Divisibility
number theory
relatively prime
IMO
IMO 1994

Answers:[/i][/u][/b] $(m,n)={(1,2),(2,1),(1,3),(3,1),(2,2),(2,5),(5,2),(3,5),(5,3)}$

[b][u][i]Solution: [/i][/u][/b] We know that $n^3$ and $mn-1$ are relatively primes, thus from $n^3(m^3+1)=(m{}^n{}^3+1)+(n^3+1)$ and we see that $n^3+1$ is divisible by $mn-1$ if and only if $m^3+1$ is divisible by $mn-1$. Therefore, if the pairs $(m,n)$ holds then $(n,m)$ also holds. Now in the future we will say that $m \geq n$.

 $1.$ If $m=n$, $\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ which gives (2,2)

 $2.$ If $m>n=1$, we have $\frac{n^3+1}{mn-1}=\frac{2}{m-1}$ we get (2,1) and (3,1)

 $3.$ If $m>n>1$, since $(m,n)$ holds, then natural $\frac{n^3+1}{mn-1}+1$ is divisible by $n$. $\frac{n^3+1}{mn-1}=kn-1$ for some $k$. Thus we get inequality $kn-1<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$,
or $(k-1)n<1+\frac{1}{n-1}$. (*)

Since $n>1$, (*) holds only when $k=1$. By the way we observe $n^3+1=(n-1)(mn-1)$ gives us $m=\frac{n^2+1}{n-1}=n+1+\frac{2}{n-1}$ which gives us (5,2) and (5,3).

davron

here we go-
suppose $ \frac{n^3\plus{}1}{mn\minus{}1}$ is a positive integer, call it k:
$ \frac{n^3\plus{}1}{mn\minus{}1}\equal{}k \equal{}> n^{3}\plus{}1\equal{}kmn\minus{}k \equal{}> 1\plus{}k \equiv 0\mod{n}$
therefore, exists integer t such that 1+k=nt:
$ n^{3}\plus{}1\equal{}(nt\minus{}1)(nm\minus{}1)\equal{}mtn^{2}\minus{}n(t\plus{}m)\plus{}1 \equal{}>(*) n^{2}\minus{}(mt)n\plus{}(t\plus{}m)\equal{}0$
if we solve this as an equation in n, we must have $ (mt)^2\minus{}4(t\plus{}m)$ be a perfect square:
$ (mt)^2\minus{}4(t\plus{}m)<(mt)^2$
$ (mt)^2\minus{}4(t\plus{}m)>(mt\minus{}1)^2 \leftrightarrow \minus{}4t\minus{}4m>\minus{}2tm\plus{}1 \leftrightarrow 2(t\minus{}2)(m\minus{}2)>9 \leftrightarrow (t\minus{}2)(m\minus{}2)>4$
so, if we want a solution, we need $ (t\minus{}2)(m\minus{}2) \leq 4$. there are few possible cases:

m=1: $ \frac{n^3\plus{}1}{n\minus{}1}\equal{}n^{2}\plus{}n\plus{}1\plus{}\frac{2}{n\minus{}1}: n\equal{}2,3\minus{}>$[b](2,1) (3,1)[/b]
when m>1:
t=6: m=2 or m=3. the solutions are not integers.
t=5: m=2 or m=3. no integer solutions.
t=4: m=2,3, or 4. no integer solutions.
t=3: m=2,...,6. m=2: n=1,5: [b](1,2) (5,2)[/b]. m=3,4,5,6: no integer solutions.
t=2: $ m^2\minus{}m\minus{}2$ should be a perfect square. for m>3: $ m^2>m^2\minus{}m\minus{}2>(m\minus{}1)^{2}$. for m=2: we get 0 ,and then n=2. [b](2,2)[/b]. for m=3: we get the solutions [b](1,3),(5,3)[/b]
t=1: $ m^2\minus{}4m\minus{}4$ should be a square. for m>6: $ (m\minus{}2)^2>m^2\minus{}4m\minus{}4>(m\minus{}3)^2$. for 1<m<7: only m=5 gives a square, and the solutions are [b](2,5),(3,5)[/b].

m=2: the same as t=2 ($ (*)$ is symmetrical with m and t) - $ t \leq 3$, and we already found the solutions for t=1,2,3.

All the solutions are: $ (2,1) (3,1) (1,2) (5,2) (2,2) (1,3) (5,3) (2,5) (3,5)$

I don't understand this part of Davron's solution:

[quote="Davron"]We know that $ n^3$ and $ mn - 1$ are relatively primes, thus from $ n^3(m^3 + 1) = (m{}^n{}^3 + 1) + (n^3 + 1)$[/quote]

n and mn-1 are relatively prime, because mn-1 is one less than a multiple of n.

It follows that $ n^3$ and mn-1 are relatively prime.

all of you right,it's an other question(it's very nice)
p is a prime number and m,n are integers.find all(m,n,p):
p=m^2+n^2 and m^3+n^3-4\p is integer

We see that
\[ \frac{n^3+1}{mn-1} = \frac{m^2n^2+mn+1+\frac{m^3+1}{mn-1}}{m^3},\]which means that $n^3+1$ is divisible by $mn-1$ iff $m^3+1$ is divisible by $mn-1$ as well. This means that if $(m,n)$ works, $(n,m)$ does as well. From now on, we can therefore assume that $m \ge n$.

If $n=1$, then $\frac{n^3+1}{mn-1} = \frac{2}{m-1}$, which means that $m=2,3$. Therefore, $(m,n)=(2,1),(3,1),(1,2),(1,3)$.

If $n=2$, then $\frac{n^3+1}{mn-1} = \frac{9}{2m-1}$, which means that $m=5,2,1$. Therefore, $(m,n)=(2,5),(2,2),(2,1),(5,2),(1,2)$.

We now know that $m \ge n > 2$. We now bring everything in terms of $n$. Since $m \ge n>1$, we have
\[ \frac{n^3+1}{mn-1} \le \frac{n^3+1}{n^2-1} = n+\frac{1}{n-1}. \]Now, note that $\frac{n^3+1}{mn-1} + 1 = \frac{n^3+mn}{mn-1}$, which divides $n$. Since $\frac{n^3+1}{mn-1}+1 \le n+1+\frac{1}{n-1}$ and $n-1>1$, we must have $\frac{n^3+1}{mn-1}+1=n$. From this, we get $m=\frac{n^2+1}{n-1} = n+1+\frac{2}{n-1}$. This means that $(m,n)=(5,2),(5,3),(2,5),(3,5)$.

Taking the union of all the solutions, we have
\[ (m,n)=(2,1),(3,1),(1,2),(1,3),(2,5),(2,2),(5,2),(5,3),(3,5). \]

Instead of $n^3+1$ and $m^3+1$, I prefer using the smaller values $n^2+m$ and $m^2+n$.

The answer is the set of $9$ pairs
\[
(1,2), \, (1,3),\, (5,2),\, (5,3),\, (2,2),\, (2,1),\,(3,1),\,(2,5),\,\text{and }(3,5).
\]
This can be seen to work; it remains to prove these are the only solutions.

Let $\tfrac{n^3+1}{mn-1} = k$, so that $n^3 + 1 = k(mn-1)$.  Taking this equality modulo $k$ yields $1\equiv -k\pmod n$, so $k\equiv -1\pmod n$.  This means there exists an integer $p$ such that $k = pn-1$, which means we now have the symmetric-looking equation
\[
n^3 + 1 = (mn-1)(pn-1).
\]
Expanding and cancelling out the $1$ yields $n^3 = mpn^2 - (m+p)n$; as $n\neq 0$, we may divide by $n$, which leaves us with the quadratic
\[
n^2 - mpn + (m+p) = 0.\qquad(*)
\]
The discriminant of this quadratic is $\Delta = (mp)^2 - 4(m+p)$; it remains to determine which ordered pairs $(m,p)$ make this a perfect square.

We case on the values of $m$ and $p$.

[list]
[*]If $m = 1$, then the discriminant is $p^2 - 4p - 4 = (p-2)^2 - 8$; this is a perfect square precisely when $p-2 = 3$, or $p = 5$.  Similarly, $p = 1$ produces $m = 5$ as the lone solution.
[*]If $m = 2$, then the discriminant is $4p^2 - 4p - 8 = (2p-1)^2 - 9$; this is a perfect square precisely when $2p-1$ is either $3$ or $5$, which yields the two solutions $p=2$ and $p=3$.  The case $p=2$ similarly yields $m=2$ and $m=3$.
[*]Finally, if both $m$ and $p$ are greater than $2$, then the inequality $(m-1)(p-1) > 2$ rewrites as $mp > m + p + 1$, and so
\[
(mp)^2 > (mp)^2 - 4(m+p) > (mp)^2 - 4(mp-1) = (mp-2)^2.
\]
This is a contradiction, as $(mp)^2 - 4(m+p)$ is a perfect square with the same parity as $(mp)^2$ and thus must be at most $(mp-2)^2$.
[/list]
The case analysis above produced three possibilities for the unordered set $\{m,p\}$: $\{1,5\}$, $\{2,2\}$, and $\{2,3\}$.  Solving for $n$ in $(*)$ for each of these three possibilities yields the nine ordered pairs shown above.

 First, let $m,n \geq 4$.
$mn-1 \mid n^3 +1 => mn-1 \mid n(n^2 +m) => mn-1 \mid n^2 +m$.
As $mn-1 \mid n^2 +m => mn-1 \mid n^2m+m^2 => mn-1 \mid mn^2+m^2 -n(mn-1) => mn-1 \mid m^2 +n$.
Now, we need $m+n^2 \geq mn-1 => n^2 +1 \geq m(n-1) => \frac{(n+1)(n-1) +2}{n-1} \geq m => n+1 + \frac2{n-1} \geq m => n+1 \geq m$. Similarly, we get $m+1 \geq n$ (Note: till here we've assumed $m,n \geq 4$). Thus, $m=n+1,n,n-1$.
Now it's just a game of substituting- put $m=n-1$,$m=n$,$m=n+1$ and you'll get no solution for $m,n \leq 4$- here's an example: If $m=n-1 => n^2 -n -1 \mid n^3 +1 => n^2 -n-1 \mid n^2 +n+1 => n^2 -n-1 \mid 2n+2 => 2n+2 \geq n^2 -n-1 => n^2  -3n-3 \leq 0$ which of course has no solutions for $n \geq 4$.
What we have left is cases when either $m$ or $n$ is lesser than $4$- we have no choice but to put in values and see which values work, and the math part is over...

Directly from the divisibility condition we get $mn-1|m^2+n, n^2+m, m^3+1$, so $WLOG  m \geq n$. We have $mn-1 \leq n^2+m$, so $m(n-1) \leq n^2-1+2$ and therefore $m \leq n+1+\frac{2}{n-1}$.

Assume $n \leq 3$. Substituting in the expression we get one-variable polynomial divisibilities and obtain the solutions $(1, 2), (2, 1), (1, 3), (3, 1), (2, 2), (2, 5), (5, 2), (3, 5), (5, 3),$ counting the ones with $n > m$ too.

Now suppose $n \geq 4$. This means $m \leq n+1+\frac{2}{n-1} < n+2$. We have the only possibilities $m=n$ and $m=n+1,$ both of which are simple one-variable polynomial divisibilities and bring no new solutions, therefore we are done $\blacksquare$.

[quote=ISL 1994 N2]Find all ordered pairs $ (m,n)$ where $ m$ and $ n$ are positive integers such that $ \frac {n^3 \plus{} 1}{mn \minus{} 1}$ is an integer.[/quote]
[b][color=#000]Solution:[/color][/b] Note $\gcd(mn-1,n)=1$
$$\implies \frac{n^3+1}{mn-1}=\left(\frac{n^3+1+mn-1}{mn-1}\right) -1 \implies \frac{n^3+mn}{mn-1}=\frac{n(n^2+m)}{mn-1} \in \mathbb{Z} \implies mn-1 \mid n^2+~m $$
Now,
$$\frac{n^2+m}{mn-1} \in \mathbb{Z} \implies \frac{n(mn-1)+m^2+n}{mn-1} \in \mathbb{Z} \implies mn-1 \mid m^2+~n \implies \begin{cases} \tfrac{m^2+1}{m-1} \geq n \\ \tfrac{n^2+1}{n-1} \geq m \end{cases}$$
Hence,
$$\begin{cases} m \le \frac{n^2+1}{n-1}=n+1 +\frac{2}{n-1} < n+2 ~ ~ \qquad [ \text{for } n>3] \\n \le \frac{m^2+1}{m-1}=m+1+\frac{2}{m-1} < m+2 ~ ~ \qquad [ \text{for } m>3] \end{cases}$$
Hence, $n<m+2<n+4 \implies m=n-1,n,n+1$ $\implies$ Yields No solution. Manually checking values when $m,n \le 3$, we get $\boxed{(m,n) = (1,2), (1,3), (2,1), (2,2), (2,5), (3,1), (3,5), (5,2), (5,3) } \qquad \blacksquare$

Here is a strange solution idea inspired by [url = https://artofproblemsolving.com/community/c1035147h2015499p14144967]this[/url].

Clearly the quotient is congruent to $-1 \pmod{n}$, so there exists $m_0$ with $n^3 + 1 = (mn - 1)(m_0n - 1)$, which becomes $n^2 = mm_0n - (m + m_0)$ or $n^2 - (m + m_0)n + mm_0$. Thus the quantities $mm_0 - n$ and $\tfrac{m + m_0}{n}$ are equal to some positive integer $n_0$.

Thus $mm_0 = n + n_0$ and $nn_0 = m + m_0$, so
\[(m - 1)(m_0 - 1) + (n - 1)(n_0 - 1) = 2.\]
Now it's easy to finish.

