quadratics
modular arithmetic
number theory
prime numbers
number theory proposed

Let $ m$ and $ n$ be given positive integers greater that 1 and $ (m,n)\equal{}1.$ Prove that there exist infinitely many prime numbers $ p$ such that $ p^{2k\plus{}1}\parallel{}m^{p\minus{}1}\minus{}n^{p\minus{}1}$ for some integer $ k\geq 0.$ ($ p^{\alpha}\parallel{}a$ iff $ p^{\alpha}|a$ and $ p^{\alpha \plus{}1}\not | a$ )

For a given prime $ p$ let $ v_p(n)$ be the highest exponent of $ p$ which divides $ n$.
It is well known that if $ p|a\minus{}b$ then $ v_p(a^n \minus{} b^n) \equal{} v_p(a\minus{}b) \plus{} v_p(n)$, we will use it later.
We will prove that if $ (m, n)\equal{}1$ and $ m>n$ then $ v_p(m^{p\minus{}1} \minus{} n^{p\minus{}1}) \equiv 1 \mod{2}$ for infinitely many primes $ p$.
So, let's suppose that there are only finitely many such primes and call them $ p_1, p_2, ..., p_k$. Denote by $ d_1, d_2, ..., d_k$ the order of $ \frac{m}{n}$ mod $ p_1, p_2, ..., p_k$ respectively. The basic property of order says that $ p_i|m^a \minus{} n^a$ iff $ d_i|a$.
Take prime $ p$ bigger then $ p_1, p_2, ..., p_k$ and not dividing $ m\minus{}n$. Let $ q$ be the arbitrary prime divisor of $ m^p \minus{} n^p$ which is not a divisor of $ m\minus{}n$. Since $ (p, d_i) \equal{} 1$ for every $ i$ $ q$ is not any of the $ p_1, ..., p_k$ and it's also clear that $ p \ne q$. It means that $ 2|v_q(m^{q\minus{}1} \minus{} n^{q\minus{}1})$, but it is easy to see that $ v_q(m^{q\minus{}1} \minus{} n^{q\minus{}1}) \equal{} v_q(m^{p} \minus{} n^{p})$. Indeed, $ p$ is an order of $ \frac{m}{n} \mod{q}$ so $ q\minus{}1 \equal{} pl$ and then $ v_q(m^{pl} \minus{} n^{pl}) \equal{} v_q(m^p \minus{} n^p) \plus{} v_q(l)$ but $ v_q(l)\equal{}0$.
Now take any prime divisor $ q$ of $ m\minus{}n$. Then $ v_q(m^p \minus{} n^p) \equal{} v_q(m\minus{}n)$ (since $ p \ne q$) so $ v_q(\frac{m^p\minus{}n^p}{m\minus{}n})\equal{}0$.
So it follows that every prime divisor of $ \frac{m^p\minus{}n^p}{m\minus{}n}$ divides it in a even exponent, so this number is a perfect square. Let $ m\minus{}n\equal{}a$ and suppose that $ a>2$.
Take a prime divisor $ q$ of $ a$ such that $ q \geq 3$ or take $ q\equal{}4$ if $ a$ is a power of two. An above reasoning holds for any sufficiently large prime $ p$, so take sufficiently big $ p$ with $ p$ being non quadratic residue mod $ q$ (we can do it, because $ q>2$). Then $ \frac{m^p\minus{}n^p}{m\minus{}n}$ is a square. But:
$ \frac{m^p\minus{}n^p}{m\minus{}n} \equal{} \frac{(n\plus{}a)^p \minus{} n^p}{a} \equal{} pn^{p\minus{}1} \plus{} \binom{p}{2}an^{p\minus{}2} \plus{} ... \plus{} a^{p\minus{}1} \equiv pn^{p\minus{}1} \mod{q}$. But $ 2|p\minus{}1$, so $ n^{p\minus{}1}$ is a square mod $ q$ and $ p$ is not a square mod $ q$, so $ pn^{p\minus{}1}$ is also not a square mod $ q$. Contradiction, since $ \frac{m^p\minus{}n^p}{m\minus{}n}$ is a square.
We are left with the cases $ a\equal{}1$ and $ a\equal{}2$. But instead of $ \frac{m^p \minus{} n^{p}}{m\minus{}n}$ we can consider numbers $ \frac{m^{pq} \minus{} n^{pq}}{m^p\minus{}n^p}$ with a fixed $ p$ and $ q$ varying. Then we can take $ x\equal{}m^p$, $ y\equal{}n^p$ then $ x\minus{}y > 2$ so the previous part of the proof works.

I know it's not very detailed so ask if something is not clear or if it is wrong.

My solution uses some useful facts about some diophantine equations:
It is well known that $ m^{2^s}+n^{2^s}$ is not a square for $ s>1$ since the equation $ x^4+y^4=z^2$ has no solutions in positive integers; also, $ x^4-y^4=z^2$ has a solution in positive integers iff $ x=y$ (Proved by Fermat). Thus for each $ s>1$ there exists a prime $ p_s$ such that $ p_s^{2k_s+1}\parallel{}m^{2^s}+n^{2^s}.$ Also, we have that $ p_s$ is an odd since the equation $ x^4+y^4=2z^2$ which is equivalent to $ z^4-(xy)^4=\left(\frac{x^4-y^4}{2}\right)^2$ has an unique solution $ x=y=z=1$ since the equation $ x^4-y^4=z^2$ is solvable in positive integers iff $ x=y.$ Thus $ m^{2^s}\equiv -n^{2^s}\pmod{p_s}$ or $ (\frac{m}{n})^{2^s}\equiv -1\pmod{p_s}.$ From here we obtain that $ 2^{s+1}$ is the order of $ m\n$ modulo $ p_s$ and therefore $ p_s=2^{s+1}l_s+1.$ Now we have that $ p_s^{2k_s+1}\parallel{}m^{2^s}+n^{2^s} | m^{2^{s+1}}-n^{2^{s+1}}$ and since $ (m^{2^s}+n^{2^s}, m^{2^s}-n^{2^s})=1$ we have that \[ p_s^{2k_s+1}\parallel{}m^{2^{s+1}}-n^{2^{s+1}} .\] Since $ p_s \not | l_s$ we get that $ p_s^{2k_s+1}\parallel{}m^{2^{s+1}l_s}-n^{2^{s+1}l_s}=m^{p_s-1}-n^{p_s-1}.$ (Here we use the fact that if $ p$ is odd and $ p^{\alpha}\parallel{}a-b$ and $ p^{\beta}\parallel{}k$ it follows that $ p^{\alpha+\beta}\parallel{}a^k-b^k.$ )
It remains to prove that $ p_s$ are infinitely many primes which follows from the fact that $ (m^{2^s}+n^{2^s},m^{2^l}+n^{2^l})=1$ for all $ k$ and $ l.$

Dear [b]TomciO[/b]
Could you post a proof or a sketch of the proof of this fact:[quote]$ v_{p}(a^{n}\minus{}b^{n}) \equal{} v_{p}(a\minus{}b)\plus{}v_{p}(n)$[/quote]
And could you explain how did you get that:[quote]$ 2|v_{q}(m^{q\minus{}1}\minus{}n^{q\minus{}1})$[/quote]
Thank you in advance.

[quote="birzhan"]Could you post a proof or a sketch of the proof of this fact:[quote]$ v_{p}(a^{n} \minus{} b^{n}) \equal{} v_{p}(a \minus{} b) \plus{} v_{p}(n)$[/quote][/quote]
http://www.mathlinks.ro/viewtopic.php?t=27757

[quote="birzhan"]Dear [b]TomciO[/b]
Could you post a proof or a sketch of the proof of this fact:[quote]$ v_{p}(a^{n} \minus{} b^{n}) \equal{} v_{p}(a \minus{} b) \plus{} v_{p}(n)$[/quote]
And could you explain how did you get that:[quote]$ 2|v_{q}(m^{q \minus{} 1} \minus{} n^{q \minus{} 1})$[/quote]
Thank you in advance.[/quote]

I am proceeding by the contradiction, I assume that $ p_1, p_2, \ldots, p_k$ are the only primes $ r$ for which $ 2 \not | v_r(m^{r\minus{}1} \minus{} n^{r\minus{}1})$. Then I take $ q$ in such a way that it is distinct from $ p_i$'s so we must have $ 2| v_q(m^{q\minus{}1} \minus{} n^{q\minus{}1})$.

[quote="bilarev"] the fact that $ (m^{2^s} \plus{} n^{2^s},m^{2^l} \plus{} n^{2^l}) \equal{} 1$ for all $ k$ and $ l.$[/quote]

 [i]  Can you prove that argument ? :maybe: [/i]

I found it in AMM 1984.May be it useful:)

[quote="nguyenvuthanhha"][quote="bilarev"] the fact that $ (m^{2^s} + n^{2^s},m^{2^l} + n^{2^l}) = 1$ for all $ k$ and $ l.$[/quote]

 [i]  Can you prove that argument ? :maybe: [/i][/quote]

I think it is easy because if $p$ is a common prime divisor of  $m^{2^s}+n^{2^s}$ and $m^{2^l}+n^{2^l}$ with $s\# l$  so
$p=2^{s+1}.k_s+1$ and $p=2^{l+1}.k_l+1$ so $s=l$
so contracdiction!


I was wondering- because there are not enough ppl in ri for a forum, would you consider "adopting" RI into this one?

I wonder if the majority will say so or an individual will. (I am fine with it.)


algebra
function
domain
real analysis
real analysis unsolved

Determine all $ x\in\mathbb{R}$ such that $ x^{x^{x^\ldots }}\in\mathbb{R}$.

There is a very large thread where the answer was determined to be $ \left[ \frac {1}{e^e}, e^{\frac {1}{e} } \right]$.  I'm trying to find it.

Edit:  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tower+power&t=77942]See here[/url].


It has been snowing heavily and it is a possibility school will be cancelled this coming wednesday for the AMC. Is it allowed to take the test at another school/university if my regular school is closed that day?


inequalities
algebra
polynomial
linear algebra

Let a,b,c,d are [b] real number [/b] such that $a \geq b \geq c \geq d$ and $a+d \geq b+c$and $a+b+c+d = 4$Find min:
$(a^{3}+b^{3}+c^{3}+d^{3})+3 (abc+bcd+cda+dab)$

 :D  :D

[hide="this is probably wrong, but..."]
$a=b=c=d=1$ so $\boxed{16}$
[/hide]

You also get 16 whenever $a=2$ and $d=0$.

[hide]
hmm... i think we have
\[4\left( a^{3}+b^{3}+c^{3}+d^{3}+3(abc+bcd+cda+dab)\right)-(a+b+c+d)^{3}\]
\[=3(a+d-b-c)(a+b-c-d)(a+c-b-d)\geq 0\]
so the minimum is indeed $\boxed{16}$. (Anything with $a+d=b+c=2$ produces equality.)

