prime factorization
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[quote="pavel25"]So all other divisors of $n^2$ are smaller than $n$ so they actually divisors of $n$. [/quote]

I don't think so [b]pavel25[/b], try compare with $n=2^5\times3^2$ and $2^6\times3$
:D

I know two sols of it and both seems boring!  :)

[quote="pavel25"]Let $\tau (n)=k$
$d_1,d_2,d_3,\ldots,d_{k-2},d_{k-1},d_k$ - divisors of $n$
${{d_k={n\over d_1},d_{k-1}={n\over d_2}},d_{k-2}={n\over d_3}},\ldots,d_3={n\over d_{k-2}},d_2={n\over d_{k-1}},d_1={n\over d_k}$
So if we are looking on the divisors of $n^2$ from the greatest one to the least one we get:
$l_1,l_2,l_3,\ldots,l_{k-2},l_{k-1},l_k$ - divisors of $n^2$
$l_1={{n^2}\over d_1},l_2={{n^2}\over d_2},l_3={{n^2}\over d_3},\ldots,l_k={{n^2}\over d_k},\ldots$
$l_k={{n^2}\over d_k}=n$
So all other divisors of $n^2$ are smaller than $n$ so they actually divisors of $n$. From here we get:
$\tau ({n^2})=2k-1$
$m={\tau ({n^2})\over \tau (n)}={{2k-1}\over k}$
$km=2k-1$
$k(2-m)=1$
The only solution in positive integers is:
$k=1,m=1$
So $m={\tau ({n^2})\over \tau (n)}$ only when $m=1,n=1$[/quote]

Why you have taken this as
${{d_k={n\over d_1},d_{k-1}={n\over d_2}},d_{k-2}={n\over d_3}},\ldots,d_3={n\over d_{k-2}},d_2={n\over d_{k-1}},d_1={n\over d_k}$ or there is any rule in such situation s=can you explain  it. Really I can not still why you have used as sequence.
Abdurashid

Yes, this is a very useful method also for example you can see an imo 2002 if i remember correctly i will check and say later.

Explanation : we know that $d_1,d_2,\dots, d_k$ are positive divisors of $n$. Then can't we say that $n=d_1\times d_k=d_2\times d_{k-1}=d_3\times d_{k-2}=\dots$ please try to understand it might be very easy.

ex: let n be 12 then we have ${1,2,3,4,6,12}$ as a postitive divisors of n, right ?
Then we can say that $1\times 12=12=2\times 6=4\times 3$, now it is yes?

davron

your solution above is obviously wrong. answer is: every odd m. you use well-known equality

if $n = \prod p_{i}^{\alpha_{i}}$, then $d(n) = \prod (\alpha_{i}+1)$ then by induction we prove that every odd m can be represented as $\frac{\prod (2\alpha_{i}+1)}{\prod (\alpha_{i}+1)}$

Let $S$ be the set of all integers $m$ representable in the form $\frac{\tau(n^2)}{\tau(n)}$. I claim that $S$ is the set of all positive odd integers. First, it is clear that $1 \in S$, as $\frac{\tau(1^2)}{\tau(1)} = 1 \in S$.

If $m \in S$, then $m$ is expressible in the form $\frac{(2e_1 + 1)(2e_2 + 1) \ldots (2e_n + 1)}{(e_1 + 1)(e_2 + 1) \ldots (e_n + 1)}$. Since the numerator of this fraction is odd, $m$ must be odd.

We now claim that every positive odd integer $m$ greater than 1 can be expressed as a product of (not necessarily distinct) numbers of the form $\frac{4a+1}{2a+1} =\frac{2(2a) + 1}{(2a) + 1}$, then $m = \frac{\tau(\prod p_i^{4a_i})}{\tau(\prod p_i^{2a_i})}$, where $p_i$ is the $i$th prime, which implies that $m \in S$. (It also follows that $m, n \in S$ implies that $mn \in S$.)

We shall prove our claim with strong induction. Our base case, $m = 3$ follows from $3 = \frac{5}{3} \cdot \frac{9}{5}$.

Now, let $m$ be any odd integer larger than 3, and suppose that all odd integers less than $p$ lie in $S$. Let $p$ be congruent to $2^{b-1} - 1$ modulo $2^b$, for some positive integer $b \geq 2$; then $p = 2^b x + 2^{b-1} - 1$ for some nonnegative integer $x$. If $b = 2$, then $p \equiv 1 \pmod{4}$, so $p = \frac{4x+1}{2x+1} \cdot (2x+1)$, which lies in $S$ since $\frac{4x+1}{2x+1} \in S$ and $2x+1 \in S$ by our inductive hypothesis.

If $b > 2$, we shall define the sequences $a_0 = 3 \cdot 2^b$, and $a_{i+1} = \frac{3a_{i+1}}{2}$, and $c_0 = 3 \cdot (2^{b-1} - 1)$ and $c_{i+1} = \frac{3c_i + 3}{2}.$ It is clear that $a_k, c_k$ are all positive integers when $0 \leq k \leq b$, $a_b = 4 \cdot 3^b$, $c_b = 2 \cdot 3^b - 1$, $\lfloor \frac{a_kx + c_k}{2} \rfloor = \frac{a_{k+1}x + c_{k+1}}{3}$ when $0 \leq k \leq b-1$. $a_kx + c_k \equiv 1 \pmod{4}$ when $0 \leq k \leq b$.

It follows from the above that each of the numbers $\frac{a_i x + c_i}{\frac{a_{i+1}x + c_{i+1}}{3}}$ for $0 \leq i \leq b-1$ and $\frac{a_b x + c_b}{2 \cdot 3^b x + 3^b}$ are of the form $\frac{4y + 1}{2y + 1}$. Hence, the product $\frac{a_0x + c_0}{\frac{a_1x + c_1}{3}} \cdot \frac{a_1x + c_1}{\frac{a_2x + c_2}{3}} \cdot \ldots \cdot \frac{a_b x + c_b}{2 \cdot 3^b x +  3^b} = \frac{a_0 x + b_0}{3(2x + 1)} = \frac{2^bx + 2^{b-1} - 1}{2x + 1} \in S$. But by our inductive hypothesis, $2x + 1 \in S$ as well, so $2^bx + 2^{b-1} - 1 \in S$, which completes our proof.

I came across that problem yesterday and after I've solved it, I saw the solutions and got confused that all the solutions I've seen where much harder. Maybe my solution is wrong, but anyway here it is:
   We will prove that each odd positive integer, greater than 1,  can be represented as $\frac{d(n^2)}{d(n)}$, for some positive integer $n$.  It's well known fact that the number of positive divisors of $P_{1}^{a_{1}}P_{2}^{a_{2}} . . . P_{k}^{a_{k}}$ is equal to  $(a_{1}+1)(a_{2}+1) . . . (a_{n}+1)$. So we have to prove that for each odd positive integer $A$ there exist such finite sequence ${a_{i}, 1\leq{i}\leq{n}}$ that $\frac{(2a_{1}+1)(2a_{2}+1)...(2a_{n}+1)}{(a_{1}+1)(a_{2}+1)...(a_{n}+1)}$ is equal to $A$. Let $S$ be the set containing all such numbers $A$. Easy to note that each element of $s$ is an odd integer.
   First, we show that if $x$ and $y$ both belong to $S$, than their product $xy$ also belong to $S$. Take $a_{i}, 1\leq{i}\leq{m}$ and $b_{j}, 1\leq{j}\leq{k}$ such that $x= \frac {(2a_{1}+1)...(2a_{m}+1)}{(a_{1}+1)...(a_{m}+1)}$ and $y=\frac{(2b_{1}+1)...(2b_{k}+1)}{(b_{1}+1)...(b_{k}+1)}$.  It means that \[xy=\frac{(2a_{1}+1)...(2a_{m}+1)(2b_{1}+1)...(2b_{k}+1)}{(a_{1}+1)...(a_{m}+1)(b_{1}+1)...(b_{k}+1)}\].  It follows that $xy$ is in $S$, as desired.
   It means that for any two $x,y$ in $S$,  $xy$ is also in $S$.    (1)
   So, if we manage to prove that each odd prime number belongs to $S$, than it will follow that each odd number is contained in $S$, which we need to prove. Apply strong induction and prove the following conjecture: If $3=p_{1}<p_{2}<...<p{n}$ are all in $S$, where $p_{i}$-s are first $n$ primes, starting from $3$, than also $p_{n+1}$ belongs to $S$, where $p_{n+1}$ is minimal prime number, greater than $p_{n}$.
   We have a basis, since $N=2^{4}3^{2}$ imply $\frac{d(n^{2})}{d(n)}=3$ and hence, $3$ is in $S$. Assume now for $3=p_{1}<...<p_{n}$. Than, from (1), each number, whose factorization into primes contains only numbers from ${p_{1},...,p_{n}}$, is contained in $S$. Consider now $L=\frac{p_{n+1}-1}{2}$. Obviously, $L$ is a positive integer and $L+1<p_{n+1}$. So, factorization of $L+1$ into primes is of the form $p_{1}^{x_{1}}...p_{n}^{x_{n}}$, where each $x_{i}$ is a nonnegative integer. By assumption, $L+1$ is in $S$.  It means that $2L+1=(L+1) \frac{2L+1}{L+1}$ is in $S$, but $2L+1=p_{n+1}$, completing the proof of our conjecture. So, each odd prime is contained in $S$ and by (1), each odd positive integer is contained in $S$, except for $1$. (Easy to prove).

[quote="lasha"]Consider now $L=\frac{p_{n+1}-1}{2}$. Obviously, $L$ is a positive integer and $L+1<p_{n+1}$. So, factorization of $L+1$ into primes is of the form $p_{1}^{x_{1}}...p_{n}^{x_{n}}$, where each $x_{i}$ is a nonnegative integer. [b]By assumption, $L+1$ is in $S$.[/b]  It means that $2L+1=(L+1) \frac{2L+1}{L+1}$ is in $S$, but $2L+1=p_{n+1}$, completing the proof of our conjecture. So, each odd prime is contained in $S$ and by (1), each odd positive integer is contained in $S$, except for $1$. (Easy to prove).[/quote]
I don't particularly understand the bolded part. If $4 | p - 3$, can't $L + 1$ be even?

I hope my solution is correct... this seemed too easy for an IMO 3/6.
Call an integer [i]good[/i] if there exists a positive integer n such that $\frac{\tau (n^{2})}{\tau (n)}=m$.

I claim that all odd positive integers are good.

First, any good integer must be expressible in the form $\dfrac {(2a_1 + 1)(2a_2 + 1) \cdots (2a_n + 1)} {(a_1 + 1) (a_2 + 1) \cdots (a_n + 1)}$. Since the numerator is odd, a good integer cannot be an even integer.

The case $m = 1$ is obvious.

Now we induct to show that all odds are good. The base case $m = 3$ can be expressed as $\dfrac {5} {3} \cdot \dfrac {9} {5}$.
Assume that all odds less than an odd number $k$ are good.

[b]Case 1:[/b] $k \equiv 1 \bmod 4$.
Then we take $k = \dfrac {k} {\dfrac {k + 1} {2}} \cdot \dfrac {k+ 1} {2}$. Since $\dfrac {k + 1} {2}$ is odd and less than $k$, it's good.

[b]Case 2:[/b] $k \equiv 3 \bmod 4$.
Let $k = x \cdot 2^a - 1$, where $x$ is odd. Let $r = k(2^a - 1) = 2^{2a} \cdot k - 2^a \cdot (1 + k) + 1$, and let $g(b) = \dfrac {b + 1} {2}$ for all integers $b$.
Then we have $\dfrac {b} {g(b)} \cdot \dfrac {g(b)} {g^2 (b)} \cdot ... \cdot \dfrac {g^{a-1} (b)} {g^a (b)} = \dfrac {(2^a \cdot x - 1)(2^a - 1)} {(x) (2^a - 1)}$. Since $x$ is odd, $x$ is good, so $\dfrac {b} {g(b)} \cdot \dfrac {g(b)} {g^2 (b)} \cdot ... \cdot \dfrac {g^{a-1} (b)} {g^a (b)} \cdot x = k$ is also good.

@cwein3

Your case of $k \equiv 1 \pmod{4}$ is correct. However, I don't understand your case of $k \equiv 3 \pmod{4}$, what is $b$? And what does $g^a(b)$ mean? It normally means $g(b)^a$, but it appears it means $g(g(g(...b))...)$ in how you are using it. Can you clarify on these two problems? Then maybe I can understand your proof :)

[quote="cwein3"]
[b]Case 2:[/b] $k \equiv 3 \bmod 4$.

Let $k = x \cdot 2^a - 1$, where $x$ is odd. Let $b = k(2^a - 1) = 2^{2a} \cdot x - 2^a \cdot (1 + x) + 1$, and let $g(b) = \dfrac {b + 1} {2}$ for all integers $b$.
Then we have $\dfrac {b} {g(b)} \cdot \dfrac {g(b)} {g^2 (b)} \cdot ... \cdot \dfrac {g^{a-1} (b)} {g^a (b)} = \dfrac {(2^a \cdot x - 1)(2^a - 1)} {(x) (2^a - 1)}$. Since $x$ is odd, $x$ is good, so $\dfrac {b} {g(b)} \cdot \dfrac {g(b)} {g^2 (b)} \cdot ... \cdot \dfrac {g^{a-1} (b)} {g^a (b)} \cdot x = k$ is also good.[/quote]

I think the solution is changed to what he meant and $g^a(b)$ means $g(g( \cdots g(a))))\cdots)$ with $a$ times taken the function $g.$

Also to understand: $g^l (b)= 2^{2a-l}x-2^{a-l}(x+1)+1$ for $l\le a$ meaning,
so $g^a(b)=2^a x -x$ and his proof is correct.

