- Precalculus Honors
- AP Statistics
- AP Biology
- AP European History
- English 4
- Student Volunteer


MATHCOUNTS
geometry
probability
number theory

I'm in Algebra 2.

Does that help me in MATHCOUNTS?

I don't think that [i]how far along you are[i] in the standard curriculum is as important as [i]how well/deeply you know[/i] the standard curriculum.Sure, being in Pre-Cal is going to give you an advantage over someone in Alg. 1, but if you are a typical Pre-Cal student(besides your age) and you're up against a thoroughly-studied Alg. 2 student, you will probably lose.

Just as a suggestion: your current topic title is quite grammatically incorrect. :diablo:

Well, i think having a better understanding of math might help.  But most of the chapter and state math problems only range from pre-algebra to geometry so bascicallly you only have to know up to geometry. :)

My class is called Geometry/Algebra 2...

Don't forget some number theory and counting.

*sigh*
what class you're in if its past geometry has no effect on how well you do.
Its how in depth you know it. I could be doing calc right now but chose not to so i could study for mathcounts, instead Im n pre-calc

My MATHCOUNTS coach/Algebra 2 teacher, really wanted us to learn geometry over the summer.
We did roughly 3 lessons per day and 5 days per week in the summer.
We spent about 5 hours per day.
2 hours on geometry and 3 hours on MATHCOUNTS.
She decided that this summer, we would review Geometry and Algebra 2 again, so we can have a better understanding.

Do you think all this was a good idea?

I think a full understanding of Algebra II would be not that helpful for MATHCOUNTS, more for AMC-12/AIME.
As for what you did over the summer, :wow: :wow:

There is not much of an advangage to being in higher level curriculum classes.  The highest you will need in Mathcounts is Geometry.  What allows you to do well is a thorough understanding of the topics taught in those lower level classes.

Several topics that are not covered in the regular curriculum that you really need to know are more complicated probability, number theory, and counting types of problems.  These can be learned through classes, by David Patrick's book, or through your mathcounts coach.  If you can master these, along with Algebra and Geometry, you will be able to do fine at Mathcounts.

I'm in algebra 2 but haven't officially taken geometry. I say officially because I've taken it outside of school.

i dont think anything bast geometry or algebra 2 will help too much


Two runners, A and B, run around a circular track in the same direction.  Runner A, who completes a lap every 78 seconds, has just passed runner B, who completes a lap every 91 seconds.  When will A again overtake B?

Thanks

[hide]Let $ n$ be the number of seconds it takes. We need $ \frac{n}{78}\minus{}\frac{n}{91}\equal{}1\implies{n\equal{}\boxed{546}}$[/hide]


Fill in the boxes below with the digits 3, 4, 7 and 9, using each digit only once, so that the resulting fraction is as close to $ \frac12$ as possible. What is the value of the fraction? \[ \frac{\Box\Box}{\Box\Box}\]

The ones we think to try are these fractions:

$ \text{a).} \frac{34}{79}, \text{b)} \frac{39}{74}, \text{c)} \frac{37}{94}, \text{and d)} \frac{47}{93}$. Comparing each fraction and their difference from $ \frac{1}{2}$, we have $ \text{a)} \equal{} \frac{1}{2} \minus{} \frac{5.5}{79}$, $ \text{b)} \equal{} \frac{1}{2} \plus{} \frac{2}{74} \equal{} \frac{1}{2} \plus{} \frac{1}{37}$, $ \text{c)} \equal{} \frac{1}{2} \minus{} \frac{10}{47}$, $ \text{d)} \equal{} \frac{1}{2} \plus{} \frac{0.5}{93}$. Now we compare their differences. We want the smallest difference, which is obvious: it is $ \frac{0.5}{93}$. That means the fraction we can create that is closest to $ \frac{1}{2}$ is $ \boxed{\frac{47}{93}}$.


limit
logarithms
inequalities
integration
real analysis
real analysis solved

What is the limit of this sequence:
$ u_n=(\sum_{i=1}^n \frac1i) - ln(n)$

Plenty of info [url=http://mathworld.wolfram.com/Euler-MascheroniConstant.html]here[/url].

