y1 = 3
y2 = 11
y3 = 24

Find y6

Find a rule for yn

that is not enough information...you have to define from when you start the sequence...and what are initial values...if i gave you the sequence 1, 4,9,16...what would you say is the next term??

I think it's too litle information since i already got for the next term 3 possibilities:
45, 578 and 17. :)

First, I moved this thread to a more appropriate forum, particularly in consideration of the point that was raised about how there is more than one possible sequence, which is certainly true.  Perhaps an Intermediate level student will come by and explain why.

Additionally, please take a look at the standards for classroom math problems.  I think any "solution" to this problem requires more math than is taught in a typical middle school (read as "not advanced math or math team") classroom.

not sure if this is a sufficient explanation, but...if you look at a function...2 points define a line, 3 points define a parabola, and so on...based on degrees...so you can have 3 points define infinitely many 3 degree equations...

See if you can find a more mathematical way to explain the polynomial concepts you are using.

Consider the following pattern:

y1=3      y2=11      y3=24

Find y6
Find a rule for yn
Find y40

if you do like this you can just draw anything on a piece of paper, find a pattern of making it grow somehow and so you got yourself a sequence, which nobody will ever guess. sorry, bu i don't find it too inteersting....more like something to do when you have too much spare time and you also wanna propose a problem nobody will solve.
suggestion: why not try something 3D, or even more? see what you get then. :)

Tarquin's right if it's quadratic. $y_x=\frac{5x^2+x}{2}$. $y_{40}=4020$


Prove that for every positive integer n there exists an n-digit number divisible by $ 5^n$ all of whose digits are odd.

We proceed by induction.

Clearly it is true for $ n\equal{}1$.

Assume it is true for $ n\equal{}k$.

Then we want to show for $ n\equal{}k\plus{}1$ there is such a number.

By the induction hypothesis we know that there exists a number $ a_{k\minus{}1}a_{k\minus{}2}\ldots a_0$ divisible by $ 5^{k}$.

For $ n\equal{}k\plus{}1$, we want one of the following to be divisible by $ 5^{k\plus{}1}:$
$ 1a_{k\minus{}1}a_{k\minus{}2}\ldots a_0\equal{}10^{k}\plus{}5^{k}\cdot m\equal{}5^{k}(2^{k}\plus{}m)$ for some integer $ m$.
$ 3a_{k\minus{}1}a_{k\minus{}2}\ldots a_0\equal{}3\cdot10^{k}\plus{}5^{k}\cdot m\equal{}5^{k}(3\cdot2^{k}\plus{}m)$
$ 5a_{k\minus{}1}a_{k\minus{}2}\ldots a_0\equal{}5\cdot10^{k}\plus{}5^{k}\cdot m\equal{}5^{k}(5\cdot2^{k}\plus{}m)$
$ 7a_{k\minus{}1}a_{k\minus{}2}\ldots a_0\equal{}7\cdot10^{k}\plus{}5^{k}\cdot m\equal{}5^{k}(7\cdot2^{k}\plus{}m)$
$ 9a_{k\minus{}1}a_{k\minus{}2}\ldots a_0\equal{}9\cdot10^{k}\plus{}5^{k}\cdot m\equal{}5^{k}(9\cdot2^{k}\plus{}m)$

We now consider the numbers $ 2^{k}, 3\cdot 2{k}. 5\cdot 2^{k}, 7\cdot 2^{k}, 9\cdot 2^{k}$ mod 5.

Clearly, these form a complete residue class, and thus at least one of them is equivalent to $ \minus{}m \mod 5$.

Then the claim follows.


calculus
integration
function
logarithms
floor function
trigonometry
factorial

Show that:
$ \int_1^N \frac{\{t\}}{t} \,dt \equal{} \frac{1}{2}\log N \plus{} C_0 \plus{} O\left(\frac{1}{N}\right)$

for some constant $ C_0$.

The Fourier series representation of the floor function together with the fact that $ \{t\} \equal{} t \minus{} \lfloor t \rfloor$ give us:
$ \frac {\{t\}}{t} \equal{} \frac {1}{2t} \minus{} 2\sum_{k \equal{} 1}^{\infty} \frac {\sin (2\pi kt)}{(2\pi kt)}$.

As akech wrote:

$ \{t\} \equal{} t \minus{} \lfloor t \rfloor$

Therefor:

$ \int ^n _ 1 \frac {\{t\} }{t} \,dt \equal{} \int ^n _ 1 \left(1 \minus{} \frac {\lfloor t \rfloor }{t} \right)\,dt \equal{} n \minus{} 1 \minus{} \int ^n _ 1 \frac {\lfloor t \rfloor }{t} \,dt$

Now, we can define:

$ \frac {\lfloor t \rfloor }{t} \equal{} \frac {n}{t}, \qquad t \in [n, n \plus{} 1\}$

Now, we have:

$ \int ^n _ 1 \frac {\lfloor t \rfloor }{t} \,dt \equal{} \sum^{\lfloor n \rfloor \minus{} 1}_{i \equal{} 1}\int^{i \plus{} 1}_i \frac {i}{t}\,dt \plus{} \int ^n_ {\lfloor n \rfloor} \frac {\lfloor n \rfloor}{t}\, dt$

$ \sum^{\lfloor n \rfloor \minus{} 1}_{i \equal{} 1}\int^{i \plus{} 1}_i \frac {i}{t}\,dt \equal{} \sum^{\lfloor n \rfloor \minus{} 1}_{i \equal{} 1} i \left( \ln(i \plus{} 1) \minus{} \ln(i)\right) \equal{} \lfloor n \rfloor \ln \lfloor n \rfloor \minus{} \ln \lfloor n \rfloor!$

$ \int ^n_ {\lfloor n \rfloor} \frac {\lfloor n \rfloor}{t}\, dt \equal{} \lfloor n \rfloor \left(\ln n \minus{} \ln \lfloor n\rfloor\right)$

Now:

$ \int ^n_ 1 \frac {\lfloor t \rfloor }{t} \,dt \equal{} \lfloor n\rfloor \ln n \minus{} \ln \lfloor n \rfloor !$

We can use Stirling approximation for the factorial in the form:

$ \ln \lfloor n \rfloor ! \approx \frac {1}{2}\ln \lfloor n \rfloor \plus{} \frac {1}{2} \ln 2\pi \plus{} \lfloor n \rfloor \ln \lfloor n \rfloor \minus{} \lfloor n \rfloor$

Now, we can put everything back in the first equation and write:

$ \int ^n _ 1 \frac {\{t\} }{t} \,dt \approx n \minus{} 1 \minus{} \lfloor n\rfloor \ln n \plus{} \frac {1}{2}\ln \lfloor n \rfloor \plus{} \frac {1}{2} \ln 2\pi \plus{} \lfloor n \rfloor \ln \lfloor n \rfloor \minus{} \lfloor n \rfloor$

Now, for large $ n$, we can write $ n \approx \lfloor n\rfloor$ and this gives us:

$ \int ^n _ 1 \frac {\{t\} }{t} \,dt \approx \frac {1}{2}\ln n \plus{} \frac {1}{2} \ln \frac {2\pi}{e^2}$

You need to be more precise with your use of Stirling's approximation - as written, you haven't shown that the error is $ O \left( \frac{1}{n} \right)$.  One needs the following more precise estimate:  for an integer $ n$,

$ \ln n! \equal{} \frac{1}{2} \ln n \plus{} \frac{1}{2} \ln 2 \pi \plus{} n \ln n \minus{} n \plus{} \lambda_n$

where $ \frac{1}{12n\plus{}1} < \lambda_n < \frac{1}{12n}$, which gives the required bound.


kk this is where all the first round matchup concerning teams plastic and dead planaria info can be found
uhhh since NO ONE SHOWED UP ON TUESDAY we will have to reschedule
[b]Time: 9:00 ET, Friday February 22[/b]
thank you

leave a comment if you've seen this topic and is on the above two teams.

Err, I showed up at 9:16...just a little late. :(


[color=cyan]2 possible approaches:
The less promising one: count up powers of 2.  I think this is much less promising than
look at the table mod 4.  That is, take every entry in the first seven or eight rows, but instead of writing them out, only look at the remainders when they are divided by 4.  Search for a pattern.  There's gonna be something nice, although it may be hard to describe exactly.  Describe it exactly.  Then prove it.  Then use it to find the answer to the question.[/color]

Paul Zeitz is known for writing extremely hard problems to do with Pascal's triangle. That question is from BAMM Test of Ingenuity 2002 either #13 or #15 (the two I didn't solve) if I remember correctly. I don't believe anyone solved that one actually. Some people talked to him after the contest, and I believe he said that there is some way to bound it, and then he wrote a computer program to check for the exact number. From his choices, only one was within his bounds. But I don't believe there is a simple pattern as Joel suggested there might be.

[color=cyan]I was working on the assumption that this problem wasn't overly difficult . . . If what ComplexZeta says is true, however, then I retract my earlier statements.  It still might be doable, though.[/color]

An easier problem is, how many numbers in the 64th row of Pascal's Triangle are not divisible by 4? And can you prove it.

Yes, this problem is much easier.  You can use the combination formula and search for non-multiples of 4.

-interesting_move


hey peeps... are we allowed to talk about the 207 school mathcounts now, or do we have to wait until March 1st?

and, if anyone has a copy of it, can you pm me immediately?

Our coach won't give it to us.(selfish)

thanks...
 :roll:

[quote="DPatrick"]We will tell you when it is OK to freely discuss the school and chapter contests; this will not be until all chapter contests in all states are concluded in late February.[/quote]

I thought the last ones were yesterday :mad:

We are still waiting to hear from the MATHCOUNTS national office regarding the conclusion of the chapter rounds.

Please do not discuss until we say otherwise.


The base 3 representation of an integer n is 21121221211212212112 what is the leading digit in the base 9 representation of n?

[quote="chronoflare"]Thats not what I got...
[hide]The leading the digit should be 7 right?[/hide][/quote]
[hide]
I also got 7.
The first two terms are $3^{19}\times2+3^{18}$, which is $3^{18}(7)$. Each two terms are paired together the same way so the first digit of the base $9$ form is $7$ since $3^{18}=9^9$ and the next term is multiplied by $9^8$ and it progresses down like that.[/hide]

oh ya, it's pi day!
does anyone know Moor Xu, currently a ninth grader who did mathcounts last year?
I wonder if he still knows past 100 places...

[quote="ShoeFactory"]Since you are going from base 3 to base 9, you can partition the number into pairs.

Like this: 21|12|12|... |12.

The first digit will be the $21_3$ converted to base 9.

$3^2 + 1 = 7$ so the answer is $7$.[/quote]

Do you mean $3*2 + 1 = 7$??

[quote="me@home"]sean, did you get to nationals yet? have you heard about the google group for that?[/quote]

I'll look for it.  By the way, who are you?  Abhi?

no. whose that? I did MCounts last year, went to nats. It was fun. I'll PM you.

[quote="Apprentice"]Sorry, I don't have the answer key yet since this problem came out on March 9th on a test.
But you're right, seven is the answer...

By the way, did anyone even notice that it's pi day???? March 14????  How did you guys celebrate the day?[/quote]I memorized a hundred digits.
Edit: I just noticed that Ryan22 has exactly 100 digits in his sig (because that's where I stopped memorizing). How coincidental.


Point A is located at (5,5) on the cartesian plane. Point B is also in the plane and has integer coordinates. If
$ 0 \le AB \le 4$ , at how many points could B be located?

I thought I made it clear that A is (5,5)

[quote="AIME15"]Hint: draw circle of radius 4 centered at (5,5) and count.[/quote]

Not a good way.

Okay, what's your way?  It's not much counting.

In this case, you need to count 48 points.

There is a general way (just do some simple calculation) to get the answer for $ 0 \le AB \le n$ .

[quote="AIME15"]What is the simple way? :)[/quote]

I won't tell you since you might be my competitor.   :rotfl:

http://mathworld.wolfram.com/CircleLatticePoints.html

This has some information pertaining to lattice points and circles.


abstract algebra
group theory
superior algebra
superior algebra unsolved

I'm having trouble understanding the following exercise from Lang:
Let G be a group, H an abelian normal subgroup, and let $ \gamma_{x}$ denote conjugation ($ \gamma_{x}(a)\equal{}xax^{\minus{}1}$).
Show that the association $ x\mapsto\gamma_{x}$ induces a homomorphism $ G/H\rightarrow\text{Aut}(H)$.

Doesn't this induce a homomorphism into $ \text{Aut}(G/H)$, not $ H$?

If $ a \in H$, then $ xax^{\minus{}1} \in H$ since $ H$ is a normal subgroup. When $ x \in H$, $ xax^{\minus{}1} \equal{} a$ since $ H$ is abelian.

So for each $ xH\in G/H$, we only conjugate elements of $ H$? I think I see how we get the homomorphism if this is the case, but how exactly does the fact that $ xax^{\minus{}1}\equal{}a$ when $ x\in H$ come into play?

If $ x \in G$, with $ a \in H$, then $ xax^{\minus{}1} \in H$ since $ H$ is a normal subgroup. So, yeah, you're only conjugating for $ a \in H$.

The fact that $ xax^{\minus{}1} \equal{} a$ is necessary to show that the homomorphism is, in fact, from $ G/H$, rather than just $ G$, because each $ x \in H$ acts trivially by conjugation.

The induced mapping you are looking for is $ xH\mapsto\gamma_x.$ The condition $ xax^{ \minus{} 1} \equal{} a$ when $ x\in H$ is necessary to show that this mapping is well defined – in other words $ xH\equal{}yH\ \Rightarrow\ \gamma_x\equal{}\gamma_y.$

Cool, I get it now, thanks for the help.

in general if G' is a normal subgroup of G one get a induced homomorphism G/Z_G(G')->Aut(G')
Patrick


Show that, for all $n \in \mathbb{N}^*$, $10^{10^n} \equiv 4 \pmod 7$.

