trigonometry
calculus
derivative
function
calculus computations

Find the derivative of $ y\equal{}\tan{^{\minus{}1}(x\minus{}\sqrt{1\plus{}x^{2}})}$ and simply where possible.

All I get is one big mess with nothing significant that will cancel, except for a $ 2$ but I can't seem to simplify it.

Please can you post how far you've got. I've just done that now and it all seems to cancel fairly quickly for me so I think you may have made a mistake somewhere

I got
\[ \frac {dy}{dx} = \frac {1}{1+(x - \sqrt {1 - x^{2}})^{2}}\frac {d}{dx}(x - \sqrt {1 + x^{2}})
\]

\[ = \frac {1}{1 + (x - \sqrt {1 + x^{2}})^{2}}\cdot{(1} - \frac {2x}{2\sqrt {1 + x^{2}}})
\]

\[ = \frac {2\sqrt {1 + x^{2}} - 2x}{(1 + (x - \sqrt {1 + x^{2}})^{2})(2\sqrt {1 + x^{2}})}
\]
I have a nasty feeling that I messed up somewhere but I checked over a ton of times. Unless I'm just not seeing some really simple cancellation...

It is correct. But you could try to transform it to the form: $ \frac{1}{2(x^2\plus{}1)}$. I think that's the final form you are after...

From the original equation we have:
1) $ \tan(y) \equal{} x \minus{} \sqrt{1 \plus{} x^2}$

2) $ \tan^{2}(y) \plus{} 1 \equal{} 2(x^2 \minus{} x\sqrt{1 \plus{} x^2} \plus{}1)$

3) Differentiating 1) we get,

 $ (\tan^{2}(y) \plus{} 1)dy \equal{} (1 \minus{} \frac{x}{\sqrt{1 \plus{} x^2}})dx \equal{} (\frac{x^2 \minus{}x\sqrt{1 \plus{}x^2} \plus{} 1}{1 \plus{} x^2})dx$.

Hence, $ \frac{dy}{dx} \equal{} \frac{(\frac{x^2 \minus{}x\sqrt{1 \plus{}x^2} \plus{} 1}{1 \plus{} x^2})}{(\tan^2(y) \plus{} 1)} \equal{} \frac{(\frac{x^2 \minus{}x\sqrt{1 \plus{}x^2} \plus{} 1}{1 \plus{} x^2})}{2(x^2 \minus{} x\sqrt{1 \plus{} x^2} \plus{}1)} \equal{} \frac{1}{2(1 \plus{}x^2)}$

So, in hindsight, how could the original function be simplified? If $ x\equal{}\tan\theta$, we're looking at $ \arctan(\tan\theta\minus{}\sec\theta)\equal{}\arctan\frac{\sin\theta\minus{}1}{\cos\theta}$. That's a version of a half-angle formula- $ \tan\frac{\alpha}{2}\equal{}\frac{1\minus{}\cos\alpha}{\sin\alpha} \equal{}\frac{\sin\alpha}{1\plus{}\cos\alpha}$. To convert this into what we need, let $ \theta\equal{}\alpha\plus{}\frac{\pi}{2}$; that means $ \sin\theta\equal{}\cos\alpha$ and $ \cos\theta\equal{}\minus{}\sin\alpha$. Our original function is $ \arctan\left(\tan\frac{\theta\minus{}\frac{\pi}{2}}{2}\right)\equal{} \frac{\theta\minus{}\frac{\pi}{2}}{2}\equal{}\frac{\arctan x}{2}\minus{}\frac{\pi}{4}$.


Is the "Trivial Inequality" referred to in AoPS as x^2 >= 0 really called the "Trivial Inequality" (the name I would use in an actual olympiad), or is it just an inequality that seems trivial?

Fierytycoon

It's actually called the Trivial Inequality.


101 ones + 101 tens + 101 hundreds is equal to what number?


First I think you intend $ 2^{\phi (n)}\minus{}1|n^{n}$  :wink:

Let $ n\equal{}2^{k}p_{1}^{k_{1}}\cdots p_{r}^{k_{r}}$   suppose for a moment that $ k\geq 1$ then $ 2^{k\plus{}r\minus{}1}|\phi(n)$
so if $ \phi(n)\equal{}2^{s}\cdot m$ then $ (2^{m}\minus{}1)(2^{m}\plus{}1)\cdots(2^{2^{s\minus{}1}m}\plus{}1)|n^{n}$
but $ n$ has $ r$ odd prime factors and so $ k\plus{}r\minus{}1\le s\le r$ so we conclude that the solutions have $ k\equal{}0$ or $ k\equal{}1$ but in both cases $ s\equal{}r$ so each $ 2^{2^{i}m}\plus{}1$ is a prime power. From here we can easily find solutions $ n\equal{}2,3,6$

[quote="TTsphn"]1)Find all n such that
$ 2^{\phi(n)\minus{}1}|n^{n}$
Some problem :
http://diendantoanhoc.net/forum/index.php?showtopic=33495
Note :So nguyen to :Prime
         So nguyen duong :Natural number
          Tim :Find all[/quote]
Obviosly all solutions are even numbers and n=1.

The way I wrote in my post the problem is from Balkan MO shortlist 2005 (Proposed by Albania, E.Pisha)


The vertices of triangle $ PQR$ have coordinates as follows: $ P(0,a),\,Q(b,0),\,R(c,d),$ where $ a,\,b,\,c$ and $ d$ are positive. The origin and point $ R$ lie on opposite sides of $ PQ$. The area of triangle $ PQR$ may be found from the expression:

$ \textbf{(A)}\ \frac{ab \plus{} ac \plus{} bc \plus{} cd}{2} \qquad
\textbf{(B)}\ \frac{ac \plus{} bd \minus{} ab}{2}\qquad
\textbf{(C)}\ \frac{ab \minus{} ac \minus{} bd}{2}\qquad
\textbf{(D)}\ \frac{ac \plus{} bd \plus{} ab}{2}\qquad
\textbf{(E)}\ \frac{ac \plus{} bd \minus{} ab \minus{} cd}{2}$

[quote="sunehra"]I think you reversed the order for shoelace, because the actual answer is $ \boxed{(B)}$.[/quote]
I did it the wrong way  :blush:


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ p$ be the smallest prime number such that $ p\mid |G|$ , where $ G$ is finite group , if $ H\leq G$ and $ [G: H] \equal{} p$ , show that $ H\triangleleft  G$.

Very classical.

http://www.mathlinks.ro/Forum/viewtopic.php?t=215134
http://www.mathlinks.ro/Forum/viewtopic.php?t=62685
http://www.mathlinks.ro/Forum/viewtopic.php?t=22672
http://www.mathlinks.ro/Forum/viewtopic.php?t=234462

  darij

A very [b]elementary[/b] proof of this result follows from this observation:

-- Let p be the smallest prime dividing |G| and let $ g \in G \minus{} H$. Then,

$ H, gH, \ldots, g^{p \minus{} 1}H$

are pairwise disjoint and

$ G \equal{} \bigcup_{k \equal{} 0}^{p \minus{} 1} g^{k}H.$

None of the proofs mentioned by [b]Darij[/b] had proceeded along these lines.  :wink:

[quote="coquitao"]A very [b]elementary[/b] proof of this result follows from this observation:

-- Let p be the smallest prime dividing |G| and let $ g \in G \minus{} H$. Then,

$ H, gH, \ldots, g^{p \minus{} 1}H$

are pairwise disjoint and

$ G \equal{} \bigcup_{k \equal{} 0}^{p \minus{} 1} g^{k}H.$

[/quote]

How do u get the normality of $ H$, from here , also how do u assure that $ g^i \ne g^j, i\ne j, 1\leq i,j<p$?

That's not clear for me . Sorry  :blush:

Can anybody post a complete proof of [u]coquitao[/u]'s hint?

That I got from his solution .

[quote="coquitao"]A very [b]elementary[/b] proof of this result follows from this observation:

-- Let p be the smallest prime dividing |G| and let $ g \in G \minus{} H$. Then,

$ H, gH, \ldots, g^{p \minus{} 1}H$

are pairwise disjoint and

$ G \equal{} \bigcup_{k \equal{} 0}^{p \minus{} 1} g^{k}H.$

[/quote]

Note that $ |g|\equal{}p$.So, $ g,g^2,\cdots,g^{p\minus{}1}$ are distinct .

[quote="coquitao"]Normality is obtained thus:

Let $ m \in G \minus{} H$ and $ n \in H$. Also, assume that $ g \equal{} mnm^{ \minus{} 1}$ doesn't belong to $ H$. My previous post allows us to ascertain that $ m \equal{} g^{k}h$  for some $ k \in \{0, \ldots, p \minus{} 1\}$ and $ h \in H$. Then, $ g \equal{} mnm^{ \minus{} 1} \equal{} (g^{k}h)n(g^{k}h)^{ \minus{} 1} \equal{} g^{k}(hnh^{ \minus{} 1})g^{ \minus{} k}.$  It follows now that $ g \equal{} hnh^{ \minus{} 1}$. The previous asseveration contradicts the assumption that $ g$ was not a member of  $ H$ and the desired conclusion follows.

 :D[/quote]

Depending on $ H$ is normal iff every element in $ G$ , we have $ ghg^{\minus{}1} \in H, \forall h\in H$.

I hope this help

[quote="Hidden Scofield"]

Note that $ |g| \equal{} p$.So, $ g,g^2,\cdots,g^{p \minus{} 1}$ are distinct .

[/quote]

Does it follow that the cosets $ g^{i}H$ are pairwise disjoint?  :blush:

If they weren't, we'd have $ g^iH\equal{}g^jH$ for some $ 0\leq i <j\leq p\minus{}1$ and hence $ g^{j\minus{}i}\in H$. Now $ g^{j\minus{}i}$ is a generator of $ \langle g\rangle$ because the exponent is positve but $ <p$ so it can't be a divisor of $ |G|$. So $ gcd(ord(g),j\minus{}i)\equal{}1 \implies \langle g\rangle\equal{}\langle g^{j\minus{}i}\rangle\subseteq H \implies g\in H$ which is a contradiction.

[b] coquitao [/b]'s proof seems not using the hypothesis that $ p$ is the [b] smallest [/b] prime divisor of $ |G|$. So does it mean that it's true for any prime divisor of $ |G|$ ?

[EDIT] I find out his error now: He assumed that $ g \equal{} m n m^{ \minus{} 1} \not\in H$ but $ g$ coincides with the original $ g$. This is the [b] notation [/b] problem. If we change the letter $ g$ to another one, the problem remains.

[quote="Gockel"]If they weren't, we'd have $ g^iH \equal{} g^jH$ for some $ 0\leq i < j\leq p \minus{} 1$ and hence $ g^{j \minus{} i}\in H$. Now $ g^{j \minus{} i}$ is a generator of $ \langle g\rangle$ because the exponent is positve but $ < p$ so it can't be a divisor of $ |G|$. So $ gcd(ord(g),j \minus{} i) \equal{} 1 \implies \langle g\rangle \equal{} \langle g^{j \minus{} i}\rangle\subseteq H \implies g\in H$ which is a contradiction.[/quote]

That was quick, esteemed Gockel! Yet, it seems like you don't use the minimality of $ p$ over there.  :(

Can any of you guys rewrite this part of Gockel's argument?

[quote="Gockel"]

Now $ g^{j \minus{} i}$ is a generator of $ \langle g\rangle$ because the exponent is positve but $ < p$ so it can't be a divisor of $ |G|$.[/quote]

Every line before this one and every line after it are crystal clear. All I need to see is what Gockel is trying to say [i]here[/i].

Thanks in advance!

It actually took me a little bit to get the details here, but the idea is that if $ g^k \in H$, $ k\in\{1,2,\ldots p\minus{}1\}$, then $ k$ must be relatively prime to $ |G|$. (using the definition of $ p$)

It follows that $ k$ is relatively prime to $ |\langle g\rangle |$, so $ g^k$ generates $ \langle g\rangle$.  Therefore $ g\in H$.

[quote="Riquelme"]TO MATHANGEL:

I see no error [i]there[/i]... Why?

[hide]
$ g$ can be any element of $ G \minus{} H$.[/hide][/quote]

First, you choose $ m$ and $ n$ and now, if $ mnm^{ \minus{} 1} \not\in H$, then it must be of the form $ g^i h'$ for some $ h' \in H$ and $ p > i > 0$, it might not be the original $ g$, am I right?. But in this case, because he choose $ g$ to identify $ mnm^{ \minus{} 1}$, his proof went smoothly.

Now if $ m n m^{ \minus{} 1} \equal{} g^i h'$ and $ m \equal{} g^j h$, then you cannot deduce anything.

Let's be clear: there is no [i]original[/i] $ g$.  You assume that $ g\equal{}mnm^{\minus{}1} \not\in H$, then you look at $ H, gH,\ldots g^{p\minus{}1} H$ and so forth.  Mathematical arguments do not always occur in the order in which they are described. :)

You are absolutely right, LydianRain!  :wink:

[quote="Hidden Scofield"]That I got from his solution ...

Note that $ |g| \equal{} p$.So, $ g,g^2,\cdots,g^{p \minus{} 1}$ are distinct .

[/quote]

I hate to revive this thread this way, but I thought I had to do it. First off: it's not valid to suppose that [b]g[/b] is always going to have order [i]p[/i]. Wonder why it was that Scofield assumed that in the first place. I have to admit that I overlooked this important point on the first perusal of his post. Needless to say, the actual remark renders meaningless most of the comments in our discussion, except for the first four of 'em and the two previous-ones...  :wallbash_red:


function
calculus
derivative
complex analysis
absolute value
topology

I would like to ask a stupid question: What is the definition of an analytic function? Some text consider an analytic function $ f$ on $ S$ as a differentiable function on $ S$, some regard it as a continuously differentiable on $ S$ (i.e. $ f\in\mathcal C^1$), and others regard it as a function which can be represented by a power series. So, which one is more popular?

