algebra
polynomial
Vieta
geometry
3D geometry
sum of roots

Is there an easier (more elegant) method to do this rather than expand out?

[b]Question 1:[/b]

Find the sum of roots:
$\frac{x}{x-a} + \frac{x}{x-b} + \frac{x}{x-c} + x-d=0$

[b]Question 2:[/b]

if $\alpha$, $\beta$, $\gamma$ are the roots of the equation $x^3 - bx^2+cx-d=0$, find an expression in terms of b, c, d for $(1+{\alpha}^3)(1+{\beta}^3)(1+{\gamma}^3)$

[quote]Find the sum of roots:
$\frac{x}{x-a} + \frac{x}{x-b} + \frac{x}{x-c} + x-d=0$[/quote]
$S=\frac{a+b+c+3d-3}{3}$

[quote="marocleverness"]
$S=\frac{a+b+c+3d-3}{3}$[/quote]

The correct answer is $a+b+c+d-3$

for the first one , only a little expansion needed ..

LHS=$\sum x(x-a)(x-b) = -(x-a)(x-b)(x-c)(x-d)$

recall vieta theorem which states that the sum of roots is actually the coefficient of $x^3$ ( since this is a degree 4 polynomial )

for the front polynomial , obiously the coefficient is 1+1+1=3 (but we have to put one negativ sign in front as in vieta). As , for the behind one , we have sum of roots = a+b+c+d . So add up both to get

$a+b+c+d-3$

Hint to the second question:
From Vieta thoerem we have:
$\alpha+\beta+\gamma=b$
$\alpha\beta+\alpha\gamma+\beta\gamma=c$
$\alpha\beta\gamma=d$

It is known, that if a polynomial is symmetrical, than it can be written as a sum and product of symmetric polynomials like:
$x+y+z, xy+yz+zx, xyz$

Polynomial $(1+\alpha^3)(1+\beta^3)(1+\gamma^3)$ is a symmetric polynomial, so we can present it as a sum and product of following polynomials:
$\alpha+\beta+\gamma$, $\alpha\beta+\alpha\gamma+\beta\gamma$, $\alpha\beta\gamma$
So it suffices to multiple the brackets and write in the abovementioned form. It's possible, so it shouldn't be difficult (but of course time-consuming :) )

[quote="calculuslover800"][quote="scorpius119"]Here's a way to do the second: [hide]If $x^3-bx^2+cx-d$ has roots $\alpha,\ \beta,\ \gamma$, then $x-bx^\frac 23 +cx^\frac 13 -d$ has roots $\alpha^3,\ \beta^3,\ \gamma^3$. The fractional powers are a problem at first, until you put the parts with fractional powers at one side of the equation and cube both sides. $cx^\frac 13 -bx^\frac 23 =d-x\Rightarrow c^3x-b^3x^2+3(bcx)(d-x)=d^3-3d^2x+3dx^2-x^3$.

After cubing (note the substitution made to get rid of the remaining fractional powers), you will get a cubic $P(x)$ with roots $\alpha^3,\ \beta^3,\ \gamma^3$; such a polynomial is $x^3-(b^3+3bc+3d)x^2+(3bcd+3d^2+c^3)x-d^3$. If $l$ is the leading coefficient of $P(x)$, then plug in $-\frac{P(-1)}{l}$ to finish.

After this, I end up with $1+3d+3b^2+3bc+b^3+c^3+d^3+3bcd$, assuming I made no stupid mistakes.[/hide][/quote]

[hide="my question"]How did you get from $cx^\frac 13 -bx^\frac 23 =d-x$ to $c^3x-b^3x^2+3(bcx)(d-x)=d^3-3d^2x+3dx^2-x^3$?  I see the RHS but not the LHS.  You said something about a substitution to get rid of the remaining fractional powers?[/hide][/quote]

$(cx^\frac 13 -bx^\frac 23 )^3=c^3x-b^3x-3c^2x^{\frac 23}bx^{\frac 23}+3cx^{\frac 13}b^2x^{\frac 43}\\=c^3x-b^3x-3bcx(cx^{\frac 13} -bx^{\frac 23})=c^3x-b^3x-3bcx(d-x)$
 ( since $cx^{\frac 13} -bx^{\frac 23} =d-x$) :)


calculus
integration
function
trigonometry
limit
calculus computations

Find the continuous function $ f(x)$ such that $ f(x)\equal{}\int_0^ x \{f(t)\cos t\minus{}\cos (t\minus{}x)\}\ dt.$

$ f(x)\equal{}\int^x_0 {f(t)cos(t)\minus{}cos(t\minus{}x)}\,dt$ $ (1)$

$ f(x)\equal{}\int^x_0 f(t)cos(t)\,dt\minus{}\int^x_0 cos(t\minus{}x)\,d(t)$

$ f(x)\equal{}\int^x_0 f(t)cos(t)\,dt\minus{}\int^x_0 [sin(t\minus{}x)]'\,dt$

$ f(x)\equal{}\int^x_0 f(t)cos(t)\,dt\minus{}sinx$

$ f(x)cos(x)\minus{}cosx\int^x_0 f(t)cos(t)\,dt\equal{}\minus{}sinxcosx$

Let $ F(x)\equal{}\int^x_0 f(t)cos(t)\,dt$ then $ F'(x)\equal{}f(x)cosx$
Thus we get $ F'(x)\minus{}cosxF(x)\equal{}\minus{}sinxcosx$
We can easily solve the linear ODE by multiplying both sides with $ e^{\minus{}sinx}$ and we find
$ F(x)\equal{}1\plus{}sin(x)\plus{}ce^{sinx} \rightarrow F'(x)\equal{}cosx\plus{}ce^{sinx}cosx
\leftrightarrow f(x)cosx\equal{}cosx(1\plus{}ce^{sinx})$

Due to $ (1)$ $ f(0)\equal{}0\leftrightarrow c\equal{}\minus{}1$. So for $ x\not\equal{}k\pi\plus{}\frac{\pi}{2}$ we have $ f(x)\equal{}1\minus{}e^{sinx}$.

$ f(x)$ is continuous in R so $ \displaystyle\lim_{x\to\ k\pi\plus{}\frac{\pi}{2}}f(x)\equal{}f(k\pi\plus{}\frac{\pi}{2})
\leftrightarrow f(k\pi\plus{}\frac{\pi}{2})\equal{}1\minus{}e^{sin(k\pi\plus{}\frac{\pi}{2})}$.


Let $ f \equal{} X^{n} \plus{} a_{n \minus{} 1}X^{n \minus{} 1} \plus{}... \plus{} a_{1}X \plus{} a_{0} \in Z[x]$, where $ n\geq 3$, such that $ a_{k} \plus{} a_{n \minus{} k}$ is even for all $ k \in \{1,2,...,n\minus{}1\}$ and $ a_{0}$ is even.

Suppose that $ f \equal{} gh$, where $ g,h$ are integer polynomials and $ \deg g \leq \deg h$ and all the coefficients of $ h$ are odd.
Prove that $ f$ has an integer root.

Let $ g(x)\equal{}\sum_{i\equal{}1}^s g_ix^i, h(x)\equal{}\sum_{i\equal{}1}^t h_ix^i$ with $ s\le t$ and suppose $ t\ge2$. Note $ g_s\equal{}h_t\equal{}1$. Then we have

$ h(x)\equiv x^t\plus{}x^{t\minus{}1}\plus{}...\plus{}1 \ (\mod2)$
and
$ g(x)\equiv x^s\plus{}g_{s\minus{}1}x^{s\minus{}1}\plus{}...\plus{}g_1x \ (\mod2)$, as $ g_0\equal{}\frac{a_0}{h_0}$ must be even.
So $ 0\equiv a_{s\plus{}1}\plus{}a_{t\minus{}1}\equiv \sum_{i\equal{}1}^{s}g_i\plus{}\sum_{i\equal{}1}^{s\minus{}1}g_i\equiv g_s\equiv 1\ (\mod2)$. Contradiction  :lol:

So $ t\le1$ and in fact, as $ n\ge3$ is requested, such a poly doesn't even exist  :P

[quote="indybar"][quote="spanferkel"]
So $ 0\equiv a_{s \plus{} 1} \plus{} a_{t \minus{} 1}\equiv \sum_{i \equal{} 1}^{s}g_i \plus{} \sum_{i \equal{} 1}^{s \minus{} 1}g_i$. [/quote]

Is this true?  :wink:[/quote]
oops, sorry, I've messed up with $ s$ and $ t$, it should be $ a_{t \plus{} 1} \plus{} a_{s \minus{} 1}$  :blush: And the initial assumption $ s\ge2$. Thus by the contradiction we have shown $ s\equal{}1$, so $ g$ is a linear factor of $ f$.  :)

[quote="indybar"][quote="spanferkel"]
oops, sorry, I've messed up with $ s$ and $ t$, it should be $ a_{t \plus{} 1} \plus{} a_{s \minus{} 1}$  :blush: And the initial assumption $ s\ge2$. Thus by the contradiction we have shown $ s \equal{} 1$, so $ g$ is a linear factor of $ f$.  :)[/quote]

How is it contradiction?[/quote]
 :?:
 :?: well, assuming $ t\ge s\ge2$, we have $ a_{s\minus{}1}\equiv \sum_{i \equal{} 1}^{s \minus{} 1}g_i$ (check it out e.g. for s=t=3, and you will see why you'll get $ 1\equiv0$).
you can also say $ a_{t \plus{} 1} \plus{} a_{s \minus{} 1}\equiv a_{t \plus{} 1} \minus{} a_{s \minus{} 1}\equal{}g_s$


calculus
integration
logarithms
search
real analysis
real analysis unsolved

i would go about doing

$ \frac{d}{da} \bigg|_{a\equal{}0}\;\int_0^\infty \;\frac{x^a} {1\plus{}x^3} \; dx$

i have shown before ( can't do any search 'cause mah computer is slow...  :D ) ,   [u]without[/u] using complex variables,

$ \int_0^\infty \;\frac{x^a} {1\plus{}x} \; dx \;\equal{}\; \pi\;  \text{cosec}\; (a\plus{}1)\pi$

hence,

$ \int_0^\infty \;\frac{x^a} {1\plus{}x^3} \; dx \;\equal{}\; \frac{\pi}{3}\;  \text{cosec}\; (a\plus{}1)\frac{\pi }{3}$

and so,
$ \frac{\pi}{3}\; \frac{d}{da}  \bigg|_{a\equal{}0} \; \text{cosec}\;(a\plus{}1) \frac{\pi }{3}  \;\equal{}\;\minus{} \left(\frac{\pi}{3} \right)^2 \;\text{cosec}\;\frac{\pi}{3}\; \cot\; \frac{\pi}{3}\; \equal{}\;\boxed{\minus{}\frac{2\pi^2}{27} }$

Thanks, that's what I thought. Could we say from the start that the value of this integral is negative?

Also, we can therefore conclude that $ \int_0^{\infty}\frac{\ln x}{x^2 \minus{} x \plus{} 1}\,dx \equal{} 0$, right?

consider :
$ I: \equal{} \int_{0}^{\infty} \frac{\ln x}{x^{3}\plus{}1}dx$
by substitution $ y \equal{} \frac{1}{x}$ we have
$ I\equal{} \int_{\infty}^{0} \frac{y\ln y}{1\plus{}y^{3}} dy \equal{}\minus{}\int_{0}^{\infty} \frac{x\ln x}{x^{3}\plus{}1}dx$
thus
\[ 0 \equal{} I\minus{}I \equal{}   \int_{0}^{\infty} \frac{\ln x}{x^{3}\plus{}1}dx \plus{} \int_{0}^{\infty} \frac{x\ln x}{x^{3}\plus{}1}dx \equal{} \int_{0}^{\infty} \frac{(1\plus{}x)\ln x}{x^{3}\plus{}1}dx \equal{}\int_{0}^{\infty} \frac{(1\plus{}x)\ln x}{(x\plus{}1)(x^{2}\minus{}x\plus{}1)}dx \equal{} \int_{0}^{\infty} \frac{\ln x}{x^{2} \minus{}x\plus{}1}\]
thus we obtain :

\[ \int_{0}^{\infty} \frac{\ln x}{x^{2} \minus{}x\plus{}1} \equal{} 0\]
this method can be used also to express  2-nd integral by 1-st.

:D

$ (2)\qquad \frac {d}{da} \bigg|_{a \equal{} 1}\;\int_0^\infty \;\frac {x^a} {1 \plus{} x^3} \; dx$

$ (3) \qquad \minus{} \frac {1}{3}\; \frac {\partial^2}{\partial a\;\partial b} \bigg|_{a \equal{} 1,\;b \equal{} 1}\;\int_0^\infty \;\frac {x^a} {b^3 \plus{} x^3} \; dx$

$ (4)\qquad \frac {d^2}{da^2} \bigg|_{a \equal{} 1}\;\int_0^\infty \;\frac {x^a} {1 \plus{} x^3} \; dx$

from

$ \int_0^\infty \;\frac {x^a} {1 \plus{} x^3} \; dx \; \equal{} \; \frac {\pi}{3}\; \text{cosec}\; (a \plus{} 1)\frac {\pi }{3}$

you can easily find $ (1)\quad\text{and}\quad(3)$.

now for $ (2)$, you most likely have to resort to    complex variables with  $ \frac {z^a}{z^3 \plus{} b^3}$ using a single  pole within a pie-shaped contour:

@ $ z \equal{} b\;e^{\frac {\imath\pi}{3}}$ with the pie slice going from $ 0\quad\text{to}\quad\frac {2\pi}{3}$

( you can equally  use a semi-circular contour if you want but take note  there is a second pole at $ z \equal{} \minus{} b$ )

$ I_1\equal{} \minus{}\frac{2\pi ^2}{27}$
$ I_2\equal{}\frac{2\pi ^2}{27}$
$ I_3\equal{}\frac{2\pi \left(\pi \minus{} 3\sqrt{3}\right)}{81}$
$ I_4\equal{}\frac{10\pi ^3}{81 \sqrt {3}}$

The methods mentioned above worked just fine, though a bit of a mess.


algorithm
calculus
algebra
polynomial
function
linear algebra
matrix

The seven problems in the millenium are the super-plus-ultra top of all mathemathical problems. If you solve one of them, you'll earn 1 million dollars as a Clay Institute's reward, and also to be remembered as one of rthe greatest mathemathicians!!! fame, money, dreams came true.......