Nice solution , cwein3!

We can obviously see that it woks for no even number .I will use induction to prove that it works for all odd.
As said earlier our target is to prove that any odd number may be represented in the form $\frac{4a_1+1}{2a_1+1}....\frac{4a_r+1}{2a_r+1}$. Assume the result is valid for all $k \le n$ Let $2^x || n+1$ and $n=2^x y-1$ where $y$ is odd .
Consider the number $(2^x y-1)(2^x -1)=2^x((2^x-1) y -1)+1$ .Now consider the sequence
$A=\frac{2^x((2^x-1) y -1)+1}{2^{x-1}((2^x-1) y -1)+1} .\frac{2^{x-1}((2^x-1) y -1)+1}{2^{x-2}((2^x-1) y -1)+1}.......
\frac{2((2^x-1) y -1)+1}{((2^x-1) y -1)+1}$.
Which is of the form we required.Above expression is equivalent to
$\frac{2^x((2^x-1) y -1)+1}{(2^x-1) y}.=\frac{(2^x y-1)(2^x -1)}{(2^x-1) y}=\frac{(2^x y-1)}{ y}$.Let $[y]$ be the representation of $y$ in the above form  .Hence  $n$ would be represented in the form $A. [y]$ .Hence the induction is complete

The answer is all the odd numbers,when k=1,3, we can just check by hand. Use induction,  suppose every odd number not bigger than 2j+1 can be expressed to the requirement. Now we working on 2j+3, if  j is even, then there are nothing to prove. If j is odd, we can do conversions on  j  endless. Finally know that k=1, by our basic inspection, it's true.

Am I missing something? :maybe:
[quote=1998 N6]For any positive integer $n$, let $\tau (n)$ denote the number of its positive divisors (including 1 and itself). Determine all positive integers $m$ for which there exists a positive integer $n$ such that $\frac{\tau (n^{2})}{\tau (n)}=m$.[/quote]

Fancy way of asking which positive integers greater than $1$ can be written as $\prod^n_{j=1} \frac{2\alpha_j+1}{\alpha_j+1}$ for $\alpha_j \geqslant 1$ and $n \geqslant 1$ integers (for $1=\tau(1^2)/ \tau(1)$ anyway). Clearly, any such number must be odd, as the numerator is odd. Note that $3$ is expressible, simply by $n=2$ and $\alpha_1=1, \alpha_2=4$.

Let $m>3$ be the smallest odd number that is not expressible. Suppose $m=2^tx-1$ where $x$ is odd and $x, t \geqslant 1$. Consider $$\frac{2^tx\ell-(2^t-1)}{\ell}=x \frac{2x\ell-1}{x\ell} \cdot \frac{2^2x\ell -3}{2x\ell-1} \dots \cdot \frac{2^tx\ell-(2^t-1)}{2^{t-1}x\ell-(2^{t-1}-1)}$$ and plug $\ell=2^t-1$ to obtain an expression for $m$, since $x<m$ is odd, hence expressible, by assumption, yielding a contradiction! So all odd numbers are expressible. $\blacksquare$

Nice problem! Here's my solution: The answer is all odd $m$. $m=1$ easily follows on taking $n=1$, so from now on assume $m>1$. First let $$n=p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$$
where the $a_i$'s are positive integers, and $p_i$'s are primes. Then the problem condition translates to $$m=\prod_{i=1}^k \left(\frac{2a_i+1}{a_i+1} \right)$$
Since numerator is odd, so for $m$ to be an integer, the denominator must also be odd (and also $m$ must be an odd integer, so we can effectively ignore even $m$ from now on). This means that $2 \nmid a_i+1$, or equivalently that all $a_i$'s are even. Letting $a_i=2b_i$, we get $$m=\prod_{i=1}^k \left(\frac{4b_i+1}{2b_i+1} \right)$$
Call all integers $m$, which can be written in the fashion above, as [i]expressible[/i] numbers. Also, for any [i]expressible[/i] number $m$, we let $g(m)$ denote its notation in the above form (note that we have $g(m)=m$ only). We proceed by strong induction to show that all odd $m$ are [i]expressible[/i]. For $m=3$, take $b_1=1$ and $b_2=2$. Now, suppose the result is true for all odd numbers less than $m$. We will show that it is true for $m$ as well. If $m \equiv 1 \pmod{4}$, then write $m=4z+1$. Thus, we have $$m=\frac{4z+1}{2z+1} \times g(2z+1)$$
where $2z+1$ is [i]expressible[/i] by induction hypothesis. So from now on let $m \not \equiv 3 \pmod{4}$. Suppose there exists some $r \geq 2$ with $m \equiv 2^r-1 \pmod{2^{r+1}}$ (we'll later prove why such an $r$ must exist). Write $m=2^{r+1}x+2^r-1$. Then we have $$(2^r-1)m=4(4^{r-1}(2x+1)-2^{r-1}(x+1))+1 \equiv 1 \pmod{4}$$
This gives $$(2^r-1)m=\frac{4(4^{r-1}(2x+1)-2^{r-1}(x+1))+1}{2(4^{r-1}(2x+1)-2^{r-1}(x+1))+1} \times \frac{4(4^{r-2}(4x+2)-2^{r-2}(x+1))+1}{2(4^{r-2}(4x+2)-2^{r-2}(x+1))+1} \times \dots \times \frac{4(2^{r-1}(2x+1)-(x+1))+1}{2(2^{r-1}(2x+1)-(x+1))+1} \times (2(2^{r-1}(2x+1)-(x+1))+1)$$
Since $2(2^{r-1}(2x+1)-(x+1))+1=(2^r-1)(2x+1)$, so we get that $$m=\frac{4(4^{r-1}(2x+1)-2^{r-1}(x+1))+1}{2(4^{r-1}(2x+1)-2^{r-1}(x+1))+1} \times \frac{4(4^{r-2}(4x+2)-2^{r-2}(x+1))+1}{2(4^{r-2}(4x+2)-2^{r-2}(x+1))+1} \times \dots \times \frac{4(2^{r-1}(2x+1)-(x+1))+1}{2(2^{r-1}(2x+1)-(x+1))+1} \times g(2x+1)$$
where $2x+1$ is [i]expressible[/i] by induction hypothesis. Thus, all odd $m$ are [i]expressible[/i] if such an $r$ exists.

Now we show that such an $r$ exists. Suppose not. Then $m \not \equiv 2^r-1 \pmod{2^{r+1}}$ for any $r \geq 1$ ($r=1$ simply gives $m \equiv 1 \pmod{4}$). This easily converts to the form that $m \not \equiv -1 \pmod{2^a}$ for any $a \geq 1$. But this means that $2 \nmid m+1$, contradicting the fact that $m$ is odd. Thus, such an $r$ exists for all odd $m$. Hence, done. $\blacksquare$

[b]REMARKS:[/b] I believe that this was a highly motivated problem, and easily follows if one works according to the direction it gives. The solution above includes the most trivial of details also (like proving $m \equiv 1 \pmod{4}$ seperately, and showing why $r$ exists), and I haven't changed anything coz this is how I discovered the solution. To be completely honest, I had a thinking that maybe just dividing into cases modulo $4$ will work. However, it soon became clear that this wasn't the case, as each case was getting divided into a sub-case. Luckily for us, the whole thing generalized easily (Or maybe the problem was designed in such a way only :P).

[quote=SnowPanda][hide = Solution]
The answer is all odd $m$. First, $\tau(n^2)$ is odd, so $m$ must be odd as well. To construct, this is equivalent to finding (not necessarily distinct) positive integers $e_1$, $e_2$, $\ldots$, $e_k$ such that \[\frac{2e_1 + 1}{e_1 + 1} \cdot \frac{2e_2 + 1}{e_2 + 1} \cdots \frac{2e_k + 1}{e_k + 1} = m.\]

Induct on $m$, with the base case $m = 1$ achieved by $n = 1$. Now assume that all odd numbers less than $m$ are achievable. Let $\nu_2(m + 1) = r \geq 1$, so $m = 2^ra - 1$ for some odd $a$. Then take $e_i = 2^{i - 1}(2^ra - a - 1)$ for $1 \leq i \leq r$, which gives \[\frac{2(2^ra - a - 1) + 1}{2^ra - a - 1 + 1} \cdot \frac{4(2^ra - a - 1) + 1}{2(2^ra - a - 1)} \cdots \frac{2^r(2^ra - a - 1) + 1}{2^{r - 1}(2^ra - a - 1) + 1} = \frac{2^ra - 1}{a}.\] Finally, multiply by a construction for $a$, which exists because $a$ is odd and less than $m$ (we always have $a \leq 2a - 1 \leq m$, and equality only holds for $m = 1$). This gives a construction for $m$.
[/hide][/quote]

Great solution. Just to prevent possible future confusion, there is a typo in the second fraction in the last big equation, it should be: \[\frac{2(2^ra - a - 1) + 1}{2^ra - a - 1 + 1} \cdot \frac{4(2^ra - a - 1) + 1}{2(2^ra - a - 1)+1} \cdots \frac{2^r(2^ra - a - 1) + 1}{2^{r - 1}(2^ra - a - 1) + 1} = \frac{2^ra - 1}{a}.\]

If $n=\prod p_i^{\alpha_i},$ then $m=\frac{\tau(n^2)}{\tau(n)}=\prod\frac{2\alpha_i+1}{\alpha_i+1}.$ Notice the numerator is always odd, so $m$ is odd. We now claim all odd $m$ work; we proceed by strong induction. Let $m=2^k\ell-1,$ letting $\ell$ be an odd integer less than $m.$ Then $\ell$ is in the form $\prod\frac{2\alpha_i+1}{\alpha_i+1},$ so $$\ell\prod_{i=0}^{k-1}\frac{2\cdot{\color{blue}2^i(2^k\ell-\ell-1)}+1}{{\color{blue}2^i(2^k\ell-\ell-1)}+1}=\ell\cdot\frac{2\cdot 2^{k-1}(2^k\ell-\ell-1)+1}{(2^k\ell-\ell-1)+1}=\ell\cdot\frac{(2^k\ell-1)(2^k-1)}{\ell(2^k-1)}=m$$ is also of the desired form. $\square$

We claim that the answer is all odd positive integers. Note $f(1)=1$, so $1$ is achievable. Let $f(n)=\tfrac{\tau(n^2)}{\tau(n)}$. Let $n=p_1^{e_1}\dots p_k^{e_k}$ be the prime factorization of $n$. Then $\tau(n)=\Pi_{i=1}^k (e_i+1)$ and $\tau(n^2)=\Pi_{i=1}^k (2e_i+1)$. In order for $f(n)$ to be an integer, it is necessary that $\tau(n)$ is odd, since $\tau(n^2)$ is odd. So $f(n)$ is always odd.

    Note that since $\tau(n)$ is multiplicative, $f(n)$ is also multiplicative. Also, note that $f(p_1^{e_1}\dots p_k^{e_k})=f(p_{k+1}^{e_1}\dots p_{2k}^{e_k})$, in other words, the primes used in the base of the prime factorization of the input "don't matter". So if $f(p_1^{e_1}\dots p_k^{e_k})=m$, and $(p_i)_{i=1}^{\infty}$ is the sequence of primes, then $$f\left(\prod_{j=0}^{\ell-1} \left(\prod_{i=1}^{k} p_{i+jk}^{e_i}\right)\right)=\prod_{j=0}^{\ell-1} f\left(\prod_{i=1}^{k} p_{i+jk}^{e_i}\right)=m^\ell,$$ for any positive integer $\ell$. It follows that if all odd primes are in the range of $f(n)$, then all odd positive integers are in the range of $f(n)$, since we can write every odd positive integer as a product of odd primes.

    Suppose for contradiction that not all primes are in the range of $f(n)$, and take the least prime $p$ that is not in the range of $f(n)$. Note that $f(2^2 3^4)=3$, so $p>3$. If $p\equiv 1\pmod{4}$ then let $e=\tfrac{p-1}{4}$, so $$f(q^{2e})=\frac{p}{\frac{p-1}{2}+1}=\frac{p}{\frac{p+1}{2}},$$ for any prime $q$. Since $\tfrac{p+1}{2}$ is odd due to $p\equiv 1\pmod{4}$ and $\tfrac{p+1}{2}<p$, there exists an $r$ such that $f(r)=\tfrac{p+1}{2}$, due to minimality of $p$. So $f(q^{2e})f(r)=p$. Since the primes in the prime factorization of $r$ "don't matter", as established earlier, we can make $\gcd(q^{2e},r)=1$, so $f(q^{2e}r)=p$, and thus $p$ must be in the range of $f(n)$, contradiction.