You have to be careful with your choice of words.  Complex functions can be differentiable (at points) without being analytic.  For example, the absolute value function is differentiable almost everywhere but it is not analytic.

Formally, a function is analytic on an open set $ D$ if for any $ x_{0}\in D$, $ f(x)\equal{}\sum_{n\equal{}0}^{\infty}a_{n}(x\minus{}x_{0})^{n}$ and the series is convergent for $ x$ in a neighborhood of $ x_{0}$.

If by differentiable in $ S$ you mean complex differentiable in an open set containing $ S$, then all three definitions are equivalent. It's completely different from the real variable setting, where differentiable, continuously differentiable and real analytic are three distinct concepts.

JRav, what do you mean? $ f(x \plus{} iy) \equal{} x^2 \plus{} y^2$ satisfies the C-R equations only at 0, and there the absolute value fails to be real differentiable, so the absolute value function has a complex derivative nowhere.

All three properties turn out to be equivalent, but most people use different words to emphasize different aspects:
-differentiable is used to mean just that (it holds meaning for reals too, in which case it is not equivalent to the others)
-holomorphic applies to complex functions only, implying that they are continuously differentiable
-analytic implies that it converges to its power series in the neighborhood of every point


quadratics
Euler
function
number theory
prime numbers
arithmetic series
algebra

I may have proof here.

-y^2 = 4 - x^5
y^2 = x^5 - 4
y = sqrt(x^5 - 4)
sqrt(x^5 - 4) must be an integer.

In order for a number to have an integer square root, each of prime factors must have an even number (i.e. 2^2 * 3^2 * 5^6) . That is because the primes must be able to be split in half.

So I don't have to keep on typing x^5, let z = x^5.

Let's assume that sqrt(z) is an integer.

Therefore, if sqrt(z-4) were to still to have an integer square root, the subtraction of 4 must have taken out two prime numbers, p. // Edit: I'm not sure, but this logic might not be sound.

Edit 2: Nope. Drats. There goes my proof.

Edit 3: Well, I think I may have another solution.
z and z-4 must both be perfect squares.
1, 4, 9, 25, etc. are perfect squares.
The differences between these squares is an arithmetic series with a cd of 2, starting at 3. Therefore, z and z-4 can never both be perfect squares.

Ignore
-----------------

z-4 = z/(p^2)
z = z/(p^2) + 4
zp^2 = z + 4p^2
zp^2 - 4p^2 - z = 0
(z-4)p^2 - z = 0
Therefore, it is a quadratic equation where a = (z-4), b = 0, and c = -z

In order for the quadratic equation to have a real solution, sqrt(b^2 - 4*a*c) must be >= 0.

sqrt(b^2 - 4*a*c) here is sqrt(0 - 4*(z-4)(-z))
sqrt(0 - (4z-16)(-z))
sqrt(0 - (-4z^2 + 16z))
sqrt(4z^2 - 16z)
If 4z^2 - 16z must be greater or equal to 0,
4z^2 >= 16z.
z^2 >= 4z
z >= 4

The whole quadratic equation for (z-4)p^2 - z would be
0 + (can't be minus) sqrt(4z^2 - 16z)/(2*(z-4))
sqrt(4z^2 - 16z)/(2z-8)
Square both the numerator and denominator:
(4z^2 - 16z)/(4z^2 - 32z + 64)
4z^2 - 16z > 4z^2 - 32z + 64
-16z > -32z + 64
-16z > 64 - 32z
16z > 32z - 64
0 > 16z - 64
64 > 16z
4 > z

Proof by contradiction.

I'm a freshman in highschool, so please forgive any problems I have with presenting a proof.

I also  hope I didn't make a silly arithmetic error, screwing my whole proof up.  :?

The only residues of x^5 (mod 11) are 0, 1, -1.

The quadratic residues modulo 11, or the possible values of y^2 (mod 11), are 0, 1, 4, 9, 5, 3.
So the possible values of y^2 + 4 (mod 11) are 4, 5, 8, 2, 9, 7.

So x^5 cannot equal y^2 + 4 (mod 11), and x^5 - y^2 = 4 (mod 11) has no integer solutions. Hence, x^5 - y^2 = 4 has no integer solutions.

I found 11 through a brute force search. Is there a general way to figure out what you can mod x^n by so that the only values are {0, 1, -1}?

Why must z-4 be z/p^2?

It might be some perfect square that has factors other than those of z.  For example, suppose 4 were 40 instead: 7^2-40=3^2.

Right. I realize that both of my solutions are wrong.

Nice solutions, yoyo.

How did you think to use mods?

Are there any budding number theorists out there who want to take a swing at yoyo's general question?

If p > 2 is prime, a^(p-1) = 1 (mod p) if (a,p) = 1 (Fermat little theorem)

So (a^(p-1)/2)^2 = 1 (mod p), so (Z/pZ is a field) a^(p-1)/2 in {1,-1}

Conclusion :

if p is prime > 2, the values of a^(p-1)/2 are in {0, 1, -1}

Set p =11

[quote="akatookey"]I am interested in yoyo's (Tim's?) question...is there a way to find a mod n where the solutions of x^n are only {-1,0,1} ???[/quote]

Take a look at i/3's post above.  In case you're a little unfamiliar with the notation, (a,p) is the greatest common divisor of a & p.

If you're not convinced, look at n^2 (mod 3), n^4 (mod 5), n^6 (mod 7), n^10 (mod 11), and so on.

Then, when you're convinced...  It's time to try proving it.

[quote="yoyo"][quote="rrusczyk"]How did you think to use mods?
[/quote]

Mods seemed to be useful on problems of a similar form, where there are no integer solutions, which I've seen before. There are also some similar ideas (and problems) in the last part of the Diophantine Equations chapter in [u]AoPS[/u] book 2. I think the advantage of using mods is that you only have to check finitely many values of x and y, instead of infinitely many. Also, if two numbers are equal, then they're equal modulo a for all a, so the contrapositive is useful (not equal (mod a) => not equal).[/quote]

Yeah, these sort of questions which involve too many variables to just solve, almost always use mods. Even if it does not solve the equation outright, it will often give some useful information. For example, it is  'wellknown'  that odd squares are 1 mod 8 (you can check this if you like.) Considering the equation mod 8, if x and y are even we get -1 == 4 mod 8, not true, and if they are both odd then x^5 must be 5 mod 8, and so x must be 5 mod 8. Usually finding the right mod (or mods) will solve the question.

Heres another problem with exactly the same idea as the above one. Prove that the equation x^3 + y^3 + z^3 = 2003 has no integer solutions. (Hint : think mods!!)

2003 mod 9 = 5

x^3  mod 9 in {0,1,-1} so x^3 + y^3 + z^3 always different from 5 mod 9.

Why 9 ???

Euler theorem :

a^phi(n) = 1 mod n if (a,n) = 1 (n prime or not).
phi being the Euler function

phi(9) = 6

so a^6 = (a^3)^2 = 1 if (a,9) = 1.

I don't have anything definite yet, but I was messing around with this in math class. Since sqrt(x^5-4) has to be an integer, you need to find an integer^5 that is 4 greater than a square. I tested this for x'es of 2 to 10 and this is what I got...
X----x^5-4---Nearest square---------Value of-------Difference between
------------number below x^5-4----that square----x^5-4 and the square
2___28________5^2_________________25_______________3
3___239_______15^2________________225_____________14
4___1020______31^2________________961_____________59
5___3121______55^2________________3025____________96
6___7772______88^2________________7749____________23
7___16803_____129^2_______________16641__________162
8___32764_____181^2_______________32761__________3
9___59045_____242^2_______________58564__________481
10__99996_____316^2______________99956___________140

I tested this for only 1,2,3, and 4 at first. I was really excited because I noticed that the differences kept on increasing. That theory was blown away when I continued the pattern. I put this up here because someone might be able to use it, perhaps there is a pattern in the differences or the squares?
Sorry about the messy underscores- the posting program deletes additional spaces. Hope that you can read this okay.

-Isaac


analytic geometry
combinatorics solved
combinatorics

It is a very very well-known problem....
You may find a proof that the answer is negative, in french, at :
http://www.animath.fr

go to 'Les cours de l'olympiade franaise de mathmatiques'
then :
'Les graphes'

Pierre.

Kuratowsky theorem:
Graph F is planar if and only if it does not contain subgraph isomorphic to K_5 or K_(3,3).

Namdung

You are obviously right, Namdung. But, in our case, we do not have to deal with the general case of Kuratowski theorem, which is difficult to prove.
To prove that K_5 and K_3,3 are not planar, you only need Euler's formula for planar graphs.

Pierre.

I hadn't heard of Kuratowsky's thm before. That's impressive.

Try this (I haven't tried it, so I don't know the answer :D ): Is K_(3,3) planar on the surface of a thorus?

Yes, both K_5 and K_3,3, are planar on the torus. It may be shown that even K_7 is, but it is too boring to describe with a keyboard. Let's try for the two others :

The torus may be seen as a square whose sides are vertically and horizontally, and for which, when going outside from one vertical side you are coming inside from the opposite side, at the same y-coordinate, and the same is for the horizontal sides (for the x-coordinate). So :

- For K_5 :
Into a big square (the torus), draw a little square ABCD where the line (AC) goes through the midpoints of opposite sides of the big square, and the same for (BD). Note that on the torus, these two midpoints represent the same point.
Let O be the center of ABCD.
We may draw each sides of ABCD and each segments connecting O to the A,B,C,D. From the orientation of ABCD, we may connect A to C 'from the outside'. And we do the same for B and D, with respect to the other direction of the torus.

- For K_3,3 :
Choose the same little square ABCD. And choose the common midpoints of pair of opposite sides, say E and F (see above), where E is collinear (for the big square) to A and C, and F with B and D.
Then draw ABCD and connect both A and C to E, and both Band D to F. It remains to connect E and F, which is easy.

Pierre.

I saw this puzzle-type question on a website:

A physicist wants to determine whether he lives in a plane or on the surface on the torus. How can he do that having only a rope?

I don't know what the answer is, but maybe they're somehow related.


Let $ a_{0}=29,a_{1}=105,a_{2}=381$ and $ a_{n+3}=3a_{n+2}+2a_{n+1}+a_{n}\forall n=0,1,2,...$

Prove that for each positive integer $ m$, there exists a number $ n$ such that $ a_{n},a_{n+1}-1,a_{n+2}-2$ are all divisible by $ m$.

It is seminar a problem in Romani Olympiad on 1998.
Consider the sequence
$ r_0 \equal{} 0,r_1 \equal{} 1,r_{2} \equal{} 3$
$ r_{n \plus{} 3} \equal{} 3r_{n \plus{} 2} \plus{} 2r_{n \plus{} 1} \plus{} r_n$
$ r_4 \equal{} a_0,r_{5} \equal{} a_1,r_{6} \equal{} a_2$ so $ r_{n \plus{} 4} \equal{} a_n$
Because $ \{r_n\}(\mod m)$ is periodic so has infinite n such that $ m|a_n,a_{n \plus{} 1} \minus{} 1,a_{n \plus{} 2} \minus{} 2$


The natural number $ n$ has the property that $ n^{2}\plus{}1$ is divisible by $ n\plus{}1$. Find $ n$.

When $ n \equal{} 1$, $ n^2 \plus{} 1 \equal{} 2$ and $ n \plus{} 1 \equal{} 2$. Clearly 2 divides itself, so the natural number is $ \boxed{n \equal{} 1}$.

Also, note that this value is the only satisfactory $ n$. Here's the brief proof:

Take $ n^2 \minus{}1$, which we can factor as $ (n \plus{} 1)(n \minus{} 1)$. We see that $ n \plus{} 1$ is a factor of $ (n^2 \plus{} 1) \minus{} 2$.

Thus, the only case for which $ n^2 \plus{} 1 \equiv [(n^2 \plus{} 1) \minus{} 2] \pmod {n \plus{} 1}$ is if $ 2 \equiv 0 \pmod {n \plus{} 1}$, so 2 is a multiple of $ n \plus{} 1$.

The only solution for $ n \in N$ is $ \boxed{n \equal{} 1}$.

Thanks goldenboy1.618.

$ n^{2}\minus{}1\equal{}(n\minus{}1)(n\plus{}1) \implies (n\plus{}1) \mid (n^{2} \minus{}1)$.

$ (n\plus{}1) \mid (n^{2} \plus{}1)$ and $ (n\plus{}1) \mid (n^{2} \minus{}1) \implies (n\plus{}1)$ divides their difference, that is, $ (n\plus{}1) \mid 2$.

But $ n$ is a natural number, so $ n\plus{}1 \ge 2$. So, $ n\plus{}1 \equal{} 2$, that is, $ n\equal{}1$.

[hide="Alternate Solution"]
If $ \dfrac{n^2 \plus{} 1}{n \plus{} 1}\in \mathbb{Z}\implies \dfrac{n^2 \plus{} 1}{n \plus{} 1} \minus{} (n \minus{} 1) \equal{} \dfrac{2}{n \plus{} 1}\in \mathbb{Z}$, from where $ n \equal{} 1$ is the only solution.
[/hide]

Edit: Beaten..

Usually for those sorts of problems a technique that (almost)always works is polynomial division.

