1.$a+b+c=1 , a,b,c>0$ . Prove the inequality :
$a^2+b^2+c^2+2.\sqrt{3abc}\leq 1$
2.$a^2+b^2+c^2=1 , a,b,c>0$ . Prove the inequality :
$a+b+c+\frac{1}{abc}\geq 4\sqrt{3}$

Replace $1$ by $(a+b+c)^2$
 It's enough to prove : $ab+ac+bc \geq 3abc(a+b+c)$ , which is obviously true

Both inequalities work with Lagrange multipliers. I have just solved the second one with this method. ;)

The second one can be solved by mixing variable .
 I'll prove : $f(a,b,c) \geq f(a,\sqrt{\frac{b^2+c^2}{2}},\sqrt{\frac{b^2+c^2}{2}})$ (1) , where $f(a,b,c)=a+b+c+\frac{1}{abc}$
 $(1) \iff (b-c)^2[\frac{1}{abc(b^2+c^2)}-\frac{1}{b+c+\sqrt{2(b^2+c^2)}}] \geq 0$ , which is true because $b+c+\sqrt{2(b^2+c^2)} \geq 2(b+c) \geq abc(b^2+c^2)$
 The rest is to prove $f(1-2t^2,t,t) \geq 4\sqrt{3}$
 It's a present for you .

[quote="nttu"] It's a present for you .[/quote]
:D   :P

[quote="nttu"] The rest is to prove $f(1-2t^{2},t,t) \geq 4\sqrt{3}$
 It's a present for you .[/quote]

Dear nttu,
I have checked your proof and I think  that we need prove

 $f(\sqrt{1-2t^{2}},t,t) \geq 4\sqrt{3}$

[quote="black_puma"]1.$a+b+c=1 , a,b,c>0$ . Prove the inequality $a^2+b^2+c^2+2\sqrt{3abc}\leq 1$[/quote]
$a^2+b^2+c^2+2\sqrt{3abc}\leq 1\iff a^2+b^2+c^2+2\sqrt{3abc}\leq (a+b+c)^2\iff$
$ bc+ca+ab\ge \sqrt{3abc}\iff  (bc+ca+ab)^2\ge 3abc(a+b+c).$


calculus
derivative
function
integration
real analysis
real analysis solved

Define the schwarzian derivative of a nonconstant meromorphic function $f: \mathbb{C}\to \mathbb{C}$ by
$D_S (f)=\frac{f^{\prime\prime\prime}(x)}{f^{\prime}(x)}-\frac 32 \left(\frac{f^{\prime\prime}(x)}{f^{\prime}(x)} \right)^2$.

Prove(or disprove, but i'm quite sure it's true) that we have $D_S(f)=D_S(g) \iff \exists\,{a,b,c,d\in \mathbb{C}, ad-bc\neq 0}\,:\ g(x)=\frac{af(x)+b}{cf(x)+d}$.

Is it really important that the functions are meromorphic?

[i]Edited by Myth[/i]

Consider $f(x)=\int \exp(\tan(x/2+a))$ and $g(x)=\int \exp(\tan(x/2+b))$.
Then $D_S(f)=D_S(g)$ and it is quite hard to believe that their are under your rational relation. But maybe I am wrong...
Of course, $f$ and $g$ are not meromorphic in $\mathbb C$, but only in each finite domain...  :?

Myth, you have destroyed a whole theory (and don't know it)...  ;)
Sorry Peter ...

Actually, we have discussed it with Peter, but he have had to go. So I constructed with example after our conversation.

...?

i thought i proved it in the last english lesson...so what's wrong with the following proof(inspired by myth's ideas in the chat)?

one direction is a trivial calculation, so assume that $D_S(f)=D_S(g)$.

let $F(x)=\frac{f^{\prime\prime}(x)}{f^{\prime}(x)}$, analogously $G(x)$. then we know that
$F'(x)-\frac 12 (F(x))^2=G'(x)-\frac 12 (G(x))^2$. if $F(x)=G(x)$, we're done.
now, what follows?

$2(F(x))^2-2F(x)G(x)-2F'(x)+2G'(x)=(F(x))^2-2F(x)G(x)+(G(x))^2$
=>$\frac{2F(x)(F(x)-G(x))-2(F'(x)-G'(x))}{(F(x)-G(x))^2}=1$
=>$\frac{d}{dx} \left( \frac{2e^{\int_0^{x} F(t) dt}}{F(x)-G(x)} \right)=e^{\int_0^{x} F(t) dt}$
=>$\frac{2e^{\int_0^{x} F(t) dt}}{F(x)-G(x)}=\int_0^{x} e^{\int_0^{s} F(t) dt} ds + C_1$

but we know that $e^{\int_0^{x} F(t) dt}=\frac{f'(x)}{f'(0)}$. this gives us that
$F(x)-2\frac{f'(x)}{f(x)+C_2}=G(x)$.
the result now follows from straightforward integration of both sides(both sides are logarithmit derivatives of something nice).

actually this counterexample doesn't seem to be one...i plotted the functions with mathematica and by taking appropriate $a,b,c,d$ the curves became quite equal...

I am verry sorry. I missed one more integration when I was performing my calculations, so right expressions are $ f(x)=\int \exp(\int\tan(x/2+a))$ and $g(x)=\int \exp(\int\tan(x/2+b))$, so it reduces to $f(x)=c_1+c_2\tan(x/2+c_3)$ and $g(x)=d_1+d_2\tan(x/2+d_3)$. And we see that, indeed, $f(x)$ and $g(x)$ are under a rational relation.

Peter! Your proof is great! (of course, since it was inspired by my ideas  :lol:  :lol: - a joke).
BTW, I am [b][color=red]M[/color][/b]yth ;)


[quote="joml88"]What is the missing number in the following sequence:
 10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, 31, 100, ---, 10000.

(Hint: The missing number is in ternary notation.)

