absolute value
combinatorics unsolved
combinatorics

Central edge: an edge in a tree that if we remove it from the tree we get a minimal difference between the number of nodes in each part of the divided tree.

Prove that if there is more then one central edge in a tree then they all( central edges) have a common node.

I think it's pretty clear that if two central edges $xy,zt$ are not adjacent, then removing any edge on the unique path linking $x$ and $z$, for isntance, achieves a better (i.e. smaller in absolute value) difference between the numbers of vertices in the two resulting groups.This means that all central edges are adjacent, so, since the graph is a tree and cannot have cycles, they must all share a common endpoint.

Is there anything wrong with this? :?

[quote="grobber"]I think it's pretty clear that if two central sides $xy,zt$ are not adjacent, then removing any side on the unique path linking $x$ and $z$, for isntance, achieves a better (i.e. smaller in absolute value) difference between the numbers of vertices in the two resulting groups.This means that all central edges are adjacent, so, since the graph is a tree and cannot have cycles, they must all share a common endpoint.

Is there anything wrong with this? :?[/quote]
what if the two sides are on this one path? or only one of them

Grobber what do you mean by removing a "side"?

I mean edge, of course :). I've edited my message.

alexrait, what if one or both of our edges lie on that path? Why would that be a problem? Simply remove another edge, different from the two, sicne the path cannot consist only of these two edges, because we assumed they weren't adjacent. I'm still not entirely sure it's Ok though.


Build a program in C that accomplishes the following task:
It reads from a file the names of some certain files and their size in MB, then it distributes these files to some folders each with a maximum of N MB. The way of distribution is such that a minimum number of folders are used and one file is not appearing in two folders. Then the program gives as output a file containing the name of each file and the number of the folder in which it was assigned.

Need an idea :) Thnx in advance!

Sounds like a [url=http://en.wikipedia.org/wiki/Bin_packing_problem]bin packing[/url] problem, which is NP-hard so there's no way to always get the optimal solution.  Here's a page containing various ways to produce a solution though: [url]http://www.cs.arizona.edu/icon/oddsends/bpack/bpack.htm[/url].

[quote="kevmo314"]NP-hard so there's no way to always get the optimal solution.[/quote]  To be precise: no [i]known, efficient[/i] way (and probably this is still true if we remove the word "known," and actually proving that is worth USD 1 million).  It's algorithmically simple to always produce the optimal solution: enumerate every possibility and then compare these to find the best.

thank you guys, but I still have difficulties understanding it... I thought maybe the bestfit algorithm is the most proper but I still can't implement it in order to build the entire program  :blush:

The best fit algorithm works as such:

[code]
for each item from 0 to i               # Attempt to place each item
     bestbin = -1                       # This is the best place to place item i
     for each bin from 0 to j           # Check each bin
          if item i + bin j <= maxsize  # If the item can be placed in the bin...
               if bin j > bin bestbin   # then check if it's a better placement
                    bestbin = j         # If it is, then store that placement
     if bestbin != -1                   # If we found a bin to place item i...
          place item i in bestbin       # then place the item
     else                               # Otherwise...
          create new bin j+1            # make a new bin
          place item i in bin j+1       # and place the item in that bin
[/code]


let $ a,b,c > 0$ prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\ge \frac {3}{8} \plus{} \frac {k}{3}$
with $ k\equal{}\frac{9(3\plus{}2\sqrt{3}}{8}$

[quote="arqady"][quote="lenhamquy"]let $ a,b,c > 0$ prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\ge \frac {3}{8} \plus{} \frac {k}{3}$
with $ k \equal{} \frac {9(3 \plus{} 2\sqrt {3}}{8}$[/quote]
I think $ k \equal{} \frac {89 \plus{} 34\sqrt7}{24}$ is a maximal value of $ k$ for which your inequality is true.[/quote]
please, detail

[quote="lenhamquy"]let $ a,b,c > 0$ prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\ge \frac {3}{8} \plus{} \frac {k}{3}$
with $ k \equal{} \frac {9(3 \plus{} 2\sqrt {3}}{8}$[/quote]
I have just seen it .  can_hang2007 had a solution by CYH for it .
And $ k \equal{} \frac {9(3 \plus{} 2\sqrt {3})}{8}$ is max value of k such that ineq is true  :wink:
Can U post it , can_hang2007 ? ( Your solution very nice , can_hang2007  :lol: )

[quote="arqady"][quote="onlylove_math"]
And $ k \equal{} \frac {9(3 \plus{} 2\sqrt {3})}{8}$ is max value of k such that ineq is true  :wink:
[/quote]
My value better!  :P  What your counterexample? Thank you![/quote]
Pk,arqady,can you try $ a \equal{} 1,b \equal{} c \equal{} \sqrt 3 \plus{} 1$with your value (I tried it and I see you are wrong,but I dont believe it :(  )I think you should check agian
The reason of your mistake,I think you wrong when you find max of
[quote="arqady"]$ \Leftrightarrow k\leq\frac {3(4a \plus{} 5)(a \plus{} 2)^2}{8(a \plus{} 1)^2}.$[/quote]
To only love math,It is not CYH,that is a technique in pqr of Can hang,a strong technique

[quote="arqady"]OK.
Let $ b \equal{} c \equal{} 1.$ Hence, $ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\ge \frac {3}{8} \plus{} \frac {k}{3}\Leftrightarrow$
$ \Leftrightarrow\frac {a^3 \plus{} 2}{2(a \plus{} 1)^2} \minus{} \frac {3}{8}\geq k\left(\frac {1}{3} \minus{} \frac {2a \plus{} 1}{(a \plus{} 2)^2}\right)\Leftrightarrow\frac {(a \minus{} 1)^2(4a \plus{} 5)}{8(a \plus{} 1)^2}\geq\frac {k(a \minus{} 1)^2}{3(a \plus{} 2)^2}\Leftrightarrow$
$ \Leftrightarrow k\leq\frac {3(4a \plus{} 5)(a \plus{} 2)^2}{8(a \plus{} 1)^2}.$
Let $ g(a) \equal{} \frac {3(4a \plus{} 5)(a \plus{} 2)^2}{8(a \plus{} 1)^2}.$ Hence, $ g'(a) \equal{} \frac {3(a \plus{} 2)(2a^2 \plus{} 2a \minus{} 3)}{4(a \plus{} 1)^2}.$
Thus, $ k\leq\min_{(0, \plus{} \infty)}g \equal{} g\left(\frac {\sqrt7 \minus{} 1}{2}\right) \equal{} \frac {89 \plus{} 34\sqrt7}{24}.$
Id est, it remains to prove your inequality for $ k \equal{} \frac {89 \plus{} 34\sqrt7}{24}.$
 Let $ f(a,b,c) \equal{} \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \minus{} \frac {3}{8} \minus{} \frac {k}{3}$  and $ a \equal{} \min\{a,b,c\}.$
Then $ b \plus{} c\geq2a$ and $ f(a,b,c) \minus{} f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right) \equal{}$
$ \equal{} (b \minus{} c)^2\left(\frac {(a \plus{} b \plus{} c)^3}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(2a \plus{} b \plus{} c)^2} \minus{} \frac {k}{4(a \plus{} b \plus{} c)^2}\right)\geq$
$ \geq(b \minus{} c)^2\left(\frac {27}{8(2a \plus{} b \plus{} c)^2} \minus{} \frac {k}{4(a \plus{} b \plus{} c)^2}\right) \equal{}$
$ \equal{} \frac {(b \minus{} c)^2(3\sqrt3(a \plus{} b \plus{} c) \plus{} \sqrt {2k}(2a \plus{} b \plus{} c)((3\sqrt3 \minus{} 2\sqrt {2k})a \plus{} (3\sqrt3 \minus{} \sqrt {2k})(b \plus{} c))}{8(a \plus{} b \plus{} c)^2(2a \plus{} b \plus{} c)^2}\geq$
$ \geq\frac {(b \minus{} c)^2(3\sqrt3(a \plus{} b \plus{} c) \plus{} \sqrt {2k}(2a \plus{} b \plus{} c)(9\sqrt3 \minus{} 4\sqrt {2k})a}{8(a \plus{} b \plus{} c)^2(2a \plus{} b \plus{} c)^2}\geq0.$
It gives $ f(a,b,c)\geq f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right).$
Hence, It remains to check $ f(a,b,b)\geq0$ but it's done!  :)[/quote]
I think you got some mistakes in your calculation, arqady. The result is $ k\equal{}\frac{9(3\plus{}2\sqrt{3})}{8}$. ;)

[quote="nguoivn"]When does equality occur ?  :maybe:[/quote]
For example, when $ a \equal{} \frac {\sqrt3 \minus{} 1}{2}$ and $ b \equal{} c \equal{} 1.$

[quote="nguoivn"]Thank you. arquady. SOS also works here.  :)[/quote]
But ,I think it is too hard (notice when equal holds),Can you post it?

[quote="Allnames"]
But ,I think it is too hard (notice when equal holds),Can you post it?[/quote]

I don't think so. See my solution here:

[quote="lenhamquy"]let $ a,b,c > 0$ prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\ge \frac {3}{8} \plus{} \frac {k}{3}$
with $ k \equal{} \frac {9(3 \plus{} 2\sqrt {3}}{8}$[/quote]
I posted it long time ago on diendantoanhoc.net. here is it: http://diendantoanhoc.net/forum/index.php?showtopic=11463&st=280 :)

[quote="arqady"][quote="Allnames"]
The reason of your mistake,I think you wrong when you find max of
[quote="arqady"]$ \Leftrightarrow k\leq\frac {3(4a \plus{} 5)(a \plus{} 2)^2}{8(a \plus{} 1)^2}.$[/quote]
[/quote]
Yes, you are right! Here is a mistake.
If $ g(a) \equal{} \frac {3(4a \plus{} 5)(a \plus{} 2)^2}{8(a \plus{} 1)^2}$ then $ g'(a) \equal{} \frac {3(a \plus{} 2)(2a^2 \plus{} 2a \minus{} 1)}{4(a \plus{} 1)^3}.$
But the method is true and easy enough.
It gives $ k\leq\min_{(0, \plus{} \infty)}g \equal{} g\left(\frac {\sqrt3 \minus{} 1}{2}\right) \equal{} \frac {9(3 \plus{} 2\sqrt {3})}{8}.$
Thank you![/quote]

Arqady, how to make your solution valid for this new value of k?

Thank you very much

[quote="Sunkern_sunflora"][b]To nguoivn:[/b]
I don't understand your solution (however easy!) because it was written entirely in your mother-tongue..... :(
Can't you provide me a solution in [b]english[/b]?? Because as per length is considered, your solution is not to big, so it ought to be nice....
Thank u.
 :)[/quote]

Indeed, I don't like to write full calculation because the coefficients are not nice (with me, only integer numbers are nice) . So, I only write the main hints. I think it won't lost the nice of the proof.
I think you can check easily that proof (as prove $ Sb \geq\ 0$ or prove $ f'(b) < 0$)
 :)

[quote="arqady"]manlio, of cause it holds! We need to check only that $ 9\sqrt3 \minus{} 4\sqrt {2k}\geq0$ is true, which is true :wink:[/quote]

You are right, Arqady. Sorry I made a mistake in my calculation :blush:


Give mechanism and predict major product  :(

The first step is [url=http://en.wikipedia.org/wiki/Favorskii_rearrangement]Favorskii Rearrangement[/url] followed by esterification :)


[quote="Golden Rule"]Prove that 2^n + 3^n is a multiple of 5 for all odd n ∈ N[/quote]
$ a\minus{}b|a^{n}\minus{}b^{n}$ for all $ a,b\in Z$
More than
Find all n such that
$ 2^{n}\plus{}3^{n}$ is a power of 5.

Barring that, is the sequence $ k_{n}$ such that $ 5^{k_{n}}| 2^{n}\plus{}3^{n}, 5^{k_{n}\plus{}1}\not | 2^{n}\plus{}3^{n}$ unbounded?

It easy
Let $ n\equal{}5^{m}$
Then $ k_{n}\geq m$

[quote="Golden Rule"]Prove that $ 2^{n}\plus{}3^{n}$ is a multiple of 5 for all odd $ n\in\mathbb{N}$[/quote]

If $ n$ is odd then: $ 2^{n}\plus{}3^{n}\equal{} (2\plus{}3)(2^{n\minus{}1}\minus{}2^{n\minus{}2}3\plus{}2^{n\minus{}3}3^{2}\minus{}........\plus{}3^{n\minus{}1})$


How many factors does the number 740 have?

Only perfect squares have odd numbers of factors and 740 isn't a perfect square  ;)
You don't have to list out all the factors.

oh yeah.. it's not

[hide]
9
[/hide]factors...

its

[hide]
12
[/hide] factors..

I forgot to count the [hide]2, 5, and 37[/hide] as a factors...

^^;;

lets have another easy one .how many factors does 2003 have ?explain and prove.


Let $ ABCD$ be a quadrilateral such that $ \angle DAC \equal{} \angle BDC \equal{} 36^{\circ}$ and $ \angle CBD \equal{} 18^{\circ}$ and $ \angle BAC \equal{} 72^{\circ}$. The diagonal $ [AC]$ meets $ [BD]$ meets at $ P$. Find $ \angle APD$.

The circumcenter of $ \triangle BCD$ lies onto the circle $ (ACD)$ [b]and[/b] onto the circle $ (ABC)$, so it is $ A$, hence the triangles $ \triangle ABC$ and $ \triangle ACD$ are isosceles at $ A$; we get $ \angle APD\equal{}108^\circ$.

Best regards,
sunken rock


$ 0 < a < 1,a_1 \equal{} 1 \plus{} a,a_n \equal{} \frac {1}{a_{n \minus{} 1}} \plus{} a(n\geq2),\Longrightarrow a_n > 1$    $ (n \in Z_ \plus{}) .$

Define $ u_0 \equal{} 1,u_{n} \equal{} \frac {1}{u_{n \minus{} 1}} \plus{} u (n\geq1),$ then

$ u_{n} \equal{} \frac {P_{n}(u)}{P_{n \minus{} 1}(u)},$

where $ P_{n} (u) \equal{} \sum_{k \equal{} 1}^{n}\left[\frac {n \plus{} k}{2}\right]u^k.$

$ u > 0\Longrightarrow P_{n} (u)> P_{n \minus{} 1}(u)\Longrightarrow u_n > 1$ for all values of $ n \equal{} 1,2,3,\cdots.$

[hide="Remark."]$ \frac {1 \plus{} v}{1 \minus{} uv \minus{} v^2} \equal{} \sum_{k \equal{} 0}^{\infty}P_k(u)v^k.$[/hide]

[quote="Ji Chen"]Define $ u_0 \equal{} 1,u_{n} \equal{} \frac {1}{u_{n \minus{} 1}} \plus{} u (n\geq1),$ then

$ u_{n} \equal{} \frac {P_{n}(u)}{P_{n \minus{} 1}(u)},$

where $ P_{n} (u) \equal{} \sum_{k \equal{} 1}^{n}\left[\frac {n \plus{} k}{2}\right]u^k.$

$ u > 0\Longrightarrow P_{n} (u) > P_{n \minus{} 1}(u)\Longrightarrow u_n > 1$ for all values of $ n \equal{} 1,2,3,\cdots.$

[hide="Remark."]$ \frac {1 \plus{} v}{1 \minus{} uv \minus{} v^2} \equal{} \sum_{k \equal{} 0}^{\infty}P_k(u)v^k.$[/hide][/quote]

How about :X(n+1)=Xn-1/Xn with x1=a,then xn=?  :maybe:


floor function
modular arithmetic
Divisibility Theory

Let $p>3$ is a prime number and $k=\lfloor\frac{2p}{3}\rfloor$. Prove that \[{p \choose 1}+{p \choose 2}+\cdots+{p \choose k}\] is divisible by $p^{2}$.