So in the case of this problem, we have $ \frac {n^2 \plus{} 1}{n \plus{} 1} \in \mathbb{N}.$ Using synthetic division, we get $ \frac {n^2 \plus{} 1}{n \plus{} 1} \equal{} n \minus{} 1 \plus{} \frac {2}{n \plus{} 1}.$ Since the RHS must be an integer, $ (n \plus{} 1) \ | \ 2,$ so the only natural number that satisfies that is $ n \equal{} 1.$


What is the hundreds digit in the following product:
$ 5 \times 6 \times 7 \times 8 \times 9 \times 10$ ?

alright: multiply the first 4 numbers, and find the tens digit. that wil be the hundreds digit b/c when you multiply by 10 you add a 0 and the 10s become hundreds. we only care abt the 10s place

5*6=3
3*7=1
1*8=8
8*9=2

so 2 is the hundreds place


I stumbled upon this recently. To my understanding, they are opening in 2009-2010 school year, enrolling 40 juniors each year for a 2 year program. Their site has an application.

I'm pretty sure I'm a year too old to apply, but I thought I'd share the word.

http://kams.fhsu.edu/

Not sure if this will be useful for me but thanks for the info!


function
calculus
derivative
calculus computations

I am having a bit of trouble understanding how to prove the basic identities of the Poisson Brackets... could someone help me out?

For two given sets of $ n$ points $ P: {p_1, p_2, \cdots, p_n}$ and $ Q: {q_1, q_2, \cdots, q_n}$ and two functions $ f, g$ which are functions of $ p_i, q_i,$ the Poisson Bracket is defined to be

\[ [f, g] \equal{} \sum{\frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i} \minus{} \frac{\partial g}{\partial q_i} \frac{\partial f}{\partial p_i}}
\]

The identity I'm having trouble with proving is

\[ [f, q_i] \equal{} \minus{}\frac{\partial f}{\partial p_i} \equal{} \frac{\partial f}{\partial q_i}
\]

Generally, I only know the definitions and minimal knowledge of partial differentiation, but no fancy manipulations. I am much better with single variable calculus. Could someone post a proof of the above identity? Thanks.  :)

Substitute $ g \equal{} q_k$ into the definition and you get
\[ [f, q_k] \equal{} \sum_i{\frac {\partial f}{\partial q_i} \frac {\partial q_k}{\partial p_i} \minus{} \frac {\partial q_k}{\partial q_i} \frac {\partial f}{\partial p_i}} \equal{}
\]
Now all the $ q_i$ and $ p_i$ are independent variables, so $ \frac {\partial q_k}{\partial p_i} \equal{} 0$ for all $ i$ and $ \frac {\partial q_k}{\partial q_i} \equal{} \begin{cases} 1 & k \equal{} i \\
0 & k\neq i \end{cases}$.  Hence,
\[ [f,q_k] \equal{} \minus{} \frac {\partial f}{\partial p_k}
\]

Ah, okay... I was just making sure that we could set $ \partial q / \partial p \equal{} 0$. As a general idea, is this allowed?:

\[ \frac{\partial f}{\partial x} \frac{\partial x}{\partial y} \equal{} \frac{\partial f}{\partial y}
\]


Solve in $x,y,z$:-
	$a = (2x + y + z)^2 -4x^2$
	$b = (2y + x + z)^2 -4y^2$
	$c = (2z + y + x)^2 -4z^2$
Under what co ndition on $a,b,c$ we get positive solutions?

[quote="rohitsingh0812"]Solve in $x,y,z$:-
	$a = (2x + y + z)^2 -4x^2$
	$b = (2y + x + z)^2 -4y^2$
	[b]$c = (2z + y + z)^2 -4z^2$[/b]
Under what co ndition on $a,b,c$ we get positive solutions?[/quote]
It's supposed to be $c = (2z + y + x)^2 -4z^2$, right?


I'm pretty positive that my answer is correct so if you want to look at the answer before posting, look at mine.

But don't look at it before you solved it :D

[hide="Solution"]

There are 6 choices on the first time and another 6 chocies on the second time.

The pairs that are LESS than 29 or GREATER than 61 are:

4-4
4-5
4-6
4-7
5-4
6-4
7-4

9-9
9-8
9-7
8-9
7-9

Oops.. There are 5*5.

But we want the ones that are in that range so subtract the 13 from total. The total, we found on the beginning is 36, so the amount on the top is 36-13 = 23.

Simplify the fraction:

\[\frac {23}{36}$[/hide]

[hide]Find what we don't want

Below 29, there are:
16
20
24
25
28

There are 5*2 below 29

Now above 61:
63
64
72
81

There are 4 above 61

There are 6*6/2 choices total, so there are 18-9=9
that work so the probability is 9/18=1/2[/hide]

Solafidefarms,

Take another look at the method you've used to count everything.

[color=green]

[hide]11/18 :D[/color][/hide]

[quote="MCrawford"]Two cubes, each number with the integers 4 through 9, are tossed.  What is the probability that the product of the two numbers rolled is greater than 29 and less than 61?  Express your answer as a common fraction.[/quote]

[hide]
We know that the product of the two numbers must be 29-61. The pairs that would satisfy that are:

4, 8
4, 9

5, 6
5, 7
5, 8
5, 9

6, 5
6, 6
6, 7
6, 8
6, 9

7, 5
7, 6
7, 7
7, 8

8, 4
8, 5
8, 6
8, 7

9, 4
9, 5

That's 22 possibilities to satisfy the requirement. There are 36 combinations in total, because 6*6 is 36.

22/36 = 11/18
[/hide]

[quote="LuCy4EvA"]I hate probability problems, i usually miss them by a very little number...is the only way to do this problem listing them?[/quote]

I don't think so. Ii usually make a 6x6 multiplication table. And just do all the multiplication to find it all the possibilities.

[quote="SoccerBrainy40"]
We know that the product of the two numbers must be 29-61. The pairs that would satisfy that are:

4, 8
4, 9

5, 6
5, 7
5, 8
5, 9

6, 5
6, 6
6, 7
6, 8
6, 9

7, 5
7, 6
7, 7
7, 8

8, 4
8, 5
8, 6
8, 7

9, 4
9, 5

That's 22 possibilities to satisfy the requirement. There are 36 combinations in total, because 6*6 is 36.

22/36 = 11/18
[/quote]
btw, you forgot 9,6... which leads to my answer... instead of yours.

[hide="solution"]find the multiple that are less than 29 or greater than 61.
(4,4)
(4,5)(5,4)
(4,6)(6,4)
(4,7)(7,4)

(8,8)
(8,9)(9,8)
(9,9)
(7,9)(9,7)

total = 13.

subtract from ALL possibilities = 36:

36-13 = 23, $\frac{23}{36}$[/hide]

oops... i understood the problem wrong...

[hide="solution"]
since they are the same kind of dice, a 4 on one of them and 6 on the other is not different from 6 on one and 4 ont he other...

meaning, there are only 21 possibilities in total.

4,4
4,5
4,6
4,7
4,8
4,9
5,5
5,6
5,7
5,8
5,9
6,6
6,7
6,8
6,9
7,7
7,8
7,9
8,8
8,9
9,9

out of those, there are 9 which does not satisfy the given property.

so, $1 - \frac{9}{21} = \frac{12}{21} = \frac{4}{7}$
[/hide]

i hope that is the right way to interpret the probelm...


[quote="omjfv"]Let in quadrilateral $ABCD$ be $\angle CAB=40, \angle CAD=30, \angle DBA=75, \angle DBC=25$. Find $\angle BDC$[/quote]

You'll find that you can determine inmediately all angles in the quadrangle except $\angle BDC$ and $\angle ACD$.  You'll find
1) Triangle $ABC$ is isosceles, base $AC$, so $AB = BC$
2) Take point $E$ on $AD$ such that $\angle BED = 110$, then $\triangle BED$ isosceles, base $BD$ hence $BE = ED$
3) Also triangle $ABE$ isosceles, base $AE$, hence $AB=BE$. Hence using 1) and 3) $B$ circumcenter of triangle $AEC$ and as $\angle EAC = 30$, you have $\angle EBC = 60 \Rightarrow \triangle BEC$ equilateral
4) $BE = BC$ and using 2) $= ED$, hence $E$ circumcenter of triangle $BCD$ with $\angle BEC = 60$ hence $\angle BDC = 30$

Daniel


Prove that if $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$ then:

$ \frac {a^3\plus{}1} {a^2\plus{}1}\plus{}\frac {b^3\plus{}1} {b^2\plus{}1}\plus{}\frac {c^3\plus{}1} {c^2\plus{}1} \ge 2$

$ \frac {a^{n\plus{}1}\plus{}1} {a^n\plus{}1}\plus{}\frac {b^{n\plus{}1}\plus{}1} {b^n\plus{}1}\plus{}\frac {c^{n\plus{}1}\plus{}1} {c^n\plus{}1} \ge 2$, where $ n\in \mathbb{N}^*$

$ \sum \frac{a^3\plus{}1}{a^2\plus{}1}\equal{}\sum a\plus{}\sum \frac{1\minus{}a}{a^2\plus{}1}\equal{}1\plus{}\sum \frac{1\minus{}a}{a^2\plus{}1}$ hence it suffices to prove that
$ \sum \frac{1\minus{}a}{a^2\plus{}1}\geq 1$
By Cauchy-Schwarz it's enough to prove that
$ 2\geq \minus{}\sum a^3\plus{}\sum a^2\iff \sum a^3\plus{}2\sum a\geq \sum a^2$ which is true.

Using $ 2(a^{n \plus{} 1} \plus{} 1)\ge (a \plus{} 1)(a^n \plus{} 1)$ which follows from $ (a \minus{} 1)(a^n \minus{} 1)\ge 0$, we get
\[ \sum\frac {a^{n \plus{} 1} \plus{} 1}{a^n \plus{} 1}\ge\sum\frac {a \plus{} 1}{2} \equal{} 2
\]

Actually the following inequality is true for $ a\plus{}b\plus{}c\equal{}1$:
\[ 3>\sum\frac {a^{n \plus{} 1} \plus{} 1}{a^n \plus{} 1}>2\]


function
topology
algebra
domain
limit
vector
complex analysis

This's very known one but I could find its solution: Prove that for all continous function $F:{\mathbb R}^{n}\to{\mathbb R}^{n}$ such that $dist(Fx,Fy)\geq Ldist(x,y)$ with $L>1$ then $F$ has a fixed point.

I also guess that in the infinitely dimesion space, the problem is quite hard, isn't it?

In an infinitely-dimensional Banach space the statement may be false: just consider twice the forward shift in $\ell^{2}$. On the other hand, in this case it may be natural to ask whether there exists an infinite-dimensional Banach space with this property.

Here's a proof for the initial problem (the finite-dimensional case):

The idea is to show that $f$ is onto (that is, $f(\mathbb R^{n})=\mathbb R^{n}$). Once we do that, it will follow that $f$ is actually a homeomorphism of $\mathbb R^{n}$ onto itself, and we'll be able to apply the Contraction Principle to $f^{-1}$ to conclude that we can find $x\in\mathbb R^{n}$ such that $f^{-1}(x)=x\Rightarrow f(x)=x$.

In order to prove that $f$ is onto, notice that, on the one hand, it's a one-to-one continuous map of $\mathbb R^{n}$ into itself and hence it's an open map, by Brouwer's [url=http://en.wikipedia.org/wiki/Invariance_of_domain]Domain Invariance Theorem[/url]. In particular, $f(\mathbb R^{n})$ is an open subset of $\mathbb R^{n}$. On the other hand, notice that the range of $f$ closed: if, say, $f(x_{n})\to y$, then $f(x_{n})$ is a Cauchy sequence and hence $x_{n}$ is a Cauchy sequence, meaning that we can find $x=\lim_{n}x_{n}$ such that $f(x)=y$. This means that the non-empty subset $f(\mathbb R^{n})$ of the connected set $\mathbb R^{n}$ is both open and closed, so it must be the whole of $\mathbb R^{n}$.

Where would grobber's proof break down in an infinite dimensional case? Note that he very much needed that $f(\mathbb{R}^{n})$ be open, and he brought in his biggest and fanciest weapon (the Brouwer domain invariance theorem) just to show that. In fedja's infinite-dimensional example of twice the right shift, $f(\ell^{2})$ is not open - it is a proper closed vector subspace, and contains no balls. However, that example does trivially contain a fixed point: zero. We'll have to make some adjustments to have an example.

In fact, I haven't yet found a counterexample on $\ell^{2}.$

There's a trivial tweak to generate a counterexample: add a constant. If $f(x_{1},x_{2},\dots)=(1,2x_{1},2x_{2},\dots)$, $f$ has no fixed points in $\ell^{2}$: a fixed point would have $x_{1}=1,x_{2}=2,x_{3}=4,\dots$, and that sequence isn't in $\ell^{2}$.

Thinking topologically, we can generalize to maps from a connected (finite-dimensional) manifold to itself; the theorem grobber quoted still works locally, since points have neighborhoods homeomorphic to $\mathbb{R}^{n}$. The distance can be any distance which generates the right topology, although there are no such maps if the space is bounded with this distance.

Ahh - somehow I'd convinced myself that if $R$ is the right shift, then $(I-2R)$ was invertible. It's not, so $Tx=2Rx+b$ does not have a fixed point as long as $b$ is not in the range of $(I-2R).$

Correcting my manifold statement- we need some extra metric/topological conditions. I think the condition that closed and bounded sets are compact does it.

Thanks  friends for useful discussings. Really it's a cool one. Agree with comments on manifolds by jmerry, but how the extra conditions play a role in here, for example let me take $H(\Omega)$ space of all holomophic function on $\Omega$. I guess this space staisfies your condition.

