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That depends on how you define the exponential operator.

Here's a question for you:  given the familiar definition for $ a^b$ in terms of repeated multiplication and square roots, we can compute $ a^b$ whenever $ b$ is rational.  So how do you compute $ a^b$ when $ b$ is irrational?

Note that the intuitive limit argument doesn't work because you have no guarantee that the limit exists.

[hide="The answer..."] Free yourself from multiplication.  One way to do this is to define $ \ln x \equal{} \int_{1}^{x} \frac {dt}{t}$, which is defined and continuous for all $ x > 0$, and show that $ \ln ab \equal{} \ln a \plus{} \ln b$, and then define $ e^x$ to be its inverse function, which is defined and continuous for all $ x \in \mathbb{R}$.  We then know that $ e^{x \plus{} y} \equal{} e^x e^y$ and that $ e^0 \equal{} 1$.  We then define

$ a^b \equal{} e^{b \ln a}$.

Note that a weakness in this definition is that it is undefined for all $ a \le 0$, but exponentials of negative numbers are very poorly defined (over the reals) anyway.  [/hide]

A simple way to view this is:

Let $ n,m\in\mathbb{R}^\plus{}: n\minus{}m\in\mathbb{R}^\plus{}$

Then $ \frac{x^{n}}{x^{m}}\equal{}x^{n\minus{}m}$

If $ n\equal{}m$, then $ 1\equal{}\frac{x^{n}}{x^{m}}\equal{}x^{n\minus{}m}\equal{}x^0$

[quote="SnipedYou"]Well the way I see it is that $ a^{n}$ means $ 1 \cdot a \cdot a \cdot ... \cdot a$, $ n$ times. So if we raise anything including $ 0$ to the $ 0$ power, then we don't multiply by anything other than $ 1$, so anything raised to the $ 0$ power equals $ 1$. I hope that makes sense.[/quote]

Yes, it does make sense.  Another way to put this is to say that an empty product equals $ 1$, i.e. "not multiplying by anything" is the same as "multiplying by one."  This also is one good explanation for why $ 0! \equal{} 1$ (it's the product of the first 0 positive integers, i.e. it's an empty product).

[quote="JZambrano201"]A simple way to view this is:

Let $ n,m\in\mathbb{R}^ \plus{} : n \minus{} m\in\mathbb{R}^ \plus{}$

Then $ \frac {x^{n}}{x^{m}} \equal{} x^{n \minus{} m}$

If $ n \equal{} m$, then $ 1 \equal{} \frac {x^{n}}{x^{m}} \equal{} x^{n \minus{} m} \equal{} x^0$[/quote]

It doesn't follow for x=0.

If x=0,  $ \frac {x^{n}}{x^{m}} \equal{} \frac00$ = undefined :wink:

I was just discussing this with a friend.  :P
L'Hôpital for the win, essentially (ignoring the exponents in this case because it does not affect the answer since $ n\equal{}m$)
$ \lim_{x \to 0} \frac{x}{x} \equal{} \lim_{x \to 0} \frac{1}{1} \equal{} 1$.

Again, this depends on your definition of exponentiation.  The property that $ \frac {x^n}{x^m} \equal{} x^{n \minus{} m}$ is a familiar property of the positive-integer-valued exponent, and we extend exponentiation to the reals such that this property continues to hold.  It is not a fundamental part of any definition of exponentiation (because as pacman mentions correctly, it doesn't hold when $ x \equal{} 0$).

The essential point of your argument (if you want to consider it an actual proof) is that the function $ f(x) \equal{} \frac {x^n}{x^n} \equal{} x^{n \minus{} n} \equal{} x^0$ has the property that $ \lim_{x \to 0} f(x) \equal{} f(0)$.  In other words, you're [b]assuming[/b] the above function is continuous at zero, which is precisely what you wanted to prove.

Again, using the definition of exponentiation I gave, the above function is not defined at zero.  It's not a removable discontinuity, either, because - again - exponentiation is very poorly defined for negative bases and it's most sensible to not allow it to be defined (over the reals) at all.

Over the integers exponetiation is defined as:

$ n,m\in\mathbb{Z}^\plus{}$
$ x\in\mathbb{Z}: x\ne0$
$ x^{m\plus{}n}\equal{}x^mx^n$
$ x^{m\minus{}n}\equal{}\frac{x^m}{x^n}$
$ x^{\minus{}n}\equal{}\frac{1}{x^n}$
$ x^0\equal{}1$
$ x^1\equal{}x$
$ x^{\minus{}1}\equal{}\frac{1}{x}$
$ (x^m)^n\equal{}x^{mn}$

Redundant.  The only thing you need to define exponentiation over integer exponents is that $ x^0 \equal{} 1$ and $ x^{n\plus{}1} \equal{} x \cdot x^n$ (which is what I meant when I said the cancellation property you talked about is not part of the definition of exponentiation.)  You can prove [b]every[/b] other property on that list from the above.

How are negative exponents defined as $ x^{\minus{}n} \equal{} \frac{1}{x^{n}}$ just using those two relations?  Sorry, I just don' t see it.  :oops:
Otherwise, nice presentation, t0rajir0u.

Also, are there other explanations besides the empty product that are valid - regardless of the math involved, including Calculus?

$ x^{\minus{}1 \plus{} 1} \equal{} x\cdot x^{\minus{}1} \implies x^0 \equal{} x\cdot x^{\minus{}1}$
$ x^{\minus{}2 \plus{} 1} \equal{} x\cdot x^{\minus{}2} \implies x^{\minus{}1} \equal{} x \cdot x^{\minus{}2}$
etc.

Yes, I know how one would get negative exponents with those relations.  :)
Maybe I was not clear, how does one know that $ x^{ \minus{} n} \equal{} \frac {1}{x^{n}}$?  Or, rather, is that something that needs to be stated?

$ x^{ \minus{} 1 \plus{} 1} \equal{} x\cdot x^{ \minus{} 1} \implies x^0 \equal{} x\cdot x^{ \minus{} 1} \implies 1 \equal{} x\cdot x^{\minus{}1} \implies \frac{1}{x} \equal{} x^{\minus{}1}$
$ x^{ \minus{} 2 \plus{} 1} \equal{} x\cdot x^{ \minus{} 2} \implies x^{ \minus{} 1} \equal{} x \cdot x^{ \minus{} 2} \implies \frac{1}{x} \equal{} x\cdot x^{\minus{}2} \implies \frac{1}{x} \cdot \frac{1}{x} \equal{} x^{\minus{}2}$
etc.

[quote="AstroPhys"]I have always been told, "Anything raised to the zero-th power is one, except for zero."  Recently, however, I have discovered that it is not necessarily true.  I have read the Wikipedia article about it and I understand most of the reasons. Are there other reasons?  How about ones that could be explain with more simplier terms geared, say, for kids that have not taken Calculus?

 :) .[/quote]

The point has probably been made, but I have found this property of exponents  easiest to understand in the following way:

Consider $ x,\,n \in \mathbb{R}$, such that $ x^n \equal{} 1$.

Taking the logarithm: $ \log(x^n) \equal{} \log 1 \equal{} 0$.

\[ \log(x^n) \equal{} n\log(x) \equal{} 0 \]

\[ n \equal{}\frac{0}{\log(x)} \equal{} 0\]

[quote="AstroPhys"]$ \log 0$ is undefined, though.[/quote]

I believe you've answered your own question, because if we have this

\[ n\cdot \underbrace{\log 0}_{\text{meaningless nonsense}} \equal{} 0\]

then what is the value of $ n$ that makes this statement true?

So, the question of what $ 0^0$ should mean usually depends on context. In several situations, different conventions are taken on to mean what people would like them to mean. I can't think of an example for $ 0^0$ to mean $ 0$, but here are two situations where it is useful for it to mean $ 1$:

1) When writing polynomials, it is often nice to write $ P(x) \equal{} \sum_{n\equal{}0}^{\deg P}a_nx^n$. But what happens when you evaluate it at $ 0$? Clearly you want it to be the constant coefficient, which only works if $ 0^0 \equal{} 1$.

2) When doing information theory, the expression $ p\ln p$ appears in entropy calculations. When $ p \equal{} 0$, we take the convention that $ p\ln p \equal{} 0$, since this coincides with $ \lim_{p \to 0^\plus{}}p\ln p$. This correlates to $ 0^0$ since $ 0\ln 0 \equal{} 0$ is essentially the same as $ 0^0 \equal{} 1$ (take the natural logarithm of both sides).

[quote="Hamster1800"]So, the question of what $ 0^0$ should mean usually depends on context. In several situations, different conventions are taken on to mean what people would like them to mean. I can't think of an example for $ 0^0$ to mean $ 0$, but here are two situations where it is useful for it to mean $ 1$:

1) When writing polynomials, it is often nice to write $ P(x) \equal{} \sum_{n \equal{} 0}^{\deg P}a_nx^n$. But what happens when you evaluate it at $ 0$? Clearly you want it to be the constant coefficient, which only works if $ 0^0 \equal{} 1$.

2) When doing information theory, the expression $ p\ln p$ appears in entropy calculations. When $ p \equal{} 0$, we take the convention that $ p\ln p \equal{} 0$, since this coincides with $ \lim_{p \to 0^ \plus{} }p\ln p$. This correlates to $ 0^0$ since $ 0\ln 0 \equal{} 0$ is essentially the same as $ 0^0 \equal{} 1$ (take the natural logarithm of both sides).[/quote]

It is possible to define $ 0^0 \equal{} 1$ if one does so [i]a priori[/i] but that leads to an inconsistency. Consider $ 0\cdot 0^x \equal{} 0^0$. This implies $ 1 \plus{} x \equal{} 0\implies x \equal{} \minus{}1$, then $ 0^x \equal{} 0^{\minus{}1}$. The number $ 0^{\minus{}1}$ does not exist.

As for the expression $ p\ln p$, this appears not only in information theory, but in thermodynamics as well, as an expression of the entropy of a state whose probability of occupation is $ p$. This is why Claude Shannon, at von Neumann's recommendation, called his information measure "entropy." That said, the information contained in a message that is received with probability zero is: $ \lim_{p\to 0} H(p) \equal{} \lim_{p\to 0} \minus{}p\log p \equal{}\lim_{n\to\infty} \minus{}\frac{1}{n}\log\left(\frac{1}{n}\right) \equal{} \lim_{n\to\infty}\frac{\log(n)}{n} \equal{} 0$ by L'Hopital's rule.  This makes sense, as there is no information in a message one cannot receive. Likewise, $ \lim_{p\to 1} H(p) \equal{}\lim_{p\to 1} \minus{}p\log p \equal{} \minus{}1\cdot\log 1 \equal{} 0$. If there is no uncertainty in a given message, said message is uninformative.

I'm not sure what you're saying in the first part. The fact that $ 0^{\minus{}1}$ doesn't exist is just recreated in your expression of $ 0\cdot 0^x \equal{} 0^0 \equal{} 1$? There's no number such that $ 0\cdot x \equal{} 1$, so clearly your $ 0^x$ can't exist...

[quote="Hamster1800"]... There's no number such that $ 0\cdot x \equal{} 1$, so clearly your $ 0^x$ can't exist...[/quote]

There is no $ x \ni 0\cdot x \equal{} 1$ and $ 0^{\minus{}1}$ does not exist are alternative ways of saying the same thing. The upshot is that $ 0^0$ is also a non-existent number; if you define it as equaling 1 then there is no way to obtain that number by multiplying factors of zero together, which is what exponentiation is all about.

[quote="t0rajir0u"][quote="Dr. No"]Consider $ 0\cdot 0^x \equal{} 0^0$. This implies $ 1 \plus{} x \equal{} 0\implies x \equal{} \minus{} 1$[/quote]
It doesn't.  $ 0^a \cdot 0^b \equal{} 0^c \equal{} 0$ holds for any positive integers $ a, b, c$.[/quote]

Not sure what your point is. Comparing your example to mine, $ a \equal{} 1,\quad b \equal{} x,\text{ and}\quad c \equal{}0$.  Neither $ b$ nor $ c$ is positive, and there is no stipulation that any of the exponents in any particular example one might dream up must be integers.

I wonder why this hasen't already been locked...

@Dr. No: You gave no reasoning, just vague guesses.
Why should $ 0 \cdot 0^x \equal{} 0^0$ imply $ 1\plus{}x\equal{}0$¿ We surely have $ 1 \cdot 1^{42} \equal{} 1^0$, but surely not $ 1\plus{}42\equal{}0$.

Additionally, your argument seems to be this:
If we do some "forbidden" operation with $ 0^0\equal{}1$ (or $ 0^0\equal{}0$), then we get something strange; thus $ 0^0$ is undefined.
This kind of argument doesn't work.

[quote="ZetaX"]I wonder why this hasen't already been locked...

@Dr. No: You gave no reasoning, just vague guesses.
Why should $ 0 \cdot 0^x \equal{} 0^0$ imply $ 1 \plus{} x \equal{} 0$¿ We surely have $ 1 \cdot 1^{42} \equal{} 1^0$, but surely not $ 1 \plus{} 42 \equal{} 0$.[/quote]

\[ 0\cdot 0^x \equal{} 0^0 \Rightarrow \log 0 \plus{} x\log 0 \equal{} 0\log 0 \]

\[ (1 \plus{} x)\log 0 \equal{} 0\log 0\]

Dividing by $ \log 0$ (legitimate because whatever $ \log 0$ might be $ \log 0 \neq 0$),

\[ 1 \plus{} x \equal{}0\]

In your example,

\[ 1\cdot 1^{42} \equal{} 1^0 \\ \Rightarrow \log 1 \plus{} 42\cdot\log 1 \equal{} 0\cdot \log 1 \\ \Rightarrow 43\cdot\log 1 \equal{} 0\].

Dividing by $ 43$,

\[ \log1 \equal{} \frac{0}{43} \equal{} 0\]

which is of course completely true.

Have you actually read our posts?  Apply the same logic to, say, $ 0 \cdot 0^{42} \equal{} 0^3$.  We get $ 1 \plus{} 42 \equal{} 3$.

Of course, having seen the way you talk about logarithms, I'm likely not going to post in this thread again.  The danger with people like you is that people who don't know any better are fooled into thinking that you know what you are talking about, so if you appear to get too convincing I may return.

I think that he's trying to say that $ 0^1\cdot0^{\minus{}1} \equal{} 0^0 \equal{} 1$, but $ 0^1 \equal{} 0$, so we have a contradiction. However, the fault in this is that $ 0^{\minus{}1}$ doesn't exist, so we can't multiply it to get $ 0^0$. Indeed, we can't make $ 0^0$ out of multiplying other powers of $ 0$, so I see no reason that your line of reasoning leads to a contradiction.

[quote="t0rajir0u"]Have you actually read our posts?  Apply the same logic to, say, $ 0 \cdot 0^{42} \equal{} 0^3$. [/quote]

I'll do better than that.