:)

[hide]

it is easy to check that $10^3\equiv -1\mod 7$ . And also $10^n-1\equiv 0\mod 3$ . Let $10^n-1=3k$ , for some positive odd integer $k$ , then

$10^{10^n}=10\cdot 10^{10^n-1}=10\cdot 10^{3k}\equiv 3\cdot (-1)\equiv 4\mod 7$

[/hide]

[hide]Really dumb way of doing it but requires absolutely nothing fancy:

$3^3 = 27 \equiv -1 \pmod{7} \Rightarrow 3^{10} = 3 \times (3{}^3){}^3 \equiv 3 \times -1 \equiv -3 \pmod{7}$

$N = 10{}^{10}{}^n = ((10^{10})^{10})^{10} \ldots (n \text{ times}) \equiv ((3^{10})^{10})^{10} \ldots (n \text{ times})$. But $-3^{10} \equiv -3 \pmod{7}$, so no matter how many times we raise to the power of $10$, $N \equiv -3 \equiv 4 \pmod{10}$.[/hide]


integration
calculus
calculus computations

How could I show that

$f_n(x)=\begin{cases}1 \quad &\text{if} \quad 0\le x \le 1/2 \\ -2^n\left (x-\tfrac{1}{2}\right )+1 \quad &\text{if} \quad \tfrac{1}{2}\le x\le 1/2+(\tfrac{1}{2})^n \\ 0 \quad &\text{if} \quad \frac{1}{2}\le x \le 1 \end{cases}$

is a Cauchy sequence in the metric

$d(f_n,f_m)=\int \limits _0 ^1 |f_n-f_m|dx \quad ?$

I tried evaluating the integral (which is very well possible), but it didn't get obvious that they will get close. Is there a quicker way to see it? Can I see it somewhere on the way? Thanks in advance for any help.

I just found a simple way to see it. Estimate!


algebra
polynomial
number theory
divisibility tests
number theory proposed

Let $n$ be a fixed natural number.
Find all $n$ tuples of natural pairwise distinct and coprime numbers like $a_1,a_2,\ldots,a_n$ such that for $1\leq i\leq n$ we have
\[ a_1+a_2+\ldots+a_n|a_1^i+a_2^i+\ldots+a_n^i  \]

here is mine ...
wlog $ a_1 < ... < a_n$
let $ s_i \equal{} \sum_{j \equal{} 1}^{n} a_j^i$ and $ P \equal{} \prod a_i$ and $ Q \equal{} \sum_{i \equal{} 1}^{n} \frac {P}{a_i}$ and it's obvious that $ \gcd(P,Q) \equal{} 1$ .
we have
$ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid (a_1 \plus{} ... \plus{} a_n)(a_1^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 1})$

$ \implies a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid \sum_{i \neq j} a_ia_j(a_i^{n \minus{} 2} \plus{} a_j^{n \minus{} 2})$

$ \implies a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid \sum_{i \neq j} a_ia_j(s_{n \minus{} 2} \minus{} a_i^{n \minus{} 2} \minus{} a_j^{n \minus{} 2})$

$ \implies a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid \sum_{i,j,k \ \text{distinct}} a_ia_ja_k(a_i^{n \minus{} 3} \plus{} a_j^{n \minus{} 3} \plus{} a_k^{n \minus{} 3})$

$ \implies a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid \sum_{i,j,k \ \text{distinct}} a_ia_ja_k(s_{n \minus{} 3} \minus{} a_i^{n \minus{} 3} \minus{} a_j^{n \minus{} 3} \minus{} a_k^{n \minus{} 3})$

$ \implies a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid \sum_{i,j,k,t \ \text{distinct}} a_ia_ja_ka_t(a_i^{n \minus{} 4} \plus{} a_j^{n \minus{} 4} \plus{} a_k^{n \minus{} 4} \plus{} a_t^{n \minus{} 4})$

well... it's easy to understand that by doing the same thing again and again we'll finally reach this divisibility :

$ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid nP$ (1)

now do exactly the same thing but this time by considering $ (a_1 \plus{} ... \plus{} a_n)(a_1^{n \minus{} 2} \plus{} ... \plus{} a_n^{n \minus{} 2})$ instead of $ (a_1 \plus{} ... \plus{} a_n)(a_1^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 1})$ at the beginning of the process ...

this time we'll finally reach this divisibility :

$ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \mid (n \minus{} 1)Q$ (2)

let me tell an example to clarify :
if $ n \equal{} 4$ and we want to find the 4-th tuples (a,b,c,d) then ...
$ a \plus{} b \plus{} c \plus{} d \mid (a \plus{} b \plus{} c \plus{} d)(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3)$

$ \implies a \plus{} b \plus{} c \plus{} d \mid ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} cd(c^2 \plus{} d^2) \plus{} da(d^2 \plus{} a^2) \plus{} db(d^2 \plus{} b^2) \plus{} ac(a^2 \plus{} c^2)$

$ \implies a \plus{} b \plus{} c \plus{} d \mid ab(c^2 \plus{} d^2) \plus{} bc ( a^2 \plus{} d^2) \plus{} ... \plus{} ac ( b^2 \plus{} d^2)$

$ \implies a \plus{} b \plus{} c \plus{} d \mid abc(a \plus{} b \plus{} c) \plus{} bcd(b \plus{} c \plus{} d) \plus{} cda(c \plus{} d \plus{} a) \plus{} dab(d \plus{} a \plus{} b)$

$ \implies a \plus{} b \plus{} c \plus{} d \mid abc(d) \plus{} bcd(a) \plus{} cda(b) \plus{} dab(c) \equal{} 4abcd$

and also  ...

$ a \plus{} b \plus{} c \plus{} d \mid (a \plus{} b \plus{} c \plus{} d)(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$
$ \implies a \plus{} b \plus{} c \plus{} d \mid ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} cd(c \plus{} d) \plus{} da(d \plus{} a) \plus{} db(d \plus{} b) \plus{} ac(a \plus{} c)$
$ \implies a \plus{} b \plus{} c \plus{} d \mid ab(c \plus{} d) \plus{} bc(a \plus{} d) \plus{} ... \plus{} ac(b \plus{} d) \equal{} 3 ( abc \plus{} bcd \plus{} cda \plus{} dab)$

back to the problem .. .
from (1) and (2) we get ...

$ a_1 \plus{} ... \plus{} a_n \mid \gcd(nP,(n \minus{} 1)Q)$ but we obviously have $ \gcd(nP,(n \minus{} 1)Q) \mid n(n \minus{} 1)$

so we must have $ a_1 \plus{} ... \plus{} a_n \mid n(n \minus{} 1)$ but since $ a_i$ are distinct we  have $ a_1 \plus{} ... \plus{} a_n \ge 1 \plus{} 2 \plus{} ... \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$ and so if $ a_1 \plus{} ... \plus{} a_n < n(n \minus{} 1)$ then since $ a_1 \plus{} ... \plus{} a_n \equal{} \frac {n(n \minus{} 1)}{k}$ for some $ k$ so we get
$ \frac {n(n \plus{} 1)}{2} \le a_1 \plus{} ... \plus{} a_n \le \frac {n(n \minus{} 1)}{2}$ which is a contradiction ...
so we must have $ a_1 \plus{} ... \plus{} a_n \equal{} n(n \minus{} 1)$
we can easily prove that $ a_1 \plus{} ... \plus{} a_n \ge 1 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1} p_{i}$ where $ p_i$ is the $ i \minus{} th$ prime and equality occurs iff $ a_1 \equal{} 1$ and $ a_i \equal{} p_{i \minus{} 1}$...
now we can easily see that for $ n \equal{} 7$ we have $ 1 \plus{} \sum_{i \equal{} 1}^{6} p_{i} \equal{} 7 (7 \minus{} 1)$ but for $ n \ge 8$ we have $ 1 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1} p_{i} > n(n \minus{} 1)$ meaning that we do not have any n tuple with the properties mentioned for $ n \ge 8$ ... it remains to check some trivial cases ...


We have got a number : ${{n} \choose k}$, n is an even number.
Is there any correlation between parity of number ${{n} \choose k}$ and number $k$
$k \in [0 ;n]$

Use Legendre theorem:
$p^l | n!$ , where $l = \sum_{k \geqslant 1}[\dfrac{n}{p^k}]$

and definition of ${n \choose k}$

I think Lucas' theorem is a better correlation than Legendre's one...
Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=lucas&t=41375

Pierre.


calculus
function
inequalities
real analysis
real analysis unsolved

Suppose f is a twice differentiable real function such that for all x, |f (x) | <=1 and |f''(x)|<=1. Prove that for all x, |f ' (x)|<=2.

here is the proof. For any x we have the Lagrange identity:

$ f(x \plus{} 2) \equal{} f(x) \plus{} 2 f'(x) \plus{} 2f''(\xi)$

therefore

$ f'(x) \equal{} \frac {f(x \plus{} 2) \minus{} f(x) }{2} \minus{} f''(\xi)$

so

$ |f'(x) | \le 1 \plus{} 1 \equal{} 2$

This inequality isn't sharp; we may conclude that $ |f'(x)|\le\sqrt{2}.$

Consider a point $ x.$ By replacing $ f(\cdot)$ by $ \minus{}f(\cdot)$ and/or $ f(\cdot)$ by $ f(2x\minus{}\cdot)$ if necessary, assume WLOG that $ f(x)\ge 0$ and $ f'(x)\ge 0.$

Now follow the same structure as thinkagain's proof, but with a different spacing:

$ f(x\plus{}\sqrt{2})\equal{}f(x)\plus{}\sqrt{2}\,f'(x)\plus{}f''(\xi).$

$ \sqrt{2}\,f'(x)\equal{}f(x\plus{}\sqrt{2})\minus{}f(x)\minus{}f''(\xi)\le 1\minus{}0\plus{}1\equal{}2.$

Hence, $ f'(x)\le\sqrt{2}.$ Remove the WLOG on the signs of $ f$ and $ f'$ and we conclude that $ |f'(x)|\le\sqrt{2}.$

The function $ f(x)\equal{}\begin{cases}\frac{x^2}2\minus{}1&\text{ on }[0,\sqrt{2}]\\ 1\minus{}\frac{(2\sqrt{2}\minus{}x)^2}{2}&\text{ on }[\sqrt{2},2\sqrt{2}]\end{cases}$ shows that the constant $ \sqrt{2}$ cannot be improved. (This $ f$ is not twice differentiable, but it can be approximated arbitrarily well by twice differentiable functions.)

The general result is this: writing $ \|f\|_{\infty}\equal{}\sup_{x\in\mathbb{R}}|f(x)|.$ if $ f$ is twice differentiable on $ \mathbb{R},$ then
\[ \|f'\|_{\infty}\le g\left(\|f\|_{\infty},\|f''\|_{\infty}\right)\]
What is the correct expression $ g\left(\|f\|_{\infty},\|f''\|_{\infty}\right)$ that makes this true, but "tight"?

kaymant sir, et tu??

I will give a link to my solution : http://www.goiit.com/posts/list/differenciation-challenge-no-2-74440.htm#367304


I was wondering if there were any awards/money associated with getting a perfect score on the AMC 10. I know you get your picture on the website, but I don't know if that's worth taking the AMC 10 instead of the AMC 12. What do you guys think or know (for those of you who have gotten perfect scores, I know you're out there)?

you could simplify the problem by acing the AMC 12  :lol:

Lol, but the point is a friend of mine who's a junior want to win the AMC 12 for the school so he can have 4 years in a row as school winner (he missed out on the 12A cuz he thinks freshman beat him =P but that freshman is taking the 10B). That's why I don't want to take the 12. Gotta support the team right?

I don't believe there are any awards for AMC10.  I've read about awards for AMC12 (usually by region...if you have a results book, it is in a section towards the front) but not for the 10.

There are plaques for state winners on the 10, but I'm not aware of book awards or the like for that contest.


There are 5 red balls; $R_1, R_2, R_3, R_4, R_5$ and 2 white balls $W_1, W_2$

The two ends must be occupied by red balls.

a) how many arrangements can be made is the white balls are not next to each other?

b) how many arrangements if $R_1, W_1$ are next to each other

c) how many arrangements to satisfy both a) and b)?

[hide="A"]Choose the order of the red balls: $5!$
There are 4 spaces between the red balls.  There are $4(3)$ ways to place the white balls.
Therefore, the answer is $5!(4)(3)=120(12)=\boxed{1440}$[/hide]

[hide="A little help"] Just use basic combinatorics and the solution follows. I don't have time now, so I let anyone to post a solution, however the problem doesn't seem difficult.[/hide]

[hide]

b) case 1) $R_1$ is at one of the end points.

Two possible locations for $R_1$, four possible choices for the other end point, of the 5 slots inbetween, 1 has to be $W_1$, so number of arrangements equals: $2\times4\times4!=192$

case 2) $R_1$ is among the middle 5 spots. For 3 of those cases, $W_1$ could then be in 2 places. For the 2nd and 6th spots, there is only one place $W_1$ could be since the end points have to be red. Then there are $4\times3$ ways to fill up the end points, and $3!$ ways to fill up the rest of the sequence. Number of arrangements equals: $(3\times2+2)\times4\times3\times3!=576$

Total number of arrangements equals $192+576=\boxed{768}$

[/hide]

How do you put a different caption/title in place of the "Click to reveal hidden content"

If you type "$\text{[hide}=\text{Solution]The answer is 100[/hide]}$", you will get

[hide="Solution"]The answer is 100[/hide]

Oh ok.
[hide="I see now"]
Thanks,  :lol:
[/hide]

[hide="A"]Take $5\choose2 * 2!$ to see which red balls occupy the ends and their order.

Take $5\choose2$ to see which slots the white balls will occupy R_ _ _ _ _R.
$4$ of them do not work because the white balls will be next to each other so the white balls can be in $6$ positions.

Multiply by  $2$ to account for the white balls' order.

Now take $3!$ to find the order of the red balls occupying the last 3 spaces.