How did I come up with that? My first attempt was to write the given expression as
\[(a+d)(a^{2}-ad+d^{2}+3bc)+(b+c)(b^{2}-bc+c^{2}+3ad)\]
and then try Chebyshev because we know $a+d\geq b+c$. Unfortunately, this doesn't work (try $a=2,b=4\sqrt{3}-6,c=d=4-2\sqrt{3}$, for instance). However, we do know that if $a+d=b+c$, then the two expressions
\[(a+d)(a^{2}-ad+d^{2}+3bc)+(b+c)(b^{2}-bc+c^{2}+3ad),\]
\[\frac{a+b+c+d}{2}(a^{2}-ad+d^{2}+3bc+b^{2}-bc+c^{2}+3bc)\]
are equal. But the second expression is just $2(a+d)^{2}+2(b+c)^{2}$ which also equals 16 if $a+d=b+c$! So we find that $a+d=b+c$ implies that the given expression is 16. This means $a+d-b-c$ better be a factor of the homogenous inequality, and since the expression is symmetric, so should $a+b-c-d$ and $a+c-b-d$. This doesn't leave too many options. :P
[/hide]

[quote="scorpius119"][hide]
hmm... i think we have
\[4\left( a^{3}+b^{3}+c^{3}+d^{3}+3(abc+bcd+cda+dab)\right)-(a+b+c+d)^{3}\]

\[=3(a+d-b-c)(a+b-c-d)(a+c-b-d)\geq 0 \]
so the minimum is indeed $\boxed{16}$. (Anything with $a+d=b+c=2$ produces equality.)

How did I come up with that? My first attempt was to write the given expression as
\[(a+d)(a^{2}-ad+d^{2}+3bc)+(b+c)(b^{2}-bc+c^{2}+3ad) \]
and then try Chebyshev because we know $a+d\geq b+c$. Unfortunately, this doesn't work (try $a=2,b=4\sqrt{3}-6,c=d=4-2\sqrt{3}$, for instance). However, we do know that if $a+d=b+c$, then the two expressions
\[(a+d)(a^{2}-ad+d^{2}+3bc)+(b+c)(b^{2}-bc+c^{2}+3ad), \]

\[\frac{a+b+c+d}{2}(a^{2}-ad+d^{2}+3bc+b^{2}-bc+c^{2}+3bc) \]
are equal. But the second expression is just $2(a+d)^{2}+2(b+c)^{2}$ which also equals 16 if $a+d=b+c$! So we find that $a+d=b+c$ implies that the given expression is 16. This means $a+d-b-c$ better be a factor of the homogenous inequality, and since the expression is symmetric, so should $a+b-c-d$ and $a+c-b-d$. This doesn't leave too many options. :P
[/hide][/quote]

  Wonderful solution. And yes, it is also my solution. In fact, I find the identity by using Linear Algebra ! But it is amazing since you can factor the polynomial by hand !


a polynomial is called good if its (1) monic (2) irreducible over the integers (3) has integer coefficients
prove if f(x) is good, and |f(0)| is not a perfect square, then f(x^2) is also good.

(edited now)

I might not understand this question , but isn't $ f(x)=x^2+4 $ a counter-example?

I saw this problem in Romanian TST 2003.
You forgot important part of statement: |f(0)| is not a perfect square. Otherwise consider f(x)=x3-4.


Given the sequence 1,2,4,5,7,8,10,11,13,14,... where every third integer is missing.

(a) Find the sum of the first hundred terms of the sequence.

(b) Find a formula, in terms of n, for the nth term of the sequence.

[hide]a. If x is even, the formula for all of terms up to the nth term is 3(n-1)

3(100-1)

3(99)

297

b. [/hide]


Let $ a,b,c,d,e$ be real numbers such that $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}0$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2>0.$ Prove that
$ \frac{ab\plus{}bc\plus{}cd\plus{}de\plus{}ea}{a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2}< \frac{\sqrt{61}\minus{}5}{9}.$

[b]Stronger[/b]
Let $ a,b,c,d,e$ be real numbers such that $ a+b+c+d+e=0$ and $ a^2+b^2+c^2+d^2+e^2>0.$ Prove that
$ \frac{ab+bc+cd+de+ea}{a^2+b^2+c^2+d^2+e^2}\leq \frac{\sqrt{5}-1}{4}.$
(Ordinarily discovered by Ky Fan)

It's just Ky Fan ineq in the condition $n=5.$

The gerenal Ky Fan ineq is that:
If $a_i\in \mathbb{R},i=1,2,...,n$ and $\sum_{i=1}^n a_i=0$ then($a_{n+1}=a_1$)
$$\sum_{i=1}^n a_ia_{i+1}\leq \cos {\frac{2\pi}{n}} \sum_{i=1}^n a_i^2.$$


counting
distinguishability
number theory
greatest common divisor
algorithm
modular arithmetic

For positive integers $ n$, find all positive integers ${ N=3n(n+1)\ }$ that have 12 devisors.

This means that we must have $ N\equal{}\{a^{11},a^5b,a^3b^2,a^2bc\}$.

If $ 3n(n\plus{}1)\equal{}a^{11}$, then $ a\equal{}3$, and there are obviously no integer solutions.

If $ 3n(n\plus{}1)\equal{}a^5b$, then we can either have $ a\equal{}3$, which results in no solutions, or $ b\equal{}3$, which also has no solutions.

If $ 3n(n\plus{}1)\equal{}a^3b^2$, then we either have $ a\equal{}3$, which has no solutions, or $ b\equal{}3$, which has no solutions.

If $ 3n(n\plus{}1)\equal{}a^2bc$ (where $ b$ and $ c$ are indistinguishable), then we either have $ a\equal{}3$, which gives solutions $ (a,b,c,n,N)\equal{}(3,5,2,5,90)$ and $ (a,b,c,n,N)\equal{}(3,7,2,6,126)$, or $ b\equal{}3$, which gives $ (a,b,c,n,N)\equal{}(2,3,5,4,60)$.

Thus, the possible values of $ N$ are $ \boxed{60,90,126}$, which are achieved by $ n\equal{}4,5,6$, respectively.

[quote="math154"]This means that we must have $ N \equal{} \{a^{11},a^5b,a^3b^2,a^2bc\}$.

If $ 3n(n \plus{} 1) \equal{} a^{11}$, then $ a \equal{} 3$, and there are obviously no integer solutions.

If $ 3n(n \plus{} 1) \equal{} a^5b$, then we can either have $ a \equal{} 3$, which results in no solutions, or $ b \equal{} 3$, which also has no solutions.

If $ 3n(n \plus{} 1) \equal{} a^3b^2$, then we either have $ a \equal{} 3$, which has no solutions, or $ b \equal{} 3$, which has no solutions.

If $ 3n(n \plus{} 1) \equal{} a^2bc$ (where $ b$ and $ c$ are indistinguishable), then we either have $ a \equal{} 3$, which gives solutions $ (a,b,c,n,N) \equal{} (3,5,2,5,90)$ and $ (a,b,c,n,N) \equal{} (3,7,2,6,126)$, or $ b \equal{} 3$, which gives $ (a,b,c,n,N) \equal{} (2,3,5,4,60)$.

Thus, the possible values of $ N$ are $ \boxed{60,90,126}$, which are achieved by $ n \equal{} 4,5,6$, respectively.[/quote]

If $ 3n(n \plus{} 1) \equal{} a^5b$, then we can either have $ a \equal{} 3$, which results in no solutions, or $ b \equal{} 3$, which also has no solutions.

If $ 3n(n \plus{} 1) \equal{} a^3b^2$, then we either have $ a \equal{} 3$, which has no solutions, or $ b \equal{} 3$, which has no solutions.

How do you know the above have no solutions?

For $ 3n(n\plus{}1)\equal{}a^5b,a\equal{}3$, we have

$ n(n\plus{}1)\equal{}81b\equal{}1\cdot81b\equal{}3\cdot27b\equal{}\cdots$.

Testing these, we see there are no solutions. The other cases are similar.

This is easy for this problem, since the numbers are relatively small. You might try modular arithmetic, but it's not worth the hassle for this problem.

[quote="math154"]For $ 3n(n \plus{} 1) \equal{} a^5b,a \equal{} 3$, we have

$ n(n \plus{} 1) \equal{} 81b \equal{} 1\cdot81b \equal{} 3\cdot27b \equal{} \cdots$.

Testing these, we see there are no solutions. The other cases are similar.

This is easy for this problem, since the numbers are relatively small. You might try modular arithmetic, but it's not worth the hassle for this problem.[/quote]

But how can you prove that n(n+1) = a^5 doesn't have integer solutions?

$ a$ is prime (by definition), but $ \text{gcd}\,(n,n\plus{}1)\equal{}1$.


function
geometric series
algebra unsolved
algebra

Let $u$ and $v$ be real numbers such that
\[ (u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8.  \]
Determine, with proof, which of the two numbers, $u$ or $v$, is larger.

[hide]
Consider $f(x) = 1+x+\cdots+x^8+10x^9$ and $g(x) = 1+x+\cdots+x^{10}+10x^{11}$.

We are given that $f(u) = g(v) = 9$.

We have $g(x)-f(x) = 10x^{11}+x^{10}-9x^9 = x^9(10x-9)(x+1)$.

$f\left(\frac{9}{10}\right) = 1+\frac{9}{10}+\cdots+\frac{9^8}{10^8}+10 \cdot \frac{9^9}{10^9}$.

By summing the geometric series,

$f\left(\frac{9}{10}\right) = \frac{1-\frac{9^9}{10^9}}{1-\frac{9}{10}} = 10-10 \cdot \frac{9^9}{10^9}+10 \cdot \frac{9^9}{10^9} = 10 > 9$.

Since $f$ is an increasing function, we have $u < \frac{9}{10}$.

Hence $g(u)-f(u) = u^9(10u-9)(u+1) < 0 \Rightarrow g(u) < f(u) = 9$.

But since $g$ is an increasing function as well, we know $g(u) < g(v) = 9 \Rightarrow u < v$.[/hide]

$12\tfrac14$ year revive!

Remark that trivially $0 < u < 1$.  Note that $(u + u^2 + \cdots + u^8) + 10u^9 = 8$ rearranges to \[\frac{u^9-1}{u-1} + 10u^9 = 9\quad\Rightarrow\quad 10u^{10} - 9u^9 - 9u + 8 = 0.\]  This rearranges to $(u^9-1)(9u-8) = u^9(1-u)$; combined with our earlier observation we actually get $u < \tfrac89$.  Furthermore, remark that the given relation further implies \[(10u^{10} - 9u^9 - 9u + 8)(u+1) = 0\quad\Rightarrow\quad 10u^{11} + u^{10} - 9u^9 = 9u^2 + u - 8 = (9u-8)(u+1) < 0.\] Thus, \[(u+u^2 + \cdots + u^{10}) + 10u^{11} = 8 - 9u^9 + u^{10} + 10u^{11} < 8.\] It follows that, upon letting $f(x) = (x+x^2+\cdots+x^{10}) + 10x^{11}-8$, we deduce that $f(u) < 0$.  But $f$ is monotonically increasing, so in fact its root $v$ must satisfy $v > u$, implying that $v$ is larger.


function
limit
irrational number
algebra unsolved
algebra

Setting $g(x)=f(x)+\frac{x^{2}}{2}$ the relation rewrites:
\[g(x)-\frac{x^{2}}{2}+g(y)-\frac{y^{2}}{2}-\left(g(x+y)-\frac{(x+y)^{2}}{2}\right)=xy\Leftrightarrow\]
\[g(x)+g(y)=g(x+y).\]
This is Cauchy's equation and, since $g$ is continuous, it is well-known that only such function are $g(x)=tx$, where $t\in\mathbb{R}$.

Thus $f(x)=tx-\frac{x^{2}}{2}$.

