Let a,b,c three integers in Z*, with a != c so that :
$\frac {a}{c} = \frac {a^2+b^2}{c^2+b^2}$
Show that :
$a^2+b^2+c^2$ isn't prime.
:)

Take the first equation:
$ac^2+ab^2=a^2c+b^2c$
$ab^2-cb^2=a^2c-c^2a$
$(a-c)b^2=(a-c)ac$
$b^2=ac$
So then we get
$a^2+b^2+c^2=a^2+ac+c^2=(a+c)^2-ac=(a+c)^2-b^2=(a+c-b)(a+c+b)$.  Therefore, it has at least those two factors (except in the special case where those two are equal to $a^2+b^2+c^2$ and $1$.  I don't think you could have that...but if you could, then we could solve that  :)  )

EDIT: Whoops, i made a typo :D thanks for catching it, mathmanman.

Thanks a lot :)
btw, on your fourth line it should be b^2  :P but, that was just a typo ;)

What about a=b=c=1.

This satisfies the conditions, and a^2 + b^2 + c^2 = 3,

which is prime.

[quote="paladin8"]I just noticed you could interpret $a!=c$ as either $a \neq c$ or $(a!) = c$. Haha. :P[/quote]
I don't think so because he wrote a !=c  not  $a!=c$. :?

a=b=c=1 is the case where $(a+b-c)=1$ and $(a+b+c)=a^2+b^2+c^2$.  But then yeah, it doesn't satisfy the original $a%Error. "neqc" is a bad command.
$.  I interpreted it as $a%Error. "neqc" is a bad command.
$ because he put a != c instead of a! = c.

Yeah sorry. I did interpret it as a factorial.

Heh, ok, sorry for that  :?
I couldn't remember the latex expression for "!=", but now, thanks to deej21, I know it's : $\ne$  (\ne)
 :)

ps: that"s understandable since : ne -> non equal hehe.

[quote="paladin8"]Yeah, I wasn't really being serious... the thought just occured to me. In this case it's pretty easy to determine what is really meant. BTW, nice solution, K81o7.[/quote]

Thanks.  Yeah i realized you were joking, but when i actually looked back at it and thought about it, i realized that it could be interpreted that way after all!  :lol:
What if we try to solve the equality with $a!=c$?  That would be more difficult...any solutions?  :D

Yeah i actualy didn't know the latex for $\neq$ before now, so when i don't know something, i just avoid try to post it...instead would write "a isn't equal to c"   :D

:rotfl: Yeah, next time, I'll do the same :P
And, indeed, it would be cool to solve it with this condition :)
Although
[quote="Gyan"][quote="paladin8"]I just noticed you could interpret $a!=c$ as either $a \neq c$ or $(a!) = c$. Haha. :P[/quote]
 :lol: b=3 and you can have a=c=1 or a=c=2! :lol:[/quote]

Well there are lots of counterexamples so we can't prove the same result...

Well for $c!=a$, for $a \neq 1,2$, this implies that $a \neq c$


Ai có cuốn này không, share cho mình với!
Recent Advances in Geometric Inequalities (Mathematics and its Applications) (Hardcover)
by Dragoslav S. Mitrinovic  , J. Pecaric , V. Volenec

[color=purple] Chu' ma`y thich' BDt hinh` a` ? Anh ma`y ti`m 2 nam nay roai` em a .Chi ko' ban thuong mai moi' dau :( [/color]

Minh cung dang tim quyen nay nhung tim mai ma khong thay. Cac thay va cac ban giup voi.

[quote="evarist"][color=purple] Nó ở đây này  :D [url]http://mathvn.vn2k.net/forum/viewthread.php?forum_id=11&thread_id=191[/url]
   Come there to solve all problem and get it :D [/color][/quote]
Bạn ơi sao tôi ko vào được vậy nhỉ?

Đây nhé anh em!
http://www.readbook5.com/book/search.php?req=Recent%20Advances%20in%20Geometric%20Inequalities&nametype=orig


Δείτε το αρχείο.

              Μπάμπης

Ίδια ιδέα με εδώ http://www.mathlinks.ro/viewtopic.php?t=232147

Η πιο συντομη λυση που μπορεσα να σκευτω:
$ a \plus{} b \equal{} \frac{1}{x \plus{} 1}, b \plus{} c \equal{} \frac{1}{y \plus{} 1}, c \plus{} a \equal{} \frac{1}{z \plus{} 1}$

τοτε: $ a \plus{} b \plus{} c \equal{} 1$, και $ x \equal{} \frac{1 \minus{} a \minus{} b}{a \plus{} b} \equal{} \frac{a \plus{} b \plus{} c \minus{} a \minus{} b}{a \plus{} b} \equal{} \frac{c}{a \plus{} b}$
και ομοιως: $ y \equal{} \frac{a}{b \plus{} c}, z \equal{} \frac{b}{c \plus{} a}$

Οποτε η ανισοτητα γραφεται:
$ 8\frac{abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \leq 1 \Leftrightarrow (a \plus{} b)(b \plus{} c)(c \plus{} a) \geq 8abc$
η οποια ισχυει, αφου προκειπτει απο κυκλικο πολλαπλασιασμο της ανισοτητας: $ a \plus{} b \geq 2\sqrt{ab}$


let $ g$ is a function : $ g: \mathbb{R}\equal{}\equal{}>\mathbb{R}$
$ x\equal{}\equal{}>g(x)$ such that  $ (\forall x\in \mathbb{R}) g\circ g(x)\equal{}\minus{}x$
$ 1$ . prove that $ g$ is $ bijective$.
$ 2$ . find $ g(0).$

1. - is obviosly.
2. Let $ x\equal{}g(0)$, then $ g(x)\equal{}0$ and $ \minus{}x\equal{}g(g(x))\equal{}g(g(g(0)))\equal{}g(\minus{}0)\equal{}g(0)\equal{}x\to x\equal{}0$.

The function is odd : $ g(g(g(x))) \equal{} g(\minus{}x) \Longrightarrow \minus{} g(x) \equal{} g( \minus{} x)$ for all $ x$ ..


geometry
trigonometry
geometric transformation
dilation
trig identities
Law of Sines

Please help.  :maybe:

(from AoPS vol. 2)

Problem:
In triangle $ ABC$ with sides $ a,b,c$, we have $ a\ge b\ge c$ and $ \frac {a^3 \plus{} b^3 \plus{} c^3}{\sin^3 A \plus{} \sin^3 B \plus{} \sin^3 C} \equal{} 7$. Find the maximum value of $ a$.

Solution:
Extended law of sines:
$ \displaystyle 7 \equal{} \frac {a^3 \plus{} b^3 \plus{} c^3}{\sin^3 A \plus{} \sin^3 B \plus{} \sin^3 C} \\
\equal{} \frac {a^3 \plus{} b^3 \plus{} c^3}{\frac {a^3 \plus{} b^3 \plus{} c^3}{8R^3}} \equal{} 8R^3 \Rightarrow R \equal{} \sqrt [3]{7} / 2 \Rightarrow a$ is a chord of a circle of radius $ \sqrt [3]{7} / 2 \Rightarrow a$ is at most the length of the diameter, or $ \boxed{\sqrt [3]{7}}$

Question of validity:
How does one know another restriction does not exist to further lower the possible length of $ a$? How does one know that such a triangle even exists?

Consider a right triangle, any right trinalge.

Now, Dilate it so that it satisfies the equation ( this is possible becuase a^3 is continuoes), and we get what we want.

The so the given solution was missing a little something?

Nevermind, everything can work backards... so it was missing the statement that for a right triangle with that R, everything will work backwards and it will all work out.


limit
floor function
trigonometry
integration
calculus
calculus computations

Let $S_{n}$ be the sum of lengths of all the sides and all diagonals of a regular $n$-gon inscribed in a circle with radius 1. Find $\lim_{n\to\infty}\frac{S_{n}}{n^{2}}.$

If $n$ is odd, $S_{n}=\sum_{k=1}^{\lfloor n/2\rfloor}2n \sin\left(\frac{k\pi}{n}\right).$

If $n$ is even, the last term of the sum needs modification, but once we divide by $n^{2}$ and take the limit, the effect of that modification will go to zero. We finish by recognizing what we have as a Riemann sum.

$\lim_{n\to\infty}\frac1{n^{2}}\sum_{k=1}^{\lfloor n/2\rfloor}2n \sin\left(\frac{k\pi}{n}\right)$

$=\lim_{n\to\infty}\frac{2}{\pi}\sum_{k=1}^{\lfloor n/2\rfloor}\sin\left(\frac{k\pi}{n}\right)\left(\frac{\pi}{n}\right)$

$=\frac2{\pi}\int_{0}^{\frac{\pi}2}\sin x\,dx=\frac{2}{\pi}.$

On second thought: count each diagonal twice, once from each end, then divide the whole thing by two. That way, we don't need special handling for $n$ even. The original sum sets up as

$\lim_{n\to\infty}\frac1{n^{2}}\cdot\frac12\sum_{k=1}^{n-1}2n \sin\left(\frac{k\pi}{n}\right)$

Leading to $\frac1{\pi}\int_{0}^{\pi}\sin x\,dx=\frac{2}{\pi}.$


calculus
integration
function
trigonometry
calculus computations

Let $0<\alpha <\frac{\pi}{2}.$ The function $F$ is defined by $F(\theta )= \int_0^{\theta } x\cos (x+\alpha ) dx.$

When $\theta$ moves in the range of $\left[0,\ \frac{\pi}{2}\right],$ find the maximum value of $F$.

$F'(\theta) = \theta \cdot \cos (\theta + \alpha) > 0 \Leftrightarrow \theta + \alpha \in (0,\pi/2)$
F is increasing on $(0,\frac{\pi}{2} - \alpha)$
F is decreasing on $(\frac{\pi}{2} - \alpha, \frac{\pi}{2})$
$F_{max} = F(\frac{\pi}{2} - \alpha) = \frac{\pi}{2} - \alpha - \cos(\alpha)$


In triangle $ ABC$, let the excircle with respect to $ A$ touch $ BC$ at $ D$, and let $ I$ be the incenter. Let $ E$ be the foot of the altitude from $ A$ to $ BC$. Prove that $ DI$ bisects $ AE$.

Let AI intersects BC at F. As D is the point of excircle on BC, CD = p-b, a, b, c and p being the sides and semiperimeter of the triangle ABC. After some easy comptutations we get DE=p(b-c)/a and FD=p(b-c)/(b+c). With AI/IF=(b+c)/a, applying Menelaos in triangle AEF with the transversal DI, we get DI passing through the middle of AE.

Best regards,
sunken rock

Dear Mathlinkers,
Let 1 the incircle of ABC, U the contact point of 1 with BC, V the antipode of U wrt 1.
UV is parallel to AE
The points A, V and D are collinear (see Coexter, Geometry revisited)
I is the midpoint of UV
then, DI bisect AE.
Sincerely
Jean-Louis

Here is my solution for this problem
[b]Solution[/b]
Let $F$ be orthogonal projection of $I$ on $BC$, $F'$ be reflection of $F$ through $I$
It's easy to that: $AF'$ passes through $D$
But: $FF'$ $\parallel$ $AE$ and $I$ is midpoint of $FF'$ so: $DI$ passes through midpoint of $AE$


Bueno, por ultimo pongo el problema 6:

Sea n la suma de todos los digitos de un entero positivo A. Decimos que A es "surtido" si cada uno de los enteros del 1, 2,..., n es suma de algunos digitos de A.
1) Demuestra que si 1, 2,..., 8 son sumas de algunos digitos de A entonces A es surtido.
2) Si 1, 2,..., 7 son sumas de algunos digitos de A, ¿es A necesariamente surtido?

Nota: El numero 117 no es surtido pues solo 1=1, 7=7, 1+1=2, 1+7=8, 1+1+7=9 se pueden escribir como suma de algunos digitos de A.

[quote="R.G.A.M."][color=blue]Tomando en cuenta el número $0$… ¿Influye el mismo en que un número sea surtido o no? ¿O lo ignoramos? ¿Es determinante que algunos de los dígitos sumen $0$, o no importa?[/color][/quote]
por la condicion del problema no influye que tengamos $0's$, ya que las sumas van desde $1$ hasta $n$.
chau, saludos, pronto posteo la solucion.

b) Creo que la parte b del problema es bastante más fácil que la a. Como con 1,2,4 se pueden escribir todos los numeros hasta el 7 (sistema binario) si ponemos un $9$ el $8$ no se puede escribir. Entonces 9421 cumple la condicion de todas las sumas hasta el 7 pero no se puede lograr el 8.

a) El argumento del anterior no funciona porque no hay digitos mayores que 9 en base 10. Usaremos inducción. Si la suma de todos los digitos de $A$ es 8 todas las sumas estan y ese caso es trivial. De lo contrario hay que garantizar primero que $9$ se puede escribir pues $n\ge 9$. Sean $a_{1}, a_{2},.. a_{9}$ la cantidad de $1,2,3,...,9$ en $A$ respectivamente. Si $a_{9}\ge 1$ ya esta pues tomamos un $9$ y ya. De lo contrario sea $i$ el menor entero tal que la representación de $8$ como suma de digitos no tiene $a_{i}$ digitos $i$. Si $i=1$ ya esta porque se lo agregamos al 8 y queda el 9.  Si $i >1$ todas las representaciones de los números de $1$ hasta $i-1$ se tienen que hacer con los digitos de la representación de $8$ pues no pueden tener una representación con numeros mayores que ellos mismos. Entonces hay un subconjunto de los digitos que suman $8$, cuya suma es $i-1$. Cambiamos estos numeros por $i$ y la suma es $9$.
Ya teniendo que todos los digitos se pueden representar, cambiemos en la demostración de que $9$ se puede el $8$ por una $m$ con $9\le m \le n-1$. El mismo argumento funciona pues todos los digitos se pueden representar.
QED


[quote="paganinio"]If I have 50 post count, just display it like this:
Posts: 50=2*5^2
or like this:
Posts: 50=$2\times5^{2}$[/quote]Yes woulnd't that add PGTimes :) Other than that it's a nice idea. We'll see what we can do about it. :P

We had discussed expressing a number of site stats in the form of prime factorizations some time ago and decided that while the idea was neat, it would unfortunately make numbers harder to read.  People like their stats and while they might enjoy the first few prime factorizations, it's going to get real old real quick.