*ternary means base 3.[/quote]
In my personal opinion, the hint trivialize the puzzle.
[hide="Answer"]121, because all the numbers are 16 in base 16, 15, 14, ... , 2[/hide]

It had the hint in [i]Mathematical Magic Show[/i] so I gave the hint also.

OK, as I said, it seems to me too easy with the hint, without it, it could be may be a little more challenging  :D

By the way, let's do something, let's not put hints on our puzzles unless they have been posted at least two days before and nobody has done them yet, what do you think (and the question is general!)


function
integration
real analysis
real analysis unsolved

If $f$ is nonnegative integrable function on a measure space $X$, prove:

(1) For any $\epsilon>0$, there is a set $B$ with $\mu(B)<\infty$,
such that $\int_{X-B}{f}\leq \epsilon$.

(2) There is a $\delta>0$, and $\int_{E}{f}\leq \epsilon$ for all $E$ with $\mu(E)<\delta$.

Also, is $f$ nonnegative necessary?

for (1), note simple integrable functions are dense in L^1.

For (2), construct a sequence of bounded functions converging to f a.e...


Calculate $ \sum_{k \equal{} 0}^n( \minus{} 1)^k\binom{n}{k}(n \minus{} 2k)^m$, for

a) $ 0\leq m\leq n \minus{} 1$

b) $ m \equal{} n$

[hide="Not sure about (a) but this is (b), I think:"][Argh! My solution was wrong!

Don’t worry, one of these days I’ll come up with a solution with no mistakes.][/hide]

Sorry, I’ve just realized something wrong with my solution was wrong. Argh! :wallbash:

I would be very interested in a solution for b) (one that's much better than what I'm about to give).

Knowing that $ a_n \equal{} \sum_{k \equal{} 0}^{n}( \minus{} 1)^k\binom{n}{k}(n \minus{} 2k)^n$

[hide="Is this relation true?"]
$ a_n \equal{} (2n)a_{n \minus{} 1}$ :maybe:?

If this recursive relation is true, then
[hide]$ a_n \equal{} \prod_{k \equal{} 1}^{n}2k \equal{} \boxed{2^n(n!)}$[/hide]
Otherwise, I'm not sure how else to tackle the problem (other than knowing that it's symmetric and the closed form is even).
[hide="Symmetry?"]$ a_n \equal{} n^n \minus{} \binom{n}{1}(n \minus{} 2)^n \plus{} \binom{n}{2}(n \minus{} 4)^n \minus{} ... \plus{} ( \minus{} 1)^n( \minus{} n)^n$
$ \equal{} 2(n^n) \minus{} 2\binom{n}{1}(n \minus{} 2)^n \plus{} ... \plus{} 2\binom{n}{l}(n \minus{} p)^n$ Folding the series like a sandwich.
Knowing that $ p \equal{} n$ if $ n$ is even and $ p\equal{}n\minus{}1$ if $ n$ is odd and $ \binom{n}{l}$ is the maximum value at the $ n$th row of pascal's triangle.[/hide][/hide]

[quote="Patterns_34"]

If this recursive relation is true, then
$ a_n \equal{} \prod_{k \equal{} 1}^{n}2k \equal{} \boxed{2^n(n!)}$[/quote]

Yes, your answer is correct. Please show how you got $ a_n \equal{} (2n)a_{n \minus{} 1}$.  :)

Well, the way I obtained that relation was more of a unsure hit-or-miss, search-for-a-pattern method.  :oops:
I just looked at it as a sequence and quickly recognized a pattern:
[hide]$ a_1 \equal{} 2$
$ a_2 \equal{} 8 \equal{} 2\times4$
$ a_3 \equal{} 48 \equal{} 2\times4\times6$
$ a_4 \equal{} 384 \equal{} 2\times4\times6\times8$
$ a_5 \equal{} 3840 \equal{} 2\times4\times6\times8\times10$
$ a_6 \equal{} 46080 \equal{} 2\times4\times6\times8\times10\times12$
etc.?[/hide]
Noticing the products and assuming it followed a recursive relation would mean
$ a_n \equal{} (2n)a_{n \minus{} 1} \equal{} 2^n(n!)$.
Even though the products shown above showed a direct relationship for a given $ n$, it was uncertain if the pattern was for real or temporary in this scenario, like [url=http://www.stmarksschool.org/data/files//MathInstituteSolutions/Microsoft_Word__Newsletter_March06_Web_Version.pdf]these[/url]. The reason I asked if the recursive relation was true was because if it was true then the solution provided [i]must[/i] be true.

Can anyone think of a direct approach with the series, showing $ \sum_{k \equal{} 0}^n( \minus{} 1)^k\binom{n}{k}(n \minus{} 2k)^n \equal{} 2n\sum_{k \equal{} 0}^{n \minus{} 1}( \minus{} 1)^k\binom{n \minus{} 1}{k}(n \minus{} 1 \minus{} 2k)^{n \minus{} 1}$ or something similar?

[hide="my solution"]
a) $ \sum^n_{k\equal{}0}(\minus{}1)^k \binom{n}{k} (n\minus{}2k)^m$
$ \equal{} \sum^n_{k\equal{}0}(\minus{}1)^k \binom{n}{k} \phi(k)$ where $ \phi(k)$ is a polynomial of degree m.

so we can write $ \phi(k) \equal{} \phi_m\binom{k}{m} \plus{} \phi_{m\minus{}1}\binom{k}{m\minus{}1} \plus{} \cdots \plus{} \phi_0\binom{k}{0}$

note that $ \binom{k}{m}\binom{n}{k} \equal{} \binom{n}{m}\binom{n\minus{}m}{k\minus{}m}$
and provided $ m<n$, $ \sum_{k\equal{}0}^n(\minus{}1)^k\binom{n}{m}\binom{n\minus{}m}{k\minus{}m} \equal{} \binom{n}{m}\sum^{n\minus{}m}_{k\equal{}0}(\minus{}1)^k\binom{n\minus{}m}{k} \equal{} 0$

we have this happen for all terms of $ \phi(k)$ so...

for part a the answer is 0.