You get $20*6*6*2 = \boxed{1440}$

edit: thanks adunakhor[/hide]

[hide="a"]
For the number of possibilities for the endpoints, order matters, so it's $20$, not $\displaystyle{5\choose 2}$

The rest of what you put is fine, though for some reason it looks like you miscopied your work into the final step. You had the 6 location possibilities for the white balls, the factor of 2 for their arrangement, and the $3!$ for the arrangment of the 3 red balls. Your final step though has a $2$ in place of the $3!$ for some reason.

number of arrangements equals: $20\times6\times2\times3!=1440$

[/hide]

what about if the white balls are not next to each other and R1 is next to W1


[u][i]  Problem [/i][/u]
 [i] Let a,b,c,d be positive real numbers such that
  $ 3(a\plus{}b\plus{}c\plus{}d) \plus{} 4(abc\plus{}bcd\plus{}cda\plus{}dab) \equal{} 8$
       Prove that
 $ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd \leq 2$[/i]
[i]   Love An forever[/i]
                     [i]  Argady, Nayel ... I am waiting for a nice solution [/i] :oops:

I think the inequality is strict, without equality.
The following is with equality:
[i]Let $ a,b,c$ be positive real numbers. If
$ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\equal{}6$,
then[/i]
(a) $ 5(a\plus{}b\plus{}c\plus{}d)\plus{}abc\plus{}bcd\plus{}cda\plus{}dab \ge 24$;
(b) $ (2\plus{}2\sqrt 2)(a\plus{}b\plus{}c\plus{}d\minus{}4)\plus{}abc\plus{}bcd\plus{}cda\plus{}dab\minus{}4 \ge 0$;

All of these inequalities belong to the class of four variable symmetric inequalities without the presence of the product $ abcd$. They are solved by using Mean Value Theorem, reducing to three numbers. For instance, after Mean Value Theorem with the substitution $ p\equal{}x\plus{}y\plus{}z$, $ q\equal{}xy\plus{}yz\plus{}zx$, $ r\equal{}xyz$, choose $ p\equal{}1$, the first by Vasile can be reduced to proving that
\[ \frac{20}9q\plus{}4r\geq\frac{8\sqrt3}3q^{3/2}.\]

[quote="nkht-tk14"][u][i]  Problem [/i][/u]
 [i] Let a,b,c,d be positive real numbers such that
  $ 3(a + b + c + d) + 4(abc + bcd + cda + dab) = 8$
       Prove that
 $ ab + ac + ad + bc + bd + cd \leq 2$[/i]
[i]   Love An forever[/i]
                     [i]  Argady, Nayel ... I am waiting for a nice solution [/i] :oops:[/quote].I think the number $ 2$ is instead of $ \frac{3}{4}$
By roll theorem have existed $ x,y,z>0$ such that :$ a+b+c+d=\frac{4}{3}(x+y+z), ab+ac+ad+bc+bd+cd=2(xy+yz+zx),abc+bcd+dca+dab=4xyz$
so the condition  will become:$ x+y+z+4xyz=2$,we must prove that $ xy+yz+zx\leq\frac{3}{4}$
we assume that $ xy+yz+zx=\frac{3}{4}$ so we need to prove that $ x+y+z+4xyz\geq 2$
we put $ f(x;y;z)=x+y+z+4xyz$ so easy to prove that  $ f(x;y;z)\geq f(x;\sqrt{(x+y)(x+z)}-x;\sqrt{(x+y)(x+z)} -x)$(*)
with ${ x=min\{x;y;z}\Rightarrow x\leq\frac{1}{2}$ thus $ (*)\Leftrightarrow (1-4x^2)(\sqrt{x+y}-\sqrt{x+z})^2\geq 0$
so we only the inequality in the case $ b=c$,we have $ 2ab+b^2=\frac{3}{4}$
$ a+2b+4a^3\geq2$
$ \Leftrightarrow 16b^2(12b^2+3)+(3-4b^2)^3\geq 256b^3$, we can expand it and prove easy :D


inequalities
trigonometry
calculus
inequalities unsolved

Actually, the second part can be proved more simply by using the inequality $\sin x \le x$ for every nonnegative x.  We have
\[\sin 20^\circ \le 20^\circ = \frac{\pi}{9} < \frac{3.15}{9} = \frac{7}{20} = \frac{21}{60}.\]


geometry
calculus
trigonometry
ratio
MATHCOUNTS
AMC 10

I'd like to open a new forum.
[u]Do mathematicians listen to music[/u]? [u]What kind of music do you listen and is it true that who loves maths doesn't like to have fun[/u]?

I start. I like music very much especially metal and punk too. I like [b]Cannibal Corpse[/b], [b]Deicide[/b], [b]Mayhem[/b], [b]Morbid Angel [/b]and also [b]Motorhead [/b]and [b]Misfits [/b]very much!!!!!!!!

Tell us what you think about these questions.
Hello!!!

I listen to music nearly all the time. (I'm currently listening to Schumann's ``Spring'' Symphony #1.)

The assertion that those who love math do not like to have fun is very false. I have plenty of fun (in my own way; it probably wouldn't be fun for some other people), and I love math.

"is it true that who loves maths doesn't like to have fun?" I think it's false but I think we are introvert.
I love music ! I listen Kyo (french group), Miossec (french too), Coldplay, Cranberries, Bob Dylan, Muse, Radiohead, Saez (french), ...
And I play acoustic guitar

Music is my life! I LOVE it! I enjoy singing it as well as listening to it. My fave singers/groups are: Josh Groban (wasn't that obvious???  ;) ), Hoobastank, Avril Lavigne, sometimes Pink (depends on the mood)...I have a rather wide range of likes when it comes to music. I have learned to appreciate each style for what it is, for the most part, but there are some I just can't get into unless it's a great song such as rap (I really like eminem's mockingbird) and various country songs.

[quote="Andreas"]is it true that who loves maths doesn't like to have fun?[/quote]

I don't really think it's possible for anyone to not want to have fun.  The only thing that is different is what is fun for that person.  For people who love math, it's probably doing math.

I love music... classic rock.  I think most math people are also involved in a band (at school or wherever), or atleast play a instrument.  I know it's true for me and all my friends

This is somewhat off-topic, but what are "maths"-- I've seen people write "maths" on numerous occasions? Is that the plural form of "math?"

geometry, algebra, calculas, trig, etc.  all fields of math thus maths.  Many other languages like spanish math (las matematicas) is plural.  this is also true for sciences.  hence 'the sciences'

Is maths really plural? Mathematics is certainly a singular noun (that just happens to end with an `s').

I return to the topic, I like music very much, especially Blind Guardian (they're the best !). But do you like to "make" music = Singing, playing? I like both (playing a bit on piano and guitar) and singing. What about you?

I love music!!!  In fact, I'm listening to classical music played by Bela Fleck on the banjo right now (yeah, I'm into the weird stuff) but I mostly listen to jazz and classical.  Probably stems from my love of the piano- I've been playing almost my whole life.  I know alot of mathematicians who are really into music- I think the two go together.   :)

My life is math, music ( and sometime sports). Often I do not know what to study Math or music.. But I think I will study mathematics. I have studied music for 6 years (i'm 16 now) and I am a percussionist, pianist and I am intersted in composing and ethnomusicology. I like all music, from classical, impressionist, rock, jazz, latin jazz latin music, cultural music from different countries, brazillian rythms and a lot more.. I also like experimental music and anything that has to do with creativity.

[quote="LynnelleYe"]As far as I can tell, maths is basically the British word for math. Correct me if I'm wrong :)[/quote]

And many people (especially Europeans) who learn English as a second language learn British English rather than American English.  (I imagine the Chinese learn American; I wonder what they do in Hong Kong nowadays.)

[quote="Danbert"]I play the piano (for ~6 years) and the viola (3 years).  I love classical music (listening to Chopin's Fantaisie-Impromptu now (and learning to play it)).
Music is basically math; it's all just ratios.[/quote]

Chopin is an pwnage machine   :D .

[quote="ComplexZeta"]The assertion that those who love math do not like to have fun is very false. I have plenty of fun (in my own way; it probably wouldn't be fun for some other people), and I love math.[/quote]

I agree with this.  My mom is always telling me that I don't have enough fun, but I believe the problem is we have different definitions of fun.

As for music, I enjoy classic rock, a.k.a. oldies.  I cannot go for a day without listening to Boston, the Beatles, Billy Joel, ELO, Def Leppard, Neil Diamond, the Righteous Brothers, and so on.  But Dave Matthews Band and Counting Crows are my favorite contemporary artists.

[size=200]Did you know that dogs is the plural of dogs![/size]
([size=59]I like music[/size]

someone asserted that maths departments have very little problem forming orchestras if they wanted (= As for me, I am learning bassoon and singing this year, but have enjoyed music for ages (=

I love music, too, especially R&B,hard rock. I like Three Doors Down, The servant, Thalia, J.Lo
More stuff:
Knowledge of mathematics is basic to understanding music.Musical pitch, tempo, timbre, and the acoustics of music all can be expressed in mathematical terms.The intangible quality of art is what distinguishes a great piece of music.Numbers enter curiously in describing what sounds are pleasing to the ear. In the 12-note tempered scale(our usual musical scale), each note is spaced in pitch by a ratio of 2^(1/12).
...and did  you know that using a special mathematical series, called a Fourier Series, it is possible to write a mathematical formula for the harmonies generated by an instrument when a single note is sounded.Thus the general form for a Fourier Series,
  :theta: n(x) = a0 + a1cosx + b1sinx + a2cos2x+ b2sin2x+…+ ancosnx + bnsinnx+…
is used to express an arbitrary periodic waveform and can therefore differentiate between,say, a clarinet and a bassoon, both playing at the same pitch.Note that the series is a sum of simple sine and cosine functions.
[color=blue]Artiona[/color]

Math is like music and the music is like math  :D

I agree with nonIMO, if you are good at math, you can't really suck at music. I think most math-people play a instrument of some sort, piano, violin, cello...  and I totally DON't agree with teh question of wheether or not mathmaticians doesn't have fun. Everyone have fun... and some people consider math fun.

[quote="LuCy4EvA"]I agree with nonIMO, if you are good at math, you can't really suck at music. [/quote]
So I'm not good at math?  :(

o, eh, nvm, no offense...:blush:  I kinda do too, but it's probably better than most people that doesn't take math at all, since music is like numbers... [size=200]SORRY[/size]

Hehe, no prob., I wasn't really offended...besides, your statement might still be true, depending on how 'good at math' is defined...I'm competent at chapter MathCounts and AMC-10 -level contests. But that's as far as I can be sure of myself, and it's not far.

[quote="jackroland"]I like "Wierd Al" Yankovic[/quote]

his different lyrics for each song is pretty...creative  :P

[quote]his different lyrics for each song is pretty...creative  :P [/quote]
To say the least. :D


Find all $ (m,n)\in N$ such that
$ m|a^n\minus{}1,\forall a\equal{}1,..,n$

Why ?Every one can see result but quite hard to prove.


trigonometry
real analysis
real analysis unsolved

Can someone help me find X(x) and Y(y) for u = XY in this PDE:

u[size=75]t[/size] = u[size=75]xx[/size] - 2u
boundary conditions: u(0,t) = 0, u(3,t) = 0
initial condition: u(x,0) = 2sin(pi*x) - sin(4*pi*x)
I know how to do the separation of variables, but again I'm having trouble finding X and Y.
Thanks for your help.

Plug $ u(x,t) \equal{} X(x)T(t)$ into the PDE to find
\[ \frac {T'}{T} \equal{} \frac {X''}{X} \minus{} 2 \equal{} \minus{} \lambda
\]
Conclude using the boundary conditions on $ X$ that solutions can only exist if $ \lambda \equal{} 2 \plus{} \left(\frac {n\pi}{3}\right)^2$ for $ n\in \mathbb{N}$, and hence by superposition find
\[ u(x,t) \equal{} 2e^{ \minus{} t\left(2 \plus{} \pi ^2\right)}\sin \pi x \minus{} e^{ \minus{} t\left(2 \plus{} 16\pi ^2\right)}\sin 4\pi x
\]


please solve

1)if a,b,c are in A.P then for equation 3a(x^2)-4bx+c=0

then

A)both roots are negative
B)both roots are of opposite sign
C)both roots lie in (0,1)
D)at least one root lies in (0,1)

the answer is (D) i.e at least one root lies in (0,1)

how do you prove it?

$ (a,b,c)$ are in A.P. then:
$ (a,b,c)\equal{}(a,a\plus{}d,a\plus{}2d)$

$ f(x)\equal{}3a(x^2)\minus{}4bx\plus{}c$

$ f(0)\equal{}c\equal{}a\plus{}2d$
$ f(1)\equal{}3a\minus{}4b\plus{}c\equal{}3a\minus{}4(a\plus{}d)\plus{}(a\plus{}2d)\equal{}3a\minus{}4a\minus{}4d\plus{}a\plus{}2d\equal{}\minus{}2d$

i know that method but please show me how to derive the answer

In order to be answer (d)

$ (a\plus{}2d,\minus{}2d)$ must have opposite signs.

however if we pick $ a\equal{}\minus{}3$ and $ d\equal{}1$
$ (a\plus{}2d,\minus{}2d)\equal{}(\minus{}1,\minus{}2)$

$ (a,b,c)\equal{}(\minus{}3,\minus{}2,\minus{}1)$
$ 3a(x^2)\minus{}4bx\plus{}c\equal{}\minus{}9x^2\plus{}8x\minus{}1\equal{}0$
$ 9x^2\minus{}8x\plus{}1\equal{}0$
with solution $ x\equal{}\frac{8\plus{}\sqrt{64\minus{}36}}{18}$ and  $ x\equal{}\frac{8\minus{}\sqrt{64\minus{}36}}{18}$ both between $ 0$ and $ 1$


O.K.this is an iitjee'96 question.Use formula of sin(A+B) and cos(A+B) to open the brackets f(x) simplifies to 5/4.Hence gof(x)=g(5/4)=1


Find all real $ x$ such that $ <(x \plus{} 1)^3> \equal{} x^3$
where $ <x> \equal{} x \minus{} [x]$ is the ffractional part of x, gretaer than 0


Let be $ a,b,c\in \mathbb{R}_\plus{}$such that $ a\plus{}b\plus{}c\equal{}3$ . Show that :

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

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}_ \plus{}$such that $ a \plus{} b \plus{} c \equal{} 3$ . Show that :

$ \frac {1}{2 \plus{} a^2b} \plus{} \frac {1}{2 \plus{} b^2c} \plus{} \frac {1}{2 \plus{} c^2a}\le \frac {a^3 \plus{} b^3 \plus{} c^3}{2 \plus{} abc}$[/quote]
I have an inequality than stronger,see it here : http://www.mathlinks.ro/forum/viewtopic.php?t=268853  :)


AwesomeMath
summer program
email
geometry

Hey everybody,

The admissions test A for the 2009 AwesomeMath summer camps is now posted. You have until [size=75]Feb. 9th 2009[/size] to work on this test. To access the test and to read more about the AMSP application, see: http://www.awesomemath.org/application.shtml .