[quote="t0rajir0u"]Of course, having seen the way you talk about logarithms, I'm likely not going to post in this thread again.  The danger with people like you is that people who don't know any better are fooled into thinking that you know what you are talking about, so if you appear to get too convincing I may return.[/quote]

Is the ad hominem really necessary?

In my previous post I implicitly assumed that $ \log 0 \in \mathbb{R}$. This time I explicitly assume this in order to prove the contrary.

Taking your example, $ 0 \cdot 0^{42} \equal{} 0^3$,

\[ \log 0 \plus{} 42\log 0 \equal{} 3\log 0 \\ \Rightarrow 40\log 0 \equal{} 0\]

Dividing by $ 40$,
\[ \log 0 \equal{} 0 \\ \Rightarrow e^{\log 0} \equal{} e^0 \\ \Rightarrow 0 \equal{} 1\]

That's a contradiction.

Therefore, $ \log 0 \not\in \mathbb{R}$.

We thus have come full circle back to my previous post:
\[ n\cdot\underbrace{\log 0}_{\text{meaningless nonsense}} \equal{} 0\]

The question remains,"What value of $ n$ makes the above statement true?"

I think the general convention for $ \ln 0$ is to say that $ \ln 0 \equal{} \minus{} \infty$. In this case, your equation ($ 43\log 0 \equal{} 3\log 0$) works perfectly fine without $ \log 0 \equal{} 0$. The question of $ 0^0 \equal{} e^{0\ln 0}$ just becomes a question of whether we want $ 0\cdot \minus{} \infty$ to equal $ 0$ or $ \minus{} \infty$ (or something else).

Again, normally $ 0^0$ is just assigned a value as a convention. It's like saying, "Hey, we want $ x^0$ to represent the same function as $ f(x) \equal{} 1$," not that $ 0^0 \equal{} 1$ in every possible meaning of the form $ 0^0$.

[quote="Hamster1800"]I think the general convention for $ \ln 0$ is to say that $ \ln 0 \equal{} \minus{} \infty$. In this case, your equation ($ 43\log 0 \equal{} 3\log 0$) works perfectly fine without $ \log 0 \equal{} 0$. The question of $ 0^0 \equal{} e^{0\ln 0}$ just becomes a question of whether we want $ 0\cdot \minus{} \infty$ to equal $ 0$ or $ \minus{} \infty$ (or something else).[/quote]

Actually, $ \log 0$ is undefined, as I showed in my previous post.

Nonetheless, it is customary to define it as, $ \log 0 \doteq \lim_{n\to\infty}\log(\frac{1}{n}) \equal{} \lim_{n\to\infty} \minus{}\log(n) \equal{} \minus{}\infty$.

However, $ \infty$ is not a real number. (It represents no actual point on the real number line.)

[quote="Hamster1800"]Again, normally $ 0^0$ is just assigned a value as a convention. It's like saying, "Hey, we want $ x^0$ to represent the same function as $ f(x) \equal{} 1$," not that $ 0^0 \equal{} 1$ in every possible meaning of the form $ 0^0$.[/quote]
Usually $ 0^0$ is defined as $ 0^0 \equal{} 1$ for the sake of convenience, but as I have repeatedly attempted to show there is no way to obtain $ 0^0$ by adding powers of zero.

As for any non-zero real number, it is not a matter of wanting $ x^0$ to represent something; $ x^0 \equal{} 1$, period. Consider,
\[ 1 \equal{} x^m\cdot x^{\minus{}m} \equal{} x^{m \minus{} m} \equal{} x^0\]

I agree with your first part. $ \log 0 \equal{} \minus{}\infty$ is really just abuse of notation meaning that $ \lim_{x\to 0^\plus{}}\log x \equal{} \minus{}\infty$.

However, I can't understand your second part. You were trying to show that $ 0^0 \equal{} 1$ leads to a contradiction, but you relied on the fact that $ 0^1\cdot 0^{\minus{}1} \equal{} 0^0$, but $ 0^{\minus{}1}$ doesn't exist, so this doesn't work. I agree, there is no way to obtain $ 0^0$ by multiplying powers of $ 0$ other than $ 0^0$.

Also, you seem to be interpreting what I said incorrectly. I said, we want the function $ x^0$ to be the same as the function $ 1$, not the function $ \begin{cases}1\text{ when }x\neq 0\\\text{undefined when }x\equal{}0\end{cases}$. I understand that $ x^0 \equal{} 1$ for nonzero $ x$ perfectly well.

[quote="Hamster1800"]I agree with your first part. $ \log 0 \equal{} \minus{} \infty$ is really just abuse of notation meaning that $ \lim_{x\to 0^ \plus{} }\log x \equal{} \minus{} \infty$.

However, I can't understand your second part. You were trying to show that $ 0^0 \equal{} 1$ leads to a contradiction, but you relied on the fact that $ 0^1\cdot 0^{ \minus{} 1} \equal{} 0^0$, but $ 0^{ \minus{} 1}$ doesn't exist, so this doesn't work. I agree, there is no way to obtain $ 0^0$ by multiplying powers of $ 0$ other than $ 0^0$.

Also, you seem to be interpreting what I said incorrectly. I said, we want the function $ x^0$ to be the same as the function $ 1$, not the function $ \begin{cases}1\text{ when }x\neq 0 \\
\text{undefined when }x \equal{} 0\end{cases}$. I understand that $ x^0 \equal{} 1$ for nonzero $ x$ perfectly well.[/quote]

I believe we agree at this point. We both understand that 1) Zero has no inverse; there is therefore no way to obtain $ 0^0$ by adding powers of zero, and 2) the expression $ \log 0 \equal{} \minus{}\infty$ is a case of symbol abuse.

It is okay as far as I'm concerned to define $ 0^0 \doteq 1$ as long as one recognizes beforehand that one can't obtain the number $ 0^0$ by algebraic manipulation (doing so requires division by zero).

This is one of those situations where it's really best to have more than one version of the same definition. If I'm talking about calculus, $ 0^0$ is "Not a Number". If I'm talking about polynomials or empty products, $ 0^0$ is $ 1$.

I'm perfectly fine with saying that $ \log 0$ is $ \minus{}\infty$. That's not a real number, but it's still something we can do some arithmetic with.

For another example: what is the degree of the zero polynomial? $ \minus{}\infty$ is a good choice.


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What does this equal? (How would you rewrite it as an integral?) I know it is a riemann sum of some kind...

$ \lim_{n\to\infty}\frac{1}{n}\left(\sqrt{\frac{1}{n}}\plus{}\sqrt{\frac{2}{n}}\plus{}...\plus{}\sqrt{\frac{n}{n}}\right)$

[quote="Kent Merryfield"]Look at the advice I gave on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=369679316&t=71992]this topic.[/url][/quote]

Just to make sure I'm doing it right, I rewrote it as $ \lim_{n\to\infty}\frac {1}{n}\sum_{i \equal{} 1}^{n}\sqrt {\frac {i}{n}}$
which if $ x_i \equal{} i/n$, then it is just $ \int^1_0\sqrt {x}\,dx$ right?


A plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?

I understand that against the wind is x - 30 and flying with the wind is x + 30. I am still having trouble solving.

I don't understand how you got r = 150 from (r -150)(r + 6).

Since $(r-150)(r+6)=0$, then either r-150=0 or r+6=0, so r=150 or -6.  But since you can't go -6 mph, the answer is 150.


Let a and b be two positive integers.

What is the smallest possible value of a+b given that 6a:^3:  = 5b:^5: ?

hehehe, got a very long number.

It's just a guess. 188956805400

Your working is really the most important thing here, not the answer. How did you get your answer?

i think he was joking or summat. i think that is. it wasnt a funny joke though. if it was

I'm not in 8th grade or below, so I can't answer this question, but to make things more clear, is it 6a^3=5b^5? I can't quite read the numbers in the exponent. They're too small for me. Perhaps other people are also having this problem.

Yep it is powers of 3 and 5, respectively.

i think the answer is 5580
because simplifying 6a^3 = 5b^5 gives:
a^3 = 5/6 (b^5) = the third root of (5b^2 /6)
that can be further simplified to b * the third root of (5b^2 / 6)
suppose x is some number that x^3 = 5b^2 / 6
simplifying gives b = root of (x^3 * 6 / 5)
this means that b has to be a multiple of 5 and that it also has to be a multiple of 6 to be able to make b a positive integer.
since the least common multiple of 5 and 6 is 30, i plugged 30 in for x and got that b = 180. after plugging that in for the original equation 6a^3 = 5b^5, i got that a was 5400. adding gives 5580.
please tell me if there is a mistake with my work and/or my thinking  :wink:

I get that too, and I'm pretty sure it's right.

Well, I must have miscounted (like I always do...)

how was that a miscount when you guessed. anywho back to my geetar.....

I did not guess!!! (even though I said it was on my first post. I think I typed it, cause I knew that big of a number would be wrong) I made it into a^3 = 5/6 x b^5. And then I figured that b = 6^n x 5^m and a = 6^x  x 5^y, and that  5n - 1 = 3x and found the smallest positive integers that worked , and then 5m + 1 = 3y, and did the same... It was about a week ago, so then I don't really remember how I came up with the answer. Anyways, cya for now.

ignore previous post!!

do a prime factor analysis and then solve 3 equations.

answer is 5580  <- not 1180

hint: only consider prime factors 2, 3, and 5.

umm barnacle you can just edit your previous post if you want to fix something... just fyi... it clears up clutter. I guess as of today I'm not an 8th grader (school's out!!!) , so I can't answer this problem... but here's a hint...
[hide]as Barnacle said... the key is the six and five multiplied by the numbers raised to their respective powers. I'm not going to say anymore before I give away the problem.[/hide]


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Hi all,
I'm new to this forum (Xevarion suggested I come in), and I was hoping for some help in a) properly articulating a curiosity of mine and, then, hopefully b) answering it.

Here goes:
I use wavelet analysis in my research (neuroscience), and have never found a good explanation for which wavelet I should use and when. I understand that there is always a tradeoff between frequency and time resolution, and that different wavelets balance these two aspects differently.
I stumbled upon a paper by Newland, 1993, about the "Harmonic Wavelet Transform," which purports to have very high frequency resolution, at the expense of low time resolution. The time-representation of the wavelet function has a classic "wavelet look" to it, but surprisingly, the frequency representation is just a step function over the frequency interval to be analyzed!
The wavelet's Fourier transform is so simple. It intuitively encapsulates why the wavelet performed as it does.

I have not been able to find a treatment of the properties of wavelets in the Fourier domain and the relations to a wavelet's time-frequency resolution tradeoff.

I have looked at a few other wavelets' frequency representations, and I believe there is a pattern to this.

Any help is much appreciated!

This is a great question.  I wish I knew the answer.

My guess is that deciding what kind of wavelet to use in what situation is difficult.  It may just be one of those issues that researchers struggle with.  Then again, I wouldn't be surprised to hear that there are good rules of thumb to help with that kind of decision.

I must confess that I know almost nothing about wavelets, and probably should not responding to your question at all.

It has been a few days now and nobody has answered your question.  So here is my question---where on the internet can one go to ask a question like this?

In a few years, lots of AoPSers will be all grown up and have jobs, so AoPS forums will probably be better for research-level applied math questions.

Lots of researchers have questions, and lots of people know the answers and would be happy to give them.  It feels like there should be a better system for matching up people with questions and people who have answers.


[b]Problem 1[/b]:Let $ a,b,c$ be non negative real numbers. Prove that

$ \frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b}\plus{}\frac{27abc}{2(a\plus{}b\plus{}c)^3}\ge2$

[b]Problem 2[/b]:Let $ a,b,c$ be non negative real numbers. Prove that

$ \frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b}\plus{}\frac{3\sqrt[3]{abc}}{2(a\plus{}b\plus{}c)}\ge2$

[quote="dduclam"][b]Problem 1[/b]:Let $ a,b,c$ be non negative real numbers. Prove that

$ \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \plus{} \frac {27abc}{2(a \plus{} b \plus{} c)^3}\ge2$

[b]Problem 2[/b]:Let $ a,b,c$ be non negative real numbers. Prove that

$ \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \plus{} \frac {3\sqrt [3]{abc}}{2(a \plus{} b \plus{} c)}\ge2$[/quote]
[hide="Problem 1"]
Use SOS method[/hide]
[hide="Problem 2"]
By Cauchy-Schwar inequality, we have:
$ \sum \frac{a}{b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\equal{}1\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)}$

So we need to prove that:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{3\sqrt[3]{abc}}{(a\plus{}b\plus{}c)} \ge 2$
It is equivalent to
$ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)\plus{}3\sqrt[3]{abc}(ab\plus{}bc\plus{}ca) \ge 2(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)$

By AM-GM inequality:
$ 3\sqrt[3]{abc}(ab\plus{}bc\plus{}ca) \ge 9abc$

We need to prove:
$ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)\plus{}9abc \ge 2(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)$
$ \Leftrightarrow a^3\plus{}b^3\plus{}c^3\plus{}3abc \ge \sum ab(a\plus{}b)$
Which is clearly true by Schur inequality.
[/hide]

We have a harder problem:

Find the best constant for inequality
$ \frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b}\plus{}\frac{3kabc}{a^3\plus{}b^3\plus{}c^3} \ge k\plus{}\frac3{2}$


[color=blue]I'm a sophmore in high school right now, and making course-selections for junior year.  I am serious about my studies, and I am interested in pursuing a career in science, probably biology.  I have not taken a high school level chemistry course yet, but did take an introductory chemistry course in eighth.  I have looked through some of the posts here and seen many have taken AP Chem.  In your opinion, should I take AP Chem or honors chem next year?[/color]

Well, it's hard to say.

Look at what most people at your school do.  Do many enter AP chemistry without honors?  At my school, that would be very effective grade suicide (and I know several who have done it).  That is to say, chemistry is a two-year course at my school.  You should probably talk to the chemistry teachers at your school.  Also, take a look at some of the work being done in the honors chem class now, and see if you can find work that was done at the beginning of AP chem.  I suppose it comes down to one question - is AP chem a continuation of honors chem, or a harder version of it?

--Standard Disclaimer--

yea...in my school, i'm a junior and taking ap chemistry now, It's not that bad ...our accelerated chem last yera, was mostly stoichiometry and equilibrium and a few other stuff (molecular geometry, history, intro, periodic trends...etc.) also....few labs.. A+'ed everything there.  And now in ap chem, it's a little different because you learn more in depth about the stuff you touched on in regular or honors chemistry, but i'm not saying not to take it, beacuse if you're serious and have an expansive mind, you'll do fine.

If you can grasp the concepts of the basic issues of chemistry, then ap chemistry is just taking it furthur...

Also, you should look through an AP chem book at a local bookstore to just skim through what you'll learn in ap chem...

hope that helps!!