    If $p\equiv 3\pmod{4}$, then take $e=\tfrac{3p-1}{4}$, so $f(q^{2e})=\tfrac{3p}{(3p+1)/2}$. Note that $\tfrac{3p+1}{2}$ is odd since $p\equiv 3\pmod{4}$. Then taking $d=\tfrac{3p-1}{8}$, we have $f(s^{2d})=\tfrac{(3p+1)/2}{(3p+3)/4}=\tfrac{(3p+1)/2}{3\cdot (p+1)/4}$, for some prime $s$. Note that $\tfrac{p+1}{4}$ is smaller than $p$, so we have some $r$ such that $f(r)=\tfrac{p+1}{4}$. So then $f(s^{2d})f(r)=\tfrac{(3p+1)/2}{3}$, and we can make $\gcd(q^{2e},s^{2d},r)=1$, so $f(s^{2d}r)=\tfrac{(3p+1)/2}{3}$. Then $$f(q^{2e}s^{2d}r)=f(q^{2e})f(s^{2d}r)=\frac{3p}{\frac{3p+1}{2}}\cdot \frac{\frac{3p+1}{2}}{3}=p,$$ so $p$ is achievable. Therefore all odd primes are in the range of $f(n)$, and we are done.

Solved together with, after gatecrashing a groupsolve by [b]mxlcv[/b] and [b]mathscrazy[/b]

We claim all odd numbers work, clearly any such number is odd so it suffices to show a construction for them. Proceed by strong induction with the base case $m = 1$ which works when $n = 1$. Now, if $m+1 = 2^k \cdot \ell$ with $\ell$ odd, then we have $$\frac{(2^k - 1)m}{\frac{(2^k - 1)m + 1}{2}} \cdot \frac{\frac{(2^k - 1)m + 1}{2}}{\frac{(2^k - 1)m+3}{4}} \cdot \frac{\frac{(2^k - 1)m + 3}{4}}{\frac{(2^k - 1)m + 7}{8}} \cdots \frac{\frac{(2^k - 1)m + 2^{k-1} - 1}{2^{k-1}}}{\frac{(2^k - 1)m + 2^k - 1}{2^k}} = \frac{m}{\ell}$$By induction, since $\ell$ can be represented, we are done.

Let $n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$ for distinct primes $p_1,p_2,\dots,p_k$ and positive integers $e_1,e_2,\dots,e_k.$ Then, we have \[\frac{\tau\left(n^2\right)}{\tau(n)}=\frac{(2e_1+1)(2e_2+1)\dots(2e_k+1)}{(e_1+1)(e_2+1)\dots(e_k+1)}.\] Since the numerator is odd, the denominator must also be odd, so all $e_i$ are even, and $m$ is odd. It just remains to solve the following equivalent problem: for which odd positive integers $m$ is it possible to find elements that multiply to $m$, not necessarily distinct, from the set $S=\left\{\tfrac{4k+1}{2k+1}\right\}$ where $k$ is a positive integer. Call rational number $m$ good if it satsifies that. Note that

(1) $1$ is good by taking no elements at all. An empty product is equal to $1$. $3$ is good by taking $5/3$ and $9/5$.
(2) If $m$ and $n$ are good rationals, not necessarily distinct, then $mn$ is good. Take all the elements from the construction for $m$ and then the ones from the one for $n$, and then take the product of them all.
(3) If $m$ is a good positive integer, then $2m-1$ is good. In this case, take the elements from the construction for $m$ and add on the element described by $k=\tfrac{m-1}{2}.$

We proceed by strong induction on $m$, for the statement for all odd $m$, $m$ is good. If $m\equiv 1\pmod 4$ then $\tfrac{m+1}{2}$ is an odd positive integer, which must be good. Thus, $2\cdot \tfrac{m+1}{2}-1=m$ is good.
\\$~$\\
Otherwise, let $m=2^a\cdot b-1$ for $b$ odd. Since $b$ is odd, $b$ must be good, so it suffices to show that $\tfrac{m}{b}$ is good. Indeed, let $c=b(2^a-1)$ then $c$ is odd, so $2c-1\equiv 1\pmod 4.$ Thus, \[\frac{2c-1}{c}\cdot \frac{4c-3}{2c-1}\cdot \dots \cdot \frac{2^a(c)-(2^a-1)}{2^{a-1}(c)-(2^{a-1}-1)}=\frac{2^ac-(2^a-1)}{c}=\frac{m}{b}\] is good as desired.

Solved with [b]GoodMorning[/b]

The answer is all odd positive integers. To see that $m$ must be odd, we have $\tau(n^2)$ is odd, so $\frac{\tau(n^2)}{\tau(n)} $ is also odd. It remains to show that all odd numbers work.

Since $\tau(n)\mid \tau(n^2)$, $\tau(n)$ is odd, so $n$ must be a perfect square. Write \[n = p_1^{2a_1} p_2 ^{2a_2} \cdots p_l ^{2a_l} \] for some positive integer $l$, distinct primes $p_1, p_2, \ldots, p_l$, and positive integers $a_1, a_2, \ldots, a_l$.

We have \[m = \frac{(4a_1 + 1)(4a_2 + 1)\cdots (4a_l + 1)}{(2a_1 + 1)(2a_2 + 1)\cdots  (2a_l + 1)} \] Note that if we can find $l$ $a_1, a_2,a_3,\ldots, a_l$ such that the expression is equal to $m$, then $m$ is a valid number in the problem.

Thus, it suffices to show that all odd positive integers can be expressed as the product of (not necessarily distinct) values in $\left\{\frac{4x+1}{2x+1}\right\}_{x\in \mathbb{N}}$

We do this by induction. The base cases $m=1,3$ work by having no $a_i$ at all, and $\frac{5}{3} \cdot \frac{9}{5}$, respectively.

Now fix an odd positive integer $c$. Suppose that all odd positive integers less than $c$ were expressible. We show that $c$ is also expressible.

Let $a = \nu_2(c+1)$ and $c = k\cdot 2^{a+1} + 2^a - 1$.

Claim: If $f(x) = \frac{x+1}{2}$ for all positive integers $x$, then we have \[f^y(c\cdot (2^a - 1)) = k\cdot 2^{a+1 - y} \cdot (2^a - 1) + 2^{2a  - y} - 2^{a+1 - y} + 1 \] for all positive integers $0\le y\le a$.
Proof: Induction (base case $y = 0$ is obvious), we see that if $y$ works, then \begin{align*}
f^{y+1}(c\cdot (2^a - 1)) = \frac{k\cdot 2^{a+1 - y} \cdot (2^a - 1) + 2^{2a  - y} - 2^{a+1 - y} +  2}{2} \\
=  k\cdot 2^{a+1 - (y+1)} \cdot (2^a - 1) + 2^{2a - (y+1)} - 2^{a+1 - (y+1)} + 1,\\
\end{align*}as desired. So the induction is complete.
$\square$

Now notice that $\frac{t}{f(t)} \in \left\{\frac{4x+1}{2x+1}\right\}_{x\in \mathbb{N}}$ for any positive integer $t\equiv 1\pmod 4$. Using the fact that $f^y(c\cdot (2^a - 1) ) \equiv 1\pmod 4$ for $0\le y\le a-1$, we get that the number \[\frac{d}{f(d)} \cdot \frac{f(d)}{f^2(d)} \cdots \frac{f^{a-1}(d)}{f^a(d)} = \frac{d}{f^a(d)} = \frac{k\cdot 2^{a+1} \cdot (2^a - 1) + (2^a - 1)^2}{(2^a - 1)(2k + 1)}  \] is expressible. Since $2k + 1$ is expressible, we can multiply by $2k + 1$ and get that \[\frac{k\cdot 2^{a+1} \cdot (2^a - 1) + (2^a - 1)^2}{2^a - 1} = c\] is also expressible, which completes the induction.

Therefore, all odds are expressible, so we are done.

Shoot... This problem I had to ACTIVELY try and reassure myself that I was not to stray away from hard problems in lieu of improving...

Note that $\frac{\tau\left(n^2\right)}{\tau(n)}=\prod\frac{2e_i+1}{e_i+1}$ (where $e_i$ are exponents in the factorization of n). From here, it's obvious evens don't work, all exponents are even, and we prove all odds work by proving you can find elements from the set $S=\left\{\frac{4k+1}{2k+1}\right\}$ (repeating allowed) that multiply to m; call these numbers [i]obtainable[/i]. We proceed by induction, manually computing base cases n=1,3; observe that if m is obtainable then 2m-1 is obtainable by also using $\frac{m-1}2$ (1), and if a and b are obtainable then ab is obtainable by combining sets (2); assume henceforth the inductive hypotheses for $k\le m$. In this way, the next minimal number $m_0>m,m_0\equiv1\pmod4$ is obtainable by using (1) and inductive hypothesis. Furthermore, if $m=2^ij-1:i\ge2$ is 3 mod 4, by inductive hypothesis j is obtainable, so by (2) it suffices to prove $\frac mj$ is obtainable. But this is evident, as $\frac{2(2^ij-j)-1}{2^ij-j}\frac{4(2^ij-j)-3}{2(2^ij-j)-1}\dots\frac{2^i(2^ij-j)-(2^i-1)}{2^{i-1}(2^ij-j)-(2^{i-1}-1)}=\frac mj$ is indeed obtained.

[b]Motivational remark.[/b] The last construction is not so easy to obtain; the way I got it was by solving for a general form in the telescope, if the first term was c, that it would be of the form $\frac{2^dc-(2^d-1)}{c}=\frac mj$ for arbitrary d (i hope i didn't get this wrong, someone pls check), and from there it was easy.

The answer is all odd $m$. Clearly $m=1$ works by taking $n=1$, so suppose $m>1$.

Clearly we are being asked to find the positive integers which can be expressed as the product of not necessarily distinct numbers of the form $\tfrac{2e-1}{e}$ for positive integers $e$. Call [i]rational[/i] numbers which can be written like this [i]good[/i]. Obviously products of good numbers are also good. Observe that each fraction always have nonpositive $\nu_2$, hence $m$ must be odd. We now show that every odd prime $p$ is good, which will finish the problem. To do this, we induct on $p$.

Let $\nu_2(k+1)=k>0$, so $p \equiv 2^k-1 \pmod{2^{k+1}}$. Then we consider the identity
$$\frac{(2^k-1)p}{\frac{(2^k-1)p+1}{2}}\cdot \frac{\frac{(2^k-1)p+1}{2}}{\frac{(2^k-1)p+3}{4}}\cdots\frac{\frac{(2^k-1)p+(2^{k-1}-1)}{2^{k-1}}}{\frac{(2^k-1)p+(2^k-1)}{2^k}}=\frac{(2^k-1)p}{\frac{(2^k-1)(p+1)}{2^k}}=\frac{2^k}{p+1}p,$$
hence $\tfrac{p+1}{2^k}p$ is good. On the other hand, by the definition of $k$, $\tfrac{p+1}{2^k}$ is odd and strictly less than $k$, hence by inductive hypothesis it is good as well. Thus $p$ is good: done. $\blacksquare$

[b]Remark (motivation):[/b] I figured out how to get $1 \pmod{4}$ primes by using $\tfrac{p+1}{2}$, but for $3 \pmod{4}$ primes this doesn't work. When I tried $p=11,19$, I found that we could "start" from $3p$, but again this doesn't work for $p \equiv 7 \pmod{8}$. But for this case "starting" from $7p$ sometimes works. At this point we figure out the solution.

[b]Remark:[/b] Oh using primes doesn't actually matter and we can generally work with odds instead. But the solution isn't actually made any harder if we only think about primes (nor is it made any easier)


geometry
3D geometry
sphere
topology
calculus
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complex analysis

I heard that the Poincare's conjecture has been proven recently.  When I read into the proof, I realized I had absolutely no idea what the conjecture is in the first place.

Yeah, I just looked at the description on Mathworld, and it looked quite..complicated. Unlike other unsolved problems, such as Fermat's Last Theorem (before Wiles, at least), it is harder for high school students to understand. Perhaps this link would help:

http://www.claymath.org/Millennium_Prize_Problems/Poincare_Conjecture/

It contains a more real-world analogy.

Fierytycoon

The statement of the Poincare Conjecture isn't really that hard to understand, it's just complicated because of some of the terminology used in stating it precisely. The first paragraph of the MathWorld reference does a pretty good job of it though: "...colloquially, the conjecture says that the three-sphere is the only type of bounded three-dimensional space possible that contains no holes."

It's basically saying that 3-dimensional spheres (or 3-spheres) are homotopically unique. That is, there aren't any 3-dimensional spaces with the same homotopy properties as the 3-sphere that AREN'T essentially 3-spheres. To understand the mathematical phrasing of the problem though, you need to understand the definitions of the terms "compact, homeomorphism, manifold, and simply connected (or homotopy class)". These are things you will learn about in a first Topology course.

The most basic of these ideas is compactness. It's essentially a generalization of the idea of a closed set (i.e. a space which contains all it's limit points).

A Manifold is something that locally (i.e. in the neighborhood of any point) looks like R^n. For example a 2-sphere (or what we just think of as a regular sphere) is an example of a manifold. To see why, just think about the Earth. When you're on it (i.e. local) it looks like the plane, R^2. That's why *some* people thought that the Earth was flat at one time - because locally it looks like a plane.

A homeomorphism is a continuous, one-to-one, and onto map between spaces which has a continuous inverse. Homeomorphisms basically preserve topological properties. So in a topological sense two spaces that are homeomorphic (i.e. have a homeomorphism between them), are the same as far as topology is concerned. This goes back to the famous "a coffee cup is the same as a donut" example.

And finally simply connectedness is the hardest one to explain. A space is said to be simply connected if any closed curve within the space can be continuously shrunken to a point within the space. For example, a 2-sphere is simply connected. You can put a rubber band around it and always shrink the rubber band to a point without breaking it or taking it off of the sphere. But a Torus (or a donut) is NOT simply connected. You can imagine how a rubber band could be put on a donut so that it could not be shrunken to a point.

