inequalities
logarithms
inequalities unsolved

Er...If I want to use latex...Do i have to download some software... :?:

Hope you can find a nice solution...

Your inequality is true for all reals $ k\geq1.$ :wink:

With the same condition,and if $ s$、$ t$are positive numbers,when $ s\geq t$,then

$ \sum{\frac{a^s}{b^s\plus{}c^s}}\geq\sum{\frac{a^t}{b^t\plus{}c^t}}$ :maybe:

[quote="Jianxing113725"]With the same condition,and if $ s$、$ t$are positive numbers,when $ s\geq t$,then

$ \sum{\frac {a^s}{b^s \plus{} c^s}}\geq\sum{\frac {a^t}{b^t \plus{} c^t}}$ :maybe:[/quote]
Of course..
If you substiute $ a \to a^t$, .., and $ k \equal{} \frac{s}{t}$ it's the same ineq :)

[quote="Jianxing113725"]With the same condition,and if $ s$、$ t$are positive numbers,when $ s\geq t$,then

$ \sum{\frac {a^s}{b^s \plus{} c^s}}\geq\sum{\frac {a^t}{b^t \plus{} c^t}}$ :maybe:[/quote]

$ a,b,c >0$ and $ x > 0$

$ f(x)\equal{} \frac {a^x}{b^x \plus{} c^x}\plus{}\frac {b^x}{c^x \plus{} a^x}\plus{}\frac {c^x}{a^x \plus{} b^x}$

We can assume :  $ a \ge b \ge c > 0$

$ f'(x)\equal{} a^xb^x\ln (\frac{a}{b})(\frac{1}{(b^x\plus{}c^x)^2}\minus{}\frac{1}{(c^x\plus{}a^x)^2})$ $ \plus{}b^xc^x\ln (\frac{b}{c})(\frac{1}{(c^x\plus{}a^x)^2}\minus{}\frac{1}{(a^x\plus{}b^x)^2})$ $ \plus{}c^xa^x\ln (\frac{c}{a})(\frac{1}{(a^x\plus{}b^x)^2}\minus{}\frac{1}{(b^x\plus{}c^x)^2}) \ge 0$

So $ f(s) \ge f(t)$ for $ s >t >0$ , (equality for $ a\equal{}b\equal{}c$ )


Να αποδείξετε ότι κάθε κυρτό $ n$- γωνο ( $ n \geq 4$) μπορεί να χωριστεί σε $ n$ εγγράψιμα τετράπλευρα !

 [i]Δεν έχω τη λύση , αλλά είναι μάλλον γνωστό θέμα , οπότε θα το κουμαντάρουμε .[/i]

[quote="stergiu"][quote="silouan"][quote="stergiu"]να χωριστεί σε $ n$ εγγράψιμα τετράπλευρα !
[/quote]

κ .Μπάμπη όταν λέτε να χωριστεί εννοείται οι κορυφές αυτών των τετραπλεύρων να είναι κορυφές του ν-γώνου ?[/quote]

OXI !

 Τα τετράπλευρα , χωρίς επικαλύψεις , μπορούν να έχουν τις κορυφές τους στο εσωτερικό  ή στις πλευρές του ν- γώνου.[/quote]
Ερώτηση στο attachment

[quote="r_boris"][quote="silouan"]Νομίζω ότι ο δρόμος που πρέπει να χρησιμοποιήσουμε είναι επαγωγή .
Αρχικά δηλ να το αποδείξουμε για ν=4 (όπως έδειξε ο κ Μπόρης ) και στην συνέχεια να θεωρήσουμε ένα ν+1 γωνο . Αυτό μπορούμε να το χωρίσουμε σε δύο μέρη όπου εκεί θα εφαρμόσουμε την επαγωγική υπόθεση . Νομίζω ότι κάπως έτσι βγαίνει ...[/quote]
Παιδιά , δεν απέδειξα τίποτε! ερώτηση έκανα.[/quote]
Εκανα μερικές σκέψεις για αυτη την όμορφη άσκηση


Slove this equation: $ sin^6x\minus{}3sin^8x\equal{}\frac{1}{256}$

could you make an equation shere sinx (x)=k and solve?
$ k^6 \minus{} 3k^8 \equal{} \frac {1}{256}$

i am ultra intuitive so $ k \equal{} \pm \frac12$
$ sin(x)\equal{} \pm 1/2$ so sinx=30+360k or 330+360k or 150+360k or 21+360k
i think
i guess it wasnt...

[hide=Kinda like above...]Set $ y=\sin^2x$. Then $ y^3-3y^4=\frac{1}{256} \iff 0=3y^4-y^3+\frac{1}{256}$. $ y=\frac{1}{4}$ is one solution to this equation. By Descartes' Rule of Signs, it is also the only positive solution. (and we can ignore negative ones since $ y$ must be positive) Therefore, $ \sin^2 x = \frac{1}{4} \Rightarrow \sin x = \pm \frac{1}{2} \Rightarrow x = \frac{\pi}{6} +n \pi, \frac{5\pi}{6} + n \pi, n \in %Error. "bbmath" is a bad command.
{Z}$.

[quote="thanhnam2902"]Slove this equation: $ sin^6x \minus{} 3sin^8x \equal{} \frac {1}{256}$[/quote]

Oh! It's nice and easy too.

$ sin^6x \minus{} 3sin^8x \equal{}sin^6x(1\minus{}3sin^2x)\equal{}sin^2x.sin^2x.sin^2x.(1\minus{}3sin^2x)\le\displaystyle \left(\frac{sin^2x\plus{}sin^2x\plus{}sin^2x\plus{}1\minus{}3sin^2x}{4}\right)^4\equal{}\frac {1}{256}$

So equation $ \Longleftrightarrow$ $ sin^2x\equal{}1\minus{}3sin^2x$ $ \Longleftrightarrow$ $ 4sin^2x\equal{}1$

$ \Longleftrightarrow$ $ 2(1\minus{}cos2x)\equal{}1$$ \Longleftrightarrow$ $ cos2x\equal{}\frac{1}{2}$. Then easy.

In your solution, we have to be careful that $ 1\minus{}3\sin^2x\ge 0$

[quote="Altheman"]In your solution, we have to be careful that $ 1 \minus{} 3\sin^2x\ge 0$[/quote]

Yes. If $ 1 \minus{} 3\sin^2x\le 0 \Longrightarrow$ equation has no root. If $ 1 \minus{} 3\sin^2x>0 \Longrightarrow$ as my solution.


I don't have any proving problems with me but I want to have some proving problems that I can try to solve as that is still and will always be my weakness...

Thanks.

Any topic is fine but I would prefer algebra as I'm struggling on that.

As Osiris posted somewhere before, prove that m + 1/m is non-integral for all rational m.

Here's another one I though of. Given that for all x, x^2 + bx + 1 > 0, prove that -2 < b < 2

Ok thx...I'll try those

Also, can you guys give me some of your advice of how you would prove algebraic problems?

After you're done with that one, you can prove the generalization:

Given that for all x, x^2 + bx + c > 0, where c > 0, prove that -2sqrt(c) < b < 2sqrt(c)

I guess the first step to solving any problem is to understand it. Now I don't mean just understand what it's asking. You should also understand how the problem "works".  You should also try to simplify the problem. For example, in my first problem, I would see that it's the same thing as asking for what values of b is the parabola x^2+bx+1 completely above the x-axis. I'd try out different values for b and see what happens. If b less than -2 or greater than 2, the parabola drops down below the x-axis, and then comes back up. If b is -2 or 2, it just touches the x-axis and then goes up. If b is between -2 and 2, the parabola never crosses the x-axis at all.

Then I'd further simplify the problem. If a parabola is completely above the x-axis then it never crosses the x-axis. If it never crosses the x-axis it has no real roots. If it has no real roots the discriminant is negative. If the discriminant is negative then:
b^2 - 4*1*1 < 0
b^2 < 4
-2 < b < 2

And there you have your proof. Now try the generalization. Hope this helps.


Does there exist an infinite set $ S$ such that for every $ a, b \in S$ we have  $ a^2 \plus{} b^2 \minus{} ab \mid (ab)^2$.

Easy problem if d=gcd(a,b)
a^2-ab+b^2|(ab)^2 iff a^2-ab+b^2|d^4
Now for number a a^2-ab+b^2<a^4 ut this holds for finite b

[quote="hossein11652"]does there exist an infinite set S such that for every a, b in S we have
a 2  + b 2  - ab | (ab) 2[/quote]
Easy problem but everyone slove problem: Let $ n$ positive integer.Prove or disprove that set $ S \equal{}\{ x_1,x_2,..,x_n\}$ such that $ gcd(x_i;x_j) \equal{} 1$ and $ x_i^2 \plus{} x_j^2 \minus{} x_ix_j|(x_ix_j)^2$

I think everyone who solved the first problem would have created problem two as well for I came up with the same problem.

The solution is very simple.
Claim: The set $ S$ can have at most one non-zero integer.
Proof of Claim: Suppose not. Let $ x_1,x_2 \in S$ with $ x_1 \neq x_2$ and $ x_1,x_2 \neq 0$.
$ gcd(x_1,x_1^2 \plus{} x_2^2 \minus{} x_1x_2) \equal{} gcd(x_1,x_2^2) \equal{} 1$
Similarly, $ gcd(x_2,x_1^2 \plus{} x_2^2 \minus{} x_1x_2) \equal{} gcd(x_2,x_1^2) \equal{} 1$
Hence, $ x_1^2 \plus{} x_2^2 \minus{} x_1x_2|(x_1x_2)^2 \Rightarrow x_1^2 \plus{} x_2^2 \minus{} x_1x_2|1$
We note that $ x_1$ and $ x_2$ should have the same sign. Otherwise, $ x_1^2 \plus{} x_2^2 \minus{} x_1x_2 \geq 3$ which is not possible.
Therefore, $ 0 < x_1x_2 \leq x_1^2 \plus{} x_2^2 \minus{}x_1x_2$ and  $ x_1^2 \plus{} x_2^2 \minus{} x_1x_2|1$.  This further implies that
$ 0 < x_1x_2 \leq x_1^2 \plus{} x_2^2 \minus{}x_1x_2 \leq 1\Rightarrow x_1x_2 \equal{} 1 \Rightarrow x_1 \equal{} x_2 \equal{} 1$ or $ x_1 \equal{} x_2 \equal{} \minus{}1$.
This contradicts $ x_1 \neq x_2$. Hence, $ S$ has at most one non-zero integer.

Therefore, $ S \equal{} \{0,a\}$ where $ a \neq 0$. Hence, for $ n>3$, there does not exist a set $ S$ with the required property.

[quote="Omid Hatami"]
Now for number a a^2-ab+b^2<a^4 ut this holds for finite b[/quote]
Why does it hold for finite b?!
Another thing:RHS isn't a^4,but d^4
Thanks.

No. Your "correction" ($d^4$ instead of $a^4$ in the RHS) would make that proof invalid. I agree Omid's proof is kind of terse, so I will make it clearer.

Take $a,b \in S$, $a\neq 0$, $a\neq b$. Take $d=\gcd(a,b)$, $a=dx$, $b=dy$, so $\gcd(x,y)=1$. We then need $x^2 + y^2 - xy \mid d^2 x^2y^2$, and since $\gcd(x^2 + y^2 - xy, x^2y^2) = 1$, this forces $x^2 + y^2 - xy \mid d^2$, so $a^2 + b^2 - ab \mid d^4$. This alone does not yet directly lead to the conclusion it can only hold for finitely many $b$, since the value of $d$ may change with $b$.

But we have $d\mid a$, so $d^4\leq a^4$. Having $a^2 + b^2 - ab \mid d^4$ forces $|a^2 + b^2 - ab| \leq d^4\leq a^4$. However, then $a_4 \geq |a^2 + b^2 - ab| = (a-b/2)^2 + 3b^2/4 \geq 3b^2/4$, so $b^2 \leq 4a^2/3$, and with $a$ fixed, it can only occur for finitely many $b$.

Thanks but i have problem in the last line.a isn't fixed so how can you suppose that it's fixed?

Why is $a$ not fixed? I took $a\in S$, $a\neq 0$. I am allowed to fix it, no?

[quote=Omid Hatami]
a^2-ab+b^2|(ab)^2 iff a^2-ab+b^2|d^4
[/quote]
Reason?


Did anyone solve a problem while washing the dishes, walking the dog, or something like this? I can't stay much time sat in front of a sheet of paper, so I like thinking while doing other things. It's difficult to concentrate, but is very nice. Well, I love geometry, so the best problems are very difficult to try to solve in that way. :P

I'm pretty sure I've never done this... maybe I should try it out sometime. I'd probably forget the problem before I solved it. :D

There was a fabulous article in the New Yorker magazine this week on insights -- unfortunately, it doesn't seem to be available on-line.  If you have access to a copy, it's [u]The Eureka Hunt[/u] by Jonah Lehrer in the July 28 issue.

sometimes i do math while surfing the web

i know that i've counted self avoiding random walks in the condensation on my shower walls

[quote="7h3.D3m0n.117"]If you take breaks every now and then while attempting a difficult problem, you'll find yourself thinking back to the problem when you're on break. I was going down the wrong path on a proof once and while I was playing CoD4 I had a sudden revelation. Of course I couldn't work out the specifics.[/quote]

That's so true. I was stuck on this physics problem (involving math, so would it count?) and I went to watch TV, and then I kept thinking about it till I got the answer..and I forgot about the TV. Haha

[quote="tjhance"]I've solved problems at school before. I suppose during lunch doesn't count, but I've solved at least one during latin class...[/quote]

The only times I do that is when I'm not paying attention in class :)

I've never done that before.
However, my math teacher said she solved a problem while sleeping.

I do it all the time...during a bastketball game, watching TV, surfing the web, playing the piano, eating ect...ect...

There was a problem on the iTest that I was thinking about one night but couldn't solve it. Then I went to bed and dreamed about punching numbers into a calculator then suddenly figured out how to do it. Then I woke up.

Unfortunately I recall that we got that problem wrong. Hmmm...

Sometimes i do math while playing on FTW...... :blush:

Or something else would be watching TV (not mathcounts!)

or while eating


It seems that if a set is measurable, then its inner measure and outer measure are the same as it measure. But I cannot construct a nonmeasurable set such that its inner measure and outer measure are the same. Any ideas?

[quote="Kalle"]A set is measurable iff the inner and outer measure are equal[/quote]

No, I don't think so  :blush: .

Some texts actually define measurable this way, and then derive the definition that I'm assuming you use.