I went right into AP chem during my junior year with little prior knowledge. My teacher was very clear and explained everything well. I got through it as well as (probably better than) those who had taken high school chemistry already.
On the other hand, I have also heard horror stories about teachers who don't cover the material clearly.
So it really depends on how it's done in your school. I would advise asking people in the class and possibly sitting in during one of the lessons, to get a good feel of how things are run.
You cover more material in AP, and in greater depth, so consider your time investments as well.

yea that's true....my teacher this year is horrible, but i read 2 other ap chem review books plus this site for help and helping others....

so check with older students about the ap chem teacher!!!


linear algebra
matrix
algorithm
linear algebra unsolved

det(AB)=det(A)det(B)

Maybe there is intuitive proving of this using the following
- $ AB \equal{} \left[Ab_1\cdots Ab_n\right]$

following is actually just def of determinant
-linear property
$ \begin{vmatrix}
{r_1}\\ {.}\\ {.}\\ {.}
\end{vmatrix}
\equal{}
\begin{vmatrix}
{k \cdot r'_1 \plus{} r''_1}\\ {.}\\ {.}\\ {.}
\end{vmatrix}
\equal{}
k
\begin{vmatrix}
{ r'_1 }\\ {.}\\ {.}\\ {.}
\end{vmatrix}
\plus{}
\begin{vmatrix}
{ r''_1 }\\ {.}\\ {.}\\ {.}
\end{vmatrix}$
-  Swapping 2 rows switches the sign of the determinant
   so Adding a scalar multiple of a row to another doesn't change the determinant
- If a single row is multiplied by a scalar r, then the determinant of the resulting matrix is r times the determinant of the original matrix.
...

Let me know some proving using $ AB \equal{} \left[Ab_1\cdots Ab_n\right]$

Decomposing into columns isn't a very informative way of looking at row operations.

Row operations correspond to multiplication on the left by elementary matrices. If you show those latter three statements, you will have proved this in the case that $ A$ is an elementary matrix of one of those three types. The next step is to find an algorithm to decompose $ A$ as a product $ E_1E_2\cdots E_m$ of elementary matrices; then $ \det(A)\equal{}\det(E_1)\det(E_2)\cdots \det(E_m)$ and $ \det(AB)\equal{}\det(E_1)\det(E_2)\cdots \det(E_m)\det(B)$.


geometry
incenter
power of a point
geometry proposed

Let $ABC$ be a triangle. $I$ is the incenter, and the incircle is tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. $P$ is the second point of intersection of $AD$ and the incircle. If $M$ is the midpoint of $EF$, show that $P$, $I$, $M$, $D$ are concyclic.

AE and AF being tangents to the circle I and being concurrent on PD, it follows that $ DEPF$ is a harmonic quad, hence the tangents at D and P to circle I concur on EF, let's call $ T$ their common point. since $ M\in AI \cap EF$, we have $ MI\perp MT$, then $ DI\perp DT$ and $ PI\perp PT$, consequently $ D, M$ and $ P$ lie on the circle of diameter $ TI$, done.

Note: This way follows that $ TI\perp AD$ (was this proof known for the issue?)

Best regards,
sunken rock

Consider The inversion center $ I$ power $ k$ ( $ k\equal{}ID^2$): $ {M}\rightarrow{A}, {P}\rightarrow{P}, {D}\rightarrow{D}$. Since we'll get the result  :)

[quote=TomciO]It is really easy. $|AP\parallel AD|=|AE|^{2}=|AM\parallel AI|$, where the first equality follows from the power of a point and the second from the similarity of $AEI$ and $AEM$. Conclusion follows.[/quote]

Well though according to u this is trivial but we need to show this before that A,M,I are collinear. Which is trivial by congruence of the triangles AFM,AEM. Then done.......by your proof

People is saying trivial when its an Ibero of 1990 and it was P4!. Show some respect to problem PLS!
Anyways...taking polars w.r.t. incircle $\mathcal P_A=EF$ thus $M$ and $A$ are inverses w.r.t. incircle and now by inversion w.r.t. incirle we have $P \to P$, $M \to A$ and $D \to D$ thus the problem is equivalent to show that $A,P,D$ are colinear and we are done since that is a condition of the problem.

@below BRUH

Trivial.
By PoP, $$AK\cdot AI=AE^2=AD\cdot AP.$$


trigonometry
inequalities
analytic geometry
absolute value
geometry proposed
geometry

[u][b]New version of Question:[/b][/u]
The circle $C_1$ with center $O_1$ and radius $R_1$, and another circle $C_2$ passing through $O_1$ with center $O_2$ and radius $R_2$ are given. $T$ is a point on $C_1$ satisfying following condition:
$O_2T \cap C_1 = \{K_1\}, O_1T \cap C_2 = \{K_2\}$ and $|O_2K_1| = |K_2T|$.
Then prove that
$O_2T$ or $K_1T$ must be equal to $R_1$

[u][b]Old version of Question:[/b][/u]
The circle $C_1$ with center $O_1$ and radius $R_1$, and another circle $C_2$ passing through $O_1$ with center $O_2$ and radius $R_2$ are given. $T$ is a point on $C_1$ satisfying following condition:
$\angle O_2O_1T < 90$, $O_2T \cap C_1 = \{K_1\}, O_1T \cap C_2 = \{K_2\}$ and $|O_2K_1| = |K_2T|$.
Then,
 $i. $ If $R_1 > R_2$, prove that $|O_1T| = |O_2T|$.
 $ii. $If  $R_1 < R_2$, find $\angle O_1TO_2$.

[color=blue][b](1)[/b][/color] Let $R_1 > R_2$. Let $A, B$ be the intersections of the center line $O_1O_2$ with the larger circle $(O_1)$, so that the points $A, O_1, O_2, B$ follow on the center line in this order. Let $C, C'$ and $D, D'$ be the intersections of normals to the center line $O_1O_2$ erected at the circle centers $O_1, O_2$ with the larger circle $(O_1)$. If the point $T$ is on the arc $CAC'$ of the circle $(O_1)$, then $K_2T > R_1 > R_2 > O_2K_1$, which is impossible. If the point $T$ is on the arc $DBD'$ of the circle $(O_1)$, then it is entirely possible to have $K_2T = O_2K_1$ while $O_1T > O_2T$. This happens when the triangle $\triangle O_1K_1T$ becomes equilateral, which is possible as long as

$R_1 \frac{\sqrt 3}{2} \le R_2 < R_1$

Therefore, I assume that there is an error in the problem conditions and that the correct condition is $\angle O_1O_2T < 90^o$, rather than $\angle O_2O_1T < 90^o$. This excludes the possibilty of the point $T$ being on the arc $DBD'$. WLOG, assume that the point $T$ is on the minor arc $CD$ of the circle $(O_1)$. Let $E$ be the intersection of a normal to the center line $O_1O_2$ dropped from the point $T$ with the circle $(O_1)$ different from the point $T$ and $F, F'$ the intersections of this normal with the circle $(O_2)$. The power of the point $O_2$ to the circle $(O_1)$ is equal to

$p(O_2, (O_1)) = -O_2T \cdot O_2K_1 = -O_2A \cdot O_2B = -(R_1 + R_2)(R_1 - R_2) = -(R_1^2 - R_2^2)$

The powers of the points $T, F$ to the circles $(O_2), (O_1)$, respectively, are equal to

$p(T, (O_2)) = TK_2 \cdot TO_1 = TF \cdot TF' = TF \cdot EF$

$p(F, (O_1)) = -FT \cdot FE = -p(T, (O_2)$

Thus the absolute value of the power of the point $F$ to the circle $(O_1)$ is the same as the absolute value of the the power of the point $T$ to the circle $(O_2)$. The points $O_2, F$ have the same power to the circle $(O_1)$ iff they are equidistant from the circle center, i.e., iff $O_1F = O_1O_2 = R_2$, i.e., iff the triangle $\triangle O_1O_2F$ is isosceles with $O_1F = O_1O_2$, which is possible only if this triangle is equilateral. If this is the case, then the triangle $\triangle O_1TO_2$ is isosceles with $O_1T = O_2T$ and

$p(O_2, (O_1) = p(F, (O_1)) = -p(T, (O_2))$

$O_2T \cdot O_2K_1 =  FT \cdot FE  = TK_2 \cdot TO_1$

and since we assumed $O_1F = O_1O_2$, which is equivalent to $O_1T = O_2T$, we proved that this implies $O_2K_1 = K_2T$. Conversely (by contradiction), while the point $T$ is on the minor arc $CD$ of the circle $(O_1)$, i.e., while the angle $\angle O_1O_2T < 90^o$, if $O_2K_1 > K_2T$, then $O_1T > O_2T$ and if $O_2K_1 < K_2T$, then $O_1T <O_2T$. If $\angle O_1O_2T \ge 90^o$ was allowed, this contradiction would not work.

[color=blue][b](2)[/b][/color] Assume $R_1 < R_2$. Denote $\phi = \angle O_1TO_2,\ \theta = \angle O_1TK_1 = 180^0 - \angle O_1K_2O_2$. The triangle $\triangle O_1K_1O_2$ is isosceles with $O_2O_1 = O_2K_2 = R_2$. Hence

$K_2T = O_1K_2 - O_1T = 2R_2 \cos \phi - R_1$

$K_2T = O_2T \cos \theta + R_2 \cos \phi = (O_2K_1 - K_1T) \cos \theta + R_2 \cos \phi =$

$= (K_2T - K_1T) \cos \theta + R_2 \cos \phi$

Substituting $\cos \phi$ from the 1st equation to the 2nd one, we get

$K_2T = (K_2T - K_1T) \cos \theta + \frac{K_2T + R_1}{2}$

$\cos \theta = \frac{K_2T - R_1}{2(K_2T - K_1T)}$

It is clear that as long as $0 < \theta < 90^o$, this equation can hold only if $\cos \theta = \frac 1 2,\ \theta = 60^o$, which implies $K_1T = R_1$. Thus the triangle $\triangle O_1TK_1$ is equilateral and the angle $\angle O_1TO_2 = 180^o - \theta = 120^o$. Assume that the smaller circle $(O_1)$ and the points $K_1, T \in (O_1)$ are fixed but $\theta = \angle O_1TK_1 \neq 60^o$. The center of the larger circle $(O_2)$ then moves on a fixed line $K_1T$. Let $R_2 \rightarrow R_1$, i.e., $O_2 \rightarrow T$. Since the isosceles triangle $\triangle O_1TK_1$ with $O_1T = O_1K_1$ is not equilateral, $K_2T = O_1T > TK_1 \equiv O_2K_1$ for $90^o > \theta > 60^o$ and $K_2T = O_1T < TK_1 \equiv O_2K_1$ for $0 < \theta < 60^o$. These inequalities are maintained as the radius $R_2$ of the larger circle $(O_2)$ grows and its center $O_2$ moves away from the point $T$ on the fixed line $K_1T$, because $K_2T > O_2K_1$ to begin with and $K_2T$ grows faster that $O_2K_1$ for $90^o > \theta > 60^o$ and similarly, $K_2T < O_2K_1$ to begin with and $K_2T$ grows slower than $O_2K_1$ for $0 < \theta < 60^o$. Hence, if $K_2T = O_2K_1$, it must be $\theta = 60^o$ and $\angle O_1TO_2 = 120^o$.

[quote="yetti"][color=blue][b](1)[/b][/color]  I assume that there is an error in the problem conditions .[/quote]

i have read your solution. but i have misunderstood some points. ( Is your $1$ equal to my $i$, why $R_1<R_2$  )i will send my solution to my question. So i and you can compare the solutions. we will find our mistakes if any.

$(\text i) \equiv$ [color=blue][b](1)[/b][/color] and $(\text {ii}) \equiv$ [color=blue][b](2)[/b][/color]. I had a typo at the beginning of [color=blue][b](1)[/b][/color], saying $R_1 < R_2$ instead of $R_1 > R_2$ (as a result of copy and paste, because I solved [color=blue][b](2)[/b][/color] first). But then I referred to the circle $(O_1)$ as the larger circle and used the inequality $R_1 > R_2$ several times in [color=blue][b](1)[/b][/color]. The typo has been edited out. Thanks.

Yetti

Let $(O_1)$ be a circle with radius $R_1$ centered at the coordinate origin $O_1$ and $\triangle O_1K_1T$ an equilateral triangle symmetrically placed with respect to the x-axis. The points $K_1, T$ have coordinates

$K_1 = \left(\frac{R_1 \sqrt 3}{2}, \frac{R_1}{2}\right),\ \ \ T = \left(\frac{R_1 \sqrt 3}{2}, -\frac{R_1}{2}\right)$

Let $P$ be the intersection of the side $K_1T$ with the x-axis. Let $(O_2)$ be a circle centered on the segment $K_1T$ and passing through the origin, for example, let the coordinates of its center be

$O_2 = \left(\frac{R_1 \sqrt 3}{2}, -d\right)$

where $0 \le d = PO_2 \le \frac{R_1}{2}$ is the distance of the point $O_2$ from the x-axis. The point $O_2$ is inside of the circle $(O_1)$ and the circle $(O_2)$ passes through the center of the circle $(O_1)$. Hence, its radius is $R_2 < R_1$ as required. The angle $\angle O_2O_1T \le \angle PO_1T = 30^o < 90^o$. The equations of the circles $(O_1), (O_2)$ are

$(O_1):\ \ x^2 + y^2 = R_1^2$

$(O_2):\ \ \left(x - \frac{R_1 \sqrt 3}{2}\right)^2 + (y + d)^2 = R_2^2 = \frac{3 R_1^2}{4} + d^2$
or

$(O_2):\ \ x^2 - R_1 x \sqrt 3 + y^2  + 2dy = 0$

The equation of the line $O_1T$ is

$O_1T:\ \ y = -x \tan{30^o} = - \frac{x \sqrt 3}{3}$

Solving the set of equations for the circle $(O_2)$ and the line $O_1T$ to get the coordinates of the point $K_2$,

$x^2 - R_1 x \sqrt 3 + \frac{x^2}{3} - \frac{2dx \sqrt 3}{3} = x \left(\frac{4x}{3} - R_1 \sqrt 3 - \frac{2d \sqrt 3}{3}\right) = 0$

The root $x = 0$ corresponds to the intersection $O_1$ of the line $O_1T$ with the circle $(O_2)$. For the other intersection $K_2$, $x \neq 0$.

$x = \frac{3R_1 \sqrt 3}{4} + \frac{d \sqrt 3}{2},\ \ \ y = -\frac{x \sqrt 3}{3} = -\frac{3R_1}{4} - \frac d 2$

Now we can calculate lengths of the segments $O_1T,\ O_2T,\ K_2T,\ O_2K_1$:

$O_1T = R_1,\ \ \ O_2T = PT - PO_2 = \frac{R_1}{2} - d < R_1$

$O_2K_1 = PK_1 + PO_2 = \frac{R_1}{2} + d$

$K_2T = \sqrt{\left(\frac{3R_1 \sqrt 3}{4} + \frac{d \sqrt 3}{2} - \frac{R_1 \sqrt 3}{2}\right)^2 + \left(-\frac{3R_1}{4} - \frac d 2 + \frac{R_1}{2}\right)^2} =$

$= \sqrt{3 \left(\frac{R_1}{4} + \frac d 2\right)^2 + \left(\frac{R_1}{4} + \frac d 2\right)^2} = \left(\frac{R_1}{4} + \frac d 2\right) \sqrt 4 = \frac{R_1}{2} + d$

Thus all problem conditions are satisfied, namely, the circle $(O_2)$ passes through the center of the circle $(O_1)$, its radius $R_2 < R_1$ is smaller than the radius of the circle $(O_1)$, the angle $\angle O_2O_1T \le 30^o < 90^o$, the segments $O_2K_1 = K_2T$ are equal, but the segments $O_1T > O_2T$ are not equal.

