A certain number of 1-cm square tiles form a rectangular region with a perimeter of 30 cm. The same number of tiles can also be used to form a square region. What is the number of centimeters in the perimeter of the square region?

Dimensions of rectangle:

8 by 7: NO
9 by 6: NO
10 by 5: NO
11 by 4: NO
12 by 3: YES

Thus, the square is 6 by 6, so there is a $ 6 \times 4\equal{}\boxed{24}$ perimeter.


I am a freshman math major, and I though it is still a few years off, I was just wondering: what is the difference between graduating with a B.S. in math, getting your M.S. in math, and getting a PhD in math? Obviously, the PhD is a much more powerful degree than a bachelor's, but where does the master's fit into all this? Is it worth getting a master's and THEN going to get a PhD?

I am thinking about double major in comp sci or physics, but if the master's is worth it, CMU has an honors program to graduate with a master's in four years. To complete that I probably wouldn't be able to double major.

But is the master's worth it?

Most math professors these days follow through a graduate Ph.D. fellowship after receiving their undergraduate degrees. A M.S. is unnecessary for math unless you are seeking to teach math (masters of education) at high schools. I see more people pursuing engineering M.S with a math BS. In fact, there are lots of successful people with Computer Science and Electrical Engineering M.S, who are eligible Ph.D. candidates, but go into industry.

Most community colleges require their faculty members to have masters degrees. My own college has an M.S. program but not a Ph.D. program; many of our graduates do go on to teach in community colleges. A significant number of our Applied Math and Statistics M.S. students are already employed in industry and their companies are supporting their further education.


superior algebra
superior algebra unsolved

If $ A$, $ B$ and $ C$ are groups such that $ A \times B$ and $ A \times C$ are isomorphic, can we infer that $ B$ and $ C$ are isomorphic ?

No, take $ A$ an infinite product of nontrivial groups $ G$ and $ B\equal{}G, C\equal{} G^2$.

[quote="ZetaX"]No, take $ A$ an infinite product of nontrivial groups $ G$ and $ B \equal{} G, C \equal{} G^2$.[/quote]

Ok, thanks. What about the case where $ A$, $ B$ and $ C$ are supposed to be finite ?


linear algebra
matrix
induction
invariant
algebra
polynomial
modular arithmetic

Let $A$ be a $nxn$ complex matrix of finite order and whose order if a power of a prime $p$. Suppose that its trace is rational. Prove that its trace is one of the numbers $n,n-p,n-2p,...,n-[n/(p-1)]p$.

[quote="harazi"]Let $A$ be a $nxn$ complex matrix of finite order and whose order if a power of a prime $p$[/quote]

What do you mean by " ...matrice of finite order ..." ?

The multiplicative group generated by this matrix is finite.

We can prove it by induction on $n$.

It's easy to check for small $n$ (like $n=1,2$, for instance). Now suppose it's true for all numbers $<n$, let $p^t$ be the order of the matrix, and let $\zeta$ be a primitive $p^t$'th root of unity. Suppose some $\alpha=\zeta^{p^uk},\ u\in\overline{0,t-1},(k,p)=1$ is an eigenvalue of our matrix. Then, since the sum of all the eigenvalues (taking multiplicity into account) is a rational number (in fact integer, since it's rational and an algebraic integer at the same time), it's invariant to all elements of the Galois group of $\mathbb Q(\zeta)$ over $\mathbb Q$, meaning that all the conjugates of $\alpha$ are eigenvalues.

Since $\alpha$ is a root of the irreducible polynomial $\varphi_{p^{t-1-u}}$ (the $p^{t-1-u}$'s cyclotomic polynomial), it has $p^{t-1-u}(p-1)$ conjugates (including itself), and this number is $0\pmod p$ if $u<t-1$ and $-1\pmod p$ if $u=t-1$, and they sum up to $0$ if $u<t-1$ and $-1$ if $u=t-1$. In other words, the sum is congruent to the number of conjugates modulo $p$. If we now eliminate these conjugates from the multiset of eigenvalues, we're left with a collection of $n-p^{t-1-u}(p-1)\ p^t$'th roots of unity on which we can apply the induction hypothesis, after which we can add back the $p^{t-1-u}(p-1)$ roots which sum up to $0$ or $-1$ depending on whether $p^{t-1-u}(p-1)$ is $0$ or $-1\pmod p$ respectively. The result follows.


Prove that for $ a,b \in (0,\infty)$ we have $ a^3\plus{}b^3\geq 3ab$

$ a \equal{} b \equal{} 1$  :roll:  (AM-GM gives $ a^3 \plus{} b^3 \plus{} 1 \geq 3ab$, and of course AM-GM has equality cases.  But there are other good ways to see that this is ridiculous.  For example, the LHS goes to 0 much faster than the RHS as $ (a, b) \to (0^ \plus{} , 0^ \plus{} )$.)


quadratics
algebra
polynomial
logarithms
induction
modular arithmetic
number theory

Hi,
In the paper by Jeffrey Shallit and Troy Vasiga: "On the iteration of certain quadratic maps over GF(p)", ( preprint version is available at: http://www.cs.uwaterloo.ca/~tmjvasig/papers/newvasiga.ps ), the Theorem 5 and Corollary 3 provide the formula for computing the number of cycles of length L for Mersenne primes for a Digraph under x^2.
With PARI/gp, the formula can be computed only up to q=127 .

Experimenting with Shallit's formula, I've been able to find that the number of cycles of length L for Mersenne primes (2^q-1) under x^2 seems independent of q and that the Sequence of numbers seems already well-known and there exists a formula that can be computed for high values of q.

With Pari/gp, I've used Shallit's formula for computing the first values for: q=5, 7, 13, 17, 19, 31, 61, 89, 107, 127. (Higher values of q for Mq primes lead to Pari/gp to end with error or to never end).
I've seen that the number of cycles of length L seems to NOT depend on q and that lengths L are [b]all[/b] the divisors of q-1.
Then, the number of cycles of length L=1..12 is: n= 1, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335 .

As an example, the number n of cycles of length L for q=127 is:
##### x^2 127
L n
------
1 1
2 1
3 2
6 9
7 18
9 56
14 1161
18 14532
21 99858
42 104715342801
63 146402730743693304
126 675163426430433459179525995420973028

Then I've searched for this Sequence in the "On-Line Encyclopedia of Integer Sequences!" and I've found that this is Sequence A001037 (same as A059966):
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A001037

With the given program: (PARI) : a(l)=if(l<1,l==0,sumdiv(l,d,moebius(d)*2^(l/d))/l)
I've been able to check that all the values I have computed with Pari and Shallit's formula are equal to the values computed with a(l).

As a consequence, I think that the "explicit" formula of the "number N of cycles of length L in the digraph G(x->x^2) for p being a Mersenne prime" (Theorem 5) is: $NC(l) = \frac{1}{l} \sum_{ d \mid l } \mu(\frac{l}{d}) 2^d$,  for all l such that $l \ \mid q-1$ .

And thus there is a link between this digraph and:
- Number of degree-n irreducible polynomials over GF(2).
- Number of n-bead necklaces with beads of 2 colors when
turning over is not allowed and with primitive period n.
- Number of binary Lyndon words of length n.

[b]Can you prove that NC(L) does provide the "number N of cycles of length L in the digraph G(x->x^2) for p being a Mersenne prime" ?[/b]

Tony

By the existence of a primitive root $\mod 2^{q}-1$, we can see it also as the following problem:
Find the number of cycles of length $l$ under the action $x \to 2x$ seen $\mod 2^{q}-2$. (this is the point where the chain/beads come into it ;) )

Now canonically lets find the number of solutions of $2^n x \equiv x \mod 2^q-2 \iff (2^n-1)x \equiv 0 \mod 2^q-2$, since this is the number of $x$ that are part of a cycle having an order dividing $n$. Since $\gcd(2^{q}-2 , 2^n-1)=2^{\gcd(q-1,n)}-1$, we have only to consider $n$ that divide $q-1$ and there are $2^n-1$ solutions then.
Let $\psi(l)$ be the number of elements of cycles of exactly length $l$.
Now we have $2^n-1 = \sum_{d|n} \psi(d)$. By the Moebius-inversion-formula, we get $\psi(n) = \sum_{d|n} \mu (\frac{n}{d}) (2^d-1) = \sum_{d|n} \mu (\frac{n}{d}) 2^d - \sum_{d|n} \mu (\frac{n}{d}) = \sum_{d|n} \mu (\frac{n}{d}) 2^d$. But by dividing through $n$ (since every cycle of length $n$ contains $n$ elements) we get the desired formula.
Now looking back, we see that this argumentation doesn't work for the cycles of length $1$, but for these we can verify it directly.

Very impressive how fast you have found an answer !
I need some time to chew your answer ... (I'm a beginner).

[quote="ZetaX"]By the existence of a primitive root $\mod 2^{q}-1$, we can see it also as the following problem: Find the number of cycles of length $l$ under the action $x \to 2x$ seen $\mod 2^{q}-2$. [/quote] How do you switch from $x \to x^2$ seen $\mod 2^{q}-1$ to $x \to 2x$ seen $\mod 2^{q}-2$ ?

Now, more complicated.
I have also computed the first values of the number of cycles of length L under $x \to x^2-2$ seen $\mod 2^{q}-1$ and the result is nearly exactly the same as the one under $x \to x^2$, but sometimes the number of cycles is different (only when L is even): it seems to be dependent on q.
Here are the data (The values of N that are equal to the case $x^2$ are highlighted by a * at first column and appear second. "* ... ???" means that the value has not been computed but is expected to be the same as with $x^2$).

[b]Can you find a formula for the number of cycles of length L under $x \to x^2-2$ seen $\mod 2^{q}-1$ and prove it ?[/b]

Tony

[code]
     L N
-----------
*   1 1
*   2 1
*   3 2
    4 1                 q=5,7,61
*   4 3                 q=17,89
*   5 6
    6 4                 q=7,19,31,127
*   6 9                 q=13,61
*   7 18
    8 14                q=89
*   8 30                q=17
*   9 56
   10 48                q=31
*  10 99                q=61
*  11 186
   12 165               q=61
*  12 ???
*  13
   14 576               q=127
*  14 ???
*  15 2182
   16 2032              q=17
*  16 ????
*  17
   18 7252              q=19
*  18 ????
*  19
   20 26163             q=61
*  20 ?????
*  21 99858
*  22 190557
   30 17894588          q=31
*  30 35790267          q=61
   42 52357646436       q=127
*  42 ???????????
*  44 399822314775
*  53 169947155749830
   60 9607679187152607  q=61
*  60 ????????????????
*  63 146402730743693304
   88 1758437555802897070185750                 q=89
*  88 ?????????????????????????
  106 382686973653805017365049507840            q=107
* 106 ??????????????????????????????
  126 337581713215216729553162315024563188      q=127
* 126 ????????????????????????????????????
[/code]

Definition:
A primitive root is a $\zeta$, such that $\zeta^k, k \in\{ 1,2,...,p-1\}$ gives all different prime residue classes $\mod p$, so $\{1,2,3,...,p-1\}$. They exist modulo every power of an odd prime and some other cases (look at http://www.mathlinks.ro/Forum/viewtopic.php?highlight=primitive+root&t=55473 for the general statement and it's prove) and also in every finite field. Especially, they exist modulo ever prime $p$.

Now fix a primitive root $\zeta \mod p$. Any prime residue class $x \mod p$ can now be seen as some power $x\equiv \zeta^k \mod p$ for suitable $k$. Since when also $y\equiv\zeta^l \mod p$, it follows that $xy\equiv\zeta^k \zeta^l = \zeta^{k+l} \equiv \zeta^{(k+l) \mod (p-1)} \mod p$ (by Fermat's theorem), so you can see multiplication $\mod p$ as addition $\mod (p-1)$ (excluding $0$). So considering these powers $k,l$ is like taking logarithm in the real numbers.

But now to the problem concerning $x^2-2$ (it's much less elementary I think):

Let $p=2^q-1$ be a prime (so $q$ is also prime).
Lets work in the field $\mathbb{F}_{p}$ or $\mathbb{F}_{p^2}$ respectivly (so the field with $p$ elements and it's quadratic extension):
Let be $a_0,a_1,a_2,...,a_n=a_0$ be a cycle (of length dividing $n$). When we can write $a_0=x+\frac{1}{x}$, we would get by induction that $a_k=x^{2^{k}}+\frac{1}{x^{2^{k}}}$ for all $k$. Such $x$ does not necessary exist in $\mathbb{F}_p$, but, since it's a quadratic equation, for sure in $\mathbb{F}_{p^2}$.
So $a=x+\frac{1}{x}$ is part of a cycle of length dividing $n$ iff
\[ x^{2^{n}}+\frac{1}{x^{2^{n}}} = a_n =a_0 = x+\frac{1}{x} \iff x^{2^{n+1}}-x^{2^n+1}-x^{2^n-1}+1=0 \iff (x^{2^n+1}-1)(x^{2^n-1}-1)=0 \]
yielding two equations to solve in $\mathbb{F}_{p^2}$ (under the additional condition of $x+\frac{1}{x} \in \mathbb{F}_p$):
a) $x^{2^n+1}=1$
b) $x^{2^n-1}=1$
Since $p^2-1=(p+1)(p-1)=2^{q+1}(2^{q-1}-1)$, all these solution are already in $\mathbb{F}_{p}$ (because of order/primitive roots again).
Now that means we are looking for $x\in \mathbb{F}_p$ with $ord(x) | 2^n+1$ or $ord(x) | 2^n-1$.

Special case: $n| p-1$ and $n$ is odd.
Now also $2n|p-1$, and (because of $2^n+1 ; 2^n-1 | 2^{2n}-1 | 2^{q-1}-1$) there are already all $2^n+1$ solutions for a) and all $2^n-1$ solutions for b) contained in $\mathbb{F}_p$ (this statement is again based on the existence of a primitive root).
The only solution of both together is $x=1$. But when $x$ is a solution, also $\frac{1}{x}$ is a solution, but $x$ and it's inverse (and only those, since a quadratic equation has just two roots) give the same $a=x+\frac{1}{x}$, and the only selfinverse $x$ are $\pm 1$ (and $-1$ is for sure not a solution, $1$ is).
So there are exactly $\frac{2^n+1+2^n-1}{2}=2^n$ different such $a$. Now again using Moebius inversion gives the result for the odd $n$ dividing $q-1$.