Let's just stick with numbers. Simple and sweet...

1, I don't think it will add PG time, because 50=$2\times5^{2}$ consists of 2 parts: the original number part(50=) and the LaTeX part($2\times5^{2}$), the number part will be loaded first. Even if you don't see the LaTeX part it doesn't matter.
2, It will not be difficult to read, the reason is the same.

[quote="Valentin Vornicu"]Sorry but this is really really not on the priority list right now :) We'll give it a try later in the summer, ok?[/quote]

Actually, we won't make any promises at all.  I personally don't like the idea at all.  It's clever and fun, but people want to read their stats.  Let's consider the subject closed.

Actually I was thinking more in the lines of implementing an option in the options or profile menu from which an user could trigger this on and off. However as I said and Mathew reinforced, this really won't be done any time soon. :)

[quote="Treething"]Posts: 2:^5: x 3:^2: x 17

I really don't think that's such a great idea :P .[/quote]

But it would be

Posts: $4896=2^5\cdot 3^2\cdot 17$

Which is immediately understandable.

Note that for most numbers, there will not be many prime factors. If I recall correctly, the normal order of the number prime factors of $n$ is $\log\log n$.

[quote="Treething"]Posts: 2:^5: x 3:^2: x 17
[/quote]

Posts: 4896=2:^5: x 3:^2: x 17

That's it!
We don't need Latex here at all!

Since Admin said this is really really not on the priority list right now, let's just forget it for the moment :lol:


AoPSwiki
articles
modular arithmetic
number theory
relatively prime

[b]Bezout Theorem[/b]

If $ a,\ b$ are positive integers, then there exists integers $ u,\ v$ such that $ au\plus{}bv\equal{}GCD(a,\ b)$.

[quote="PowerOfPi"]I was looking for theorems and formulas related to divisibility and modular arithmetic, more specifically, including powers. I found two, the Chinese Remainder Theorem and Wilson's Theorem. Could anyone provide a name or explanation for these kinds of theorems?[/quote]

Fermat's Little Theorem and Euler's Theorem...the AoPSWiki articles explain much better than I can.

There's the divisibility rules...don't know if you know them.

1: always divisible
2: last digit divisible by 2
3: sum of digits divisible by 3
4: last two digits divisible by 4
5. last digit divisible by 5
6: divisible by both 2 and 3
7: multiply the last digit by 2, subtract it from the rest of the number. You can do this multiple times if the number is too big. If you end up with a number divisible by 7, so is the first number. For example, let's try 14: 4x2=8, 1-8=-7, which is divisible by 7, so 14 is also.
8. Last 3 digits divisible by 8.
9. Sum of digits divisible by 9.
10. Last digit is 0.
11. Subtract all of the even digits (relative to the first digit) and add the odd digits. If the result is divisible by 11, so is the number you started with. For example, in 12345, you do 1-2+3-4+5, which is 3, and not divisible by 11, so neither is 12345.
12. Divisible by 3 and 4
13. Same as 7, except you multiply by 9.
14. Divisible by 2 and 7.
15. Divisible by 3 and 5.
16. last 4 digits divisible by 16. You can do this type of thing for all powers of 2; 2^n|x if the last n digits of x are divisible by 2^n.
17. Same as 7, but multiply by 5.
18. Divisible by 2 and 9.
19. Same as 7, but multiply by 17 (or multiply by 2 and add)
20: divisible by 4 and 5
21: Divisible by 3 and 7
22: divisible by 2 and 11
23: Multiply the last digit by 7, add to the rest of the number. ADD, don't subtract.
24. Divisible by 3 and 8.
25. Last two digits divisible by 25. You can use the same rules as numbers of the form 2^n for those of the form 5^n.
26. Divisible by 2 and 13.
27. Hard to do this one. Group the number into 3-digit intervals, and add them together; repeat as necessary. If the result is divisible by 27, so is the first number. For example, let's try 123456. We group it as 123, 456; 123+456=579, which is not divisible, so neither is 27.
28. divisible by 4 and 7.
29. 3 times the last digit plus the rest of the number.
30. Divisible by 5 and 6 (or 2, 3, and 5).

I'll stop there.

Well there USED to be a page on Wikipedia that had Raina's Little Theorem, which says that $ a^n \equiv 1 \pmod{a \minus{} 1}$ for all non-negative integers $ a$ and $ n$ but someone deleted it :mad:

FLT is probably the most useful (?) and it says that if $ p$ is a prime number, then:
$ a^p \equiv a \pmod p$
for all integers $ a$

And then theres a variation of it which says that

$ a^{p \minus{} 1} \equiv 1 \pmod p$

And obviously for this variation $ a$ and $ p$ have to be relatively prime, because if they are not then $ a^{p \minus{} 1} \equiv 0 \pmod p$


AMC
AMC 12
AIME
USAMTS
number theory
modular arithmetic

When did this website begin? I wish I had known about it earlier because I have learned so much since joining this site. One year ago, I could barely do some of the easiest AMC12 questions and I didn't even know what easy stuff like modular arithmetic was-- no I am not kidding. Now, I can even handle some of the easy/intermediate AIME problems and I've been able to do USAMTS and stuff. Also, I never knew that there were so many people in this world who think exactly like me. Honestly, sometimes I see a message and I know that I could have been the one writing it. At school, this is never the case. I can't believe that I didn't know-- well, I still don't really "know" -- the people here at AoPS for the past decade and a half of my life. I love AoPS!!!

The earliest AoPSers signed up in late April or May, so that's around when it "started", although it was first set up in March or so, I believe.  Over the summer, AoPS started to expand more, and by the time school had started, had built up quite a member list.  Classes began soon after that, and even more people came along.

The site was released publicly near the end of May, 2003.

Glad we've been able to help, and we hope to release plenty more useful materials in the next few years.

AOPS sent me a letter in May. But I went to China during the summer. So I joined after I came back. But I had often visited it. So it should say May for me.


A,C are nxn matrices with real entries, C is any invertible matrix.

Show that if $ C^{ \minus{} 1}AC \equal{} A$ for all invertible matrices C, then A is a scalar multiple of the identity.

Since $ A$ is similar to its transpose, let $ S$ be an invertible matrix such that $ A \equal{} S^{ \minus{} 1}A^TS$. Now,

$ A \equal{} C^{ \minus{} 1}AC \equal{} C^{ \minus{} 1} (S^{ \minus{} 1}A^TS) C \equal{} (SC)^{ \minus{} 1} A^T (SC)$.

So, if we choose $ C \equal{} S^{ \minus{} 1}$, we get $ A \equal{} A^T$, and $ A$ is a symmetric matrix. Therefore, there exists an invertible matrix $ Q$ and a diagonal matrix $ D$ such that $ A \equal{} Q^{ \minus{} 1}DQ$. Again,

$ A \equal{} C^{ \minus{} 1}AC \equal{} C^{ \minus{} 1} (Q^{ \minus{} 1}DQ) C \equal{} (QC)^{ \minus{} 1} D (QC)$,

and choosing $ C \equal{} Q^{ \minus{} 1}$, we may conclude that $ A \equal{} D$: $ A$ is a diagonal matrix. Finally, if we choose for $ C$ the permutation matrix $ P_{ij}$ [obtained from I by interchanging rows (or columns) i and j], then $ CA \equal{} AC$ implies that $ a_{ii} \equal{} a_{jj}$, and so all diagonal elements of $ A$ are equal. This, together with the conclusion that $ A$ is diagonal, shows that $ A \equal{} k I, k \in \mathbb{R}$. $ \Diamond$

This argument, however, looks too lenghty, and there is certainly a simpler proof.

Why is A similar to its transpose? Isn't the problem statement that A is similar to itself?

The statement could be rephrased as $ AC \equal{} CA$ for all invertible $ C,$ or $ A$ commutes with every invertible matrix.

It's pretty easy to prove that if $ A$ commutes with all matrices, then it is a scalar multiple of the identity. But what makes that proof easy is that you get to use matrices with one nonzero element - which are very much not invertible.

But we can get back to that. Let $ E_{jk}$ be the matrix that has a $ 1$ in the $ (j,k)$ position and zeros elsewhere. (We're assuming that $ j\ne k.$) $ E_{jk}$ is not invertible, but $ I \plus{} E_{j,k}$ is. Then $ (I \plus{} E_{jk})A \equal{} A \plus{} E_{jk}A \equal{} A(I \plus{} E_{jk}) \equal{} A \plus{} AE_{jk}.$ So by subtraction, we have that $ E_{jk}A \equal{} AE_{jk}.$

If we multiply that out and equate it, we find that $ a_{jk} \equal{} a_{kj} \equal{} 0$ and $ a_{jj} \equal{} a_{kk}.$ Do that for all $ j$ and $ k,$ and we find that $ A$ is zero off the main diagonal and the entries on the main diagonal are all equal.

[quote]Why is A similar to its transpose?[/quote]

Because every matrix is similar to its transpose. Look, for instance, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317900]here[/url].


John travels to work at 45 mph. On his way back from work he goes 36 mph on a road that is 3 miles shorter. He took 5 minutes more. How long is the road on the way to work? Please show your work.

[hide]Please PM me the answer and post the answer here[/hide]

[hide]Set up a system of equations using $ d \equal{} rt$.

$ d \equal{} 45t$
$ d \minus{} 3 \equal{} 36(t \plus{} \frac {1}{12})$

Substituting, $ 45t \minus{} 3 \equal{} 36t \plus{} 3 \implies 9t \equal{} 6 \implies t \equal{} \frac {2}{3}$.

Plugging this in the first equation,
$ d \equal{} 45\times{\frac {2}{3}} \equal{} \boxed{30}$.[/hide]


For positive real numbers $a,b,c$ satisfying $a+b+c=1,$

Prove that we have $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leqq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).$ Note that you don't need to state for the condition
for which the equality holds.

This one has already been posted in the forum:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=japanese+inequality&t=25780

I see. Could you please delete my post, moderator.

Clearing the denominator, we get

$6a^2b^2c^2+6S(3,2,1) \leq 2(a^4b^2+b^4c^2+c^4a^2)+6S(3,3,0)$

From AM-GM inequality,
$6a^2b^2c^2 \leq 2(a^4b^2+b^4c^2+c^4a^2)$

From Muirhead inequality,
$S(3,2,1) \leq S(3,3,0)$

So the proof is completed. :)

Can this problem be solved without using Muirhead?

$\frac{1+a}{1-a} = \frac{b+c+2a}{b+c} = 1+\frac{2a}{b+c}$ .
Now we need to prove :
$$\frac{3}{2} \leq \sum \frac{bc}{a(c + a)}$$
Now we can suppose $abc=1$ . So we have $p,q,r$ such as $p =\frac{1}{a} ,....$

Now we need to prove :
$$\frac{3}{2}\leq \sum \frac{p^2}{q(r + p)}$$ ( :first: )
( :first: ) is easy by $T-2$ lemma

[quote=Kunihiko_Chikaya]For positive real numbers $a,b,c$ satisfying $a+b+c=1,$

Prove that we have $$\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leqq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).$$ Note that you don't need to state for the condition
for which the equality holds.[/quote]
[url]https://artofproblemsolving.com/community/c6h25780p161786[/url]
[url]https://artofproblemsolving.com/community/c6h2045685p14495540[/url]

[b]Generalization 1[/b]
Let $a,b,c,k$ be positive reals such that $a+b+c=\lambda$ ise

$$\dfrac{\lambda k+a}{\lambda-a}+\dfrac{\lambda k+b}{\lambda-b}+\dfrac{\lambda k+c}{\lambda-c}+3k-3\leq \left(k+1\right)\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)$$


pigeonhole principle
email
summer program
MathPath
AwesomeMath

[quote="mdk"]Given any 9 integers whose prime factors lie in the set {3,7,11} prove that there must be two whose product is a
square.[/quote]

[hide]There are only 7 different such integers so there has to be 2 pairs that are the same whose product will be a square.[/hide]

Johnny, there's more than seven different such integers. Every single number of the form $ 3^{k}$, $ k$ being an integer, is in that set.

That being said, we use pigeonhole and mod two. We can express every number in this set in the form $ 3^{j}\cdot 7^{k}\cdot 11^{l}$. If j, k, and l are the same mod 2 for two different integers in the set, then the product will be a square. However, there are only eight different possibilities mod 2 for j, k, and l, so we can pigeonhole the ninth integer in.

oh yeah... forgot about that.

how do u know my name :maybe:


Determine a vector equation for the following line:
With the same x-int as [x,y,z]=[3,0,0]+t[4,-4,1]
and the same z-intercept as [x,y,z]=[6,-2,-3]+t[3,-1,-2]

*****if it helps the answer is [x,y,z]=[0,0,-3]+t[-3,0,-3]*****I just don't know how to get it!!!!

Well, you've been given two points on the line (though those points are described in a rather convoluted way).  So, probably you know how to find the vector equation once you've found the two points explicitly (and if you don't, you should reread that section of your textbook) and so all you need to do is figure out the explicit equations of those two points.  One of them is very easy ;)

(A side note: I'm assuming here that "$ x$-intercept" means "place where the line intersects the $ x$-axis."  This is not a very useful notion in three dimensions, since most lines don't intersect the $ x$-axis at all and are instead skew to it, and something special happens here because someone constructed the problem in order to make it happen.  This is quite different than in the case of two dimensions.  What happens in three dimensions that's similar to what happens in two dimensions is that most lines intersect the $ xy$-, $ yz$- and $ xz$-[i]planes[/i], and so the "$ yz$-intercept" (for example) is a meaningful idea.)

Edit: the answer you've given is not right, at least if my interpretation of "$ x$-intercept" is the intended one.


It's equivalent to
\[ 0\leq (a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\left( \frac{a\plus{}b\plus{}c}{2abc}\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2}\minus{}\frac{1}{2(ab\plus{}bc\plus{}ca)}\right)\]
which is true, since
\[ \frac{a\plus{}b\plus{}c}{abc}\geq \frac{9}{ab\plus{}bc\plus{}ca}\geq \frac{9}{a^2\plus{}b^2\plus{}c^2}\]


function
algebra
functional equation
algebra unsolved

Find all continous functions $f: \mathbb R \to \mathbb R$ which satisfy
\[ f(x\plus{}y)\equal{}f(x\plus{}f(y))\]
for all real numbers $x$ and $y$.