The question has been corrected. Please look at the first message.


limit
absolute value
linear algebra
linear algebra unsolved

Let $ A\in\mathcal{M}_m(\mathbb C) , m\in\mathbb N^*$, so that $ | detA |>1$.
Prove that  $ \lim_{n\to\infty}|det(2I_m\plus{}A^n)|\equal{}\infty$

Triangularize it. Now the determinant comes from the diagonal, and the diagonal entries are of the form $ \lambda^n\plus{}2$. If $ |\lambda|>1$, this entry goes to $ \infty$ (in absolute value); if $ |\lambda|\le 1$, it is always at least $ 1$ in magnitude.
At least one $ |\lambda|$ is greater than one, so the product of these goes to $ \infty$.


modular arithmetic
function
algorithm
number theory
totient function
Euclidean algorithm

i'm kinda new with mods so i was wondering how you would go about finding $x$ in
\[3^{3^{3^{3}}}\equiv x \pmod{25}\]

[b]Definition:[/b]  Let $\varphi(n)$, the totient function, be defined as the number of positive integers $a \le n$ such that $gcd(n, a) = 1$.

[b]Euler's Totient Theorem:[/b]  For any positive integer $a$ such that $gcd(n, a) = 1$,

$a^{ \varphi(n)}\equiv 1 \bmod n$

Now see if you can solve the problem.

hmm ok so did i do this right?

I let $n=25$ and got that
\[3^{3^{20}}\equiv 1 \pmod{25}\]
and since $3^{3}\equiv 2 \pmod{25}$,
\[3^{3^{27}}\equiv 2^{7}\equiv 3 \pmod{25}\ \Rightarrow \ \boxed{x=3}\]

Sorry, there are two problems with that.

First, the first statement should be $3^{20}\equiv 1 \bmod 25$.

Second, $3{}^{3{}^{27}}\neq (3^{3})^{27}$.

Now, from the first statement we would need to [hide="find..."] $3{}^{3}{}^{3}\bmod 20$; this becomes our new exponent. [/hide]

hmm ok how so how about this now?

\[3^{20}\equiv 1 \pmod{25}\ \Rightarrow \ 3^{27}\equiv 3^{7}\equiv 12 \pmod{25}\ \Rightarrow \ 3^{27^{3}}\equiv 12^{3}\equiv 3 \pmod{25}\]

Is the last step of raising $3^{27}$ valid? Im still getting $x=3$.

No, $x=3$ is not right; everyone is saying that you're doing the order of exponentiation wrong.
$3^{27^{3}}$ means $3^{(27^{3})}$, and it is not the same as $(3^{27})^{3}$.

[hide="solution"]$3^{\varphi (25)}\equiv 1\pmod{25}$, where $\varphi(25)=20$.
Then ${3^{3^{3^{3}}}}\equiv 3^{y}\pmod{25}$, where $y\equiv 3^{3^{3}}\pmod{20}$. Since $\gcd(3,20)=1$, $3^{\varphi(20)}\equiv 1\pmod{20}$. $\varphi(20)=8$.
Then $y\equiv 3^{z}\pmod{20}$, where $z\equiv 3^{3}\equiv 3\pmod 8$.
Back-substituting, $y\equiv 3^{3}\equiv 7\pmod{20}$, $x\equiv 3^{7}\equiv 27\cdot 27\cdot 3\equiv 12\pmod{25}$.[/hide]

[quote="NeverOddOrEven"]so the answer i got, $x=3$, is correct?[/quote]
no , the right answer is $12$
first , notice that $3^{20}\equiv 1 (mod 25)$
so we go on to find $y$ , where $3{}^{3}{}^{3}\equiv y(mod 20)$
notice that $3^{4}\equiv 1 (mod 80)$
and now we continue finding z where $3^{3}\equiv z (mod 4)$ , and clearly $z=3$
now , we have $3^{4k+3}\equiv y (mod 20)$ , clearly $y =7$
then $3^{20k+7}\equiv x (mod 25)$ , so it's equivalent to $3^{7}\equiv x(mod 25)$
now you easily got the answer

hmm i've never seen these methods before. could someone give me a explanation of how this works?

I guess I shouldn't have been so patchy. I'm not sure what methods you're referring to, but essentially, this was the theorem I was using.
For $a,m,s,t\in\mathbb{N},\gcd(a,m)=1$, we have $a^{s}\equiv a^{t}\pmod m$ iff $s\equiv t\pmod{\varphi(m)}$. This follows from Euler's Totient theorem and the definition of mod.
Extension to negative $s$ and $t$ is easy with the definition $a^{-1}\equiv b\pmod m$, where $b$ is such that $ab\equiv 1\pmod m$, which exists since $\gcd(a,m)=1$. (Use the Euclidean algorithm to find it.)

By the way, x\pmod{25}, x\bmod{25}, and x\mod{25} gives $x\pmod{25}$, $x\bmod{25}$, and $x\mod{25}$, respectively. Also, it's 3^{3^3} for $3^{3^{3}}$, not 3^{3}^{3}.

oh ok i think i get it now thanks guys. one more question.

Bictor, in your solution why did you have to state that $3^{\varphi(25)}\equiv 1 \pmod{25}$. It seems like you used it nowhere in your solution.

If $3^{20}\equiv 1 \bmod 25$, then for any exponent $a$, if $a \equiv b \bmod 20$ then

$3^{a}\equiv 3^{b}\bmod 25$

So since we have $a = 3^{27}$ in this problem, we want to find $b$ where

$3^{27}\equiv b \bmod 20$

This is the essence of the problem.

According the Bictor's second statement, if i found that
\[3^{3^{3^{3}}}\equiv 3^{-17}\pmod{125}\]
Could i simplify that to
\[3^{3^{3^{3}}}\equiv 3^{-17}\equiv 3^{17}\pmod{125}\]

... No.  You would have to find the number $b$ such that $3b \equiv 1 \bmod 125$ (why are you taking $\bmod 125$?), and then

$3^{-17}\equiv (3^{-1})^{17}\equiv b^{17}\bmod 125$

In fact, it is easy to find that $b = 42$, since

$3 \cdot 42 = 126 \equiv 1 \bmod 125$

[quote="Bictor717"]For $a,m,s,t\in\mathbb{N},\gcd(a,m)=1$, we have $a^{s}\equiv a^{t}\pmod m$ iff $s\equiv t\pmod{\varphi(m)}$. [/quote]

Careful; the implication works from right to left only.  The implication from left to right is trivially false for $a = 1$.


I know the inductive proof of rearrangement, but are there any other proofs, especially by AM-GM, Chebyshev, or Cauchy?

You seem to have it backwards.  Rearrangement is a very basic inequality.  It's used to prove Chebyshev, not the other way around.


Question:

What runs all over town but never comes inside?

mustafa,

well, there ARE bus garages where certain buses go inside.

I enjoy your answer.

However, I did not make up the question/answer.

Officially, no, it is not correct according to the Popsicle stick.

Hint...

Find the hint below...
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.
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.
.
.
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.
.
Think of a homonym for "Rhodes" that was in my question in my subject heading.

Answer below...

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as in roads ---> "the street"


here are the local olympiad problems from Bucharest, Romania 2004.
in .dvi and .pdf format.

I must say that I have modified some of the problems, because in the actual contest there were many mistakes. That's what happens when one person decides he can do it all alone.

look at pb3 10th grade

Let (fn) be the Fibo sequence! where do u use it? should it be a_n=sum
(k=1,n) (1/(f_n*f_(n+2)))?

cheers!

indeed (although I can hardly call that a problem!).

how should we interpret the grades? what is 9th-12th grade exactly?

If I am not wrong
12th = final year in high school
11th = the year before the final year in high school
10th = two years before the final year in H.S.
9th = three years before the final year in H.S.

you are not wrong :)  in US english :D
12th grade = senior
11th grade = junior
10th grade = sophmore (is this the correct spelling?)
9th grade = freshman

In China:
7~9th grade = Junior 1~3
10~12th grade = Senior 1~3

oic

so they're starting pretty young with those olympiads there

here, since 2 years ago, 10th grades have their own competition, before it were only 12th graders and some 11th ones...

i thought sophomore and freshman were terms for the collegians

[quote="Peter VDD"]oic

so they're starting pretty young with those olympiads there

.[/quote] if this feels young imagine we are having an balkan math olympiad for students UNDER 15 years :)
also olympiads start from grade 4th, which is usually 10 years for the student :D

wished things like that existed here :(

I've been begging for something like this until I was in 10th grade, and then they finally came up with a junior olympiad that year :D whiiiii finally some cool maths to do

( not the stuff like solve the equitation 126x²-163x+487=0 )


Where can I get the old Mathcounts State Competitions, like the ones from the time Mathcounts started?

you can find some at
http://www.yashar.biz/mathcounts/
or at the mathcounts website(students :arrow: prior competitions)
there are some in the [u]All time Greatest mathcounts problems[/u] book
and as alanchou said, try asking your coach. Be sure to save all the tests that you get! :)
EDIT: for the old ones, the only resource I know is the all time greatest mathcounts  problems
if you have to, you can order off of the Mathcounts store via http://www.mathcounts.org/webarticles/anmviewer.asp?a=183&z=6

There's a sticky at the top of the page called "MATHCOUNTS Resources HERE!"...

the only tests that you can get like from the 80's are 1. very hard to get and only the oldest of veteran coaches may still have them and 2. the tests from the 80's are outdated with each one having 40 questions + the fact that the curriculum has changed since then, for example many of the tests back then required that you knew basic inputs and computer programming. I suggest you use aops Volume 1 the Basics, order from the Mathcounts online store and find a local coach/teacher who has lots of experience with mathcounts


Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length $s$, is circumscribed around the balls. Then $s$ equals

$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)} \ 1+2\sqrt 6 \qquad \text{(E)} \ 2+2\sqrt 6$


[b]4991(din nr.1/2008):[/b]
Demonstrati ca intr-un triunghi oarecare ABC, are loc inegalitatea: $ a^2\plus{}b^2\plus{}c^2\minus{}\frac{1}{2}(|a\minus{}b|\plus{}|b\minus{}c|\plus{}|c\minus{}a|)^2\ge 4.\sqrt{3}.S.$ ([i]S.G. Abdullaev[/i])
[b]OBSERVATIE:[/b]
Inegalitatea lui Abdullaev este mai tare ca inegalitatea lui Finsler-Hadwiger: $ a^2\plus{}b^2\plus{}c^2\minus{}(a\minus{}b)^2\minus{}(b\minus{}c)^2\minus{}(c\minus{}a)^2\ge 4.\sqrt{3}.S.$

[b]5019(din nr.6/2008):[/b]
Fie ABC un triunghi oarecare si sa notam cu M, N si cu P mijloacele laturilor [BC], [CA] si respectiv [AB]; iar cu G centrul sau de greutate.
Daca $ r_1,\ r_2, \ r_3, \ r_4, \ r_5, \ r_6$ sunt razele cercurilor inscrise in triunghiurile GBM, GCN, GAP, GCM, GAN si respectiv GBP, artunci avem inegalitatea:
$ \frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}\frac{1}{r_3}\equal{}\frac{1}{r_4}\plus{}\frac{1}{r_5}\plus{}\frac{1}{r_6}.$  (enunt usor modificat) ([i]I.I. Voronovici[/i]-Minsk)
[b]OBSERVATIE:[/b]
Daca $ R_1, \ R_2, \ R_3, \ R_4, \ R_5, \ R_6$ sunt razele cercurilor circumscrise acelorasi triunghiuri, atunci are loc urmatoarea relatie [i]JAPONEZA[/i]: $ R_1R_2R_3\equal{}R_4R_5R_6.$

4991. Here my solution:  Lets assume  $ a\ge b\ge c$
Then  $ a^2\plus{}b^2\plus{}c^2\minus{}\frac{1}{2}(\lvert\ a\minus{}b \rvert\plus{}\lvert\ b\minus{}c \rvert\plus{}\lvert\ c\minus{}a\rvert)^2\equal{}a^2\plus{}b^2\plus{}c^2\minus{}\frac{1}{2}
\cdot 4(a\minus{}c)^2\equal{}b^2\minus{}a^2\minus{}c^2\plus{}4ac$
It known $ b^2\equal{}a^2\plus{}c^2\minus{}2ac\cos\beta$
$ S\equal{}\frac{1}{2}ac\sin\beta$
Hence $ b^2\minus{}a^2\minus{}c^2\plus{}4ac\minus{}4\sqrt{3}S\equal{}2ac(2\minus{}\cos\beta)\minus{}2\sqrt{3}ac\sin\beta\equal{}4ac(1\minus{}\frac{1}{2}\cos\beta\minus{}\frac{\sqrt{3}}{2}\sin\beta)\equal{}4ac[1\minus{}\cos(\beta\minus{}60^{\circ})]\geq0$  q.e.d !   :lol:

 _____________________________________
Azerbaijan Land of Fire    :ninja:  :D


LaTeX
modular arithmetic
number theory
relatively prime
number theory unsolved

Find all non negative integers m and n such that 6 in the power of m + 2 in the power of n +2 is a perfect square. 6m+2n+2=k2

Please, try to use $ \text{\LaTeX}$  :oops:

Find all non negative integers $ m$ and $ n$ such that $ 6^m+2^n+2$ is a perfect square.

Hint: It's very simple :blush:

Ok, let's kill this problem.

If both $ m$ and $ n$ are greater than or equal to $ 2$, then the expression $ 6^m+2^n+2$ is even but it can't be a square since $ 6^m+2^n+2\equiv 2 \pmod{4}$, while an even square has to be $ \equiv 0 \pmod{4}$. Thus, let's analyze the following cases:

a) $ m=n=0$,      $ 6^0+2^0+2=4=2^$     OK,
b) $ m=1,\,n=0$,     $ 6^1+2^0+2=9=3^2$     OK,
c) $ m=0,\,n=1$,     $ 6^0+2^1+2=5$  NO,
d) $ m=n=1$,     $ 6^1+2^1+2=10$  NO.

Now, the following general cases remain:

1) $ m\geq 2,\, n=0$, $ 6^m+3=a^2$. This has no solutions since for $ a$ can't be $ \equiv 3  \pmod{4}$ for $ a$ odd,
2) $ m\geq 2,\, n=1$, $ 6^m+4=a^2$. We will analyze this case later,
3) $ n\geq 2,\, m=0$, $ 2^n+3=a^2$. This has no solutions since for $ a$ can't be $ \equiv 3  \pmod{4}$ for $ a$ odd,
4) $ n\geq 2,\, m=1$, $ 2^n+8=a^2$. We will analyze this case later.