Let $ a,b,c$ be real numbers and let $ M$ be the maximum of the function $ y\equal{}|4x^3\plus{}ax^2\plus{}bx\plus{}c|$ in the interval $ [\minus{}1,1]$. Show that $ M \ge 1$ and find all cases where equality occurs.


in triangle $ABC$ let $G$ be the centroid, and $A'$ be inersection point of $G$-symmedian of $\triangle GBC$ with circumcircle of $GBC$,$B',C'$ are defined like this.
prove that 3 circles $AGA',BGB',CGC'$ are coaxal.

Inversion with center G carries the $\triangle ABC$ into a $\triangle PQR$ and the $\triangle A'B'C'$ into a $\triangle P'Q'R'$ inscribed in the $\triangle PQR,$ so that $P' \in QR,\ Q' \in RP,\ R' \in PQ.$ Circumcircles of $\triangle AGA',\ \triangle BGB',\ \triangle CGC'$ go into the lines PP', QQ', RR' and we have to show that these 3 lines are concurrent. By the basic properties of inversion, $\triangle GRQ \sim \triangle GBC,\ \triangle GPR \sim \triangle GCA,\ \triangle GQP \sim \triangle GAB$ are oppositely similar, hence the G-symmedians $GA' \equiv GP',\ GB' \equiv GQ',\ GC' \equiv GR'$ of $\triangle GBC,\ \triangle GCA,\ \triangle GAB$ are the G-medians of $\triangle GRQ,\ \triangle GPR,\ \triangle GQP$ and P', Q', R' the midpoints of QR, RP, PQ. Thus PP', QQ', RR' are medians of $\triangle PQR$ concurrent at its centroid.

Dear Mathlinkers,
this situation is a particular case of

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=616641

Sincerely
Jean-Louis


integration
calculus
trigonometry
trig identities
calculus computations

Calculate the arc length of the indicated portion of the curve r(t).

r(t) = (7 - 2t)i + (5 + 6t)j + (9t - 7)k,   -10 ≤ t ≤ -3

I started from here:
x(t)=(7-2t)  ===> x'(t)=-2
y(t)=(5+6t) ===>y'(t)= 6
z(t)=(9t-7) ===>z'(t)=9
 from the formula of arec length I got
from t=-10 to t=-3 square root of 121 which is 11 d(t)
and I stopped !

 if somebody can explain to me this very clear step by step to see what I missed  I will appreciate this

There's no mistake - you just didn't trust yourself enough to finish.

You have arrived at $ \int_{\minus{}10}^{\minus{}3}11\,dt.$

The integral of a constant is simply that constant times the length of the integral: $ \int_a^bc\,dt\equal{}c(b\minus{}a).$

Thus in this case we get an arclength of $ 11\cdot7\equal{}77.$

Of course, we didn't need calculus for this. This is a line segment. It starts at $ (7\minus{}2(\minus{}10),5\plus{}6(\minus{}10),5(\minus{}10)\minus{}7)\equal{}(27,\minus{}55,\minus{}57)$ and end at $ (7\minus{}2(\minus{}3),5\plus{}6(\minus{}3),5(\minus{}3)\minus{}7)\equal{}(13,\minus{}13,\minus{}22).$ The arclength of this segment is the same as the distance between its endpoints.

Thanks a lot I did not know of the intergral of constant was the constant times the lenght of the integral. LOL

Please take a look at the attached image, I have to find the arc length of different thing and I got to the point
of square root of 16+9sin^2(3/5t) +9cos^2(3/5t)

... which simplifies a lot (in fact, it's another constant: you should be familiar with trig identities such as dealing with $ \sin^2 \theta\plus{}\cos^2 \theta$).

That should be:

$ \int_{\minus{}1}^{10} \sqrt{16 \plus{} 9\left(\sin^2(\frac{3}{5}t)\plus{}\cos^2(\frac{3}{5}t)\right)} dt$

which boils down to

$ \int_{\minus{}1}^{10} \sqrt{25} \ dt \equal{} \int_{\minus{}1}^{10} 5 \ dt \equal{} 5(10 \minus{} \minus{}1) \equal{} 55$

yes that's right it should be 55, forgot the square root.

this is a solved thread you can go head and close it


a) For which nonnegative integers a, b, c is $4^a + 4^b + 4^c$ a perfect square?
b) For which nonnegative integers n is $n2^{n - 1} + 1$ a perfect square?

well i think b) is quite easy

n*2^n = x^2-1 = (x-1)(x+1)

and both have to be even, but one is 2 mod 4...so one have one 2 in its factorization and the other 2^(n-1) and it is 2^(n-1)*k . Then the other is >= 2^(n-1) - 2 but it is less of equal to n but for n>=4 it isn´t true cause the left side growns a lot more than n

so trying n=0,1, 2 and 3                : 0 1 and 3 are the answers

am i right?

Well, you have some nice ideas, but you will need to improve them.

[quote] but for n>=4 it isn´t true[/quote]
  How about n=5 when LHS is 81 ..
(Also check your post,  you may have a typo/error wehn n=1)


calculus
integration
trigonometry
logarithms
calculus computations

Let $ f(x)\equal{}\frac{1}{\sin x\sqrt{1\minus{}\cos x}}\ (0<x<\pi)$.

(1) Find the local minimum value of $ f(x)$.

(2) Evaluate $ \int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} f(x)\ dx$.

(a) is the answer $ f\left(\frac{1}{2}\tan^{\minus{}1}\sqrt{2}\right)\equal{}\frac{3\sqrt{3}}{4\sqrt{2}}$? :(

Forgive me for my lazyness but i wrote reply in greek so consider cosx instead of συνx and sinx instead of ημx
for 2)

$ \begin{array}{l}
 \int\limits_{\frac{\pi }{2}}^{\frac{{2\pi }}{3}} {f(x)\;dx}  \equal{} \int\limits_{\frac{\pi }{2}}^{\frac{{2\pi }}{3}} {\;\frac{1}{{\eta \mu x\sqrt {1 \minus{} \sigma \upsilon \nu x} }}dx} \; \equal{} \int\limits_{\frac{\pi }{2}}^{\frac{{2\pi }}{3}} {\;\frac{{\sqrt {1 \plus{} \sigma \upsilon \nu x} }}{{\eta \mu x\sqrt {1 \minus{} \sigma \upsilon {\nu ^2}x} }}dx}  \equal{}  \\
  \equal{} \int\limits_{\frac{\pi }{2}}^{\frac{{2\pi }}{3}} {\;\frac{{\sqrt {1 \plus{} \sigma \upsilon \nu x} }}{{\eta {\mu ^2}x}}dx}  \equal{} \int\limits_{\frac{\pi }{2}}^{\frac{{2\pi }}{3}} {\;\frac{{\sqrt {2\sigma \upsilon {\nu ^2}\frac{x}{2}} }}{{4\eta {\mu ^2}\frac{x}{2}\sigma \upsilon {\nu ^2}\frac{x}{2}}}dx}  \equal{} \frac{{\sqrt 2 }}{2}\int\limits_{\frac{\pi }{2}}^{\frac{{2\pi }}{3}} {\;\frac{{\sigma \upsilon \nu \frac{x}{2}}}{{2\eta {\mu ^2}\frac{x}{2}\sigma \upsilon {\nu ^2}\frac{x}{2}}}dx}  \equal{}  \\
 \mathop  \equal{} \limits_{\frac{1}{2}\sigma \upsilon \nu \frac{x}{2}dx \equal{} du}^{\eta \mu \frac{x}{2} \equal{} u} \frac{{\sqrt 2 }}{2}\int\limits_{\frac{{\sqrt 2 }}{2}}^{\frac{{\sqrt 3 }}{2}} {\;\frac{1}{{{u^2}\left( {1 \minus{} {u^2}} \right)}}dx}  \equal{} \frac{{\sqrt 2 }}{2}\int\limits_{\frac{{\sqrt 2 }}{2}}^{\frac{{\sqrt 3 }}{2}} {\;\left( {\frac{1}{{{u^2}}} \plus{} \frac{1}{{1 \minus{} {u^2}}}} \right)dx}  \equal{}  \\
 \end{array}$
$ \begin{array}{l}
  \equal{} \frac{{\sqrt 2 }}{2}\int\limits_{\frac{{\sqrt 2 }}{2}}^{\frac{{\sqrt 3 }}{2}} {\;\left( {\frac{1}{{{u^2}}} \plus{} \frac{1}{{2\left( {1 \minus{} u} \right)}} \plus{} \frac{1}{{2\left( {1 \plus{} u} \right)}}} \right)dx}  \equal{} \frac{{\sqrt 2 }}{2}\left[ { \minus{} \frac{1}{u} \plus{} \frac{1}{2}\ln \frac{{1 \plus{} u}}{{1 \minus{} u}}} \right]_{\frac{{\sqrt 2 }}{2}}^{\frac{{\sqrt 3 }}{2}} \equal{}  \\
  \equal{} \frac{{\sqrt 2 }}{2}\left( {\frac{{2\left( {\sqrt 3  \minus{} \sqrt 2 } \right)}}{{\sqrt 6 }} \plus{} \frac{1}{2}\ln \frac{{\left( {2 \plus{} \sqrt 3 } \right)\left( {2 \minus{} \sqrt 2 } \right)}}{{\left( {2 \minus{} \sqrt 3 } \right)\left( {2 \plus{} \sqrt 2 } \right)}}} \right) \\
 \end{array}$

That's correct. :)

You can summarize it as $ 1\minus{}\sqrt{\frac{2}{3}}\plus{}\frac{1}{\sqrt{2}}\ln (2\plus{}\sqrt{3})(\sqrt{2}\minus{}1)$


Assume that the length of the Earth's equator is exactly 25100 miles and the earth is the perfect sphere. A city is at 45 degrees latitude (middle of north pole and equator). What is the distance of the circumfence of the circle on Earth parallel to the equator and passing through the city? Round to the nearest hundred miles.

[hide="Answer"]$25100\cdot \frac{\sqrt{2}}{2}=12550\sqrt{2}=12550\cdot 1.414=17645.7$ miles, and since the problem asks for the answer rounded to the nearest hundred, we have $\approx 17600$ miles.[/hide]


Every point on a plane is colored in two colors, black and white. Must there be three points of the same color forming an equaliteral triangle of length $ 1$?

No.
Color the plane by parallel monochromatic strips of width $ \frac {\sqrt 3}2$, that is $ M(x,y)$ is red if $ [\frac {2y}{\sqrt 3} ]$ is even, and blue otherwise.

Pierre.

Every point on a plane is colored in two colors, black and white. [b]PROVE THAT THERE ARE [/b]three points of the same color forming an equaliteral triangle of length 1 [b]OR 3^1/2[/b]?

Please post your solution . Good luck
 :ninja:


If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?

[hide]
I know the prime factorization of a number's exponents + 1 is the number of divisors, but I'm not sure where to go from there.  Can someone help?
[/hide]

This url contains 2 solutions
http://mathcentral.uregina.ca/QQ/database/QQ.09.97/james2.html

[quote="Jimmy"]Can you give an example of what you mean by "the prime factorization of a number's exponents +1 = the number of divisors"?
Thanks[/quote]

$2^{a_1}\cdot 3^{a_2}\cdot 5^{a_3}\cdots\text{ has }(a_1+1)(a_2+1)(a_3+1)\ldots\text{ divisors.}$

Thanks, do you know where I can find a proof of this?


geometry perimeter ARML calculus derivative

Given any closed curve, show that there must exist four points on the perimeter of the curve which are the vertices of a square.

Was this unsolved until recently, or am I being stupid? Also, does the curve have to enclose a convex region?

At ARML I remembered spending a lot of time thinking about it in 7th grade (the only redeeming quality of my geometry textbook was that it mentioned this theorem (conjecture?) on some obscure page) and I tried, once again without success, to think of a solution.  I was quite interested in finding a solution so Richard then suggested that I post the problem on this forum.  I think the book may have mentioned that the problem was unsolved -- but that memory could be faulty and the problem may have been solved since then (book was quite old).  For all I know, however, the problem is unsolved to this day.

I guess people should be aware that I dont have much of an idea of the difficulty of this problem (except that I am fairly sure it deserves a spot in the advanced forum) before they start working on it.  You have been warned!

And to answer your other question -- the closed curve does not need to be convex.

I believe that the problem is still open.  I know it has been proven for smooth curves, but the general case is still open.

Interesting -- is the definition of a smooth curve just that the derivative must be continuous?

[quote="JBL"]"Smooth curve" I think means just differentiable, not C1 -- that is, it needs to have a derivative everywhere, but the derivative does not need to be continuous.[/quote]

I believe a curve, r(t), is smooth iff it is differentiable and r'(t) != 0

I think the term ``smooth'' is not very standardized. Sometimes it is used to mean C1; sometimes is it used to mean [tex]C^\infty[/tex]. It might also be used to mean just differentiable (as opposed to continuously differentiable), as Joel said. A smooth curve [tex]\gamma(t)[/tex] need not have [tex]\gamma'(t)\not=0[/tex] everywhere; such a curve is called a regular curve.


Let $ f$ be a function satisfying $ f(xy) \equal{} f(x)/y$ for all positive real numbers $ x$ and $ y$. If $ f(500) \equal{} 3$, what is the value of $ f(600)$?

$ \textbf{(A)} \ 1 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ \displaystyle \frac {5}{2} \qquad \textbf{(D)} \ 3 \qquad \textbf{(E)} \ \displaystyle \frac {18}{5}$

[quote="236factorial"][quote="dangvh"][hide] We can rewrite $f(500)$ as $f(100\cdot 5)$. So we see that $f(xy)=f(100\cdot 5)=\frac{f(100)}{5}=3$. So, we can conclude that $f(100)=15$. Now, we can express $f(600)$ as $f(100\cdot 6)$. So we have $f(600)=\frac{f(100)}{6}=\frac{15}{6}=\frac{5}{2}$ So our answer is $\boxed{C}$[/hide][/quote]

Why can't we write $f(500)$ as $f(250\cdot2)$?[/quote]

You can. I think dangvh used $f(100\cdot5)$ because 100 is a factor of 600 where 250 isn't. [hide="But it still works..."]$f(500)=f(250\cdot2)=\frac{f(250)}{2}=3\Rightarrow\ f(250)=6\Rightarrow\ f(600)=f(250\cdot2.4)=\frac{f(250)}{2.4}=\frac{6}{2.4}=\frac{5}{2}$[/hide]
I personally like what JesusFreak did - it does the same thing without separating the 500.

yep, i just picked a factor of both $500$ and $600$. I figured the problem would be easier that way, and $100$ was the first factor that popped into my head so i used that. My solution isn't as elegant as JesusFreak's though, look at his  :D

At first I was going to do it your way, but then it just popped into my head, "Why not do it this way?" So I did. :lol:

It's basically already been said, but we have:

$xy=500$

$\frac{f\left(x\right)}{y}=3$

Then we have:

$f\left(x\right)=\frac{1500}{x}\implies f\left(xy\right)=\frac{1500}{xy}$

$xy=600=100\cdot 6\implies f\left(600\right)=\frac{1500}{600}=\frac{5}{2}$


real analysis
function
real analysis unsolved

Suppose that $ f: [0,1]\to \mathbb{R}$ is a Lebesugue measurable function. Then there exists a function $ g$ such that $ g\equal{}f$ a.e. and $ g$ maps every closed subset of $ [0,1]$ to a $ F_\sigma$ set.