Let $$\frac{n^3 + 1}{mn - 1} = k \Longleftrightarrow n^3+ 1 = mnk - k$$Clearly, taking modulo $n$ from both sides, we see that $k \equiv -1 \mod n$. Therefore, $k$ can be written as $an - 1$ for some $a \in \mathbb{Z}$. We can re-write what we have as:
$$\frac{n^3 + 1}{mn - 1} = k \Longleftrightarrow \frac{n^3 + 1}{mn - 1} = an - 1 \Longleftrightarrow n^3 + 1=amn^2 - n(a+m) + 1 \Longleftrightarrow n^2 = amn - (a+m)$$Solving, we see that the discriminant is $(pm)^2 - 4(p+m)$, which we seek to make a perfect square.
Claim:  $(am)^2 - 4(a+m) > (am)^2$
Assume by sake of contradiction that $(am)^2 - 4(a+m) \le (pm)^2$. That would imply that $a+m$ is nonnegative which clearly isn’t true.
Case 1: $(am)^2 - 4(a+m) = (am-k)^2$, where $k$ is odd.
If $am$ is odd, then the LHS is odd while the RHS is even. On the other hand, if $am$ is even, the LHS is even, while the RHS is odd. This is a contradiction.
Case 2: $(am)^2 - 4(a+m) = (am - 2)^2$
This is the same as saying that $(am)^2 - 4(a+m) = (am)^2 - 4am + 4 \Longleftrightarrow  a+m = am - 1 \Longleftrightarrow (a-1)(m-1) = 2 \Longleftrightarrow (a,m) = (3,2),(2,3)$.
Case 3: $(am)^2 - 4(a+m) = (am-k)^2$ where $k \ge 4$
Assume by sake of contradiction that $(am)^2 - 4(a+m) = (am-k)^2 \le (am-4)^2 \Longleftrightarrow (2a-1)(2m-1) \le 9$. Here the only solutions are of the form $(a, m) = (4,1),(1,4),(5,1),(1,5),(1,3),(3,1),(2,1),$ and $(1,2),(2,2)$. Checking, we see that the only $(a,m)$ that actually produce a perfect square are $(5,1)$ and $(1,5)$ and $(2,2)$.
Putting this all together and included what we found in the previous cases, we see that the final answer is $$\boxed{(1,2), \, (1,3),\, (5,2),\, (5,3),\, (2,2),\, (2,1),\,(3,1),\,(2,5),\,\text{and }(3,5)}.$$

Vieta jumping. We claim the solutions are $(1,2),(1,3),(2,2),(2,5),(3,5)$ and permutations.

We first take care of the case $m=n.$ Note that this implies $n^2-1\mid n^3+1,$ or $n^2-1\mid n+1,$ or $n-1\mid 1,$ implying $n=2.$ For the rest of the proof, we assume $m\neq n.$

Claim (Symmetry): For $m,n,$ $\frac{n^3+1}{mn-1}$ is integer if and only if $\frac{m^3+1}{mn-1}$ is integer.

Proof: Obviously this only needs to be proven in one direction. Note that
\[mn-1\mid n^3+1\implies\]
\[mn-1\mid mn^3+m\implies\]
\[mn-1\mid n^2+m\implies\]
\[mn-1\mid mn^2+m^2\implies\]
\[mn-1\mid n+m^2\implies\]
\[mn-1\mid nm+m^3\implies\]
\[mn-1\mid m^3+1.\]

Now take $mn-1$ and $n^3+1$ mod $n,$ and note that $\frac{n^3+1}{mn-1}\equiv -1\pmod{n},$ so we can express it as $m'n-1.$

Claim (Bounding): We claim that for $n\geq 4,$ both $m$ and $m'$ must be less than $n.$
\end{fact}

Proof: Note that $(mn-1)(m'n-1)=n^3+1.$ We cannot have $m=1$ since $n-1\mid n^3+1\implies n-1\mid 2,$ and $n-1\geq 3.$ Since $(2n-1)(n^2-1)>n^3+1$ for all $n\geq 4,$ we must have $m<n.$

Claim: We cannot have $m,n\geq 4.$

Proof: Put Symmetry and Bounding together.

Note that Bounding implies $n<m,$ but Symmetry implies that $(n,m)$ being a valid pair implies $(m,n)$ is a valid pair, which requires $m<n,$ contradiction.

Now we only need to manually check $n=1,n=2,n=3,$ and then we can permute.

Given $mn - 1 | n^3 + 1 \implies mn-1|n(n^2 + m) \implies mn-1|n^2 + m$ (since $mn-1, n$ are coprime). We have obtained $mn - 1 | n^2 + m \qquad (1)$. Continuing this, we get $mn-1|m^2(n^2 + m) - m^2n^2 + mn \implies mn - 1 | m^3 + mn \implies mn-1|m(m^2 + n) \implies mn-1|m^2 + n \qquad (2)$. From $(1)$ and $(2)$, we have both $mn - 1 \leq m^2 + n \implies n(m - 1) \leq m^2 + 1$ and $m(n-1) \leq n^2 + 1$.

Suppose $m < n$. Then we have from $n(m-1) \leq m^2 + 1$ that $n=m+1$ or $n = m+2$. If $n = m+2$, then $m \leq 3$ gives the solutions $(1, 3), (3, 5), (1, 2), (2, 5)$. Suppose $n = m+1$. Then we require $m^2 + m - 1 | m^3 + 3m^2 + 3m + 2$ in the original question. This gives $$m^2 + m - 1 |  m^3 + 3m^2 + 3m + 2 - m(m^2 + m - 1) \implies m^2 + m - 1 | 2m^2 + 4m + 2$$ But $m^2 + m - 1$ is always odd, so $m^2 + m - 1 | m^2 + 2m + 1 \implies m^2 + m - 1 | m+2 \implies m = 1$ which gives the solutions $(1, 2), (1, 3)$ which we already had.

Now suppose $m > n$. From $m(n-1) \leq n^2 + 1$, we get that $m = n+1$ or $m = n+2$. If $m = n+2$, we easily see $n \leq 3$ which gives the new solutions $(5, 3), (3, 1), (2, 1)$.
Therefore, $m = n+1$. Plugging it in the original question gives us $n^2 + n - 1 | n^3 + 1 \implies n^2 + n - 1 | n^3 + 1 - n(n^2 + n - 1) \implies n^2 + n - 1 | n^2 - n -1$. This is only possible if $n = 1, m = 2$ which gives a solution that we already have.

Finally, if $m = n$, it is trivial to see that $m = n = 2$.

So, our obtained solutions are $\boxed{(m, n) = (1, 3), (3, 5), (1, 2), (2, 5), (5, 3), (3, 1), (2, 1), (2, 2)}$ and we are done.

We have:
$$mn-1\mid n^3+1\Rightarrow mn-1\mid m(n^3+1)-n^2(mn-1)=n^2+m$$
and
$$mn-1\mid m(n^2+m)-n(mn-1)=m^2+n.$$
We will find all solutions to these two divisibility conditions. WLOG $m\ge n$, and suppose $n\ge4$. Let $f(x)=\frac{x^2+1}{x-1}$. Note that $x+2-f(x)=\frac{x-3}{x-1}$, so $f(x)<x+2$ for $x\ge4$. Since $mn-1\le n^2+m$, we have $m\le f(n)<n+2$, so $m\le n+1$. Then either $m=n$ or $m=n+1$.
If $m=n$, then $n^2-1\mid n^2+n$, so $n^2-1\mid n+1$. From $n^2-1\le n+1$, we derive $n\le2$ which is a contradiction.
If $m=n+1$, then $n^2+n-1\mid n^2+3n+1$, so $n^2+n-1\mid 3n+2$. From $n^2+n-1\le3n+2$, we derive $n\le3$ which is a contradiction.

Now, we check the cases $n\le3$.
[list]
[*] $n=1$: Then $m-1\mid m+1$, so $m-1\mid 2$. This gives $m=2,3$.
[*] $n=2$: Then $2m-1\mid m+4$, so $2m-1\le m+4$ or $m\le5$. Checking each of these gives that only $m=2,5$ work.
[*] $n=3$: Then $3m-1\mid m+9$, so $3m-1\le m+9$ or $m\le5$. Checking each of these gives that only $m=5$ works.
[/list]
This gives the only solutions:
$$(m,n)=(1,2),(1,3),(2,2),(2,5),(3,5)$$
along with their reverses:
$$(m,n)=(2,1),(3,1),(5,2),(5,3).$$

If mn-1 is divisible by n^3+1 then mn-1 is divisible by m(n^3+1)-n^2(mn-1)=n^2+m as it is the same. simillarly m(n^2+m)-n(mn-1)=m^2+n. hence, $mn-1,$ is divisible by n^2+m, and m^2+n. then rest is just casework an modular arithmetic


I do realize this is a trivial answer, but take $ m = -a$ and $ n = 0$. Then you have:

$ a+m=0$
$ a\cdot n=0$

and

$ a+n=a$
$ a\cdot m=-a^2$

Would this work?

[quote="gntca"]Prove that for every irrational real number a, there are irrational real numbers m and n such that a+m and a*n are both rational while a*m and a+n are both irrational[/quote]
one of  $ m \equal{} \minus{} a$ or $ m \equal{} 1 \minus{} a$ work!otherwise we must have $ \minus{} a^2$ and $ a(1 \minus{} a)$ are both rational,and therefore $ a \equal{} a(1 \minus{} a) \minus{} ( \minus{} a^2)$ is rational...contradiction!
and also one of  $ n \equal{} \frac {1}{a}$ or $ n \equal{} \frac {2}{a}$ work!otherwise we must have $ a \plus{} \frac {1}{a}$ and $ a \plus{} \frac {2}{a}$ are both rational,and therefore $ a \equal{} \frac {1}{a \plus{} \frac {2}{a} \minus{} (a \plus{} \frac {1}{a})}$ is rational...contradiction!


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ginger

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[u]Μπάμπης[/u]

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Μη περιμένεις IMO training φίλε Κώστα !

Δες το σαν εκδρομή..

Αν θέλει κάποιος να πάει ας το πει... να κανονίσουμε να πάμε όλοι μαζί παρέα... !  :)

Από ό,τι μπόρεσα να μάθω, τα μαθήματα θα γίνονται σε δύο επίπεδα και το Αναλυτικό Πρόγραμμα θα περιέχει Στοιχειώδη Συνδυαστική, Ευκλείδεια Γεωμετρία, Στοιχειώδη Θεωρία Αριθμών, Συναρτησιακές Εξισώσεις, Πολυώνυμα και Ανισότητες. Στο τέλος ίσως υπάρξει και "a mock Math Olympiad Session" with Prizes and everything. Yπάρχει όριο συμμετοχής: 120 άτομα. Το ξενοδοχείο δεν χωράει άλλους. Οι διδάσκοντες είναι γνωστοί στο χώρο των Διαγωνισμών και έχουν χρηματίσει αρχηγοί, υπαρχηγοί, trainers etc. των Ελληνικών ΒΜΟ και ΙΜΟ ομάδων. Δυστυχώς δεν μπορώ να παρευρίσκομαι γιατί θα είμαι στην Αμερική. Το μέρος όμως είναι υπέροχο και χρησιμοποιήθηκε για την επιλογή των θεμάτων από το Συμβούλιο Κριτών της Βαλκανικής Ολυμπιάδας Νέων (JBMO JURY) το 2005.

ginger

[color=blue]Προς όλους τους γονείς![/color]

  Σας προτρέπω να στείλλετε τα παιδιά σας  , αλλά  να πάτε και σεις μαζί τους αν μπορείτε , στην ΗΜΑΘΙΑ. Εδώ τα έξοδα θα πιάσουν τόπο. Μπορείτε να κάνετε  διακοπές και να δώσετε συγχρόνως στα παιδιά σας μια μοναδική ευκαιρία . Με τη μαθηματική παιδεία θα ζήσουν και πιο καλά και πιο πλούσια και πιο ευτυχισμένα.Είναι ώρα να βγούμε όλοι από την αφάνεια και να πάρουμε τη ζωή στα χέρια μας.

  Δώστε στο παιδί σας την ευκαιρία να γνωρίσει  το χαμόγελο και τη χαρά που μόνο μια μαθηματική απόδειξη μπορεί να χαρίσει  !

[u] Μπάμπης[/u]
 (αυτό το τελευταίο μπορεί ο αγαπητός φίλος Κώστας Παπ. από τη  Βέροια να το κάνει σλογκαν.Δεν έχει copy right!)

Στέλιο σκέφτομαι να πάω. Ακόμα όμως δεν είμαι σίγουρος. Βρίσκω ενδιαφέρον το πρόγραμμα και αν πάω θα ρθει και ο αδερφός μου που είναι Α Γυμνασίου για να αρχίσει σιγά σιγά να μπαίνει στο κλίμα. :wink:  Αν πάτε πολλοί και δεν είμαστε μόνοι μας  :P (που μάλλον αποκλείεται) θα το σκεφτώ σοβαρά.

Άξιζουν συγχαρητήρια σε όσους διοργανώνουν αυτό το καλοκαιρινό πρόγραμμα για μαθηματικά.Παπ σκέφτομαι να έρθω και εγώ.Ξέρει κανείς που μπορούμε να δηλώσουμε συμμετοχή :?:


find all injective functions from the set of positive integers into itself such that : f(f(n)) <= (n+f(n))/2

http://www.mathlinks.ro/Forum/viewtopic.php?t=32562    :?  ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=5449   ;)

Pierre.


geometry
inequalities
inequalities proposed

Prove that in any triangle we have:

$ \frac{s_{a}}{m_{a}} + \frac{s_{b}}{m_{b}} + \frac{s_{c}}{m_{c}} \geq \frac{p^{2}(p^{2}+2r^{2}+8Rr)+r^{2}(r^{2}+8R^{2}+8Rr)}{p^{2}(p^{2}+4Rr)-r^{2}(16R^{2}+8Rr+1)}$
(standard notations)!

I hope this is original and even more: correct! :D

What is symmedian? And why is there 1 in your rhs?

And I got one from the computations I made, I hope they are corect, I'll check. If they are not corect I will rewrite the ineq!

cheers! :D :D


function
inequalities unsolved
inequalities

find maximum of expression P
$P= \frac{a}{4+bc} + \frac{b}{4+ac} + \frac{c}{4+ba}$ where $a,b,c \in$ [$0;2$]

$a=b=c= 0 \Longrightarrow P=0$.

