inequalities
Pythagorean Theorem
geometry
triangle inequality

[hide]
12. I think, $\frac{-bc}{d}$
13. 0, but it seems too simple...[/hide]

[hide="13"]Assume that $a^{2}+b^{2}=c^{2}$ and $a+b=c$ are both true.

Substitute $c$ into  the first equation to get $a^{2}+b^{2}=a^{2}+2ab+b^{2}$.

This is only true if $ab=0$, which is impossible because $a$ and $b$ must be positive integers.

James will flunk his PT quiz, too.[/hide]

[quote="i_like_pie"][hide="13"]Assume that $a^{2}+b^{2}=c^{2}$ and $a+b=c$ are both true.

Substitute $c$ into  the first equation to get $a^{2}+b^{2}=a^{2}+2ab+b^{2}$.

This is only true is $ab=0$, which is impossible because $a$ and $b$ must be positive integers.

James will flunk this PT quiz, too.[/hide][/quote]
[hide="alternatively"]
$a+b$ must be >c by the triangle inequality, so 0.[/hide]

[quote="diophantient"]
[hide="alternatively"]
$a+b$ must be >c by the triangle inequality, so 0.[/hide][/quote]

That was my method. Good job. :)

Oooh, thought of a good variation of 13.

James is moved to the lowest math class there is, where, somehow, they are also covering the Pythagorean theorem. They're just about to take a test on manipulating Pythagorean triples-that is, multiplying by a scale factor and still being left with a Pythagorean triple. However, James is confused(shocking), and thinks he can simply add a constant to each side and be left with one. There qill be a similar format-2007 questions, lengths of the same range as below, but a pyth. triple is given, and they must supply another based on it. How many could he possibly get right?

[quote="perfect628"] However, James is confused(shocking), and thinks he can simply add a constant to each side and be left with one.[/quote]

Please elaborate

For example, he thinks with the triple $3-4-5$ he can add 1 to each side and be left with a new one, $4-5-6$, which is clearly not one.

I think it's 0 because I just can't think of any off the top of my head, and I guess you could prove it with algebra, too.

a^2+b^2=c^2

(a-x)^2+(b-x)^2=(c-x)^2

Expand, reduce, and something else I can't think of.

[quote="13375P34K43V312"]12. James is trying to learn how to multiply binomials; he wants to expand expressions such as $(2x+3)(3x+1)$. However, he learns incorrectly; instead of using the FOIL method, in which one multiplies the first terms $(2x,3x)$, the outside terms $(2x,1)$, the inside terms $(3,3x)$, and the last terms $(3,1)$, and adds these products together, James only multiplies the first terms and last terms, and adds those together. His teacher gives one of this type of exercise, to expand $(ax+b)(cx+d)$, where $a,b,c,d$ are real constants. Miraculously, James uses his method and ends up with the correct answer. Find the value of $a$ in terms of $b$, $c$, and $d$. Express your answer as a fraction.

13. After flunking the binomial multiplication test, James’ math class moves onto a new topic: The Pythagorean Theorem. He learns that in any right triangle with legs of length $a$ and $b$, and hypotenuse of length $c$, $a^{2}+b^{2}=c^{2}$. Being the ignorant student that he is, James thinks he can take the square root of each individual term, and reduces the Pythagorean Theorem to $a+b=c$. One day, James’ teacher gives a pop quiz on the Pythagorean Theorem, with 2007 questions. In each problem, the students must find the length of the hypotenuse of a right triangle given the lengths of the two legs, which have distinct integer lengths between 1 and 2007 inclusive. James uses his (incorrect) method on the pop quiz for every problem. If all the problems are different (they have different legs), what is the maximum number of problems that James can get right on the quiz?[/quote]
[hide="Solution for 12"]
$acx^{2}+bd=acx^{2}+(ad+bc)x+bd$
$0=(ad+bc)x$
$ad=-bc$ or $x=0$
$a=\frac{-bc}{d}$
[/hide]

[hide="Solution for 13"]
$a^{2}+b^{2}=c^{2}$
$a+b=c$
$a^{2}+2ab+b^{2}=c^{2}$
$a^{2}+b^{2}=a^{2}+2ab+b^{2}$
$0=2ab$
$ab=0$
$a$ or $b$ has to be 0.
That's impossible since the lengths are between 1 and 2007 (inclusive), so he will not get any questions right.

For a triangle, in order for it to have the length of one side be equal to the sum of the lengths of two other sides, it has to break the rule that the length of the longest side has to be less than the sum of the lengths of the two other sides (triangle inequality).
[/hide]


function
real analysis
real analysis unsolved

Let $ A$ be the set of all continuous functions on $ R$. Prove that there is a bijection between $ A$ and $ R$.

A real continuous function is completely determined by its values on rationals.

But $ \mathbb{R}^{\aleph_0} \sim (2^{\aleph_0})^{\aleph_0} \sim 2^{\aleph_0}  \sim \mathbb{R}$.


Hello,

I'm using TeXnicCenter could someone tell me how can I add a picture  :?:

No problem with the server here. Alternatively AoPSWiki has help with pictures at [[LaTeX:Pictures]]

Thanks and for the site I don't it says "The website is too busy to show the webpage" anyway I know how to do it now thanks a lot  :)


For $ a,b,c$ positive real numbers, prove that
$ \left.\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 4\left[\frac{ab}{(b\plus{}\sqrt{ca})^2}\plus{}\frac{bc}{(c\plus{}\sqrt{ab})^2}\plus{}\frac{ca}{(a\plus{}\sqrt{bc})^2}\right]\right.$

$ \frac {4ab}{(b + \sqrt {ca})^2}\leq \frac {4ab}{4b\sqrt {ca}} = \sqrt {\frac {a}{c}}$ e.t.c.

$ \therefore 4\left[\frac {ab}{(b + \sqrt {ca})^2} + \frac {bc}{(c + \sqrt {ab})^2} + \frac {ca}{(a + \sqrt {bc})^2}\right]
\leq \sqrt {\frac {a}{c}} + \sqrt {\frac {b}{a}} + \sqrt {\frac {c}{b}}\leq \frac {a}{b} + \frac {b}{c} + \frac {c}{a}$

$ \because \sqrt {\frac {a}{b}}\sqrt {\frac {b}{c}} + \sqrt {\frac {b}{c}}\sqrt {\frac {c}{a}} + \sqrt {\frac {c}{a}}\sqrt {\frac {a}{b}}\leq \frac {a}{b} + \frac {b}{c} + \frac {c}{a}$

Thanks, [b]kunny[/b]! Nice solution. Shorter than mine. Congratulations!


probability
trigonometry
floor function
search
limit
logarithms

A particular botched coin is controlled by a [i]God-sequence[/i], $ \{ G_n \} _{n\equal{}1}^{\infty}$; the probability of the coin landing heads on the $ n^{\underline{\text{th}}}$ flip is $ G_n$.

Prove or disprove the existence of a God-sequence with the following properties:

$ \blacktriangleright$ There is no integer $ k$ such that the coin will certainly land heads on the $ k^{\underline{\text{th}}}$ flip
$ \blacktriangleright$ The coin will almost certainly land heads at least once
$ \blacktriangleright  \; \left( \sum_{n\equal{}1}^{\infty} G_n \right) \in \mathbb{R}$

[hide="is this right?"]Take $ G_n\equal{}\frac{1}{n^2}$.  Then $ \sum_{n\equal{}1}^\infty G_n\equal{}\frac{\pi^2}{6}$ and

$ \prod_{n\equal{}1}^\infty\left(1\minus{}G_n\right)$

$ \equal{}\prod_{n\equal{}1}^\infty\left(1\minus{}\frac{\pi^2}{n^2\pi^2}\right)$

$ \equal{}\sin\pi\equal{}0$[/hide]

Clever, but there's an oversight in there  :wink:

You have $ G_1 = 1$, which violates the first condition. If you leave out that term, then the probability of having at least one heads is only $ 1/2$.

The first condition is in there because otherwise, sequences like $ G_n = \left\{\begin{array}{cc} 1 & n = k & 0 & n \neq k\end{array} \right$ would work trivially.

Are the coin flips required to be independent?

[hide="how about this?"]We know that if $ G_n\le\frac {1}{n \plus{} 1}$ then

$ \prod_{n \equal{} 1}^\infty \left(1 \minus{} G_n\right)\ge\prod_{n \equal{} 1}^\infty \left(1 \minus{} \frac {1}{n \plus{} 1}\right)$

$ \equal{} \prod_{n \equal{} 1}^\infty \left(\frac {n}{n \plus{} 1}\right) \equal{} 1$

Thus, we must have $ G_n\ge \frac {1}{n \plus{} 1}$. But then $ \sum_{n \equal{} 1}^\infty G_n\ge \minus{} 1 \plus{} \sum_{n \equal{} 1}^\infty\frac {1}{n} \equal{} \infty$.[/hide]

edit: this is wrong, don't read it

[b]@ Ravi B:[/b]

They are only dependent in the sense that $ n$ is increasing.

The probability on a particular coinflip is determined solely by the value of the corresponding term of the God-sequence.

If you want, you might formulate the problem algebraically

[hide="Rephrase with Algebra"]
Does there exist a sequence $ G_n$ such that:

$ \blacktriangleright \qquad\forall n \in \mathbb{N}, \; \;\lfloor G_n \rfloor \equal{} 0$

$ \blacktriangleright \qquad \left( \sum_{n \equal{} 1}^{\infty} G_n \right) \in \mathbb{R}$

$ \blacktriangleright \qquad \prod_{n \equal{} 1}^{\infty} (1 \minus{} G_n) \equal{} 0$
[/hide]

[hide="Progress"]I suspect there isn't one.

If $ \sum G_n$ converges, $ \prod e^{G_n}$ also does, so the second condition is equivalent to

$ \prod_{n\equal{}1}^{\infty} e^{G_n} \equal{} r$

for some $ r \in \mathbb{R}$. By using some lemmas about convergence, we can determine that for all $ \epsilon > 0$, there is an $ M$ such that

$ \prod_{n\equal{}M}^{\infty} e^{G_n} < 1 \plus{} \epsilon.$

In particular, $ e^{G_n} < 1 \plus{} \epsilon$ for all $ n \geq M$, so $ G_n$ gets very close to $ 0$. If we take $ \epsilon$ small enough, I think we can just pretend that $ e^{G_n}$ is $ 1 \plus{} G_n$ by Taylor series.

Since $ 0 \leq G_n < 1$, we must also have $ \prod_{n\equal{}M}^{\infty} (1 \minus{} G_n) \equal{} 0$ for every $ M$.

So now we simultaneously require that for all $ \epsilon > 0$, there is an $ M$ such that

$ \prod_{n\equal{}M}^{\infty} (1 \plus{} G_n) < 1 \plus{} \epsilon$

$ \prod_{n\equal{}M}^{\infty} (1 \minus{} G_n) \equal{} 0$

This seems contradictory to me, but I can't finish it.

Also, my analysis is pretty poor, so the chance of a logical mistake is pretty high.[/hide]

Kent Merryfield's posts [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1839316794&t=204755]here[/url] might be relevant.

[hide="Thoughts (?)"]Condition I & III suggest that $ \lim_{n \rightarrow \infty} G_{n} \equal{} 1$, but that would mean that $ \sum_{n \equal{} 1}^{\infty} G_{n}$ diverges, if $ \{G_{n}\}^{\infty}_{n \equal{} 1}$ is to be considered monotonic.[/hide]

... maybe I shouldn't trust my own thoughts at 2 a.m. :P

[quote="metafor"]Condition I & III suggest that $ \lim_{n \rightarrow \infty} G_{n} \equal{} 1$[/quote]

Not exactly.  We can take $ G_n \equal{} \frac{1}{n\plus{}1}$, for example, although it's still true that the sum diverges.