If you are familiar with Lusin theorem, then apply it here, to construct suitable compact exhaustion of our interval $ [0,1]$    :wink:


[quote="now a ranger"]If the vertex angle of an isosceles triangle measures $19$ degrees $30'$, then each base angle measures:[/quote]

 :|  what does $19$ degrees $30'$ mean? specifically the 30'.. is that 30 NOT?

$d^\circ m's''$ <-- This is the angular notation, where $d,m,s$ each stand for degrees, minutes, seconds, respectively. So $19^\circ30'=19.5^\circ$.

Wow.

So does that imply that the measure of a base angle is $\frac{321}{4}$ because the base angles of an isosceles triangle are congruent?

I didn't know this meant degrees, minutes, and seconds.  :blush:

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[quote="now a ranger"]If the vertex angle of an isosceles triangle measures $19$ degrees $30'$, then each base angle measures:[/quote]
[hide]$\alpha=\frac{180-19.5}2=80.25^{\circ}$[/hide]


induction
number theory open
number theory

Are there or not an infinity of prime numbers in the Fibonacci sequence?

[i]unknown source[/i]

..i've no idea about this...
but we can prove if n is prime, F_n is a prime(there is only one exception)

If it wasn't a question, liyi, please post a proof to your last statement

very very sorry  :( .... i missed some words....  :(
i meant to say
"if n is NOT a prime, then F_n is NOT a prime either"

First of all I didn't solve the problem. [b]Because I think this problem is realy hard, I chalange you to a little brainstorming in order to come up with a proof.[/b]
What I want is to gather up enough information concerning Fibonacci numbers and prime numbers.
So let's see what we know:
1. What liyi said: If n is not a prime, then F_n is not a prime either.
Proof: This uses the fact that F_n is something we call a Fibonacci-Mersenne. This means F_gcd(m,n)=gcd(F_m,F_n) for all positive integers m,n.
To prove this, prove first the relation: F_(m+n)=F_(n-1)F_m+F_nF_(m+1). (Hint: use induction)

2. This is mine: For every prime p, there is an integer n such that p|F_n.

Proof: This is going to look ugly and not quite elementary.
The first idea is to consider the explicit form of F_n.
This is F_n=(1/ \sqrt 5)(((1+ \sqrt 5)/2) n-((1- \sqrt 5)/2) n). This writing is hard to work with so let's make some notations:
a=  \sqrt 5
b=(1+a)/2 and c=(1-a)/2.
Then F_n=(1/a)(b n-c n) [b][1][/b]. For the ones of you that didn't kwon that check that b and c are the roots of x 2 -x-1=0, so F_n+2=F_n+1+F_n and F_0=0 and F_1=1. This shows that F_n is indeed the Fibonacci sequence.

Now I'm going to do something nasty again and compute F_n mod p in [1].

[b]I am going to use square residues theory so I recommend you read something on Legendre symbol, the Law of cuadratic reciprocity([b]LCR[/b]) (hope I got the name wright).[/b]
Denote by (r p)=r (p-1)/2 (mod p) (chose the residue that is 1 or -1), the Legendre symbol.

Case0: p=5, chose n=5 and then F_5=5.
Case1: p=(1 or 4) mod 5  this means (p 5)=1 mod 5 (0 1 and 4 are the square residues mod 5). By LCR, (p 5)(5 p)=(-1) (p-1)(5-1)/4 =1
This means that x 2 -5=0 has 2 solutions mod 5, so modulo p, 5 is a square residue, so mod p, the equation x 2=5 has a solution.
I did this to prove that  \sqrt 5 is an element (or an element and it's opposite) of Z_p (the set of residues mod p). Chose d in Z_p such that d 2 =5 mod p.
F_p-1=(1/a)(b p-1-c p-1)=
(1/a)( (1+a) p/(2 p-1(1+a))-(1-a) p/(2 p-1(1-a)))=(1/a)( (1+a p)/(1+a)-(1-a p )/(1-a))=2(1/a)(a p-a)/(1-a 2 )=(1-a p-1)/2. Remember a=  \sqrt 5 is in Z_p, so 1-a p-1=0.
So p|F_p-1
I used (1+a) p=1+a p (mod p of course) and 2 p-1=1.
It can be shown in a similar manner that F_p=1 mod p.

Case2: p=2 or 3 mod 5, then (5 p)=-1, or 5 (p-1)/2=-1.
Repeat the computations made earlier and show that F_p-1=(1-a p-1)/2=(1-5 (p-1)/2)/2=1 mod p. I used a p-1=5 (p-1)/2.
So F_p-1=1 mod p.
F_p=...=(1/a)(a p)=a p-1=-1.
So F_p=-1. Then F_p+1=F_p-1+F_p=0, so p|F_p+1.

If you can prove this easier, by using another idea, like an identity for Fibonacci sequence, please post it. That is because now I think highly of myself for proving what I think is something quite hard :D.
I think this exhausting post makes up for all my past little useless posts. So post spam and prosper! :D  :D
Feel free to extend the brainstorming to the other problems. That is what [b]mathlinks[/b] is all about wright?  :D

okay my self confidence decreased as Pliviu found a simpler proof to my previous statement:
Take F_n mod p. The sequence F_n is infinite and takes finitely many values mod p, so there are i<j such that F_i=F_j and F_i+1=F_j+1 mod p then F_i-1=F_i+1-F_i=F_j+1-F_j=F_j-1 and going down we get to 0=F_0=F_j-i mod p of course. Isn't this simple?  :(

But the good part about my previous proof, and the reason it's worth reading it  :D , is that it found a vaue that was n=p-1 or p+1 for which p|F_n. It is not the smallest though.

Remember tha brainstorming.

I don't know why you said that was your "theorem". I saw lots of problems of this kind in olympiads and mathematical magazines. So this is quite a very well known result.

I only wanted to say that [b]the first proof[/b] was mine.
Is it known that p|F_p \pm 1 or just that p|F_n for some n?
I saw lots of problems too, but just not this one, and I don't think you will meet with such proofs in Olympiad problems. They require time to prove. Sory if it is simple for you, for me it ment 2 days of hard work.

Anyway don't forget about the initial problem here.

[quote=amfulger]I forgot to ask for this before. Is this problem solved? Or is it a conjecture?
I ask for this because there is a small resemblance between this problem and the question wether there are or not infinetely many Mersene primes.[/quote]

this is an unsolved problem
[url]https://www.quora.com/Are-there-infinitely-many-prime-numbers-in-the-Fibonacci-sequence-If-there-are-how-do-you-prove-it[/url]


Can moderators see your ip when you post or is that reserved to administrators?

admins. [size=0]this message size limit is bullshit[/size]


prove that thre exist infinitely many numbers of the form n(n+1)/2 that can be represented as sum of squares of 2 natural numbers in two distinct ways.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=122564 and use $2n=a^{2}+b^{2}\implies n=\left(\frac{a+b}{2}\right)^{2}+\left(\frac{a-b}{2}\right)^{2}$.

And you don't need to tell us that it is number theory, we all can read it in the forum's title. Especially you don't need to show this again in the source.
It is much more usefull to use a title like "n(n+1)/2 as sum of two squares".


inequalities
Euler
function
limit
number theory proposed
number theory

If $\phi$ denotes the Euler function (see e.g.
[url]http://mathworld.wolfram.com/TotientFunction.html[/url])
then
\[ \sum_{k=1}^n\frac{\phi(k)}{k}\ge\frac{n+1}{2} \]
for all $n\ge 1$.

Sketch of a solution:

Take the intervals $\left[0, \frac{1}{n+1}\right]$ and $\left[1 - \frac{1}{(n+1)}, 1\right]$
and for each fraction $a/b$ having $0<a<b\le n$ with $b>1$, take the interval
\[ \left[\frac{a}{b}-\frac{1}{(n+1)b},\frac{a}{b}+\frac{1}{(n+1)b}\right] \]
The Dirichlet approximation theorem
[url]http://planetmath.org/encyclopedia/DirichletsApproximationTheorem.html[/url]
tells us that these intervals *cover* the interval $[0,1]$. Therefore the sum of their sizes is $\ge 1$. The number of intervals of size $2/b(n+1)$ is $\phi(b)$. Grouping the summands this way, and adding, we get the desired inequality.

Write
\[ S(n) = \sum_{k=1}^n\frac{\phi(k)}{k}  \]
for $n\ge 1$. It seems that
\[ \lim_{n\to\infty} \frac{S(n)}{n} = \frac{6}{\pi^2} = \frac{1}{\zeta(2)} \]
and I have heard from an expert that he has proved this.


Solve in integers:

(1) 32^x - y:^3: = 1

(2) 32^x - y:^2: = 1

And a hint to go along with each of them, which you shouldn't read until you've already tried the problems!

(1)[hide]32^x has a very nice factorization.  What can you do with that?[/hide]

(2)[hide]Gotta love them mods.  What mod makes 32 and y:^2: look nice?[/hide]

JBL wrote:Solve in integers:
(1) 32^x - y:^3: = 1
(2) 32^x - y:^2: = 1

Easy!

Sol'n:[hide]x=0, y=0 for both problems[/hide]

You didn't say "non-trivial" and you didn't say "positive integers"

Um I think you assume that it's positive...

I'm having a hard time with this...is there anymore hints?

Err...you have to prove that those are all the solutions. For example, the solution to #2:

[hide]32^x - y^2 = 1 (mod 4)

y^2 = -1 (mod 4) for x > 0

But y^2 must be 0 or 1 mod 4, a contradiction.

Now for x < 0

32^x < 1, but since y^2 must be a non-negative integer since y is an integer, 32^x - y^2 cannot equal 1, so that is also a contradiction. Therefore, x = 0, y = 0 is the only solution.[/hide]

Fierytycoon is correct -- "solve" means "find all solutions," not just find one.  Which implies that you must prove that your solution is unique.  Tare -- for these particular questions, it doesn't really matter if you take positive integers or all integers -- it turns out the real meat of the problem (that is, the proving part) is the same either way.  I'm not really sure if "Diophantine Equations" in general implies all integers or just positive integers -- probably one should assume the most general solution left open by the problem statement, which in this case would include negative integers and 0 as well.

I'm pretty sure Diophantine Equations are for all integers, not just positive ones.

Well, if Simon says so, it must be true :D

Number 1 is still up for grabs, and there's the hint available at the top for those who want.

Note that (y + 1)(y - y + 1) = 25x.

So y + 1 and y - y + 1 must be powers of 2.

So we write y = 2k - 1.

From here on it is straightforward.


If the minmum number of "Friday the 13th's" that can occur in a calendar year is m and the maximum number of "Friday the 13th's" that can occur in a calendar year is M, what's the ordered pair (m,M)

[quote="Nawahd Werdna"]Hmm... Old Math League problem.... I believe the answer is (1,3) - You just analyze the 7 cases for January 1st and everything can be determined from there. :D[/quote]
Or you can use mods to analyze them all at once. You should also examine leap years (even though it doesn't end up making a difference)


Find all the values of a and b for which the system:

\[
\left\{ \begin{array}{l}
 \left| {\frac{{x^y  - 1}}{{x^y  + 1}}} \right| = a \\
 x^2  + y^2  = b \\
 \end{array} \right.
\]

has only one solution with $ (a,b,x,y \in R \  \ and \  \ x > 0) $.

First we will prove that y=0.
Suppose y is a solution to the system then so does -y.
So we must have y=0.
This gives a=0,b>0.x= :sqrt: b
Hey man this is not quite true.If a=0 and b>1 then we have (x,y)=(1; :sqrt: (b-1)),(1;- :sqrt: (b-1))
so we must have b <= 1,indeed if a=0 and b <= 1 gives us y=0 and x= :sqrt: (b)


AMC USA(J)MO USAMO geometry AIME \/closed

If you are already a USAMO qualifier, I would assume that you should take Olympiad classes... unless you don't feel too confident about the AMC-12 (which is somewhat unlikely).

I'd suggest also looking at intermediate level courses in subject areas in which you don't feel completely confident.

[quote="bubala"]If I'm a USAMO qualifier, when I start prepping for the AMC 12, would it be useful to take the AMC 12 Problem series or not?[/quote]

Some USAMO qualifiers have enjoyed the AIME Problem Series including one of the members of last year's IMO team.  If you want to solidify your chances of returning to the USAMO, that's a better course for you.  Other than that, Intermediate and Olympiad subject classes will help you more.


Let$ ABC$ be a triangle and denote $ l_a,l_b,l_c$ the angle-bisectors of the angle
$ A,B,C$ respectively.Put$ BC \equal{} a,CA \equal{} b,AB \equal{} c$.Prove that:
$ \frac {a \plus{} b}{2l_a \plus{} 2l_b \plus{} l_c} \plus{} \frac {b \plus{} c}{2l_b \plus{} 2l_c \plus{} l_a} \plus{} \frac {c \plus{} a}{2l_c \plus{} 2l_a \plus{} l_b} \ge \frac {4\sqrt {3}}{5}$
It similar the inequality of evarist :)

[quote="mathVNpro"]Always inequality,... There is a non- stop increasing amount of students in VN crazy about inequality nowadays :lol:[/quote]
I known :wink: .But i post it because i have a nice solution with this inequality,and Can_hang 2007 post it in this forum, regards:1 months if you solve it :wink:
What about the following?
$ \frac{a\plus{}b}{m_a\plus{}m_b}\plus{}\frac{b\plus{}c}{m_b\plus{}m_c}\plus{}\frac{c\plus{}a}{m_c\plus{}m_a} \ge 2\sqrt{3}$


The first and thirteenth terms of an arithmetic sequence are $ \frac79$ and $ \frac45$, respectively. What is the seventh term?

k this is an arithmetic, or linear sequence.
You get $ x\plus{}c\equal{}\frac79$ and $ 13x\plus{}c\equal{}\frac45$
So add them up and divide by 2   and you get $ 7x\plus{}c$, which is 7th term.
$ \frac{\frac79\plus{}\frac45}{2}\equal{}\frac{\frac{35\plus{}36}{45}}{2}\equal{}\frac{71}{90}$


Prove that $(p-1)!+1$ is not power of $p$ where $p$ is aprime number.(exept $2, 3, 5$)

[quote="amirhtlusa"]i might be wrong but isn't this just a generalization of Wilson's theorem? ;)[/quote]

i don't think so. wilson's theorem just states that $(p-1)! \equiv -1 \mod{p}$ so we know $(p-1)!+1$ is always at least divisible by $p$. i'm not sure how to prove it can't be a power of $p$ though.

aha ...
i got it.(and as i suspected, i was previously wrong and this had not much to do with the Wilson's Thm. :) )
let me restate the problem
we want to find all $(p,r)$ in which $(p-1)!=p^r-1$

let $p-1=q$,
then we have $q!=(q+1)^r-1$.
now suppose $r>1$ and $q>4$, then if $q+1$ is not a prime, then $gcd(q!,q+1)\not =1\Rightarrow q+1$ is prime. also we have:
$(q-1)!= q^{r-1}+rq^{r-2}+...+\frac{r(r-1)}{2}q+r$. from  here we see that $q$ must be a prime. because if its not aprime then $q|(q-1)!\Rightarrow q|r \Rightarrow r\geq q\Rightarrow q!\geq (q+1)^q-1$ Contradiction $\Rightarrow q $ is a prime.$\Rightarrow$, contradiction because we proved that both $q$ and $q+1$ must be primes.
so now we must look at the cases in which $q\le 4$. which is very easy.($p=2,3,5$).

hope this is right.