[quote="alekk"]is it rational  :D   :D[/quote]
 :D  :D  :D  :D

Kvant magazine, problem [b]597[/b].
Let $x_n=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$
[b]1)[/b] Prove that there exists $\gamma=\lim_{n\to\infty}(x_n-\ln n)$.
[b]2)[/b] Prove that $\gamma<x_n+x_m-x_{nm}\leq 1$ holds for all $n,m\in\mathbb{N}$.
[b]3)[/b] Find $\gamma$ accurate to 0.01.

Oh really? By the way,what is Kvant?

I will be able to show $u_{n}>\frac{1}{2}$ elementarily.

kunny

Kvant is a famous russian mathematical journal for students (11 - 18+ years :) ).

I think the first one has been proven on the forum before. Actually, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18230]here[/url]'s a generalization posted by Moubinool.

The first inequality from the second one follows from the fact that, after fixing $n$, $a_m=\frac 1{m+1}+\frac 1{m+2}+\ldots+\frac 1{mn}\to \ln n$ and $a_m$ is increasing, so $a_m\le \ln n,\ \forall m$, which means that $x_n+x_m-x_{mn}=x_n-a_m>x_n-\ln n,\ \forall n$, and $x_n-\ln n$ is decreasing and goes to $\gamma$.

As for the second inequality, we use again the fact that $a_m$ is increasing for fixed $n$, so $x_n-a_m\le x_n-a_1=1$.

Thank you, Myth! :)

I will show you the following inequality.

$\frac{1}{n+1}<\int^{n+1}_n \frac{1}{x} dx<\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)  (n=1,2,3,\cdot\cdot\cdot)$

kunny

It is particular case of Karamata inequality for $\frac{1}{x}$, since
\[\frac{1}{n}+\frac{1}{n+1}>\frac{1}{n+t}+\frac{1}{n+1-t},\]
where $t\in(0,1)$.


trigonometry
limit
calculus
derivative
analytic geometry
geometry
inequalities

How do you find the limit of the problems below without using L'Hôpital's rule?  Thank you.
1.  $ \lim_{x \to0} \frac{1-\cos x}{x} $

2. $ \lim_{x \to0} \frac{\sin x}{x} $

Note that these are the derivatives of sine and cosine, evaluated at 0. (Well, the negative of the cosine.) The most typical textbook derivations of the derivatives of sine and cosine in general start by first establishing these two limits.

So: you need to know these limits in order to derive the derivatives of sine and cosine. And you would need to know the derivatives of sine and cosine in order to use L'Hôpital's rule. Therefore, to use L'Hôpital's rule on these problems is either a dishonest piece of circular reasoning, or a lazy man's way to remember the result of a hard-won theorem.

Here's one possible path:

Define $\cos x$ and $\sin x$ in terms of coordinates of a point on the unit circle. As part of this definition you must establish what you mean by the radian measure of an angle. Whether by definition or derivation, note that the area of the circular wedge subtended by angle $x$ is $\frac x2.$

Find a way to argue that both sine and cosine are continuous functions.

Assume $x$ lies in the first quadrant. Let $O$ be the origin, $A = (1,0),$ $B=(\cos x,\sin x),$ and $C = (1,\tan x)$ so that $O,A,$ and $C$ are collinear.

The wedge of angle $x$ contains $\triangle OAB$ is contained in $\triangle OAC.$ Turn those containments into a chain of inequalities for the areas, and we have

$\frac12\,\sin x<\frac12\,x<\frac12\,\tan x.$

So $\frac{\sin x}x<1$ and $\cos x<\frac{\sin x}x.$

These inequalities are valid for first quadrant $x$, but since $\frac{\sin x}x$ is an even function, we have for $0<|x|<\frac{\pi}2$ that

$\cos x<\frac{\sin x}x<1.$

Continuity implies that $\lim_{x\to 1}\cos x=1,$ so by the squeeze theorem,

$\lim_{x\to 1}\frac{\sin x}x=1.$

As for the other case:

$\frac{1-\cos x}x=\frac{1-\cos x}x\cdot\frac{1+\cos x}{1+\cos x}= \frac{\sin^2x}{x(1+\cos x)}=\frac{\sin x}x\cdot\frac{\sin x}{1+\cos x}$

As $x\to 0$ this tends to $1\cdot\frac02=0.$

[This can all be found in many textbooks.]


invariant
number theory proposed
number theory

Start with the set {14, 35}. At each step you can replace only one of the elements of the previous set with either the sum or the difference of the elements of that set. For example from {14, 35} you can reach either {14, 49} or {14, 21} or {49, 35} or {21, 35}. Can you ever reach {1323, 1519}?