*Conditions (sorry about the typo) ...

Wait, but, if $ G_{n} \equal{} (n \plus{} 1)^{\minus{}1}$, wouldn't $ \prod^{\infty}_{n\equal{}1} \left(1 \minus{} \frac{1}{n \plus{} 1} \right) \equal{} \prod^{\infty}_{n \equal{} 1} \frac{n}{n\plus{}1}$ be nonzero?

(I considered that definition of the sequence, but I didn't give much thought to the possibility that the infinite product of $ n (n \plus{} 1)^{\minus{}1}$ was zero.)

${ \prod_{n=1}}^{N} \frac{n}{n+1} = \frac{1}{N}$, so the product "diverges to zero" as per Kent Merryfield's definitions.  (This is behavior that you don't find in sums - it is equivalent to saying that the sum of the logarithms diverges to $ -\infty$.)  In any case,

[hide="The solution"] [b]Case:[/b]  $ \sum |G_n|^2$ converges.  Then by the lemma given in the link, $ \prod (1 - G_n) = 0 \Leftrightarrow \sum G_n = \infty$; contradiction.

[b]Case:[/b]  $ \sum |G_n|^2$ diverges.  Then obviously $ \sum G_n$ diverges. [/hide]

Yes, I already understood the concept of "diverging to zero."

It is interesting that ${ \prod_{n=1}}^{\infty} \frac{n}{n + 1}$ is the 'inverse' (though this is somewhat inaccurate) of $ \prod^{\infty}_{n = 1} \left(1 + \frac{1}{n}\right) = \prod^{\infty}_{n = 1} \frac{n + 1}{n}$, which diverges to infinity.

Anyway, thanks for the clarification. :)


rotation
probability
analytic geometry
ratio
geometry
3D geometry
floor function

1. (1999 Team #4) In a triangle whose sides have lengths 6 cm, 50cm, and 52cm, what is the number of centimeters in the length of the altitude to the shortest side of the triangle?

2. (2000 Team #1) Right triange ABC with legs AB = 3 cm and Cb = 4 cm is rotated about one of its legs. What is the greatest possible number of cubic centimeters in the volume of the resulting solid? Express answer to nearest whole number.

3. (2000 Team #2) Erica flips a fair two-sided coin. Nate then flips the coin and wins if it matches Erica's flip. If it doesn't match, then Noah flips the coin and wins if it matches Nate's flip. If Noah doesn't win, then Erica flips and wins if it matches Noah's flip. The game continues until there is a winner. What is the probability that Nate wins the game? Express answer as a common fraction.

4. (2000 team #3) what is the sum of the digits needed to write al the whole numbers from 0 to 10000, inclusive?

5. (2002 Team #4) A circle of radius 5 with its center at (0,0) is drawn on a Cartesian coordinate system. How many lattice points (points with integer coordinates) lie within or on the circle?

Answers:
1. 48
2. 50
3. 4/7
4. 180001
5. 81

Solution to #3

Nate has a $ 1/2$ chance of winning. $ 1/8$ of the time, when everyone misses, Nate has a $ 1/2$ chance again. Through similar logic:

$ 1/2 \plus{} 1/(2*8) \plus{} 1/(2*8^2)...$

$ 1/2 \plus{} 1/16 \plus{} 1/128...$

That's an infinite geometric sequence with starting term $ 1/2$ and common ratio $ 1/8$

thus, its $ (1/2)/(1 \minus{} 1/8) \equal{} (1/2)/(7/8) \equal{} 8/14 \equal{} 4/7$

stop spamming the freakin forum mewto!!!

Solution to #1 [hide]using heron's formula you can find that the area of the triangle is 144. The altitude to the shortest is therefor 288/6 which equals 48.[/hide]

solution to #2 [hide]the greatest volume when a right triangle is rotated about one of its legs is the cone with radius of the longer leg. The volume of a cone is $ \frac {1}{3}h\pi{r^2}$ just substitute 4 for $ r$ and 3 for $ h$ and you get $ 16\pi$ which rounds to $ 50$
[/hide]

for #3, mewto's solution is correct

for number #4, i give a hint, find the sum of all the ones digits from 1-10000. That sum is the same for tens, hundreds and thousands digits!

BTW, is there a formula for number 5? I just did it by counting every single lattice point

I'm not even good at the lattice ones. I can't see how there is possibly 81 points. Could you possibly list them?

[quote="tinytim"]stop spamming the freakin forum mewto!!!

Solution to #1 [hide]using heron's formula you can find that the area of the triangle is 144. The altitude to the shortest is therefor 288/6 which equals 48.[/hide]

BTW, is there a formula for number 5? I just did it by counting every single lattice point[/quote]

Wait why do we divided 288 by 6 to get the shortest altitude?

[hide="hint"]the formula for the circle is $ x^2\plus{}y^2\equal{}25$[/hide]

#4 see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1056127&search_id=1069183997#1056127[/url]

#5
[hide]Draw quadrant I of the coordinate plane, and make a box that is 5 x 5 ($ \frac {1}{4}$ of the total area).  Since our radius is 5, we can have 3-4-5 triangles as shown in the diagram (blue and brown).  If we connect the points that are 5 units from the origin (red dots, red line), all lattice points inside will be separated from the out side.

Let's look at the lattice points outside the circle.  By counting we find there are 10 lattice points outside the circle.  Since this is $ \frac {1}{4}$ of the circle, we multiply this by four getting 40.

We should notice that the entire figure contains 121 lattice points.

Now subtract 40 from 121 and get 81, which, of course, is the answer.[/hide]

[quote="kl2836"]there's a thing called Gauss's Circle Problem that can calculate the number of lattice points, but i think its a little bit too advanced for mathcounts. is there a generalization of the formula?[/quote]

If you look it up on mathworld, the equation $ N(r)\equal{}1\plus{}4\lfloor r\rfloor\plus{}4\sum_{i\equal{}1}^{\lfloor r\rfloor}\left\lfloor \sqrt{r^2\minus{}i^2}\right\rfloor$ is pretty easy to both memorize and understand :)


[quote="314.math"]Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole
number of dollars to spend, and together they had $ \$56$. The absolute difference
between the amounts Ashley and Betty had to spend was $ \$19$. The absolute
difference between the amounts Betty and Carlos had was $ \$7$, between Carlos
and Dick was $ \$5$, between Dick and Elgin was $ \$4$, and between Elgin and
Ashley was $ \$11$. How much did Elgin have?

(A) $ \$6$ (B) $ \$7$ (C) $ \$8$ (D) $ \$9$ (E) $ \$10$

Thanks in advance![/quote]

[hide]Let $ A,B,C,D,E$ be the amount of money they have each. Notice that $ (A\minus{}B)\plus{}(B\minus{}C)\plus{}(C\minus{}D)\plus{}(D\minus{}E)\plus{}(E\minus{}A)\equal{}0$ and we know that $ A\minus{}B \equal{} \pm 19$, $ B\minus{}C \equal{} \pm 7$, $ C\minus{}D \equal{} \pm 5$, $ D\minus{}E \equal{} \pm 4$, and $ E\minus{}A \equal{} \pm 11$. Now note that $ A\minus{}B$ and $ E\minus{}A$ must have different signs (if they didn't, then their sum is $ \pm 30$, and that has too big of an absolute value for the other terms to compensate). So $ E \minus{} B \equal{} A\minus{}B \plus{} E\minus{}A \equal{} \pm 8$. Now it's not too hard to compute the values that $ \pm 7 \pm 5 \pm 4$ can take on (just 8 calculations). We get:

$ 7\plus{}5\minus{}4 \equal{} 8$ and $ \minus{}7\minus{}5\plus{}4 \equal{} \minus{}8$.

So we have two possibilities:
$ A\minus{}B \equal{} 19$, $ B\minus{}C \equal{} \minus{}7$, $ C\minus{}D \equal{} \minus{}5$, $ D\minus{}E \equal{} 4$, $ E\minus{}A \equal{} \minus{}11$
or the same thing with signs switched.

Now we use the fact that their sum is $ 56$. We can get everything in terms of $ E$:

$ A \equal{} E \plus{} 11$, $ B \equal{} E \minus{} 8$, $ C \equal{} E \minus{} 1$, $ D \equal{} E\plus{}4$ $ E \equal{} E$ (first case)

So their sum would be $ 5E \plus{} 6 \equal{} 56 \implies E \equal{} 10 \implies \textbf{E}$. The other case doesn't work, we get $ E$ to be non-integral.[/hide]

[hide="Initial Thoughts"]The first thing we notice with this problem is that we have two possibilities for each difference.  This implies that we will probably have to do case work.  Also, because we have a sum of 5 numbers, we will try to express 4 of them in terms of the last one.[/hide]
[hide="Solution"]For simplicity of the solution we will abbreviate each of the names with the first letter.  Because we want to find out what E is equal to, we will try to express the other 4 numbers in terms of E.  Let us first look at $ A\minus{}B$ and $ E\minus{}A$.  The difference between B and E is either 30 or 8.  If $ B\minus{}E\equal{}\pm30$ then $ B\minus{}C\plus{}C\minus{}D\plus{}D\minus{}E\equal{}\pm30$.  The maximum of $ (B\minus{}C)\plus{}(C\minus{}D)\plus{}(D\minus{}E)$ is $ \pm7\plus{}\pm5\plus{}\pm4\equal{}\pm16$ and therefore, $ B\minus{}E\not\equal{}\pm30$.  This implies $ B\minus{}E\equal{}\pm8$.
We know that $ B\minus{}C\equal{}\pm7, C\minus{}D\equal{}\pm5, D\minus{}E\equal{}\pm4$ and through simple calculations we see $ 7\plus{}5\minus{}4\equal{}8$ and $ \minus{}7\minus{}5\plus{}4\equal{}\minus{}8$.  From this we get that
$ A\minus{}B\equal{}19, B\minus{}C\equal{}\minus{}7, C\minus{}D\equal{}\minus{}5, D\minus{}E\equal{}4, E\minus{}A\equal{}\minus{}11$ or $ A\minus{}B\equal{}\minus{}19, B\minus{}C\equal{}7, C\minus{}D\equal{}5, D\minus{}E\equal{}\minus{}4, E\minus{}A\equal{}11$.
From this we get two equations in E.
$ A\equal{}E\plus{}11, B\equal{}E\minus{}8, C\equal{}E\minus{}1, D\equal{}E\plus{}4, E\equal{}E$ or $ A\equal{}E\minus{}11, B\equal{}E\plus{}8, C\equal{}E\plus{}1, D\equal{}E\minus{}4, E\equal{}E$
Adding these up we get $ 5E\plus{}6\equal{}56$ and $ 5E\minus{}6\equal{}56$.
From the first equation $ E\equal{}10$ and from the second $ E\equal{}12.4$.
Because E has to be a whole number, $ E\equal{}10 \implies \boxed{E}$[/hide]


Hello

I am taking an introductory Chinese class. It is safe to assume  I know close to nothing about mandarin Chinese.

When I found out tokenadult's post on the college forum that he knew Chinese, I pmed him a few questions but he said it would be better for me to post them here:

tokenadult suggested I listen to Internet radio from China and Taiwan for listening exposure. I have no money... Where can I listen to it for free? Also, I know there are different dialects... will that be a problem when I'm searching?

What advice would you have for someone in my position learning a new language?

What methods did you use to learn writing and speaking Chinese words? When I write a word in mandarin, I can't spell it out like in English... the words are like pictures to me.

What advice would you have for someone having trouble distinguishing things like:
the second tone vs. the third tone
in the pinyin writing, Zh vs. J, Q vs. ch, and so on...
I'm having trouble with these because I don't recognize English equivalents to these differences.

