modular arithmetic
algebra
polynomial
number theory
Additive combinatorics
combinatorics proposed
combinatorics

Find all the pairs of integers $ (a,b)$ satisfying $ ab(a \minus{} b)\not \equal{} 0$ such that there exists a subset $ Z_{0}$ of set of integers $ Z,$ for any integer $ n$, exactly one among three integers $ n,n \plus{} a,n \plus{} b$ belongs to $ Z_{0}$.

It's my first time here, so sorry for any glaring deficiencies in my proof (most notably, how I cannot express how m and n are relatively prime). If anyone could correct them or give tips, that would be helpful. :)

Label $ x$ as the middle integer of $ n$, $ n \plus{} a$, $ n \plus{} b$. $ x \minus{} m$ will be the smallest integer and $ x \plus{} o$ will be the greatest.

If $ m \equal{} o$, clearly the condition will be satisfied with $ Z_0$ containing all integers $ 3km$, where k is an integer. So from now on we will work with $ m \neq o$.

If $ m$ and $ o$ are not relatively prime, $ Z$ will be split into $ (m, o)$ sets? fields? rings? groups? whatever
But each of these "bunches" of integers will only affect integers inside it, and shifting and dividing by $ (m, o)$ will result in our original problem with $ (m, o) \equal{} 1$. So assume m and n are relatively prime WLOG, as every non-relatively prime pair will only work if their relatively prime cousins work.

There must be at least one integer in $ Z_0$. Call it $ f$.

Take $ x \equal{} f$. $ f \plus{} o$ cannot be in $ Z_0$.
Take $ x \equal{} f \plus{} m$. $ f \plus{} m$ and $ f \plus{} m \plus{} o$ cannot be in $ Z_0$.
Take $ x \equal{} f \plus{} m \plus{} o$. $ f \plus{} m \plus{} 2o$ must be in $ Z_0$.
Going left instead of right, we get that $ f \minus{} 2m \minus{} o$ must also be in $ Z_0$.
If $ f$ is in $ Z_0$, so is $ f \plus{} m \plus{} 2o \minus{} 2m \minus{} o \equal{} f \plus{} o \minus{} m$, and therefore $ f \plus{} k (o \minus{} m)$ is in $ Z_0$, where k is an integer.

$ (o \minus{} m, m \plus{} 2o) \equal{} (o \minus{} m, \minus{}2m \minus{} o) \equal{} (m \minus{} o, 3o) \equal{} (m \minus{} o, 3m) \equal{} (3o, 3m) \equal{} 3$

Assume one of $ m$, $ o$ is divisible by 3. (make it o because it does not matter)
$ (o \minus{} m, 3o) \equal{} 3$. Thus, $ m$ must be divisible by three. However, this gives us $ (o, m) \equal{} 3$. Contradiction!

Assume $ m \plus{} o \equal{} 0 \pmod{3}$. $ (m \minus{} o, 3o) \equal{} (m \plus{} 2o, 3o) \equal{} 3$, so that means m + 2o must be a multiple of 3. However, our assumption gives us that $ o$ is a multiple of 3, which by the lemma above cannot satisfy the given condition. Contradiction!

So we have that neither $ m$, $ o$, nor $ m \plus{} o$ is divisible by 3. These are the three distances between the three integers $ x$, $ x \minus{} m$, $ x \plus{} n$, so none of them can have the same residue mod 3.

We put every third integer into $ Z_0$, and as $ x \minus{} m$, $ x$, and $ x \plus{} n$ are $ 0$, $ 1$, and $ 2 \pmod{3}$, not necessarily in that order, this set will satisfy all m and n outlined above.

This means that all integers $ a$, $ b$ such that neither $ a$, $ b$, nor $ a \minus{} b$ (the three distances) is a multiple of 3 work.

To undo the relatively prime restriction, we multiply a and b by k, where k can be a multiple of three. Then $ Z_0$ can contain $ 3kc, 3kc \plus{} 1, 3kc \plus{} 2, \cdots 3kc \plus{} k \minus{} 1$.

Thus, our answer is all integers $ a \equal{} ky$, $ b \equal{} kz$, where neither $ y$, $ z$, nor $ y \minus{} z$ is a multiple of 3.

I'm sorry! it should be Chinese TST 2009 2nd quiz P2!

Let $f(n)=1$ if $n\in Z_0$ and $0$ otherwise, so $f(n)+f(n+a)+f(n+b)=1$ for all $n\in\mathbb{Z}$.

Then $f(n+b+1)-f(n+b)+f(n+a+1)-f(n+a)+f(n+1)-f(n)=0$ for all integers $n$, so by the [url=http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/]Skolem-Mahler-Lech theorem[/url] the set of zeroes of $f$ is eventually periodic for sufficiently large $n>N$ and consequently purely periodic with period $\ell>0$ (easily seen by "reversing" the recurrence). Hence
\[P(x)=\sum_{k=0}^{\ell-1}f(k)x^k\implies P(x)(1+x^{-a}+x^{-b}) \equiv 1+x+\cdots+x^{\ell-1} \pmod{x^\ell-1}.\]Modulo $x-1$ forces $3P(1)=\ell$, whence $\ell=3^r s$ for some $r\ge1$ and $\gcd(s,3)=1$. If $\gcd(x^{3^r}-1,1+x^{-a}+x^{-b})=1$, then
\[\frac{x^{3^r}-1}{x-1} \bigg{|} P(x)\implies 3^r|P(1)=\frac{\ell}{3}=3^{r-1}s,\]a contradiction, so there exists some $1\le t\le r$ such that $\Phi_{3^t}(x) = 1+x^{3^{t-1}}+x^{2\cdot3^{t-1}} | 1+x^{-a}+x^{-b}$ (since cyclotomic polynomials are irreducible in $\mathbb{Q}[x]$). Now working in $\mathbb{Q}(\omega)[x]$ for $\omega=e^{2\pi i/3}$, we have $x^{3^{t-1}}-\omega | 1+x^{-a}+x^{-b}$, so if $-a=3^{t-1}a_1+a_2$ and $-b=3^{t-1}b_1+b_2$ for $0\le a_2,b_2<3^{t-1}$, then it's easy to see that $a_2=b_2=0$ and $\{a_1,b_1\}\equiv\{1,2\}\pmod{3}$.

For the construction, taking $\ell=3^t$ and $P(x)=1+x+\cdots+x^{3^{t-1}-1}$ works.

Also note that in general, if $X\subseteq\mathbb{Z}$ and $X+a = \{x+a | x \in X\}$, then assuming integers $a_i$ exist such that $X+a_1, \ldots , X+a_n$ partition $\mathbb{Z}$, there must also exist an integer $N$ such that $X = X+N$. I think this allows us to generalize the original problem to $f(n+a_1)+f(n+a_2)+\cdots+f(n+a_p)=1$ if $p$ is prime.

Note that 2012 China TST Day 2 #3 actually reduces to a specific case of this problem (you are also given exactly one of $n$, $n-a$, and $n-b$ is in the set).

Awesome solution of math154 to the problem! I wish to remark here
[quote=math154]Then $f(n+b+1)-f(n+b)+f(n+a+1)-f(n+a)+f(n+1)-f(n)=0$ for all integers $n$, so by the [url=http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/]Skolem-Mahler-Lech theorem[/url] the set of zeroes of $f$ is eventually periodic for sufficiently large $n>N$ and consequently purely periodic with period $\ell>0$ (easily seen by "reversing" the recurrence).[/quote]
as $f$ is given by a recurrence relation and takes values in a finite set, it’s automatically eventually periodic (the recurrence relation even doesn’t need to be linear), so we don’t need Skolem-Mahler-Lech.

[quote=math154]I think this allows us to generalize the original problem to $f(n+a_1)+f(n+a_2)+\cdots+f(n+a_p)=1$ if $p$ is prime.[/quote]
Also let me put down a small step needed to run math154’s arguments for $p$, namely the fact that if $p$ powers of a $p^n$-th root of unity add up to $0$, then they divide the unit circle evenly. WLOG we may assume arguments of these powers are between $0$ and $2(p-1)\pi/p$, then since the minimal polynomial of a primitive $p^n$-th root of unity over $\mathbb{Q}$, i.e. the $p^n$-th cyclotomic polynomial has degree $p^{n-1}(p-1)$, these powers must correspond to terms of the cyclotomic polynomial and equal the $p$-th roots of unity.

It's easy to check that if $n$ appears in the set, all integers congruent to $n$ (mod $a+b$) will be in set too. Suppose $S$ is the set of remainders mod $a+b$ wich appeared in the set; then the sets $S+a$ and $S+b$ and $S$ are disjoint and covering all remainders mod $a+b$. Thus $a+b$ is divisable by 3 and the rest is easy by claiming that if (a,b) is an answer, (a/3,b/3) is an answer too.

This problem is very similar to [url=https://artofproblemsolving.com/community/c6h19789]IMO ShortList 1999, C6[/url]. One may see [url=https://artofproblemsolving.com/community/c6h19789p3375556]#8 there[/url] for a solution, using the Discrete Fourier Transform (DFT). The same approach can be applied here and it's even easier in my opinion. I decided  to write something on DFT in [url=https://dgrozev.wordpress.com/2020/08/16/application-of-discrete-fourier-transform-on-olympiad-problems/]my blog,[/url] and as an illustration to see how it works on both problems.


calculus
integration
algebra unsolved
algebra

Let $x,y,z$ be randomly (and independently) chosen from $[0,1]$.

Find the average value of $max(x,y,z) - min(x,y,z)$.

It is easy to calculate average max and min: $M(max(x_1,x_2,\dots,x_n))=n/(n+1),M(min(x_1,x_2,\dots,x_n)=1/(n+1)$.
Therefore $M(max-min)=\frac{n-1}{n+1}$. In your case 1/2.

How can you prove those formulas whitout using probablistic calculus? I will be happy to see an elementary proof. thank you!


Given that $a^2+b^2+c^2=1$ and $n$ is a FIXED natural number, what is the maximum value of $ a^n(b+c)+b^n(c+a)+c^n(a+b)$.

Of course, the inequality $a^n(b+c)+b^n(c+a)+c^n(a+b)\leq 2$ is not difficult to establish, but what is the maximal value?

I don't understand you harazi... the maximal value is clearly 2 and equality occurs when, for example, $a=b=c=\frac{1}{\sqrt{3}}$ and $n=1$.

I am sorry, but for every $n$ you must find the maximal value. This will probably depend on $n$, nut it is not always $2$.

Because (a^2)+(b^2)+(c^2)=1 we have the max value when n=1. so,We have
2(ab+ac+bc)<=2[(a^2)+(b^2)+(c^2)]=2

Hi Harazi,

Maybe you should reformulate the question so there is no more misunderstanding.

Oops sorry, you already done it...........

It is posted in the UNSOLVED section. :)

Well, i didnt solved it but is this a bit helpfull , isnt?

$\sum_{cyc}^{a,b,c} a^n(b+c)=\sum_{cyc}^{a,b,c} a^nb+\sum_{cyc}^{a,b,c} a^nc$ ,
but i found a lemma inequality which is proved by muierhead(and as harazi wishes chebyshev :lol: ) and it is this :
for $m,n \in \mathbb{N}$ we have :

$\sum_{cyc}^{a,b,c} a^mb^n \le \sum_{cyc}^{a,b,c} a^{m+n}  $

so using this inequality twice at our problem we will have :
$\sum_{cyc}^{a,b,c} a^n(b+c) \le 2\sum_{cyc}^{a,b,c} a^{n+1}$ so the maximum of our state occures  when this lemma takes its euality form .

Unfortunately, it doesn't work, since the maximal value of the last expression is attained in another point than the equality in the lemma. By the way, to prove the lemma one does not need Muierhead, just Chebyshev (I say this just because I hate from all my heart Muierhead).

Because I am 15 years please harazi tell me if it is correct

I regarded just your final answer and I realized that it is not helpful, since the upper boudn you find is far from optimal for sufficiently large $n$ (for example, if $n$ is large the number you indicated will be much larger than 6, which is clearly an upper bound for our expression). The problem is again the fact that the two inequalities used have different cases of equality. To be sincere, I think the only good way would be to use mixing variables and tons of computations. For $n=3$ I have smart solution.

I think I haven't made me understood. No, it is not good and I gave the explanation above. Your result is far from the maximal value of the expression. This is why at first I did not regard all solution. Now I regarded it and my expections were right: although the inequalities you used are correct, the final result is far from optimal simply because the inequalities you used have different cases of equality. Did I make myself clear now?

Could you post me a solution with a private message if you do not want the others see yout solution. Please reply yes,

I really don't know what to say. I think I speak french. I posted it in the unsolved section, since I don't have a solution for it. It bothers me for more than a year, but I never managed to solve it.

Sorry but i have not had understand you. Sorry again.

Now I have an another good solution.Please Mr Harazi tell me if it is correct

I would say you have another wrong solution. You cannot say that $a^n+b^n+c^n\leq 3(\frac{\sqrt{3}}{3})^n$. On the contrary.

Yes i would say this! We know if (x^2)+(y^2)=1 we have that (x^n)+(y^n)=[(x^n)+(1-x^2)^1/2]^n=f(x)
In an onother way we know tht when a sum is constant the product is max when x=y=z

Silouan, so you want to say that if $a^2+b^2+c^2=1$ then $a^n+b^n+c^n$ is maximal when all variables are equal? Very bad for you, but this holds only for $n=1$. Actually, for $n>2$ the sum $a^n+b^n+c^n$ is MINIMAL when all variables are equal. Sorry, your solution is wrong.

harazi, I guess this was posted by hungkhtn in proposed problem section before?

I never saw a solution to this problem, so if someone presented a link with a complete solution (not observations of the type "it must follow from") I would be very grateful. Thanks again.

My second solution are correct in positive reals?

No, silouan, sorry, but it is not correct. Can't you understand my explanations or what?

http://www.mathlinks.ro/Forum/viewtopic.php?t=30732&postorder=asc
it's a little bit different, but maybe it can inspire someone.

Exactly as I thought! There is no proof, absolutely nothing! :(


By using Muirheads inequality can we say that $[3,3,0]\geq[3,2,1]$ is true...?