Write $p-1=q$.  It is clear that $q$ is composite for $p>3$

We have $q!=(q+1)^k-1$.  Say that $q$ is not a perfect square of the form $l^2$ where $l$ is prime.  Then $q^2|q!$, so $q^2|(q+1)^k-1$ and hence $q|k$, so that

$q!=(q+1)^{aq}-1$ for some $a\geq 1$, which is clearly impossible.

Now, say $q=l^2$  Then $l^2+1=p$.  If $p$ is odd, then $l$ is even.  But then $l=2$, which implies that $p\leq 5$.  If $p$ is even, then it is easy to analyze.

oh..., thanks
Actually my problem was that prove there are infintiely many $n$ such that $n!+1$ is divisible by at least two distinct primes.
And I used tihs lemma and proved it . ;)

$(p-1)! \equiv -1 \mod{p}$ so we know $(p-1)!+1$ is divisible by $ p $.
But because p is a prime number $ p^2 $ divides  $(p-1)!+1$. So $ p^2 <= (p-1)!+1 $
But the equality does not holds obviously so and we are done

[quote="silouan"] $ p^2 <= (p-1)!+1 $
But the equality does not holds obviously so and we are done[/quote]
Are you sure :?  :huh:
consider $p>3$ then $p^2 \leq (p-1)!+1$


let $a;b;c;d>0$ prove that:
$\sqrt[6]{\frac{(a+b)(b+c)(c+a)(a+d)(d+b)(d+c)}{64}} \geq \sqrt[3]{\frac{abc+bcd+cda+adb}{4}}$

Great problem! I still remember the trick Namdung used to solve a problem from Crux: we have the marvelous inequality $ (x+y)(y+z)(z+t)(t+x)\geq (x+y+z+t)(xyz+yzt+ztx+txy)$. Now, write this inequality for all permutations of $a,b,c,d$ multiply all the inequalties of the above form and use the fact that $ (a+b+c+d)^3\geq 16(abc+bcd+cda+dab)$ You will get your inequality.

[quote="harazi"] $ (x+y)(y+z)(z+t)(t+x)\geq (x+y+z+t)(xyz+yzt+ztx+txy)$[/quote]
[b]harazi[/b] can you help me to prove this one :( .many thanks :P

All you have to do is to break everything and you will see that there is almost nothing which remains to be proved.  :P

Or it is simply Mac-Laurin inequality. No computations! ;)

Harazi, do you think it is possible to generalize it to n variables? :?

Who can prove the following inequality:
 $\sqrt[6]{\frac{(a+b)(b+c)(c+a)(a+d)(d+b)(d+c)}{64}} \geq\frac{\sqrt{ab}+\sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}+\sqrt{cd}}{6}$

No, I know it is true, but I cannot prove it. I saw it in a magazine for the general case.

It is not true. See the case $a=b=0$. :P

Doesn't matter, we make two variables very close to 0. Well, at least it is nice is false, we don't have to prove it anymore. I adore false inequalities! :P

[quote="manlio"]This is true by AM-GM

$\displaystyle (a+b+c+d)^3\geq 16(abc+bcd+cda+dab)$

as it is equivalent to

$6\sum a^3 +18\sum ab(a+b) \geq 60\sum abc$[/quote]

We can also homogonize it as follows $a+b+c+d=4$, then we need to prove $abc+bcd+cda+dab\leq4$. By AM-GM we have
\[ ab(c+d)+cd(b+a)\leq\left(\frac{a+b}2\right)^2(c+d)+\left(\frac{c+d}2\right)^2(a+b)\\ =\frac14(a+b)(c+d)(a+b+c+d)\leq 4. \]


Ali, Bonnie, Carlo and Dianna are going to drive together to
a nearby theme park.  The car they are using has four seats: one
driver's seat, one front passenger seat and two different back seats. Bonnie
and Carlo are the only two who can drive the car. How many
possible seating arrangements are there?

Let's take care of our drivers seat first.  Since only Bonnie and Carlo can drive, we have $ 2$ choices.  Since the rest of the seats have no restrictions, we have $ 3!$.  In total, we have $ 2\times 3!\equal{}2\times6\equal{}\boxed{12}$

[hide="Clearer Explanation"] Only 2 people can drive the first seat so they are 2 choices. There are 3 choices for the other front seat. There are 2 choices for the 1st back seat and 1 choice for the last back seat. 3*2*2*1=6*2=12(Like jxl28 said, there are factorials.)[/hide]

Usually A, B, C, and D can drive, but in this problem only 2 of 4 can drive, B, C.

Answer: $\dfrac{4!}{2}=12$.


[quote="silouan"]Suppose that in the exterior of a convex quadrilateral ABCD equilateral
triangles XAB, YBC, ZCD, WDA with centroids $S_{1}, S_{2}, S_{3}, S_{4}$ respectively
are constructed. Prove that $S_{1}S_{3}\bot S_{2}S_{4}$ if and only if $AC \bot BD$.[/quote]

Just in order to close(?) an old topic:
http://www.mathlinks.ro/Forum/viewtopic.php?t=17861
http://www.mathlinks.ro/Forum/viewtopic.php?t=1370
Note that your above problem should read $AC=BD$ instead of $AC\bot BD$. Maybe this was the reason nobody answered your posts.

  Darij


real analysis
calculus
integration
function
real analysis unsolved

Let f and g be two Lebesgue integrable extended real valued measurable functions on Lebesgue measurable set D. Show that if  f is bounded on D, then f*g is  Lebesgue integrable

$ \left\Vert fg\right\Vert _{1}\le\left\Vert f\right\Vert _{\infty}\left\Vert g\right\Vert _{1}<\infty$


LaTeX
inequalities
inequalities proposed

Resend your problem with proper use of $ LATEX$. :wink:

[quote="Math pro"]a,b,c >0 such that abc=1 prove that

$ \sum \frac {\sqrt a}{\sqrt {a^3 \plus{} 1}} \geq \frac {2 \sqrt 2}{\sum \sqrt {ab}}$[/quote]
Is your problem my friend?

[quote="mitdac123"]It's old problem[/quote]
yes! It is old problem but my proof is too long. :D
Can you post your solution,mitdac?Thanks!

do you have any nice solutions for it,everybody? :blush:

It is equivalent to prove $ (a^3\plus{}1)(b^3\plus{}1)(c^3\plus{}1)\leq \frac{3^6}{8^3}.$      :wink:

[quote="Sunkern_sunflora"]It is equivalent to prove $ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\leq \frac {3^6}{8^3}.$      :wink:[/quote]
hello Sunkern_sunflora how can this come can you show it?? i dont think this problem can turn to that form  :)

Of course it can!

$ (\sum \frac {\sqrt a}{\sqrt {a^3 \plus{} 1}})^2 \geq\sum \frac {a}{\sqrt {a^3 \plus{} 1}}\geq\frac{9\sqrt[3]{abc}}{\sqrt[3]{(a^3\plus{}1)(b^3\plus{}1)(c^3\plus{}1)}}$

Thus it only remains to prove: $ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\leq \frac {3^6}{8^3}.$ :wink:

But does the inequality holds?? :maybe: If it does, still I can't prove it. Can you, Rose_joker?

[quote="Sunkern_sunflora"]Of course it can!

$ (\sum \frac {\sqrt a}{\sqrt {a^3 \plus{} 1}})^2 \geq\sum \frac {a}{\sqrt {a^3 \plus{} 1}}\geq\frac {9\sqrt [3]{abc}}{\sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)}}$

Thus it only remains to prove: $ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\leq \frac {3^6}{8^3}.$ :wink:

But does the inequality holds?? :maybe: If it does, still I can't prove it. Can you, Rose_joker?[/quote]
Byt the last ineq is wrong,try $ a\equal{}1,b\equal{}2,c\equal{}\frac{1}{2}$ (or easy find since $ AM\minus{}GM$

yes this ineq is fairly strong,I think you have to use some stronger method! :lol:


Cauchy Schwarz Engel's form and $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ can solve easily.

I don't think as you, post all solution please!


Hi
Well the matter is: I need help ,because I have to practise Games theory and i don´t have exercices

If someone post some problems I`ll be very thaknfull

Well that`s all , and anyway thanks.

 :maybe:  :ewpu:  :stretcher:  :weightlift:

i don't think there is a such thing as that...
it's just strategy... and thinking. :):P:(

[quote="james4l"]Yes, there is something called game theory.

Unfortunately I don't know too many game theory questions...

Find the winning domain for the following (if landing on 1 = you lose) takeaway sets:
{1,2,3}
{1,3,4}
{1,3,5,7,9,...}
{1,2,4,8,16,32,...}
{1,2,3,5,8,13,21,...} (I actually don't know how to solve this one, guidance may be helpful)
among others.

Other questions:
Prove that White cannot lose if he plays optimally in double-chess (every turn you get two moves)

I'm out, sorry.[/quote]
[quote="cgyao15"]i don't think there is a such thing as that...
it's just strategy... and thinking. SmileRazzSad
[/quote]
Thank you ; Anyway i`ll  post some topics about Games theory but rightnow i don`t have much time so may be later

And again thanks.
 :)  :lol:  :thumbup:

Also, the problems posted so far are not game theory, they are COMBINATORIAL game theory, which is a seperate field from game theory.

Example game theory problem: Prisoner's Dilemma (If you dont know what it is, google it...)


function
limit
real analysis
real analysis unsolved

Let $f$ be the real valued continous function on $[0, \infty)$ such that $\lim_{n \rightarrow \infty}f(nx)$ exist for every $x \in [0 \infty)$. Prove that $\lim_{x \rightarrow \infty}f(x)$ exist.

This is essentially the same as [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70016[/url] (at least the solution given there works after an obvious modification).


search
number theory unsolved
number theory

find all nonnegative integer$m,n$ such that $mn(m+1)(n+1)$ be perfect square.

well because i dont know very good english can you explain me what integer is??and what do you mean by "perfect square"?thk you..

An integer is a number $n$ which belongs to $\{...,-3,-2,-1,0,1,2,3,...\}$.

The nonnegative integers are 0, 1, 2, 3, 4, 5, ...

A perfect square is the square ($n^2$) of an integer $n$.

it must be n=x and m=x+2 and for every x m*n(m+1)(n+1) will be a perfect square.. oh i forgot that x must be a nonnegetive integer, and not a number like 1/2.1/3e.g(i dont know how to say it)

Well there are infinitely many such numbers actually. Observe that the problem is the same as finding two triangular numbers whose product is a square. And there are infinitely many such numbers. Also m = n is a solution of course.

Yes, DK, we know that there are infinitely many such triples, but the problems is to describe all those triples, to find all those solutions, even if there are infinitely many :)

I think that m=n is the only solution there are only a couple possiblitys for mn(m+1)(n+1) to be a perfect square m(m+1) nad n(n+1) have to botht be a perfect square which is obviously impossible or n(m+1) and m(n+1) have to but be perfect squares which is impossible because m and n+1 would have to both be perfect squares or the same number and then m+1 and n would have to be a perfect square or the same aswell which is obviously impossible.

Well that's not quite true. For example you might have n = 1, m = 8.
Here is an interesting page about triangular numbers

http://www.shyamsundergupta.com/triangle.htm

Some of my results for a simplified task:
One can observe that if we simplify the task by saying m = 1 which means that we're looking for numbers n for which 2n(n+1) is a square then every solution implies an infinite number of other solutions: if 2i(i+1) is a complete square then so is 2j(j+1) if we say j = 4i(i+1).  So the solution (1,8) implies (1,288) etc... Not all solutions can be obtained that way: the other class of solutions for this simplified problem is given n = (2i+1)(2i+1) with i an arbitraty integer for which 2i²+2i+1 is a perfect square.
So in this simplified case (m=1) we will find all solutions by finding all integers i for which 2i²+2i+1 is a square, then the solutions are n=(2i+1)² [ and then 4n(n+1) is a solution and so on].  This way I got
(1,1) -> (1,8) -> (1,288) -> (1,332928) -> ...
(1,49) -> (1,9800) -> (1,384199200) -> ...
(1,1681)-> (1,11309768) -> ...

I found the numbers 1,49,1681 by performing a computer search on the numbers <100  e.g. 1681 = (2*20+1)² because 2*20²+2*20+1 is a square found by the computer.

this is of course just a very small part of the solutions but it may help...

**edit** I just noticed that the problem of finding integers i such that 2i²+2i+1 is a square is the same as finding integers k such that 2k²-1 is a square (2i²+2i+1=k²).


Find all pairs m, n of positive integers such that (m^2 - n) divides (m + n^2) and (n^2 - m) divides (m^2 + n).

If $m^{2}-n|m+n^{2}$ then $m^{2}-n\leq m+n^{2}\Rightarrow m-n\leq 1$.
If $n^{2}-m|m^{2}+n$ then $n^{2}-m\leq m^{2}+n\Rightarrow m-n\geq-1$.
So $-1\leq m-n\leq 1\Rightarrow m-n\in\{-1,0,1\}$.
If $m-n=0\Leftrightarrow m=n\Rightarrow n(n-1)|n(n+1)\Rightarrow n-1|n+1\Rightarrow n\in\{2,3\}$. So we have the solutions $(2,2), \ (3,3)$.
If $m=n+1\Rightarrow n^{2}+n+1|n^{2}+n+1$ and $n^{2}-n-1|n^{2}+3n+1$. Then $ n^{2}-n-1\leq\frac{n^{2}+3n+1}{2}\Rightarrow n\in\{0,1,2\}$ So we have the solutions $(0,1), \ (1,2), \ (2,3)$.
If $m=n-1\Rightarrow n^{2}+n-1|n^{2}-3n+1$ and $n^{2}-n+1|n^{2}-n+1$. Then $ n^{2}+n-1\leq\frac{n^{2}-3n+1}{2}\Rightarrow n=0\Rightarrow m=-1$ and $m$ is not positive.