[quote="Peter"]Let $ p > 3$ is a prime number and $ k \equal{} \lfloor\frac {2p}{3}\rfloor$. Prove that
\[ {p \choose 1} \plus{} {p \choose 2} \plus{} \cdots \plus{} {p \choose k}
\]
is divisible by $ p^{2}$.[/quote]

I will assume paragraphs 1. and 2. of http://www.mathlinks.ro/Forum/viewtopic.php?t=150391 (or http://www.problem-solving.be/pen/viewtopic.php?t=25 ) post #4 as known.

We have to prove that $ p^{2}\mid\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\binom{p}{i}$.

In http://www.mathlinks.ro/Forum/viewtopic.php?t=150400 (or http://www.problem-solving.be/pen/viewtopic.php?t=34 ) post #2, Theorem 1, I showed that $ v_{p}\left(\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}\right)\geq 1$. This rewrites as $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}\equiv 0\mod p$, where we are working with fractions modulo $ p$ (what is allowed in our case because the denominators are integers $ i$ satisfying $ 1\leq i\leq\left\lfloor 2p/3\right\rfloor$, and these integers are not divisible by $ p$). In other words, $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\left( \minus{} 1\right)^{i \minus{} 1}\equiv 0\mod p$.

Now we prove a simple lemma:

[color=blue][b]Lemma 1.[/b] If $ p$ is a prime and $ i$ is an integer satisfying $ 0\leq i\leq p \minus{} 1$, then $ \binom{p \minus{} 1}{i}\equiv\left( \minus{} 1\right)^{i}\mod p$.[/color]

[i]Proof of Lemma 1.[/i] We have $ \binom{p \minus{} 1}{i} \equal{} \frac {\left(p \minus{} 1\right)\cdot\left(p \minus{} 2\right)\cdot ...\cdot\left(p \minus{} i\right)}{i!}$, and thus

$ i!\cdot\binom{p \minus{} 1}{i} \equal{} \left(p \minus{} 1\right)\cdot\left(p \minus{} 2\right)\cdot ...\cdot\left(p \minus{} i\right)\equiv\left( \minus{} 1\right)\cdot\left( \minus{} 2\right)\cdot ...\cdot\left( \minus{} i\right)$
$ \equal{} \left( \minus{} 1\right)^{i}\cdot\left(1\cdot 2\cdot ...\cdot i\right) \equal{} \left( \minus{} 1\right)^{i}\cdot i!\mod p$.

Now, $ i!$ is coprime with $ p$ (since $ i! \equal{} 1\cdot 2\cdot ...\cdot i$, and all the numbers $ 1$, $ 2$, ..., $ i$ are coprime with $ p$ since $ p$ is a prime and $ i < p$); hence, we can divide this congruence by $ i!$, and obtain $ \binom{p \minus{} 1}{i}\equiv\left( \minus{} 1\right)^{i}\mod p$. Lemma 1 is proven.

Now, for every integer $ i$ satisfying $ 1\leq i\leq\left\lfloor 2p/3\right\rfloor$, Lemma 1 yields $ \binom{p \minus{} 1}{i \minus{} 1}\equiv\left( \minus{} 1\right)^{i \minus{} 1}\mod p$, so that $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\left( \minus{} 1\right)^{i \minus{} 1}\equiv 0\mod p$ becomes $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\binom{p \minus{} 1}{i \minus{} 1}\equiv 0\mod p$. In other words,

$ v_{p}\left(\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\binom{p \minus{} 1}{i \minus{} 1}\right)\geq 1$.

Together with $ v_{p}\left(p\right) \equal{} 1$, this yields

$ v_{p}\left(p\cdot\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\binom{p \minus{} 1}{i \minus{} 1}\right) \equal{} v_{p}\left(p\right) \plus{} v_{p}\left(\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\binom{p \minus{} 1}{i \minus{} 1}\right)\geq 1 \plus{} 1 \equal{} 2$.

But

$ p\cdot\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\binom{p \minus{} 1}{i \minus{} 1} \equal{} p\cdot\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\cdot\frac {\left(p \minus{} 1\right)!}{\left(i \minus{} 1\right)!\cdot\left(\left(p \minus{} 1\right) \minus{} \left(i \minus{} 1\right)\right)!}$
$ \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {p\cdot\left(p \minus{} 1\right)!}{\left(i\cdot\left(i \minus{} 1\right)!\right)\cdot\left(\left(p \minus{} 1\right) \minus{} \left(i \minus{} 1\right)\right)!} \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {p!}{i!\cdot\left(p \minus{} i\right)!} \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\binom{p}{i}$.

Hence, we get $ v_{p}\left(\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\binom{p}{i}\right)\geq 2$. Since $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\binom{p}{i}$ is an integer, this means $ p^{2}\mid \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\binom{p}{i}$, qed.

Note that this problem also appeared as question 2 in the 11th Annual Vojtech Jarnik International Mathematical Competition 2001, Category I. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=151379 .

  Darij

Where's the lemma no 2? :D, Mr [b]darij grinberg[/b]?

We haven't any exact and simple solutions, example: the Darij's solution! I think that this isn't a hard problem and we can solve it more simply and purer! I cann't understand the termes, example: "p-ardic"... Help me, please!

[quote="ricardokaka"]I think that this isn't a hard problem and we can solve it more simply and purer![/quote]Then post your solution, please! :)

I also believe p-adic numbers can be avoided.

Sorry, but p-adic numbers were never used in any of the proofs (even not in the linked ones).

[quote="darij grinberg"]
Now, for every integer $ i$ satisfying $ 1\leq i\leq\left\lfloor 2p/3\right\rfloor$, Lemma 1 yields $ \binom{p \minus{} 1}{i \minus{} 1}\equiv\left( \minus{} 1\right)^{i \minus{} 1}\mod p$, so that $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\left( \minus{} 1\right)^{i \minus{} 1}\equiv 0\mod p$ becomes $ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i}\binom{p \minus{} 1}{i \minus{} 1}\equiv 0\mod p$. In other words,
  Darij[/quote]
I don't see prove. I give full solution at least 2 times.
Let $ S(a,b)\equal{}\sum_{a<i<b}\frac 1i \mod p$. Then
$ \sum_{0<i<2p/3}\frac{(\minus{}1)^{i\minus{}1}}{i}\equal{}\sum_{0<i<2p/3, i\minus{}odd}\frac 1i \minus{}\frac 12 S(0,\frac p3) \equal{}\minus{}\sum_{p/3<i<p, p\minus{}even}\frac 1i\minus{}\frac 12 S(0,\frac p3)\equal{}\minus{}\frac 12 (S(\frac p6 ,\frac p2 )\plus{}S(0,\frac p3 )).$
Let $ S_i\equal{}S(\frac{ip}{6},\frac{(i\plus{}1)p}{6})\mod p,i\equal{}0,1,2,3,4,5$. Obviosly $ S_i\plus{}S_{5\minus{}i}\equal{}0$.
Consider
$ T_a\equal{}\frac{a^{p\minus{}1}\minus{}1}{p}\sum_{x\equal{}1}^{p\minus{}1}x^{p\minus{}1}\equal{}\frac 1p \sum_{x\equal{}1}^{p\minus{}1}(ax)^{p\minus{}1}\minus{}x^{p\minus{}1}\equal{}\sum_{j\equal{}1}^{p\minus{}1}j(p\minus{}1)\sum_{jp/a<x<(j\plus{}1)p/a}(ax\minus{}p[ax/p])^{p\minus{}2} \mod p\equal{}\sum_{j\equal{}1}^{a\minus{}1}\frac{\minus{}j}{a} \sum_{jp/a<x<(j\plus{}1)p/a}\frac 1x \mod p.$
It give $ \sum_{i\equal{}0}^{[(a\minus{}2)/2]}(a\minus{}1\minus{}2i)S_{i,a}\equal{}aT_a\mod p$, were $ S_{i,a}\equal{}\sum_{ip/a<x<(i\plus{}1)p/a}\frac 1x \mod p$.
Because $ T_6\equal{}T_2\plus{}T_3$, we get
$ 5S_0\plus{}3S_1\plus{}S_2\minus{}3(S_0\plus{}S_1\plus{}S_2)\minus{}2(2S_0\plus{}2S_1)\equal{}6T_6\minus{}3*(2T_2)\minus{}2*(3T_3)\equal{}0\mod p.$
It give $ 0\equal{}S_0\plus{}2S_1\plus{}S_2\equal{}S(\frac p6,\frac p2 )\plus{}S(0,\frac p3 )\mod p$

I have derived a shorter solution here, wonder if it's valid. :)

[b]Solution:[/b]
Re-write it as $p(\frac{(p-1)!}{1!(p-1)!}+\frac{(p-1)!}{2!(p-2)!}+...+\frac{(p-1)!}{k!(p-k!)})$ , factoring out $p$, and taking modulo $p$ gives,
\[\ \equiv 1!-\frac{1!}{2!}+\frac{2!}{3!}+...+(-1)^{k-1}\frac{(k-1)!}{k!} \]
Simplifying, the above expression is equivalent to,
\[\ 1+\frac{1}{2}+...+\frac{1}{k}-2(1+...+\frac{1}{\lfloor \frac{k}{2} \rfloor}) = \frac{1}{\lfloor \frac{k}{2} \rfloor +1 } +...+ \frac{1}{k} \]

Now we consider the cases when $p=6n+1,6n-1$.
[b]case 1:[/b] $p=6n+1$
We have, $k=4n$, the expression becomes,
\[\ \frac{1}{2n+1} +...+ \frac{1}{4n} \equiv 0 \pmod {p} \]
Since we can pair opposite reciprocals, such that their numerator is $6n+1$. i.e $(4n)+(1+2n)=6n+1$. And note that there are exactly, $(4n)-(2n+1)+1=2n$ terms, which is an even number. Hence, the pairing is possible.

[b]case 2: [/b]$p=6n-1$,
we have $k=4n-1$, the expression becomes,
\[\ \frac{1}{2n} +...+ \frac{1}{4n-1} \equiv 0 \pmod {p} \]
which can be solved in the exact same way as [b]case 1[/b].
$\Box$

EDIT: edited some typos :p

$\dbinom{p}{r}\equiv p\times\frac{(-1)^{k-1}}{k} \mod p^2 \implies$

now we should check two cases $p=6n+1$ and $p=6n+5$

[color=#f00]case 1:[/color] let $p=6n+1\implies \sum_{i=1}^{\lfloor\frac{2p}{3}\rfloor} \dbinom{p}{i} \div p \equiv \frac{1}{1}-\frac{1}{2}+...-\frac{1}{4n} \equiv$ $\frac{1}{1}+\frac{1}{2}+...+\frac{1}{4n} - 2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4n})$

$\equiv \frac{1}{2n+1}+...+\frac{1}{4n} \mod p$ which is obvious as we take pairs $\frac{1}{2n+1+i}, \frac{1}{4n-i}$

[color=#f00]case 2:[/color] let $p=6n+5 \implies \sum_{i=1}^{\lfloor\frac{2p}{3}\rfloor} \dbinom{p}{i} \div p \equiv \frac{1}{1}-\frac{1}{2}+...-\frac{1}{4n+2}+\frac{1}{4n+3} \equiv$

$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{4n+2}+\frac{1}{4n+3} -2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4n+2}) \equiv \frac{1}{2n+2}+...+\frac{1}{4n+3} \mod p$

which again obvious by taking pairs


Hi.

um, what is ACoPS?

i heard some saying its art and craft of problem solving.

is it a book?

could someone please explain?

It is the book The Art and Craft of Problem Solving, by Paul Zeitz.

It is sort of teh next step after aops vol.2.

Having just finished it, I highly, highly recommend it.

But I heard the solutions to a lot of problems in the book aren't provided?


Solve the following trig equations:

1.
\[ \sin(x\minus{}12^\circ)\equal{}2\cos(6^\circ)\cos(x)
\]

2.
\[ \frac{\sin(x)}{\sin(40^\circ \minus{}x)} \equal{}1\minus{}\frac{\sin(40^\circ)}{\sin(80^\circ)}
\]

1) $ sinxcos12^0\minus{}cosxsin12^0\minus{}2cos6^0cosx\equal{}0\Leftrightarrow tanx\equal{}\frac{sin12^0\plus{}2cos6^0}{cos12^0}$  .OK! (Especially $ \Leftrightarrow tanx\equal{}tan66^0$)
2) Similarly!

How did you evaluate that fraction?...I got to the same point, but then I couldn't evaluate it.


Find the minimum value of $ K$ such that

$ \frac {a\plus{}b\plus{}c\minus{}3}{1\minus{}abc} \le K$

for all nonnegative real numbers $ a,b,c \le 2$ such that  $ abc<1$ and $ ab\plus{}bc\plus{}ca\equal{}3$.