The original hypothesis is concerns metric properties. What's your metric on $H(\Omega)?$

Then again, if $\Delta$ is the unit disc, and we map $H(\Delta)$ to $H(\Delta)$ by $T(f)(z)=1+2zf(z),$ what do we have? (Look earlier on this thread, and you'll understand the inspiration for the example.)

The space of holomorphic functions is infinite-dimensional; we should have counterexamples in any sensible metric. My statement about manifolds did not leave the finite-dimensional world.

ah I just don't understand some English words. In fact i mean $H(\Omega)$ with metric generate from semi-norm squences (as usual). This topo on $H(\Omega)$  satisfying the condition which jmerry want.
By the way, there is a same question: If $X$ is Banach space; $f: X\to X$ with the above condition, ie... $|f(x)-f(y)|\geq L|x-y|$ with $L>1$. Assume that   $f$ has a fixed point. Does $f$ must be $1-1$ from $X$ onto $X$?

Certainly not. The counterexamples already given in this thread are in Banach spaces. I don't think anything infinite-dimensional works.

It doesn't make sense to ask the question unless our space is a complete metric space; if we require that it's a vector space as well and that the metric is a norm, we have a Banach space. That isn't enough.

The correct manifold extension: If $X$ is a topological manifold (locally homeomorphic to $\mathbb{R}^{n}$ for some fixed $n$) which is connected and complete in some metric $d$, then any $f$ which satisfies $d(f(x),f(y))\ge L(d(x,y))$ with $L>1$ is a homeomorphism from $X$ to itself, and has a unique fixed point.

[quote="goldhp"]ah I just don't understand some English words. In fact i mean $H(\Omega)$ with metric generate from semi-norm squences (as usual). This topo on $H(\Omega)$  satisfying the condition which jmerry want.[/quote]
Then jmerry and I did both understand you correctly, and the answer is no, because it's infinite-dimensional. jmerry's conditions started with the set being a finite-dimensional manifold. The example I gave, or something like it, should be a counterexample.

[quote]By the way, there is a same question: If $X$ is Banach space; $f: X\to X$ with the above condition, ie... $|f(x)-f(y)|\geq L|x-y|$ with $L>1$. Assume that   $f$ has a fixed point. Does $f$ must be $1-1$ from $X$ onto $X$?[/quote]
False. Just go back to the first thing fedja said, twice the right shift on $\ell^{2}.$ This mapping satisfies your expansion condition, and it has a fixed point: zero. It's 1-1, as any mapping satisfying the expansion condition trivially must be, but it does not map $\ell^{2}$ onto $\ell^{2}.$

hi. I am interested in the original problem proposed by goldhp. I went through grobber's solution , is there any other way to solve this problem? ( i dont know brouwer's domain invariance theorem)  also when i first saw the problem, i thought that the problem is false. the distance between 2 points keeps increasing on applying f , so how come there is a fixed point?

Here's how you construct a fixed point: apply $f^{-1}$ repeatedly. We need powerful theorems to show that $f^{-1}$ is defined everywhere, but it is certainly a contraction if it is.

[quote="kouboy"] is there any other way to solve this problem? ( i dont know brouwer's domain invariance theorem)  [/quote] Something like Brouwer's theorem should be used if your function is assumed to be just continuous. On the other hand, if you know that it is differentiable, everything is easy: just consider the point where $|f(x)-x|$ attains its minimum. I played for a while with the idea of approximating continuous expansions by differentiable ones but couldn't figure out how to do that :(. If somebody can do such approximation by elementary means, it will result in another solution.


inequalities
function
integration
real analysis
real analysis unsolved

Assume that $f(x)$ is an increasing function and $n\in \mathbb{N}$.
Prove the following inequality.

\[0\leqq \frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right) -\int_0^1 f(x)dx\leqq \frac{1}{n}\{f(1)-f(0)\}\]

Since $f$ is increasing,
\[\frac{1}{n}f(\frac{k-1}{n})\int_\frac{k-1}{n}^\frac{k}{n} f(x)dx \leq \frac{1}{n}f(\frac{k}{n}),\]
thus
\[0 \leq \frac{1}{n}f(\frac{k}{n}) - \int_\frac{k-1}{n}^\frac{k}{n} f(x)dx \leq \frac{1}{n}\left(f(\frac{k}{n})-f(\frac{k-1}{n})\right)\]
Sum them up for $k=1,...,n$ we obtain the inequality as desired.

A monotonous function on [0,1] is always Riemann-integrable (because it is continuous except at (at most) a countable number of points); for the two inequalities, draw a graph and see what happens (it's basic majoration / minoration).


This is actually a somewhat tricky problem so here's a hint: The only thing keeping you from solving the problem is that we don't know the height of the cone.  However, we can find it.  Since the sphere has radius 5, from the center of the sphere to one of its edges is 5 (the center of the sphere is somewhere along the center axis of the cone).  And we know the radius of the cone is 3 so we can set up a triangle and solve for the distance from the center of the sphere to the base of the cone.  Then we can add this to the radius of the sphere which would give us the total height.


calculus
integration
function
real analysis
real analysis theorems

Let $ f$ and $ g$ be summable over the interval [a,b], and suppose $ F(x) \equal{} F(a) \plus{} \int_a^x f(y)dy$, and $ G(x) \equal{} G(a) \plus{} \int_a^x g(y)dy$, then

$ \int_a^b F(x)g(x)dx \equal{} F(b)G(b) \minus{} F(a)G(a) \minus{} \int_a^b G(x)f(x)dx$.

-For the sake of example, I want to come up with a concrete $ F$ and $ g$ and carry out the integration. In fact, i'm having difficulty finding functions which are Lebesgue-integrable but not Riemann integrable (I know that statement must sound really funny since they are plenty of them). My text doesn't provide any concrete examples  :|  Any good webpages that can be recommended?

Los

You mean that you'd like to see [i]non Riemann integrables[/i] f and g for which integration by part applies?

As for a non Riemann integrable but Lebesgue integrable function, the typical example is the characteristic function of the rationals in [0,1], $ \mathbb{I}_{\mathbb{Q}\cap[0,1]}$.

What is the question? Anyway, for the formula to be true it is sufficient (and nearly necessary) that $ g$ be lebesgue integrable and $ F$ to be absolutely continuous. If $ F$ is not absolutely continuous then it has a singular part which is detected in the LHS, but not the RHS.


For $ a,b,c \ge 0$ and $ a\plus{}b\plus{}c\equal{}1$ then prove that
\[ \frac{{1 \plus{} a}}{{1 \minus{} a}} \plus{} \frac{{1 \plus{} b}}{{1 \minus{} b}} \plus{} \frac{{1 \plus{} c}}{{1 \minus{} c}} \le 2\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right)\]
\[ \Leftrightarrow \frac{{2a \plus{} b \plus{} c}}{{b \plus{} c}} \plus{} \frac{{a \plus{} 2b \plus{} c}}{{a \plus{} c}} \plus{} \frac{{a \plus{} b \plus{} 2c}}{{a \plus{} b}} \le 2\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right)
\]
\[ \Leftrightarrow 3 \plus{} 2\left( {\frac{a}{{b \plus{} c}} \plus{} \frac{b}{{c \plus{} a}} \plus{} \frac{c}{{a \plus{} b}}} \right) \le 2\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right)
\]

which is the inequality of mathlinks contest
\[ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \minus{} \frac{3}{2} \ge \frac{a}{{b \plus{} c}} \plus{} \frac{b}{{c \plus{} a}} \plus{} \frac{c}{{a \plus{} b}}
\]
:)

[quote="frankvista"]Did you install MathPlayer?

Here's a question:
If a>0,b>0,c>0,a+b+c=1
Prove that:(1+a)/(1-a)+(1+b)/(1-b)+(1+c)/(1-c)≤2(a/b+b/c+c/a)。

$ (1 \plus{} a)/(1 \minus{} a) \plus{} (1 \plus{} b)/(1 \minus{} b) \plus{} (1 \plus{} c)/(1 \minus{} c) < \equal{} 2(a/b \plus{} b/c \plus{} c/a)$

Unluckily,I can't solve it.

You're clever,can you solve it?[/quote]
Let $ c \equal{} \min\{a,b,c\}.$ We obtain:
$ 2\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)\geq\frac {1 \plus{} a}{1 \minus{} a} \plus{} \frac {1 \plus{} b}{1 \minus{} b} \plus{} \frac {1 \plus{} c}{1 \minus{} c}\Leftrightarrow$
$ \Leftrightarrow2\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right)\geq\sum_{cyc}\left(\frac {2a}{b \plus{} c} \minus{} 1\right)\Leftrightarrow$
$ \Leftrightarrow2\left(\frac {a}{b} \plus{} \frac {b}{a} \minus{} 2 \plus{} \frac {b}{c} \minus{} \frac {b}{a} \plus{} \frac {c}{a} \minus{} 1\right)\geq\frac {2a^3 \plus{} 2b^3 \plus{} 2c^3 \minus{} a^2b \minus{} a^2c \minus{} b^2a \minus{} b^2c \minus{} c^2a \minus{} c^2b}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}\Leftrightarrow$
$ \Leftrightarrow2\left(\frac {(a \minus{} b)^2}{ab} \plus{} \frac {(c \minus{} a)(c \minus{} b)}{ac}\right)\geq\frac {2(a \plus{} b)(a \minus{} b)^2 \plus{} (a \plus{} b \plus{} 2c)(c \minus{} a)(c \minus{} b)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}\Leftrightarrow$
$ \Leftrightarrow2(a \minus{} b)^2\left(\frac {1}{ab} \minus{} \frac {1}{(a \plus{} c)(b \plus{} c)}\right) \plus{}$
$ \plus{} (c \minus{} a)(c \minus{} b)\left(\frac {2}{ac} \minus{} \frac {a \plus{} b \plus{} 2c}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}\right)\geq0,$ which is obviously true.

[quote="frankvista"][quote]
which is the inequality of mathlinks contest
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \frac {3}{2} \ge \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}}
\]
:)[/quote]

Can you tell me the details of the proof?[/quote]

See here frankvista, http://www.mathlinks.ro/viewtopic.php?t=206917 :)

[quote="frankvista"][quote]
which is the inequality of mathlinks contest
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \frac {3}{2} \ge \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}}
\]
:)[/quote]

Can you tell me the details of the proof?[/quote]

Inequality <=> $ \sum a(\frac1{b}\minus{}\frac1{b\plus{}c})\ge\frac3{2}$  <=> $ \sum\frac{ca}{b(b\plus{}c)}\ge\frac3{2}$
By Cauchy-Schwarz and AM-GM we have
$ \sum\frac{ca}{b(b\plus{}c)}\ge\frac{(ab\plus{}bc\plus{}ca)^2}{2abc(a\plus{}b\plus{}c)}\ge \frac{3abc(a\plus{}b\plus{}c)}{2abc(a\plus{}b\plus{}c)}\equal{}\frac3{2}$

We have done!  :roll:

[quote="arqady"]Nice proof, akai! :lol:[/quote]
Thanks,arqady   :lol:  Nother proof (dduclam tell me):

Because $ \sum\frac {a}b\ge\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca}$
thus we need prove only  $ \frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca} \minus{} 3\ge \frac a{b \plus{} c} \minus{} \frac3{2}$
<=>$ \sum\frac {(b \minus{} c)^2}{2(ab \plus{} bc \plus{} ca)}\ge\sum\frac {(b \minus{} c)^2}{2(a \plus{} b)(a \plus{} c)}$
<=> $ \sum\frac {a^2(b \minus{} c)^2}{(a \plus{} b)(a \plus{} c)}\ge0$
so we are done  :roll:

@arqady: Your proof also nice!  :lol:

PS: It's Japan TST 2004  :)


algebra
polynomial
group theory
abstract algebra
topology
vector
function

Miklós Schweitzer Competition 2009 will be on 30 October - 9 November 2009.
The problems will be announced in the competition web-sites on this Friday, 30 October 2009 at 12:00.
The guidelines of 2009 Competition: http://www.cs.elte.hu/students/schweitzer/kiiras2009.pdf (in Hungarian)

http://www.cs.elte.hu/students/schweitzer/
http://www.math.klte.hu/schw09/
http://www.bolyai.hu/schweitzer.html

Problems [b]#3,4,5,8,9[/b] are translated into English.
(Disclaimer. This is unofficial translation. Moreover at the moment it has not been checked by native Hungarian speaker yet.)

[i]Attachment deleted.[/i]

[quote="dmitin"]Current version of the translation. (5 and 8 are fixed, 12 added.)

Please watch the updates at Miklós Maróti's (aka [b]mmaroti[/b]) site:
http://www.math.u-szeged.hu/~mmaroti/schweitzer/[/quote]

All problems are now translated. Please let me know if you find any typos of problems.

Best,
Miklos

[size=150][b]Problems of the Miklós Schweitzer Memorial Competition 2009[/b][/size]

[url=http://www.mathlinks.ro/viewtopic.php?t=311746][b]1.[/b][/url] On every card of a deck of cards a regular 17-gon is displayed with all sides and diagonals, and the vertices are numbered from 1 through 17. On every card all edges (sides and diagonals) are colored with a color 1,2,...,105 such that the following property holds: for every 15 vertices of the 17-gon the 105 edges connecting these vertices are colored with different colors on at least one of the cards. What is the minimum number of cards in the deck?