But back to reality what dou you think about them???

If someone solves the Riemann Hypothesis the world economy will probably collapse, I guess that's a revolution

To be honest I think not many of us are substantially familiar with these problems to truly appreciate their difficulty and consequences.

Side note: Does anyone watch numbe3rs? What's with some guys daughter getting kidnapped because he was doing stuff on the Riemann Hypothesis?

The Riemann Hypothesis is not close to being proven (according to Sarnak who is an expert). He seems to think that the Twin Prime Conjecture will be settled soon (in the affirmative) though.

I went to a talk about the Birch & Swinnerton-Dyer conjecture one time. Christopher Skinner gave the talk. The way he described it made it seem within reach, but maybe it was just the way he talked. This was a while ago anyway so I don't really remember.

I don't know anything about the Hodge Conjecture or the Navier-Stokes equations. Yang-Mills theory isn't a problem, it's more of a broad goal, so it's hard to compare it to the other ones.

I know a bit about P vs NP but not enough to comment on how hard it is.

For myself, I'm interested in some topics in analysis and their applications to number theory, so probably Riemann is the only one I'd have a chance at myself (as in I might be in that field). But I am not going to set myself on proving Riemann, which means it probably won't happen.

I don't think anyone has tools to grapple with problems like $ P$ vs $ NP$.  This is actually a member of a much larger family of problems, those of the form, "It takes me time $ t$ to solve an algorithm in this model of computation; how long should it take in this other model?"  (The two models for $ P$ vs $ NP$ are deterministic and nondeterministic computation -- nondeterminism is powerful in theory, but no implementation is really possible, so we want to know whether we can solve questions deterministically that we already have the nondeterministic solutions to without causing exponential expansion in the time necessary.)  As far as l know, no one has solved any substantial question of this form roughly since the Turing Machine and $ \lambda$-calculus were shown equivalent, and no one has any tools which solve anything that looks like one of these problems.  In particular, the best known lower bounds for computing problems in $ NP$ are linear.  In other words, $ P \neq NP$ conjectures that no polynomial grows fast enough to solve a given NP problem, and we've managed to prove that you need to grow at least as fast as a linear function if you want to have a chance.  There's a pretty substantial gulf, there :)

Navier-Stokes is the set of (partial differential) equations that govern fluid flows.  The problem is to come up with some sort of family of solutions (maybe a boundary-independent family?  I'm not sure).  Particular instances are either solvable or modelable, but I haven't heard anything to suggest that the more general problem is close to being solved.  (I also don't know if anyone is really trying to solve it in general in the way that some people definitely have work that they hope might touch on the RH.)

One interesting thing about the RH is that we have results that look a lot like it ... the Prime Number Theorem is essentially a special case.  Solving the RH would not, I think, destroy the world economy.  I can only assume that comment had something to do with encryption using primes (although what, I'm not sure -- I don't think it has any implications for factoring (although it definitely has implications for primality testing)), but that's going to become problematic because of quantum computers (and essentially simultaneously irrelevant because of quantum encryption) long before anyone solves RH.

[quote="Pul de Algodoncito"]Why did you say that world economy would collapse if Riemman Hypothesis is proved????[/quote]

So much of the world economy is dependent upon good cryptography. Whenever you send data over the internet, say to browse this website even, everything has to be encrypted and decrypted to protect your security. The encryption scheme used most is called RSA, and the reason it works is because information is encoded as a product of two very large prime numbers, 250 or so digits. If you know one of the prime factors it's no problem to decode the information, IE it takes very little computing time, but if you don't know one of them it takes forever to factor the number. So it takes very little computer time to encode and decode information but the time taken to hack a message would be so large as to prohibit anyone from doing it. If the Riemann hypothesis were proven, there would be big implications for figuring out how to factor very large numbers, so it might be easier to crack RSA.

[quote="hadjimurad"]If the Riemann hypothesis were proven, there would be big implications for figuring out how to factor very large numbers, so it might be easier to crack RSA.[/quote]

Anything in particular?  :)

I have no idea what the preceding comment means, but (as I mentioned above) the Riemann Hypothesis is definitely not the future discovery likely to have a major effect on cryptosystems.  In particular, no efficient algorithm for factoring is known even with the assumption that RH is true, while very efficient algorithms (Shor's algorithm, in particular) are known for factoring given a large enough quantum computer.  (In fact, they can already factor 15. :) )  If P vs. NP were resolved in the affirmative (i.e. P = NP), that would have definite theoretical implications for the security of RSA (and all other conventional cryptosystems: the hypothesis that secure cryptography (in the "traditional" sense, i.e. not quantum cryptography) is possible is actually stronger than the assumption that $ P \neq NP$) far beyond anything the RH is known to imply.  Of course, hardly anyone believes that $ P \equal{} NP$, while a great many people believe that the Riemann Hypothesis is true.

I have my doubts that $ P \equal{} NP$ has practical implications, and I think there are more interesting specific questions within theoretical computer science. Suppose you found an algorithm that could solve a problem in $ O(N^{20})$ time. Then that would place the problem firmly within class $ P,$ but you still couldn't practically solve it for $ N$ in the hundreds. On the other hand, the theoretical placement of linear programming within class $ P$ has not killed off the usefulness of the non-$ P$ simplex algorithm.

Anyone know how to compute the permanent of a matrix? (Defined like determinant, but without the alternating sign - all positives.)

[quote="Kent Merryfield"]
Anyone know how to compute the permanent of a matrix? (Defined like determinant, but without the alternating sign - all positives.)[/quote]

$ \operatorname{perm}\begin{bmatrix}A\end{bmatrix}\equal{}\sum_{\sigma\in S_n}\prod_{i\equal{}1}^n a_{i,\sigma(i)}$

Like $ \operatorname{perm}\begin{bmatrix}1&2\\3&4\end{bmatrix}\equal{}(1\cdot4)\plus{}(2\cdot3)\equal{}10$.

Yes, that's the definition, but can you actually compute the permanent of a concretely given $ 50\times 50$ matrix - say, with integer values?

[quote="Kent Merryfield"]I have my doubts that $ P \equal{} NP$ has practical implications[/quote]
If (as is likely) $ P \neq NP$ then of course this conversation is meaningless.  If $ P \equal{} NP$, then that fact alone may not have any important practical implications, but a proof of that fact almost certainly will.

You guys know the Poincare conjecture has been proved, right? So we're down to 6 unsolved millennium problems.

[quote="Kent Merryfield"]
Anyone know how to compute the permanent of a matrix? (Defined like determinant, but without the alternating sign - all positives.)[/quote]
There is no subexponentially fast deterministic approach known but, as far as I remember, there is a randomized algorithm that allows you to compute the logarithm of the permanent with any prescribed absolute error in polynomial time when all the matrix entries are positive. If you are really interested in that, look at [url]http://www.cs.berkeley.edu/~sinclair/perm2.ps[/url]. Unfortunately, the power of $ n$ in the running time estimate is about $ 25$, so the algorithm is not really practical.

As to Perelman, it is true that he worked alone but it is false to say that he intentionally isolated himself from the outside world to work on Poincare.

[quote="bos1234"]Is it true that Pereleman went into 5 years of isolation in order to solve the Ponicaire conjecture?[/quote]
No, he didn't pull a Wiles.

[quote="Pul de Algodoncito"]What is the importance of these problems? Are these the more difficult, the more intersting, or do they have a huge application like to be called the "millenium problems"?

¿What would every of these problems imply?[/quote]

Some recommended reading: http://www.amazon.com/Millennium-Problems-Greatest-Unsolved-Mathematical/dp/0465017290

Good discussion for the amateur.  The discussion of the Hodge Conjecture is somewhat involved, since it is one of the more abstract of the problems, but I still found it very readable.

Or, more simply, read the brief descriptions provided by CMI: http://www.claymath.org/millennium/

Poincare Conjecture was already solved by Grigori Perelman (Russian) at 2006 and he received for that the Fields Medal.

[quote="kurt.math"]Which is the hardest millennium problem, from hardest to easiest?[/quote]

Well isn't that a bit subjective? Many people would probably say the Riemann Hypothesis because it is Analytical Number Theory (you really need to know the relevant theorems/lemmas basically you need to know "everything"). But I can't see how we can rank these problems. They are all "hard" in different ways which is why they are still unsolved.


Well, there have been threads about the AMC12 and AIME getting easier/harder, but how about the USAMO? I personally think that the USAMO has been harder in recent years than those of a couple decades ago.

I think there isn't much of a debate that the USAMO problems are harder now than they were even ten years ago. But that doesn't mean that the USAMO is harder. The USAMO used to be 5 problems in 3 and a half hours. Now it's 3 problems in 4 and a half hours. That's an enormous difference in time.

The USAMO even back in the 80s jumped around quite a bit in difficulty.  Take a look at '87.  Top score was below 60/100.  Only 6 people cleared 40, only around 25 cleared 20.

So USAMO questions used to be 20 points each? Interesting.


Let $ \triangle ABC$ with incenter $ I$ and $ A$ - excenter $ I_a$ . The $ B$ - excircle touches

$ AC$ at $ X$. Denote $ Y\in IX\cap BC$ . Prove that $ A\equal{}90^{\circ}\ \Longleftrightarrow\ I_aY\perp AC$ .

A-excircle $(I_a)$ touches $AC$ at $D$ and $I_aD$ cuts $BC$ at $Z.$ Then $I_aY \perp AC$ $\Longleftrightarrow$ $Y \equiv Z.$ Let $V \equiv AI \cap BC.$ By Menelaus' theorem for $\triangle ACV$ cut by $\overline{ZDI_a}$ and $\overline{YXI},$ we get

$\frac{ZC}{ZV}=\frac{I_aA}{VI_a} \cdot \frac{DC}{AD} \ \ , \ \ \frac{YC}{YV}=\frac{IA}{VI} \cdot \frac{XC}{AX}$

$Y \equiv Z\Longleftrightarrow \ \frac{ZC}{ZV}=\frac{YC}{YV} \Longleftrightarrow \ \frac{I_aA}{I_aV} \cdot \frac{DC}{AD}=\frac{IA}{IV} \cdot \frac{XC}{AX}$

Since $(A,V,I,I_a)=-1,$ then $\frac{I_aA}{I_aV}=\frac{IA}{IV} \ \Longrightarrow$

$Y \equiv Z   \Longleftrightarrow \ \frac{DC}{AD}=\frac{XC}{AX} \Longleftrightarrow \ \frac{s-b}{s}=\frac{s-a}{s-c} \Longleftrightarrow \ a^2=b^2+c^2.$


Euler
parameterization
algebra
system of equations
number theory
Diophantine equation
number theory open

this is a well-known open problem:

"

An Euler Brick is just a cuboid, or a rectangular box, in which all of the edges (length, depth, and height) have integer dimensions; and in which the diagonals on all three sides are also integers.

So if the length, depth and height are a, b, and c respectively, then a, b, and c are integers, as are the quantities $ \sqrt{{a}^2 \plus{} {b}^2}$ and $ \sqrt{{b}^2 \plus{} {c}^2}$ and $ \sqrt{{a}^2 \plus{} {c}^2}$ .

The problem is to find a perfect cuboid, which is an Euler Brick in which the space diagonal, that is, the distance from any corner to its opposite corner, given by the formula $ \sqrt{{a}^2 \plus{} {b}^2 \plus{} {c}^2}$, is also an integer, or prove that such a cuboid cannot exist .

"
or alternatively stated:

find a postitive integer solution to the following system of equations or show that such solution doesn't exist:

$ {a}^2 \plus{} {b}^2 \equal{} {x}^2$
$ {b}^2 \plus{} {c}^2 \equal{} {y}^2$
$ {a}^2 \plus{} {c}^2 \equal{} {z}^2$
$ {a}^2 \plus{} {b}^2 \plus{} {c}^2 \equal{} {t}^2$

...

my question is,"Why" this question is still open? i mean why this is so hard...
(please i don't mean "why you people (mankind!) can not solve it",meaning that i don't want answers like "so if you think it's not that hard,why don't *you* just solve it?! " ... i just wonder what makes it this hard...)

Solving any of those equations individually is easy; there are well-known parameterizations that will give integer Pythagorean triples or quadruples.  The difficulty is in the fact that each variable is [b]simultaneously[/b] a member of two Pythagorean triples, and to top it off must also be a member of a Pythagorean quadruple.

In general, Diophantine equations can become very difficult without looking that way.  For example, the study of cubic Diophantines in two variables spawned the massive field of elliptic curve theory.

thanks t0rajir0u :)
if someone ask me "what do you love the most in mathematics? " i would say : "Diophantine Equations" ...yes that's a truth,those really make my body "creep"!
and if someone ask me "what scare you in mathematics the most?" i would say again: "Diophantine Equations!"...yeah i think i've got a "Phobia" or something about that...you look at a diophantine equation,and you understand it well,it is as easy as eating a cake! ,but some,are solvable during a one-hour'n-half ,and some are like a 400 years nightmare...!
so i wouldn't let my babies ask me to solve "any diophantine equations they want"!! (i don't have babies right now,i imagined 15-20 years later or more!)


geometry
Pythagorean Theorem
perpendicular bisector

Consider a semicircle with radius 6. Two semicircles are drawn inside it with a diameter of 6. A circle is drawn insde the bigger semicircle such that is tangent to the three semicircles. Find the radius of the circle.