[quote="Algadin"]Find all continous function $ f: R\to R$ which satisfy:
$ f(x \plus{} y) \equal{} f(x \plus{} f(y)) (1)$[/quote]

Since (1), put x = 0 gives $ f(f(y)) \equal{} f(y)$ for all real number y, so $ f(x) \equal{} x$ for all x in the value set of f
Let $ a$ and $ b$, respectively, be the minimum and maximum of $ f$ ($ a$ and $ b$ might be infinitive)
- If $ a \equal{} b$, then f is constant.
- If $ a \neq b$:
  + If $ a$ is finite, then $ a \equal{} lim_{x \rightarrow a} f(x) \equal{} f(lim_{x \rightarrow a} x) \equal{} f(a)$
So $ f(a) \equal{} a$, and since $ f$ is continuous, so there exists a real number $ d$ such that
$ f(x) \equal{} x \forall x \in [a; a \plus{} 2d]$
Now we prove that $ f(a \minus{} s) \ge a \plus{} d \forall s \in (0;d]$
Indeed, suppose there exists $ s$ such that $ f(a \minus{} s) \equal{} a \plus{} q < a \plus{} d$, we have
$ a \equal{} f(a \minus{} s \plus{} s) \equal{} f(f(a \minus{} s) \plus{} s) \equal{} f(a \plus{} q \plus{} s) \equal{} a \plus{} q \plus{} s$ (since $ q \plus{} s < 2d$)
$ \rightarrow q \plus{} s \equal{} 0$, which is obviously wrong

So $ f(a \minus{} s) \ge a \plus{} d \forall x \in (0; d]$. But $ f(a) \equal{} a$, which contradicts the Intermediate Value Theorem. So our suppose is wrong and $ a$ is infinite. Similarity, $ b$ is infinite, which means $ f(x) \equal{} x$ for all real number x

Conclusion:
There are two functions satisfying the functional equation is $ f(x) \equal{} c$ and $ f(x) \equal{} x$

[quote="Algadin"]Find all continous function $ f: R\to R$ which satisfy:
$ f(x \plus{} y) \equal{} f(x \plus{} f(y))$[/quote]

Let $ g(x)\equal{}f(x)\minus{}x$. The equation becomes $ f(x)\equal{}f(x\plus{}g(y))$

If $ g(x)\equal{}c$ (constant), then $ f(x)\equal{}x\plus{}c$ and so, plugging back in original equation, $ c\equal{}0$ and $ f(x)\equal{}x$ which, indeed, is a solution.

If $ g(x)$ is non constant, and continuous, $ \exists [a,b]\subset g(\mathbb R)$ with $ b>a$. So $ f(x)\equal{}f(x\plus{}g(y))$ implies $ f(x)$ constant over $ [x\plus{}a,x\plus{}b]$ $ \forall x$ and so $ f(x)$ constant over $ \mathbb R$, which indeed is a solution.

Hence the two solutions :
$ f(x)\equal{}c$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$

Nice solution, pco. And here is another solution of my friend:

It's easy to check that $ f(x) \equal{} x$ and $ f(x) \equal{} c$ satisfying the equation
Suppose that $ f(x) \neq x$, then there exists a real number $ y_0$ such that $ f(y_0) \minus{} y_0 \equal{} k \neq 0$
And we have
$ f(x \plus{} y_0) \equal{} f(x \plus{} f(y_0) \rightarrow f(x) \equal{} f(x \plus{} k)$, implying $ f$ is periodic.
Since $ f$ is continuous and periodic, so there exists a real number $ T > 0$ be the smallest period of $ f$

Since $ f(x \plus{} y) \equal{} f(x \plus{} f(y)$, it implies that $ f$ is periodic, and $ f(y) \minus{} y$ is the period. So $ |f(y) \minus{} y| \equal{} T.h$, where $ h \in N$
Because $ f$ is continuous, so $ lim_{y \rightarrow y_0} [f(y) \minus{} y] \equal{} f(y_0) \minus{} y_0 \equal{} T.s$, implying there exists $ d > 0$ such that
$ |[f(y) \minus{} y] \minus{} [f(y_0) \minus{} y_0]| < T \forall y \in (y_0 \minus{} d; y_0 \plus{} d)$, implying $ f(y) \minus{} y \equal{} f(y_0) \minus{} y_0$
Finally, we have $ f(x) \equal{} x \plus{} c$. Putting back into the first equation implies $ c \equal{} 0$

Conclusion:
There are two functions satisfying the functional equation, namely $ f(x) \equal{} x$ and $ f(x) \equal{} c$


logarithms
inequalities
real analysis
real analysis unsolved

Hi, we know that if $ p \leq q$ then $ \ell_{p} \subset \ell_{q}$,

 But i cant prove that $ \| f \|_{q} \leq \| f \|_{p}$ for all $ f\in \ell_{p}$.

Can someone give me a hint, I have tried Holder and no luck

Thx

How didi you know ? [quote="Keenan"]Hi, we know that if $ p \leq q$ then $ \ell_{p} \subset \ell_{q}$,

 [/quote]  :lol:
ok
suppose $ f \in \ell_{p}$ and we need to prove :
$ \parallel{}f\parallel{}_{q}\leq \parallel{}f\parallel{}_{p}$ for $ p \leq q$

since : $ \parallel{}f\parallel{}_{p} \equal{} \left( \sum_{k \equal{} 0}^{\infty}|f_{k}|^{p} \right) ^{1/p} < \infty$
hence $ |f_{k}| \to 0$ as $ k \to \infty$ thus there exist $ N$ such as $ |f_{n}| < 1$ for all $ n \geq N$
suppose $ g \equal{} \max\{f_{0},f_{1},...,f_{N},1\}$ then :
$ \parallel{}f\parallel{}_{p} \equal{} \left( \sum_{k \equal{} 0}^{\infty}|f_{k}|^{p} \right) ^{1/p} \equal{} |g|\left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/p}$
thus we need to prove :
$ |g|\left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{q} \right) ^{1/q}\leq |g|\left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/p}$
or
$ \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{q} \right) ^{1/q}\leq \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/p}$
$ \left( \frac {|f_{k}|}{|g|}\right)^{q} \leq \left( \frac {|f_{k}|}{|g|}\right)^{p}$
since $ \frac {|f_{k}|}{|g|}\leq 1$ and $ p \leq q$
hence
$ \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{q} \right) ^{1/q}\leq \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/q}$
and
$ \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/q} \leq \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/p}$
since
$ \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \geq 1$ because
there exist $ k$ such as
$ \frac {|f_{k}|}{|g|} \equal{} 1$
done.

Here is a direct approach:

[b]Lemma.[/b] Let $ a_1, \cdots, a_n$ be positive numbers, $ f(x) \equal{} a_1^x \plus{} \cdots \plus{} a_n^x$ and $ g(x) \equal{} f(x)^{1/x}$. Then $ g$ is decreasing for $ x > 0$.

[i]proof.[/i] We instead show that $ \log g$ is decreasing. Differentiating,

$ (\log g(x))' \equal{} \minus{} \frac {1}{x f(x)} \sum_{k \equal{} 1}^{n} a_k^x (\log g(x) \minus{} \log a_k).$

Since $ g(x) > a_k$ for each $ k$, we have $ (\log g(x))' < 0$. ////

Now suppose $ 1 \leq p \leq q < \infty$ and $ f \in \ell^p$. Then we know that $ (|f_1|^q \plus{} \cdots \plus{} |f_n|^q)^{1/q} \leq (|f_1|^p \plus{} \cdots \plus{} |f_n|^p)^{1/p}$.

Taking $ n \to \infty$, we have $ \| f \|_q \leq \| f \|_p$.

[quote="Extremal"] because
there exist $ k$ such as
$ \frac {|f_{k}|}{|g|} \equal{} 1$
done.[/quote]
Why is it so and why do we need it?

We shouldn't be dividing by a maximum value, as Extremal has done. Instead, the argument goes like this:

[b]Case 1:[/b] Suppose $ \|f\|_p\equal{}1.$ Then $ \sum_k|f_k|^p\equal{}1.$ It follows that $ |f_k|\le 1$ for all $ k$ and hence $ |f_k|^q\le |f_k|^p.$

Then $ \|f\|_q^q\equal{}\sum_k|f_k|^q\le\sum_k|f_k|^p\equal{}1$ and $ \|f\|_q\le 1.$

[b]Case 2:[/b] If $ \|f\|_p\equal{}0,$ then $ f$ is identically zero and $ \|f\|_q\equal{}0.$

[b]General proof:[/b] WLOG, we may assume that $ \|f\|_p\ne 0.$ Let $ g\equal{}\frac1{\|f\|_p}\,f.$ Then $ \|g\|_p\equal{}\frac{\|f\|_p}{\|f\|_p}\equal{}1$ and $ \|g\|_q\equal{}\frac{\|f\|_q}{\|f\|_p}.$

By case 1, $ \|g\|_q\le 1$ so $ \frac{\|f\|_q}{\|f\|_p}\le1$ or $ \|f\|_q\le\|f\|_p.$

By careful examination of the proof, we can find the criterion for equality in this inequality: the two norms are equal if and only if $ f$ is identically zero or $ f$ is zero except for one element.

at first I absolutely agree with sos440. in some sense it's same what I have done.
I absolutely agree with Kent, it's full proof which also consider case when $ g \equal{} 0$ then statement is obvius.
I
My last inequality is :

$ \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/q} \leq \left( \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \right) ^{1/p}$
we need to check it.
suppose :
$ \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \equal{} A$
then $ A^{1/q}\leq A^{1/p}$ if and only if when $ A\geq 1$ since $ q \geq p$
but why  $ A > 1$???
since $ A \equal{} \sum_{k \equal{} 0}^{\infty}\left( \frac {|f_{k}|}{|g|}\right)^{p} \equal{} 1 \plus{} \alpha$ where $ \alpha \geq 0$ since there exist $ k$ such as $ \left( \frac {|f_{k}|}{|g|}\right)^{p} \equal{} 1$ and all other terms is more or equal to $ 0$.

Here's a variant, posed mostly to test understanding of the proof.

Let $ (w_k)$ be a sequence of positive numbers (a set of weights) that is both bounded above and bounded away from zero.

For $ 1\le p < \infty,$ define $ f\in\ell_w^p$ iff $ \sum_kw_k|f_k|^p < \infty$ and define $ \|f\|_{p,w} \equal{} \left(\sum_kw_k|f_k|^p\right)^{\frac1p}.$

Prove that if $ p < q,$ then $ \ell_w^p\subset\ell_w^q.$

Furthermore, there is a constant $ C$ depending on $ p$ and $ q$ such that $ \|f\|_{q,w}\le C \|f\|_{p,w}$ for all $ f\in \ell_w^p.$ Find $ C$ and prove this.

Would this still be true if we removed the hypothesis that $ w_k$ is bounded above? Would this still be true if we removed the hypothesis that $ w_k$ is bounded away from zero?

I don't agree with you .
I suppose you mean $ \ell_w^p\subset\ell_w^q.$

[quote="Kent Merryfield"]

Prove that if $ p < q,$ then $ \ell_w^q\subset\ell_w^p.$

[/quote]

condition 1): both bounded above and bounded away from zero - I think it's means there exist $ \max w_{k}$ and $ \min w_{k}$ which is not equal to $ \infty,0$
hence :
$ \parallel{}f\parallel{}_{q,w} \leq \max w^{1/q}_{k}\parallel{}f\parallel{}_{q} \leq \max w^{1/q}_{k}\parallel{}f\parallel{}_{p} \equal{} \frac {\max w^{1/q}_{k}}{\min w^{1/p}_{k} } \min w^{1/p}_{k} \parallel{}f\parallel{}_{p}\leq \frac {\max w^{1/q}_{k}}{\min w^{1/p}_{k} }\parallel{}f\parallel{}_{p,w}$

If we have no condition 1), then there exist such $ C$ for which :
$ \parallel{}f\parallel{}_{q,w} \leq C\parallel{} f\parallel{}_{p,w}$ still is true.
but, now I have no idea how to calculate the best $ C$for fixed weights.

[quote="limsup"][quote="Extremal"] because
there exist $ k$ such as
$ \frac {|f_{k}|}{|g|} \equal{} 1$
done.[/quote]
Why is it so ...?[/quote]
Oh, I have forgotten to answer...
since $ g \equal{} \max\{ |f_{0}|,|f_{1}|,....,|f_{N}|\}$  hence  there exist $ k$ such as $ \frac {|f_{k}|}{g} \equal{} 1$

But you defined your $ g$ as    $ g \equal{} \max\{f_{0},f_{1},...,f_{N},1\}$

oh, there should be $ g \equal{} \max\{|f_{0}|,|f_{1}|,...,|f_{N}|\}$,
but there is one bad  moment . if $ g \equal{} 0$ then my solution don't works.
therefore let's $ g \equal{} \max \{ |f_{0}|,|f_{1}|,...,|f_{k}|,....\}$
maximum exist since $ |f_{k}| \to 0$ as $ k \to \infty$
if $ f \equal{} 0$ then statement is obvius.
if $ f \neq 0$ then $ \infty > g > 0$
after you can repeat all my next steps. :)

ok. thank you Extremal. Dr. Kent's proof is also nice. Could you answer my questions in the topic titled "measure zero set"?


inequalities
LaTeX
inequalities unsolved

The side lengths condition is useful because if $x^2, y^2, z^2$ are not side lenghts then I am pretty sure the ineq does not always hold.

Consider replacing $x^2, y^2, z^2$ with $sinA * 2R, sinB * 2R, sinC *2R$; that could help.

I don't see how you can solve the ineq just using the fact that $x^2 \leq y^2+z^2$. :(

hi .l am new in this forum..
did you try to solve it with Muirhead and the AM-GM?!!!! i think it can solve it !! :?