If you are interested in the maximal value of $P$ then note that if we fix two variables then we have a convex function of the third one. Hence maximum is attained when $a,\ b,\ c$ are at the endpoints of an interval. Easy checking gives that this maximum equals $2$, e.g. take $a=b=2,\ c=0$.

can yo y help me ?
 i want to  a so  lution    with  ineq  Cauchy  or AM -GM ineq    . Thank  you.

can you help me ?   the more basical solution do  the better i like .

If we consider the left hand side as a function of any one of the three varialbes $a,b,c$, say $a$. The resulting function is sum of a convex and a linear one. Thus, it is a convex function. We have
\[ f(a)\leq \max\{f(0),f(2)\}. \]

The rest is for you!


Positive real numbers are arranged in the form:
$ 1 \ \ \ 3 \ \ \ 6 \ \ \ 10 \ \ \ 15 ...$
$ 2 \ \ \ 5 \ \ \ 9 \ \ \ 14 ...$
$ 4 \ \ \ 8 \ \ \ 13 ...$
$ 7 \ \ \ 12 ...$
$ 11 ...$
Find the number of the line and column where the number 2002 stays.

[hide]We turn the arrays about 45 degrees clockwise to get
[code]    1
   2 3
  4 5 6
 7 8 9 10
...[/code]

The first number of row $ n\plus{}1$ is greater than $ t_n$ (the $ n$-th triangular number) by 1, i.e., $ a_n\equal{}\frac{n(n\minus{}1)}2\plus{}1$.

We call the numbers that are greater than triangular numbers by 1 "cool numbers". Note that $ a_{63}\equal{}1954$ is the greatest cool numbers which is less than 2002.

Hence 2002 is in the 63-th row. Since 2002-1954=48, then 2002 is in the 49-th column.

Now we turn the array back. It is easy to see that 2002 is in the 15-th row and in the 49-th column.[/hide]


Cuenca, you are fantastic, but we don't know the results anyway.

[quote="cuenca"]Hi, just for sake of curiosity: does somebody know who are the favorites to win? I heard that Renan Henrique Finder (BRA2) is sure to get the first place and become mr. IMO, but I dont know anything about the favorite girls...[/quote]

that's true... though i voted for RUS5..

And the winners are:

MISS IMO: Carmela Acevedo (VEN1)

MISTER IMO: Renan Finder (BRA2)

It seems that Latin America rules!!!!


The John wrote five positive integers, not necessarily distinct, such that
their product is equal to the sum. What may have been the numbers written by John?

Do you want all possible combinations?

[hide="One Combination"]$ 1, 1, 2, 2, 2$.[/hide]

Yes, how many combinations ?
Please, send me all of them, if you have.


calculus
integration
geometry
parameterization
conics
hyperbola
calculus computations

Let $ x,\ y$ and $ t$ be real numbers such that

$ \{ \begin{array}{ll} t^3 - (2y - 1)t^2 - t + 2y - 1 = 0 & \quad \\
xt - 2t + 2y = 0 & \quad \end{array} $

In $ x - y$ plane find the area of the figure bounded by the part of curves above with $ x\geq 0$ and the line $ y = 2.$

I tried to start from the fact that $ 0 \equal{} t^3 \minus{} (2y \minus{} 1)t^2 \minus{} t \plus{} 2y \minus{} 1 \equal{} (t^2 \minus{} 1)(t \minus{} 2y \plus{} 1) \equal{} (t \minus{} 1)(t \plus{} 1)(t \minus{} 2y \plus{} 1)$, so that $ t \equal{} \pm 1$ or $ t \equal{} 2y \minus{} 1$. This result is not very clear to me. Does this mean that the whole curve is composed by three different curves, one for each value of the parameter $ t$? :maybe: (I realized such curves would turn out to be a hyperbola and two straight lines). Also, after trying to sketch the curve I found no way to define the area to be computed since I misunderstood the statement, I'm afraid :( Any suggestions, please? :)


Until this morning, my profile said that I had 49 posts, which is true (this is my 50th post)
Also, when I logged on I had a rating of 5 (I think two sixes, one five, two fours)
Now, I see that I have 27 posts and that I have not yet been rated.
Why is this?I couldn't find this topic anywhere else and this is why i posted it here.
The system that I am using is windows and the browser internet explorer 7

Thank you in advance for your help,
salonikios

me too
$ 1045 \rightarrow 987$ :o
Navier - Stokes $ \rightarrow$ Yang - mills :rotfl:

My post count has also decreased: $ 1380\Rightarrow 875$ ?!?! :o

I think it is because the posts in community forums are no longer counted.

[quote="rem"]

I think it is because the posts in community forums are no longer counted.[/quote]
no not atleast the asian section

[quote="rem"]I think it is because the posts in community forums are no longer counted.[/quote]You've got it. The community forums are not counted towards post count and rating so people abuse less :P

is it that if u disable ratings the post count also disables  :?:
i think the problem was with ratings like many people did random bad ratings....but what's the problem with post counts

[quote="salonikios"]I see now...But I think I also had some ratings outside community forums (like in Number Theory and Calculus) and these also do not count towards my rating.
Is this correct?
What ratings do actually count?[/quote]Ratings in Number Theory and Calculus count. It's probably because you don't have enough rated posts yet. The ratings now require a certain amount of rated posts before you can have a rating (so a new user with 1 rated post 6 will not show up as with rating 6).

Thanks, admins!  :)  Was this because if ilp/my suggestion, or...?

[quote="Temperal"] Was this because if ilp/my suggestion, or...?[/quote]No, it was because my basketball court light switch broke, so I couldn't play basketball last night so I had some unwanted free time off.

[quote="Xevarion"]I've noticed that some people now are rated 0. Perhaps they were rated before the change, and now no longer have enough posts to be rated, and instead of being reset to "not rated yet" somehow they are reset to 0? (The people I noticed it for mostly have relatively few total posts.) Regardless of the reason, it might be worthwhile to remedy this so that certain people don't feel their posts are completely worthless...  :wink:[/quote]Well if you are rated 0 it's like you are not rated yet (you can't rate anyone 0). The minimal rating (when you are rated) is obviously 1 :D

[quote="Valentin Vornicu"]Well if you are rated 0 it's like you are not rated yet (you can't rate anyone 0). The minimal rating (when you are rated) is obviously 1 :D[/quote]
Right. But for some people it actually says "not yet rated" and for others it says "Rating: 0"

It says "Not Yet Rated" at first page load. After you've rated something on the page, the script automatically updates the ratings of the users and (here's a small glitch) the number is displayed instead of the language variable.

There are some dynamic community forums (like Greek and Romanian), and I see they always post and discuss about maths, so maybe we should still count post in community forums  :?:  Just my opinion.

I find the new decision absolutely okay. If post counts are to measure contributions to MathLinks, then posts in any language other than English should be rated, if at all, then lower.

  darij

but as April said,for example in Iran's forum we don't disscus about anything other than math,you can check it if you want,and also many Iranian people here in ML can't speak or write their questions in english,so it is a great opportunity for them to post their problems in Iranian language in it's section.but now according to what you said,these posts are not being counted,it just doesnt make any sense,why would you do this?I know that the number of posts isn't that important but at least I myself would respect one with more posts more,so in Iran's forum we would respect peoples with more posts more than others,which would be 0 now... :maybe:
so if a member only posts in a community forum,and is really active in there then there would be no difference between him and a new member!? :huh:

Guys I know that some local communities people only discuss math and they are serious forums, the question was a programming question (simpler to get all the forums under national communities than to handpick all of the bad ones). For now it will be like this, but I will think up a better solution soon. Remember that when the post count rules change, the post counts can be updated (so if say today I add Iran Forum back to the list, then all your post counts will be updated - nothing is lost it's just the way it's displayed).


Find the sum of the 4 smallest different integers such that each integer is $\equiv1\pmod{x^2}$, for any integer $x$, and $\equiv2\pmod{y^3}$, for any integer, $y$.

[hide="hint"]The first 3 are all 10 or less.[/hide]

I'm guessing you mean for SOME integer x and y--otherwise it would mean it has to be true for all.
Are we assuming that x,y are greater than one or something similar? Because all $z\equiv 1\equiv 2 \bmod{1}$, which means that all numbers could be expressed that way.

Good point.  I guess I should say find the sum of the smallest four different integers that are 1 more than a perfect square, and 2 more than a perfect cube.

Making it positive integers eliminates the two easiest solutions (hint, hint).

In that case, it would be
[hide][b]1[/b] = 0^2 + 1 = (-1)^3 + 2, [b]2[/b] = 1^2 + 1 = 0^3 + 2, [b]10[/b] = 3^2 + 1 = 2^3 + 2, and... either there's no last one or it's pretty big... (larger than 40000), and I'm too lazy to find it  :P .[/hide]

[quote="b-flat"]Very good, you got the entire solution correct  :lol:  :lol:  :D

I just wanted to see if anyone would even bother to just guess that it was rediculously hard to find and wasn't worth the time for the 4th one.[/quote]

Wait... so does that mean there's no 4th one?

I don't know  :D

I will try to figure that out, though (and waste time looking for something I meant just to waste time  :roll: ).


topology
geometry
vector
advanced fields
advanced fields unsolved

I'm trying to get myself ready for a differential geometry qual sequence, and unfortunately running into a lot of difficulty.  I can seem to grasp the overall mechanics, but when it comes the small details I get lost.  For example, Lee says in passing that for a vector space $ V$ of dimension $ n$:

$ E : \mathbb{R}^n \rightarrow V$ defined by $ E(x) \equal{} \sum_{i\equal{}1}^n x^iE_i$

is a homeomorphism.  I went to go prove this, and I can get bijective and continuous sure enough, but I seem to be stuck on proving that it is open.

Could anyone help me out?

It's linear. You can just find the inverse.

Yes, the inverse is quite obvious.  But I would still need to prove it is continuous.

So far, what I've been doing was going off the norm and proving things are open with the same approach I would in a metric space (rather than a topological space).  So letting $ U \in \mathbb{R}^n$ be an open set, $ x \in U$ hence $ E(x) \in E(U)$, what I want to show is there exists $ \varepsilon > 0$ such that $ \lVert E(x) \minus{} E(y) \rVert < \varepsilon$ implies that $ E(y) \in E(U)$, or rather, $ y \in U$.

However the norm only gives me larger approximations with the triangle inequality.  This worked just fine going in the "reverse" direction (for continuity rather than openness), but I am not sure what to do from here.

There's something you've omitted: any mention of how we define the topology of $ V$. Why can we mostly ignore it? Because any two norms on a finite-dimensional real vector space are equivalent and induce the same topology.

Try showing this: any linear map from a finite-dimensional real vector space to another is continuous.

Then $ E$ and its inverse are continuous, and we are done.

Ok, so I think what I was really missing was that every linear map between finite dimensional vector spaces is bounded.  I knew that, but forgot it.

In any case, this is my proof, and it seems to be rather straightforward.  What I'm wondering, is there a better way?  Or is this more of the standard was to prove this result?

Let $ T$ be a linear map between finite dimensional vector spaces $ V$ and $ W$.  Then $ T$ is bounded, so there exists a $ C \in \mathbb{R}$ such that $ \lVert T(v) \rVert_W \leq C \cdot \lVert v \rVert_V$ for all $ v \in V$.

Let $ U \subset W$ be open.  We will show that $ T^{\minus{}1}(U)$ is open in $ V$.  Let $ v_1 \in T^{\minus{}1}(U)$ be arbitrary.  Then since $ U$ is open, there exists an $ \varepsilon > 0$ such that $ B_\varepsilon(T(v_1)) \subset U$.  To prove that $ T^{\minus{}1}(U)$ is open, we shall show that $ B_{\frac{\varepsilon}{C}}(v_1) \subset T^{\minus{}1}(U)$.  Thus, it suffices to prove that $ \lVert v_1 \minus{} v_2 \rVert_V < \frac{\varepsilon}{C}$ implies that $ \lVert T(v_1) \minus{} T(v_2) \rVert_W < \varepsilon$.  Indeed, we have:

$ \lVert T(v_1) \minus{} T(v_2) \rVert_W \equal{} \lVert T(v_1 \minus{} v_2) \rVert_W \leq C \cdot \lVert v_1 \minus{} v_2 \rVert_V < C \cdot \frac{\varepsilon}{C} \equal{} \varepsilon$

As required.


parameterization
logarithms
inequalities
algebra proposed
algebra

Find the value of real parameter $ a$ such that $ 2$ is the smallest integer solution of \[ \frac{x\plus{}\log_2 (2^x\minus{}3a)}{1\plus{}\log_2 a} >2.\]

[quote="Differ"]$ x \plus{} \log_2(2^x \minus{} 3a) > 2(1 \plus{} \log_2a)$
$ x \plus{} \log_2(2^x \minus{} 3a) > \log_2(4a^2)$
$ 2^{2x} \minus{} 3a2^x > 4a^2$
$ 2^{2x} \minus{} 3a2^x \minus{} 4a^2 > 0$
$ (2^x \minus{} 4a)(2^x \plus{} a) > 0$

We need all $ 2^x \leq 2$ not to work. That gives us $ a \leq \minus{} 2$, $ a \geq \frac {1}{2}$.[/quote]

Your method works but you made certain mistakes in your solution.
Mistake 1: $ 1\plus{}log_2 a$ can be negative, so the first row is wrong.
Mistake 2: $ a$ cannot be negative or $ log_2 a$ undefined. So $ a \le \minus{}2$ is impossible.

[hide]First of all, note that $ x \equal{} \frac{1}{2}$ or $ a \le 0$ yield an undefined expression.

Case I: $ 1 \plus{} \log_2a > 0 \Rightarrow a > \frac{1}{2}$

$ x \plus{} \log_2(2^x \minus{} 3a) > 2(1 \plus{} \log_2a)$
$ x \plus{} \log_2(2^x \minus{} 3a) > \log_2(4a^2)$
$ 2^{2x} \minus{} 3a2^x > 4a^2$
$ 2^{2x} \minus{} 3a2^x \minus{} 4a^2 > 0$
$ (2^x \minus{} 4a)(2^x \plus{} a) > 0$

Both factors are positive if $ 2^x > 4a$ and $ 2^x > \minus{}a$, or equivalently $ 2^x > 4a$.
Since $ x \le 1$ does not fit this condition, and $ x\equal{}2$ does, we have $ 2^1 \le 4a < 2^2 \Rightarrow a \in \left(\frac{1}{2}, 1\right)$.