[geogebra]82b7c64b8dd10e845d9b8061458fd27e25e8950f[/geogebra]

Please provide a complete solution
Where can I find many geometry problems of this level of difficulty?

Thanks! :lol:

Hint

[hide]Apply Pythagoras,[/hide]

Solution

[hide]$ (r \plus{} 3)^2 \equal{} 3^2 \plus{} (6 \minus{} r)^2$, then $ r \equal{} 2$.[/hide]

Poincare,for using Pythagorean Theorem you must prove that the origin of the bigger semicircle, the origin of the little circle and the point of tangency between the little circle and the big semicircle are colinear and the line passing through the three of them is perpendicular to the diameter of the bigger semicircle.

Why make it so complicated?
It is so easy to see it is just assumed.
Lets see, I connect the origin of the full circle to origins og the small semicircle.
It creates an isoceles triangle, and the median of an isoceles triangle creates a right angle by some proof we learnt in geometry.
Good Enough?  :lol:

Let $ O$, $ O_1$ and $ O_2$ be the centers of the largest semicircle and the smaller ones.
It's clear that $ O$ is the midpoint of $ O_1O_2$.
Let $ O_3$ be the center of the small circle (unknown radius).
Then triangle $ O_1O_3O_2$ is isosceles, therefore $ O_3O_1$ is the perpendicular bisector of $ O_1O_2$.
Then apply Pyth in right triangle $ O_1OO_3$.

Sorry RobRoobiks I didnt see your post...


algebra
polynomial
superior algebra
superior algebra unsolved

Consider homogeneous polynomials

$ F(x_0,x_1)\equal{}a_mx_0^m\plus{}...\plus{}a_0x_1^m,   G(x_0,x_1)\equal{}b_nx_0^n\plus{}...\plus{}b_0x_1^n \in k[x_0,x_1]$.

Set $ \hat F(x_0,x_1)\equal{}F(x_1,x_0),    \hat G(x_0,x_1)\equal{}G(x_1,x_0)$

and show that $ Res(\hat F,\hat G)\equal{}(\minus{}1)^{\varepsilon (m,n)}Res(F,G)$, $ \varepsilon (m,n)\equal{}\frac{m(m\minus{}1)\plus{}n(n\minus{}1)\plus{}(m\plus{}n)(m\plus{}n\minus{}1)}{2}$

All you have to prove is $ \varepsilon(m,n)\equiv mn (mod 2)$.


geometry
number theory
algebra proposed
algebra

Let the sides be $n-1,n,n+1$, thus $s=\frac{3}{2} n$.
Then we want
$A=\sqrt{\frac{3}{2} n \cdot (\frac{3}{2} n - n +1) \cdot (\frac{3}{2} n - n ) \cdot (\frac{3}{2} n - n -1) }=$
$=\sqrt{\frac{3}{2} n \cdot (\frac{1}{2}n +1) \cdot (\frac{1}{2} n ) \cdot (\frac{1}{2} n -1) }=$
$=\sqrt{\frac{3}{16} n^2 \cdot (n^2 -4) }$
to be an integer.
Set $n=2k$, then $3 k^2 \cdot (k^2 -1)$ must be a perfect square, so $3 \cdot (k^2 -1)=m^2$ must be.
This is equivalent to $m^2-3 k^2 = -3$, which has $\infty$ many solutions by Pell.

PS: Please move it (back) to Number Theory.

[quote="ZetaX"]Set $n=2k$, then $3 k^2 \cdot (k^2 -1)$ must be a perfect square, so $3 \cdot (k^2 -1)=m^2$ must be.
This is equivalent to $m^2-3 k^2 = -3$, which has $\infty$ many solutions by Pell.

PS: Please move it (back) to Number Theory.[/quote]
Solutions are  supposed to be found by Pell's equation yes.
But problem is actually a mixture of various math fields.
Dedicated to Heron.

Well, as good as everything that wants integers is number theory ;)
There is no real geometry involved, and especially no Algebra.
But a lot people think that Algebra = everything you can't totally fit into anything else. But that point of view is wrong.

But another famous problem:
Call a rational number "congruent" if it is the area of a right-angled triangle with rational sides
Find all (well thats open) that are open or at least prove or disprove for some integers $n$ that they are (not).
For example, (when I'm not mistaken) $1,2,3,4$ are not but $5,6,7$ are congruent.

[quote="esti"]Find [b]at least[/b] six triangles with sides that are subsequent natural numbers with condition that triangles' areas
are natural numbers as well.

Example :
(a=[b]3[/b],b=[b]4[/b],c=[b]5[/b])$\Longrightarrow A=6$[/quote]
First 6 triangles listed in increasing order are:
[b](a,b,c)[/b]
 (3,4,5)
(13,14,15)
(51,52,53)
(193,194,195)
(723,724,725)
(2701,2702,2703)
...


Bueno,
aqui estoy colgando la parte inicial del Libro de la Editorial Mir,
(en español)

consideramos, el indice del libro....

Carlos Bravo ;)
Lima - Peru

Disculpen...

habia reimpreso este pdf de manera equivocada ..
pero ya esta .

akii lo tienen..

ire colgando este libro en partes ..

debido a q existe un tope para las cargas es este forum ...

Carlos Bravo

Una pergunta, hablando de e-books: Alguien tiene el libro "Retorno a la Geometría"?? Porque yo lo tengo en inglés, pero me gustaría tenerlo en castellano. Y otra cosita: algún libro de geometría no euclidiana????. Si alguien sabe de donde los puedo conseguir en la red o alguna información, por favor mándenme un mensaje privado, o simplemente contesten este mensaje en el foro... Gracias!!!!!!!!

[color=blue]
 Buenos dias,

les informo que por restricciones en mi
cuota de uploading...

quitare estos pdf,
por tanto, descarguenlos  antes del lunes en la noche,

Gracias[/color]


geometry
geometric transformation
reflection
function
trigonometry
geometric series

Can anyone tell me where can I find Ramanujan's proof that $\sum\limits_{i=1}^\infty i =-1/12$?
Thanks :)

[quote="jb05"]$1+2+3+4+...=-\frac 1{12}$[/quote]

Sorry??????? Is this as the proof where $2=1$ and you are dividing by $0$, so the proof is not correct????

If you analytically continue $\zeta(s)$ into $\mathbb{C}\setminus\{1\}$, use the reflection formula, and some easily calculated values of the $\Gamma$ function, you can get the result.

Sorry, but I don't know anything about what you say... but I still don't undertand how $1+2+3+4+...=-\frac{1}{12}$

[quote="jb05"]Consider the sum $-\sum_{i=1}^\infty i(-r)^i$. Let this sum equal S.[...]
[/quote]
Thanks!

I couldn't understand only this passage:
[quote="jb05"]
$S+Sr=r-r^2+r^3-r^4+...$
$S=\frac r{(r+1)^2}$
[/quote]

Use the sum of a geometric series:
$r-r^2+r^3-...=r(1+(-r)+(-r)^2+(-r)^3+...)=\frac r{r+1}$ and divide by r+1 to isolate S.

No, this is not like the proof that 1=2. Since we add the $r^i$ to the summation and take the limit as r approaches 1, this is Abel Summation, a method of computing otherwise divergent sums. It is probably not appropriate that this has gained enough acceptance that people actually say that 1+2+3+...=-1/12, but with Able Summation, that is true.

Oh, yes it is easy to show that $r-r^2+r^3-r^4+r^5-...=r/(r+1)$  ...

It's not true, obviously, but it is true in a normalized sense, e.g, as a value of the analytically continued zeta function.

Yes, and you can find the developement of what blahblahblah is talking about on this forum :)

No, I don't think so.  Since when you subtract $2S$ from $S$, 2S always has one more term than $S/\{1\}$. i.e, we it is finite and it goes up to 8, we have $(1+2+4+8)-(2+4+8+16)=-15\neq -1$.  So that is not a valid proof. ;)

Consider the function $f(a)=1+a+a^2+\ldots$.  Note that for all $|a|\leq 1$, $f(a)=\frac{1}{1-a}$ and $\frac{1}{1-a}$ is defined for $\mathbb{C}\setminus\{1\}$.  Note also that there is a pole at $a=1$, so we can't analytically continue it any further.

Thus we can say that $1+2+4+8+\ldots=\frac{1}{1-2}=-1$, in a sense.

Sorry for late post,but i have worked on this thing and trough research,maybe i have found a proof.

This proof is based on this exstension : \[ \zeta(s)=2^s\pi^{s-1}sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s) \],where $ \Gamma(t+1) = t! $

For $ s = -1 $ we have \[ \zeta(-1) = \frac{1}{2} \cdot \frac{1}{\pi^2} \sin(\frac{-\pi}{2})\Gamma(2)\zeta(2)=\frac{1}{2\pi^2} \cdot -1 \cdot 1! \cdot \frac{\pi^2}{6} = -\frac{1}{2\pi^2}\frac{\pi^2}{6}=-\frac{1}{12}\]
Therefore \[ \sum_{n=1}^{\infty} n = 1+2+3+4+...+{\infty} = \zeta(-1) = -\frac{1}{12}\]

[quote="blahblahblah"]Thus we can say that $1+2+4+8+\ldots=\frac{1}{1-2}=-1$, in a sense.[/quote]
Also note that this and the proof are completely fine in $2$-adic numbers.


limit
number theory unsolved
number theory

Given a sequence $ n_1,n_2,n_3,...$in natural number，let

$ x_1 \equal{} \frac {1}{n_1}, x_2 \equal{} \frac {1}{n_1 \plus{} \frac {1}{n_2}}, x_3 \equal{} \frac {1}{n_1 \plus{} \frac {1}{n_2 \plus{} \frac {1}{n_3}}}...$
,show that $ \lim_{n\to\infty}{x_n} \equal{} x$ exists and is an irrational number.

thanks~ :lol:

Prove that $ x\equal{}[0;n_1,n_2,\dots]$ and use the fact that rational numbers have finite continued fractions.


ceiling function
calculus
derivative
logarithms
geometry
3D geometry
trigonometry

Could somebody give me any reference for the following queston:
given $ \rho \in (0,1)$, what is the asymptotic behavior of
$ \frac{\sum_{k\equal{}\lceil N \rho \rceil}^N \binom{N}{k} (\minus{}1)^{k}(k\minus{}N \rho)^{N\minus{}1}k}{N\sum_{k\equal{}\lceil N \rho \rceil}^N \binom{N}{k} (\minus{}1)^{k}(k\minus{}N \rho)^{N\minus{}1}}$
and
$ \frac{\sum_{k\equal{}\lceil N \rho \rceil}^N \binom{N}{k} (\minus{}1)^{k}(k\minus{}N \rho)^{N\minus{}1}k^2}{N^3\sum_{k\equal{}\lceil N \rho \rceil}^N \binom{N}{k} (\minus{}1)^{k}(k\minus{}N \rho)^{N\minus{}1}}$
for large N

This is just an absolutely terrible reformulation of the general question about the asymptotics of the average value of $ \Bigl(\sum_{j\equal{}1}^N x_j\minus{}N\rho\Bigr)^p$ with fixed $ p>0$ over $ Q(\rho)\equal{}\{x_j\in[0,1], \textstyle{\sum_j x_j\ge N\rho}\}$. This is non-trivial only for $ \rho>\frac 12$ when it reduces to finding the asymptotics of the derivative of the logarithm of the volume of the $ N\minus{}1$-dimensional section of the cube $ [0,1]^N$ by the hyperplane $ \sum_j x_j\equal{} N\rho$ with respect to $ \rho$. That, in turn, reduces to finding the saddle point of $ \left|\frac{\sin z}z e^{i\lambda z}\right|$ for $ 0<\lambda<1$ on the imaginary axis, i.e. to the minimization of $ \frac{e^{y}\minus{}e^{\minus{}y}}y e^{\minus{}\lambda y}$ over $ y\in(0,\plus{}\infty)$. The answer is the unique root of some transcendental equation. You can easily compute it with arbitrary precision but there is no nice formula for it.

As to a reference, I doubt it is written anywhere: it is just a routine exercise in Fourier analysis and the mountain pass lemma with no particularly nice answer.

Thanks for you reply, but I'm not sure I got how is it a reformulation of getting the asymptotics of that expression. Also I don't get how it is only non-trivial for $ \rho > \frac 12$.

Oops, I misread the problem a bit (took $ N \minus{} 1$ for $ N$ in hurry) :blush:. But the correct version is almost the same. Let's talk about the first quotient.

If you have any function $ f$ of one variable, then
\[ \begin{aligned} \sum_{k \equal{} 0}^N \binom{N}{k} ( \minus{} 1)^{k}f(k) & \equal{} ( \minus{} 1)^N \int_{[0,1]^N}f^{(N)}(x_1 \plus{} \dots \plus{} x_N)\,dx_1\dots dx_N \\
& \equal{} \int_0^N f^{(N)}(t)S(t)\,dt \end{aligned}\]
where $ S(t)$ is the $ N \minus{} 1$-dimensional volume of the cross-section of $ [0,1]^N$ by the plane $ x_1 \plus{} \dots x_N \equal{} t$. In the denominator, we have just $ (N \minus{} 1)!S(N\rho)$ because the $ N$-th derivative of $ (k \minus{} \rho N)_ \plus{} ^{N \minus{} 1}$ is just the $ \delta$-function with that factor. In the numerator, we can wtite $ k \equal{} (k \minus{} N\rho) \plus{} N\rho$ and see that the numerator is $ (N \minus{} 1)! N\rho F(N\rho)$, which cancels with the denominator nicely, plus $ N!\int_{N\rho}^N F(t)\,dt$. But, if $ \rho > \frac 12$, then $ F(t)\le F(N\rho)e^{ \minus{} c(t \minus{} N\rho)}$ with $ c \equal{} \minus{} \frac d{dt}\log F(t)_{t \equal{} N\rho} > 0$, so only an interval of constant length is interesting for the asymptotics, on which $ F(t)$ is almost the same as $ F()e^{ \minus{} c(t \minus{} N\rho)}$ and you only need to know the asymptotics of $ c$ to find the answer (which I showed).