Case 2) $ m\geq 2,\, n=1$, $ 6^m+4=a^2\to 2^m3^m=(a-2)(a+2)$. Since $ a$ must be even, we put $ a=2s$, then $ 2^m3^m=4(s-1)(s+1)\to 2^{m-2}3^m=(s-1)(s+1)$. Putting $ s-1=k$, finally have $ 2^{m-2}3^m=k(k+2)$. If $ m=2$ this equation has non solution. Then, for $ m>2$, $ k$ has to be even, $ k=2n$ and  $ 2^{m-2}3^m=4n(n+1)\to 2^{m-4}3^m=n(n+1)$. Again, for $ m=4$ there is not a solution, so $ m>4$. Moreover $ n$ and $ n+1$ are relatively prime so one must be equal to $ 2^{m-4}$ and the other to $ 3^m$. We see that we necessarily must have $ n+1=3^m$ and $ n=2^{m-4}$ ($ 3^m>2^{m-4}$), but this implies $ 1=3^m-2^{m-4}$ which can't hold for $ m>4$. No solutin.

Case 4) $ n\geq 2,\, m=1$, $ 2^n+8=a^2$. If $ n=2$, we have $ 2^2+8=12\ne a^2$, thus $ n\geq 3$. Now, $ 2^3(2^{n-3}+1)=a^2$. If $ n>3$ the factor $ (2^{n-3}+1)$ is odd but $ 2^3$ isn't an even square. So we must have $ n=3$, which gives $ 2^3+8=16=4^2$. Finish.

So, the solutions are in terms of $ (m,n)$: $ (0,0),\,(1,0),\,(1,3)$ :yup:

thank a lot. also i found as solution the same numbers but your way is good and well organised.


Εστω κυκλος κεντρου Ο, σημειο Ρ εκτος αυτου και οι εφαπτομενες ΡΑ, ΡΒ. Εστω το Α' αντιδιαμετρικο του Α και η ΒΗ καθετη στην ΑΑ' οπου Η σημειο της ΑΑ'.
Εστω Μ το σημειο τομης των ΒΗ και ΡΑ' και Q σημειο τομης των ΒΗ και ΡΟ. Εστω S η προβολη του Ρ στην ΒΗ.
ΝΔΟ:  i) ΜΒ=ΜΗ
         ii) SΒ=ΗQ

Nice Saki.
Η λύση μου για το 1:
Έστω η $ AB$ να τέμνει την $ PA'$ sto $ K$.Από εύκολο angle chasing βγάζουμε ότι $ BA$ είναι διχοτόμος της $ \angle PBH$ και η $ BA'$ της εξωτερικής της.Επομένως $ (P,M,K,A')\equal{}\minus{}1$.Άρα η $ A(P,K,M,A')$ είναι αρμονική δέσμη και η ΒΗ παράλληλη στην PA.Από γνωστό Lemma ξέρουμε ότι όταν μία ευθεία είναι π\λ σε ένα radii της αρμονικής δέσμης τα άλλα τρία radii την χωρίζουν σε ίσα τμήματα :wink:
Θα κοιτάξω και το δεύτερο όταν βρω χρόνο.....

Hmmm νομίζω έβγαλα και μία τελείως elementary λύση για το δεύτερο αλλά μπορεί να έχω κάπου λάθος γιατί είναι και βράδυ :P .

Για να είναι $ QH\equal{}SB$ αρκεί τα $ \triangle AHQ\equal{}\triangle PSB$
Aλλά $ AH\equal{}PS$.Επομένως αρκεί $ PB\equal{}AQ$.
Αρκεί λοιπόν το $ \triangle APQ$ να είναι ισοσκελές.
Έστω το σημείο τομής των $ PQ$ και $ AB$ να είναι το $ W$.
Τότε $ AW\bot PQ$,αρα αρκεί η $ AW$ να είναι διχοτόμος της $ \angle PAQ$.Έστω η $ AQ$ να τέμνει τον κύκλο στο $ Z$.Για νδο  η $ AW$ να είναι διχοτόμος της $ \angle PAQ$ αρκεί νδο $ BA\equal{}BZ$.To οποίο όμως ισχύει γιατί η $ BO$ είναι μεσοκάθετος της $ AZ$.(note that $ O$ is the orthocenter of the $ \triangle ABQ$ :) ).

Dimitris X

Ας βάλω και γω τη λύση μου στα 2 ερωτήματα, χρησιμοποιώντας και στα 2 αρμονικές δέσμες :)

Αν η $ PB$ τέμνει την $ AA'$ στο $ R$, έχουμε ότι: $ A,H,A',R$ αρμονικά συζηγή (στο τρίγωνο $ ABA'$, η $ ΒR$ είναι εφαπτομένη του περιγεγραμμένου ενώ η $ BH$ συμμετροδιάμεσος). Ετσι, $ P(A,H,M,B)$ αρμονική δέσμη κι αφού $ BH//PA$ έχουμε το 1ο ζητούμενο!

Για το 2ο, έχουμε ότι $ P(A,O,A',S)$ είναι αρμονική δέσμη γιατί $ PS//AA'$ και $ OA=OA'$ (το αντίστροφο του προηγούμενου λήμματος). Ετσι, P(A,Q,M,S) είναι αρμονική δέσμη κι αφού $ PA//SQ$ έπεται το 2ο ζητούμενο!

ego prospathousa na grapso ti lusi apo xthes alla de me afine na kano submit giati evgaze "no post mode specified"- kserei kaneis ti einai auto?

tespa..polu oraies luseis. oles pano kato sto idio pneuma (suberilamvanomenis kai tis dikias mou)

gia to proto theoro tin tomi $ T$ tis efaptomenis tou kuklou sto $ A'$ me tin poliki tou $ P$, xrisimopoio theorima La Hire apo opou exo oti i poliki tou $ T$ einai h $ PA'$. Ara h desmh $ A'(A,M,B,T)$ einai armoniki, q.e.d.

gia to deutero einai paneukolo na deiksei kaneis oti ta trigonakia $ AQH$ kai $ PBS$ einai isa, opote telos.

ps. DimitrisX nomizo oti i veltiosi sou se tetoiou eidous geometrika themata einai aisthiti! carry on.

Και μια elementary λυση για το 1)
Εστω Ν η τομη της ΑΒ με την ΡΟ, ευκολα βλεπουμε οτι Ν ειναι μεσο της ΑΒ, οποτε αρκει ΜΝ//ΑΑ', η αρκει ΟΝ/ΝΡ=ΗΜ/ΜΣ. Ομως ΟΝ/ΝΡ=(ΟΑ/ΑΡ)^2 και ΗΜ/ΜΣ = ΗΑ'/ΣΡ = Α'Η/ΗΑ = (Α'Β/ΒΑ)^2 οποτε αρκει τα ορθογωνια τριγωνα ΡΑΟ και ΑΒΑ' να ειναι ομοια με <ΟΑΝ = <ΑΡΟ, πραγμα που ευκολα βλεπουμε οτι ισχυει.

Το 2ο λυνετε elementary χρησημοποιοντας ισοτητες τριγωνων οπως στις λυσεις του Δημητρη και του Νικου.

Κατι αλλο τωρα: Nick λες οτι επηδη η Α'Ρ ειναι πολικη του Τ, η δεσμη Α'(Τ,Β,Η,Α) ειναι αρμονικη. Επηδη δεν εχω δει και πολλες ασκησεις που να λυνονται με αρμονικοτητα και πολικες, αν δεν σου ειναι κοπος μπορεις να εξηγησεις τι ακριβως λεει το θεωρημα χρησημοποιεις στο συγκεκριμενο σημειο?

Ολες οι λυσεις ηταν εξαιρετικες.
H λυσεις που ειχα στο νου ειναι αυτες του Δημητρη.

[quote="Nick Rapanos"][quote="NickNafplio"]

Κατι αλλο τωρα: Nick λες οτι επηδη η Α'Ρ ειναι πολικη του Τ, η δεσμη Α'(Τ,Β,Η,Α) ειναι αρμονικη. Επηδη δεν εχω δει και πολλες ασκησεις που να λυνονται με αρμονικοτητα και πολικες, αν δεν σου ειναι κοπος μπορεις να εξηγησεις τι ακριβως λεει το θεωρημα χρησημοποιεις στο συγκεκριμενο σημειο?[/quote]

loipon leo oti apo to theorima La Hire syberainoume oti i $ A'P$ einai poliki tou $ T$. Auto simainei oti an paroume ta simeia tomis me ton kuklo kai me tin poliki tou $ T$ mias opoiasdipote eutheias pou dierxetai apo to $ T$ mazi me to simeio $ T$ apoteloun armoniki tetrada.
Auto kano kai tipota parapano.
den exo sxhma brosta mou, gi auto an thelei kapoios perissoteres leptomereies to ksanasuzhtame.[/quote]

ok, thanx, me kalipses


Let $ \bigtriangleup B'AC,$ $ \bigtriangleup C'AB$ be, two similar isosceles triangles, erected on the side segments $ AC,$ $ AB$ respectively, of a given triangle $ \bigtriangleup ABC,$ outwardly to it. We denote as $ B'',$ $ C'',$ the reflexions of $ B',$ $ C',$ with respect to $ AC,$ $ AB$ respectively. Prove that the line connecting the points $ K_{1}\equiv BB'\cap CC'$ and $ K_{2}\equiv BB''\cap CC'',$ passes through the point $ K,$ as the [b][size=100]Lemoine point[/size][/b], of $ \bigtriangleup ABC.$

Kostas vittas.

PS. I apologize if this problem has already been posted again in this Forum.

I've edited your message, Kostas. Wonderful problem by the way! A straightforward solution is by using homogenous barycentric coordinates, i.e.

It is known that $ K_{1}\equal{}\left(\frac{1}{S_{A}\plus{}S_{\omega}} : \frac{1}{S_{B}\plus{}S_{\omega}} : \frac{1}{S_{C}\plus{}S_{\omega}}\right)$ and similarly obtain that $ K_{2}\equal{}\left(\frac{1}{S_{A}\minus{}S_{\omega}} : \frac{1}{S_{B}\minus{}S_{\omega}} : \frac{1}{S_{C}\minus{}S_{\omega}}\right)$, where $ S_{\alpha}\equal{}2[ABC] \cdot \cot{\alpha}$.

Since $ K\equal{} \left(a^{2}: b^{2}: c^{2}\right)\equal{}\left(S_{B}\plus{}S_{C} : S_{C}\plus{}S_{A}: S_{A}\plus{}S_{B}\right)$, it is enough to prove that

\[ \left|\begin{array}{ccc}\frac{1}{S_{A}\plus{}S_{\omega}} & \frac{1}{S_{B}\plus{}S_{\omega}} & \frac{1}{S_{C}\plus{}S_{\omega}} \\
\ \frac{1}{S_{A}\minus{}S_{\omega}} & \frac{1}{S_{B}\minus{}S_{\omega}} & \frac{1}{S_{C}\minus{}S_{\omega}} \\
\ S_{B}\plus{}S_{C} & S_{C}\plus{}S_{A} & S_{A}\plus{}S_{B}\end{array}\right| \equal{} 0,\]
which is clear.

Best regards,
Cosmin

$ B'B'', C'C''$ cuts $ BC$ at $ B_o,C_o$. ( I label other points like [b]Kostas Vittas[/b] in his figure  :D )
Consider the rotation $ f \equal{} Q_A^{(\overrightarrow{AB},\overrightarrow{AC})}$, we have:
$ f: \ [AC') \to [AB'')$
$ f: \ [AC'') \to [AB')$
$ f: \ [AC_1) \to [AB_o)$
$ f: \ [AC_o) \to [AB_1)$
( Proving them is easy, only simple angle chase )
So $ A(C'C''C_1C_o) \equal{} A(B''B'B_oB_1)$
$ \Rightarrow C(C'C''C_1C_o) \equal{} B(B''B'B_oB_1) \equal{} B(B'B''B_1B_o)$
$ \Rightarrow (CC',CC'',CB_1,CB) \equal{} (BB',BB'',BC_1,BC)$
Then, $ K_1 \ (CC' \cap BB') , K_2 (CC'' \cap BB''),$ and $ K \ (CB_1 \cap BC_1)$ are collinear  :wink:

Dear [b][size=100]Pohoatza[/size][/b], [b][size=100]PDatK40SP[/size][/b] and all my friends.

Thank you for your interest and solutions. The only one proof of this problem I have in mind, based on the [b][size=100]double ratio[/size][/b] argument, as present by [b][size=100]PDtK40SP[/size][/b].

Best regards, Kostas Vittas.

Actually, we have some interesting properties of Kiepert point as following: :)

Let $ D,E,F $ be the point out of triangle $ ABC $ satisfy
$ \angle DBC=\angle DCB =\angle ECA=\angle EAC=\angle FAB=\angle FBA=a $
Let $ K(a)=AD\cap BE\cap CF $ and $ K'(a) $ be the isogonal conjugate of $ K(a) $ WRT $ \triangle ABC $

(1) $ K(x), K'(-x), X(2) $ are collinear.
(2) $ K(x), K(90-x), X(3) $ are collinear.
(3) $ K(x), K'(90-x), X(4) $ are collinear.
(4) $ K(x), K(90+x), X(5) $ are collinear.
(5) $ K(x), K(-x), X(6) $ are collinear.

Dear Friends, What is the name of this problem? who is the first person found this? I want using the problem put in to my paper?


limit
real analysis
real analysis unsolved

d_n is  the general term of a divergent series , that diverge to infinity.d_n is positif.
WHAT is the nature of the following series defined by their general terms
1.d_n/(1+nd_n)
2. d_n / (1+sqr d_n)

[hide="Possible argument for 1"]If $ d_n$ diverges then also $ nd_n$ diverges. Let's now compare series 1 with the harmonic series:

$ \lim_{n \to \infty} \frac{\frac{d_n}{1 \plus{} nd_n}}{\frac{1}{n}} \equal{} \lim_{n \to \infty} \frac{nd_n}{1 \plus{} nd_n} \equal{} \lim_{n \to \infty} \frac{1}{1 \plus{} \frac{1}{nd_n}} \equal{} 1$,

and so both series have the same nature.[/hide]


Find  f:$\mathbb{R}^{+}\Rightarrow \mathbb{R}^{+}$ such that:
$xf(xf(y))=f(f(y))\forall x,y \in \mathbb{R}^{+}$

Let $f(1)=c$. Then for all $x$, $xf(cx)=f(c)$, and therefore $f(x)=\frac{c\cdot f(c)}{x}=\frac{\alpha}{x}$. It's easy to check that for all $\alpha\in\mathbb R^{+}$ function $f(x)=\frac{\alpha}{x}$, $x\in \mathbb R^{+}$, satisfies given equation.