Nope, as $\gcd(a,b)$ is invariant under $(a,b)\mapsto (a,|a\pm b|)$ or $(a,b)\mapsto(|a\pm b|,b)$, but $\gcd(14,35)=7$ whilst $\gcd(1323,1519)=49$


Woot we came in 6th

daine better see this

Yes I see this. :)
This topic seems a bit late.

If none of us made any stupid mistakes.  We definitely would've gotten 5th.

What are you guys doing on our forum?  :D
I'm going to post in the Michigan forum now :P

Aw man, your public forum is dead and the other one is private.

[quote="kyyuanmathcount"]Michigan messed up the team round, and still got 13th, w00t.  yeah.......If I hadn't blown target (and sprint), we probably would have made top 10.[/quote]

Dude, you didn't blow sprint. I blew sprint. So did Sam, too, kinda, seeing that he only beat me by 1.

i resent that.(not really, but it seemed like the thing 2 say)

Get out of our forum or...

I'll tell one of the admins to make mods for the local community forums, I'll become one of the mods, and....well I still won't have enough power to throw you out...

well anyway you have your own private forum, Michiganders are just jealous that we beat them by 7 places at MC... :wink:

Fine, Connecticuters out of our forumm too.

[quote="Phelpedo"][quote="Ignite168"]CT definitely would've gotten into top 10 if we had Neal Wu instead of Dan Tobias (2nd highest scorer on our team)[/quote]

32nd to 10th? Neal only scored like 20 points higher than Dan...[/quote]

that's only a 5 point team score difference where a 1/4 points can mean the difference between 1st place and 2nd

@SplashD: just because .25 separated 1st from 2nd doesn't mean it separates any two consecutive places, and even if it did that's still only 12th place.

Yeah, PA told us that when they were in our room the night before we left.

[quote="13375P34K43V312"]California came in 10th, and this is my estimate to their scores:

Rolland Wu: 38 (know for a fact)
Kevin Yang: 32 (pretty sure)
Jim Li: 31 (ditto)
Tony: 25 or so, don't know

Team: 14 (know for a fact)

45.5 in total. So that was the cutoff, approximately. I am really just guessing Tony's score. I know for a fact MA got 46.75 (6th), MD took 7th with 46.5. PA said they came in 8th with 46 even I think, and a NCer told me they were 9th with 45.something. So the cutoff was at least 45, probably for the top 10.

So I remember Ignite saying that the CT scores were like 30-22-20-17 or something like that. With Neal Wu, that's 42-30-20-17. With Neal, let's say CT gets a 14 on the team round, since Neal told me LA got 12 on the team round. So that comes to a 41.25, missing the cutoff by at least 4.[/quote]

I'm pretty sure Kevin Yang got a 30 considering he tied with me.

[quote="13375P34K43V312"]I'm not exactly sure...but I did hear that Kevin Yang got was the highest of some scoring group (might have been 30, might have been 31)[/quote]

Yes it was 30.

44th-57th were all 30.

I was 56th. Kevin Yang was 44th.

We all know that neither of them got a 40, because neither of them was in the countdown. Moogra is probably just a little bit confused.

Talk about off topic...to TRY to get back on, I'm just happy we beat Maryland.

Michigan got owned because of me and my target death, and my team dumb mistakes.

[quote="perfect628"]MA could have done better hadf Carl and I not bombed the sprint.[/quote]

I believe without my six careless sprint errors and one target, if you had the same, we would have come in 4th, and I would have like 30th.

[quote="Ignite168"][quote="moogra"]jim li got 40 something, kevin yang also got 40 something. im sure about taht[/quote]

Then why did none of them win written round? Also, how was a 40 something 44th and 45th a 30?[/quote]

He means the place tehy got not the score...

Kevin Yang got 44th, Jim Li got 47th...


solve equation in R:
 [img]http://www.gnux.be/latex/data/7956473dc5336a5288d01587211c6584.png[/img]

determine the domain of definition
$ x\equal{}9$it's the only solution

yes, it is the only solution but it is needed show him, point out that only one case of equality is:
[img]http://www.gnux.be/latex/data/21a86d16f86d0c3a2c77500f44bf3c0b.png[/img]


Prove that for each integer $m$ and each integer  $k>0$ , there exist integers $a,\ b>0$ relative prime wiht $m$ and such that $a-b=2k$.