Now disregard what I said about $ \rho < \frac 12$. If $ N \minus{} 1$ were $ N$, we would be dealing with the upper portion of the cube, which, in this case, would be the entire cube. Unfortunately, with $ N \minus{} 1$, we need to be more careful. The integral becomes trivial, indeed (just $ 1$) because almost all of the volume comes from $ t\approx \frac N2$, but now it dominates the poor cross-section volume term and we need to know the asymtotics of $ F(N\rho)$ rather than just that of $ \log F(N\rho)$, which requires more accurate computations in the mountain pass lemma. Anyway, here the ratio is huge, not approaching a constant, so I do not know if you really need to know it rather than just its logarithm and I told you how to estimate the latter.

The analysis of the remaining exceptional case $ \rho \equal{} \frac 12$ is completely covered by the central limit theorem, so it shouldn't present any difficulties.

The second quotient can be treated similarly.

Please, accept my apologies for misreading the statement. Feel free to ask as many more questions as you need :).

thanks again for your reply, but I realized I should have stated the problem differently, since the term I want to get the asymptotic behavior for is
$ \frac{(N(1\minus{}2\rho)\plus{}1\plus{}2\rho)\sum_{k\equal{}\lceil N\rho \rceil}^N \binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}k\minus{}2\sum_{k\equal{}\lceil N\rho \rceil}^N \binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}k^2}{N^3\sum_{k\equal{}\lceil N\rho \rceil}^N \binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}}$
which I realized can be restated as
$ \frac{(N(1\minus{}2\rho)\plus{}1\plus{}2\rho)\sum_{k\equal{}0}^{\lceil N\rho \rceil\minus{}1} \binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}k\minus{}2\sum_{k\equal{}0}^{\lceil N\rho \rceil\minus{}1} \binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}k^2}{N^3\sum_{k\equal{}0}^{\lceil N\rho \rceil\minus{}1} \binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}}$
for this, I expect to have a finite constant this term approaches and I will try to use the same idea you have mentioned to work it out.

Are you absolutely sure that you have $ N^3$, not $ N^2$ as a factor in the denominator? I'm asking because, from what I said above, it follows that the expression you have now tends to $ 0$ for $ \rho>\frac 12$... :? Also, it should blow up exponentially for $ \rho<\frac 12$, which you will, probably, not like in the slightest (that is because the the numerator represents some integral over the main portion of the cube but the denominator is still just the cross-section volume). Are you sure you do not have $ N$ instead of $ N\minus{}1$, i.e., that my misreading wasn't actually the truth?

I'm sorry, I forgot an N in the numerator, now it's
$ \frac{(N^2(1\minus{}2\rho)\plus{}N(1\plus{}2\rho))\sum_{k\equal{}0}^{\lceil N\rho\rceil\minus{}1}\binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}k\minus{}2\sum_{k\equal{}0}^{\lceil N\rho\rceil\minus{}1}\binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}k^{2}}{N^{3}\sum_{k\equal{}0}^{\lceil N\rho\rceil\minus{}1}\binom{N}{k}(\minus{}1)^{N\minus{}k}(k\minus{}N\rho)^{N\minus{}1}}$
but I am sure that I have the power as (N-1) rather than N

Are you sure that you didn't forget $ N$ in the sum with $ k^2$ too?

Anyway, as long as you have $ N\minus{}1$, it blows up for $ \rho<\frac 12$ (but for $ \rho>\frac 12$, you have a finite non-zero limit, indeed).

I think there are no N's I'm missing now, but are you sure it still blows up even after making the sum from $ 0$ to $ N\rho \minus{} 1$ rather than from $ N\rho$ to $ N$ ?

Oops, missed that. Let me think a bit then :)


geometry
incenter
geometric transformation
reflection
angle bisector
geometry solved

I once saw this property(possibly from Yufei) but I can not remember how to prove this. I do remember that the proof was very simple and short.
Let M and N be points in the triangle ABC. Show that angle MCB= angle NCA, if angle MAB=NAC and angle MBA= angle NBC.
Is this something to do with incentre?

[quote="pleurestique"]M and N are reflections of each other in both the angle bisector of angle A and the angle bisector of angle B (two distinct lines).  the only way this can happen is if M = N = I, the incenter.  and the result follows.[/quote]
M and N is not necessarily reflections. But AN is reflection of AM by bisector of angle A, BN is reflection of BM by bisector of angle B(straight line, not segment).

The problem follows easily from the angle version of Ceva's theorem.

M and N are then said to be isogonal conjugates:
[url=http://mathworld.wolfram.com/IsogonalConjugate.html]mathworld.wolfram.com/IsogonalConjugate.html[/url]

Here's another approach:

Let $M_a,M_b,M_c$ be the projections of $M$ on $BC,CA,AB$ respectively. Let $\ell_a,\ell_b,\ell_c$ be the isogonal conjugates of the lines $AM,BM,CM$ respectively. Show (very simple angle chase) that $\ell_a\perp M_bM_c$, and similar relations hold for $\ell_b,_c$.

The perpendiculars from $M_a,M_b,M_c$ to $BC,CA,AB$ respectively are concurrent in $M$, which means that $ABC,M_aM_bM_c$ are orthologic. This means that the perpendiculars from $A,B,C$ to $M_bM_c,M_cM_a,M_aM_b$ respectively are also concurrent. According to the above paragraph, these perpendicularsd are precisely $\ell_a,_b,_c$, so we have shown that $\ell_a,_b,_c$ are concurrent, Q.E.D.


LaTeX
trigonometry
geometry theorems
geometry

ABC is triangle $\angle ABC=x$. Then altitude of $BC$ from $A$ is $h$ and $h$ intersect $BC$ at $E$
$h^{2}-h.(BC).cotx=BE*EC$

excuse me ...

could u plz use latex  ;)

Now I forgot you. Please, learn and write in [b]LaTeX[/b] !
Here is the posted Osmantelli's problem:

[u]The hypothesis:[/u] $\triangle ABC$: $E\in BC$, $AE\perp BC\ (AE=h_a).$

[u]The conclusion:[/u] $EB\cdot EC=h^2_a-ah_a\cot B.$

[i]Osmantelli[/i], is it correctly ?. I am waiting your .... and the answer ! $\blacksquare .$

[b]Remark.  [/b] $(1)\ \ h^2_a+EB\cdot EC=ah_a\cdot cot B,\ \underline {E\in (BC)}\Longrightarrow B=45^{\circ}.$

[u]A solution.[/u] From the relations $h_a=c\sin B$, $a=b\cos C +c\cos B$,

$EB=c\cos B$, $EC=b\cos C$ results:

$(1)\Longleftrightarrow c^2\sin ^2B+bc\cos B\cos C=ac\sin B\cdot \frac{\cos B}{\sin B}\Longleftrightarrow$

$c\sin ^2 B+b\cos B\cos C=a\cos B\Longrightarrow$

$c\sin ^2B=\cos B(a-b\cos C)\Longrightarrow c\sin ^2B=c\cos^2 B\Longrightarrow B=45^{\circ}.$

Osmantelli's problem cannot be correct as it is written now, since one of the equality members is symmetrical in $B$ and $C$, while the other isn't ($cot B \neq cot C$)

EDIT : it is in fact correct if we replace $cot B$ by $cot A$, giving :

\[ EB . EC = h_a^2 - a h_a cot A  \]


What is the maximum number of non-intersecting diagonals in a convex $n$-gon?

[hide="hint"] Induction/recursion works well. For a convex $n$-gon, draw any diagonal and you have reduced it to a case of two smaller polygons. [/hide]

Alternatively, is it not just clearly $n-3$? This is achievable, and using the fact that one cannot draw a diagonal within a triangle, it must be maximal.


function
induction
algorithm
algebra
domain
number theory
relatively prime

Find all function $f:Q_+ \rightarrow Q_+$ such that :

1) $f(x) = f( \frac{1}{x} ) $ for all $x \in Q_+$.
2) $( \frac{1}{x} +1)f(x)=f(x+1)$ for all $x \in Q_+$.

Let f(1) = y.

$f:Q_+ \rightarrow Q_+ => y \in Q_+$.

$( \frac{1}{x} +1)f(x)=f(x+1) => f(n) = ny$ for all $n \in N_$.
(This is an induction.  $( \frac{1}{1} +1)f(1)=f(2) => f(2)=2y$
and we are already given that $( \frac{1+x}{x} )f(x)=f(x+1)$ so $( \frac{1+n}{n} )n*y=f(n+1)$.)

$f(x) = f( \frac{1}{x} ) => f( \frac{1}{n} )=ny$ for all $n \in N_$.

Note that 2) may be rewritten as $f(x)=( \frac{x}{x-1} )*f(x-1)$ for x>1.

If n > m*k for $k \in N_$ then $f( \frac{n}{m} )= \frac{n}{n-m*k} *f( \frac{n-m*k}{m} )$.
(This is an induction as well.  $f( \frac{n}{m} )= \frac{n/m}{(n-m)/m} *f( \frac{n}{m} -1)=\frac{n}{n-m} *f( \frac{n-m}{m} )$
and $\frac{n}{n-m*(p-1)} *f( \frac{n-m*(p-1)}{m} )$
=$ \frac{n}{n-m*(p-1)} *( \frac{ \frac{n-m*(p-1)}{m} }{ \frac{n-m*(p-1)}{m} -1} )*f( \frac{n-m*(p-1)}{m} -1)$
=$ \frac{n}{n-m*(p-1)} *( \frac{ \frac{n-m*(p-1)}{m} }{ \frac{n-m*p}{m} } )*f( \frac{n-m*p}{m} )$
=$ \frac{n}{n-m*p}*f( \frac{n-m*p}{m} )$.)

There exist $m,n \in N_$ for all $x \in Q_+$ such that $x = \frac{m}{n}$ and m and n are relatively prime.

If m<n then there exist $q,r \in N_$ such that r < m and n = m*q +r

Then, using 1) and the result of the second induction, $f( \frac{m}{n} )=f( \frac{n}{m} )= \frac{n}{r} *f( \frac{r}{m} )$

This relates the value of each rational number in (0,1) to another rational number in (0,1) with the property that both the numerator and denominator are less than that of the original.  Furthermore, because m and n are relatively prime the Euclidean Algorithm tells us that iterating this relation will always eventually lead to a relation to a rational number of the form 1/n, which we have already found the values of f for.  We can now construct f for any rational number a in (0,1) and for each rational number b in (1, infinity) there exists a rational number a in (0,1) such that a = 1/b and so f(a) = f(b).   Therefore, once we have a value of f for 1 we may construct the value of f for any other number in the domain.  To put it another way, for each value of f for 1 there is only one value of f that each given number in the domain may have.  So, for each y there exists only one possible f.

There exist $m,n \in N_$ for all $x \in Q_+$ such that $x = \frac{m}{n}$ and m and n are relatively prime.

Let g(x)=g(m/n)= m*n*y where y=g(1)

Then 1) $g(x)=m*n*y=g( \frac{n}{m} )=g( \frac{1}{x} )$ for all $x \in Q_+$

And 2) $( \frac{1}{x} +1)g(x)= \frac{n+m}{m} *m*n*y=(n+m)*n*y=g( \frac{n+m}{n})=g( \frac{m}{n} +1)=g(x+1)$ for all $x \in Q_+$.

Therefore, for each value of f(1) there exists at most one function that satisfy criteria 1) and 2) and g(x) satisfies these criteria, so the solution is {f| f(x) = m*n*y where $x,y \in Q_+$, $x = \frac{m}{n}$, and m and n are relatively prime}.

To derive this set of functions directly takes a few more steps, but can be found by showing a correlation between the numbers that pop up in the chain formed by relating a fraction of the form m/n to one of the form 1/n and the numbers that occur in the colums of the table for the Extended Euclidian Algorithm of n and m.  For example:

f(8/13)=f(13/8)=13/5*f(5/8)=13/5*f(8/5)=13/5*8/3*f(3/5)=13/5*8/3*f(5/3)=13/5*8/3*5/2*f(2/3)=13/5*8/3*5/2*f(3/2)
=13/5*8/3*5/2*3/1*f(1/2)=13/5*8/3*5/2*3/1*f(2/1)= $ \frac{13}{5} * \frac{8}{3} * \frac{5}{2} * \frac{3}{1} * \frac{2}{1} *y$=13*8*y

The first numerator in this process is always the denominator of the original fraction, the second is always the numerator.  The first- and second-to-last denominators are always 1, and the n+2 numerator always equals the n denominator.  This all results in cancellation of everything but n, m, and y.

[i]Edited because some browsers don't automatically break text strings for wrapping at punctuation like Internet Explorer.[/i]


calculus
derivative
function
calculus computations

Look for the derivative of the function

$ f(x)\equal{}\prod_{i\equal{}1}^{n}(x\minus{}a_i)^{a_i}$


[b][size=100][color=DarkBlue]Δίνεται κύκλος $ (O)$ χορδής $ AB$ και έστω $ M,$ το σημείο τομής του από τη μεσοκάθετη του $ AB.$ Αποδείξτε ότι για κάθε σημείο $ P\neq M$ του τόξου $ AB$ του $ (O)$ που περιέχει το $ M,$ ισχύει η ανισότητα $ PA\cdot PB < MA\cdot MB \equal{} \kappa^{2}$[/color][/size][/b].

Κώστας Βήττας.