[quote="bodom"]let \[ x,y,z\ge 0  \] such that \[ x^2,y^2,z^2  \] are the side lenghts of a triangle.prove or disprove the following:
\[ x^8+y^8+z^8+x^6yz+xy^6z+xyz^6+2x^3y^3z^2+2x^3y^2z^3+2x^2y^3z^3\ge x^5y^2z+y^5z^2x+z^5x^2y+x^5yz^2+y^5zx^2+z^5xy^2+2x^4y^4+2y^4z^4+2z^4x^4  \]
i'm not sure that the condition with the side lenghts is usefull  :([/quote]
Nope. Because you have on RS $2x^4y^4+2y^4z^4+2z^4x^4$ and only half of this can be "eliminated" with $x^8+y^8+z^8$ and the other half cannot be eliminated by anything. ;)


Find the sum of all distinct positive divisors of the number $104060401$.


calculus
integration
function
logarithms
real analysis
real analysis unsolved

For a given positive real r, does there exist a piecewise-continuous function f such that:
$ \int_{\minus{}\infty}^\infty \! f(x)x^a \, dx$ exists and is finite for $ a\equal{}0,r$,
and $ \int_{\minus{}\infty}^\infty \! f(x)x^a \, dx \equal{} \infty$ for all $ a>r$?

of course: you can play around with $ f(x)\equal{}x^{\minus{}\beta}1_{x>1}$

I tried something similar, but the integral will diverge iff $ a \ge \beta\minus{}1$. We want $ >$ instead of $ \ge$.
Tell me If I am missing something trivial...

Then how about $ \frac1{x^{\beta}\ln^{\gamma}x}\,\mathbf{1}_{\{x>2\}}.$ You may have to play around with $ \gamma$ a little to get it to do what you want.


geometry
circumcircle
inradius
incenter
Euler
inequalities
area of a triangle

$ABC$ is an acute-angled triangle and $P$ a point inside or on the boundary. The feet of the perpendiculars from $P$ to $BC, CA, AB$ are $A', B', C'$ respectively. Show that if $ABC$ is equilateral, then $\frac{AC'+BA'+CB'}{PA'+PB'+PC'}$ is the same for all positions of $P$, but that for any other triangle it is not.

Take $P\in OI$, then $AC'+BA'+CB'=s/2$. The conclusion is trivial then.

[quote="amir2"]$ABC$ is an acute-angled triangle and $P$ a point inside or on the boundary. The feet of the perpendiculars from $P$ to $BC, CA, AB$ are $A', B', C'$ respectively. Show that if $ABC$ is equilateral, then $\frac{AC' + BA' + CB'}{PA' + PB' + PC'}$ is the same for all positions of $P$, but that for any other triangle it is not.

[color=red][Moderator edit: This problem is 12th Austrian-Polish MO 1989 problem 8.][/color][/quote]

This is a rather straightforward one, but probably Kalva was too lazy to write down the solution, so let me do it here:

We have to prove two assertions:

[color=blue][b]Assertion 1.[/b] If ABC is an equilateral triangle, and P is a point inside it or on the boundary, and if A', B', C' are the feet of the perpendiculars from this point P to the sides BC, CA, AB of triangle ABC, then the value of $\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}$ is the same for all positions of P.

[b]Assertion 2.[/b] If ABC is an acute-angled triangle, and for all points P inside it or on the boundary, denoting by A', B', C' the feet of the perpendiculars from the point P to the sides BC, CA, AB of triangle ABC, the value of $\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}$ is the same for all positions of P, then the triangle ABC is equilateral.[/color]

[i]Proof of Assertion 1.[/i] Let a = BC = CA = AB be the sidelength of the equilateral triangle ABC. Then, the area of the triangle ABC is $\frac{\sqrt3}{4}a^2$.

Now, consider the triangle PBC. Its side BC has the length a, and the altitude to this side is PA'. Since the area of a triangle equals $\frac12\cdot$ side $\cdot$ corresponding altitude, we thus see that the area of triangle PBC equals $\frac12\cdot a\cdot PA^{\prime}$. Similarly, the areas of triangles PCA and PAB are $\frac12\cdot a\cdot PB^{\prime}$ and $\frac12\cdot a\cdot PC^{\prime}$, respectively. But since the triangles PBC, PCA and PAB, combined, form the triangle ABC, their areas, added up together, yield the area of triangle ABC, which is $\frac{\sqrt3}{4}a^2$, as we know. Thus, we have $\frac12\cdot a\cdot PA^{\prime}+\frac12\cdot a\cdot PB^{\prime}+\frac12\cdot a\cdot PC^{\prime}=\frac{\sqrt3}{4}a^2$. Upon multiplication by $\frac{2}{a}$, this becomes $PA^{\prime}+PB^{\prime}+PC^{\prime}=\frac{\sqrt3}{2}a$.

On the other hand, by the Pythagoras theorem, applied to the right-angled triangles BA'P and CA'P, we have $BP^2=BA^{\prime}\ ^2+A^{\prime}P^2$ and $CP^2=CA^{\prime}\ ^2+A^{\prime}P^2$, so that $BP^2-CP^2=BA^{\prime}\ ^2-CA^{\prime}\ ^2$. In other words, $BP^2-CP^2=\left(BA^{\prime}+CA^{\prime}\right)\left(BA^{\prime}-CA^{\prime}\right)$. But BA' + CA' = BC = a, and $BA^{\prime}-CA^{\prime}=2\cdot BA^{\prime}-\left(BA^{\prime}+CA^{\prime}\right)=2\cdot BA^{\prime}-a$. Thus, $BP^2-CP^2=a\left(2\cdot BA^{\prime}-a\right)$. Similarly, $CP^2-AP^2=a\left(2\cdot CB^{\prime}-a\right)$ and $AP^2-BP^2=a\left(2\cdot AC^{\prime}-a\right)$. Thus,

$0=\left(BP^2-CP^2\right)+\left(CP^2-AP^2\right)+\left(AP^2-BP^2\right)$
$=a\left(2\cdot BA^{\prime}-a\right)+a\left(2\cdot CB^{\prime}-a\right)+a\left(2\cdot AC^{\prime}-a\right)$
$=a\left(2\cdot\left(AC^{\prime}+BA^{\prime}+CB^{\prime}\right)-3a\right)$.

Since $a\neq 0$, we thus have $2\cdot\left(AC^{\prime}+BA^{\prime}+CB^{\prime}\right)-3a=0$, so that $AC^{\prime}+BA^{\prime}+CB^{\prime}=\frac32a$. Hence,

$\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}=\frac{\frac32a}{\frac{\sqrt3}{2}a}=\sqrt3$.

Clearly, this value is indeed the same for all positions of P, and thus, Assertion 1 is proven.

[i]Proof of Assertion 2.[/i] We have assumed that the value of $\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}$ is the same for all positions of the point P inside or on the boundary of triangle ABC; denote this value by k. Then, we have $k=\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}$ for every point P inside or on the boundary of triangle ABC.

First let's consider the case when the point P is the circumcenter of triangle ABC (the circumcenter lies inside the triangle ABC, since the triangle ABC is acute-angled). In this case, the point P is the circumcenter of triangle ABC and thus lies on the perpendicular bisectors of its sides BC, CA, AB; thus, the feet A', B', C' of the perpendiculars from the point P to these sides BC, CA, AB are the midpoints of these sides, respectively. Hence, $AC^{\prime}=\frac{AB}{2}=\frac{c}{2}$, $BA^{\prime}=\frac{BC}{2}=\frac{a}{2}$ and $CB^{\prime}=\frac{CA}{2}=\frac{b}{2}$, so that $AC^{\prime}+BA^{\prime}+CB^{\prime}=\frac{c}{2}+\frac{a}{2}+\frac{b}{2}=\frac{a+b+c}{2}=s$, where we have denoted by $s=\frac{a+b+c}{2}$ the semiperimeter of triangle ABC. On the other hand, by [url=http://www.cut-the-knot.org/proofs/carnot.shtml]the Carnot theorem[/url], since the point P is the circumcenter of the acute-angled triangle ABC, we have PA' + PB' + PC' = R + r, where R is the circumradius and r is the inradius of triangle ABC. Thus, $k=\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}=\frac{s}{R+r}$.

On the other hand, we can also consider the case when the point P is the incenter of triangle ABC (it is clear that the incenter will always lie inside the triangle ABC). In this case, the feet A', B', C' of the perpendiculars from the point P to the sides BC, CA, AB of triangle ABC are the points where the incircle of triangle ABC touches these sides. Thus, on the one hand, PA' = PB' = PC' = r (since the points A', B', C' lie on the incircle of triangle ABC, which has center P and radius r), while on the other hand, AC' = s - a, BA' = s - b and CB' = s - c. Thus, PA' + PB' + PC' = r + r + r = 3r and AC' + BA' + CB' = (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s. Hence, $k=\frac{AC^{\prime}+BA^{\prime}+CB^{\prime}}{PA^{\prime}+PB^{\prime}+PC^{\prime}}=\frac{s}{3r}$.

Comparing the expressions $k=\frac{s}{R+r}$ and $k=\frac{s}{3r}$ obtained now, we get R + r = 3r, so that R = 2r. But there is the famous Euler inequality, which states that $R\geq 2r$, and equality occurs only in the case when the triangle ABC is equilateral. Hence, the triangle ABC must be equilateral, and Assertion 2 is proven.

  Darij


prove that for any $a\in N$ and any $M\in N$ there exists $n\in N$ s.t some of the first digits of $a^n$ equals $M$ for instance for$M= 137226374840$ we have $n\in N$ s.t:$2^n=137226374840\underbrace{.....}_{other digits}$

Well, this is really well-known, byt you must ask for a not to be a power of 10. It's also a problem of mathlinks contest and an old russian problem. It's a direct consequence of Kronecker theorem.


AMC
AIME
analytic geometry
geometry
geometric transformation
rotation
trigonometry

I'm stumped on this problem.

A particle is located on the coordinate plane at $ (5,0)$. Define a move for the particle as a counterclockwise rotation of $ \frac{\pi}{4}$ radians about the origin followed by a translation of $ 10$ units in the positive $ x$-direction. Given that the particle's position after 150 moves is $ (p,q)$, find the greatest integer less than or equal to $ |p|\plus{}|q|$.

I think this is a bit too advanced for me, so can someone explain?

A solution is given [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2008&p=1088469]here[/url].

Sadly, that's the same solution I didn't understand, sorry :( . I was wondering if someone could explain it? :)

Suppose the particle is at $ (x,y)$. To rotate the particle by $ \theta$, we use:

$ (x',y')\equal{}(x \cos \theta \minus{} y \sin \theta, x \sin \theta \plus{} y \cos \theta)$

(Unfortunately, the only explanation I saw of this was in my linear algebra class, which uses vectors and matrices, so you may have to take this by faith if you haven't seen it.)

Therefore, our particle's movements can be mapped as:

$ x_{n\plus{}1}\equal{}x_n \cos \frac{\pi}{4} \minus{} y_n \sin \frac{\pi}{4}\plus{}10\equal{}\left(\frac{\sqrt{2}}{2}\right)(x_n\minus{}y_n)\plus{}10$
$ y_{n\plus{}1}\equal{}x_n \sin \frac{\pi}{4} \plus{} y_n \cos \frac{\pi}{4}\equal{}\left(\frac{\sqrt{2}}{2}\right)(x_n\plus{}y_n)$

From here, the solution uses this to relate $ (x_{n\plus{}2},y_{n\plus{}2})$ to $ (x_n,y_n)$, and then shows that $ x_{n\plus{}8}\plus{}y_{n\plus{}8}\equal{}x_n\plus{}y_n$ and $ x_{n\plus{}8}\minus{}y_{n\plus{}8}\equal{}x_n\minus{}y_n$, for all non-negative $ n$ by recursion.  The problem asks for $ |x_{150}|\plus{}|y_{150}|$, which reduces to finding $ |x_6|\plus{}|y_6|$, which is done by solving the two-equation system given by the values of $ x_6\plus{}y_6$ and $ x_6\minus{}y_6$.

You can also think about rotations using polar coordinates. For example, if you wanted to rotate the point $ (a,b)$ by $ \frac {\pi}{4}$, you can convert $ (a,b)$ to polar form: $ z \equal{} r\text{ cis } \theta , r \equal{} \sqrt {a^2 \plus{} b^2}$, $ \theta \equal{} \arctan\left(\frac {b}{a}\right)$. Then multiply $ z$ by $ \text { cis } \frac {\pi}{4}$ because $ (r\text{ cis } \theta)\cdot\left(\text{cis } \frac {\pi}{4}\right) \equal{} r \text{ cis } \left(\theta \plus{} \frac {\pi}{4}\right)$. Then convert back to cartesian coordinates.

I would suggest reading about polar coordinates if you're not comfortable with this.

Solutions using complex numbers can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088469#p1088469


inequalities
trigonometry
search
AMC
USA(J)MO
USAMO
triangle inequality

Will someone explain this to me please. I know the proof and everything, but I can't figure how to use it efficiently.

I guess that's where practice comes in.

The inequalities section of this book has some problems:
http://www.thiel.edu/mathproject/atps/tbloc.htm

A common exercise:

What is the distance from the origin to the line $5x+12y = 60$?

OMG... do you have any more questions that can be solved using Cauchy? I think I need the practice

You should probably tell them $\sum_{i=1}^{\infty}\frac{1}{i^{2}}=\frac{\pi^{2}}{6}$.

Yes, that was the intended solution.

There are plenty of inequality problems in the Olympiad Inequalities section, although they are fairly hard...

Edit:  Well, here's another one.  Find the amplitude of $y = A \cos x+B \sin x$.

I posted a solution before to this, I used trigonometric identities, but I think this is how you do it with Cauchy.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118178

[hide]
By Cauchy, we have $(\cos^{2}x+\sin^{2}x) (A^{2}+B^{2}) \ge (A\cos x+B\sin x)^{2}$

So, $\sqrt{A^{2}+B^{2}}\ge (A\cos x+B\sin x)$

$\boxed{\sqrt{A^{2}+B^{2}}}$
[/hide]

Similar:

Prove that for $0 \le x \le \frac{\pi}{2}$ we have

$\frac{\sin^{2}x}{\cos x}+\frac{\cos^{2}x}{\sin x}\ge 1$

Is it really simple? I can't seem to get it :(

I proved it, but it's not 100% Cauchy.