General case:
Let $k_n$ be the number of solutions of a).
Let $l_n$ be the number of solutions of b).
Then the number of cycles of length dividing $n$ is $\frac{k_n+l_n}{2}$.
Now by the same reasons as before, we get that $k_n = \gcd(2^n+1,2^{q-1}-1)$ and $l_n = \gcd(2^n-1,2^{q-1}-1)$, thus, the problem is solved after using Moebius again.

Note that most of this arguing works for all primes, not just that of Mersenen type.

I hope all is correct :?

OK. Seems I have to buy several books in order to understand everything ;) (I'm a beginner) and it will take some time.
Since it is too late now for doing my home work (I was at cinema before ...), I'll just provide what I've found before your post.

The Sequence b(n): 1, 0, 1, 1, 3, 4, 9, 14, 28, 48, 93, 165, 315, 576 ... is built by:

$\ b(n) = \big( \ \sum_{d|n \ odd} \mu(d) 2^{n/d} \ + \sum_{d|n \ even} \mu(d) 2^{n/d-1} \ \ \big) / l$

but I was not able to link this result to q (since the number of cycles of length n depends on q).

I'll try tomorrow to link your nice work with my (experimental) work.

Thanks,
Tony

Ask me if something is not clear to you :)
I checked my formula on $q=5;7$ and it worked.

But to simplify the formula one more time:
We just have to consider divisors of $q-1$ as length of cycles again, so let $n$ be a divisor of $q-1$ from now on.
Let $\psi (n)$ denote the number of elements that are part of a cycle of a length dividing $n$.
Claim:
\[ \psi (n) = \left\{ \begin{array}{l} 2^n \text{ iff } 2n|q-1 \\ 2^{n-1} \text{ otherwise} \\ \end{array} \right.  \]

Prove: since $n|q-1$, we have $\gcd(2^n-1,2^{q-1}-1) = 2^n-1$, so there are $2^n-1$ solutions for b).
Since $2^n+1$ and $2^n-1$ are coprime, we get $\gcd (2^n+1, 2^{q-1}-1) =\frac{\gcd (2^n-1, 2^{q-1}-1) \gcd (2^n+1, 2^{q-1}-1)}{\gcd (2^n-1, 2^{q-1}-1)}= \frac{\gcd (2^{2n}-1, 2^{q-1}-1)}{\gcd (2^n-1, 2^{q-1}-1)}$.
Now if $2n | q-1$, we get $\frac{\gcd (2^{2n}-1, 2^{q-1}-1)}{\gcd (2^n-1, 2^{q-1}-1)} = \frac{2^{2n}-1}{2^n-1}=2^n+1$ and there are $\frac{2^n-1+2^n+1}{2}=2^n$ solutions then.
If otherwise $2n \nmid q-1$, we get $\frac{\gcd (2^{2n}-1, 2^{q-1}-1)}{\gcd (2^n-1, 2^{q-1}-1)} = \frac{2^n-1}{2^n-1}=1$ and there are $\frac{2^n-1+1}{2}=2^{n-1}$ solutions then.

Now when $\varsigma (n)$ describes the number of cycles of length exactly $n$, we get that $n \cdot \varsigma (n)$ is the number of elements that are part of such a cycle and by Moebius we reach $n \cdot \varsigma (n) = \sum_{d|n} \mu \left( \frac{n}{d} \right) \psi(d)$ or equivalently \[ \varsigma (n) = \frac{1}{n} \sum_{d|n} \mu \left( \frac{n}{d} \right) \psi(d)  \]

To write it without that cases, let $q-1=2^s u$, where $u$ is odd.
Then the formula reduces to \[ \varsigma (n) = \frac{1}{n} \left( \sum_{2^s \nmid d|n} \mu \left( \frac{n}{d} \right) 2^d + \sum_{2^s | d|n} \mu \left( \frac{n}{d} \right) 2^{d-1} \right) = \frac{1}{n} \left( \sum_{d|n} \mu \left( \frac{n}{d} \right) 2^d - \sum_{2^s | d|n} \mu \left( \frac{n}{d} \right) 2^{d-1} \right)  \]

Not that for the divisors of $\frac{q-1}{2}$ this is simply the formula from the $x^2$ case!

I now calculated and got:

$q=19$:
1: 2
2: 0
3: 2
6: 4
9: 56
18: 7252

$q=107$:
1: 2
53: 169947155749830
106: 382686973653805017365049507840

$q=521$:
1: 2
2: 1
4: 3
5: 6
8: 14
10: 99
13: 630
20: 52377
26: 2580795
40: 13743869130
52: 86607683851185
65: 567592125344909154
104: 97511584632940679888211521850
130: 10470226674490414260127730319842974161
260: 7125667029988688949142829846688486637119071635317558733759491696897244000323
520: 3300383490447408516818221538019900585225690048145819908362722816916126530479
        638769917715602127378580016469113156431850364472762710961015011478214497983630

Check them if you want.

[quote="ZetaX"]Now we have $2^n-1 = \sum_{d|n} \psi(d)$.
By the Moebius-inversion-formula, we get $\psi(l) = \sum_{d|n} \mu (d) (2^d-1)$ [/quote] I did not know the "Moebius Inversion Formula". Looking at: [url=http://mathworld.wolfram.com/MoebiusInversionFormula.html]MathWorld[/url], it does not look the same you do.
Instead of $2^d-1$ I would expect $2^{n/d}-1$ . And, instead of $\phi(l)$, I would expect $\phi(n)$ because $l$ is the length of a cycle and $l$ must be a divisor of $q-1$, which is $n$ here.
Am I correct ?
Tony

Yes, that was a stupid mistake of me. It should be correct now.

[quote="ZetaX"]....
Check them if you want.[/quote] That's perfectly correct !
I've written a PARI program, and I get with it the same results I had computed with Shallit's formula. About q=521, I get the same result as you get.
(In the following, I compute $1$ as the number of cycles of length $1$ under $x^2-2$ because Shallit considers that $2 \to 2$ is the terminator of the tree and must not be counted as a true cycle.)

I am still chewing your explanations, which are great ! Thanks !
I was looking at this problem since years, and I realize now that it was REALLY far over my very limited skills in Number Theory. I would really like to be 15 years old and to discover Number Theory ...

Don't you think all of that should make a nice paper ?

Tony
[code]
NC(q)=
{
	s=0; while((q-1)%2^s==0, s++); s--;
	print("q= ", q, " = 1 + 2^", s, ".", (q-1)/2^s, "\n");
	dq=divisors(q-1);
	lq=length(dq);

	print("n= ", 1, " -> 1");
	for(j=2,lq,
		n =dq[j];
		dn=divisors(n);
		ln=length(dn);
		S=0;
		for(i=1,ln,
			ddn=dn[i];
				S += moebius(n/ddn)*2^(ddn);
\\				print(dn[i], " + ", ddn, " ", moebius(ddn), " ", n/ddn);
			if(ddn%2^s==0,
				S -= moebius(n/ddn)*2^(ddn-1);
\\				print(dn[i], " - ", ddn, " ", moebius(ddn), " ", n/ddn-1);
			);
		);
		if(S!=0, print("n= ", n, " -> ", S/n););
	);
	print("\n");
}

[/code]


We have Fibonacci and Lucas sequences defined like so:
F_0=0;F_1=1 and F_(n+2)=F_(n+1)+F_n
L_0=2;L_1=1 and L_(n+2)=L_(n+1)+L_n.
Compute:

lim(n-->oo) (L_n,F_(n-1));
where (n,k)=n!/[k!(n-k)!]

cheers! :D :D

This is probably the hardest limit I have seen in my life so far! Please help! :D

Perhaps this can help Anyone Of Us :) :
F_(2n)=F_n*L_n
L_n=F_(n+1)+F_(n-1)
using these two relations we can transform the limit only in Fibonacci!! :D

anyway part b) of this problem is:
compute the limit of: lim(n-->oo) (L_n,F_(n-1))/3^n;

cheers! :D :D

when I saw a new post in this topic I thought u solved it! :(
damn... I am still so anxious for a solution to this problem!
:D :D :D

Since Ln=Fn-1+Fn+1 we get that C(Ln,Fn-1)=C(Fn-1+Fn+1,Fn-1) which is obviously bigger than Fn+1+Fn-1->oo
results that the first limit is oo.


Time for comedy. Results of 10th National Science Olympiad are out.  http://sofworld.org/10nso/2lresult/

Ranks:
Cyber-Shreyas (ME) - 99  :rotfl:  :rotfl:  :rotfl:

Ultimate comedy. My performance was so pathetic that I expected nothing but Last rank and I end up with 99.

Rangarajan (Valeriummaximum) - 130's
Albert Einstein (Srinivas) - 130's
Prasad (Physici Olympiaci) - 73
Jothi - 23
Rakesh - 40's (not sure)

Anit Sahu - Probably All india 1.

How pathetic. I give a poor performance and end up getting 99. I wish this happened in fiitjee.

Congrats da :D What rewards do you get??? :D :D

I am 41 da shre :rotfl:  :rotfl:  :rotfl:

Cha vethu da :D Atleast below 50, there shd be some recognition :)

:furious:  :bomb:  :censored: 11-250 ranks all Rs.250 this is very very very irritating

My transport charges for travelling from my home to the centre (to and fro) was almost Rs.250. So overall its a neutralisation reaction.  :rotfl:


Can you replace the question marks with maths symbols (+-/x) and make the equation correct:
(9 ? 1) ? (7 ? 6) ? (8 ? 4) = 3


limit
real analysis
real analysis unsolved

$a_{n+1}=2^{1-a_n}$
 So $0 <a_n <2 \forall n \geq 2$
  If $a_n >1$ then $a_{n+1} <1$
  Put $G(x) = f(f(x)) -x$ , where $f(x) =2^{1-x}$
  We always have : $G'(x) < 0$
  So $(a_{2n})$ is increasing and $a_{2n+1}$ is decreasing
   $\lim a_{2n} = \lim a_{2n+1} = 1$


In how many ways can a person enter a room with 5 doorways
and leave through a different doorway?

The person has $ \ 5$ ways to enter; since he can't go back the way he went, he has $ \ 4$ ways now. Find the number of ways he has by multiplying $ \ 5 \cdot 4 \equal{} \boxed {20}$


The tangents at points $ B,C$ to the circumcircle of a triangle $ ABC$ meet at $ T$. Tha points $ L,M,N$ are the midpoints of sides

$ BC,CA,AB$ respectively. The lines $ AT$ and $ NL$ meet at $ X$ and the lines $ AT , LM$ meet at $ Y$. Prove that $ BX \parallel CY$.

Babis

Dear Mathlinkers,
for a proof in english, see
http://perso.orange.fr/jl.ayme/  vol. 1  Another unlikely concurrence
Sincerely
Jean-Louis

Since $ \angle BNX \equal{} \angle LMC$, we have to show that $ \triangle BNX \sim \triangle YMC$
But this is the same as showing $ \frac {BN}{NX} \equal{} \frac {YM}{MC}$, or $ \frac {NA}{NX} \equal{} \frac {MY}{MA}$, or $ \triangle ANX\sim \triangle AMY$.
But since $ NX \parallel AM$, $ \angle NXA \equal{} \angle MAY$, similarly $ \angle NAX \equal{} \angle MYA$, so they are similar.

In fact, the problem is true for any ray AX passing through A, not just the symmedian.
 [geogebra]fe205c162abfbd3da4d97f98c88ae499d69f31a4[/geogebra]

[quote][color=darkred]Let $ ABC$ be a triangle. For a point $ L\in (BC)$ define the points $ \left\|\ \begin{array}{c} M\in AC\ ,\ LM\parallel AB \\
\ N\in AB\ ,\ LN\parallel AC\end{array}\ \right\|$ .

For another point $ S\in (BC)$ define the points $ X\in AS\cap LN$ and $ Y\in AS\cap LM$ . Prove that $ BX\parallel CY$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ \left\|\ \begin{array}{ccc} XL\parallel AC & \implies & \frac {SX}{SA} \equal{} \frac {SL}{SC} \\
 \\
YL\parallel AB & \implies & \frac {SA}{SY} \equal{} \frac {SB}{SL}\end{array}\ \right\|\ \bigodot\implies$ $ \frac {SX}{SY} \equal{} \frac {SB}{SC}$ $ \implies BX\parallel CY$ .[/color]

$AT$ and $AL$ are symmetrical w.r.t. bisector of $\widehat{BAC}$, hence $\triangle ALM \sim \triangle YAM$, so $ML\cdot MY = AM^2=MC^2$, that is, $AC$ is a tangent to the circle $(CLY)$, similarly $AB$ to the circle $(BLX)$, done.

Best regards,
sunken rock


Let n>=1.  Show that S=1/3 +1/5 +1/7 + ... + 1/(2n+1)  is not an integer.

Please explain to me with steps.. THZ A LOT!!

put $ B=lcm (3,5,...,2n+1)$ so we have:
$ A=\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n+1}=\frac{a_{1}+...+an}{B}$
where $ a_{i}=\frac{B}{2i+1}$...
we`ll show that $ 3 \mid B$ but $ 3 \not\mid a_{1}+...+a_{n}$

[hide="why?"]every natural number $ n$ can be represented in the form $ 3^{k}r$ where $ \gcd (r,3)=1$
now if we write the numbers $ \{3,5,...,2n+1\}$ in the above form we get that:
$ \{3,5,...,2n+1\}=\{3^{k_{1}}r_{1},...,3^{k_{n}}r_{n}\}$
now one can see that among the numbers $ r_{1},...,r_{n}$ only one of them has the maximum value...
let it be $ k_{t}$,
now its trivial that the greatest power of $ 3$ which devides $ B$ is $ 3^{k_{t}}$,now according to what we said above we get that for every $ i \not= t$,$ a_{i}$ is divisible by $ 3$ because $ r_{i}<r_{t}$...
but in the other hand $ a_{t}$ is not divisible by $ 3$,so $ a_{1}+...+a_{n}$ is not divisible by $ 3$,but $ B$ is divisible,hence $ A$ cant be an integer...[/hide]


I am stuck on this problem on divisibility of integers:

Let a1,a2,a3,...,an be n distinct integers. Show that the product of all fractions of form (ak-al)/(k-l) , where n>=k>=l, is an integer.