A sketch of proof for determining $g$ is as follows:
$\textbf{1}$
Set $g(1)=t$. Find $g(x)=tx$, for $x\in\mathbb{Z}$.
$\textbf{2}$
From $k\cdot g\left(\frac{1}{k}\right)=g\left(\frac{1}{k}+\dots+\frac{1}{k}\right)=g(1)=t$, we get $g\left(\frac{1}{k}\right)=\frac{t}{k}$, for $k\in\mathbb{Z}$, nonzero
$\textbf{3}$
Get $g\left(\frac{n}{k}\right)=ng\left(\frac{1}{k}\right)=t\cdot\frac{n}{k}$.
So $g(x)=tx$, for $x\in\mathbb{Q}$.
$\textbf{4}$
Using the fact that $g$ is continuous we "approximate" any irrational number by two sequences of rational numbers, deducing $f(x)=tx$, $\forall x\in\mathbb{R}$.

show that $\lim_{x\to0}\frac{f(x)}{x}=0$
and see $\lim_{x\to0}\frac{f(y+x)-f(x)}{x}=\lim_{x\to0}-y+\frac{f(x)}{x}=-y$.  so  $f(y)^{\prime}=-y$ or $f(x)^{\prime}=-x$
so $f(x)=\frac{-1}{2}x^{2}+c$  and using the condition u will get
$f(x)=\frac{-1}{2}x^{2}$. :idea:
verry nice freemind :oops:

Sorry for revival :P but f(x)=x(2c+1-x)/2 also works.

It is just freemind's solution with $t = c + \dfrac{1}{2}$.

[quote=maths314]Find all continuous functions f:R to R such that
             $$f(x)+f(y)-f(x+y)=f(xy)$$[/quote]

We claim that $f(x)=0$ is then only solution.
Actually you should post this as a new topic.

[quote=pco][quote=IMO2019][We claim that $f(x)=0$ is then only solution.[/quote]
$f(x)=17$ $\forall x$ is also a solution  :blush:[/quote]

Sorry, for the incovinience but I didn;t do that since the op posted hapharzadly. OFC, $f(x)=c$ is a solution. :blush:

[hide=#1][url=aops.com/community/user/28466][b]p.v.ninh[/b][/url] · Mar 21, 2007, 9:52 AM [url=aops.com/community/p790953](view)[/url][color=transparent]helo[/color]
Find all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that \[f(x)+f(y)-f(x+y)=xy\;\forall x,y \in\mathbb{R}.\]

-----------
[color=#5b7083][aops]x[/aops] 1[color=transparent]hellloolo[/color] [aops]Y[/aops] 1 [color=transparent]hellloolo[/color] [/hide]: $g(x)=f(x)+\frac{x^2}2\Rightarrow g(x+y)=g(x)+g(y)\Rightarrow g(x)=cx\Rightarrow f(x)=-\frac12x^2+cx,c\in\mathbb R$, which all work.

[hide=#9][url=aops.com/community/user/415688][b]maths314[/b][/url] · Jun 4, 2018, 8:27 AM [url=aops.com/community/p10466758](view)[/url][color=transparent]helo[/color]
Find all continuous functions f:R to R such that
             f(x)+f(y)-f(x+y)=f(xy)

-----------
[color=#5b7083][aops]x[/aops] 9[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]: $P(x-1,1)\Rightarrow f(x)=c,c\in\mathbb R$, works.

[quote=pco][quote=maths314]Find all continuous functions f:R to R such that
             f(x)+f(y)-f(x+y)=f(xy)[/quote]
Let $P(x,y)$ be the assertion $f(x)+f(y)-f(x+y)=f(xy)$
Let $a=f(1)$

$P(x-1,1)$ $\implies$ $\boxed{f(x)=a\quad\forall x}$ which indeed is a solution[/quote]

Nice


(i) Let $ ABC$ be a triangle. Let $ M$ be the mid-point of the side $ AC$. Show that the areas of triangles $ ABM$ and $ MBC$ are equal.

(ii) Let $ ABC$ be a triangle. Let $ N$ be the point of intersection of the side $ AC$ with the bisector of angle $ B$. Show that $ AB: BC \equal{} AN: NC$.

[hide="1"]
They have the same height, so the ratio of their areas is the ratio of their bases, which is $ 1$ because the bases are both one half of $ AC$.
[/hide]

[hide="2"]
This is known as the Angle Bisector Theorem.

Draw a line parallel to $ AB$ passing through $ C$. Extend $ BN$ so that it hits the parallel line at $ D$. Then $ \angle D$ is clearly the same as $ \angle ABN$ and $ \angle NBC$. $ \triangle BCD$ is isosceles, so $ BC\equal{}CD$. $ \triangle ABN\sim \triangle CND$, so $ AB/DC\equal{}AN/NC$. Substitute $ BC\equal{}CD$ and the conclusion follows.
[/hide]


calculus
integration
algebra
polynomial
analytic geometry
calculus computations

Use the given transformation to evaluate the integral:

$ \iint_R(x\minus{}3y)dA$, where $ R$ is the triangular region with vertices $ (0,0)$, $ (2,1)$, and $ (1,2)$; $ x\equal{}2u\plus{}v$, $ y\equal{}u\plus{}2v$

The book claims it's -3, but the first time I got 10.5, and now I keep getting -17.5.  It's clearly negative, because $ 3y\geq\frac{3}{2}x\in R\implies x\minus{}3y\leq 0\in R$, but am I right or is the book?  I'm pretty sure I haven't made any mistakes in the integrals (they're just polynomials after all), but I've gone over it about 4 times and can't get the same answer as the book.  -17.5 seems to make more sense, because $ x\minus{}3y\equal{}\minus{}u\minus{}5v$, and at $ (2,1)$, $ \minus{}u\minus{}5v\equal{}\minus{}11$, and at $ (1,2)$, $ \minus{}u\minus{}5v\equal{}\minus{}7$.  Is my book just screwing me over?

Those corners are given as $ (x,y)$ pairs. What are their $ (u,v)$ coordinates?

Thanks.  Now I feel stupid  :blush:   Must have been because I was doing stuff with finding images of regions given in the uv-plane right before that...


[quote="GameBot"]A spherical scoop of vanilla ice cream with radius of 2 inches is dropped onto the surface of a dish of hot chocolate sauce. As it melts, the ice cream spreads out uniformly forming a cylindrical region 8 inches in radius. Assuming the density of the ice cream remains constant, how many inches deep is the melted ice cream? Express your answer as a common fraction.[/quote]

The volume of the sphere is $ \frac {4}{3}\pir^3 = \frac43\pi8 = \frac {32}{3}\pi$.  The area of the circle it will make is $ \pir^2 = \pi64 = 64\pi$  The depth will thus be then $ \frac {\frac {32}{3}}{64} = \boxed{\frac {1}{6}}$.

I hope I'm not misunderstanding the question or or doing it wrong.


[quote="pascal12"]Should it be
$ ( \minus{} 3x^2)(4xy^2( \minus{} \frac {2}{3}x^2))$?[/quote][b]yes, this is my question please help me.[/b]

Are calculators allowed? If so, then you can just plug it into the calculator. Everywhere you see a $ x$, replace it with a $ \frac{3}{4}$. Everywhere you see a $ y$, replace it with a $ 3$.

Have you tried plugging in the values for x and y?

$ (\minus{}3\left(\frac{3}{4}\right)^{2})(4\left(\frac{3}{4}\right)(3)^{2}(\minus{}\frac{2}{3}\left(\frac{3}{4}\right)^{2}))$

Well, you substitute
And I get an answer of [hide]2187/128?[/hide]

We could begin by simplifying the given expression to get

$ 8x^{5}y^{2}$

and then substitute to get

$ 8 \cdot \frac{3^{5}}{4^{5}} \cdot 3^{2}$

$ \equal{} \frac{2^{3}\cdot 3^{7}}{2^{10}}$

$ \equal{}\frac{3^{7}}{2^{7}}$

$ \equal{}\frac{2187}{128}$.

[quote="samuelkim0814"]They seriously need to improve the game I got banned for no reason?!
I mean seriously it's getting annoying...not the first time btw[/quote]

I'm not surprised that you got banned... seeing as you made a completely random post in a random thread unrelated to any game that you could be talking about.

What's this all about? As this is your first post as samuelkimo814, I'm guessing you've changed your identity.

[quote="samuelkim0814"]They seriously need to improve the game I got banned for no reason?!
I mean seriously it's getting annoying...not the first time btw[/quote]

Hmmm... this user just swore at me in FTW after I entered his open game and got banned for 15 mins.... I thought people with less than 11 posts were muted.


Greetings all Indianians and others who visit this board.

We've had some reports of inappropriate behavior on this board.

Please be mature while using this board and any part of the AoPS site.  We will ban people (either by name or IP address) found not to be using the board appropriately.

Name calling seems to be the most pervasive problem.  Using words like retard, idiot, making implications about a sexual orientation, race, etc. is absolutely, positively, inappropriate and extremely immature.

The second largest problem is people using more than one account.  Please use one account and stick with it.  If you have more than one account currently, please provide a list of those accounts we should deactivate.

If you see any inappropriate messages, please send me a private message so that we can take care of it as soon as possible.

Hmm guess that would be me...
but the kid im refering to knows its a joke...

hes also got an account here but idont know his name on here so i was trying to lure him  into posting...

anyways, sorry to anyone that was offended

[quote="mathcrazed"]Thank you!! At last! I have a question. What does it take to apply for the position of Indiana moderator? Because we apparently have a completely inactive one.[/quote]
me too  :D


There are twelve dancers in   the musical ``42nd Street.'' Six of the
dancers wear silver costumes  and six of them wear gold costumes. How
many ways are there for the   dancers to arrange their dancing line by
only considering the color of their costumes?

If they were all distinguishable there would be 12! possibilities

Since they are not all distinguishable (6 silver, 6 gold) we would divide by 6! for silver and 6! for gold

So Ans = 12!/(6!6!)=[b]924[/b]

Note: to simplify $ \frac{12!}{6!6!}$, you can do the following.

First, write it out to see if you can cancel anything.

$ \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Hmm...let's get rid of those pesky 1's.

$ \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}$

Note we can cancel out a $ 6 \cdot 5 \cdot \dots 2$.

$ \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}$

Hey, we know what $ \frac{12}{6}$ is - that's two!

$ 2 \times \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2}$

And there's more two's we can take out.

$ 2^3 \times \frac{11 \cdot 9 \cdot 7}{3 \cdot 2}$

And a 3 as well...

$ 2^3 \times 3 \times \frac{11 \cdot 7}{2}$

$ 8 \times 3 \times \frac{77}{2}$

$ 24 \times \frac{77}{2}$

$ 24 \times 77 \equal{} 4 \cdot 6 \cdot 7 \cdot 11 \equal{} 44 \cdot 6 \cdot 7 \equal{} 264 \cdot 7 \equal{} 1848$

$ 1848 \div 2 \equal{} \boxed{924}$

Ta-da!

We can optimize a couple of steps:

$ \frac {12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

At the beginning: We can just cancel the last six numbers:

$ \frac {12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Hmmm, 4*3=12. Also, 5*2=10. Cancel!

$ \frac {11 \cdot 9 \cdot 8 \cdot 7}{6}$

And 6=2*3.

$ \frac {11 \cdot 9 \cdot 8 \cdot 7}{3 \cdot 2}$

9/3=3, 8/2=4

$ {11 \cdot 3 \cdot 4 \cdot 7} \equal{} {84 \cdot 11}\equal{}840\plus{}84\equal{}\boxed{924}$

[quote=DRTFROG2]how do u do that in 15 sec? ftw[/quote]

You get good at mental math with practice.