[hide]
$\left(\frac{1}{\sin^{2}x}+\frac{1}{\cos^{2}x}\right)\left( \sin^{6}x+\cos^{6}x\right) \ge 1$

$2\sin x \cos x \ge 0$ when $0\le x < \frac{\pi}{2}$

So, if we add the two inequalities, we get $\frac{\sin^{2}x}{\cos x}+\frac{\cos^{2}x}{\sin x}\ge 1$[/hide]

The Cauchy-Schwarz inequality is:
$ \left(\sum a^{2}\right)\left(\sum b^{2}\right)\ge \left(\sum ab\right)^{2}$
$ \left(a_{1}^{2}+a_{2}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}\right)\ge \left(a_{1}b_{1}+a_{2}b_{2}\right)^{2}$
If you set $a_{1}= \frac{\sin{x}}{\sqrt{\cos{x}}}$, $a_{2}= \frac{\cos{x}}{\sqrt{\sin{x}}}$, $b_{1}= \sqrt{\cos{x}}$, $b_{2}= \sqrt{\sin{x}}$, we have
$\left(\frac{\sin^{2}{x}}{\cos{x}}+\frac{\cos^{2}{x}}{\sin{x}}\right)\left(\cos{x}+\sin{x}\right) \ge \left(\sin{x}+\cos{x}\right)^{2}$ and $\cos{x}+\sin{x}> 1$ by the triangle inequality, so $\frac{\sin^{2}{x}}{\cos{x}}+\frac{\cos^{2}{x}}{\sin{x}}\ge 1$

How would you do the following problem with Cauchy?

Prove:

$\frac{a^{2}}{x}+\frac{b^{2}}{y}\ge \frac{\left(a^{2}+b^{2}\right)}{\left(x+y\right)}$

$\frac{1}{a^{3}\left(b+c\right)}+\frac{1}{b^{3}\left(a+c\right)}+\frac{1}{c^{3}\left(a+b\right)}\ge \frac{3}{2}$

This was from Mathematical Olympiad Threasures if anyone wondered.

#1

[hide]

Assuming they're positive, cauchy's gives you that it's greater than (a+b)^2, but that's obviously greater than a^2+b^2.

[/hide]

#2

[hide]

This is tricky unless you first think to do  and so on, and then force the resulting expression into Cauchy's (it looks terrible, but works).

[/hide]

[quote="Max300"][quote="computer_nuke"]How would you do the following problem with Cauchy?

Prove:

$\frac{1}{a^{3}\left(b+c\right)}+\frac{1}{b^{3}\left(a+c\right)}+\frac{1}{c^{3}\left(a+b\right)}\ge \frac{3}{2}$

[/quote]

I think you are supposed to be given that $abc=1$
If so...

[hide]First we rewrite the left side as $\frac{\frac{1}{a^{2}}}{a(b+c)}+\frac{\frac{1}{b^{2}}}{b(a+c)}+\frac{\frac{1}{c^{2}}}{c(a+b)}$. By Cauchy-Schwarz, $\frac{\frac{1}{a^{2}}}{a(b+c)}+\frac{\frac{1}{b^{2}}}{b(a+c)}+\frac{\frac{1}{c^{2}}}{c(a+b)}\ge \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{a(b+c)+b(a+c)+c(a+b)}$. So if we prove that $\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{a(b+c)+b(a+c)+c(a+b)}\ge \frac{3}{2}$ then we have proven the inequality. Simplifying the left hand side gives us $\frac{(\frac{ab+bc+ca}{abc})^{2}}{2ab+2bc+2ca}\ge \frac{3}{2}$. Substituting $abc$ with 1 and canceling like terms gives us $ab+bc+ca \ge 3$ which is true by AM-GM.[/hide][/quote]
yeah it would have to include abc=1 or else its not true
it would approach 0 as a b and c got bigger

I think USAMO problems are a little too hard, but USAMO 2002 #2 becomes easy if you use cauchy I think.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=337847

[quote="PenguinIntegral"]The old inequality marathon had bunch of cauchy problems.

(hint: search button)[/quote]here is the link: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40857]Inequality lessons and maybe a marathon[/url]

[quote="computer_nuke"]$\frac{a^{2}}{x}+\frac{b^{2}}{y}\ge \frac{\left(a^{2}+b^{2}\right)}{\left(x+y\right)}$[/quote]

This is (edit: weaker than) the two-variable case of another form of Cauchy, known as Engel form, which follows directly from the original form:

$\sum \frac{a_{k}^{2}}{b_{k}}\ge \frac{ \left( \sum a_{k}\right)^{2}}{\sum b_{k}}$

[quote="t0rajir0u"][quote="computer_nuke"]$\frac{a^{2}}{x}+\frac{b^{2}}{y}\ge \frac{\left(a^{2}+b^{2}\right)}{\left(x+y\right)}$[/quote]

This is the two-variable case of another form of Cauchy, known as Engel form, which follows directly from the original form:

$\sum \frac{a_{k}^{2}}{b_{k}}\ge \frac{ \sum a_{k}^{2}}{\sum b_{k}}$[/quote]

I think this inequality doesn't hold. :maybe:

[quote="t0rajir0u"][quote="computer_nuke"]$\frac{a^{2}}{x}+\frac{b^{2}}{y}\ge \frac{\left(a^{2}+b^{2}\right)}{\left(x+y\right)}$[/quote]

This is the two-variable case of another form of Cauchy, known as Engel form, which follows directly from the original form:

$\sum \frac{a_{k}^{2}}{b_{k}}\ge \frac{ \sum a_{k}^{2}}{\sum b_{k}}$[/quote]

You mean

$\sum \frac{a_{k}^{2}}{b_{k}}\ge \frac{ \left(\sum a_{k}\right)^{2}}{\sum b_{k}}$.

Although the original one does hold for positive reals since it is weaker.

Yes! but we should add some condition.
Anyway Cauchy-Scwarz inequality can be used in case that we can see the form such as [b]the product of the squared of distance $\geq$ the squared of inner product.[/b]

The lemma: $x_{i},y_{i}\in\mathbb{R_{+}}$
\[\frac{x_{1}}{y_{1}}+\frac{x_{2}}{y_{2}}+\cdots+\frac{x_{n}}{y_{n}}\geq \frac{(x_{1}+x_{2}+\cdots x_{n})^{2}}{x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}}\]

Problem:

For positive numbers $a,b,c$ the inequality holds:

\[\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1\]

To solve it apply the lemma and th inequality: $(a+b+c)^{2}\geq 3(ab+bc+ca)$

that follows directly form what paladin8 posted. :wink:

[quote="Max300"][quote="computer_nuke"]How would you do the following problem with Cauchy?

Prove:

$\frac{1}{a^{3}\left(b+c\right)}+\frac{1}{b^{3}\left(a+c\right)}+\frac{1}{c^{3}\left(a+b\right)}\ge \frac{3}{2}$

[/quote]

I think you are supposed to be given that $abc=1$
If so...

[hide]First we rewrite the left side as $\frac{\frac{1}{a^{2}}}{a(b+c)}+\frac{\frac{1}{b^{2}}}{b(a+c)}+\frac{\frac{1}{c^{2}}}{c(a+b)}$. By Cauchy-Schwarz, $\frac{\frac{1}{a^{2}}}{a(b+c)}+\frac{\frac{1}{b^{2}}}{b(a+c)}+\frac{\frac{1}{c^{2}}}{c(a+b)}\ge \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{a(b+c)+b(a+c)+c(a+b)}$. So if we prove that $\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{a(b+c)+b(a+c)+c(a+b)}\ge \frac{3}{2}$ then we have proven the inequality. Simplifying the left hand side gives us $\frac{(\frac{ab+bc+ca}{abc})^{2}}{2ab+2bc+2ca}\ge \frac{3}{2}$. Substituting $abc$ with 1 and canceling like terms gives us $ab+bc+ca \ge 3$ which is true by AM-GM.[/hide][/quote]

Yes it is abc=1.

But I see what you're doing, I'm just confused on how it fits into the original equation: $\left(\sum a^{2}\right)\left(\sum b^{2}\right)\ge \left(\sum ab\right)^{2}$

I mean that I cannot see which are the a_n and which are the b_n, I cannot see the correlation between the two.

On the first problem, I can see how cauchy could prove it and then it could be generalized and used on the second, but is there a more direct result from Cauchy?

[quote="t0rajir0u"][quote="computer_nuke"]I see what you're doing, I'm just confused on how it fits into the original equation: $\left(\sum a^{2}\right)\left(\sum b^{2}\right)\ge \left(\sum ab\right)^{2}$[/quote]

His only use of Cauchy-Schwarz follows from the [b]Engel form[/b] that I gave, which follows from the original statement of Cauchy thus:

$\left( \sum \frac{a_{k}^{2}}{b_{k}}\right) \left( \sum b_{k}\right) \ge \left( \sum a_{k}\right)^{2}$[/quote]

How do you get this form from the original form? :maybe:

[quote="paladin8"]Let $x_{k}= \frac{a_{k}}{\sqrt{b_{k}}}$ and $y_{k}= \sqrt{b_{k}}$. It follows from Cauchy on the sequences $x$ and $y$.[/quote]

Hmmm, Nice manipulation, I see it now.

Thanks alot to everyone that helped  :lol:

Hmm.  So the condition for Engel form is stronger than I thought.  Each $b_{k}$ must be positive.

Are there any other interesting forms of the Cauchy inequality?


analytic geometry
combinatorics unsolved
combinatorics

Let be the set $ M\equal{}{(x,y,z) such that 1\leq x,y,z\leq 4n }$ .We colour all poits of this set by $ 4n$ points . Prove that there exist a triangle with three vertices are monochromatic and three sides are parallel to three axises.

There is no triangle in space, each of whose sides is parallel to one of the coordinate axes.  Do you perhaps mean the $ xy$-, $ yz$- and $ zx$-planes?


ratio
geometry
perimeter
3D geometry
pyramid
rotation
area of a triangle

These are the problems I got wrong on my first practice test. Help me if you can. Sorry there are so many.

1. How many ways can you rearrange the letters in "ALGEBRA" ?

2. A password is created by randomly selecting 3 digits to form a 3-digit number. How many passwords are possible?

3. The ratio of the perimeters of two similar polygons is 2:3. What is the ratio of their aeas.

4. A boy ran up a hill at 3/2 mph and came down it at 9/2 mph. The trip took him 6 hours. How many miles is it to the top of the hill?

5. A ream has 2000 sheets of paper. A box has nine reams. How many 200 page books can be printed with one box?

6. As Sarah continues preparing for her backpack trip she found that lard has 120 calories/13 grams. She wants to know how many calories are in each ounce of lard. Use the following information she found on the package to help her out. Each package is one pound. Each package has 35 servings. Each serving is 13 grams.

7. What is the volume pf a pyramid with base area 18 and height 27?

8. What is the distance from the point (33,17) to the point (33, -51)

9. 500 cars in a lot using a character followed by a single digit. How many cars will have the same catalog code as another car?

10. What is the area of a triangle with vertices at points (0,0), (5,0), and (2,4)?

11. A hubcap with radius 24 inches falls off the Math Team's land rover and rolls down the street. How many times will it rotate if the street is 200 feet long?

1. There are 7 letters in ALGEBRA, but there are 2 A's, so there are a total of $ \frac{7!}{2!}\equal{}2520$ unique words.

2. Since it doesn't say the numbers have to be unique, there are $ 10^{3}\equal{}1000$ different passwords.

8. They have same $ x$ so the distance will be $ |\minus{}51\minus{}17|\equal{}68$

10. Use Heron's formula, or just turn the triangle into 2 right triangles.

onlyrevolutions, I believe that you posted this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245864]here[/url] in the middle school classroom math forum. In the future, please do not post topics in multiple places. :maybe:


cho tam giac ABC voi I la tam duong tron bang tiep goc C. Goi D,E thu tu la cac tiep diem cua cac canh AB,BC.Goi AH la duong cao cua tam giac ABC .DE cat AH tai K .CMR AK=r

bainay co ban .k0 co ai giai thi minh giai vay .
<AKD=90+<B\2,<ADK=<B\2.
Ap dung dinh ly sin cho tam giac AKD ta duoc
AK=AD.tanB\2.tinh duoc AD=p-b .vay ta co dpcm

Xin lỗi vì spam nhưng hôm nọ thứ 7,thầy mới chữa bài này,cả lớp mỗi mình thằng Thế Anh (and me) làm được mà :P

Ai bao the .Hom day tao cung lam duoc bai nay.Thang the anh chep cua Khanh thi co .Ma may len mang it thoi ko lai bi thay hung nhac nho .Tao len mang đoc viet bai vo van thoi ,trinh tieng anh thi thap


[quote="Kurt Gödel"]Impressive  :o How did you find that? By trying to find an explicit formula before proving it?[/quote]

No, by referring to my own short compendium on recurrences :D ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107173]this topic[/url], fifth post, third case). (Starting by taking the reciprocals was kinda obvious, though.)


In how many ways can the letters of the word BANANA be rearranged such that the new word does not begin with a B?

I believe this is
$ \frac{5!}{3!2!}\equal{}10$

since it cannot begin with a B, there are

$ 5 \times 5 \times 4 \times 3 \times 2 \equal{} 600$

but then we have to divide by $ 3!$ and $ 2!$ because of the number of N's and A's

so we get the answer of $ \frac {600} {3!2!} \equal{} \frac {600} {6 \times 2} \equal{} \frac {600} {12} \equal{} \boxed {50}$

Alternate solution:

Total number= $ \frac{6!}{3!2!}\equal{}60$

NUmber that start with B=$ \frac{5!}{3!2!}\equal{}10$

$ 60\minus{}10\equal{}50$


[quote="math12061992"]FIND ALL SOLUTIONS TO THE EQUATION IN POSITIVE INTEGERS
a!b!=a!+b!+c![/quote]
WLOG, $ a \le b$
[b]Case 1:[/b] $ c \ge b > a$
$ a!\equal{}\frac{a!}{b!}\plus{}\frac{c!}{b!}$, L.H.S. is an integer but R.H.S. is not , contradiction, so no solution in this case.