b) if n=m then we have

$ \sum^n_{k\equal{}0}(\minus{}1)^k\binom{n}{k}(n\minus{}2k)^n$
$ \equal{}\sum^n_{k\equal{}0}(\minus{}1)^k\binom{n}{k}\phi_n\binom{k}{n}$

The $ k^n$ coefficient of $ \binom{k}{n}$ is $ 1/n!$
The $ k^n$ coefficient of $ (n\minus{}2k)^n$ is $ (\minus{}2)^n$
so $ \phi_n \equal{} n!(\minus{}2)^n$

so our sum above is:
$ \equal{}n!(\minus{}2)^n\binom{n}{n} \sum^n_{k\equal{}0}(\minus{}1)^k\binom{n\minus{}m}{k\minus{}m}$
$ \equal{}n!(\minus{}2)^n\sum^0_{k\equal{}0}\binom{0}{k}(\minus{}1)^{k\plus{}n}$
$ \equal{}\boxed{2^nn!}$
[/hide]


In how many ways can 7 people sit around a circular table? Seating arrangements are considered different only if at least one person has at least one different neighbor.  Btw, the answer isnt 6!

Yeah. You can flip the table, so the answer is $ \frac{6!}{2}\equal{}360$


[hide]If every odd chair is opposite another odd chair, there must be an even number of odd chairs.  Call the number of odd chairs $2k$, where $k$ is a positive integer.  Since there are an even number of chairs, the number of odd chairs equals the number of even chairs.  So there are $2k$ even chairs, giving a total of $4k$ chairs, clearly a multiple of 4.[/hide]

[quote="buzzer11"]Chairs are equally sapced around a table and numbered from 1 to 2n. Prove that if every odd chair is directly opposite another odd numbered chair, then the number of chairs is a multiple of 4.[/quote]

[hide]If every odd chair is directly opposite another odd numbered chair, then every even chair is directly opposite another even numbered chair. So number of odds=number of evens.
The total number of chairs is x. The number of odd and even chairs are equal. so there are $\frac{x}{2}$ odds and $\frac{x}{2}$ evens.
Since equal amounts of each kind are on each side of a table,  there are $\frac{x}{4}$ odds or evens on a side of the table. Then that means x is a multiple of 4.[/hide]

both good proofs!

i maybe will put up the given proof in a few days.


graph theory
combinatorics unsolved
combinatorics

What is the maximum number of edges in a bipartite graph that contains no matching with $k$ edges and no vertex of degree $d$?

(A matching is a set of independet edges)


you could just notice that 15(6h+12y) is simply 9z, so 9z=xz and x=9

[quote="mathnerd314"][quote="i_like_pie"][quote="mr. math"]How would you do this problem:

Solve for x: If 5(2h+4y)=z then 15(6h+12y)=xz[/quote]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=115701[/url][/quote]

"The topic or post you requested does not exist." :P[/quote]
It used to. :huh:

[quote="bpms"][quote="i_like_pie"][quote="k8reindeer"]wait! what if h=-2y?[/quote]
Then both equations turn into $0=0,$ which is true.[/quote]
But then x could be anything.[/quote]
How many times do I have to say this? If you come to a true statement while solving an equation, that does not mean the variable can be anything. If you multiply both sides of an equation by 0 (yes, it is legal), that doesn't do you any good, but it doesn't mean the variable can be anything.

[hide="Solution"]$5(2h+4y)=z;\,15(6h+12y)=xz$

$10(h+2y)=z;\,90(h+2y)=xz$

Clearly, $x=9.$[/hide]

[quote="tim1234133"]I'm a little unsure here, but when $h=-2y$ then the equations are
$5\cdot0=z$
$15\cdot0=xz$
Clearly $z=0$ from the first equation, and $x=1$,$x=2$,$x=7.4632$ etc. all fit the second equation: both equations are satisfied by $x=$ anything.
We haven't multiplied by $0$ anywhere.  :maybe:[/quote]
If $h=-2y,$ then $5(2h+4y)=z$ becomes $z=0.$ How can you say that $x=\text{anything?}$

Again, assuming $h=-2y,$ the second equation is $xz=0,$ and $x=\frac{0}{z}.$ Since we know that $z=0,$ it becomes undefined and we can't say anything about $x.$

I can say that $x=$ anything because (as you say) $xz=0$ and $z=0$. If I substitute any number I like for $x$ into that equation, then the equation is satisfied.
In fact, if I substitue any number I like for $x$ into the original equations, then they are also satisfied. Hence any value of $x$ solves the equations. (in the case $h=-2y$)


Let $ X \subset \mathbb{R}^{2}$ be a connected graph that is the union of a finite number of straight line segments. Show that $ \pi_{1}(X)$ is free with a basis consisting of loops formed by the boundaries of the bounded complementary regions of $ X$, joined to a basepoint by suitably chosen paths in X.

you can prove that finite graph $ X$ is homotopic to wedge sum of finite circles by some kind of collapsing , say n , hence the fundamental groups must be free group with $ n$ generator. Furthermore $ n \equal{} 1 \minus{} \chi(X)$

But I am wondering why you are mentioning Jordan in this case ?

because his statement is more precise, and (probably) its proof requires jct.


[quote="grobber"]I know stronger results hold (I think I can remember sam-n proving that if $S$ is infinite, then $S$ and $S\times S$ have the same cardinality :?; I can't prove that, though :)).

Given an infinite subset $S$, show that we can find $A,B\subseteq S,\ A\cap B=\emptyset$ s.t. both $A,B$ have the same cardinality as $S$.[/quote]

 Are we allowed to use the Zorn lemma? :)

I don't think it can be done without something equivalent to the axiom of choice, so yes :).

Actually, I think my proof my flawed.. :( I don't think I can solve it, infortunately (unless I'm blind or something :().