[quote="Vasc"]Find the minimum value of $ K$ such that

$ \frac {a \plus{} b \plus{} c \minus{} 3}{1 \minus{} abc} \le K$

for all nonnegative real numbers $ a,b,c \le 2$ such that  $ abc < 1$ and $ ab \plus{} bc \plus{} ca \equal{} 3$.[/quote]
$ k\equal{}1\plus{}\frac{2}{\sqrt{7}}$ ;) :)

[quote="can_hang2007"][quote="Vasc"]Find the minimum value of $ K$ such that

$ \frac {a \plus{} b \plus{} c \minus{} 3}{1 \minus{} abc} \le K$

for all nonnegative real numbers $ a,b,c \le 2$ such that  $ abc < 1$ and $ ab \plus{} bc \plus{} ca \equal{} 3$.[/quote]
$ k \equal{} 1 \plus{} \frac {2}{\sqrt {7}}$ ;) :)[/quote]
I found the same value. :wink:

hello, i have also $ K_{max}\equal{}\frac{1}{7}(7\plus{}2\sqrt{7})$, this will be obtained for $ a\equal{}2,b\equal{}\minus{}2\plus{}\sqrt{7},c\equal{}\minus{}2\plus{}\sqrt{7}$.
Sonnhard.


geometry
perimeter
geometric transformation
rotation
projective geometry
trigonometry
calculus

Between all the triangles with fixed perimeter find the triangle that generates the maximum volume of the solid or revolution whose axis of rotation is a sideline of the given triangle

By the Theorem of Pappus, the volume of the solid of revolution is the area times the circumference traveled by the center of mass of the planar object. Draw a triangle with base $ b$ along the axis of rotation and altitude $ h$ perpendicular to that axis. The are is $ \frac12bh$ and the center of mass lies $ \frac13h$ from the axis - a third of the altitude. Hence the Theorem of Pappus gives the volume as $ V\equal{}\frac{\pi}3bh^2.$

Among all triangles with base $ b$ and altitude $ h,$ the one of least perimeter is the isosceles triangle for which that particular altitude is the line of symmetry. Hence, we may safely assume that our maximal triangle is just such an isosceles triangle. For such a triangle, the perimeter is $ b\plus{}2\sqrt{h^2\plus{}\frac{b^2}4}\equal{}b\plus{}\sqrt{4h^2\plus{}b^2}.$

Our problem has now become this:

Maximize $ V\equal{} \frac{\pi}3bh^2$ subject to the constraint $ b\plus{}\sqrt{4h^2\plus{}b^2}\equal{}P$ for some fixed $ P.$

(So far, I'm pretty confident of that much. What follows depends on me having done some fairly messy algebra correctly while solving this by Lagrange multipliers; I have less confidence in that.)

I get that the shape of the answer is that $ h\equal{}\sqrt{2}\,b.$ This describes an isosceles triangle whose apex angle is $ \cos^{\minus{}1}\left(\frac79\right),$ which is about $ 38.9^{\circ}.$ The two matching base angles are then about $ 70.5^{\circ}.$

Thanks for your fast reply Kent. The given answer is an isosceles triangle as u said such that the length of its legs is $ \frac {3p}{8}$ and the length of the base is $ \frac {p}{4}$ , so u are right  and the apex angle is about 38,9 degrees as u said.  :)

Ah. My first method was a little ugly. It would have been better if I'd worked with the slant height instead of the altitude. Assume a base of $ b$ and a slant height of $ a.$ Then the altitude is $ h \equal{} \sqrt {a^2 \minus{} \frac {b^2}4}.$ Note that the formula for volume involves just the square of $ h,$ which is nice. The problem is:

Maximize $ V \equal{} \frac {\pi}3b\left(a^2 \minus{} \frac {b^2}4\right) \equal{} \frac {\pi}{12}(4a^2b \minus{} b^3)$ subject to the constraint $ b \plus{} 2a \equal{} P.$

Now the Lagrange multipliers work out much cleaner. We fairly quickly come to the equation

$ 4a^2 \minus{} 3b^2 \equal{} 4ab$ or $ 4a^2 \minus{} 4ab \minus{} 3b^2 \equal{} 0.$

This factors as $ (2a \plus{} b)(2a \minus{} 3b) \equal{} 0$

The negative numbers that would result from $ 2a \plus{} b \equal{} 0$ make no physical sense, so we must have $ 2a \equal{} 3b.$ Put that into $ b \plus{} 2a \equal{} b \plus{} 3b \equal{} P$ to get $ b \equal{} \frac {P}4$ and $ a \equal{} \frac {3P}8.$


number theory
least common multiple
algorithm
algebra
polynomial
induction
relatively prime

I don't know if I can solve this one (I have an idea but I don't know if it works), and that's why I'm not posting it under "Proposed Problems":

Let p be an odd prime. Show that there exists 1<a<p s.t. a is a primitive root modulo p^n for any n in N*.

for all non-Romanian members N* = {1,2,3 ...} :)

The proof for this fact is pretty involved, I will try to post some parts to it but it is too long to write out in one sitting.  The basic idea is as follows:

Definition U(p) = unit group mod p (group of invertible elements)

steps to the proof:

1) show that if a and b are elements in U(p) and that if a has order m and if b has order n, then there exists an element u of order lcm (m,n).

2) first prove that U(p) is cyclic
  - done by showing that if n is the max orderof all the elements in U(p), then for all w in U(p), wn is congruent to 1 mod p

3) show that because U(p) is cyclic, U(Pn) has an element of order p-1 and an element of order pn-1 and this should establish the result by 1)

for 1)

Lemma: if a and b have orders m and n mod t, and (m,n) = 1 then a*b has order m*n.

Proof:

we have a with order m and b with order n, and (m,n) = 1 and suppose that the order of a*b = L.

Now, (a*b)m*n = am*n*bm*n = (am)^n*(bn)^m = 1n*1m = 1, so L|m*n

also, 1 = (a*b)m*L = bm*L so n|m*L, and by a similar method, we can show that m|n*L, and since (m,n) = 1, then n|L and m|L so m*n|L (since they are relatively prime), and since L|m*n, and m*n|L, L = m*n.

Now onto the proof of 1),

so suppose we have an element a with order m and b with order n mod t, we wish to show that there is an element with order lcm (m,n).  Now if (m,n) = 1 we showed above that a*b has order lcm (m,n) = m*n, if WLOG m|n, then the lcm (m,n) = n and and b has order n.  For the last case:

consider the algorithm of finding the lcm of two positive integers, you write them out in "complete" factorization and you compare exponents of primes and take the larger one for the lcm, for example for the lcm of 15 and 24, 15 = 31*51*20and 24 = 23*31*50, so we go through the primes and we see that the lcm would be 23*31*51 = 120.

Now, suppose we write the "complete" prime factorizations of m and n
m = p_1^e_1*p_2^e_2*...*p_k^e_k and n = p_1^f_1*...*p_k^f_k
then let g_i = e_i for e_i  \geq f_i and 0 otherwise
and  let h_i = f_i for f_i > e_i and 0 otherwise

then let m' = p_1^g_1*...*p_k^g_k and let n' = p_1^h_1*...*p_k^h_k, and we quickly see that m'*n' = lcm(m,n) and that (m',n') = 1 and m'|m and n'|n

Now, consider the element a^(m/m')*b^(n/n') now the order of a^(m/m') = m' and the order of b^(n/n') = n', and since (m',n') = 1, by the lemma above a^(m/m')*b^(n/n') has order m'*n' = lcm(m,n)

for U(p) is cyclic:

lemma: if n is the max order of all elements in U(p), then for all u, in U(p) u^n = 1

proof: let n = max order or all elements in U(p), and suppose there exists some g in U(p) s.t. order of g = k and k does not divide n.   Then g^n != 1, but if this were the case, then by the proof of 1), we can find an element of U(p) with order lcm (k, n) > n, contradiction.

now, U(p) has p-1 elemtns, and according the lemma above, if n = max order in U(p), then all elements raised to this power are 1.  then consider the polynomial xn - 1, this polynomial has p - 1 roots, so n  \geq p-1, but we also know that by fermat's little theorem, n|p-1 since all elements raised to the p-1 yield 1.  Then since n|p-1, n  \leq p-1, and at the same time n  \geq p-1, so n = p-1, and the max order is p-1, meaning that there is a primitive root (a generator) for U(p), U(p) is cyclic.

sorry for the delay in post, I was out of town for the last few days

lemma I: a = 1 mod pm  iff ap = 1 mod p m+1 for m a positve integer and some integer a and odd prime

-> suppose we have a = 1 mod pm, then there exists some integer r s.t a = pm*r + 1, then ap = (pm*r + 1)p =  \sum pi=0 C(p,i)*(pm*r)i = 1 + pm+1*(some other terms) = 1 mod pm+1

<- suppose ap = 1 mod pm+1, then ap = 1 mod p, and by Fermat's little Theorem, a = 1 mod p, then there exists some k so that a = 1 + pk, and ap = (1 + pk)p = 1 mod pm+1 or pm+1|(1 + pk)p - 1, and we can expand this to see that pm+1|p2*(k + C(P,2)*k2 + ... + pp-2*kp) or pm-1|k*(1 + C(p,2)*k + ... + pp-2*kp-1), and each term in (1 + C(p,2)*k + ... + pp-2*kp-1),  is divisible by p except for the 1, so (1 + C(p,2)*k + ... + pp-2*kp-1),  = 1 mod p and (1 + C(p,2)*k + ... + pp-2*kp-1),  is relatively prime to p, implying that (1 + C(p,2)*k + ... + pp-2*kp-1),  is relatively prime to pm-1, so therefore pm-1|k so there exists some integer t such that pm-1*t = k, then a = 1 + pk = 1 + pm*t = 1 mod pm

we can now use Lemma I to prove 3)

an easy induction combined with lemma I shows us that if a = 1 mod p, then a(p^m) = 1 mod pm+1 (exercise left to reader)

then since we know that U(p) is cyclic, there exists some g element of U(p) so that the order of g is p-1, gp-1 = 1 mod p.

Then gp-1^(p^(m-1)) = 1 mod p^m, so that g(p^(m-1))^(p-1) = 1 mod p^m, so that if the order of g(p^(m-1)) is L, then L|(p-1) or L \leq p-1, but at the same time if g(p^(m-1))^L = 1 mod p^m, then g(p^(m-1))^L = 1 mod p, and by Fermat's little Theorem gL = 1 mod p, so (p-1)|L, and p-1 \leq L therefore p-1 = L, and we have found an element of order p-1

now, we show that the element p+1 has order p^(m-1) in U(p^m) by induction

base case: p + 1 = 1 mod p, so by lemma I, (p+1)^p = 1 mod p^2 so the order of p + 1 is either 1 or p, but it isn't 1 becase p<p+1<p^2, so its order is p^1

Induction Hypothesis: suppose that the order of (p+1) is p^(m-1) mod p^m

Inductive Step: by the induction hypothesis and Lemma I, (p+1)^(p^m) = 1 mod p^(m+1), so the order of (p+1) is p^k, k \leq m, but then by lemma 1 and the induction hypothesis k-1 = m-1 so k = m, and the order of (p+1) = p^m, proving our claim

Now, in U(p^m) we have found an element of order p-1 and an element of order p^(m-1), so by 1), there is an element of lcm(p-1,p^(m-1)) and since (p-1,p^(m-1)) = 1, there is an element c, with order (p-1)*p^(m-1), and since the number of elements of U(p^m) is phi(p^m) = (p-1)*p^(m-1), c is a generator of U(p^m), it is cyclic, it has a primitive root

QED

Here's a nice result. I read it a few weeks ago, and only now have I realized that it fits perfectly here:

If $x$ is a primitive root mod $p$ and $p^2$ then it's a primitive root mod $p^n,\ \forall n\in N^*$.

I't proven by induction. Let's show it for $p^3$. This will illustrate the induction step.

Let's assume the order of $x$ mod $p^3$ is $u$. Since $x$ is a primitive root of $p$, we find that $p-1|u$. $x$ is also a primitive root of $p^2$, so $p(p-1)|u$. What we want to show is that $u\neq p(p-1)$, or, in other words, $p^3$ doesn't divide $x^{p(p-1)}-1$.

Let's assume the opposite. It's easy to see that this is equivalent to $p^2|\frac{x^{p(p-1)}-1}{x^{p-1}-1}=(x^{p-1})^{p-1}+(x^{p-1})^{p-2}+\cdots+x{p-1}+1$. However, $x^{p-1}=kp+1\Rightarrow p^2|((p-1)kp+1)+((p-2)kp+1)+\cdots+(kp+1)+1\mbox{ (by expanding the powers of the binomials in (*))}=M_{p^2}+p$, which is obviously false. This shows that $x$ is a primitive root of $p^3$. The induction step $n\rightarrow n+1$ is basicly identical to the proof for $p^3$.

This means that it suffices to show that there exists a primitive root of $p$ and $p^2$ betqeen $1$ and $p$.

By a similar method you can show if g is a primitive root modulo $p^n$, then either g is a primitive root mod $p^{n+1}$ or (g+p) is a primitive root mod $p^{n+1}$.


conics
ellipse
rotation
trigonometry
trig identities
Law of Cosines

You have an $m$ by $n$ (assume $m\geq n$) piece of paper.  The sides are horizontal and drawn on it is the largest ellipse that will fit in it, with the sides tangent to it at the endpoints of the major and minor axes.
You put a pencil on the center of the paper and a pen on the ellipse so that it is positioned to the right of the pencil, at an endpoint of the major axis, and the tangent line through the pen is vertical.
Then, you move the pencil and the paper moves with it and rotate the paper around the pencil such that the pen stays in the same position, traces arround the allipse, and always the tangent through the pen is always vertical.
The pen returns to its starting position on the ellipse.
Describe the movement of the pencil.

It is possible to do as follows (although this is probably not the nicest method):

Property of an elipse: given any point $P$ on an elipse, the angles made by the segments joining $P$ to the foci $F_1$ and $F_2$ with the tangent at $P$ are equal.  $m$ is the sum of the distances from each point on the elipse to $F_1$ and $F_2$ and $c = \frac12\sqrt{m^2 - n^2}$ is the distance $F_1F_2$.  Thus, if distance $PF_1$ is $x$, we can determine by the law of cosines the cosine of angle $F_1PF_2$ from which we can determine the angles between $F_1P$ and $F_2P$ and the tangent line, from which we can determine the coefficients of $F_1$ and $F_2$ (taking $P$, the pen, as the origin and the fixed tangent line as an axis) and then the center of the elipse is their midpoint and you have a nice parametric expression for it.  From that, you should be able to figure out and prove what the shape actually is (an elipse of some sort, it appears, but it would be nice to pin down its center, shape and size).


What does <==> mean? (double arrow) I keep seeing it in proofs.

It means that each statements implies one another.


What is the radius of the biggest circle that can be drawn inside a quarter circle of radius 1cm?

Now how to solve this?

[hide="hint"]
Find the distance from the center of the desired circle to the center of the quarter circle in terms of the radius of the circle.[/hide]

[hide="Generalized"]
For a $ k^{\circ}$ degree slice of a unit circle, $ k\le 180$, we'll find the largest circle which can be inscribed.  The distance from the center of the desired circle to the center of the unit circle we cut a slice from is $ r\csc{\frac{k}{2}}$.  Then $ 1 \equal{} r(1\plus{}\csc{\frac{k}{2}})\implies r \equal{}\frac{1}{1\plus{}\csc{\frac{k}{2}}}$.[/hide]

And this doesn't go in the Olympiad section...


linear algebra
matrix
induction
algebra
polynomial
modular arithmetic
algorithm

Let $ a_nx \plus{} b_n\ (n \equal{} 1,\ 2,\ \cdots)$ be the residue when $ x^{n \plus{} 1}$ is divide by $ x^2 \minus{} x \minus{} 1.$ Prove that $ a_n,\ b_n$ are coprime.

[hide="In fact..."] $ a_n, b_n$ are consecutive Fibonacci numbers.  This result is equivalent to Binet's formula. [/hide]

The simplest proof is by induction.  One can also explicitly compute the behavior of multiplication by $ x$ on the basis $ \{ 1, x \}$ of $ \mathbb{Z}[x]/(x^2 \minus{} x \minus{} 1)$, which is in general the companion matrix of the corresponding polynomial.  Companion matrices are diagonalized by Vandermonde matrices, and the result follows.