Silouan: what are you talking about?! I think that you missed some details ... don't you?? I mean please let us learn any relation between k and m either the problem seems too easy!

davron

if i'm not doing any mistake it's not difficult indeed!
just show it for m prime (easy) and finish it of with chinese remainder theorem... :wink:

Yes Albanian Eagle that's the main idea  :wink:


Gauss
floor function
geometric series
algebra unsolved
algebra

What is the last two digit of the integer $\lfloor{\frac{10^{1995}}{10^{95}+3}}\rfloor$

 :)

$\begin{array}{lll}\left\lfloor\frac{10^{1995}}{10^{95}+3}\right\rfloor&=&\left\lfloor\frac{{10^{95}}^{21}}{10^{95}+3}\right\rfloor\\&&\\&=&\left\lfloor{10^{95}}^{20}-3.{10^{95}}^{19}+9.{10^{95}}^{19}\cdots\right\rfloor\\&&\\&=&\left\lfloor\displaystyle\sum_{i=0}^\infty3^i{10^{95}}^{20-i}\right\rfloor\\&&\\&=&\displaystyle\sum_{i=0}^{20}3^i{10^{95}}^{20-i}+\left\lfloor\displaystyle\sum_{i=1}^\infty3^{20+i}{10^{95}}^{-i}\right\rfloor\end{array}$

Now it is easy to see that the geometric series in the floor brackets has a sum less that one, so the required floor must be $\sum_{i=0}^{20}3^i{10^{95}}^{20-i}$.  But the last two digits of this must simply be the last two digits of $3^{20}$, since all other terms have a factor of $10^{95}$, so the two required digits are given by $3^{20}\equiv3^4\equiv81\mod{100}$ (since $\phi(100)=16$).

Hmmm... so it is.  :blush:

The $10^{95}$ should really be in brackets (i.e. $\left(10^{95}\right)^{20}$), but I got lazy.


Consider three points in plane: $ A(0,1)$, $ B(1,2)$, $ C(2,3)$. The coordinates of the centroid of triangle $ \triangle ABC$ are:

$ (A)$ $ (3,6)$
$ (B)$ $ (6,3)$
$ (C)$ $ (1,2)$
$ (D)$ $ (2,1)$
$ (E)$ $ (3,2)$

[hide="Solution"]
The centroid of a triangle with vertices $ (x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $ \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).$ The given coordinates give us \begin{align*}\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)&=\left(\frac{0+1+2}{3},\frac{1+2+3}{3}\right)\\&=(1,2),\end{align*} which is the choice $ \boxed{\textbf{(C)}}.$[/hide]

It's not really a triangle, more like a straight line.

Oh wow didn't notice that. But the centroid is still $ (1,2)$ even though the triangle is degenerate.


[i]source: some econ question by[/i] RLE

Prove that  :sqrt:(n-1) + :sqrt:(n+1) is irrational for every integer n >= 1

[quote="Dang Hop"]Let's have a look at:
http://www.mathlinks.ro/Forum/viewtopic.php?p=169040&highlight=#169040[/quote]
thanks Dang for the refferal but am still stuck :( .  anyone else have an opinion on it?

$\sqrt{n-1}+\sqrt{n+1}=\frac{p}{q}$
so $\frac{2}{\sqrt{n+1}-\sqrt{n-1}}=\frac{p}{q}$
so $\sqrt{n+1}-\sqrt{n-1}=\frac{2q}{p}$
so $\sqrt{n+1}=\frac{1}{2}(\frac{p}{q}+\frac{2q}{p})$ is rational,
so $n+1$ is the square of an integer, so $n-1$ isn't.


AMC
USA(J)MO
USAMO
geometry
inequalities
number theory
\/closed

I was just wondering how advanced the classes will get?  Will there be one on USAMO, IMO problems?  Thanks

I believe this was discussed earlier on in the year. The answer was that when we have a membership mass (and interest) that would make such a class actually worth the effort, it would definitely be a possibility. (That's the way I rmr it) I think it would be best if you got an answer from the instructors though.

A couple of our classes already include olympiad material (the intermediate counting class, for example, has a few problems of that difficulty).

We will be trying a couple olympiad-level Problem Series next spring (start in February, probably).  Each course would cover a specific subject (say number theory or geometry or inequalities, etc).


modular arithmetic
combinatorics unsolved
combinatorics

A graph with $ n$ vertices has degree of each vertex equal to $ k$. What can be said about $ n$?(what values can take $ n$?)