[color=red]
Edit: The deadline for the admissions test A has been extended to Feb. 14th[/color]

for the letter of recommendation that is "from someone who is familiar with the student’s activities and accomplishments outside the world of mathematics," there is still something on there that asks a "comment on the student’s skills, problem solving, and reasoning ability, willingness to accept the challenge of difficult problems and concepts". Am I correct in assuming that this person should know me as a math student as well?

[quote="hurdler"] Am I correct in assuming that this person should know me as a math student as well?[/quote]

You should ask AMSP directly by sending an email to titu@awesomemath.org, Dr. Titu Andreescu returns email very quickly  :thumbup:

Do returning students get in automatically and are allowed to take whatever classes they want or do they only get admitted or are allowed to take some classes if their admission test and other application materials are good? I want to know this because I have WOOT lectures next week and might want to take olympiad geometry and have other things so I may not have that much time to spend doing admission test problems. I heard that they allowed returning students to come back to the camp and take whatever classes they wanted last year but I want to clarify for this year.

[quote="hurdler"]It says that we will know of whether we are accepted by the 18th...what if my application does not get received till the 16th or 17th? (I mailed). Will they really go through the applications that quickly?[/quote]
You should ask Dr. Titu Andreescu at titu@awesomemath.org

[quote="hurdler"]It says that we will know of whether we are accepted by the 18th...what if my application does not get received till the 16th or 17th? (I mailed). Will they really go through the applications that quickly?[/quote]

I wouldn't worry about it. Your application will get a fair look.


floor function
continued fraction
number theory unsolved
number theory

can anyone explain what the most appropriate way is to solve the pells equation?

Develope the continued fraction of $ \{\sqrt n\} : \equal{} \sqrt n \minus{} \lfloor \sqrt n \rfloor$. It's periodic, and if it's period is $ d$, then take the $ d$-th approximation $ \frac xy$. It's the smallest solution to $ x^2\minus{}ny^2 \equal{} \pm 1$. All other solutions are up to sign a power of that one, so of type $ x^\prime \plus{} y^\prime \sqrt n \equal{} \pm (x\plus{}y \sqrt n)^k$.
Also take a look at http://en.wikipedia.org/wiki/Pell%27s_equation .

aadil you could post these topics in the open section ok. Please post only problems here.

The "Open Problems" section is for open problems and not for questions. You can post this kind of question either here ("Unsolved Problems") or in "Theorems and Formulas", the latter may be the better one.


How can I factor the following,

x^4 .(y-z) + y^4 .(x-z) + z^4 .(x-y)

I feel like it shouldn't be very difficult but I'm completely at sea.

${(y-z) x^{4}+(x-y) z^{4}+y^{4}(x-z)=L}$

${y x^{4}-z x^{4}+y^{4}x+z^{4}x-y z^{4}-y^{4}z=L}$
${\left(z^{4}x-z^{4}y\right)+\left(x^{4}y-x^{4}z\right)+\left(y^{4}x-y^{4}z\right)=L}$
${(y-z) x^{4}+(x-y) z^{4}+y^{4}(x-z)=L}$

It means that we cannot factor it in a more simply way.On possible way is

${(x-z) \left(y^{4}+z^{3}y+x (y-z) z^{2}+x^{3}(y-z)+x^{2}(y-z) z\right)=L}$

But it is more hard that the first.

Another possible is

${(y-z) x^{4}+(x-y) z^{4}+y^{4}(x-z)}$

The expression you posted isn't cyclic -- was that intentional, or an error?  As Thales notes, the current expression has a factor of $x-z$ in it, but the remaining part doesn't seem to factor.  However, if you were going for the cyclic form, then it has factors of $x-y$, $y-z$ and $z-x$.

It was intentional, and thanks, bear with just another question please.
If it were cyclic, as,
L=x^4 . (y-z) + y^4 .(z-x) + z^4 .(x-y)

then,
L=(x-y)(z-x)(y-z).F

Where F is a factor. How would I find F?

So I return to the subject of cyclic sums yesterday, the non cyclic expression here put me off for a bit but...

There's a bunch of facts that you can assume about the factors of cyclics,
for example,
E = a^3 .(b-c) + b^3.(c-a) + c^3.(a-b)
treating this as a polynomial in a is..well pretty basic, but note how,
E= a^3 (b-c) + a(c^3 - b^3) + bc(b^2 - c^2)
  = (b-c)[a^3 -(b^2 +bc +c^2)a + bc(b+c)]
Now the expression equivalent to E/(b-c) admits as roots(in a of course) only,
+/-b , +/-c , +/-bc , +/-(b+c) , +/-c(b+c) , +/-b(b+c)

so now we're frowning at the possibilities, but note how a whole bunch of possibilities are eliminated, eg- the possible factors
(a-bc) and (a-b-c) are not symmetric.
The remaining factors are just those required, we conclude,
E= (b-c)(a-b)(a-c)(a+b+c)

Thanks for listening to the rant, this seemed more elegant than the textbook solution, at least in this case. :wink:

Yes that is the best way to deal with the given form of cyclic expressions for n>3,
However the first expression in question was not cyclic and has an inconvenient factorisation as shown by others.


show that the only numbers, $ n$, that have the property that either $ 5n^2\minus{}4$ or $ 5n^2\plus{}4$ are perfect squares are the fibonacci numbers (the converse is also true).

[quote="jinmadaxia"]the only numbers  is  1.
5n^2-4=a^2,5n^2+4=b^2,(0<a<b)we have b^2-a^2=(b+a)(b-a)=8.
because  b+a>b-a>0,so b+a=8,b-a=1 or b+a=4,b-a=1,so only b=3,a=1,n=1.
the converse is also true[/code][/quote]
No it's definitely wrong since the problem just states "either...or", which means we just need 1 out of 2 to be perfect square. Another counter example for your solution is n=2, which also gives 5.2^2-4=16 is perfect square.
I temporarily can prove the converse(if it's true) by using the general formula of the Fibonacci numbers with algebraic manipulation. But the main problem is hard for me, especially when I don't see why the density of those numbers that give either 5n^2-4 or 5n^2+4 are always Fibonacci numbers. :o


superior algebra
superior algebra unsolved

Hi
How can I show that 2 is a prime in Z_5

Thanks

Indeed. In every standard interpretation I can think of (and there's not just one), $ 2$ is invertible.


Mikhail currently has a balance of $ \$5400$ in his bank account. This is $ 60 \%$ of the balance he had when he opened the account. How much money will be in his account when the balance is down to $ 35\%$ of the balance he had when he opened the account?

Our first step is to find the amount of money Mikhail had when he originally opened his account. We know he currently has $ \$5400$ and that this is $ 60\%$ of the balance when he opened the account. Let $ x$ = the amount of money he originally had in his account when he first opened it.

$ 0.6x \equal{} 5400$

$ \frac{0.6x}{0.6} \equal{} \frac{5400}{0.6}$

$ x \equal{} 9000$

(Note: Generally, when you know that a certain number $ a$ is $ x\%$ of another number $ b$, just divide $ a$ by $ \frac{x}{100}$. So say $ 5$ is $ 50\%$ of $ b$. Divide $ 5$ by $ 0.5$ and you'll get $ 10$. So $ b$ in our example would be $ 10$.)

Now that we know what the balance was when Mikhail opened the account, we can find $ 35\%$ of this to get our answer.

$ 9000 \times 0.35 \equal{} 3150$

So our answer is $ \boxed{\$3150}$


linear algebra
matrix
trigonometry
advanced fields
advanced fields unsolved

How to show that every $ P\in SO(2)$ is of the form
\[ \begin{bmatrix}\cos\varphi &\minus{}\sin\varphi\\ \sin\varphi &\cos\varphi\end{bmatrix}?\]
I found a proof but it requires 23 lines of calculations so I was thinking if there is some other proofs than brute force.

brute force, 23 lines?!

just take (a,b;c,d) in SO(2), thus a,b,c,d are real numbers satisfying ad-bc=1, ab+cd=0, a^2+c^2=1=b^2+d^2. put these equations together and conclude a=d, b=-c.


It was posted before in "Number Theory" but it is still unsolved.

For rationals a,b:
S(a,b)={a(x^2)+b(y^2): x,y are rationals}.
let a,b,c,d be rational numbers different form 0. Show that:

S(a,b)=S(c,d) iff ab/(cd) is the square of a rational number
and there is q (different form 0) in S(a,b) and in S(c,d).

[quote]and there is q (different form 0) in S(a,b) and in S(c,d).[/quote]
i don't understand what this means :maybe: ....what is q and what is meant by 'there is q in S(a,b)'.....please clarify


Consider the equation x^2 + y^5= z^3 where x, y, and z, are positive integers.

A) Can you give at least three solutions to it ?

B) Determine whether or not there is an infinite number of solutions.

see http://www.kalva.demon.co.uk/short/soln/sh976.html  :wink:


prove that a square cannot be divided into odd number of congruent triangles

You can find a solution [url=http://www.mathlinks.ro/viewtopic.php?t=80752]here[/url].

My problem is a special case of that problem. I think the solution of my problem might be much simpler. I haven't solved this though.


function
integration
calculus
derivative
algebra
functional equation
calculus computations

Let the function $ f$ have the property

$ f(x)f'(\minus{}x)\equal{}f(\minus{}x)f'(x)$  for all $ x$
  and $ f(0)\equal{}3$

Proove that

[a] $ f(x)f(\minus{}x)\equal{}9$

[b] $ f(x)>0$ for all $ x$

[c] Find $ \int_{\minus{}2000}^{2000}\frac{1}{3\plus{}f(x)}dx$

[quote="giannis"]Let the function $ f$ have the property

$ f(x)f'( \minus{} x) \equal{} f( \minus{} x)f'(x)$  for all $ x$
  and $ f(0) \equal{} 3$

Proove that

[a] $ f(x)f( \minus{} x) \equal{} 9$

[b] $ f(x) > 0$ for all $ x$

[c] Find $ \int_{ \minus{} 2000}^{2000}\frac {1}{3 \plus{} f(x)}dx$[/quote]

[a]$ g(x) \equal{} f(x)f(\minus{}x) \Rightarrow g'(x) \equal{} f'(x)f(\minus{}x) \minus{} f'(\minus{}x)f(x) \equal{} 0 \Rightarrow g(x) \equal{} c, c\in \mathbb{R}$
To get c, substitute x=0 to get $ f(0)^2 \equal{} 9 \equal{} c$

[quote="isomorphism"]I could not get [b], but...[/quote]
$ f$ is differentiable, hence continuous. The functional equation $ f(x)f(\minus{}x)\equal{}9$ implies that $ f(x)$ is never zero. There is one point at which $ f(x)>0.$ Therefore, by the Intermediate Value Theorem, $ f(x)>0$ everywhere.

You need this in order to justify the manipulations you did in computing the answer to c). In fact, you need this just to claim that that integral exists.

[quote="Kent Merryfield"]$ f$ is differentiable, hence continuous. The functional equation $ f(x)f( \minus{} x) \equal{} 9$ implies that $ f(x)$ is never zero. There is one point at which $ f(x) > 0.$ Therefore, by the Intermediate Value Theorem, $ f(x) > 0$ everywhere.[/quote]
I had thought about this at first but soon got carried away. At the end I was trying to manipulate derivatives and f(x)f(-x) = 9 and I lost track :(

[quote="Kent Merryfield"]You need this in order to justify the manipulations you did in computing the answer to c). In fact, you need this just to claim that that integral exists.[/quote]

Actually I am not familiar with analysis beyond sequences and sums. So could you please explain where I will be needing f(x) > 0 for the integral?

Thanks again,
Spai

What you really need is $ f(x)>\minus{}3,$ so that the denominators in the integrals don't go to zero and so that the integrals (of continuous functions on a closed bounded interval) converge. Also at one point, you replaced $ f(\minus{}x)$ by $ \frac{9}{f(x)}$ - it helps to know that that fraction exists. Once you know the integral exists, you can then flip it around, add pieces together, and so on.

[quote="Kent Merryfield"]What you really need is $ f(x) > \minus{} 3,$ so that the denominators in the integrals don't go to zero and so that the integrals (of continuous functions on a closed bounded interval) converge. Also at one point, you replaced $ f( \minus{} x)$ by $ \frac {9}{f(x)}$ - it helps to know that that fraction exists. Once you know the integral exists, you can then flip it around, add pieces together, and so on.[/quote]

Oh so essentially I needed $ f(x) \neq 0, \minus{}3$. $ f(x) \neq 0$ is quite obvious, but $ f(x) \neq \minus{}3$ is unfortunately not. I see where I needed f(x) > 0.

Thanks,
Spai


There is the finite set M with m elements und 1986 further sets M_1, M_2, M_3, ... , M_1986. Each of them contains more than m/2 elements from M.