What's wrong with induction?  Have you actually done the computation?  The key is that $ x^{n\plus{}2} \equiv x^{n\plus{}1} \plus{} x^n \bmod (x^2 \minus{} x \minus{} 1)$.

[hide="From the two hints"] If you accept that "polynomial modular arithmetic" makes sense, the following is straightforward:  $ x^{n \plus{} 1} \equiv a_n x \plus{} b_n \bmod (x^2 \minus{} x \minus{} 1)$.  On the other hand, $ x^{n \plus{} 3} \equiv x^{n \plus{} 2} \plus{} x^{n \plus{} 1} \bmod (x^2 \minus{} x \minus{} 1)$.  It follows that

$ a_{n \plus{} 2} x \plus{} b_{n \plus{} 2} \equiv (a_{n \plus{} 1} \plus{} a_n) x \plus{} (b_{n \plus{} 1} \plus{} b_n) \bmod (x^2 \minus{} x \minus{} 1)$

from which it follows that

$ a_{n \plus{} 2} \equal{} a_{n \plus{} 1} \plus{} a_n$
$ b_{n \plus{} 2} \equal{} b_{n \plus{} 1} \plus{} b_n$

and now one only has to check the initial conditions.  Since $ x \equiv x \bmod (x^2 \minus{} x \minus{} 1)$ and $ x^2 \equiv x \plus{} 1 \bmod (x^2 \minus{} x \minus{} 1)$, those initial conditions are $ a_0 \equal{} 1, b_0 \equal{} 0, a_1 \equal{} 1, b_1 \equal{} 1$ and the result follows.

More precisely, one finds that $ a_n \equal{} F_{n\plus{}1}, b_n \equal{} F_n$.  This is equivalent to the pair of identities

$ \phi^{n\plus{}1} \equal{} F_{n\plus{}1} \phi \plus{} F_n$
$ \varphi^{n\plus{}1} \equal{} F_{n\plus{}1} \varphi \plus{} F_n$

and subtracting the two identities and dividing we obtain Binet's formula,

$ F_{n\plus{}1} \equal{} \frac{\phi^{n\plus{}1} \minus{} \varphi^{n\plus{}1}}{\phi \minus{} \varphi}$.  [/hide]

I use complicated terminology, but one shouldn't take it as an indication that the concepts themselves are complicated.  I mention the general solution because it is general.

Why do you think the residue when $ x^{n\plus{}2}$ is divided by $ x^2\minus{}x\minus{}1$?

Really? Even I can follow t0rajir0u's line of reasoning.
[hide]
$ x^2 \minus{} x \minus{} 1 \equiv 0 \pmod{x^2 \minus{} x \minus{} 1}$
$ x^2 \equiv x \plus{} 1 \pmod{x^2 \minus{} x \minus{} 1}$
$ x^{n \plus{} 1} \equiv x^{n} \plus{} x^{n \minus{} 1} \pmod{x^2 \minus{} x \minus{} 1}$

Let $ F_0 \equal{} 0$, $ F_1 \equal{} 1$, and $ F_{n \plus{} 2} \equal{} F_{n \plus{} 1} \plus{} F_{n}$.

Now proving that $ x^{n \plus{} 1} \equiv F_{n \plus{} 1}x \plus{} F_{n} \pmod{x^2 \minus{} x \minus{} 1}$ using induction:

For $ n \equal{} 0$ we have:
$ x^1 \equiv x \equiv F_{1}x \plus{} F_{0} \pmod{x^2 \minus{} x \minus{} 1}$

For $ n \equal{} 1$ we have:
$ x^2 \equiv x \plus{} 1 \equiv F_{2}x \plus{} F_{1} \pmod{x^2 \minus{} x \minus{} 1}$

If $ x^{n \minus{} 1} \equiv F_{n \minus{} 1}x \plus{} F_{n \minus{} 2}\pmod{x^2 \minus{} x \minus{} 1}$ and $ x^{n} \equiv F_{n}x \plus{} F_{n \minus{} 1}\pmod{x^2 \minus{} x \minus{} 1}$ then:

$ x^{n \plus{} 1} \equiv x^{n} \plus{} x^{n \minus{} 1} \pmod{x^2 \minus{} x \minus{} 1}$
$ x^{n \plus{} 1} \equiv F_{n}x \plus{} F_{n \minus{} 1} \plus{} F_{n \minus{} 1}x \plus{} F_{n \minus{} 2} \pmod{x^2 \minus{} x \minus{} 1}$
$ x^{n \plus{} 1} \equiv (F_{n} \plus{} F_{n \minus{} 1})x \plus{} (F_{n \minus{} 1} \plus{} F_{n \minus{} 2}) \pmod{x^2 \minus{} x \minus{} 1}$
$ x^{n \plus{} 1} \equiv F_{n \plus{} 1}x \plus{} F_{n} \pmod{x^2 \minus{} x \minus{} 1}$ as desired.

Therefore, $ a_n$ and $ b_n$ are consecutive Fibonacci numbers. It is easy to prove by induction using the Euclidean Algorithm that $ \gcd(a_n,b_n) \equal{} 1$. QED. [/hide]

EDIT: Sorry, didn't see the last few posts.

[quote="mathwizarddude"]Is there an elementary solution to this problem?[/quote]

First, I dare to say, the problem has been posed in Tokyo University entrance examination.

[color=blue]You can use mathematical induction and contradiction.[/color]

Generally speaking, Japanese high school students don't study modular arithmetic.

In my opinion, I don't like modular arithemetic, because just remembering the property gives no sense for residue.

I should have the problem in the intermidate Section.

Here is my solution of the half:

We can write $ x^{n \plus{} 1} \equal{} (x^2 \minus{} x \minus{} 1)Q(x) \plus{} a_nx \plus{} b_n, Q(x):$ polynomial.
From $ x^{n \plus{} 2} \equal{} (x^2 \minus{} x \minus{} 1)xQ(x) \plus{} a_nx^2 \plus{} b_nx$
$ \equal{} (x^2 \minus{} x \minus{} 1)\{xQ(x) \plus{} a_n\} \plus{} (a_n \plus{} b_n)x \plus{} a_n$, we have $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n,\ b_{n \plus{} 1} \equal{} a_n.$

[quote="kunny"]We can write $ x^{n \plus{} 1} \equal{} (x^2 \minus{} x \minus{} 1)Q(x) \plus{} a_nx \plus{} b_n, Q(x):$ polynomial. [/quote]
But this is exactly what the modular notation means.

[quote="kunny"]From $ x^{n \plus{} 2} \equal{} (x^2 \minus{} x \minus{} 1)xQ(x) \plus{} a_nx^2 \plus{} b_nx$
$ \equal{} (x^2 \minus{} x \minus{} 1)\{xQ(x) \plus{} a_n\} \plus{} (a_n \plus{} b_n)x \plus{} a_n$, we have $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n,\ b_{n \plus{} 1} \equal{} a_n.$[/quote]
Yes, this is exactly what I meant when I said that we can write down the action of multiplication by $ x$ on $ \mathbb{Z}[x]/(x^2 \minus{} x \minus{} 1)$.

[quote="t0rajir0u"]
[quote="kunny"]From $ x^{n \plus{} 2} \equal{} (x^2 \minus{} x \minus{} 1)xQ(x) \plus{} a_nx^2 \plus{} b_nx$
$ \equal{} (x^2 \minus{} x \minus{} 1)\{xQ(x) \plus{} a_n\} \plus{} (a_n \plus{} b_n)x \plus{} a_n$, we have $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n,\ b_{n \plus{} 1} \equal{} a_n.$[/quote]
Yes, this is exactly what I meant when I said that we can write down the action of multiplication by $ x$ on $ \mathbb{Z}[x]/(x^2 \minus{} x \minus{} 1)$.[/quote]

How do you continue after that without using Fibonacci Sequence directly?

Hmm...

Proceed by induction. $ n\equal{}1$ gives $ (1,0)\equal{}1$. Assume that $ x^k\equal{}q(x)(x^2\minus{}x\minus{}1)\plus{}ax\plus{}b$ where $ (a,b)\equal{}1$.

Then $ x^{k\plus{}1}\equal{}(x q(x))(x^2\minus{}x\minus{}1)\plus{}ax^2\plus{}bx\equal{}(x q(x)\minus{}a)\plus{}(a\plus{}b)x\plus{}a$. So obviously $ (a\plus{}b,a)\equal{}(a,b)\equal{}1$ so we are done by induction.

[quote="kunny"]How do you continue after that without using Fibonacci Sequence directly?[/quote]
Take the determinant of $ \left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]$.

[quote="kunny"][quote="Altheman"]

Then $ x^{k \plus{} 1} \equal{} (x q(x))(x^2 \minus{} x \minus{} 1) \plus{} ax^2 \plus{} bx \equal{} (x q(x) \minus{} a) \plus{} (a \plus{} b)x \plus{} a$.[/quote]

I can't understand the part of $ (x q(x))(x^2 \minus{} x \minus{} 1) \plus{} ax^2 \plus{} bx \equal{} (x q(x) \minus{} a) \plus{} (a \plus{} b)x \plus{} a$.[/quote]
This is [b]exactly[/b] what you did.

Alternately, you can simply plug in the two points and find the equation of the line. It'll work for any degree polynomial although you might have trouble Lagranging it once it grows too large. :P

[quote="t0rajir0u"][quote="Altheman"]

Then $ x^{k \plus{} 1} \equal{} (x q(x))(x^2 \minus{} x \minus{} 1) \plus{} ax^2 \plus{} bx \equal{} (x q(x) \minus{} a) \plus{} (a \plus{} b)x \plus{} a$.[/quote]

[quote="kunny"]I can't understand the part of $ (x q(x))(x^2 \minus{} x \minus{} 1) \plus{} ax^2 \plus{} bx \equal{} (x q(x) \minus{} a) \plus{} (a \plus{} b)x \plus{} a$.[/quote]
This is [b]exactly[/b] what you did.[/quote]

It's all Greek to me!

[quote="kunny"]From $ x^{n \plus{} 2} \equal{} (x^2 \minus{} x \minus{} 1)xQ(x) \plus{} a_nx^2 \plus{} b_nx$
$ \equal{} (x^2 \minus{} x \minus{} 1)\{xQ(x) \plus{} a_n\} \plus{} (a_n \plus{} b_n)x \plus{} a_n$, we have $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n,\ b_{n \plus{} 1} \equal{} a_n.$[/quote]
Barring typos, Altheman did exactly what you did here.

$ \left\|\ \begin{array}{ccccc}
\left(X^2\minus{}X\minus{}1\right) & | & \left(X^{n\plus{}1}\minus{}a_nX\minus{}b_n\right) & \equiv & f_n\\\\
\left(X^2\minus{}X\minus{}1\right) & | & \left(X^{n\plus{}2}\minus{}a_{n\plus{}1}X\minus{}b_{n\plus{}1}\right) & \equiv & f_{n\plus{}1}\end{array}\ \right\|\ \implies$ $ \left(X^2\minus{}X\minus{}1\right)\ |\left(f_{n\plus{}1}\minus{}X\cdot f_n\right)$ $ \implies$

$ \left(X^2\minus{}X\minus{}1\right)\ |\ \left[a_nX^2\plus{}\left(b_n\minus{}a_{n\plus{}1}\right)X\minus{}b_{n\plus{}1}\right]\implies$ $ \frac {a_n}{1}\equal{}\frac {b_n\minus{}a_{n\plus{}1}}{\minus{}1}\equal{}\frac {\minus{}b_{n\plus{}1}}{\minus{}1}\implies$

for any $ n\in\mathbb N$ we have $ \left\|\ \begin{array}{c}
a_{n\plus{}1}\equal{}a_n\plus{}b_n\\\\
b_{n\plus{}1}\equal{}a_n\end{array}\ \right\|$ , where $ a_0\equal{}1$ and $ b_0\equal{}0$ . In conclusion,

for any $ n\in\mathbb N$ we have $ a_{n\plus{}2}\equal{}a_{n\plus{}1}\plus{}a_n$ and $ b_{n\plus{}1}\equal{}a_n$ , where $ a_0\equal{}a_1\equal{}1$ and $ b_0\equal{}0$ a.s.o.


In $\triangle ABC$, $AX=XY=YB=BC$ and $\angle ABC=120$. Find $\angle BAC$.

Please show your work.

[color=blue][size=75]I believe this was 1993 National target #7, check the greatest problems book, I'll change it..[/size][/color]

[color=red][size=75]Oh, I wonder what I put...[/size][/color]

[hide="Solution"]
ln $e^15$ = $\boxed{15}$

too lazy to provide a full explanation without a diagram
[/hide]

Ahh this was fun, I just did this yesterday! I ended up getting the answer 2 minutes out of time along with #8 but I set up a couple of equations and used substitution using the fact that some of the angles were supplementary and all 3 of the triangles were isosceles.

Hmm just found my work, here are all of the equations I ended up making (Note: Most are not necessary and unsimplified)
y+180-2x=x+180-2y+z
x+180-2y+z=179
y+180-2z=120
y+180-2x=180
y+60-2z=x
3y+60=3z+x
3z+x=3x+z+60
3y=3x+z
2x+60=2z

x is the first triangle's base angle measure and the one you need to find, y the middle one, and z the last one

[hide]
All you do is set one of the angles to x and solve using the sum of triangles angles formula (sum=180) and mark congruent angles and use supplementary angles... not too hard
$15$[/hide]


Does anybody have the folowing documents or related resources:-

1) INMO (Indian National Mathematical Olympiad) - 1989 and 1990
2) RMO (Regional Mathematical Olympiad) - 1989 , 1990 , 2005
3) Postal Coaching-  1989 - 1995 , 1999 - 2003
4) IMO Training Camp - 1989 - 1995, 1999 -  2003

If you have the above resource and their problems, please post them here. If you have any other matter related to the Indian MO System - like other regional/national olympiad level papers, online scans/pdfs of books recommended by INMO at http://www.iisc.ernet.in/mocell , etc. please post them here and share for everone's benefit!!!!

Thanks!!!

Anbody have some resources (similar to ones above ) for INMO!?!?!?

Klava has got just INMO (some ) problems which I am already posting in the resources Section here! Any more suggestions?

Do visit the Resources section for all the problems regarding the Indian MO system that I have. Please post if you have other resources like problems and recommended books besides the ones that I have already posted.

Come on Rushil! There are tons of books with plenty of problems for INMO and the works.
 All the books by C.R. Pranesachar and his colleagues are all extremely helpful and useful.
         Ashwath

There are only 2 - "not TONS!!!' - besides I dont have those 'tons' , probably you have! SO , go ahead and post the problems!