*cough cough u know that bro*


21. In how many ways can 9 be written as the sum of three (not necessarily distinct) positive integers if order is not important?

I tried Brute Force and got it right, but I think there is a shortcut out of this... Would Brute Force be the best way to solve it?

[hide="Answer"]7 ways[/hide]

Listing the partitions of 9 into three positive integers

the combinations won't work as you count somethings two or three times.

[quote="abacadaea"][hide]let the first digit be 1
ways to make 8 is 8
let the first digit be 2
ways to make 7 is 7, and so on
36 total ways.
order doesnt matter so we divide by 6(number of combinations of 3 numbers)
36/6=6[/hide]

erm, did i forget something?[/quote]

I believe you thought that every sequence has 3 distinct numbers, and therefore is counted 6 times.  In reality, sequences like 225 are counted 3 times, not 6, and 333 is counted once.

Last number doesn't matter.
[code]
Ways
1 1
1 2
1 3
1 4
2 2
2 3
3 3
[/code]

I, for one believe brute force would help most so you would not have to account for combinations that appear only once. Also, there are only 7 combinations so brute force would not be that tedious to carry out.
you proceed by listing combinations containing a 1
117 126 135 144
then combinations with a 2 (remember there can be no 1)
225 234
lastly, 333 for a total of 7 possibilities...

[quote="CircleSquared"]The best way to do it is with balls and urns.
Since each number is positive, we can subtract 3 from both sides of a+b+c=9 to get d+e+f=6, where d, e, and f are nonnegative.
Now we can use balls and urns - there are 2 dividers (between d and e and between e and f) ,which gives $ \binom{8}{2}$[/quote]
I did exactly that. But the answer is 7, not 28... I thought I did it wrong. Weird..

[quote="stevenmeow"]Also, there are only 7 combinations so brute force would not be that tedious to carry out.  [/quote]
Hmm.. What if the answer is bigger? It would certainly devour our time to do it with Brute Force. I think what Professional ProblemSolvers will do is to skip and come back to it, right?

brute force will not be that bad and i dont think MATHCOUNTS would really put anything THAT hard but anyways, i will post a solution if the 9 is really 20

with 1 being one of the numbers, the 2nd number may be 1-9

with 2 being one of the numbers, the 2nd number may be 2-9

it will contine with 3-8, 4-8, 5-7, and 6-7 this means there are
9, 8, 6, 5, 3, and 2 possibilities for the 2nd number. these numbers would continue to decrease by 1 and 2 alternatively had there been a bigger number then it would be fairly easy to sum up the numbers.


calculus
integration
function
calculus computations

Let $ f(x)$ be a continuous function defined in the interval $ 0\leq x\leq 1.$

Prove that $ \int_0^1 xf(x)f(1\minus{}x)\ dx\leq \frac{1}{4}\int_0^1 \{f(x)^2\plus{}f(1\minus{}x)^2\}\ dx.$

$ 0\le [f(x)\minus{}f(1\minus{}x)]^2\equal{} f(x)^2\minus{}2f(x)f(1\minus{}x)\plus{}f(1\minus{}x)^2$

$ \int_0^1 xf(x)f(1\minus{}x)dx\equal{}\int_0^1 (1\minus{}x)f(x)f(1\minus{}x)dx$
so $ \int_0^1 f(x)f(1\minus{}x)dx\equal{}\int_0^1(x\plus{}(1\minus{}x))f(x)f(1\minus{}x)dx\equal{}2\int_0^1 xf(x)f(1\minus{}x)dx$

and so

$ \int_0^1 f(x)^2 \plus{}f(1\minus{}x)^2 dx \ge 2\int_0^1 f(x)f(1\minus{}x)dx\equal{}4\int_0^1 xf(x)f(1\minus{}x)dx$

I have nothing to add more than this. :lol:


geometry
parallelogram
rectangle
similar triangles
Pythagorean Theorem
geometry unsolved

Let ABC be a triangle, with AB (NO=) AC.  Drop a perpendicular from A to BC, meeting at O.  Let AD be the median joining A to BC.  If <OAB = <CAD, show that <CAB is a right angle.

$O$ is a foot of $A$-symmedian, so $\frac{BO}{OC}=\frac{AB^{2}}{AC^{2}}=\frac{c^{2}}{b^{2}}$ (you can consider similar triangles to get it). Now we want to prove:

(*) $BC^{2}=(BO+OC)^{2}=(\frac{c^{2}+b^{2}}{b^{2}}OC)^{2}=c^{2}+b^{2}.$

From similar triangles $AOB$ and $AOC$ plus Pythagorean theorem in $AOC$ we have:

$\frac{c}{b}=\frac{AO}{OC}$ so
$\frac{c^{2}}{b^{2}}=\frac{AO^{2}}{OC^{2}}=\frac{b^{2}-OC^{2}}{OC^{2}}$

which is equivalent to:

$\frac{c^{2}+b^{2}}{b^{4}}OC^{2}=1$ which is plainly (*). QED

[i] Probably I have sent a solution to this problem in the past , but here we are :[/i]

If  $K , L$  are symmetric points of  $A$  with respect to  $O,D$ ,

  then  $ACLB$  is a parallelogram and so :

$\angle BKA = \angle BAK = \angle CAL = \angle ALB$.

 This means that $ABKL$  is inscribed , so

  $\angle ABL= \angle AKL = 90$ ,

since $DO // KL$ and $AO \perp OD$.

   Thus $ABLC$  is a rectangle and hence $\angle A = 90$.

[u] Babis [/u]


What is the ones digit of $ 7^{35}$ when written as an integer?

All arithmetic is done in mod ten.
\begin{align*}
7^{35} &= 7^{32} \cdot 7^3 \\
&= (7^4)^8 \cdot 7^3 \\
&\equiv 1^8 \cdot 7^3 \\
&= 343 \\
&\equiv \boxed{3}
\end{align*}


number theory
prime numbers
number theory open

Hi,

I put new on the title because I did not find the sequence in Sloane Encyclopedia. But I don't know if it is a new kind of prime or not.

Let a integer>0

Let i varying from a-1 to 1

Let c(i) any number coprime to a and <a

Let b the product of all the coprime c(i)

Let k=a+b

Example :

a=7

6,5,4,3,2 are coprime to 7

So b=6*5*4*3*2=720

k=7+720=727 is prime

Another example

a=9

8,7,5,4,2 are coprime to 9

b=8*7*5*4*2=2240

k=9+2240=2249  is not prime

So the first primes of the kind of k are :

2,3,5,7,29,11,727,113,199,3628811,397,19319,85103

Do someone has any idea about this kind of prime (documentation or references)?

Thank you for your help.

Well, you can't really call it a new kind of prime if they're not, in fact, all prime.  Like the other sequence you've conjured up, this sequence is good at avoiding small prime factors, but there are no controls on having large prime factors.

One of the major problems of trying to find a quick formula that gives you a lot of primes is that it is very easy to construct sequences - one of the most notorious being $ a_n \equal{} n^2 \plus{} n \plus{} 41$ - that [i]appear[/i] to give a lot of primes for small values of $ n$, but what you really have to do is carry out the computation much, much further than the first twenty or even the first hundred values to figure out if you're really onto something because the first few values can be [i]very[/i] deceptive.

I have found a formula (I can improve it) which gives me 2002 prime numbers out of 4620 (43%).
It require lot of tests to improve it.
I do not know if 43% is a good performance.
Here is the file attached to the post.
It is easy to understand.
I used a sequence of 12 values repeated indefinitly.

The asymptotic distribution of primes in most types of sequences tends to get worse and worse the more terms you test.  Try testing ten times more terms and see what happens.


prove or disprove the following statement:

If $ E$ is an open subset of real line, containing $ \mathbb{Q}$, then $ E\equal{}\mathbb{R}$.


A square is divided into 9 congruent squares. The distinct integers $ a, b, c$ occupy the first row, $ d, e, f$ the second row and $ g, h, i$ the third row (all 9 integers are distinct). The elements in each row, diagonal and column have the same product $ N$. What is the least possible value of $ N$?

Notice that the "center" number $ e$ has to be the cube root of $ N$
Product of numbers = $ N^3$
$ N$ can have one or more distinct prime factors.
If it has more than one distinct prime factor, every prime factor in $ N$ has to occur in a power atleast 3.Further all integers are distinct and product of "opposite" numbers has to be a perfect square, i.e, $ e^2$. this forces all numbers to have the same prime factors.
Least prime factor is 2. So, every number is a power of 2. Sum of powers=3k. All powers are distinct so, minimum $ N^3=2^{1+2+3....+9}\Rightarrow N=\boxed {2^{15}}$.
The numbers can be easily found. One solution is $ {2^8,2^1,2^6,2^3,2^5,2^7,2^4,2^9,2^2}$

But $ N \equal{} 216$ works.. place 6 in the middle and the other factors can fit nicely around.


abstract algebra
integration
Functional Analysis
advanced fields
advanced fields unsolved

$X$is a Hilbert space,$A\in \mathcal{L}(X)$ a bounded linear operator.
If $A$ is on-to, its ad-joint $A^*$ (i.e $<Ax,y> = <x,A^{*}y>$)surely be injective.
How about the converse: if $A$ is injective,must $A^*$ be on-to ?
I think the answer should be negative,but I haven't got a counterexample.Any comment?

The kernel of $A^*$ is precisely the orthogonal complement of the range $\mathcal R(A)$ of $A$: $A^*y=0\iff (x,A^*y)=0,\ \forall x$, which is equivalent to $(Ax,y)=0,\ \forall x\iff y\in\mathcal R(A)^\perp$. This means that $A^*$ is one-to-one iff the range of $A$ is dense. It is possible for $A$ to have dense range which is not the entire $H$, and in such a situation $A$ will not be onto, but $A^*$ will still be one-to-one.

Try to find such an operator (with dense range but which is not onto). It shouldn't be too difficult.

Here are two examples of such operators:

1. On $L^2([0,1]),$ let $Tf(x)=\int_0^xf(t)\,dt.$

2. On $\ell^2,$ let $(Tx)_n=\frac1n\cdot x_n.$

Note that in both cases the inverse operator, which is densely defined because it is defined on its range, is not a bounded operator.


I've  got a problem solving situation with Melelaos.
Arbitrary triangle ABC. [url=http://www.wageningse-methode.nl/images/puzgif/puz14.gif]triangle[/url]
Intersection of straight line from point A to BC is E, intersection of line from point B to AC is F, intersection of line from point C to AB is D.
Lines AE, BF en CD are completely arbitrary, no 60 degrees, or other criteria that would ik make easier.
The second triangle in the middle must have some relations to the triangle ABC. In area of length of the lines.
Starting is:
|AB| : |BD| = p : 1
|BE| : |CE| = q : 1
|CD| : |AF| = r : 1
This is not quite the theorem of Menelaos but should be helpful.
Anyone who just loves geometry?

Sammi,
the result you're looking for is known as Routh's theorem:
[url=http://mathworld.wolfram.com/RouthsTheorem.html]mathworld.wolfram.com/RouthsTheorem.html[/url]

[quote="jhaussmann5"]Sammi,
the result you're looking for is known as Routh's theorem:
[/quote]
You're my hero. Can't thank you enough.  :) !!!


I believe you will like it. I'm looking for earlier stages of this olympiad...

Thank you very much.
Could someone please translate the solutions into english?
Does anybody knows the problems and results of the last open Chinese olympiad,which took place in the end of Januray in Harbine(sorry,if made a typo in writing the name of the city.).