MATHCOUNTS
AMC
AIME
geometry
3D geometry
probability
conics

Here's what you've all been waiting for... The Pythagorus! :)  :)  :)

So anyway, The Pythagorus is the contest I've been working on (sorry for lack of originality in the name) in case you haven't read the thread explaining the contest:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32638]Thread[/url]

Most of the following rules are copied from the thread, but I've added some stuff, so it might be worthwhile to scim through it.

There will be 5 questions in each of the 3 groups A, B, and C. The group A problems will be on the level of the harder (around National) Mathcounts questions, and could be solved by a relatively gifted middle-schooler. All the group A questions will have numerical answers. You do not have to show work for any of the group A questions.

The group B problems will be slightly harder than the group A questions, and contain both proofs and numerical answer questions. You will have to show your work for the group B questions. These questions will be challenging, but I've seen harder on this forum.

The group C problems will be the hardest, and probably the level of the harder AIME questions, but will include proofs. Again, you'll have to show your work for group C problems.

You can choose one of the 3 groups of questions to complete, and PM the answers/solutions to me.
Part A will be marked out of 5 (you get 1 mark for a right answer, 0 marks for a wrong),

Part B and C will be marked out of 25 (you may get partial marks)

This is a PM contest so you send your answers/solutions to me instead of posting them on the thread.

You may also do more than one group, if you have the free time. I will post the top people and their scores after the contest finishes.

Calculators are allowed, but no help from other people/math tutors. If you have any other questions, post them on this thread.

You have 1 week to complete this contest.
GROUP A:

1. David has a 4 by 4 by 4 3-dimensional tic toe board. As usual, one can win by placing 4 counters in horizontal, vertical, or up-down rows. One can also win diagonally, whether it is the diagonal of the cube, or in a line parallel to a diagonal of one of the faces. How many different winning rows are there? (For example, in normal 3 by 3 tic-tac-toe, there are 8)

2. A circle has center A. Two lines are drawn tangent to the circle at points B and D. They intersect at point C in an angle of 120 degrees. Line BD is drawn and has length 6rt3. Find the length of BC.

3. John tosses 10 coins. What is the probability that the amount of heads is an even number?

4. Let N be the number of positive integers less than or equal to 1000 which are multiples of 2, 3, or 5. Find N.

5. Let x/y = 8, and xy^3 = 2. Find (x^5)(y^5)

GROUP B:

1. [n] is defined as the greatest integer less than or equal to n. For example, [3.141] = 3,  [19] = 19, and [-6.7] = -7. Find the sum of all n[sqrt(n)] as n ranges from 1 to 50 inclusively, and n is an integer. (In other words, find: $\displaystyle\sum_{n=1}^{50}n[\sqrt{n}]$)

2. A line with equation y=mx+b is tangent to the two parabolas $y=x^2 + 5$ and $y = -x^2 + 2$. If m is positive, find $m^2 + 4b^2$. (NOTE: Yes, there is a non-calculus solution).

3. A circle is drawn with radius one and tangent to the two axes. A smaller circle is drawn tangent to the large circle and the two axes. Another circle is drawn tangent to the smaller circle and tangent to the two axes. This process is continued on infinitely, to an infinite amount of circles. Find the total area of all the circles.

4. Inside a regular hexagon of side length 1, a point P is chosen. An arbitrary side of the hexagon is chosen, and the endpoints of this side are connected to P. The endpoints opposite side of the hexagon are also connected to P. This act forms two triangles. Prove that the sum of these two triangles is constant, and find the value that it is equal to.

5. Given seven distinct positive integers, all less than 127, show that there exists two positive integers satisfying $1 < \frac{y}{x} \le 2$.

GROUP C:

1. Squares are constructed on the sides of a 3-4-5 right triangle. The vertices of these squares are connected as shown in the diagram. Find the area of the large hexagon.

2. Prove that, for positive a, b, and c, that:
$(a+b)^2 + (a+c)^2 + (b+c)^2 \ge \frac{(a^2+b^2+c^2)\sqrt[3]{abc}}{(a+b+c)}$

3. A man is standing in a line. The fraction a/b of the line is in front of him, and the fraction c/d of the line is behind him, where a, b, c, and d are integers and the two fractions are in lowest terms. Prove that the amount of people in the line is the LCM of b and d.

4. Call a triangle graphed in the Cartesian plane dk2ian if it has integral side lengths and integral coordinates. Prove that there are no dk2ian triangles with perimeter 43.

5. Let ABC be the right-angled isosceles triangle whose equal sides have length 1. P is a point on the hypotenuse, and the feet of the perpendiculars from P to the other sides are Q and R. Consider the areas of the triangles APQ and PBR, and the area of the rectangle QCRP. Prove that regardless of how P is chosen, the largest of these three areas is at least 2/9.

GOOD LUCK  :lol:

Just randomly looking at problems, C2's inequality doesn't seem to be true.

id agree with probability1.01
plugging in 100 for a b and c gives us something grossly off

[quote="probability1.01"]Perhaps you meant $\sqrt[3]{abc}$ on the RHS instead of abc.  That would work.  Although I'm not sure if it should be in Getting Started.[/quote]

Yes, thats what I did mean, but I think the LATEX messed up.

And I did want to make the group C questions challenging, and the question is easier than it looks.

On another note, Zyloch has sent in his answers for group B. I just received them, so I haven't marked them yet, but they look good  ;).

Remember, you can still write the contest (you didn't need to sign up), just get the questions to me by next sunday (anytime during sunday is OK).

On the one with the line of people, I think all multiples of the LCM are possible given two fractions.

[edit. Ignore this post. I made a really dumb mistake]

[quote="EFuzzy"]On the one with the line of people, I think all multiples of the LCM are possible given two fractions.[/quote]

Really? Take this for an example. (P is the person in consideration, L are all other people):

LLLPLL.

In this case, 1/3 of the line is behind P, and 1/2 of the line is in front of P.
LCM(3, 2)=6.

EFuzzy, I think your saying that all multiples of the LCM are possible? Then 12 is a multiple of the LCM.

So let's say we have a line with 12 people that satisfies the condition.
If we let 1/2 of the line be in front of P, that is 6 people. If we let 1/3 of the line be behind P, that is 4 people. That gives us a total of 1+ 6 +4=11 people in the line, and this is a contradiction.

But maybe I misunderstood your statement, although I'm quite sure the question is correct.

I haven't gotten many responses to this contest yet, so remember, you don't have to send in solutions to all three groups. You may send in solutions to all three groups, or you could just send solutions to one or two groups. I rank the groups separately anyways (e.g, I give the top scorers in Group A, group B, and Group C).

[quote="white_horse_king88"]By the way, posting C group in the beginner forum won't get you too many responses, so don't be surprised. It is much more Intermediate oriented, as probability1.01 mentioned. Perhaps you should make a different contest with those problems on the Intermediate forum.[/quote]

Actually, thats probably a pretty good idea. I'm still keeping these questions here, but I'll post links to this contest on the Mathcounts (for group A), and Intermediate (for group C, like you mentioned). But I tried to make the group B questions do-able for this forum, so at least people that regularly post on this forum are challenged (but can actually solve the question).

[quote="Scrambled"][quote="mathgeek2006"]Pythagorus?  More like Pythagoras.  Spelling mistakes do annoy me...[/quote]
maybe he wanted a name not yet taken (i doubt anyone has named a contest pythagorus yet, and i know of an obscure contest called pythagoras)[/quote]

 :blush: Oops  :blush:

Oh well, I'll keep the name as Pythagorus. It makes it... 'unique'.

Make sure you are on all of Sunday because you will probably get most Group C answers then. They do not seem to be so difficult. But doing all of them by Sunday may be too much for the Getting Started forum. Try to get it into the Intermediate forum as soon as possible. You could still get responses from them by Sunday.

[quote="dangvh"]i just finished group B and i dont think i am able to do all of C, so can i do A for fun and send it to you?? :D[/quote]

Of course.

[size=150]The contest is over now[/size]

I'm working on correcting the solutions I have and will post the results either tonight or tommorow.

Will you post the solutions (the calculations, not just "the answer on A4 is 734 (I solved this one, I think)) or can people who solved the problems do that? I would like to see some of them (all of them :D ). Honestly, it would be very kind :)


function trigonometry calculus integration Euler

Is there a formula that allows to compute the values of the zeta function?
I know this:
$\zeta(2)=\frac{\pi^{2}}{6}$ (I know how to prove this)
$\zeta(4)=\frac{\pi^{4}}{90}$ (can anyone prove that for me?)

What about other values, like $\zeta(3)$, and $\zeta(5)$?

Odd values don't have a closed form.  The rational coefficients of the even values are related to the Bernoulli numbers.  Look it up on Mathworld sometime  :)

Edit:  There is a nice proof of $\zeta(2)$ using Fourier series that I think can be readily extended to finding $\zeta(4)$.

Yes those Fourier series based calculations are rather elegant.

[url]http://en.wikipedia.org/wiki/Riemann_zeta_function[/url]
[url]http://mathworld.wolfram.com/RiemannZetaFunction.html[/url]

Easiest is to compare the expression of $\frac{\sin x}x$ as function of $x^{2}$
as $1-x^{2}/6+x^{4}/5!-\dots$
(or $1-y/6+y^{2}/120-\dots$
Sum of  reciprocals of roots of y = (1/6)
Sum of squares of reciprocals of the roots = (1/6)^2 -2(1/120) = 1/90
Since the roots of the equations are at $(1)^{2}(\pi)^{2},(2^{2})(\pi)^{2},(3^{2})(\pi^{2}) \dots$
we have  $(1/\pi^{2}) (1+1/(2^{2})+1/(3^{2})+\dots) = 1/6$
and $((1/\pi^{4}) (\zeta (4)) = 1/90$

One of the  easiest way to approximate $\zeta (3)$ etc is to just sum first 3-4 values and then use integral (+ correction according to Euler method) ..3 or 4 terms will give 5-6 figure accuracy..


trigonometry
algebra
polynomial
Vieta
complex numbers

(a) Solve for $\theta\in\mathbb{R}$: $\cos(4\theta) = \cos(3\theta)$

(b) $\cos\left(\frac{2\pi}{7}\right)$, $\cos\left(\frac{4\pi}{7}\right)$ and $\cos\left(\frac{6\pi}{7}\right)$ are the roots of an equation of the form $ax^3+bx^2+cx+d = 0$ where $a, b, c, d$ are integers. Determine $a, b, c$ and $d$.

[quote="Peter VDD"](a) trivial
(b) $1,\frac12,-\frac12,-\frac18$.[/quote]

Yeah, $a$, $b$, $c$ and $d$ are integers, but that's right  :)

b) Its well known that 8n^3-4n^2-4n+1=0, for cos(pi/7, 3pi/7, 5pi/7) and the negatives of those are (cos 6pi/7, cos 4pi/7, cos 2pi/7) so the desired equation is 8n^3+4n^2-4n-1. So a, b, c, d=8, 4, -4, -1 correspondingly

[quote="ISTormrage"]So a, b, c, d=8, 4, -4, -1 correspondingly[/quote]Or a multiple of that.

For the exercise $(a)$ you can use the following essential relations :

$S.\blacktriangleright \sin x=\sin y\Longleftrightarrow (x-y)\in 2\pi Z\ \ \vee\ \ (x+y)\in 2\pi Z+\pi\ .$

$C.\blacktriangleright \cos x=\cos y\Longleftrightarrow (x\pm y)\in 2\pi Z\ .$

$T.\blacktriangleright \tan x=\tan y\Longleftrightarrow x-y\in \pi Z\ .$

so for (a) we have: $2\pi Z_1 +3\theta= 2\pi Z_2 +4\theta$ so $\theta = 2\pi Z$ (where $Z, Z_1, Z_2$ are any whole number)
correct, right?

me@home, not any whole number, but any [i]integer[/i],

also how did #1 get there, that is so trivial

$\cos 4\theta-\cos 3\theta=0$
$\sin\frac{7}{2}\theta\sin\frac{1}{2}\theta=0$
$\theta= \frac{2}{7}*(k\pi), 2*(kpi)$ for $k\in\mathbb{Z}$

part 1 was simply a hint for part 2. ;)

Now there's already only 1 pupil who solved part b, what wouldn't it have been without hint...

Some other participants made some progress towards the solution of 1b...
You don't actually *need* the hint, complex numbers work nicely as well.

Let $x^3-ax^2+bx-c=0$ be the appreciated equation
with the roots $\cos\frac{2\pi}{7}$, $\cos\frac{4\pi}{7}$, $\cos\frac{6\pi}{7}\ .$

$1.\blacktriangleright 2b=2\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+2\cos\frac{4\pi}{7}\cos \frac{6\pi}{7}+2\cos\frac{6\pi}{7}\cos\frac{2\pi}{7}=$ $\cos \frac{2\pi}{7}+\cos \frac{6\pi}{7}+\cos \frac{2\pi}{7}+\cos \frac{10\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=$ $\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=$ $2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)=2a\Longrightarrow \boxed {\ b\ =\ a\ }\ .$

$2.\blacktriangleright 4c=4\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}\cos\frac{6\pi}{7}=2\left(\cos \frac{2\pi}{7}+\cos\frac{6\pi}{7}\right)\cos\frac{6\pi}{7}=$
$2\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+2\cos ^2\frac{6\pi}{7}=$ $\cos \frac{4\pi}{7}+\cos\frac{8\pi}{7}+1+\cos \frac{12\pi}{7}=$
$\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+1+\cos\frac{2\pi}{7}=$ $1+a\Longrightarrow \boxed {\ 4c\ =\ a\ +\ 1\ }\ .$

$3.\blacktriangleright 2a\sin \frac{2\pi}{7}=2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{2\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{2\pi}{7}\cos\frac{6\pi}{7}=$ $\sin\frac{4\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}+\sin\frac{8\pi}{7}-\sin \frac{4\pi}{7}=$ $\sin\frac{\pi}{7}-\sin\frac{2\pi}{7}-\sin\frac{\pi}{7}=-\sin\frac{2\pi}{7}\Longrightarrow$
$\boxed {\ a= -\frac 12\ }\ .$ The proof of this $3^{th}\ -$ relation is similarly with the [u]Shyong's[/u] .