Only 2. Please!
There are ever so many, just 2 are mentioned on the website. most of them are actually available if you write to the publishers.
   Ashwath

P.S I will post a few problems here and there if you like. Most of them have already been posted on the ML forum however.

You can see the posted indian problems at http://www.mathlinks.ro/Forum/resources.php?c=78 ;)

Ashwanth, please give the links of the poblems you have posted and which competition they are from IF they are not already posted in the Resources section - Valentin has given the link above!

I don't quite understand your question.Could you be more specific?
    Ashwath

Please post the links for the problems you intend to post. Now that you understand :?:

well for starts you could try
 [url]http://vyasa.math.iisc.ernet.in/problems/[/url]
   Ashwath
The real majority are in books actually. Not just by the
indian authors but also by ones from abroad.
 I'll try to give you a list of books pretty soon...
   Ashwath

P.S. I'm Ashwath not Ashwanth

Hello,
 Regarding IMO books,there are actually very little compared to places like Russia and USA(and China most definitely). However, here are a few books considered standard
1)Challenges and Thrills of Pre College Mathematics Pranesachar,KN Ranganthan, Venkatachala and Krishnamurthy
2)Number Theory Hardy and Wright
3)Inequalities Polya
4)Problem Primer(Elementary level)
5)Excursion in Mathematics(Published by those inimitable Pune people)
Some books are availble if you contact publishers directly
You can see my post on Good Books in the same Forum for more.
Sathej

Could someone tell me where to get Indian physics olympiads?

By the way,I think you guys must be knowing 123iitjee.com.


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:
\[ \frac{1}{a^3\plus{}3abc\plus{}b^3}\plus{}\frac{1}{a^3\plus{}3abc\plus{}c^3}\plus{}\frac{1}{b^3\plus{}3abc\plus{}c^3}\geq\frac{81}{5(a\plus{}b\plus{}c)^3}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {1}{a^3 \plus{} 3abc \plus{} b^3} \plus{} \frac {1}{a^3 \plus{} 3abc \plus{} c^3} \plus{} \frac {1}{b^3 \plus{} 3abc \plus{} c^3}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}
\]
[/quote]

  Such a nice inequality. n-1 EV does not work. But expanding  works , Argady.  I think the orginal inequality should deserve a nice solution.   Do you use $ a\plus{}b\plus{}c\equal{}3u$ .... to solve it or a completely  new approach  ?

[quote="Gibbenergy"]

But expanding  works , Argady. [/quote]
After expanding we obtain the following inequality:
$ \sum_{sym}(5a^9\plus{}30a^8b\plus{}30a^7b^2\minus{}32a^6b^3\plus{}90a^5b^4\minus{}153a^7bc\plus{}390a^6b^2c\plus{}$
$ \plus{}450a^5b^3c\minus{}963a^5b^2c^2\minus{}519a^4b^4c\plus{}1170a^4b^3c^2\minus{}498a^3b^3c^3)\geq0.$
Here I did not find nothing.  :(
[quote="Gibbenergy"] I think the orginal inequality should deserve a nice solution. [/quote]
Thank you! I have found a nice solution! :P
[quote="Gibbenergy"]  Do you use $ a \plus{} b \plus{} c \equal{} 3u$ .... to solve it or a completely  new approach  ?[/quote]
No and no!  :)

[quote="arqady"][quote="Gibbenergy"]

But expanding  works , Argady. [/quote]
After expanding we obtain the following inequality:
$ \sum_{sym}(5a^9 \plus{} 30a^8b \plus{} 30a^7b^2 \minus{} 32a^6b^3 \plus{} 90a^5b^4 \minus{} 153a^7bc \plus{} 390a^6b^2c \plus{}$
$ \plus{} 450a^5b^3c \minus{} 963a^5b^2c^2 \minus{} 519a^4b^4c \plus{} 1170a^4b^3c^2 \minus{} 498a^3b^3c^3)\geq0.$
Here I did not find nothing.  :(
[quote="Gibbenergy"] I think the orginal inequality should deserve a nice solution. [/quote]
Thank you! I have found a nice solution! :P
[quote="Gibbenergy"]  Do you use $ a \plus{} b \plus{} c \equal{} 3u$ .... to solve it or a completely  new approach  ?[/quote]
No and no!  :)[/quote]
I have already proved it by CYH technique. :)

[quote="Gibbenergy"][quote="arqady"][quote="Gibbenergy"]

But expanding  works , Argady. [/quote]
After expanding we obtain the following inequality:
$ \sum_{sym}(5a^9 \plus{} 30a^8b \plus{} 30a^7b^2 \minus{} 32a^6b^3 \plus{} 90a^5b^4 \minus{} 153a^7bc \plus{} 390a^6b^2c \plus{}$
$ \plus{} 450a^5b^3c \minus{} 963a^5b^2c^2 \minus{} 519a^4b^4c \plus{} 1170a^4b^3c^2 \minus{} 498a^3b^3c^3)\geq0.$
[/quote]

Honestly, this is the one that my expanding technique completely fail in term of breaking the numerator  into Schur term and sum of square terms. And I have to use the dirty trick " assume $ a \equal{}$  min { a,b,c}"

If $ a \leq b \leq c$  then $ b \equal{} a \plus{} x$ , $ c \equal{} a \plus{} x \plus{} y$, expand (computer's works)
[hide]
$ 135a^7x^2 \plus{} 678a^6x^3 \plus{} 1875a^5x^4 \plus{} 135a^7xy \plus{} 1017a^6x^2y \plus{} 873a^6xy^2 \plus{} 3750a^5x^3y \plus{} 3960a^5x^2y^2 \plus{} 2085a^5xy^3 \plus{} 7545a^4x^4y \plus{} 10158a^4x^3y^2 \plus{} 7692a^4x^2y^3 \plus{} 8472a^3x^5y \plus{} 14238a^3x^4y^2 \plus{} 14356a^3x^3y^3 \plus{} 5187a^2x^6y \plus{} 10737a^2x^5y^2 \plus{} 13875a^2x^4y^3 \plus{} 38x^9 \plus{} 3018a^4x^5 \plus{} 2824a^3x^6 \plus{} 135a^7y^2 \plus{} 267a^6y^3 \plus{} 1482a^2x^7 \plus{} 387ax^8 \plus{} 171x^8y \plus{} 552x^7y^2 \plus{} 1134x^6y^3 \plus{} 1548ax^7y \plus{} 4002ax^6y^2 \plus{} 6588ax^5y^3 \plus{} 3885a^4xy^4 \plus{} 9591a^3x^2y^4 \plus{} 11769a^2x^3y^4 \plus{} 642a^5y^4 \plus{} 1560x^5y^4 \plus{} 7026ax^4y^4 \plus{} 3825a^3xy^5 \plus{} 912a^4y^5 \plus{} 1464x^4y^5 \plus{} 6372a^2x^2y^5 \plus{} 4878ax^3y^5 \plus{} 667a^3y^6 \plus{} 2040a^2xy^6 \plus{} 2196ax^2y^6 \plus{} 884x^3y^6 \plus{} 297a^2y^7 \plus{} 597ay^7x \plus{} 75ay^8 \plus{} 315y^7x^2 \plus{} 60y^8x \plus{} 5y^9 \geq 0$
[/hide]
If $ a \leq c \leq b$ then $ c \equal{} a \plus{} x$ ,$ b \equal{} a \plus{} x \plus{} y$
expand and get all terms positive

So, I am curious to see the solution involved no expanding  :D[/quote]

Why are all terms positive? What if $ a < 0$ ?

Dear Can_hang, did U meant
$ \sum\ (3a\plus{}b\plus{}c)^2(b^3\plus{}3abc\plus{}c^3) \leq\ 225$
with $ a\plus{}b\plus{}c\equal{}3$
 :)

[quote="Mathias_DK"]  Why are all terms positive? [/quote]

The problem says that a ,b ,c are nonegative.

[quote="can_hang2007"] I have already proved it by CYH technique.  :) Smile [/quote] Can I see your proof ? Thanks

[quote="nguoivn"]Dear Can_hang, did U meant
$ \sum\ (3a \plus{} b \plus{} c)^2(b^3 \plus{} 3abc \plus{} c^3) \leq\ 225$
with $ a \plus{} b \plus{} c \equal{} 3$
 :)[/quote]
Try $ a \equal{} b \equal{} c \equal{} 1.$ :wink:
Maybe you mean $ \sum (3a \plus{} b \plus{} c)^2(b^3 \plus{} 3abc \plus{} c^3) \leq 375$ $ ?$
If so, try $ a\rightarrow3$ and $ b \equal{} c\rightarrow0.$ :wink:

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {1}{a^3 \plus{} 3abc \plus{} b^3} \plus{} \frac {1}{a^3 \plus{} 3abc \plus{} c^3} \plus{} \frac {1}{b^3 \plus{} 3abc \plus{} c^3}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}
\]
[/quote]
A more clearly hint:
\[ \left( \sum \frac{1}{b^3\plus{}3abc\plus{}c^3}\right) \left( \sum \frac{b^3\plus{}3abc\plus{}c^3}{(b\plus{}c)^2}\right) \ge \left( \sum \frac{1}{b\plus{}c}\right)^2\]
:)

what is CYH method????please send your complete solution

[quote="Gibbenergy"][quote="can_hang2007"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {1}{a^3 \plus{} 3abc \plus{} b^3} \plus{} \frac {1}{a^3 \plus{} 3abc \plus{} c^3} \plus{} \frac {1}{b^3 \plus{} 3abc \plus{} c^3}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}
\]
[/quote]
A more clearly hint:
\[ \left( \sum \frac {1}{b^3 \plus{} 3abc \plus{} c^3}\right) \left( \sum \frac {b^3 \plus{} 3abc \plus{} c^3}{(b \plus{} c)^2}\right) \ge \left( \sum \frac {1}{b \plus{} c}\right)^2
\]
:)[/quote]

Yes, your solution is correct. Thanks  :D[/quote]

Gibbenergy, do you use pqr method to solve

$ 5(\sum a)^3(\sum a^2\plus{}3\sum ab) \geq 81\sum(b^3\plus{}3abc\plus{}c^3)(b^2 \plus{}\sum ab)^2$

Thank you very much.

[quote="gilcu3"]Please can_hang2007, can you post a complete solution to this? Thanks[/quote]
See my full proof at: http://canhang2007.wordpress.com/2008/12/09/michael-rozenbergs-inequality/ :)
What is your proof, arqady? Thank you :)

Here is my proof:
$ \sum_{cyc}\frac {1}{a^3 \plus{} 3abc \plus{} b^3} \equal{} \sum_{cyc}\frac {(4c \plus{} a \plus{} b)^2}{(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}\geq$
$ \geq\frac {36(a \plus{} b \plus{} c)^2}{\sum(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}.$
Hence, it remains to prove that $ \frac {36(a \plus{} b \plus{} c)^2}{\sum(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}.$
But $ \frac {36(a \plus{} b \plus{} c)^2}{\sum(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}\Leftrightarrow$
$ \sum_{cyc}(2a^5 \plus{} 10a^4b \plus{} 10a^4c \plus{} 47a^3b^2 \plus{} 47a^3c^2 \minus{} 230a^3bc \plus{} 114a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5 \minus{} a^4b \minus{} ab^4 \plus{} b^5) \plus{} 11\sum_{cyc}(a^4b \minus{} a^3b^2 \minus{} a^2b^3 \plus{} ab^4) \plus{}$
$ \plus{} 58\sum_{cyc}(c^3a^2 \minus{} 2c^3ab \plus{} c^3b^2) \minus{} 114\sum_{cyc}(a^3bc \minus{} a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(58c^3 \minus{} 57abc \plus{} a^3 \plus{} 12a^2b \plus{} 12ab^2 \plus{} b^3)\geq0.$
$ a^3\plus{}12a^2b \plus{} 12ab^2 \plus{} b^3\geq26\left(\sqrt {ab}\right)^3.$ Let $ c \equal{} t\sqrt {ab}.$
Hence, it remains to prove that $ 58t^3 \minus{} 57t \plus{} 26\geq0,$ which is true because $ 3\sqrt [3]{58\cdot13^2} > 57.$

[quote="arqady"]Here is my proof:
$ \sum_{cyc}\frac {1}{a^3 \plus{} 3abc \plus{} b^3} \equal{} \sum_{cyc}\frac {(4c \plus{} a \plus{} b)^2}{(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}\geq$
$ \geq\frac {36(a \plus{} b \plus{} c)^2}{\sum(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}.$
Hence, it remains to prove that $ \frac {36(a \plus{} b \plus{} c)^2}{\sum(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}.$
But $ \frac {36(a \plus{} b \plus{} c)^2}{\sum(4c \plus{} a \plus{} b)^2(a^3 \plus{} 3abc \plus{} b^3)}\geq\frac {81}{5(a \plus{} b \plus{} c)^3}\Leftrightarrow$
$ \sum_{cyc}(2a^5 \plus{} 10a^4b \plus{} 10a^4c \plus{} 47a^3b^2 \plus{} 47a^3c^2 \minus{} 230a^3bc \plus{} 114a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5 \minus{} a^4b \minus{} ab^4 \plus{} b^5) \plus{} 11\sum_{cyc}(a^4b \minus{} a^3b^2 \minus{} a^2b^3 \plus{} ab^4) \plus{}$
$ \plus{} 58\sum_{cyc}(c^3a^2 \minus{} 2c^3ab \plus{} c^3b^2) \minus{} 114\sum_{cyc}(a^3bc \minus{} a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(58c^3 \minus{} 57abc \plus{} a^3 \plus{} 12a^2b \plus{} 12ab^2 \plus{} b^3)\geq0.$
$ a^3 \plus{} 12a^2b \plus{} 12ab^2 \plus{} b^3\geq26\left(\sqrt {ab}\right)^3.$ Let $ c \equal{} t\sqrt {ab}.$
Hence, it remains to prove that $ 58t^3 \minus{} 57t \plus{} 26\geq0,$ which is true because $ 3\sqrt [3]{58\cdot13^2} > 57.$[/quote]
Nice. So we have 2 ways to prove this inequality by Cauchy Schwarz. :)  :lol:

[size=200][b]Cauchy Schwarz is forever![/b][/size]

[quote="can_hang2007"]
[size=200][b]Cauchy Schwarz is forever![/b][/size][/quote]
Yes! :lol:


How do I make a commented bibliography?

I imagine something like:

[Gar59] Gardner, Martin. The Scientific American Book of Mathematical Puzzles and Diversions. Simon and Schuster, 1959.
This book contains many puzzles and games of interest to the mathematician.

Thanks a bunch

do you use bibtex (with a file *.bib)?  In that case, you can add the field NOTE.

Ledurt

[quote="Ledurt"]do you use bibtex (with a file *.bib)?  In that case, you can add the field NOTE.[/quote]

I do, thanks, that's just what I need.

Can I modify the bibliography style to insert a linebreak before the note? (Well, how?)