How can I type Chinese characters?

What general advice would you have about learning a new language? What general advice would you have to me, an English speaker in Virginia, about opening up my mind and learning to learn new grammar rules, dealing with no alphabet, and being converse in Chinese without translating from Chinese to English to meaning back to Chinese...?

I don't know about listening to Chinese radio. It's very easy to feel discouraged about listening if you don't even understand what you're listening to.

But here's a link to a Taiwanese soap opera, they speak in Standard Mandarin. And there are subtitles. Although it's still very easy to feel discouraged if you don't understand what they 're saying, at least you can kind of use the subtitles to kind of...get a feel for what's going on.

http://www.youtube.com/profile?user=aranelcharis

for learning characters,

your best bet is repetitions of writing the same character over and over again. Of course, it's important that with each repetition you stick to the correct strokes, otherwise it's counterproductive. Also, that's usually how characters are taught to native Chinese students.

for tones,

You should get a CD or some audio file of a native CHinese speaker speaking. And trying to duplicate those sounds. Once again repetition applies here as well. The point is to build "muscle memory".

typing chinese characters,

I only know how to do this for windows. click on "start" and go to "Control Panel". You should find "Date, Time, Language, and Regional Options". Then depending on what windows you ( I think) you should be able to customize the language on your computer to "Chinese(PRC)" which is the standard in China. You ll be able to switch back and forth between Chinese and English when writing a text document ( with some button clicking). Sorry I m being so nonspecific, I just don't know what version you have and whether the procedures differ between versions.

I hope this helps some.

good luck.


Solve the equation: $x=1+ln(\frac{x^{2}}{1+lnx})$

really? how did you solve it? show me what you did? :wink:


Given the positive integers ordered increasingly.

$ 5;\; 13;\; a;\; 24;\; 30$

Calculate the $ \overline {MA} \plus{} a$ if the variance of these data is $ 74,8$

Greetings  :D

What does $ \overline {MA}$ mean?
How can the variance be two numbers, or do you mean $ 74.8$?

$ \overline{MA}$ refers to the arithmetic mean.

Greetings  :D


the circles O and Q intersect at A and B. Q is a point on circle O and C is a point on circle Q. Lines AC and BC meet circle O at points E and D, respectively. Prove that lines QC and DE are perpendicular.


Let $ a,\ b,\ c$ be side lengths of a triangle. Prove that $ 3(a\plus{}b\plus{}c)^2\plus{}8(ab\plus{}bc\plus{}ca)>4(a\plus{}b\plus{}c)(\sqrt{ab\plus{}bc}\plus{}\sqrt{bc\plus{}ca}\plus{}\sqrt{ca\plus{}ab}).$

[quote="kunny"]Let $ a,\ b,\ c$ be side lengths of a triangle. Prove that $ 3(a \plus{} b \plus{} c)^2 \plus{} 8(ab \plus{} bc \plus{} ca) > 4(a \plus{} b \plus{} c)(\sqrt {ab \plus{} bc} \plus{} \sqrt {bc \plus{} ca} \plus{} \sqrt {ca \plus{} ab}).$[/quote]
If $ a,b,c$ are not side lengths of triangle then problem is true.
Because $ (\sqrt {ab \plus{} bc} \plus{} \sqrt {bc \plus{} ca} \plus{} \sqrt {ca \plus{} ab})\leq \sqrt {6\sum ab}$.
We need prove:$ 3X^2 \plus{} 8Y^2 \minus{} 4\sqrt {6} XY\geq 0$. Where $ X \equal{} a \plus{} b \plus{} c ;Y \equal{} \sqrt {\sum ab}$
Ineq is true.

[quote="kunny"]When does the equality hold?[/quote]
I do not think this result have to explain.
Because:equallity where ineq :$ (\sqrt {ab \plus{} bc} \plus{} \sqrt {bc \plus{} ca} \plus{} \sqrt {ca \plus{} ab})\leq\sqrt {6\sum ab}$ be $ a \equal{} b \equal{} c$.
When :$ 3X^{2} \plus{} 8Y^{2} \minus{} 4\sqrt {6}XY > 0$.

[quote="kunny"]Here is my solution:

$ \sum (\sqrt {a \plus{} b} \minus{} \sqrt {c})^4 > 0.$ Done![/quote]

It don't need a;b;c  be side lengths of a triangle


I was going to post a solution by induction, and then I saw your post KWMAN... nice solution :)

haha much brilliance. Fantastic simplicity is always nice.

:huuh:

$\sum_{i=0}^{n} i^3$ is the number of rectangles in a n*n square which is $(\frac{n(n+1)}{2})^2$. also $\binom{n+1}{2}=\frac{n(n+1)}{2}\mathbb{QED}$

they have the visual proof on the sidebar


floor function
real analysis
real analysis unsolved

Prove that there exists a positive integer $n$ such that
$$\left\lfloor \frac{n+2^{\sqrt{2}}}{\sqrt{2}}\right\rfloor-\left\lfloor \frac{n-3}{\sqrt{2}}\right\rfloor=5,$$
and find the smallest value of $n$.


integration
limit
algebra unsolved
algebra

Let $ (x_n)_n$ be the real sequence $ x_0\equal{}1,x_1\equal{}a,x_{n\plus{}2}\equal{}x_{n\plus{}1}\plus{}\dfrac{x_n}{n\plus{}1}$.
Find $ a$ s.t. $ x_n\rightarrow{0}$.

Well, I got $ a \equal{} 2 \minus{} e$. This is what I did:

Denote:

$ F(z) \equal{} \sum^{\infty}_{n \equal{} 0}x_n z^n$

also, suppose that $ F^{(n)}(0)$ exists, that $ F$ is differentiable (and integrable) and bounded (not pretty sure if this is the right term).

Multiplying the recursive equation with $ z^n$ and summing up to $ \infty$ gives us three parts that can be rewritten as:

$ \sum^{\infty}_{n \equal{} 0}x_{n \plus{} 2} z^n \equal{} \frac {1}{z^2}F(z) \minus{} \frac {1}{z^2} \minus{} \frac {a}{z}$

$ \sum^{\infty}_{n \equal{} 0}x_{n \plus{} 1} z^n \equal{} \frac {1}{z}F(z) \minus{} \frac {1}{z}$

$ \sum^{\infty}_{n \equal{} 0}\frac {x_n}{n \plus{} 1} z^n \equal{} \frac {1}{z}\int^z_b F(\psi)\, d\psi$

Plugging that in the equation, and rewriting it we get:

$ \frac {1}{z}F(z) \minus{} \frac {1}{z} \minus{} a \equal{} F(z) \minus{} 1 \plus{} \int^z_b F(\psi)\, d\psi$

Now, after we derivate both sides with respect to $ z$ and find the solution to the differential equation (we have $ f'(0) \equal{} a$ as a beginning condition), we get:

$ F(z) \equal{} \frac {1}{(1 \minus{} z)^2} \plus{} (a \minus{} 2)\frac {z e^{ \minus{} z}}{(1 \minus{} z)^2}$

Now, from the definition of $ F(z)$ we can denote the following:

$ \frac {1}{n!}\frac {d^n F(0)}{dz^n} \equal{} x_n$

Also, can evaluate that:

$ \frac {d^n}{dz^n}\left(\frac {1}{(1 \minus{} z)^2}\right) \equal{} (n \plus{} 1)!$

and also:

$ \frac {d^n}{dz^n}\left(\frac {z e^{ \minus{} z}}{(1 \minus{} z)^2}\right) \equal{} (a \minus{} 2)(n \plus{} 1)!\sum^{n \plus{} 1}_{k \equal{} 0}\frac {( \minus{} 1)^k}{k!}$

Now, from this we have:

$ x_n \equal{} n \left(1 \plus{} (a \minus{} 2)\sum^{n \plus{} 1}_{k \equal{} 0}\frac {( \minus{} 1)^k}{k!}\right)$

also, we have:

$ \lim_{n \rightarrow \infty} x_n \equal{} 0$

so for large enough $ n$ we have:

$ 1 \plus{} \frac {(a \minus{} 2)}{e} \equal{} 0$

and from here:

$ a \equal{} 2 \minus{} e$

Hi Milin,
 a little mistake: $ x_n\equal{}(n\plus{}1)(\cdots)$.


superior algebra
superior algebra solved

How do you describe $ Spec(\mathbb{Z}[x])$

[hide]I got
$ (0)\cup \{(f)\}$, f is irreducible in $ \mathbb{Z}[x]\cup \{(g,p)\},$ $ p$ prime and $ g$ irreducible in $ \mathbb{Z}/\mathbb{Z}p[x]$
I'm waiting for a confirmation.[/hide]

Yes, see http://www.mathlinks.ro/Forum/viewtopic.php?t=72993 .


[quote="xzlbq"][size=150]Let $ x,y,z \geq 0$,prove that:

$ (\frac {z}{x} \plus{} 1)(1 \plus{} \frac {y}{x}) \geq (\sqrt {\frac {y}{x}} \plus{} \sqrt {\frac {z}{x}})^2$

BQ[/size][/quote]

$ \iff (x\minus{}\sqrt{yz})^2 \ge 0.$  :wink:

[quote="xzlbq"][size=150]So easy,I have to replace it:

Let ABC be a triangle,prove that:

$ r_a \plus{} r_b \plus{} r_c\geq w_aw_bw_c(\frac {1}{w_a^2} \plus{} \frac {1}{w_b^2} \plus{} \frac {1}{w_c^2}) \geq \sqrt {3}s.$[/size][/quote]

$ r_a \plus{} r_b \plus{} r_c\geq w_aw_bw_c(\frac {1}{w_a^2} \plus{} \frac {1}{w_b^2} \plus{} \frac {1}{w_c^2})$

$ \Longleftrightarrow (r_a \plus{} r_b \plus{} r_c)^2\geq (w_aw_bw_c(\frac {1}{w_a^2} \plus{} \frac {1}{w_b^2} \plus{} \frac {1}{w_c^2}))^2...(*)$

substitution $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$

$ (*)\Longleftrightarrow \sum_{sym} (x^3y^2 \minus{} x^2y^2z)\geq 0$

from Muirhead theorem the inequality holds

$ w_aw_bw_c(\frac {1}{w_a^2} \plus{} \frac {1}{w_b^2} \plus{} \frac {1}{w_c^2}) \geq \sqrt {3}s$

I can prove it by Muirhead theorem.
 :D


Let $ a,b,c$ be positive real numbers. For $ p\ge \frac{15+\sqrt{97}}{8}$ (the greatest root of the equation $ 4p^{2}-15p+8=0$), prove that

\[ \frac{1}{pa^{2}+bc}+\frac{1}{pb^{2}+ca}+\frac{1}{pc^{2}+ab}\ge \frac{27}{(p+1)(a+b+c)^{2}}.\]

This solution is really hard to understand,  :(

Is there another solution? THis problem is the general problem of many hard inequalities VASC post here.


[quote="Harry Potter"]There are 2n point on the circle .How many the way to add 2n point by n chord don't cut ?[/quote]
Based on what I understood from the problem, the only answer I can think of is "Yes!"

[b]Solution.[/b] Suppose that you have a graph with $2n$ vertices satisfying the conditions of the problem, that is, $n$ non-intersecting chords connecting pairs of points.  Go around this circular graph and label the first end of all chords by $r$ and the second end by $u$.  Then you have a bijection between a so-called "good" graph and Catalan numbers.  The answer is thus $\frac{1}{n+1}\binom{2n}{n}$. $\Box$


integration
trigonometry
analytic geometry
function
limit
calculus
complex analysis

$\int_{-\infty}^\infty \frac{x^{4}}{1+x^{8}}dx$. I was trying to use the theorem that says this is just $2\pi i$ times the sum of the residues in the upper half plane (of which there are four). This is some hefty factoring and residue calculating for prb. 1 in the section on "application of the residue theorem."