Can anyone please give an approach to this problem. By the way i started with the fact that when a number M is divided by any other number it leaves any of the remainders 0.1. 2 ... or M-1 but i am not getting any further with that.


Let show a rational coefficient polynomial that its minimum value is
a)$ \sqrt 2$
b)$ \minus{} \sqrt2$

Its very easy and exist many many polynomials for example see the polynomial: P(x)=x^3 -6x at x=sqrt(2) has local minimum its value is=-sqrt(2)
and so one....
But about absolute minimum value .......

No ideal?
I guess the polynomial must be four-degree one,but I can find at least example,It is too hard for me


Two  diagonally opposite squares are cut out from a checkerboard (with 64 squares). The checkerboard now only has 62 squares. You are supplied  with 31 identical dominoes with dimensions of 2 squares by 1 square. Is it possible to cover the modified  board completely  with non-overlapping dominoes?  Explain.

Thanks.

[hide="hint"]
Each domino would cover a black square and a red square.
[/hide]


Let a,b,c be positive real number.
Prove that $ \frac{a^3}{2a^2\plus{}b^2}\plus{}\frac{b^3}{2b^2\plus{}c^2} \plus{}\frac{c^3}{2c^2\plus{}a^2}  \geq \frac{a\plus{}b\plus{}c}{3}$
I have a solution, it's quite good, but I want to see the best solution :)  :lol:


function
floor function
inequalities
arithmetic series
algebra

Define a digit sum to be the sum of the digits of a certain base ten positive integer.  Prove that in every arithmetic series we can find two numbers whose digital sum is the same.  Can you generalize?

[hide]

Call the first term $a$ and the common difference $d$. Now, choose some arbitrary integer $n$. For now, we want to restrict the sequence to all the terms that are less than or equal to $10^n$. How many such terms are there, in terms of $a$, $d$ and $n$? Well, call that number $k$. Then we know that $k$ is the largest integer such that

$a + d(k-1) \leq 10^n$
$k \leq \frac{10^n - a}{d} + 1$

So we can express $k$ in terms of the floor function as

$k = \left\lfloor \frac{10^n - a}{d} + 1 \right\rfloor$

Since no term in this finite sequence of $k$ terms exceeds $n$ digits (possibly with the exception of the term $10^n$), the maximum digital sum in this sequence is $9n$. Since digital sums must be natural numbers, that means that there are $9n$ possible values of digital sums in this sequence.

Now, if the number of terms in this sequence exceeds the number of possible digital sums, then we know that some digital sums must appear more than once. So to prove that at least one digital sum must appear more than once, the following is sufficient:

$\left\lfloor \frac{10^n - a}{d} + 1 \right\rfloor > 9n$

...for at least one natural $n$. Now, remember, we get to choose whatever $n$ we want. The expression on the left, which contains $10^n$, obviously increases a lot faster than $9n$ after a while. So we will be able to find an $n$ big enough such that the inequality is true, which means that at least one digital sum must appear more than once in the first $\left\lfloor \frac{10^n - a}{d} + 1 \right\rfloor$ terms of the sequence, and consequently in the infinite sequence itself.

This is easily extendable to other bases. Just switch $10^n$ to $b^n$, and switch $9n$ to $(b-1)n$. The inequality still has to be true for a big enough value of $n$, so the sequence in that base still holds the property that a digital sum must be repeated somewhere.

[/hide]


Let $b$ be a positive integer. Determine all $2002$-tuples of non-negative integers $(a_{1}, a_{2}, \cdots, a_{2002})$ satisfying \[\sum^{2002}_{j=1}{a_{j}}^{a_{j}}=2002{b}^{b}.\]

how to solve this problem? Is there any nice solution?

There should be.

I can prove that when $ b\ge 2002$, then $ a_1 \equal{} ... \equal{} a_{2002} \equal{} b$ is the only solution (simply by inequalities), but I have no idea how to handle $ b < 2002$.

Yimin Ge, do you know more? You're from Autstria...

Let $ c\equal{}max_j a_j$. Then $ c^c\le S\equal{}\sum_j a_j^{a_j}\le 2002 c^c$.
If $ c\ge 737$, then $ 2002(c\minus{}1)^{c\minus{}1}<c^c$, therefore $ b\ge c\to a_j\equal{}c \ \forall j$.
When $ b<737$ may be another solutions. For example
$ b\equal{}1,a_1\equal{}a_2\equal{}...a_k\equal{}1,a_{k\plus{}1}\equal{}...\equal{}a_{2002}\equal{}0.$

In the competition were this problem appeared, there was the additional condition $ b>800$.  :wink:

[quote="FelixD"]In the competition were this problem appeared, there was the additional condition $ b > 800$.  :wink:[/quote]That is interesting information! Do you have a solution?


Solve the equation $ 3^n\equal{}2^{n\plus{}1}\plus{}1$ where $ n$ is a positive integer

[hide="Solution"]The only solutions to $ 3^n \equal{} 2^{n\plus{}1} \plus{} 1$ are the solutions to $ 3^n \minus{} 2^{n\plus{}1} \equal{} 1$. By Catalan's conjecture, the only solution is then $ \boxed{n \equal{} 2}$.[/hide]

I didn't worked by Catalan's conjecture before but can you use a conjuncture in proofing something.

Thanks  :D

Observing in mod 3, we see that $ n$ must be even. After testing some values, we see that $ n\equal{}2$ is a solution and as $ n$ increases, the LHS seems to be increasing much faster than RHS, from which we can conclude that $ \boxed{n\equal{}2}$ is the only positive integer solution.

*Note: Not a valid proof for a competition though. But it's good if the test just cares about the answer.

Catalan's conjecture is more of a misnomer now, since it was actually proven. It states that the only natural number solutions to $ a^b \minus{} c^d \equal{} 1$ are $ a \equal{} d \equal{} 3$ and $ b \equal{} c \equal{} 2$.


yea i read that earlier. China is way over doing "looking good for the world". It's like wearing too much make up to the prom. (That was a terrible analogy but it was all I could come up with.)

I think its ridiculous... most people wouldnt judge your country on the cuteness of your 6 year olds

[quote="you-know-who"]The Olympics is about sending your best out there. I think it's perfectly okay to do that. I'm sure the USA has done so many times, but nobody cares. But when China does it, it suddenly becomes a huge crime! We should all take a look at ourselves before criticizing others. Yang Peiyi doesn't seem to mind much too; why should we be freaking out over this when the person who is not seen is okay with it? This is the entertainment business, and people in Hollywood get fired based on their looks. This is nothing new![/quote]

The general idea of you-know-who's statement is correct, but the claim that China is innocent is false. That kind of faking is a crime, no doubt. It is much like anabolic steroids and illegal drugs. It is definitely a crime, but a human growth hormone-using athlete has no right to level accusations at another user of human growth hormone, though the IOC does. The US as a whole has no right to complain, but maybe some other countries might (though I doubt it)

I think that something like that must never happen. This isn't something new, but try now to be in the skin of that girl. What do you think how does she feel(the girl who was singing)?

I would like to echo you-know-who's statement, it's no question that people are judged based on their looks, and china wanted to put their best out there.  is it a little bit dishonest?  yes, i think so, but, it's not some high crime, which is what the media is making it out to be

This is also in Round Table: http://www.mathlinks.ro/viewtopic.php?p=1223729

[quote="zephyredx"][quote="you-know-who"]Yang Peiyi doesn't seem to mind much too; why should we be freaking out over this when the person who is not seen is okay with it? [/quote]

I agree. In fact, I admire her ability to not mind being backstage; if it were me, I'd have to make sure everyone in the room knew I was the singer.

Somehow I like feel this topic should also belong in the Round Table.[/quote]

umm...shes probably been told wat to say to the media, so i wouldnt believe that claim.


$ABC$ is an equilateral triangle.

It is folded such that point $A$ now lies on $BC$. The fold meets the erstwhile $AB$ and $AC$ at $D$ and $E$ respectively.

The area of $DAB$ is $36cm^2$, and that of $EAC$ is $16cm^2$.

What is the area of the original triangle $ABC$?

[quote="TheAmazingOne"][hide="Hint"]Triangles EAC and ADB are similar.[/hide]

[hide]
I got 150 cm², but I don't feel like writing out a whole solution. The hint should be enough.
[/hide][/quote]

[hide]I could get the same answer using the calculations... but not able to prove that the triangle EAC and ADB are similar? How dp you get that?[/hide]

I can tell you why they are similar but kindly give me the complete soln.
I am not getting 150 as the answer....
something wrong somewhere, plzzzz help

similarity proof is given in the attached diagram
 :roll:

[quote="TheAmazingOne"][hide]
Since the ratio between the areas of triangles $EA'C$ and $A'DB$ is $36: 16$, the ratio between the sides is $6: 4 = 3: 2$. So set $A'D = 3a$. Therefore $A'E = 2a$. Since $ADE$ and $A'DE$ are congruent, we have $AD = 3a$ and $AE = 2a$. Now set $BD = 3b$. Therefore $A'C = 2b$. Since $AB = BC$, we have $3a+3b = 2b + A'B$ and thus $A'B = 3a + b$. And for the same reason, $CE = a + 3b$. But because of the similarity, we know that $2A'B = 3CE$. That gives us

$2(3a+b) = 3(a+3b)$
$6a + 2b = 3a + 9b$
$3a = 7b$

From this we can deduce that the side length of the equilateral triangle is $10b$. That means that the area in terms of $b$ is $\left(\frac{1}{2}\right)(10b)^2\sin 60^\circ = 25b^2\sqrt{3}$.

We know that $A'B = 3a + b = 8b$ and $BD = 3b$. So the area of $A'BD$ is

$\left(\frac{1}{2}\right)(3b)(8b)\sin 60^\circ = 6b^2\sqrt{3} = 36$

Multiply both sides of the equation by $\frac{25}{6}$:

$25b^2\sqrt{3} = 150$

This is the area of $ABC$, as we noted above.
[/hide][/quote]

Thats nice...


There is a man in a park going with constant velocity $ v$ in a straight line. THere is a dog with $ l$ initial distance to the man. the dog starts to move perpendicular to the line of the man with constant velocity $ v$. During the motion the dog always runs towards the man. What will be the final distance between the dog and the man?

Consider frame of man. After time $ t$, assume dog is at $ (r,\theta)$.
Dog has velocity $ v$ downwards and $ v$ radially inwards. Thus we have pair of differential equations
$ \frac{dr}{dt}\equal{}v_0(\sin\theta \minus{} 1)$ and $ r\frac{d\theta}{dt}\equal{}v_0\cos\theta$
Solving simultaneously,
$ \ln\frac{l}{l_0}\equal{}\minus{}\int_{\minus{}\frac{\pi}{2}}^{0} \frac{\sin\theta \minus{} 1}{\cos\theta}d\theta$

I got final $ l\equal{}2l_0$

Could you explain, please, what is $ v_0$ and $ l_0$ ?

Sorry about that. I have used $ v_0$ and $ l_0$ for the original $ v$ and $ l$.

So, $ l$ should be the final (asked) distance, right?

But, in that case the distance has been increased? Wrong, I think... (notice the sign of $ {{dr}\over{dt}}$ in your own expression).

Seems to be just the opposite: the final distance is half the initial one.


inequalities
induction
linear algebra
matrix
triangle inequality
inequalities unsolved

Let $ (a_1,a_2,...a_n)\succ (b_1,b_2,...,b_2)$ and let $ x_1\geq x_2\geq ...\geq x_n$
Prove
$ [(a_1x_1 \plus{} a_2x_2 \plus{} ... \plus{} a_nx_n),...,(a_1x_{\pi(1)} \plus{} a_2x_{\pi(2)} \plus{} ... \plus{} a_nx_{\pi(n)}),...,(a_1x_n \plus{} a_2x_{n \minus{} 1} \plus{} ... \plus{} a_nx_1)]\succ$
$ \succ[(b_1x_1 \plus{} b_2x_2 \plus{} ... \plus{} b_nx_n),...,(b_1x_{\pi(1)} \plus{} b_2x_{\pi(2)} \plus{} ... \plus{} b_nx_{\pi(n)}),...,(b_1x_n \plus{} b_2x_{n \minus{} 1} \plus{} ... \plus{} b_nx_1)]$

1. From rearrangement we have :
$ (a_1x_1 \plus{} a_2x_2 \plus{} ... \plus{} a_nx_n)\geq ...\geq (a_1x_{\pi(1)} \plus{} a_2x_{\pi(2)} \plus{} ... \plus{} a_nx_{\pi(n)})\geq ...\geq (a_1x_n \plus{} a_2x_{n \minus{} 1} \plus{} ... \plus{} a_nx_1)$
and
$ (b_1x_1 \plus{} b_2x_2 \plus{} ... \plus{} b_nx_n)\geq ...\geq(b_1x_{\pi(1)} \plus{} b_2x_{\pi(2)} \plus{} ... \plus{} b_nx_{\pi(n)})\geq ...\geq (b_1x_n \plus{} b_2x_{n \minus{} 1} \plus{} ... \plus{} b_nx_1)$

2. ???

3. $ \sum^n_{i \equal{} 1}a_i.[(n \minus{} 1)!(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)] \equal{} \sum^n_{i \equal{} 1}b_i.[(n \minus{} 1)!(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)]$ <=> $ \sum^n_{i \equal{} 1}a_i \equal{} \sum^n_{i \equal{} 1}b_i$ which is true because $ (a_1,a_2,...a_n)\succ (b_1,b_2,...,b_2)$

when we have that it's obvious that we have [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=140365#140365]Muirhead revisited[/url] using Karamata for $ f(x) \equal{} ln x$
and The original Muirhead using Karamata for $ f(x) \equal{} e^x$.
Maybe that's been proven yet , but I don't know where exactly  :huh:

True, indeed, but I would prove it differently.