I'm not quiet sure,but at leat $ nk$ must be even,and also $ n\geq k \plus{} 1$

And $ n$ being a multiple of $ k\plus{}1$ is also good :P

[quote="sunnykim"]And $ n$ being a multiple of $ k \plus{} 1$ is also good :P[/quote]
I don't think so... :maybe:

Why don't? Consider "groups" of (k+1)-complet graphs with no edges among the groups...

yes,those kinds of graphs are desired to,but note that in this problem we want some necessary conditions.

Yes, I didn't pretend I have done the problem(found all $ n$), I just remarked some possible values :P

Well..

For small k, we have:

$ k \equal{} 2 \Rightarrow n \geq 3$, but no other conditions (consider $ C_n$)
$ k \equal{} 3 \Rightarrow n \geq 4$ and $ n \equiv 0 \pmod{2}$
I believe
$ k \equal{} 4 \Rightarrow n \geq 5$ and no other conditions as well... so I think that Babak's conditions in the second post are both necessary and sufficient, but I'm not entirely sure how to prove it.

For k even, $ n \ge k\plus{}1$: consider in $ \mathbb{Z}_n$ the relation $ |a\minus{}b| \le \frac k2$.
For k odd, n even, $ n \ge k\plus{}1$: consider in $ \mathbb{Z}_n$ the relation $ |a\minus{}b| \le \frac{k\minus{}1}2$ or $ a\minus{}b \equal{} \frac n2$.

So jour conjectures seem to be true.  :)

[quote="edriv"]For k even, $ n \ge k \plus{} 1$: consider in $ \mathbb{Z}_n$ the relation $ |a \minus{} b| \le \frac k2$.
For k odd, n even, $ n \ge k \plus{} 1$: consider in $ \mathbb{Z}_n$ the relation $ |a \minus{} b| \le \frac {k \minus{} 1}2$ or $ a \minus{} b \equal{} \frac n2$.
[/quote]

I don't understand your solution...Who are $ a$ and $ b$? :maybe:

$ \mathbb Z _n$ is the set of vertices.
If $ a,b \in \mathbb Z_n$, we put an edge between a and b if and only if $ \minus{} \frac k2 \le a \minus{} b \le \frac k2$.
(the order relation we use is obtained identifying $ \mathbb Z_n$ with $ \{1,2,\ldots,n\}$)

And the same for the second example.


How many factors of 2^5 multiplied by 3^6 are perfect squares?

[quote="andersonw"][hide]
$ 3^6$ is already a perfect square (since it's raised to an even power), so we just have to look for factors of $ 2^5$ that are squares, since a perfect square multiplied by another perfect square is also a perfect square.

There are 3: $ 2^2$,$ 2^4$, and $ 1$.
[/hide][/quote]

wrong the answer is 12

How many even natural-number factors does[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/c/7/bc782dbe9df7f4539e717279f906b71aef6fb215.gif[/img]have?

Oh, I misinterpreted your question, thought it meant how many of the factors of $ 2^5$, when those factors multiplied by $ 3^6$, are perfect squares.


Alcumus
AMC
AIME
probability
pigeonhole principle
calculus
Support

What level does Alcumus go up to? The end of AOPS's intro to C and P? intermediate? AIME level?

[quote="Alcumus FAQ"]Problems are based on Art of Problem Solving's  Introduction to Counting and Probability book.[/quote]
Through end of Intro Counting and Probability level.

Could the admin please kindly make the problems' difficulty level a bit higher (like all the way to intermediate and maybe beyond)? Perhaps it would be better for users beyond intro and give them more challenge (franky i don't have any aops textbook and i only learn from this website)...  :D

You may be interested in [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting & Probability[/url].

Yeah I know that textbook but it's kinda expensive to buy (like 50 USD)... plus right now there are tons of problems on this site that I haven't even seen yet... so i don't have any free time to actually read the book, plus i could find almost all those topics on the internet (like PIE, pigeonhole, etc)... :lol:

Come on, the free program (Alcumus) is pretty good already; you shouldn't be hesitant to spend 50 bucks on a great-quality math book. Don't be so demanding. Knowledge > 50 bucks.

And also, I don't believe it's actually that expensive, is it? Or am I mistaken? I have the book, and it was only like 35 bucks.

To be precise, the book costs 47 dollars including the solutions manual.