:lol: You are welcome. I like beautiful problems. I like especially the Olympiads of the following countries: Vietnam, China, Russia, USA, Poland, Canada, Iran, India, UK (and Bulgaria of course! )
(I don't know there may be some other great math olympiads but I cannot know all possible problems from all over the world for all years.)
and I collect problems from these countries.

I want to ask for the rules of Chinese Math Olympiad and Vietnamese Math Olympiad and some examples for problems for early stages of these Olympiads (1995-1999 or if are not available for other years).
I also saw there are not complete set of Iranian Math Olympiad on the Net. So if it is possible - someone to translate all stages of that olympiad for all years - there are great problems I believe.

Thank You Very Much!
They are really very nice problems :!:
I hope that somebody can say how can we get the solutions(in English)  :maybe:

[quote="Fang-jh"]Problem 1:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=184264[/url]
Problem 2:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=184417[/url]
Problem 3:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=184281[/url]
Problem 4:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=184433[/url]
Problem 5:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=184424[/url]
Problem 6:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=184291[/url][/quote]
Do you know the results of this olympaid?
Best regards,Erken.

Thanks.How many points was each problem estimated?And tell me please medal cut-offs?

Each problem is 21 points. the perfect grade is 126 points

Excellent problems! Please, ask to my questions. I took a look at the site you gave a link but I don't know Chinese. I know only Bulgarian, English and Russian - when China become the next Superpower I probably must learn Chinese...

(I want to ask for the rules of Chinese Math Olympiad and some examples for problems for early stages/rounds of that Olympiad (1995-1999 or if are not available for other years).

P.S. I have an idea - moderators - please put all problems from this topic in the Resources Section. In this way it will be easier for all people to prepare themself for math competitions in high level.

Gold medal: 84 points
Silver medal: 42 points
Bronze: all participants except for gold and silver medals

Actually we have translated the old CMOs here: http://www.mathlinks.ro/viewtopic.php?t=66332

But due to copyright problems, they can not be put to the resources section.

What about the rules and the earlier stages of the olympiad?

I can translate them sometime if anyone else wants (solutions are sorta long...if I have time i'll do it. Rules sure, but I don't see them anywhere and I don't know where you can find them. My computer is wierd, can't type chinese, so google isn't an option).

Are there China 2007 solutions? I don't see them in that folder.


topology
real analysis
real analysis unsolved

Let $\mathbb{R}$ be a metric space of real numbers with the standard metric. For any subset $A$ of $\mathbb{R}$, denote the set of limit points of $A$ by $L(A)$. Then is it true that $A \backslash L(A)$ is always countable?

I think I have made a proof for this, but I hope someone would assure me that I'm going right.

[hide="Yes"]because each point $x\in A\setminus L(A)$ is contained in an open interval with rational endpoints that is disjoint from $A\setminus\{x\}$.[/hide]


trigonometry
algebra
polynomial
calculus
calculus computations

Let $ x,\ y$ be variables such that $ x^2\plus{}y^2\equal{}1,\ x\geq 0,\ y\geq 0$. Let $ u\equal{}x\plus{}y,\ v\equal{}x^3\plus{}y^3$.

(1) Find the range of $ u$.

(2) Express $ v$ in terms of $ u$.

(3) Find the range of $ v$.

Same bit with circular functions... I get

(1) $ 1\le u\le\sqrt {2}$

I used polynomial curve fitting on $ v\equal{}au^3\plus{}bu^2\plus{}cu$ to get

(2) $ v \equal{} \minus{} {\scriptstyle \frac {1}{2}}(u^3 \minus{} 3u)$

v is now perameterized via (1) so its range is

(3) $ {\scriptstyle \frac {1}{\sqrt {2}}}\le v\le 1$

[quote="kunny"]Your answer to (1),(3) is incorrect.[/quote]
Thanks! A stupid mistake  :lol:
I corrected the solution.

[quote="M@ri@n"]If the curve $ c$ is given by $ x^2 + y^2 = 1$ and $ x\ge 0$, $ y\ge 0$, then for every point $ P = [x_P,y_P]$ on $ c$ there exists an angle $ \varphi_P\in\left [0,\frac {\pi}{2}\right ]$ such that $ P = [\cos\varphi _P,\sin\varphi _P]$. Hence $ u(x,y): = x + y$ can be written in the transformated form as $ u(\varphi) = \cos\varphi + \sin\varphi$. From here the range for $ u$, namely, if $ P\in c$ then $ u(\varphi _P)\in [1,\sqrt {2}]$.

To express $ v(x,y): = x^3 + y^3$ in terms of $ u(x,y)$ we can write
\[ v(x,y) \& = x^3 + y^3 = (x + y)(x^2 + y^2 - xy) = u(x,y)\cdot (1 - xy).\]
But $ u^2(x,y) = x^2 + y^2 + 2xy = 1 + 2xy$. Thus $ xy = \frac {1}{2}(u^2 - 1)$. From this we deduce that $ v(x,y) = \frac {u(x,y)}{2}\cdot (3 - u^2(x,y))$.

The range of $ v(x,y)$ can be found easy. We get the result $ v(x,y)\in\left [\frac {1}{\sqrt {2}},1\right ]$ for every $ P\in c$.[/quote]

Now the problem is completed. :lol:


linear algebra
matrix
linear algebra unsolved

$ A,B \in M(n,C)$

Show there exist $ P,Q \in GL(n,C)$ such that

$ PAQ$ and $ PBQ$ are triangular superior

Already asked, no answer yet
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252477


$ x^2 \log x \equal{} (2x)^x$

How is this solved?

hello, i think your equation has no real solutions.
Sonnhard.


assume ABCD a convex quadrilatral. P and Q are on BC and DC respectively such that angle BAP= angle DAQ .prove that [ADQ]=[ABP]
([ABC] means its area ) iff the line which crosses through the orthocenters of these traingles , is perpendicular to AC.

It's a bit boring to write it all. Let Hb and Hd be the orthocenters of ABP and ADQ respectively. Let B' and P' be the feet of the perpendiculars from B and P respectively in ABP and let D' and Q' be defined similarly for triangle ADQ. It's easy to show that if <BAP=<DAQ and their areas are equal then AP*AB'=AQ*AD'.

Now consider the inversion of pole A and power AP*AB'=AQ*AD'=AHb*AP'=AHd*AQ'. This inversion turns HbHd into the circle which passes through A, P' and Q', and since CP' perpendicular to AP' and CQ' perpendicular to AQ', it means that the circle AP'Q' has AC as its diameter, and a line is always turned into a circle which passes through the pole s.t. the diameter of the circle which passes through the pole is perpendicular to the initial line, so, in our case, AC perpendicular to HbHd.

This is only one of the implications, but the other one isn't hard to derive from this one:

We know that the lines are perpendicular if the areas are equal, and if we move one of P and Q a bit, say P, the orthocenter Hb moves along AP', which is fixed, Hd is fixed, AC is fixed, so AC and HbHd can't be perpendicular anymore.

I have an inversion free solution.

Let H1, H2 be the orthocenters of DAQ, BAP. Let R1, R2 be the circumradi of DAQ and BAP.
Let DAQ=BAP=a.
Let C1 be the angle ACD and C2 be the angle ACB.
Let F be a point such as CF and BC are perpendicular and AF is paralel to DC.
Let E be a point such as CE and CD are perpendicular and AE and BC are paralel.
It is easy to see that C is the orthocenter of AEF and CFE=C2 and CEF=C1. We get CE/CF=sinC2/sinC1

It is well known that AH1=2R1cosa and AH2=2R2cosa.

AH1H2 and CEF have AH1||CE and AH2||CF. C is the orthocenter of AEF, so AC and EF are perpendicular.

It is clear that H1H2 and AC are perpendicular iff H1H2||EF iff AH1H2 and CEF are similar iff CE/CF=AH1/AH2 iff sinC2/sinC1=R1/R2 iff R1sinC1=R2sinC2.

Now Area[DAQ]=Area[BAP] iff R1*AD*sina*sinD=R2*AB*sina*sinB iff
R1*(AC*sinC1/sinD)*sina*sinD=R2*(AC*sinC2/sinB)*sina*sinB iff R1*sinC1=R2*sinC2.

Problem solved. I'll do my best to offer you a picture.

If C=900 then the construction of E and F doesn't work.

In this case, take M to be AH1 \cap CD and N=AH2 \cap BC. Then AMCN is a ractangle and H1AC=C2 and H2AC=C1.

As before, we have [DAQ]=[BAP] iff R1sinC1=R2sinC2.

H1H2 is perpendicular to AC iff AC is an altitude in AH1H2, iff AH1H2=C1 and AH2H1=C2, iff R1/R2=AH1/AH2=sinC2/sinC1 iff R1sinC1=R2sinC2.


#1. LynnelleYe

[img]http://img.photobucket.com/albums/v189/bigsister/monkey.jpg[/img]

#2. sky9073

[img]http://www.artofproblemsolving.com/Forum/files/thumbs/t_monkey_683.gif[/img]

You have one week to vote. Next Sunday we will find out the winner! I will vote in case of a tie.

NOTE: The competitors CAN vote for themselves.

I hypocritically voted for myself despite actually preferring sky's monkey...heh.

Hmm...who else voted for me? They're crazy.


vector
complex numbers
superior algebra
superior algebra unsolved

group of complex numbers from unit circle
and
group of nonzero complex numbers

- isomorphic or not?

Yes.  $ \mathbb{R}$ is a vector space over $ \mathbb{Q}$, so choose a basis $ a_j, j \in J$ where $ a_1 \equal{} 1$.  Now, the circle group is isomorphic to $ \mathbb{R}/\mathbb{Z}$ whereas the group of nonzero complex numbers is isomorphic to $ \mathbb{R} \times \mathbb{R}/\mathbb{Z}$.  It follows that both groups are isomorphic to $ \mathbb{Q}/\mathbb{Z} \times \mathbb{Q}^J$ (where the latter is the infinite direct sum, not the infinite direct product).

Note that the isomorphism does not hold in the topological category, since one is compact and the other is not.


number theory
prime factorization
divisibility tests

does anybody know how to find the largest prime factor of a huge number without using prime factorization??? :?:

well, you know the prime factors are less than the square root of that number, or else times one less than that. for example 9889 has a squareroot less than  100. so start at 99 and go down. thats what i do

that's what i usually do and it slows me down a lot.
My friend can find the largest prime factor in less than 5 seconds...it's crazy.
More replys would be appreciated

How big are the numbers involved?  And are they randomly chosen or are you thinking of a particular problem?

usually...they are in the 3 digits.........better if you can tell me how to do it for four digit problems

All the primes lower than the square root of the number.

Such numbers are fairly small, and I'd go with prime factorization to find their largest prime factors (although if you're testing small primes you can disregard a factor as soon as you find it - the last prime you use will be the largest prime factor).

yea...i know........but how do people do it in like less than 5 seconds???...that's crazy

if no one knows....ill like cry cause i have to know

The key is fast mental arithmetic.  The first thing to do is divisibility tests for $ 2, 3, 5, 11$ (the test for $ 7$ isn't that fast).  For the rest, I have a little trick I use, but it's difficult to explain generally so I'll use an example.