Prove that one has not to mark more than ten elements from M so that each set M_i (i = 1,...,1986) has at least one marked element.

[i]Bundeswettbewerb (BWM), 1986, stage *, problem *[/i]

Ok, let's solve another old unsolved problem....

For clarity, let's denotes the sets A_i instead of M_i....

Let M = {x_1,...,x_m}. For each i, let n_i be the number of sets A_j such that x_i belongs to A_j.
Then   \sum {1 \leq i \leq m} n_i =  \sum {1 \leq j \leq 1986} |A_j| > 1986*m/2 = 993m.
It follows that t least one of the n_i's is greater than 993. Say n_1  \geq 994, and pick x_1.

Clearly, if n_1 = 1986, we are done. So, assume that n_1 < 1986.

Let p_1 = 1986.
Wlog, we may assume that A_1,....,A_(p_2) are those of the A_j's which do not contains x_1.
Then p_2 = p_1 - n_1 = 1986 - n_1  \leq 992.
Note that each of the sets A_1,...,A_(p_2) contains more than the half of the elements of M-{x_1}. Thus, we may use the same reasoning as above, and assume that n_2 > (p_2)/2. So we pick x_2, and assume that A_1,...,A_(p_3) are those of the A_i's which do not contain x_1 nor x_2 (if there is no such A_i, we are done).
Note that, since they are integers, we have n_2 \geq [(p_2)/2] + 1, where [.] denotes the integer part.
And p_3 = p_2 - n_2 \leq p_2 - [(p_2)/2] - 1.

By this reasoning, we construct a sequence p_i of nonegative integers such that, for each i :
p_(i+1)  \leq p_i - [(p_i)/2] - 1.

If p_i = 0 for some i \leq 10 then we are done.
So assume that it is not the case. Thus p_1 = 1986, p_2  \leq 992, p_3  \leq 495, p_4  \leq 247, p_5 \leq 123, p_6 \leq 61, p_7 \leq 30, p_8 \leq 14, p_9  \leq 6 and p_10 \leq 2.
At each of the first nine steps, we picked an element of M, say x_1,...,x_9. And, each of the A_i's with i > p_10 has at least one element in {x_1,...,x_9}.

Now, since 1 \leq p_10  \leq 2, we have two cases :
_ If p_10 = 1, then select any element from A_1 to be x_10.
_ If p_10 = 2, then there are still two sets A_1 and A_2 for which we do not have select any element. But |A_1| > m/2 and |A_2| > m/2, then A_1 and A_2 do have at least one element in common. So select x_10 to be such a common element. And we are done.

Pierre.


Prove that
$\sum_{n=1}^{\infty}\frac{x^{2n-1}}{1-x^{2n}}\equiv \sum_{n=1}^{\infty}\left(\frac{1}{1-x^{2n-1}}-\frac{1}{1-x^{2n}}\right)$

where x is arbitrary fixed number different than $\pm 1$

This equality is taken from the book, but I think it's wrong (typo ?).
When you write out the first partial sums for the left hand side and the right hand side and plug in some values (say x=2) the sums are not the same.
The book says that the n-th partial sum equals to $\frac{1}{1-x}-\frac{1}{1-x^{2n}}$, but the terms don't cancel out in the RHS.

Sorry. Bad copy and paste case. Now it's correct.


Let $a,b,c>0,a+b+c=3$ prove that $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\geqslant ab+bc+ca$ thanks.

I think is similar with:
http://www.mathlinks.ro/Forum/viewtopic.php?t=105823

Vasc generalized this: For all $a,b,c\in\mathbb{R}^{+}$ such that $a+b+c=3$ we have \[a^{r}+b^{r}+c^{r}\geq ab+bc+ca\] where \[r\geq\frac{\log9-\log8}{\log3-\log2}\approx0.29048870864854522302381\]


search probability AMC AIME USA(J)MO USAMO USAJMO

So I am working through the Art and Craft of Problem Solving. I'll label the problems as they are found in the book. If I don't find the problem using the search feature and I think its interesting, I will post it (unless told otherwise).

This is a good intermediate level problem:

When $ n$ standard six -sided dice are rolled, the probability of obtaining a sum of $ 1994$ is greater than zero and is the same as the probability of obtaining a sum of $ S$. What is the smallest possible value of $ S$?

Interesting problem, veezbo! Could you please post some more such problems (and maybe we can start a marathon)?  :lol:

Suppose there are $ a_1,a_2,...,a_6$ such that
$ \sum_{k\equal{}1}^6 k*a_k\equal{} 1994$ and $ \sum a_k\equal{}n$

Since the probability is > 0, we must have $ 6n\ge1994\implies n\ge333$ and $ n\le1994$. So $ 1994\ge n\ge 333$.

For $ S$ to have equal probability, we must have

$ \sum_{k\equal{}1}^6 (7\minus{}k)*a_k\equal{}S$.

So $ 1994\plus{}S\equal{}7n\implies 7*1994\ge1994\plus{}S\ge7*333\implies S\ge7*333\minus{}1994\equal{}337$

Nice solution. Much more legit than mine, which just deals with the fact that there is a bijection between adding 4 to 333 1's and subtracting 4 from 333 6's.

As for the problems, I guess I could post them. It might be of more help to people if I put them in separate threads (but only if the moderator is okay with this). I posted this one because the 1994 AHSME (which this problem is from) isn't available on AoPS.

Buuuut, we could have a marathon with the contest problems that are available here. From chapter 1, it seems like there are some pretty good problems that we could start off with (AMC, AIME, and USAMO level).

This is a tiny bit unrelated, but what I have noticed is this:

In ACoPS, many of the earlier chapters have problems that aren't easily solved, but if you come back to them after finishing a relevant chapter, then you will definitely see a new path. This was indicated in the introduction of the book and is slowly proving itself.

Elefont, which part do you not understand?
Veezbo, yes I highly agree to post problems that are not available in the resource section (like the early AHSME and ARML)... since I don't see the point of reposting the archived problems with their solutions available already...
So does anyone mind to start a marathon on problems around intermediate level? :?:

[quote="Elefont"]can somebody explain mathwizarddude's explanation a little more? thanks!

also, veezbo, why is USAMO harder to qualify for this year?[/quote]

I do not think it would be easiest for me to explain mathwizarddude's solution.

USAMO is harder to make this year because of the implementation of the USAJMO (which I have no interest of making), cutting the number of qualifiers drastically. The index will be raised most likely to somewhere between 210-220, so my goal is to have a 220 index somehow.


[quote="james4l"]I believe there are 21.

there are 3 rhombi for every equilateral triangle nxn with even n, there are 6 2x2 triangles, 1 4x4 triangle, so 7 triangles, so 7*3=[b]21[/b] rhombi[/quote]
There's some rhombi with longer ones.

If there's longer ones, it has to be more than 21, therefore being more than 20...


calculus
integration
logarithms
calculus computations

Evaluate $ \int_0^1 {\left( {\frac{{\ln u}}{{9 \minus{} u}} \minus{} \frac{{\ln u}}{{8 \plus{} u}}} \right)\,du} .$

For a positive real number $ a$ define $ F(x)\equal{}\frac{x\, \ln x}{a\plus{}x}\minus{}\ln(a\plus{}x)$

then $ F'(x)\equal{}\frac{\ln(x)\plus{}1}{a\plus{}x}\minus{}\frac{x\ln x}{(a\plus{}x)^2}\minus{}\frac{1}{a\plus{}x}\equal{}\frac{a\, \ln x}{(a\plus{}x)^2}$

hence $ a\int_0^1 \frac{\ln x}{(a\plus{}x)^2}\, dx\equal{}F(1)\minus{}F(0)\equal{}\ln(a)\minus{}\ln(a\plus{}1) \qquad *$

Similarly define $ G(x)\equal{}\frac{x\, \ln x}{a\plus{}1\minus{}x}\plus{}\ln(a\plus{}1\minus{}x)$

then $ G'(x)\equal{}\frac{\ln(x)\plus{}1}{a\plus{}1\minus{}x}\plus{}\frac{x\, \ln x}{(a\plus{}1\minus{}x)^2}\minus{}\frac{1}{a\plus{}1\minus{}x}\equal{}\frac{(a\plus{}1)\, \ln x}{(a\plus{}1\minus{}x)^2}$

hence $ (a\plus{}1)\int_0^1 \frac{\ln x}{(a\plus{}1\minus{}x)^2} \, dx\equal{}G(1)\minus{}G(0)\equal{}\ln(a)\minus{}\ln(a\plus{}1) \qquad **$

With $ *$ and $ **$ you obtain

$ \int_0^1 \left[\frac{\ln x}{(a\plus{}x)^2}\minus{}\frac{\ln x}{(a\plus{}1\minus{}x)^2}\right] dx\equal{}\Big[\ln(a)\minus{}\ln(a\plus{}1)\Big]\, \left(\frac{1}{a}\minus{}\frac{1}{a\plus{}1}\right)$

After that integrate both sides with respect to $ a$ :

$ \int_0^1 \left(\frac{\ln x}{a\plus{}1\minus{}x}\minus{}\frac{\ln x}{a\plus{}x}\right) dx\equal{}\frac12 \Big[\ln(a)\minus{}\ln(a\plus{}1)\Big]^2\plus{}C$

whereat $ C\equal{}0$ . Concerning this let $ a$ tend to $ \infty$ .


Hello
Let's say we have a system with The Sun, Planet X and Planet Y. Planet's X and Y mass is very smaller than the mass of the Sun, so we consider that the Sun is is "fixed".

So Planet X and Planet Y will be influenced by The Sun and eachother, so for each planet:

F_Sun + F_other_planet = m_planet a_planet

Where F_ is the gravitacional force. The question is regarding the movement of these planets. Is the trajectory fixed, Or are there small variations in the trajectory?

Thanks.

The trajectories are fixed.  However, if one views a planet traectory without consideration of the the planet to planet interaction, one will find that the trajectories have variations.  If the planets are large/close enough the effect of a close interaction can be very significant.

I note that some of the recently discovered rings actually twist about themselves.  If one were looking for an interesting orbit and not an interesting place to live, one could make up a system where the planets were linked and the linked planets orbited the sun, sort of like the earth/ moon system, where the center of the system is close to the surface of the earth, not nearly the center of the earth, as most of earth's inhabitents "know." (Obscure reference to MIB movie, what people know.)

[quote="wizard2378"]The trajectories are fixed.  However, if one views a planet traectory without consideration of the the planet to planet interaction, one will find that the trajectories have variations.  If the planets are large/close enough the effect of a close interaction can be very significant.
[/quote]

Hello, The reason I asked is this: in a simulation program based on the Gravitacional Force one of the planet's orbits didn't make a fixed eliptic orbit, when each "loop" ends the new elipse isn't drawn above the previous one. The other planet does maintain a fixed orbit. So I wanted to know if that makes some sense.Thanks.


function
real analysis
real analysis unsolved

Show that there is no continuous function $f : R\to R$ such that for every $c \in R$ there exists exactly two solutions to the equation $f(x)=c$.

Assume such a function exists, and write $x_{0},x_{1}$ the solutions of the equation $f(x)=0$ (we assume$x_{0}<x_{1}$). From the hypothesis, without loss of generality, we can assume that $\forall x < x_{0}, f(x)<0$ and $\forall x >x_{1}, f(x)>1$ (otherwise $f$ has a lower bound or an upper bound, which can't be). Also, $f$ is either positive or negative on $(x_{0},x_{1})$. We can assume it's positive. In this case, it is easy to show that $f(x)=\epsilon >0$ has at least $3$ solutions for some $\epsilon>0$ small enough.


LaTeX
inequalities
inequalities proposed

yes, I have some problems about latex . thank you.

It looks like there will be a lot of expansion, no nice solution
is this true?

\[4\sum_{cyc}\frac{(3a+b)a^{2}}{b+c}\geq 11\sum_{}a^{2}-3\sum{}bc\]  \[4\sum_{cyc}\frac{(3a+b)a^{2}}{b+c}=4\sum_{cyc}\frac{((3a+b)a)^{2}}{(b+c)(3a+b)}\geq 4\frac{(\sum 3a^{2}+ab)^{2}}{\sum b^{2}+7\sum ab}\geq 11\sum_{}a^{2}-3\sum{}bc\] multiply out and you've got it
see thats what i meant by messy proof (lol probly nicer way, can you post it?)

[quote="me@home"]
$4\frac{(\sum 3a^{2}+ab)^{2}}{\sum b^{2}+7\sum ab}\geq 11\sum_{}a^{2}-3\sum{}bc$[/quote]
This kind of thing isn't actually so bad -- you just have to get used to abbreviations. If you let $x = \sum a^{2}$ and $y = \sum ab$, we're just proving
$4\frac{(3x+y)^{2}}{x+7y}\ge 11x-3y$
$36x^{2}+24xy+4y^{2}\ge 11x^{2}+74xy-21y^{2}$
$25x^{2}+25y^{2}\ge 50xy$
AM-GM :D You don't even have to use that $x \ge y$.

Nice solution, I'll have to use that more often (naming things variables to make the problem seem trivial :D ) For all these types of "dumbassing" solutions can you always get an AM-GM solution? (Or at least a solution that is part AM-GM and part x>=y)

[quote="me@home"] For all these types of "dumbassing" solutions can you always get an AM-GM solution? (Or at least a solution that is part AM-GM and part x>=y)[/quote]
I think "dumbassing" by definition means expanding and solving an inequality using only AM-GM (perhaps in the form of Muirhead) and Schur. So to answer your question, no, sometimes you might need Schur.

My solution is here.
I create this ineq from expanding the cyclic sum of (b+c-2a)^2 (3a+c)/(b+c)
sorry for inconvenient i'm trying to use tex.


In $ \triangle ABC$, point $ E$ is on $ AB$ and $ F$ is on $ BC$.

$ EC$ and $ AF$ intersect at $ D$.