Davron

[quote="Davron"]By using Muirheads inequality can we say that $[3,3,0]\geq[3,2,1]$ is true...?[/quote]

You mean that

$2\left(b^3c^3+c^3a^3+a^3b^3\right)\geq a^3b^2c+a^3c^2b+b^3c^2a+b^3a^2c+c^3a^2b+c^3b^2a$ ?

Yes, we can say that. But we can also say that this follows by summing up the three not really hard inequalities

$a^3\left(b^3+c^3\right)\geq a^3\left(b^2c+c^2b\right)$;
$b^3\left(c^3+a^3\right)\geq b^3\left(c^2a+a^2c\right)$;
$c^3\left(a^3+b^3\right)\geq c^3\left(a^2b+b^2a\right)$,

and thus applying Muirhead here is a bit of unnecessary.

  darij

A bacha why not. You can say every think you want!!! :D

 :arrow: Because you are Davron!!!!!!!!!!!!! not only Davron but Davron Latipov!!!!!!!!!!!!!!!!! :oops:


Find all increasing functions $ f:\mathbb{R}\to\mathbb{R}$ which satisfy
\[f(xf(y))=yf(2x), \quad \forall x,y \in \mathbb R.\]

[quote="quangpbc"]Find all $ f:\mathbb{R}\to\mathbb{R}$ so that :

$ 1,f$ is an increasing function.

$ 2,f(xf(y)) \equal{} yf(2x)$

 :)[/quote]

Let $ P(x,y)$ be the property $ f(xf(y)) \equal{} yf(2x)$

$ P(1,x)$  $ \implies$ $ f(f(x))\equal{}f(2)x$. $ f(2)\neq 0$ since $ f(x)$ is strictly increasing, so $ f(x)$ is a bijective function.
$ P(x,0)$ $ \implies$ $ f(xf(0))\equal{}0$ and so $ f(0)\equal{}0$ since $ f(x)$ is an injective function.
$ P(f(x),y)$ $ \implies$ $ f(f(x)f(y))\equal{}yf(2f(x))$
$ P(f(y),x)$ $ \implies$ $ f(f(x)f(y))\equal{}xf(2f(y))$
So $ yf(2f(x))\equal{}xf(2f(y))$ and so $ \frac{f(2f(x))}{x}\equal{}\frac{f(2f(y))}{y}\equal{}a$ $ \forall x,y\neq 0$
So $ f(2f(x))\equal{}ax$ $ \forall x$ with $ a\neq 0$
So we have a new property $ Q(x,y)$ : $ f(f(x)f(y))\equal{}axy$
Let then $ b\neq 0$ such that $ f(b)\equal{}1$ ($ b$ exists since $ f(x)$ is a surjective function)
$ Q(x,b)$ $ \implies$ $ f(f(x))\equal{}abx$ and so $ f(abx)\equal{}abf(x)$
$ Q(f(\frac{x}{ab}),f(\frac{y}{ab}))$ $ \implies$ $ f(xy)\equal{}\frac{1}{ab^{2}}f(x)f(y)$

And this is a well known Cauchy equation which gives $ f(x)\equal{}ux$ for some $ u$ and $ \forall x\in\mathbb{Q}$ and then $ f(x)\equal{}ux$ $ \forall x$ since $ f(x)$ is an increasing function.

Putting back this solution in the original equation, we find $ u\equal{}2$

And the only solution is $ f(x)\equal{}2x$


greatest common divisor
number theory
least common multiple

If $x$ is an integer such that $\text{gcf}(x,24)=8$ and $\text{lcm}(x,24)=312$, find $x$.
Eli, you better not answer.

hmm...i actually didnt notice that Eli dont answer.  :(

but heres a hint: (a really big hint, that basically gets you the answer, but anyways....

[hide]lcm times gcf=product of the 2 numbers[/hide]

[quote="nutz_for2.718281828"][hide="solution"]24x=2496, x=104

other way:

We divide 312 by 8, and get 39. we divide by 3 to get 13. So the divisors of x are 13 and 8. Multiply: x=104[/hide][/quote]
Where did the three come from?

[hide] using the hint of Walk Around The River, we now know that the LCM times the GCF of two numbers equals the product. Why this works i have yet to find out, but anyways. we get
$24x=312*8$. (24*x comes from the fact that the problem gives you them when it tells you about GCF/LCM.)
anyways...moving on.
$24x=2496$, which becomes
$x=104$
$\boxed{104}$. [/hide]

[hide]since lcm*gcf=product of numbers, x=8*312/24=104[/hide]


[quote="pascal_1623"]What is the sum of $\frac{1}{2x}*\frac{2}{3}*\frac{3}{4}*...\frac{99y}{100}$?[/quote]

are you sure that's a sum? It looks more like a product...

[hide]many terms cancel leaving the product as y/100x i believe.[/hide]

[hide]
$\displaystyle \frac{1}{2x}*\frac{2}{3}*\frac{3}{4}*...\frac{99y}{100}$

as you can see the 2 in 2x will cancle out with the 2 in 2/3
also the 3 in 3/4 will cancel with the 3 in 2/3
this goes on for a while.........
you are left with $\frac{y}{100x}$[/hide]


search
inequalities unsolved
inequalities

let a,b,c  be reals and such that

\[\begin{array}{l}a^{2}+b^{2}+c^{2}= 9 \\ \\ then prove that : 2(a+b+c)-abc \le 10 \\ \end{array}\]


vector
inequalities
triangle inequality
linear algebra
linear algebra theorems

How to show that x -> (|x1|^p + |x2|^p + ... + |xn|^p)^(1/p) defines a vector norm on R^n for 1<=p<infinity ?

Hope someone can enlighten me.. thanks![/code]

To prove that it is a norm, you need to prove three things: that it is positive definite, that it scales properly with scalar multiplication, and that it satisfies the triangle inequality. The first two items are straightforward; the triangle inequality is not.

$ p\equal{}1$ and $ p\equal{}\infty$ are special cases with their own, easier, proofs. So assume that $ 1<p<\infty.$

The standard intermediate step is an important result in its own right: [b]Hölder's Inequality:[/b]

Suppose that $ \frac1p\plus{}\frac1q\equal{}1.$ Then

$ \left|\sum_{j\equal{}1}^nx_jy_j\right|\le \left(\sum_{j\equal{}1}^n|x_j|^p\right)^{\frac1p} \left(\sum_{j\equal{}1}^n|y_j|^q\right)^{\frac1q}$

I'll let you think about proving that (or looking it up). (Note that if $ p\equal{}2,$ then we would have $ q\equal{}2$ and Hölder's Inequality would become Cauchy's Inequality.)


Given real numbers $x,y,z$ such that $x^2+y^2+z^2=k$, $k>0$, prove the inequality
$\frac{2}{k}xyz-\sqrt{2k}\le x+y+z\le \frac{2}{k}xyz+\sqrt{2k}$.
[hide]When k=2, see poland 1991 http://www.artofproblemsolving.com/Forum/resources.php?c=137&cid=39&year=1991 [/hide]

This inequality can be viewed also at Mathscope at

http://www.imo.org.yu

or at

http://www.math4u.de

Thuan

[quote="Beat"]Given real numbers $ x,y,z$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} k(k > 0)$, prove that

$ \frac {2}{k}xyz \minus{} \sqrt {2k}\le x \plus{} y \plus{} z\le \frac {2}{k}xyz \plus{} \sqrt {2k}$.

When $ k \equal{} 2$, see Poland 1991 http://www.artofproblemsolving.com/Forum/resources.php?c=137&cid=39&year=1991
[/quote]
$ 2\left(x^2 \plus{} y^2 \plus{} z^2\right)^3 \minus{} \left[(x \plus{} y \plus{} z)\left(x^2 \plus{} y^2 \plus{} z^2\right) \minus{} 2xyz\right]^2$

$ \equal{} 2\left(x^2 \plus{} y^2 \plus{} z^2\right)^3 \minus{} \left[x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(yz \plus{} zx \plus{} xy)\right]\left(x^2 \plus{} y^2 \plus{} z^2\right)^2$

$ \plus{} 4xyz(x \plus{} y \plus{} z)\left(x^2 \plus{} y^2 \plus{} z^2\right) \minus{} 4x^2y^2z^2$

$ \equal{} \left(x^2 \plus{} y^2 \plus{} z^2 \minus{} 2yz\right)\left(y^2 \plus{} z^2 \plus{} x^2 \minus{} 2zx\right)\left(z^2 \plus{} x^2 \plus{} y^2 \minus{} 2xy\right) \plus{} 4x^2y^2z^2$

$ \geq 5x^2y^2z^2\geq0$.


Rico can walk 3 miles in the same amount of time that Donna can walk 2 miles. Rico walks a rate 2 miles per hour faster than Donna. At that rate, what is the number of miles that Rico walks in 2 hours and 10 minutes?

$ \frac{3}{t}\equal{}\frac{2}{t}\plus{}2$.
$ \therefore \ t\equal{}\frac{1}{2}$[hour].
So, Rico walks with $ 6$ miles per an hour.
Thus, $ 6\times\left(2\plus{}\frac{10}{60}\right) \equal{} 13$[miles].


logarithms
analytic geometry
geometry
calculus
integration
calculus computations

Sketch the graphs of $ y \equal{} \ln x$ and $ y \equal{} 2\ln (x \minus{} 6)$.

The curves meet at the point $ P$ with $ x$-coordinate $ a$.

State the pair of transformations which maps $ y \equal{} \ln x$ to $ y \equal{} 2\ln (x \minus{} 6)$.

Solve an equation to find $ a$.

Use Simpson's rule with (i) two strips and (ii) four strips to find approximations to the area enclosed by $ x \equal{} a$, $ y \equal{} 2\ln (x \minus{} 6)$ and the point where $ y \equal{} 2\ln (x \minus{} 6)$ meets the $ x$-axis.

[I make $ a\equal{}9$ (because $ 4<7$) and the area $ 2.580800674$ and $ 2.590643337$.

Using integration by parts I get $ 6\ln 3 \minus{}4 \equal{} 2.591673732...$.]

Disappointed that no-one replied but I'm fairly sure I got it right. Just wanted to check to be sure.

hello, yes i have got $ 2\int_{7}^9\ln(x\minus{}6)\,dx\equal{}6\ln(3)\minus{}4$.
Sonnhard.


[quote="kunny"]For $ 0\leq x < 2\pi$, find the number of $ x$ which satisfies the equation : $ 2\sqrt {2}(\sin ^ 3 x \plus{} \cos ^ 3 x) \plus{} 3\sin x\cos x \equal{} 0$.[/quote]

[hide="partial solution"]$ \sin ^ 3 x \plus{} \cos ^ 3 x\equal{}(\sin x\plus{}\cos x)(\sin^2{x}\minus{}\sin{x}\cos{x}\plus{}\cos^2{x})\equal{}(\sin{x}\plus{}\cos{x})(1\minus{}\sin{x}\cos{x})$

$ 2\sqrt{2}(\sin{x}\plus{}\cos{x})(1\minus{}\sin{x}\cos{x})\plus{}3\sin{x}\cos{x}\equal{}0$[/hide]

There I am shtuck.

Put sinx+cosx = t

then sinxcosx = (t^2-1)/2


The real numbers x, y, z satisfy:

0 \leq x, y, z \leq 1
3x + 2y + z \leq 4

Find the greatest possible value of 3x2 + 2y2 +z2.

I'm not sure this is the right statement, since the problem is too simple. We obviously have 3x^2+2y^2+z^2 \leq  3x+2y+z \leq 4 and this is attained when x=1 and z=1 and y=0.


I have 18 distinguishable socks in my drawer: 8 white, 6 brown, and 4 blue.  In how many ways can I choose a pair of socks, provided that I get two socks of the same color?

To get a pair of white socks, there are $ 8C2$ or $ 28$ ways.

To get a pair of brown socks, there are $ 6C2$ or $ 15$ ways.

To get a pair of blue socks, there are $ 4C2$ or $ 6$ ways.

$ 28 \plus{} 15 \plus{} 6 \equal{} \boxed{49}$


complex numbers
geometry unsolved
geometry

A regular $ n\minus{}$gon $ A_1A_2A_3 \cdots A_k \cdots A_n$ inscribed in a circle of radius $ R$ is given. If $ S$ is a point on the circle, calculate \[ T \equal{} \sum^n_{k\equal{}1} SA^2_k.\]

Let's assume WLOG that $A_{1}A_{2}A_{3}\cdots A_{k}\cdots A_{n}$ are in counterclockwise order
Let the origin of the complex plane be at the center of the $n-$gon. Let $a_k$ be the complex number representing $A_k$, and let $s$ be the complex number representing $S$.
Then $a_k = wRe^{i2k\pi /n}$ and $s = vwR$ for some unitary complex numbers $w, v$.
$T =\sum^{n}_{k=1}SA^{2}_{k} = \sum^{n}_{k=1}|s-a_k|^2 = R^2|w^2|\sum^{n}_{k=1}|v-e^{i2k\pi /n}|^2$
$= R^2\sum^{n}_{k=1}(v-e^{i2k\pi /n})\overline{(v-e^{i2k\pi /n})} = R^2\sum^{n}_{k=1}(v-e^{i2k\pi /n})(1/v-1/e^{i2k\pi /n})$
$= R^2(2n+v\sum^{n}_{k=1}e^{i2k\pi /n}+1/v\sum^{n}_{k=1}1/e^{i2k\pi /n}) = 2nR^2$.


[b]I'm trying to take old AHSMEs  under time limits to practice for the next AMC. I expect to answer less than half of the questions and get a few wrong.

How would you go about taking them if you were me? [/b]
[hide="More Specific Questions"]
Should I make sure I read all the problems before the time is up so when I go over the answers, I have some idea of what the solutions mean?
OR
Should I just answer the questions I can, check over them with the solutions, then try the remaining problems afterward without looking at their solutions?

After I go over the answers to one contest should I...
1. Take it immediatly afterward to see if I actually understood the solutions I read?
2. Wait until later and take the competition again?
3. Do something else?

Is there any point in guessing while taking the previous competition?

Should I get one of those scary looking scantrons to fill in the bubbles with?