So the Poincare Conjecture basically states that any 3-dimensional space with this property has to be homeomorphic to the 3-sphere. On one hand it seems intuitively obvious, but then lots of weird things have been found to happen in dimensions 3 and 4. And it has, obviously, been enormously difficult to prove, as are lots of simple statements in topology. For example, just try and prove the Jordan Curve Theorem that says a circle in the plane divives it into precisely two components of which it is the common boundary. Sounds simple enough, but not so easy to prove.

A fun little book to get aquinted with the ideas of thinking in higher dimensions is a classic one called [u]Flatland[/u] by Edwin Abbott. A bit sexist and a bit outdated, but still fun. Then a good book with a chapter that describes the Poincare Conjecture in layman's terms is [u]The Millennium Problems[/u] by Keith Devlin.

And if you're really interested in learning topology, a nice introduction is the Dover book called [u]Elementary Topology[/u] by Michael Gemingnani. It's really cheap, like under 10 bucks. But to really learn topology you can't do much better than the holy grail for undergraduate students, which is a book called [u]Topology[/u] by Munkres. It's ridiculously expensive though. Hope that helps a little.

actually, from what I've heard, Flatland is not sexist. The author was actually satirizing the sexist views held by others of his time.

I'd just like to say the only reason I know anything at all about this is because of [u]The Mathematical Tourist[/u] by Ivars Peterson, which, from my point of view, does a really good job of introducing people to such subjects in mathematics.

About the Jordan Curve Theorem, it doesn't say that just any circle divides the plane into two regions. It says that any simple closed curve divides the plane into two regions. This is significantly more difficult.

By the Poincare Conjecture (Theorem) for n = 1, any simple closed curve is homeomorphic to the unit circle. So I think the problems are essentially the same - though still very difficult.

I still don't understand it completely, but the website has aided comprehension significantly.  Thx

You have dishonored me by replying to your own message with a reply that is not worthy of being a reply to your own message.

This is just sad...
The Idiot without a village

[quote="ComplexZeta"]I was thinking the same thing, but then that shouldn't be too hard to prove with a contour integral I think. So then perhaps you can't do that for some reason. Perhaps some theorems about contour integration rely on the Jordan Curve Theorem (which is extremely likely).[/quote]

Most books in complex analysis glide around the issue of seperating R^2~C into connected components. The theorems on contour integration described are usually only proven to apply to C^1 curves but they apply to any continuous, not merely differentiable, closed curve. For that, one needs the Jordan Curve Theorem. (BTW, the standard proof of the Poincare Conjecture for n=1 uses enough homological algebra that the Jordan Curve Theorem is a trivial corallary of it).


function calculus derivative algebra polynomial integration logarithms

If $ f: [a,b] \to \mathbb{R}$ is a continous function with derivative in $ (a,b)$, the mean value theorem guarantees that there exists a $ c \in (a,b)$ such that $ f(b)\minus{}f(a)\equal{}f'(c)(b\minus{}a)$.

Which conditions on a real function $ f$ are enough so we can get that $ c$ is a differentiable function of $ b$, taking $ a$ fixed and $ b$ variable in $ (a,\plus{}\infty)$?

As an example, if $ f(x)\equal{}x^2$, then $ b^2\minus{}a^2\equal{}2c(b\minus{}a)$ with $ c\equal{}\frac{a\plus{}b}{2}$, hence here $ c$ is a polynomial of degree 1 respect to the variable $ b$, thus differentiable.

Can we assure that this is always true?

My gut reaction is that there's trouble here, even for smooth $ f.$ The reason for that reaction: $ c$ is not uniquely defined. There may be several $ c$ in the same interval that satisfy this equation. Perhaps as we let $ b$ change, we might find ourselves obligated to jump from one definition to another.

I don't have an example yet to illustrate this idea, but it's something to look for.

Here's an example.  Probably it's a good idea to draw the pictures to make sense of the following:

Take some polynomial like $ f(x) \equal{} x^5 \minus{} 2x^3 \plus{} x$.  One can show that $ f( \minus{} 1) \equal{} 0$ and that there exists a unique $ b > \minus{} 1$ such that $ f(b) \equal{} b \plus{} 1$.  (In particular, $ b \approx 1.5128\ldots$.)  We have $ f'(0) \equal{} 1$ and so $ f(b) \minus{} f( \minus{} 1) \equal{} b \plus{} 1 \equal{} f'(0)(b \minus{} ( \minus{} 1))$, so on the interval $ [ \minus{} 1, b]$ the Mean Value Theorem lets us select $ c \equal{} 0$.  However, there is no way to continuously extend this solution to $ b' > b$: we have that $ f'(x)$ achieves a local maximum at $ x \equal{} 0$ but that $ \frac {f(b') \minus{} f( \minus{} 1)}{b' \minus{} ( \minus{} 1)} > 1$ for all $ b' > b$.

(The other root is $ c' \equal{} \frac {\sqrt {30}}{5}$, and you can climb continuously up from there.  However, you can't continuously extend that root to all smaller values of $ b$: at some point, you'll begin falling outside of the interval.)

It should work for, say, twice continuously differentiable $ f$ satisfying $ f''(x)\neq 0$ for all $ x$.

Dear friends, thank you for your answers.

I would like it would be true for $ f(x) \equal{} \int_a^x {g(t)dt}$ being $ g(x) \equal{} \frac{1}{{\ln \displaystyle \frac{{b \plus{} x}}{{a \plus{} x}}}}$ or $ g(x) \equal{}  \frac{1}{{\ln \displaystyle \frac{{b(a \plus{} x)}}{{a(b \plus{} x)}}}},$ both giving $ f''(x) \neq 0$.

Why $ f''(x) \neq 0$ solves the problem?

It avoids the problem of double roots.  Any particular root "wants to" behave nicely, but sometimes they will pop into or out of existence (e.g., when the function has a bunch of wiggles, as in the example I gave); a nonzero second derivative means that the first derivative is monotonic, and so injective; this ensures that there's always a unique root.  (I actually haven't tried to prove JoeBlow's assertion, but this is my heuristic for why it should be true.)

[quote="pacoga"]Dear friends, thank you for your answers.

I would like it would be true for $ f(x) \equal{} \int_a^x {g(t)dt}$ being $ g(x) \equal{} \frac {1}{{\ln \displaystyle \frac {{b \plus{} x}}{{a \plus{} x}}}}$ or $ g(x) \equal{} \frac {1}{{\ln \displaystyle \frac {{b(a \plus{} x)}}{{a(b \plus{} x)}}}},$ both giving $ f''(x) \neq 0$.

Why $ f''(x) \neq 0$ solves the problem?[/quote]

We have $ f(x) \minus{} f(a) \equal{} f'(c_x)(x \minus{} a)$, which gives  $ (f(x) \minus{} f(a))/(x \minus{} a) \equal{} f'(c_x)$. Without worrying about it yet, let's assume $ f'$ has a well defined inverse. We then get  $ (f')^{ \minus{} 1}[(f(x) \minus{} f(a))/(x \minus{} a)] \equal{} c_x$. So if $ f'$ is $ 1 \minus{} 1$ and $ (f')^{ \minus{} 1}$ is differentiable, then $ c_x$ is a differentiable function of $ x$. The condition $ f''(x)\neq 0$ is the perfect assumption to guarantee both (and I don't think we need the continuity of $ f''$ here).


We have that $ B$ is a collection of closed balls in $ R^n$ with positive radii that are bounded above. Suppose $ A$ is a set such that each point $ a$ in $ A$ is the center of arbitrarily small balls in $ B$.
Prove that if $ \mu$ is finite Borel regular measure on $ R^n$, then there is a sequence $ B_{k}$ of disjoint balls in $ B$ that cover $ \mu$-almost all of A.

Note that im only looking for a hint for the problem, i was already given that i should find a finite disjoint collection of balls in $ B$ that cover at least $ \frac{1}{Q\plus{}1}$ of $ A$. However ive been plowing away for a while and have made little progress. Also we're using the following formulation of the besicovitch covering theorem:
 Given a collection of closed balls in $ R^n$ with positive radii that are bounded above there exists a number $ Q\equal{}Q(n)$ depending only on n and that there are collections $ B_{1},B_{2},...,B_{Q}$ of $ B$ such that (1) each collection $ B_{i}$ is disjoint (i.e. for every $ B(x_{k},r)\in B_{i}$, $ B(x_{k},r)$ intersect $ B(x_{j},r)$ for all $ j\ne k$ (2) the balls in the union of the collections cover A.

[quote]arbitrarily small balls[/quote]
This suggests using (perhaps the ideas of the proof of) [url=http://en.wikipedia.org/wiki/Vitali_covering_lemma#Vitali_covering_theorem]Vitali covering theorem[/url], rather than Besicovitch's.

Addition, should $ \mu$ have doubling property ?

Im not quite sure what you mean by the doubling property

A measure $ \mu$ has the doubling property if there is a constant $ C$ such that $ \mu(2B)\le C\mu(B)$ for any ball $ B$. (Here $ 2B$ is the ball with the same center as $ B$ and twice the radius). This property plays well with covering lemmas, but I don't know if it's needed here or not.


Marcy bought 120 apples for $ \$24$. When she got home, she discovered that $ \frac{1}{5}$ of the apples were rotten. If she figures she spent the $ \$24$ on only the good apples how many cents did each good apple actually cost?

[quote="GameBot"]Marcy bought 120 apples for $ \$24$. When she got home, she discovered that $ \frac {1}{5}$ of the apples were rotten. If she figures she spent the $ \$24$ on only the good apples how many cents did each good apple actually cost?[/quote]
1/5 of the apples is 24 apples, so there are 120-24=96 good apples. Therefore she spent 24/96=1/4 of a dollar for each good apple, which is 25 cents.

YAY 1337th post! :D  :D  :D


linear algebra
matrix
algebra
polynomial
complex numbers
superior algebra
superior algebra solved

Let A matrix in M_3(R) such that trA=trA=0
Prove that det(A+I_3)=(detA)+1
Radu Gologan ONM 2003
-----------------------------------------------------------
Here my solution, if you have a different one i 'll take it

See A as matrix in M_3(C), A is similar to a triangular with diagonal
(a,b,c). A is real coefficient c=b* where b* is the complex conjugate of b
(1) trA=a+b+b*=0
(2) trA=a+b+b*=0
(1) & (2) => (3) ab+ab*+bb*=0
det(A+I)=(a+1)(b+1)(b*+1)=
(detA)+1+a+b+b*+ab+ab*+bb*
=(detA)+1+ ab+ab*+bb*

(3)=>0=(ab+ab*+bb*)=ab+ab*+bb*+2abb*(a+b+b*)
det(A+I)=(detA)+1

M.Omarjee

Let f(x)=x3+ax-d, g(x)=x3+bx-d2 be the characteristical polynomials for A, respectively A2. We have to prove that
f(i)f(-i)=d2+1 <=>  (-i+ai-d)(i-ai-d)=d2+1 <=> a \in {0,2}
We have A^6=(dI-aA)2=d2I-bA2 <=> A2(b+a2)-2adA=0. If d<>0 then A is invertible and A(b+a2)=2adI thus (b+a2) tr A=4ad => a=0 and we're ok.

Otherwise, if d=0 we have A2(b+a2)=0, and if A2=0 then A has all the eigenvalues equal with 0, thus f(x)=x3, which also means a=0. Therefore b+a2=0 and thus g(x)=x3-a2x=x(x-a)(x+a), and f(x)=x3+ax, so, from f(A)=0 => A2(A2+aI)=0 which means that the eigenvalues of A can only be 0 or -a, and thus g(x) has only 0 or a as roots, which means that a=0 and again we are done.


[quote="Hamster1800"]All I know about $\mathbb{Z, Q, R, C}$ is that they mean Integers, Rationals, Reals, and Complex. I also know that $\mathbb{R}-\mathbb{Q}$ is the set of irrationals, but how else do you use them? For example the book countdown about some IMO people mentions $\mathbb{Z}\mod{2\mathbb{Z}}$ and says it separates the integers into evens and odds. How does it do this and what do things such as $\mathbb{C}^2$ mean? I saw that one in the puzzles & brainteasers forum. I'm sorry if this is too easy/difficult for this forum but I often see these things and I really have no idea what they mean fully. If it's the wrong forum, please move it, but I see it used in both this forum and pre-olympiad a lot.[/quote]

$\mathbb{Z} \bmod 2 \mathbb{Z}$ simply refers to the integers $\bmod 2$ - that is, $0, 1$.  More specifically, it refers to the ring that contains the integers $\bmod 2$ and the associated modular operations.  $0$ is the even integers, $1$ is the odd integers.  In a problem where only evenness/oddness matters, you can reduce the problem from $\mathbb{Z}$ to this much easier ring.

$\mathbb{C}^2$ refers to the set of all ordered triples $(x, y)$ with $x, y \in \mathbb{C}$.

For now, $\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$ these are the sets that you probably ever need to know. If you ask about something like $\mathbb{R}^2$, this means the Cartesian product $\mathbb{R}\times\mathbb{R}$. For example, points in the real plane like $(x,y)$ they are elements of $\mathbb{R}^2$; similarly, you can deduce that $(x,y,z)\in\mathbb{R}^3$. These are all convenient notations in mathematics.