We can use Zorn lemma to decompose $S$ into a family $\{F_\alpha\}_{\alpha \in I}$ of countable subsets.
 Then we decompose each countable subset $F_{\alpha}$ into two other countable ones, namely $F_{\alpha A} $ and $F_{\alpha B}$
 Let $A = \cup_{\alpha \in I}F_{\alpha A}$, $B = = \cup_{\alpha \in I}F_{\alpha B}$
 That's my solution :)


calculus
integration
quadratics
algebra
quadratic formula

Solve the equation
$(x^{2}+y)(x+y^{2})= (x-y)^{3}$
 on the set of integers.

[hide="meh..."]
Not particularly elegant; i bet there's a better approach. I think I got 'em all though...

First, we can expand this to achieve:
$xy+3x^{2}y-3xy^{2}+x^{2}y^{2}+2y^{3}=0$
$y(x+3x^{2}-3xy+x^{2}y+2y^{2})=0$
$y=0\qquad \underline{\text{OR}}\qquad x+3x^{2}-3xy+x^{2}y+2y^{2}=0$

The former produces the solution set $(x,0)$ for all integral $x$.

The latter rearranges to:
$(2)y^{2}+(x^{2}-3x)y+(x+3x^{2})=0$
Which, by the quadratic formula, yields:
$y=\frac{(3x-x^{2}\pm (x+1)\sqrt{x^{2}-8x})}{4}$

Clearly from the above, $x^{2}-8x$ must be a perfect square, or $x+1=0 \to x=-1$. The latter produces $y=-1$, so $(-1,-1)$ is a solution.

For the former case, we know $x^{2}-8x+16=(x-4)^{2}$ is a perfect square, so we're looking for two perfect squares that differ by 16. The two such are $0^{2},4^{2}$ and $3^{2},5^{2}$.
The former produces $x-4=4 \to x=8$ or $x-4=-4 \to x=0$
The latter produces $x-4=4 \to x=9$ and $x-4=-5 \to x=-1$.

$x=8$ causes $y=-10$ and $x=9$ causes $y=-6$ and $y=-21$.

$x=0$ causes $y^{3}=-y^{3}\to y=0$, which has already been covered, and $x=-1$ was covered when we considered setting $x+1$ equal to $0$.

[b]Solutions:[/b]
$(x,0)$
$(-1,-1)$
$(8,-10)$
$(9,-6)$
$(9,-21)$
[/hide]


Factor and Simplify:

(csc x) ^3 - (csc x) ^2 - (csc x) + 1

[hide]Use factor by grouping. $ (\csc^2{x} \minus{} 1)(\csc{x} \minus{} 1) \equal{} (\csc{x} \minus{} 1)^2(\csc{x} \plus{} 1)$

Thus, $ \sin{x} \equal{} \pm{1}$, so $ x \equal{} \displaystyle\frac {\pi}{2} \plus{} n\pi$.[/hide]

Wait, you are trying to get the zeroes of that?
I think it's just simplify?


number theory
prime numbers
combinatorics unsolved
combinatorics

A 7-digit telephone number cannot start with a 9, 0, or 1, and must contain exactly two distinct prime digits.  How many telephone numbers are possible?

By “two distinct prime digits”, do you mean that a number like, say, 1223334 is allowed? Or do you mean only two prime digits in the whole number, both prime digits distinct? :huh:

In what follows, I will assume the latter. ;)

Two prime numbers can be chosen from {2,3,5,7} in $ ^4\mathrm{C}_2$ ways.

Suppose one of the prime digits is the first number. Then the other prime digit can occupy any of the other 6 places, and the 5 remaining places can be filled by any of 6 digits (0, 1, 4, 6, 8, 9), giving $ 6^5$ permutations. The two prime digits can also swap over. Hence there are $ ^4\mathrm{C}_2\times6\times6^5\times2$ numbers in which the first digit is prime.

Now suppose the first digit is not prime. Then the two prime digits can be placed in the other 6 places in $ ^6\mathrm{P}_2$ ways (taking into account the order in which they appear). The first digit can only be 4, 6 or 8, so there are 3 ways for that. The remaining 4 digits can be from {0,1,4,6,8,9}, giving $ 6^4$ permutations. Thus there are $ ^4\mathrm{C}_2\times^6\mathrm{P}_2\times3\times6^4$ numbers in which the first digit is not prime.

If I can use a calculator correctly, the answer you got is 1259712, which is actually one of the foils on this problem.

It says the correct answer is 1679616.


Which 2-digit numbers are divisible by both the sum and the product of their digits?

h_s_potter2002 wrote:[hide]Any number divisble by 12.[/hide]

48? 60? 72? 84? 96?

Any number divisible by 10

That and 36, as well as 24.

There might be more...

[quote="Treething"]Any number divisible by 10[/quote]

It said any the number is divisible by the sum and THE PRODUCT
and u cant have a 0 in the denominator cause any number that is a multiple of 10, it's digits product equals 0.
I dont think it counts if u divide it and there is a undefined answer for the product

[hide]You can't have odd numbers because if the ten's digit and the one's digit are both odd, then the sum would be even and that wouldn't work. If the ten's digit is even, then the product would be even and then it wouldn't work. If also can't be divisible by 10 because of what dinoman said, so I just crossed the rest off one by one and I got 12, 24, 36, and 84[/hide]


You like gambits? What about the "other" choice on the poll?

Sorry for the confusion. I meant "Out of the choices below".

Queens Gambit accepted and declined are up to the opponent, not you.

Which one do you favor? And what if you play black?

I like playing the Benko Gambit with black if white opens Queen's Gambit. Black builds up a tremendous attack


geometry
geometric transformation
number theory
relatively prime
IMO
IMO 2005

At the imo official page I saw that two contestants of the Israel team , one with 8 points and the other with 3 points , and they got bronze medal and honorable mention respectively. Why this did happen ? Could someone tells us about that story ?

See the footnote:
[i]§ Contestant was awarded by the decision of the Jury due to special circumstances.[/i]

The Hebrew translation of the problem texts had a mistake that made a problem trivial.

  darij

I had seen the foot note , but I wanted to know about these "special circumstances"  :)
Thanks for the explanation

[quote="darij grinberg"]The Hebrew translation of the problem texts had a mistake that made a problem trivial.[/quote]

This was problem IMO 2005/4.  The correct formulation was:
Determine all positive integers relatively prime to all the terms of the infinite sequence  $ a_n\equal{}2^n\plus{}3^n\plus{}6^n \minus{}1$.