First, a random odd three-digit number:  say, $ 691$.  We're going to test the primes up to $ \sqrt {691} < 27$.  Clearly not divisible by $ 2, 3, 5$.  Now,

$ 691 \equal{} 630 \plus{} 61 \equal{} 90 \cdot 7 \plus{} 61$

so $ 691$ is not divisible by $ 7$ or $ 61$.  Next is $ 11$.  But

$ 691 \equal{} 680 \plus{} 11 \equal{} 40 \cdot 17 \plus{} 11$

so $ 691$ is not divisible by $ 11$ or $ 17$.  (We could've used the divisibility test for $ 11$, but this way I find out information about two primes at the same time.)  Next to try is $ 13$.  But

$ 691 \equal{} 650 \plus{} 41 \equal{} 50 \cdot 13 \plus{} 41$

so $ 691$ is not divisible by $ 13$ or $ 41$.  Next is $ 19$.  It helps to know that $ 57$ is a multiple of $ 19$ (it might not be obvious initially).  So:

$ 691 \equal{} 570 \plus{} 121 \equal{} 30 \cdot 19 \plus{} 11^2$

so $ 691$ is not divisible by $ 19$.  Next is $ 23$.  But of course

$ 691 \equal{} 690 \plus{} 1 \equal{} 30 \cdot 23 \plus{} 1$

so $ 691$ is not divisible by $ 23$.  We conclude that $ 691$ is prime.

In my head, that wouldn't have taken longer than $ 20$ seconds, and if I were faster at mental arithmetic it would take less time still.  :)

Um, yeah, if it's like a target round question involving prime factors, just make a calculator primality or prime factorization program and use that to solve the question.  It's actually pretty easy to make.  In the case that it's non-calculator, just do what abadadaea does: Find prime factors less than its square root.  It really doesn't take that long.  And Mathloser (no 1337), problem solving doesn't always involve using some magic formula to solve everything.  Just work your problems out systematically and accurately and you won't need to do the problem in "less than 5 seconds."
Hope that helps! :D


conics ellipse geometry trapezoid ARML trigonometry quadratics

I really lke number 6.  It seemed hard at first but then I got it and it became easy, and not many people got it.

Was 6 the one with the circle in the ellipse?  If so, that was my favorite as well, for the same reasons you listed.

Yeah, that was the one.  It really told you if you could really do well, by noticing how to do the problem, and also sort of with 8, and with 7

I love #8 now that I know how to do it, and I see how easy it is. :lol: :P

i liked #3, the clock problem cuz i could do it by using skeches. i drew like 7 clocks on my paper. i probably should have gotten quicker. o well.

[quote="Sly Si"]I liked #4.  I'm always a fan of the type of problem (usually geometry) that allows you to get a really quick solution by shortcutting it with a key assumption--in this case, that the trapezoid had a right angle.  ARML seems to like this type of problem.  #8 was neat, too, just because it was so straightforward if you got on the right track, yet still so inexplicably hard.[/quote]
Yeah, I assumed that one of the base angles was right, and I got the right answer!!

I actually started with the more obvious assumption that the trapezoid was isosceles, since I knew I could use a couple of theorems about trapezoids and their diagonals to get a brute-force solution.  But I didn't like my answer that I got that way, so with about 30 seconds left, I decided to use the other assumption because it would reduce the number of points and lines.  I got it at the last second! :)

#6. It was the easiest one that i did NOT get.

[quote="shake9991"]i liked umber seven. I think it was the trig. one (I may be mistaken however).
but I liked the the trig question. It was extremely messy at first, but reduced simply[/quote]

Yes, that was number 7.  I took the geometric approach which worked out beautifully.

[quote="joml88"]I took the geometric approach which worked out beautifully.[/quote]

Do you mean the circle of radius 2 and circle of radius 1, automatically making a distance of 3? If so, I did that too, but then I forgot to switch the $\sin$ and $\cos$ for the coordinates. >_<

[quote="JesusFreak197"]Do you mean the circle of radius 2 and circle of radius 1, automatically making a distance of 3? If so, I did that too, but then I forgot to switch the $\sin$ and $\cos$ for the coordinates. >_<[/quote]
I did that too, and I did not forget to switch the sin and the cos.
however, i did even worse mistake on #6 - failed to solve quadratic equation correctly.


algebra
polynomial
limit
real analysis
real analysis unsolved

Let $ P(t) \equal{} \sum_{i \equal{} 1}^nc_je^{b_jt}$, where $ b_j \in \mathbb{R}$ and $ c_j\in\mathbb{C}$.
Prove that if $ P(t) \equal{} 0$ for all integers $ t$ then $ P(x) \equal{} 0$ for all real numbers $ x$.

Suppose $ P$ is not identically $ 0$, then we can assume without loss of generality that all $ c_j\neq 0$, all $ b_j$ are distinct, $ b_1 \equal{} \min_{1\leq j\leq n}\{b_j\}$, and $ b_{n} \equal{} \max_{1\leq j\leq n}\{b_j \}$.  Then $ \lim_{x\to \infty}P(x)e^{ \minus{} b_nx} \equal{} c_n$ and $ \lim_{x\to \minus{} \infty}P(x)e^{ \minus{} b_1x} \equal{} c_1$, but since $ P(t) \equal{} 0$ for each integer $ t$ the limits can only be $ 0$ and hence $ c_1 \equal{} c_n \equal{} 0$, a contradiction.


Find all triplets of positive integers (a,b,c) such that there exists infinitely many positive integers n satisfying: a^n | b^n - c^n.

Have a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=4611&highlight=divisibility

Pierre.


How many real solns does the eqn $ x^4\minus{}10x^2\plus{}x\plus{}20\equal{}0$ have?

Solution:

Let $ x^4\minus{}10x^2\plus{}x\plus{}20\equal{}f(x)$

$ f(0)\equal{}20$
$ f(1)\equal{}12$
$ f(2)\equal{}\minus{}2$
$ f(3)\equal{}14$
$ f(\minus{}1)\equal{}10$
$ f(\minus{}2)\equal{}\minus{}6$
$ f(\minus{}3)\equal{}8$
So there exist real roots in $ (1,2), (2,3), (\minus{}1,\minus{}2), (\minus{}2,\minus{}3)$.

Ans: 4 real roots.

Does descartes rule say there are surely real roots? I dont think so! I think it states that there is a max of so many real roots if im not wrong? And also the roots neednt be real?
The soln im looking for is different atleast and even i did with descartes rule but im not sure that that cann justify the answer.

The answer is 4 only though  :)

No see the polynomial function is continuous... and so if it takes a postive value at 1 and a negative value at 2, we can surely say that there is a "real root" between 1 and 2, because a crude explanation would be- if you plot the function, it would take a value above the x-axis at 1 and a value below the x-axis at 2 , we know the function is continuous, so it must cut the x-axis at some point between 1 and 2 and that becomes a real root.

yeah i get that but in many cases ive heard that we cant ascertain with descartes rule of signs bcos it can only talk abt the max no. of possible roots roots  :maybe:
Anyways ill post the other soln which i had in mind

By that method we can also get the roots  :D

[hide]
Put a = 5
Then the eqn transforms into a quadratic in a as:
$ a^2 \minus{} a \cdot (2 \cdot x^2 \plus{} 1) \plus{} x^4 \plus{} x \equal{} 0$
Discriminant of this eqn is: $ (2 \cdot x \minus{} 1)^2$ , a perfect square
Hence get the roots of the eqn as: $ a \equal{} x^2 \plus{} x, x^2 \minus{} x \plus{} 1$
Then get the roots  :)

By this method we can also get the roots  :D
    [/hide]

Brilliant method no? I read it in a magazine

yes that was very nice n it would take something to assign a as 5! I wonder how to think of that during an exam or is it impossible?  :maybe:
Also problem 96 is also very ingenious try it  :D

i think wat thealchemist posted was the intermediate mean values theorme and not the descartes rule, which merely speaks abt the sign of the real roots (if any)

the mean value theorem as thealchemist said, gaurentees the existence of a real root as ever polynomial with real coeffs is continuos

Descartes rule ONLY gives the maximum no of a certain type of roots. i mean, if
f(x) has say some t changes of sign, then it can have UTMOST t positive roots. in fact the claim is that it can have (t- (an even no)) of roots. Next the no of changes in sign in f(-x) gives us a similar bound on the no of negative roots. Further, we can give a lower bound to the no of non-real roots using the fact that no. of such roots= degree of the polynomial - (no. of positive roots+negative roots).

actually, even though descartes rle may sound utterly useless, it is employed in many jee objective questions. an inmo 86 or 87 problem, which involved a quintic or 6th degree polynomial or sth merely reqd the knowledge of this rule to solve it


modular arithmetic
number theory proposed
number theory

[hide]This problem becomes very easy when it is considered that $ m^n\equiv m^{n \pmod 4\plus{}4} \pmod {10}$ for natural numbers, which can be proven by checking each individual possibility of $ m$.

Since we are just working mod $ 10$, we can reduce that whole sum to $ 1\plus{}2^2\plus{}3^3\plus{}4^4\plus{}5^1\plus{}6^4\plus{}7^3\plus{}8^4\plus{}9^1$, equivalent to $ 1\plus{}4\plus{}9\plus{}6\plus{}5\plus{}6\plus{}1\plus{}6\plus{}9$ or $ 7$.[/hide]


vector
calculus
derivative
conics
ellipse
calculus computations

In $f(x,y)=ax^2 + by^2 + 2hxy + 2gx + 2fy +c =0$.
Why do we get the cente of the curve by solving
${\frac{\partial f}{\partial x}} {=} {0}$
and
${\frac{\partial f}{\partial y}} {=} {0}$

[quote="TheBB"]I don't know what you mean by the "centre" of the curve, but what you get are the [i]critical points[/i] of the function.

Critical points are where $\nabla f = 0$. In two dimension this reduces to the two equations you posed:

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

$\nabla f$ is a vector pointing in the direction where the derivative of $f$ is greatest. If $\nabla f = 0$, what you have is a critical point, a point where the derivative is zero in all directions. This probably also characterise what you call the "centre of the curve".[/quote]

By centre I meant that solving the partial derivative equatioins for an ellipse gives us the centre of ellipse. For a pair of straight lines gives us the point of intersection of the lines represented by the equation.
Thanks for replying.
Bye :)

Let $(x\,,y)=\vec{R}$, The curve is symmetric about the center $\vec{R}_0$, i.e. after we do the transformation $\vec{R}=\vec{R}_0+\vec{r}$, the new equation should have no linear term. But we know the new linear term is $\nabla f\left(\vec{R}_0\right) \cdot \vec{r}$,  thus $\nabla f\left(\vec{R}_0\right)=0$.


In last year's discussion of USAMO selection, I proposed an index that subtracted about 20 points from AMC 10 scores (to be exact, the difference between the AIME qualifying minimum scores). My best guess would be that that would have almost exactly the same correlation with USAMO scores as index without that adjustment - but I am curious as to whether it would be tweaked in either direction enough to see.

Uh, not that it's a big deal or anything, but the USAMO qualifiers page states that [quote]The USAMO is a slightly better predictor according to that analysis.[/quote]
which is missing the word "index" after "USAMO".

Question: Could USAMO scores be factored into the index?  This would make it more likely that students who have already done well on the exam that the index is trying to qualify them for would be more likely to get in, which is what the AMC is looking for, right?

Then again, I'm sour because I missed qualifying for the USAMO by one point this year, having scored 14 on it last year.  I also, from the looks of things, took the harder of the two AMC12s and the harder of the 2 AIMEs.

[quote="AMCDirector"]As a statistical aside, we did a correlation between indices and USAMO score using the 2007 USAMO data.