[color=blue][b]Theorem 1.[/b] Let $ (a_1,a_2,...,a_n)$, $ (b_1,b_2,...,b_n)$, $ (x_1,x_2,...,x_n)$ be three number arrays such that $ (a_1,a_2,...,a_n)\succ(b_1,b_2,...,b_n)$. Then, $ (\sum_{k=1}^n a_kx_{\pi(k)})_{\pi\in S_n}\succ(\sum_{k=1}^n b_kx_{\pi(k)})_{\pi\in S_n}$.[/color]

Here, $ (\sum_{k=1}^n a_kx_{\pi(k)})_{\pi\in S_n}$ means the number array consisting of $ n!$ numbers - namely, of the numbers $ \sum_{k=1}^n a_kx_{\pi(k)}$ for every permutation $ \pi$ of the set $ \{1,2,...,n\}$. Similarly, the number array $ (\sum_{k=1}^n b_kx_{\pi(k)})_{\pi\in S_n}$ has to be understood.

For the proof, we use [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5849]a known lemma[/url]:

[color=blue][b]Theorem 2.[/b] Let $ (a_1,a_2,...,a_n)$ and $ (b_1,b_2,...,b_n)$ be two number arrays such that $ (a_1,a_2,...,a_n)\succ(b_1,b_2,...,b_n)$. Then, we can assign a nonnegative real $ \lambda_{\pi}$ to every permutation $ \pi\in S_n$ such that $ \sum_{\pi\in S_n}\lambda_{\pi}=1$ and

$ b_i=\sum_{\pi\in S_n}\lambda_{\pi}a_{\pi(i)}$ for every $ i\in\{1,2,...,n\}$.

In words: If a number array majorizes another one, then the second array is a convex combination of the arrays obtained from the first array by permuting its coordinates.[/color]

Apart from the proof given in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5849]above link[/url], this theorem can be proven in two other ways:
- by induction, using the fact that the array $ (b_1,b_2,...,b_n)$ can be obtained from the array $ (a_1,a_2,...,a_n)$ through a finite number of compressions (see http://www.mathlinks.ro/Forum/viewtopic.php?t=17288 - maybe I am going to post a proof for this fact when time allows);
- using the Birkhoff theorem on bistochastic matrices and the bistochastic matrix criterion for majorization.

Now to the [i]proof of Theorem 1[/i]: According to Theorem 2, we can assign a nonnegative real $ \lambda_{\pi}$ to every permutation $ \pi\in S_n$ such that $ \sum_{\pi\in S_n}\lambda_{\pi}=1$ and

$ b_i=\sum_{\pi\in S_n}\lambda_{\pi}a_{\pi(i)}$ for every $ i\in\{1,2,...,n\}$.

Hence, for any real $ t$, we have

$ \sum_{\phi\in S_n}|\sum_{k=1}^n b_kx_{\phi(k)}-t|=\sum_{\phi\in S_n}|\sum_{k=1}^n \sum_{\pi\in S_n}\lambda_{\pi}a_{\pi(k)}x_{\phi(k)}-t|=\sum_{\phi\in S_n}|\sum_{\pi\in S_n}\lambda_{\pi}\sum_{k=1}^n a_{\pi(k)}x_{\phi(k)}-t|$
$ =\sum_{\phi\in S_n}|\sum_{\pi\in S_n}\lambda_{\pi}\sum_{k=1}^n a_{\pi(k)}x_{\phi(k)}-\sum_{\pi\in S_n}\lambda_{\pi}t|$ (since $ \sum_{\pi\in S_n}\lambda_{\pi}=1$)
$ =\sum_{\phi\in S_n}|\sum_{\pi\in S_n}\lambda_{\pi}(\sum_{k=1}^n a_{\pi(k)}x_{\phi(k)}-t)|$
$ \leq \sum_{\phi\in S_n}\sum_{\pi\in S_n}\lambda_{\pi}|\sum_{k=1}^n a_{\pi(k)}x_{\phi(k)}-t|$ (in fact, $ |\sum_{\pi\in S_n}\lambda_{\pi}(\sum_{k=1}^n a_{\pi(k)}x_{\phi(k)}-t)|\leq \sum_{\pi\in S_n}\lambda_{\pi}|\sum_{k=1}^n a_{\pi(k)}x_{\phi(k)}-t|$ by the triangle inequality, since the $ \lambda_{\pi}$ are all nonnegative)
$ = \sum_{\phi\in S_n}\sum_{\pi\in S_n}\lambda_{\pi}|\sum_{k=1}^n a_{\pi(k)}x_{(\phi\circ\pi^{-1})(\pi(k))}-t|$
$ = \sum_{\phi\in S_n}\sum_{\pi\in S_n}\lambda_{\pi}|\sum_{m=1}^n a_{m}x_{(\phi\circ\pi^{-1})(m)}-t|$ (at this step we have set $ m=\pi(k)$; since $ \pi$ is a permutation of $ \{1,2,...,n\}$, we were able to replace summation over $ k$ by summation over $ m$)
$ = \sum_{\pi\in S_n}\lambda_{\pi}\sum_{\phi\in S_n}|\sum_{m=1}^n a_{m}x_{(\phi\circ\pi^{-1})(m)}-t|$
$ = \sum_{\pi\in S_n}\lambda_{\pi}\sum_{\theta\in S_n}|\sum_{m=1}^n a_{m}x_{\theta(m)}-t|$ (at this step we have set $ \theta=\phi\circ\pi^{-1}$, and since $ x\mapsto x\circ\pi^{-1}$ is a permutation of the set $ S_n$, we were able to replace summation over $ \phi$ by summation over $ \theta$)
$ = \sum_{\pi\in S_n}\lambda_{\pi}\cdot \sum_{\theta\in S_n}|\sum_{m=1}^n a_{m}x_{\theta(m)}-t|$ (since the sum $ \sum_{\theta\in S_n}|\sum_{m=1}^n a_{m}x_{\theta(m)}-t|$ does not depend on $ \pi$)
$ =\sum_{\theta\in S_n}|\sum_{m=1}^n a_{m}x_{\theta(m)}-t|$ (since $ \sum_{\pi\in S_n}\lambda_{\pi}=1$)
$ =\sum_{\phi\in S_n}|\sum_{k=1}^n a_{k}x_{\phi(k)}-t|$ (at this step, we have just renamed the indices $ m$ and $ \theta$ as $ k$ and $ \phi$, respectively);

in other words,

$ \sum_{\phi\in S_n}|\sum_{k=1}^n a_{k}x_{\phi(k)}-t|\geq\sum_{\phi\in S_n}|\sum_{k=1}^n b_kx_{\phi(k)}-t|$.

Hence, according to the symmetric majorization criterion ( http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 post #11 Lemma 1), we have $ (\sum_{k=1}^n a_kx_{\pi(k)})_{\pi\in S_n}\succ(\sum_{k=1}^n b_kx_{\pi(k)})_{\pi\in S_n}$. This proves Theorem 1.

  darij

[quote="BG Yoda"]thank you very much darij , but i'm in class 11 and we haven't already learned what means :
$ \phi \circ \pi ^{ \minus{} 1}$
can you give me a link of something where i can read about that operation ??
i guess $ \circ$ is a concatenation isn't it[/quote]

Yes. If $ f$ and $ g$ are two transformations, then $ f\circ g$ is the transformation defined by $ \left(f\circ g\right)\left(x\right)\equal{}f\left(g\left(x\right)\right)$ for every $ x$.

  Darij


given is $\lim_{x\rightarrow0}\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$ calculate this well: I take  $u=\frac{1}{x}$ then we get $\lim_{u\rightarrow \infty}\frac{1-u}{1+u}$ then we get $\lim_{u\rightarrow \infty}\frac{u(\frac{1}{u}-1)}{u(\frac{1}{u}+1)}$ so we get $\lim_{u\rightarrow \infty}\frac{(\frac{1}{u}-1)}{(\frac{1}{u}+1)}$ then $\lim_{u\rightarrow \infty}\frac{(0-1)}{(0+1)}=-1$

So my question is this methode correct? We need to use this without using l'hopital.

Thanks a lot.

Almost correct. As $x\to0^{+},$ then $u\to\infty.$ Hence, technically you have only computed the one-sided limit at zero. To get the other side, you have to also consider what happens when $u\to-\infty.$

Of course, for this particular problem, you did too much work. Isn't there a simple algebraic step you can do right at the beginning that would make your life easier?

$\frac{1-\frac1x}{1+\frac1x}=\frac{1-\frac1x}{1+\frac1x}\cdot\frac{x}{x}=\frac{x-1}{x+1}$

In fact, you did this exact step, but later, after you had switched to $u.$

oh great I find an other way so $\frac{\frac{x}{x}-\frac{1}{x}}{\frac{x}{x}+\frac{1}{x}}$then we get $\frac{\frac{x-1}{x}}{\frac{x+1}{x}}$ finaly $\frac{x-1}{x}\frac{x}{x+1}$ and then we will come.

Thanks.


what is the 42nd fibonnacci number?

is there a fast way to do this

yeah there's a formula, but it's terribly complicated. you would need a calculator.

You could use [url=http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html][b]Binet's Formula[/b][/url], which states that for the term $ n$ in the fibonacci recurrence, $ F_{n}=\frac{{(1+\sqrt{5})}^{n}-{(1-\sqrt{5})}^{n}}{{2}^{n}\sqrt{5}}$.

Using binet's formula, we get 267914296 as our answer.

Or you could use this: $ F_{n}\approx\frac{\phi^{n}}{\sqrt 5}$, which is just a consequence of the above formula. Here $ \phi=\frac{1+\sqrt 5}{2}$ is the golden ratio.


Given are positive integers $ n>1$ and $ a$ so that $ a>n^2$, and among the integers $ a\plus{}1, a\plus{}2, \ldots, a\plus{}n$ one can find a multiple of each of the numbers $ n^2\plus{}1, n^2\plus{}2, \ldots, n^2\plus{}n$. Prove that $ a>n^4\minus{}n^3$.

[quote="April"]Given are positive integers $ n > 1$ and $ a$ so that $ a > n^2$, and among the integers $ a \plus{} 1, a \plus{} 2, \ldots, a \plus{} n$ one can find a multiple of each of the numbers $ n^2 \plus{} 1, n^2 \plus{} 2, \ldots, n^2 \plus{} n$. Prove that $ a > n^4 \minus{} n^3$.[/quote]

Let $ k_1 (n^2\plus{}1)\equal{}a\plus{}y_1$, $ k_2 (n^2\plus{}2)\equal{}a\plus{}y_2$

Then $ k_1 (n^2\plus{}1)\minus{}n \le a \le k_1 (n^2\plus{}1)\minus{}1$  and  $ k_2 (n^2\plus{}2)\minus{}n \le a \le k_2 (n^2\plus{}2)\minus{}1$

$ \Rightarrow \minus{}(n\minus{}1) \le k_2 (n^2\plus{}2)\minus{}k_1 (n^2\plus{}1)\equal{}(k_2\minus{}k_1)(n^2\plus{}2)\plus{}k_1 \le n\minus{}1$

Similarly we have $ \minus{}(n\minus{}1) \le (k_{i\plus{}1}\minus{}k_i)(n^2\plus{} (i\plus{}1))\plus{}k_i \le n\minus{}1$

So either $ k_i$ is a constant for all $ i$ , which is clearly impossible or

We can find $ i$ so that $ \minus{}(n\minus{}1) \le (k_{i\plus{}1}\minus{}k_i)(n^2\plus{} (i\plus{}1))\plus{}k_i \le \minus{}(n^2\plus{}(i\plus{}1))\plus{}k_i$

So $ n^2\minus{}n\plus{}(i\plus{}2) \le k_i$

$ a\plus{}n \ge a\plus{}y_i\equal{}k_i(n^2\plus{}i) \ge [n^2\minus{}n\plus{}(i\plus{}2)] (n^2\plus{}i) > n^4\minus{}n^3\plus{}(i\plus{}2)n^2$

$ \Rightarrow a>n^4\minus{}3n^3$

i don't understand  why $ \minus{} (n \minus{} 1) \leq (k_2 \minus{} k_1)(n^2 \plus{} 2) \plus{} k_1 \leq n \minus{} 1$

Explain me, plz!!!
  i understood! sorry for my spam !

Let $n^2 + i \mid a + x_{i}$ and $n^2 + i+1 \mid a + x_{i+1}$ then we have
$$(n^2+i)(n^2+i+1) \mid (a+x_i)(n^2 + i+1) - (a+x_{i+1})(n^2 + i) = a+x_i + (n^2 + i)(x_i - x_{i+1})$$
hence
$$(n^2 + i)^2 + (n^2 + i) \le |a + x_i + (n^2+i)(x_i - x_{i+1})|$$
If the quantity inside the modulus is positive for some $i$ then
$$a \ge (n^2+i)^2 + (n^2+i)(1 + x_{i+1} - x_i) - x_i \ge (n^2+i)^2 + (n^2 + i)(1 -n + 1) + n = n^4 - n^3 + 2n^2(i+1) + 2i - ni$$
but the quantity $2n^2(i+1) + 2i - ni$ is most definitely positive and we are done. [In the above inequality we placed $x_i = n$ and $x_{i+1}=1$ to get the maximum negative value of $-(n-1)$ which is to be multiplied with $n^2+i$]. Henceforth assume that for all $i$ we have the quantity $a + x_i + (n^2+i)(x_i - x_{i+1}) < 0$ which simplifies to $a \le -(n^2+i)(x_i - x_{i+1}) - x_i \le -(n^2 + i) - x_i < n^2$ if $x_i - x_{i+1} \ge 1$ which is a contradiction, otherwise we have $x_i - x_{i+1} \le -1$ for all $i$ which after summing all $i$ yields $x_1 - x_n \le -(n-1)$ implying $x_i - x_{i+1} =-1$ (due to equality everywhere) $\implies x_i = i$ but then
$$0 < a - n^2 = a+x_i + (n^2+i)(x_i - x_{i+1}) < 0$$
which is another contradiction and we are done.


group theory
abstract algebra
linear algebra
linear algebra unsolved

Let $ p$ be a prime number, and $ G$ a subgroup of $ \mbox{GL}_{n}(\mathbb{Z}/p \mathbb{Z})$, s.t. $ |G|=p^{n(n-1)/2}$.
Show that all elements of $ G$ are trigonalizable in a common basis.