Αν κατάλαβα καλά το πρόβλημα:

Έχουμε ότι το πρόβλημα είναι ισοδύναμο με το ότι Εμβαδό(ΜΑΒ)>Εμβαδό(ΡΑΒ) [αφού ΜΑ*ΜΒ=Εμβαδό(ΜΑΒ)/ημφ και ΡΑ*ΡΒ=Εμβαδό(ΡΑΒ)/ημφ]. Αλλά Εμβαδό(ΜΑΒ)>Εμβαδό(ΡΑΒ)<=>d(M,AB)>d(P,AB), που ισχύει, αφού το Μ είναι το μέσο του τόξου ΑΒ

Kώστα μια διανυσματική λύση
ΡΑ.ΡΒ=(δ.ΡΑ.δ.ΡΒ)/συνΡ=[(ΡΚ)^2-(ΚΒ)^2]/συνΡ<[(ΜΚ)^2-(ΚΒ)^2]/συνΜ=(δ.ΜΑ)(δ.ΜΒ)/συνΜ=(ΜΑ)(ΜΒ) όπου Κ το μέσο του ΑΒ

Ηλία και Χρήστο σας ευχαριστώ.

Η λύση που έχω υπόψη μου, βασίζεται στην ιδέα ότι για κάθε $ P\neq M,$ ισχύει επίσης ότι $ PA \plus{} PB < MA \plus{} MB \equal{} 2\kappa$ $ ,(1)$

Επειδή τώρα, το γινόμενο δύο άνισων ευθύγραμμων τμημάτων, είναι μικρότερο από το τετράγωνο του ημιαθροίσματός τους ( αποδεικνύεται εύκολα αλγεβρικά αλλά και γεωμετρικά ), προκύπτει άμεσα το ζητούμενο.

Για την απόδειξη της $ (1),$ έχω υπόψη μου μία γεωμετρική λύση χωρίς πολύ θεωρία, αλλά ίσως με τριγωνομετρία και διανύσματα ( όπου δυστυχώς πολύ λίγα θυμάμαι ), να είναι πάλι πιο απλά τα πράγματα.

Κώστας Βήττας.

Esto $ x$ i gonia ton $ AB,MB$ kai $ y$ i gonia ton $ AB,BM$
Isxii oti $ x \plus{} y$=sta8ero
kai episis $ PA*PB \equal{} 4R^2 *sinx*siny$
kai  epidi $ sinx*siny \equal{} \frac {cos(x \minus{} y) \minus{} cos(x \plus{} y)}{2}$,i arxiki posostita ginete megisti otan
$ cos(x \minus{} y)$ megistopiite (afu cos(x+y)=sta8ero)
diladi otan $ x \equal{} y$

Edit:
Me entelos omia silogistiki mpori na vgi oti k to a8risma ine mikrotero...dld $ PA \plus{} PB < MA \plus{} MB \equal{} 2\kappa$


For any positive integer x, let S(x) be the sum of the digits of x, and let T(x) be abs(S(x+2)-S(x)) How many values T(x) do not exceed 1999?

[hide="(edited) Solution"]
$T(x)$ takes on the value $2$ and values $\equiv -2 \bmod 9$ (easy to prove).  It's easy to construct examples where it takes on all of them, depending on how many digits need to be carried by the addition $x + 2$.  Therefore, it takes on $\lfloor \frac{1999 - 7}{9} \rfloor + 1 = \boxed{223}$ values. [/hide]

[hide]Do you mean $\equiv 7 \bmod 9$ ... ? (remember abs)

(and 2 of course)[/hide]
Also, you seem to have made an arthimetic error in the last step. ;)


Solve in $ \mathbb{R}$ the equation: $ 3^{|x\plus{}1|}\minus{}2|3^x\minus{}1|\equal{}3^x\plus{}2$ .

Divide in to three case in order to solve it.
very easy.

Let's look at when $ x \plus{} 1$ and $ 3^x \minus{} 1$ are nonnegative/negative.

[hide]$ x \plus{} 1 \geq 0 \rightarrow x \geq \minus{} 1$

$ x \plus{} 1 < 0 \rightarrow x < \minus{} 1$

$ 3^x \minus{} 1 \geq 0 \rightarrow x \geq 0$

$ 3^x \minus{} 1 < 0 \rightarrow x < 0$[/hide]
Case 1 : $ 0 > x \geq \minus{} 1$

[hide]$ 3^{x \plus{} 1} \plus{} 2(3^x \minus{} 1) \equal{} 3^x \plus{} 2$

$ 4 \cdot 3^x \minus{} 4 \equal{} 0$

$ x \equal{} 0$[/hide]

Case 2: $ x \geq 0$

[hide]$ 3^{x \plus{} 1} \minus{} 2(3^x \minus{} 1) \equal{} 3^x \plus{} 2$

$ 0 \equal{} 0$
This means infinite solutions for all nonnegative $ x$.[/hide]

Case 3: $ x < \minus{} 1$

[hide]$ 3^{ \minus{} x \minus{} 1} \plus{} 2(3^x \minus{} 1) \equal{} 3^x \plus{} 2$

Solving gives $ x \equal{} \log_{3} (2 \pm \frac {\sqrt {33}}{3})$[/hide]

PS: Tell me if you see any mistkes.


analytic geometry
combinatorics unsolved
combinatorics

[quote="ehsan2004"]Let $A_1,A_2,...,A_n$ be the sets each consisting of $n$ segments on a given line.Prove that the intersection $A_1\cap A_2\cap...\cap A_n$ contains not more than $n^2-n+1$ disjoint segments.[/quote]
For $n^2$ is easy. But I don't see how to solve for $n^2-n+1$.

for each point calculate the coordinate of it right endpoint, and let $r_1<r_2<r_3....<r_{n^2}$ This order allows us to show the desired maximun, just think about it....it is a big clue


geometry
geometric transformation
rotation

Comment on the following statement:

"The spoked wheels of a stage coach on a TV Western film often appear to be rotating backwards ,but similar spoked wheels on a farm cart in the same film seldom do"

I closely observed a fastly moving cycle wheel and I feel that its not the same spoke which we observe to move backwards but the task of our eye to change the picture in 1/16th of a second makes us look towards the next coming spoke in continuation with the first picture .Thus we are made to look towards the 'backward motion' by the unstrained eye.This is
not the case certainly  for a slow motion.
Please give your views.


At 3:00 pm while driving on highway US 2 towards Oakton and Trenary, Karela sees the sign:  “Oakton 37 miles;  Trenary 57 miles.”  At 3:15 pm, Karela sees another sign and notices that Trenary is now twice as far ahead as Oakton.   What was Karela’s average speed between 3:00 pm and 3:15 pm?

A.   60 mph  B.   62 mph   C.   64 mph  D.   65 mph   E.   68 mph

[hide] Let $ x$ be the number of miles she traveled in that time. Then,

$ 2(37\minus{}x)\equal{}(57\minus{}x) \implies x\equal{}17$.

Thus, she traveled $ 17$ miles in $ \frac14$ of an hour, and using $ d\equal{}rt$, her speed was $ (17)(4)\equal{}\boxed{68}$. Thus, the answer is E.[/hide]


trigonometry
vector
geometry solved
geometry

$ OA, OB, OC, OD$ are 4 rays in space such that the angle between any two is the same. Show that for a variable ray $ OX,$ the sum of the cosines of the angles $ XOA, XOB, XOC, XOD$ is constant and the sum of the squares of the cosines is also constant.

Let A,B,C,D,X be points on OA,OB,... such that OA=OB=OC=OD=OX=1.
Then cos(XOA)=(2-XA 2 )/2...
I use vectors to prove that the sums (XA 2n+XB2n+XC2n+XD2n) are constant for each nonnegative integer.
From now on xa means the vector XA (not the segment XA).
xa=xo+oa.
Apply Newton formula for (a+b)k using the scalar product for vectors.
In that formula use xa2 =XA2 , and oa+ob+oc+od=0. Because those sums are constant it should easily follow the fact that the sums cosn(XOA)+... are constant.


function
induction
algebra proposed
algebra

Find all the continuous functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that :
$\forall(x,y) \in \mathbb{R}^2$,
$f(x+y)\cdot f(x-y) = [f(x)\cdot f(y)]^2$

EDIT : Sorry, maybe this post should rather go into the "proposed and own problems" section  :oops:

A variation on the old Cauchy theme.

If $f(x)=0$ for any $x$ then $f$ must be uniformly 0.

Otherwise, it's easier to coinsider $g(x)=\log|f(x)|$.  Clearly $g(x+y)+g(x-y)=2g(x)+2g(y)$.  By induction $g(nx)=n^2g(x)$ for all integers $n$, and so $g(\frac{n}{m}x)=n^2g(\frac{1}{m}x)=(\frac{n}{m})^2g(x)$ for all rational numbers $\frac{n}{m}$.  By continuity, then, $g(x)=kx^2$, so for any $x$, $f(x)=\pm a^{x^2}$, where $a$ is some positive constant.

Substituting $x=r+s$, $y=r-s$, we see that $f(r)f(s)=[f(r+s)f(r-s)]^2>0$ for all $r$, $s$, so either $f(x)=a^{x^2}$ for all $x$ or $f(x)=-a^{x^2}$ for all $x$.


The numerator of a fraction is $ 6x \plus{} 1$, then denominator is $ 7 \minus{} 4x$, and $ x$ can have any value between $ \minus{}2$ and $ 2$, both included. The values of $ x$ for which the numerator is greater than the denominator are:

$ \textbf{(A)}\ \frac{3}{5} < x \le 2\qquad
\textbf{(B)}\ \frac{3}{5} \le x \le 2\qquad
\textbf{(C)}\ 0 < x \le 2\qquad \\
\textbf{(D)}\ 0 \le x \le 2\qquad
\textbf{(E)}\ \minus{}2 \le x \le 2$

We just get $ 6x\plus{}1>7\minus{}4x$. Solving:

$ 6x\plus{}1>7\minus{}4x$
$ 10x\plus{}1>7$
$ 10x>6$
$ x>\frac{3}{5}$

But we also have a limit of 2. So:

$ \frac{3}{5}<x\le2$

This is choice $ \mathbf{(A)}$.


algebra
polynomial
function
algebra proposed

Let $f(x)=a_0+a_1x+\ldots +a_nx^n $ and for $k\in \{0,1,...,n\} $ denote $b_k=\displaystyle \sum\limits_{j=0}^{k}\frac{{k\choose j}}{{n\choose j}}a_j\; . $ Prove that
\[
f\left(\frac{X}{1+X}\right)(1+X)^n=\sum\limits_{k=0}^{n}{n\choose
k}b_kX^k\; .
\]

i think ,there is a mistake ,
if we assume that the result is tru , by using the inversion formula we have:
$ b_nx^n\equal{}\sum_{k\equal{}0}^{n}(\minus{}1)^{n\minus{}k}\dbinom{n}{k}f(\frac{x}{x\plus{}1})(1\plus{}x)^k\equal{}f(\frac{x}{x\plus{}1})x^n$ so $ b_n\equal{}f(1)\equal{}f(\frac{x}{x\plus{}1})$ wich is clearly false

in my mind there is some uncorrect arguments.
because of the [url=http://www.akifaltundal.net]Principles and Techniques in Combinatorics[/url].

i'm sorry i put there a wrong url. The true one will come.


#36

This is not too hard question and more close to MC question..

A man engages in a shooting contest. Each time he hits a target he receives 10 cents, while each time he misses he pays 5 cents. If after 20 shots, he has lost 10 cents, how many times has he hit the target?

[hide]In order for him to lose 10 cents he must miss the target $2x + 2$ times where $x$ is the number of times he hits it because $10$ is twice 5, and in order to "balance" it out, he must miss twice as much as he hits, then to lose 10 cents, must miss twice. $2x + 2 + x = 3x + 2 = 20$ so $x = 6$ and he hits the target 6 times.[/hide]

If $A$ is the number of times he hits the target, and $B$ is the number of times he misses, then
\begin{eqnarray*}
10A-5B&=&-10\\
A+B&=&20\\
\end{eqnarray*}
Now we see that
\begin{eqnarray*}
10A-5B&=&-10\\
5A+5B&=&100\\
15A&=&90\\
A&=&6
\end{eqnarray*}
So he hit the target 6 times.

quick in your head method

too lazy to use caps

two misses + one hit = evens out[

12 misses + 6 hits = evens out

14 misses + 6 hits = 10 cents loss

WIN


conics
parabola
analytic geometry
calculus
function
geometry
geometric transformation

A side ways parabola shown here: [url]http://www.analyzemath.com/calculus_questions/analytical/functions.html[/url]
is not a function. It is understandable that there is no equation for it, and multiple y-values for each corresponding x-value, but is it justifiable?

Basically, I am wondering there is an equation (that can be proved), be it cartesian-coordinate based or not, for this graph.
(I am guessing you need calculus)

Thank you very much in advance.
This topic has been incessantly bugging me!
:spider:

I might be reading you wrong, but I think you want a function for sideways parabolas.
You can take a parabola and parameterize it, then apply a rotation matrix, or I guess you just take the inverse function of the parabola and plot the two branches of it as separate functions.

Why would I need calculus?

It has the form $ x\equal{}\minus{}(y\minus{}c)^{2}$, where $ c$ is some constant.

So wait... are you guys saying in essence it is a parabolic cylinder???

I am confused.

If you rotate the diagram shown in the website you refer to, through $ 90$ degrees anticlockwise so that the $ x$-axis is vertical and the $ y$-axis points to the left and then reflect in the vertical $ x$-axis, you get a graph which is that of $ y\equal{}\minus{}x^{2} \plus{} 1$ but with the roles of $ x$ and $ y$ interchanged. That is you have a curve whose equation is $ x\equal{}\minus{}y^{2} \plus{}1$. That is $ y{^2} \equal{} 1\minus{}x$ (which is only defined for $ x\le 1$). That is $ y \equal{} \pm(\sqrt{1\minus{}x})$. This consists of two functions: $ y \equal{} \sqrt{1\minus{}x}$, the branch above the x-axis and $ y \equal{} \minus{}\sqrt{1\minus{}x}$, the branch below the $ x$-axis. I hope this helps.

ok.. so pretty much the normal parabolas that we see every day (that are functions)

are of the form $ y\equal{} ax^2\plus{}b$ (these go up, or down depending on the value of "a")
the one that u showed, and others are of the form $ x\equal{} ay^2\plus{}b$ (these go sideways left and right depending on "a")

basically if "x" is squared, then the graph goes up and down... if "y" is squared, then the graph goes sideways...

the graph that u showed is not a function because "y" is being squared, so there are 2 values of "y" for each "x"...

and plus it is not a function because the vertical line test says so...