Both factors are negative if $ 2^x < 4a$ and $ 2^x < \minus{}a$, or equivalently $ 2^x < \minus{}a$.
However, this is impossible since $ a > \frac{1}{2} \Rightarrow 2^x < \minus{}\frac{1}{2}$.

Case II: $ 1 \plus{} \log_2a < 0 \Rightarrow 0 < a < \frac{1}{2}$

$ x \plus{} \log_2(2^x \minus{} 3a) < 2(1 \plus{} \log_2a)$
$ x \plus{} \log_2(2^x \minus{} 3a) < \log_2(4a^2)$
$ 2^{2x} \minus{} 3a2^x  < 4a^2$
$ 2^{2x} \minus{} 3a2^x \minus{} 4a^2 < 0$
$ (2^x \minus{} 4a)(2^x \plus{} a) < 0$

Since $ 2^x\plus{}a > 2^x \minus{} 4a$, we must have $ 2^x\plus{}a > 0$ and $ 2^x \minus{} 4a < 0$.
This is equivalent to $ \minus{}a < 2^x < 4a$. However, since $ a>0$ and $ 2^x > 0$, if $ x \equal{} 2$ satisfies this inequality, then any $ x \le 1$ also satisfies this inequality.

Thus, there are no permissible values of $ a$ in this case.

Therefore, the only values of $ a$ such that the smallest integer solution is $ x \equal{} 2$ is $ a \in \left( \frac{1}{2}, 1 \right)$.[/hide]

[quote="stephencheng"][quote="Differ"]$ x \plus{} \log_2(2^x \minus{} 3a) > 2(1 \plus{} \log_2a)$
$ x \plus{} \log_2(2^x \minus{} 3a) > \log_2(4a^2)$
$ 2^{2x} \minus{} 3a2^x > 4a^2$
$ 2^{2x} \minus{} 3a2^x \minus{} 4a^2 > 0$
$ (2^x \minus{} 4a)(2^x \plus{} a) > 0$

We need all $ 2^x \leq 2$ not to work. That gives us $ a \leq \minus{} 2$, $ a \geq \frac {1}{2}$.[/quote]

Your method works but you made certain mistakes in your solution.
Mistake 1: $ 1 \plus{} log_2 a$ can be negative, so the first row is wrong.
Mistake 2: $ a$ cannot be negative or $ log_2 a$ undefined. So $ a \le \minus{} 2$ is impossible.[/quote]

And that's why I got bronze on USAMTS. :P


How many revolutions must be made by a $ \frac1{11}$ cm diameter wheel in order for it to travel one kilometer? Use $ \frac{22}7$ as an approximation for $ \pi$.

That means the circumfrence is:

$ \frac {1}{11} \times \frac {22}{7} \equal{} \frac {2}{7}$ Centimeters

Since there are 100000 cm in a kilometer we simply do:

$ \frac {100000}{\frac {2}{7}} \equal{} \boxed{350000}$


this problem's REALLY been bugging me...see if anyone can solve it

$x^4+x^3+x^2+x+1$ has a root in the form $a+bi$.

If $ar+bs=\cos{3}$ and $r=\cos{75}$, what is $r+s$

this problem's confusing, but I'm sure the average *joe* can solve it :P

[quote="Iversonfan2005"]
this problem's confusing, but I'm sure the average *joe* can solve it :P[/quote]

 :roll:

[hide="solution"]
Ok, so $x^4+x^3+x^2+x+1$ cries ROOTS OF UNITY out.  So we convert it to $\frac{x^5-1}{x-1}$ (you could also see this by using the formula for a geometric series).  So the polynomial has the fifth roots of unity, excluding $x=1$, as its roots.  The fifth roots of unity are $\text{cis }\left(\frac{360ni}5\right)$ for $n=0,1,2,3,4,5.$  So this gives $\text{cis }\theta$ for $\theta=72, 144, 216, 288$ (I threw out $0^\circ$ since it corresponds to $x=1$).

Ok, from here I just kinda looked at what they give you (which seems kinda random) and saw 3 and 75, whose difference is 72, so I just let $a+bi=\text{cis }72 = \cos 72 + i \sin 72\Leftrightarrow a=\cos 72, b=\sin 72$.

This makes $ar+bs$ change into $\cos 72\cos 75 + \sin 72 s=\cos 3$.  If we let $s=\sin 75$ this is just the cosine difference angle formula.  So $r+s=\cos 75+\sin 75$.

Perhaps this is simple enough.  I dunno, but it can be simplified even further so...

\[ \cos 75=\cos(45+30)=\cos 45\cos 30 - \sin 45\sin 30 = \frac{\sqrt{6}-\sqrt{2}}4  \]

\[ \sin 75 =\sin(45+30) = \sin 45\cos 30 +\sin 30 \cos 45 = \frac{\sqrt{6}+\sqrt{2}}4 \]

So $r+s=\frac{\sqrt{6}}2$.[/hide]

[quote="Iversonfan2005"]hmmm...looking at the answer $r+s=\cos{75}+\sin{75}$ :?

hmmm...i think you made a careless mistake or something[/quote]

oh, um, yeah :oops:

That makes the simplifications afterwards QUITE A BIT EASIER!!


ABCD yek motavaziolazla . E rooye emtedade AB betorike BE=AD  B beyne A,E . az E bar AB , az C bar BD amood mikonim ta in 2 khat dar F yekdigar ra ghat konand sabet konid AF nimsaze  zaviyeye A ast.

bebinid vaseye esbate hokme masale bayad sabet konim 3 khatte [b]amood bar $ AB$ dar E[/b] va [b]amood bar $ BD$ az $ C$[/b] va [b]nimsaze zavie $ \angle A$[/b] hamresan dge...
khob vase inam man miam inkaro mikonam,migam [b]farz konid amood bar $ AB$ dar $ E$ va nimsaze zavie $ \angle A$ hamdigaro too $ F$ ghat konan,oonmoghe sabet mikonam $ FC\perp BD$...[/b]
ghaboole?

khob pas masalamoon be in tabdil mishe:

[b]motavaziolazlae $ ABCD$ mafrooz ast,noghte $ E$ ra bar emtedade $ AB$ tori dar nazar begirid ke avvalan $ B$ beine $ A,E$ bashad va sanian $ AD\equal{}BE$ hal az $ E$ bar $ AB$ amood konid ta nimsaze zavie $ \angle DAB$ ro dar $ F$ ghat konad,sabet konid $ FC\perp BD$.[/b]
---------------------------------------------------------
khob vase inke sabet konim $ FC\perp BD$ tebghe karno kafie sabet konim $ \boxed{FD^2\minus{}FB^2\equal{}CD^2\minus{}CB^2}$

khob gharar bedid $ AB\equal{}DC\equal{}b$ va $ AD\equal{}BC\equal{}BE\equal{}a$...

khob hala noghte $ F$ rooye nimsaze zavie $ \angle DAB$ gharar dare pas az do zele in zavie be yek faselas,yeni age ma az $ F$ bar $ AD$ amood konimo esme paye amoodo bezarim $ H$ oonmoghe $ FH\equal{}FE$...

khob hala too 4zelie $ AEFH$ do zavie $ \angle H,\angle E$ barabare $ 90^\circ$ hastand va hamchenin hamtor ke didim $ FH\equal{}FE$ pas in 4zeli kayte,pas $ AE\equal{}AH$ amma $ AE\equal{}b\plus{}a$ pas $ AH\equal{}b\plus{}a$ amma $ AH\equal{}AD\plus{}DH\equal{}a\plus{}DH$ pas $ DH\equal{}B$...

khob hala dar mosalase ghaemozavie $ \triangle FHD$ darim:

$ FD^2\equal{}FH^2\plus{}HD^2\equal{}FH^2\plus{}b^2$ (1)

hamchenin too mosalase ghaemozavie $ \triangle FEB$ darim:

$ FB^2\equal{}FE^2\plus{}BE^2\equal{}FH^2\plus{}a^2$ (2)

hala ba tavvajoh be (1),(2) darim:

$ \boxed{FD^2\minus{}FB^2\equal{}b^2\minus{}a^2} (I)$

hamchenin darim:

$ CB\equal{}a,CD\equal{}b\Rightarrow \boxed{CD^2\minus{}CB^2\equal{}b^2\minus{}a^2} (II)$

hala ba tavajoh be (I),(II) darim:

$ FD^2\minus{}FB^2\equal{}CD^2\minus{}CB^2$

yeni hamoon chizi ke mikhastim...


Is there integer $n$ such that $n!$ begins with $2005$ ?

Have a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=23417

And http://www.mathlinks.ro/Forum/viewtopic.php?t=46302
And http://www.mathlinks.ro/Forum/viewtopic.php?t=44064

The exact answer can also be found in the first topic.


Προέκυψε σε μία συζήτηση που είχα με ένα φίλο μου, και είπα να το βάλω σαν (πολύ εύκολη) ασκησούλα επειδή μου φαίνεται cute :)

Η συνάρτηση του Euler $ \phi(n)$ ορίζεται ως ο αριθμός των θετικών ακεραίων $ \leq n$ που είναι σχετικά πρώτοι ως προς το $ n$ (π.χ. $ \phi(9)\equal{}6$ επειδή οι $ 3,6,9$ δεν είναι σχετικά πρώτοι ως προς το $ 9$).

Είναι προφανές (?) ότι $ \lim \phi(n) \equal{} \infty$. Τί μπορείτε να πείτε για το όριο $ \lim\frac{\phi(n)}{n}$?

Cheerio,

Durandal 1707

PS: Οι ιδιότητες της $ \phi$ είναι γνωστές, οπότε όσοι το ξέρετε αφήστε το για όσους πιθανώς δεν το έχουν δει ;)

Το κρύβω για να μην χαλάσω τη δουλειά των άλλων . Παναγιώτη τόσο βγαίνει ?
[hide]$ \sim\sup \frac {\phi(n)}{n}\equal{}1$  και $ \sim\inf \frac {\phi(n)}{n}\equal{}0$

Το πρώτο είναι προφανές αν πάρουμε $ n\equal{}p$ πρώτο και το άλλο βγαίνει εύκολα με την αρμονική σειρά [/hide]

[hide]Πολύ σωστά Monsieur Silouan :)[/hide]
Anyone else?

Durandal 1707


Let be given a convex polyhedron whose all faces are triangular. The vertices of the polyhedron are colored using three colors. Prove that the number of faces with vertices in all the three colors is even.

Assuming the colors are red, green, and blue, count the red-green sides of each face. The number of such edges of a face is odd iff the triangle is red-green-blue. However, the number we get by summing all these numbers must be even, since each side is counted twice, being the side of two faces. This means that there must be an even number of red-green-blue triangles, as desired.

a solution similar to grobber's:
on each edge, write $0$ iff the vertices of this edge are the same color. number of a triangle is sum of the numbers of it's three edges. now sum up all of triangle's numbers. the rest is like gobber's solution.


let $ g$ is a function such that :
$ g: R\equal{}\equal{}>R$
 $ x\equal{}\equal{}>g(x)$
and $ (\forall x \in R) g(x\plus{}1)g(x)\plus{}g(x\plus{}1)\plus{}1 \equal{} g(x)$

1. prove that $ (\forall x \in R) g(x)(g(x)\plus{}1)\neq 0$
2. prove that $ (\forall x \in R) g(x\plus{}2)\equal{} \minus{} \frac{1}{g(x)}$
3. prove that $ g$ is $ periodic$.

Rewrite the given condition as $ (g(x\plus{}1)\plus{}1)(g(x)\plus{}1)\equal{}2g(x)$ for all $ x$.

1) If $ g(x)\equal{}\minus{}1$ for some $ x$ then, from above, it follows that $ g(x)\equal{}0$. A contradiction.
Therefore, $ g(x) \neq \minus{}1$ for all $ x$.
Moreover, neither $ g(x\plus{}1)\plus{}1$ nor $ g(x)\plus{}1$ is zero, so that $ g(x) \neq 0$ for all $ x$.

2) Frome the condition, we deduce that $ g(x\plus{}1) \equal{} \frac {g(x)\minus{}1}{g(x)\plus{}1}$.
Using the same relation between $ g(x\plus{}2)$ and $ g(x\plus{}1)$ and substituing leads to the desired result.

3) From 2) it is easy to verify that $ g$ is $ 4$-periodic.

Pierre.


calculus
derivative
function
calculus computations

$f''(x)=0$ and $f'''(x)\neq 0$ implies that $x$ is a flex point of $f$

Standardly, this means the function is never increasing or decreasing i.e constant.  Maybe this is too elementary an answer  :blush:


What is the least natural number, greater than 1, that is a factor of $ 11000\plus{}1100\plus{}11$?


How many integers $ n$ satisfy $ \sqrt8 < n < \sqrt{80}$


How do you systematically solve this -
Find the least value of x such that
$4x + 2$ [color=red]MOD[/color] $3 = 0$

$4x+2 \equiv 0 \mod {3}$
$3x+x \equiv -2 \mod {3}$
$0+x \equiv 1 \mod {3}$
$x \equiv 1 \mod {3}$

Clearly the smallest value is $\fbox {1}$.

[quote="amirhtlusa"]$x\in\mathbb{N}$??
if yes,
$x=1$[/quote]
thats not a systematic way of solving... its just a numeric answer!
when you have $a\mod b=c$ it means that $a=bq+c$ where $q$ is an integer you dont need to care about except you are finding the [b]quotient[/b].
now for your problem, we can translate it into easier equation:

from this $4x+2\mod 3=0$ you get $4x+2=3q+0=3q$. so what number makes $4x+2$ divisible by 3? $x=1$ because $4\cdot 1+2=6$ is divisible by 3. try to do more divisibility problems with mods and soon you will find mod is actually easier than you think it is  :)

edit: please look above at chess64's solution, that is one short way of doing it  :)

Ooh...I just learned mods today  :D and I solved this problem correctly, since if $x=1$ then $MOD3 of 6$ would be $0$ since there are no remainder left over  :D

It can, but we assumed, as amirhtlusa said, that $x \in \mathbb {N}.$

Well , your assumption was no problem , for $\displaystyle x \in \mathbb {N}$ is indeed true in the problem I wanted to solve.