Also, the answer is given as $\frac{\pi}{4}\left(\sin\left(\frac{3}{8}\pi\right)\right)^{-1}$. I don't see how a sine is going to figure in doing it my way. Is there some nice substitution I'm missing? Just want to make sure before I go through the trouble of all that calculation.

That $\sin$ comes from where the poles are; their coordinates are trig functions of multiples of $\frac{\pi}8$.

The poles are very similar to each other ($x^{4}$ varies only in sign), which makes the computation easier.

My first approach would be to divide cut the interval in half (integrate from 0 to infinite).  Then, I'd let x = exp(y).  You should be able to use rectangular contours to figure it out.  Just a guess, but I've used that substitution frequently for integrals of that sort.

There's no need to do that. We have rapid decay at $ \infty$, so closing things off with a big semicircle works and is simpler.

Divide the integral in half (even function). Then integrate over a "pie slice" contour: out on the real axis, partway around a large circular arc, then back in on the line $ x\equal{}y,$ which we parameterize (backwards) as $ z\equal{}e^{\pi i/4}t.$ Note that on that last line, $ dz\equal{}e^{\pi i/4}\,dt,$ $ z^4\equal{}\minus{}t^4,$ and $ 1\plus{}z^8\equal{}1\plus{}t^8.$

The computational advantage of this is that we only need to use the first of Kouichi Nakagawa's four residue computations. (I'll just quote that computation.)

We get that

$ \left(1\plus{}e^{\pi i/4}\right)\int_0^{\infty}\frac{x^4}{1\plus{}x^8}\,dx\equal{} 2\pi i\cdot \frac18\,e^{3\pi i/8}.$

Finishing from there is an algebraic computation (which will still take some work).


$e$=$lim (1+\frac{1}{n})^{n}$=$lim\underbrace{(1+\frac{1}{n})(1+\frac{1}{n})\cdots(1+\frac{1}{n})}_{n times}$
=$\underbrace{lim(1+\frac{1}{n})lim(1+\frac{1}{n})\cdots\lim(1+\frac{1}{n})}_{n times}$

=$\underbrace{1.1\cdots1}_{n times}=1$

what is wrong?

What if $n = \pi$.. how can you multiply something by itself $\pi$ times?


I made a similar post last year:

I am a U.S. citizen in Taiwan.  My AMC/AIME scores are as follows:

AMC 12A: 145.5
AIME II: 14

Index: 285.5.

My name was not posted on the USAMO qualifier website.  I have already emailed and PMmed the AMC staff, but I have received no responses.  What do I do?

It's probably an issue relating to your residence.  If you had not already emailed, I would tell you to do that at some point (though maybe not until Friday).

It may depend on the time that you emailed them.  If I were working for the AMC, I wouldn't try to respond to a 12:00 am email after a long day.

Thanks. Last year, I remember that all my school arrangements were last minute since I couldn't confirm my qualification for a while.

Try calling them directly. They can't ignore that. =D
Also, it might have to do with the fact that you aren't a US resident, though you are a citizen.


$ABC$ is a triangle such that point $D$ and $E$ lie on $AB$ and $AC$ respectively . If $DE$ is parallel to $BC$ and the incircle is also tangent to $DE$ , prove that

$AB+BC+AC \geq 8DE$

[hide="Here is a proof of this nice and elementary geometrical inequality."]$2S=\boxed {ah_a=2pr}$, $\frac{DE}{BC}=\frac{h_a-2r}{h_a}$ $\Longrightarrow$ $DE=\frac{ah_a-2ar}{h_a}=$ $\frac{a(ah_a-2ar)}{ah_a}=$
$\frac{a(2pr-2ar)}{2pr}=$ $\frac{2ar(p-a)}{2pr}=\frac{a(p-a)}{p}$ $\Longrightarrow$ $\boxed {DE=\frac{a(p-a)}{p}}$. Therefore,
$\boxed {AB+BC+CA\ge 8\cdot DE}$ $\Longleftrightarrow$ $p\ge \frac{4a(p-a)}{p}$ $\Longleftrightarrow$
$(a+b+c)^2\ge 8a(b+c-a)$ $\Longleftrightarrow$ $\left[(b+c-a)+2a\right]^2\ge 8a(b+c-a)$ $\Longleftrightarrow$
$\left[ (b+c-a)-2a\right]^2\ge 0$ $\Longleftrightarrow$ $\boxed {(b+c-3a)^2\ge 0}$ (with equality iff $b+c=3a$).

[b]Remark.[/b] Denote the semiperimeter $p(XYZ)$ of the triangle $XYZ$. If the incircle of the triangle $ABC$ touches the side $AB$ in the point $F$, then $AF=p(ADE)$ and $\triangle ADE\sim\triangle ABC$ $\Longrightarrow$ $\frac{DE}{a}=$ $\frac{p(ADE)}{p(ABC)}=$ $\frac{AF}{p}=$ $\frac{p-a}{p}$ $\Longrightarrow$ $\boxed {DE=\frac{a(p-a)}{p}}\ .$[/hide]

thanks for ur beautiful solution Sir Virgil Nicula .

I just wanna ask how to prove that $AF=p(ADE)$  :)


trigonometry
calculus
integration
calculus computations

let

$\frac{dx}{dt}= 3+\cos{t^{2}}$

at $t=2$, the $x=1$

find $x$ when $t=4$

$\cos (t^{2})$?

Are you sure it's so simple?

I just need a numerical method of solving this. It doesn't need to be exact.
I'm not sure how to go after it.

Would Euler's method come into play here?

It certainly would.  $x$ is not expressible in terms of elementary functions.  See [url=http://mathworld.wolfram.com/FresnelIntegrals.html]Fresnel Integrals[/url].

Notice that as differential equations go, this is just an antiderivative problem.

We can write $x(t)=1+\int_{2}^{4}\cos t^{2}\,dt$

Now, that's an integral with no closed-form antiderivative, but you can certainly use any method of numerical integration that you know.

For my own part, I note that that's an oscillatory integrand. To improve my accuracy, I think I'd like to integrate by parts first:

$\int_{2}^{4}\cos t^{2}\,dt=\left.\frac1{2t}\sin t^{2}\right|_{2}^{4}-\int_{2}^{4}\frac1{2t^{2}}\sin t^{2}\,dt$

$=\left.\frac1{2t}\sin t^{2}\right|_{2}^{4}+\left.\frac1{4t^{2}}\cos t^{2}\right|_{2}^{4}+\int_{2}^{4}\frac1{2t^{3}}\cos t^{2}\,dt$

Maybe even go another round or two on the integration by parts, then use your method of numerical integration on the integral that is left.

----

Using Euler's method on this differential equation that's actually an integration problem would be the same as using left-endpoint Riemann sums. As numerical methods, they both have what is called a first-order local truncation error. In practice, that means that to get twice the accuracy, you need about twice as many steps, and to gain one more decimal place, you need 10 times the steps.

It's worth investing the time to learn about approximating methods with higher-order local truncation errors. (And if you apply fourth-order Runge Kutta to this problem which is actually an integral, that would amount to using Simpson's Rule on the integral)

----

Tinkering around on a spreadsheet (using Romberg integration, if you have to ask) I get

$x(4)\approx 1.13299887.$


$ M,N,P$ are the midpoints of the sides $ AB,BC,CA$ of the triangle $ ABC.$ Consider $ E$ on $ NP$ so that $ NP\equal{}PE;$ ($ P$ is between $ N$ and $ E$) and the point $ D$ on $ CM$ so that $ CM\equal{}MD$; ($ M$ is between $ C$ and $ D$). Prove that: $ AD\equal{}2AE.$

:rotfl: we have $ AE\equal{}NC\equal{}1/2BC$ $ and$ $ AD\equal{}BC.$ :rotfl:

we can see that ADBC and AECN are parellelograms and since 2NC=BC hence 2AE=AD


${f=\sqrt{x z+\sqrt{x y}}+\sqrt{y z+\sqrt{x y}}+\sqrt{x y+\sqrt{x z}}+}$
${\sqrt{y z+\sqrt{x z}}+\sqrt{x y+\sqrt{y z}}+\sqrt{x z+\sqrt{y z}}}$

${x+y+z=\theta }$
${(x,y,z,\theta )>0}$

proove that

${f\leq 2 \sqrt{\theta (\theta+3)}}$

[quote="Thales418"]yes yes

but solve this[/quote]
Ok, We have $f^{2}\leq 6(2(xy+yz+zx)+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}))$ and $xy+yz+zx\leq\frac{(x+y+z)^{2}}{3}=\frac{\theta^{3}}{3}$ and $(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})^{2}\leq 3(xy+yz+zx)\leq (x+y+z)^{2}=\theta^{2}$. Done :D

[quote="Thales418"]u did not proove that

${f\leq 2 \sqrt{\theta (\theta+3)}}$[/quote]
Why you not try read carefully my post?


Mine is

The Raven
[hide]
Once upon a midnight dreary, while I pondered, weak and weary,
Over many a quaint and curious volume of forgotten lore,
While I nodded, nearly napping, suddenly there came a tapping,
As of someone gently rapping, rapping at my chamber door.
" 'Tis some visitor," I muttered, "tapping at my chamber door;
Only this, and nothing more."

Ah, distinctly I remember, it was in the bleak December,
And each separate dying ember wrought its ghost upon the floor.
Eagerly I wished the morrow; vainly I had sought to borrow
From my books surcease of sorrow, sorrow for the lost Lenore,.
For the rare and radiant maiden whom the angels name Lenore,
Nameless here forevermore.

And the silken sad uncertain rustling of each purple curtain
Thrilled me---filled me with fantastic terrors never felt before;
So that now, to still the beating of my heart, I stood repeating,
" 'Tis some visitor entreating entrance at my chamber door,
Some late visitor entreating entrance at my chamber door.
This it is, and nothing more."

Presently my soul grew stronger; hesitating then no longer,
"Sir," said I, "or madam, truly your forgiveness I implore;
But the fact is, I was napping, and so gently you came rapping,
And so faintly you came tapping, tapping at my chamber door,
That I scarce was sure I heard you." Here I opened wide the door;---
Darkness there, and nothing more.

Deep into the darkness peering, long I stood there, wondering, fearing
Doubting, dreaming dreams no mortals ever dared to dream before;
But the silence was unbroken, and the stillness gave no token,
And the only word there spoken was the whispered word,
Lenore?, This I whispered, and an echo murmured back the word,
"Lenore!" Merely this, and nothing more.

Back into the chamber turning, all my soul within me burning,
Soon again I heard a tapping, something louder than before,
"Surely," said I, "surely, that is something at my window lattice.
Let me see, then, what thereat is, and this mystery explore.
Let my heart be still a moment, and this mystery explore.
" 'Tis the wind, and nothing more."

Open here I flung the shutter, when, with many a flirt and flutter,
In there stepped a stately raven, of the saintly days of yore.
Not the least obeisance made he; not a minute stopped or stayed he;
But with mien of lord or lady, perched above my chamber door.
Perched upon a bust of Pallas, just above my chamber door,
Perched, and sat, and nothing more.

Then this ebony bird beguiling my sad fancy into smiling,
By the grave and stern decorum of the countenance it wore,
"Though thy crest be shorn and shaven thou," I said, "art sure no craven,
Ghastly, grim, and ancient raven, wandering from the nightly shore.
Tell me what the lordly name is on the Night's Plutonian shore."
Quoth the raven, "Nevermore."

Much I marvelled this ungainly fowl to hear discourse so plainly,
Though its answer little meaning, little relevancy bore;
For we cannot help agreeing that no living human being
Ever yet was blessed with seeing bird above his chamber door,
Bird or beast upon the sculptured bust above his chamber door,
With such name as "Nevermore."