[quote="MCmaniac"]It is understandable that there is no equation for it[/quote]
No.  There's no equation of the form $ y \equal{} f(x)$, but there [b]is[/b] an equation; see JRav's post.  Are you familiar with the idea of an [url=http://en.wikipedia.org/wiki/Implicit_function]implicit function[/url]?

I have read somewhere that if a straight vertical line crosses a curve more than once, then the curve can not be represented as a function. So it wouldn't be a function, rather an equation with a $ \pm$ in it somewhere. According to the link, it would be $ y\equal{}\pm\sqrt{c\minus{}x}$ where in the specific case, $ c\equal{}1$.

[quote="PowerOfPi"]So it wouldn't be a function[/quote]
No, it wouldn't.  Why are the only interesting graphs functions?  Circles aren't functions, either.

Also, you should consider reading about [url=http://en.wikipedia.org/wiki/Curve]curves[/url].  A circle is an example of a curve, parametrized by $ \textbf{X}(t)\equal{}(\cos{t},\sin{t})$.  Curves can do all sorts of interesting things that functions can't, for example, the [url=http://en.wikipedia.org/wiki/Peano_curve]Peano space-filling curves[/url].


I have a certain ring of mass $ m$ and radius $ r$ rotating at $ \omega$ angular velocity in the horizontal about an axis passing through the vertical. If it is thrown upwards with a velocity $ v_0$ and is observed at it's maximum height to have the same orientation as it had when it was being launched. Find the minimum value of $ v_0$ such that this is possible.

if i am not mistaken then the ring rotates in the horizontal plane only so even after going up it will rotate in the horizontal plane.so the orientation remains the same provided all parts are indistinguishable or do u mean u have coloured all the parts and u want the samw orientation of colour?

Guessing what means "rotating  in the horizontal" "about an axis passing through the vertical" and "same orientation",
\[ v_0 = {{2\pi g} \over \omega
}\] .
Could you, pleaase, confirm...


Open MATHCOUNTS Topics.  Talk about whatever you want as long as it pertains to MATHCOUNTS.  Also, if you have any secret tips, tell them to me, and I will tell you if they are worth anything. :huh:

[quote="perfect628"]What about the WV forum?[/quote]

Where is the WV Forum???

I'm not sure there is one (although there is one for almost all other states including VA). :)

Yeah, when I said that I assumed that all states had one. Apparently not, though.

what do they think were too much like hillbillies to own computers.  we have technology here in wv.  i think ill just start my own wv forum


abstract algebra
group theory
superior algebra
superior algebra unsolved

let a(n) denotes no of abelian groups of order n.show a(n) is no. theoretically multiplicative.
 p.s. are there any good bouns on a(n)?

You can calculate $a(n)$ in terms of $s(d)$, where we define $s(d)$ as the number of ways to write $d$ as sum of positive integers (without ordering, thus $1+2$ and $2+1$ are the same).
Then:

Write $n=\prod p_{i}^{v_{i}}$ with different primes $p_{i}$. Then $a(n) = \prod s(v_{i})$.

Proof: use the structure theorem on finite abelian groups and write every finite abelian group as product of groups of type $\mathbb{Z}/(p^{k}) \mathbb{Z}$, $p$ prime (the abelian groups correspond then bijectively with the (multi)sets consisting of the $p^{k}$).
By grouping the terms with the same $p$, we get the desired multiplicativity, and then it just runs out to prove it for $n=p^{k}$, being true more or less by definition.


geometry
combinatorics open
combinatorics

Consider a following operation with n$\in\mathbb{N}$ lines on a plane so that no three lines intersect at one point. These lines divide this plane into several areas. Same of them we paint black and the rest are white, so that every two areas that have a common side have different colours. For given n find the maximum of black areas.

I proved that it is at least (for $n\geq 3$):
for even $n$: $a_n=\frac{n(n+4)}{4}-2$
for odd $n$: $a_n=\frac{(n+1)(n+3)}{4}-2$
It means that for $n\geq 5$, $a_n=a_{n-2}+n+1$.


Calculate the following, and express your answer as a common fraction:
$ \frac{\frac23}{0.\overline6} \div \frac35$

$ 0.\overline6\equal{}\frac{2}{3}$.  We then have $ \frac{\frac{2}{3}}{\frac{2}{3}}\div\frac{3}{5}$.  The $ \frac{2}{3}'\text{s}$ cancel out, so we get $ 1\div\frac{3}{5}$.  When dividing by a fraction, we multiply by the recipricol, so the answer is $ 1\times\frac{5}{3}\equal{}\boxed{\frac{5}{3}}$

i don't really think this problem should be on there because the the bar on the .6 looked like part of the fraction :(

That is EXACTLY what I thought, therefore getting the answer of $\boxed{\frac{25}{9}}$.


So does $0$.  It's not hard to conjecture that there are no other solutions, but a proof is elusive.

How about this:
Find all integers n where $4^{n}+3^{n}$ is divisible by $2^{n}+1^{n}$


integration
trigonometry
logarithms
limit
calculus
calculus computations

Please help me to check the solution, thanks.
I dont know why Im wrong?

\[ I\equal{}\int{{{e}^{\minus{}ax}}.{{\left( \cos bx \right)}^{2}}.dx}\equal{}\frac{1}{2}\left( \minus{}\frac{{{e}^{\minus{}ax}}}{a}\plus{}\frac{\minus{}a.{{e}^{\minus{}ax}}\sin \left( 2bx \right)\minus{}2b.{{e}^{\minus{}ax}}\cos \left( 2bx \right)}{{{a}^{2}}\plus{}4{{b}^{2}}} \right)\]

\[ \Rightarrow \int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.{{\left( \cos bx \right)}^{2}}.dx}\equal{}\minus{}\frac{1}{2}\left( \minus{}\frac{1}{a}\plus{}\frac{\minus{}a.{{e}^{\minus{}ax}}.\sin 0\minus{}2b.\cos 0}{{{a}^{2}}\plus{}4{{b}^{2}}} \right)\equal{}\frac{1}{2}\left( \frac{1}{a}\plus{}\frac{2b}{{{a}^{2}}\plus{}4{{b}^{2}}} \right)\]

\[ \Rightarrow \int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.dx}\int\limits_{0}^{1}{{{\left( \cos bx \right)}^{2}}.da}\equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.dx}.\int\limits_{0}^{1}{\frac{\cos \left( 2bx \right)\plus{}1}{2}.db}\equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.dx}.\left[ \frac{\sin \left( 2bx \right)}{4x}\plus{}\frac{b}{2} \right]_{b\equal{}0}^{b\equal{}1}\]

\[ \equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\left( \frac{\sin \left( 2x \right)}{4x}\plus{}\frac{1}{2} \right)dx}\equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{4x}.dx}\plus{}\frac{1}{2}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.dx}\]

\[ \equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{4x}.dx}\minus{}\frac{1}{2a}\left[ {{e}^{\minus{}ax}} \right]_{0}^{\plus{}\infty }\equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{4x}.dx}\plus{}\frac{1}{2a}\]

\[ \Rightarrow \int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.dx}\int\limits_{0}^{1}{{{\left( \cos bx \right)}^{2}}.db}\equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{4x}.dx}\plus{}\frac{1}{2a}\equal{}\frac{1}{2}\int\limits_{0}^{1}{\left( \frac{1}{a}\plus{}\frac{2b}{{{a}^{2}}\plus{}4{{b}^{2}}} \right).db}\]

\[ \equal{}\left[ \frac{b}{2a} \right]_{0}^{1}\plus{}\frac{1}{2}.\frac{1}{4}\int\limits_{0}^{1}{\frac{d\left( {{b}^{2}} \right)}{{{b}^{2}}\plus{}{{a}^{2}}/4}}\equal{}\frac{1}{2}\plus{}\frac{1}{8}.\left[ \ln \left| {{b}^{2}}\plus{}\frac{{{a}^{2}}}{4} \right| \right]_{b\equal{}0}^{b\equal{}1}\equal{}\frac{1}{2}\plus{}\frac{1}{8}.\left( \ln \left| \frac{{{a}^{2}}\plus{}5}{4} \right|\minus{}\ln \left| \frac{{{a}^{2}}}{4} \right| \right)\]

\[ \equal{}\frac{1}{2a}\plus{}\frac{1}{8}.\ln \left| \frac{{{a}^{2}}\plus{}5}{{{a}^{2}}} \right|\Rightarrow \int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{8x}.d\left( 2x \right)}\plus{}\frac{1}{2a}\equal{}\frac{1}{2a}\plus{}\frac{1}{8}.\ln \left| \frac{{{a}^{2}}\plus{}5}{{{a}^{2}}} \right|\]

\[ \Leftrightarrow \int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{2x}.d\left( 2x \right)}\equal{}\frac{1}{2}.\ln \left| \frac{{{a}^{2}}\plus{}5}{{{a}^{2}}} \right|\]

\[ \Rightarrow \underset{a\to \plus{}\infty }{\mathop{\lim }}\,\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}ax}}.\frac{\sin \left( 2x \right)}{2x}.d\left( 2x \right)}\equal{}\int\limits_{0}^{\plus{}\infty }{\frac{\sin x}{x}.dx}\equal{}\underset{a\to \plus{}\infty }{\mathop{\lim }}\,\frac{1}{2}.\ln \left| \frac{{{a}^{2}}\plus{}5}{{{a}^{2}}} \right|\equal{}0\]

Hi chuong01.

I've tried the integral twice (using integration by parts three times) and get the same answer each time, which is different to yours. I'll checkit again if I get time beofre someone else sorts it out. I get

$ \frac{e^{\minus{}ax}}{a} (\minus{}(\cos bx)^{2} \plus{} \frac{(ab \sin 2bx \plus{} 2b^{2} \cos 2bx)}{(a^{2} \plus{} 4b^{2})})$.

Hi chuong01. Originally your integrand contained $ e^{\minus{}ax}$ and not $ e^{ax}$.

All the best.

[quote="AndrewTom"]Hi chuong01. Originally your integrand contained $ e^{ \minus{} ax}$ and not $ e^{ax}$.

All the best.[/quote]
Oh thanks, but transforming from e^(ax) into e^(– ax) is easy.

hello, i think we have
$ \int e^{\minus{}ax}(\cos(bx))^2\,dx\equal{}\minus{}\frac{e^{ax}(a^2\plus{}4b^2\plus{}a^2\cos(2bx)\minus{}2ab\sin(bx)}{2(a^3\plus{}4ab^2)}\plus{}C$
Sonnhard.


absolute value
real analysis
real analysis unsolved

Let $ \mathbf{U}$ be a unitary operator of form $ \mathbf{I}\plus{}\mathbf{C}$, $ \mathbf{C}$ compact. Show that $ \mathbf{U}$ has a complete set of orthonormal eigenvectors, and that all eigenvalues have absolute value 1 (in a book due to P. Lax).

The second part is easy, just do as following. Since
\[ \left( {x,x} \right) \equal{} \left( {{\mathbf{U}}x,{\mathbf{U}}x} \right) \equal{} \left( {\lambda x,\lambda x} \right) \equal{} \lambda \overline \lambda  \left( {x,x} \right) \equal{} {\left| \lambda  \right|^2}\left( {x,x} \right)\]then all eigenvalues of $ \mathbf{U}$ have absolute value 1. But I still do not know how to prove the existence of "a complete set of orthonormal eigenvectors".

 :oops:

Since $ U \equal{} I \plus{} C$, $ U$ has a discrete spectrum that is actually the set of its eigenvalues, $ \{\lambda_n\}_{n\in\mathbf N}$. The spectrum of $ C$ is $ \{\lambda_n \minus{} 1\}_{n \in\mathbf N}$. Let $ H_0$ denote the linear subspace of $ H$ spanned by the eigenvectors of $ U$ corresponding to $ \lambda_n$'s; $ H_0$ is a closed subspace of $ H$. The claim now reads: Prove that $ H_0 \equal{} H$.

Assume to the contrary that $ H_0 \neq H$. Since $ H_0$ is closed, we then can find another nontrivial closed linear subspace $ H_1$ of $ H$ such that $ H \equal{} H_0 \oplus H_1$. But then $ U \colon H_1 \to H_1$, so $ U$ must have an eigenvector in $ H_1$ (while we know that all the eigenvectors of $ U$ are in $ H_0$). This contradiction proves the claim.


function
advanced fields
advanced fields unsolved

I'm a bit clueless here on how to start this exercise.

Let $X$ be a nonempty set with the discrete metric and show that $C(X,\mathbb{R})$ is separable $\iff X$ is finite.

Here $C(X,\mathbb{R})$ is the set of bounded continuous mappings from $X$ into $\mathbb{R}$.

[quote="jmerry"]Show that the only dense set in the discrete metric is the whole space.[/quote]

Thank you for the help, but I'm afraid I'm still unable to see a solution.

If $X$ is infinite, then by a previous exercise we can take $X=\mathbb{N}^+$ and define $f(n)=0$ if $|f_n(n)|\ge 1$ and $f(n)=|f_n(n)|+1$ if $|f_n(n)|<1$ where $(f_n)$ is a sequence in $C(X,\mathbb{R})$, for then $\|f-f_n \|\ge 1$ for all $n$. Hence $C(X,\mathbb{R})$ isn't separable since it doesn't even have an everywhere dense subset.