Please post here rumours about the results of your team at IMO. :wink:
P.S:
Here are Kazakhstan team results(that's what we think it is gonna be):
1.+,+,-
2.+,-,-
3.+,+,-
4.+,+/-,-
5.+,+/-,-
6.+,-,-.

i`m so stupid
i solved 2(a) but  i don´t  slolve the (b)
i made a mistace in discriminant

i`m Kaz 5

 :(  :(  :(

Kazakhstan contestants:

1 Egor
2 Asset
3 Erken
4 Nursultan
5 Sanzhar (i)
6 Iskander

What about 2nd day!!!
I see that result is normal, but what happened with Asset?

[color=green]I was wondering if Valentin (or somebody else) was going to post the partial results of the imo, just like he did last year...

I'm sure we all want to know :P [/color]

Ok, [b]baysa[/b] wrote:
Mongolia, Problem 1
Mong 1 - 6 points
Mong 2 - 7 points
Mong 3 - 7 points
Mong 4 - 7 points
Mong 5 - 5 points
Mong 6 - 7 points

Please, post more results....
For example: Romania, Turkey and Moldavian couple Frimu-Iliasenco...

[quote="cafer tayyar"]here is turkey team results
melih: + + + + + -
umut: + + + + + -
tekin: + + + + + -
fehmi: + + + + - -
alper: + + - + - -
semih: + + - + - -
 - not means that 0[/quote]

And + not means 7 ?
If it means 7, i think that the cutoff for gold will be 35-36

What about Moldova ?
Meatman, salut,de ce nu raspunzi la mesajul privat?

[quote="boggart0803"]can someone get scores of Taiwan team?
Or maybe China or Japan[/quote]

I would like to know the data of Japan Team. :)

And + not means 7 ?
If it means 7, i think that the cutoff for gold will be 35-36

expect two of them + means 7 most probably

Brazilian's possible results:

Pondé : ++-++-
Finder : ++-++-
Régis   : +(+/-)-++-
Davi :     +(+/-)-++-
Tupynamba : +(+/-)-++-
Gauy : +--+--

PLEASE, if any of you is attending the coordinations and have some official scores to share with us, post them.

(Oh Valentin, we miss you so much...)

Last information about Mongolia (from [b]baysa[/b]):

[size=150]Mong 1    6 - - 7 7 -
Mong 2    7 - - 7 - -
Mong 3    7 - 1 7 7 -
Mong 4    7 - - - - -
Mong 5    5 - - 3 - -
Mong 6    7 - - 4 - -[/size]
- means no information available

Could somebody post Canada and India's results please?  :D

[quote="Fedor Petrov"]As I know about Russian team, Babichev Dmitri and Volkov Vladislav have 35, Boiky Roman and Kudyk Nikita have 32, Gorinov Evgeni has 31 and Bazhov Ivan 29.

Quite poor.[/quote]

I hope this is not exactly the same as those problems of the second day. :D Do you know about Chineese team?

I've just heard that two of them solved six problems.

I want to get the result of Chinese team.

[quote="Quick"]The (imcomplete) situation of China before problem 6 is pretty simple to describe, and pretty impressive too : except for one student who got 0 on problem 5, they all have 7 on all five first questions...[/quote]

I heard rumor say Chinese team got 41/42 on Problem 2...

Does anyone hear anything about Vietnamese result?

Any further update about India? I'm curious to hear of the cutoffs...

Official IMO 2008 results are posted here: http://www.imo-official.org/year_country_r.aspx?year=2008  :)

[quote="Fedor Petrov"]As I know about Russian team, Babichev Dmitri and Volkov Vladislav have 35, Boiky Roman and Kudyk Nikita have 32, Gorinov Evgeni has 31 and Bazhov Ivan 29.

Quite poor.[/quote]

Fedor gave lower bounds on the results of Russia contestants which are only tight for half of the contestants: Nikita Kudyk (32 points), Roman Boykiy (32 points), Evgeny Gorinov (31 points):

- http://www.imo-official.org/team_r.aspx?code=RUS&year=2008

And I know inofficial team rank 2 after default winner China is so bad...  :D

- http://www.imo-official.org/year_country_r.aspx?year=2008

That are six gold medals which is one gold medal more than China:

- http://www.imo-official.org/team_r.aspx?code=CHN&year=2008

Though of course Ivan Bazhov and Evgeny Gorinov who scored 31 were lucky to have the cut-off for gold at 31!  :)

[quote="ktl4ik"]kaakhstanski resultat napiwite!!! :mad:[/quote]
Tebe nado ty i pishi :rotfl:

I am sorry, the results I posted as final were results without problem 6.


[quote="mr.danh"][quote="thanhnam2902"][color=darkblue]Let $ ABC$ is a triangle, known that $ \frac {cosA}{5} \plus{} \frac {cosB}{12} \plus{} \frac {cosC}{13} \equal{} \frac {119}{780}$. Prove that $ ABC$ is right triangle.[/color][/quote]
You made a mistake in your post. $ \frac {cosA}{5} \plus{} \frac {cosB}{12} \plus{} \frac {cosC}{13} \equal{} \frac {169}{780}$
See at [url]http://www.mathlinks.ro/viewtopic.php?t=206374[/url]
Apply (1) with x=5,y=12,z=13, hence result.[/quote]
thank you very much.


You got what? OK, farmertim - if you're reading this, you can't have Brut3Forc3. He's ours, not yours. (I don't send letters - we run more of a word of mouth operation down here.)

[quote="Ubemaya"]Hello,
Was every USAMO qualifier supposed to receive an ARML letter? Everyone else in my school got one except me, which made me very sad :'( .[/quote]
I didn't get one, nor did anyone at my school to the best of my knowledge. However last year I think everyone got one or something, but that happened later in the year.

i'm glad to see the letters are getting out.  the primary purpose is to recruit USAMO qualifiers who aren't already involved in the ARML team.  if you went to ARML last year you probably didn't get a letter.  there were also some people whose home addresses didn't show up properly on the AMC list, so they won't have received letters either.  admittedly i'm not getting 100% coverage here, but that's where you all come in.  i can sure use your help to make sure all our USAMO qualifiers join the team this year..

BTW Kent, i start with the CA USAMO qualifier list and remove ZIP codes that are obviously in southern california.  if any of them slipped through and i end up recruiting the next Andrew Wiles for you because your word of mouth didn't reach him/her, you can thank me later..  :)

[quote="Brut3Forc3"]I think so, considering that I got one...[/quote]

Yay me too. The drive is a bit long, though.


geometry
circumcircle
cyclic quadrilateral
power of a point
radical axis
geometry unsolved

Let $PQRS$ be a cyclic quadrilateral such that the segments $PQ$ and $RS$ are not parallel.  Consider the set of circles through $P$ and $Q$, and the set of circles through $R$ and $S$.  Determine the set $A$ of points of tangency of circles in these two sets.

Let (O) be the quadrilateral circumcircle and X the intersection of PQ and RS. Let $\sigma_1$ be an arbitrary circle passing through P, Q tangent at A to a circle $\sigma_2$ passing through R, S. PQ is the radical axis of $(O), \sigma_1$ and RS is the radical axis of $(O), \sigma_2$. Hence, X is the radical center of $(O), \sigma_1, \sigma_2$ independent of the circles $\sigma_1, \sigma_2$. The single common internal tangent (resp. the single common external tangent) at A of the circles $\sigma_1, \sigma_2$ is their radical axis, if they touch externally (resp. if they touch internally). Thus this common tangent passes through the radical center X. Since all possible circles $\sigma_1$ belong to the same pencil with the base points P, Q and since X is on the radical axis of this pencil, the tangent length XA is constant and equal to the square root of the power of the point X to any circle $\sigma_1$, i.e., $r = XA = \sqrt{XP \cdot XQ} = \sqrt{XR \cdot XS}$. Consequently, the locus of the tangency points A of the circles $\sigma_1, \sigma_2$ is a circle centered at X and with radius $r$.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1995Problem3

Vo Duc Dien


Prove(or disprove) that 666 is the largest triangular number with all of its digits the same.

(i have no idea)

I haven't found a counterexample yet, but it could be that the next number like that is much larger (I believe over $ 100000$).  Anyway, to restate your question: Find $ (k,a)$ with $ k \ge 3$ such that there exists $ n \in \mathbb{Z}$ such that $ n(n \plus{} 1)/2 \equal{} a(10^k \minus{} 1)/9$. (it is easy to verify that $ 777$, $ 888$, and $ 999$ are not triangular numbers)

Reference:
Bellew, D. W. and Weger, R. C. "Repdigit Triangular Numbers."
J. Recr.Math. 8, 96-97, 1975-76

Some of us don't have access to the article... But here's what I found:

[quote="[url]http://mathserver.sdu.edu.cn/mathency/math/b/b082.htm[/url]"]666 is a Repdigit, and is also a Triangular Number.

In fact, it is the largest Repdigit Triangular Number (Bellew and Weger 1975-76)[/quote]


Question:

Prove that if $ a$ is an odd integer, then

$ a^{2^{n}}\equiv 1 (mod\,2^{n+2})\ ,\forall n \in N$

I think you meant $ a$ is an odd integer.

I don't know if there's an elementary way to prove this.  It amounts to showing that $ \mathbb{Z}_{2^{k}}$ doesn't have a primitive root for $ k > 2$, which is not an easy result as far as I remember.

Edit:  [hide="Oh, but of course..."] Naively, let $ a = 2b+1$.  We then have

$ (2b+1)^{2^{n}}= 1+2^{n}(2b)+{2^{n}\choose 2}(2b)^{2}+{2^{n}\choose 3}(2b)^{3}+...$

Which can be rewritten

$ 1+2^{n+1}b+2^{n+1}(2^{n}-1) b^{2}+2^{n+2}( \text{stuff})$

Now, $ b+(2^{n}-1) b^{2}\equiv b^{2}+b \equiv 0 \bmod 2$, so it follows that

$ (2b+1)^{2^{n}}\equiv 1 \bmod 2^{n+2}$ [/hide]
The hard result is characterizing the moduli that do have primitive roots.  Righty-o.

I assume you meant $ a$ is an odd integer instead of $ n$.
[hide="Hint"]We need to prove that $ a^{2^{n}}-1$ has at least $ n+2$ even divisors. So factor $ a^{2^{n}}-1$.[/hide]

[hide="Solution"]$ a^{2^{n}}-1=(a^{2^{n-1}}+1)(a^{2^{n-2}}+1) \cdots (a^{2^{1}}+1)(a^{2}-1)$. Now note that the first $ n-1$ factors are even. Also we have $ a^{2}-1\equiv 0\pmod{2^{3}}$, since $ a$ is odd. Thus $ a^{2^{n}}-1$ has $ n-1+3=n+2$ even factors.  [/hide]

EDIT: t0rajir0u beat me to it.

My solution is:

Lets see if the conjecture holds for $ n = 1$:

$ LHS = a^{2}$
$ RHS \equiv 1(mod\, 2^{3}) \equiv 1(mod\, 8)$

For $ a \equiv \pm 1(mod\,8)$, $ a^{2}\equiv 1(mod\,8)$
For $ a \equiv \pm 3(mod\,8)$, $ a^{2}\equiv 9(mod\,8) \equiv 1(mod\,8)$

$ \Longrightarrow$ conjecture holds for $ n = 1$

Lets assume the conjecture holds for $ n = k$

$ \Longrightarrow a^{2^{k}}\equiv 1 (mod\, 2^{k+2})$

$ \Longrightarrow a^{2^{k}}\equiv 1 (mod\, 2^{k+3})$ or $ a^{2^{k}}\equiv 2^{k+2}+1 (mod\, 2^{k+3})$

But $ {a^{2^{k+1}}}={a^{2\cdot2^{k}}}= (a^{2^{k}})^{2}$

$ \Longrightarrow{a^{2^{k+1}}}\equiv 1 (mod\, 2^{k+3})$ or $ {a^{2^{k+1}}}\equiv (2^{k+2}+1)^{2}(mod\, 2^{k+3})$

But $ (2^{k+2}+1)^{2}= 2^{2k+4}+2^{k+3}+1 = 2^{k+3}(2^{k+1}+1)+1$

$ \Longrightarrow (2^{k+2}+1)^{2}(mod\, 2^{k+3}) \equiv 1 (mod\, 2^{k+3})$

$ \Longrightarrow a^{2^{k+1}}\equiv 1 (mod\, 2^{k+3})$ completing the induction step.

Question 2:

Solve the following linear congruence:

$ 140x \equiv 133 (mod\,301)$

[quote="sludgethrower"]
Solve the following linear congruence:

$ 140x \equiv 133 (mod\,301)$[/quote]

[hide="solution"]Since $ \gcd(140,133,301)=7$, dividing by $ 7$ we get $ 20x\equiv 19\pmod{43}\equiv 19+7\cdot 43\pmod{43}\equiv 320\pmod{43}$. Thus $ x\equiv 16\pmod{43}$[/hide]

nayel...isn't it customary to refer to the solution of the linear congruence $ ax \equiv b(mod\,n)$ as the set of solutions which are incongruent modulo $ n$?

Maybe, but I'm not sure, to be honest! :P
In that case we could describe the solutions $ x\equiv 16\pmod{43}$ as $ x\equiv 16,16+43,16+2\cdot43,\dots,16+6\cdot 43\pmod{301}$ which is equivalent to $ x\equiv 16,43,102,145,188,231,\text{ or }274\pmod{301}$.

Or in abbreviated form:

$ x = 16+43t (mod\,301)$ where $ t = 0, 1,\ldots,6$

It means exactly the same thing -- why should it matter which you say?
Also, nayel, you had a small typo: $ 16+43 = 59 \neq 43$.

It doesn't matter what you say. It is just convenient to treat solutions of $ ax \equiv b(mod\,n)$ which are congruent modulo $ n$ as being "equal" even though they are not equal in the usual sense. In short: when we refer to the number of solutions of $ ax \equiv b(mod\,n)$, we mean the number of incongruent integers satisfying this congruence. For this reason, it is sometimes useful to make the incongruent solutions distinct.

Question 4: I am really stuck with this one. I don't know how to begin tackling it.

Find all the solutions of the linear congruence $ 3x-7y \equiv 11 (mod\,13)$.


For $a,b,c$ positive reals prove that

$\displaystyle 4\sum\frac{a^2+bc}{(b+c)^2} \ge 3+\frac{(a+b+c)^2}{ab+bc+ca}$

Nobody is interested in this inequality?

Here is my proof:

By the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]Cauchy-Schwarz inequality in the Engel form[/url], we have

$\frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)}\geq\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(a+b+c)^2}{2(bc+ca+ab)}$.

In other words,

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq\frac{(a+b+c)^2}{2(bc+ca+ab)}$.

Multiplication with 2 yields

$2(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq\frac{(a+b+c)^2}{bc+ca+ab}$.