But the raven, sitting lonely on that placid bust, spoke only
That one word, as if his soul in that one word he did outpour.
Nothing further then he uttered; not a feather then he fluttered;
Till I scarcely more than muttered, "Other friends have flown before;
On the morrow he will leave me, as my hopes have flown before."
Then the bird said, "Nevermore."

Startled at the stillness broken by reply so aptly spoken,
"Doubtless," said I, "what it utters is its only stock and store,
Caught from some unhappy master, whom unmerciful disaster
Followed fast and followed faster, till his songs one burden bore,---
Till the dirges of his hope that melancholy burden bore
Of "Never---nevermore."

But the raven still beguiling all my sad soul into smiling,
Straight I wheeled a cushioned seat in front of bird, and bust and door;
Then, upon the velvet sinking, I betook myself to linking
Fancy unto fancy, thinking what this ominous bird of yore --
What this grim, ungainly, ghastly, gaunt and ominous bird of yore
Meant in croaking "Nevermore."

Thus I sat engaged in guessing, but no syllable expressing
To the fowl, whose fiery eyes now burned into my bosom's core;
This and more I sat divining, with my head at ease reclining
On the cushion's velvet lining that the lamplight gloated o'er,
But whose velvet violet lining with the lamplight gloating o'er
She shall press, ah, nevermore!

Then, methought, the air grew denser, perfumed from an unseen censer
Swung by seraphim whose footfalls tinkled on the tufted floor.
"Wretch," I cried, "thy God hath lent thee -- by these angels he hath
Sent thee respite---respite and nepenthe from thy memories of Lenore!
Quaff, O quaff this kind nepenthe, and forget this lost Lenore!"
Quoth the raven, "Nevermore!"

"Prophet!" said I, "thing of evil!--prophet still, if bird or devil!
Whether tempter sent, or whether tempest tossed thee here ashore,
Desolate, yet all undaunted, on this desert land enchanted--
On this home by horror haunted--tell me truly, I implore:
Is there--is there balm in Gilead?--tell me--tell me I implore!"
Quoth the raven, "Nevermore."

"Prophet!" said I, "thing of evil--prophet still, if bird or devil!
By that heaven that bends above us--by that God we both adore--
Tell this soul with sorrow laden, if, within the distant Aidenn,
It shall clasp a sainted maiden, whom the angels name Lenore---
Clasp a rare and radiant maiden, whom the angels name Lenore?
Quoth the raven, "Nevermore."

"Be that word our sign of parting, bird or fiend!" I shrieked, upstarting--
"Get thee back into the tempest and the Night's Plutonian shore!
Leave no black plume as a token of that lie thy soul hath spoken!
Leave my loneliness unbroken! -- quit the bust above my door!
Take thy beak from out my heart, and take thy form from off my door!"
Quoth the raven, "Nevermore."

And the raven, never flitting, still is sitting, still is sitting
On the pallid bust of Pallas just above my chamber door;
And his eyes have all the seeming of a demon's that is dreaming.
And the lamplight o'er him streaming throws his shadow on the floor;
And my soul from out that shadow that lies floating on the floor
Shall be lifted---nevermore![/hide]

Sorry if this post is too long... I hope you don't mind... I'll edit it if you want me to.

[color=green][size=42]Klebian's note to all: Please put the poems in hide tags.[/size][/color]

Here's one by Lewis Carol

The Mad Gardener's Song

[hide]He thought he saw an Elephant,
That practised on a fife:
He looked again, and found it was
A letter from his wife.
'At length I realise,' he said,
The bitterness of Life!'

He thought he saw a Buffalo
Upon the chimney-piece:
He looked again, and found it was
His Sister's Husband's Niece.
'Unless you leave this house,' he said,
"I'll send for the Police!'

He thought he saw a Rattlesnake
That questioned him in Greek:
He looked again, and found it was
The Middle of Next Week.
'The one thing I regret,' he said,
'Is that it cannot speak!'

He thought he saw a Banker's Clerk
Descending from the bus:
He looked again, and found it was
A Hippopotamus.
'If this should stay to dine,' he said,
'There won't be much for us!'

He thought he saw a Kangaroo
That worked a coffee-mill:
He looked again, and found it was
A Vegetable-Pill.
'Were I to swallow this,' he said,
'I should be very ill!'

He thought he saw a Coach-and-Four
That stood beside his bed:
He looked again, and found it was
A Bear without a Head.
'Poor thing,' he said, 'poor silly thing!
It's waiting to be fed!'

He thought he saw an Albatross
That fluttered round the lamp:
He looked again, and found it was
A Penny-Postage Stamp.
'You'd best be getting home,' he said:
'The nights are very damp!'

He thought he saw a Garden-Door
That opened with a key:
He looked again, and found it was
A Double Rule of Three:
'And all its mystery,' he said,
'Is clear as day to me!'

He thought he saw a Argument
That proved he was the Pope:
He looked again, and found it was
A Bar of Mottled Soap.
'A fact so dread,' he faintly said,
'Extinguishes all hope!'[/hide]

:rotfl:  :rotfl:

Where did you find that one?

It's a poem by my favorite poet.
He wrote a bunch of really weird/funny poems.
Here's another one:

The Walrus and the Carpenter

[hide]The sun was shining on the sea,
Shining with all his might:
He did his very best to make
The billows smooth and bright--
And this was odd, because it was
The middle of the night.

The moon was shining sulkily,
Because she thought the sun
Had got no business to be there
After the day was done--
"It's very rude of him," she said,
"To come and spoil the fun!"

The sea was wet as wet could be,
The sands were dry as dry.
You could not see a cloud, because
No cloud was in the sky:
No birds were flying overhead--
There were no birds to fly.

The Walrus and the Carpenter
Were walking close at hand;
They wept like anything to see
Such quantities of sand:
"If this were only cleared away,"
They said, "it would be grand!"

"If seven maids with seven mops
Swept it for half a year.
Do you suppose," the Walrus said,
"That they could get it clear?"
"I doubt it," said the Carpenter,
And shed a bitter tear.

"O Oysters, come and walk with us!"
The Walrus did beseech.
"A pleasant walk, a pleasant talk,
Along the briny beach:
We cannot do with more than four,
To give a hand to each."

The eldest Oyster looked at him,
But never a word he said:
The eldest Oyster winked his eye,
And shook his heavy head--
Meaning to say he did not choose
To leave the oyster-bed.

But four young Oysters hurried up,
All eager for the treat:
Their coats were brushed, their faces washed,
Their shoes were clean and neat--
And this was odd, because, you know,
They hadn't any feet.

Four other Oysters followed them,
And yet another four;
And thick and fast they came at last,
And more, and more, and more--
All hopping through the frothy waves,
And scrambling to the shore.

The Walrus and the Carpenter
Walked on a mile or so,
And then they rested on a rock
Conveniently low:
And all the little Oysters stood
And waited in a row.

"The time has come," the Walrus said,
"To talk of many things:
Of shoes--and ships--and sealing-wax--
Of cabbages--and kings--
And why the sea is boiling hot--
And whether pigs have wings."

"But wait a bit," the Oysters cried,
"Before we have our chat;
For some of us are out of breath,
And all of us are fat!"
"No hurry!" said the Carpenter.
They thanked him much for that.

"A loaf of bread," the Walrus said,
"Is what we chiefly need:
Pepper and vinegar besides
Are very good indeed--
Now if you're ready, Oysters dear,
We can begin to feed."

"But not on us!" the Oysters cried,
Turning a little blue.
"After such kindness, that would be
A dismal thing to do!"
"The night is fine," the Walrus said.
"Do you admire the view?

"It was so kind of you to come!
And you are very nice!"
The Carpenter said nothing but
"Cut us another slice:
I wish you were not quite so deaf--
I've had to ask you twice!"

"It seems a shame," the Walrus said,
"To play them such a trick,
After we've brought them out so far,
And made them trot so quick!"
The Carpenter said nothing but
"The butter's spread too thick!"

"I weep for you," the Walrus said:
"I deeply sympathize."
With sobs and tears he sorted out
Those of the largest size,
Holding his pocket-handkerchief
Before his streaming eyes.

"O Oysters," said the Carpenter,
"You've had a pleasant run!
Shall we be trotting home again?'
But answer came there none--
And this was scarcely odd, because
They'd eaten every one.[/hide]

My favorite Carroll poem is "Jabberwocky."  My favorite poem of all time, however, is probably [url=http://etext.virginia.edu/stc/Coleridge/poems/Rime_Ancient_Mariner.html]The Rime of the Ancient Mariner[/url], by Samuel Taylor Coleridge (Author of [url=http://etext.virginia.edu/stc/Coleridge/ascii_files/geometry_poem_letter.html]A Mathematical Proposition.[/url]

[hide]Homework! Oh, Homework!
I hate you! You stink!
I wish I could wash you away in the sink,
if only a bomb
would explode you to bits.
Homework! Oh, homework!
You're giving me fits.

I'd rather take baths
with a man-eating shark,
or wrestle a lion
alone in the dark,
eat spinach and liver,
pet ten porcupines,
than tackle the homework,
my teacher assigns.

Homework! Oh, homework!
you're last on my list,
I simple can't see
why you even exist,
if you just disappeared
it would tickle me pink.
Homework! Oh, homework!
I hate you! You stink!
[/hide]
Jack Prelutsky

hahaha!!! Homework....

do you like any of the classics, like the Bells, The Lady of Shalott, etc.

I like the poems by other people...(1/2 pixieman mostly) on english.unitecnology.ac.nz. It is sad though because everyone who used to be on disappeared and i'm not active anymore...


trigonometry calculus function quadratics inequalities algebra polynomial

College Algebra
Plane Trig
Finite Math with application
Elements of Calculus with application
Calculus 1
Calculus 2
Calculus 3
Fundamental Math

What on earth is College Algebra? And fundamental math?  :maybe:

You'll have to say a bit more about the classes so that we can understand what they are.

Clearly, the three calculus classes are in order. Elements of calculus sounds like it is either a replacement for one of the earlier calc classes or something to do before or after calc 1 (again, I can't tell because the name could mean anything). You need to know plane trig before you can do calculus.

If College algebra is referring to Algebra II or precalc, take it before calculus.

I think College Algebra refers to algebra of the abstract type :wink: Not the high school Algebra II. I haven't a clue what Finite or Fundamental Math are. Plane trig sounds simple. My guess is

Plane Trig
Calculus I, II, III
College Algebra

The other three courses sound like fillers for liberal arts people  who want to have a little background but don't want to get too involved.

College Algebra is a course that falls between Algebra II and Precalculus. It is usually a course that re-iterates a lot of the material of Algebra II (linear functions, quadratic functions, inequalities, factoring, polynomials, rules of exponents, exponential functions) and introduces some of the material of precalculus (logarithms, graphs of functions, and sometimes some trigonometry).

As to the original question, difficulty is relative to where [i]you[/i] are in your mathematical development. None of the classes are intrinsically harder than any other. To determine which is the appropriate class for you, you would need to talk to someone who understands your math proficiency up to this point and knows your reasons for wanting to study mathematics.

By the way, Finite Math is basically a non-calculus math class for business and social science majors. It typically covers matrices, linear inequalities, linear programming, sets and counting, probability, statistics, finance, game theory, and boolean logic.

The "difficulty" of a math course is largely based on the things such as the teacher, one's own learning style, what types of math one is better at, etc. For example, I found calculus to be easier than algebra II (not that I found that difficult in the least). The point is, you can't really generalize as to which one is "harder" than the others.