[url=http://www.mathlinks.ro/viewtopic.php?t=311748][b]2.[/b][/url] Let $ p_1,\dots,p_k$ be prime numbers, and let $ S$ be the set of those integers whose all prime divisors are among $ p_1,\dots,p_k$. For a finite subset $ A$ of the integers let us denote by $ \mathcal G(A)$ the graph whose vertices are the elements of $ A$, and the edges are those pairs $ a,b\in A$ for which $ a \minus{} b\in S$. Does there exist for all $ m\geq 3$ an $ m$-element subset $ A$ of the integers such that
(i) $ \mathcal G(A)$ is complete?
(ii) $ \mathcal G(A)$ is connected, but all vertices have degree at most 2?

[url=http://www.mathlinks.ro/viewtopic.php?t=311749][b]3.[/b][/url] Prove that there exist positive constants $ c$ and $ n_0$ with the following property. If $ A$ is a finite set of integers, $ |A| \equal{} n > n_0$, then
\[ |A \minus{} A| \minus{} |A \plus{} A| \leq n^2 \minus{} c n^{8/5}.\]
[url=http://www.mathlinks.ro/viewtopic.php?t=311750][b]4.[/b][/url] Prove that the polynomial
\[ f(x) \equal{} \frac {x^n \plus{} x^m \minus{} 2}{x^{\gcd(m,n)} \minus{} 1}\]
is irreducible over $ \mathbb{Q}$ for all integers $ n > m > 0$.

[url=http://www.mathlinks.ro/viewtopic.php?t=311751][b]5.[/b][/url] Let  $ G$ be a finite non-commutative group of order $ t \equal{} 2^nm$, where $ n, m$ are positive and $ m$ is odd. Prove, that if the group contains an element of order $ 2^n$, then
(i) $ G$ is not simple;
(ii) $ G$ contains a normal subgroup of order $ m$.

[url=http://www.mathlinks.ro/viewtopic.php?t=311753][b]6.[/b][/url] A set system $ (S,L)$ is called a Steiner triple system, if $ L\neq\emptyset$, any pair $ x,y\in S$, $ x\neq y$ of points lie on a unique line $ \ell\in L$, and every line $ \ell\in L$ contains exactly three points. Let $ (S,L)$ be a Steiner triple system, and let us denote by $ xy$ the thrid point on a line determined by the points $ x\neq y$. Let $ A$ be a group whose factor by its center $ C(A)$ is of prime power order. Let $ f,h: S\to A$ be maps, such that $ C(A)$ contains the range of $ f$, and the range of $ h$ generates $ A$.
Show, that if
\[ f(x) \equal{} h(x)h(y)h(x)h(xy)\]
holds for all pairs $ x\neq y$ of points, then $ A$ is commutative, and there exists an element $ k\in A$, such that $ f(x) \equal{} kh(x)$ for all $ x\in S$.

[url=http://www.mathlinks.ro/viewtopic.php?t=311754][b]7.[/b][/url] Let $ H$ be an arbitrary subgroup of the diffeomorphism group $ \mathsf{Diff}^\infty(M)$ of a differentiable manifold $ M$. We say that an $ \mathcal C^\infty$-vector field $ X$ is [i]weakly tangent[/i] to the group $ H$, if there exists a positive integer $ k$ and a $ \mathcal C^\infty$-differentiable map $ \varphi \mathrel{: } \mathord{]} \minus{} \varepsilon,\varepsilon\mathord{[}^k\times M\to M$ such that
(i) for fixed $ t_1,\dots,t_k$ the map
\[ \varphi_{t_1,\dots,t_k} : x\in M\mapsto \varphi(t_1,\dots,t_k,x)\]
is a diffeomorphism of $ M$, and $ \varphi_{t_1,\dots,t_k}\in H$;
(ii) $ \varphi_{t_1,\dots,t_k}\in H \equal{} \mathsf{Id}$ whenever $ t_j \equal{} 0$ for some $ 1\leq j\leq k$;
(iii) for any $ \mathcal C^\infty$-function $ f: M\to \mathbb R$
\[ X f \equal{} \left.\frac {\partial^k(f\circ\varphi_{t_1,\dots,t_k})}{\partial t_1\dots\partial t_k}\right|_{(t_1,\dots,t_k) \equal{} (0,\dots,0)}.\]
Prove, that the commutators of $ \mathcal C^\infty$-vector fields that are weakly tangent to $ H\subset \textsf{Diff}^\infty(M)$ are also weakly tangent to $ H$.

[url=http://www.mathlinks.ro/viewtopic.php?t=311755][b]8.[/b][/url] Let $ \{A_n\}_{n \in \mathbb{N}}$ be a sequence of measurable subsets of the real line which covers almost every point infinitely often. Prove, that there exists a set $ B \subset \mathbb{N}$ of zero density, such that $ \{A_n\}_{n \in B}$ also covers almost every point infinitely often. (The set $ B \subset \mathbb{N}$ is of zero density if $ \lim_{n \to \infty} \frac {\#\{B \cap \{0, \dots, n \minus{} 1\}\}}{n} \equal{} 0$.)

[url=http://www.mathlinks.ro/viewtopic.php?t=311756][b]9.[/b][/url] Let $ P\subseteq \mathbb{R}^m$ be a non-empty compact convex set and $ f: P\rightarrow \mathbb{R}_{ \plus{} }$ be a concave function. Prove, that for every $ \xi\in \mathbb{R}^m$
\[ \int_{P}\langle \xi,x \rangle f(x)dx\leq \left[\frac {m \plus{} 1}{m \plus{} 2}\sup_{x\in P}{\langle\xi,x\rangle} \plus{} \frac {1}{m \plus{} 2}\inf_{x\in P}{\langle\xi,x\rangle}\right] \cdot\int_{P}f(x)dx.\]
[url=http://www.mathlinks.ro/viewtopic.php?t=311757][b]10.[/b][/url] Let $ U\subset\mathbb R^n$ be an open set, and let $ L: U\times\mathbb R^n\to\mathbb R$ be a continuous, in its second variable first order positive homogeneous, positive over $ U\times (\mathbb R^n\setminus\{0\})$ and of $ C^2$-class Langrange function, such that for all $ p\in U$ the Gauss-curvature of the hyper surface
\[ \{ v\in\mathbb R^n \mid L(p,v) \equal{} 1 \}\]
is nowhere zero. Determine the extremals of $ L$ if it satisfies the following system
\[ \sum_{k \equal{} 1}^n y^k\partial_k\partial_{n \plus{} i}L \equal{} \sum_{k \equal{} 1}^n y^k\partial_i\partial_{n \plus{} k} L \qquad (i\in\{1,\dots,n\})\]
of partial differetial equations, where $ y^k(u,v) : \equal{} v^k$ for $ (u,v)\in U\times\mathbb R^k$, $ v \equal{} (v^1,\dots,v^k)$.

[url=http://www.mathlinks.ro/viewtopic.php?t=311758][b]11.[/b][/url] Denote by $ H_n$ the linear space of $ n\times n$ self-adjoint complex matrices, and by $ P_n$ the cone of positive-semidefinite matrices in this space. Let us consider the usual inner product on $ H_n$
\[ \langle A,B\rangle \equal{} {\rm tr} AB\qquad (A,B\in H_n)\]
and its derived metric. Show that every $ \phi: P_n\to P_n$ isometry (that is a not necessarily surjective, distance preserving map with respect to the above metric) can be expressed as
\[ \phi(A) \equal{} UAU^* \plus{} X\qquad (A\in H_n)\]
or
\[ \phi(A) \equal{} UA^TU^* \plus{} X\qquad (A\in H_n)\]
where $ U$ is an $ n\times n$ unitary matrix, $ X$ is a positive-semidefinite matrix, and $ ^T$ and $ ^*$ denote taking the transpose and the adjoint, respectively.

[url=http://www.mathlinks.ro/viewtopic.php?t=311760][b]12.[/b][/url] Let $ Z_1,\,Z_2\dots,\,Z_n$ be $ d$-dimensional independent random (column) vectors with standard normal distribution, $ n \minus{} 1 > d$. Furthermore let
\[ \overline Z \equal{} \frac {1}{n}\sum_{i \equal{} 1}^n Z_i,\quad S_n \equal{} \frac {1}{n \minus{} 1}\sum_{i \equal{} 1}^n(Z_i \minus{} \overline Z)(Z_i \minus{} \overline Z)^\top\]
be the sample mean and corrected empirical covariance matrix. Consider the standardized samples $ Y_i \equal{} S_n^{ \minus{} 1/2}(Z_i \minus{} \overline Z)$, $ i \equal{} 1,2,\dots,n$. Show that
\[ \frac {E|Y_1 \minus{} Y_2|}{E|Z_1 \minus{} Z_2|} > 1,\]
and that the ratio does not depend on $ d$, only on $ n$.

http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2009-eng.pdf

Can anyone post solution of Miklós Schweitzer Memorial Competition 2000- 2009 ( in Hungarian)?
Many thank


What is the conjugate of a real number $a$? of a pure imaginary number $bi$?

Please explain.  :lol:

[hide="first question"]let $a=a+0i$. you can see $\overline{a}=\overline{a+0i}=a-0i=a$[/hide]
[hide="second question"]$bi=0+bi$ so $\overline{bi}=\overline{0+bi}=0-bi=-bi$ [/hide]

oh snap.

I guess I was thinking too much.


How do we derive the formula tan(x/2)=tan(x)sin(x)/(tan(x)+sin(x))

multiply numerator and denominator by cot x and we get sin/(1+cos) which we get by taking the normal tan half-angle and multplying top and bottom by sqrt(1+cos)


linear algebra
matrix
abstract algebra
group theory
linear algebra unsolved

I will admit that this question is related to one of my previous investigations, but I have cleaned the problem up into a much nicer version. The basic idea is this.

THE CAYLEY TABLE MATRIX
Construct the Cayley table for the multiplicative group (mod m).
Assign to each residue class a real positive finite value according to a specific formula.
Substitute these values into the Cayley table in their corresponding positions.
Now treat the table as a matrix.

THE PROBLEM
Given knowledge about the distribution of the variables assigned to the residue classes, what is the easiest way to prove that the determinant of this matrix is nonzero (in a situation where this fact is true)?

UPDATE: I have an idea. I wonder if squarefree factorization helps.

The "group determinant" of a finite group turns out to be a product over terms that have to do with its irreducible finite-dimensional representations; this was in fact Frobenius' original motivation for studying representations of finite groups ([url=http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.22.3825]reference[/url]).  For abelian groups the story is a little simpler and the determinant splits into linear factors, so it will be nonzero if and only if all of these factors are nonzero.  See:

[url=http://en.wikipedia.org/wiki/Discrete_Fourier_transform]Discrete Fourier transform[/url]
[url=http://en.wikipedia.org/wiki/Pontryagin_duality]Pontryagin duality[/url]
[url=http://en.wikipedia.org/wiki/Regular_representation]Regular representation[/url]

In particular, for a finite cyclic group one recovers a classical formula for the eigenvalues and determinant of a [url=http://en.wikipedia.org/wiki/Circulant_matrix]circulant matrix[/url].  The general case for finite abelian groups is essentially repeated tensor products of this case.

t0rajir0u, your response has helped me a great deal. The information about the group determinant being expressible as a product of linear factors has turned out to be especially valuable, allowing me to see certain connections in a whole new way. Thank you! If anything unusual comes from this, I'll update the topic for anyone interested.


The pages on the site have been coming up really slow today....please help!

Possibilities (from the most likely to least likely):
a1.) Server Trouble
a2.) Server Maintainance
a3.) Switching server
b1.) Switching to independent server
b2.) Forest fire disrupted the server :) (Although I'm not sure if there [b]is[/b] any forest fire left :lol:)
b3.) Construction in/around server
c.) major change somewhere on AoPS
d.) Human error on the site/in the server whether it's turning the server off accidentally or tripping over the cord :) or whatever
e.) phpBB programming problem

j.) Problems on the user side

Pick your reason :)

The server went a little batty last night for some reason.  If all goes as planned, we're changing servers in 2-4 weeks and hopefully a lot of these little problems will go away.

[quote="mathfanatic"]It could also be that the large number of users are chewing up the server's bandwidth, or something to that effect.  Each time you view or refresh a page, you're using some of the file transfer capacity of the site up.  Multiply that by 688, factor in Tare, and you have a big number.[/quote]
What??? My computer's no better/faster than yours so it probably won't matter who the user is, that's why I put this at j instead of f.

Remember file transfer capacity has little to do with the available memory in the server, and correct me if I'm wrong.

Also since this problem also happens when there's only a couple of people here, I doubt that's the problem.

Thanks for your input Mr. Rusczyk.

But that's the thing.  I don't think server memory is necessarily a problem; rather, it's with many computers trying to access the information in the server at the same time.

And I meant "factor in Tare" because you spend a lot of time on AoPS, therefore you view a lot of pages.  Therefore your computer requests more pages from the server than most people's computers do.  The number of users only affects the number that are likely to be requesting pages from the server.

Mmm, although I spend a lot of time here not all the time I'm logged on I'm "active"...I wonder if you just being there helps to clog up the connection...I mean if you are in artofproblemsolving.com but are doing absolutely nothing I doubt that takes up any memory or bandwidth.

You do have a point there, hmm...

[quote="Tare"]Mmm, although I spend a lot of time here not all the time I'm logged on I'm "active"...I wonder if you just being there helps to clog up the connection...I mean if you are in artofproblemsolving.com but are doing absolutely nothing I doubt that takes up any memory or bandwidth.[/quote]

I don't think it does.  Your computer only transmits/recieves signals to/from AoPS when you're submitting something, trying to view a page, or in a virtual classroom.  If the page sits idle, nothing.


vector
analytic geometry
calculus
absolute value

I was trying to get what vectors are in the Encarta Encyclopedia but I can't. I don't think they're AP. The Encarta has a strange way to explain them(to me.) Does anyone have a simpler way to explain? Or do they have the Encarta and think that is the simplest way to explain?