$ G$ is a Sylow $ p$-subgroup of $ \mbox{Gl}_{n}(\mathbb Z/p\mathbb Z)$. All Sylow $ p$-subgroups of a finite group belong to the same conjugacy class, so there is an inner automorphism of $ \mbox{Gl}_{n}(\mathbb Z/p\mathbb Z)$ which sends $ G$ onto the group consisting of matrices with $ 1$ on and $ 0$ below the main diagonal.


linear algebra
matrix
algebra
polynomial
LaTeX
real analysis
real analysis theorems

Hello everyone :)

I need rigourous justifications for the continuity of the following applications :
A---------det(A) where A is square matrix
A---------(characteristic polynomial of A)

Thank you and sorry for not using latex

Both are polynomial in the entries of $A$. Polynomials are continuous in any reasonable topology.


Dear Valentin,

Surely this is overkill? ;)

[quote="Valentin Vornicu"]I just remembered that in 2001, almost all balkan countries had at least a girl in their team (we can also include Hungary here), and I remember the organizers throwing a "Girls in Maths" banquet for that at some point :)

I was curious how many will be there now :) What do you think ?[/quote]

I think there's a good reason that didn't get repeated...  it was the girls, plus the guys of the countries that went, plus ice-cream (the only decent part of the evening)... and the talk was as if we didn't know anything (Sierpinski triangle I think? For the first half an hour??)... Although. They had an official games event thing where they got lots of games and put them in a room and forced us to go there... that was good...


I made a signature in my profile, but I don't know how to post it...
I need help plz.  :lol:

If you have saved your signature, then all you have to do i select "Attach signature" below the text box while posting, or go to Profile >> Preferences >> Posting a message, where there is an option "Always attach signature".


[quote="Allan Z"]Let $x_{1}$, $x_{2}$, $x_{3}$, ..., $x_{n}$ are positive.
Prove:
${\frac{{x_{1}}^{2}}{x_{2}}+\frac{{x_{2}}^{2}}{x_{3}}+...+\frac{{x_{n-1}}^{2}}{x_{n}}+\frac{{x_{n}}^{2}}{x^{1}}}\geq{x_{1}+x_{2}+x_{3}+...+x_{n}}$[/quote]

[hide="rearangement"]
WLOG, let $x_{1}\leq x_{2}\leq \dotsm \leq x_{n}$

Then $x_{1}+x_{2}+\dotsm+x_{n}= (x_{1}^{2})(1/x_{1})+(x_{2}^{2})(1/x_{2})+\dotsm+(x_{n}^{2})(1/x_{n})$

Notice on the RHS that the $(1/x_{1})$ is the smallest of the reciprocal terms, and $x_{1}^{2}$ is the largest of the other terms.  Therefore, the RHS is oppositely sorted, so by the rearangement inequality, every other permutation of the RHS terms is going to be more than or equal to that permutaion.  QED.

[/hide]

LHS $\geq$ (sum)^2/sum = RHS by cauchy in engel form.


Sorry, the exact problem is
Prove that
$(2n+1)^{(}2n+1)\leq (2n+1)!!$$\pi^{n}$
for every natural n, where $(2n+1)!!=(2n+1)(2n-1)(2n-3)...1$

Rewrite it in a readable form:

[quote="nmt_2006"]Prove that: \[(2n+1)^{(2n+1)}\leq \pi^{n}\cdot(2n+1)!!\] for every natural $n$, where $(2n+1)!!$ is factorial of odd numbers from 1 to 2n+1.[/quote]

Well $k(2n-k)\pi=(2n+1)^{2}$ for $01 \le k \le 2n-1$, and $k$ is odd.

You can make an even stronger bound with not $\pi$, but with $4$, it seems.

Please don't post any solution here. It is a Problem in TH&TT magazine too.
It's very easy, just Induction. :D


calculus
integration
calculus computations

I can't seem to find an appropriate change of variables to evalute the following integral:

 The Double integral : ${{\int\int \exp(\frac{x^3+y^3}{xy})}} dxdy$

Over the region D , defined as: $D$: { ${{y^2 <= 2px }}$ ${{; x^2 <= 2py}}$  ${{,(x,y)}\in R}$ }

Is there some general method, or even general guidelines when trying to find a change of variables when evaluating integrals such as this.

Thanks in advance,

Sharper

The substitution that leaps out at me is $u=\frac{x^2}y,\,v=\frac{y^2}x$.
That makes the region nice, and the integrand is easily expressed. Note that the region in question is entirely first quadrant; both $x$ and $y$ must be positive.

We are lucky and the Jacobian comes out nicely; $dx\times dy=\frac13 du\times dv$. I used the inverse form of the transformation to calculate it, since we don't have any extra factors in the original integral.

The transformed integral is then $\int_0^{2p}\int_0^{2p}\frac13e^{u+v}\,du\,dv$.


Fie n>=4 ; Aratati ca orice patrulater convex poate fii impartit in n patrulatere inscriptibile.

Fie $ ABCD$ patrulaterul convex.
1). Pentru $ {n} \equal{} 4$:  luam un punct $ O$ in interiorul sau astfel incit proiectiile sale, punctele $ M$, $ N$, $ P$ si $ Q$, pe dreptele $ AB$, $ BC$, $ CD$ si $ DA$
suport ale laturilor patrulaterului, sa se gaseasca pe laturile respective. Patrulaterele $ AMOQ$, $ BNOM$, $ CPON$ si $ DQOP$ sunt inscriptibile.
2). Presupunem ca stim imparti un patrulater in $ {n} > 4$ patrulater inscriptibile. In acest caz il putem imparti si in $ {n \plus{} 1}$ patrulatere inscriptibile. Iata cum procedam: Prin varfurile  A si B ducem un cerc care sa intersecteze laturile ($ AD$) si respectiv ($ BC$) in punctele $ X$ si $ Y$. Patrulaterul $ ABYX$ este inscriptibil, iar patrulaterul $ CDXY$ prin ipoteza (de inductie) stim deja sa-l impartim in $ {n}$ patrulater inscriptibile! Obtinem astfel o impartire a patrulaterului $ ABCD$ in $ {n \plus{} 1}$ patrulatere inscriptibile!


geometry
3D geometry
sphere
trigonometry

[quote="sran"]/*This topic also posted on the "Other Problem-Solving Topics" forum, please delete if illegal, apologize if illegal*/

Consider two points on a perfect sphere. Let the radius of the sphere be r, and the latitudes and longitudes of the two points be (a, b) and (x, y) (Or whatever you prefer to call it). It has been discovered that the points are relatively close together (Say, the are in the same octant). What is the distance between the points, 1) along the surface of the sphere, and 2) through Euclidean space?

For 1), I'm thinking you apply $\theta * r = s$, but I'm nt sure how to get the $\theta$.

For 2), its an SAS trig problem if you can find $\theta$, but again, I'm having trouble with $\theta$.

Any help would be appreciated. Thanks![/quote]
first, is that you should avoid double posting the same topic in different forums even if it is important. i know this maybe urgent to you, but your urgency is not the forum's responsibility. in other words, the forum is immune to your urgency for help.

second, there are 'imperfect spheres'?

third, for 1) you can use $s=r\theta$. for more detailed info, here is a [url=http://mathworld.wolfram.com/LawofCosines.html]link to mathworld on law of cosine[/url]. using this law of cosine allows you to find $\theta$ because you have the latitudes/longitudes of the two given points.

fourth, 2) is related to 1) can be done the same way with law of cosine.

fifth, here is the difference between the two problems:
1) apply law of cosine on spherical triangle
2) apply law of cosine on Euclidean (plane) triangle


geometry
circumcircle
geometric transformation
rotation
trigonometry
projective geometry
geometry unsolved

On the circumcircle of the equilateral triangle ABC, choose an arbitrary point P on arc AB. Mark the points Q and R on arc BC and the points S and T on arc CA for which AP=BQ=CR=CS=AT holds. Let lines BQ and PR meet at U and let lines AS and PT meet at V. Finally, let lines AU and BV meet at W. Show that each two of lines AU, BV and PW enclose an angle of 60.

Now, I just can prove the degree of angle AWF=60(F is the intersection between BW and the circumcircle(O)).aND i TRY TO FIND THE WAY TO PROVE PW//AF but I'm still unsuccessfull! Can anybody help me prove this property?  :(

[quote="PDatK40SP"]$ O$ is the center of $ \Delta ABC$
$ QT$ cuts $ CS$ at $ U'$
Use the trigonometric form of Céva's theorem with $ \Delta SBT$ and the concurrence of:
+ $ SA,BV,TP$, we'll have a condition
+ $ SC,BU',TQ$, we'll have a condition
From these two conditions, we have $ \overrightarrow{BU'} \uparrow \uparrow \overrightarrow{BV}$, or $ B,V,U'$ are collinear    $ (*)$
Consider the rotation with center $ O$ and angle $ \frac { 2 \pi}{3}$ we have:
+ $ A \to B, B \to C, C \to A$
+ $ P \to Q, Q \to S, S \to P, R \to T$
So we have $ U \to U'$, it means $ AU \to BV$, so $ (\overrightarrow{AU}, \overrightarrow{BV}) \equiv \frac {2 \pi}{3} \ (mod \ 2 \pi)$
From $ (*)$ and Pascal's theorem, we have $ PC,AQ,BV$ concurrent at a point $ X$
Cause $ (PX,PA) \equiv (PC,PA) \equiv \frac {\pi}{3} \equiv (BV,AU) \equiv (WX,WA) \ (mod \ \pi)$ we have $ P,A,W,X$ are concyclic.
And so, $ (BV,PW) \equiv (WX,WP) \equiv (AX,AP) \equiv (AQ,AP) \equiv \frac { \minus{} \pi}{3} \ (mod \ \pi)$[/quote]
What do you meant by Pascal's theorem  since I know that Pascal just apply for hexagon which is inscribed in the circle! :( And I proved the second angle just by using inscribed quadrilaterals+add the measure of the chord! :) Anyway, please explain more clearly about how to use Pascal's theorem and your solution is extremely short+perfect!Thx for your help so much,PDKATSP! :)

Suppose that $ AQ$ cuts $ PC$ at $ X$
Apply Pascal's theorem with the hexagon $ CPTQAS$ we have $ X,V U'$ are collinear. From this and $ (*)$ we have $ X \in BV$, or $ PC,QA,BV$ are concurrent at a point $ X$  :)

[quote="ghjk"][quote="PDatK40SP"]$ O$ is the center of $ \Delta ABC$
$ QT$ cuts $ CS$ at $ U'$
Use the trigonometric form of Céva's theorem with $ \Delta SBT$ and the concurrence of:
+ $ SA,BV,TP$, we'll have a condition
+ $ SC,BU',TQ$, we'll have a condition
From these two conditions, we have $ \overrightarrow{BU'} \uparrow \uparrow \overrightarrow{BV}$, or $ B,V,U'$ are collinear    $ (*)$[/quote]
Sorry but how you can apply Ceva's theorem in triangle SBT even when SA,BV,TP belongs to triangle ATB(I meant concurrence),PDKATSP?I think we have nothing to do with triangle SBT,right? :) And I also misunderstand your idea about SC,BU' and TQ since the simillar reason as above![/quote]
Can you explain more details ? My English is so poor, so I can't understand your words  :oops:
Use Trig-Céva:
From $ SA,BV,TP$ are concurrent, we have:
$ \frac{ sin ( \overrightarrow{SA} , \overrightarrow{SB} ) }{sin (\overrightarrow{SA}, \overrightarrow{ST} ) }. \frac{ sin ( \overrightarrow{BV} , \overrightarrow{BT}) }{ sin (\overrightarrow{BV}, \overrightarrow{BS} )}. \frac{ sin ( \overrightarrow{TP} , \overrightarrow{TS} ) }{sin (\overrightarrow{TP}, \overrightarrow{TB} ) } \equal{} \minus{}1$
From $ SC,BU', TQ$ are concurrent, we have:
$ \frac{ sin ( \overrightarrow{SC} , \overrightarrow{SB} ) }{sin (\overrightarrow{SC}, \overrightarrow{ST} ) }. \frac{ sin ( \overrightarrow{BU'} , \overrightarrow{BT}) }{ sin (\overrightarrow{BU'}, \overrightarrow{BS} )}. \frac{ sin ( \overrightarrow{TQ} , \overrightarrow{TS} ) }{sin (\overrightarrow{TQ}, \overrightarrow{TB} ) } \equal{} \minus{}1$
Note that:
$ (\overrightarrow{SA}, \overrightarrow{SB} ) \equiv \minus{} (\overrightarrow{SC}, \overrightarrow{SB}
) \ \text{ (mod } 2 \pi \text{ ) } \\ (\overrightarrow{SA}, \overrightarrow{ST} ) \equiv  (\overrightarrow{TQ}, \overrightarrow{TB} ) \ \text{ (mod } 2 \pi \text{ ) } \\ (\overrightarrow{TP}, \overrightarrow{TS} ) \equiv \pi \minus{} (\overrightarrow{TQ}, \overrightarrow{TS} ) \ \text{ (mod } 2 \pi \text{ ) } \\ (\overrightarrow{TP}, \overrightarrow{TB} ) \equiv  (\overrightarrow{SC}, \overrightarrow{ST} ) \minus{} \pi \ \text{ (mod } 2 \pi \text{ ) }$
So $ \frac{ sin ( \overrightarrow{BV} , \overrightarrow{BT}) }{ sin (\overrightarrow{BV}, \overrightarrow{BS} )} \equal{} \frac{ sin ( \overrightarrow{BU'} , \overrightarrow{BT}) }{ sin (\overrightarrow{BU'}, \overrightarrow{BS} )}$, which means that $ \overrightarrow{BV} \uparrow \uparrow \overrightarrow{BU'}$