Should I leisurely work on the problem at my own pace?
______________________________________________________________[/hide][b]Let's say you just took a practice test and you're going over it. When you get a question wrong or left it blank and you want to go over the problem, what is the best way to understand and learn the solution?[/b]

[hide=" Viewing solutions as more than just tricks"]
Sometimes when I see the solution, I say to myself, "Oh! That was the trick... " and move on before I feel confident that I can answer any similar question with changed numbers. It feels a little disgusting when I tell myself that I didn't get a problem correctly just because I didn't see "a trick." I want to say to myself: "Chris, if you were more experienced or understood this-and-that better, than you would have gotten the problem correct." ... [/hide]
[b]What is the correct response when you see that the solution required no math that you didn't know?[/b] You can't just say, "Oh I knew that. Maybe I'll see the trick for the next problem!"
It just doesn't feel right.

Yes, it is quite frustrating when you go over a solution and find that it required a non-trivial step using concepts that you are familiar with. I would say to go back a few weeks later and do the problem again. Also, it's always better to think about the problem after you take the test rather than simply reading the solution after taking the test. Make sure you give all problems a good deal of thought, even if you take more time than the exam allots. This will probably be the case if you only expect to solve half of the problems.

Your hide tag does not work because of the quotes ("") in it.

Yeah, thanks for telling about the hide. Also, you're right about what you said. It's not good to read a solution to a problem you haven't thought about.

OK this is what i did to go from 13 on AMC 10 to 8 on AIME...when you're done with an AMC or AHSME, then go over the problems you did wrong/didn't do and spend time on them. The amount of time doesn't matter. Even when you start off, DO NOT time your work, just focus on accuracy. Speed comes naturally after that. :)

Its totally true that any AMC problem is solved by a 'trick' or two. You should be under the assumption that there is a reasonable solution. Think like a test writer. What ideas do they usually use? Ultimately, everything will boil down to a "trick" or two. But saying that it is just a trick means that you don't understand how to solve the problem on your own. There is going to be a well-motivated solution and you have to work towards finding it yourself.

Go back to the questions that you missed because of a "trick" after a week or so to see if you remember what the trick was. There are a lot of tricks, but question writers tend to reuse some tricks, so once you get familiar with a certain set of tricks that's that many more problems you can do.

For example, one trick that comes up often on combinatorial questions is the bijection on the balls-and-urns argument. While the questions that could be asked about this are very diverse, doing a few questions involving this "trick" will help you see this trick on a different question in the future.


It is Bertsnaider theorem (Cosine law for quadrilateral)

Prove that $ {d_1}^2{d_2}^2\equal{}a^2c^2\plus{}b^2d^2\minus{}2abcd.\cos\left(\alpha \plus{}\gamma \right )$,

where d1 and d2 are diagonals of ABCD, a, b, c, d are its sides α and γ are angles А and С.

I think it is not an easy problem. Do you want a hint?
Here is the complete solution: [hide]http://www.math10.com/forumbg/viewtopic.php?t=2588.[/hide]


For a certain triangle all of its altitudes are integers whose sum is less than 20. If its Inradius is also an integer Find all possible values of area of the triangle.

[hide="Very big hint"]

$ S\equal{}rs\equal{}\frac{ah_a}2\equal{}\frac{bh_b}2\equal{}\frac{ch_c}2\rightarrow \frac1h_a\plus{}\frac1h_b\plus{}\frac1h_c\equal{}\frac1r$

[/hide]

YEs Bugi it is a very big hint for asuch an easy problem.
More over thi sthing simplifies it further, that is
$ \sum {\triangle}{a} < 10$. $ h \equal{} \frac{2\triangle}{side}$
$ r \in \mathbb{N}$. THis implies $ \triangle \equal{} ks$ where $ k$ is a postive integer and $ s$ is the semiperimeter.
We can easily say now that $ \frac {9}{a \plus{} b \plus{} c} \le \sum \frac {1}{a} < \frac {10}{\triangle}$.The first part is by Cauchy Schwarz.
this implies that $ 9\triangle < 20 s$ THis implies that $ k < \frac {20}{9}$. Thus $ k \equal{} 1$ or $ 2$.
From now on, there is nothing interesting in the problem as it is so so easy. Sad that India is asking problems like these in the TSTs.


calculus
integration
trigonometry
linear algebra
matrix
calculus computations

Compute:

1) $ \int \int_D e^{\frac{y\minus{}x}{y\plus{}x}}\,dx\,dy$, where D is the triangle limitated by the lines x = 0, y = 0, and x + y = 2.

2) $ \int \int_D(x\minus{}y)^2 \sin^2(x\plus{}y) \,dx\,dy$, where D is the polygon with vertices at the points $ (\pi, 0)$, $ (2 \pi, \pi)$, $ (\pi, 2 \pi)$, and $ (0, \pi)$.

3) $ \int \int_D xy\,dx\,dy$, where $ D \equal{} \{(x,y) \in \mathbb{R} ^2: (x^2\plus{}y^2 \leq 4 \,\wedge \, x \geq 0 \, \wedge \, y \geq 0) \, \vee$

$ \vee \, (4x^2\plus{}y^2 \leq 4 \,\wedge \, x \leq 0 \, \wedge \, y \geq 0) \}$.

I'll take (2). The region $ D$ is bounded by $ x\plus{}y\equal{}\pi,x\plus{}y\equal{}3\pi,x\minus{}y\equal{}\pi,$ and $ x\minus{}y\equal{}\minus{}\pi.$ Make the change of variables $ u\equal{}x\minus{}y,v\equal{}x\plus{}y.$ This linear transformation has matrix $ \begin{bmatrix}1&1\\ \minus{}1&1\end{bmatrix}$ (or its transpose, depending on your convention). The Jacobian matrix of a linear transformation is simply its matrix, so the Jacobian determinant $ \frac{\partial (u,v)}{\partial (x,y)}$ is $ 2.$ But that's the reciprocal of what we want, which is that $ \frac{\partial (x,y)}{\partial (u,v)}\equal{}\frac12$ and $ dx\,dy\equal{}\frac12\,du\,dv.$ The integral becomes:

$ \int_{\minus{}\pi}^{\pi}\int_{\pi}^{3\pi}\frac12u^2\sin^2v\,du\,dv\equal{}\int_{0}^{\pi}\int_{\pi}^{3\pi}u^2\sin^2v\,du\,dv$

$ \equal{}\frac{\pi^3}3\int_{\pi}^{3\pi}\sin^2v\,dv\equal{}\frac{\pi^4}{3}$

In the last step, I used the "yin-yang lemma": the integral of $ \sin^2$ or $ \cos^2$ of a linear quantity taken over an interval that starts and stops at the ends of quadrants is half the length of the interval.

(I'm teaching exactly this course this semester. But the semester hasn't started yet - it starts Monday - and these questions won't be until my third of four midterm exams.)

4) Compute $ \int_0^1 \int_0^{1\minus{}x} e^{\frac{y}{x\plus{}y}}\,dy\,dx$.


Be $\alpha(G)$ the size of the largest independent set in $G$ and $\chi(G)$ its chromatic number. Prove that :

$\alpha(G)\gneq\frac{n}{\chi(G)}$

where $n$ is the number of vertices of $G$.

I tried and got to some points that don't lead me nowhere. Using induction on the vertices of $G$ by supressing on vertex $v$ we have :

$\alpha(W)-1\lneq\alpha(G)\lneq\alpha(W)$
$\chi(W)-1\lneq\chi(G)\lneq\chi(W)$

so $n-1\lneq\alpha(W)\chi(W)\lneq\alpha(G)\chi(G)$

where $W$ is the graph obtained of supressing $v$ from $G$.

$\alpha(G)\lneq n-\chi(G)+1$

and some other things that don't lead me anywhere too...

Any hints....

Thx :(

First, strict inequality fails: just consider a complete graph or its complement.

Actually, though, this question is easier than you're making it -- no need for strong inductions.  My hint is to consider what a vertex-coloring really is.

Well the definition of chromatic number is clear, the minimun number of colors needed to colored a graph $G$, such $(v, w)$ will never have the same color for every vertices $v,w$.

What I'm thinking is to work with the set of independent vertices separately and with the rest of the vertices as well. But still can go any further than

$\alpha(G)\chi(G)\gneq n-1$

and still can't reduce that from $n-1$ to $n$

[quote="Andy"]What I'm thinking is to work with the set of independent vertices separately and with the rest of the vertices as well. But still can go any further than

$\alpha(G)\chi(G)\gneq n-1$

and still can't reduce that from $n-1$ to $n$[/quote]
I'm not quite sure what you mean by this, but as I said, it's not in fact possible to prove $\alpha(G)\chi(G) > n$ because it's not in general true.

[quote="Andy"]Well the definition of chromatic number is clear, the minimun number of colors needed to colored a graph $G$, such $(v, w)$ will never have the same color for every vertices $v,w$.
[/quote]
Yes, right.  Now, my second hint: suppose we have a coloring of the graph $G$ with $\chi(G)$ colors, and we use the color red.  What can you tell me about the set of red vertices?

We can say that the red vertices can be considered as an independent set of vertices of G.

'cause coloring the graph is like finding a set of independent set of vertices of G.

Color red is an independent set.
Color blue is an independent set.

uhmmm I think I got it...I'll work on it later is too late...thx

Well, yeah...I know how to prove that coloring $G$ is like doing the following...

Getting the maximun independent set of vertices from $G$.
Now from the graph that's left of removing this independent set with all edges incident to each vertex we can have the maximum independent set of vertices from it... and so on......

So this like dividing $G$ into a minimum of independent set of vertices...


trigonometry
geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry solved

i'm sorry for the disobedience, but when i see a problem like this i just can't think on the trigonometric form of ceva's theorem, anyway there's my solution.

Since $\angle B+\angle C=140$, we have $\angle A=40$, then $\angle OAB=30$, and because of $\angle OCA=10$, we have $\angle OCD=70$. Now, let $\angle OBC=\alpha$, so $\angle OBA=60-\alpha$. By the trigonometric form of Ceva, we have $\sin\angle OCA\cdot\sin\angle OAB\cdot\sin\angle OBC=\sin\angle OAC\cdot\sin\angle OBA\cdot\sin\angle OCB \\\Rightarrow \sin10\cdot\sin30\cdot\sin\alpha=\sin10\cdot\sin(60-\alpha)\cdot\sin70 \\\Rightarrow \sin\alpha=2\sin(60-\alpha)\cdot\cos20\\ \Rightarrow \sin\alpha=\sin(40-\alpha)+\sin(80-\alpha)$
Now, if $\alpha>40$ we have $LHS>RHS$, and if $\alpha<40$ we get $LHS<RHS$, so we must have $\alpha=40 \Rightarrow \angle COD=\angle OCD=70 \Rightarrow BO=BC$.

I've been searching for some ingenious geometric construction but I didn't immediately find one. I'll think some more about it. I hope there is a nice solution.

Since you want a proof without trigonometry, here is one, although I am almost sure there are much easier ones:

We show a more general fact:

[color=blue][b]Theorem 1.[/b] Let ABC be a nondegenerate triangle, O an arbitrary point not lying on any of its sidelines, and t an angle such that < OAB = 30°, < CAO = < OCA = t and < BCO = 60° + t, where we use directed angles modulo 180°. Then, BO = BC.[/color]

[i]Proof of Theorem 1.[/i] Let E be the reflection of the point O in the line AB. Then, $OE\perp AB$ and < EAB = - < OAB, < AEO = - < AOE and BE = BO. Since < OAB = 30°, we have < EAB = - < OAB = -30°.

From $OE\perp AB$, we conclude < (OE; AB) = 90°, so that < EOA = < (OE; AO) = < (OE; AB) - < (AO; AB) = 90° - < OAB = 90° - 30° = 60°. Since < AEO = - < AOE = < EOA, we thus have < AEO = 60°. So the triangle AEO has two 60° angles; therefore, it is equilateral, so that AO = EO. Also, since < CAO = < OCA, the triangle AOC is isosceles, and thus AO = CO. Hence, AO = CO = EO. Thus, the points A, C and E lie on a circle with center O. Hence, by the central angle theorem for directed angles modulo 180°, we have < OEC = 90° - < CAE. Thus, < CAE = 90° - < OEC.

Now,

   < CAE = < CAO + < OAB + < BAE = < CAO + < OAB - < EAB = t + 30° - (-30°) = 60° + t = < BCO.

Hence, < CAE = 90° - < OEC becomes < BCO = 90° - < OEC.

On the other hand, if B' is the circumcenter of triangle CEO, then, since B' is the center of a circle passing through the points C, E and O, the central angle theorem yields < B'CO = 90° - < OEC. Hence, < BCO = < B'CO.

Now, assume that the points B and B' are distinct. Then, < BCO = < B'CO yields that the line BB' passes through the point C. On the other hand, the point B' lies on the perpendicular bisector of the segment OE (since it is the circumcenter of triangle CEO). But since the point E is the reflection of the point O in the line AB, the perpendicular bisector of the segment OE is the line AB, and thus the point B' lies on the line AB. Hence, the line BB' is simply the line AB, and therefore it cannot pass through the point C, since the triangle ABC is nondegenerate. Hence, we get a contradiction.

Thus, the assumption that the points B and B' are distinct was wrong, i. e. the points B and B' coincide. Since the point B' was defined as the circumcenter of triangle CEO, it follows that the point B is the circumcenter of triangle CEO, and thus BO = BC, and Theorem 1 is proven.

Theorem 1 quickly yields your initial problem: Let's work with non-directed angles. In your figure, by the sum of angles in triangle ABC, we have < BAC = A = 180° - B - C = 180° - 60° - 80° = 40°, so that < OAB = < BAC - < OAC = 40° - 10° = 30°, < CAO = < OCA = 10° and < BCO = < BCA - < OCA = C - < OCA = 80° - 10° = 70° = 60° + 10°. Hence, your problem follows from Theorem 1 for t = 10°.

  darij

[quote="stergiu"]Let a triangle  $\triangle  ABC $    with  $\angle B = 60 $ and  $\angle C = 80 $. In the interior of  $ \triangle ABC  $ we take point  $ O $  such that   $\angle OAC = \angle OCA = 10$. Prove that   $BO = BC $ .