Therefore, $a=b=-\frac 12\ \ \wedge\ \ c=\frac 18\Longrightarrow \boxed {\ 8x^3+4x^2-4x-1=0\ }\ .$

[b]Remark (two remarkable relations).[/b] $n\in N^*\ ,\ \alpha_{k+1}=\alpha _k+r\ ,\ k\in \overline {1,n}\Longrightarrow$

$\sum^n_{k=1}\cos\alpha_k=\frac{\cos\frac{\alpha_1+\alpha_n}{2}\sin\frac{nr}{2}}{\sin\frac{r}{2}}\ \ \wedge\ \ \sum^n_{k=1}\sin\alpha_k=\frac{\sin\frac{\alpha_1+\alpha_n}{2}\sin\frac{nr}{2}}{\sin\frac r2}\ .$
In the our case, $a=\frac{\cos\frac{4\pi}{7}\sin\frac{3\pi}{7}}{\sin \frac{\pi}{7}}=\frac{-\cos\frac{3\pi}{7}\sin\frac{3\pi}{7}}{\sin \frac{\pi}{7}}=-\frac 12\ .$

it is kind of odd that i derived part b on my own before seeing the problem...consider

$x^7-1=0$, the roots are $e^{\frac{2\pi i k}{7}}$ for $k=-3,-2,-1,0,1,2,3$, note that the roots form conjugate pairs

$\text{cis} \, k*\text{cis} -k=1$ and $\text{cis}\, k+\text{cis} -k=2\cos k$, so we have that the polynomial factors into

$(x-1)(x^2-ax+1)(x^2-bx+1)(x^2-cx+1)$ where $(a,b,c)=(2\cos\frac{2\pi}{7},2\cos\frac{4\pi}{7},2\cos\frac{6\pi}{7})$

expanding and comparing coefficients, we have

$a+b+c=-1$, $ab+bc+ca=-2$, and $abc=1$, using vieta, a polynomial with those roots is:

$r^3+r^2-2r-1=0$, and getting rid of the 2, making the roots cos...

$8x^3+4x^2-4x-1=0$

[color=red][b]Very nice, Altheman ! Thanks.[/b][/color] I applied the your method for obtaining the equation

$x^5-ax^4+bx^3-cx^2+dx-e=0$ with the roots $t_k=\cos\frac{2k\pi}{11}\ ,\ k\in \overline {1,5}\ .$

$\blacktriangleright\sum^{10}_{k=0}x^k=\prod^5_{i=1}\left(x^2-2t_ix+1\right)\Longrightarrow \sum^{10}_{k=0}x^k=(x^2+1)^5-2ax(x^2+1)^4+$

$+4bx^2(x^2+1)^3-8cx^3(x^2+1)^2+16dx^4(x^2+1)-32ex^5\Longrightarrow$

$\frac{-a}{16}=\frac{b}{-32}=\frac{-c}{-12}=\frac d6=\frac{-e}{1}=\frac{1}{32}\ ,$ i.e. the in demand equation is\[ \boxed {\ 32x^5+16x^4-32x^3-12x^2+6x+1=0\ }\ . \][color=blue][b]Remark.[/b] Try a generalization : use the [u]Altheman's method[/u] to obtain

the equation with the roots $t_k=\cos\frac{2k\pi}{n}\ ,\ k\in \overline {1,p}\ ,$ where $n=2p+1\ ,\ p\in N^*\ .$[/color]

[quote="Altheman"]$\theta= \frac{2}{7}*(k\pi), 2*(kpi)$ for $k\in\mathbb{Z}$[/quote]

Actually, this is not correct. $\frac{6\pi}{7}$, for instance, does not work.

I am pretty sure that $\dfrac{6\pi}{7}$ works.

Also, I revived because I have a kind of different solution:

[hide="sol"]
Note that $\dfrac{2\pi}{7},\dfrac{4\pi}{7}, \dfrac{6\pi}{7}$ all satisfy $\cos 4\theta = \cos 3\theta$.

That equation in equivalent to $$8\cos^4\theta-8\cos^2\theta+1=4\cos^3\theta-3\cos\theta$$ or $$8\cos^4\theta-4\cos^3\theta-8\cos^2\theta+3\cos\theta+1=0$$

Thus, the roots $\cos \left(\dfrac{2\pi}{7}\right), \cos \left(\dfrac{4\pi}{7}\right), \cos \left(\dfrac{6\pi}{7}\right)$ satisfy the polynomial $$8x^4-4x^3-8x^2+3x+1=0$$

Since $1$ is a trivial root of this polynomial (created from the trivial solution $2\pi n$ in part a) we do synthetic division of the polynomial by $x-1$ to get the polynomial $$8x^3+4x^2-4x-1=0$$ and we're done.

[/hide]


in a circular loop of wire carrying a current i, is the magnetic induction [b]B[/b] uniform everywhere within the loop?

No . It Isn't.

It would be uniform inside an ideal infinite long solenoid.


Use the Euclidian Algorithm to find all integers p and q such that 599p + 53q = 1

Express your answers in terms of the integer parameter "t".

Ok but explain how you were lead to those expressions of $t$. Justify that that gives all the answers and not just some.

Just saying sludgethrower, you should have 1 topic for all the problems you put on at once.


prove that     (a\a+b)^n  +(b\b+c)^n  +(c\c+a)^n   >=  3\2^n  +(a-b)(a-c)(b-c)\(a+b)(b+c)(c+a)2^n        for a b c positives number
 :ninja:

I guess you mean in Latex:[quote="fermat3"]Prove that  $\bigl(\frac a{a+b}\bigr)^n +\bigl(\frac b{b+c}\bigr)^n + \bigl(\frac c{c+a}\bigr)^n \geq  \bigl(\frac 32\bigr)^n +\frac{(a-b)(a-c)(b-c)}{(a+b)(b+c)(c+a)2^n }$
for $a, b, c$ positive numbers  :ninja:[/quote]

the inequality above is wrong!
$c\rightarrow 0 a\rightarrow infinity$

Oops, certainly he meant $\frac3{2^n}.$ Truly, $(\frac32)^n$ doesn't make sense.  :?:
Fermat3, you should really learn some Latex, it isn't hard  :)

So:[quote="fermat3"]Prove that  $\bigl(\frac a{a+b}\bigr)^n +\bigl(\frac b{b+c}\bigr)^n + \bigl(\frac c{c+a}\bigr)^n \geq \frac3{2^n} +\frac{(a-b)(a-c)(b-c)}{(a+b)(b+c)(c+a)2^n }$
for $a, b, c$ positive numbers  :ninja:[/quote]

Anyway this problem is wrong :) ,take $n=1$ you will easy see that it can't be true,I have cheked even for $n=2$ :D ,it was wrong again.I don't think that this ineq. can be true for $n\geq 3$,so I won't advise you to try this one :lol: .

you  can  use   bernoulli     and  try if  the  inegualitie  false or no :rotfl:  ;)

[quote="fermat3"]we  have   (1+a-b\a+b)^n  >=  1+n(a-b\a+b)

then adding we find   A >= 3+n[  a-b\a+b  + b-c\b+c  +c-a\c+a ]                                                                                or it s easy to  prove that   a-b\a+b  +b-c\b+c  +c-a \c+a = (a-b)(b-c)(a-c)\(a+b)(a+c)(b+c)              and  1+a-b\a+b =2a\a+b :ninja:[/quote]
 :o  :rotfl:
Have you seen what have you proved? :rotfl:
You proved that
(a\a+b)^n +(b\b+c)^n +(c\c+a)^n >= 3\2^n +$n*$(a-b)(a-c)(b-c)\(a+b)(b+c)(c+a)2^n, aren't you?
Now look at your first post :!:
[quote="fermat3"]prove that (a\a+b)^n +(b\b+c)^n +(c\c+a)^n >= 3\2^n +(a-b)(a-c)(b-c)\(a+b)(b+c)(c+a)2^n for a b c positives number [/quote]
wher is your $n$?  ;)

[quote="aviateurpilot"][quote]we  have   (1+a-b\a+b)^n  >=  1+n(a-b\a+b)

then adding we find   A >= 3+n[  a-b\a+b  + b-c\b+c  +c-a\c+a ]                                                                                or it s easy to  prove that   a-b\a+b  +b-c\b+c  +c-a \c+a = (a-b)(b-c)(a-c)\(a+b)(a+c)(b+c)              and  1+a-b\a+b =2a\a+b :ninja:[/quote]

[b]fermat3[/b] found:
$(\frac{a}{a+b})^{n}+(\frac{b}{c+b})^{n}+(\frac{c}{a+c})^{n}\geq \frac{2}{3^{n}}+n\frac{(a-b)(b-c)(a-c)}{2^{n}(a+b)(a+c)(b+c)}\geq \frac{2}{3^{n}}+\frac{(a-b)(b-c)(a-c)}{2^{n}(a+b)(a+c)(b+c)}$
where is the probleme [b]Tiks[/b].[/quote]
The problem is that you can't show me that $(a-b)(b-c)(a-c)\geq 0$ :P


inequalities
trigonometry
calculus
inequalities unsolved

$0< \alpha <\dfrac{\pi}{2}$
Prove that:
$\sin{\alpha}+\dfrac{1}{2}\sin{2\alpha}+\dfrac{1}{3}\sin{3\alpha}>0$

I can do this solution by using caculus. In fact if $f(x)=\frac1nsin (nx)$  then
$f'(x)=cos(nx)$ .
  Just  use caculus here and u can easily find the root of $g(x)=cos(x)+cos(2x)+cos(3x)$
 :)

let me have a try...
by the trig addition formulas we have that
$\sin2\alpha=2\sin\alpha\cos\alpha$ and $\sin3\alpha=\sin\alpha(4\cos^2\alpha-1)$, substituting in the problem we have
$\sin{\alpha}+\dfrac{1}{2}\sin{2\alpha}+\dfrac{1}{3}\sin{3\alpha}=\dfrac{1}{3}\sin \alpha(3+3\cos\alpha+4\cos^2\alpha-1)=\\\dfrac{1}{3}\sin \alpha(4\cos^2\alpha+3\cos\alpha+2)$
which is clearly positive.

Indeed , we have :
 $F_n(x)=\sinh +\dfrac{1}{2}\sin2x +\dfrac{1}{3}\sin3x+...+\dfrac{1}{n}\sin nx >0 \forall x$

Not for all x, but for all x from (0, Pi)

If I am not wrong, then one calls it "Fejer inequality"

Yes $x \in [0,\pi]$ . Sometimes this ineq is also called Jackson's ineq

Hi nttu and teacher Nam Dung!
   Could you please show me how to solve that problem in the general case please?
$F_n(x)=\sinh +\dfrac{1}{2}\sin2x +\dfrac{1}{3}\sin3x+...+\dfrac{1}{n}\sin nx >0$
x from (0, Pi)

 I have done it by using caculus but I wonder about "Fejer inequality"
   What is it?

$ \sin{\alpha} \plus{} \frac{\sin{2\alpha}}{2} \plus{} \frac{\sin{3\alpha}}{3}\equal{}\frac{2t\left(3t^4\minus{}4t^2\plus{}9\right)}{3(t^2\plus{}1)^3}>0,$

where $ t\equal{}\tan\frac{\alpha}{2}\left(0<\alpha<\pi\right).$

Stronger inequality $ \sin{\alpha} \plus{} \frac {\sin{2\alpha}}{2} \plus{} \frac {\sin{3\alpha}}{3} >\frac{\alpha}{6}(\pi\minus{}\alpha)$ holds for $ 0 < \alpha < \pi.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=114058


How many prime numbers are between 30 and 50?

Let's see:

31

37

41

43

47

There are $ 5$ primes.

[hide="answer"]i just listed them out and there were 5 [/hide]

There would be 5 prime numbers between 30 and 50.


Slove the system of equations
$ y(x\plus{}y)^2\equal{}9$
$ y(x^3\minus{}y^3)\equal{}7$

Solving  $ y(x \plus{} y)^2 \equal{} 9$ in $ x,\ y > 0$ and

$ x \equal{} \pm\frac {3}{\sqrt {y}} \minus{} y\ \cdots\ (1)$

Solving  $ y(x^3 \minus{} y^3) \equal{} 7$  in $ x,\ y\neq0$ and

$ x^3 \equal{} \frac {7}{y} \plus{} y^{2}\ \cdots\ (2).$

Since $ y > 0,\ x > 0$ and (1) must $ x \equal{} \frac {3}{\sqrt {y}} \minus{} y$ and $ 0 < y < 3^{\frac {2}{3}}.$

Let $ y \equal{} t^2,$ then

 (1) is:

$ x \equal{} \frac {3}{t} \minus{} t^{2}$

$ xt \equal{} 3 \minus{} t^{3}$

$ (xt)^{3} \equal{} (3 \minus{} t^{3})^{3}$

and (2) is:

$ x^3 \equal{} \frac {7}{t^{2}} \plus{} t^{4}$

$ (xt)^3 \equal{} 7t \plus{} t^{7}.$

So solving the equation on $ t$ with $ 0 < t < 3^{\frac {1}{3}}$,

$ 7t \plus{} t^{7} \minus{} (3 \minus{} t^{3})^{3} \equal{} 0\ \cdots\ (3)$

will give the solution.

(3) is

$ f(t) \equal{} t^9 \plus{} t^7 \minus{} 9t^6 \plus{} 27t^3 \plus{} 7t \minus{} 27 \equal{} 0.$

$ f(t) \equal{} (t \minus{} 1)(t^8 \plus{} t^7 \plus{} 2 t^6 \minus{} 7 t^5 \minus{} 7 t^4 \minus{} 7 t^3 \plus{} 20 t^2 \plus{} 20 t \plus{} 27) \equal{} (t \minus{} 1)g(t).$

$ g(t) \equal{} t^8 \plus{} t^7 \plus{} 2 t^6 \minus{} 7 t^5 \minus{} 7 t^4 \minus{} 7 t^3 \plus{} 20 t^2 \plus{} 20 t \plus{} 27$
$ \equal{} (t^6 \minus{} 7t^3 \plus{} 20)(t^2 \plus{} t \plus{} 1) \plus{} t^6 \plus{} 7 > 2t^6 \minus{} 7t^3 \plus{} 27$

and the determinant of RSH is negative , so $ g(t) > 0$ on $ 0 < t < 3^{\frac {1}{3}}.$

So the solution of $ f(t) \equal{} 0$ on $ 0 < t < 3^{\frac {1}{3}}$ is $ t \equal{} 1$ and so $ y \equal{} 1,x \equal{} 2.$


function
limit
geometry
real analysis
real analysis unsolved

Let the function $ g^{[1]}(x) \equal{} \left\{{\begin{array}{*{20}c} {{x \minus{} x\,\ln|{x}|}\hspace{0.1cm},\hspace{0.2cm}x\in\mathbb{R}^{*}}\hfill\vspace{0.2cm} \\
{{0}\hspace{0.1cm},\hspace{1.9cm}x \equal{} 0}\hfill \\
\end{array}}\right.$

and  the functions $ g^{[n]}(x) \equal{} \left\{{\begin{array}{*{20}c} {{\underbrace{g^{[1]}\circ{g^{[1]}}\circ\dots\circ{g^{[1]}}}_{n \minus{} times}}\hspace{0.1cm},\hspace{0.2cm}x\in\mathbb{R} \minus{} \left\lbrace{0,\pm{e}}\right\rbrace}\hfill\vspace{0.2cm} \\
{{0}\hspace{0.1cm},\hspace{3.1cm}x\in\left\lbrace{0,\pm{e}}\right\rbrace}\hfill \\
\end{array}}\right.$, $ n\in\mathbb{N}^{*},\ n\geq2$.