The correlation between AIME score and USAMO score is 0.40

The correlation between USAMO Index and USAMO score is 0.45

The USAMO index is a slightly better predictor of USAMO score according to that analysis.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

There has to be a ton of bias here. Namely you are throwing out people with good AIME scores and bad USAMO Index scores who failed to qualify for USAMO. These people would significantly outperform the demographic that has a monster score on the AMC-10/12 and barely gets in with a weak AIME score (these are often the USAMO first-timers who are excited to get their name on the list and might not even sit there for the whole exam -- my first time I left early).

What does the correlation between AIME and USAMO scores mean that the AMC Director is talking about?

[url=http://en.wikipedia.org/wiki/Correlation_coefficient]Wikipedia for Correlation Coefficient[/url]

Basically, he's giving a quantitative measure of how good an indicator the AMC Index is of USAMO performance -- or more fundamentally, how useful of a measure is the AMC Index.

Correlation coefficients generally occupy a chapter in a statistics book, so explaining it here is not appropriate.   It would be a good start to read such a chapter.  I've always liked the book Statistics by Freedman, Pisani and Purves.  It has a lot on correlation presented in an interesting fashion.

Correlation has the advantage that it is well-understood and already present as a procedure in spreadsheet programs.  It may or may not be a good way to measure what I am interested in measuring.  I would be interested in having others on this forum suggest other ways of objectively  measuring the association of the Index with the USAMO score.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="Hamster1800"]Of course, there's also me with a 6 AIME, 204 Index, and an 18 USAMO.

I think no matter what you're graphing against, I'm a bit off the main line :).[/quote]

You're also an example of the reason we sponsored the expansion of USAMO qualification.  Congrats.

I think the only reason that index correlates better is that it's simply more data. The sample size on AIME is very, very small.

If the goal is to identify potential of success in the USAMO, you might might want to see what happens to the correlations if the data is limited to USAMO scores $ \geq 10$, or some number in that ballpark. Then, you can calculate the correlations in that limited set to see what is a better indicator for people who are doing well.

There will always be lots of people who are better at finding solutions than at proving those solutions. But the goal is to make sure that as many potential successes also shine in whatever such method.


function
induction
real analysis
real analysis unsolved

Find all functions $f: [0,\infty)\to [0,\infty)$, differentiable at $x=1$ and satisfying $f(x^{3})+f(x^{2})+f(x)=x^{3}+x^{2}+x$, for all $x\ge{0}$.

  [i]Mihai Piticari[/i]


Prove that all prime =1 mod 3 is expressible as form a^2+3b^2.

Bomb :D

Do it the same way you proof it for $p=x^2+y^2$, only change conditions


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let D be the point on the side BC of triangle ABC so that AD bisects the angle <BAC. A circle passes thru A and is tangent to the side BC at D. Assume M be the 2nd intersection of the side AC with the circle and P be the 2nd intersection of the circle with the segment BM.
Show that P lies on a median of triangle ABD

Call $N$ the second intersection of circle with $AB$. Then it's easy to see that $MN \parallel BC$ (for example consider the center $O$ of the circle, then $OD$ is perpendicular to both $MN$ and $BC$). So $\angle BAP = \angle PMN = \angle MBC$. But this implies that a circumcircle of triangle $ABP$ is tangent to $BC$ and $B$. Call $X$ the intersection of $AP$ with $BC$. Since $X$ lies on radical axis of this two circles we have $|XB|^{2}= |XD|^{2}$, which finishes the proof.


A sphere of radius R filled with water hangs from the ceiling by a  thread of length l.
find the ratio of small oscillations before & after the water freezes.neglect viscosity & change in vol.
(thread dose not go to the  center, only till the surface.)

                                      -------------------------
                                             /
                                            /
                                           /
                                          /
                                   (sphere)

LET ME TELL YOU
BOTH ARE NOT THE SAME

there ratio is square root of (1 + (l/R)^2)

SEE I.E. Irodov
section 4
its frm that book.
very tough  indeed


Hi, I got informed by my teacher about the AMC 10 test. I'm new to the stuff, si I'm gonna need suggestions. :) I have about three months to prepare, what specific books are the best?

There are plenty of resources in this forum right here.  Look around and you ll find them.  The Mock contests are great.  Keep an eye out because beta is working on one and it will be administered probably before the new year.  As for books: The Art of Problem Solving Volume 1 and 2 (volume 1 moreso for AMC 10).  There are other books that may help but I don't see them as necessary.  Just using the AoPS books and taking old AMC's/AHSME's is great practice.  I would also look into taking a class on this website if your parents would allow you to.

[url=http://www.unl.edu/amc/d-publication/d6-books/testsoln.html#1]Old AMC 10 tests[/url] are by far the best way to prepare for the AMC 10. Take them under timed conditions. Be sure to score your practice tests according to the official scoring rules: learn when NOT to guess.

Another very good way to prepare for the AMC 10 is to sign up for the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#amc10]AoPS online course about the AMC 10[/url], which has already begun but which you can still sign up for. (Transcripts of the classes you missed will be available to people who officially enroll.)

Other books that are good to read for preparing for the AMC 10 include [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674&highlight=readable+books]readable books[/url] about algebra and geometry, as mentioned in earlier threads. Work all the hardest problems in the books you can find about algebra and geometry, for example [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]The Art of Problem Solving[/url] volumes 1 and 2.

Hope this helps! Good luck on the test. Don't guess.

[quote="joml88"] The Mock contests are great.  Keep an eye out because beta is working on one and it will be administered probably before the new year.[/quote]

Actually you don't have to wait that long. It should be ready by next week.   And AoPS books are great for preparing for the AMCs!


1. I just want to register myself, but the thing at mandelbrot.org says i have to register a team with a coach. could i just have my mom or dad be my coach, and do i have to register a team?

2. who is competing next year.

You need at least three other people to make up a team, you can not compete with less.

Shoot an email to Sam Vandervelde at info@mandelbrot.org.  I don't know what Sam's policies are on this.  Back when I helped him run the contest, we allowed Armies of One, but usually they were affiliated with schools.  However, I'm pretty sure he allows homeschool and area teams (like the excellent ICAE in Michigan) to participate, so you might qualify under that.

You're at a disadvantage in the Team Round, of course, but you'll still enjoy it & learn a ton.  You might even be fine on your own - we had a few students who could tackle team tests alone back when I did the contest.  Of course, they had names like Lenny Ng or Mathew Crawford.

Come on guys.. A guy named Lenny? He must have been good at math... I always regret my parents not naming me Lenny... leonard... leonardo... the Len-meister... lentasitc the list goes on... I bow to a mathlete named Lenny.

IW.OAV

thx for the feedback. ill email him in a few minutes. Lenny Ng....that guy's good. he got top scores probably.

[color=cyan]I actually have a friend who did the team round by himself, although he did get three other people to take individual tests with him.  Actaully, on round 4, he tied our 4-person team on the team portion.  I was very impressed.[/color]


i wrote it last Tuesday in my school...
it's harder than any other Canadian math contest i think...
and my teacher said if u got 10 questions, u would be the toppest students in the country and get scholarship from your  school...
is that true?
it seems to be easy for gr12's

here are my answers:
1.084
2.901
3.725
4.124
5.936
6.360
7.408
8.000(which is absolutely wrong)
9.046
10.211(i got the answer 1 day later and this one is wrong)
11.896(guessed...the possibility of doing it right is no more than 1/999)
12.552(same as #11)
13.900(wt* i forgot to minus 1....)
14.031
15.000(the official answer is 27.and i think it's 003.what did u get?)

8.  089
10.  065
11.  458
12.  906
13.  899
14.  183
15.  027 (definitely not 003, been proven by AoPS members repeatedly)

Making your final score 8.  Pretty good.

(Scholarship?  I got 11 and I haven't seen a scholarship so far :P )

he was just kidding i think
but 8 is not a good mark really..
the contest should be easy for gr12's
and it's kinda unfair to me bcz i havent learnt about complex number and other stuffs

I don't think it should be 'easy'.

I'm in grade 12, and I didn't find it easy.  Out of everyone in my school who took it, there's one kid who found it easy.  Maybe two, but I highly doubt that.

What's your definition of easy? No contest is easy, I hate the word easy >:(

Well it's outta 15, so if it is 'easy' for you, then you should get full marks, MAYBE you can't get one of them and you make a maximum of 1 stupid mistake.  So that makes it 13-15 outta 15.  But even 13 would be pushin it to call it 'easy'.

because i am gr 10 and because of the floor value rule, getting 9 or 10 on aime is not too bad for me. Though I think it's pretty horrible. Oh well.

[quote="nirvanatear"]
the contest should be easy for gr12's
and it's kinda unfair to me bcz i havent learnt about complex number and other stuffs[/quote]
It actually is fair if you are in grade 10 or below. then, if your AIME score is above the lowest AIME score of the top 240 grade 12s, you are automatically in. You can find this out somewhere on the amc website.
however, if you are in grade 11 and don't know complex numbers, you are in trouble, because you have pretty much no chance at winning USAMO.

well...
im actually gr8...
and i just meant it's not that hard is it?
olympiad is way harder than it

[quote="lowest_reynolds"][quote="rem"]You got 8, so you might actually get in, the projected floor value is 7 or 8.
But the USAMo is insanely hard. But you will get experience, that's good.
However, you still need a decent AMC score. For a score of 8 on AIME, it should be 120 - 130.[/quote]

No he doesn't.  All he needs is the floor value.[/quote]
That's not true. If more than 120 students in gr 10 or lower get above the floor value, then the organizors will look at the highest 120 indexes of those students.

It's all over the AMC boards.  217.0, which is more or less what people were predicting.

[quote="YawningLion"]What is this year's USAMO cutoff for grade 12s (i.e. the combined points from AMC and AIME required to qualify for USAMO)? I've heard it's insanely low.

I wouldn't call AIME easy, though it's not hard in the usual problem-toughness sense, but rather in the tedious answer-checking sense. It's funny that this contest is used for USAMO selection, since it requires completely different skills.[/quote]

I agree.

Grumble grumble... :roll:

[quote="rem"]Is 217 actually the mark released by organizors or is it just the average of all the predictions?
If i got higher than that, but i am in gr 10, will they count me as gr 12 and let me in without the floor value thing?[/quote]

$217.0$ is the value I got in my official USAMO invitation.  It doesn't matter what grade you are if you get higher than the index.


[quote="NaZAf"]How to solve the following inequality:
$\frac{1}{1- \left | \frac{x}{1-x} \right |}<1-\frac{x}{1- \frac{1}{|1-x|}}$[/quote]
Don't forget inequality's condition, it's also important.
$x=0$.

[quote]
Don't forget inequality's condition, it's also important.
$x=0$.[/quote]

$x \ne \frac{1}{2}, \ x \ne 0, \ x \ne 2, \ x \ne 1$


When will the AMC 8 stats be available on the AMC website?

Thank you.

The AMC 8 statistics are available through the top link under What's New on our web page:
http://www.unl.edu/amc/

That's one month after more than 147,000 students at more than 2230 schools in over 20 countries world-wide participated.  Hats off to the AMC staff who made it happen!

Steve Dunbar
AMC Director


calculus
derivative
function
calculus computations

Hi,

I have tried to graph an equation, dN/dt = N (1-(N/50)) - 9N/(5+N). It is attached with this post? Does it look right? Also... i am asked to take a derivative of it. I tried it, i just want someone to see where i am going right/wrong?

the derivative is dN/dt = 1-(N/25) - 45/(5+N)^2

Thanks so much :blush:  :oops:

You just graphed $ N'$ against $ N$. I think you want to graph $ N$ against $ t$.