Now add this inequality to the inequality

$2(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq 3$,

which follows from $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac32$, what is just Nesbitt's inequality. We get

$4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq 3+\frac{(a+b+c)^2}{bc+ca+ab}$.

Finally, after Lemma 2 of http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9, we have

$\frac{( a-b) ( a-c) }{( b+c) ^{2}}+\frac{( b-c) ( b-a) }{( c+a) ^{2}}+\frac{( c-a) ( c-b) }{( a+b) ^{2}}\geq 0$.

Since

$\frac{a^2+bc}{(b+c)^2}=\frac{( a-b) ( a-c) }{( b+c) ^{2}}+\frac{a}{b+c}$;
$\frac{b^2+ca}{(c+a)^2}=\frac{( b-c) ( b-a) }{( c+a) ^{2}}+\frac{b}{c+a}$;
$\frac{c^2+ab}{(a+b)^2}=\frac{( c-a) ( c-b) }{( a+b) ^{2}}+\frac{c}{a+b}$,

we have

$4(\frac{a^2+bc}{(b+c)^2}+\frac{b^2+ca}{(c+a)^2}+\frac{c^2+ab}{(a+b)^2})$
$=4((\frac{( a-b) ( a-c) }{( b+c) ^{2}}+\frac{a}{b+c})+(\frac{( b-c) ( b-a) }{( c+a) ^{2}}+\frac{b}{c+a})+(\frac{( c-a) ( c-b) }{( a+b) ^{2}}+\frac{c}{a+b}))$
$=4(\frac{( a-b) ( a-c) }{( b+c) ^{2}}+\frac{( b-c) ( b-a) }{( c+a) ^{2}}+\frac{( c-a) ( c-b) }{( a+b) ^{2}})+4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$
$\geq 4\cdot 0+(3+\frac{(a+b+c)^2}{bc+ca+ab})=3+\frac{(a+b+c)^2}{bc+ca+ab}$.

This proves your inequality.

  Darij

[quote="Vasc"]The following stronger inequality holds:
$\displaystyle \sum\frac{a^2+bc}{(b+c)^2} \ge \frac{(a+b+c)^2}{2(ab+bc+ca)}$.[/quote]

Of course, it follows by adding together

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq\frac{(a+b+c)^2}{2(bc+ca+ab)}$

and

$\frac{( a-b) ( a-c) }{( b+c) ^{2}}+\frac{( b-c) ( b-a) }{( c+a) ^{2}}+\frac{( c-a) ( c-b) }{( a+b) ^{2}}\geq 0$.

So we see that Hungkhtn's inequality is not too strong.

  Darij


inequalities
geometry
circumcircle
triangle inequality

let triangle ABC, AB=AC. $ P\in$ triangle ABC
prove that:
        $ PA^2 \plus{} PB.PC  \leq AB^2$

[hide="If P is on BC"]
Let $ x$ be the smaller of $ PB$ and $ PC$, and $ y \equal{} BC \minus{} x$. Then it is true that

$ PA^2 \plus{} PB\cdot PC \equal{} PA^2 \plus{} xy \leq AB(PA) \plus{} AB(y)$

So, we only need to show

$ AB(PA \plus{} y)\leq AB^2$

which is equivalent to showing $ PA \plus{} y\leq AB$, which his true by the triangle inequality.[/hide]

[hide="If P is on AB or AC"]
Without loss of generality, let $ P$ be on $ AB$. Then note $ PA \leq AB$ and $ PB \leq AB$. So,

$ PA^2 \plus{} PB\cdot PC \leq AB(PA \plus{} PC) \leq AB^2$

by the triangle inequality on $ \triangle PAC$.[/hide]

Unfinished business
--------------------------

Does this problem not also include showing the inequality is true
if point P belongs to the interior of triangle ABC as well,
(still with AB = AC)?

[hide="Hint"]
Draw the circle centered at A with radius AP, then $ AB^2\minus{}AP^2$ is the power of either B or C with respect to the circle, we only have to show one of them is $ \ge PB \cdot PC$.[/hide]

dgreen can you tell me what you mean by power of B and C?

Power of a point
http://en.wikipedia.org/wiki/Power_of_a_point

Let $ M$ be the midpoint of $ BC$ and WLOG assume that $ P$ lies inside $ \triangle ABM.$  Let $ Q \in \overrightarrow{CP}$ such that $ PB \equal{} PQ.$ The circumcenter $ O'$ of $ \triangle QBC$ lies on $ AM$ such that $ \angle BO'C \equal{} 2\angle BQC \equal{} \angle BPC.$  Thus the power of $ P$ WRT $ (O')$ is $ PC \cdot PQ \equal{} PC \cdot PB \equal{} O'C^2 \minus{} O'P^2.$

$ S \equiv PC \cap AM$ and $ N$ is the midpoint of $ CP.$ For each $ P$ lying on $ AB,$ the midpoint of the cevian $ CP$ lies on the C-midline. Therefore, for each $ P$ inside $ \triangle ABM,$ the midpoint $ N$ of $ CP$ lies inside $ \triangle AMC$ $ \Longrightarrow$ $ S$ is between $ P,N.$

On the other hand, since $ O'$ is the midpoint of the arc $ BPC,$  the ray $ \overrightarrow{CP}$ is internal to $ \angle O'CB$ $ \Longrightarrow$ $ O'$ is between $ S$ and $ A.$  Then it follows that the orthogonal projection $ X$ of $ O'$ onto $ CP$ lies between $ C$ and the orthogonal projection $ Y$ of $ A$ onto $ CP.$ Thus

$ AC^2 \minus{} AP^2 \equal{} YC^2 \minus{} YP^2 \equal{} 2PC \cdot NY $

$O'C^2 \minus{} O'P^2 \equal{} XC^2 \minus{} XP^2 \equal{} 2PC \cdot NX$

$ NY \ge NX$ $ \Longrightarrow$ $ AC^2 \minus{} AP^2 \ge O'C^2 \minus{} O'P^2$ $ \Longrightarrow$ $ AB^2 \minus{} PA^2 \ge PB \cdot PC.$


quadratics
inequalities unsolved
inequalities

let $ a, b, c, m$ be four positive real numbers, prove that
$ (a^2 \plus{} m)(b^2 \plus{} m)(c^2 \plus{} m) \geq \frac {3}{4} m^2 (a \plus{} b \plus{} c)^2$

the answer in the book was a bit too long, can somebodyu give a shorter and more beautiful solution? thanks!

Can you prove it for m=2?
It is a quadratic in a, you could solve it that way but it is very ugly.

[quote="tmy1018"]let $ a, b, c, m$ be four positive real numbers, prove that
$ (a^2 \plus{} m)(b^2 \plus{} m)(c^2 \plus{} m) \geq \frac {3}{4} m^2 (a \plus{} b \plus{} c)^2$

[/quote]
Let $ a \equal{} x\sqrt {\frac {m}{2}},$ $ b \equal{} y\sqrt {\frac {m}{2}}$ and $ c \equal{} z \sqrt {\frac {m}{2}}.$
Hence, $ (a^2 \plus{} m)(b^2 \plus{} m)(c^2 \plus{} m) \geq \frac {3}{4} m^2 (a \plus{} b \plus{} c)^2\Leftrightarrow$
$ \Leftrightarrow(x^2 \plus{} 2)(y^2 \plus{} 2)(z^2 \plus{} 2)\geq3(x \plus{} y \plus{} z)^2$ and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4830
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76508


abstract algebra
group theory
superior algebra
superior algebra unsolved

this is small case of the main theorem due to Reidei:
if $G$ is abelian of order $p^{2}$ and $G=AB$ such that $A$ and $B$ contain identity and $|A|=|B|=p$ (p is prime)then $A$ or $B$ is subgroup of $G$.

uhm...
don't we need to ask for something more about $G$?
otherwise $D_{3}$ (or $S_{3}$, as you prefer to call it) seems to provide a counterexample: $A = B = \{e, r, s\}$ should work...

uhm.. i think it's not enough yet: take $A=B=\{0,1,3\}\subset \mathbb{Z}/4\mathbb{Z}$..
maybe $p$ should divide the order of $G$?


function
limit
trigonometry
real analysis
real analysis unsolved

Let $f$ be a differentiable function from the positive reals to the positive reals with the property that $f(x)<x$ for all $x$. Suppose that $x_{1}>0$, and for $n>1$ let $x_{n}=f(x_{n-1})$. Suppose further that $\lim_{n\rightarrow \infty}x_{n}=0$ and that there exists positive numbers $a$ and $k$ such that

$\lim_{x\rightarrow 0}\frac{x^{a}-(f(x))^{a}}{x^{a}(f(x))^{a}}=\frac{1}{k^{a}}$.

Prove that:

a) $\lim_{n\rightarrow \infty}n^{1/a}x_{n}=k$.

b) Suppose that $0<x_{1}<1$, and specialize to the case where $f$ is given by $f(x)=\sin x$ if $x<\pi/2$ and $f(x)=1$ if $x\geq \pi/2$. Show that $\lim_{n\rightarrow \infty}x_{n}\sqrt{n}=\sqrt{3}$.

c) Finally, suppose instead that $0<x_{1}<1$ and that $f(x)=1-e^{-x}$. Show that, in this case, $\lim_{n\rightarrow \infty}nx_{n}=2$.


calculus
integration
limit
symmetry
complex analysis
complex analysis unsolved

Let $ z \in \mathbb{C},|z| \equal{} 1$, $ B_n(x) \equal{} \prod_{i \equal{} 1}^{n} \frac {x \minus{} \alpha_i}{1 \minus{} \overline{\alpha_i}x}$, $ |\alpha_i| < 1, 1\leq i \leq n$.

$ R_n(\xi,z) \equal{} \frac {1}{\xi \minus{} z}\left( \xi \frac {B_n(\xi)}{B_n(z)} \minus{} z \frac {B_n(z)}{B_n(\xi)} \right)$.

Prove that $ \displaystyle \lim_{n \to \infty} \int_{S^ \plus{} } R_n(\xi,z)d\xi \equal{} \lim_{n \to \infty} \int_{S^ \minus{} } R_n(\xi,z)d\xi$ for any $ |z| \equal{} 1$   (*).

First i thought, that it is even true for every $ n$, without limit, but then substitutng simply $ n \equal{} 1$ i seemingly found lots of $ \alpha_1$ for which two integrals are different.

I would be grateful if you could possibly impose some conditions on numbers $ \alpha_i$ in order for (*) to be true, in case it is not true as stated.

What are $ S^{\pm}$? :? Probably, they are the upper and the lower semicircles, but in which directions are they travelled?

Sorry. $ S^\minus{} \equal{}  \{ z \in \mathbb{C}, z\equal{} e^{it},  \minus{}\pi < t < 0\}, S^\plus{} \equal{}  \{ z \in \mathbb{C}, z\equal{} e^{it}, 0 < t < \pi \}$

So both are travelled counterclockwise, right? And $ \alpha_j$ are assumed to satisfy the Blaschke condition or they are just completely arbitrary numbers?

To be honest, I do not see any reason whatsoever for the equality to hold with or without limit unless $ \alpha_k$ have some special symmetry (though I may be wrong here). What are you really aiming at, eugene?

[quote="fedja"]So both are travelled counterclockwise, right? [/quote]

Exactly.

[quote="fedja"]
And $ \alpha_j$ are assumed to satisfy the Blaschke condition or they are just completely arbitrary numbers? [/quote]

Actually, it is vice versa supposed that $ \sum (1 \minus{} |\alpha_k|) \equal{} \infty$. But i'm not sure the Blaschke condition should be imposed on the $ \alpha_k$'s.

[quote="fedja"]To be honest, I do not see any reason whatsoever for the equality to hold with or without limit unless $ \alpha_k$ have some special symmetry (though I may be wrong here). What are you really aiming at, eugene?[/quote]

First, i also thought that at least the condition that $ \alpha_k$ are pairwise symmetric wrt to the origin(or something like this with conjugacy) should be imposed, at least intuitively.
If you see any proof/idea using any condition on $ \alpha_k$, i would be glad to see it.

For now, i have some ideas, which prove this is true(even without the limit of $ n$ ) for any $ \alpha_k$, but it should be wrong, because as i said, when i checked both integrals for $ n \equal{} 1$, they turned our to be distinc in general.

Well, what can be more symmetric than the sequence $ \alpha_j \equal{} 0$? Still, we have $ R_n(\xi,z) \equal{} \xi^{n}z^{ \minus{} n} \plus{} \dots \plus{} 1 \plus{} \dots \plus{} \xi^{ \minus{} n}z^{n}$. The integrals of $ 1$ are certainly not the same (they are $ 2$ and $ \minus{} 2$) and everything else is some series depending on $ z$ in such a way that the average over $ z$ of each term is $ 0$. So, for this sequence, you'll never have what you want. Actually, I don't know any sequence $ \alpha_j$, for which the statement holds. I'm too lazy to try to figure out right now what exactly is wrong with the attached text (maybe I'll do it later if you don't find a mistake yourself) but regardless of anything, my advice would be to forget completely about trying to prove this obviously wrong statement and to see if you can do what you need without it.


Let $ P \equal{} \frac {ax^2 \plus{} by^2}{ax \plus{} by} \plus{} \frac {az^2 \plus{} bt^2}{az \plus{} bt}$
Know that: $ x \plus{} z \equal{} y \plus{} t \equal{} 1$ and $ x,y,z,t \ge 0$ and $ a, b$ are constan.
Find Max of $ P$ and Min of $ P$.

[quote="thanhnam2902"]Let $ P = \frac {ax^2 + by^2}{ax + by} + \frac {az^2 + bt^2}{az + bt}$
Know that: $ x + z = y + t = 1$ and $ x,y,z,t \ge 0$ and $ a, b$ are constan.
Find Max of $ P$ and Min of $ P$.[/quote]
${{ P  \geq  \frac{(ax+by)^2}{(a+b)(ax+by)}}+\frac{(az+bt)^2}{(a+b)(az+bt)}}  = \frac{ax+by}{a+b} + \frac{az+bt}{a+b} =1$
 occur  equal  when and only when :$ x=y=z=t=\frac{1}{2}$

puppose  that  $ x \geq y$  then $ t \geq z$
  we have : $ \frac {ax^2 \plus{} by^2}{ax \plus{} by} \leq \frac{ax^2}{ax} \equal{}x \leq 1$(  because of $ xy(x\minus{}y) \geq 0$)
and other hand, we have :$ \frac {az^2 \plus{} bt^2}{az \plus{} bt} \leq t \leq 1$
 hence $ P \leq 2$
Max$ P \equal{}2$  and  hold  equal if  and only if :$ x\equal{}t\equal{}1,y\equal{}z\equal{}0$


Hint: Try writing the radicals as rational exponents, then simplify from there. Should help a bit.

I did that before but can't see how to continue. Another hint please  :D

[quote="Totally Zealous"][color=red]removed by b-flat[/color]
[/quote]
I hope you realize that this is from the Mandelbrot Competition and can't be discussed until the Mandelbrot Competition gives us permission to discuss these questions.