Funcia :o

Certain quantities in physics, such as the velocity of an object, or the acceleration of an object, or the force that is being exerted on an object, have not only a magnitude but also a direction.  Such quantities are called vectors.  Intuitively, a vector is a "quantity that has both a magnitude and a direction."

Can you visualize a vector?  Well... A vector $\vec{v}$ can be "represented" by a "directed line segment" (in other words, an arrow) whose length is the magnitude of $\vec{v}$ and whose direction is the direction of $\vec{v}$.  However, there are many (infinitely many) such directed line segments, and any one of them "represents" the same vector $\vec{v}$.  When I think of a vector, I visualize one of these directed line segments... But I remind myself that there are other directed line segments (parallel to the one I'm visualizing, and having the same length) which also represent the same vector.

Vectors can be added.  For example, suppose that a cruise ship is traveling 3 meters/sec to the North.  And suppose that on board that cruise ship, a boy is riding a skateboard 4 meters/sec to the West.  (That's his velocity with respect to the cruise ship.)  Can you find his velocity relative to the ocean surface?  That's a little puzzle.  Turns out the boy's velocity relative to the ocean surface is 5 meters/sec Northwest.  Therefore, the sum of the vectors "3 m/s to the North" and "4 m/s to the West" is the vector "5 m/s Northwest".  You could draw a little picture, with arrows to represent each of the vectors, to illustrate this fact.

Vectors can also be multiplied by real numbers.  For example, if your velocity is 5 m/s Northwest, and my velocity is five times your velocity, then what's my velocity?  25 m/s Northwest.

If you introduce a coordinate system, then you can associate with each vector $\vec{v}$ an ordered triple of real numbers $(x,y,z)$.  To do that, visualize the directed line segment that represents $\vec{v}$ and which is based at the origin.  Look at the point at the tip of that directed line segment.  The coordinates of that point are $x$, $y$, and $z$.  The numbers $x$, $y$ and $z$ are called the "components" of $\vec{v}$ with respect to the coordinate system you are using.  There's a certain simple formula for adding vectors in terms of the components of those vectors.  There's a similar formula for multiplying a vector by a real number.  And there's a formula for the magnitude of a vector in terms of that vector's components.

Since every vector can be associated in this way with an ordered triple of real numbers $(x,y,z)$ (assuming you have introduced a coordinate system), an ordered triple of real numbers may itself also be called a "vector".  But I believe it's important to remember that there's actually a difference between the geometric/physical vector $\vec{v}$ and the ordered triple $(x,y,z)$ that tells you the components of $\vec{v}$, just as there is a difference between a geometric point in space and the ordered triple that tells you the coordinates of that point.

That's a very short introduction to vectors... I'm sure there are plenty of much more thorough explanations of vectors out there on the internet.  Also, freshman level college physics books, or even high school physics books, tend to have decent explanations of vectors.  Lots of math books do too; for example, a Pre-Calculus book, or perhaps an Algebra II book, might have a chapter or a section on vectors.  You could also try Calculus books or Vector Calculus books.

Wow! Thanks! I guess I'll go research... :coolspeak:

The person above me provided a good mathematical explanation of what vectors are, but i will explain it in layman's terms just so you can have several ways to think about it. Like in the post above, a vector is a quantity with both magnitude and direction. Now, some people might be lost already at this point. What does it mean for a quantity to have "direction"? Well, it is quite simple actually. Here is an example to show this concept of "direction":

First of all, note that distance is a scalar quantity (no direction is involved), but it's vectoral counterpart, displacement, does have direction. Now, to the example.

Say if you are standing at a given point. You walk 2 meters to the right, then 2 meters down, then 2 meters back, then 2 meters up (it might help to draw this situation, since i can't type out a diagram). Anyways, the person just ends up back where they started. Now, the person traveled a total DISTANCE of 8 meters (scalar addtion = 2 + 2 + 2 + 2 = 8, since which way they traveled is negligible). HOWEVER, the displacement of this person is ZERO, since they ended up where they started. To understand why this is in a more mathematical way, let us go back an analyze each step.

First of all, lets call traveling "right" (or forward) and "back" (or left) the X direction and call traveling up and down the Y direction. Also let's call going forward/right the positive X direction and going up the positive Y direction. The opposites of these are the negative direction of each respective axis or "direction".

The person traveled 2 meters to the right (+2x), then 2 meters down (-2y), then 2 meters to the left (-2x), and then up 2 meters (+2y). Notice that we have employed positive and negative signs to change direction. Doing vector addition yields:

(+2x) + (-2y) + (-2x) + (+2y)

now i am just going to rearrange the terms to match up the x's and y's to make it easier to see that it equals zero.

(+2x) + (-2x) + (2y) + (-2y) = 0x + 0y = 0

Thus, the displacement is zero.

However, notice that if you do vector addition but take the absolute values of the quantities, you still end up with 8 meters. This is how scalars and vectors are mathematically related. A scalar is the absolute value of it's respective vector. This is because, by taking the absolute value, you make all the quantities "positive", thus it doesn't matter what their initial sign or "direction" was. This mathematically shows that sign or "direction" is negligible for scalar quantities.

OK, so there was a basic idea for you to help you differentiate between, yet relate both scalars and vectors. I hope this helps  :)


geometry
Putnam
complex numbers
real analysis
real analysis solved

hi all,

 here is a nice pb:
Let z_n be a sequence of non nul complex numbers so that
|z_i - z_j|>=1 for every i and j. Let a>=3 a real.
Prove (or disprove) that the serie
  Sn=sum( (1/z_i)^a, i=1..n) converges.

Could you improve the bound p>=a ??

Such a nice problem! The series is absolutely convergent for any $ a>2 $. To see this, think about the points having affixes $ z_k$, let them $ P_k$. Then the sum of the terms $\frac{1}{(OP_i)^a} $ where $ n\leq OP_i<n+1$ is at most $\frac{N}{n^a} $, where $N$ is the number of indexes $j$ for which $ n\leq OP_j<n+1$. Now, if we take a disk a radius $\frac{1}{2} $ around each point, then those disks are disjoint and a simple comparison of areas shows that $N\leq 4[(n+\frac{3}{2})^2-(n-\frac{1}{2})^2]<AN $ where A is an absolute constant. Thus, the series having general term $\frac{1}{(OP-i)^a}$ is smaller than the series having general term $\frac{An}{n^a} $ and the latter is convergent for $a>2$.

Compare to problem A3 on the 1949 Putnam:
[url]http://www.kalva.demon.co.uk/putnam/putn49.html[/url]


Just wondering if someone could help me out with this question. It has me completely stumped. the question is

Prove that the equation x^3 + 2(y^3) = 4(z^3) has no solution in positive integers x, y and z.

Or you can find the proper question at:

http://members.optusnet.com.au/gujju_style/maths2.JPG

Thanks.

If we look at the equation, we note $2|x^{3}\Rightarrow 2|x \Rightarrow x=2x_{1}$ for some positive integer $x_{1}.$

So, we have $4x_{1}^{3}+y^{3}= 2z^{3}$. Again, we note $2|y^{3}\Rightarrow 2|y \Rightarrow y = 2y_{1}$ for some positive integer $y_{1}$.

So, we have $2x_{1}^{3}+4y_{1}^{3}= z^{3}.$ Note again $2|z^{3}\Rightarrow 2|z \Rightarrow z=2z_{1}$ for some positive integer $z_{1}.$

And, now we get $x_{1}^{3}+2y_{1}^{3}= 4z_{1}^{3}.$ This obviously has the same form as our orginal equation. And, now we can repeat our argument again, leading to an infinite descent. Thus, there are no solutions in positive integers $x,y$ and $z.$

I don't know if this is a very good argument, but... :maybe:
[hide="Here it is"]Take mod two of the entire equation:
$x^{3}\equiv0\pmod2$
$x\equiv0\pmod2$
Let $x=2p$
$8p^{3}+2y^{2}=4z^{2}$
$4p^{3}+y^{3}=2z^{3}$
Similarly, $y=2q$:
$4p^{3}+8q^{3}=2z^{3}$
$2p^{3}+4q^{3}=z^{3}$
Similarly, $z=2r$:
$2p^{3}+4q^{3}=8r^{3}$
$p^{3}+2q^{3}=4r^{3}$
Similarly, $p=2{p}_{1}$

We can continue like this forever. $x$, $y$, and $z$ will have to have an infinite factors of $2$. This means that everything will be infinity. So there are no solutions. [/hide]

Did I oversimplify the problem?

Thanks for the reply guys, very much appreciated. I'm just a little confused as to which one to look at?
Are both solutions correct?

Look at either one. Each of them has basically the same type of argument.  :P

[quote="Kurt Gödel"][quote="FieryHydra"]And, now we can repeat our argument again, leading to an infinite descent.[/quote]
Or just note that if $\gcd(x,y,z)>1$ we can divide the given equation by $\gcd(x,y,z)$, so we can assume that $\gcd(x,y,z)=1$, which will lead to a contradiction, as we proved that $x\equiv y\equiv z\mod{2}$.[/quote]
In the last sentence, $x\equiv y \pmod 2$ is not the same as $x\equiv 0 \pmod 2.$

[quote="Ikarus"][quote="FieryHydra"]And, now we can repeat our argument again, leading to an [b]infinite descent[/b]. Thus, there are no solutions in positive integers $ x,y$ and $ z.$[/quote]
Can someone explain this? How can I really show by infinite descent that there are no solutions?

So how can I prove this implication ("And, now we can repeat our argument again, leading to an infinite descent") $ \rightarrow$ ("Thus, there are no solutions in positive integers $ x,y$ and $ z.$") mathematically?[/quote]

The most rigorous way to do it is using the Well Ordering Principle.  You agree that the set of solutions is a subset of the set of ordered triplets of positive integers, and so by the Well Ordering Principle, there exists a solution which has a smallest value of x (WOP states that any subset of the naturals has a smallest element).  However, we can use any solution to produce a smaller solution, so a smallest solution cannot exist, and we have a contradiction.


hi,
prove the identity
\[ 1^{2}\binom{n}{1} \plus{} 3^{2}\binom{n}{3} \plus{} ... \equal{} n(n\plus{}1)2^{n\minus{}3}.\]

The LHS counts the number of ways to form a committee consisting of an odd number of people out of $ n$ people along with a president and vice president of that committee (who are allowed to be the same).  Pick the president and vice president first:  if they're the same person (which occurs in $ n$ cases) then select an even subset of the other $ n \minus{} 1$ people, which can be done in $ 2^{n \minus{} 2}$ ways.  Otherwise (which occurs in $ n(n \minus{} 1)$ cases) select an odd subset of the other $ n \minus{} 2$ people, which can be done in $ 2^{n \minus{} 3}$ ways.  This gives $ (2n \plus{} n(n \minus{} 1)) 2^{n \minus{} 3}$ as desired.


Find $ f$ where $ f(x)\cdot f(y) \equal{} f(x\plus{}y) \plus{}xy$ and $ x, y \in \mathbb{R}$

[quote="AndrewTom"]Find $ f$ where $ f(x)\cdot f(y) \equal{} f(x \plus{} y) \plus{} xy$ and $ x, y \in \mathbb{R}$[/quote]
$ f(0)\cdot f(y) \equal{} f(y)$ so $ f(0) \equal{} 1$

$ f(1)\cdot f(\minus{}1) \equal{} f(0) \minus{} 1 \equal{} 0$ so $ f(1) \equal{} 0$ or $ f(\minus{}1) \equal{} 0$

If $ f(1) \equal{} 0$ then $ f(x\minus{}1)\cdot f(1) \equal{} f(x) \plus{} x \minus{} 1$ so $ f(x) \equal{} 1 \minus{} x$

If $ f(\minus{}1) \equal{} 0$ then $ f(x\plus{}1)\cdot f(\minus{}1) \equal{} f(x) \minus{} x \minus{} 1$ so $ f(x) \equal{} 1 \plus{} x$

[quote="AndrewTom"]Find $ f$ where $ f(x)\cdot f(y) \equal{} f(x \plus{} y) \plus{} xy$ and $ x, y \in \mathbb{R}$[/quote]

Let $ P(x,y)$ the equation $ f(x)\cdot f(y) \equal{} f(x \plus{} y) \plus{} xy$ with $ x, y \in \mathbb{R}$

$ P(0,0)$  $ \implies f^2(0) \equal{} f(0)$ and so either $ f(0)\equal{}0$ or $ f(0)\equal{}1.$

$ f(0)\equal{}0$ $ \implies f(x)\equal{}0$ $ \forall x \in  \mathbb{R}$ and it's easy to check that this is not a solution.

So $ f(0)\equal{}1.$

$ P(x,\minus{}x)$ $ \implies f(x)f(\minus{}x)\equal{}1\minus{}x^2\equal{}(1\minus{}x)(1\plus{}x)$.

So the first solution is $ f_1(x)\equal{}(1\minus{}x)$  and $ f_1(\minus{}x)\equal{}(1\plus{}x)$ $ \forall x \in  \mathbb{R}.$

The second solution is $ f_2(x)\equal{}(1\plus{}x)$  and $ f_2(\minus{}x)\equal{}(1\minus{}x)$ $ \forall x \in  \mathbb{R}.$

It is simple to check that both functions are solution of the original equation.