Just as a sidenote, is \[4x+2\text{ mod }3=0\] really a valid way to write this? I always understood that $\text{mod}$ is written after the congruence in parentheses so you don't have to write "... in modulo $n$." There isn't really a "mod function." Sorry I'm always so picky about notation.. and correct me if I'm wrong.

[quote="karthik_k"]Hey AntonioMainenti ,
Thanks a real lot for that ... I just got begun with the MOD function , and could someone tell me how exactly you write an expression ? And if something in the way the question is written is wrong , could you tell me what it is and how it is to be written the correct way ?
P.S {I'm used to programming where x mod y is a very commonly used statement.....}[/quote]

You write it like this:

$4x+2$=$0(Mod 3)$

But I don't know how to make the congruent sign which is 3 dashes, so the = sign should have 3 dashes instead of 2, okay.

$4x+2 \equiv 0 \pmod 3$

[code]4x+2 \equiv 0 \pmod 3[/code]

$x \equiv y \mod {n}$ means that $x$ leaves a remainder of $y$ when divided by $n$. For example, $6 \equiv 0 \mod {3}$. Also, $5 \equiv 1 \mod {4}$.

When you have $-y$, as in $9 \equiv -1 \mod {5}$, it's just the same as $9 \equiv 4 \mod {5}$. That's why I find it easier to think of it as $x \equiv y \mod {n}$ means $x-y$ is divisible by $n$. So in the $-y$ case, you would have $x+y \equiv 0 \mod {n}$. You can also think of it as $x \equiv n-y \mod {n}$.

I hope that's not confusing.

[quote="karthik_k"]Hmmm...So I need to use the $\equiv$ sign - But how do you write this : 4 mod 3 = 1 ?? :oops:
By the way , could someone give a detailed explanation on the use of the $\equiv$ sign ??  :?[/quote]

you don't write it like that...you write it as:

$1(mod 3)\equiv\4$ as in when $4$ is divided by $3$, it leaves a remainder of $1$, it's almost like $3$ times whatever plus $1$ is equivalent to $4$.

By the way, you should always use parenthesis with mods because it is much easier to read and understand. :juggle:

[quote="karthik_k"]Hmmm...So I need to use the $\equiv$ sign - But how do you write this : 4 mod 3 = 1 ?? :oops:
By the way , could someone give a detailed explanation on the use of the $\equiv$ sign ??  :?[/quote]You would write that as $4\equiv1\pmod{3}$. You're basically saying $4\equiv1$, and the $\pmod{3}$ means that it's in modulo $3$. Equality with the $=$ symbol is actually just a special case of congruence with the $\equiv$ symbol, since $x=y\Leftrightarrow{x}\equiv{y}\pmod{0}$. As for the congruence symbol itself.. I believe Gauss introduced that in Disquisitiones Arithmeticae. The definition I know for congruence is just $a\equiv{b}\pmod{m}\Leftrightarrow\exists{r}\in\mathbf{Z}\text{ such that }a=b+rm$.

[quote="Iversonfan2005"]$1(mod 3)\equiv\4$[/quote]

$\pmod {n}$ is usually written as $\$\pmod {n}$\$ if you want paranthesis. If not, just $\$\mod {n}$\$.  ;)


geometric sequence
algebra solved
algebra

Define f on the positive integers by f(n)=k 2 +k+1 where 2^k is the highest power of 2 dividing n. Find the smallest n such that:
f(1)+f(2)+...+f(n) \geq 123456

Hey, Maverick, have you solved this one? (I mean is it proposed or are you asking for help?)

I'm asking you because I think I found n to be 24710 and I was wondering if it was OK. If you had solved it and told me that it wasn't correct I wouldn't have bothered to post my 'soln' (which is kind of long, non-elegant and I don't even know if it's correct, I'm not sure).

So what do you say?

P.S. for n=24710 I think that sum is exactly 123456, so this would be a strong argument in favor of the correctness of my soln (I'll post it a bit later if you want me to).

Well I didn't solve it yet cause I was kind of busy and that's why I proposed it to be solved. I'll think about it(if I have the time)

Have a nice day! :)

Is this a mathematical problem?

Simple little computer program gives an answer:  24710

It is really f(1)+...+f(24710)=123456

maybe there is a mathematical solution - i mean 123456 is not just another ordinar number ... :D

I guess I have a mathematical soln, just not a very nice one.

We can calculate f(1)+f(2)+..+f(n) like this: the number of odd numbers <=n is n-[n/2]. The number of numbers <=n divisible by 2 but not divisible by 4 is [n/2]-[n/4] and so on. Then f(1)+..+f(n)=n+ \sum k>=1 [n/2^k]*(g(k)-g(k-1)) where g(k)=k^2+k+1. We see that g(k)-g(k-1)=2k, so our sum is S(n)=f(1)+..+f(n)=n+  \sum k>=1 2k*[n/2^k] (*). If S(n) >= 123456 then A(n)=n+  \sum k>=1 2k*n/2^k >= 123456. If we calculate A(n) (we use the formula for the infinite geometric progression) we find 5n. This means that n>= [123456/5]= 24691. I calculated S(n) for 24692 (it's not that hard, I used the formula (*) and I just had to make 14 divisions, because the greatest power of 2 <= 24692 is 2^14) and I got 123456-108. Then I just started increasing n until I reached 123456. This didn't take long, since 108 is rather small and 24692+12=24704 is divisible by 128 = 2^7, so when I got to 24704 I added f(7) = 57, which is rather large, and for each new n we add at least f(0)=1. Actually, you can see that I made 24710-24692=18 steps. This isn?t nice at all though..

You can also see that I didn?t use any particularities of 123456.


Prove that: $ a_{n}\equal{}(\minus{}1)^{n}cos(n\plus{}\frac{1}{n})\Pi$ is convergent.

[hide="Hint"] Compute a limit for $ n\equal{}2k$ and $ n\equal{}2k\plus{}1$ for $ k\in\mathbb{Z}$ and remember that $ \cos (\alpha\plus{}2m\pi)\equal{}\cos\alpha$ for $ m\in\mathbb{Z}$[/hide]


trigonometry
algebra
polynomial
induction
number theory
relatively prime
algebra proposed

Let $n\geq 3$ be odd integer,and $ \{\ a_1,a_2,...,a_{\phi (n)}\}\ $  be the set of positive integers less than n and relatively prime to n.prove that:
$|\prod_{k=1}^{\phi (n)}\cos (\frac{a_k \pi}{n})|=\frac{1}{2^{\phi (n)}}$

Let $P_n(x)$ be the $n^{th}$ cyclotomic polynomial.
That is $P_n(x) = \prod_{gcd(k,n)=1} (x - e^{ \frac {2ik \pi} n} ).$

Thus :
$P_n(-1)= \prod (-1-e^ {\frac {2i \alpha_k \pi} n}) = (-2)^{ \phi (n)} \prod [e^ {\frac {i \alpha_k \pi} n}) cos(\frac {\alpha_k \pi} n) ]$

from which it follows that :
$|P_n(-1)| = 2^{ \phi (n)}| \prod cos(\frac {\alpha_k \pi} n)|$ (1)

In another hand, it is well-known that for each $d$, $P_d(X)$ is a polynomial with integer coefficients and that $X^n - 1 = \prod _{d|n} P_d(X) $.
Since $n$ is supposed to be odd, we then have :
$-2 = \prod _{d|n} P_d(-1) = P_n(-1) \cdot P_1(-1) \cdot \prod _{d|n,1<d<n} P_d(-1)$ and each term of the RHS is an integer, with $P_1(-1) = -2$.
Thus $|P_n(-1)| = 1$, (2)

From (1) and (2), the result follows.

Pierre.

let d be a divisor of n and [tex]A_d={0<=i<=n/ gcd(i,n)d=n}[/tex]
we can proove the identity by induction !
let's assume that is true for all positive integers less than n
we know that the product of |cos(k :pi: /n)| where k in{1,2,....,n} equals [tex]2^{-n}[/tex]( by using the polynome [tex]X^n-1[/tex] and his factorisation and by taking X=-1
so the set {1,2,....,n} = :cup:  [tex]A_d[/tex] when d|n
the [tex]A_d[/tex]'s are disjoints!
so the cos(k :pi: /n)=cos(i :pi: /r)  where i=k/gcd(n,k) and r=n/gcd(n,k)
so the product which is equals to 2^(n- :Sigma:  :phi: (d))  where d|n.
we have just to proove that  n=:Sigma:  :phi: (d)  where d|n
it's quit simple if we use the cardinal of the union!

Well, it is the same proof but with different words...

Pierre.

Exercices d'arithmetic Sup Monier!
 :Sigma:  :phi: (d)=n where d|n
;)


Kindly post a non-brute force solution. :D

[quote="jhredsox"][quote="i_like_pie"]For what value of $n$ is the five-digit number $7n867$ divisible by 33?[/quote]

[hide]6[/hide]
-jorian[/quote]
Incorrect.

[quote="SplashD"]I think the answer is 5, just use the divisibility rule for 3 and 11.[/quote]
Correct! :D

[quote="jhredsox"]i had forgotten rule for 11, so i was doing it with brute force :maybe:
jorian[/quote]

Well I assume you know the divisibility rule for 3? because 6 doesn't fit that either.

no, i know it now (i knew it but forgot, so i looked it up)

ty for everyone's help
-jorian

[quote="anirudh"]11 is
abcde:

If abcde is divisible by 11:
then a+c+e/b+d must equal {0, or a multiple of 11}.[/quote]
Um... sorry, but no, that's not the correct rule.  :wink: (If you're going to post a formula, don't post an incorrect one.  :roll: )
The correct one is actually
$a-b+c-d+e$ is a multiple of $11$.
(In other words, the alternating sum of the digits must be divisible by 11.)

If you want to try to figure out why this works, here's a hint: 11, 1001, 100001, etc., are multiples of 11, and 99, 9999, 999999, etc., are multiples of 11 also.

Another hint is to look at the equation it in mod 11. That is how AoPS vol. 1 does it...

[hide]
5, because if n=5, then (7+8+7)-(5+6)=11, so 75867 is divisible by 11
7+5+8+6+7=33, which is divisible by 3.[/hide]

[hide="solution"]we test for divisibility by 3, and see that the umbers can be only 2, 5, or 8. We test for divisibility for 11, and see that only 5 works.[/hide]

[hide]The digits at first add to 28. To make it divisible by 3, n must be 2,5, or 8. The one the makes it divisible by 11 is the one that satisfies the equation 22-6-n= 0 mod 11. It's pretty obvious that this must be 11. 16-n=11, n=[i][b]5[/b][/i][/hide]


[quote="Interval"]Light travels about 186,000 miles per second. Jesse figured that it takes approximately 8 minutes for light from the sun to reach one of the planets on the chart. To which planet is Jesse referring?

Earth 93

Venus 67

Jupiter 484

Saturn 888

All of the numbers next to the planets represents the average distance from the sun (in millions of miles).[/quote]
[hide]Since the $d=vt$, the distance is $186,000\cdot8\cdot60=89280000$ (changing minutes into seconds)
Earth ( 93 mil.) is the closest to 89 mil. miles away.[/hide]

I played around with this question for about 10 minutes before giving up.

I knew that 8 minutes x 60 seconds = 480 seconds.

What part of the question indicated to multiply 186,000 x 60 seconds x 8 minutes?

How did you know to do that in your reply?

[quote="Interval"]I played around with this question for about 10 minutes before giving up.

I knew that 8 minutes x 60 seconds = 480 seconds.

What part of the question indiacted to multiply 186,000 x 60 seconds x 8 minutes?

How did you know to do that in your reply?[/quote]
To go from a rate to a distance you have $\frac{\text{miles}}{\text{sec}}*\text{sec}=\text{miles}$ (${r}\cdot{t}=d$

But, assuming that all these facts are even REMOTELY accurate, you could use a simple astronomical approximation that it takes ~8 min. for light to get from the sun to here. (Come on, you MUST have heard that SOMETIME in your life)...(sorry if you havent)

Yeah it's supposed to be a widely known fact that it takes 8 mins for sunlight to reach earth!! :D
Or if u really want to calculate...speed=186,000 and time=8  mins or 480 seconds,$Distance=Speed*Time=18600*480=8928000$ which is closest to Earth...


I recently graduated from undergrad (University of Michigan) with a double-major in Economics and History. However, recently I have been extremely interested in Mathematics and I'm seriously considering a Master's degree in it. The problem is that the most math that I took during my undergrad was Multivariate Calculus, and I don't have the necessary coursework to get into good programs yet.

So, my question you guys - hopefully, a few of you are pursuing math in higher education - is whether I should go back to undergrad, finish a major in math, and then apply for a Master's Degree.... or should I just fulfill those required courses at other colleges (or even through online distance-learning) and then apply for a Master's once I'm set? I really want to get into a good school though (one of the top 10, by rankings).

Also, do most college offer placement exams so that I could skip ahead of those required classes if I teach myself the material? And suppose I apply to Master's programs now, is it possible for me to make up for all these required classes early on in the program or do I need to have it all done before applying?

Any general advice that you can offer me would be greatly appreciated. Thanks ahead.

~ Siam

At my college I was exempted from a lot of courses by passing final exams. Maybe you have the same situation. I highly recommend that you take a full courseload in undergrad math before you move onto a master's degree.

Siam, if you are able to use your Michigan connections to gain hassle-free admission to the masters program there, then that would be one way to go.  Although you are not yet ready for graduate work in math, you could perhaps start with an aggressive load of undergrad courses in year one and then move on to basic grad courses the next year.  You'd have to look into how many graduate level courses you'd need for the masters degree.  Alternatively, you could just go back to undergrad at Michigan (I'm not sure how that works, but I assume you don't have to get admitted), and just take as much math as you can.  Even under this route, you could plausibly end up with enough credits to get a masters.