[b]Case 2:[/b] $ c \ge b\equal{}a$, the equation $ \Leftrightarrow (a!)^2\equal{}2(a!)\plus{}c! \Leftrightarrow a!\equal{}2\plus{}\frac{c!}{a!}$
[b](i)[/b]$ c \minus{} a \ge 4$, then $ 4 \nmid$ R.H.S. $ \Rightarrow a \le 3$. Clearly only $ a\equal{}3$ is possible , $ \Rightarrow (a,b,c)\equal{}(3,3,4)$
[b](ii)[/b]$ c\equal{}a\plus{}3 \Rightarrow a!\equal{}2\plus{}(a\plus{}3)(a\plus{}2)(a\plus{}1) \Rightarrow a|8 \Rightarrow$ no solution.
[b](iii)[/b]$ c\equal{}a\plus{}2 \Rightarrow a!\equal{}2\plus{}(a\plus{}2)(a\plus{}1) \Rightarrow a|4 \Rightarrow$ no solution.
[b](iv)[/b]$ c\equal{}a\plus{}1 \Rightarrow a!\equal{}2\plus{}(a\plus{}1) \Rightarrow a|3 \Rightarrow (a,b,c)\equal{}(3,3,4)$
[b](v)[/b]$ c\equal{}a \Rightarrow a!\equal{}3 \Rightarrow$ no solution.

[b]Case 3:[/b] $ c <b$, then $ b!<a!b!<3b! \Rightarrow a\equal{}2 \Rightarrow b!\equal{}2\plus{}c!$, no solution.

So $ (a,b,c)\equal{}(3,3,4)$ is the only solution.

VERY VERY THANK U FOR UR FANTASTIC SOLUTION STEPHENCHENG.


geometry
geometric transformation
rotation
analytic geometry
graphing lines
slope
trigonometry

Find the moment of inertia of a square with side $a$ The axis of rotation is making angle $\alpha$ with the $x$ axis.The mass of the square is $M$.

The answer must be $\frac{1}{12}Ma^{2}$. :(

[hide]
Let us cut the square into four smaller squares. If the unknown moment of inertia of the big square is $I$, than the moment of inertia of each of the small squares (with the axis of rotation making angle $\alpha$ with the $x$ axis of the smaller square) is $\frac{I}{16}$.
Using the parallel axis theorem we get:
$I=2(\frac{I}{16}+\frac{m}{4}d^{2}+\frac{I}{16}+\frac{m}{4}D^{2})$
$d$ and $D$ are the distances between the axis of rotation of the big square and the centres of the small squares.
From some geometrical relations we get:
$D=\frac{a}{4}(1+\tan\alpha)\cos\alpha=\frac{a}{4}(\cos\alpha+\sin\alpha)$
and $d=\frac{a}{4}(1-\tan\alpha)\cos\alpha=\frac{a}{4}(\cos\alpha-\sin\alpha)$
so $D^{2}+d^{2}=\frac{a^{2}}{8}$
Finally:
$I=\frac{I}{4}+\frac{m}{2}(d^{2}+D^{2})$
$I=\frac{ma^{2}}{12}$
[/hide]

[quote="minsoens"][hide="not completely sure"]the line making an angle $\alpha$ has slope $\tan\alpha$, so the equation of the line is $y=(\tan\alpha)x$.
then we use the point-to-line distance formula:
\begin{eqnarray*}d(x_{0},y_{0})&=&\frac{ax_{0}+by_{0}+c}{\sqrt{a^{2}+b^{2}}}\\ &=&\frac{(\tan\alpha)x_{0}-y_{0}}{\sqrt{1^{2}+(\tan\alpha)^{2}}}\\ &=&x_{0}\sin\alpha-y_{0}\cos\alpha \end{eqnarray*}
\begin{eqnarray*}\int_{-a}^{a}\int_{-a}^{a}\frac{M}{a^{2}}d(x,y)^{2}\,dx\,dy&=&\int_{-a}^{a}\int_{-a}^{a}\frac{M}{a^{2}}(x\sin\alpha-y\cos\alpha)^{2}\,dx\,dy\\ &=&\frac{M}{a^{2}}\int_{-a}^{a}\int_{-a}^{a}x^{2}\sin^{2}\alpha-2xy\sin\alpha\cos\alpha+y^{2}\cos^{2}\alpha\,dx\,dy\\ &=&\frac{M}{a^{2}}\int_{-a}^{a}\frac{2a^{3}\sin^{2}\alpha}{3}+2ay^{2}\cos^{2}\alpha\,dy\\ &=&\frac{M}{a^{2}}\left(\frac{2a^{3}\sin^{2}\alpha}{3}(2a)+\frac{2a}{3}(2a^{3})\cos^{2}\alpha\right)\\ &=&\frac{4Ma^{2}}{3}\end{eqnarray*}
[/hide][/quote] :huh: i cannot edit my above post..
anyway.. i make mistake in integrating from $-a$ to $a$..it's supposed to be $\frac{-a}{2}$ to $\frac{a}{2}$ :blush:
\begin{eqnarray*}\int_{-\frac{a}{2}}^{\frac{a}{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}}\frac{M}{a^{2}}d(x,y)^{2}\,dx\,dy&=&\int_{-\frac{a}{2}}^{\frac{a}{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}}\frac{M}{a^{2}}(x\sin\alpha-y\cos\alpha)^{2}\,dx\,dy\\ &=&\frac{M}{a^{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}}x^{2}\sin^{2}\alpha-2xy\sin\alpha\cos\alpha+y^{2}\cos^{2}\alpha\,dx\,dy\\ &=&\frac{M}{a^{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}}\frac{a^{3}\sin^{2}\alpha}{12}+ay^{2}\cos^{2}\alpha\,dy\\ &=&\frac{M}{a^{2}}\left(\frac{a^{3}\sin^{2}\alpha}{12}(a)+a\left(\frac{a^{3}}{12}\right)\cos^{2}\alpha\right)\\ &=&\frac{Ma^{2}}{12}\end{eqnarray*}


[quote="Iversonfan2007"]22-23 i think depending on 2b[/quote]
Same here  :( . How come we can't get under 10? :P

I got 2b wrong, so probably a 22-23 also.

[quote="SplashD"]Do they consider your grade at all when grading?[/quote]
Most likely not.  :maybe:

[quote="AdorableAddict"]Probably 25 if nothing goes wrong.  :maybe:[/quote]

lol of course you did rui ;)

Well looking over the Math Jam transcript I got all the answers right, and my proofs were overly-rigorous and explanatory (my solution file was 13 pages- probably overkill, but it can't hurt to be as clear as possible), so I'm gonna be arrogant and say 25.

I also had 13 pages. :D
My number 1 solution was way too complicated.  :lol:
I did messy casework on 2b, and included a diagram in 4 (which took almost a page). And another diagram in 5, along with a generalization.

13 pages seems like not too much, I thought I was conservative and wrote 11... thats only like 2 per problem, some problems needed a minimum of three no matter how stingy you were

I sure as **** better get a 25 and several commendations too. I tried really hard to make my answers comprehensive but elegant. Each of my answers fits on a single page.

I checked and all my answers are correct, so I probably did get a 25 (but last year in one of the rounds I got all the answers right but still didn't get full credit). I'm really expecting a commendation on 2 because I managed to avoid doing casework. My other solutions were nice too, although 1 was pretty straightforward and my solution on 4 was pretty much the exact same as the one given in the math jam. We'll see.

I'll go with 25 because I haven't caught any mistakes yet, but my solutions were pretty badly explained and not as thorough as they could have been, so actually I'll probably get a few style points docked at the very least. I only had 7 pages--two of which are almost entirely white space (my explanation for #4 only took up a small paragraph). And my solution for #1 is weird. Not nearly as nice as the official one.

I'm hoping for a 25, because I believe I did everything right.  I think my style is okay, not perfect, but hopefully good enough to not take points off.

i'll probably get 25, even though i had an extremely minor error on 2b (i added 2 and 2 when i should have multiplied  :lol:  :huh: ).

25...unless Samso pwns me.  Grar...have mercy! :wink:

[quote="joml88"]How many pages you have doesn't really say anything about what score you will get.  One person could write out every step and get 4 pages for a solution that another person condenses into 2.

Last year, for example, my solutions went like this:

Round 1 - 6 pages - 25
Round 2 - 8 pages - 25
Round 3 - 17 pages (9 pages of total brute force on problem #5 which contained a ton of repetitive diagrams) - 25
Round 4 - 6 pages - 20

I would shoot for simplicity.  Don't go through and prove the Pythagorean theorem every time you use it (i.e. if a theorem is famous you can just cite it and move on with the proof).  If you know how to brute force it and can't find an elegant approach, just brute force it out.[/quote]

Wow so you missed gold by 1 point. :(

Ugh!
I got them all right, but I finished my solutions like 2 weeks ago and forgot to send them in.  Lol.  Dang it.

17. Because.

I finished #5 six hours too late, and gave up on #2b - I forbade myself from using messy casework. I think the rest were fine - my #1 has the correct value, but I didn't use the same method as everyone else, but it should be fine. My lemma for #3 and all of #4 are rather hastily written, and thus the phrasing and use of symbols is sloppy.

I think I'll start the next round >2 days before it's due. It was in some ways the most interesting two days in the past month, but not really worth the (mental [i]and[/i] physical) exhaustion.

[quote="bpms"]There should've been a 30 option to see how many liars there are in AoPS.[/quote]

LMAO fourteen 25s eh? ;)

[quote="bookaholic"][quote="Iversonfan2007"]LMAO fourteen 25s eh? ;)[/quote]
That's not that unbelievable. This round seemed a lot easier to me than any of last year's rounds, presumably because the admins wanted to get more people to participate, and even last year there were some 30 perfects per round and many more than that on the first round.[/quote]
But 40 % of the people having 25s is pretty unbelievable...

[quote="lotrgreengrapes7926"][quote="bpms"]But 40 % of the people having 25s is pretty unbelievable...[/quote] 40% of [b] AoPS members [/b] is not that unbelievable.[/quote]

um first of all it's not 40% of all aopsers...not all the ones who did round 1 voted....

and another thing just b/c you got all the answers right doesn't mean you'll definitely get a 25.

[quote="bpms"][quote="bookaholic"]Well, that's another thing. I would expect that people with lower scores are less likely to vote, especially if they look at the results first. So the percentage of voters with very high scores can be much higher than is representative without anyone lying.[/quote]
Heh, I can't vote because I think I got a lower score than the minimum in the pole. If I got what I think I got, my goal is to triple my score :P[/quote]
I meant triple my score on the first round in the second but I can't edit it. Can a mod or someone in power combine these?

[quote="13375P34K43V312"]I like how so many people think they got 25. I'm curious to see how many people actually do.[/quote]

a [i]lot[/i] of people do get 25s on the first round

[i]2 messages were deleted[/i]

Please do not discuss any aspect of Round 2 while it is in progress.

People who don't think they did well don't vote.  Like me, for instance.


conics
parabola
calculus
geometry
geometric transformation
reflection

[quote="mathgeniuse^ln(x)"][quote="cincodemayo5590"]Preferably without calculus  :P , find the parabola in the form $x^{2}+ax+b$ which is tangent to both lines $x+4y=4$ and $y-2x=1$.[/quote]

[hide]Hint

Equate it with both equations so that when you take the discriminant both are equal to 0.[/hide][/quote]

Hey, if you wanted the hide to show up as "Hint" instead of "Click to blah blah blah," you'd type in this:
[code] [hide=Hint] stuff [/hide] [/code]
and it'll show up as: [hide="Hint"] stuff [/hide]

[quote="mathgeniuse^ln(x)"][quote="cincodemayo5590"]Preferably without calculus  :P , find the parabola in the form $x^{2}+ax+b$ which is tangent to both lines $x+4y=4$ and $y-2x=1$.[/quote]

[hide]Hint

Equate it with both equations so that when you take the discriminant both are equal to 0.[/hide][/quote]

why does this work?

Because if the discriminant is equal to zero, there is one solution, meaning one intersection of the graphs, which is the definition of tangent.

[quote="cincodemayo5590"]Preferably without calculus  :P , find the parabola in the form $x^{2}+ax+b$ which is tangent to both lines $x+4y=4$ and $y-2x=1$.[/quote]

if we wanted to be really tricky, we would use the reflection properties of the parabola to find the equation...

[quote="Ubemaya"]But don't the parabola and the tangents never touch?[/quote]

Ok, when you say tangent, usually it means to touch something.  So there you go.


function
induction
inequalities
algebra unsolved
algebra

Does there exist a pair $ (f; g)$ of strictly monotonic functions, both from $ \mathbb{N}$ to $ \mathbb{N}$, such that \[ f(g(g(n))) < g(f(n))\] for every $ n \in\mathbb{N}$?

Assume that such functions exist. Since the functions are increasing, by simple induction we get $ f(n),g(n)\ge n$ for all $ n\in\mathbb{N}$. If $ f(n)\equal{}n$ for some $ n\in\mathbb{N}$, subsituting to the given inequality, we get $ g(g(n))<g(n)$, contradicting the hypothesis that $ g$ is increasing. Similarly, if $ g(n)\equal{}n$ for some $ n\in\mathbb{N}$, we get $ f(n)<f(n)$ which is impossible. Therefore $ f(n),g(n)>n$.

We now have $ f(g(g(n)))<g(f(n))<f(g(f(n)))$. We get $ f(g(g(n)))<f(g(f(n)))$, so $ g(n)<f(n)$. Now, $ f(g(g(n)))<g(f(n))<f(f(n))$, so $ g(g(n))<f(n)$. By induction, we get that $ f(n)>g(g(...(n)...))$. This is impossible because $ g(n)<g(g(n))<...<g(g(...(n)...))$. Therefore no such functions exist.


for $a, b, c > 0$

prove that

$\sum_{cyclic} a(a-b)(a-c) \leq {1 \over 2} \sum_{cyclic} {ab \over c}(a-b)^2$

This is equivalent to Peter Scholze's inequality

$a^3+b^3+c^3+3abc-b^2c-c^2b-c^2a-a^2c-a^2b-b^2a$
$\leq\frac12\left(\frac{bc}{a}\left(b-c\right)^2+\frac{ca}{b}\left(c-a\right)^2+\frac{ab}{c}\left(a-b\right)^2\right)$,

which was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=6563 .