If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is
[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0);
draw(P--C--D^^A--B^^Circle(Q,1));
label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, N);
label("$D$", D, S);
label("$P$", P, S);
label("$Q$", Q, SE);
label("$60^\circ$", P+0.0.5*dir(30), dir(30));[/asy]
$ \textbf{(A)} \ \frac{\sqrt{3}}{2} \qquad \textbf{(B)} \ \frac{\sqrt{3}}{3} \qquad \textbf{(C)} \ \frac{\sqrt{2}}{2} \qquad \textbf{(D)} \ \frac12 \qquad \textbf{(E)} \ \frac23 $

[hide]AQ=CQ (because they're both radii). Triangle PQC is a 30-60-90 triangle, so $\frac{PQ}{CQ}=\frac{1}{\sqrt{3}}=\frac{PQ}{AQ}$, which makes the answer $\boxed{B}$, $\frac{\sqrt{3}}{3}$. [/hide]

-BLt

[quote="4everwise"]If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is
$\text{(A)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=464[/img][/quote]

[hide](B)
we are looking for $\frac{PQ}{AQ}$ so if CQ=X, then PQ=$\frac{X}{\sqrt{3}}$, and AQ=X, so $\frac{PQ}{AQ}$=$\frac{\frac{X}{\sqrt{3}}}{X}$=$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$[/hide]
edit: same thing  :P

[hide]We find that $m\angle PCQ=30^\circ$. Because it is a $30^\circ-60^\circ-90^\circ$ right triangle, we can let $PQ=x$ and $CQ=AQ=x\sqrt{3}$. Thus, $\frac{PQ}{AQ}=\frac{x}{x\sqrt{3}}=\frac{\sqrt{3}}{3}\Rightarrow\boxed{B}$.[/hide]

[hide="Solution"]$\text{Answer: (B)}$

We first express this ratio in terms of things we know, then solve. Because $AQ=CQ$ as radii of the circle,

$\frac{PQ}{AQ}=\frac{PQ}{CQ}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$[/hide]

[hide]Set the radius to equal $\sqrt{3}PQ$

$\frac{PQ}{\sqrt{3}PQ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} = \boxed{B}$[/hide]


Determine the function f(x) such that f(x) is a quadratic polynomial and satisfies the condition f(x+2)=f(x)+x+2 for all x, and f(2)=3

[hide]
Let $ f(x)=ax^{2}+bx+c$
We have $ ax^{2}+(4a+b)x+4a+2b=ax^{2}+(b+1)x+2$
$ \Rightarrow\left\{\begin{array}{ll}4a=1\\4a+2b=2\end{array}\right.\Rightarrow a=\frac{1}{4},b=\frac{1}{2}$
$ f(2)=3\Rightarrow c=1$
[/hide]


MATHCOUNTS
ratio
geometry
3D geometry
sphere

[hide]let x = radius of ballon, y = radius of hemi

(4 :pi: x^3)/3 = (4:pi:y^3)/6

24:pi:x^3 = 12:pi:y^3

2x^3 = y^3

x/y = 1:(cuberoot)2[/hide]

no, find your mistake and justify every step you took. youll kick yourself for what you did wrong too.

Theres a mistake? (I assume : was a typo for / but..)

Not much of a mistake. Teaches me to read the last sentence! AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

he didnt put it in simplest radical form, and on mathcounts his answer would be wrong. but he still has not justified or corrected his answer

Syntax Error wrote:a water filled spherical balloon lands on a sidewalk, momentarily flattening to a hemisphere.  what is the ratio of the radius of the spherical balloon to the radius of the hemisphere? express your answer as a common fraction in simplest radical form.

[hide]Volume of a sphere = (4/3):pi:*r^3

Volume of a hemisphere = (1/2)Volume of sphere = (2/3):pi:*r^3

Let the spherical radius of the ballon be x and the hemispherical radius be y.

(4/3):pi:*x^3 = (2/3):pi:*y^3

2*x^3 = y^3

21/3x = y

x/y = x/21/3x = 1/21/3

x/y = 22/3/2[/hide]

edit:[hide]

Or if you want 22/3/2 in other forms:

(1/2)*(cuberoot2)^2

(cuberoot(4))/2[/hide]


Given a_1,a_2,...,a_n real numbers. We denote with S=a_1+a_2+...+a_n.
If \sum {k=1,n} a_k/(S-a_k)=1 and S-a_k=/=0 for all k \in {1,2,...,n} prove that \sum {k=1,n}ak^2/(S-a_k)=0.

S=S* \sum a_k/(S-a_k)= \sum a_k^2/(S-a_k)+S (check it!) :D


Find the sum of the squares of the solutions of $ x^2\minus{}13x\plus{}4\equal{}0$.

Notice that $ a^2 \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 2ab$. We can use Vieta's formulas to find the sum and the product. The sum is $ \minus{}\frac{\minus{}13}{1} \equal{} 13$ and the product $ \frac{4}{1} \equal{} 4$, so our answer is $ 13^2 \minus{} 2(4) \equal{} 169 \minus{} 8 \equal{} \boxed{161}$.

The formula for the sum of the squares of the roots of a quadratic is:
(-b/a)^2 -2ac

------------------
Sorry for the lack of LATEX.


First, of all, is this network traceable? If a graph is traceable, then you can go over all of the lines without going over a line more than once, and you can hit a point more than one time.
From which points can you start tracing?
And lastly, if you start a certain point, at which point must you end at?

Hopefully your geometry class has gone over this :)

Shoot, there's a way to solve this...and I don't have my geometry textbook with me. Crud!

I believe this discovery is attributed to Euler.

[quote="Monkey"][hide] Denote each point as a "node".

If a path is to enter one "node", there must also be a path out of that node. Thus, the number of line segments which intersect at any node must be even. Exception: You can have two nodes with an odd # of lines intersecting there, since it means those are the start and end points (there's only one line required to start, and only one to end).

This path shown is traceable because it has two odd nodes and everything else even nodes.

The path must start at 8 or 11 ("odd" nodes), since otherwise, there's no way to exit once you enter them).

The path must therefore end at either 8 or 11 (depending on where you started). [/hide][/quote]
Oh, so thatswhy. I was always wondering why the cross in rectangle picture couldn't do it...

[hide]I had to count this thing twice.  I forgot to count all the nodes.  Yes, it's traceable but you have to start and end on the numbers with odd nodes.[/hide]
Did you find this figure in a book or did you make it yourself?

[quote="kingof21penguins"]i traced it hehe :blush: me and my incorrect skills[/quote]

Better if you do things the slow way at Nats because then it gives everyone else a chance :P (even though you are on my team)

hehe...just kidding

I find 2^323729792749702 of these problems in math puzzle books. It always asks you to draw a path in some manner without any path run over twice


Prove : $ \frac{(b\plus{}c)(c\plus{}a)}{c}\plus{}\frac{(c\plus{}a)(a\plus{}b)}{a}\plus{}\frac{(a\plus{}b)(b\plus{}c)}{b}\plus{}\frac{1}{2}\equal{}\frac{1}{2a}\plus{}\frac{1}{2b}\plus{}\frac{1}{2c}$,
for $ a\plus{}b\plus{}c\equal{}1$, and $ a,b,c$ real positive.

really? if this is in the inequalities section, where's the inequality sign?  :)

[quote="wangsacl"]Prove : $ \frac {(b \plus{} c)(c \plus{} a)}{c} \plus{} \frac {(c \plus{} a)(a \plus{} b)}{a} \plus{} \frac {(a \plus{} b)(b \plus{} c)}{b} \plus{} \frac {1}{2} \equal{} \frac {1}{2a} \plus{} \frac {1}{2b} \plus{} \frac {1}{2c}$,
for $ a \plus{} b \plus{} c \equal{} 1$, and $ a,b,c$ real positive.[/quote]
I think, it needs to be
\[ \frac {(b \plus{} c)(c \plus{} a)}{c} \plus{} \frac {(c \plus{} a)(a \plus{} b)}{a} \plus{} \frac {(a \plus{} b)(b \plus{} c)}{b} \plus{} \frac {1}{2} \leq \frac {1}{2a} \plus{} \frac {1}{2b} \plus{} \frac {1}{2c}
\]
It's true and easy enough.

The sign of the inequality is as arqady says, since
\[ \frac {1}{2a} \minus{} \frac {\left(c \plus{} a\right)\left(a \plus{} b\right)}{a} \equal{} \frac {\left(a \plus{} b \plus{} c\right)^2 \minus{} 2a^2 \minus{} 2\left(ab \plus{} bc \plus{} ca\right)}{2a} \equal{} \frac {b^2 \plus{} c^2}{2a} \minus{} \frac {a}{2},
\]
and similarly for the cyclic permutations, leaving the equivalent inequality to the one proposed:
\[ \frac {b^2 \plus{} c^2}{2a} \plus{} \frac {c^2 \plus{} a^2}{2b} \plus{} \frac {a^2 \plus{} b^2}{2c}\geq1.
\]
Assume now that $ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\geq1$.  Then, using the AM-GM, $ \frac {b^2 \plus{} c^2}{2a}\geq\frac {bc}{a}$, and similarly for its cyclic permutations, making the proposed inequality true.

Assume now that $ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\leq1$.  Then, it suffices to prove that
\[ \frac {\left(b \plus{} c\right)^2}{2a} \plus{} \frac {\left(c \plus{} a\right)^2}{2b} \plus{} \frac {\left(a \plus{} b\right)^2}{2c}\geq2.
\]
But $ \frac {\left(1 \minus{} a\right)^2}{2a} \equal{} \frac {1}{2a} \plus{} \frac {a}{2} \minus{} 1$, or the inequality that we need to prove is equivalent to
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq9.
\]
But this is trivially true using the inequality between arithmetic and harmonic means, and we are done.  Note that, trivially, equality is achieved iff $ a\equal{}b\equal{}c$, since we need either $ b\equal{}c$ and its cyclic permutations in the first case (making trivially $ \frac{bc}{a}\plus{}\frac{ca}{b}\plus{}\frac{ab}{c}\equal{}a\plus{}b\plus{}c\equal{}1$), and $ a\equal{}b\equal{}c$ in the second case (ibidem).

I don't understand the solution above.
Yes, the inequality is right, as arqandy says.
But, I don't understand the solution.
Please give a complete solution, so I can study the solution.

The complete solution is there...  $ 1 \equal{} \left(a \plus{} b \plus{} c\right)^2$ and $ \left(a \plus{} b\right)\left(c \plus{} a\right) \equal{} a^2 \plus{} \left(ab \plus{} bc \plus{} ca\right)$, use these two to show that the first line of calculations is correct.
Substitute this first line in the given equation to show that the inequality proposed by arqady is equivalent to
\[ \frac {b^2 \plus{} c^2}{2a} \plus{} \frac {c^2 \plus{} a^2}{2b} \plus{} \frac {a^2 \plus{} b^2}{2c}\geq1,
\]
and if we can prove this, the problem is finished.  We analyze this in two possible cases, depending on whether $ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}$ is larger than or equal to, or less than or equal to, 1.  In the first case, and since the inequality between arithmetic and geometric means ensures that $ b^2 \plus{} c^2\geq2bc$, with equality if and only if $ b \equal{} c$, and similarly for the other three fractions, finding that the inequality holds is immediate.  In the second case, if it was true that
\[ \frac {\left(b \plus{} c\right)^2}{2a} \plus{} \frac {\left(b \plus{} c\right)^2}{2a} \plus{} \frac {\left(b \plus{} c\right)^2}{2a}\geq2,
\]
we can see that the LHS may be written as the sum of $ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}$, which is less than 1, and $ \frac {b^2 \plus{} c^2}{2a} \plus{} \frac {c^2 \plus{} a^2}{2b} \plus{} \frac {a^2 \plus{} b^2}{2c}$, which would then be larger than or equal to 1 (which is what we need to show).  So we just need to prove that $ \frac {\left(b \plus{} c\right)^2}{2a} \plus{} \frac {\left(b \plus{} c\right)^2}{2a} \plus{} \frac {\left(b \plus{} c\right)^2}{2a}$ is indeed larger than or equal to 2.  Substitute then $ b \plus{} c \equal{} 1 \minus{} a$, and similarly for $ c \plus{} a \equal{} 1 \minus{} b$ and $ a \plus{} b \equal{} 1 \minus{} c$, expand the squares and the fractions as shown, gather together constant terms and terms with $ a,b,c$ (these latter ones will add up to $ \frac {1}{2}$), and the inequality that we need to prove is then shown to be equivalent to $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq9$.  The LHS of this inequality is 3 divided by the harmonic mean of $ a,b,c$, and the harmonic mean is always less than or equal to the arithmetic mean ($ a,b,c$ are positive), and using this it is trivial that the LHS is indeed larger than or equal to 9, with equality if and only if $ a \equal{} b \equal{} c \equal{} \frac{1}{3}$, and the problem is complete in both cases considered.