Assuming all goes according to plan and you do well, you may be able to convince Michigan to let you into their PhD program (which is a good program), or if you do really well, you could transfer into a PhD program at some other good school.  But if you do, even if you have a masters, it might be best to start as a first-year grad student, depending on how far along you get at Michigan.

Keep in mind that a masters is not important at all if you plan to get a PhD, but I emphasize its importance to *you* because at this point, you really don't know if a math PhD is right for you because of your inexperience.  You may decide not to go through with it, and if that happens, at least you can come away with a masters degree in exchange for the time and work you put in.

There are various other issues that might be important, like money, but I don't know your situation.

I don't recommend trying to apply to a top school right away.  First, unless there are people who will vouch that you are a genius, your math background will not get you in.  Second, if you did get in, you'd be way behind your peers.

Last piece of advice:  Talk to someone on the math faculty at Michigan about your plans, especially if you're thinking of going back to school there.

That is your decision. I don't even know you, therefore I'm not about to give you advice on what to do since the decision will affect the rest of your life most likely. However here's my advice, do what you feel will lead to a better future for you.


geometry
circumcircle
trigonometry
geometric transformation
geometry proposed

[b]Hypothesis.[/b] $\triangle ABC: \ D\in (BC),\ AC=DC;$ circumcircle $w=C(O,R);\ \{ A,A'\}\subset AD\cap w.$

[b]Conclusion.[/b] $\underline {\overline {\left|\ m(\angle BAD)=30^{\circ}\Longleftrightarrow DA'=R.\ \right| }}$

See my last reply at http://www.mathlinks.ro/Forum/viewtopic.php?t=51025

Write the power of the point D in two ways:
     AD . DA' = BD . DC = BD . AC => $\frac{AD}{BD} = \frac{AC}{DA'}$
   Write the sine law in triangle ABD:
    $\frac{BD}{sin 30} = \frac{AD}{sin B}$ <=> $\frac{AD}{BD} = \frac{sin B}{0.5} = \frac{AC}{R}$
   Comparing the relations above, we obtain DA' = R.

Please use Romanian language or provide a translation in the Romanian forums. This topic is moved into the Geometry forums.


geometry
3D geometry
tetrahedron
inequalities solved
inequalities

try this ppl
the 4 medians of a tetrahedron are the possible sides of a quadrilateral

This is trivial 'cause  [u]m_a[/u] + [u]m_b[/u] + [u]m_c[/u] + [u]m_d[/u] = [u]0[/u], is'nt it?

==========
Namdung


:rotfl:  :rotfl:  :rotfl:
that tells us SO MUCH ABOUT THE AZNESS of mathcounts...


AMC
AIME
MATHCOUNTS
blogs
USA(J)MO
USAMO
AMC 10

Hello,

People are probably going to "computer-yell" at me because I am, in a sense, making a double-post. However, I need to know what to do to improve my math skills.

Last year, I learned a lot. Now, I am only slightly above the level of my peers.

Many people only said do more problems. My question is where? Where am I supposed to get the problems? And problems aren't enough; where am I going to learn new concepts and formulas without a coach (I can't afford the AoPS texts and classes)? The internet? Not a very good teacher (and I don't know any websites).

My point is: I have nothing, but I want to get better.

I will probably come up with even more questions so keep looking at this post.

Uh.. This is going to start another LONG thread. But I will not shout at you. I once wanted to know the answer to that question. But the answer is: There is no correct answer. Everyone has there own study skills. But my study skills are just AoPS, Past MC tests, and AMC10's/AMC12's. This is pretty much what I do.

But if MC stuff don't work for you.. IDK...  :oops:

okay. i tend to do okay in math (144 amc 10, 2nd place michigan state mc with a 44, failed aime with a 5, etc.), so i'll give you my tips.

1. do a lot of old problems. it's like the classic "how do i lose wait?" well....you exercise. you have to do old mathcounts stuff, old amcs, old problems of any sort. try to accumulate a vast amount of resources, and USE THEM.

2. read. read, read, read. coming from someone who doesn't particularly like to read, i may not be the best influence, but do it. even on aops, look at blogs that have math (aime's blog, izzy's, bogtro's, pythags [although pythag is a bit of a freak...]), and other such...things.

3. alcumus. ftw. these two things have helped me a ton. i don't have any of the aops books (not because i can't afford them, i just...don't). utilize aops to it's fullest:its free, its fun, and its educational. don't just look at the mathcounts/moems forums, look at the amc's, the HSB and intermediate (pre olymp might be a bit hard.)

4. have fun. math is just like everything else, practice makes perfect. you have to have fun when you're doing math; i know that the thrill of cracking a difficult problem for me is pretty great. and if math isn't your thing...so what? become a beast at writing (MY SISTER! doesn't like math too much, but has won national essay competitions.), become the next mozart, just do what you like. math pwnage comes from within.

wow. i sound like a frickin' psychologist. hope that helps.

[quote="nikeballa96"]pythags [although pythag is a bit of a freak...])[/quote]

Ahem?

Anyways, doing problems is pretty important... don't worry about the advanced techniques, mathcounts usually doesn't have that many advanced techniques. Don't look at the solutions on problems unless you've been working on them for 5/10 minutes (since this is mathcounts... usamo its more like a few days at least) but the solutions are usually really useful. I found most of the techniques i know by looking at the solutions to online problems.

[quote="jonathanchou711"][quote="pythag011"][quote="nikeballa96"]pythags [although pythag is a bit of a freak...])[/quote]

Ahem?
.[/quote]

LOLLOLLOLLOL.... Kind of aggressive there eh nike?[/quote]

pythag, realize that nike calling you a freak is more of a compliment than anything.
i said freak because it was in context of blogs, while you go around posting IMPOSSIBLE questions and labeling them "Triviality!"

=] =] =]

by the way, agreed with what pythag said regarding the topic of this thread.

At nikeballa, how am I supposed to do the AMC/Beginning/Intermediate ones when I didn't even make MC nationals? The point is [b]I suck[/b].

pythag didn't make nats last year. he made usamo.
you're from texas. mathcounts isn't the end of anything. in fact, it's not even the beginning.
just read through them. a lot of HSB/AMC's are easier (well...maybe.) then some mc material.

The main issue with mc is time, which is why mc problems often seem harder than amc probs. (Though time is really important also in amc, aime, and even usamo (though it might not seem like it, time is pretty important in usamo...))

You can improve a lot just by solving the problems that people post on aops.

About nikeballa's post; how much time should I spend doing each? Right now, I basically spend all my time reading on the forums.


trigonometry
real analysis
real analysis unsolved

for wich $ m$ and $ n$ this series will converg?
$ \sum_{k \equal{} 1}\frac{\tan^{m}k }{k^{n}}$
for first  find it for
$ \sum\frac{\tan{n}}{n^{6}}$?

Let $ M \equal{}\mbox{max}_{k\in\mathbb{N}}\left|\tan^{m}k\right|$, then
\[ \sum_{k\equal{}1}^{\infty}\frac{\tan^{m}k}{k^{n}}\leq (M\plus{}1)\sum_{k\equal{}1}^{\infty}\frac{1}{k^{n}}\]
which converges for all $ n > 1$ so for all real $ m$ and $ n>1$.

$ M$ is infinite.

This is about how well $ \pi$ can be approximated by rational numbers. The series definitely diverges when $ n\le 1$ or $ 2m\ge n$. It definitely converges when $ n>1$ and $ n>9m$, but that requires a pretty advanced theorem to prove. The exact threshold, and whether $ (6,1)$ works or not, depends on an open question.


Ok, I'll give you some simple Flemish olympiad problems :

____________________________
(1) (FMO1991) Determine whether the following numbers are prime :
* 111...111 (1991 digits "1")
* 64000...00027 (10000 digits "0")
____________________________
(2) (FMO2002) Determine all functions which satisfy
xf(x/2) - f(2/x) = 1
for any real number x (different from 0).
____________________________
(3) (FMO2000) A unknown number N ...
* consists of precisely 7 distinct digits ;
* is divisible by each of its digits.
Which digits cannot appear in the decimal representation of N ?
____________________________
(4) (FMO2001) Prove that for any positive integer n :
(n - 1)2 | nn - 1 - 1
____________________________
(5) (FMO1997) Determine the greatest possible product of positive integers with sum 1997.
____________________________
(6) (FMO1994) Suppose that a, b and c are the lengths of the sides of a triangle ABC for which  = 90 (and, as usual, a = BC, b = CA and c = AB).
Solve for real numbers x : ax > bx + cx.
____________________________

A hint for number 2:

[hide]Consider the given functional equation for x = 2y and for x = 2/y.[/hide]

4) [hide]This is equivalent to proving that

n - 1 | nn - 2 + nn - 3 + ... + n + 1.

R.S. = nn - 2 + nn - 3 + ... + n - (n - 2) + (n - 1)

= (nn - 2 + nn - 3 + ... + n - (n - 2)) + (n - 1)

Recall that polynomial (x - 1) divides f (x), a polynomial function, iff the coefficients of f (x) add up to 0. Hence

n - 1 | nn - 2 + nn - 3 + ... + n - (n - 2)

if we treat (n - 2) as a constant term.

QED.[/hide]

Partial solution for 5)

[hide]Let x1 ... n be the sequence of the desired numbers. Then, applying AM-GM, we find

(:pi: is the product symbol)

:Sigma:k = 1 ... n xk :ge: n:sqrt:(:pi:k = 1 ... n xk)

So: :pi: k = 1 ... n xk :le: (1997 / n)n.

What now?[/hide]

using calculus will give you an idea what to do

Actually it is much simpler. No calculus !

Number 6:[hide]

For all x :ge: 2, it will work.  We know that we have a right triangle.  Thus, we can call one of its angles y, and we will have that

b = a*cos(y), c = a*sin(y), so we are now trying to solve the equation

1 > cos(y)^x + sin(y)^x for all y in [0, :pi:/2].

For x = 2, that gives us a constant value of 1 on the right side.  If x > 2, cos(y)^x + sin(y)^x = cos(y)^2*cos(y)^(positive number) + sin(y)^2*sin(y)^(postiive number).  cos(y)^(postiive number) :le: 1, and sin(y)^(positive number) :le: 1, and the two equalities cannot both be satisfied at the same time.  Thus,

cos(y)^x + sin(y)^x < cos(y)^2 + sin(y)^2 = 1.

A similar argument shows that for x < 1, we will have our expression greater than 1.  All done.[/hide]

Arne wrote:Actually it is much simpler. No AM GM or calculus !

are you referring to an [hide]algorithmic argument? that namely, all numbers should be replaces with 3's, and then some 2's left over?[/hide]

that works, but it's not obvious. that's why calc gets you a general idea.

About 4. Here's something that might come in handy for some other problems: we can prove that, generally,                                     ((a^m-1)/(a-1), a-1) = (m, a-1).

In our case a=n, m=n-1, so ((n^(n-1)-1)/(n-1), n-1) = (n-1, n-1) = n-1, so n-1 | (n^(n-1)-1)/(n-1), which is what we wanted in the first place.


Breaking News: A teacher in Britain was arrested. On him they found a pencil, eraser, ruler, protractor and compass. They said he was part of the Al Gebra network, and that he used weapons of math instruction.

lol now you say that ivashka? This was posted 6 yrs ago!

Some people take longer to get a joke :)

hey it didn't take me that long to get the joke I joined this year


Prove that the recurrence $n=x_{n}.(x_{n-1}+x_{n}+x_{n+1})$, $n=1,2,...,$ $x_{0}=0$ has exactly one nonnegative solution.

This appeared as Problem 6 in the Miklós Schweitzer Competition 2003. See [url=http://artofproblemsolving.com/community/c6h410109]here [/url]for a sketch of the solution.


Suppose we have a solution of transparent alcooholic drink. To become more easy it can be sake.

What is the solution, in small quantities, that we can put on the sake to turn it in "wine", the color of the wine, just putting after a single cristal of the solution?!

Example, we can put Phenolphthalein and make the PH of the wine below 10, and after add a small quantitie of some basic element to PH be above 10, and it turn reddish purple.

But, there is another better way to turn "Water" in "Wine"?

PS: The result solution cannot be toxic, it need to be drinkeable, the problem is with this. hehe

I think there's more than just a toxicity problem here. Wine and other alcoholic drinks are quite acidic, with the pH of wine generally being from around three to around four. One would need to put in a significant amount of base for the liquid to reach a pH of ten, and actually, the liquid doesn't need to have a pH of ten. It just needs to be at least over seven--very slightly basic.

Also, using phenolphthalein to color the "wine" would be a bad idea because the indicator is poisonous when ingested and has a very un-winelike magenta hue, rather than the deep reddish purple you're suggesting.

It's a good idea; there are just a few kinks to work out. :)

Well, I think phenolphthalein has a pKa somewhere around 9... 9.3 maybe?  Anyway, a good rule of thumb for indicators is that the color change is detectable for pHs around their $pKa\pm 1$.  This means that you should first get the some pink around a pH of 8.5 or so...

Might I ask what this reaction is going to be used for?  It sounds like an interesting idea; I am intrigued.

Hmm...that's interesting about the phenolphthalein, gregx007. I was told by my chem teacher back when we were doing titrations that the indicator would change color when the reaction reached equilibrium (and passed it a little to become slightly basic), and I guess I assumed that meant it was effective for solutions with a pH that was simply over 7. However, I did some research, and phenolphthalein has a pKa of 9.3 and is effective through pH levels of 8.2 through 10.0. Sometimes I wish my teachers would explain a little more... :roll:

Anyway, to do this "water to wine" trick with our ethanol and our phenolphthalein--which we probably shouldn't be using, but whatever--we need to neutralize the acid with a base to make a non-toxic salt. The base would have to be non-toxic also if we want to be able to drink it when we're done because there will be some base still in solution. So, to start, what bases do we know of that will react with CH[size=75]3[/size]CH[size=75]2[/size]OH and make a salt that isn't poisonous?