I still can't findthe suitable triangle for your equations above! :mad: And I think you mistyposince using trig-Ceva always give the result1, not -1! Please mention as clearas you can about:"Which triangle your equation applies for?".Pleasereply meas soon as you can.Thx so much for your help

No, the result $ 1$ only comes when you use normal angle, and it's not correct. If you use direct angle, the result will be $ - 1$
I use trig-Céva with $ \triangle{SBT}$
And I don't understand what "belong" means in your above post  :D
[quote="ghjk"]
Sorry but how you can apply Ceva's theorem in triangle SBT even when SA,BV,TP belongs to triangle ATB(I meant concurrence),PDKATSP?I think we have nothing to do with triangle SBT,right? :) And I also misunderstand your idea about SC,BU' and TQ since the simillar reason as above![/quote]


How many inches long is the path $ ABCDECA$ if $ \widehat{ABC}$ is a semicircle with a $ 3\frac13$ inch diameter and $ \widehat{CDE}$ is a semicircle with an $ 8\frac23$ inch diameter? Express your answer in terms of $ \pi$.
[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.9));
draw(Arc(origin,5/3,0,180)); draw(Arc((6,0),13/3,180,360));
draw((-5/3,0)--(31/3,0));
label("$A$",(-5/3,0),W); label("$B$",(0,5/3),N); label("$C$",(5/3,0),dir(45)); label("$D$",(6,-13/3),S); label("$E$",(31/3,0),E);[/asy]

Use the diameters to find circumference of each segment. ABC is 5/3 pi and CED is 13/3 pi. Then add the circumferences to the diameters so it should be 12 + 6pi


vector
function
algebra
polynomial
real analysis
real analysis unsolved

Consider the function $ F$ which transforms arbitrary $ (x_1, x_2, \ldots, x_n)$ from $ [0, 1]^n$ to the coefficients $ (c_1, c_2, \ldots, c_n)$ of polynomial
\[ f (x) \equal{} (x \minus{} x_1)(x \minus{} x_2)\cdot \ldots \cdot (x \minus{} x_n) \equal{} x^n \plus{} c_1 x^{n \minus{} 1} \plus{} \ldots \plus{} c_{n \minus{} 1} x \plus{} c_n.\]
Is this vector function $ F$ continuous?

Why wouldn't it be?  Every component is polynomial in every variable.


[quote="pvthuan"]Determine the greatest value of $k$ such that the following inequality holds for positive $a,b,c,d$ \[\frac{ab}c+\frac{bc}d+\frac{cd}a+\frac{da}b\geq a^{k}+b^{k}+c^{k}+d^{k}.\] [/quote]
If $k>1$ then your inequality is wrong: $a=b=c=d$ and $a>1.$
If $k<1$ then it is wrong for $a=b=c=d$ and $0<a<1.$
If $k=1$ then it is wrong for $a=91,$ $b=3,$ $c=5$ and $d=1.$ :)

I am sorry. I missed the condition $a+b+c+d=4$.

Determine the greatest value of $k$ such that the following inequality holds for positive $a,b,c,d$ such that $a+b+c+d=4$ \[\frac{ab}c+\frac{bc}d+\frac{cd}a+\frac{da}b\geq a^{k}+b^{k}+c^{k}+d^{k}.\]

Let $a=c=2-x, b=d=x$ and $x\to 0$, then we will have $k\leq 1$

But $k=1$ is not correct, because of this case $27(a,b,c,d)=(100,2,5,1)$


One thing we seem to be lacking in the mock contest area is a mock AMC 8.  With that in mind, I have compiled a mock AMC 8 from old problems and newly written ones.

There is no narrow window of time that you must meet, as I believe that causes more inconvenience than benefit.  Try and take it in the standard time limit (40 minutes), but I'll leave it up to you to practice with it in whatever way you think will help best.  If you do take it in the time limit, send me your answers and I'll keep a scoreboard.

So without further adieu, here is the exam.  Remember that the standard rules are:

- 40 minutes
- non-QWERTY calculators are allowed
- scoring is the number of questions right
- a large print edition is available for disabled students*

Enjoy!  If you're curious, problem sources are listed in the TeX file.

[size=75]*Instructions: in IE, go to View --> Text Size --> Largest; in Mozilla, press [CTRL][+][/size]

[Edit]: Fixed two problems.

[Edit 2]: Actual fix.

I have a new incentive for those in 8th grade or less taking this test:  whoever scores the highest will get a gmail invite from me.  Good luck!

Oops!  There were two mistakes in the first version of the test (#16 and #25).   Thanks to soilent green for pointing them out.  I guess this complicates the scoring process a little - if you've already taken the test, give yourself a few minutes to do #16 and #25 again.  You can still submit your answers.

And if you downloaded the test before today (6:00 PST), make sure to redownload it to get the corrected version.

Oops again.  Everything should be better now.  Redownload again.  Sorry.

I'm confused, what's the practice AMC for?

No, Why is there a practice AMC8 and a regualr one?  Couldn't the second be called AMC8B instead?

And who should we send the answers to for joml88's contest?

[quote="h_s_potter2002"]And who should we send the answers to for joml88's contest?[/quote]

[url=http://www.artofproblemsolving.com/Forum/privmsg.php?mode=post]Me.[/url]

It's not my contest. It is Mathfanatic's.  So please don't send them to me.  I am just offering a gmail invite to whoever does the best and is in <= 8th grade.

[quote="h_s_potter2002"]No, Why is there a practice AMC8 and a regualr one?  Couldn't the second be called AMC8B instead?
[/quote]

They're the same, just one is a .tex file with pictures and the other is a PDF

[quote="mathfanatic"]Sorry about the confusion.  RC-7th is right; the two files are for the same test, but in different formats.

Please send your answers in by this coming friday.  Only three people have responded so far.  (There must be more than 3 middle-schoolers here, right?!)[/quote]

Um... are you implying elementary-schoolers can't participate in the Gmail contest :?  :(

Anyone in 8th or under.  I don't see many elementary school students floating around here.  If there are they should fell free to entry if they want to.

Is the contest over now or has the deadline been extended?

What is the deadline, I can't seem to find it. :blush:

I don't think there is a deadline.  I don't think there was meant to be one too, but since joml88 brought up the contest idea, either him or mathfanatic should make one.

Uh, the deadline is...now.  It was supposed to be last friday, as I mentioned in the previous post, but I was both lenient and lazy.

You can still send your answers in...but I won't accept them.  (Insert evil laugh.)

Here are the final answers: (confirmed means I checked the original problem listing to make sure the answer was correct; unconfirmed means, well, that I didn't.)

1. B confirmed
2. E confirmed
3. C confirmed
4. C confirmed
5. B confirmed
6. C confirmed
7. D confirmed
8. B confirmed
9. E unconfirmed, but checked
10. B confirmed
11. B confirmed
12. C confirmed
13. E confirmed
14. E confirmed
15. C confirmed
16. E uncomfirmed, but checked
17. B unconfirmed, but checked
18. C confirmed
19. C unconfirmed, but checked
20. D confirmed
21. D confirmed
22. C confirmed
23. B confirmed
24. E confirmed
25. E

I'll have a scoreboard up in a moment.

Alright, the final results are in:

RC-7th won with a score of 24/25.  soilent green and h_s_potter2002 tied for second with a score of 23/25.  No other middle-schoolers (or below) entered.

Only 3 people entered, and all of them basically got the same score, with a 1 point difference.  That's really suprising.

it's really great of u(mathfanatic) to compile the problems

I think it's really a good idea..

I've time traveled ten years into the future! :)

Anyway. Am I allowed to do the test?

Sure, you did above.

You can do what ever you want.

Beware it might be a little easier since the earlier tests are generally easier than the 2010- tests.


L is on a fixed line which is perpendicular on MN ?!?!?!?!?  :D  :roll:

I corrected: $not\ A\in BC$. I am sorry. The locus is not perpendicular on a mobile line, by all means !!!???!!!

[quote="Virgil Nicula"]Let the line $BC$ and the points $D,E,F\in BC,\ A\not \in BC$. A mobile line which pass through $D$ cuts the lines $AB$, $AC$ in the points $M$, $N$ respectively. Ascertain the geometrical locus of the point $L\in ME\cap NF.$[/quote]

This problem didn't solved two months !

Let next mobile line which pass through $D$ cuts the lines $AB,AC$ in the points
$M',N'$ respectively and $L'\in{M'E}\cap{N'F}$.

We know
[color=red]Dezarg’s theorem:[/color]If $ABC$ and $A'B'C'$ be two triangle in the plene that
$A_1\in{A'B'}\cap{AB},B_1\in{B'C'}\cap{BC},C_1\in{CA}\cap{C'A'}$and $A_1,B_1,C_1$  lie in same line $\Rightarrow$ ,$AA',BB'$and $CC'$ intersecting in same point.

We see that $D\in{N'M'}\cap{MN},E\in{M'L'}\cap{ML},F\in{NL}\cap{N'L'}$ but $D,E,F$lie in $AB$=>
Line$MM',NN'$and$LL'$intersecting in same point ,but $A\in {MM'}\cap{NN'}$=>
$A\in{LL'}$=>
That locus is a line pass through $A$ ;)


Does there exist any a Enumerable set A, that for any non-trivil decidable set C, $ A\bigcap C\neq \Phi$. Non-trivil set B means both B and $ B^c$ are infinite.

Yes, it does. Every [i]simple set[/i] ( http://en.wikipedia.org/wiki/Simple_set ) has a non-empty intersection with every r.e. infinite set and, in particular, with every infinite decidable set.

Here is an example of a simple set. Let $ T_i$ be a Turing machine that enumerates the elements of the $ i$-th recursively enumerable set (and the numbering of $ T_i$ is effective). Then
\[ \{x: x \text{ is the first number greater than } 2i \text{ that is output by } T_i \}\]
is a simple set.

P.S. I assume that you are looking for a non-trivial set $ A$ and that $ \Phi \equal{} \varnothing$.

Thanks.
Well, how to enumerate the set you propose? It seems to require $ A_{TM}$.

What is $ A_{TM}$?

Here is how we enumerate the set. Run all machines $ T_i$ simultaneously [hide="(how?)."] First, execute one step of program $ T_1$. Then execute one step of $ T_1$ and one step of $ T_2$, then one step of $ T_1$, $ T_2$, and $ T_3$ etc. [/hide] Once $ T_i$ prints a number $ x$ greater than $ 2i$ output $ x$ (i.e., add it to the set) and stop $ T_i$.

It may (and will) happen that some $ T_i$-s don't print any numbers greater than $ 2i$ -- we will run those $ T_i$-s forever. But this is OK, since those $ T_i$-s enumerate [i]finite[/i] sets.

$ A_{TM}$ is the acceptance problem for TMs.  And mathiq probably was missing the "run all machines simultaneously" part.

Thanks a lot. I mistake "output" by "accept" that's why I can't construct an example.


Have they posted the mathletes who have placed in the top 30 or so yet in the 2009 Mathcounts Nationals? Just out of curiosity.

i don't think so... i was on the MathCounts website today, because i wanted to know the top 57... nothing so far. They should have it up pretty soon, though.

Here are the results you are looking for.

frost

What?? Darryl Wu got 33th????????? How did that happen?  :o  :o  :o

[quote="nikeballa96"]Last years top three written were Bobby, Jason Hyun, and Darryl Wu. Two didn't get top 30 this year; the other won.

Mathcounts is so easy to mess up.[/quote]

I think people improved and some people didn't.

It's okay Darryl. You have next year.

[quote="echen"]Like it has been said in a lot of places, one competition cannot judge all the competitors perfectly. Some people have really good days and some people really bad. We cannot say that whoever got 1st place in mathcounts is definitely the best and whoever is 70th is definitely 70th. Unfortunately, we will never have a competition that can place everyone perfectly in order consistently.[/quote]

Well, you could run the competition about 50 times, and you would get much more accurate results, but that's not practical.

Mieh, /agree that people have bad days and good days. There was a quote from JBoyd in some other thread that I can't currently find about this...

[quote="nikeballa96"]my teammate got 59th.  :P[/quote]

lol that was me

[quote="echen"]Like it has been said in a lot of places, one competition cannot judge all the competitors perfectly. Some people have really good days and some people really bad. We cannot say that whoever got 1st place in mathcounts is definitely the best and whoever is 70th is definitely 70th. Unfortunately, we will never have a competition that can place everyone perfectly in order consistently.[/quote]

Yes; for example, I had a very good day and got a far higher score than I normally would in practice, just like I did on the AIME.

[quote]does anyone know the cut off this year for top.... 58?[/quote]

I know 57th was 18/12; presumably 58th was the same breakdown.

Also what's up with 58 and not 57? And 22 and not 20?

"Yes; for example, I had a very good day and got a far higher score than I normally would in practice, just like I did on the AIME. "

Eh, there are people who do better on competitions than practices. This is a very useful skill.

For the past two years, John Zhang has been from Washington.
2008-#38
[url]http://mathcounts.org/Document.Doc?id=221[/url]
2009-Last name on page 8
[url]http://mathcounts.org/Document.Doc?id=341[/url]

Perhaps you didn't read the previous posts, all they wanted to say was that MATHCOUNTS placed him in MI for
https://mathcounts.org/Page.aspx?pid=1336

I met him once-upon-a-long-time-ago...pretty funny person.

Haha yeah, I know he's from WA. That's why I put the rofl's there, 'cause they messed up.  :rotfl:  :rotfl:

Hope John Zhang doesn't see that. Michigan is getting credit. LOL.

Well, they occasionally make mistakes, last year they had one person's name in the yearbook twice

Didn't they put one of the CD's from Maryland last year in 7th grade when he was really in 8th grade? Also, didn't they spell Delaware wrong?

[quote="gauss1181"]Last year, they also spelled Wisconsin wrong.[/quote]

Let me guess, they spelled like Wicsoncin?      :rotfl:

lol, I was that 58th person.  I'm lucky to even make it on that list.  I epicly failed because I didn't study the week of the nats.