Perhaps in my previous post I just indicated steps that seemed trivial to me.  I hope that this more fully explained post will help.

[quote="daniel73"]  [quote]We analyze this in two possible cases, depending on whether $ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}$ is larger than or equal to, or less than or equal to, 1.[/quote][/quote]
we have $ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\ge a\plus{}b\plus{}c\equal{}1$ for all positive reals numbers $ a,b$ and $ c$.

Thanks to Cauchy-Schwarz inequality.  Call $ x \equal{} \sqrt {\frac {bc}{a}}$, $ y \equal{} \sqrt {\frac {ca}{b}}$, $ z \equal{} \sqrt {\frac {ab}{c}}$, and then
\[ \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c} \equal{} x^2 \plus{} y^2 \plus{} z^2\geq xy \plus{} yz \plus{} zx \equal{} c \plus{} a \plus{} b \equal{} 1,
\]
with equality iff $ x \equal{} y \equal{} z$, or equivalently $ a \equal{} b \equal{} c$.  I didn't realize it...   :blush:


If we have $ a_n^2$, if and how can we make the 2 be 'after'/'behind' the n?


So the problem is to find the 3 digits number N such as [tex]180\ |\ 10^N -N [/tex]?

[tex]180 = 2^2\cdot 3^2\cdot 5[/tex]

Looking mod 4 and 5 , we must have 20 | N
Looking mod 9 we must have N = 1 [9]

This solves to the numbers 20*N with 2N=1 [9] and to 100, 280, 460, 640, 820


An inequality, I believe. Try searching Muirhead inequality on wolfram or here on the forum.

[quote="neelnal"][quote="boxedexe"]You can easily kill it with muirhead. :P[/quote]

What is muirhead?[/quote]
[url]http://en.wikipedia.org/wiki/Muirhead_inequality[/url]

[quote="neelnal"]What is muirhead?[/quote]
Here is the answer:
[quote="harazi"]Muirhead inequality is for me one of the worst things that ever happened since the invention of the sign $\geq$.[/quote]

[quote="Ilthigore"]How can you justify such an opinion? It provides a routine method for the solution of many homogeneous algebraic inequalities which, sure, might make them less valid IMO questions, but really is an important and highly useful breakthrough in this field of mathematics. To me, Muirhead epitimises the ideal generalised theorem, and ever since I first read an article about it a few months ago I have absolutely loved it.[/quote]
Read my post [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=647853#647853]here[/url] and the links I attached there.


[quote="Peter"]If an integer $n$ is such that $7n$ is the form $a^2 +3b^2$, prove that $n$ is also of that form.[/quote]

In fact, let $A$ and $B$ be two integers satisfying $7n=A^2+3B^2$. Then, $7\mid A^2+3B^2$, so that also $7\mid\left(A^2+3B^2\right)-7B^2=A^2-4B^2=\left(A+2B\right)\left(A-2B\right)$. Thus, since $7$ is prime, we have $7\mid A+2B$ or $7\mid A-2B$.

If $7\mid A+2B$, then $\frac{A+2B}{7}$ is an integer; then, by writing

$\left(2\cdot\frac{A+2B}{7}-B\right)^2+3\left(\frac{A+2B}{7}\right)^2=\frac{A^2+3B^2}{7}=\frac{7n}{7}=n$,

we obtain a representation of $n$ in the form $a^2+3b^2$.

If $7\mid A-2B$, then $\frac{A-2B}{7}$ is an integer; then, by writing

$\left(2\cdot\frac{A-2B}{7}+B\right)^2+3\left(\frac{A-2B}{7}\right)^2=\frac{A^2+3B^2}{7}=\frac{7n}{7}=n$,

we obtain a representation of $n$ in the form $a^2+3b^2$.

In both cases, we have seen that $n$ can be represented in the form $a^2+3b^2$. This completes the solution.

  Darij


define  positive integers   $ x, y$, such that      $ x^2 \minus{} y^3 \equal{} 1$


Compute without using a calculator: $ \dfrac{9!}{6!3!}$

The $ 6!$ cancels to give $ \frac{9\cdot8\cdot7}{6}\equal{}\boxed{84}$


What are the requirements for a graph to an Eulerian path? What about a directed graph?

i think it'd be useful to provide some background info on those kinds of graphs
people who know what they mean probably know the answer to this, as it is classic
and people who haven't been exposed to graph theory before, probably don't

http://en.wikipedia.org/wiki/Eulerian_graph
an Eulerian path is a path in a graph which visits each edge exactly once.

http://en.wikipedia.org/wiki/Directed_graph#Directed_graph
the explanation for directed grpah isn't so good there imo... but mathworlds is even more O_O


function
real analysis
real analysis theorems

Apologies if this is the wrong place to ask these basic question; I'm new to this board so please correct me if this post should have been made elsewhere.  How do I solve the following basic functions?

f(x)=(x-1)(x+5)      for f(a+1)
I try to solve as follows but get it wrong.
  f(a+1)= ( (a+1) - 1 ) ( (a+1) + 5 )
           = (a+1)^2 + 5(a+1) - (a+1) - 5
           = a^2 + 1 + 4(a+1) - 5
           = a^2 + 4a
           = a(a+4)

can anyone help?

[quote="KSt"]Apologies if this is the wrong place to ask these basic question; I'm new to this board so please correct me if this post should have been made elsewhere.  How do I solve the following basic functions?

f(x)=(x-1)(x+5)      for f(a+1)
I try to solve as follows but get it wrong.
  f(a+1)= ( (a+1) - 1 ) ( (a+1) + 5 )
           = [color=red](a+1)^2[/color] + 5(a+1) - (a+1) - 5
           = [color=red]a^2 + 1 [/color]+ 4(a+1) - 5
           = a^2 + 4a
           = a(a+4)

can anyone help?[/quote]

  = (a+1)^2 + 5(a+1) - (a+1) - 5
           = a^2 + 1+2a + 4(a+1) - 5 = a(a+6).
Of course you could avoid all these simplifications , since  " (a+1) - 1 = a "
 Please write in the section << getting start >> or <<intermediate >>


factorial
number theory
greatest common divisor
analytic geometry
graphing lines
slope

$1$. You have a rectangular solid with dimensions of $230$, $231$ and $232$ mm. Grid lines are drawn on each face at intervals of $1$ mm, thereby forming tiles of $1$mm $\times$ $1$mm on the faces.

If you draw diagonals across the faces, on which face will the diagonal cross the maximum number of tiles?

$2.$ use the digits $1, 2, 3, 4, 5$ and by using the mathematical operations and signs like  $+$ ( addition ), $-$ ( subtraction ), $*$ ( multiplication ), $/$ ( division ), ^ ( power ) and $!$ ( factorial ), how close can you come to the final figure of $67890$. you can use brackets ( ) for clarity.

concatenation (i mean merging in chain such as $123$ by joining / merging $1, 2, 3$ )  of digits is permissible.

for example, you can write

$12^{4}\times 3+5! = 62328$

the number $62328$ is $91.80$% of $67890$. so deviation is $8.20$ %. Come as close as possible...

For the second one, I got $(2^{3+1})^{4}+5=65541$.

EDIT: Got a better one: $23 \times 4! \times 5!+1=66241$.

[quote="AdorableAddict"]$(3^{2}+\frac14)^{5}=67718.7$[/quote]

any one to better this?

[quote="xxxyyyy"][quote="Xantos C. Guin"][hide="Problem 1"]The number of squares a diagonal in an $a \times b$ array will go through is $a+b-gcd(a,b)$.

Using this formula:
$230+231-gcd(230,231)=230+231-1=640$
$231+232-gcd(231,232)=231+232-1=642$
$232+230-gcd(232,230)=232+230-2=640$

Therefore, the face that the diagonal will cross the maximum number of tiles is one of the $231 \times 232$ faces. [/hide][/quote]

can you explain...
[hide]How do you prove that the number of squares a diagonal in an $a \times b$ array will go through is $a+b-gcd(a,b)$.[/hide][/quote] That's just Principle of Inclusion-Exclusion. I don't know exactly how to prove it.  :|

[quote="xxxyyyy"][quote="Xantos C. Guin"][hide="Problem 1"]The number of squares a diagonal in an $a \times b$ array will go through is $a+b-gcd(a,b)$.

Using this formula:
$230+231-gcd(230,231)=230+231-1=640$
$231+232-gcd(231,232)=231+232-1=642$
$232+230-gcd(232,230)=232+230-2=640$

Therefore, the face that the diagonal will cross the maximum number of tiles is one of the $231 \times 232$ faces. [/hide][/quote]

can you explain...
[hide]How do you prove that the number of squares a diagonal in an $a \times b$ array will go through is $a+b-gcd(a,b)$.[/hide][/quote]

The diagonal will have to cross $a$ squares horizontally, and $b$ squares vertically to get to the opposite vertex. From this, you have to subtract the number of vertices it crosses since only square is crossed in that case. This occurs every $GCD(a, b)$ tiles.

I can't see the GCD relationship. I seems it is a grid that is x by y. Then the number of vertices should be the number of integer points on the line with slope x/y.

How is the GCD the solution to this?

if you look at it, it can go through a squares vertically and b squares horizontally, but every time it goes through an intersection of two grid lines, it subtracts 1 from the total amount of squares it passes through (do you see why?). it's pretty easy to show that the gcd is the number of times it goes through a lattice point (excluding the very first one).

i know my explanation sucks but it's 10:45 and i'm never fully lucid at this time

Sorry, that was a poor explanation of mine. Every case where the diagonal crosses the vertex of a square is double counted since at that moment, it increased both the horizontal and vertical distance, while still only crossing one square. This occurs every gcd(a, b) squares, which is what you subtract from a+b.

Can you please show a proof of the GCD being the interval for integer solutions?


[quote="eagerkill"][img]http://img153.imageshack.us/img153/1773/untitled9bu.jpg[/img][/quote]
It's wrong: $x=0.4,$ $y=0.59$ and $z=0.01.$ :wink:


blogs
function
number theory
prime numbers

i just saw this one posted somewhere.
I'll find it.

[quote="IntrepidMath"][quote="mgao"]How about this?

I am doing a fun thingy over here, sort of like a math generator.

This is how it works, you have to contribute a problem and then somebody else

answers it, and the same somebody else contirbutes another problem and I would

like it very much if they include a formula to go with it.  Doesn't have to be related

to the problem.  I'll pop in every now and then to see how you're doing.