Also,
$ BF\plus{}FD\equal{}10$
$ BE\plus{}ED\equal{}10$
$ BC\plus{}BD\equal{}20\times\sqrt(3)$

what is value of $ BA\plus{}AD$?

[quote="Farenhajt"]Just a question: is it really $ BC\plus{}BD$? No typo? (Perhaps it should be $ BC\plus{}CD$?)[/quote]

my mistake... I have just edited it. Sorry for the mistake.

[quote="xxxyyyy"][quote="Farenhajt"]Just a question: is it really $ BC\plus{}BD$? No typo? (Perhaps it should be $ BC\plus{}CD$?)[/quote]

my mistake... I will edit it. Sorry for the mistake.[/quote]

I am not able to edit it... can any mod help?

[quote="Farenhajt"]Just type the correct statement here, and a mod will paste it back up there.[/quote]

$ BC\plus{}CD \equal{} 20\times\sqrt(3)$


A lucky integer is a positive integer which is divisible by the sum of its digits. What is the least positive multiple of 9 that is not a lucky integer?

let's start listing numbers multiples of 9

9
18
27
36
45
54
63
72
81
90
99

we see that ever multiply of 9 up to 90 is a lucky number
however, 99 is not divisible by the sum of its numbers, so our answer is $ \boxed{99}$

You notice that the multiples of 9 from 9 to 90 always their digits always add up to 9. So you starts with the multiples of 9 after 90.
And the first number is the answer which is 99.


For postive constant $ r$, denote $ M$ be the all sets of complex $ z$ that satisfy $ |z \minus{} 4 \minus{} 3i| \equal{} r$.

(1) Find the value of $ r$ such that there is only one real number which is belong to $ M$.

(2) If $ r$ takes the value found in (1), find the maximum value $ k$ of $ |z|$ for complex number $ z$ which is belong to $ M$.

[hide]Let $ z \equal{} a \plus{} bi$. Then $ (a \minus{} 4)^2 \plus{} (b \minus{} 3)^2 \equal{} r^2$ and we must have only one solution when $ b \equal{} 0$, which means $ r \equal{} 3,a \equal{} 4$ since $ (a \minus{} 4)^2 \equal{} r^2 \minus{} 3^2$. To find the maximum of $ \sqrt{a^2 \plus{} b^2}$, or the max of the distance of $ (a,b)$ from the origin, we must have the point $ (a,b)$ being collinear with the origin and the center of the circle $ (a \minus{} 4)^2 \plus{} (b \minus{} 3)^2 \equal{} 3^2$ by the triangle inequality. So $ \max(\sqrt{a^2 \plus{} b^2}) \equal{} \sqrt {3^2 \plus{} 4^2} \plus{} 3 \equal{} 8$.[/hide]


inequalities
function
logarithms
inequalities proposed

Let $ a,b,c\ge 0$. Prove the following inequality:

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge \frac{2}{3}(ab\plus{}bc\plus{}ca)\left(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\right)$

[quote="Mashimaru"]Let $ a,b,c\ge 0$. Prove the following inequality:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac {2}{3}(ab \plus{} bc \plus{} ca)\left(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right)$[/quote]
It's equivalent to $ \sum_{sym}(a^5b\plus{}2a^4b^2\plus{}a^4bc\minus{}2a^3b^3\minus{}2a^3b^2c)\geq0,$ which is obviously true.

Nice ineq,Mashimaru!
And it could be solve directly using the Nesbit inequality:
Note that:$ \sum \frac {a}{b \plus{} c} \equal{} \sum \frac {ab \plus{} ac}{(b \plus{} c)^2}$
So $ \frac {3}{2}(LHS \minus{} RHS) \equal{} \frac {1}{2}(\sum \frac {a}{b \plus{} c}) \minus{} \sum \frac {bc}{(b \plus{} c)^2}$
By Nesbit ineq we have:$ \frac {1}{2}(\sum \frac {a}{b \plus{} c}) \geq \frac {3}{4}$
And by AM-GM we have $ \frac{3}{4} \geq \sum \frac{bc}{(b\plus{}c)^2}$
Q.E.D

[quote="Mashimaru"]Let $ a,b,c\ge 0$. Prove the following inequality:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac {2}{3}(ab \plus{} bc \plus{} ca)\left(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right)$[/quote]

It's my Inequality on MS forum, here http://forum.mathscope.org/showthread.php?t=4794   :wink:

Intersting coincidence, [b]dduclam[/b] :lol:

More weighted problem:

Let $ a,b,c\ge 0$. Find the greatest positive real number $ k$ such that:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge$
$ \ge \frac {2}{3}(ab \plus{} bc \plus{} ca)\left(\frac {1}{(a \plus{} b)^{2}} \plus{} \frac {1}{(b \plus{} c)^{2}} \plus{} \frac {1}{(c \plus{} a)^{2}}\right) \plus{} k\frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$

@[b]Mathias_DK[/b]: Yes, I meant the above inequality. Sorry for mistyping :oops:

[quote="Mashimaru"]Intersting coincidence, [b]dduclam[/b] :lol:

More weighted problem:

Let $ a,b,c\ge 0$. Find the greatest positive real number $ k$ such that:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge$
$ \ge \frac {2}{3}(ab \plus{} bc \plus{} ca)\left(\frac {1}{(a \plus{} b)^{2}} \plus{} \frac {1}{(b \plus{} c)^{2}} \plus{} \frac {1}{(c \plus{} a)^{2}} \plus{} k\frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}\right)$[/quote]
$ k \equal{} 0$ ;)

I suppose you meant:
$ \ge \frac {2}{3}(ab \plus{} bc \plus{} ca)\left(\frac {1}{(a \plus{} b)^{2}} \plus{} \frac {1}{(b \plus{} c)^{2}} \plus{} \frac {1}{(c \plus{} a)^{2}}\right) \plus{} k\frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$ which is a bit harder :)

[quote="Stephen"]if a, b, c>0 and a, b, c are real numbers, prove that
(a/(b+c))+(b/(c+a))+(c/(a+b))≤(a²/(b²+c²))+(b²/(c²+a²))+(c²/(a²+b²)).[/quote]

[b]Stephen's problem:[/b] Let $ a,b,c \ge 0$. Prove that:
$ \frac{a^2}{b^2\plus{}c^2}\plus{}\frac{b^2}{c^2\plus{}a^2}\plus{}\frac{c^2}{a^2\plus{}b^2} \ge \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}$

[b]Generalized problem:[/b] Let $ a,b,c \ge 0$ and $ r \ge s >0$. Prove that:

$ \frac{a^r}{b^r\plus{}c^r}\plus{}\frac{b^r}{c^r\plus{}a^r}\plus{}\frac{c^r}{a^r\plus{}b^r} \ge \frac{a^s}{b^s\plus{}c^s}\plus{}\frac{b^s}{c^s\plus{}a^s}\plus{}\frac{c^s}{a^s\plus{}b^s}$

[b]Solution:[/b]

It is enough to prove that $ f(k)\equal{}\frac{a^k}{b^k\plus{}c^k}\plus{}\frac{b^k}{c^k\plus{}a^k}\plus{}\frac{c^k}{a^k\plus{}b^k}$ is an increasing function on $ (0;\plus{}\infty)$.

WLOG, assume that $ a \ge b \ge c$. We have:

$ f'(k)\equal{}\sum_{cyc}^{a,b,c} \frac{a^k(b^k\plus{}c^k)\ln{a}\minus{}a^k(b^k\ln{b}\plus{}c^k\ln{c})}{(b^k\plus{}c^k)^2}$
$ \equal{}\sum_{cyc}^{a,b,c} a^kb^k(\ln{a}\minus{}\ln{b})\left(\frac{1}{(c^k\plus{}b^k)^2}\minus{}\frac{1}{(c^k\plus{}b^k)^2}\right) \ge 0$

Moreover, $ f'(k)\equal{}0 \Leftrightarrow a\equal{}b\equal{}c$, this implies that in general, $ f(k)$ is an increasing function on $ (0;\plus{}\infty)$ and we are done!

Besides, for all $ a,b,c\ge 0$ we have following Inequality
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\le \frac {2}{3}(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right)
\]


Let $a,b,c$ be the side lengths of a triangle. Show that
\[\frac{a^{2}+b^{2}}{2c}+\frac{b^{2}+c^{2}}{2a}+\frac{c^{2}+a^{2}}{2b}\leq\frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}\]

Notice that \[\sum\frac{bc}{a}\leq \frac{a^{2}+b^{2}}{2c}+\frac{b^{2}+c^{2}}{2a}+\frac{c^{2}+a^{2}}{2b}\leq\frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}\]
So its a pretty novel expression. But other than that it is just too easy :P

Or for another nice thing to try: \[LHS=\sum_{s}\frac{a^{2}}{2c}\geq \frac{(\sum a)^{2}}{\sum a}=\sum a\]
to give an even weaker lower bound than before.

I also think there's a cauchy to prove the ineq. but I don't see it right now.

[quote="Jan"]The inequality is equivalent with:
\[\sum_{sym}{a^{4}}\geq \sum_{sym}{a^{3}b}\]
which is obviously true.

Note that $a,b,c$ don't have to be the sides of a triangle.[/quote]

I don't think they are equivalent and it's not that obvious to me. :wink:

[quote="me@home"]Notice that
\[\sum\frac{bc}{a}\leq \frac{a^{2}+b^{2}}{2c}+\frac{b^{2}+c^{2}}{2a}+\frac{c^{2}+a^{2}}{2b}\leq\frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}\]
So its a pretty novel expression. But other than that it is just too easy :P

[/quote]

I didn't see any proof there. :huh:

For Jan's, just multiply through by $2abc$.

And I'm not sure what me@home was doing; it seems that he is using the result of this problem to make weaker inequalities?

[quote="paladin8"]For Jan's, just multiply through by $2abc$.

[/quote]

Multiply through by $2abc$ I thought it would be:
\[2\sum_{sym}{a^{4}}\geq \sum_{sym}{a^{3}b}\]

, which is not "obviously" true. :roll:

Sorry for being not clear. I was just playing around with weaker inequalities and the result of the problem, like paladin said.

The result of mine jelly quoted follows really easily... I thought I could leave out proof. It is just noticing that $\frac{a^{2}+b^{2}}{2}\geq ab$ so that $\frac{a^{2}+b^{2}}{2c}\geq \frac{ab}{c}$... :huh:

Also in your response to paladin, jelly, notice that you have a typo; it should be
\[2\sum_{cyc}{a^{4}}\geq \sum_{sym}{a^{3}b}\]
instead of
\[2\sum_{sym}{a^{4}}\geq \sum_{sym}{a^{3}b}\]
And then notice that $2\sum_{cyc}{a^{4}}=\sum_{sym}a^{4}$.

Of course, it is obviously true by Muirhead; $[4,0,0]$ maximizes $[3,1,0]$.

EDIT: didn't see post above me

Um here's a slightly nicer rearrangement proof if Am-Gm isn't your fancy

\[\iff \frac{a^{3}}{bc}+\frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}+\frac{c^{3}}{ab}\geq \frac{a^{3}}{ac}+\frac{c^{3}}{ac}+\frac{b^{3}}{bc}+\frac{c^{3}}{bc}+\frac{a^{3}}{ab}+\frac{b^{3}}{ab}\]
For the sequences: \[A=(a^{3},a^{3},b^{3},b^{3},c^{3},c^{3}) \qquad B=(\frac{1}{bc},\frac{1}{bc},\frac{1}{ca},\frac{1}{ca},\frac{1}{ab},\frac{1}{ab})\]
We can tell that the left hand side is similarly sorted because:
\[a\geq b\geq c\implies a^{3}\geq b^{3}\geq c^{3}\implies \frac{1}{bc}\geq \frac{1}{ac}\geq \frac{1}{ab}\]

Yes that's exactly what I meant. I didn't remember the symbol name. So $\succ$ I guess short for succeeds? Then what is the other symbol? Is it something like $%Error. "lecc" is a bad command.
$

[quote="me@home"]

The result of mine jelly quoted follows really easily... I thought I could leave out proof. It is just noticing that $\frac{a^{2}+b^{2}}{2}\geq ab$ so that $\frac{a^{2}+b^{2}}{2c}\geq \frac{ab}{c}$... :huh:

[/quote]

I meant I didn't see any proof in the right part of your expression:
\[\sum\frac{bc}{a}\leq \frac{a^{2}+b^{2}}{2c}+\frac{b^{2}+c^{2}}{2a}+\frac{c^{2}+a^{2}}{2b}\leq\frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}\]
The left part is obviously true as you described. But how is the right part obviously true?? :huh:

Maybe it's really "obvious"....I got this in no obvious way. :maybe:

Multiply through by $2abc$, it's equivalent to prove
\[ab(a^{2}+b^{2})+bc(b^{2}+c^{2})+ca(c^{2}+a^{2})\leq 2a^{4}+2b^{4}+2c^{4}\]
By Cauchy and AM-GM,respectively, we have
\[a^{4}+b^{4}\geq \frac{1}{2}(a^{2}+b^{2})^{2}\geq ab(a^{2}+b^{2}) \]

\[b^{4}+c^{4}\geq \frac{1}{2}(b^{2}+c^{2})^{2}\geq bc(b^{2}+c^{2}) \]

\[c^{4}+a^{4}\geq \frac{1}{2}(c^{2}+a^{2})^{2}\geq ca(c^{2}+a^{2}) \]
Adding all these together and we're done!

As I said, I "assumed" that one without proof, just adopting Jan's basic Muirhead proof. We already went over how Muirhead is "obvious" :P

Your proof for this was nice since using cauchy is a "pretty" way of proving the muirhead.
However, the muirhead style proof using only Am-Gm is just
\[\frac34 a^{4}+\frac14 b^{4}\geq a^{3}b \]
by Am-Gm
after cyclically summing, one finds the inequality.