Your alleged derivative is nonsense. Ask yourself these questions: What is the variable? What is the function?


Find all functions $ f : \mathbb{R} \ \to \ \mathbb{R}$ such that $0< f(0)$ and \[ f(x\plus{}y) \geq f(x) \plus{} yf(f(x)) , \quad \forall x, y \in \mathbb{R}.\]

[quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R} \ \to \ \mathbb{R}$  such that  :

   $ 0 < f(0)$ and $ f(x \plus{} y) \ \geq \ f(x) \plus{} yf(f(x)) \ \ \forall x; y \ \in \ \mathbb{R}$[/i][/quote]
[hide="Intermediate result"]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\geq f(x)\plus{}yf(f(x))$

$ P(x\plus{}y,\minus{}y)$ $ \implies$ $ f(x)\geq f(x\plus{}y)\minus{}yf(f(x\plus{}y))$ and so $ f(x\plus{}y)\minus{}yf(f(x))\geq f(x)\geq f(x\plus{}y)\minus{}yf(f(x\plus{}y))$

And so $ yf(f(x\plus{}y))\geq f(x\plus{}y)\minus{}f(x)\geq yf(f(x))$

A first conclusion is that $ yf(f(x\plus{}y))\geq yf(f(x))$ and so that $ f(f(x))$ is a non decreasing function.

A second conclusion is that $ f(x)$ is a continuous function (just set $ y\to 0$ and remember that $ f(f(x))$ is non decreasing)

A third conclusion is then that, for $ y>0$ : $ f(f(x\plus{}y))\geq \frac{f(x\plus{}y)\minus{}f(x)}y\geq f(f(x))$ and so $ f'(x)$ exists and $ f'(x)\equal{}f(f(x))$. And since $ f(f(x))$ is non decreasing, it means that $ f(x)$ is a convex function.

So $ f(x)$ solution $ \implies$ $ f(x)$ is a convex function such that $ f(0)>0$ and $ f'(x)\equal{}f(f(x))$

And it is easy to see that these necessary conditions are enough and so :

$ f(x)$ solution $ \iff$ $ f(x)$ is a convex function such that $ f(0)>0$ and $ f'(x)\equal{}f(f(x))$
[/hide]

[quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R} \ \to \ \mathbb{R}$  such that  :

   $ 0 < f(0)$ and $ f(x \plus{} y) \ \geq \ f(x) \plus{} yf(f(x)) \ \ \forall x; y \ \in \ \mathbb{R}$[/i][/quote]
There are no solution to this equation
Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\geq f(x)\plus{}yf(f(x))$

1) [b][u]$ f(x)$ is $ C_1$, convex and such that $ f'(x)\equal{}f(f(x))$[/u][/b]

$ P(x\plus{}y,\minus{}y)$ $ \implies$ $ f(x)\geq f(x\plus{}y)\minus{}yf(f(x\plus{}y))$ and so $ f(x\plus{}y)\minus{}yf(f(x))\geq f(x)\geq f(x\plus{}y)\minus{}yf(f(x\plus{}y))$

And so $ yf(f(x\plus{}y))\geq f(x\plus{}y)\minus{}f(x)\geq yf(f(x))$

A first conclusion is that $ yf(f(x\plus{}y))\geq yf(f(x))$ and so that $ f(f(x))$ is a non decreasing function.

A second conclusion is that $ f(x)$ is a continuous function (just set $ y\to 0$ and remember that $ f(f(x))$ is monotonous)

A third conclusion is then that, for $ y>0$ : $ f(f(x\plus{}y))\geq \frac{f(x\plus{}y)\minus{}f(x)}y\geq f(f(x))$ and so $ f'(x)$ exists and $ f'(x)\equal{}f(f(x))$. And since $ f(f(x))$ is non decreasing, it means that $ f(x)$ is a convex function.

So $ f(x)$ solution $ \implies$ $ f(x)$ is a convex function such that $ f(0)>0$ and $ f'(x)\equal{}f(f(x))$

Q.E.D

2)[b] [u]$ \forall x>0$ : $ f(x)>0$ and $ f(f(x))>0$[/u][/b]
If $ f(u)<0$ for some $ u>0$, and since $ f(0)>0$ and $ f(x)$ continuous, $ \exists v>0$ such that $ f(v)\equal{}0$
Let then $ w\equal{}\inf\{v>0$ such that $ f(v)\equal{}0\}$. Continuity implies that $ f(w)\equal{}0$ and (definition of $ w$) $ f(x)>0$ $ \forall x\in[0,w)$
But then $ f'(w)\equal{}f(f(w))\equal{}f(0)>0$, which is in contradiction with $ f(x)>0$ $ \forall x\in[0,w)$.
So $ f(x)>0$ $ \forall x\geq 0$
Then $ f'(0)\equal{}f(f(0))>0$ since $ f(0)>0$ and, since $ f(x)$ is convex, $ f(f(x))\equal{}f'(x)\geq f'(0)>0$ $ \forall x>0$
Q.E.D.

3) [b][u]The contradiction[/u][/b]
Let $ x>0$ : $ P(x,f(x)\minus{}x)$ $ \implies$ $ f(f(x))\geq f(x)\plus{}(f(x)\minus{}x)f(f(x))$ and, since $ f(f(x))>0$ : $ f(x)\minus{}x\leq 1\minus{}\frac{f(x)}{f(f(x))}$
And so, since $ f(x)>0$ and $ f(f(x))>0$ : $ f(x)<x\plus{}1$ $ \forall x>0$
And, since $ f(x)$ is convex, this implies $ f'(x)\leq 1$ $ \forall x$ and so $ f(f(x))\leq 1$ $ \forall x$
But $ f(x)$ convex and $ f'(0)>0$ imply $ \lim_{x\to \plus{}\infty}f(x)\equal{}\plus{}\infty$
So contradiction.

Hence the conclusion : no such function exists.

[i]Oh , Pco , you are really patient . I don't have time to read your solution But it seems to take many time :lol:

   Here is the solution :

  Assume $ f$ satisfies all conditions of the problem

    If $ f(f(x)) \ \leq \ 0 \ \forall \ x \in \ \mathbb{R} \ \ (1)$ then with all real number $ y \ \leq \ 0$ , we have : $ yf(f(x))   \ \geq \ 0\ \forall \ x \in \ \mathbb{R}$

  $ \rightarrow f(x\plus{}y) \ \geq \ f(x) \plus{} yf(f(x))   \ \geq \ f(x)  \ \forall \ x \in \ \mathbb{R}$

  $ f$ is non-increasing function .
  $ \rightarrow f(0) > 0 \ \geq \  f(f(x)) \ \forall \ x \in \ \mathbb{R}  \rightarrow f(x) >0 \ \forall \ x \in \ \mathbb{R}$ , contradict with $ (1)$

    $ \rightarrow$ there is $ z \ \in \  \mathbb{R} : f(f(z)) >0$

  Due to the fact that : $ f(x\plus{}z) \ \geq \ f(z) \plus{} xf(f(z))   \ \forall \ x \in \ \mathbb{R}$

   We can easily prove that $ \lim_{x \ \to \ \plus{} \infty} f(x) \equal{} \ \ \plus{}  \infty  \rightarrow \lim_{x \ \to \ \plus{} \infty} f(f(x)) \equal{} \ \ \plus{} \infty$

  So we can choose positive real numbers $ x ; y$ such that

$ f(x) \ \geq \ 0 ;  f(f(x)) > 1 ; y \ \geq \ \frac{x\plus{}1}{f(f(x)) \minus{}1 } ; f(f(x\plus{}y\plus{}1)) \ \geq \ 0$

 We have : $ f(x\plus{}y) \ \geq \ f(x) \plus{} yf(f(x))   \ \geq \   yf(f(x)) \equal{} y(f(f(x)) \minus{}1) \plus{} y \ \geq \ x\plus{}y\plus{}1 \ \ (2)$
   Using $ (2)$ :
  $ f(f(x\plus{}y)) \equal{} f(  (x\plus{}y\plus{}1) \plus{} (f(x\plus{}y) \minus{} ( x\plus{}y\plus{}1) ) )$

$ \ \geq \   f(x\plus{}y\plus{}1) \plus{}   (f(x\plus{}y) \minus{} ( x\plus{}y\plus{}1) ) f(f(x\plus{}y\plus{}1)) \ \geq f(x\plus{}y\plus{}1)$

$ \geq \ \ f(x\plus{}y) \plus{} f(f(x\plus{}y))  \geq \ \ f(x) \plus{} yf(f(x) \plus{} f(f(x\plus{}y)) > f(f(x\plus{}y))$ , contradiction

   And our initial assumation is wrong .So , there is no function satisfies the condition  :lol: [/i]


geometry
3D geometry
tetrahedron
puzzles

What are the next terms in the sequence which begins $1, \infty, 5, 6, 3$ ?

Hmm, [hide]does it have anything to do with the number of (convex) regular polyhedra in dimensions 1, 2, 3, etc.?  I don't know anything beyond the 3rd dimension.[/hide]

Thanks, but that doesn't mean I know the next terms. :)  I guess I could google the answer, but that wouldn't be sporting.  So this puzzle is still open...

As far as I know, all the next terms are $3$.

Yes, Ravi and grobber are both correct.  All the remaining terms in the sequence are 3.  It is the number of regular polyhedra in dimension n.

[quote="Devendra Patel"]please explain
I am lost or let me know a site where I can check this out
This is really interesting because of the Infinite in the sequence[/quote]No... because of the geometrical interpretation.

Now you've done it - you rubbed the lamp...

Anyway the nth entry in the sequence is the number of "regular shapes" in n dimensions.

In 1 dimension the line segment is the only regular shape.

In 2 dimensions any n-sided regular polygon (equilateral triangle, square, regular pentagon, etc.) is a regular shape.

In 3 dimensions there are only five shape - the five platonic solids (with sides 4, 6, 8, 12, and 20)

In 4 dimentions there are six shapes. The proof of this is surprisingly not that hard, but does require a little imagination for figuring out how to "flatten" four dimensional shapes into their three dimensional nets.

In higher dimensions the only regular shapes are the Simplex, the Hypercube and the dual of the Hypercube, which is why the remaining terms of the sequence are all 3.

Hope this helps.

Hey, well since we're on the topic of n-dimensions.  I had countless hours of fun playing around with this site's Java applet.

[url=http://dogfeathers.com/java/hyperstar.html]Hyper Cubes etc[/url]

 Though in general people consider "time" the 4th dimension, it's a little more than that, but since we don't have 4 dimensional drawing boards... I suppose that time is probably the nicest way to show the sections of the dimensions, as this site has so beautifully done.

How does a regular solid $12$ look like?

Cute problem !
FWIW - Book Geometry and the Imagination (By D. Hilbert and Cohn-Vossen) is fun book  and has a whole chapter on Regular polyhedra in Three, Four and higher dimension....
For n>=5 , only possible figures are:
(n+1) cell (with n+1 vertices - self dual and like a tetrahedron )
2n -cell  (with 2n vertices )
2^n - cell (with 2^n vertices)
(Above two are mutually dual)

Another fun problem some one asked me had something to do with knot - apparently only 3-d you can tie one..

That's really weird. How do you mathematically tie a knot? Things seem so out of reach in $n$-D, where $n \geq 4$.