[quote]plus i could find almost all those topics on the internet (like PIE, pigeonhole, etc)[/quote]

Sure.  As in-depth as in the AoPS book?  Doubt it.

The book [i]is[/i] expensive, but worth it. If you can't afford it (or just want a cheaper price anyway  :P ), you can always try getting a used copy on Amazon. I once got a fantastic calculus book for about 40 cents, and it was in relatively good condition, too. Even better, if you live near a math circle, see if you can borrow the book!


geometry
calculus
integration
floor function

I'm not sure what you mean by "an integer in terms of $ n$ and $ m$."
[hide="The locus is"]the circle with diameter $ OP$.[/hide]

as in, how many points in the locus are such that $ QP$ is an integer? Express your answer in terms of $ n$ and $ m$.

Are m and n integers themselves, or does that matter?

[hide="EDIT:"]We split into two cases:

Case 1: m is an integer

We first draw the perpendicular of OP at P. That chord has QP equal to 0. We then have every value from 1 to m-1 on the left and right side, and an m(on the diameter). That's 2m.

Case 2: m is non-integral

We draw the perpendicular again with QP=0. We then have 1 through $ \lfloor m \rfloor$ on both sides. We have $ 2\lfloor m \rfloor \plus{}1$[/hide]

Why don't we need to know n?!

Sorry, but your solution is incorrect.

[quote]We then have every value from 1 to m-1 on the left and right side[/quote]

is a rather problematic statement.

[quote="Temperal"]Consider a circle $ O$ of radius $ n$. A point $ P$ is $ m$ units from the center such that $ m < n$ is selected. Consider the locus $ \Gamma$ of the midpoints of all the chords through $ P$. Find the number of points $ Q$ in $ \Gamma$ such that $ QP$ is an integer in terms of $ n$ and $ m$.[/quote]
I am assuming that $ m$ and $ n$ are integers.  Otherwise, the solution can be easily modified.
[hide="Solution"]
First, we find what $ \Gamma$ is.  Consider chord $ APB$ with midpoint $ M$.  Now, notice that $ \angle OMB \equal{} 90\implies \angle OMP \equal{} 90$.  Hence, for all midpoints of the chords through $ P$, say $ M_i$, we have that $ \angle OM_iP \equal{} 90$ and thus the locus of $ M_i$ is the circle with diameter $ OP$.  Simiarly, we can show that any point $ X$ on the circle with diameter $ OP$ is the midpoint of a chord through $ P$ as follows.  Let $ PX$ intersect the circle at $ Y$ and $ Z$.  Thus, we have that $ \angle OXY \equal{} \angle OXP \equal{} 90\implies OX\perp XY\implies OX\perp YZ$, so $ X$ is the midpoint of $ YZ$.

Now, with this in mind, we find the number of points on $ \Gamma$, $ Q$, so that $ QP$ is an integer.  Now, notice that $ QP\le PO \equal{} m$.  Hence, $ QP$ has $ m \plus{} 1$ values between $ 0$ and $ m$.  Realize, however, that there is only one point $ Q$ so that $ QP \equal{} 0$, namely $ Q \equal{} P$, and one point $ P$ so that $ QP \equal{} m$, namely $ Q \equal{} O$.  For all of the other $ m \minus{} 1$ integers (say $ x_i$), there are $ 2$ points such that $ QP \equal{} x_i$.  Hence, the number of points $ Q$ is $ 2(m \minus{} 1) \plus{} 1 \plus{} 1 \equal{} \boxed{2m}$. [/hide]

EDIT: Dear, [b]Bictor717[/b], my apologies.  I meant that $ QP\le PO$.  Solution edited.  That's odd though; my answer is still different than yours...

EDIT 2: Thanks again, [b]Bictor717[/b].  I misread the problem, but I edited my solution.  Still, however, I get a different answer than you.  What am I doing wrong now?

Temperal, I'm not sure why I found that wording confusing. :blush:
[quote="The QuattoMaster 6000"][hide]
Now, notice that $ QP\le QO \equal{} n \minus{} m$.[/hide][/quote]
[hide="Your solution is incorrect here"]$ OP$ is a diameter. There are certainly points on $ \Gamma$ that are closer to $ O$ than they are to $ P$.[/hide][hide="The rest of my informal solution"]
Consider concentric circles centered at $ P$ with integer radii. If the radius is less than $ m$, the circle intersects $ \Gamma$ twice. If the radius is $ m$, the circle is tangent to $ \Gamma$. These intersections and $ P$ form the locus of points $ Q$ where $ QP$ is an integer and $ Q$ lies on $ \Gamma$. Thus, there are $ 2m$ points if $ m$ is an integer, and $ 2\lfloor m\rfloor \plus{}1$ points if $ m$ is not an integer.[/hide]

I'm not sure where the mistake is in your solution, Bictor, but I'm fairly sure that Quatto is correct now...