  Darij

Oh, It's a wonderful solution  :D  :D

$ ( ab(a-b)+bc(b-c)+ca(c-a))^2\geq 0 $

Thank you for information. ^^


ARML
function
geometry
geometric transformation
rotation

$f$ is a continuous complex-valued function satisfying:

i) $|f(z)| = |z|$
ii) $|f(z)-z| = |z|$

Find $f(f(f(2007)))$

[hide="Solution"] We know two things about $f$:  One, that it maps the unit circle to itself; another, that the origin, $z$, and $f(z)$ form an equilateral triangle.

We conclude that $f(z)$ must be a rotation by $60^{\circ}$ in either direction; in other words, that $f(z) = z e^{ \pm i \frac{\pi}{3}}$.

Hence $f(f(f(2007))) = 2007 e^{\pm i \pi}= \boxed{-2007 }$. [/hide]

[quote="gotztahbeazn"]why does a rotation of $60^\circ$ imply a third root of unity? wouldn't it be sixth root[/quote]

Sorry, you're right.  I've fixed it.

[quote="AstroPhys"]:maybe: It makes a triangle such that the angles are each $60^{\circ}$.
Not a rotation though.[/quote]

One leg of this triangle is $z$, and the other is $f(z)$.  Hence $f(z)$ is a $60^{\circ}$ rotation of $z$.

In that sense of the meaning of rotation, yes.  :)

I really dont think you should post this a day before it is due, do your own homework! And even if you are just posting this because you thought it was interesting, you should still wait until after it is due.

As far as I know the ARML coaches have always advocated kids working together and bouncing ideas off of each other.

I think they advocate working together and bouncing ideas of eachother, as in with other members of the team or with kids from your school, not over the internet, and especially not on a site where people post full solutions to problems posed.

but perhaps I am wrong.

heh it does you no good to post a question and copy the solution word for word and i agree that that would be cheating

however seeing as how i even corrected his solution, that shows that at least i understand his solution and i kno w where he's coming from with his solution and isn't that the point of hw?

anyways if the only reason you're posting this stuff is because you're worried that this might offset the proper team selection process, then take into account that the coaches consider your individual/power/relay performance during practice as well, in addition to AIME/AMC and ICTM stuff that you do

Hey, kind of off topic, but I've completely lost my ARML schedule... the next practice is tomorrow? When are the others?

[quote="gotztahbeazn"]May 17th is the last one[/quote]

Ok, thanks.  :)

NOOOO THEY HAVE MADE POST RATINGS NOOOOO

anyway...i came up with and posted an almost identical problem before...so w/e

i.e. let g(x)=[1/x]f(x)


geometry
inequalities
search
triangle inequality

Two of the altitudes of the scalene triangle $ ABC$ have length $ 4$ and $ 12$.  If the elngth of the third altitude is also an integer, what is the biggest it can be?

(A) 4
(B) 5
(C) 6
(D) 7
(E) none of these

Thanks in advance!

[hide]Let $ A \equal{} \frac 42a \equal{} \frac {12}2b \equal{} \frac h2 c$ be the area, where the sides are $ a,b,c$, with altitudes respectively $ 4,12,h$; then by the triangle inequality, we have $ a < b\plus{}c \Longrightarrow \frac{A}{2} < \frac{A}{6} \plus{} \frac{2A}{h} \Longrightarrow h < 6$. Hence $ \mathrm{(B)}$.[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=935421804&t=219040]Search[/url]

[hide]just use triangle inequality

Let area Be A
then
2A(1/12+1/4)>2A/x
=>1/3>1/x
gives x>3

also 2A(1/12+1/x)>2A/4
gives 1/x>1/6
x<6

so only value is x=5 [/hide]


[quote="konstantine"]Γεια σας παιδιά πριν από λίγο γράφτηκα στο φόρουμ και αυτη είναι η πρώτη άσκηση που ποστάρω.Συγχωρέστε με αν είναι πολύ κάτω των δυνατοτήτων σας.
Αν για την εξίσωση λχ2+χ+μ=0 ειναι |λ+μ+1|<|λ| νδο αυτη εχει μια τουλάχιστον ρίζα στο (0,2) (λ όχι μηδέν)[/quote]
Brika mia lusi alla mono gia λ>0.
Sigoura kati tha mou diafeugei.....
Thetoume λ=a kai μ=m gia eukolia sto Latex.
Exoume loipon oti:
$ a^2 > a^2 \plus{} m^2 \plus{} 1 \plus{} 2am \plus{} 2a \plus{} 2m \Leftrightarrow m^2 \plus{} 1 \plus{} 2am \plus{} 2a \plus{} 2m < 0$
Alla $ m^2 \plus{} 1 > 2m$,ara $ 2a(m \plus{} 1) < 0$
Gia $ a > 0$,o $ m < \minus{} 1$
Estw h $ f(x) \equal{} ax^2 \plus{} x \plus{} m$.
Tote $ f(0) \equal{} m$ kai $ f(2) \equal{} 4a \plus{} 2 \plus{} m \equal{} a \plus{} m \plus{} 1 \plus{} a \plus{} 2a \plus{} 1$
O $ a \plus{} m \plus{} 1 \plus{} a > 0$ (afou $ |a| > |a \plus{} m \plus{} 1|$ kai $ a > 0$)
Epomenws $ f(0) < 0,f(2) > 0$ ara apo theorima $ Bolzano$ i $ f(x)$ tha exei toulaxiston mia riza sto $ (0,2)$

Διαίρεσε και τα δύο μέλη της δοσμένης ανισότητας με |λ| και μετα τύπους Vieta για αθροισμα και γινόμενο ριζών...

Xmmm wraia askisi....
$ |a \plus{} m \plus{} 1| < |a| \Leftrightarrow |1 \plus{} \frac {m}{a} \plus{} \frac {1}{a}| \le 1 \Leftrightarrow \minus{} 2 < \frac {m}{a} \plus{} \frac {1}{a} < 0$
Alla $ \frac {1}{a} \equal{} \minus{} x_1 \minus{} x_2$ kai $ \frac {m}{a} \equal{} x_1x_2$.
Ara $ \minus{} 2 < \minus{} x_1 \minus{} x_2 \plus{} x_1x_2 < 0 \Leftrightarrow 0 < x_1 \plus{} x_2 \minus{} x_1x_2 < 2$
H opoia mporei na ksanagrafei ws:
$ \minus{} 1 < (x_1 \minus{} 1)(x_2 \minus{} 1) < 1$,kai me atopo pernoume eukola tis zitoumenes.....


function
parameterization
trigonometry
topology
real analysis
real analysis unsolved

how can we prove that the set of local extrem(strictly) points of a function is at most a numerable set?

Each extremal point is contained in an open interval with rational endpoints which contains no other such points.

Yes, that's what I've been thinking too, but I had a moment of hesitation, thinking that there could alway be another extrem point between those rationals...but I realized this was imposible and everything is clear now.
Thanks :)

Trying to write down the detailed solution I realized that my hesitation did not dissaper  :(  If $x_{0}$ is a local extrem(let it be maximal) point and if there exists a sequence $x_{n}\to x_{0}$ with $x_{n}$ extrem point for every n, then we cannot find 2 rationals such that there is no other local extrem point between them but $x_{0}$. I thought i could contradict this by saying that, in the case of existance of such a sequence, there will be $x_{0}$ and another maximal point,$x_{k}$, in the same interval and then we would had both $f(x_{0})>f(x_{k})$ and $f(x_{k})>f(x_{0})$ true..but we can find another interval, which does not contain $x_{0}$ and on which $x_{k}$ be local maximal point... Where am I wrong? :maybe: I still think that this is the idea but i can't write it down...

It is better to keep track of another parameter: the (rational) radius $r>0$ of the ball $B(x,r)$ around an extremal point $x$ such that $f(y)<f(x)$ for all $y\in B(x,r)$, $y\ne x$. For each fixed $r>0$, the set of possible $x$ with this $r$ is separated (by the same $r$) and, therefore, countable (unless you work in a non-separable metric space, but then the statement can be false). The rest should be clear.

I don't know whether mlok's argument can be repaired: for instance, $|x|\left(2+\sin \frac{1}{|x|}\right)$ has a global minimum at $0$ and infinitely many local minima in every neighborhood of $0$.


function
integration
analytic geometry
Gauss
logarithms
limit
calculus

I think I put my original post in the wrong forum:S

Please help me with this question or at least give me a hint or a starting point so I know what direction I can take :

1) Show that the constant A from the free-space Green's function in 2d
$ G \equal{} Alog(x\minus{}x_0)^2\[ \plus{}(y\minus{}y_0)^2\] )^{1/2}$

is given by $ A\equal{}\minus{}\frac{1}{2\pi}$, if G statisfies the [i]impulse condition[/i]

$ \int\int_D\[ [\nabla^2\] G\plus{} \delta(x\minus{}x_0$$ ,y\minus{}y_0)]dxdy\equal{}0$,

where D is a circular disc or arbitrary radius R centered in $ (x_0,y_0)$

$ Hint : You can use that \int\int_{R^2}\delta(x\minus{}x_0,y\minus{}y_0)\equal{}1$ and then use polar coordinates and the Gauss theorem

I started using divergence theorm and got nowhere :(

please help

thanks in advance

Let $ \varepsilon > 0$ and $ B_r(\mathrm{x})$ be an open ball of radius $ r$ centered at $ \mathrm{x}$.

Suppose $ g$ is a $ C^1$ function with compact support. That is, there is a positive number $ M$ such that $ g$ vanishes outside $ b_M(0)$. Then we can choose $ R$ sufficiently large so that $ g = 0$ outside $ B_R(\mathrm{x_0})$.

Applying divergence theorem, we have

\begin{eqnarray*} & & \int_{\mathbb{R}^2 \backslash B_{\varepsilon}(\mathrm{x_0})} \nabla^2 \log | \mathrm{x} - \mathrm{x_0} | \, g(\mathrm{x})\, d\mathrm{x} \\
& = & \int_{B_{R}(\mathrm{x_0}) \backslash B_{\varepsilon}(\mathrm{x_0})} \nabla^2 \log | \mathrm{x} - \mathrm{x_0} | \, g(\mathrm{x}) \, d\mathrm{x} \\
& = & \int_{\partial B_{R}(\mathrm{x_0})} \nabla \log | \mathrm{x} - \mathrm{x_0} | \, g(\mathrm{x}) \cdot \nu \, dS_{\mathrm{x}} + \int_{\partial B_{\varepsilon}(\mathrm{x_0})} \nabla \log | \mathrm{x} - \mathrm{x_0} | \, g(\mathrm{x}) \cdot \nu \, dS_{\mathrm{x}} \\
& = & \int_{\partial B_{\varepsilon}(\mathrm{x_0})} \nabla \log | \mathrm{x} - \mathrm{x_0} | \, g(\mathrm{x}) \cdot \nu \, dS_{\mathrm{x}} \\
& = & \int_{\partial B_{\varepsilon}(\mathrm{x_0})} \frac {\mathrm{x} - \mathrm{x_0}}{| \mathrm{x} - \mathrm{x_0} |^2} \, g(\mathrm{x}) \cdot \left( - \frac {\mathrm{x} - \mathrm{x_0}}{\varepsilon} \right) \, dS_{\mathrm{x}} \\
& = & - \frac {1}{\varepsilon} \int_{\partial B_{\varepsilon}(\mathrm{x_0})} \, g(\mathrm{x}) \, dS_{\mathrm{x}} \end{eqnarray*}

where $ \nu$ is outward unit normal to the region $ B_{R}(0) \backslash B_{\varepsilon}(0)$. Taking limit $ \varepsilon \to 0^ +$, we find that

$ \lim_{\varepsilon \to 0^ + } - \frac {1}{\varepsilon} \int_{\partial B_{\varepsilon}(\mathrm{x_0})} \, g(\mathrm{x}) \, dS_{\mathrm{x}} = - 2\pi g(\mathrm{x_0})$.

Thus we find that

$ \int_{\mathbb{R}^2} \nabla^2 \log | \mathrm{x} - \mathrm{x_0} | g(\mathrm{x})\, d\mathrm{x} = - 2\pi g(x_0)$

for all such $ g$. Formally, this indicates that $ \nabla^2 \log | \mathrm{x} | = - 2\pi \delta(\mathrm{x})$. Therefore, $ A = - \frac {1}{2\pi}$.

thanks a lot, it's amazing how you managed to solve it

I am however doing an applied math course meaning this answer will not be accepted

Thanks again for your insightful information


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that if Theta is homorphism form G onto H, and N is a normal subgroup of G then Theta(N) is a normal subgroup of H.

Please Help!!

Consider $g \in G$, $n \in N$.
For $x \in \theta(N)$, $x = \theta (n)$ for some $n \in N$, then since
$\theta(g) \theta(n) \theta(g^{-1}) = \theta(gng^{-1})$ and $gng^{-1} \in N$, we are done.


probability
geometry
geometric transformation
rotation

Hello,

Could you guys help me solve this problem without writing down all of the cases?

Problem: How many ways can 3 dimes, 3 nickels, and 3 quarters be arranged in a circle if rotations do not matter?

In case of clarification, these 9 coins form a ring.

Thanks.

attempt?

Fix one of the 9 spots to be a quarter. Now you have 8 spots to choose from and 3 D's, 3 N's, and 2 Q's to put in these spots. The number of distinct ways to arrange these 8 coins is the same as the number of distinct "words" you can make out of the letters $ DDDNNNQQ$ which is $ \frac{8!}{3!3!2!}\equal{}560$ ways

Good attempt, but I'm not too sure if that is correct.

For example, if we only had to arrange 3 pennies and 3 dimes in a circle, there would only be 4 ways (did the cases).

So in this case, that method will not work (i think, unless i am applying it incorrectly) because it would be
$ \frac{5!}{2!3!} \equal{} 10$.

Thanks for trying though.  :)


[quote="cylor"]Without using calculator, show that 7^(sqr5) > 5^(sqr7)
(^: to the power of ;sqr : squareroot)[/quote]

Just to make it easier to read:

$7^{\sqrt{5}} > 5^{\sqrt{7}} $

Without trying it, the first thing i would do would be to to replace sqr5 with a smaller rational number and replace sqr7 with a larger rational number, and work it out from there...

according to [b]seamusoboyle[/b], the first thing I tried was 7 2 > 5 3, however, it's wrong
I remember that there is a way of showing 57>75, but I can't recall it.