 [i]A solution without trigonometry is preferable [/i]

Babis[/quote]
What I came up  with is a solution based on really simple trigonometry.Namely, denote angle OBA=x, we`ll show that x=20.After applying sine theorem for triangle OAB and OBC we get the equation sin30/sinx=sin70/sin(60-x).
sin70/sin30=sin(60-x)/sinx.Clearly, as we increase x, sin(60-x) decreases and sinx increases(since x<60), so there can be at most 1 value of x that suutisfies the equation.We show it is 20.Namely, we should prove the identity 2sin20sin70=2sin40sin30
LHS=cos50=sin40  RHS=cos10-cos70=sin80-sin20
What is to be proved now is that sin40+sin20=sin80, which follows directly from the formula sinA+sinB=2sin((A+B/2))cos((A-B)/2)

Well , I think it's OK with this problem and the solutions are good. Darij has encouranged me to try again for a pure and a litle simpler geometric solution. If I find something , I ' ll come back again.

Thanks to all of you!

Babis


rotation
analytic geometry
geometry
geometric transformation
reflection
linear algebra
matrix

The point $ P \equal{} (1,2,3)$ is reflected in the $ xy$-plane, then its image $ Q$ is rotated by $ 180^\circ$ about the $ x$-axis to produce $ R$, and finally, $ R$ is translated by 5 units in the positive-$ y$ direction to produce $ S$. What are the coordinates of $ S$?

$ \textbf{(A)}\  (1,7, \minus{} 3) \qquad \textbf{(B)}\  ( \minus{} 1,7, \minus{} 3) \qquad \textbf{(C)}\  ( \minus{} 1, \minus{} 2,8) \qquad \textbf{(D)}\  ( \minus{} 1,3,3) \qquad \textbf{(E)}\  (1,3,3)$

[hide="Answer"]Reflecting across the $xy$ plane gives $(1,2,-3)$, $180^\circ$ about the x-axis flips the $y$ and $z$ signs and gives $(1,-2,3)$, and adding $5$ to $y$ gives $(1,3,3)\Rightarrow \boxed{E}$.[/hide]

[hide]$(1, 2, 3) \Rightarrow$ across $xy \Rightarrow (1, 2, -3)$
$\Rightarrow 180^\circ x-axis \Rightarrow (1, -2, 3)$
$\Rightarrow +5 units y-axis \Rightarrow (1, 3, 3) = \boxed{E}$[/hide]

What does it really mean to reflect in the xy plane anyway?

Suppose you want to reflect a point $ P$ across the $ xy$-plane. Suppose the image of $ P$ upon reflection is $ P'$. Then by definition, $ \overline{PP'}$ is perpendicular to the plane $ xy$ and the plane bisects the line segment.

Oh I see so it's like flipping the entire plane over and where P is after that is P'.

[quote="TheMaskedMagician"]Can someone explain how to rotate in the xy plane?[/quote]

Rotating by multiples of $90$ degrees counterclockwise is fairly intuitive.  It turns $(x, y)$ into $(-y, x)$.  Switch the coordinates, and negate the second.  If you do this a second time, you get $(-x, -y)$, which is a $180$ degree rotation.  Doing it four ties gets you $(x, y)$ - coming literally full circle.

For some fun, you can prove that $(-y,x)$ is the rotation of $(x,y)$ - all you have to prove is that they're the same distance from the origin, and that they form a right angle.  (Hint:  When you see "right angle" in coordinate geometry, you should immediately think of the number $-1$.)

Reflecting $(x,y)$ across the x-axis gives $(x, -y)$ - instead of above (or below), it's below (or above) the x-axis.  Reflecting the original point across the y-axis gives $(-x, y)$.  Doing the two reflections together gives $(-x, -y)$ - so rotating by $180$ or reflecting twice gives you the same point.

Now, what if it's not $90$ degrees?  A rotation matrix makes it much easier, but I hate $\LaTeX$ing matrices.  If you rotate $(x,y)$ by $\theta$, then your new point is $(x\cos \theta - y\sin \theta,  x\sin \theta+ y\cos \theta )$.  Don't memorize the formula in this form - go to Wikipedia and search for the rotation matrix article - you'll find it there.

For fun, you might try to prove that the two points are the same distance from the origin, and that they form an angle of $\theta$.

Oh I thought the problem was talking in terms of 3D so I was trying to visualize the point being rotated on the xyz plane. Rotation matrices look very useful and interesting but I would need to learn some things prior to tackling them. Thanks for the explanation.

Note also that rotating in the x-y plane is (sort of) the same thing as "rotating around the z-axis".  You have to specify a direction:  clockwise or counterclockwise.  In this problem, you're rotating around the y-axis, so you're rotating within the x-z plane.  The clockwise/counterclockwise issue is alleviated by the fact that it's a 180 degree turn.

In fact, that might cause you to ask a very good question:  which way would be clockwise or counterclockwise?  Well, in the x-y plane, we assume that we're looking "from above" (or from the positive direction) when we're talking about clockwise/counterclockwise.  To see what I mean, find a floor fan and turn it on slowly.  Notice which way the blades go - whether it's clockwise or counterclockwise.  Now, go behind the fan.  The blades go in the opposite direction!

In physics, there's actually something called the "right hand rule" that gives the direction.  If you point your right thumb up, your fingers curl clockwise.  It basically gives the orientation of the three axes.

Luckily I had a water bottle next to me, and I imagined the point being the corner of the water bottle (my water bottle is in the shape of a rectangular prism, not a cylinder). It really helped me visualize the rotations and reflections :D .


induction
algorithm
number theory proposed
number theory

Prove that for every positive integer $n$ can be represented uniquely in the form \[n=\sum_{k=1}^{\infty}a_{k}\cdot k!,\]
where \[a_{k}\in\{0,1,...,k\}.\]

could anyone give a proof for this interesting result? or Rust  explain his post

A generalization:  Given a sequence $b_{k}$ of positive integers, every positive integer $n$ can be represented uniquely in the form

$n = \sum_{k=1}^{\infty}a_{k}(b_{1}b_{2}... b_{k})$

Where $0 \le a_{k}\le b_{k+1}-1$.  For $b_{k}= b$ it gives the familiar base-$b$ expansion, and for $b_{k}= k$ it gives this result.

The proof is not difficult by induction or a counting argument, and relies on the identity

$\sum_{k=1}^{M}(b_{k+1}-1)(b_{1}b_{2}... b_{k}) = b_{1}b_{2}... b_{k+1}-1$

Which has familiar analogues in the cases $b_{k}= b, b_{k}= k$.

Another generalization: Let $m_{0},m_{1},...$ be a strictly increasing sequence of positive integers such that $m_{0}=1$ and $m_{i}|m_{i+1}\forall i\geq 0.$ Prove that for every positive integer $n$ can be represented uniquely in the form \[n=\sum_{i=0}^{\infty}a_{i}\cdot m_{i},\] where \[0\leq a_{i}\leq\frac{m_{i+1}}{m_{i}}-1\forall i\geq 0\] and $m_{i}=0$ for all but finitely many $i$.

Oh, I think I understand.  Rust is showing that $a_{1}$ is unique because it is the unique residue of $n \bmod 2$ (since $2$ divides $k!$ for all $k \ge 2$).  Once the uniqueness of $a_{1}$ has been established, we can subtract it out and then $a_{2}$ is the unique residue of $\frac{n-a_{1}\cdot 1!}{2}\bmod 3$, and so forth.  By induction it follows that $a_{k}$ is unique for a given $n$.  Very nice!  :)

This also provides a fast proof of my generalization.  Impressive.

yes I think it's clear. by division algorithm $n=2k_{1}+a_{1}$,then sets $n_{2}=\frac{n-a_{1}1!}{2}$ and $n_{2}=3k_{2}+a_{2}$ and then he  finds $n_{3}$ etc.
it is a very nice idea  :)
but shouldn't it be  $n_{3}=n_{2}-a_{2}2!/3$ ?


inequalities
rearrangement inequality
inequalities proposed

It's not trues.Try a = b = 1; c = 0 :wink:

[quote="math_Tutor"](a+b+c)^3= a^3+b^3+c^3+2ab+2bc+2ca :lol:[/quote]

 :rotfl:  :rotfl:  :rotfl:

Does it hold for the flipped one?? :maybe:

[quote="Gibbenergy"]Let $ a,b,c$ are positive such that not 3 variables are zero simultaneously. Prove that
$ \frac {(a \plus{} b \plus{} c)^3}{2(ab \plus{} bc \plus{} ca)} \geq 3 \left ( \frac {a^3}{a^2 \plus{} bc} \plus{} \frac {b^3}{b^2 \plus{} ac} \plus{} \frac {c^3}{c^2 \plus{} ab} \right)$

This is my version for the inequality posted here http://www.mathlinks.ro/viewtopic.php?t=217556
Can anyone prove the inequality above without using Trebusep (or rearrangement inequality) ?[/quote]
wrong try $ (2,1,1)$  :wink:

I asked whether $ \frac {(a \plus{} b \plus{} c)^3}{2(ab \plus{} bc \plus{} ca)} \leq 3 \left ( \frac {a^3}{a^2 \plus{} bc} \plus{} \frac {b^3}{b^2 \plus{} ac} \plus{} \frac {c^3}{c^2 \plus{} ab} \right)$ holds?????

@ sunkern_sunflora : no :wink:  try $ (1,\frac{1}{5}, \frac{1}{5})$


trigonometry
inequalities
inequalities proposed

Prove that in any triangle we have:

${ \sum ( \frac{h_{a}}{r_{a}}+ \frac{\cos(B)-\cos(C)}{p-a}}) \geq \frac{ (\sum \sin(A))^{2}}{ \sum \cos^{2}( \frac{A}{2})}$.

I like this one a lot, I hope it is my creation and nobody else thought of it before me and I hope it is correct! :D :cool:

Lagrangia, how did you create a so hard inequality?

Did you make a conjecture about  its truth, by your mathematical intuition, and then you  tested it?

Or did you start from the truth of a more simple expression and then you worked out it to deduce your final inequality?

I am very interested in your answer.

Thank you, very much.

i am glad you like it! :) i just come up with it yesterday! Indeed by  conjecturing it and then I proved it! :D

I hope you like the other ineqs I created! :D

I don't know why, but I find it very hard to believe that someone could conjecture such a thing. But maybe I'm wrong and a new Ramamujan lives here. Who knows?

[quote="harazi"] A new Ramamujan lives here. Who knows?[/quote]

 :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

I found it hard myself to believe that I could conjecture such a thing! If I didn't believe it, why would I ask for someone to believe it! :D :D I am amazed of this inequality myself!

 :cool:


integration
calculus
derivative
real analysis
real analysis unsolved

How to prove the following:
Let $(X,M,\mu)$ $\sigma$-finite measurable space and $\nu: M\to[-\infty,\infty]$ $\sigma$-finite $\pm$-measure. If $\nu\ll\mu$, there exists $f\in L^{1}(X)$ (with respect to $\mu$) such that
\[\nu (A)=\int_{A}f\,d\mu \]
for all $A\in M$. If $g: X\to [-\infty,\infty]$ is $\mu$-measurable and
\[\nu(A)=\int_{A}g\,d\mu \]
for all $A\in M$, then $g=f$ $\mu$-almost everywhere.

there is an approach with differentiation of measure, another approach with the Riesz representation theorem, but I do not know very straightforward proof for this strong theorem. Anyone ?

Folland's [i]Real Analysis[/i]* has a lot of slick proofs, but even there Radon-Nikodym takes a page (after preliminary work).

(*) Recommended reading

It looks like my university's library does not have this Folland's book. Are there any other sources where I can find the proof?

You can find it in any textbook on measure theory.

I was wondering is my proof valid:

By Hahn decomposition theorem, for a $\pm$-measure $\mu: A\to [0,\infty]$ we can find measures $\mu^{+},\mu^{-}$ such that $\mu (A)=\mu^{+}(A)-\mu^{-}(A)$. Let us choose a measure $\mu_{1}$ such that  $\mu^{+}\ll \mu_{1}$ and $\mu^{-}\ll \mu_{1}$. It follows from the standard Radon-Nikodym theorem that there are $f,g\in L^{1}(X,\mu)$ such that
\[\mu^{+}(A)=\int_{A}f\,d\mu_{1}\quad \text{and}\quad \mu^{-}(A)=\int_{A}g\,d\mu_{1}. \]
Thus we have
\[\mu (A)=\mu^{+}(A)-\mu^{-}(A)=\int_{A}f\,d\mu_{1}-\int_{A}g\,d\mu_{1}=\int_{A}(f-g)\,d\mu_{1}. \]
But $f,g\in L^{1}(X,\mu)$ implies $f-g\in L^{1}(X,\mu)$. Uniqueness follows by the uniqueness of $f$ and $g$ in standard Radon-Nikodym theorem.

I need more help if this is not a valid proof.


calculus
limit
function
trigonometry
inequalities
algebra
floor function

If $ \lim_{x\to a}g(x)$ exists then is it true that
$ \lim_{x\to a}f(g(x))\equal{}f(\lim_{x\to a}g(x))$,provided f(x) is continuous at x=a

I read this somewhere but do not believe it to be authentic.Can someone give a rigorous proof for this.

Is this rule applicable in the following two cases:
(i)f(x)=[x] , where [x] denotes the greatest integer function and $ g(x)\equal{}\frac{sinx}{x}$.
   Is $ \lim_{x\to 0}f(g(x))\equal{}f(\lim_{x\to 0}g(x))$ in this case.

(ii)f(x)=[x] , where [x] denotes the greatest integer function and $ g(x)\equal{}\frac{tanx}{x}$.
   Is $ \lim_{x\to 0}f(g(x))\equal{}f(\lim_{x\to 0}g(x))$ in this case

The condition should be, "provided $ f$ is continuous at $ x \equal{} L$, where $ L \equal{} \lim_{x \to a} g(x)$."  To see if the rule is applicable, see if $ f$ and $ g$ satisfy all the required conditions.

Your trig functions will look better if you use \sin, \cos, \tan, etc. (and be sure to leave a space, i.e. "\sin x" instead of "\sinx").

The proof you are looking for follows directly from the definition of continuity.