I prefer the style alpha.

ok, in the main style file of bibtex, there's no style that can do that.  But, on this site [url=http://www-math.cudenver.edu/~billups/courses/ma5779/annotated_bibliography.html]www-math.cudenver.edu/~billups/courses/ma5779/annotated_bibliography.html[/url] you can get a style file of bibtex to create annoted bibliography.  I tried it and it worked so if you have an trouble to add this file, just tell me.  In this case, you use the field "annote" instead of "note".

By the way, do you know how to change "References" to "Bibliogrphy" in an article?

Ledurt

[quote="Ledurt"]By the way, do you know how to change "References" to "Bibliogrphy" in an article?[/quote]Try \renewcommand{\bibname}{Bibliography}, where you can replace "Bibliography" with whatever you want to call it.

If I put this command in the preambule, I have an error because latex doesn't recongnize the command \bibname.  If I put it in the document, I don't have an error but the name stays "references".  Any clues?

ledurt

Great, this works.

You wouldn't happen to know of an annote style based on alpha rather than plain?

I'm afraid I can't help with your Ref->Bib.

Thanks

[quote="Ledurt"]If I put this command in the preambule, I have an error because latex doesn't recongnize the command \bibname.  If I put it in the document, I don't have an error but the name stays "references".  Any clues?[/quote]Oops, I forgot that I was using the [b]cite[/b] package.  Sorry -- it may not work in standard LaTeX.  On the other hand, my editor seems to recognize \bibname as a standard LaTeX command.

In other words, I don't have a clue.  :wacko:

[quote="Xevarion"]Great, this works.
You wouldn't happen to know of an annote style based on alpha rather than plain?
[/quote]

No, I don't.  I found this while looking on google so maybe you can find one or edit a new one.

ledurt


function
integration
inequalities
limit
real analysis
real analysis solved

Let [tex]f:[0,1]\to[0,\infty)[/tex] be a continous function such that [tex]\int_0^xf(t)dt\leq x\int_0^1f\left(\frac{t}{2}\right)dt,\;\forall x\in[0,1][/tex]. Prove that [tex]\int_0^xf(t)dt\leq f\left(\frac{1}{2}\right),\;\forall x\in[0,1][/tex].

Put $F(x)=\int^x_0 f(t)dt$. The hypothesis simply tells us that $\frac{F(x)}x$ reaches its maximum in $x=\frac 12$. We have to show that $F(1)\le F'(\frac 12)$ (this suffices, since for all $x\in[0,1]$ we have $F(x)\le F(1)$).

It's easy to show now that if $x<\frac 12$, then $\frac{F(x)-F(\frac 12)}{x-\frac 12}\le\frac{F(\frac 12)}{\frac 12}$, and the inequality is reversed if $x>\frac 12$. This means that $F'(\frac 12)=\frac{F(\frac 12)}{\frac 12}$, and the conclusion follows from here.

Another way could be:

$x \int_0^1f\left(\frac{t}{2}\right)dt= x\int_0^{2x}f\left(\frac{t}{2}\right)dt + x\int_{2x}^1 f\left(\frac{t}{2}\right)dt \\
= 2x\int_0^x f(t)dt + x(1-2x) f\left(\xi\right),$

where $\xi\in (2x, 1)$ and $x < \frac{1}{2}$.

Therefore

$(1- 2x)\int_0^x f(t)dt \leq x(1-2x)f(\xi)$

where $\xi \in (2x,1)$ and $x\in [0,\frac{1}{2})$,

hence

$\int_0^x f(t)dt \leq x f(\xi)$

and now taking $\lim_{x \rightarrow \frac{1}{2}, x < \frac{1}{2}}$ from both sides gives the result.

Sorry, $\xi \in (x, \frac{1}{2})$ in my previous posting.


limit
trigonometry
calculus
calculus computations

Hi. I wonder how you can find the following limit:

$ \lim_{x\to\infty}\left[x\sin\left(\frac{1}{x}\right)\right]$

I see that $ 1/x$ goes towards $ 0$, and that $ \sin(1/x)$ therefore also goes towards $ 0$. But $ x$ goes towards infinity. Mathematica gave me the answer: $ \pi$. But why is it so..?

hello, are yor sure, that Mathematica gives the answer $ \pi$?
Write your limit in the form $ \lim_{x \to \infty}\frac{sin(\frac{1}{x})}{\frac{1}{x}}$ and substitute $ t\equal{}\frac{1}{x}$.
Sonnhard.

Thanks a lot Sonnhard! That was a smart trick. You are of course right, the correct answer is 1. My original problem was $ \sin(\pi/x)$, not $ \sin(1/x)$, but I forgot the $ \pi$ when I entered it into the forum
 :oops:


A hockey team consists of: 1 goalkeeper, 4 defenders, 4 midfielders and 2 forwards. There are 4 substitutes: 1 goalkeeper, 1 defender, 1 midfielder and 1 forward.

Given a maximum of three substitutions can be made, 11 players end the game, and the team formation is never changed (ie. a midfielder can only be replaced by another midfielder), how many possible teams could finish the match?

Note a player that has been substituted off can be still be substituted back on again.

[hide]
No substitutions are made: $1$ team (starting team)

$1$ substitution is made: $11$ different teams

$2$ substitutions are made:

The different pairs of substituions (goalie/def, goalie/mid, goalie/for, def/mid, def/for, mid/for):

$1(4+4+2) + 4(4+2) + 4(2) = 42$ different teams

$3$ substitutions are made:

The different triplets (goalie/def/mid, goalie/def/for, goalie/mid/for, def/mid/for)

$1\times 4 \times 4 + 1\times 4 \times 4 + 1\times 4 \times 2 + 4\times 4 \times 4 = 64$ different teams that are not included in the one substitution group

Total number of teams is then $118$

[/hide]


calculus
integration
calculus computations

Let $ K$ be the figure bounded by the curve $ y\equal{}e^x$ and 3 lines $ x\equal{}0,\ x\equal{}1,\ y\equal{}0$ in the $ xy$ plane.

(1) Find the volume of the solid formed by revolving $ K$ about the $ x$ axis.

(2) Find the volume of the solid formed by revolving $ K$ about the $ y$ axis.

hello kunny, to (1), we have $ V_x\equal{}\pi\cdot\int_{0}^1(e^x)^2\,dx\equal{}2\pi(e\minus{}1)$.
Sonnhard.

hello, to (2), we have $ V_y\equal{}\pi\int_{1}^e(\ln(x))^2\,dx\equal{}\pi(e\minus{}2)$.
Sonnhard.


Alice leaves Denver, driving at a constant speed. After a while, she passes a mile marker displaying a two-digit number. One hour later, she passes a second marker with the same two digits in reverse order. In another hour, she passes a third marker with the same two digits separated by a zero. What is the rate of Alice's car in miles per hour?

[hide]We know that in the first number, the first digit must be a 1, as otherwise the 2nd and 3rd numbers wouldn't correspond.

A bit of guess and check leads to the answer 45 miles per hour (16, 61, 106)[/hide]

[hide="hint"]

How do you put into words " the same two digits in reverse order?"[/hide]

[hide="answer"]

The first number we can call 10a+b, the second 10b+a and the third 100a+b

We have the speed being x.

10a+b+x=10b+a

100a+b=10b+a+x

Simplifying these, you can get x=45a.  Since we are dealing with speeds (legally), "a" would have to be one (or two if your in montana or something) .

Just try the two possibilities, the only restriction is a and b must be single digit.

The speed is thus 45 mph.  [/hide]

who said that they were travelling legal speeds? :D

[hide]Supposing the first digit is a and the second b;
The first marker is shown as 10a+b.
The second marker is shown as 10b+a.
The third marker is shown as 100a+b.

Supposing X is the velocity of the car(mph);
X=9(b-a)
X=9(11a-b)
X=45a

Using the last equation, without exeeding the speed limit, "a" must be one concluding the velocity of the car is 45mph
a=1, X=45
a=2, X=90
a=3, X=135

[/hide]

remember this from the 150 greatest problems
[hide]45 miles per hour yo

[color=blue]Moderator kyyuanmathcount's note: AGAIN, PLEASE, and PLEASE give us reasons, we already know the answer is 45, but we really need to know better methods.[/color]
[/hide]


I got into a discussion with my professor about $ (NH_{4})_{2}S$. He claimed that it was soluble, but that it did not break up into ions. Since it is a soluble ionic compound, I don't understand why it wouldn't break up into ions (just like NaCl). I didn't understand his explanation.

Anyway, help would be much appreciated. I don't want to lose any more test points in the future.

Just read it up on wiki (it is a stub). According to wiki, that salt does not exist.

If an ionic compound is soluble (in water), then in aqueous solution it will split into solvated ions. If a compound is soluble but does not split into ions that means only one thing: the compound has a considerable degree of covalent character.


arithmetic sequence
number theory proposed
number theory

Show that there exist strictly increasing infinite arithmetic sequence of integers which has no numbers in common with the Fibonacci sequence.

[hide]Consider Fibbonaci sequence $ \mod 8$[/hide]

The Fibonacci sequence is defined by $f_1 = 0, f_2 = 1$, and $f_{n+2} = f_{n+1}+f_n$ for $n\ge 1$. Prove that there is a strictly increasing arithmetic progression whose no term is in the Fibonacci sequence.

Just stating the problem as it was in the contest for the sake of the Resources section. BTW: The problem is from the 2004 contest.

The $n$-th Pisano period for $n = F_{2k}+F_{2k+2}$ is small, namely precisely $4k+2$ long. So for $k>1$, all these numbers $n$ work ($11,29,76,\ldots$), by missing residues in the Fibonacci sequence. See [url]http://en.wikipedia.org/wiki/Pisano_period[/url].


limit
integration
calculus
calculus computations

evaluate limite of the sequence , $ U_n\equal{}\sum_{k\equal{}1}^{n}(\frac{k}{n})^n$

Is your problem $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \left(\frac{k}{n}\right)^n$ or $ \lim_{n\to\infty} \frac{1}{n} \sum_{k\equal{}1}^n \left(\frac{k}{n}\right)^n$?

$ \lim_{n\rightarrow\infty}U_{n}\equal{}\lim_{n\rightarrow\infty}\left[\left(\frac{n}{n}\right)^{n}\plus{}\left(\frac{n\minus{}1}{n}\right)^{n}\plus{}\left(\frac{n\minus{}2}{n}\right)^{n}\plus{}\cdots\right]\equal{}\sum_{k\equal{}0}^{\infty}e^{\minus{}k}\equal{}\frac{e}{e\minus{}1}$ :)
And also $ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k\equal{}1}^{\infty}\left(\frac{k}{n}\right)^{r}\equal{}\int_{0}^{1}x^{r}dx\equal{}\frac{1}{1\plus{}r},\forall r\in\mathbb{R}^{\plus{}}$ so your second limit is zero right kunny?

the probleme is without $ \frac{1}{n}$ kunny ,


limit
integration
trigonometry
calculus
real analysis
real analysis unsolved

Prove that $\lim_{n\to\infty}\int^\frac{\pi}{2}_{0}\frac{\sin 2nx}{\sin x}\,dx = \frac{\pi}{2}.$

$\frac{\sin 2nx}{\sin x}-2\cos (2n-1)x= \frac{\sin x\cos (2n-1)x+\sin (2n-1)x\cos x}{\sin x}-\frac{2\sin x\cos (2n-1)x}{\sin x}$
$=\frac{\sin (2n-1)x\cos x-\cos (2n-1)x\sin x}{\sin x}=\frac{\sin ((2n-1)x-x)}{\sin x}=\frac{\sin (2n-2)x}{\sin x}$.
Summing this identity, $\frac{\sin 2nx}{\sin x}=2(\cos x+\cos 3x+\cdots+\cos (2n-1)x)$. The integral is $2\left(1-\frac13+\frac15-\cdots+\frac{(-1)^{n-1}}{2n-1}\right)$. As $n\to\infty$, this series is well-known to converge to $\frac{\pi}2$.

This is not the only proof, of course. We don't have to calculate the sequence to find its limit.

$\int_{0}^{\frac{\pi}2}\frac{\sin((2n+1)x)}{\sin x}\,dx=\frac{\pi}2,$ exactly.

Who can figure out how to use this?

Let's use Kent's nice idea:

We have

$|\int_{0}^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx-\int_{0}^{\pi/2}\frac{\sin(2nx)}{\sin(x)}dx|=|\int_{0}^{\pi/2}\frac{\cos(2n+1/2)x}{\cos x/2}dx|$

$\leq \frac{\sqrt{2}}{2n+1/2}$

The result follows.

Another idea: substitute $t=2nx$. The integral becomes
$\int_{0}^{n\pi}\frac{\sin t}{\sin\frac{t}{2n}}\cdot\frac1{2n}\,dt= \int_{0}^{n\pi}\frac{\sin t}{2n\sin\frac{t}{2n}}\,dt$

The new integrand converges to $\frac{\sin t}{t}$; with appropriate convergence arguments, the limit is $\int_{0}^\infty \frac{\sin t}{t}\,dt=\frac{\pi}{2}$.

[quote="jmerry"]...with appropriate convergence arguments...[/quote]
I looked at that and decided that the arguments were messier and more delicate than I really wanted to write out.

[quote="didilica"]
$|\int_{0}^{\pi/2}\frac{\cos(2n+1/2)x}{\cos x/2}dx|\leq \frac{\sqrt{2}}{2n+1/2}$

The result follows.[/quote]

 Didilica, can you prove that?

[quote="kunny"]Let $I_{n}=\int^\frac{\pi}{2}_{0}\frac{\sin 2nx}{\sin x}\,dx$,

$I_{n}-I_{n-1}=....$[/quote]

What do you mean with "..."?
I`m looking for a solution without improper integrals or using the series. In fact, my problem is to find the limit of that series:$2(1-\frac{1}{3}+\frac{1}{5}-...+\frac{(-1)^{n-1}}{2n-1})$ by calculating the limit of $I_{2n}$. This is a posible subject for the graduation exam in my country, which supposes to be at an intermediate level...


induction
number theory
algebra
binomial theorem

Prove Fermats theorem by expanding (1+1+...+1)^p where p is a prime.
 :maybe:  :maybe:

You mean his little theorem? Fermat has kinda many theorems :)

Hi,

If you mean Fermat's Little Theorem, then my attempt follows.

[hide]a^p = (1+1+... +1)^p , where 1 is summed a times

By Multinomial expansion, we get

a^p = sum of the multinomial coefficients C(p ; m_1,m_2,...m_a), where sum of m_i = p.

Since p is prime, if all m_i are less than p, then the multinomial coefficient is divisible by p and hence is 0 mod p.

But, if exactly one of m_i = p, and the rest are zeros, then the coefficient is 1. These are the only non-zero terms in the expansion mod p.

Thus, working modulo p,

a^p = 1 + 1 + ... + 1 (a times) = a (mod p)

Thus, for all integers a,  a^p = a (mod p); and if gcd(a,p)=1, then we can cancel out a from both sides to get a^(p-1) = 1 (mod p) -----> Fermat's Little Theorem

(I think I saw this question in 'Concrete Mathematics' by Knuth, Graham and Patashnik; but I didnt try the problems at that time.)