Prove that   $ \mathop{\lim}\limits_{n\rightarrow{ \plus{} \infty}}\left. {g^{[n]}(x)}\right|_{x\in\left[{0,e}\right]} \equal{} \left. {g(x)}\right|_{x\in\left[{0,e}\right]} \equal{} \left\{{\begin{array}{*{20}c} {{1}\hspace{0.1cm},\hspace{0.2cm}x\in\left({0,e}\right)}\hfill\vspace{0.2cm} \\
{{0}\hspace{0.1cm},\hspace{0.2cm}x\in\left\lbrace{0,\pm{e}}\right\rbrace}\hfill \\
\end{array}}\right.$.

____________________________________________________________________________________________

unfortunately the graph in the attachment does not help to much

The equality clearly holds for $ x\in \{0,1,e\}$.  Let us consider the remaining intervals $ (0,1)$, $ (1,e)$ seperately.

Fix an $ x\in (0,1)$ and define the sequence $ a_k \equal{} g^{[k]}(x)$, $ k\geq 1$.  Note that $ x < g^{[1]}(x) < 1$, so $ \{a_k\}$ is a bounded increasing sequence, and therefore convergent.  Convergence of $ \{a_k\}$ implies $ \lim_{k\to \infty}g^{[1]}(a_k) \minus{} a_k \equal{} 0$, and therefore $ \lim_{k\to \infty}a_k \equal{} L$ where $ g^{[1]}(L) \equal{} L$.  The only fixed points of $ g^{[1]}$ are $ 0$ and $ 1$, but $ 0 < x$ so $ L \equal{} 1$.  Thus, $ \lim_{n\to \infty}g^{[n]}(x) \equal{} 1$ for all $ x\in (0,1)$.

Now fix an $ x\in (1,e)$.  $ g^{[1]}(x)\in (0,1)$, so we apply the result above to $ g^{[1]}(x)$ to obtain $ \lim_{n\to \infty}g^{[n]}(x) \equal{} \lim_{n\to \infty}g^{[n]}(g^{[1]}(x)) \equal{} 1$.

First wish to thank JoeBlow for the solution. In my efforts  I failed  to recognize that $ \mathop{\lim}\limits_{n\rightarrow{ \plus{} \infty}}\left[{g^{[n \plus{} 1]}(x) \minus{} g^{[n]}(x)}\right] \equal{} 0$ leads to the fixed point of $ g^{[1]}$.
Though it is not so obvious, the specific limit is the core $ (\star)$ of a problem I posted a few days ago in Mathlinks,
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=231132]common points of graphs and limit of area[/url] .


geometry
number theory
greatest common divisor
rotation
trigonometry
number theory proposed

Let $ ABC$ be a triangle with sides $ a \equal{} BC$, $ b \equal{} AC$ and $ c \equal{} AB$ and area $ S$. It is known that $ a^2, b^2, c^2$ and $ 2S$ are integers satisfying $ \text{gcd}(a^2, b^2, c^2) \equal{} 1$. Prove that we can translate and rotate $ ABC$ in such a way that $ A, B$ and $ C$ become lattice points (i.e., points with integer coordinates).

$ bc\sin A\equal{}2S\equal{}n$ and $ 2bc\cos A\equal{}b^2\plus{}c^2\minus{}a^2$ are integers, hence $ \tan A$ is rational. $ \tan A\equal{}\sin A/\cos A\equal{}n/\sqrt{b^2c^2\minus{}n^2}$. So $ b^2c^2\equal{}m^2\plus{}n^2$, where $ m$ is an integer. We call an integer [i]good[/i] if it can be represented as a sum of two squares. $ b^2c^2$ is good. Similarly we can show that $ c^2a^2$ and $ a^2b^2$ are good. We know the lemma that an integer is good iff the exponent of each prime of the form $ 4k\plus{}3$ in its canonical representation is even. It follows that $ [a^2b^2,b^2c^2,c^2a^2]$ is good. Since $ a^2b^2c^2$ $ \equal{}(a^2,b^2,c^2)[a^2b^2,b^2c^2,c^2a^2]$ $ \equal{}[a^2b^2,b^2c^2,c^2a^2]$, we conclude that $ a^2b^2c^2$ is good. Therefore by the lemma $ a^2b^2c^2/b^2c^2\equal{}a^2$ is good. Similarly $ b^2$ and $ c^2$ are also good. It follows that triangle $ ABC$ can be translated and rotated such that $ A,B,C$ become lattice points.


inequalities
logarithms
inequalities proposed

maybe someone will find it is equal to some problem which have been posted.
but I still want to post it because it is so nice.as following:
$a,b,c,d$ are all positive numbers.and $a+b+c+d=1$
prove $(1-\sqrt a)(1-\sqrt b)(1-\sqrt c)(1-\sqrt d) \geq \sqrt{(abcd)}$.

[quote="yptsoi"]This inequality seems to be quite strange.  :? Zhaobin, did you make any typos?[/quote]
 ;)  ;) I have edited it.

Nice problem: let $A=\sum_{sym}\sqrt{a}$, $B=\sum_{sym}\sqrt{ab}$, $C=\sum_{sym}\sqrt{abc}$.

The inequality is equivalent to $1+B\geq A+C$

We easily prove: $A^2=1+2B$ and also $\frac{36AC}{16}\leq B^2$, and also $A\geq 4C$ ( the last one can be done mltiplying the rhs by $a+b+c+d$ and using muirheads)

Summing the last three inequalities, we get: $(B+1)^2\geq A^2+\frac{36AC}{16}\geq (A+C)^2$ like we wanted.

Guys, it suffices for someone to replace a number by its square root and nobody see the classical problem, so much discussed on the forum, of Vasc:
   If $a^2+b^2+c^2+d^2=1$ then $^(1-a)(1-b)(1-c)(1-d)\geq abcd$? Come on! :D
  And since we're speaking about this one, I will propose again the last problem from Old and New Inequalities, which generalizes this one. I posted it before, but as usual is lying somewhere on the forum, forgotten:
  Find the best $k_n$ such that if $a_1^2+...+a_n^2=1$ then $(1-a_1)...(1-a_n)\geq k_n a_1...a_n$. Hope next time we wont't replace the variables with their 4-th root!

Pascual2005 and harazi were quicker than me, but anyway, here is my solution.

Suppose WLOG that $a\leq b\leq c\leq d$. Then we have:

$\sqrt a\leq \sqrt{\frac 1 4}=\frac 1 2\Rightarrow 1-\sqrt a\geq \sqrt a$

$\sqrt a+\sqrt b\leq2\sqrt{\frac{a+b}2}\leq2\sqrt{\frac{\frac 1 2}2}=1$ (by RSM-AM inequality) $\Rightarrow 1-\sqrt a+1-\sqrt b\geq\sqrt a+\sqrt b$

$\sqrt a+\sqrt b+\sqrt c\leq 3\sqrt{\frac{a+b+c}3}\leq 3\sqrt{\frac{\frac 3 4} 3}=\frac 3 2\Rightarrow$
$\Rightarrow 1-\sqrt{a}+1-\sqrt{b}+1-\sqrt{c}\geq\sqrt{a}+\sqrt{b}+\sqrt{c}$

$1-\sqrt{a}+1-\sqrt{b}+1-\sqrt{c}+1-\sqrt{d}=\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+4-2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)$

Now we have
$\left(1-\sqrt{a},1-\sqrt{b},1-\sqrt{c},1-\sqrt{d}\right)\succ\left(\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d}+4-2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)\right)$.

Also note that $4-2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)\geq 4-8\sqrt{\frac{a+b+c+d}4}=0$ (RSM-AM inequality again).

Now we apply Karamata's inequality for $f(x)=\ln x$:

$\ln\left(1-\sqrt{a}\right)+\ln\left(1-\sqrt{b}\right)+\ln\left(1-\sqrt{c}\right)+\ln\left(1-\sqrt{d}\right)\geq$
$\geq\ln\sqrt{a}+\ln\sqrt{b}+\ln\sqrt{c}+\ln\left(\sqrt{d}+4-2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)\right)\geq$
$\geq\ln\sqrt{a}+\ln\sqrt{b}+\ln\sqrt{c}+\ln\sqrt{d}$

$\ln\left(1-\sqrt{a}\right)\left(1-\sqrt{b}\right)\left(1-\sqrt{c}\right)\left(1-\sqrt{d}\right)\geq\ln\sqrt{abcd}$

$\left(1-\sqrt{a}\right)\left(1-\sqrt{b}\right)\left(1-\sqrt{c}\right)\left(1-\sqrt{d}\right)\geq\sqrt{abcd}$

QED

[quote="harazi"]No, no, you two still have some work on the general case, too.  :D  ;)[/quote]
I believe that my idea could be useful, but right now I don't have time, I'll try to solve it later.

I don't think your solution is correct.
$\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}$ and $1-\sqrt{a},1-\sqrt{b},1-\sqrt{c},1-\sqrt{d}$ are in different order, so you haven't proved the arrays majorized the other.

Unfortunately, my solution is absolutely wrong, besides siuhochung's comment another mistake is that $\ln x$ is concave and not convex. Sorry, I will try to correct it.

harazi,can you give me the link of your general problem :)
sorry,I can't find it.maybe my searching skill is too poor ;)


Fie $x, y, z$, numere reale pozitive. Demonstrati inegalitatea:
\[\frac{1}{x^2+y}+\frac{1}{y^2+z}+\frac{1}{z^2+x}\leq\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}.\]

Se vede treaba ca nu prezinta interes inegalitatea.
Cumva e prea simpla?
Mi-ar placea sa vad si alte solutii diferite de a mea.

Te credeam om serios. Dar daca incepi si tu cu vaicareli de genul "s-ar parea ca nu intereseaza pe nimeni...", atunci meriti genul de raspunsuri pe care ti le da Cezar Lupu.

Imi cer scuze daca am deranjat pe cineva cu maniera mea de a da "refresh" subiectului postat. Asa am vazut ca fac altii si am vrut sa fiu la moda. In legatura cu Cezar Lupu, eu personal il apreciez foarte mult; e problema lui daca  ma apreciaza sau nu.

poti totusi sa postezi solutia ta te rog frumos ?

[u]Definitie.[/u] $X.a.s.Y\Longleftrightarrow sgnX=sgnY\Longleftrightarrow XY>0\ \vee\ X=Y=0$
(X,Y au [u][b]a[/u][/b]celasi  [u][b]s[/b][/u]emn, adica X,Y au aceeasi pozitie relativa fata de 0).

Fie numerele pozitive $x_1\le x_2\le x_3\le \ldots \le x_n,\ n\in N^*$.
Pentru o permutare $p: \overline {1,n}\rightarrow \overline {1,n}$ notam $S_p =\sum \limits_{k=1}^n \frac{1}{x^2_k+x_{p(k)}}$. Fie transpozitia $t=(i,j)$.
Se arata usor ca $\left [S_{p\cdot t}-S_p \right] .a.s.[(x_j -x_i)(x_{p(i)} -x_{p(j)})].a.s.[(j-i)[p(i)-p(j)]$.

Daca permutarea $p$ are cel putin o inversiune, fie aceasta perechea $(i,j)$, atunci $S_{p\cdot t}\le S_p$,

ceea ce inseamna ca $\max\limits_{p} S_p =S_e$, unde $e$ este permutarea identica. In concluzie, $\left( \forall \right)\ p,\ S_p\le S_e$, adica
\[ \underline {\overline {\left| \bigstar\left( \forall \right)\ p,\ \sum\limits_{k=1}^n \frac{1}{x^2_k +x_{p(k)}}\le \sum\limits_{k=1}^n \frac{1}{x^2_k +x_k}\ \bigstar\right| }}.  \] [u]Generalizare[/u] (imediata). Fie functiile strict crescatoare (sau strict descrescatoare) $f,g\ : \ R^*_+\longrightarrow R^*_+$.
Atunci exista inegalitatea:\[ \underline {\overline {\left| \bigstar\left( \forall \right)\ p,\ \sum \limits_{k=1}^n \frac{1}{f(x_k)+g(x_{p(k)})}\le \sum\limits_{k=1}^n \frac{1}{f(x_k)+g(x_k)}\ \bigstar \right| }}. \] [u]Recomandare.[/u] Aceasta generalizare mi se pare ridicola. Nu incercati sa "saturati" (prin generalizare extrema) inegalitatea propusa. Va sfatuiesc sa renuntati deoarece nu va va oferi vreo satisfactie deosebita. Bucurati-va de inegalitatea propusa care contine in sine orice dezvoltare ulterioara precum un pui de caprioara: are nevoie doar de protectie.


I'm not sure if this is the right topic to post this, but I need help!  What is the sum of the coefficients of (x-2)^2007?  Is there some cool formula for this? :D

[hide="Hint"] You're thinking too hard.  Look at small examples. [/hide]
[hide="Spoiler"] $ (1-2)^{2007}$. [/hide]

Try changing the 2007 to some smaller  number (1, 2, etc.) and you'll see a pattern very quickly.

[quote="Math Geek"]Huh?  I replaced it with some small numbers, but I still don't understand.  The question asks for the coefficient when x is an unknown variable (this might be a really dumb thing to say).  I'm sorry for my slowness, but I still don't get it![/quote]

[hide="Well,"]
What does the value come out to be each time, when you add up the coefficients? When $ n$ is even? When $ n$ is odd?[/hide]
[hide="Explanation"]
What are you doing when you add up the coefficients of a polynomial, but simply setting $ x=1$ and adding?