[quote="paganinio"]I got it!! just 30 sec later
 :D

$
since 7^{\frac{1}{5}} and 5^{\frac{1}{7}} are both greater than 1,
and (7^{\frac{1}{5}})^{35}=7^{7}, (5^{\frac{1}{7}})^{35}=5^{5},
obviously 7^{7}>5^{5},
therefore 7^{\frac{1}{5}} > 5^{\frac{1}{7}}
$
[/quote]

SORRY that was another problem, I was just confused
what we should prove is actually
$7^{5^{\frac{1}{2}}}>5^{7^{\frac{1}{2}}}$

i know something that if a was greater than b, but if their difference is less than one, then a powered to b is greater than b powered to a.

if that statement makes sense in any way, then i can prove it.

say sqrt(5)=b, sqrt(7)=a.

then the equation becomes something like this.

(a^2)^b is greater than b^2^(a)

rearrange the exponents than we have

(a^b)^2 is greater than (b^a)^2.

squareroot both sides

a^b is greater than b^a.

since a is greater than b, but their difference is not greater than 1, above makes sence.

Not sure if this will work, i didnt do it out fully.
Round $\sqrt{5}$ down to something minutely smaller than it, say $\frac{2236}{1000}$ and round $\sqrt{7}$ up to $\frac{2646}{1000}$, thereby reducing the problem to proving
that $7^{2236}>5^{2646}$, still not nice but easier!

Can one log both sides of an inequality, like an equation, with it still remaining true?

$7^\sqrt{5}$$>5^\sqrt{7}$
$\sqrt{5}\log 7>\sqrt{7}\log 5$
$\log_5 7>\sqrt{\frac{7}{5}}$
$\left(\log_5 \frac{7}{5}+\log_5 5\right)^2>\frac{7}{5}$
$\left(\log_5 \frac{7}{5}\right)^2+2\left(\log_5 \frac{7}{5}\right)+1>\frac{7}{5}$
neglecting the first term on the left side,
$2\log_5\frac{7}{5}>\frac{2}{5}$
$\log_5\frac{7}{5}>\frac{1}{5}$
$7^5>5^6$
Q.E.D.
You might even have those exponents memorized already.


Let a;b;c be positive real numbers. Prove that:
$\large \sum \frac{1}{a} (b+c-a)^3 \geq  \sum a^2$


what is the angle between the hands of a clock if it is 3:23

(I hope this isn't too hard)

[hide="a formula that will give the problem away"]$|30H-5.5M|$ :D  I hope its right[/hide]

[hide]The hour hand moves at 1/2 degree per min.
The min. hand moves at 6 degrees per min.

1/2*203=101.5
6*23     =138

138-101.5= [b]36.5[/b][/hide]

[quote="hockeyadi23"][quote="jli"][quote="hockeyadi23"]I'm not too sure, but i think your formula may not apply to all cases. If you do the absolute valuse of .5*(total minutes off of 12:00) - 6(minutes off the hour), you should always arrive at your answer.[/quote]

the ".5*(total minutes off of 12:00) " is the same as 30(hours)   but oh well

Your formula doesn't work... for example 6:30
in fact, im pretty sure that it only works when it is _:00

oh! and nice job GoBraves[/quote]

O yeah... i didn't realize that... but the .5 method is much easier than 30H when you have minutes in the time... (you end up running into fractions and such because otherwise it applys only to times on the hour)

My formula [b]does work[/b], even with your example too jli(6:30)... we agree that 6:30 would have an angle of 15 degrees (as the hour hand moves 15 degrees off the six).... and $(6*60+30)/2 = 390/2 = 195$. When you subtract $6*30 = 180$ from 195, you get $195-180=15$[/quote]

yeah..it does work.  I didn't know that you had to add the minutes to hour*60. sorry :blush:  nice job

Sok.. it's my fault actually, I should have made it clearer... then again, that's the beauty of Math... there are a million ways to do one thing!

the answer to this problem is 36.5 because if you do 23/60(30)=11.5   this is the amount that the hour hand has already moved from 3 so you do 30-11.5=18.5  ok then you do 3/5(30)=18  this is the amount the minute hand has already moved from 4 because it is 3 minutes more than 20 which is exactly at fouyr.  Therefore 18.5+18=36.5 which is the answer :D

In my experience, the formula  abs(30h - 5.5m) for hours(h) and minutes(m) always works although if the answer you get is greater than 180 degrees, then you also need to subtract 360 and again take the absolute value again.

Note that using this formula 12:00 for is silly as the hands are clearly in the same place and the angle is 0.  You could think of 12:00 as 0:00, but the abs{[12(30)-5.5(0)]-360}=0

Give it a try, anyone find any that don't work???

frost

Thanks for that... it clears up the confusion...  :)

--Adi

[quote="jli"][quote="hockeyadi23"][quote="DON FAVOR"][11d-60s]/2

d-its minute
s-hour

thiis is the forumul of it if u want i vindicate it to u![/quote]

What if the time is 12:00? And how does your method take into account the motion the hour hand takes every minute. Sorry if i have a formula, i want to be sure it works, so I might seem a bit over-interrogative...[/quote]

actually it does... 360 degrees is 0 degress[/quote]
 :blush:  :blush:  :blush:  :blush:
lol... good point...


I wonder if it is just something wrong with my eyes/browser or the $ \text{\LaTeX}$ formulae really became a bit blurrier recently with all those shadows and shades of gray instead of crisp black symbols on the blueish background (Firefox, Windows XP, AoPS skin). They are still readable, of course, but I believe they were better before...

No setting has changed server side (I see no blurriness). Might be something wrong on your end ...

Interesting... They are, indeed, fine on IE even on my computer but my FF makes them rather smeared. So, you are right, that is something on my end. Does anybody have idea of what it might be? :?

Perhaps you could try Firefox in safe mode - maybe an add-on is resizing the images? Have you tried images on other sites to see if you get the same effect? Googling for [i]firefox blurry images[/i] gives a lot of discussion about bilinear filters in Firefox 3, but I don't know if that's relevant or if you have upgraded Firefox recently.

[quote="levans"]You may have firefox magnified. Try control-0 (zero)[/quote]

I had the same problem.

Ctrl-0 worked!!!!!

Thanks a ton .....!!!!!


Is not hard,WLOG ,we assume that b is the second greatest number in the set $ {a,b,c}$ ,then $ c(a\minus{}b)(c\minus{}b)\le 0$
$ \Leftrightarrow c^2a\plus{}b^2c\le bc^2\plus{}abc$
$ \Leftrightarrow a^2b\plus{}b^2c\plus{}c^2a\plus{}abc\le bc^2\plus{}abc\plus{}b^2c\plus{}abc\equal{}$
$ \equal{}b(a\plus{}c)^2\le \frac{1}{2}(\frac{2b\plus{}a\plus{}c\plus{}a\plus{}c}{3})^3\equal{}4$

[quote=Inequalities Master]Let $ a,b,c\geq0$ s.t. $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ ab^2\plus{}bc^2\plus{}ca^2\plus{}abc\leq4$[/quote]

https://artofproblemsolving.com/community/c6h1900506_inequality_with_three_variables


calculus
integration
real analysis
real analysis theorems

Just wondering if anyone can tell me if the right triangle is the only  method for solving quotient integrals besides substitution

I have no idea what you just said. Can you give an example of what you're talking about?


Define the half-factorial of a positive integer n as $ \displaystyle\prod_{i \equal{} 2}^{2n}{\frac {i}{2}}$.  For what n is the half-factorial of n an integer?

The condition $ \Leftrightarrow 2^{2n \minus{} 1}|1 \cdot 2 \cdot 3 .... \cdot (2n)$

The highest power of $ 2$ that divides $ (2n)!$  $ \equal{} \lfloor \frac {2n}{2} \rfloor \plus{} \lfloor \frac {2n}{2^2} \rfloor \plus{} ... \le 2n \minus{} 1$
and equality holds iff $ 2n$ is a power of $ 2$, i.e. for $ n$ being a power of $ 2$, the half-factorial of $ n$ is an integer.

Somewhat trivial extension: What if it starts at $ i\equal{}1$?

[quote="Temperal"]Somewhat trivial extension: What if it starts at $ i \equal{} 1$?[/quote]

Then no such $ n$ exist as $ \lfloor \frac {2n}{2} \rfloor \plus{} \lfloor \frac {2n}{2^2} \rfloor \plus{} ...<2n$


reddevil, actually I think the easy one is...
[hide]$\frac{1}{1}$[/hide]

Oops. How shallow are my thinking levels...

[hide="Medium"]4
$\frac 15 + \frac 12 + \frac 1{20} + \frac 14$[/hide]

[quote="Hexaditidom"]I was looking at a pizza and wondering about different ways to divide it... then I came up with this:

A set of numbers can be called a "pizza" if: Every element is a distinct unit fraction (a unit fraction is 1/n where n is an integer), and all of its terms add up to 1.

Easy: What pizza set has the fewest number of terms?

Medium: How many elements are in the smallest pizza set where 1/5 is the largest term?

Hard: Prove (or disprove) that as n gets bigger, there always exists a pizza set with largest term 1/n. (Of course, the slices would be too small to care about)[/quote]

Um, 1/2 is greater than 1/5 o_o ...

[hide="Medium"]
1/5

1/5 + 1/20 = 1/4
1/4 + 1/12 = 1/3
1/3 + 1/6 = 1/2

1/5 + 1/6 + 1/12 + 1/20 = 1/2 = k

...so we need to find more stuff that adds to 1/2 >_< .

1/7 + x = 1/2
x = 5/14

1/14 + x = 5/14
x = 2/7

1/21 + x = 2/7
x = 5/21

wtf... I don't see a pattern.
[/hide]

[hide="Hard - a start"]
1/n + 1/((n)(n-1) = 1/(n-1)
[/hide]

Ok fixed hopefully.
[hide="Medium"]11
$\frac 15 + \frac 1{10} + \frac 1{20} + \frac 1{15} + \frac 1{12} + \frac 19 + \frac 16 + \frac 1{18} + \frac 1{36} + \frac 18 + \frac 1{72}$
Should work. [/hide]

I think...
[hide="Hard"]That the elements of a pizza set with largest element $\frac 1n$ can always be a geometric sequence with a common ratio of $r$ where $r$ can be expressed as a unit fraction. Of course, the geometric sequence will have a sum of 1.
-edit- never mind. Doesn't work for sure.[/hide]

treethings thing for hard is all you need, and it also gives that there are infinitly many ways to cut a pizza with largest term 1/n by further dissecting the other


[quote="GameBot"]The ratio of youths to adults at a skating rink is 13:7. What percent
of these people are youths?[/quote]

$ \frac{13}{20}\equal{}65 \%$.


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Since the other was great here is another one. Some rules:

1. No rap. I know some of you will complain and say "WTFS RAP EZ TEH BESTES!", but no one really needs to hear it.

I shall start with Motorhead.
[youtube]vqtNGkSzh1o[/youtube]

[quote="hunter34"]Some rules:

1. No rap. I know some of you will complain and say "WTFS RAP EZ TEH BESTES!", but no one really needs to hear it.[/quote]

I'm not saying that I'm a fan of rap, but seriously, that's just lame.
Like I can say

3. Death metal is too loud. No death metal.
4. No bands beside {insert band name}, all the rest suck, and we don't need to here them.
5. No songs with a piano, it sounds lame.
6. No songs without a guitar in it, guitars are awesome.

See my point? As long as the songs are appropriate, why shouldn't rap be allowed?

Please. Just look at the first one and go through it all. Geez. And post a song.

Dang where penguin when you need him.

Fine. Might as well quote myself:
[quote="Nerd_of_the_Ages"]dude, this song is so awesome, it can keep me listening the entire day.[youtube]tGkr6PxV1MI[/youtube][/quote]

I'll see your Barney theme song and raise you Autechre. :P

[youtube]atjByPSKTRA[/youtube]

The Barney theme song?!  WTF.

[youtube]ECA8yqSWCwE[/youtube]

Hmm. Political music. I better challenge it with cheesy pop.

[youtube]n36IqITO76g[/youtube]

I'm bored, so even though I don't really listen to it, I'll post a Dragonforce song.
[youtube]q3M2X4ZAH_w[/youtube]

Edit:Elemnop is right.

[youtube]Tpy_pYXSpPA[/youtube]

Enjoy some "OVER 9000!!!!". That's pretty awesome stepmania song, lol
[youtube]RzW09jX0xwg[/youtube]

[quote="hunter34"]Boy, think before you post. Ever listen to them live. They sound like BS.
[/quote]

yea, i have, have you? it's not like you go to a dragonforce concert to simply sit and listne to the music, who cares if it doesn't sound studio perfect

better than whatever you posted nayways

Chill out, guys. How 'bout some... uh.. Radiohead?
[youtube]NCPDiEz-GcE[/youtube]

Hello, I'm the first one to post this kind :

[youtube]xgHmSdpjEIk[/youtube]

Here's some Weird Al:

[youtube]-xEzGIuY7kw[/youtube]

VERY NICE....I needed a band that I didn't hear until now....can you tell me more heavy metal bands....not so known

[youtube]KyTOQxzL97o[/youtube]

this song is pretty beast

A better version of a previous song:

[youtube]24SoPihLdq4[/youtube]

Oops, needed the beast acoustic solo in my previous video.

[youtube]MZV146LVNfE[/youtube]

METALLICA camed in ROMANIA thie week....FANTASTIC CONCERT

oh meh.

More BLS
[youtube]c48oMTmYcz4[/youtube]

Did I post this already?

[youtube]8S6Fj_aCd8Y[/youtube]

Music start: 1:37
Music end: 3:38

OH NO. MJ IS NOT BRITISH. (actually no one is really sure what he/she is)

[youtube]0Xd8ykpZkwA[/youtube]

oh boy this clip is going to jory
[youtube]BUNDfyy2f5M[/youtube]

Since this is turning into something usable by trolls, I am stopping this thread. On that note this should be the last song.[youtube]4xdP2dJnGHM[/youtube]

I was looking back at this. The winner is junggi's painkiler-judas priest.

They don't make em' like they used too.=/