[/hide]

*****************************************************************

Sambit

Yeah, I mean the little theorem.

What are the multinomial coefficients and what do m_i's represent in yor solution, Sambit?

Hi,

I will try to explain what multinomial coefficients are.

Consider $n$ objects (not necessarily distinct) out of which $m_1$ are of type $1$, $m_2$ are of type $2$, and so on upto $m_k$ of type $k$.
Here, $m_1 + m_2 + ... + m_k = n$.
Then the no. of possible distinct arrangements of these objects is
$\left( \begin{array}{c} n \\ m_1, m_2,..., m_k \end{array} \right) = \frac{n!}{m_1! m_2! .... m_k!}$ ---------- (1)

This is also the number of ways one can distribute $n$ distinct balls in $k$ distinct cells such that $1$st cell gets $m_1$ balls, $2$nd cell gets $m_2$ balls, and so on upto $k$th cell gets $m_k$ balls.

It is called the multinomial coefficient because
\[{ (a_1 + a_2 + ... + a_k)^n = \sum_{\sum_{i=1}^k m_i = n}} \left( \begin{array}{c} n \\ m_1, m_2,..., m_k \end{array} \right) a_1^{m_1} a_2^{m_2} ... a_k^{m_k} \] --- (2)

(2) is a generalization of the binomial theorem, and it is called the multinomial theorem.

Now, one can see that if $n$ is prime and if all $m_i$'s are less than $n$, then the multinomial coefficient will be divisible by $n$, since in the numerator of (1), a factor of $n$ is present, and it doesn't get cancelled out with any of the terms in the denominator as $n$ is prime and all $m_i$'s are less than it. But if one of the $m_i$'s is equal to $n$, then the others are zeros, and hence the multinomial coefficient is $1$.

Now, I guess you can follow the proof posted before.

PS : Hi all, if I have erred at some point in my explanation of multinomial coefficients, kindly correct me.

*********************************************************************************
Sambit

to avoid multinomials we can use induction: it is very easy to show via binomial expansion that $(a+b)^p\equiv a^p+b^p\; mod\; p$. now, it is clear that $1^p\equiv 1\; mod\; p$. but $(a+1)^p\equiv a^p+1^p\equiv a+1\; mod\; p$. just for those who want a very simple solution :)

Peter


calculus
integration
trigonometry
calculus computations

ı couldn't find this question of jacobien so that. ı can't solve. Can you help me??

you can revrite this integral as
$ \int_{0}^{1}dx\int_{0}^{(1\minus{}\sqrt{x})^{2}}\sqrt{\sqrt{x}\plus{}\sqrt{y}}dy$

here is another way.
After substituting $ x \equal{} r\cos^{4}\alpha$
 and $ y \equal{} r\sin^{4}\alpha$ we have Jacobien
$ |J| \equal{} 4r\sin^{3}\alpha\cos^{3}\alpha$
and $ \sqrt {\sqrt {x} \plus{} \sqrt {y}} \equal{} r^{1/4}$
finally
since $ x\geq 0$ ,$ y \geq 0$ and $ \sqrt {x} \plus{} \sqrt {y}\leq 1$ we have $ 0\leq r \leq 1$ and $ 0 \leq \alpha \leq 2\pi$ hence:
$ \int_{0}^{1}dr\int_{0}^{2\pi}r^{\frac {5}{4}}\cos^{3}\alpha \sin^{3}\alpha d\alpha$ after yourself.
to find last integral you can use $ \sin 2\alpha \equal{} 2\cos\alpha \sin \alpha$


induction
inequalities proposed
inequalities

Let $ a_1,a_2,a_3,......,a_n$ be $ n$ non-negative real numbers, such that $ a_1\plus{}a_2\plus{}.....a_n\equal{}1$. Prove that,
\[ a_1a_2\plus{}a_2a_3\plus{}a_3a_4\plus{}...........a_{n\minus{}1}a_n\leq{\frac{1}{4}}\]

It does not work when $ n\equal{}1$. You need $ n\ge2$.

When $ n\equal{}2$, and $ a_1\plus{}a_2\equal{}1$, we do have $ a_1a_2\le\frac{1}{4}$. The proof is easy and I’ll omit it here.

Assume that for some $ n\ge2$, $ a_1a_2\plus{}\cdots\plus{}a_{n\minus{}1}a_n\le\frac{1}{4}$ whenever $ a_1\plus{}\cdots\plus{}a_n\equal{}1$.

Let $ a_1\plus{}a_2\plus{}\cdots\plus{}a_n\plus{}a_{n\plus{}1}\equal{}1$. WLOG, assume that $ a_n\le a_{n\minus{}1}$.

Then if $ A\equal{}a_n\plus{}a_{n\plus{}1}$ we have $ a_1\plus{}\cdots\plus{}a_{n\minus{}1}\plus{}A\equal{}1$; hence $ a_1a_2\plus{}\cdots\plus{}a_{n\minus{}1}A\le\frac{1}{4}$.

$ \therefore\ a_1a_2\plus{}\cdots\plus{}a_{n\minus{}1}a_n\plus{}a_na_{n\plus{}1}\le a_1a_2\plus{}\cdots\plus{}a_{n\minus{}1}\left(a_n\plus{}a_{n\plus{}1}\right)\le\frac{1}{4}$.

Hence the result is true by induction.

[quote="Jane_Bennet"]It does not work when $ n\equal{}1$. You need $ n\geq{2}$[/quote]
You are right. I didn't mention it since it is very obvious.
Besides, I have already solved the problem with induction. So, solve it without induction! :P

Brilliant! Thanks mitdac123!

You really are a "STUPID BOY"!!! :P  :P  :P


trigonometry
calculus
integration
real analysis
real analysis unsolved

Consider $ f_n : \left[ \minus{} \pi, \pi \right] \to \left[ \minus{} \pi, \pi \right]$, $ f_n (x) \equal{} \sin \left( nx \right)$.
Prove that $ \{ f_n : n  \geq 1 \}$ is closed and bounded in $ L^2$, but not compact.

How do you tackle this? For the first part, I think we should prove that the set has no limit points (haven't yet managed, but I bet it's pretty trivial), but for the second part I have absolutely no idea...

you can find the solution in Rudin's Principles book, page 143  :) ( The section named : Equicontinuous families of fuctions)

Of course, this is an infinite discrete set, and $ \|f_n\minus{}f_m\|_2^2\equal{}1$ for all $ n\ne m.$

Choose the collection of open balls of radius $ \frac12$ centered at each $ f_n.$ This is an open cover of $ \{f_n\},$ but it has no finite subcover, since each of those balls covers only one point and you can't leave any of them out.

(Note: I was normalizing my integrals, saying $ \|f\|_2^2\equal{}\frac1{2\pi}\int_{\minus{}\pi}^{\pi}|f(x)|^2\,dx.$ If you don't divide by $ 2\pi$ in your definition, just make the appropriate constant multiple corrections to what I said above.)


So you are in a maze.

[url]http://www.mafiascum.net/forum/viewtopic.php?t=9427[/url]
for the idea + rules

Maximum 10 players.

EDIT:

I enter the maze.  I turn eastward.

[i][u]AIME15[/u]: You are in a cave. You can see the features of a corner. Wait... so that means I can go in only two directions? Yes, I suppose so. Go South or East.[/i]

james4l, what's your account on MS?
just wondering...
Enter, go north

[i][u]AIME15[/u]: You get gobbled up by this black blob of doom! You frantically try to attack this mysterious blob's stomach, but to no avail...
[b]AIME15 has died, resurrected as AIME16.[/b][/i]

[i][u]pacman2812[/u]: Ouch, that must hurt to bump into a wall. S/E

(to answer your question I don't have a MS account)
[/i]

[i][u]1=2[/u]: It is nice and airy here. You can see remnants of a dinosaur. Maybe you'll get big money... nah, you have to escape from this labyrinth first. N/S/E [/i]

Go east.
Then how did you get that game?

[i][u]AIME16[/u]: S/E [/i]

[i][u]pacman2812[/u]:A voice sounds: "You were dropped from above. Unfortunately there are no ways of getting out of here."

Another blob comes up and kills you!

In your death dream, you hear the same mysterious voice, "Don't repeat the same silly mistakes..."

[b]pacman2812 has died, resurrected as Pacman II.[/b]

S/E
[/i]

[i][u]1=2[/u]: N/S/E[/i]

[i][u]dragon96[/u]: Somehow the dinosaur bones are less shiny than before. You spot a note on the ground! N/S/E[/i]

To pick up an object, just say, "Pick up xxx"

I'll be james for a while.

cf: The wall is solid. The wall hurts. You get hurt. S/E

Dude just wait. If james wants this to die, let it die.

I'm almost done with the map, now onto the enemies and objects XD

It looks like this game is going to continue, but unfortunately, I'm going to have to drop out of it.  Sorry.  :(


Hey,

I'm a noobie at math, so I was hoping I could get some general help. =]

First of all, does practicing olympiad problems improve one's performance on short-answer problems?

Second, I've almost completed AoPS volume 2. However, I've often  had trouble retaining a lot of the information and trouble applying knowledge to contests.

Any advice?

Thanks so much

As long as you are practicing problems that are not too easy (you know the solution before you start) and not too hard (you always have to look up the solution), you will improve your math in general. So practicing olympiad problems may not give as many direct benefits to short answer problems, but they do help.

As for AoPS 2, if you have done all the practice problems there should be no problem retaining the knowledge. If you haven't, do them. This might take a while, but it'll help you remember.


i recently deleted my blog and am now unable to create a new one, do i have to create a new account to make a blog or is there a way to revive my old one???

Thanks

I deleted my blog about half a year ago, and it is still there. :maybe:

I have the same problem. Unfortunately, as far as I know, there is no current fix for this problem. There is however, a post in the Help and Support Forum that may offer you a little information on the topic. If anything does happen, that will probably be the one to watch.


Find all $f: Z\rightarrow Z$ such that

$f(x+y+f(y))=f(x)+2y$.

Let $c=y+f(y)$, then $f(x+c)=f(x)+2y$. Therefore $f(x)$ is surektive and $g(x)=f(x)-\frac{2y}{c}x$ satisfyed $g(x+c)=g(x)$. It give $g(x)=const$ or $f(x)=x$.

It posted here and solved for $f: R \rightarrow R$:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120585


The equation with roots $ 3 \plus{} \sqrt{2}, 3 \minus{} \sqrt{2}, \minus{}3 \plus{} i\sqrt{2}, \minus{}3 \minus{} i\sqrt{2}$ is in the form $ x^4 \plus{} Ax^3 \plus{} Bx^2 \plus{} Cx \plus{} D \equal{} 0$. Find A + B + C + D.

[hide="my way"] I eventually found it by muliplying $ (x\minus{}(3 \plus{} \sqrt{2}))...$ etc., but the solution in the book states that $ f(1) \equal{} A \plus{} B \plus{} C \plus{} D \plus{} 1$... I used that and found the answer shortly just like the rest of the solution, but how did it come to that conclusion? [/hide]

What happens if you replace all the $ x$s with $ 1$s in the equation you're given?

Oh.... wow... :blush: Oh... I thought it had to with the symmetric sums ( because the problem was in that section).. Thanks.  :rotfl:


Let $x,y,z>0$ with  $x+y+z=1$

Prove that

$\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x}+9xyz\geq x^2+y^2+z^2+1$

[quote="pvthuan"]From this I think of the following:
Let $x,y,z$ be positive real numbers such that $x+y+z=1$, prove that
\[ \frac{xy}z+\frac{zx}y+\frac{zy}x\geq(x+y+z)^2.  \][/quote]
And  $9xyz+2(xy+xz+yz)\geq1?$ But this is not true! ;)

[quote="silouan"]You mean that the initial ineq is wrong ?[/quote]
I meant, that on mine pvthuan was mistaken. Your inequality is true, I have proved it, but I can not find the nice proof.

[quote="silouan"]Could you post your solution?[/quote]
$\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x}+9xyz\geq x^{2}+y^{2}+z^{2}+1\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{4}y^{2}+x^{4}z^{2}-2x^{4}yz)+2\cdot\sum_{cyc}(x^{3}y^{3}-x^{3}y^{2}z-x^{3}z^{2}y+x^{2}y^{2}z^{2})\geq0,$
which true by Muirhead and Schur.


Is there any thread for Geometry where one can start from beginner level and then reach up to Olympiad level. We have one such thread in Pre-Olympiad for Inequalities. If one can start it for Geometry also then it would be very helpful for many students and me too.

Till then solve my following doubts:
1.	Equilateral triangles $ABX$ and $ACY$ are described on sides $AB$,$AC$ of a triangle $ABC$ externally to triangle $ABC$. Prove that $CX=BY$.
2.	Prove that the lines joining the mid-points of opposite sides of a quadrilateral and the line joining the mid-points of diagonals are concurrent.

[quote="dangvh"]what does "described" mean?  :huh:[/quote]
Described means drawn.

[hide="Hint for 1"]
Look for a pair of congruent triangles.
If you notice it, it's pretty straight forward.
[/hide]


inequalities
trigonometry
inequalities unsolved

Hm.. I'm sorry but I think that doesn't work in this problem...
I have an idea to rephrase the problem with a trigonometry substitution but I can't finish the outcome.. Can anybody help ?  :maybe:

[quote="RDeepMath91"]Let $a,b,c$ be nonnegative real numbers, so that $ab+bc+ca = 1$
Prove that $\sum_{cyc}\frac{1}{b+c}\geq \frac{5}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476

Another way to prove this Inequality is homogenizing :
$ (ab\plus{}bc\plus{}ca)(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a})^2\ge \frac{25}{4}$
This Ineq go to Schur form..

:rotfl: [quote="BttfUS"][quote="Ji Chen"][quote="RDeepMath91"]Let $ x,y,z$ be nonnegative numbers such that $ yz \plus{} zx \plus{} xy \equal{} 1.$ Prove that

$ \frac {1}{y \plus{} z}\plus{}\frac {1}{z \plus{} x}\plus{}\frac {1}{x \plus{} y}\geq \frac {5}{2},$

with equality if $ x\equal{}0,y\equal{}z\equal{}1.$[/quote]$ \sum{\frac {1}{y \plus{} z}}\minus{}\frac {5}{2}\equal{}\frac{4(x\plus{}y\plus{}z\minus{}2)^2\plus{}xyz\plus{}3\sum{x(y\plus{}z\minus{}1)^2}}{4(y\plus{}z)(z\plus{}x)(x\plus{}y)}\geq0.$[/quote]

That is crazy！[/quote]
it's product by computer program....  :rotfl:

[quote=pohoatza]i think we could use iran 96 here
$\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\geq \frac{9}{4(xy+yz+zx)}$[/quote]

But how to prove this  inequality?