After working out certain things I have come to the conclusion that
$ \lim_{x\to 0}[\frac{\sin x}{x}]\equal{}0$  and that
$ [\lim_{x\to 0}\frac{\sin x}{x}]\equal{}1$
and also
$ \lim_{x\to 0}[\frac{\tan x}{x}]\equal{}1$  and that
$ [\lim_{x\to 0}\frac{\tan x}{x}]\equal{}1$
Am I correct.Would appreciate if somebody solves  $ [\lim_{x\to 0}\frac{\sin x}{x}]\equal{}1$ and  $ [\lim_{x\to 0}\frac{\tan x}{x}]\equal{}1$
Please check my answer

I only checked your first two since the other two are similarly checked. The first is wrong but the second is correct (nothing really to check there but the limit since it is an integer).

Try breaking the first into left and right handed limits. Your theorem does not apply since no limit exists for [ ] at integer values.

It appears to me all are correct since $ 0<\frac{\sin x}{x}<1$ in the neighbourhood of 0 therefore  $ \lim_{x\to 0}[\frac{\sin x}{x}]\equal{}0$  and since $ \lim_{x\to 0}\frac{\sin x}{x}\equal{}1$ ,we must have  $ [\lim_{x\to 0}\frac{\sin x}{x}]\equal{}1$

You are right about the first. I should have changed the inequality sign after dividing by a negative real number but I was doing it in my head  :oops:


Dati exemplu de un sir cu un punct de acumulare care nu are nici un subsir convergent la acel punct de acumulare. :)

chestia asta a fost de fapt discutata recent aici: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79850

[hide="Numai pentru Cezar."]Ce este acela "punct de acumulare" asa pur si simplu ? Dupa parerea mea, notiunea de [u]punct de acumulare[/u] este relativ la (in raport cu) o multime pe care trebuie sa o precizezi ![/hide][hide="Cateva elemente de topologie generala."][color=red][b]Definitie.[/b] Un punct $x_0\in \overline R \equiv [-\infty,\infty]$ este [b]punct de acumulare in raport cu multimea $A\subset R$[/b] daca si numai daca exista cel putin un sir $x_n,\ n\in N^*$ cu urmatoarele trei proprietati:
$1.\blacktriangleright$ pentru orice $n\in N^*$, $x_n\ne x_0\ ;$
$2.\blacktriangleright$ pentru orice $n\in N^*$, $x_n\in A\ ;$
$3.\blacktriangleright x_n\rightarrow x_0\ .$
[b]Definitie.[/b] Un punct $x_0\in \overline R$ este [b]punct de aderenta in raport cu multimea[/b] $A$ daca si numai daca exista un sir $x_n,\ n\in N^*$ cu proprietatile $(2)$ si $(3)$ mentionate mai sus.
[u]Observatie[/u]. Multimea punctelor de acumulare in raport cu multimea $A$ se noteaza $A'$ si se numeste [b]derivata multimii[/b] $A\ .$ Multimea punctelor de aderenta in raport cu multimea $A$ se noteaza $\overline A$ si se numeste [b]inchiderea multimii[/b] $A\ .$ Se arata usor ca $\overline A=A\cup A'\ .$[/color][/hide][hide="Tot numai pentru Cezar."]Si acum, te rog, enunta corect problema propusa. Cred ca in capul tau este putina ceata. Confunzi denumiri asemanatoare din diferite limbi straine cu cele din limba romana, cred.[/hide]


geometry
trigonometry
geometric transformation
reflection
trig identities
Law of Cosines
Law of Sines

I will use the law of cosines to the triangle AFE . We will  have

$ FE^2\equal{}AH^2\plus{}R^2\minus{}2AH\cdot R\cdot\cos A$ (1) . But $ AH\equal{}2OM\equal{}2R\cos A$ so from (1) we have $ FE\equal{}R$ and we are done . :)

Very nice and simple solution silouan :wink: It can also be proven by law of sines:

Note that $ \frac {\sin{AEF}}{\sin{AFE}} \equal{} 2\cos{A}$. Then it's really simple to prove by elementary trigonometry that $ \tan{A} \equal{} \tan{AFE}$ and the result follows :)

Take $D$ the reflection of $O$ about $BC$, so, it is well known, $OD=AH=AF$; with $AE=AO=BO$ and $\angle BOD=\angle EAF$ we get $\triangle EAF\equiv\triangle BOD$ (s.a.s.), but $\triangle BOD$ is isosceles, and so must be $\triangle EAF$, done.

Best regards,
sunken rock


Seven people each have a single water balloon, and each of the seven people choose one person out of the other six to throw her/his waterballoon at.

How many scenarios are there such that no two people throw water balloons at eachother?

PIE that i hope will not fail
define a GRAHAMOSER to be a group of 2 ppl who throw their waterballoons at eachother
we want 0 GRAHAMOSER's
we do PIE with respet to GRAHAMOSERs
there are at most 3 of these such pairs
yeah ok...
so $ 6^7$ (every1 is free to choose) total cases to consider
suppose the first 2 ppl form a GRAHAMOSER, then there are $ 6^5$ possibilities for this case (the last 5 ppl are free to choose)
now suppose the first and third form a GRAHAMOSER, then there are still $ 6^5$ possibilities (it will always be $ 6^5$ no matter which pair is picked)
there are 21 distinct pairs of 2 ppl, so there are 21 pairs times $ 6^5$ possibilities that must be subtracted
now suppose the first 2 and the next 2 form 2 GRAHAMOSER's (first with second, third with fourth)
there are now $ 6^3$ (3 ppl can freely choose) for this case (there are 6^3 for every such case with 2 GRAHAMOSER's)
there are 35 ways to pick a group of 4 ppl and 3 ways to split them up into 2 GRAHAMOSER's, so we need to add back 105 times 216
and if there are 3 GRAHAMOSER's, the last person has 6 ppl to choose from
there are 21 ways to pick the first pair, 10 ways to pick the nest, and 3 ways to pick the last BUT we must divide by 6 since there are 6 orders for these pairs to be picked so we must subtract 105 times 6 possibilities so our answer is
$ 6^7\minus{}21 \cdot 6^5 \plus{} 105 \cdot 216 \minus{} 105 \cdot 6 \equal{} 138690$

question: why do we choose PIE over complementary counting?

What is wrong with this argument:

we count the ways no two pair splashes each other. so the first guy has 6 choices, the second has 5 choices... so we have 6! ways that no two people splash each other.

i looked over your idea, and it gives 6*5*4*3*2*1*0 = 0! but this is incorrect (i do not want to tell why, it gets sonfusing). but as long as a method gets the right answer, it shouldnt matter which method you use, with in reasonability (regarding amount of work) but my method does not use too much work

the main reason I had this post was to ask why the factorial is incorrect

also, why is it true that 6*5*4*3*2*1*0 = 0!
?

i meant 0 exclamation mark, as in stating my explanation at the supposed answer
soory for confusion

i think the factorial idea is right, and we get answer as 0 possibilities.
steven meow gave a solution with 138690 possibilities, but do you think there can so many possibilities in such a small situation
secondly, if u think this solution is incorrect post a counterexample from ur 138690 possibilities.

not sure if you are serious. well i haven't checked the number but the aops book says use PIE and that you get alot of casework if you try complementary counting.

Could someone tell me what the casework would be for complementary counting?

there are definitely not 0 possibilities
call the people a b c d e f g
x-y denotes the first person throws their balloon at y
a-b
b-c
c-d
d-e
e-f
f-g
g-a
there thats 1 and there are definitely not 0

That's because it should have been 6*5*4*3*2*1*1.

And I think even that is incorrect. It said no two people should throw water balloons at each other, but it didn't say that each person should get hit once. If a person can get hit by multiple balloons, or 3 people throw 'em at each other in one place, and the other 4 throw 'em in a different group somewhere else, there are suddenly a lot more possibilities.

thats where pie comes in and the answer is 138690


geometry
ratio
AMC
AIME
Asymptote
MATHCOUNTS
ez game

A point $P$ is chosen in the interior of $\triangle ABC$ so that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles, $t_1$, $t_2$, and $t_3$ in the figure, have areas 4, 9, and 49, respectively. Find the area of $\triangle ABC$.
[asy]
size(200);
pathpen=black+linewidth(0.65);pointpen=black;
pair A=(0,0),B=(12,0),C=(4,5);
D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12);
MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */
MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N);
MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW);
MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE);[/asy]

[hide="solution"]First, we see that $t_1$, $t_2$, and $t_3$, are similar to the original triangle $ABC$, because all three lines were drawn parallel to the sides of the original triangle. Additionally, the sum of three corresponding lengths of the smaller triangles is equal to the corresponding side of triangle $ABC$. (If this is confusing, notice how the dashed lines on the diagram slide into one another.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=797&sid=1e38719b0f9e9839980011166e8a879f[/img]
Because the areas of the three small triangles are in the ratio $4: 9: 49$, their sides are in the ratio $2: 3: 7$. By adding these together, the ratio of their respective sides to that of triangle $ABC$ is $2: 3: 7: 12$. Now we square this to obtain the answer, $144$.[/hide]

[hide="Answer"]WLOG, assume all triangles are isosceles right triangles. We have triangles with sides $2\sqrt{2}$, $3\sqrt{2}$, and $7\sqrt{2}$, respectively. Drawing a diagram, we see that a side of $\triangle ABC$ has length equal to the sum of the lengths of the smaller triangles, so we have $\frac{1}{2}(12\sqrt{2})^2=\boxed{144}$.[/hide]

don't you wish they still made AIME problems like these...

Because all triangles are similar
$ c_{1}\equal{}kc \Rightarrow S_{1}\equal{}k^2S$
$ c_{2}\equal{}lc \Rightarrow S_{2}\equal{}l^2S$
$ c_{3}\equal{}mc \Rightarrow S_{3}\equal{}m^2S$

Adding first 3 equations we get $ c_{1} \plus{} c_{2} \plus{} c_{3}\equal{}c(k\plus{}l\plus{}m)$
But beacuase $ c_{1} \plus{} c_{2} \plus{} c_{3}\equal{}c$ we get $ k\plus{}l\plus{}m\equal{}1$

Now we add square root of other 3 equations we get $ \sqrt {S_{1}}\plus{}\sqrt {S_{2}}\plus{}\sqrt {S_{3}}\equal{}\sqrt {S}(k\plus{}l\plus{}m)$
Therefore  $ \sqrt {S_{1}}\plus{}\sqrt {S_{2}}\plus{}\sqrt {S_{3}}\equal{}\sqrt {S}$
Squaring this eqatuion we get $ (\sqrt {S_{1}}\plus{}\sqrt {S_{2}}\plus{}\sqrt {S_{3}})^2\equal{}S$
Now we plug in the numbers $ S\equal{}(2\plus{}3\plus{}7)^2\equal{}12^2\equal{}144$

[hide="Solution"]
Note that if we have $\triangle ABC\sim \triangle DGF$, and $\frac{AB}{DE}=k$, then $\frac{[ABC]}{[DEF]}=k^2$.  We use this to get the following ratios:
[asy]
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pen zzttqq = rgb(0.6,0.2,0);

draw((3.42,4.04)--(0,0)--(6,0)--cycle, zzttqq);
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */
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draw((3.42,4.04)--(0,0), zzttqq);
draw((0,0)--(6,0), zzttqq);
draw((6,0)--(3.42,4.04), zzttqq);
draw((xmin, 0*xmin + 2.06)--(xmax, 0*xmax + 2.06)); /* line */
draw((xmin, 1.18*xmin-2.19)--(xmax, 1.18*xmax-2.19)); /* line */
draw((xmin, -1.57*xmin + 7.7)--(xmax, -1.57*xmax + 7.7)); /* line */
label("x",(2.36,2.24),SE*labelscalefactor);
label("$\frac{7}{2}x$",(3.06,0.18),SE*labelscalefactor);
label("$\frac{3}{2}x$",(4.1,2.5),SE*labelscalefactor);
 /* dots and labels */
dot((3.42,4.04),dotstyle);
label("$A$", (3.5,4.16), NE * labelscalefactor);
dot((0,0),dotstyle);
label("$B$", (0.08,0.12), NE * labelscalefactor);
dot((6,0),dotstyle);
label("$C$", (6.08,0.12), NE * labelscalefactor);
dot((3.6,2.06),dotstyle);
label("$D$", (3.68,2.18), NE * labelscalefactor);
dot((2.8,3.31),dotstyle);
label("$E$", (2.88,3.42), NE * labelscalefactor);
dot((4.92,0),dotstyle);
label("$F$", (5,0.12), NE * labelscalefactor);
dot((1.86,0),dotstyle);
label("$G$", (1.94,0.12), NE * labelscalefactor);
dot((4.22,2.79),dotstyle);
label("$H$", (4.3,2.92), NE * labelscalefactor);
dot((1.76,2.07),dotstyle);
label("$I$", (1.84,2.2), NE * labelscalefactor);
dot((4.64,2.12),dotstyle);
label("$J$", (4.72,2.24), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */ [/asy]

Using this, we get $\frac{[EBF]}{[EID]}=(\frac{9}{2})^2\implies [EBF]=4*\frac{81}{4}=81$.  Therefore, $[BGID]=28$.  Similarly, we get $[AIJ]=25\implies [AEDH]=12$, and $[HGC]=100\implies [JDFC]=42$.  Sum up all these areas in the picture to give us $\boxed{144}$.  [/hide]

[quote="Binomial-theorem"][hide="Solution"]
Note that if we have $\triangle ABC\sim \triangle DEF$, and $\frac{AB}{DE}=k$, then $\frac{[ABC]}{[DEF]}=k^2$.  We use this to get the following ratios:
[asy]
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real labelscalefactor = 0.5; /* changes label-to-point distance */
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pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 15.52, ymin = -8.94, ymax = 6.3;  /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0);

draw((3.42,4.04)--(0,0)--(6,0)--cycle, zzttqq);
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */
 /* draw figures */
draw((3.42,4.04)--(0,0), zzttqq);
draw((0,0)--(6,0), zzttqq);
draw((6,0)--(3.42,4.04), zzttqq);
draw((xmin, 0*xmin + 2.06)--(xmax, 0*xmax + 2.06)); /* line */
draw((xmin, 1.18*xmin-2.19)--(xmax, 1.18*xmax-2.19)); /* line */
draw((xmin, -1.57*xmin + 7.7)--(xmax, -1.57*xmax + 7.7)); /* line */
label("x",(2.36,2.24),SE*labelscalefactor);
label("$\frac{7}{2}x$",(3.06,0.18),SE*labelscalefactor);
label("$\frac{3}{2}x$",(4.1,2.5),SE*labelscalefactor);
 /* dots and labels */
dot((3.42,4.04),dotstyle);
label("$A$", (3.5,4.16), NE * labelscalefactor);
dot((0,0),dotstyle);
label("$B$", (0.08,0.12), NE * labelscalefactor);
dot((6,0),dotstyle);
label("$C$", (6.08,0.12), NE * labelscalefactor);
dot((3.6,2.06),dotstyle);
label("$D$", (3.68,2.18), NE * labelscalefactor);
dot((2.8,3.31),dotstyle);
label("$E$", (2.88,3.42), NE * labelscalefactor);
dot((4.92,0),dotstyle);
label("$F$", (5,0.12), NE * labelscalefactor);
dot((1.86,0),dotstyle);
label("$G$", (1.94,0.12), NE * labelscalefactor);
dot((4.22,2.79),dotstyle);
label("$H$", (4.3,2.92), NE * labelscalefactor);
dot((1.76,2.07),dotstyle);
label("$I$", (1.84,2.2), NE * labelscalefactor);
dot((4.64,2.12),dotstyle);
label("$J$", (4.72,2.24), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */ [/asy]

Using this, we get $\frac{[EBF]}{[EID]}=(\frac{9}{2})^2\implies [EBF]=4*\frac{81}{4}=81$.  Therefore, $[BGID]=28$.  Similarly, we get $[AIJ]=25\implies [AEDH]=12$, and $[HGC]=100\implies [JDFC]=42$.  Sum up all these areas in the picture to give us $\boxed{144}$.  [/hide][/quote]

A troll point:

$\frac{HJ}{AC} + \frac{IE}{AB} + \frac{GF}{BC} = 1$

[quote="Binomial-theorem"][hide="Solution"]
Note that if we have $\triangle ABC\sim \triangle DEF$, [/hide][/quote]
Am I missing something?  $\triangle DEF$ is not a triangle...