In the Hebrew version, a typo had turned one of the plusses into a minus:
Determine all positive integers relatively prime to all the terms of the infinite sequence  $ a_n\equal{}2^n\plus{}3^n\minus{}6^n \minus{}1$.


Explain why $ 143\times927 \equal{} 132561$ can be done at sight.

[hide=" Hint"]

$ 143\times 7 \equal{} 1001$.

[/hide]

[quote="Nlakkamraju"]...... well because........143*7 = 1001 and143*20 = 2860 and 143*900 = 128700, add the whole thing and you get 132561[/quote]

How exactly is this all supposed to be done on sight?? You'd have to be able to hold a lot of numbers in your head...

Still, the adding in your head is a bit confuzzling  :)


vector
linear algebra
linear algebra unsolved

LET V =R^n equipped with the usual inner product (-,-)
Prove that f : V-->R  if only if there exists z such that f(x) =(x,z) foe all x belong to V

In other words: the dual of a finite-dimensional vector space is isomorphic to the original space, by any choice of basis to construct an inner product.

You don't exactly have a clear or correct statement there...


If I'm not wrong, this problem appeared for the first time in Kvant, in 1974 or 1978. It can be proved observing that (a+b-c)^2*(c+a-b)^2>=(a^2+b^2-c^2)(c^2+a^2-b^2), which is easy to prove.

Ok, (a+c-b)^2(b+a-c)^2=(a^2-(b-c)^2)^2=a^4-2a^2(b-c)^2+(b-c)^4. But (a^2+c^2-b^2)(b^2+a^2-c^2)=a^4-(b^2-c^2)^2. So, it is enough to prove that (b^2-c^2)^2+(b-c)^4>=2a^2(b-c)^2 or (b+c)^2+(b-c)^2>=2a^2, or b^2+c^2-a^2>=0. We can assume that b^2+c^2-a^2>=0, c^2+a^2-b^2>=0, a^2+b^2-c^2>=0 (only one of them could be negative and then it's trivial), so these inequalities hold. Multiply them.


Take a right angles triangle with integral sides. Locate its centroid
and connect it to the three vertices of the rightt-angled-triangle to
form three smaller triangles. Let the areas of these three triangles
be $ x$, $ y$ and $ z$.

What can you say about the numbers $ x$, $ y$ and $ z$? List all properties that you can deduce.

[quote="Altheman"]They are all equal...it is the centroid. I don't see the point of having a right angled triangle with integer side lengths.[/quote]

there are two other properties...

I'm guessing you want that $ x$, $ y$, and $ z$ all have even integer areas, since the areas of all Pythagorean triple triangles have areas divisible by 6. One more property?

err...x=y=z=Area of the triangle divided by 3...

using the whole primitive pythagorean triples, it follows that they are integers...

$ 3|\frac{1}{2}*(2kmn)(2k(m-n)(m+n))$


I saw a problem about paths along grids, and I remembered a clever problem similar to that one.

Consider a 2*n array of circles. On each turn you can move down, to the right, or up and to the right. Given n=7 (as an example), how many ways can you go from the upper left to the bottom right. Generalize.

[hide]the nth fibonacci number
the ways to get to the second one is the ways to get to the first
the ways to get to the third =ways to get to second+ways to get to 1st(which is juust 1)
ways to get to 4th=ways to get to 3rd+ways to get to second
this is how the fibonacci series are generated[/hide]

An+1=An+An-1
there fore the answer is 13.


calculus
integration
induction
number theory unsolved
number theory

Let $B$ be an integer greater than 10 such that everyone of its digits belongs to the set $\{1,3,7,9\}$. Show that $B$ has a [b]prime divisor[/b]  greater than or equal to 11.

Maybe you mean the prime factor,or it's very very easy(just itself).

Yes, it should be "prime divisor" indeed.

Yes..... you're right...  Now, it's corrected ;)

and thanks .

Carlos Bravo ;)
Lima -PERU

It's not that difficult anyway.

Since none of the digits can be even, $2$ cannot divide $B$.  Also, since none of the digits can be $0$ or $5$, we know that $5$ cannot divide $B$.

Now we proceed by contradiction.  If the supposition is false, then $\exists n\in\mathbb{N}$ such that all of its digits belong to the set $\{1,3,7,9\}$ and $n=3^a7^b$ for some non-negative, integral $a$ and $b$.

Now, note that $3^07^0=1$, and thus the ones digit is in $\{1,3,7,9\}$ and the tens digit is even.

Now proceed by induction.  If $n=100x+10y+z$ where $x\ge 0$, $y$ is even, and $z\in\{1,3,7,9\}$, then both $3n$ and $7n$ will contain an even tens-digit and either $1,3,7,$ or $9$ as a ones-digit.  So by induction we have that for all $a,b\ge 0$ that $3^a7^b$ will contain an even-tens digit.  But this violates the fact that none of the digits can be even; contradiction.

Therefore the theorem must hold.

Arguing by contradiction, we have $n = 3^a 7^b$ for some $a, b$. Now, consider $n \pmod {20}.$ We know by casework that
$$\{k, 3k, 7k, 9k \} \equiv \{1, 3, 7, 9\} \pmod {20}$$
for $k = 3, 7$. Thus by induction $3^a 7^b \in \{1, 3, 7, 9\} \pmod {20}$, so $n$ has even tens digit, contradiction.

[hide=Solution]
Similar to the solution above, let us assume that the statement is false, then there exists $B=3^a7^b$. To make it easier set $b=0$, if $a \geq 3$ the ten's digit is even, to prove this just note that $3^a$'s last digit is one of ${1,3, 7, 9}$ and in the usual multiplication algorithm they all carry out an even number to the ten's digit and since multiplication by an even number turns out to be even and inductive argument assures us that the tens digit will always be even.We can use a similar argument for the case $B=7^b$ and from these two the general case follows, a contradiction. Thus the result follows.