Hm...apparently the top 58 posted by someone on the first page of this thread (eg. the one shown at nats) and the one on the website is different.

they forgot to put someone on the list...Edward Park of GA at 41st place.

[quote="Urc"]

they forgot to put someone on the list...Edward Park of GA at 41st place.[/quote]

lol...you just made Edward Park famous...

[quote="PowerOfPi"][quote="Urc"][quote="PowerOfPi"]It is interesting how the one on the MathCounts website and the one on the first page of this thread are exactly the same except for $ 41^{st}$ place down and at the bottom, the newer one says "modified 5/18/2009 - Georgia/Park"[/quote]

oh snap, i didn't see that..[/quote]
I didn't say you were wrong. I meant that you were right, but MathCounts made the mistake.[/quote]

Yeah, but it's just that I took the time to actually find who was missing, when I could have just scrolled down...


inequalities
trigonometry
complex analysis
complex analysis unsolved

Let  $ a > 0\ \text{and}\ k > 1$. Prove that
\[ |\, \text{arg}\ z|\leq 2\arccos\frac {1}{k}\Rightarrow |\, a \plus{} z\,|\geq \frac {a \plus{} |\,z\,|}{k}\]

So $ k>1$ implies that $ 0<1/k<1$ so we have $ \cos^{\minus{}1}$ well defined. Also $ \cos^{\minus{}1}$ is in $ 0,\frac{\pi}{2}$ so $ |arg z|<\pi$.

Note that cosine is decreasing on $ 0,\pi$, so if we take cosine of both sides, we reverse the inequality. Let $ \theta$ be the argument of $ z$. We can WLOG say that that argument is $ >0$ because the inequality is unchanged under conjugation.

So $ \cos\theta\ge \cos 2\cos^{\minus{}1}\frac{1}{k}\equal{}\frac{2}{k^2}\minus{}1$.

Now in the inequality that we must prove, if we square and expand it, we get

$ a^2\plus{}|z|^2\plus{}2a(z\plus{}\overline{z})\ge \frac{a^2\plus{}|z|^2\plus{}2a|z|}{k^2}$

Now observe that $ z\plus{}\overline{z}\equal{}2|z|\cos\theta$. Now for the inequality to hold, it is strongest when $ \cos\theta$ is minimized because $ a,|z|$ are independent of $ \theta$. So the inequality that we must prove is true iff the following is true:

$ a^2\plus{}b^2\plus{}\frac{4ab}{k^2}\minus{}2ab\minus{}\frac{a^2\plus{}b^2\plus{}2ab}{k^2}\ge 0$
where $ b\equal{}|z|$.

Then we rewrite this as

$ (a\minus{}b)^2\plus{}\frac{4ab}{k^2}\minus{}\frac{(a\minus{}b)^2\plus{}4ab}{k^2}\ge 0$
$ (a\minus{}b)^2(1\minus{}1/k^2)\ge 0$ which is true YAY


Zé Roberto and Humberto are playing the Millenium Game!
There are 30 empty boxes in a queue, and each box have a capacity of one blue stome.
Each player takes a blue stone and places it in a box (and it is a [i]move[/i]).
The winner is who, in its move, obtain three full consecutive boxes.

If Zé Roberto is the first player, who has the winner strategy?

what about playing symetrically with respect to the line between the $15$th and $16$th box? I think it gives a winning strategy for player 2!

Unless the first player decides to start by putting a stone in the $15$'th box :).

well and then what if first player starts in the 15th stone and then plays simetrically unless he gets a winning position? :D

Maybe I am wrong, but I think that the number of boxes is even, so by putting a stone in the 15th box, you can't play simetrically afterwards   :?
I actually have an idea that might lead to the solution:
Let us notice that any stone in the $3\ldots 28$ range deletes 5 boxes where stones can be put (otherwhise, the player that puts a stone there loses)...A similar calculation for the other stones.....Plus some cases....Some stones eliminate other specific places...etc....I think that with a small effort (which I am not ready to make :oops: ) someone could notice the solution....Anyway....this all might be wrong :(

Ok, I think that this problem can actually be solved with an extension of my idea.....
So....consider the sequence $a_n$, so that $a_n=1$ if for $n$ stones, the $1^{st}$ player wins and 0 otherwise.
Now, for small values we calculate it by hand.
To find out what is $a_n$ if we know $a_i,\any i <n$, we must notice that if there exists a $k$ so that $a_k=0$ and $a_n-k-5=0$, then $a_n=1$.
I hope it is clear that the recurence came from putting the stone in position $k+2$ and considering the game in the other halves.......
Ok, we would have to make about $450$ elementary operations to find out $a_{30}$ but I think it is "achievable" :D
I hope it is a correct idea....

"simetrically" because if player $B$ plays in square $16$ we have wining position for us before or after that ;)  So it is like a forbidden square for player $B$ and we therefore dont use it unless it becomes a winning square.

Ok, maybe I don't understand something, but wouldn't it follow that the first player always wins (at least for even $n$).
But for $n=6$ he looses, I think.....

I never said it worked! I was just explaining what i thought could work, and BTW I just found it doesnt work at all...

This version of the game might be easier to think about:

Zé Roberto and Humberto alternately place 1x3 blocks on a 1x32 board, so that they don't overlap; Zé Roberto goes first and the player who can't place a block loses. It's easy to see why this is analogous to the original game. Replacing $n$ by $32$, we can check looking at this that Humberto wins for $n=1, 2$, Zé Roberto wins for $n=3, 4, ..., 7$, Humberto wins for $n=8$ and Zé Roberto wins for $n=9, 10, ..., 13$.

Now, if Humberto wins for the 1xn board Zé Roberto wins for 1x(n+3), 1x(n+4) and 1x(n+5), but there doesn't seem to be a similar strategy for Humberto. Also if we want to find something about 1x(n+r) by looking at 1xn, with r>5, it's harder because when the board is divided into two parts, it's not obvious at all (at least to me) that the best strategies for the players are the same as they would be if they were playing on each part separately.

Actually, this is what I have been trying to say with that recurence and $a_n$, but the problem is the block are not $1\times 3$ but $1\times 5$. And also we should take care of the "edge" situations....Again, to find the answer, we should make about 450 operations, I think......

I computedthe nim numbers of the games for $n$ up to $30$, and, if there's no mistake, I think the second player should win. I couldn't find any pattern though :(. I think that for numbers up to $30$, the second player wins when $n=6,12,22,30$ (and $0$, of course :)).

Hi!
I am sorry if I am going to say something wrong, but I think $12$ can't be a winning position for the $2^{nd}$ player, since the $1^{st}$ puts a stone in box nr.$3$, so he has "eaten" 5 boxes.....And there are only 7 left and obviously the player at move for $7$ loses.....(Btw, should $7$ be in your list) ??????
Sorry if I'm wrong :oops:

Whenever the number of boxes is odd, the first player puts a stone in the middle, and then playes symmetrically, so the first player wins when there's an odd number of bozes. In particular, the first player ("the player at move", as you call him :)) wins when there are $7$ boxes.

The blocks are actually 1x3, not 1x5. If box $k$ has a stone, the player to move can't place a stone in $k+1, k+2, k-1, k-2$. Indeed, if we use 1x3 non-overlapping blocks, if the center square of a block is $k$ it takes $k-1, k, k+1$; then the player to move can't place a block with its center there, nor can he in $k+2, k-2$ as the ends would overlap. 1x5 non-overlapping blocks would imply that the player to move can't play any of $k+1, k+2, k+3, k+4, k-1, k-2, k-3, k-4$ which is obvious overkill... ;)

What do you think about this ?

Let's note a "losing" box such a box that if one player puts a stone into it, then the next player can make a winning move. Let's say a box "cuts out k boxes" if after having put a stone into that box, there are k-1 losing boxes, so altogether k new boxes where one can not put a stone.
Now, when the first player puts a stone into a box, which cuts out an odd number k boxes, then as we assume that both are playing to win, we can "eliminate" these k boxes. So we are left an odd number of boxes and obiviously the second player who starts to play these boxes, has a winning strategy. Hence, the first player needs to cut out an even number k boxes. The only possibility how the first player can cut out even number of boxes is putting a stone into the second or 29. box. But now the second player can do this for the other 26 non-losing boxes (as we can again eliminate the first or last 4 boxes), so as none of the players wants to cut out odd number of boxes the result comes down to who is winning the came when there are 6 boxes. It is clear that then the second player is winner. So 14,22,30 are also winning standings for the second player.


$ \text{Prove that there are no integers } m,n\ge 1 \text{ such that}$

\begin{eqnarray}
\sqrt{m+\sqrt{m+\sqrt{m+...+\sqrt{m}}}}=n \nonumber
\end{eqnarray}

$ \text{where there are 2006 square root signs.}$

[hide]If we repeatedly square and rearrange, we have
\[ \sqrt {m \plus{} \sqrt {m}} \equal{} \left(\left(n^2 \minus{} m\right)^2 \minus{} m\right)^2 \minus{} m \cdots.
\]
To simplify, we know the RHS is an integer, so let it be $ k$. Then
\[ \sqrt {m \plus{} \sqrt {m}} \equal{} k \implies k^2 \equal{} m \plus{} \sqrt {m}.
\]
Then $ \sqrt {m}$ is an integer. Let $ \sqrt {m} \equal{} x$. Then we have
\[ k^2 \equal{} m \plus{} \sqrt {m} \implies k^2 \equal{} x^2 \plus{} x,
\]
and then the RHS must be a perfect square. However,
\[ x^2 < x^2 \plus{} x < \left(x \plus{} 1\right)^2 \equal{} x^2 \plus{} 2x \plus{} 1,
\]
so the RHS cannot be a perfect square. By contradiction, we have proved that $ m$ and $ n$ cannot both be integers.

So I guess that part about the 2006 square root signs was just to make the problem-solver think more?[/hide]

:) Perfect solution.
Yeah, the 2006 square roots is not really relevant, just keep opening up the square roots until you reach the contradiction. I mean, it's the year, and usually such random numbers don't really contribute to the problem mathematically.  :P


[quote="Catalan"]There is no common difference or common ratio.[/quote]

And so the sequence in question is not arithmetic nor geometric.

[quote="catalan"]Hello there,

For the following two sequences, I would like to know if I can find the general term by using the formulas for Arithmetic and Geometric Sequences.

Thank you very much!

---

1. $ 1, \frac {1}{4}, \frac {1}{9}, \frac {1}{16} ...$

$ a_n \equal{} \frac {1}{n^2}$

The general term is rather obvious, but how would I find it using the formulas for Arithmetic and Geometric Sequences? There is no common difference or common ratio.

2. $ \frac {2}{1 \times 3}, \frac {4}{3 \times 5}, \frac {6}{5 \times 7} ...$

The general term is more difficult to see in one shot. How could I find it using the formulas?[/quote]
don't even attempt to solve the first one using geometric terms/arithmetic series or any such formulas. let me tell you the answer right now, as it was proven by euler, because using your methods you won't be able to get the answer
$ \frac {1}{1^2} \plus{} \frac{1}{2^2}... \equal{} \frac {\pi^2}{6}$
i bet your thoroughly confused, but the proof requires some calculus :-)

The question never asked to sum the terms up, he only asked how the $ n^{th}$ term would look like, and as Carcul said, for the first one, it's not either an arithmetic or geometric sequence.

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

Just try looking for similarities in the sequence. I don't think you can do it using formulas for geometric or arithmetic sequences as neither are a geometric or arithmetic sequence.

I would break 2) up into three different arithmetic progressions.  The way the progression is written, we can clearly write

$ a_n \equal{} \frac{b_n}{c_n \times d_n}$

where

- $ b_1 \equal{} 2, b_2 \equal{} 4, b_3 \equal{} 6, ... b_n \equal{} 2n$,
- $ c_1 \equal{} 1, c_2 \equal{} 3, c_3 \equal{} 5, ... c_n \equal{} 2n\minus{}1$,
- $ d_1 \equal{} 3, d_2 \equal{} 5, d_3 \equal{} 7, ... d_n \equal{} 2n\plus{}1$,

and it should be fairly obvious by inspection that these are all arithmetic sequences.  This is the kind of reasoning we need to use to analyze more exotic sequences.  There is [b]no foolproof method[/b] for finding the general term of a sequence (in fact, an infinite number of "sequences" can be constructed with any set of starting terms); almost by definition, the problem of finding the general term of a sequence is a matter of guesswork and an intuition of the problem-creator's intent.


This terrible thing :D is equivalent to:
[tex]\frac{a^8+b^8+c^8}{a^2 b^2 c^2}+\frac{4((p-a)^8+(p-b)^8+(p-c)^8)}{(p-a)^2(p-a)^2(p-a)^2}\geq p^2+5\frac{S^2}{p^2}+5\frac{abc}p[/tex]
If we take x=(p-a)^2 ... then
[tex]\frac{4(x^4+y^4+z^4)}{xyz}\geq 4(x+y+z)\geq\frac43 p^2[/tex],
[tex]\frac{a^8+b^8+c^8}{a^2 b^2 c^2}\geq\frac43 p^2[/tex].
It is obvious that  40p^2/27>=5abc/p  and  5p^2/27>=5S^2/p^2=5(p-a)(p-b)(p-c)/p.
Summing we get the result.

[quote="harazi"]Bleah![/quote]

what is this?  :?  :?

I get it!!, why harazi made that interjection, he is jealous that he can't solve it !!!!!!!!!!!!  :cool:

You didn't deduce he right thing. Come on, I could solve this problem!