However, I must be missing something here for I cannot see this fails just because $X$ is finite.

[quote="Jimmy_K"]However, I must be missing something here for I cannot see this fails just because $X$ is finite.[/quote]Think about what it is you're doing. Let $\{f_n\}_{n \in \mathbf{N}}$ be a countable subset of $\mathcal{C}(\mathbf{N},\mathbf{R})$. You want to construct a function $f$ which is 'far away' enough from every $f_n$. In order to guarantee this, you pick the value of $f(1)$ so that $|f - f_1| > 1$, the value of $f(2)$ so that $|f - f_2| > 1$, etc. In order for this process to not terminate, you need to have $f$ defined on some infinite set -- otherwise you'd run out of degrees of freedom for your function $f$.

To prove that $\mathcal{C}(X,\mathbf{R})$ is separable when $X$ is finite, just use that $\mathbf{R}$ is separable.

Thank you. I realized it later on and its pretty obvious now. It's very similar to the diagonal construction in the proof of the uncountability of $\mathbb{R}$.


calculate the amount in M of HCO-3 that is added to blood for the blood PH to rise to 8.0

and then calculate the amount of H+ in M that is removed when this amount HCO-3 of is added.

What equation can i use, i dont want the answer i just want an equation or a certain way i can do this problem

In this problem we are considering that the pH of blood is controled solely by the buffer system represented by the equilibrium

$H_{2}CO_{3}\rightleftharpoons H^{+}+HCO_{3}^{-}$

where the carbonic acid arises from the equilibrium

$CO_{2}+H_{2}O \rightleftharpoons H_{2}CO_{3}$.

I don't know what is data that you are given to solves this problem, but perhaps the Henderson-Hasselbach equation may help you.

Carcul


[hide]since 4x^2-9 is the difference of two squares, it can be factored down to (2x+3)(2x-3)[/hide]

In general, the binomial theorem can be used.
$(x+y)^{n}=\binom{n}{0}x^{n}+\binom{n}{1}x^{n-1}y+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}y^{n}$

[quote="AstroPhys"][quote="Temperal"]Better yet,

$(x+y)^{n}= \sum^{4}_{r=0}\binom{4}{r}x^{4-r}y^{r}$[/quote]

Should not it be $(x+y)^{4}$?[/quote]

Or $(x+y)^{n}= \sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^{r}$.

[quote="alanchou"]Why are we discussing the binomial theorem?

[hide="anyways"]
$4x^{2}-9$ can be factored as $(2x-3)(2x+3)$.
[/hide]

The question is pretty much answered and the thread is getting off topic.[/quote]

Well the thread isn't really "off topic" because we're presenting generalizations that jamesstine will probably find very helpful later.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Suppose $ \det(A)\neq 0$, then $ A$ is invertible, so

\begin{eqnarray*}
AB & = & A+BA\\
 & = & (I+B)A \\
\det(AB) & = & \det((I+B)A) \\
 \det(A)\det(B) & = & \det(I+B)\det(A) \\
 \det(B) & = & \det(I+B) \\
\end{eqnarray*}

A contradiction.

[quote="Beginner"]Prove that if $ AB \minus{} BA \equal{} A$, then $ \det(A) \equal{} 0$.[/quote]
Actually , having these conditions we can prove that $ A^n\equal{}O_n$

if $ \det (A) \not\equal{}0$ and $ AB\minus{}BA\equal{}A$ then $ B\minus{}A^{\minus{}1}BA\equal{}I$ which is impossible since $ \text{tr}(B)\equal{}\text{tr}(A^{\minus{}1}BA)$.

@Svejk
Nice result, do you mean that $ \text{Rank}(A)\equal{}1$ ?

@HilbertThm90:
$ \det(B) \equal{} \det(I \plus{} B)$ is not a contradiction. There exist (a lot of) polynomials $ P$ such that $ P(0) \equal{} P(1)$

@Beginner:
$ A^n \equal{} 0$ does not imply that $ Rank(A) \equal{} 1$ (google: Jordan block)

Beginner's argument works over a field of characteristic zero. Over a field of finite characteristic, the original result may not hold.

Over a field of characteristic $ 2,$ consider $ A\equal{}\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}1&0\\0&0\end{bmatrix}.$

Proof of Svejk's claim, assuming we're working over a field of characteristic zero.

We have $ A \equal{} AB \minus{} BA,$ which we can also write as $ AB\equal{}A \plus{} BA.$

$ A^2 \equal{} A(AB \minus{} BA) \equal{} A^2B \minus{} ABA \equal{} A^2B \minus{} (A \plus{} BA)A \equal{} A^2B \minus{} A \minus{} BA^2.$

Hence, $ 2A^2 \equal{} A^2B \minus{} BA^2$

Assume $ kA^k \equal{} A^kB \minus{} BA^k.$ Then

$ kA^{k \plus{} 1} \equal{} A^{k \plus{} 1}B \minus{} ABA^k \equal{} A^{k \plus{} 1}B \minus{} (A \plus{} BA)A^k$ yielding

$ (k \plus{} 1)A^{k \plus{} 1} \equal{} A^{k \plus{} 1}B \minus{} BA^{k \plus{} 1}.$

Thus $ kA^k \equal{} A^kB \minus{} BA^k$ for all $ K\in\mathbb{N}.$

Take the trace and divide by $ k$ to get that $ \text{Tr}(A^k) \equal{} 0\ \forall\,k\in\mathbb{N}.$

Now, all we need is a proof of the fact that if all of the powers of a matrix have trace zero, then the matrix is nilpotent.

[quote="Kent Merryfield"]

Now, all we need is a proof of the fact that if all of the powers of a matrix have trace zero, then the matrix is nilpotent.[/quote]

[url=http://www.mathlinks.ro/viewtopic.php?t=205969]Here[/url] it is  :wink:

An alternate version of the same argument:

Let's try this. Let $ p(x)$ be the characteristic polynomial of $ A.$ By Cayley-Hamilton, $ P(A)\equal{}0.$ Take the trace of $ P(A)\equal{}0$ and we get that $ \text{Tr}(a_0I)\equal{}0,$ from which $ a_0,$ the constant term of the characteristic polynomial, must be 0.

Let $ x$ be an eigenvector such that $ Ax\equal{}0.$ Build an invertible matrix $ P$ whose first column is $ x.$ Then $ P^{\minus{}1}AP\equal{}\begin{bmatrix}0&*\\0&A'\end{bmatrix}.$ But if we take powers of this, we see that $ \text{Tr}(A'^k)\equal{}0$ for all $ k\in\mathbb{N}.$ So the same process can be applied to $ A'.$

After $ j$ steps of this process we have that $ A$ is similar to $ \begin{bmatrix}N&*\\0&B\end{bmatrix},$ where $ N$ is strictly upper triangular (upper triangular with zeros on the main diagonal). But then $ \text{Tr}(A^k)\equal{}\text{Tr}(B^k)\equal{}0,$ so the argument applies to $ B$ and we can take another zero off. This recursive process can only end when $ B$ vanishes altogether. We conclude that $ A$ is similar to a strictly upper triangular matrix, and hence $ A$ is nilpotent.

(Essentially, this is just a minor modification of the proof of Schur's Lemma.)


(I'd PMed this to you, Mr. Rusczyk, if you were wondering about an email notification, but I thought it might help others as a forum topic)

I'd enrolled in the Calculus class, but today's Math Jam has left me with a lot of doubt as to how useful it will be for me, since I've already taken Calculus.

I'd like to drop the class and enroll in WOOT instead. Is it too late? I noticed the "real" problem solving sessions haven't started yet.

[quote="My Classes"]Click the 'Drop Class' button if you wish to drop the course. A full refund will be issued within 30 days.[/quote]
If I drop the Calculus class, how will it be refunded? Will I receive a check, or a money transfer directly to my dad's bank account? Also, exactly how fast will the process be? My mom seems worried that we might be a bit short on money this month if we don't receive the refund before I pay for and enroll in WOOT.

I hope my late enrollment won't cause too much trouble for you guys. Thanks!

It is definitely not too late to enroll in WOOT.  This week's classes are just "orientation"; the mathematical content of WOOT doesn't start until next week.

Normally, if you drop a class, we issue the refund via the original form of payment (for example, if you paid by credit card, we'll issue a refund to your credit card).  On occasion, though, we're unable to do this (for reasons I don't want to go into), in which case we'll mail you a check.

However, if you want to drop one class and immediately enroll in another, we can instead issue you a non-refundable AoPS gift certificate for the dropped class that you can immediately apply to the new class.

Email us at classes@artofproblemsolving.com with your information (essentially, with what you've said in this post), and we can set up the details.

By the way, PM'ing an individual is usually the least-effective way to contact us.  Emailing us at one the addresses on the "Contact Us" page (there's a link at the bottom of every web page) is by far the better way to get in touch with us.

Thank you for the info, Dr. Patrick! I will use the classes email address.


Let $ a,b,c,d$ be non-nagative real numbers such that: $ a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}$. Find the min value of the following expression:
\[ N=abc+bcd+cda+dab \]

to my brother, you should try it  :P Don't ask me  :P  Actually, it is very easy.

[quote="hungkhtn"]Hi my younger brother  :lol: I am not spamming ! I want to ask if you have an typo in "minimum" instead of "maximum". The minimum is clearly $ 0$ as in my example.[/quote]
okie, I am always understand you  :D But  the minimum value of $ N$ is not equal $ 0$. It means you are wrong  :D I haven't ever tried to find out the maximum value but I will try now  :P

I have 2 solutions for above problem. The proof is using Cauchy-Schawrz what I like best  :D

[quote="zaizai-hoang"][quote="hungkhtn"]Hi my younger brother  :lol: I am not spamming ! I want to ask if you have an typo in "minimum" instead of "maximum". The minimum is clearly $ 0$ as in my example.[/quote]
But  the minimum value of $ N$ is not equal $ 0$. [/quote]

Why the minimum is not equal to $ 0$? For $ a=1,b=c=d=0$ then $ a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}$ and
$ abc+bcd+cda+dab=0$. Even if the problem is given for positive real numbers, the infimum is $ 0$, too.


Hi there!

Let $x_{1}\, , \, x_{2}\, \, \ldots \, , \, x_{N}$ be $N \geqslant 2$ (real) numbers and consider the range of them, that is, the greatest possible absolute value of the differences between any 2 of them: \[range = \,\,\mathop{\max }_{1\, \leqslant \,u,v\, \leqslant \,N}\,\left|{x_{u}-x_{v}}\right|\]

If $\mu$ is the average of the $N$ given numbers, is it possible to have $\left|{x_{j}-\mu }\right| > \,\frac{{range}}{2}$ , where $j \in \left\{{1,2, \ldots ,N}\right\}\, \,$ ?

It is certainly possible, consider the data set $\{1,99,99,99,99,99,99\}$.  The range is $98$ and $\mu = 85$. And $|1-85|> \frac{98}{2}$.

EDIT: Yes sorry, it is fixed.

Correct, illcrowflu. You certainly mean that the range is 98 and $\left|{1-85}\right| > \,\,\frac{{98}}{2}$

On "the same subject", considering the standard deviation, defined as:

\[\sigma = \sqrt{\frac{1}{N}\cdot \sum_{j = 1}^{N}{\left({x_{j}-\mu }\right)^{2}}}\]

is it possible to have $\, \, \, \sigma > \frac{{range}}{2}$ ?

I´ve posted another (related-) problem at my last message. It´s a nice one, I guess!  Have a try!

Small technical hint:
[hide]Without loss of generality, you may consider $x_{1}\leqslant x_{2}\leqslant \ldots \leqslant x_{N}$ and, therefore, $range = x_{N}-x_{1}$[/hide]


In a circle with center M, two perpendicular chords AC and BD are given. The circle with diameter AM intersects the circle with diameter BM in M and P. The circle with diameter BM intersects the circle with diameter CM in M and Q. The circle with diameter CM intersects the circle with diameter DM in M and R. The circle with diameter DM intersects the circle with diameter AM in M and S.
Show that PQRS is a rectangle.

Because angles MQB and MQC are right we can see that Q is, in fact, the projection of M on BC, so it's the midpt of BC in the same way we find that P, R, S are the midpts of AB, CD, DA respectively. This means that PQ and RS are parallel to AC and QR and PS are parallel to BD, and from here we derive the conclusion.


induction
combinatorics unsolved
combinatorics

Let M ={ 1,2,....,40} .Find the smallest positive integer n for which it is possible to partition M into n disjoint subsets such that whenever a, b, and c ( not necessarily distinct ) are in the same subset ,then a is not equal to (b+c)

This is related to the so-called Schur's numbers.

Schur's original construction proves that the set $\{1, \cdots , \frac {3^n - 1} 2 \}$ can be splitted into $n$ pairwise disjoint subsets such that none of them contains a solution of the equation $x+y=z$ (in non necessarly distincts integers).
Therefore if $ 40 \leq  \frac {3^n - 1} 2$ that is $n \geq 4$ such a partition does exist.

In another hand it is a classical problem to prove that $\{ 1, 2 , \cdots, 14 \}$ cannot be partitioned into three subsets such that none of them contains a solution of the equation $x+y=z.$

Thus the desired minimum is $n = 4$.

Pierre.

If n =4 , how can we divide the set into 4 subsets that satisfy the condition ?

you can prove by induction that (3^n-1)/2 numbers can be divided into such n subsets.


There's a fair, handing out 218 prizes. 9 groups try to win the prizes.
3 groups do better compared to the others;
Group A is 15% of the population and gets 46 prizes
Group B is 19%, and gets 47
Group C is 6% and gets 19
What are the odds that someone in each group can win a prize?

[hide]Group A: 86/23

Group B: 171/47

Group C: 199/19

Other : 53/56[/hide]


geometry
circumcircle
incenter
geometry unsolved

The circumcircle of triangle ABC has center O and diameter AD . Let I be the incenter of triangle ABC . The lines AI, DI cut again the circumcircle at H, K respectively. Draw the line IJ perpendicular to BC at J . Prove that H, K, J are collinear .