Synthesize $ \text{1,2,3\minus{}trichlorobenzene}$ from $ \text{Aniline}$ by a multi-step route which involves a nitration.

[hide="Solution"]1) aniline + $ \ce{CH3COCl}/pyridine$ -----> acetanilide

2) acetanilide + $ \ce{HNO3}/\ce{H2SO4}$ -----> p-nitroacetanilide

3) p-nitroacetanilide + hydrolysis -----> p-nitroaniline

4) p-nitroaniline + $ \ce{Cl2}$/water -----> 2,6-dichloro-4-nitroaniline

5) 2,6-dichloro-4-nitroaniline + $ t\minus{}BuONO/CuCl$ -----> 1,2,3-trichloro-5-nitrobenzene

6) 1,2,3-trichloro-5-nitrobenzene + reduction -----> 3,4,5-trichloroaniline

7) 3,4,5-trichloroaniline + $ \ce{NaNO2}/\ce{H3PO2}$ -----> 1,2,3-trichlorobenzene[/hide]

holy crap carcul.... thats a beastly mechanism....how long did that take you?

I didn't post a mechanism, but instead a synthetic scheme, which are very different things.


search
superior algebra
superior algebra unsolved

Let $ I$ be an ideal of $ R[x_1,...,x_n]$.
If for any $ p \in I$, each homogeneous component of $ p$ is also in $ I$,
then $ I$ is generated by homogeneous polynomials.

[quote="t0rajir0u"]This problem was answered the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=166093037&t=213365]last time[/url] you posted it.[/quote]

Answer of my last time question is 'clear'.
Why is this clear?

Please don't post the same thing in two places.


geometry
geometric transformation
rotation
3D geometry
tetrahedron
symmetry
search

How many inequivalent colorings using k colors are there to paint the vertices of a polyhedron of n sides.  I thought about using Burnsides' theorem of polygons.  The rotation of the vertices of a polyhedron still forms a group, but we have more than one dimension that I can rotate it now.  So I guess Burnsides' theorem still applies?  Just that it could be very hard (?) to count the number of fixed colorings.

http://en.wikipedia.org/wiki/Burnside%27s_lemma

Yes, you can use Burnside's theorem. For example, let's compute how many ways there are to color tetrahedron by three colors. Totally we have $ 3^4 \equal{} 81$ ways to color it. Let's see how many of the are the same up to rotational symmetry. The rotational symmetry group of tetrahedron is $ G \equal{} A_4$. Let $ \Omega \equal{} \{T_1,\ldots,T_{81}\}$ be the set of $ 81$ tetrahedrons. Now $ G$ acts if $ \Omega$ in natural way so it maps every $ T_i$ to some $ T_j$. Two tetrahedron $ S$ and $ T$ are colored the same way if there exist $ g\in G$ such that   $ S*G \equal{} T$. Thus one can count the coloring of tetrahedron the same way as counting how many $ G$-orbits there are in $ \Omega$.

Now I can use Burnside's theorem. The neutral element fixes everything so $ \chi(1) \equal{} 81$. Three cycle $ \theta$ has axis as a vertex of the tetrahedron and it goes through the middle point of other side. Thus $ \theta$ fixes a coloring where every side has the same color. I got $ 9$ colorings of this kind and the bottom color can be arbitrary.

An element $ \phi$ whose order is two acts as $ (12)(34)$. So if I want it to fix a coloring, I have to make sure that sides $ 1$ and $ 2$ are the same color and $ 3$ and $ 4$ are the same color.

There are eight three cycles $ \theta\in A_3$ and $ \chi(\theta) \equal{} 9$ for every one of them. There are three elements of order two and $ \chi (\phi) \equal{} 9$. Now Burnside's theorem gives total
\[ \frac {1}{|A_4|}\sum_{g\in A_4}|\chi (g)| \equal{} \frac {1}{12}(81 \plus{} 8\cdot 9 \plus{} 3\cdot 9) \equal{} 15\]
colorings.

Can anyone solve http://www.mathlinks.ro/viewtopic.php?search_id=1650611433&t=228658 using Burnside's theorem?

But in general a polyhedron of $ n$ sides is not necessary regular so you have to compute how many positions looks the same. I think this makes computing trickier. And I think that there can be several polyhedrons which has the same number of faces but they are not similar.


I've just installed the mhchem package for you Chemistry wiz people out there. Let's give it a run:

$ \ce{1/2H2O}$

Documentation for this package is attached in the .pdf file.

However, because of the way the LaTeX interpreter works in this site some of the formulas will not have the expected out come.

brilliant work and thanks a lot  :clap:  :omighty:

I really comment on the new package.
Thank you very much.  :D

I can't seem to get it to align properly sometimes.

For instance, consider the unbalanced chemical equation H + O2 -> H2O.  This is what I get when I use the new package:

[code]
$\ce{H + O2 -> H2O}$[/code]

$\ce{H + O2 -> H2O}$

Upon further inspection, I see that when the $\LaTeX$ parses, it gets rid of many of the spaces.  However, these spaces are vital to the parsing of the code.  Is there any way to fix this?  Thank you!

You can use the new code:
[code]
$\ce{CO2} + \ce{C} \ce{->} \ce{2CO}$[/code]
              OR
[code]$CO_{2} + C \ce{->} 2CO$[/code]
$\ce{CO2} + \ce{C} \ce{->} \ce{2CO}$


Don't remember which year it is, but the question is:

How many three element sets of positive integers $a,b,c$ are there such that $abc=2310$?

The correct answer is $40$, but I don't understand what I did wrong...