So, we can take the polynomial $ (x-2)^{2007}$, and set $ x=1$, to obtain $ (-1)^{2007}= \boxed{-1}$.[/hide]

Oohhhhhhhh....  Darn, I got that question wrong, then.  Oh well, thanks!!   :lol:


geometry
rectangle
analytic geometry
graphing lines
slope
vector
AMC

In rectangle $ ABCD$, we have $ A \equal{} (6, \minus{} 22)$, $ B \equal{} (2006,178)$, and $ D \equal{} (8,y)$, for some integer $ y$. What is the area of rectangle $ ABCD$?

$ \textbf{(A) } 4000 \qquad \textbf{(B) } 4040 \qquad \textbf{(C) } 4400 \qquad \textbf{(D) } 40,000 \qquad \textbf{(E) } 40,400$

[hide="Better yet"]Translate everything by the vector $<-6,22>$, and we have the three points $A(0,0)$, $B(2000,200)$ and $D(2,z)$, where $z=y+22$. Since $\overline{AB}$ and $\overline{AD}$ are perpendicular, their slopes will be inverse reciprocals of each other. So $\frac{200}{2000}=-\frac{2}{z}\implies z=-20$. By the distance formula, $AB=\sqrt{2000^2+200^2}=200\sqrt{101}$, and $AD=\sqrt{20^2+2^2}=2\sqrt{101}$, so $[ABCD]=AB\cdot AD=400\cdot101=40400$, $\text{(E)}$.[/hide]


What is the smallest integer (in degrees) angle, which can be constructed with straightedge and compass?

for gauss it can be constructed with straightedge and compass a poligon with N side if $ N \equal{} 2^{{k}_{0}}\cdot P_{1}^{{k}_{2}}\cdots P_{s}^{{k}_{s}}$ where $ k_{i}$ are non-negative integer and $ p_{j}$ Fermat's primes. $ \left ( F_{n}\equal{} 2^{{2}^{n}}\plus{}1\right )$
and so we can costruct a poligon with $ N \equal{} 2^{3}\cdot 3^{2}\cdot 5^{1}\equal{} 360$ and then a angle of 1 degrees


calculus
derivative
inequalities
algebra
polynomial
inequalities unsolved

Polynom from degree n > 1 (deg P(x)=n) has n different solutions $ e_1$,$ e_2$, ... , $ e_n$. The roots of his dericative are $ y_1$, $ y_2$, ... , $ y_n\minus{}1$. Prove the Inequality.

$ \frac{e_1^2\plus{}e_2^2\plus{}...\plus{}e_n^2}{n} > \frac{y_1^2\plus{}y_2^2\plus{}...\plus{}y_{n\minus{}1}^2}{n\minus{}1}$

P(x) = a_n*x^n+ a_(n-1)*x^(n-1)...
derivative = n*a_n*x^(n-1)...
sum of squares of roots for P(X) =(a_(n-1)/a_n)^2 - 2*a_(n-2)/a_n
sum of squares of roots for derivative =
(n-1)*a_n-1/(n*a_n))^2 - 2*a_(n-2)*(n-2)/(a_n*n)
let a = a_n
b = a_(n-1)
c = a_(n-2)

((b/a)^2 - 2*c/a)n > (((n-1)*b/(n*a))^2 -2*x*(n-2)/(a*n))/(n-1)
this simplifies to (n-1)*b^2 > 2*n*a*c

I don't know if that is provable
for n = 2 it is discriminant > 0

Wlog assume $ P(x)$ is monic, call $ P(x) \equal{} x^n \minus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} a_{n \minus{} 2}x^{n \minus{} 2} \plus{} ...$.  Using Cardano-Viete relations, the LHS is $ \frac {a_{n \minus{} 1}^2 \minus{} 2a_{n \minus{} 2}}{n}$.  Also, $ P'(x) \equal{} nx^{n \minus{} 1} \minus{} (n \minus{} 1)a_{n \minus{} 1}x^{n \minus{} 2} \plus{} (n \minus{} 2)a_{n \minus{} 2}x^{n \minus{} 3} \plus{} ...$, so using again Cardano-Viete relations, the RHS is $ \frac {(n \minus{} 1)^2a_{n \minus{} 1}^2 \minus{} 2(n \minus{} 2)a_{n \minus{} 2}}{n^2(n \minus{} 1)}$.  The given inequality is then equivalent to $ a_{n \minus{} 1}^2 \minus{} 2a_{n \minus{} 2} > 0$, which is trivially true since the LHS is the sum of the squares of the roots of $ P(x)$, and not all roots may be 0 since they are different.

Note however that I have assumed (the original statement didn't say so) that all roots are real.  Otherwise, the result is not necessarily true: consider e.g. polynomial $ P(x) \equal{} x^2 \plus{} 1$, and you will find LHS equal to $ \minus{} 1$, RHS equal to 0.

Note that the problem could have been stated with a $ \geq$ sign and without different roots for $ P(x)$, and asked for the case where equality holds.  Obviously, this case is $ P(x) \equal{} x^n$.


A steel pipe of $27$ feet long is being carried down a hallway $8$ feet wide. At the end of the hall there is a left-angled turn into another hallway.
[b]What is the minimum width of the hallway so that the pipe can be carried horizontally around the corner[/b]?

the longest ladder tha can be carried between 2 halls of width $a$ and $b$ is

$(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$

if you would like a proof just let me know.

That's correct ! :)

How about this:

The upper right-hand corner of a piece of paper, $12$ in. by $8$ in., as in the figure below, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold?


limit
induction
real analysis
real analysis solved

The sequencces ${a_n},{b_n}$,whose terms are all positive respectively, are  defined as follows.

\[a_1=b_1=1\ \ \ a_{n+1}-a_n=\frac{1}{a_n},\ b_{n+1}-b_{n}=\frac{1}{b_{n+1}}\ (n=1,2,\cdots)\]
Evaluate $\lim_{n\to\infty} \frac{b_n}{a_n}$

It is easy to prove by induction that
1) $\sqrt{2n}\leq a_n\leq\sqrt{2n}+\frac{1}{2}$;
2) $\sqrt{2n-4}-\frac{1}{2}\leq b_n\leq \sqrt{2n-4}$.

Thus $\lim\frac{b_n}{a_n}=1$.

[quote="kunny"]You are right. My solution is longer than yours. ;)[/quote]
Of course, because, actually, I didn't write solution, but only its main part ;)

I think we can make it with Cesaro Stolz.. If we denote by $L=\lim_{n\to\infty}\frac{a_n}{b_n}$ then $L=\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{b_{n+1}}{a_{n+1}}=\frac 1L$. So $L=1$

Hmm... At least, you use that limit exists, don't you?

Yes, I used that $\lim_{n\to\infty}\frac{a_n}{b_n}$ exists.


algebra
polynomial
modular arithmetic
arithmetic sequence
algebra proposed

Here's a cute quick problem. Sorry if you solve it for 30 seconds  :D :

Does there exist a polynomial $f \in \mathbb{Q}[x]$ with rational coefficients such that $f(1) \neq -1$, and $x^nf(x) + 1$ is a reducible polynomial for every $n \in \mathbb{N}$?

( :) ) Enjoy!  ;)

We shouldn't say about problems of Schinzel that they are solvable in 30 seconds. ;)

:D  Yes, but this problem is really quite simple  ;) . It is just relevant to some deeper results of Schinzel.  :)
Notice the similarity with the classical problem of Selfridge that there is a $k$ such that $k2^n + 1$ is composite for all $n$.

I must confess I am idiot. In fact, I knew that before, but now it's clear. I cannot find such a polynomial. And here is an amusing result: if we replace $ x^n f(x)+1$ with $ x^n+f(x)$ then such a monster is $(3x^9+8x^8+6x^7+9x^6+8x^4+3x^3+6x+5)/12$. Nice, isn't it?

Great!  :) . I knew it. In fact, I got it to, after posting this.

From what I can tell, the idea to solve this one is very similar to that used to solve the one mentioned by vess, concerning $k2^n+1$. In fact, there's another similarity between the two situations: Pierre (I think) posted, a long time ago, a problem of Sierpinski, I believe, stating that there's an odd $k$ s.t. $2^n+k$ is composite for all $n$.

If we find distinct $a_1,\ldots,a_t$ s.t. there are $b_1,\ldots,b_t$ with the property that the progressions $(a_ix+b_i)$ cover all the integers, then we can freely choose the residues our polynomial takes when divided by $\phi_{a_i}$ (the cyclotomic polynomials), s.t. $f(x)+x^n$ (or $x^nf(x)+1$) is divisible by $\phi_{a_i}$ whenever $n\equiv b_i\pmod{a_i}$. Harazi found $(a_1,b_1),(a_2,b_2),(a_3,b_3),(a_4,b_4),(a_5,b_5)$ equal to $(2,1),(4,2),(3,2),(6,4),(12,0)$.

Yes. That's why I said I got the idea :)
Actually, I also tried it, but I stoped for three sequences, because I haven't general insight  :blush:

Exactly. It is a useful thing to know that there exist covering sequences with distinct moduli $>1$. Another such sequence (with the same moduli) is $0 \pmod{2}, 0 \pmod{3}, 1 \pmod{4}, 1 \pmod{6}, 11 \pmod{12}$. There also exists a covering system with moduli $\{2,3,4,8,12,24\}$, which was used in Erdos' proof of the existence of an infinite arithmetic progression of odd integers that cannot be represented as the sum of a prime and a power of $2$. Two unsolved problems about covering sets are: 1) for each $k$, is there a finite covering system with distinct moduli, each $>k$?; and 2) does there exist a covering system with odd distinct moduli $>1$?

Concerning this problem, it is not known whether there exists an [b]integer[/b] non-trivial $f$ such that $x^nf(x) + 1$ (resp. $f(x) + x^n$) is reducible for all $n$. It is shown by Schinzel  :)  that the existence of such polynomial would imply 2).


counting
distinguishability
geometry
3D geometry
tetrahedron
geometric transformation
rotation

How many distinguishable tetrahedral dice are there where each face has one of the numbers 1, 2, 3, or 4 on it? Two such dice are said to be indistinguishable if one can be obtained from the other by a rotation.

[hide="solution"]Let's count the number of distinct tetrahedra on one face.

WLOG, assume that 1 is on the bottom. Then 2, 3, and 4 are visible. There are 6 arrangements of those:

*minor note: Imagine that you are looking at the top of the tetrahedron*
[code]2 3
 4[/code]

[code]3 2
 4[/code]

[code]2 4
 3[/code]

[code]4 2
 3[/code]

[code]4 3
 2[/code]

[code]3 4
 2[/code]

But we have to account for rotations, as stated in the problem. So there are only two distinct arrangements:

[code]2 3
 4[/code]

and

[code]3 2
 4[/code].

Now we count the ones with a 2 on the bottom, taking rotations in mind.

[code]3 1
 4[/code]

[code]1 3
 4[/code]

But those can be rotated to get

[code]2 3
 4[/code]

[code]3 2
 4[/code]

With 1 on the bottom. We have already counted those.

The same argument holds for 3 and 4 on the bottom, so there are only 2 distinct tetrahedra.
[/hide]

[quote="i_like_pie"]Your solution is correct, but there is a much shorter one that does not require brute force. Anybody else?[/quote]
[hide]
$ 1$ can be next to $ 2,3,4$ (clockwise), or $ 2,4,3$ (clockwise; basically $ 4,3,2$ counterclockwise)
that's $ 2$ ways.
It's the same when you look at it from any other side, because you cant rearrange it so that $ 1$ can be next to those $ 2$ combinations in the same ways and be different tetrahedrons, so its $ 2$ ways...

By the way, does the orrientation of the number written down make a difference?
(like the bottom of the $ 1$ could be facing a vertex, or a side, and the side could be either $ 2$, $ 3$, or $ 4$, or the vertex could be either between $ 2$ and $ 3$, $ 3$ and $ 4$, or $ 4$ and $ 2$)
If it does make a difference, then I think there's $ 6^4$ times as many ways, or $ 2592$ ways in all.
[/hide]

The orientation of the number doesn't matter, and your solution is closer to the optimal one.

[hide]
The 1 and 2 will always be next to each other because every face of a tetrahedron touches every other.
Position the tetrahedron such that the 1 is on the left, and the 2 is on the right, and the 3 and 4 sides converge at the edge farthest from you.  Either the 3 is on top, or the 4 is on top, so there are 2 distinct tetrahedra.
[/hide]

That's the one I was looking for. :)

[hide="Short Version"]Numbers 1 and 2 will always be adjacent, and there are $ 2! \equal{} 2$ permutations of the other two.[/hide]


[quote="fesha3t"]Please help me(hard)
let $ a,b,c\geq 0$
Prove that
a) Easy $ \sum \sqrt {\frac {a}{4a \plus{} b \plus{} c}} \leq \sqrt {1.5}$
b) hard $ 0.5 \geq \sum \sqrt {\frac {a}{4a \plus{} b \plus{} c}}$[/quote]
Are you sure for the second???
Take $ a\equal{}b\equal{}c\equal{}1$ :wink:

[quote="fesha3t"]Please help me(hard)
let $ a,b,c\geq 0$
Prove that
a) Easy $ \sum \sqrt {\frac {a}{4a \plus{} b \plus{} c}} \leq \sqrt {1.5}$
b) hard $ 0.5 \geq \sum \sqrt {\frac {a}{4a \plus{} b \plus{} c}}$[/quote]
a) $ \sum_{cyc} \sqrt {\frac {a}{4a \plus{} b \plus{} c}} \leq \sqrt {1.5} \equal{} \frac {\sqrt {6}}{2}$
Assume wlog $ a \plus{} b \plus{} c \equal{} 3$. $ \sqrt {\frac {a}{4a \plus{} b \plus{} c}} \equal{} \sqrt {\frac {a}{3(a \plus{} 1)}} \le \frac {a \plus{} 3}{4\sqrt {6}} \iff (a \minus{} 1)^2(3a \plus{} 27) \ge 0$
So $ \sum_{cyc} \sqrt {\frac {a}{4a \plus{} b \plus{} c}} \leq \sum_{cyc} \frac {a \plus{} 3}{4\sqrt {6}} \equal{} \frac {12}{4\sqrt {6}} \equal{} \frac {\sqrt {6}}{2}$ :)

[quote="Mathias_DK"][quote="FelixD"]Did you mean $ \frac {1}{2} \le \sum_{cyc} \sqrt {\frac {a}{4a \plus{} b \plus{} c}}$?
Then assume again $ a \plus{} b \plus{} c \equal{} 3$ and use $ \sqrt {\frac {a}{3a \plus{} 3}} \ge \frac {a}{6}$.[/quote]
Nice !  :D
Tried to solve it, but it never occured that we could use the same method :P[/quote]

= When
$ a\equal{}b\equal{}0, c\in R$
but i don't know