You also asked about $\mathbb{Z}/p\mathbb{Z}$. This is the set of elements in a p-modular arithmetic. For example, let $p=2$ be simplest example. Because $\mod$ is an equivalence relation, we can partition the set of integers $\mathbb{Z}$ into equivalence classes. For this particular $p=2$, there are only 2 equivalence classes, $[0]$ and $[1]$. To put everything into symbols you can get a quick glance of it, check out below:

$\mathbb{Z}=\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$

$[0]=\{\ldots,-6,-4,-2,0,2,4,6,\ldots\}$

$[1]=\{\ldots,-5,-3,-1,1,3,5,7,\ldots\}$

$\mathbb{Z}/2\mathbb{Z}=\{[0],[1]\}$

PS: Many books uses the following notation $\mathbb{Z}/p\mathbb{Z}=\mathbb{Z}_p$
PSS: Like t0rajir0u said, this constitutes an algebraic structure known as a ring (commutative ring)

[quote="Hamster1800"]OK thanks. That made a lot of things a lot clearer. Also, am I likely to learn this in school? If so, when? My precalculus book sometimes uses things such as $n\in\mathbb{Z}$ but nowhere in any of my previous books or this one has it been explained.

Also, I knew about $\mathbb{Z}_p$ but I didn't know that it was the same as $\mathbb{Z}/p\mathbb{Z}$. Thanks for letting me know that.[/quote]It's normal that you probably never heard of the $\in$ symbol because that's certainly introduced in basic set theory. You probably have seen this inverted A $\forall$ which stands for 'for all'. You will learn many many more new things, and this makes mathematics much more interesting.

About $\mathbb{Z}/p\mathbb{Z}$, this notation has a certain use in higher algebra. The only thing I know why they write it like a fraction is that it denotes some sort quotient system. But other than that, I don't know yet.

Yes, as 10000th user pointed out, this has its reason in a quotient, namely the group or ring of quotients (taking quotients $A/B$ in general, not only for this case, means descripte to see all the elements of $B \subset A$ as "the same", and everything else "the same" that is necessary for not loosing the prior structure).
The notation $\mathbb{Z}_p$ for the residues $\mod p$ is really junk when seen from an algebraic point of view:
- it conflicts with the notation for the $p$-adic integers
- there is $\mathbb{Z}/p\mathbb{Z}$ for the integral residues $\mod p$, $C_p$ (rarely used and not very common) for the cyclic group with $p$ elements and $\mathbb{F}_p$ or $\mathrm{GF}(p)$ for the field with $p$ elements
The only reason to use it I see is to create confusion ( ;) ) and to shorten the writting (by the very important number of two letters).

Another question: In $\mathbb{Z}/p\mathbb{Z}$, why is there the second $\mathbb{Z}$? Would things like $\mathbb{Z}/p\mathbb{Q}$ make any sense to anybody?

The second set must be a subset of the first... I don't know if it is used, but maybe one could interpret $\mathbb{Q}/\mathbb{Z}$ as the rationals in $[0,1[$?

Yes, $\mathbb{Q} / \mathbb{Z}$ is used (have seen it in  paper some time ago).
And when for example the second set is "generated" by some element $p$ (here by the multiples of $p$), then you could also write $\mathbb{Z}/p$ (but only when it is clear what "generated" means; for integers this should be clear everytime).

[quote="quantum leap"][quote="t0rajir0u"]$\mathbb{C}^2$ refers to the set of all ordered triples $(x, y)$ with $x, y \in \mathbb{C}$.[/quote]
Um, is it ordered triples or pairs?[/quote]
Pairs ;)

[quote="ZetaX"]Yes, $\mathbb{Q} / \mathbb{Z}$ is used (have seen it in  paper some time ago).[/quote]

Just consider it as a quotient group? (I used this as a counterexample for some statement on my algebra exam)

I think I understand. So $\mathbb{Z}/p\mathbb{Z}$ is all the integers on the interval $[0,p)$? Or at least it makes them equivalent to one in that interval, right? Would changing the first one make it all the rationals/reals/whatever on the same interval?

[quote="Hamster1800"]So $\mathbb{Z}/p\mathbb{Z}$ is all the integers on the interval $[0,p)$? [/quote]

I think it makes more sense to think of it as the integers $\bmod p$, but yes, that is correct.  Although with your definition it's not clear how to define addition, multiplication, etc. in such a ring.

Example:  Working $\in \mathbb{Z}$ we have such statements as $7 + 13 = 20, 7 \times 13 = 91$, etc.

Working $\in \mathbb{Z} /17 \mathbb{Z}$ (as an example) those statements become $7 + 13 = 3, 7 \times 13 = 6, \frac{7}{13} = 11$.


algebra
polynomial
Ring Theory
number theory
relatively prime
algebra unsolved

Find a fifth degree polynomial $ p(x)$, such that $ p(x) \plus{} 1$ is divisible by $ (x \minus{} 1)^3$ and $ p(x) \minus{} 1$ is divisible by $ (x\plus{} 1)^3$.

[quote="t0rajir0u"]Are you familiar with the generalization of the Chinese Remainder Theorem to PIDs?  This is a simple example.[/quote]
Sorry, I don't know that generalization. I was looking for some elementary solution.

The construction is perfectly elementary; the CRT just tells you that $ p$ is unique.  Since $ (x \plus{} 1)^3$ and $ (x \minus{} 1)^3$ are relatively prime, there exist unique polynomials such that

$ p(x) \equal{} q(x) (x \minus{} 1)^3 \plus{} r(x) (x \plus{} 1)^3$

and we can take $ q, r$ to be quadratic.  The given conditions are equivalent to $ p(1) \plus{} 1 \equal{} p'(1) \equal{} p''(1) \equal{} 0$ and $ p(\minus{}1) \minus{} 1 \equal{} p'(\minus{}1) \equal{} p''(\minus{}1) \equal{} 0$, and it's easy to solve for $ q, r$ from here.

you $ (x \plus{} 1)^3 / P(x) \minus{} 1$ so $ (x \plus{} 1)^2 / P'(x)$

and  $ (x \minus{} 1)^3 / P(x) \plus{} 1$ so $ (x \minus{} 1)^2 / P'(x)$  so $ (x \minus{} 1)^2(x \plus{} 1)^2 /P'(x)$ and deg(P')=4 so

$ P'(x) \equal{} a(x^4 \minus{} 2x^2 \plus{} 1) so P(x) \equal{} a(\frac {x^5}{5} \minus{} \frac {2x^3}{3} \plus{} x \plus{} b)$

and you can detemine a and b ....


I couldn't get this '03 National Sprint.

What is the greatest 3-digit number "abc" such that 4,a,b forms a geometric sequence and b,c,5 forms a arithmetic sequence?

you know that a,b, and c have to be single-digit integers, and that leads you to the geometric sequence. Since the problem gives you that 4b=a2. The only combination that works here is a=2, b=1. With this, you can plug into the arithmatic sequence, you get c=3. Thus, the answer is 213.

Actually that isn't the answer. :(  :(  :? Way bigger.

Oops.. I didn't read the question fully. The answer is 697. Set up the proportion 4/a=a/b. In addition to my previous solution, six and nine work (a=6, b=9). If you substitute nine into the question, you get c=7, giving you 697.


Problem 2 (Group 1). Considering the structures of graphite and diamond, and the following data - $ \Delta H_{f}^{o}[C(s), diamond] = 1.9 kJ/mol, \Delta H_{f}^{o}[C(s), graphite] = 0 kJ/mol$ -, select from the following the correct alternative:

A - The standard enthalpy variation for the formation of 1 mol of diamond from graphite is -1.9 kJ/mol.
B - The energy of the covalent C-C bond in diamond is smaller than the energy of the C-C covalent bond in graphite.
C - The geometrical arrangement of the carbon atoms, in diamond and in graphite, corresponds to the same hybridization of the corresponding atomic orbitals.
D - The great hardness of diamond compared to graphite is due to the greater C-C bond lenght in diamond.

i think it should be D beacuse i don't find others right

I think it's $ D$ as well.

$ A,B$ are obviously wrong (unless I am much mistaken) and $ C$ I think grahite is $ sp^{2}$ hybridization while diamond is $ sp^{3}$ hybridization.

How does the energy of a bond varies with bond order?

Bond energy is inversely propotional to bond order for similarly bonded atoms

So, for example, a C-C triple bond has less energy than a C-C simple bond?

includin the negative sgn yes considring absolute value carbon carbon triple bond has more energy than carbob carbon simple bond

[quote="Pardesi"]considring absolute value carbon carbon triple bond has more energy than carbob carbon simple bond[/quote]

So, what is the conclusion regarding the exam problem?

Does that really make sense? Diamond is harder just because of bond lenght?

well as i said in my first post i think it is D because others are not true for sure if there would have been a none of thee option i would have gone with that


If we launch a satellite to a circular orbit around the Earth at height 357.1 km, to find the velocity needed at launch, do we just set the energies equal?:
\[ \minus{}\frac{\mu}{2\left(r_E\plus{}h\right)} \equal{} \frac{v_L^2}{2} \minus{} \frac{\mu}{r_E}\]
and then solve for $ v_L$?

Can you tell me what these quantities are and where they are coming from?  I would have an easier time answering your question if I could understand what you have there and then retrace your steps to figure out whether or not it makes sense.

First of all, I forgot to say that $ \mu \equal{} GM$, where $ M$ is the mass of the Earth.
Then, in space, the mechanical energy is
\[ \frac {v^2}{2} \minus{} \frac {\mu}{r}.\]
Using centripetal force, we have
\[ \frac {m v^2}{r} \equal{} \frac {\mu m}{r^2} \implies v^2 \equal{} \frac {\mu}{r},\]
so the mechanical energy is
\[ \minus{} \frac {\mu}{2r}.\]
Since the mechanical energy is conserved at the surface, we set it equal to that mechanical energy to find the launch velocity. Is my reasoning for the launch velocity correct?

Thanks; I was just wondering because some other person said to just let the velocity at the top stage to be $ 0$.


limit
logarithms
calculus
derivative
function
real analysis
real analysis unsolved

I often see in a book something like this:

$\lim_{h \rightarrow a}f(g(h)) = \lim_{k \rightarrow b}f(k)$ since $k=g(h) \rightarrow b$ as $h \rightarrow a$.

For instance, in a book that develops the logarithm first, in computing the derivative of $E(x)=e^{x}$, it says something such as

$\lim_{h \rightarrow 0}\frac{E(h)-1}{h}= \lim_{k \rightarrow 0}\frac{k}{\L (k+1)}$, where $k = k(h) = E(h)-1$, $L = \log$; and $\lim_{k \rightarrow 0}\frac{L(k+1)}{k}= L'(1) = 1$, so the original limit is $1$.

In this case, isn't this only justified because $k(h)$ is near $0$ but not equal to $0$ when $h$ is near $0$ and not equal to $0$, because $E$ is strictly increasing.  So in general, a limit change like in the beginning is only justified when, in addition to $k \rightarrow b$, we have $k(h) \neq b$ for $h$ in some open interval containing $a$ and $h \neq a$.

If your $g$ and $h$ functions are continous I can't see anything unclear as for a continous function $\lim_{x\to a}f(x)=f(a)$.
If they are not continous I am not so sure that holds but I guess that some stronger conditions than continuity may exist.

Also, wrong section I believe.

oops, you're right, so that's a bad example.  I meant to pick a $g$ that's not continuous.  So is what i was saying the necessary condition in the case that $g$ is not continuous?

Ok so let me continue with this question?

say all we know about a function $ u$ is something like

$ u = u(x) =-x/2+x^{2}/3+o(x^{2})$, so $ u \rightarrow 0$ as $ x \rightarrow 0$.

Then is it necessarily true that we can find an approximation as $ x \rightarrow 0$ of $ e^{u(x)}$ by saying

$ e^{u}= \sum_{i = 0}^{n}\frac{u^{i}}{i!}+o(u^{n})$ as $ x \rightarrow 0$.

What I mean is, how can we say that $ e^{u}$ necessarily acts like that if we don't know that as $ x \rightarrow 0$, in addition to $ u \rightarrow 0$, ALSO $ u$ is not $ 0$ for small enough $ x$.  I guess one way to justify the change in practice, where $ u(x)$ would be completely specified and from which the asymptotic formula would be derived, is to look at the complete function $ u(x)$ and make sure it's nice at $ 0$.  But is there a justification when $ u$ is not exactly known, or is too complicated to want to verify this.

These kind of arguments seem to be used very often, as in the previous case I mentioned, or in something like this, an approximation of a function, but i never see a justification for the variable change in the limit.

I don't see this last example so much as a variable change as an example of being careful with the notation.

We know that $ e^{u}=1+u+\frac{u^{2}}2+o(u^{2}),$ and we have your representation of $ u.$

First, note that since $ u=O(x)$ as $ x\to0,$ it follows that $ u^{3}=O(x^{3})=o(x^{2}).$

So $ e^{u}=1+u+\frac{u^{2}}2+o(x^{2}).$

Let's take a particular look at $ u^{2}:$

$ \left(-\frac{x}{2}+\frac{x^{2}}3+o(x^{2})\right)^{2}=\frac{x^{2}}4-\frac{x^{3}}3+\frac{x^{4}}9+o(x\cdot x^{2})+o(x^{2}\cdot x^{2})+o(x^{2})\cdot o(x^{2})$

$ =\frac{x^{2}}4+o(x^{2})$ since [b]all[/b] of those other terms are $ o(x^{2}).$

So $ e^{u}=1-\frac{x}2+\frac{x^{2}}3+\frac{x^{2}}4+o(x^{2})= 1-\frac{x}2+\frac{7x^{2}}{12}+o(x^{2}).$

Note also that we cannot do any more than that with the information you gave me: the very particular $ x^{3}$ and $ x^{4}$ terms that appear in my intermediate calculations are added to some nonspecific $ o(x^{2})$ terms, which means that $ o(x^{2})$ is the most accurate information that we have.