[hide="my work"]Since $2310=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$, we can split the numbers into subsets which only contain three elements ${a,b,c}$, and we can split it up in four different ways:

1) $(0,2,3)$ in which the numbers can be chosen $\binom{5}{0} \cdot \binom{5}{2} \cdot \binom{3}{3}=10$

2) $(1,1,3)$ in which it will be $\binom{5}{1} \cdot \binom{4}{1} \cdot \binom{3}{3}=20$

3) $(1,2,2)$ in which it will be $\binom{5}{1} \cdot \binom{4}{2} \cdot \binom{2}{2}=30$

4) $(0,1,4)$ in which it will be $\binom{5}{0} \cdot \binom{5}{1} \cdot \binom {4}{4}=5$

Adding everything together, we have $65$ total ways...which is far from the correct answer :o [/hide]

[quote="chess64"][hide="Start of a possible solution"]
Each of a, b, and c must be in the form $2^{p}3^{q}5^r7^s11^t$, where all the exponents are either 0 or 1. Then \[ abc=2^{p_a+p_b+p_c}3^{q_a+q_b+q_c}5^{r_a+r_b+r_c}7^{s_a+s_b+s_c}11^{t_a+t_b+t_c}.  \] Hence $p_a+p_b+p_c=1$, etc. This means that the possibilities for the powers on each factor are (1,0,0), (0,1,0), or (0,0,1).
[/hide][/quote]

EDIT:  i understand this, but my approach was very similar but it didn't work..

My ideas:  One of $a,b,c$ must have a factor of $2$, one must have a factor of $3$, and so on.  We need to "assign" each of the five factors to one of $a,b,c$.  Therefore it seems to me that the answer should be $3^{5}=243$.  Is there anything wrong with this reasoning?  (I clearly didn't get the right answer)

[quote="matt276eagles"]One of $a,b,c$ must have a factor of $2$, one must have a factor of $3$, and so on.  We need to "assign" each of the five factors to one of $a,b,c$.  Therefore it seems to me that the answer should be $3^{5}=243$.  Is there anything wrong with this reasoning?  (I clearly didn't get the right answer)[/quote]

I think you're overcounting and might be missing the factor of $1$.

uhh....Jesusfreak...I don't think I want to waste 10 minutes just listing out possibilities during the AMC, especially when it's so easy to miscount :P

IversonFan2005, I think your counting method doubled the # of actual results for certain cases.

[hide]A $(1,1,3)$ group could be $2\times3\times(5\times7\times11)$ I think the 20 is doubled because we counted the above as separate from $3\times2\times(5\times7\times11)$ so therefore we would have twice as many. Same with a $(1,2,2)$ group. [/hide]

Good work Jesusfreak  :lol: ...But I still dont understand what was wrong with Iversonfan's solution or my solution.  Can anyone help?

[quote="matt276eagles"]Also, you assumed that $a\le b\le c$.  Is that what the original problem said?[/quote]

You can assume, since they aren't ordered pairs.

It actually only took me about 5 minutes to list them out. Once you have the pairs beginning with 1, you can refer back to that. 5 minutes doesn't seem bad to me for a #29, when the average is 3 minutes per. ;)

[quote="matt276eagles"]The factor of 1 is accounted for, but it is only used when all factors are assigned to the other variables.

In your solution, Iversonfan, I think you missed the case $(0,0,5)$.  Also, you assumed that $a\le b\le c$.  Is that what the original problem said?[/quote]

The original problem never said the integers have to be distinct

[quote="plokoon51"]IversonFan2005, I think your counting method doubled the # of actual results for certain cases.

[hide]A $(1,1,3)$ group could be $2*3*(5*7*11)$ I think the 20 is doubled because we counted the above as separate from $3*2*(5*7*11)$ so therefore we would have twice as many. Same with a $(1,2,2)$ group. [/hide][/quote]

Ahah...that makes sense now, so we need to divide the second and third by $2!$ and I also forgot to add in $(0,0,5)$ which would have $5$ choices so then it would add up to $\boxed{40}$


conics
geometry
geometric transformation
reflection
circumcircle
angle bisector
geometry unsolved

Reflections $P,Q,R$ of $A,B,C$ about $O$ obviously lie on the wanted conic $\mathcal{H}.$ Isogonal conjugation WRT $\triangle ABC$ takes circumcircle $\mathcal{K}$ of $\triangle ABC$ into the line at infinity and $\mathcal{H}$ into a line $\eta$ passing through the isogonal conjugates $P^{-1},Q^{-1}$ of $P,Q$ WRT $\triangle ABC.$ So, the fourth intersection $U$ of $\mathcal{K} ,\mathcal{H}$ is the isogonal conjugate of the infinite point of $\eta,$ this is: Draw the parallel $\tau$ from $A$ to $\eta,$ the reflection of $\tau$ across the angle bisector of $\angle bc$ cuts $\mathcal{K}$ again at $U.$ If $D \equiv BC \cap AU,$ then angle bisectors $u,v$ of $\angle (AD,BC)$ are parallel to the axes of the conic $\Longrightarrow$ parallels from $O$ to $u,v$ are the corresponding focal axis $f$ and minor axis $e.$

Construct a tangent line to $\mathcal{H}$ by using Pascal theorem. For instance, by Pascal theorem for the non-convex hexagon $PPABCR,$ the intersections $X \equiv PP \cap BC,$  $O \equiv PA \cap CR$ and $Y \equiv AB \cap RP$ are collinear $\Longrightarrow$ $\lambda \equiv XP$ is tangent to $\mathcal{H}$ at $P,$ where $X \equiv OY \cap BC.$

Denote by $F,F_0$  the foci of $\mathcal{H}.$ The tangent line $\lambda$ and its normal $\lambda' \perp \lambda$  through $P$ are angle bisectors of $\angle (PF,PF_0).$ Hence if $\lambda,\lambda'$ cut $f$ at $M,N,$ then $(F,F_0,M,N)=-1$ $\Longrightarrow$ $\overline{OF}^2={\overline{OF_0}}^2= \overline{OM} \cdot \overline{ON}$ $\Longrightarrow$ circle $\omega$ centered at $O$ whose radius equals the geometric mean of the known segments $OM,ON$ cuts $f$ at $F,F_0.$


How many odd numbers between $ 100$ and $ 999$ have distinct digits?