$ OP\equal{}m$ because $ P$ is $ m$ units from the center. $ n$ has nothing to do with $ \Gamma$.


advanced fields
advanced fields unsolved

Let $X$ be a seperated metric space prothat all $M\subset X$ is seperated metric space, too.

As I know every  metric space is seperated, in virtute of rational (ray) balls. Am I wrong?

yes, you're wrong. Example, $L^{\infty}(\mathbb{R})$ is not separated.

That depends on what "separated" [url=http://en.wikipedia.org/wiki/Hausdorff_space]means[/url].


Given $a,b\in{R}$  and the equation $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least one real root, determine the minimum value of $a^{2}+b^{2}$.

[quote="cats..."]you posted it here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104868[/quote]

I thought I've posted it in the wrong place ...but I couldn't move it. :oops:


To proof,that if for  $ n\in \aleph$, $ 3^{n+1}\2^{3}^{n}+1$, but not divide $ 3^{n+2}$.

Hi Yavor Tsvetanov 52. Can you retype your question? Thanks.

Yes,sory.
If $ n$ is a natural number,then proof that $ 3^{n\plus{}1}|2^{{3}^{n}}\plus{}1$ but $ 3^{n\plus{}2}$ not divide $ 2^{{3}^{n}}\plus{}1$.

The question is a direct application of lifting the exponent .
Take a look at this:
http://docs.google.com/gview?a=v&q=cache:93o6nb9DW4gJ:reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf+exponent+lifting&hl=en&gl=in&sig=AFQjCNGzcOHCHRG9vZP3eoAlmAXjqwv3Ww

this lemma solves imo 90's third problem  :wink:

I still can't understand the solution.Can you post the solution using the lemma?

$ p^{k}|n$ but $ p^{k\plus{}1}\nmid n$  $ \implies$  $ p^{k}\parallel{} n$   :wink:

Well,here it's my attemp:
We will use an induction.
$ 1) n \equal{} 1$ , then $ a_{1} \equal{} 9$ and $ 3^{n \plus{} 1} \equal{} 9$ and $ 3^{n \plus{} 2} \equal{} 27$.
$ \equal{} > 9|9$ but $ 27\nmid 9$ , so we have a base.
Let when $ n \equal{} k$ the statement is right,so:
$ 3^{k \plus{} 1}|2^{{3}^k} \plus{} 1$ and  $ 3^{k \plus{} 2}\nmid 2^{{3}^k} \plus{} 1$.
Now we must proof that the statement is right and when $ n \equal{} k \plus{} 1$.
$ \equal{} > a_{k \plus{} 1} \equal{} 2^{{3}^k \plus{} 1} \plus{} 1 \equal{} 2^{3k.3} \plus{} 1 \equal{} (2^{{3}^k})^3 \plus{} 1 \equal{} (2^{{3}^k} \plus{} 1)(2^{3^{k}.2} \plus{} 1 \minus{} 2^{{3}^k}) \equal{} a_{k}(2^{3^{k}.2} \plus{} 1 \minus{} 2^{{3}^k})$
Now we have that $ 3^{k \plus{} 1}|2^{{3}^k} \plus{} 1$,so we have $ 3^{k \plus{} 2}|3(2^{{3}^k} \plus{} 1)$.
So now we have to proof that $ 3|2^{3^{k}.2} \plus{} 1 \minus{} 2^{{3}^k}$
$ 1\equiv 1(mod 3) , 2^{3^{k}.2}\equiv 1(mod 3)$ and $ 2^{3^{k}}\equiv 2(mod 3)$
So $ 3|2^{3^{k}.2} \plus{} 1 \minus{} 2^{{3}^k}$
$ \equal{} > 3^{k \plus{} 2}|2^{{3}^{k \plus{} 1}} \plus{} 1$

But the problem is that i don't know how to continue.

I think this belongs in intermediate (HSI); just a tip.  :)