Let K' be the intersection of HJ and (O). K'I meets (O) at K' and D'.
$ \hat{BK'H}\equal{}\hat{HBJ} \equal{}> BHJ\sim K'HB \equal{}> HJ.HK'\equal{}HB^2\equal{}HI^2 \equal{}> IHJ\sim K'HI$
$ \equal{}> \hat{JIH}\equal{}\hat{HK'I}\equal{}\hat{HK'D'}\equal{}\hat{HAD'}$
IJ parallel to OH => $ \hat{JIH}\equal{}\hat{IHO}\equal{}\hat{HAO}$
Hence, $ \hat{HAO}\equal{}\hat{HAD'}$, so $ A,O,D'$ are collinear.
$ \equal{}> D'\equiv D \equal{}> K'\equiv K$, hence $ H,K,J$ are collinear.

Dear Mathlinkers,
let 0 be the circumcircle of ABC, U the orthogonal projection of I on the A-altitude of ABC.
It is knowed that U, J and H are collinear.
Let Th the tangent to 0 at H and 1 the circle with diameter AI ; it goes through K and E .
We have Th // UI
Applying the Reim's theorem to 0 and 1, it follows that K, U and H are collinear...
Sincerely
Jean-Louis

[quote="conan_naruto236"] [color=darkblue]Let $ [AD]$ be the diameter of the circumcircle $ w \equal{} C(O,R)$ of $ \triangle ABC$ and let $ C(I,r)$ be the incircle of $ \triangle ABC$ . The

lines $ AI$ , $ DI$ cut again the circle $ w$ at $ H$ , $ K$ respectively. Denote $ J\in BC$ for which $ IJ\perp BC$ Prove that $ J\in HK$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] This nice problem is one from the celebrated old properties of a triangle. Observe easily that

$ \widehat {IAD}\equiv\widehat {JIH}$ and $ \frac {AI}{AD} \equal{} \frac {IJ}{IH}$ because the power of $ I$ w.r.t. $ w$ is $ p_w(I)\equiv\overline {IA}\cdot \overline {IH} \stackrel {(*)}{\ \equal{}\ } \minus{} 2Rr$ and

$ AD\cdot IJ \equal{} 2Rr$ . Therefore $ \triangle IAD\sim\triangle JIH$ $ \implies$ $ \widehat {JHI}\equiv\widehat {IDA}$ $ \implies$ $ \widehat {JHI}\equiv\widehat {KDA}$ , i.e. $ J\in HK$ .   [/color]

========================================================================

$ (*)\ \blacktriangleright$ Denote $ R\in AC$ so that $ IR\perp AC$ and the diameter $ [NS]$ of $ w$ so that $ NS\perp BC$

and the line $ BC$ separates $ A$ , $ S$ . Observe easily that $ \triangle AIR\sim\triangle NSC$ $ \implies$ $ \frac {AI}{NS}\equal{}\frac {IR}{SC}$ $ \implies$

$ IA\cdot SC\equal{}2Rr$ $ \stackrel {(**)}{\ \implies\ }$ $ IA\cdot IS\equal{}2Rr$ , i.e. the power of $ I$ w.r.t. $ w$ is $ p_w(I)\equal{}\minus{}2Rr$ .

========================================================================

$ (**)\ \blacktriangleright\ m(\widehat {SIC})\equal{}m(\widehat {SCI})\equal{}\frac {A\plus{}C}{2}\equal{}90^{\circ}\minus{}\frac B2$ $ \implies$ $ SB\equal{}SC\equal{}SI$ .


Let $ S$ be the set of all points with coordinates $ (x,y,z)$, where $ x, y,$ and $ z$ are each chosen from the set $ \{ 0, 1, 2\}$. How many equilateral triangles have all their vertices in $ S$?

$ \textbf{(A)}\  72 \qquad \textbf{(B)}\  76 \qquad \textbf{(C)}\  80 \qquad \textbf{(D)}\  84 \qquad \textbf{(E)}\  88$

I don't think I missed any...

[hide="Answer"]There are $9$ distinct cubes with edges parallel to the axes. Each of these cubes has $8$ distinct equilateral triangles formed by three face diagonals of the cube, so the answer is $8\cdot 9=72\Rightarrow \boxed{A}$.[/hide]

hmm..
i dont see how you got 9 distinct cubes. ;)

How can we be sure (or prove) there are no more equilateral triangles?

There can't be any more equilateral triangles per cube. If there were, it wouldn't fit any of the answer choices. ;) The only thing I'm not entirely sure of is that there aren't any more cubes, but I am [b]fairly[/b] sure that there aren't any others.

Anyway, to be a little more thorough, there can only be eight equilateral triangles per cube because I covered all where there are two points on top and two on bottom, and that automatically covers the sides, as well. It's a little hard to explain, but I'm sure there aren't any more. ;)

your statement about the equalateral triangles is correct.

There is also a cube with verticies on the points $(1,1,0)$ ,$(0,0,1)$, $(2,0,1)$, $(0,2,1)$, $(2,2,1)$, and $(1,1,2)$.  This cube has eight new triangles, so the final answer is $80$.

[quote="JesusFreak197"]Cubes have eight vertices. :huh:[/quote]
looks 6 to me. ;)

hmm..
weird, Problem 25 on the
[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2005]Contest Resorces[/url]
page is different from this one.! ;)

(its actually a duplicate of the problem 24)

[quote="amirhtlusa"]looks 6 to me. ;)[/quote]

Exactly my point. ;)

I made that same mistake at first as well, but I corrected it.

[quote="JesusFreak197"]Cubes have eight vertices. :huh:[/quote]

Oops sorry about that. :)

I checked the official answer and it's 80, but I can't think of any triangles we missed...

http://www.unl.edu/amc/mathclub/problems/H-problems/H-web/2005web/ha05-25-ia.html

That would mean that we missed a cube, but I don't see how you can possibly make another cube with the given points. :huh:

Oh, wow, I completely didn't think of using other points in the larger cube. :oops:

i got 80
i think the ones you are missing are on the planes in the middle of the 3x3x3 cube
there are 2 planes and each plane has 4 triangles on it

Could someone maybe post a rigorous solution for this since the transcripts to the mathjam are not available and everyone else in this thread were not able to?

The case that was missed is when the side length is $\sqrt{1^2+1^2+2^2}=\sqrt{6}$, which gives 8 more triangles. This happens when we take 3 midpoints of the edges of the 2x2x2 cube (that are also non-parallel and nonadjacent). The first edge can be any of 12, WLOG let it be parallel to the x-axis. Then, of the 4 edges parallel the the y-axis, we can choose 2 (the other 2 are adjacent). The edge parallel to the z-axis is then fixed. Since we have counted each triangle 3 times, we divide by 3 to get 12*2/3 = 8 more triangles.

I dont rrly get the solutions on the wiki, or here, so can someone please explain it to me if they get how to do it?


inequalities
number theory unsolved
number theory

I found this problem in some old papers, from some class

PROBLEM: Let $1=d_1<d_2<...<d_k=n$ be positive divisors of $n$. Prove that
\[k\sqrt{n}\leq d_1+d_2+...+d_k\].
First solve this then awnser my question...
I am asking when does equality holds (if it holds at all)??
[hide]
This is my solution $d_1+d_k\geq 2\sqrt{d_1d_k}=2\sqrt{n}, d_2+d_{k-1}\geq
2\sqrt{d_2d_{k-1}}=2\sqrt{n}... d_k+d_1\geq 2\sqrt{d_kd_1}=2\sqrt{n}$ So when we sum
all this inequalities we obtain $2(d_1+d_2+...+d_k)\geq2k\sqrt{n}$ But as one can
see EQUALITY doesn't stand for any $n$, so where do I make mistake, or there is
mistake in text.[/hide]

I'm tired, so read at youre own risk ;)
[What a pity I made an awfull mistake, I'm sorry :(]

I have a solution for the inegality:

$\frac{d_1+d_2+...+d_k}{k}>= \sqrt[k]{d_1d_2...d_k}$  (AM-GM)

Now, I put $d_i$ with $d_{k-i}$ because $d_id_{k-i}=n$

FIRST CASE: n is even:

$\frac{d_1+d_2+...+d_k}{k}>= \sqrt[k]{n^{\frac{n}{2}-1}*n/2} >= \sqrt[k]{n^{\frac{k}{2}}}$ because k is the number of divisor.
and we are done

SECOND CASE : n is odd:

$\frac{d_1+d_2+...+d_k}{k}>= \sqrt[k]{n^{\frac{n}{2}-1}} >= \sqrt[k]{n^{\frac{k}{2}}}$ because k is the number of divisor.
and we are done

I forgot to add, in the text it goes that $n$ is natural and $n>1$ so equaltiy doesn't hold at all

Thank you for the solution libillau but that part wasn't relly my problem, look at my questions :D :D D

P.S. You have typo, it isn't $d_id_{k-i}=n$, its $d_id_{k-i+1}=n$ because $d_1$ goes with $d_k$, $d_2$ goes with $d_{k-1}$ and so on...
And you really don't need cases but never mind your solution is in basics correct and almost same as mine

I will post now second part of that problem so if you are interestef look at it, but it $wierder$ then this one


Define $ n \in \mathbb{N}$ as the minimum positive integer such that in every set of $ n$ consecutive positive integers there is at least one element with digital sum multiple of 11. Show that for all $ m \in \mathbb{N}$ there exist a infinite number of sets of $ n\plus{}7$ consecutive positive integers such that all begin with $ m$ (in the usual decimal representation) and exactly one element has its digital sum multiple of 11.

(i.e. with "begin" i mean that 676498 begins with 67, or also with 6764..)

I hope you'll like it  :lol:

[b]1) We claim that the minimum positive integer $ n$ as described in the problem post equals $ 39$.[/b]

[i]Proof[/i]: Consideration of the series $ \{999981,999991,\ldots,1000018\}$ yields $ n > 38$.

For any positive integer $ N = \sum_{k = 0}^Ma_k10^k$ let $ S(N)$ denote the set of digital sums corresponding to the numbers $ N,N + 1,\ldots,N + 9$. If $ 10|N\Longleftrightarrow a_0 = 0$ we have $ S(N) = \{\sum_{k = 1}^Ma_k,1 + \sum_{k = 1}^Ma_k,\ldots,9 + \sum_{k = 1}^Ma_k\}$. Therefore, for such $ N$ we have $ S(N)\cap 11\cdot\mathbb{Z} = \emptyset$, iff $ \sum_{k = 1}^Ma_k\equiv 1\bmod{11}$ holds. Moreover, if additionally $ S(N + 10)\cap 11\cdot\mathbb{Z} = \emptyset$ holds, we must have $ a_1 = 9$ or equivalently $ N\equiv 90\bmod{100}$. In particular, if $ 10|N$ holds, at least one of the sets $ S(N),S(N + 10),S(N + 20)$ contains a multiple of $ 11$.
Now consider an arbitrary series $ N,N + 1,\ldots,N + 38$ of $ 39$ consecutive numbers. Let $ N\equiv r\bmod{10}$ with $ 0<r\leq 10$, then the series has $ N+10-r,\ldots,N+39-r$ as a subseries and therefore, application of the previous remark to $ S(N+10-r),S(N+20-r),S(N+30-r)$ completes the proof.

[b]2) We claim that for any $ m\in\mathbb{N}$ there are infinitely many series of $ 47$ consecutive numbers, such that their decimal representation starts with $ m$ and only one of their corresponding digital sums is divisible by $ 11$.[/b] (Note, that this is stronger than the claim of $ 46= n + 7$ in the initial post)

[i]Proof[/i]: For arbitrary $ m\in\mathbb{N}$ let $ d(m)$ denotes its digital sum. Choose $ 0\leq B\leq 9,0\leq A < 3$ such that $ A + B + d(m)\equiv 0\bmod{11}$ holds. Now, for arbitrary $ T$ with $ T\equiv 5\bmod{11}$, we define $ N_T = m10^{T + 3} + B10^{T + 2} + A10^{T + 1} + 9\sum_{k = 2}^T10^k+81$ with digital sum $ A + B + d(m) + 9T\equiv 1\bmod{11}$. The digital sums corresponding to the series $ N_T,N_T + 1,\ldots,N_T + 46$ are then determined by:

$ {\cal R} = \underbrace{d(m) + A + B + 9T,\ldots,d(m) + A + B + 9T + 8}_{\mbox{first}\;9},\underbrace{d(m) + A + B +9T,\ldots,d(m) + A + B + 9T+9}_{\mbox{next}\;10},$

$ \underbrace{d(m) + A + B + 1,\ldots,d(m) + A + B + 10}_{\mbox{next}\;10},\underbrace{d(m) + A + B + 2,\ldots,d(m) + A + B + 11}_{\mbox{next}\;10},$ $ \underbrace{d(m) + A + B + 3,\ldots,d(m) + A + B + 10}_{\mbox{final}\;8}\\[0.2cm]$

Further transformation modulo $ 11$ shows $ {\cal R} = \underbrace{1,\ldots,9}_{\mbox{first}\;9},\underbrace{1,\ldots,10}_{\mbox{next}\;10} ,\underbrace{1,\ldots,10}_{\mbox{next}\;10} ,\underbrace{2,\ldots,11}_{\mbox{next}\;10},\underbrace{3,\ldots,10}_{\mbox{final}\;8}$ and only the digital sum corresponding to $ N_T + 38 = m10^{T + 3} + B10^{T + 2} + (A + 1)10^{T + 1} + 19$ is a multiple of $ 11$.

i don't understand: you get the right proof with the wrong solution  :D  :D

NB. consider {999981,...} :wink:

Thx, for pointing it out. Actually, my proof was flawed. I edited my original post and now it is fine.