Hence the solution
$ \boxed{
\begin{array}{ll}
f(x)\equal{}1\minus{}x & \forall x \in  \mathbb{R} \\
f(x)\equal{}1\plus{}x & \forall x \in  \mathbb{R} \end{array}}$


=================================
Prove that the set $ \{1, 2, \ldots , 1989\}$ can be expressed as the disjoint union of subsets
$ A_1, A_2, \ldots , A_{117}$ in such a way that each $ A_i$ contains 17 elements and the sum of
the elements in each $ A_i$ is the same.
=================================

This was the first problem posed at IMO'1989, and it has been proposed by the Philippines.

Does anybody in this forum know the name of the composer of this problem?
It should be added to
http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions

I believe it's Jose Marasigan.

I'll confirm with him next time I see him.


geometry
trigonometry
angle bisector
geometry proposed

Given triangle $ ABC$ of area 1. Let $ BM$ be the perpendicular from $ B$ to the bisector of angle $ C$. Determine the area of triangle $ AMC$.

$ S_{ABC} \equal{} \frac {a \cdot b \sin C} {2} \equal{} 1$ ( 1 )
$ CM \equal{} a \cdot \cos \frac {C} {2}$ ( 2 )
$ S_{ACM} \equal{} \frac {AC \cdot CM \cdot \sin \frac {C} {2}} {2}$ ( 3 )
After substituting (2) in (3), taking into account (1), we get $ S_{ACM} \equal{} \frac {1} {2}$.

(Here a and b means the lengths of BC and AC respectively), while $ S_{XYZ}$ means the area of $ \Delta XYZ$).

Best regards,
sunken rock

Extend $ BM$ to meet $ CA$ at $ D$. Then as $ CM$ is both an angle bisector and an altitude, $ DM \equal{} MB$. So the perpendicular from $ M$ to $ DC$ will be half the perpendicular from $ B$ to $ DC$, so $ [AMC] \equal{} \frac {1}{2}$.


MOP=MOSP=MOsP

And uncyclopedia is a joke...

[quote="AIME15USAMO"][quote="abacadaea"]this site also has a lot of good information
http://uncyclopedia.wikia.com/wiki/MOP[/quote]

Wow, really?[/quote]

uncyclopedia has given me the clearest definition of MOP I have ever read/heard. Do not criticize it. :rotfl:

[quote="topofmath"][quote="PowerOfPi"]Can someone give me more information on MOP, like who qualifies and how, what happens there, where is it, and anything else I should know.

This is all I know about MOP:

- Also called MOSP
- Stands for Mathematical Olympiad (Summer) Program
- There are three levels, red, blue and black (I don't know the difference, though)

Thanks![/quote]

Black MOP is top 12 USAMOers.
Blue MOP is top 12 that are not in Black MOP, are not seniors
Red MOP is top 12 freshman not in Black or Blue MOP[/quote]

Actually, I believe there is a qualifying score range for each of the respectful "MOPs".

Yes, and this qualifying score range is chosen such that there are a certain number of MOPpers in each group.

Everything in topofmath's post is true except that the number of red MOPpers is much greater than 12 - usually around 25, and that the numbers are approximate - there were more than 12 people in Blue in 2009, for example.

[quote="not_trig"][quote="Brut3Forc3"]Blue is supposed to be as close to 18 without exceeding it, I believe.[/quote]

Something like this.

Black: Top 12 overall.
Blue: Next ~12-18 non-seniors.
Red: Top 20-30 freshmen. Note that USAJMO is now an option for qualifying for this...

In recent years, potential qualifiers for the China Girls' Math Olympiad (CGMO) have also trained at MOP (though not necessarily every year). The top-scoring ~8-10 girls on the USAMO [USAJMO may allow one to qualify, but not sure??] qualify.

It's 3 weeks of math and getting pwned by Zuming. Pretty awesome.[/quote]

What's usually the average scores for red Moppers?

On the name, I believe it was once officially MOP- but if so, that was before my time. The silent S has lasted over ten years, pretty impressive when you consider how much turnover there is in everybody involved.

[quote="PowerOfPi"]In the AoPS Wiki page it said that exceptional seventh and eighth graders have sometimes been accepted.[/quote] In Blue (and now Black MOP.) Red MOP is for 9th graders only
@marc I think it's somewhere around 10.

[quote="PowerOfPi"]In the AoPS Wiki page it said that exceptional seventh and eighth graders have sometimes been accepted.[/quote]

I think thats for blue mop, not red mop. As far as I know, nobody below 9th grade has ever been accepted for red mop.

Wait, a score of 10 on what? USAMO I guess?

[quote="PowerOfPi"]Are you being sarcastic? So, if I get a 150 on the AMC10, 8 on the AIME, qualify for USAMO because I am first in the state, and get a 12 (2 points per problem), I will make Red MOP?[/quote]

...And getting 2 points per problem is obviously very easy.

BTW, that was sarcastic.

Getting 2 points per problem is impossible almost. Its very, very hard to do.

Look at some RECENT usamo problems to see why they are considered hard and why getting 2 points on each problem might be... not easy.

Mode and Median score in 2009: 4 on usamo.

[quote="PowerOfPi"]Are you being sarcastic? So, if I get a 150 on the AMC10, 8 on the AIME, qualify for USAMO because I am first in the state, and get a 12 (2 points per problem), I will make Red MOP?[/quote]

What pythag said. I think that getting a 2 on each problem is almost as hard as getting a 20+ (3 complete/nearly complete solutions + maybe a bit inroad on others) The most common way of making Red MOP is either two complete solutions or 1 with a few more semi-completed.

[quote="dysfunctionalequations"][quote="pythag011"]Mode and Median score in 2009: 4 on usamo.[/quote]

If I'm not mistaken, the mode score was actually a $ 2$.[/quote]
It was. But you could get 2 points for giving the (very obvious) answer to number 2 without proof, or so I have heard.


The triangle $\triangle ABC.$ The bisectrix $AA ', BB', CC '.$ Circles $(I), (O), (E)$ to turn be inscribed circle, circumsribed circle of the triangle $\triangle ABC$ and escribed circle to the angle $\angle A.$ Since $I,$ draw $IH$ perpendicular $B'C '.$ Prove that:
$R \cdot r = OE \cdot IH$

Let $ (I)$ touch $ BC, CA, AB$ at $ U, V, W.$ $ V$ is pole of $ CA$ and direction $ \perp BIB'$ is pole of $ BIB'$ WRT $ (I)$ $ \Longrightarrow$ $ b' \perp BIB'$ through $ V$ is polar of $ B'$ and likewise, $ c' \perp CIC'$ through $ W$ is polar of $ C'$ WRT $ (I)$ $ \Longrightarrow$ $ U' \equiv b' \cap c'$ is pole of $ B'C'$ WRT $ (I).$ Therefore $ \overline{IH} \cdot \overline {IU'} \equal{} r^2$ (1). Let $ V', W'$ be poles of $ C'A', A'B'$ WRT $ (I),$ obtained in a similar way. $ \triangle U'V'W'$ is anticomplementary to $ \triangle UVW.$ Let $ E, F, G$ be excenters of $ \triangle ABC$ against $ A, B, C,$ respectively. $ \triangle EFG \sim \triangle U'V'W'$ are similar, having parallel sides, $ \perp AIA', BIB', CIC'.$ $ (O), (I)$ are their respective 9-point circles $ \Longrightarrow$ $ \frac {\overline{OE}}{\overline{IU'}} \equal{} \minus{} \frac {R}{r}$ (2). Combining (1) & (2), $ \overline{OE} \cdot \overline{IH} \equal{} \minus{}R \cdot r.$


trigonometry
trig identities
Law of Cosines

Equilateral triangles $ ABF$ and $ CAG$ are erected externally on the side $ CA$ and hypotenuse $ AB$ of right triangle $ ABC$. $ M$ is the midpoint of $ [BC]$. $ |MF|\equal{}11$ $ |MG|\equal{}7$ then find $ |BC|$

Does anybody know the solution ?, please share it   :(

By law of cosines on CGM and AC=GC,
$ AC^2 \plus{} MC^2 \plus{} \sqrt {3}AC \cdot CM \equal{} 49$
By law of cosines on MBF and MB=MC,
$ CM^2 \plus{} AB^2 \minus{} 2 \cdot CM \cdot AB \cdot cos (60 \plus{} ABC) \equal{} 121$
$ CM^2 \plus{} AB^2 \minus{} 2 \cdot CM \cdot AB(\frac {1}{2} \cdot \frac {2CM}{AB} \minus{} \frac {\sqrt {3}}{2} \cdot \frac {AC}{AB})$
$ AB^2 \minus{} CM^2 \plus{} \sqrt {3}AC\cdot CM \equal{} 121$
$ AC^2 \plus{} 3MC^2 \plus{} \sqrt {3}AC \cdot CM \equal{} 121$
Subtracting from the first
$ 2MC^2 \equal{} 72$
$ MC \equal{} 6$
$ BC \equal{} 12$


Is it true that if there is a $ c \in \mathbb{C}$ such that $ D_wf(z)\equal{}cw$ for each $ w$ such that $ |w| \equal{} 1$ then $ f$ is $ \mathbb{C}$-differentiable at $ z$?
The above $ D$ means the directional derivative in the $ w$ direction.

Yes, because $ f$ is $ \mathbb C$-differentiable at $ z_0$ iff the total differential of $ f$ at $ z_0$ (in the sense of [b]real[/b] calculus) is $ \mathbb C$-linear.


I posted this at "Unsolved Problems" but got no reply. :(

 If $ a_1,a_2,...,a_n$ are positive real numbers then prove

$ \frac {a_1}{a_2 \plus{} a_3} \plus{} \frac {a_2}{a_3 \plus{} a_4} \plus{} ... \plus{} \frac {a_{n \minus{} 2}}{a_{n \minus{} 1} \plus{} a_n } \plus{} \frac {a_{n \minus{} 1}}{a_n \plus{} a_1} \plus{} \frac {a_n}{a_1 \plus{} a_2} \geq \frac {n}{4}$

Let $ b_1\equal{}max  a_i$; $ b_2$ is largest of two numbers in denominator of fraction, where $ b_1$ is numerator; ... ; $ b_k$ is largest of two numbers in denominator of fraction, where $ b_{k\minus{}1}$ is numerator; $ b_1$ is largest of two numbers in denominator of fraction, where $ b_k$ is numerator. obviously. $ k \geq \frac{n}{2}$


[color=olive]Show that if a and b are integers with (a,b)=1, then (a+b,a-b)=1 or 2 [/color]

Should I use the fact that (a+cb,b)=(a,b)? I've also tried to start with (a,b)=1 => 1=xa+yb, but then I was stuck, don't know how to begin. Thanks for helping me out.

http://www.mathlinks.ro/Forum/viewtopic.php?t=111291

It wasn't necessary to repost in the correct forum: the original thread was moved here.


Prime number is known by me to be defined as a number whose factors are only itself and one. The lowest prime number i previously have been tought was 2. But with a spark of random thinking on my part, i thought that [b]-1[/b]'s factors are only -1 and one. By the definition of prime numbers above, all negative numbers are composite. Example: -7's factors are 1,7,-7,-1. But negative one's factors are 1,-1. No others. Please explain.

A prime number is defined to be natural. This stuff doesn't work well with negatives.

So a negative number can't be prime? The onlynegative number that might possibly be prime is -1.

It really depends on your definition of prime, since primes are not well defined in the negatives.

hmm. We need to ask a mod. They would know

Look [url=http://en.wikipedia.org/wiki/Prime_number]here[/url].

A factor is defined to be positive, and prime numbers have to be greater than $ 1$.


geometry
3D geometry
calculus
MATHCOUNTS

I found a piece of baumkuchen, about 1/8 of the original. The outer part of the piece was about 5 and a half cm. long and the inner part was about 1 and a half cm. long. It's thickness (height) was 5 cm. If my sister ate the rest of the baumkuchen, how much did she eat? (just round off the radius)

(Note: baumkuchen is a German dessert/pastry. It looks like a donut except [b]all sides are flat[/b] Reference:
[img]http://www.ja-inaba.or.jp/tokusan/img/t_27.jpg[/img]
[img]http://www.yosanet.com/gensen/shop/003/img/item003-l.jpg[/img][img]http://www.okashi.co.jp/suzuya/suzuya08.gif[/img]
[img]http://www.shunkado.co.jp/images/bamukuheni.jpg[/img]

To all the German people: Yes, I know that traditional Baumkuchen differ from the Baumkuchen I'm talking about (My version of baumkuchen is one of the Japanese versions of baumkuchen, not the traditional which can have a round outside like a donut, shaped slightly like a cone instead of a perfect circle, has creme in the middle, etc.), but I don't think calculus is in MATHCOUNTS :)

My answer might be diffrent from yours, since I rounded diffrently. The volume for a roll of toilet paper, or Baumkuchen, is V= :pi: R^2H- :pi: r^2H. R=radius of outer circle, r=radius of inner circle, H=height. Since the problem gives 1/8 of the circumferences,
1/8C=5.5
C=44
44=2 :pi: R
22= :pi: R
7=R
This also gives 2 for r. So
V=7^2 :pi: 5-2^2 :pi: 5
V=245 :pi: -20 :pi:
V=225 :pi:
V=707
So the 7/8 the sister ate's volume would be 619 cm^3.

Yup yup...er, toilet paper? :lol:

 :cry: So happy someone solved this one, all the picture-finding didn't go to waste after all...it would really be great if you could solve all (50 :lol: ) unsolved problems I made...

There were 50 unsolvable problems?  Where?

Oh, wait.  Unsolved.  Slightly different.  Oh, well.  Let the fanatic at them!