[quote=mihaig][quote=RDeepMath91]Let $a,b,c$ be nonnegative real numbers, so that $ab+bc+ca = 1$
Prove that $\sum_{cyc}\frac{1}{b+c}\geq \frac{5}{2}$[/quote]

This is the professor's Laurentiu Panaitopol problem.
Classic and classy one.[/quote]

In addition, I give a solution using the pqr method.

$q = 1$. We want to prove that $2p^2-5pq+2q+5r\ge 0$, or $2p^2 - 5p + 2 + 5r\ge 0$, or $(2p-1)(p-2) + 5r \ge 0$.

1) $p\ge 2$: Clearly, it is true.

2) $p < 2$: Since $p^3 + 9r\ge 4pq$ (Schur's inequality), we have $r \ge \frac{4pq-p^3}{9} = \frac{4p-p^3}{9}$. It suffices to prove that
$2p^2 - 5p + 2 + \frac{5(4p-p^3)}{9} \ge 0$, or $(2-p)(5p^2-8p+9) \ge 0.$ It is true.

Done.

$ \frac {1}{a + b}+\frac {1}{b + c}+\frac {1}{c + a}\geq \frac {5}{2},$
We put ab+bc+ca=1
We get

$ \frac{ab+bc+ca}{a+b}+\frac{ab+bc+ca}{c+a}+\frac{ab+bc+ca}{a+b}\geq\frac{5}{2},$

[quote=Simonbilth]iran 96 is really enough[/quote]
for example,square the both side,use the iran 96,and reduction of fractions to a common denominator,then obviously true.


Given the following:

$ \ce{ZnSO4 . 7H2O}_{(s)} \rightleftharpoons \ce{ZnSO4 . 6H2O}_{(s)} \plus{} \ce{H2O}_{(l)}$

$ \Delta G \equal{} 1.48 kJ$ and $ T\equal{}291K$.

Given that the vapour pressure of pure water is $ 2.06kPa$ at $ 291K$. What is the vapour pressure of the above reaction?

Please, someone otherwise I can't sleep tonight (troubled by the problem)  :( .

[quote="BanishedTraitor"]Confirming something first. If the activity of a pure substance is one. Then the equilibrium constant $ K \equal{} 1$ so $ \Delta G^{\circ} \equal{} 0$. What's wrong? :([/quote]

I dont think the amount of water stays constant in this case. So its activity isn't 1.  :maybe:
Find $ K_p$ which is nothing but the vapour pressure of pure water.
Using that, get $ K_c$ and then use :
$ \Delta G \equal{} \Delta G^{\circ} \plus{} RT ln K_c$

And as for the sum in general:
what do you mean by vapour pressure of a reaction??  :maybe:  u mean, at equilibrium, what is the vapour pressure?


function limit integration calculus derivative trigonometry inequalities

1) Let $ f : (0, \plus{} \infty) \to \mathbb{R}$ be a coutinuous function such that : $ \lim_{x \to \plus{} \infty}{f(x)} \equal{} 0$
 Prove that : $ \lim_{x \to \plus{} \infty}{e^{ \minus{} x}\int_{0}^{x}{e^{t}f(t)dt}} \equal{} 0$
 2) Let $ f : [0, \plus{} \infty) \to \mathbb{R}$ be an $ n$-time differentiable function with the $ n$-derivative continuos such that
there exists : $ \lim_{x \to \plus{} \infty}{\sum_{k \equal{} 0}^{n}{C_{n}^{k}f^{(k)}(x)}} \equal{} A$
 Prove that : $ \lim_{x \to \plus{} \infty}{f(x)} \equal{} A$

For the first one, if I'm not mistaken, you can just use L'Hospital: $ \lim_{x\to \plus{}\infty}e^{\minus{}x}\int_0^x e^tf(t)dt\equal{}\lim_{x\to \plus{}\infty}\frac{\int_0^x e^tf(t)dt}{e^x}$. By L'Hospital, that limit is the same as $ \lim_{x\to \plus{}\infty}\frac{e^xf(x)}{e^x}\equal{}\lim_{x\to \plus{}\infty} f(x)\equal{}0$.

By L'Hospital rule
$ \lim_{x\to\plus{}\infty} f(x) \equal{} \lim_{x\to\plus{}\infty} \frac{e^xf(x)}{e^x} \equal{}\lim_{x\to\plus{}\infty} \frac{e^xf(x) \plus{} e^xf'(x)}{e^x} \equal{}\lim_{x\to\plus{}\infty} \frac{e^xf(x) \plus{} 2e^x f'(x) \plus{} e^xf''(x)}{e^x} \equal{} ... \equal{} \lim_{x\to\plus{}\infty} \frac{e^x \sum_{k \equal{} 0}^{n}{C_{n}^{k}f^{(k)}(x)} }{e^x} \equal{} \lim \sum_{k \equal{} 0}^{n}{C_{n}^{k}f^{(k)}(x)} \equal{} A$

@anik_andrew and mornik,
How can you make sure that the use of L'Hospital's valid? (Whether the numerator really diverges to infinity?)

if denumenator is goes to infty, than  L'Hospital rule is valid...

Not true, L'Hospital's rule is not applicable if $ f$ is, say, $ f(t) \equal{} \frac {\sin t}{t}$.  I think proving this statement directly from the definition of the limit is cleaner and more revealing.

For any $ \epsilon > 0$, there is a $ \xi$ such that $ |f(x)| < \epsilon$ for all $ x > \xi$, so setting $ c \equal{} \left \vert \int_0^\xi e^tf(t)dt\right \vert$, we have by the triangle inequality:
\[ \lim_{x\to \infty}\left \vert e^{ \minus{} x}\int_0^x e^tf(t)dt\right \vert \leq \lim_{x\to \infty} e^{\minus{}x}\left(c \plus{} \int_{\xi}^x \epsilon e^tdt \right) \equal{} \epsilon\]
Taking $ \epsilon \to 0$ we obtain the desired result.

[quote="JoeBlow"]Not true, L'Hospital's rule is not applicable if $ f$ is, say, $ f(t) \equal{} \frac {\sin t}{t}$. [/quote]
but $ \lim_{t\to \plus{} \infty} \frac {\cos t}{1}$ is not exists... so your example is not the disproof

What I am saying is that if we take $ f(t) \equal{} \frac {\sin t}{t}$, L'Hospital's rule is not applicable to $ \lim_{x\to \infty}\frac {\int_0^xe^t f(t)dt}{e^x}$ because the numerator does not have a limit.

[quote="mitelman"]if denomenator is goes to infty, than  L'Hospital rule is valid...[/quote] so we dont need to check the numerator
am i wrong??  :?:

Yes, that statement is wrong both generally, as well as in the context of problem (1) in this post, which was the point of the example.

Slight tangent: If no restrictions are placed on functions $ f,g$ other than $ f,g\in C^1$, $ f(x)\to \infty$, and $ g(x)\to \infty$ as $ x\to \infty$, then L'Hospital's rule is valid only if [u]both[/u] limits $ \lim_{x\to \infty}\frac {f(x)}{g(x)}$ and $ \lim_{x\to \infty}\frac {f'(x)}{g'(x)}$ exist.  Therefore L'Hospitals rule is only good for determining a value of a limit when one has already proven that the limit exists.  Using it to prove that a limit exists is, essentially, circular reasoning.

[quote="JoeBlow"] If no restrictions are placed on functions $ f,g$ other than $ f,g\in C^1$, $ f(x)\to \infty$, and $ g(x)\to \infty$ as $ x\to \infty$, then L'Hospital's rule is valid only if [u]both[/u] limits $ \lim_{x\to \infty}\frac {f(x)}{g(x)}$ and $ \lim_{x\to \infty}\frac {f'(x)}{g'(x)}$ exist. [/quote]
Can you give an example of $ f,g$ such that $ f,g\to\infty$, and exists $ \lim\frac{f'}{g'}$ but $ \lim\frac{f}{g}$ is not exists ???
in all source that i found (and as i know)
"If

$ \lim_{x\to\infty}\frac{f'(x)}{g'(x)}$

has a finite value or if the limit is , then

$ \lim_{x\to\infty}\frac{f(x)}{g(x)} \equal{} \lim_{x\to\infty}\frac{f'(x)}{g'(x)}$

"

Your source is wrong (can you cite the text, so I can make sure to avoid that author? ;)).

Take $ f(x) \equal{} x \plus{} \cos x \sin x$, $ g(x) \equal{} e^{\sin x}f(x)$ (this particular example is from [url=http://mathworld.wolfram.com/LHospitalsRule.html]Mathworld[/url], but it's not hard to construct similar examples -- the main point is that derivatives of $ f$ and $ g$ switch signs infinitely often beyond arbitrarily large $ x$).

Thanks JoeBlow. My source is
http://mathworld.wolfram.com/LHospitalsRule.html  :)
Ok )) Lets add  one more  assumption
"that  $ g'$ does not change sign infinitely often in a neighborhood of $ \infty$".
Then the L'Hospitals rule is valid. I mean if $ \lim \frac {f'}{g'}$ is exists  [b]then[/b] $ \lim \frac {f}{g}$ exists too. Right?

Yes, If both functions are eventually just strictly increasing or decreasing, then it's true.  We can invert everything and linearize near the origin.  But this is a pretty serious restriction on the space of functions to which L'Hospital's rule is applicable, and I think underlines why the rule is really not a useful tool in analysis.  Besides, in any situation where L'Hospital's rule is applicable, one can always evaluate the limit just as easily by making some first order estimates.

in my proof $ g(x)\equal{}e^x$. So $ g'>0$ for all $ x$.

So is there a version of L'Hospital rule that does not require existence of limit in numerator? And can anyone give an example of differentiable functions $ f,g$ such that $ \lim_{x \rightarrow \infty} g(x) \equal{} \infty$, $ g'(x) \neq 0$ for sufficiently large $ x$, $ \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} \equal{} L$ but $ \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}$ does not exist or is not $ L$?

[quote="anik_andrew"]in our university course we learn such L'ospitals rule.
I found just one link.
http://www.imc-math.org.uk/imc2005/day1_solutions.pdf
problem 5 there.[/quote]

That's also what I learnt, but suddenly I am uncertain about everything.  :maybe:

[quote="anik_andrew"]
Can you give an example of $ f,g$ such that $ f,g\to\infty$, and exists $ \lim\frac {f'}{g'}$ but $ \lim\frac {f}{g}$ is not exists ???

"[/quote]

I would say there is no such example, because implicit in the existence of $ \lim\frac {f'}{g'}$ is the assumption that $ \frac {f'}{g'}$ is defined on some interval $ [a,\infty)$, which means of course that we are not dividing by $ 0$ on that interval, i.e., $ g'\neq0$ on that interval.


[quote="GameBot"]On the planet Fraltramod, volume is measured in units referred to as Kwarts and Kaffos. There are five Kwarts to one Kaffo. How many Kwarts, expressed as a mixed number, are there in $ \frac25$ Kaffo $ \plus{}$ $ \frac52$ Kwart?[/quote]
Let $ x$ be a Kwart and $ y$ be a Kaffo:
$ 5x \equal{} y$
$ y \equal{} \frac{1}{5}x$
Plug in the value:
$ \frac{2}{5} \times \frac{1}{5}x \plus{} \frac{5}{2}x$
$ \frac{2}{25}x \plus{} \frac{5}{2}x \equal{} 2 \frac{29}{50} x$
Thus there are $ 2\frac{29}{50}$ Kwarts in the container given.

[quote="FreeForAll"]
$ 5x = y$
$ y = \frac {1}{5}x$
[/quote]
You were right the first time: $ 5x=y$
Plugging that in, we have
\begin{align*}
\frac{2}{5}y+\frac{5}{2}x&=\frac{2}{5}\cdot 5x+\frac{5}{2}x\\
&=2x+\Big(2\frac{1}{2}\Big)x\\
&=\Big(4\frac{1}{2}\Big)x\end{align*}
There are thus $ \boxed{4\frac{1}{2}}$ Kwarts in $ \frac{2}{5}$ Kaffo $ +$ $ \frac{5}{2}$ Kwart

[quote="jxl28"][quote="FreeForAll"]
$ 5x = y$
$ y = \frac {1}{5}x$
[/quote]
You were right the first time: $ 5x = y$
Plugging that in, we have
\begin{align*} \frac {2}{5}y + \frac {5}{2}x & = \frac {2}{5}\cdot 5x + \frac {5}{2}x \\
& = 2x + \Big(2\frac {1}{2}\Big)x \\
& = \Big(4\frac {1}{2}\Big)x\end{align*}
There are thus $ \boxed{4\frac {1}{2}}$ Kwarts in $ \frac {2}{5}$ Kaffo $ +$ $ \frac {5}{2}$ Kwart[/quote]
Yes, I input the values incorrectly.


a.b.c>0 prouv that
5(a+b+c)+3/abc>=18 we know that a^2+b^2+c^2=3
please i want clear demonstration

I think it is not correct. Where is the condition of problem? It means $ a+b+c=k,ab+bc+ca=p,...$ etc ... What is the value of $ k,p$
It is not homogeneous  :D

okie. It is not hard to prove  :wink:
Let $ p=a+b+c,q=ab+bc+ca,r=abc\to q=\frac{p^{2}-3}{2}$. We have:
\[ \frac{1}{r}\ge \frac{3p}{q^{2}}\]
The inequality is equivalent to:
\[ 5p+\frac{3}{r}-18\ge 5p+\frac{9p}{q^{2}}-18\ge 0\]
\[ \iff 5p+\frac{36p}{(p^{2}-3)^{2}}-18\ge 0\]
\[ \iff \frac{(p-3)^{2}(5p^{3}+12p^{2}-3p-18)}{(p^{2}-3)^{2}}\ge 0\]
which is clearly correct because:
\[ (5p^{3}+12p^{2}-3p-18)=(5p+12)(p^{2}-3)+6(2p+3)\ge 0\]
Note that: $ p=a+b+c\ge \sqrt{3}\iff 2q\ge 0$
we are done. Equality occurs if and only if $ a=b=c=1$


Let $ABC$ be a non equilateral triangle with $O$ as the circumcentre and $H$ as the incentre.Then prove that
                     $OH > 3 OB$

Please correct the problem. It's obviously false.

soory actually the reverse is to be proved i.e
$OH$ < $3OB$


[quote="rotfl"][quote="isabella2296"][quote="GameBot"]What is the simplified value of $ (7^{4})(8 \minus{} 2^{3}) \plus{} (11^{4(8) \minus{} 32}))$?[/quote]

$ 8 \equal{} 2^3$, so that's 0. Anything times 0 is 0, so now we only care about the 11 part. However, 4(8) = 32, thus our answer is $ 11^0 \equal{} \boxed{1}$.[/quote]
I was stupid, I answered 0 instead of 1.[/quote]

O. K.

[quote="davidkim2106"]
O. K.[/quote]

Please don't post stuff like this. It contributes nothing to the thread.