This topic should be deleted...

I wish I did Mandelbrot, It sounds like fun. I have the book.

I do not generally delete topics, so I will just lock this topic and remove the parts from posts that need to be removed.


Find all non-negative integer $ n$ such that $ 0! \plus{} 1! \plus{} 2! \plus{} \cdots \plus{} n!$ is a perfect square.(we define $ 0!\equal{}1$)

[hide="solution"]Let us consider mod 8. A square can only be 0, 1, or 4 modulo 8. When $ n\geq 3$, the sum is 2 modulo 8, so we must have $ n<3$. $ n\equal{}0$ yields 1, $ n\equal{}1$ yields 2, and $ n\equal{}2$ yields 4. Only $ n\equal{}0,2$ work.[/hide]

[hide="nice problem"]
When $ n\equal{}3$, the sum is $ 2*5$.

Let the sum for a number $ n$ be $ 2*k$ where $ k$ is odd. Then adding $ (n\plus{}1)!$ will yield $ 2(k\plus{}\frac{(n\plus{}1)!}{2}$. If $ n\plus{}1\ge4$, then $ \frac{(n\plus{}1)!}{2}$ is even, so we get a new number $ 2*l$ where $ l$ is odd. So, for $ n\ge4$, $ 0!\plus{}1!\plus{}2!\plus{}...\plus{}n!$ cannot be a perfect square.

So the only answers are when $ n\le3$. Perfect squares occur when $ n\equal{}0$ and $ n\equal{}2$.

[/hide]


Find the smallest natural number a such that a-1 is the sum of 7  positive divisors (not necessarily different) of a. This is related to a problem published in Kvant, which says that for any n there are only finitely many values of a for which a-1 is the sum of n positive divisors of a.

This very old post by Harazi hasn't received any replies yet, after almost 2 years.

I thought it might be a good idea to move it from pre-olympiad to this forum. Anyone?

Doesn't a = 8 work?  Because 8 - 1 can be expressed as 7 ones.

Maybe the problem was supposed to be "largest", not "smallest"?

You've got a point there :D Harazi, is that the question you meant to ask?

Sorry, it is the largest of course. Probably this is the reason for which nobody answered... :D

I think the question boils down to this:  If 8 (positive) unit fractions add up to 1, what is the largest possible denominator?  I think I have seen that question before, so I will let others finish.


Αν $ a,b,c$ είναι θετικοί πραγματικοί αριθμοί τέτοιοι ώστε $ a\plus{}b\plus{}c\plus{}1\equal{}4abc$ να αποδείξετε ότι

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq3\geq\frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}$

Η παρακάτω λύση είναι στο πόδι οπότε μπορεί να μην έχω δει κάτι προφανές ή να έχω κάνει λάθος ... Please check
H συνθήκη  $ a\plus{}b\plus{}c\plus{}1\equal{}4abc$ είναι ισοδύναμη με $ \frac{1}{1\plus{}2a}\plus{}\frac{1}{1\plus{}2b}\plus{}\frac{1}{1\plus{}2c}\equal{}1$

Θέτουμε $ \frac{1}{1\plus{}2a}\equal{}x$ .Τότε $ x\plus{}y\plus{}z\equal{}1$ kai  η προς απόδειξη γίνεται :

$ \sum\frac{1\minus{}x}{2x}\geq 3\geq\sum\frac{2\sqrt{xy}}{\sqrt{(1\minus{}x)(1\minus{}y)}}$

Η αριστερή είναι ισοδύναμη με $ \sum\frac{1}{x}\geq 9$ που ισχύει αφού $ (x\plus{}y\plus{}z)(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})\geq 9$

Τώρα για τη δειξιά από την ανισότητα AM-GM  έχουμε :
$ \frac{2\sqrt{xy}}{\sqrt{(1\minus{}x)(1\minus{}y)}}\equal{}\frac{2\sqrt{xy}}{\sqrt{(y\plus{}z)(z\plus{}x)}}\leq\frac{y}{y\plus{}z}\plus{}\frac{x}{x\plus{}z}$

Προσθέτωντας παίρνουμε τη ζητούμενη .

[quote="galois01"]Αν $ a,b,c$ είναι θετικοί πραγματικοί αριθμοί τέτοιοι ώστε $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$ να αποδείξετε ότι

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq3\geq\frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {bc}} \plus{} \frac {1}{\sqrt {ca}}$[/quote]

Εχω μια λυση αλλα η αποδειξη του δεξιου σκελους με δυσκολευε και την εκανα με πολλαπλασιαστες Lagrange, διωρθωστε με αν εχω κανει πουθενα κανενα λαθος.

[hide]
Απο τη συνθηκη εχουμε:

$ 1)$ $ \frac {1}{ab} \plus{} \frac {1}{ac} \plus{} \frac {1}{cb} \equal{} 4 \minus{} \frac {1}{abc}$

$ 2)$ Απο AM-GM: $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1 \geq 4\sqrt [4]{abc} \Leftrightarrow abc \geq 1 \Leftrightarrow \minus{} \frac {1}{abc} \geq \minus{} 1$

Για το αριστερο σκελος της ανισοτητας εχουμε:

$ (LHS)^2 \geq 3\left(\frac {1}{ab} \plus{} \frac {1}{ac} \plus{} \frac {1}{cb}\right) \equal{} 3\left(4 \minus{} \frac {1}{abc}\right) \geq 3(4 \minus{} 1) \equal{} 9 \Leftrightarrow LHS \geq 3$

Τωρα για το δεξι σκελος της ανισοτητας λυνουμε τη συνθηκη ως προς $ c$ και παιρνουμε:

$ c \equal{} \frac {a \plus{} b \plus{} 1}{4ab \minus{} 1}$

Θετουμε $ f(a,b,c) \equal{} \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {ac}} \plus{} \frac {1}{\sqrt {cb}}$, οποτε:

$ f(a,b,c) \equal{} f(a,b) \equal{} \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {\frac {a^2 \plus{} ab \plus{} a}{4ab \minus{} 1}}} \plus{} \frac {1}{\sqrt {\frac {b^2 \plus{} ab \plus{} b}{4ab \minus{} 1}}}$

Προφανος το πεδιο ορισμου της $ f(a)$ ειναι το διαστημα $ (\frac {1}{4b}, \plus{} \infty)$

Το δεξι ακρο του συνολου τιμων της $ f$ θα ειναι ο μεγαλυτερος αριθμος απο τις ακρεες τιμες της $ f$   και απο τα ακροτατα που αυτη παρουσιαζει στο ανοιχτο διαστημα στο οποιο οριζεται

Εχουμε τωρα $ \lim_{a \rightarrow \plus{} \infty}{f(a)} \equal{} 2$ και $ \lim_{a \rightarrow \frac {1}{4b}}{f(a)} \equal{} 2$

Εστω τωρα οτι η $ f$ παρουσιαζει καποιο ακροτατο στο ανοιχτο διαστημα στο οποιο οριζεται, για ευκολια θετουμε $ \frac {1}{\sqrt {ab}} \equal{} x, \frac {1}{\sqrt {ac}} \equal{} y, \frac {1}{\sqrt {cb}} \equal{} z$, οποτε ειναι $ f(x,y,z) \equal{} x \plus{} y \plus{} z$, ενω η συνθηκη γινεται $ x^2 \plus{} y^2 \plus{} z^2 \plus{} xyz \equal{} 4$, θετουμε ακομα $ g(x,y,z) \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} xyz \minus{} 4$, και αν $ (x_0, y_0, z_0)$ ειναι η θεση του ακροτατου τοτε υπαρχει αριθμος $ k$ τετοιος ωστε οι μερικες παραγωγοι της $ F(x,y,z) \equal{} f(x,y,z) \plus{} kg(x,y,z)$ στο σημειο $ (x_0, y_0, z_0)$ να ειναι ολες ισες με 0

Εχουμε τωρα να λυσουμε το συστημα: $ \nabla F(x_0, y_0, z_0) \equal{} < 0, 0, 0 >$, δηλαδη το:
$ k(2x_0 \plus{} y_0z_0) \plus{} 1 \equal{} 0$

$ k(2y_0 \plus{} x_0z_0) \plus{} 1 \equal{} 0$

$ k(2z_0 \plus{} x_0y_0) \plus{} 1 \equal{} 0$

$ {x_0}^2 \plus{} {y_0}^2 \plus{} {z_0}^2 \plus{} x_0y_0z_0 \minus{} 4 \equal{} 0$

Απο τις 2 πρωτες εξισωσεις παιρνουμε: $ 2x_0 \plus{} y_0z_0 \equal{} 2y_0 \plus{} x_0z_0 \Leftrightarrow x_0(2 \minus{} z_0) \equal{} y_0(2 \minus{} z_0) \Leftrightarrow z_0 \equal{} 2$ η $ x_0 \equal{} y_0$. Ομοια απο τους υπολοιπους συνδιασμους των 3 πρωτων εξισωσεων παιρνουμαι: $ x_0 \equal{} y_0 \equal{} z_0$ η $ x_0 \equal{} y_0 \equal{} z_0 \equal{} 2$, και λογο της 4ης εξισωσης τελικα εχουμε $ (x_0, y_0, z_0) \equal{} (1,1,1)$. Αρα η $ f$ στο ανοιχτο διαστημα στο οποιο οριζεται παρουσιαζει ακροτατο μονο για $ x \equal{} y \equal{} z \equal{} 1$ το οποιο ειναι ο αριθμος $ f(1,1,1) \equal{} 3$, και επηδη $ 3 > 2 \equal{} \lim_{a \rightarrow \plus{} \infty}{f(a)} \equal{} \lim_{a \rightarrow \frac {1}{4b}}{f(a)}$ εχουμε οτι το μεγιστο της $ f$ ειναι ο αριθμος $ 3$ οποτε η αποδειξη του δεξιου σκελους της ανισοτητας ολοκληρωθηκε.[/hide]

Για το δεξί μέλος κάντε το μετασχηματισμό $ x\equal{}\frac{1}{2\sqrt {ab}}$ και τα λοιπά.
Από την δοσμένη ισότητα έχουμε:
$ \frac{1}{4ab}\plus{}\frac{1}{4bc}\plus{}\frac{1}{4ca}\plus{}\frac{1}{4abc}\equal{}1 \Longrightarrow$

$ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$

Συνεπώς τα $ x,y,z$ είναι συνημίτονα γωνιών ενός τριγώνου.

Τώρα η δεξιά ανισότητα ισοδυναμεί με:

$ x\plus{}y\plus{}z\leq \frac{3}{2}$ που είναι γνωστό  :wink:


calculus
integration
geometry
inradius
algebra
system of equations

Find all positive integral solutions of the system of equations: $\begin{cases}a^{2}=b^{2}+c^{2}\\ a+b+c=\frac{1}{2}bc\end{cases}$

Eliminating $a$ we obtain th equatin: $(b-4)(c-4)=8$
We see that $b-4=1, 2, 4$ or $8$, whence $b=5, 6, 8$ or 12,...

[hide="Remark"] This is equivalent to finding an integer right triangle $(b, c, a)$ with inradius $r = 2$.  This can be generalized; if the inradius is written in the form

$r = kn(m-n)$

Then we have the familiar

$b = 2kmn$
$c = k(m^{2}-n^{2})$
$a = k(m^{2}+n^{2})$  [/hide]


projective geometry
geometry unsolved
geometry

Is this China TST? I think it's too easy!
Let the tangency of (O) which passes through E cut BC at Q.
We have $ (QDBC)\equal{}\minus{}1$ and $ \angle QED\equal{}\angle QDE\equal{}\angle DFC$ then $ EQ//CG$
$ \Rightarrow FG\equal{}FC$

[color=darkblue][b]I have a solution same livetolove212[/b].

Let $ (d)$ is a line tangent with inscribed of $ \triangle ABC$ at point $ E$. Let $ (d)$ meet $ BC$ at point $ S$.

We have $ \angle{SED} \equal{} \angle{SDE} \equal{} \angle{CDF} \equal{} \angle{CFD} \Longrightarrow ES // CF$

You known that $ (ES, EG, EF, EC) \equal{} \minus{} 1 \Longrightarrow F$ is midpoint of $ CG$.[/color]

Perhaps my solution is complicated  :oops: :draw line EK that is tangent to inscribed circle .The line meets BC at K, and meet AB and AC at M' and M .Obviously, EK is parallel to CF ,then we get DE/EF=KD/CK~~~~(#); then join BM and CM';
By Newton Theorem, we know that AD , CM' and BM meet at one point ;
then we use Ceva theorem:(BD/DC)*(CM/MA)*(AM'/M'B)=1~~~①;
Then we use Menelaus theorem:(BK/CK)*(CM/MA)*(AM'/M'B)=1~~~②;
 ①/②→BD/CD=BK/CK→DB*CK=CD*KB~~③;
On segment BK ;By Ptolemy Theorem ,we get DC*BK+BD*CK=BC*DK; By③,WE can easily know that :1/2(BC*DK)=DB*KC=CD*BK;
In △FDC ;we use Menelaus again to get:(CB/BD)*(DE/FE)*(FG/CG)=1;
By  (#), we can get :(CB/BD)*(KD/CK)*(FG/CG)=1.
∵2BD*KC=BC*KD;∴BC/BD*DK/KC=2;∴FG/CG=1/2;
Then the answer follows.
 :P

[color=darkblue]This is a full solution of me for this problem!

Let $ AB, AC$ tangents with circle $ (O)$ at points $ M, N$ respectively. Let tangent of $ (O)$ at point $ E$ meet $ BC$ at point $ S$.

We have $ S, M, N$ are lies on polar line of point $ A$ with circle $ (O) \Longrightarrow$ $ S, M, N$ are collinear.

Let $ AD$ meet $ MN$ at point $ H \Longrightarrow (SHMN) \equal{} \minus{}1$, (property of polar – polar line).

$ \Longrightarrow (AS, AH, AM, AN) \equal{} \minus{}1 \Longrightarrow (SDBC) \equal{} \minus{}1$

$ \Longrightarrow (ES, ED, EB, EC) \equal{} \minus{}1$, (*).

But $ ES // CG$, (**) (because $ \angle {SED}\equal{}\angle{SDE}\equal{}\angle{FDC}$).

Since (*) and (**) $ \Longrightarrow FC \equal{} FG$. [/color]

[quote="CCMath1"]The inscribed circle I of △ABC tagent side BC at point D join AD meet the circle at point  E . join BE . F is point on AD ,and CF = CD;line CF meet BE at G. Prove that :FG=CF.[/quote]
this problem is China TST 2008 not 2009 and easy

This is quite a nice problem! It seems that the Chinese coaches like projective geometry.

[quote="livetolove212"]Is this China TST? I think it's too easy!
[/quote]
It was absolutely a problem in Chinese TST 2008 . See some other proofs : http://www.mathlinks.ro/viewtopic.php?p=1088276#1088276.
I don't think it is  too easy,livetolove212!.The step "Draw the tangency of $ (O)$ " is not obvious .