[quote="topper"]My geometry teacher called this kind of problem a "toilet paper problem".[/quote]
 :lol:, I understand where he/she's getting it from. They kind of look the same too except for the shape and color (they both have layers)

I doubt I can think up of a problem where it'll be unsolvable to you, mathfanatic :)

That's not true.. Just ask whether there are infinitely many primes of a certain form. Chances are it will be unsolvable to everyone in the world.

Yeah, I guess so, but then again I would've never thought of that :)


I saw the phrase "mass points" in relation to triangles and looked it up on the Wiki, but there was nothing other than a link to a nonexistent article.  What is it?

Mass points are also known as [url=http://www.cut-the-knot.org/triangle/barycenter.shtml]barycentric coordinates[/url].  If you learn to use them properly, they can be very useful.

a semi-old topic but still...
Thanks for your reply, but I don't know how to find the barycenter of two points, which is involved in the link description of how to find the barycenter of three points. Hopefully that info. will help in understanding the rest of the article, because I'm pretty confused about the rest of it, too.  :o

The barycenter is also known as the [url=http://en.wikipedia.org/wiki/Center_of_mass]center of mass[/url].


AMC
USA(J)MO
USAMO
modular arithmetic
geometry
algorithm
induction

This is where you may talk about the problems and solutions and where the results will be posted in 4 days time.

2 graders gave a score to each solution, without knowing who the solution was from, and then the two scores were averaged. That is why half-points are given to some solutions.

Best solutions in the near future.

Overall Winners:
[hide]31 - rem
39 - JSteinhart
18 - SplashD
19 - pkerichang
25 - 101101101[/hide]

Day 1 Winners:
[hide]18-SplashD
19-pkerichang
31 - rem
37 - Xantos C. Guin
39 - JSteinhart
[/hide]
Most Perfects: (tie)SplashD and Xantos C. Guin

Scores:
[hide]
The average is the average score of the solutions, not the average score of the participants. \begin{tabular}{c|l|l|l|r\parallel l|l|l|r}ID&P1& P2&P3&Day 1&P4&P5&P6&Total\\ \hline Average&5.08&5.85&3.69&12.25&2.88&1.67&3.5&14.43\\ 1&3.5&5.5&0&9&&&1&10\\ 8&3&5.5&4&12.5&&&&12.5\\ 14&&6.5&&6.5&.5&&&7\\ 15&6&5&0&11&&&&11\\ 18&7&6&7&20&&&&20\\ 19&6&7&6.5&19.5&&&&19.5\\ 22&5&&&5&&&&5\\ 25&4.5&5&1.5&11&3&.5&4.5&19\\ 28&5.5&6&2&13.5&&&&13.5\\ 31&7&6&4.5&17.5&3&3&4&27.5\\ 33&3.5&6.5&&10&&&&10\\ 37&7&7&&14&1.5&&&15.5\\ 39&4&5.5&4&13.5&3.5&1.5&4.5&23\\ junggi&4&4.5&&9.5&&&&9.5\\ \end{tabular} [/hide]

I managed to solve everything but maybe made some stupid mistakes, but I don't remember my answers...
I really liked the problems :)

[quote="lotrgreengrapes7926"][hide="2"] For 2, I got 279.  :maybe: Is that right? [/hide][/quote]
I think so.

OMG I made an arithmetic error on this quetsion. I am so dumb!  :wallbash_red: I solved the question but messed up in my calculations. hopefully the graders will take only a few marks off...

[quote="chess64"]#1 was the only one I had time to attempt, and it was just Euclidean algorithm. The answer was all $x$ such that $x\equiv 3\pmod{5}$, $x\equiv 4\pmod{17}$, or $x\equiv 38\pmod{85}$.[/quote]

$x\equiv 38\pmod{85}$ is already covered by both $x\equiv 3\pmod{5}$ and $x\equiv 4\pmod{17}$, so I don't think it needs to be listed again.

For number 2:
[hide]Since $5$ rocks in a box is equal to $7$ rocks in the previous box, the arrangement of rocks is the base $\frac{7}{5}$ representation of the number of rocks in base $10$. So the answer is the largest integer that has $10$ digits in base $\frac{7}{5}$ which after using a formula is $279_{10}=5362324266_{\frac{7}{5}}$.
[/hide]
This is why it is solafidefarms' favorite problem.

[quote="Xantos C. Guin"]For number 2:
[hide]Since $5$ rocks in a box is equal to $7$ rocks in the previous box, the arrangement of rocks is the base $\frac{7}{5}$ representation of the number of rocks in base $10$. So the answer is the largest integer that has $10$ digits in base $\frac{7}{5}$ which after using a formula is $279_{10}=5362324266_{\frac{7}{5}}$.
[/hide]
This is why it is solafidefarms' favorite problem.[/quote]

Wow, and me with my brute force...

[quote="Altheman"]I spent a while grading and I have a few comments:

Of the 12 sets of 3 problems that I graded, only one set had adequate spacing. All you have to do is press enter a few times between each line.

For the third problem...if you say that the orthric triangles are homothetic wrt the orthocenter of ABC, that does not mean that the larger triangles are as well. You have to justify that. There was only one person who gave a satisfactory argument to follow. I am also suprised that only one other person recognized the 9 point circle, which was my intent of the problem.[/quote]

I don't remember personally saying anything about orthic triangles, but it seems to me that showing that the altitudes of the two triangles are homothetic is sufficient, as a triangle is uniquely determined by its altitudes. (In fact, it is also uniquely determined by its orthic triangle, as the excenters of the orthic triangle are the vertices of the original triangle.) It therefore follows that if the orthic triangles are homothetic, so are the original triangles.

I also seem to remember hearing at one point that showing that any three corresponding points on a triangle are homothetic about a point is sufficient...can anyone confirm this?

This is not an attempt to get the graders to change anything, in fact I hope they won't, as otherwise the integrity of the contest would be ruined, I am just seeking to make sure that what I think is true is actually true, as I am personally not so good at geometry.

Btw, did you guys get $\binom{n}{2}(n-1)^{4}$ for problem 4?

[quote="Xantos C. Guin"][quote="JSteinhardt"]Btw, did you guys get $\binom{n}{2}(n-1)^{4}$ for problem 4?[/quote]

That's probably right. I messed that one up. My answer was an integer $\text{ iff }n\not\equiv 3 \bmod{4}$[/quote]

I used lots of casework :roll: and it was an integer $\text{ iff }n\equiv0\mod2$.  :blush:

[quote="lotrgreengrapes7926"]What's the criteria for getting a 7? Does it have to be brilliant and ingenious  :idea: ? Or just correct and well-explained?[/quote]

Yes. Most people got the right answer for 1 and 2, but the majority of people got less than a 5 on the first and no more than a 5.5 on the second. 7 was for well-explained pretty solution.

Btw Day 1 seemed significantly easier than a normal USAMO (I would be ecstatic if a USAMO number 3 were that easy, on a geometry problem no less), whereas Day 2 seemed about on par with normal difficulty, (question 4 was a bit easier than most question 4's, question 5 was a bit harder than most question 5's, question 6 was about as hard as most question 6's.

[quote="chess64"]#1 was the only one I had time to attempt, and it was just Euclidean algorithm. The answer was all $x$ such that $x\equiv 3\pmod{5}$, $x\equiv 4\pmod{17}$, or $x\equiv 38\pmod{85}$.[/quote]

If only I had waited a bit to take the first part of the test....
I learned Euclidean algorithm in interm. num theory class the day after  :lol:

Part of it was you were late :P. Secondly I thought it wasn't as pretty as I liked. Thirdly, I was the only grader for your solution, because the other #1s had already been graded. So maybe it should have gotten a 6.5.

\gcd(x^{2}+1,2x+9) by the Euclidean Algorithm. Applying the algorithm again, we can say that

2(x^{2}+1)-x(2x+9)=-9x+2 \implies \gcd(f(x),g(x))=\gcd(2x+9,-9x+2).

It seems like you claim gcd(a, b) = gcd(b, 2a - kb). Try this with b = 4, a = 2, k = 1. We get 2 = gcd(2, 4) = gcd(4, 4 - 4) = 4 OH GOD HAVE WE DISPROVEN MATH

[quote="JSteinhardt"][quote="solafidefarms"][b]Secondly I thought it wasn't as pretty as I liked.[/b][/quote]

I believe this has been said three times in this thread, but Alex and I are being ignored.

[b]Style does not affect your score.[/b][/quote]

Even if you overly repeat and write tons of pages when it can be done with only one?

Yes, even in this case style does not affect your score. If someone writes 25 pages, but they are 25 pages of rigorous statements that eventually prove the desired result, they should get 7 points. This is how it works on the actual USAMO.

It should also be noted the chess64's solution was less than 1 page...

Edit: Just so things don't get confused, I'm not complaining or anything, I'm just attempting to make this point clear for the purpose of future Mock USAMO's.

[quote="chess64"][quote="MysticTerminator"][quote]$\gcd(x^{2}+1,2x+9)$ by the Euclidean Algorithm. Applying the algorithm again, we can say that \[2(x^{2}+1)-x(2x+9)=-9x+2 \implies \gcd(f(x),g(x))=\gcd(2x+9,-9x+2).\] [/quote]

It seems like you claim gcd(a, b) = gcd(b, 2a - kb). Try this with b = 4, a = 2, k = 1. We get 2 = gcd(2, 4) = gcd(4, 4 - 4) = 4 OH GOD HAVE WE DISPROVEN MATH[/quote]

Umm Arnav...

THEY'RE POLYNOMIALS[/quote]

...ok? so let's do this with polynomials

2x = gcd(2x, 4x) = gcd(4x, 4x - 4x) = gcd(4x, 0) = 4x

hmm try using the euclidean algorithm arnav

$\frac{x^{2}+1}{2x+9}=\frac{1}{2}x-\frac{9}{4}+\frac{85/4}{2x+9}$

$\iff 4(x^{2}+1)=(2x+9)(2x-9)+85$

$\therefore \gcd(x^{2}+1,2x+9)=\gcd(2x+9,85)$

which is actually faster than my first way

but whatever

arnav is still wrong :P

[quote="solafidefarms"][quote="JSteinhardt"][quote="solafidefarms"][b]Secondly I thought it wasn't as pretty as I liked.[/b][/quote]

I believe this has been said three times in this thread, but Alex and I are being ignored.

[b]Style does not affect your score.[/b][/quote]

Style did affect your score in this case, because I think it [i]should[/i] affect score. Maybe it doesn't matter on the real thing, but I'd rather consider my work nearly a Book proof or, if I had to bruteforce it, as elegant bruteforce as possible, then just to hack out a correct but not-wonderfully-readable solution. Also, in How to Write a Proof it puts a bit of emphasis on niceness. Thus it did matter, even if it doesn't usually matter. When you do a mock USAMO you don't have to grade by style, and, judging from the outcry, I may not grade on style next time either.[/quote]

eh, it's cool either way (as USAMO is indeed independent of style while MOP tests have style), just give a heads up beforehand so people know what to expect

in fact, I think the mathlinks contest on this site had a rubric that gave 6 for correct proof that was not very innovative, so it's up to the individual contest-maker's discretion

[quote="chess64"]hmm try using the euclidean algorithm arnav

$\frac{x^{2}+1}{2x+9}=\frac{1}{2}x-\frac{9}{4}+\frac{85/4}{2x+9}$

$\iff 4(x^{2}+1)=(2x+9)(2x-9)+85$

$\therefore \gcd(x^{2}+1,2x+9)=\gcd(2x+9,85)$

which is actually faster than my first way

but whatever

haha I'm an idiot :P[/quote]

erm this is definitely not what you did the first time. I'm not disputing that every step in your line of equations is correct. I'm just saying your reasoning is flawed. Try justifying each step.

Yeah I agree with Arnav, if you want to grade on style you can, but the title Mock USAMO indicates by default that it would be graded identically to a USAMO. So next time it would b a good idea to let people know beforehand. In fact, if you want to do it the way it works at MOP, you could assign a score solely based on Math, 1-7, and another based solely on Style, 0.0-1.0. At MOP, a well-written proof gets 0.8, sloppy proof or "dumbass/brute force" solution gets 0.7, and stuff like Lagrange multipliers or mindless coordinates gets 0.5-0.6; 0.9 is reserved for elegant solutions, and 1.0 for extremely elegant, etc. etc.; also proofs with non-trivial generalizations might get higher style. You don't have to do that, it's just an idea. I think that might be easier to grade that way because you could concentrate just on whether it's correct, then just on whether it's nice. (Btw, your "score" would then be style * math.) It would also prevent people from asking why they lost points, because they would know exactly why -> if they lose math points, it's because they did something that wasn't rigorous or just plain wrong. If they lose style points, it's because they had bad style.

I'm glad to see that the organizers are keeping an open mind, and I apologize if I was a bit harsh before, basically I wanted to get a point across for future reference and it seemed that everyone was ignoring it.

[quote="MysticTerminator"]

[quote="chess64"]hmm try using the euclidean algorithm arnav

$\frac{x^{2}+1}{2x+9}=\frac{1}{2}x-\frac{9}{4}+\frac{85/4}{2x+9}$

$\iff 4(x^{2}+1)=(2x+9)(2x-9)+85$

$\therefore \gcd(x^{2}+1,2x+9)=\gcd(2x+9,85)$

which is actually faster than my first way

but whatever

haha I'm an idiot :P[/quote]

erm this is definitely not what you did the first time. I'm not disputing that every step in your line of equations is correct. I'm just saying your reasoning is flawed. Try justifying each step.[/quote]

huh huh any response :P

Shouldn't multiplying by 2 be allowed here, since the other term (2x + 9) is never divisible by 2?