EDIT: oops, I meant symmetric summing lol

It seems that I'd better learn more about Muirhead. :blush:


algebra
polynomial
complex numbers
algebra unsolved

find x if

$ x^3\minus{}33x^2\plus{}27x\minus{}3\equal{}0$

[hide="hint"]Graph it, and if their are other imaginary roots use polynomial division.[/hide]

[quote="aktyw19"]find x if

$ x^3 \minus{} 33x^2 \plus{} 27x \minus{} 3 \equal{} 0$[/quote]

Newton gives :
0,704088191...
0,132474331...
32,16343748...

None is a rational
None is in the form $ p\plus{}q\sqrt r$
Cardan will give ugly formulas with complex numbers

Where is this problem from ?

[quote="aktyw19"]Is simplest  solution maybe?[/quote]

What do you mean ?

And, please, could you give us the origin of the problem, so we can maybe give you some directions ?

hello, one solution is of the form
$ x\equal{}11\plus{}\frac{28}{\sqrt[3]{\frac{1}{2} \left(37\plus{}i
   \sqrt{3}\right)}}\plus{}2 2^{2/3} \sqrt[3]{37\plus{}i \sqrt{3}}$
Sonnhard.

[quote="Dumel"]it's well known equation
there are 3 real roots:
$ tan^2 20^o$
$ tan^2 40^o$
$ tan^2 80^o$[/quote]

Nice :)

But I think it's a crazy olympiad problem since it's nearly impossible to find these values if you have never seen the equation before  :(

@sonnhard : Cardan gives very ugly formulas ....

[quote="Dumel"]it's well known equation
there are 3 real roots:
$ tan^2 20^o$
$ tan^2 40^o$
$ tan^2 80^o$[/quote]

Why such solution?

[i]Actually , this problem is very well - known

   It's idea is followed the formula :

   $ tg 3x \equal{} \frac{3tgx \minus{} tg^3 x }{1 \minus{} 3 tg^2 x}$  :)

   Just square two side
   [/i]

[quote="nguyenvuthanhha"]Actually , this problem is very well - known[/quote]
Not by me  :(  :blush:

the original problem was:
prove that $ tan^220^o$ [...]
are roots of equation [...]  :lol:


hello, by writing your equation in the form
$ (1\plus{}3\cos(2x))^2(257\plus{}132\cos(2x)\plus{}27\cos(4x))\equal{}0$
and solving this we find
$ x\equal{}\frac{1}{2}\left(\minus{}\text{arccos}(\minus{}\frac{1}{3})\plus{}2k\pi\right)$
$ x\equal{}\frac{1}{2}\left(\text{arccos}(\minus{}\frac{1}{3})\plus{}2k\pi\right)$
with $ k \in \mathbb{Z}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, by writing your equation in the form
$ (1 \plus{} 3\cos(2x))^2(257 \plus{} 132\cos(2x) \plus{} 27\cos(4x)) \equal{} 0$
and solving this we find
$ x \equal{} \frac {1}{2}\left( \minus{} \text{arccos}( \minus{} \frac {1}{3}) \plus{} 2k\pi\right)$
$ x \equal{} \frac {1}{2}\left(\text{arccos}( \minus{} \frac {1}{3}) \plus{} 2k\pi\right)$
with $ k \in \mathbb{Z}$.
Sonnhard.[/quote]

[hide]Thank you very much![/hide]

Hi! every One. Who can give a other solution?

If $ u\equal{}cos^2x$, then we have

$ u^4\plus{}\frac{1}{8}\cdot (1\minus{}u)^4\minus{}\frac{1}{27}\equal{}\frac{1}{216}\cdot (\minus{}1 \plus{} 3 u)^2 (19 \plus{} 6 u \plus{} 27 u^2)$


system of equations
algebra
Diophantine Equations
IMO
Cauchy-Schwarz inequality

Determine all real numbers a for which there exists positive reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $ \sum_{k=1}^{5} kx_{k}=a,$ $ \sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $ \sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$

Let $A$ the first sum, $B$ the second and $C$ the third.

from cauchy, we have $AC \ge B^2$. but, $AC = B^2$.

the equality in cauchy occur if the terms are proportional.

so, $jx_j/j^5x_j = ix_i/i^5x_i$ for all $i$,$j$ $\Longrightarrow i = j$, implossible.  :)

E.L

i believe the original question is non-negative reals..

also, i think there are six solns but some sources say there are only five..
n=0,1,4,9,16,25

Yes, the original question is.. "non-negative reals"...
I just added a solution on the main page

$a^2\sum_{k=1}^{5}{kx\textsubscript{k}}-2a\sum_{k=1}^{5}{k^3x\textsubscript{k}}+\sum_{k=1}^{5}{k^5x\textsubscript{k}}=a^3-2a^3+a^3$
$\sum_{k=1}^{5}{(a^2kx\textsubscript{k}-2ak^3x\textsubscript{k}+k^5x\textsubscript{k})}=0$
$\sum_{k=1}^{5}{k(a^2-2ak^2+(k^2)^2)x\textsubscript{k}}=0$
$\sum_{k=1}^{5}{k(a-k^2)^2x\textsubscript{k}}=0$
Because $x\textsubscript{k}\gneq0$:
* all $x\textsubscript{k}=0$ --> $a=0$
* one $x\textsubscript{k}\neq0$ and the others $x\textsubscript{i}=0$ --> $a-k^2=0 \iff a=k^2$
--> $a=\{0,1,2^2,3^2,4^2,5^2\}$

C-S directly gives:
$$\left(\sum_{k=1}^5kx_k\right)\left(\sum_{k=1}^5k^5x_k\right)^2\ge\left(\sum_{k=1}^5k^3x_k\right)^2.$$
But both sides are actually equal, so by analyzing the equality case of C-S, we have $k^5x_k=kx_k\cdot c$ for some constant $c$ and all $k\in\{1,\ldots,5\}$. Then $c=k^4$ for each $k\in\{1,\ldots,5\}$ which is absurd, so no such sequence exists.


Plz help me to solve the following questions:

Let $ f: \mathbb R^n\to\mathbb R$ so that $ f(x)\equal{}\exp\left(\frac{1}{|x|^2\minus{}1}\right)$   if $ |x|<1$ and $ f(x)\equal{}0$ if $ |x|\geq1$.

(a) Show that $ f\in C^\infty(\mathbb R^n)$.

(b) Does $ f$ is analytic on $ \mathbb R^n$ ?.


inequalities
induction
geometry
rectangle
combinatorics proposed
combinatorics

Four squares of $25 \cdot 25$ chessboard are called a quartet if their centers form a rectangular with sides parallel to the sides of the board. What is the maximum number of quartets which do not have any common squares ?

:D does anyone know where i can find a good website with notes on inequalities  :?:

[quote="manuel"]Count the number of edge disjoint 4-cycles in a complete bipartite graph each  partition having 25 vertices.[/quote]

Can u tell me the rest ??

Consider a bipartite graph. One partition represents the columns and the other represents the rows. If we label each partiton with the numbers $1,2,...,n$ an edge $ij$ represents the square that is in the intersection of column $i$ and row $j$. Quartets are clearly 4-cycles.

I think you should prove by induction that the number of edge disjoint 4-cycles in a $K_{n,n}$ is $[\frac{n}{2}]^{2}$ where $[x]$ denotes the integer part of $x$.

Note: I am assuming that "common sides" in the problem means a common side of equal length.
In other words, I am asumming that rectangles like these do not have common sides:
 _____
|_____|____
|__________|


I was just wondering what times everyone has.  I got 7 seconds this morning when I saw one for the first time

um
its been a long time
i think it was 16 sec or something using a 3x3x3 method
thats probably wrong

mybest is 26 :(

i use the 3 by 3 method too


algebra
polynomial
calculus
calculus computations

$ \frac{{\left( \begin{array}{l}
 n \\
 k \\
 \end{array} \right)}}{{q^n }}$

k is a fixed natural number and q > 1.

How can we show that this limit tends to 0 as n tends to infinity?

$ \binom{n}{k}$ is a $ k$th degree polynomial in $ n$ with fixed coefficients. This limit is simply a matter of exponentials growing faster than powers of $ n.$

I am not very strong with limits so tell me if this is essentially whether this is a rigorous argument.

$ \frac {{n^{\frac {k}{n}} }}{q} \to \frac {1}{q}$

and

$ \frac {1}{{q^n }} \to 0$

Hence

$ \frac {{n^k }}{{q^n }} \to 0$

A different argument is that

$ \frac{n}{{\sqrt[k]{q}^n }} \to 0$

so

$ \frac{{n^k }}{{q^n }} \to 0$


A parliament has $n$ senators. The senators form 10 parties and 10 committees, such that any senator belongs to exactly one party and one committee. Find the least possible $n$ for which it is possible to label the parties and the committees with numbers from 1 to 10, such that there are at least 11 senators for which the numbers of the corresponding party and committee are equal.

Let $ n_0$  senators have (#party)=(#committee)
$ n_1$  senators have (#party)-(#committee)$ \equiv 1(mod10)$
....
$ n_9$  senators have (#party)-(#committee)$ \equiv 9(mod10)$
$ n\equal{}n_0\plus{}n_1\plus{}..\plus{}n_9$
If n=101, exist $ n_k\geq 11$. Then re-label parties $ p \to (p\minus{}k)(mod10)$, and that $ n_k$ senators will have equal #party and #committee. Let also #10:=#0
If n=100 senators : one senator for each pair ((#party),(#committee))-we can't label


geometry
ratio
symmetry
trigonometry
Putnam
geometry proposed

In triangle ABC

K,L,M are points on BC,CA,AB respectively such that

AB/MB = BC/KC = CA/LA   =r

Lines AK,BL ,CM  intersect inside triangle ABC to form a second triangle PQR such that Q and P lie on AK with Q being the nearer point to A.

Find in terms of r the values of{CPK}/{CAB} and {PQR}/{CAB} where the notation {CPK} denotes the area of triangle CPK and so on for the other triangles

If the point $Q$ was closer to the vertex $A$ than the point $P$ on the line $AK$, it would be ridiculous to find the area ratio for the triangle $\triangle CPK$, because it is not one of the triangles in the figure. Hence, I assume that the point $P$ is closer to the vertex $A$. It does not matter, the interesting problem is to find the area ratio for the triangle $\triangle PQR$.

Any triangle $\triangle ABC$ can be transformed by a parallel projection (or by an affine transformation) of the plane into an equilateral triangle. Ratio of areas of any plane figures are preserved in this transformation. Ratios of line segments on the same lines are also preserved. Therefore, it is sufficient to find the desired ratio of areas for an equilateral triangle $\triangle ABC$.

Let $a = AB = BC = CA$ be the side length of the equilateral triangle $\triangle ABC$. Because of symmetry of the equilateral triangle, the triangles $\triangle ABK \cong \triangle BCL \cong \triangle CAM$ are congruent and the triangles $\triangle APM \cong \triangle BQK \cong \triangle CRL$ are also congruent. Hence, the angles $\measuredangle AKB = \measuredangle BLC = \measuredangle CMA$ are equal and the angles $\measuredangle KAB = \measuredangle LBC = \measuredangle MCA$ are also equal. In addition, the triangles $\triangle ABK \sim \triangle AQM$ are similar (the angle $\measuredangle KAB \equiv \measuredangle PAM$ is common and the angles $\measuredangle AKB = \measuredangle CMA \equiv \measuredangle PMA$ are equal).

Using the cosine theorem for the triangle $\triangle AKB$,

$AK^2 = AB^2 + BK^2 - 2 AB \cdot BK \cos{\widehat{ABK}}$

Since $\frac{BK}{BC} = \frac{AM}{AB} = 1 - \frac{MB}{AB} = \frac{r - 1}{r}$ and $\measuredangle ABK = 60^o$, we have

$AK^2 = a^2 + a^2 \frac{(r - 1)^2}{r^2} - a^2 \frac{r - 1}{r} = a^2 \frac{r^2 - r + 1}{r^2}$

$AK = a \frac{\sqrt{r^2 - r + 1}}{r}$

From similarity of the triangles $\triangle AKB, \triangle AMP$,

$\frac{AP}{AM} = \frac{AB}{AK}$

$AP =  a \frac{r - 1}{r} \frac {r}{\sqrt{r^2 - r + 1}} = a \frac{r - 1}{\sqrt{r^2 - r + 1}}$

$\frac{PK}{AP} = \frac{AK}{AP} - 1 = \frac{\sqrt{r^2 - r + 1}}{r} \frac{\sqrt{r^2 - r + 1}}{r - 1} - 1 = \frac{1}{r(r - 1)}$

The areas $|\triangle AKB|, |\triangle AMP|$ of the triangles $\triangle AKB, \triangle AMP$ are in the ratio equal to the square of their coefficient of similarity:

$\frac{|\triangle AKB|}{|\triangle AMP|} = \frac{AB^2}{AP^2} = \frac{r^2 - r + 1}{(r - 1)^2}$

The triangles $\triangle APC, \triangle PKC$ have a common altitude from the common vertex $C$ to the line $AK$. Hence, their areas $|\triangle APC|, |\triangle PKC|$ are in the ratio of their bases $AP, PK$:

$\frac{|\triangle APC|}{|\triangle PKC|} = \frac{AP}{PK} = r(r - 1)$

Denote the areas of the triangles $\triangle ABC, \triangle PQR, \triangle ABK, \triangle AMP$ as $\triangle_0 = |\triangle ABC|$, $\triangle_1 = |\triangle PQR|$, $\triangle_2 = |\triangle ABK|$, $\triangle_3 = |\triangle AMP|$. Because of symmetry of the equilateral triangle $\triangle ABC$, we obtain the following relation for these areas:

$\frac{|\triangle APC|}{|\triangle PKC|} = \frac{\triangle_2 - \triangle_3}{\triangle_1 + \triangle_2 - 2 \triangle_3} = r(r - 1)$

$\triangle_1 = \frac{\triangle_2 - \triangle_3}{r(r - 1)} - \triangle_2 + 2 \triangle_3$

Calculating the triangle areas $\triangle_0, \triangle_2, \triangle_3$ and their ratios:

$\triangle_0 = \frac{a^2 \sqrt 3}{4}$

$\triangle_2 = \frac 1 2 AB \cdot BK \sin{\widehat{ABK}} = \frac{a^2 \sqrt 3}{4} \frac{r - 1}{r}$

$\triangle_3 = \triangle_2\ \frac{(r - 1)^2}{r^2 - r + 1} =  \frac{a^2 \sqrt 3}{4} \frac{(r - 1)^3}{r(r^2 - r + 1)}$

$\frac{\triangle_2}{\triangle_0} = \frac{r - 1}{r}$

$\frac{\triangle_3}{\triangle_0} = \frac{(r - 1)^3}{r(r^2 - r + 1)}$

$\frac{\triangle_1}{\triangle_0} = \frac{|\triangle PQR|}{|\triangle ABC|} = \frac{1}{r^2} - \frac{(r - 1)^2}{r^2(r^2 - r + 1)} - \frac{r - 1}{r} + \frac{2(r - 1)^3}{r(r^2 - r + 1)}$

After some dull algebra, we get a simple expression

$\frac{|\triangle PQR|}{|\triangle ABC|} = \frac{(r - 2)^2}{r^2 - r + 1}$

which checks with the Sketchpad. The ratio of areas of the triangles $\triangle CPK, \triangle ABC$ can be obtained in a similar way:

$\frac{|\triangle CPK|}{|\triangle ABC|} = \frac{\triangle_1 + \triangle_2 - 2 \triangle_3}{\triangle_0} = \frac{(r - 2)^2}{r^2 - r + 1} + \frac{r - 1}{r} - \frac{2(r - 1)^3}{r(r^2 - r + 1)} = \frac{1}{r(r^2 - r + 1)}$

Again, this simple expression checks with the Sketchpad.