The last digit is $1$, $3$, $5$, $7$, or $9$. So there are $5$ choices. Nine choices exist for the hundreds digit ($1$, $2$, $\ldots$ , $9$), but we've used one of these, so since our digits must be different. We subtract one for $8$ total choices. Finally the tens can be anything from $0$ to $9$ except the two digits we've used already, leaving $8$ choices for the tens digit. There are $5\cdot 8 \cdot 8 = \boxed{320}$ numbers.

I believe that jamesshin97's solution is correct, for the following:

[hide="Explanation"]The most limited digit is the last one (ones digit), since it would be much more difficult to find the number of ways to fill in the first 2 digits and then making cases. As the ones digit must be odd, there are 5 choices {1,3,5,7,9}. The next limiting digit is the hundreds place. This cannot be 0, nor can it be the same digit as the last one. We then have 10-2=8 possible digits here. Finally, the tens digit has no limitations other than being distinct from the hundreds and ones digits, so this has 10-2=8 choices. Thus, our total number of numbers is $5\times8\times8= \boxed{320}$.[/hide]

Hope this helps.

I actually found casework easier.

[hide="Solution"]
Let the 3 digit number be $abc$.
[hide="Case 1(a is even)"]
[b]Subcase 1: b is odd[/b]
$4\times 5\times 4=80$
[b]Subcase 2: b is even[/b]
$4\times 4\times 5=80$
[/hide]

[hide="Case 2(a is odd)"]
[b]Subcase 1: b is odd[/b]
$5\times 4\times 3=60$
[b]Subcase 2: b is even[/b]
$5\times 5\times 4=100$
[/hide]

So we have $80+80+60+100=\boxed{320\text{ ways}}$.
[/hide]

[quote=$LaTeX$][quote="1ntegration1"]Why do you get different answers? Why must you do the last digit, first digit, then the second?[/quote]

The reason why jamesshin and PoP got different answers was because PoP assumed that the second digit was not a $0$ (which it can be), which resulted in him missing a few cases.[/quote]

No, he/she had it right because he said that there will be 9 choices.  There are 10 numbers from 0-9 inclusive.  $10-1=9$.

[quote=1ntegration1]Why do you get different answers? Why must you do the last digit, first digit, then the second?[/quote]
[quote=Eulalia]So what's wrong with his/her answer which was this:
[quote=PowerOfPi]We will start with the last digit. It has 5 choices (1, 3, 5, 7, 9). The second digit can be anything but the last, so it has 9 choices. Finally, the first can be anything other than 0, the second one, or the last one, so it has 7 choices. Multiplying:

$ 5 \cdot 9 \cdot 7\equal{}\boxed{315}$[/quote][/quote]

The second solution finds more, and is correct. Therefore, it is the correct answer. The more restricted, the more to the front it goes.

I found the phrasing of this problem to be confusing. There's an ambiguity of distinct digits (not having all 3 be the same) vs. all digits being distinct (no two digits are the same). I had to waste a guess figuring out which of these forms was the intention of the question.

No, it's obviously the second, just by logic.

I'll admit that the "all distinct" interpretation makes more sense, since the "not all the same" interpretation is quite trivial, but I wouldn't say it's entirely obvious. I've seen other problems where the logically obvious interpretation is not what is being asked, and without wording to clarify, it could be interpreted either way.

My main point is that is that it could be been worded more specifically.

What did I do wrong?

We can use complimetry counting.

So clearly there are $450$ numbers in total. Then for the amount of cases that don't work we know that at least one of the digits must be the same as one of the other two so clearly there are $10 \cdot 1 \cdot 5$ ways to make a number like this. So we get $450 - 50 = \boxed{400}$.

@above

The problem with this is you don't count the "repeated digits case" correctly. First of all you have to order the digits. Second of all even if you implement that you are overcounting the cases where the number is, say, $333.$ See what I'm getting at here?

I don't think complementary counting would work for this problem as it's too much of a hassle.

[quote]The problem with this is you don't count the "repeated digits case" correctly. First of all you have to order the digits. Second of all even if you implement that you are overcounting the cases where the number is, say, $333.$ See what I'm getting at here?
[/quote]

1: I did order the digits, right?

2: I don't really see tbh how I'm counting $333$ multiple times...

Sorry I suck at C+P :P

How did you order the digits? You assumed that the [10] place and the [1] place are in those positions, but you must order them! What if one of the numbers that's repeated is the units digit? Then that must be odd.

I'm just saying it's hard to do with complementary counting :)

anyways ima go sleep bai

@below

no problem

[hide=Solution]
Deal with the restrictions first; there can only be five choices for the ones digit (odd numbers, or $1, 3, 5, 7, 9$). The hundreds digit cannot be $0$ or the last digit, so there are $8$ choices for it. Finally, the tens digit cannot be either the hundreds or ones digits, so there are $10-2=8$ choices for it. As such, the total will be $8 \times 8 \times 5 = \boxed{320}$.
[/hide]

[quote name="1ntegration1" url="/community/p2351202"]
Why do you get different answers? Why must you do the last digit, first digit, then the second?
[/quote]

Cuz otherwise you must do casework. If you do the first digit first, you have to deal with whether you chose an even or odd digit. Same goes will the other orders

The final digit has 5 choices, the second has 9 choices, and the third has 7 choices.

5*9*7=315

The problem with the $315$ solution is that it undercounts all cases where the second digit is $0$. If the second digit is $0$, then there are $8$ choices for the first digit. This adds $5$ solutions, which means the answer is indeed $\boxed{320}$.


Let ABCD be a quadrilatercal with inscribed in cirde (O) .Point E,F on the sides AB,CD respectively such that $ \frac{EA}{EB}\equal{}\frac{FD}{FC}$ ;Point K,L on the side EF such that $ \hat{AKB}\equal{}\hat{DLC}$ .Prove that $ \hat{BKC}\equal{}\hat{BLC}$

A question to clarify the statement: ABCD is inscribed in a circle (O), or ABCD has an inscribed circle in it? Thanks