[quote]PS: The result solution cannot be toxic, it need to be drinkeable, the problem is with this. hehe[/quote]

I just make a easily example. because phenolphthalein is the most well-know example of indicator.

Like someone said, we need a acid indicator, not a basic.
But we want a "wine" color.

And the solution we add cant modify (a lot) the taste of the "wine"

To be more convincent; The water can be transparent, dont have taste and smell. and the Wine be colored, have taste of wine and smell of wine.

with this the problem become toooo hard and quasi-impossible.

But it will be so wonderfull!! with a little cristal of the solution we "magically" turn water in wine.

with a list of Indicators
http://en.wikipedia.org/wiki/PH_indicator

If I have a indicator that turn Red in Acid solution and another one that turn Blue in acid solution, the result is purple?

can we work with 2, or more indicators to make the color we want? i never tried this.
we can use buffer solution and anothers solutions.

But I really dont know a good way to make this experience without killing the person who drinks. hehe

And this experience can be used in ClassRooms, this is a good project to students make calculations.

because the best in chemistry is make the theory and after see the theory in action

I will try to make someting in Crocodile Chemistry after.

thanks

So, you're saying you don't want to make the acid into a base, you just want a red/purple acid indicator? Sorry, I didn't understand that you were just using phenolphthalein for an example.

Our problem now is that most indicators are toxic. I was looking at methyl red because it seems to give somewhat of a "wine" color, but it's also very poisonous. I wonder if we could just use some food coloring dye. We could still try to find a way to make whatever substance into ethanol, but we wouldn't need to worry about trying to find a nontoxic indicator.

Of course, we could still look for one, but we could focus on other aspects of making the "wine" and come back to the color later.

Before continue we need to make a convention.

what is the wine solutions? Is not pure ethanol!!

I don't know the percentage of alcool.
But we can suppose that the wine have $x+y=100\%$,. with $x$ the percentage of alcool and $y$ the percentage of wine.

Since in brazil we have a drink that is more alcool than water, we can suppose $x>y$

and, how much moles will have the solution?! with the percentage we can calculate the moles. but. how much?!

this is just arbitrary. our supposition is: Wine = Ethanol + Water.

And we want to "hide" the ethanol.

we can make reversible reation with ethanol.

the color we can add after. but. since with the reversible reation, the result can be the "water" we want.

but the 'general' form of indicators are "acid = Blue"  and "basic = Red"

If I try to solve this problem in a simulator I cannot find the tast and the smell of the wine, just the color.

hmm....


The first round of invitations have been sent.  If you have done well on either OCTM or AMC, you can expect an invitation arriving at your home or your school's Math team coach.

Alan

Hm do you know where I can get homework?

[quote="ayang"]Send me an e-mail, and I'll send you the homework.

I have some idea who you are, but I want to be sure.

Alan[/quote]

Yep I just emailed you

If you have received invitation from me to join Ohio ARML, and
if you've sent me an e-mail, and
if you have not heard from me so far,

then please send me another e-mail or send me a snail mail, because sometimes e-mail can be lost in the cyber space.

Alan

This year's photos are up; check them out
http://www.cscc.edu/docs/math/faculty/ARML/2009%20Photos/

If you want a copy of any of them, shoot me an e-mail.

Alan

Wow, I'm actually in some this year. How'd that happen? :P

lol...
Some people I know are on there! *cough* some people*cough*


complex numbers
inequalities proposed
inequalities
geometric inequality

Let $ ABC$ be a triangle of sidelengths $ a,b,c$ and $ X$ a point in plane. We denote the distances of $ X$ from the vertices $ A,B,C$ with $ x,y,z$. Prove that

$ ayz\plus{}bzx\plus{}cxy \geq abc$

WITHOUT USING COMPLEX NUMBERS OR POLAR INVERSION !!!!! :wink:

Does it exist such a solution?  I cannot ind it. :(

[quote="mr.danh"]We proceed from the inequality $ mXA^2 \plus{} nXB^2 \plus{} pXC^2\ge \frac {mnc^2 \plus{} npa^2 \plus{} pmb^2}{m \plus{} n \plus{} p}$ for every triples m,n,p of real numbers.
Apply that ineq with $ m \equal{} \frac {a}{XA},n \equal{} \frac {b}{XB},p \equal{} \frac {c}{XC}$, hence result.[/quote]

Thank you very much, mr. danh.

I am so stupid that I didn't see it.  :blush: . Sorry.


A bridge is to be built in the shape of a parabolic arch and it to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

I do NOT just want the answer. I want you to explain how to find it and what you did to find the answer.

Try find the equation of the graph (quadratic) such that $ f(50) \equal{} f( \minus{} 50) \equal{} 0$  and also that $ f(40) \equal{} 10$

then $ f(0)$ is the answer.

This is a double-post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1468433257&t=172759 is the original

... and it also has appeared earlier in the Indiana forum: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=549457695&t=143371


[quote="Cezar Lupu"]Let $x, y, z$ three nonnegative reals. Prove that \[\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq2\left(1-\frac{2xyz}{(x+y)(y+z)(z+x)}\right).\] [/quote]

Isn't it just Schur when expanded?

I  have solved a more general one [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112118]here[/url]!

[quote="SplashD"]

Isn't it just Schur when expanded?[/quote]

yes, it is :) :blush:


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let $ [E: F] <\infty$ and $ S \equal{}\{ K | F\leq K\leq E\}$ does it mean $ |S| <\infty$.

This question has been posted here before. Standard example is something like: take an infinite field $ k$ of characteristic $ p>0$, $ E\equal{}k(T,U)$, $ F\equal{}k(T^{p},U^{p})$, $ K\equal{}F(T\plus{}aU)$ for $ a\in k$. owk

Actually, a finite field extension $ F\subseteq E$ has finitely many intermediate extensions iff it's simple (that is, $ E\equal{}F(\alpha)$ for some $ \alpha\in E$).

[quote="owk"]This question has been posted here before. Standard example is something like: take an infinite field $ k$ of characteristic $ p > 0$, $ E \equal{} k(T,U)$, $ F \equal{} k(T^{p},U^{p})$, $ K \equal{} F(T\plus{}aU)$ for $ a\in k$. owk[/quote]
Sorry, I am not familar with this notation. Can you explain?
(Or perhaps a more elementary example).

$ E$ is the quotient field of the polynomial ring in two variables $ T,U$ and coefficients in $ k$. the extension $ E/F$ is finite (it is not hard to see, that it has degree $ p^{2}$), but the $ F(T\plus{}aU)$ yield infinitely many intermediate fields.

[quote="grobber"]Actually, a finite field extension $ F\subseteq E$ has finitely many intermediate extensions iff it's simple (that is, $ E \equal{} F(\alpha)$ for some $ \alpha\in E$).[/quote]
Can you post a proof or a link, please?

(The case $ F$ having charachteristic $ 0$ is trivial. But otherwise it gets harder.)

http://www.mathreference.com/fld-sep,pet.html
it is not trivial for char 0.


Would anyone recommend against independently studying chemistry? I'm considering this for my Junior year for a few reasons; the primary being that the consensus among every chem student at my high school is that the current teacher is incapable of facilitating an actual understanding of the subject. If I did study independently, I would have to forego the lab portion of the course. In terms of college admissions, would not having a lab component make doing this not worth it?

Why would you not just take the class and read the book on your own to make up for the teacher's lackings?  It seems unlikely to me that the teacher could make it more difficult to learn chemistry than an absence of teacher.

It's not just that the teacher can't teach. It's that she spams you with busywork -- stupid assignments that require time that could be used on real classes such as AP Calc, Psych and Euro.

Does this make sense?

Missing the lab experience is high risk, not least because you don't really understand how science is done until you've spent substantial time in a lab. Albert Einstein, as I recall from an autobiographical statement of his, used to look for lab experience outside of his school (which he didn't like either), and he did lots of independent reading in libraries, for sure. If the teacher is demonstrably bad and time-wasting, and the only teacher in your school, you should do your utmost to find a different lab-based way to learn chemistry next year.

My local homeschooling group hunted all over town for lab space, and got a teacher recommended by our local university. We had to pay out of pocket for the teacher and for the lab equipment, but that was better learning than just reading a book. Maybe you can hunt up a group like that in your town.

Best wishes. I feel for you, as I had HUGE science interest when I was a kid, and few school experiences in science that encouraged that interest.

Thanks for the response tokenadult. I'll try to work something like that out for sure. Maybe I could even do the experiments with the class but study on my own; we'll see. Thanks again.


Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3.$

Prove that: $ \frac {(b \plus{} c)^5 \plus{} 32}{a^3 \plus{} 1} \plus{} \frac {(c \plus{} a)^5 \plus{} 32}{b^3 \plus{} 1} \plus{} \frac {(a \plus{} b)^5 \plus{} 32}{c^3 \plus{} 1} \geq\ 96$

 :)

[quote="nguoivn"][quote="Mathias_DK"][quote="Flakky"]Can we use Jensen ?[/quote]
Maybe ;)

$ f(x) \equal{} \frac {(3 \minus{} x)^5 \plus{} 32}{x^3 \plus{} 1}$, then $ f''(x) \equal{} \frac { \minus{} 2(x^9 \minus{} 268x^6 \plus{} 1260x^5 \minus{} 2190x^4 \plus{} 1900x^3 \minus{} 2520x^2 \plus{} 1095x \minus{} 270)}{(x^3 \plus{} 1)^3}$, so $ f''(x) > 0 \forall x \in [0;3]$.... And we are done :)[/quote]
Wow, nice work, my friend. Althought we only need to use Am-Gm (my proof is only 2 lines)  :)[/quote]
I'll see if I can find the nice proof by AM-GM then. :)

[quote="Potla"]Remember, AM GM works for POSITIVE reals only, but here the condition given is $ a, b, c\ge0$......
AM GM won't directly work here. :|[/quote]
Am-Gm works for non-negative reals... ;)

We could try proving $ \frac{{(3\minus{}x)}^{5}\plus{}32}{{x}^{3}\plus{}1}\minus{}120\plus{}88x\geq 0$ for $ 0\leq x\leq 3$
But this is equivalent with $ {(x\minus{}1)}^{2}(\minus{}{x}^{3}\plus{}101{x}^{2}\minus{}7x\plus{}155)\geq 0$ So all we have to do is summing over a,b and c and we have our inequality!


Find the number of ordered triple $(a,b,c)$ where $a$, $b$, and $c$ are positive integers, $a$ is a factor of $b$, $a$ is a factor of $c$, and $a+b+c=100$.

did you use the hockey-stick identity for this one?

[hide]Create x and y such that $b=ax$ and $c=ay$.

Now you have $a+ax+ay=100$, factor out an a $a(1+x+y)=100$. Realize a can be any factor of 100, and after you divide a out on both sides, you subtract y. This leaves you with $x+y=0, 1, 3, 4, 9, 19, 24, 49, 99$. Since x and y are both positive integers, the number of possible solutions for x and y are always 1 less than what they equal so you get $0+0+2+3+8+18+23+48+98=\boxed{200}$.[/hide]

oh if only 0 was a positive integer   :(

[quote="AznChristmas"]oh if only 0 was a positive integer   :([/quote]

OMG, seriously!!! did u get 217? that's what i got (i counted 0's too  :( )

[quote="AznChristmas"]217. my method was perfect... except for the common sense.[/quote]

i know what you mean. i'd say that's a kinda important part in solving a problem.

phdman, you don't have to insult people.  Also, your solution is very unclear.  Also, the real answer is $ 200$.

See [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_II_Problems/Problem_2]here[/url] for solution.

[hide=Is there a faster way?]
Let $b=ak, c=aw$ where $k,w$ are both positive integers.
$a(1+k+w)=100.$
Pair the factors of $100$:
$1, 100$
$2,50$
$4,25$
$5,20$
$10,10$
and alternate the order accordingly.
For each case we need $k$ and $w$ to be positive integers.
Doing all this casework, we get
$8+98+48+23+2+18+3=200.$
[/hide]

This seems like a very slow way. if the number was huge like $3600$, how would I go about solving it.

It's the AIME, the questions will take a few minutes

^okay.

But can you explain what you would do if the RHS was a number with many many factors like 3600, instead of 100?


A paper cone is to be made from a three-quarter circle having radius 4 inches (shaded). What is the length of the arc on the discarded quarter-circle (dotted portion)? Express your answer in terms of $ \pi$.

[asy]size(80);
import graph;
draw(Arc((0,0),1,0,90),linetype("2 4"));
filldraw((0,0)--(0,1)--Arc((0,0),1,90,360)--cycle,gray(.6),linewidth(1));
label("$4''$",(.5,0),S,fontsize(12pt));[/asy]

[hide]We are looking for 1/4 th the circumference. The diameter is 8, so

1/4(8 pi)= [b]2pi inches[/b][/hide]


vector
calculus
integration
algebra unsolved
algebra

Prove that for any irrational $\alpha$ there exists $n$ so that {$\alpha ^n$} $\in [\frac{1}{3},\frac{2}{3}]$

[quote="cyberjoe"]I've seen it already on the forum ...[/quote]

You might be right because I have a feeling like I've seen it too, though I searched and found nothing  :?
So I'm waiting for a solution (or link if problem already appeared somewhere)

Megus, I think some of the problems you've been posting lately might come from a file containing problems which appear to have been used in the Romanian IMO training in 2000. I would upload it, but apparently, there are big problems with my internet connection, so uploading it would take like two hours :). This problem is in that file, and so are the two with unit vectors which you posted in the "Combinatorics" section last night.

If this is indeed the case, then please be careful. Don't just post everything you see, as there are many mistakes and unclear statements in that file.

For instance, I'm almost sure this problem is supposed to be like this:

Prove that there is a positive irrational $\alpha$ s.t. for all positive integers $n$ we have $\{\alpha^n\}\in\left[\frac 13,\frac 23\right]$.

If in the problem you posted $n$ takes positive integral values, then think about the case $0<\alpha<\frac 13$. It would clearly be a counterexample.