Harazi, really, don't bother to answer my message, because I was angry as Monaco lost last night in the Champions League and I picked on you, just a wave of madness!


inequalities
AMC
USA(J)MO
USAMO
induction
geometry
geometric transformation

Reading through the following paper has made me realize extactly how much I suck at inequalities. While I'm great at stuffing pigeons into holes, inequalities fry my brain with their giant number of terms. So while I'm off learning schur, why don't you take at look at the following mock contest?

http://www.tjhsst.edu/~tmildorf/math/2004TJUSAMO3.pdf

[hide="#0"]For any real $x$ \[x^{4}-x^{2}-2x+2\ge 0\] [hide="solution"]$x^{4}-x^{2}-2x+2\ge 0$
$\Leftrightarrow (x-1)^{2}(x^{2}+2x+2)\ge 0$
$\Leftrightarrow (x-1)^{2}((x+1)^{2}+1)\ge 0$

which is true by the Trivial Inequality[/hide][/hide]

Most of them get easier when you know rearrangement. I'd take a look at that first.

could you prove number 1 by induction as well?

[quote="Peter VDD"][quote="PenguinIntegral"]How so?[/quote]I think the most well-known is number 2 (=the third one), where you can rearrange the denominator: $a^{3}+b^{3}\ge a^{2}b+b^{2}a$, but then the whole LHS reduces to $\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$.[/quote]

Oh, very nice! I guess I'd need to practice it to get the hang of finding places to use it, but it looks like a very powerful tool.

I don't think the document lists it in the front, so I'm guessing there's another way to do it. Suggestions?

We have answers for the first three? Anyone want to type up more, or should I go searching through kalva?

No, you should solve them, not lookup the answers. And if you can't solve them, start with a set of easier problems first.

It's your right to do what you want, but that's the way you'll learn it best. Get a place that gives the theorems and a range of exercices from trivial to USAMO level. You will see the ideas are not so strange or genius as you probably think they are.

Remember that you learn a lot more from seriously attempting a problem and failing than just looking at the solution.

[quote="Peter VDD"]Get a place that gives the theorems and a range of exercices from trivial to USAMO level. [/quote]

Could you suggest a place? I can't seem to find one, so I thought it would be best to learn through example. :blush:

[quote="Phelpedo"]The IMO compendium:

http://www.imo.org.yu/index.php?options=mbb|Bfa|f|Alg|Ine&p=ww200[/quote]
How does that give a range of difficulty?

Old IMOs are easier. Very old IMOs are much easier. To the extent that I can do some of the problems. :D

edit: sorry, I didn't look at the page. That particular one might not help so much. Try the resources on this site. (Old IMOs.)

I don't know about a good English place that's free at the same time.

Maybe this helps: it's the first chapters of a document I'm writing. It's completely in dutch, but I guess 90% of the problems are readable in english, as it's mainly mathematical language. The theory explainings are not readable, but I assume you already know what a permutation is and what $\sum$ means and such.

Chapter 1 = sums of squares
Chapter 2 = rearrangement
Chapter 3 = AM-GM (and alikes) & Cauchy

[hide="short list of useful translations (even though you do not really need them)"] Zij ... = Let ...
Toon aan dat = Show that
Bewijs dat = Prove that
Vind = Find
maximale waarde = maximal value
minimale waarde = minimal value
de waarde van = the value of
gelijkheid = equality
als en slechts als = iff
Wanneer (treedt) gelijkheid (op)? = When (is) equality (attained)?[/hide]

Ok the URL got screwed up somehow...there is a page under Theorems&Concepts that gives theorems.

I have a tex file with all imo problems from 1959-2003...if you need that

[quote="Phelpedo"]:w00t: Are you serious?[/quote]

Ever heard of [url=http://www.kalva.demon.co.uk]Kalva[/url]?
This is the URL to the .tex-file:
http://www.kalva.demon.co.uk/imo/tex/imo-all.tex

yeah downloaded it from kalva long ago...

[quote="pianoforte"][hide="#0"]$(x^{2}-1)^{2}=x^{4}-2x^{2}+1\ge0$ and
$(x-1)^{2}=x^{2}-2x+1\ge0$, from Trivial inequality.
Adding the two gives $x^{4}-x^{2}-2x+2\ge0$.[/hide][/quote]
yay SoS thats how i did it

[quote="Peter VDD"][quote="PenguinIntegral"]I don't think the document lists it in the front, so I'm guessing there's another way to do it. Suggestions?[/quote]You don't need the theorem, it's just the idea that helps you finding this. You can also find it by $a^{3}+b^{3}\ge a^{2}b+b^{2}a\Leftrightarrow (a+b)(a-b)^{2}\ge0$, but the idea is just impossible to come up with in that way.[/quote]
[hide="Here is another easy solution without that idea:"]
$c\geq b\geq a$. It's acute so (by law of cosines) $a^{2}+b^{2}>c^{2}\implies a^{2}+b^{2}+c^{2}>2c^{2}$. Now triangle inequality: $a+b>c\implies a+b+c>2c$. Multiply those: $(a+b+c)(a^{2}+b^{2}+c^{2})>4c^{3}$. It remains to prove that $4c^{3}(a^{3}+b^{3}+c^{3})\geq 4(a^{6}+b^{6}+c^{6}) \text{ or }(ac)^{3}+(bc)^{3}\geq a^{6}+b^{6}$ .
But we know that $c\geq b,a$ so $(ac)^{3}\geq a^{6}, (bc)^{3}\geq b^{6}$, and we're done... no need for any rearrangement ideas![/hide]

one more thing:
[quote="shobber"]Problem 2:
$\frac{1}{(a+b)(a^{2}-ab+b^{2})+abc}=\frac{1}{(a+b)((a-b)^{2}+ab)+abc}\leq$

$\leq \frac{1}{(a+b)ab+abc}= \frac{1}{ab(a+b+c)}$.

Hence take sum we get:

$\sum\frac{1}{a^{3}+b^{3}+abc}\leq \frac{1}{a+b+c}\cdot \sum\frac{1}{bc}= \frac{1}{a+b+c}\cdot \frac{a+b+c}{abc}= \frac{1}{abc}$.[/quote]
Shobber, is that a fairly common ineqaulity in the first step or did you notice that after working backwards?

Shobber used the same rearrangement idea, and used a simple calculation to justify it without mentioning rearrangement (and probably without thinking of it as rearrangement).

How's the mildorf packet in conjunction with AoPS 2 and the learning inequalities thread? Will it give me enough practice problems with the harder inequalities?


Prove the uniqueness part of the fundamental theorem of arithmetic without using the theorem that if a prime number p divides ab, then it divides a or b.

Suppose there are two different representations of the same number:
$ k \equal{} p_{1} p_{2} p_{3}... p_n \equal{} q_{1} q_{2} q_{3}... q_{m}$,
where $ p_{1} < p_{2} < ...p_{n}$ and $ q_{1} < q_{2} < ... < q_{m}$
and let k be the MINIMAL of such numbers.

If $ p_{x} \equal{} q_{y}$, then we can divide by it on both sides, contradicting minimality.

Let $ p_{1} < q_{1}$.

Then let

$ q_{i} \equal{} a_{i}p_{1} \plus{} r_{i}$, where $ 1\leq i \leq m$ and $ 0 \leq r < p$

Then,

$ q_{1} q_{2} q_{3}... q_{m} \equal{} bp_{1} \plus{} r_{1} r_{2} r_{3}... r_{m} \equal{} n$

It follows that $ r_{1} r_{2} r_{3}... r_{m}$ is a multiple of $ p_{1}$, but that means that
$ r_{1} r_{2} r_{3}... r_{m} \equal{} 0$ due to the restriction we set ($ 0 \leq r < p$)

However, this cannot be the case:

Let $ r_{c} \equal{} 0$; then:
$ q_{c} \equal{} p_{1}d$
But since $ q_{c}$ is prime, $ d \equal{} 1$, so
$ q_{c} \equal{} p_{1}$. Contradiction.

Therefore there is a unique representation.


Find the best constant $k$ such that the following inequality satisfy for all $x,y,z \ge 0$ and $x+y+z=3$ :

$x^4+y^4+x^4 \ge 3xyz + k(x-y)(y-z)(z-x)$

_____________

Edit.  :(  Oh, sorry. I post it in wrong subforum, because I had solution.

[quote="arqady"]$k_{max}=\sqrt8$ :)[/quote]
I think you may have a small mistake when mutli $x+y+z$.I find $k_{max}=3 \sqrt8$ :lol:
with a very ugly solution.

You are right, Zhaobin: $-3\sqrt8\leq k\leq3\sqrt8.$

Any solution here?

My solution only uses AM-GM.

Ok. Here 's the solution.

Notice that the inequality's only solve in case $x \ge y \ge z=0$. Then :

\[ x^4+y^4=(x^2-y^2)+2x^2y^2 \ge 2\sqrt{2}xy(x^2-y^2)=6\sqrt{2}(x-y)(y-z)(z-x) \]

The equal takes if :

$z=0, x^2-y^2=\sqrt{2}xy$ and $x+y+z=3$.

[quote="hungkhtn"]Ok. Here 's the solution.

Notice that the inequality's only solve in case $x \ge y \ge z=0$.
.[/quote]
Why??

[quote="hungkhtn"]Now, You decresed $x,y,z$ by a number $t\le z$.[/quote]
But then you lose the condition $x+y+z=3$.

No,spanferkel. I meant that the expression

$f(a,b,c)=\frac{a^4+b^4+c^4}{a+b+c}-3abc$

Decreased when we decrease all $a,b,c$ by $t$. You can check it.


Derive the formula for the roots of $P(x)=ax^{2}+bx+c$.

[hide="Solution(Standard, Textbook Style)"]
$ax^{2}+bx+c = 0$
$ax^{2}+bx =-c$
$x^{2}+\frac{b}{a}x =-\frac{c}{a}$
Complete the square:
$x^{2}+\frac{b}{a}x+\left(\frac{b}{2a}\right)^{2}=-\frac{c}{a}+\left(\frac{b}{2a}\right)^{2}$
$(x+\frac{b}{2a})^{2}=-\frac{c}{a}+\left(\frac{b}{2a}\right)^{2}$
$(x+\frac{b}{2a})^{2}=-\frac{4ac}{4a^{2}}+\frac{b^{2}}{4a^{2}}$
$(x+\frac{b}{2a})^{2}= \frac{b^{2}-4ac}{4a^{2}}$
Take the square root of both sides:
$(x+\frac{b}{2a}) = \frac{\pm\sqrt{b^{2}-4ac}}{2a}$
Subtract $\frac{b}{2a}$ from both sides:
$x = \frac{-b \pm\sqrt{{b^{2}-4ac}}}{2a}$
[/hide]


Heres a problem i always had trouble with. Can anyone solve it?

6. An ammonia and water mixture fills a five-gallon container. Eighty percent of the mixture is ammonia but some of the mixture will be drained and replaced with pure water. If a five-gallon mixture of fifty percent ammonia is desired, how many quarts of the mixture need to be drained before the water is added? (A quart is one-fourth of a gallon.) Express your answer as a decimal to the nearest tenth.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=303690642&t=193449
Please search before posting.

also note that the solutions for the 2008 school, chapter, and state rounds is up on the mathcounts site.

It is also available on the Minnesota MathCounts site, along with 2006 and 2007.


This is based off an AoPS problem, which I don't completely understand.

If there are six people A, B, C, D, E, and F and six chairs. If A can't sit in chair A, B can't sit in chair B, ..., F can't sit in chair F. How many seating arrangements are there?

I understand the answer, but I don't really have a good intuitive grasp as to why you can't just say that the number of seating arrangements is 5 x 5 x 4 x 3 x 2 x 1. For chair A, you have 5 choices. For chair B, you have 5 choices, etc.

It has to do with the fact that A can't sit in A, not only that there are 5 choices.  I think.  I don't know exactly how to do it though.

Because those options you wrote down count impossible possiblilities.   In particular, after A sits down, there probably won't be 5 seats available to B.  If A sits in seat C, then B can sit in seats A, D, E, or F -- only 4 possibilities.  There are only 5 choices for B if A sits in seat B.

Also, if A sits in B, B in A, C in D, D in E, and E in C, that leaves F nowhere to sit at all -- another possibility that your expression counts.

this is just derangements right?  Which is just the result of applying Principle of inclusion-exclusion.

It's really easy to say "apply inclusion-exclusion."  Show a solution :).

well once you do it once do you really want to do it every time?

|A| is # of arrangements with A in his spot, |A^B| with A and B in their spots, etc.  Plus because of symmetry, |A|=|B|=...=|F|, |A^B|=|A^C|, etc.

6!= |A|+|B|+|C|+...|F| - (|A^B|+|A^C|+...|E^F|)+taken 3 at a time - taken 4 at a time + taken 5 at a time - 6 taken at a time + (the choices where none are in their right spot)

6!= 6*5!-C(6,2)4!+C(6,3)*3!-C(6,4)*2!+C(6,5)-1+(the choices where none are in their right spot)

(the choices where none are in their right spot) = 6! -  6*5!-C(6,2)4!+C(6,3)*3!-C(6,4)*2!+C(6,5)-1 which we can play with to get  6!(1-1/1!+1/2!-1/3!+1/4!-1/5!+1/6!)

.

When someone gives a challenge and you respond with an answer and a brief solution (such as a sentance saying "inclusion-exclusion and what you apply it to,") that satisfies me.  But if someone asks for help, that really isn't sufficient -- I mean, if there were very familiar with using inclusion-exclusion, they wouldn't have needed help on the problem.

and when he says he understands the answer, i'm assuming that he understands the answer.  (and if you want to differentiate between answer and solution, well, i don't know how you can get that answer without having a solution).  My interpretation was that he came up with a different argument and knew that it was wrong but couldn't understand why.  I do that all the time... making a simple counting argument that over/undercounts but i can't see why.  And I completely agree with you - if someone wants help a crappy one-sentence outline probablly isn't any good.  Being mostly on the receiving end of help, i know how that is.

Ooops, my apologies, I didn't see that.  Touchee.

For calculating derangements, I have read somewhere that the answer is always equal to n!/e rounded to the nearest whole number. I don't have a proof of this, but it does work for this problem.

6!/e = 720/e :approx: 264.87319...(which rounds to 265, which is the answer)

Does anyone know why this works?

probably because when you take the 1-1/2!+1/3!-1/4!.... to infinity its sum is e.

Actually it sums to 1/e. That's basically the reason that the derangement of n is the nearest integer to n!/e. It takes a bit of work to show it because you have to make sure that the series is converging quickly enough.