[hide]
Label the diagram as shown. Let the length of $FG$ be $x$. Since $t_1\sim t_2\sim t_3\sim \triangle ABC$, $\frac{DP}{FG}=\sqrt{\frac{4}{49}}=\frac27$, so $DP=\frac{2x}7$. Similarly, $\frac{PE}{FG}=\sqrt{\frac9{49}}=\frac37$, so $PE=\frac{3x}7$. Since $DP=BF$ and $PE=GC$, $BC=DP+FG+PE=\frac{12x}7$. Therefore, $\frac{BC}{FG}=\frac{12}7$, so $\frac{area[ABC]}{area[PFG]}=\frac{144}{49}$. Hence, $area[ABC]=144$.

[asy]/* File unicodetex not found. */

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draw((4,1)--(-1,5));
draw((-1,5)--(-2,1));
draw((xmin, 4*xmin + 5.96)--(xmax, 4*xmax + 5.96)); /* line */
draw((xmin, -0.8*xmin + 2.89)--(xmax, -0.8*xmax + 2.89)); /* line */
draw((xmin, 0*xmin + 3.4)--(xmax, 0*xmax + 3.4)); /* line */
label("$ x $",(0.26,1.1),SE*labelscalefactor);
label("$ A $",(-1.2,5.84),SE*labelscalefactor);
label("$ B $",(-2.32,1.12),SE*labelscalefactor);
label("$ C $",(4.26,1.24),SE*labelscalefactor);
label("$ D $",(-1.88,3.54),SE*labelscalefactor);
label("$ P $",(-0.7,3.36),SE*labelscalefactor);
label("$ E $",(1.12,4.02),SE*labelscalefactor);
label("$ F $",(-1.2,1.06),SE*labelscalefactor);
label("$ G $",(2.04,1.1),SE*labelscalefactor);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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darn geogebra -> asy not working well.
[/hide]

Whoops, that should have said $\triangle DGF$ instead of $\triangle DEF$ :).  Thanks for the fix.

This is the exact same problem as the 2013 MathCounts National Sprint #16. There may be a couple different numbers, but the set up is literally the exact same. Goes to show how much more difficult MathCounts has become, and how easy the AIME was back then.

[hide="Easiest Solution"]The ratio of one side of t1 to one side of t2 to one sideof t3 is 2:3:7. So we can say that the ratio is 2x:3x:7x. If we use the triangle where t1+t2 is the hypotenuse, we have the hypotenuse=5x. So the height can be expressed as 5x. The height of triangle t3 can be expressed as 7x. This means that the height of triangle ABC can be expressed as 5x+7x=12x. This means that the area of triangle ABC is (12)^2=144.[/hide]

[quote=Math_007][hide="Easiest Solution"]The ratio of one side of $t1$ to one side of $t2$ to one sideof $t3$ is $2:3:7$. So we can say that the ratio is $2x:3x:7x$. If we use the triangle where $t1+t2$ is the hypotenuse, we have the hypotenuse=5x. So the height can be expressed as 5x. The height of triangle $t3$ can be expressed as $7x$. This means that the height of $\triangle ABC$ can be expressed as $5x+7x=12x$. This means that the area of
$\triangle ABC$ is $(12)^2=144$.[/hide][/quote]
FTFY :)

This is very similar to another Mathcounts National problem I have encountered. Since all of the triangles are similar to each other and the entire triangle, and we can find the ratio of their sides, the answer is (2+3+7)^2 = 144.

That MATHCOUNTS problem was in turn ripped from an AIME from the 1980s.

[hide = The Best Solution (Quick, Elegant, and Simple Method to Solve)]
Let A be the area of triangle ABC and let the bottom side of triangles $t_1, t_2,$ and $t_3$ be $a_1, a_2,$ and $a_3$. Then we have $BC = a_1 + a_2 + a_3$ (notice the 2 parallelograms to the left and to the right of triangle $t_3$.) Therefore, since we have similar triangles, the ratio of areas is equal to ratio of corresponding side lengths squared of the similar triangles, so the ratio of corresponding side lengths is equal to the square root of the ratio of the areas of the similar triangles. Thus, we obtain the following equations:

$\frac{a_1}{a_1+a_2+a_3} = \sqrt{\frac{4}{A}}$
$\frac{a_2}{a_1+a_2+a_3} = \sqrt{\frac{9}{A}}$
$\frac{a_3}{a_1+a_2+a_3} = \sqrt{\frac{49}{A}}$

Therefore adding the 3 equations the left side becomes $\frac{a_1+a_2+a_3}{a_1+a_2+a_3} = 1$ and therefore substituting we obtain:

$1 = \frac{2}{\sqrt{A}} + \frac{3}{\sqrt{A}} + \frac{7}{\sqrt{A}}= \frac{12}{\sqrt{A}}$.  Solving for A we have $\sqrt{A} = 12,$ and therefore,
$A = \boxed{144}$

[hide=Solution]Let the longest side of triangle $t_1$ be $x$. Then the longest side of triangle $t_2$ is $\frac{3}{2}x$ and the longest side of triangle $t_3$ is $\frac{7}{2}x$. Therefore, the longest side of the largest triangle is $6x$. Then its area is $\frac{6}{1} \cdot \frac{6}{1} \cdot 4 = \boxed{144}$.[/hide]

[hide=Solution]The ratio of the sides are $2:3:7:12$ (by similar triangles) so the area of the big one is $12^2=\boxed{144}.$[/hide]


[quote="arkhammedos"]let $u_{n}=\sum_{k=o\to n}\frac{1}{2^{k}}\cos(kx)$
evaluate  $limu_{n}$.[/quote]
Find this sum $\sum\frac{z^{n}}{2^{n}}$ where $z$ is complex $|z|<1$ and use Abelia theorem.

No need for the theorem. This is simply $\text{Re}\left(\sum_{k=0}^{\infty}z^{k}\right)= \text{Re}\left(\frac1{1-z}\right)$, where $z=\frac12e^{ix}$. That's strictly inside the radius of convergence, so only basic tools are needed- we can even evaluate the partial sums explicitly for this function.

The rest is just algebra.


In the addition problem below,
A,H, and E represent different numbers.
AH is a two digit number and HEE is a three digit number.
What number does HEE represent?

AH + A = HEE

Since we have 2-digit + 1 digit, and the tens and ones is the same, A=9.  Also, H=1.  We then have:

9H + 9 = 1EE

EE are the same, but even of H was 9, we would get 98 + 9 = 107, which is not the same.  E must equal 0 to get:

91 + 9 = 100

Therefore, HEE = 100


geometry
trigonometry
area of a triangle
trig identities
Law of Cosines
Law of Sines

You are given one angle measure, one side measure, and the area of a triangle where the side is opposite the angle.  Does this uniquely determine a triangle?

If the side is one of the two that form the given angle, yes, since, by using the area law

$\sin A (bc)/2$

we have the other side, and using the law of cosines, we have all three.

I don't believe that if the opposite side from the angle is given we have enough information to uniquely determine the triangle, but I haven't worked on that case yet.

[hide]in $\triangle ABC$, let $a$ be the side we know, let $\angle A$ be the angle we know, let $K$ be the area

using the extended law of sines, $R=\frac{a}{2\sin{\angle A}}$

using $K=\frac{abc}{4R}$, we find $bc=\frac{4KR}{a}=\frac{2K}{\sin{\angle A}}$

then using the law of cosines, we have

$a^2=b^2+c^2-2bc\cos{\angle A}$

substituting, we get

$0=b^2+\left( \frac{2K}{b\sin{\angle A}} \right)^2-\frac{4K\cos{\angle A}}{\sin{\angle A}}-a^2$

$0=b^4 -b^2\left( \frac{4K\cos{\angle A}}{\sin{\angle A}}+a^2\right) +\left( \frac{2K}{\sin{\angle A}} \right)^2$

here we have a quartic about $b$ (but the roots come in opposite pairs), so we can either have 0,1,2 positive answers, using the discriminant (and simplifying):

$\left( \frac{4K\cos{\angle A}}{\sin{\angle A}}+a^2 \right)^2-\left( \frac{4K}{\sin{\angle A}} \right)^2$

now simplifying is a lot of algebra, but you see where this is going, if the triangle is defined or not is dependant on the exact numbers that are given[/hide]


prove that every prime number of the form 4n+1 can be represented as the sum of 2 perfect squares

see here:
http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares

also , see proofs here:
http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares


limit
function
calculus
calculus computations

Hello out there.  I hope everyone ids doing well.

I could use a little guidance on this:

suppose $f$ is continuous for all $x$, and

$\lim_{x\rightarrow -\infty}f(x) = -1$ and $\lim_{x\rightarrow +\infty}f(x) = 10$

Show that $f(x) = 0$ for at least one $x$

I know I need to use the Intermediate Value Theorem and the definition of the limit...but I'm not really sure how to apply them.

Thanks,
Bailey

<edit>multiple typos...sorry

Let $a$ and $b$ such as $f(x)$ is a continuous function on the closed interval $[a,b]$, with

$f(a) < 0 < f(b)$.

We have to show that there is a real number c, between a and b, such that f(c) = 0.

First of all we have to see the Sign-Preserving property of Continuous Functions:
If f is continuous at $a$, and $f(a) < 0$, then there is an open interval $I$ containing a such that $f(x) < 0$ for every $x$ in $I$. (this is quite easy to see, but you can ask if you want).

So, to proof that $f(c)=0$ for $c$ in $[a,b]$ (that is called Bolzano's theorem):

Let S be the set of numbers $x$ within the closed interval from $a$ to $b$ where $f(x)<0$.

Since $S$ is not empty (it contains $a$) and $S$ is bounded, the Least Upper Bound axiom asserts the existence of a least upper bound, say $c$, for $S$.
To show that this number $c$ satisfies the requirements of Bolzano's theorem:

There are three possibilities for $f(c)$: either $f(c) < 0$, $f(c) > 0$, or $f(c) = 0$. We show the first two choices lead to contradictions.

Suppose $f(c) < 0$, so that $c$ is a member of $S$. Then the Sign Preserving Property of Continuous Functions asserts the existence of an open interval $I$ containing $c$ where f takes only negative values -- that is, $I$ is a subset of $S$. But $I$ (and thus $S$) contains points greater than $c$, so $c$ cannot be an upper bound for $S$. This contradiction forces us to reject the possibility that $f(c)<0$.

Suppose instead that $f(c) > 0$. The argument is similar to the previous case.

We conclude that $f(c) = 0$

I know another proof that i like more:
let $a$ and $b$ the same as before. let $x=(a+b)/2$ if $f(x)<0$ we call $x=a_2$ if $f(x)>0$ we call $x=b_2$ and if $f(x)=0$ we call $x=c$ and that's it  :P .

The next step is to do the same but in the interval $[a,b_2]$ or $[a_2,b]$ depending on $f(x)$ obtaining either $[a_2,b_2]$ or $[a,b_3]$ for the first case or $[a_3,b]$ or $[a_2,b_2]$ in the second case.

If we continue this process infinitely we are getting two successions: $a_n$ and $b_n$ were $a_n$ is monotone and increasing, and $b_n$ is monotone and decreasing $dist(a_n,b_n)$ tends to zero, so both are a pair of convergent monotone successions (i'm not sure about this concept in english) so $\displaystyle\lim_{n\to\infty} a_n=\displaystyle\lim_{n\to\infty}\b_n=c$ :D. $f(c)=0$  :lol:

SOL you type it so fast  :D
By definition of limit, $\lim_{x \rightarrow - \infty} f(x) = -1$,  we take $\epsilon= 2$ then there exist $M \in \mathbb{R}$ so that $(\forall x)$, $\left( x < M \Rightarrow -3 < f(x)< 1 \right)$.
and for the second limit, $\lim_{x \rightarrow \infty} f(x) = 10$
taking $\epsilon = 2$ then there exist $N \in \mathbb{R}$ so that $(\forall x)$, $\left( x > N \Rightarrow 8 < f(x) < 12 \right )$.
so we have seen that  $-3 <f(x)<12$. (Edit: this for M>N or M<N)
Now apply the Intermediate Value Theorem then there exist $x$ so that $f(x)=0$

Thanks SOL and Beginner...you have both shed a lot of light on my question.  Thanks a bunch...I understand it now.