Kent,

I understand everything you wrote, but i don't understand how it answers my question.  (note: there's a small typo in the calculation of $ u^{2}$, $ -2$ should be replaced with $ -x/2$)

I don't know if i made the question clear though.  When we write $ e^{u}= 1+u+\frac{u^{2}}{2}+o(u^{2})$ as $ u \rightarrow 0$, we are asserting that the statement

$ \lim_{u \rightarrow 0}\frac{e^{u}-(1+u+u^{2}/2)}{u^{2}}= 0$

holds.  However, in our case, we let $ u$ be a function of $ x$, so that the implicit assertion made as you replace $ u$ with $ u(x)$ is now that

$ \lim_{x \rightarrow 0}\frac{e^{u(x)}-(1+u(x)+u^{2}(x)/2)}{u^{2}(x)}= 0$.

for $ u(x)$ defined as was given.  Even though we know $ u \rightarrow 0$ as $ x \rightarrow 0$, I don't see why the limit necessarily holds, since we don't know from what was given that $ u(x)$ is nice around $ 0$.  (this, i think, makes the question almost exactly the same as the first example/question i asked about in the thread)


For more information see [url=http://mersenne.zapto.org/FTW/Forum/viewtopic.php?f=3&t=16]this[/url].

It isn't supposed to.  #H34N1 has resigned.  Until someone else wants to to run/help run it, I will take his place.

cool. is everyone here just like kids and stuff?

Minicon, don't worry about having a website.  It is just a HUGE distraction and takes away from things getting done.  This isn't going to be fun if we just worry about the stuff on the website.  Hopefully, you agree.

The February tournament is cancelled, but the March is open for registration at http://aopsftw.freehostia.com/home/

Your team hasn't been registered yet, so you can go ahead and do it.


$p$ is a prime number. first prove that $p|(p-2)!+1$ then prove that it can not be power of $p$ if $p>5$.
see here if you cannot solve it
[hide]Do you know wilson $(p-1)! \equiv 1 \mod p$[/hide]

you have the sign flipped on [hide]Wilson[/hide]; thus, you probably mean (p-2)! - 1 ?

yes MystickTerminator you are right it is $(p-2)!-1$ :blush:  :blush:


Na deixtei oti  $ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\geq 3$ an $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$
 a,b,c pragmatikoi thetikoi. Silouan perimenoume lysi.

[quote="silouan"]Ρε δε σας έχω πει να μην κοντραρίζεστε με τους άλλους γιατί πάντα χάνουμε  :P  (αφού δε  σας βαλαν τίποτα m-dimensional cube πάλι , πάλι καλα)
Λοιπόν να η λύση . Από Ηοlder exoume :
$ (\sum\frac {a^2}{b})(\sum\frac {a^2}{b})(\sum a^2b^2)\geq (\sum a^2)^3$

Άρα αρκεί νδο $ (\sum a^2)^3\geq 9\sum a^2b^2$ ή αν το γράψουμε σε ομογενή μορφή :

$ x^3\geq 3y\sqrt {3(x^2 \minus{} 2y)}$ όπου $ x \equal{} a^2 \plus{} b^2 \plus{} c^2$ και $ y \equal{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$
Αν πάμε την τελευταία ισοδύναμα αρκεί νδο
$ x^6 \minus{} 27x^2y^2 \plus{} 54y^3\geq 0$ ή $ (x^2 \minus{} 3y)^2(x^2 \plus{} 6y)\geq 0$

QED  :D ('Αμα τον ξερω αυτόν που πρότεινε την ανισότητα δώστε του χαιρετίσματα  :P )[/quote]

m-dimensional cube in an n-dimensional cube klp klp e???
pantos kalo einai na kontrarizomaste akomi ki an xanoume...

[quote="silouan"]Nick ευτιχώς πέρσυ τους έδωσες κάποια στιγμή τη γωμετριάρα σου και βλέπαν λάθος και το γεωμετρικό τόπο στον υπολογιστή  :P  kai τους ξεφορτοθήκαμε (ευτυχώς δεν ήταν Μολδαβοι αυτή τη φορά) γιατί αλλιώς μέρι τότε μας είχαν σκίσει με τις άλλες geom  :maybe:[/quote]

sil etsi einai...opos ta les!na sou po tin alitheia ta eixa ksexasei kai mou ta thimises. to thema einai oti panta uparxei kapoios pou einai kaluteros kai tha se skisei. sou thimizo oti tin idia geometria eixa dosei kai sto jurie...kai hmouna polu confident. to ti egine to thumasai..sosta?

[quote]sou thimizo oti tin idia geometria eixa dosei kai sto jurie[/quote]

να ελεγα οτι την εδινες και σε κανεναν ασχετο βρε Νικο...

Απο περιεργεια τι εγινε?

[quote="filosofimenos"]Απο περιεργεια τι εγινε?[/quote]

egine oti einai o monadikos pou mou exei lusei auti tin askisi me synthetiki geometria..i askisi xorizetai se duo merh. sto proto meros edose lusi diaforetiki apo ti dikia mou kai eksairetiki eno sto deutero meros edose tin idia akrivos lush me ton upofainomeno kataskeuasti! i plaka einai se auto pou mou eipe...tin askisi tou tin eixa pei ston aera..xoris xarti kai molyvi. me ton idio tropo prospathouse na ti lusei, alla ekseplagei dioti den borouse na kataferei tpt. telika pige spiti, kathise ena tetarto kai telos...tora gia to tetarto den eimai kai polu sigouros gt mou tin eipe tin epomeni mera, alla opos kai na exei ego perimena oti oute oi 5 meres tis bmo tha tou itan arketes-opote opos katalavaineis me tapose kanonika!!

[quote="Dimitris X"][quote="Nick Rapanos"]
egine oti einai o monadikos pou mou exei lusei auti tin askisi me synthetiki geometria..[/quote]
Τι ακριβώς είναι η συνθετική γεωμετρία????[/quote]

ela sorry an xoris na to thelo fanike kati san ufo i "synthetiki geometria"...auto pou ithela na po einai me sunthetiko tropo (dld oxi me analytiki, migadikous klp)

[quote="Dimitris X"]Την έχεις ποστάρει στο forum αυτήν Nick???[/quote]

kapou uparxei sto aggliko forum (i auti i meros tis-kata pasa pithanotita olokliri) kai exei dosei mia lush o yetti me polikes suntetagmenes an den kano lathos.


It looks like all figures in the those AIME PDF files, such as question #15 in http://www.artofproblemsolving.com/Forum/resources/files/usa/45/USA-AIME-2008-45-182.pdf, are corrupted. Can it be fixed?

Thanks,
-zhidong

It's not a bug, it's just a missing feature for now. It will be corrected in the future.

Any idea how far into the future? just wonder when I should check back.

Thanks,
-Zhidong


Here is the paper and solutions to the recent Pan African Mathematics Olympiad held in Dakar Senegal. 31 July-8 August


Oh , It really is my own  :mad:   :wink:

[quote="chien than"][quote="Harry Potter"]Oh , It really is my own  :mad:   :wink:[/quote]
To Harry Potter
[hide]Anh mathboy nay,anh dung nhu vay chu,sao lai post bai tren THTT len vay,anh lam em that vong ve anh qua![/hide][/quote]
Oh My god
i am very sad   :( and despair
To moderator : Let's Close this topic . Please
To everyone : Mathlinks is a biggest mathForum ( which are the forum i know ) , Very Much member forum around the world and some member are from Viet Nam , And some time they post the problem in mathematices and youth _ A big magazin in viet Nam  .
But this really is my own
You should remember that : A lost of Problem in M & Y are not New Problem
I'm Very Sad  :(

They not same problems:
	[quote="thesunflower"]
thesunflower
Consider 6 positive numbers $ a,b,c,d,e,f$ satisfy: $ a \plus{} c \plus{} e \equal{} b \plus{} d \plus{} f.$
Prove that: \[ \frac {af \plus{} be}{(a \plus{} b)(e \plus{} f)} \plus{} \frac {ad \plus{} bc}{(a \plus{} b)(c \plus{} d)} \plus{} \frac {cf \plus{} de}{(c \plus{} d)(e \plus{} f)} \geq \frac {3}{2}\]
[/quote]
and
[quote="Harry Potter"]For $ x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}$ are the positive real number such that :
$ x_{1} \plus{} y_{1} \plus{} z_{1} \equal{} x_{2} \plus{} y_{2} \plus{} z_{2}$
Find the minimum : $ \frac {x_{1}y_{1} \plus{} x_{2}y_{2}}{(x_{1} \plus{} x_{2})(y_{1} \plus{} y_{2})} \plus{} \frac {z_{1}y_{1} \plus{} z_{2}y_{2}}{(y_{1} \plus{} y_{2})(z_{1} \plus{} z_{2})} \plus{} \frac {x_{1}z_{1} \plus{} x_{2}z_{2}}{(x_{1} \plus{} x_{2})(z_{1} \plus{} z_{2})}$[/quote] :wink:

argady, you are right  :) they are rather like each other but they are not same form  :wink:  Harry, Are you sure it is really hard? If not, I think that you should change the tittle of this topic.

[quote="arqady"][quote="Harry Potter"]For $ x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}$ are the positive real number such that :
$ x_{1} \plus{} y_{1} \plus{} z_{1} \equal{} x_{2} \plus{} y_{2} \plus{} z_{2}$
Find the minimum : $ \frac {x_{1}y_{1} \plus{} x_{2}y_{2}}{(x_{1} \plus{} x_{2})(y_{1} \plus{} y_{2})} \plus{} \frac {z_{1}y_{1} \plus{} z_{2}y_{2}}{(y_{1} \plus{} y_{2})(z_{1} \plus{} z_{2})} \plus{} \frac {x_{1}z_{1} \plus{} x_{2}z_{2}}{(x_{1} \plus{} x_{2})(z_{1} \plus{} z_{2})}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170845[/quote]
This inequality
[quote="thesunflower"]
Consider 6 positive numbers $a,b,c,d,e,f$ satisfy: $a+c+e=b+d+f$.
Prove that: $\frac{af+be}{(a+b)(e+f)}+\frac{ad+bc}{(a+b)(c+d)}+\frac{cf+de}{(c+d)(e+f)} \geq \frac{3}{2}$
[/quote]
is wrong! Try $a=3$, $c=4$, $e=3$, $b=8$, $d=1$ and $f=1$.


geometry
geometric transformation
rotation

Sunlight is known to spend 8.33 minutes on reaching Earth, and 43.3 minutes on reaching Jupiter.

a) What is the rotation period of Jupiter around the Sun?
b) Compute the mass of the Sun.

Thanks for your help! :D

(a) [hide="Answer"]From the times given and using the speed of light we can calculate the distance between each planet ($ R_{J}$, the distance of Jupiter, and $ R_{E}$, the distance of Earth) and the Sun:

$ c = \frac{R_{J}}{\Delta t_{J}}= \frac{R_{E}}{\Delta t_{E}}\Rightarrow  R_{J}= c \Delta t_{J}$ and $ R_{E}= c \Delta t_{E}$.

Now we can use the third law of Kepler (and use the known period of rotation of Earth) to derive an expression for the period of Jupiter:

$ \frac{T^{2}_{J}}{R^{3}_{J}}= \frac{T^{2}_{E}}{R^{3}_{E}}\Rightarrow  T_{J}= T_{E}\sqrt{\left(\frac{R_{J}}{R_{E}}\right)^{3}}$

$ \Rightarrow  T_{J}= T_{E}\sqrt{\left(\frac{\Delta t_{J}}{\Delta t_{E}}\right)^{3}}$.

Using the given data and $ T_{E}= 1\, year$ we get $ T_{J}= 11.85\, years$.
[/hide]

(b) [hide="Answer"]The centripetal force necessary for the curvilinear motion of a planet is identified with the gravitational force exerted by the Sun. Applying this to Jupiter:

$ M_{J}\cdot \frac{v^{2}_{J}}{R_{J}}= \frac{G M_{S}M_{J}}{R^{2}_{J}}\Rightarrow  \omega^{2}_{J}R^{2}_{J}= \frac{G M_{S}}{R_{J}}$ and $ \omega_{J}= \frac{2 \pi}{T_{J}}$

$ \Rightarrow  \frac{4 \pi^{2}R^{3}_{J}}{T^{2}_{J}}= G M_{S}\Rightarrow  M_{S}= \frac{4 \pi^{2}(c \Delta t_{J})^{3}}{G T^{2}_{J}}$.

Substituting we get $ M_{S}= 2.0 \times 10^{30}\, kg$.

[/hide]


yeah, i see what you're sayng.

maybe aops could make a filter or something.  that would be super! :D

I think, as Mr. Ruscsyk (splling?) has already said, Interval has a right to post as many problems as he would like. However, my personal opinion is that most, if not all of them, should go into classroom math.

If they MUST reside on the HSB forums, he should attempt to collect them all into ONE post with all of the problems.

[quote="now a ranger"]I used to come to HSB to look at some alluring problems. Now I  never come anymore since all I see is Interval's bombardment ( if that's even a word ) of Interval's Impaired Posts.

It is quite appalling.

Quite chancy and caustic.

What do you think?