I would like to add that if $r = \frac{1 + \sqrt 5}{2}$ is equal to the golden ratio, then $\frac {AP}{PK} = r(r - 1) = 1$, i.e., the points $P, Q, R$ are the midpoints of the segments $AK, BL, CM$. The triangles $\triangle PQR, \triangle AMP$ then have the same areas:

$\frac{|\triangle AMP|}{|\triangle PQR|} = \frac{\triangle_3}{\triangle_1} = \frac{(r - 1)^3}{r(r - 2)^2} = 1$

If the problem condition was that the points $P, Q, R$ are the midpoints of the segments $AK, BL, CM$, rather than $\frac{MB}{AB} = r = \frac{1 + \sqrt 5}{2}$, then it would be one of the Putnam 2001 problems. This Putnam problem permits a more elegant solution (without using the cosine theorem), precisely because the areas of the triangles $\triangle PQR, \triangle AMP$ are equal.


induction
modular arithmetic
number theory

Prove Fermat's little theorem by induction on a.
Fermat's little theorem :

\[a^{p-1} \equiv 1 \pmod {p}\]

what I have so far is

the theorem holds true for a=1.

Assume it is true for a=m

then, we'd have to show that \[(m+1)^{p-1} \equiv 1 \pmod {p}\]

using binomial theorem.

\[(m+1)^{p-1}= \binom{p-1}{0}m^{p-1}+\binom{p-1}{1}m^{p-2}+\cdots +\binom{p-1}{p-1}m^{0}\]

this is as far as I can get.

Obviously, $\binom{p-1}{0}m^{p-1} \equiv 1 pmod{p}$ from the assumption
and $\binom{p-1}{p-1}m^{0} \equiv 1 \pmod{p}$

and I do not know how to deal with $\binom{p-1}{1}m^{p-2}$

cuz if we do the division thing
we will have $\binom{p-1}{1}m^{p-2} \equiv \binom{p-1}{1} \frac{1}{m} \pmod {p}$ ???
this wouldn't make sense

so, can somebody solve the mystery?
 :P

Instead of inducting on $m^{p-1} \equiv 1 \pmod{p}$.

We can induct on $m^p \equiv m \pmod{p}$, which if we do the case $m+1$, we have $(m+1)^p \equiv (m+1) \pmod{p}$, which is a little easier to work with.

After binomial expansion, we get $\binom{p}{0}m^p + \binom{p}{1}m^{p-1} + \ldots + \binom{p}{p-1}m^1 + \binom{p}{p}m^0 \\ \equiv m + pm^{p-1} + \ldots + pm + 1$. If we bracket it, the middle expressions are all divisible by p so are thus equivalent to 0 modulo p. So, we are left with $(m+1)^p \equiv (m+1) \pmod{p}$. And our inductive step is done.

thre are manyways to prove it.
also you can use the Wilson's theorem to prove that.


here are some more (quite easy) problems:

1. In a mathematical competition consisting of 12 problems, 8 marks are given for a correct response, 0 marks for an incorrect response and no response is given 3 marks. Vicki scored 35 marks. What is the largest number of incorrect responses she could have had?

2. Let l1 be the line the y-intercept 16 and x-intercept 8. Let l2 be the line with y-intercept 6 and x-intercept 18. Find the area of the region enclosed by l1, l2 the y-axis and the x-axis.

3. Three circles and three straight lines lie in a plane. What is the largest number of points in which they may intersect one another?

4. A regular hexagon can be formed by 3 congruent pentagons (in the trivial way). What is the minimum number of these pentagons needed to form a regular hexagon larger than the one given?

5. Steve has a broken calculator. When just turned on, it displays 0, If the + key is pressed, it adds 51. If the - key is pressed it subtracts 51. If the * key is pressed it adds 85. If the / key is pressed it subtracts 85. The other keys do not function. What is the closest attainable number to 2003?

6. A four digit numbers leaves remainder 10 when divided by 21, remainder 11 when divided by 23 and remainder 12 when divided by 25. What is the sum of the digits of this number?

7. How many numbers less than 10000 have the product of their digits equal to 84?

8. The POM system is a base three system, with the digits P, O and M representing 1,0 and -1 respectively. For example PMOMP represents P*3^4 + M*3^3 + ... = 52. When 2003 is represented in the POM system, what are the last 2 digits?

9. A 3*3 square is divided into 9 1*1 squares. Different integers from 1 to 9 are placed in these 9 squares. For each two squares sharing a common edge, the sum of the integers in them is calculated. What is the minimum possitlbe number of different sums?

10. For each vertex of a cube, a plane is constructed through the three neighbouring vertices. Into how many parts do these 8 planes dissect the cube?

is the answer [hide]33 [/hide]parts for number 10?

#9 is [hide]4[/hide].  The center square makes 4 the lower bound for the minimum and  [hide]                   9 1 8

                                                             2 7 3

                                                             6 4 5[/hide]

is an example.

I think #7 is [hide] 60 [/hide].  The combinations of digits are [hide]2, 2, 3, 7 & 1, 2, 6, 7 & 1,3,4, 7.[/hide]

question 10 has already been at mathlinks and i gave answer 21. i'm quite sure that it's true.
#1: 8
#3: 21
#5: 2006
#6: 16
#8: MM
didn't yet think of the others, but they don't look difficult. however, i don't want to bother with any more problems until monday(first IMO paper :D ), but sure i can't let it be(f***ing drugs ;) ).

Peter

ok, thx.
I see my error now.  The answer is 1 + 8 + 12 and not 1 + 8 + 24.

Peter -- try using the spoiler button so that other people can do the questions without having seen your answers already.

I'm only going to try a couple

1.[hide]

Let x be the number of correct problems, y be the number of problems Vickie didn't answer, and z be the number of incorrect problems.

x+y+z = 12

8x+3y=35

We want the smallest possible way that 8x+3y can equal 35.  Since 8 is bigger than 3, there will be less 8's in 35 than three.  We can't go over though, since there is no way to get negative points.

The smallest number divisible by 8 is 32.  8*4=32, so we know x has to be 4.

32+3y=35

3y=3

y=1

Now we know that y=1

4+1+z=12

z=7

Vickie answered 7 problems incorrectly.[/hide]

3.[hide]

I know the circles are aranged in a vendiagram.

So that would make it 3 (for each circle touching the other) and 3 (for all the veritices in the little triangle thing) which eaqual 6.

Then you can arrange each line, so that it touches two sides of the triangle thing, two sides of a place occupied by only two triangles, and the edge of two traingles, which makes 6 for each line, which maks a grand total of 24 so far.

All of the lines should also cross at the same point, adding one, making the number 25.[/hide]

I included a picture, in case some of you didn't understand what I said, but It's very confusing too.

If anyone needs a better picture just pm me or something.

Oops.  Is there any way to hide the picture so other people can't see?

3) [hide]All possible intersections: 2 lines intersect, 2 circles intersect, a line and a circle intersect.

a line and a circle intersect in at most 2 points:

3(3)=9=total number of ways a line can intersect a circle 9*2=18.

2circles intersect in at most 2 points,

3C2=3=total number of ways two circle intersect. 2*3=6

2 lines intersect in one point at most:

3C2=3 ways of 2 lines intersect. 3*1=3

18+6+3=27[/hide]

[hide=@#9][url=aops.com/community/user/586791][b]mathpro12345[/b][/url] · 24 minutes ago [url=aops.com/community/p21022128](view)[/url][color=transparent]helo[/color]
\bumpity \bumpity \bump

(16 year old bump!!!!!!!)

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[color=#5b7083][aops]x[/aops] 9[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
Stop bumping. These are nice problems tho - imma check them :)

3. Post #6 only gave 25, and Post #8 doesn't show equality for 27, I was only able to get 26:
[img]https://i.ibb.co/F7hc6nd/Screen-Shot-2021-03-19-at-7-09-31-AM.png[/img]


inequalities
trigonometry
inequalities proposed

[quote="pvthuan"]i) Prove that if $a,b,c\geq0$, then
\[\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}+\sqrt{c^{2}-ca+a^{2}}\geq\sqrt{3(a^{2}+b^{2}+c^{2})}. \]
ii) Prove that if $a,b,c\geq0$, then
\[\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}+\sqrt{c^{2}-ca+a^{2}}\geq\sqrt{4(a^{2}+b^{2}+c^{2})-(ab+bc+ca)}. \]
[/quote]
The first one is immediately shown from the posted inequality $\sum\sqrt{a^{2}-ab+b^{2}}\sqrt{b^{2}-bc+c^{2}}\geq a^{2}+b^{2}+c^{2}.$

[quote="Lovasz"]
The first one is immediately shown from the posted inequality $\sum\sqrt{a^{2}-ab+b^{2}}\sqrt{b^{2}-bc+c^{2}}\geq a^{2}+b^{2}+c^{2}.$[/quote]

So is the second one.

Yes, that's right Duc.

The second one is proposed by Nguyen Anh Tuan inspired by mine. He then sent it to VMEO III.

[quote="zaizai-hoang"][quote="Lovasz"]
The first one is immediately shown from the posted inequality $\sum\sqrt{a^{2}-ab+b^{2}}\sqrt{b^{2}-bc+c^{2}}\geq a^{2}+b^{2}+c^{2}.$[/quote]
how to solve it?  :D This problem come from VMEO contest in Diendantoanhoc.net[/quote]
To prove it, you only need to prove
\[\left(\sum a\right)\sqrt{(a^{2}-ab+b^{2})(b^{2}-bc+c^{2})(c^{2}-ca+a^{2})}\ge \left(\sum ab \right) \left(\sum a^{2}\right)-\left(\sum a^{2}b^{2}\right) \]
:)


linear algebra
matrix
combinatorics unsolved
combinatorics

We have a group of 24 student, where are 8 girls (=> 16 boys).
In how many ways we can arrange the students in a classroom of 24 places, which are distributed like a matrix (4 x 6), such that there are no, two neighbour girls (on the line and on the column) ?

[hide="a restating of the question in a different form"]
Let $ A \equal{} \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & 0 \\
0 & a_{11} & a_{12} & \cdots & a_{16} & 0 \\
0 & a_{21} & a_{22} & \cdots & a_{26} & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & a_{41} & a_{42} & \cdots & a_{46} & 0 \\
0 & 0 & 0 & \cdots & 0 & 0 \\
\end{pmatrix}$
such that 8 of the $ a_{ij} \equal{} 1$ and the rest equal 0.

Then the problem is equivalent to counting the number of ways such that $ (\forall i\in [1,4], j\in[1,6])\ a_{ij} \plus{} a_{(i \plus{} 1)j} \plus{} a_{(i \minus{} 1)j} \plus{} a_{(i \minus{} 1)j} \plus{} a_{i(j \plus{} 1)} \plus{} a_{i(j \minus{} 1)} \equal{} 1$
[/hide]


I am the type of person who is curious.

Am I allowed to make a poll on religion?

I am asking this first so that I would not get in trouble...

Please don't get mad at me. I have good intentions...

It is perfectly fine to make a poll as long as it doesn't contain anything that could be offensive.

However, when making a serious poll, I suggest you post it [url=http://www.artofproblemsolving.com/Forum/index.php?f=19]here[/url], in the official polls forum.

In G&FF, not as many people will take it seriously and there is a very high chance that it will get spammed/flamed.

Thanks i_like_pie. It seems you are everywhere! (I mean that in a good way)

Using my newly acquired [b]skillz[/b] in searching, I have found out that there are already 2 polls on religion, so i wouldn't bother making another one.

[size=200]DELETE THIS THREAD PLEASE[/size]

I won't be anywhere once school starts. :(

But if I am, I'll be glad to help anytime. :)


[hide=" if you want An x,y"]If you just want an x,y such that it satisfies the equations, guessing they were integers, caused the 1/(x+y) term to equal 1/3, from there, (x,y)=(2,1)[/hide]


Find all $f$:$GL_{n}(C)\rightarrow M_{n}(C)$
such that for any $M\in GL_{n}(C)$ we have $M^{-1}f(-M)+f(M^{-1})=M$

Simple algebra gives $f(M)=\frac12(M^2+M^{-1})$.

This question has nothing to do with linear algebra.