Best wishes,
bailey

Bailey, I've edited my post. I forget the condition $N<M$ for the last inequality. but nervermind, for $N>M$ we can take $(- \infty,\infty)$ for the interval ;)


Today, Tuesday, November 15, 2005 is the official day for the 2005 AMC 8.  Some schools will be administering the contest through Tuesday, November 22.    So please, no discussion of the problems or solutions on the AMC 8 until after Tuesday, November 22.

I know that this forum usually discusses the AMC 10, AMC 12, AIME and USAMO, but if you know of other forums where AMC 8 is discussed, please pass the word along, or let us know at the AMC office.

Thanks for your cooperation.

Steve Dunbar
AMC Director

It's ok If I unannounce this right?

It's well past November 15.


Can anyone please show me why
[tex] cos 2x = -sqrt3/2 [/tex]
has four solutions?

I feel kind of dumb, this is a regular high school Precalc textbook and I'm stuck on an easy problem :(

This problem was moved from Getting Started.

Trig isn't a getting started.

Also, question is:

$\cos 2x = \frac {- \sqrt 3}{2}$

I'm not expert on the trig but I assume it has to do with the quadrant, I think. ;)

Actually it has infinite solutions .. we do know that  cos (pi-pi/6) and cos (pi+pi/6) both are equal to  - (:sqrt: 3)/(2)  so   there are two values.. not to mention you can add   2*n*pi (n = 0,1,2,3...) to any of the values ..

(Generally  one wants solution only between 0 and 2pi range.. but there are two solutions  even in this range)...

phatso -- what possible values of x are you considering?  There are in fact infinitely many solutions, since from any solution you can just keep adding $2\pi$ (or 360 degrees) as many times as you want and not change the cosine.  So, the problem must ask for all the solutions that fit some other condition such as $0 \leq x \leq 2\pi$.  After that, Silverfalcon's suggestion is a small first step, perhaps.

First,
\[\cos 2x=-\frac{\sqrt{3}}{2}\]

has infinitely many restrictions if you don't put a restriction on $x.$

I assume you meant that $0\leq x<2\pi$ or something equivalent.
(Ok I see that two ppl have already pointed this out but i'll leave it here anyways )

Now let's go back to the equation that I said had infinitely many.  If you can understand it, then you can understand the other question.

Cosine is negative in quadrants II and III, so we know our angles will be in there.  Next, what angle gives a cosine of $\frac{\sqrt{3}}{2}$.  Well, you should recognize $\sqrt{3}$, and $2$ from a 30-60-90 triangle.  So it follows that the $30^\circ$ (or $\frac{\pi}{6}$ in radians) angle gives a cosine of $\frac{\sqrt{3}}{2}$.

So since the angle has to be in quadrant II or III the angle will be $30^\circ$ above or below $180^\circ$ (or any angle coincident with $180$).

We can write this a little more neatly by saying that any angle, $\theta$, will satisfy $\cos \theta=-\frac{\sqrt{3}}{2}$ iff it is of the form $\theta = 360k^\circ + 150^\circ$ or $\theta=360k^\circ + 210^\circ$ where $k$ is any integer (even negatives).

Or if you prefer radians, as I do, they would look like: $\theta=2\pi k+\frac{5\pi}{6}$ or $\theta=2\pi k+\frac{7\pi}{6}$

Now try using this knowledge to complete your problem (with that restriction on $x$ of course).

I hope this helps.

ur confusing by the way ur stating the question. The question is actually for what values of x is [tex]cos2x = -\frac{\sqrt{3}}{2}[/tex]


algebra
polynomial
absolute value
superior algebra
superior algebra unsolved

Hi,

does anyone of you know whether there exists, for
an arbitrary fixed even number $ m$, a nonhomogeneous
polynomial with real and alternating coefficients which has
no real roots at all (all real roots must be strictly positive
in this case, but there can be no real roots at all,
see the example below) and furthermore has
exactly $ m$ roots of least absolute value?

For $ m\equal{}2$ I find such a thing, namely
$ g : \equal{} x^2 \minus{} 2x \plus{} 2 \equal{} (x\minus{}(1\plus{}i))\cdot (x\minus{}(1\minus{}i)).$

The requirements are met and there are
two roots of least absolute value, which, here, is $ \sqrt{2}$.

Salut,
Sébastien

For any $ \alpha_1,\ldots,\alpha_m,\beta\in\mathbb{R}_\plus{}$ with $ \alpha_i^2<4\beta$ for $ i\equal{}1,\ldots,m$
we have $ f(x)\equal{}(x^2\plus{}\alpha_1 x\plus{}\beta)\cdots(x^2\plus{}\alpha_m x\plus{}\beta)\in\mathbb{R}[x]$ with non-negative coefficients.
So $ f(\minus{}x)$ has alternating coefficients and only non-real roots of absolute value $ \sqrt{\beta}$.


How many different ways can change for $ \$1.00$ be made using nickels, dimes, and/or quarters?

[i]Case 1: We have no quarters.[/i]
If we have no quarters, we can have from $ 0\minus{}10$ dimes, leaving us $ 11$ ways.

[i]Case 2: We have one quarter.[/i]
If we have one quarter, we can have from $ 0\minus{}7$ dimes, leaving us $ 8$ ways.

[i]Case 3: We have two quarters.[/i]
If we have two quarters, we can have from $ 0\minus{}5$ dimes, leaving us with $ 6$ ways.

[i]Case 4: We have three quarters.[/i]
If we have three quarters, we can have from $ 0\minus{}2$ dimes, leaving us with $ 3$ ways.

[i]Case 5: We have four quarters.[/i]
If we have four quarters, we can't have any dimes, leaving us with $ 1$ way.

Adding these yields $ 11\plus{}8\plus{}6\plus{}3\plus{}1\equal{}\boxed{29}$.

Once you know how many quarters and dimes are used, there's only one choice for nickels.


inequalities induction function probability

For $k=n-1$ or $k=1$, the first inequality becomes:
$n^{n}<n(n+1)((n-1)^{n-1})$, or:
$(1+\frac{1}{n-1})^{n-1}<n+1$. But LHS$<e<3\leq n$.
The second inequality becomes:
$n<\frac{n^{n}}{(n-1)^{n-1}}$, or:
$1<(1+\frac{1}{n-1})^{n-1}$, evidenty true.
Now return to the initial problem.
For $n=3$, the inequality is true, since we proved it for $k=1, 2=3-1$.
Consider the inequalities true for all $n,k$ such that $k<n\leq t-1$. We'll prove the inequality for $t$.
Consider $k>1$.
The second inequality becomes:
$\frac{t!}{k!(t-k)!}<\frac{t^{t}}{k^{k}(t-k)^{t-k}} \Leftrightarrow$,
$\frac{(t-1)!}{(k-1)!(t-k)!)}<\frac{t^{t-1}}{k^{k-1}(t-k)^{t-k}}$.
By the induction hypothesis:
$\frac{(t-1)!}{(k-1)!(t-k)!}<\frac{(t-1)^{t-1}}{(k-1)^{k-1}(t-k)^{t-k}}$.
So it suffices to prove that:
$\frac{(t-1)^{t-1}}{(k-1)^{k-1}(t-k)^{t-k}}<\frac{t^{t-1}}{k^{k-1}(t-k)^{t-k}}$, or:
$(1+\frac{1}{k-1})^{k-1}<(1+\frac{1}{t-1})^{t-1}$. But the function $(1+\frac{1}{x})^{x}$ is increasing.
Unfortunately, the first inequality doesn't result in a similar way :(

[quote="shobber"]Let $n,k$ be given positive integers with $n>k$. Prove that:
\[ \frac{1}{n+1} \cdot \frac{n^n}{k^k (n-k)^{n-k}} < \frac{n!}{k! (n-k)!} < \frac{n^n}{k^k(n-k)^{n-k}} \][/quote]

[b]Lemma[/b]: Let $a$ and $b$ be any positive integers. For any integer $i$ between 0 and $a+b$, $\binom{a+b}{i} a^i b^{a+b-i} \leq \binom{a+b}{a} a^a b^b$, with equality not holding when $i=0$ or when $i = a+b.$
[i]Proof[/i]:
Let $f(i) = \binom{a+b}{i} a^i b^{a+b-i}$. We observe that $f(a) = \binom{a+b}{a} a^a b^b$. We will show that $\frac{f(i)}{f(i+1)} \geq 1$ whenever $i \geq a$ and that $\frac{f(i)}{f(i+1)} \leq 1$ whenever $i \leq a$.

We have
\[ \frac{f(i)}{f(i+1)} = \frac{\binom{a+b}{i} a^i b^{a+b-i}}{\binom{a+b}{i+1} a^{i+1} b^{a+b-i-1}} = \frac{\frac{(a+b)!}{i! (a+b-i)!}}{\frac{(a+b)!}{(i+1)! (a+b-i-1)!}} \cdot \frac{b}{a} = \frac{i}{a+b-i} \cdot \frac{b}{a}.\]

If $i \geq a$, then
\[ \frac{i}{a+b-i} \cdot \frac{b}{a} = \frac{1}{\frac{a+b}{i} - 1} \cdot \frac{b}{a} \geq \frac{1}{\frac{a+b}{a} - 1} \cdot \frac{b}{a} = 1, \]
and if $i \leq a$, then
\[ \frac{i}{a+b-i} \cdot \frac{b}{a} = \frac{1}{\frac{a+b}{i} - 1} \cdot \frac{b}{a} \leq \frac{1}{\frac{a+b}{a} - 1} \cdot \frac{b}{a} = 1,\]
as desired.

To show that $f(0) < f(a)$ or $f(a+b) < f(a)$, we suppose without loss of generality $a \geq b$. Then
\[ \frac{f(a)}{f(0)} = \frac{a^a b^b \binom{a+b}{a}}{b^{a+b}} = \left(\frac{a}{b}\right)^a \binom{a+b}{a} > 1.\]

--------------------------------------------------------------------------------------------------------------------------

Let $j = n-k$. We wish to show that
\[ \frac{1}{j+k+1}{\cdot \frac{(j+k)^{j+k}}{j^j k^k} < \frac{(j+k)!}{j! k!} < \frac{(j+k)^{j+k}}{j^j k^k} }\]
We have
\[ (j+k)^{j+k} = \sum_{i=0}^{j+k} \binom{j+k}{i} j^i k^{j+k-i} \&> \binom{j+k}{j} j^j k^k \]
which rearranges to
\[ \frac{(j+k)!}{j! k!} < \frac{(j+k)^{j+k}}{j^j k^k}. \]

In addition, by the lemma, we have
\[ (j+k)^{j+k} = \sum_{i=0}^{j+k} \binom{j+k}{i} j^i k^{j+k-i} > \sum_{i=0}^{j+k} \binom{j+k}{j} j^j k^k = (j+k+1) \binom{j+k}{j} j^j k^k, \]
which rearranges to
\[ \frac{1}{j+k+1}{\cdot \frac{(j+k)^{j+k}}{j^j k^k} < \frac{(j+k)!}{j! k!}, }\]
so we are done.

well, let k/n = p. Then, it is pretty easy to see that the claim is equivalent to showing that

1/(n+1) < (nCk)(p^k*(1-p)^(n-k) < 1

however, simply considering a binomial distribution with n trials, a probability of success p, we are nearly instantly done (a binomial distribution is maximized at the mean, np = k.)

Lets consider $n^n=(k+n-k)^n.$ The expansion of $n^n=\sum _{i=0}^n\binom{n}{i}k^k(n-k)^{n-k}.$ One of the term of the expansion is $n\binom{n}{i}k^k{(n-k)}^{n-k}$ which is strictly less than $n^n$. Dividing by $k^k\times (n-k)^{n-k}$ we get the second inequality.
For the first inequality, we have to show that the highest term of the expansion of $n^n$ is $\binom{n}{k}k^k{(n-k)}^{n-k}$.
For this let $T_r=\binom{n}{r}k^r{(n-k)}^{n-r}$ and $T_{r+1}=\binom{n}{r+1}k^{r+1}{(n-k)}^{n-r-1}$. Notice that $T_r>T_{r+1}$ for $r\geq k$ and smaller when $r<k$. Therefore we get $(n+1)\binom{n}{k}k^k{(n-k)}^{n-k}>n^n$. Dividing both side by $k^k\times (n-k)^{n-k}$ and rearranging  we get our desired first inequality.

[quote=Abdullahil_Kafi]Lets consider $n^n=(k+n-k)^n.$ The expansion of $n^n=\sum _{i=0}^n\binom{n}{i}k^k(n-k)^{n-k}.$ One of the term of the expansion is $n\binom{n}{i}k^k{(n-k)}^{n-k}$ which is strictly less than $n^n$. Dividing by $k^k\times (n-k)^{n-k}$ we get the second inequality.
For the first inequality, we have to show that the highest term of the expansion of $n^n$ is $\binom{n}{k}k^k{(n-k)}^{n-k}$.
For this let $T_r=\binom{n}{r}k^r{(n-k)}^{n-r}$ and $T_{r+1}=\binom{n}{r+1}k^k{(n-k)}^{n-r-1}$. Notice that $T_r>T_{r+1}$ for $r>m$ and smaller when $r<m$. Therefore we get $(n+1)\binom{n}{k}k^k{(n-k)}^{n-k}>n^n$. Dividing both side by $k^k\times (n-k)^{n-k}$ and rearranging  we get our desired first inequality.[/quote]

can u explain how did you prove the first ineq it is unclear to me

[quote=TFIRSTMGMEDALIST]
can u explain how did you prove the first ineq it is unclear to me
[/quote]

In any $n$ power binomial, there are exactly $n+1$ terms, and all the terms are not equal. Therefore if we sum the biggest term $n+1$ times then it is obviously strictly bigger than the original and hence done.

[quote=TFIRSTMGMEDALIST]yes but before this how did you prove that the highest term of the expansion is .....
you didn't define m and there's a mistake in  the expansion of n up  to n[/quote]

Ok, $m$ is actually $k$ (I edited).