That is my opinion. Please feel free to share your thoughts.[/quote]
I think someone should start posting more questions. I think I'll start sometime now

I must say that interval's posts might be derogatory to AoPS itself.
Consider this case:
What if you are a 9th grader who got 80 on the AMC 10, so you know that you need help with the basics in High School. You are eager to do math, and want to find a place where you can do non-routine math problems that you won't find in school. You go to the High School Basics forum, but all you see are trivial school
textbook questions. When you turn to the Intermediate Topics section, the questions are WAY too hard for you. With no other places to turn to, you leave aops. This almost happened to me, because I was scared of the difficulty of the mathcounts forum.

So, I suggest an alternative (it may be implausible because I have no idea as to how much trouble it would take). A new forum where people such as interval who want help with homework problems can go.

[quote="DiscreetFourierTransform"]Or just have a mod collect all of his posts into one large post...

Or make a sticky for problems like that...

Or just delete them all and break out the "modly shotgun of banning?" [\sarcasm][/quote]
Okay, a sticky would be more practical I suppose than my idea.

[quote="jli"]heres an idea

every week make one big topic called interval's questions or something. whenever he needs help, he can post them in there.  that way it might be easier  for him to keep track of everything and also it wont spam our forum[/quote]

I think this is a better idea than creating a new forum, since other than Interval I really haven't seen [i]too[/i] many other ppl posting textbook questions; and idk about everyone else, but I would hardly ever visit that forum. People who do need this kind of help would probably only be heard by other people who need help, and threads wouldn't really go anywhere.

Can a mod create a sticky thread in this forum, called something like "Interval's Questions" or w/e, and PM him to post his questions in it?  (Or a weekly thread would work as well, but I don't know to what degree he'd comply...)

Thanks.  :)


topology
real analysis
real analysis unsolved

Assume $G$ is an open set in $\mathbb{R}^{n}$. How to show that
1. $x\in \mathbb{R}^{n}\setminus G$ might be a density point of $G$?
2. $x\in\partial G$ might be a density point of $\mathbb{R}^{n}\setminus G$?

1/take the center of a punctured disk $G={x: 0<|x|<1}$

2/ for example, in $\mathbb{R}^{2}$, if you consider $G = \{(x,y) \in \mathbb{R}^{2}: |y|<x^{2}\}$, then $0$ is on the boundary of $G$ and is a density point of $\mathbb{R}^{2}\setminus G$, I think.

[quote="alekk"]2/ for example, in $\mathbb{R}^{2}$, if you consider $G = \{(x,y) \in \mathbb{R}^{2}: |y|<x^{2}\}$, then $0$ is on the boundary of $G$ and is a density point of $\mathbb{R}^{2}\setminus G$, I think.[/quote]
I think you are right. I just don't see the equation for the general case $n$.


Let $ K$ denote the intersection of the two diagonals of a quadrilateral $ ABCD$ where $ \hat{ABC}\equal{}\hat{ADC}\equal{}90$ and $ AC\equal{}AB\plus{}AD$. Prove that the radi of the inscribed of the triangles $ ABK$ and $ ADK$ are equal.

Stop! This is Problem T12/374 of Math&Youth Magazine, see at [url]http://www.mathlinks.ro/viewtopic.php?t=168836&search_id=79333260&start=20[/url]


Let $ f(x)=\sqrt [3] {e^x-1-x-\frac{x^2}{2}}$,prove exist ${ f'(0}$ and find it
Have funs,my friend,Nam :lol:

[quote="Hong Quy"][quote="Allnames"]Let $ f(x) = \sqrt [3] {e^x - 1 - x - \frac {x^2}{2}}$,prove exist ${ f'(0}$ and find it
Have funs,my friend,Nam :lol:[/quote]
 We have:  $ e^x = 1 + x + \frac {x^2}{2} + \frac {x^3}{3!} + ...$
$ f(x) = \sqrt [3]{e^x - 1 - x - \frac {x^2}{2}} = \sqrt [3] {\frac {x^3}{3!} + ...}$
$ f(x) = \frac {x}{6}.\sqrt [3] {1 + x/4 + ...} = \frac {x}{6}(1 + x/12 + ...)$
Hence: $ f'(0) = \frac {1}{6}$[/quote]
Can you check it agian,my result is $ \frac{1}{\sqrt [3] 6}$ but my main idea is similar to yours


A train traveling at 60 miles per hour reaches a tunnel. The front of the train enters the tunnel, and 40 seconds later the front of the engine exits the tunnel, How many feet are in the length of the tunnel?

Assuming that the engine is located at the front of the train...
The train goes $ 60$ miles per hour, or $ 5280$ feet per minute. It takes $ 40$ seconds for the train to go through the tunnel, and $ 40$ seconds is $ 2/3$ of a minute. Thus, the tunnel is $ 5280 \times \frac {2}{3} \equal{} 3520$ feet long.


Santa Claus has a triangular area fenced in on a square lattice
on the xy-plane.  Each lattice point has one post at it.  There
are only posts at the lattice points.  He has a fence going from
location (0,0) to (200,10) to (10,200) back to (0,0).  (The
lattice points are 1 unit apart vertically and horizontally from
their nearest neighbors.)  He wants to know how many posts
are strictly inside the triangular fenced in area.  He said you
may use a combination of arithmetic and elementary algebra only.

Even if you decide not to attempt this, please listen to him here
(with your sound turned on):
http://www.birthdaycards.com/pickup?ID=A222-KT8H-92CP-ZWS

[hide="hint"]
Pick's Theorem
[/hide]

[hide="solution"]
We have a triangle with vertices at (0,0), (10,200), and (200, 10).
The determinant (shoelace) method for finding areas of polygons gives the area as 19950.
There are 210 lattice points on the boundary.
By Pick's Theorem, A=i+b/2-1, where b is the number of lattice points on the boundary and i is the number of lattice points in the interior.
19950=i+210/2-1
19950=i+105-1
19846=i
[/hide]

Previous poster,

please check your home for an extra Christmas present.

Also, for an alternate way of doing part of the hint:
[hide]This is an isosceles triangle; the line y = x is the
line of symmetry.

Look at the midpoint of the line segment containing
(200,10) and (10,200).  It is (105,105).  The height
of the triangle is the distance from (0,0) to
(105,105), or 105sqrt(2) units.  The length of the base
of the triangle is the distance from (10,200) to (200,10)**,
or 190sqrt(2) units.

The area of the triangle = (1/2)bh =
(1/2)[190sqrt(2)][105sqrt(2)] = (1/2)(2)(190)(105) =
19950 square units.

**Or it is equal to twice the distance from the midpoint
(105,105) to one of the endpoints on that line segment.
[/hide]


probability
function
integration
calculus
real analysis
real analysis unsolved

Suppose $ p(x)$ is a continuous probability density function which
satisfying the following condition (where $ M$ and $ K$ are positive
constants, t is time and no less than zero):
\[ \frac {\int_{t}^{\infty}(x \minus{} t)p(x)dx}{\int_{t}^{\infty}p(x)dx} \equal{} Kt \plus{} M.\]
Then through computation, it is clear that $ p(x) \equal{} (K \plus{} 1)M^{1 \plus{} \frac {1}{K}}(Kx \plus{} M)^{ \minus{} 2 \minus{} \frac {1}{K}}$ is one solution.
Then my question is: Is the solution is unique? If it is, then how to prove the uniqueness?
Thanks a lot! :)

Multiply both sides by $ \int_{t}^\infty p(x)dx$ and differentiate twice with respect to $ t$ to obtain the ODE
\[ p(t) \equal{} \minus{} 2Kp(t) \minus{} (M \plus{} Kt)p'(t)\]
The family of functions $ p(t) \equal{} c(M \plus{} Kt)^{ \minus{} 2 \minus{} \frac {1}{k}}$ for any constant $ c$ satisfies the ODE (though of course the normalization constraint if $ p$ is a probability density fixes $ c$).  Further, these are the [b]only[/b] solutions to the ODE as guaranteed by the usual uniqueness results from the theory of differential equations.

Then, the condition that this is a probability density function fixes the integral (from $ 0$ to $ \infty$) at $ 1$, nailing down $ c\equal{}(K\plus{}1)M^{1\plus{}\frac1K}$. The solution is unique.


geometry
incenter
inradius
perimeter
area of a triangle

Prove that in triangle ABC, Ra = K / (s-a), where Ra is the exradius of excircle opposite vertex A, K is the area, s the semiperimeter, and a the length of side BC.

[hide="Solution"]
Draw the incircle, and then draw the excircle, let $ I$ be the incenter and $ E_a$ be the excircles center, extend line $ AC$ and draw the radius of each circle and call the tangent points on the line $ AC$ for the incircle and excircle points $ P_1$ and $ P_2$ respectively.

Let the exradius and the inradius be $ R_a$ and $ r$ respectively.

Note that $ AP_2 \equal{} s$, now $ \triangle AIP_1$ is similar to $ \triangle AE_aP_2$, which means that $ \dfrac{E_aP_2}{IP_1} \equal{} \dfrac{s}{s \minus{} a}\implies R_a \equal{} \dfrac{s\cdot r}{s \minus{} a}$. It is well-known that the area of a triangle is the inradius times the semi-perimeter $ \implies$ Q.E.D.

- I'm making a drawing so that it looks clearer.
[/hide]


limit
trigonometry
calculus
calculus computations

1. $ \lim_{x\rightarrow0}\frac{x}{\sqrt{1\minus{}cos4x}}\equal{}?$

2. $ \lim_{x\rightarrow0}\frac{2sinx\minus{}sin2x}{x^3}\equal{}?$

(without L'Hospital)

Since we have $ \cos 2x \equal{} 1\minus{} 2\sin^2 x$, we find that
\[ \lim_{x \to 0} \frac{x}{\sqrt{1\minus{}\cos 4x}}\equal{}\lim_{x \to 0} \frac{x}{\sqrt{2\sin^2 2x}}\equal{}\lim_{x \to 0} \frac{x}{\sqrt{2} \cdot \sin 2x}.\]
Manipulating, we have
\[ \lim_{x \to 0} \frac{1}{2\sqrt{2}} \cdot \frac{2x}{\sin 2x} \equal{} \frac{1}{2\sqrt{2}} \equal{} \boxed{\frac{\sqrt{2}}{4}}.\]

Don't post the problem related to Calculus in this forum.

You should post such problems in Calculus Computations and Tutrials next time.

Japanese Communities Moderator

kunny

Sorry! I thought that restriction is only for intermediate topics :blush:

[quote="queen"]1. $ \lim_{x\rightarrow0}\frac {x}{\sqrt {1 \minus{} cos4x}} \equal{} ?$

2. $ \lim_{x\rightarrow0}\frac {2sinx \minus{} sin2x}{x^3} \equal{} ?$

(without L'Hospital)[/quote]Thx kunny. @OP+AoPSers: Please read the subforum's guidelines BEFORE posting, limits is not HSB topic. I did not PM so that other users of this forum should know this fact as well.
2. Recall that $ \sin2\theta \equal{} 2\sin\theta\cos\theta$ and remember $ \lim_{x\rightarrow0}\frac {\sin x}{x} \equal{} 1$ and $ \lim_{x\rightarrow0}\frac {1 \minus{} \cos x}{x^2} \equal{} \frac12$. The rest is straightforward.


calculus
vector
integration
geometry
probability
email
complex numbers

This thread is created for discussions about this year's training camp for IMO.

if i do get a chance at the camp,
i will post the partial results of various students at the camp. :maybe:

[b]best of luck [/b]to the juniors who i think must have already reached the camp.

senior batch camp starts from 10th May,2007and continues till 31st May , 2007.

hope india does wonders at IMO 2007 to be held in Hanoi, Vietnam
and regains back lost glory.

http://www.iisc.ernet.in/mocell/indiainimo/

the above link gives india's rank at IMO till 2001.
i will write the rest

IMO 2002
[b]total score[/b]-156
[b]rank[/b] - 9
[b]1[/b] gold
[b]3[/b] silvers
[b]2[/b] bronze

IMO 2003
[b]total score [/b]-115
[b]rank[/b]-15
[b]4[/b] silvers
[b]1[/b] bronze
1 HM

IMO 2004
[b]total score[/b]-151
[b]rank[/b] -14
[b]4[/b] silvers
[b]2[/b] bronze

the nightmare starts:
IMO 2005
[b]total score[/b]-81
[b] rank[/b] -36
[b]1[/b] silver
[b]1[/b] bronze
3 HM's

IMO 2006
[b]total score[/b]-92
[b]rank[/b]-35
[b]5[/b] bronze
1 HM

http://en.wikipedia.org/wiki/India_at_the_IMO

what will happen this year?

it seems indians are showing declining interest in the IMOs.

i think there are three possible reasons:
1)they are more serious about their IIT preparations and IMO is just another competition for them. :o

2)the training programme is turning into a farce.
last year when i went to the camp as a junior, everything was so obsolete and almost like i was in a regional level camp. :mad:

the class taken on geometrical transformations by Professor K.Ranaganathan was the one of the few relevant and interesting classes.
but bad luck he took only one class.
rest was the same old problems and tricks which the profs must have taught ten years back.

they are not changing with times. very few people in the indian camp know about modern methods of geometry like inversive, projective, affine,etcetera and they profs also dont teach these.