Edit:

What do you know! :) Apparently, I've managed to upload it after all! :)

Thanks for clarifying Grobber, this is indeed the file which I had and took problems from - some statements are indeed very unclear -  do you know whether there are solutions available to this paper anywhere on the Internet ? Otherwise, I'll keep on posting some of the problems (unless they were already discussed) as I found them interesting.

Also do you have a proof to fixed version of the problem above?

No, I've never seen a file with solutions, and although I have worked on this problem before (a very long time ago, I can't remember if I solved it :)), I don't have the time to look at it right now. Anyway, the idea is to construct some closed nested intervals which must contain every number with the proeprties we want, and to conclude that the number must exist, since the intersection of every descending sequence of compact intervals is non-void.


$ \frac{1\plus{}a^{2}}{\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}}\plus{}\frac{1\plus{}b^{2}}{\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}}\plus{}\frac{1\plus{}c^{2}}{\sqrt{2c^{2}\plus{}3ac\minus{}b^{2}}}\equal{}$
$ \equal{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}a^{2}}{\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}}\plus{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}b^{2}}{\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}}\plus{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}c^{2}}{\sqrt{2c^{2}\plus{}3ac\minus{}b^{2}}}\geq\frac{(4a\plus{}4b\plus{}4c)^{2}}{4(\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}\plus{}\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}\plus{}\sqrt{2c^{2}\plus{}3ac\minus{}b^{2}})}$
by Cauchy-Schwarz

$ \sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}\plus{}\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}\plus{}\sqrt{2c^{2}\plus{}3ac\minus{}b^{2}}\leq 3\sqrt{\frac{2a^{2}\plus{}3ab\minus{}c^{2}\plus{}2b^{2}\plus{}3bc\minus{}a^{2}\plus{}2c^{2}\plus{}3ac\minus{}b^{2}}{3}}\equal{}3\sqrt{\frac{1\plus{}3(ab\plus{}bc\plus{}ac)}{3}}$
by AM-QM
So
$ \frac{(4a\plus{}4b\plus{}4c)^{2}}{4(\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}\plus{}\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}\plus{}\sqrt{2c^{2}\plus{}3ac\minus{}b^{2}})}\geq \frac{4(a\plus{}b\plus{}c)^{2}}{3\sqrt{\frac{1\plus{}3(ab\plus{}bc\plus{}ac)}{3}}}$
Now we should prove that
$ 2(a\plus{}b\plus{}c)\geq  3\sqrt{\frac{1\plus{}3(ab\plus{}bc\plus{}ac)}{3}}$
$ 4(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2(ab\plus{}bc\plus{}ac))\geq 3(1\plus{}3(ab\plus{}bc\plus{}ac))$
$ 1\geq ab\plus{}bc\plus{}ac$, which obviously true, because $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}1$
So
$ \frac{4(a\plus{}b\plus{}c)^{2}}{3\sqrt{\frac{1\plus{}3(ab\plus{}bc\plus{}ac)}{3}}}\geq 2(a\plus{}b\plus{}c)$ :)
Hope that's right. And what is your proof, Michael?

It is problem from Ukranian TST 2007 number 1.
Here is  my solution. It is also author's solution.
We use inequality $ \frac{a^{2}}{b}\ge 2a\minus{}b$
$ \sum_{cyc}\sqrt{\frac{(1\plus{}a^{2})^{2}}{2a^{2}\plus{}3ab\minus{}c^{2}}}\ge\sum_{cyc}\sqrt{2\plus{}c^{2}\minus{}3ab}\ge\sum_{cyc}\sqrt{\frac{a^{2}\plus{}b^{2}\plus{}6c^{2}}{2}}\ge\sum_{cyc}\frac{a\plus{}b\plus{}6c}{4}\equal{} 2(a\plus{}b\plus{}c)$

[quote="Shishkin"]It is problem from Ukranian TST 2007 number 1.
[/quote]

Can you post all the problems from this TST?

[quote="arqady"]$ \{a,b,c\}\subset\left(\frac{1}{\sqrt6},\plus{}\infty\right)$ such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1.$ Prove that
$ \frac{1\plus{}a^{2}}{\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}}\plus{}\frac{1\plus{}b^{2}}{\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}}\plus{}\frac{1\plus{}c^{2}}{\sqrt{2c^{2}\plus{}3ca\minus{}b^{2}}}\ge2(a\plus{}b\plus{}c).$[/quote]

Another possible solution is with Holder, i.e.

\[ \left(\frac{1\plus{}a^{2}}{\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}}\plus{}\frac{1\plus{}b^{2}}{\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}}\plus{}\frac{1\plus{}c^{2}}{\sqrt{2c^{2}\plus{}3ca\minus{}b^{2}}}\right)^{2}\geq\frac{4^{3}}{\sum{(1\plus{}a^{2})(2a^{2}\plus{}3ab\minus{}c^{2})}}.\]

Hence it remains to prove that $ 16\geq (\sum{a})^{2}\sum{(1\plus{}a^{2})(2a^{2}\plus{}3ab\minus{}c^{2})}$, which is true.

[quote="arqady"]$ \{a,b,c\}\subset\left(\frac {1}{\sqrt6}, \plus{} \infty\right)$ such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1.$ Prove that
$ \frac {1 \plus{} a^{2}}{\sqrt {2a^{2} \plus{} 3ab \minus{} c^{2}}} \plus{} \frac {1 \plus{} b^{2}}{\sqrt {2b^{2} \plus{} 3bc \minus{} a^{2}}} \plus{} \frac {1 \plus{} c^{2}}{\sqrt {2c^{2} \plus{} 3ca \minus{} b^{2}}}\ge2(a \plus{} b \plus{} c).$[/quote]
By the Cauchy Schwarz Inequality, we get
\[ \sum \frac{a^2}{\sqrt{2a^2\plus{}3ab\minus{}c^2}} \ge \frac{(a\plus{}b\plus{}c)^2}{\sum \sqrt{2a^2\plus{}3ab\minus{}c^2}} \ge \frac{(a\plus{}b\plus{}c)^2}{\sqrt{3(\sum a^2\plus{}3\sum ab)}} \ge \frac{(a\plus{}b\plus{}c)^2}{\sqrt{3 \cdot \frac{4}{3}(a\plus{}b\plus{}c)^2}} \equal{}\frac{1}{2}(a\plus{}b\plus{}c)\]
and
\[ \sum \frac{1}{\sqrt{2a^2\plus{}3ab\minus{}c^2}} \ge \frac{9}{\sum \sqrt{2a^2\plus{}3ab\minus{}c^2}} \ge \frac{9}{\sqrt{3(\sum a^2\plus{}3\sum ab)}} \ge \frac{9}{\sqrt{3\cdot \frac{4}{3}(a\plus{}b\plus{}c)^2}} \equal{}\frac{9}{2(a\plus{}b\plus{}c)} \equal{}\frac{9(a\plus{}b\plus{}c)}{2(a\plus{}b\plus{}c)^2} \ge \frac{9(a\plus{}b\plus{}c)}{2\cdot 3(a^2\plus{}b^2\plus{}c^2)} \equal{}\frac{3}{2}(a\plus{}b\plus{}c)\]
Adding up these inequalities, we can get the result. :)

[quote="micf"][quote="Shishkin"]
$ \sum_{cyc}\sqrt {\frac {a^{2} \plus{} b^{2} \plus{} 6c^{2}}{2}}\ge\sum_{cyc}\frac {a \plus{} b \plus{} 6c}{4}$[/quote]
Can anybody explain in to me?[/quote]
$ (a^{2} \plus{} b^{2} \plus{} 6c^{2})(1\plus{}1\plus{}6)\ge\ (a\plus{}b\plus{}6c)^2$

$ \rightarrow \sum_{cyc}\sqrt {\frac {a^{2} \plus{} b^{2} \plus{} 6c^{2}}{2}}\ge\sum_{cyc}\frac {a \plus{} b \plus{} 6c}{4}$

I solve it by using Carlson inequality first,and then
by a series of complicated calculation,and using:
3$\sum-{cyc}a^{3}b\le\ (a^{2}+b^{2}+c^{2})^2$
the problem can be solved


I am really interested in taking the Intermediate Trig. class... do any classes run through the summer?

Yes, we will have classes in the summer.  [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]Two of them have already been announced[/url], and we hope to announce our full summer schedule fairly soon.


[img]http://i47.photobucket.com/albums/f180/HawkG/Problem007.jpg[/img]

We know that AB=BC; AD - diameter; Question is to find angle 1 which is BAC.

I was wondering if it was even possible to find angle BAC with given information  :(  :?:

So can anyone solve it?

[quote="Centy"]There is not enough information to find $\angle BAC$ directly, although we can get it in terms of $\angle DAC$.[/quote]

Yes,  I know about DAC. (It's twice angle 1). So it's impossible to solve with given information. I guess so, there must be a typo in the book.

Thanks for the help.

they could have meant this:
find angle 1 if $AB=BC=CD$

[quote="Centy"]In which case its a simple angle case using Sine Rule.[/quote]
We don't even need to use Law of Sines.
If $AB=BC=CD$, then $\triangle OAB$, $\triangle OBC$ and $\triangle OCD$ are congruent equilateral triangles ;)


trigonometry
geometry
angle bisector
geometry proposed

$\Rightarrow$ follows pretty quickly from $(AP,AC;AB,AM)=(AQ,AB;AC,AM)=-1$. The other part does too, but we need just a tiny bit of trigonometry.

Sure, but I'll leave that pleasure to someone else for now.. :)

For $\Rightarrow$:

The lines AM and AA' are symmetric to each other with respect to the angle bisector of the angle C'AB' (in fact, both lines AM and AA' coincide with this angle bisector, since < MAB = < MAC).

Now apply the problem version 1 in my first post in the thread http://mathlinks.ro/Forum/viewtopic.php?t=6082 to the triangle B'C'A and the points M and A'. Remember, of course, that the lines AM and AA' are symmetric to each other with respect to the angle bisector of the angle C'AB'. The lines B'A' and C'A' meet the lines C'M and B'M at the points Q and P, respectively. Then, from that problem, it follows that the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle C'AB'. In other words, < MAP = < MAQ.

For $\Leftarrow$:

The proof is similar to the above for the $\Rightarrow$ direction, but this time you will have to apply the problem version 1 in my first post in the thread http://mathlinks.ro/Forum/viewtopic.php?t=6082 to the triangle PQA and the points M and A'.

Problem solved.

  Darij

[quote="nttu"]Let a triangle ABC .  M is inside ABC . MA intersects BC at A' . MB intersects AC at B' . MC intersects AB at C' . MB intersects A'C' at P . MC intersects A'B' at Q . Prove that : <MAB = <MAC <==> <MAP = <MAQ[/quote]
Very nice problem.

It is associated in my mind with the following problem:
Let $AH$ be altitude, $H\in BC$, $M\in AH$. Line $BM$ intersect $AC$ at point $B'$, $CM$ intersects $BC$ at point $C'$. Prove that $\angle C'HA=\angle B'AH$.

Is there some kind if inverse statement?

[quote="Myth"]It is associated in my mind with the following problem:
Let $AH$ be altitude, $H\in BC$, $M\in AH$. Line $BM$ intersect $AC$ at point $B'$, $CM$ intersects $BC$ at point $C'$. Prove that $\angle C'HA=\angle B'AH$.[/quote]

You probably meant $\angle C'HA = \angle B'HA$ here.

[quote="Myth"]Is there some kind if inverse statement?[/quote]

Of course, there is: If H is an arbitrary point on the line BC (not necessarily the foot of the altitude from A), and if < C'HA = < B'HA, then the line AH is the altitude from A in triangle ABC.

  Darij

[quote]Let $ABC$ be a triangle, and let $M$ be a point inside triangle $ABC$. The line $MA$ intersects the line $BC$ at $A^{\prime}$. The line $MB$ intersects the line $CA$ at $B^{\prime}$. The line $MC$ intersects the line $AB$ at $C^{\prime}$. The lines $MB$ and $C^{\prime}A^{\prime}$ intersect each other at $P$. The lines $MC$ and $A^{\prime}B^{\prime}$ intersect each other at $Q$. Prove that $\measuredangle MAB=\measuredangle MAC\Longleftrightarrow\measuredangle MAP=\measuredangle MAQ$.[/quote]
Nice problem [b]nttu[/b]. I'll prove only one side, second side can be easily proven in the same way. Here is my solution.
Our target is to prove that if $\angle{MAB}=\angle{MAC}\Rightarrow \angle{MAP}=\angle{MAQ}$.Denote $\angle{MAP}=\alpha,\angle{MAQ}=\beta,\angle{PAB}=x,\angle{QAC}=y.$
It is sufficient to prove that $\frac{\sin{\alpha}}{\sin{x}}=\frac{\sin{\beta}}{\sin{y}}$(obvious). Now from triangles ACP'and APA' we get that $\frac{\sin{\alpha}}{\sin{x}}=\frac{AC'\cdot{A'P}}{AA'\cdot{PC'}}$.From triangles AA'Q and AQB' we get  $\frac{\sin{\beta}}{\sin{y}}=\frac{AB'\cdot{A'Q}}{AA'\cdot{B'Q}}$. So we have to prove that $\frac{AC'\cdot{A'P}}{AA'\cdot{PC'}}=\frac{AB'\cdot{A'Q}}{AA'\cdot{B'Q}}$ or$\frac{AC'}{AB'}\cdot{\frac{A'P}{PC'}}\cdot{\frac{QB'}{QA'}}=1$. But using triangles BPA' and BPC',A'QC and QB'C we get that our target is to prove that $\frac{A'C\cdot{AB'}\cdot{BC'}}{B'C\cdot{AC'}\cdot{BA'}}\cdot{\frac{\sin{MCB}\cdot{\sin{MBA}}}{\sin{MCA}\cdot{\sin{MAC}}}=1}$
And this follows immediatelly follows from Cheva (simple) and Cheva (trigonometric). I know, you think that my solution is ugly but in fact it is very simple.  :)

Oh ,yeah . You have a nice solution , Armo !! :)  :)

Thank you very much. :)
Can you please post your solution [b]nttu[/b]?