Using Stirling's approximation (the [hide=stronger version]\[\sqrt {2\pi n}\left(\frac{n}{e}\right)^{n}e^{\frac{1}{12n+1}}<n!<\sqrt {2\pi n} \left(\frac{n}{e}\right)^{n}e^{\frac {1}{12n}}\][/hide] being an exact bound rather than an asymptotic formula) we have:
\begin{align*}\frac{n!}{k! (n-k)!}&>\frac{\sqrt {2\pi n}\left(\frac{n}{e}\right)^{n}e^{\frac{1}{12n+1}}}{\left(\sqrt {2\pi k}\left(\frac{k}{e}\right)^{k}e^{\frac{1}{12k}}\right)\cdot \left(\sqrt {2\pi (n-k)}\left(\frac{n-k}{e}\right)^{n-k}e^{\frac{1}{12(n-k)}}\right)} \\
&=\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{\sqrt{n}}{\sqrt{k(n-k)}}\cdot\frac{1}{\sqrt{2 \pi}} \cdot e^{\frac{1}{12n+1}-\frac{1}{12k}-\frac{1}{12n-12k}}\\
&>\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{\sqrt{n}}{\sqrt{k(n-k)}}\cdot\frac{1}{\sqrt{2 \pi}} \cdot e^{-1/6}\\
&>\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{4}{n+1}\cdot \frac{1}{3}\\
&>\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{1}{n+1}\\
\end{align*}
where we've used the lower bound approximation for $n!$ and the upper bound for $k!$ and $(n-k)!$ as they're in the denominator. Also, to carry out the inequality we've used the following (far from strict) bounds:
\[\frac{1}{12n+1}-\frac{1}{12k}-\frac{1}{12n-12k}>-\frac{1}{12k}-\frac{1}{12(n-k)}\geq-\frac{1}{12}-\frac{1}{12}=-\frac{1}{6}\]
\[\frac{e^{-1/6}}{\sqrt{2 \pi}}=\frac{1}{\sqrt[6]{8\pi^3e}}>\frac{1}{3}\]
\[\frac{n}{2}=\frac{k+(n-k)}{2}\geq \sqrt{k(n-k)}\Longrightarrow \frac{\sqrt{n}}{\sqrt{k(n-k)}}\geq\frac{\sqrt{n}}{n/2}=\frac{2}{\sqrt{n}}\geq \frac{2}{\frac{n+1}{2}}=\frac{4}{n+1}\]
We prove the right-hand inequality in the same manner:
\begin{align*}\frac{n!}{k! (n-k)!}&<\frac{\sqrt {2\pi n}\left(\frac{n}{e}\right)^{n}e^{\frac{1}{12n}}}{\left(\sqrt {2\pi k}\left(\frac{k}{e}\right)^{k}e^{\frac{1}{12k-1}}\right)\cdot \left(\sqrt {2\pi (n-k)}\left(\frac{n-k}{e}\right)^{n-k}e^{\frac{1}{12(n-k)+1}}\right)} \\
&=\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{\sqrt{n}}{\sqrt{k(n-k)}}\cdot\frac{1}{\sqrt{2 \pi}} \cdot e^{\frac{1}{12n}-\frac{1}{12k+1}-\frac{1}{12n-12k+1}}\\
&<\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{\sqrt{n}}{\sqrt{k(n-k)}}\cdot\frac{1}{\sqrt{2 \pi}} \cdot 1\\
&<\frac{n^{n}}{k^{k}(n-k)^{n-k}}\cdot \frac{\sqrt{n}}{\sqrt{n-1}}\cdot \frac{1}{2}\\
&<\frac{n^{n}}{k^{k}(n-k)^{n-k}}\\
\end{align*}
where we've used the upper bound approximation for $n!$ and the lower bound for $k!$ and $(n-k)!$ as they're in the denominator. Also, to carry out the inequality we've used the following (far from strict) bounds:
\[\frac{1}{12n}-\frac{1}{12k+1}-\frac{1}{12n-12k+1}<\frac{1}{12n}-\frac{1}{12k+1}\leq \frac{1}{12n}-\frac{1}{12n-11}<0\]
\[\frac{1}{\sqrt{2 \pi}}<\frac{1}{2}\]
\[\frac{\sqrt{n}}{2\sqrt{k(n-k)}}=\frac{\sqrt{n}}{2\sqrt{\left(\frac{n}{2}\right)^2-\left(\frac{n}{2}-k\right)^2}}\leq \frac{\sqrt{n}}{2\sqrt{\left(\frac{n}{2}\right)^2-\left(\frac{n}{2}-1\right)^2}}=\frac{\sqrt{n}}{2\sqrt{n-1}}=\sqrt{\frac{n}{4n-4}}<1\]
[hide=Remark]I assume there exists a slick combinatorial argument that proves the two inequalities, but I'm not aware of such a solution.[/hide]


quadratics
algebra
system of equations
linear algebra
linear algebra theorems

We know how to solve system of equations by using matrices, but, is there a way to solve quadratic ones by using matrices as well?

NO (unless only squares of variables appear, in which case the system is linear in the squares of variables  :) ).


Let $ a,b,c$ be nonnegative reals such that $ 3\leq a\plus{}b\plus{}c\leq 6$.

Prove that $ \sqrt{1\plus{}a}\plus{}\sqrt{1\plus{}b}\plus{}\sqrt{1\plus{}c}\geq \sqrt{15\plus{}ab\plus{}bc\plus{}ca}$

[quote="nguyentrang"]Let $ a,b,c$ be nonnegative reals such that $ 3\leq a \plus{} b \plus{} c\leq 6$.

Prove that $ \sqrt {1 \plus{} a} \plus{} \sqrt {1 \plus{} b} \plus{} \sqrt {1 \plus{} c}\geq \sqrt {15 \plus{} ab \plus{} bc \plus{} ca}$[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=4900

  darij


Let $ a$ and $b$ be fixed real numbers which are not zero at the same time. Find all functions $ f: \mathbb R\to \mathbb R$ such that
\[ af(xy)\plus{}bf(xz)\minus{}f(x)f(yz)\geq \frac{1}{4}\cdot(a\plus{}b)^2\]
holds for all $ x,y,z\in \mathbb R$.

[quote="marin.bancos"]Let $ a,b$ real fixed numbers, no both of them are zero.
Find all functions $ f: R\longrightarrow R$ such that
$ af(xy) \plus{} bf(xz) \minus{} f(x)f(yz)\geq \frac {1}{4}\cdot(a \plus{} b)^2$
holds for all $ x,y,z\in R$[/quote]

Let $ s \equal{} a \plus{} b$
Let $ P(x,y,z)$ be the assertion $ af(xy) \plus{} bf(xz) \minus{} f(x)f(yz)\ge \frac {s^2}4$

$ P(0,0,0)$ $ \implies$ $ f(0)(s \minus{} f(0))\ge \frac {s^2}4$ $ \iff$ $ 4f(0)s \minus{} 4f(0)^2\ge s^2$ $ \iff$ $ 0\ge (s \minus{} 2f(0))^2$ and so $ f(0) \equal{} \frac s2$

$ P(1,1,1)$ $ \implies$ $ f(1)(s \minus{} f(1))\ge \frac {s^2}4$ $ \iff$ $ 4f(1)s \minus{} 4f(1)^2\ge s^2$ $ \iff$ $ 0\ge (s \minus{} 2f(1))^2$ and so $ f(1) \equal{} \frac s2$

$ P(x,0,0)$ $ \implies$ $ a\frac s2 \plus{} b\frac s2 \minus{} f(x)\frac s2\ge \frac {s^2}4$ $ \iff$ $ 2sf(x)\le s^2$

$ P(x,1,1)$ $ \implies$ $ af(x) \plus{} bf(x) \minus{} f(x)\frac s2\ge \frac {s^2}4$ $ \iff$ $ 2sf(x)\ge s^2$

So $ 2sf(x) \equal{} s^2$

If $ s\ne 0$, we get $ f(x) \equal{} \frac s2$ $ \forall x$ and this indeed is a solution.

If $ s \equal{} 0$, we get $ b \equal{} \minus{} a\ne 0$ and $ f(0) \equal{} f(1) \equal{} 0$ and then :
$ P(1,x,1)$ $ \implies$ $ af(x)\ge 0$
$ P(1,1,x)$ $ \implies$ $ \minus{} af(x)\ge 0$
And so $ f(x) \equal{} 0 \equal{} \frac s2$ $ \forall x$ which indeed is a solution.

Hence the answer : $ \boxed{f(x) \equal{} \frac {a\plus{}b}2}$ $ \forall x$

Dear Patrick,
Well done! Excellent! This is the solution to my functional inecuation. I hope you've enjoyed this problem.

Bravo Patrick   :lol:  super solution  !!!!!!!!!!!!!!!!!!!!!!


For  positive reals x,y,z  with  xyz=1 , show that
Σ1/(1+x+y) <=1
18  th  tournament  of towns  spring  1997
Thanks!

Write x=a^3, y=b^3 and z=c^3. Then x+y+1=a^3+b^3+1>=ab(a+b)+1=(a+b+c)/c. Thus, 1/(x+y+1)<=c/(a+b+c). Sum up these inequalities.


calculus
integration
trigonometry
logarithms
calculus computations

hello, i have found
$ {1/6\,{\frac { \left( \sin \left( x \right)  \right) ^{5}}{ \left( \cos
 \left( x \right)  \right) ^{6}}}+1/24\,{\frac { \left( \sin \left( x
 \right)  \right) ^{5}}{ \left( \cos \left( x \right)  \right) ^{4}}}-
1/48\,{\frac { \left( \sin \left( x \right)  \right) ^{5}}{ \left(
\cos \left( x \right)  \right) ^{2}}}-1/48\, \left( \sin \left( x
 \right)  \right) ^{3}-1/16\,\sin \left( x \right) +1/16\,\ln  \left(
\sec \left( x \right) +\tan \left( x \right)  \right)+C}$
Sonnhard.

I want the solution too
And what is omnipotence formula after all ?

By writing $ \tan^2x\equal{}\sec^2x\minus{}1,$ we can write this integral as

$ \int\sec^7x\,dx\minus{}2\int\sec^5x\,dx\plus{}\int\sec^3x\,dx.$

Those can be attacked by the reduction formula for powers of secant:

$ \int \sec^nx\,dx\equal{}\frac1{n\minus{}1}\sec^{n\minus{}2} x\tan x\plus{}\frac{n\minus{}2}{n\minus{}1}\int \sec^{n\minus{}2}x\,dx.$

You can prove the reduction formula by a combination of an integration by parts, a trig identity, and solving an equation that has the unknown integral on both sides.


someone got a geometrical demostration for:
\[ senx+cosy = 2sen(\frac{x+y}{2})cos(\frac{x-y}{2}) \]

I think you're asking for the sum and difference to product formula.

$ \sin(x)+\cos(y) = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$

I'm not sure what you mean by geometrical demonstration. Can you elaborate?

by geometrical i mean a graphical ,don't know if you get it

I got one.

[hide="With Complex Number"]
Show that,
$ e^{ia}+e^{ib}=2\cos \left( \frac{a-b}{2}\right) e^{i\left( \frac{a+b}{2}\right)}$

This is done by the law of cosines/sines because $ e^{ia}+e^{ib}$ forms a triangle by vector additions. Finds its norm and its angle. To find its norm, you need to use law of cosines. To finds its angle you need to use law of sines.

Now the norm is $ 2\cos \left( \frac{a-b}{2}\right)$ and the angle is $ \frac{a+b}{2}$. Which establishes equality. I can do the details but they are trivial.
[/hide]


combinatorial geometry
inequalities proposed
inequalities

Let $(x_{n})_{n\in\mathbb{N}}$ be a sequence of points in ${\mathbb{R}^{2}}$ such that the convex hull of $\{x_{n},x_{n+1},x_{n+2}\}$ contains the point $x_{n+3}$, for every $n$. Show that the subsequence $(x_{3n})$ is convergent.

(The convex hull of a three-element set $\{A,B,C\}$ is the set of all points
\[xA+yB+zC\]
such that $x,y,z\ge 0$ and $x+y+z=1$.)

[quote="Larry Hammick"]Let $(x_{n})_{n\in\mathbb{N}}$ be a sequence of points in ${\mathbb{R}^{2}}$ such that the convex hull of $\{x_{n},x_{n+1},x_{n+2}\}$ contains the point $x_{n+3}$, for every $n$. Show that the subsequence $(x_{3n})$ is convergent.[/quote]
Sorry, the claim is wrong. E.g. take $x_{2n}= (0,0)$ and $x_{2n+1}=(0,1)$ for all $n$. I will try to fix it.

(edit) Maybe the subsequences $(x_{6n+a})_{n}$ converge. And the above statement is probably okay if add this hypothesis: The intersection of the interiors of all the triangles
x_{n} x_{n+1} x_{n+2}
is nonempty.
What I'm trying to find is some 2-dimensional version of the Leibniz convergence test for series.
http://en.wikipedia.org/wiki/Alternating_series_test


Determine all positive integer solutions $(x, y, z, t)$ of the equation \[(x+y)(y+z)(z+x)=xyzt\] for which $\gcd(x, y)=\gcd(y, z)=\gcd(z, x)=1$.

Condition $\gcd(x, y)=\gcd(y, z)=\gcd(z, x)=1$ gives us:

$x|y+z$, $y|x+z$, $z|x+y$ and hence $xyz|x+y+z$ from which we easily get $(1,1,1)$ and $(1,1,2)$ and permutations as the only solution (and $t$ can be easily determined)

Very well remarked, yunxiu! Can you provide a correct, complete solution?

From  $ k\equal{}\frac{x\plus{}y\plus{}z}{xyz}\in \mathbb{Z}_\plus{}$ , WLOG $ x\le y \le z$
then $ x\plus{}y\plus{}z\le 3z$ then $ k \le \frac{3z}{xyz}$ then $ xy \le \frac{3}{k}$
then $ xy \le 3$ then possible solutions are:
$ (x,y)\equal{}(1,1)$ with $ k\equal{}\frac{2\plus{}z}{z}$ with $ (x,y,z)\equal{}(1,1,1)$ or $ (1,1,2)$

$ (x,y)\equal{}(1,2)$ with $ k\equal{}\frac{3\plus{}z}{z}$ with $ (x,y,z)\equal{}(1,2,3)$

$ (x,y)\equal{}(1,3)$ with $ k\equal{}\frac{4\plus{}z}{z}$ with candidate $ (x,y,z)\equal{}(1,3,4)$ however does not fullfilled for $ y$