[i]a welcome change[/i]:
last year Prof R.Bapat was proposing the inclusion of previous years senior students to spare some time in the summers for the IMOTC. :roll:
only these young people know what it takes to appear for an olympiad and they will be better teachers (i hope!!)
([b]these steps have been taken by other good countries long ago.[/b])

3)people like abhishek dang have lost interest in the IMO and are drawn devotedly towards 'mathematics'.  :(
see he got a bronze in 9th, a silver in 10th (really commendable) and we expected him to get a gold this year!!

let us take the case of Moldova's TSTs.
India's TSTs are tougher than theirs.

But Moldova could produce an iurie boreico (1 silver, three golds, 2 perfect scores) running for the 4th gold and third perfect score. :first:
hats off to him and his patience.(doesn't he get bored of olympiads?)

dang scored 128/150 in the indian TST and there's no doubting that he is a genius and well-deserved a gold.but i think he went more inclined towards deeper maths.(which generally leads to loss of touch in techniques and skills of solving probs in 4 and half hours)
(maybe like terence tao who gave up many of his IMO chances for building up a proper career in mathematics.)

so we never know. :maybe:

we cannot conclude that indians are getting dumber or sth.
its that there is a gradual decrease in scope, training and interest.

i seriously hope all these are revived again and indians come into the top 10 this year.

so
enough of wailing and crying and reflecting. :P

next i will be posting things relevant to IMOTC 2007 only. :huh:

A well written account,Soumyashant.A good analysis of the causes behind India's debacle in the last few years.Not quite a debacle,but it could turn into a seriious one if proper measures are not taken immediately.
Good luck at the camp.
Sathej

thanks

and [color=indigo][b]best of luck[/b][/color]
to all the IMOTC participants this year.

P.S.
one of the reasons of posting this is to make my number of posts 14.
[i]Triskaidekaphobia[/i] they call it. :P

[b]List of INMO-2007 Awardees[/b]
Students from Class XII
No. Name Class Region
1 U.Abhiman Acharya              XII KARNATAKA
2 Ankit Garg                             XII PUNJAB
3 Harsh H.Pareek                    XII MUMBAI
4 Shah Nisarg Kirtikumar         XII GUJARAT
5 P.N.Bhargav                         XII TAMILNADU
6 Ankesh Agarwal                   XII RAJASTHAN

Students from Class          XI and below
No. Name Class Region
7 Anuj Kumar                             XI DELHI
8 Saurabh Goyal                        XI RAJASTHAN
9 Rudradev Basak                     XI WEST BENGAL
10 Aditya Mehta                         XI MUMBAI
11 Rik Sengupta                        XI WEST BENGAL
12 Abhinav Garg                        XI PUNJAB
13 Prashant Sunil Sohani          XI MAHARASHTRA
14 Modi Aniket Diliprao              XI MAHARASHTRA
15 Aravindh Mahendran             XI MUMBAI
16 Nishant Totla                         XI MAHARASHTRA
17 Abhimanyu-Banerjee           XI KARNATAKA
18 Arijit Banerjee                       X KARNATAKA
19 Sajiv Chandra(KV)               XI TAMILNADU
20 Soudeep Deb                       XI WEST BENGAL
21 Rushabh. B. Shah                XI MUMBAI
22 Karthik.C.S                           XI KARNATAKA
23 M.Prasanna(KV)                   XI TAMILNADU
24 Ananth Shankar(KV)             X TAMILNADU
25 Anuj Tawari                          XI RAJASTHAN
26 Ninad Lalit Sancheti             XI MAHARASHTRA
27 Ashwath Rabindranath          X TAMILNADU
28 Himalay Senapati                 IX ORISSA
29 Moumanti Podder                 XI WEST BENGAL
30 Pradeep George Mathias    XI KARNATAKA
31 M.N.Ravi Chandra               XI ANDHRA PRADESH
32 George Sebastian(KV)         XI KARNATAKA
33 Utkarsh Tripathi(KV)            XI UTTAR PRADESH
34 Priyanka C Mehta(?)             XI GUJARATH

the above people form the [i]junior batch [/i]at the camp.

dont know the complete list of [i]senior batch [/i]students who will be attending the camp this year.
 :|

so this data can only be provided when i reach the camp.

most probably the first practice test is on 11th May, 2007.  :no:

The link below is the yahoo group for IMOTC ers.

http://groups.yahoo.com/group/Yo_IMOTC/

as per the information from the link above
some of the seniors are:

1)G. Vignesh      _________________      Chennai  XI
2)Rohit Singh     __________________       Dehradun  XII
3)Santhosh Ananthakrishnan  __________Mumbai   XII
4)Manas Joglekar    ________________  Mumbai  XII  (bronze-IMO 2006)
5)Soumyashant Nayak(me) ________  Bhubaneswar XI
6)Udbhav Singh___________________  Allahabad    XI
7)Arpit Amar Goel___________________ New Delhi    XII
8)Rahul Singh ____________________       New Delhi  XII
9)Gaurav       ____________                              "
10)Ashish       ____________                             "
11)Vihang Mehta   ___________                     Pune   XI
12)Sameer Wagh     _________                   Pune  IX
13)Aswin Deopurkar   _________                 Pune   XII
14)Ankit Baraskar     __________                 Pune    XI
15)Gautam          _____________                   Mumbai XII
16) Venkata        ______________                    Jamshedpur  XII
17)Abhishek Dang*  ________________          Pune XII   (bronze IMO-2004
                                                              silver IMO - 2005
                                                               HM IMO - 2006)
* means i have no sources to confirm this but i have no reason to doubt that also.

im typing from IMOTC Bombay.
Correction: First practice test is on 8th.

P.S.
Find all positive integral solutions of:
a^3+ 81b^6= c^12

why ,aren't we seniors allowed to give the first practice test? :maybe:

then why did they call us on 10th.

they should have called us on 8th maybe.

 :lol:

i think this year there will be 3 practice tests.
or maybe there will be 6 selection tests.

the latter will be a better idea. :wink:

:thumbup:
Best of Luck to all (Belated though to juniors :( )
Have a nice Camp.

[i]It seems they are going to reduce the selection tests to 4 this year........... [/i] :o

i hope this turns out to be a rumour.

5 was good enough..

some years back even we had 6 selection tests and we were doing well at the IMO (disclaimer-any resemblance with fact is coincidental  :lol: )

personally i believe more number of tests reduce the luck factor and omit the probability of a bad day affecting an able person's chance for selection. :ninja:

IMOTC results are out

the six member team consists of

1)abhishek dang (pune)  72/84    (that's obvious)  super-super senior
2)manas joglekar  (mumbai)  67/84  [b]one best solution award[/b]  senior
3)prashant sohani  (pune)    56/84  (not very sure about the place)  junior
  [b]got two best solution awards[/b]
4)utkarsh tripathy  (uttar pradesh) 53/84  [b]one best solution award[/b]  junior
5)atul shekhar   (darbhanga, bihar)  48/84   senior
6)P.N. Bhargav    (Chennai)    44/84   junior
[i]he came to know about regional maths olympiad this year and this was his last chance and he managed to make it to the team.great performance.[/i]

Arpit amar goel from delhi got a best solution award also.

[b]best wishes[/b] to the indian team for IMO 2007 to be held at Hanoi, Vietnam from 19th july to 31st july.

abhishek dang has already won a bronze, a silver and an HM in 2004,2005,2006 respectively.
manas joglekar won a bronze in 2006.

so four new faces.

Lanchka  :coolspeak:  :thumbup: .(P.N.Bhargav i suppose)
Great....

[b]P.S.[/b]soumyashant post some problems in their respective sections

Well Prashant is originally from Mumbai but he has been in Pune from the last year.

I think four ST's was a bad idea.
And yeah Bhargav is a really great person ... he deserved what he achieved.

Congrats Bhargav! I heard you have got around AIR 250 in JEE. I would strongly advise you to go in for a Mathematics career after a UG Course at CMI or ISI. A person who makes it to the team in the first attempt without a significant amount of coaching would make a really good mathematician. You could ask Vipul for suggestions.Wonderful performance for a debutant.After a long time, a contetsant from Chennai for the IMO.Really great!All the best for the IMO.
Sathej

Thank you for saying that I deserved the spot...I messed up the last test and hence the very lower score...It was a great camp and we will all do our best to make India proud!!!

Cheers
Bhargav

I (told you Bhargav but will say it once again) - had an amazing time at IMOTC and would like to congratulate all the team members - particularly Bhargav as this is really awesome for someone from Madras. Good luck for the IMO to all six of them. Come back with golds!!

I agree with those people who just dont care for imo, since if someone just solves problems all day, he will lose his imagination and just become a problemsolvingmachine without any thinking further, about new things.
The other thing is the time, who cares if one can solve 999 problems in x hours, when he becomes a mathematician there will be no time limit.
So what the imo organizers should do, is not a contest where you have to solve problems in a given time, but  like find something new  discover a new proof for a theorem etc. :wink:  :read:

True.Problem Solving is not a pre requisite to become a good mathematician.There are some excellent problem solvers who otherwise have made no significant contribution to realms of higher mathematics.And there are top-notch mathematicians who are novices at problem solving.Solving problems,that too,within a time bound is not the best judge of a person. Nevertheless,the IMO has admirably stood the test of time and to change the mode would require some thought.
Sathej

Hi everybody,

Yes Bhargav, well done, though I - for one - was expecting you to get a better score on your selection tests. All the best for IMO to all six members of the team...nothing short of gold will do, so no pressure...

And yes, Vihang or Udi or any of you out there, send me a full list of the seniors' emails at rik_sengupta@yahoo.co.in, so that I can send out some of the photos.

Bhargav, we want you back next year on the camp with a gold medal (obviously as a medallist). Aren't Prashant and Utkarsh on this forum?

Cheers, and all the best to make India proud!

Rik

agreed. :D

IMO has no excuse for leaving mainstream subjects like calculus.They actually removed complex numbers few years back from the syllabus.

when IMO started ,calculus was not taught in high school but now it is.

the course should develop with time and not stagnate with guessing type of problems.

Yes, including Calculus would not be a bad idea. But, that would make it more like Physics and Chemistry Olympiads where students below Class XI literally have no chance. A further problem would be the level of Calculus. I know certain countries that have even Vector Calculus and Multiple Integrals in High School.
Sathej

i think the problems in olympiad without even being included calculus can be made a bit less "guessing type" which appear almost everyear in INMO....should be replaced by beautiful applications of  certain fundamental theorems taught at the basic level giving a chance to young bright students.
having guessing type questions for ex.
[b]INMO '07-2
INMO '06-6(ii),2(somewhat)
INMO '05-i think 4
INMO '04-5[/b]
.....
should be replaced by better problems.
the questions have to change with time.well,calculus is great but olympiad like calculus problems can get messy..

True, problems whose solutions can be guessed may be avoided as they do not really stretch the proof-writing capabilities of a contestant. However,hardly any IMO problems are of this type.INMO-well, they would do well to avoid such problems.
Sathej

yes IMO problems are probably could be a good model  to set problems they are well balanced or rather we should say well balanced problems are IMO problems .they could be used as standards while setting questions...
actuallly the entire thing  lies on the fact that in math olympiad unlike other olympiads technique ,experience plays a huge role determining your sucess.Experience though may not help you always but technique surely does...

Experience and problem solving techniques go a long way in cracking such a time-based contest.Discounting the rarely seen show of genius,experience comes in handy. Developing problem solving techniques at an early age also helps.
Sathej

i think the level of Indian ST paper is going down.  :(
past year papers were way better than the present situation.

a person who sits on AOPS for an hour everyday will know atleast five of the 12 questions beforehand.

think about the less fortunate people who dont even know that sth like aops exists and about people like me who do not frequent AOPS throughout the year.

they should realise that the ISL questions can leak by these kind of forums (i am no way blaming AOPS  :lol: ).

The professors do part-time research and their main job is in the MO cell. so cant they prepare 12 good original questions in a year? :mad:
(anyways they dont even do that for the postal tests which are generally taken from other countries TST like romania)

this is not a good situation as some good people might miss the chance just because sb knew the quetions before.

 :ninja:

personally i feel india will as a team get 1 gold,1 silver,4 bronzes(or 3 B + 1HM) this time without reference to any individual performance. :wink:

Very true. They must come up with original problems!
Sathej

This thread deserves a pin lol. Is there someway we could get the INMO organisers, TC people to read this. Maybe those of us who make it can do something to change the system. I'm highly unlikely to make it  :rotfl: I can only do questions where no formulae are involved, or if they are from PNC :p But guys like keshav prolly will make it.

Just read this thread, and I realised it should be made an announcement.
[hide=Comment]This thread reflects upon what can be done for the mathematical olympiad programme in India in an insightful manner. Also, it is a good example of how discussions can be done in a productive manner, in view of the recent happenings on the forum. [/hide]
Edit : 2000th post.

This thread will help more $9^{\text{th}}$ & $10^{\text{th}}$ grader Indian students to send petitions to HBCSE about the faults in the selection procedure in RMO, INMO & IMO Team. :D

Nice thread BTW. :)


Could you guys help me with these questions. For the drawings part could you give me a link where I could find it. ty

1. Make a drawing to show how the gyri help increase the surface area of the brain.
Write a short paragraph explaining the drawing and telling why increased surface area is important

2. make a labelled drawing showing what you think a coronal cross-section cut through the middle of your brain model would look like.

3. write a short paragraph explaining
a. what makes your play-doh creation a useful model.
b. why your play doh model is not a perfect model.

we made models of brains with clay. we had the cerebellum, amygdala, etc. in it.

THANKS FOR YOUR HELP

I don't get it. Can someone help me? I can't seem to make any equations to solve this problem.

plz put it in the biology section??????????????