You can also put mathematical paradoxes along w/ the formula. {optional}

I'll start by giving you a problem:

Find the next row:

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
13211321321112131221212311311222113111221131221

 :P  :P
 :?  :!:[/quote]
[hide]
111312211312111312311211131122111211121321132132211331222113112211
[/hide]
Problem:
Find the next term of the sequence:
O,T,T,F,F,S,S,E,N,....[/quote]

T, then E, T... they stand for one, two, three, ect.

this is posted somewhere else, too.
11, 31, 71, 91, 32, 92, 13, 73...

hey, mgao, arent u the one that posted both of these sequences? haha.

[hide]14, 34, 74,... they're prime numbers switched backwards.[/hide]

1, 2, -1, 8, -19, 62...

[hide="hint"] it goes by a function thing, xn+y. (n being the number before it)[/hide]

sorry ignite, you're wrong. it goes by a function, as y=mx+b, where x is the previous number. good idea though! :)

edit: ignite's wrong about the answer, not his new sequence.

edit #2: never mind, ignite's right, but he didn't use the function. his way was good,though. i used the function y=-3x+5 to get the sequence. maybe we should remember that, that's kind of cool! :lol:


Round 1 problems of Year 18 of the USA Mathematical Talent Search are now available at [url]http://www.usamts.org[/url].  The postmark deadline for solutions is October 10, 2006.  A copy of the rules is available with the problems; please read them carefully.

If you have not already registered online for Year 18, please do so at http://www.usamts.org.  If you participated in Year 17, all you have to do to register for Year 18 is log in to [url]http://www.usamts.org[/url], then set your grade in the Profile section of the My USAMTS pages.

Do not discuss the problems with others or on this message board until October 14.  This includes posting questions on this board about wording issues in the questions.

Is there a featured article about a specific topic this year like last year(Expected value)?

When will there be an updated solution form, since there is only one for Year 17.

Oops, we forgot to update it.  It's now updated for Year 18.

[quote="junggi"]um.. I kind of need clarification on a question.
How do I ask?
pm one of the administrators?[/quote]You can email usamts@usamts.org, but I should warn you that it is [i]extremely[/i] unlikely that we will "clarify" any question.  However, if you believe that there is an [i]error[/i] in a problem statement, then please email and let us know.

This is my first year doing USAMTS actually using $\LaTeX$ and I have a few questions.

In the solutions template, it says to enter your AoPs username, do they mean your USAMTS username?

I just want to be clear on this. What files do you actually sumbit through email and in what format?

[quote="rrusczyk"]Yes, enter your USAMTS username.

When you email files, email us the .tex file and the .pdf file if you use LaTeX.[/quote]

what about the usamts.tex file?

What should we write in our email, besides the attachments?
Is there a specific name we should title our solution files as?


quadratics geometry trigonometry algebra polynomial Vieta AMC

What do you guys/girls think are the 100 most important concepts/formulas/theorems/etc as far as the AMC 12 is concerned?
Of course this excludes common knowledge like Pythagorean Theorem, Quadratic Formula, etc.

Organize them by the subject:(algebra, geometry, counting/probability, number theory, trig, other)

Feel free to post as or few as you want :)[/list]

Viete's Formulas, though not necessary, can definitely shave a minute or two off the easier problems.

If you know the idea of the solution, you don't need any formula. Everything can be derived.
Mathematics competition's problems are more like based on basic things, just added a little tweak to it. Math's competition doesn't require you to know master level maths, it requires you to think. Of course if you do know all those formulas, it 'll be easier, but I'd say that you only need basic knowledge about those different areas you pointed out.

The AoPS books exist more or less to answer this question.

In AMC, the topics that you exclude [i]are[/i] the important topics which you need to know to solve the problems. It's just that you need to apply them in a more creative way than what you're used to (and of course there are some things which aren't covered in classrooms like Vietas but its almost never necessary to solve the problem)

well, on amc10s there are geometry problems at the end that i fail at. amc12, ive never taken. everything on amc8 can be solved with stuff from aops volume 1 generally. aime, i dont know cuz ive never taken it :( .

I have no experience with 12 but with, 10 I think that the test is designed around applying the things you learn in the standard 9th and 10th grade curriculum on a higher level. I'm making this guess on what I read about the test and because in 8th grade, I thought I did good on the 2007 AMC 8 (23, maybe I just got lucky), but I got demolished on the 2007 AMC 10 (14/25). After a year of geometry, the a lot of the old AMC problems are getting easier for me.

Maybe it's just that I'm so new to these kind of problems, it just seems like (from reading solutions) there are a lot of theorems, formulas, etc that I had never heard of.  Actually its kind of funny, because we did learn some of Viete's formulas in my pre-cal class, but like you guys said we never learned to apply it very well.

law of consine and sine
If the people in my school knows it, I won't be the only one from my school qualified for AIME though 1st AMC.
There's usually some thing like law of cos in AIME also, but never on USAMO.

yes, and I'll add
- basic trig identities
- sequences and series (sums, averages, nth term, products)
- combinations and permutations

[quote="MellowMelon"][quote="moplam"]There's usually some thing like law of cos in AIME also, but never on USAMO.[/quote]
Please look at the solution to 2 [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2007-s.pdf]here[/url].

I don't know of anything that appears on easier tests but not the USAMO.[/quote]

I'll take that back then.

[quote="MellowMelon"]Please look at the solution to 2 [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2007-s.pdf]here[/url].[/quote]
where did you get that link? Also where can I find previous usamo problems that are not in the resource section? Thanks.

It's at the AMC website [url=http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.shtml]here[/url].

The resources section contains more questions than the archive, though.

but why does the usamo problems in [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27]the resource section[/url] only starts at year 1989? All the previous years are missing... :(

can I possibly find the missing problems on the internet? If so, where?

If someone bought me a copy of the USAMO CD, I'd be glad to start posting them, but as is I am too lazy.

Anyway, this is off-topic. If no one has anything else to say on the actual subject of the topic, I'm going to go ahead and lock this.

Eh. I got a 144 AMC10A this year, and the one I got wrong was just me overcounting and being dumb. The last question required common sense, and the extent of geometry that I used was ... coordinate geo. I'd say that coordinate geo can be a decent replacement for similar triangles, only if you fail with those (like me!). Shoelace is definitely helpful; it can save a ton of time, as can mass point. Most AMC10 algebra concepts are a multitude of basic concepts clumped together in one question. AMC12, on the other hand (121.5 for me), can require some more advanced concepts for the last five questions.

[quote="CEM_College_Champ"]What about Lagrange multipliers, partial derivatives or Jacobians?[/quote]

This is clearly [i]not[/i] tested in AMC 10 (or any of AMC for high schoolers for that matter).

Occasionally you'll see something on which single-variable calculus-based optimization may be a useful shortcut, but Lagrange multipliers will never be.

Usually on AMC's the earlier problems don't require any certain conepts.  On the later ones, I've seen complex roots of unity used often, some Stewart's Thm problems for geometry, and Hockey Stick or balls-and-urns used for combinatorics.  Alot of these carry through to AIME's as well.

Make sure you [i]understand[/i] sums and roots very well. Also, pascal's identity is quite useful.

I don't quite agree with what most of the people here have said... While it is true that there are no calculus or any other university topics in math contests, the questions still do require forumlas and such. Off all the questions that I couldn't do on the AMCs and AIMEs, over 95% of them contained formulas I never seen before.

That said, assuming you mastered all of the formulas in the AoPS books vol. I and II, how far would that get you? Assuming you could use all the formulas and tricks there on any question, but you can't derive anything on your own? Because I've seen IMO questions in there, so is it up to IMO level?

If you've truly mastered the concepts in the AoPS books, you should be able to solve most olympiad-level problems. Most problems are about using stuff you already know creatively.

Ah I see, thanks. I dont' want to make this thread about me, and I don't want to make a new thread, so beside the vol. I and II books, are there any other books that are very good for AIME-->olympiad level thinking? Any formulas that are too specific to a certain topic that are not covered in the vols. I and II?

Because you see, I've never actually gone through the school system, so what I'm "expected" to know is not necessarily what I do know.

Now on topic, I'll put some formulas I think are crucial.

"- basic trig identities
- sequences and series (sums, averages, nth term, products)
- combinations and permutations"

perhaps quadratic formula, completeing the square, binomial theorem

surface area/volume/areas of basic shapes (circles, triangles, rectangles, cone, cylinder, sphere etc)

angles (parallel lines with a line through it, triangles, x-gons, etc)

all the Log I-don't-know-what-they're-called's

a+bi

[quote="U§ername"]beside the vol. I and II books, are there any other books that are very good for AIME-->olympiad level thinking?[/quote]

The Art and Craft of Problem Solving is a great book for studying the Olympiad level topics.  It covers things from around AIME level up to even some Putnam problems.  I think it goes into alot harder topics than the AoPS vol. I and II books, as well.

Hmmm, so to add another formula, Hook-length Formula is really cool although relatively useless (ive never seen a problem that uses it).

You guys are focusing too much on formulas and tricks. See [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php[/url]. I'd say as a general rule, if you can't rederive the formula on the spot, you shouldn't be using it.

REderiving is completely not the same as deriving it for the first time. It's very hard to find a formula yourself... but if you've at least seen it before it makes it a whole lot easier to understand. Sure, you need to be very familiar with the formula to use it, and I"m not going for just a pure memorization of formulas, but there's a big difference between knowing the formulas before, and trying to derive a formula yourself on the spot for that single question.

It's like doing all the math that has been done in the 4 or 5 thousand years again yourself.

I don't think that worthawholebean was implying that you should rederive every formula you will use during a contest.  The point is that if you can't rederive a formula on the spot if necessary, then you don't understand the concepts behind the formula, and don't exactly know what the proper methods of using it are.

I didn't say that. I said that you do need to be familiar with the formulas to be able to use them, but you still need to have known them in the first place

I agree completely.  All I was getting at is that the ability to rederive a formula is a good indication that you are familiar with it.

Well, it doesn't hurt to memorize them. Wastes you less time.

Again, I agree completely.

[quote="I"]
All I was getting at is that the ability to rederive a formula is a good indication that you are familiar with it.
[/quote]

Having the [i]ability[/i] to rederive a formula is different than actually rederiving it every time.  It only shows understanding.

But yes, of course you should have it memorized.

[quote="qntty"]In AMC, the topics that you exclude [i]are[/i] the important topics which you need to know to solve the problems.[/quote]

Crap, I exclude Topology. :(

In my experience, All the harder problems that I've solved I solved just by having a little insight on the problem. If you know a few advanced formulas, good for you; you can bash some of the harder questions. But I prefer the questions that really make me think (this is what AIME and USAMO are for :) ), and late AMC 12 does this.

Also, the way to get good at the problems quicker is if you are routinely exposed to them at an early age. I officially took the AMC 12 the first time when I was supposed to be in fifth grade (99.5 wah :( ) (I skipped a grade (I was held back again for purely political reasons)), and was practicing them once a week half the year, and I've been gaining insights on the harder problems since (126 in 7th grade, 133.5 in 8th). Don't be afraid to take the harder test.

Also the only reason that I took the AMC 12 was that my group didn't offer the AMC 10. I was the highest scoring person in my grade in my state since 2006.


Support
function
analytic geometry
advanced fields
advanced fields unsolved

Let $T: L^{2}(\mathbb{R}^{n}) \to L^{2}(\mathbb{R}^{n})$ a continuous linear operateur satisfying: \[Supp(Tu) \subset Supp(u)\] Show that there exists a function $\phi \in L^{\infty}(\mathbb{R}^{n})$ such that \[Tu=\phi u\]

Write $\chi_{E}$ for the characteristic function of a rectangular box $E\subset\mathbb{R}^{n}$ (the boxes are assumed oriented along the coordinate axes). Using the assumption $\mathrm{supp}\ T(\chi_{E})\subset E$ and the linearity of $T$, one can prove that $T(\chi_{E})=T(\chi_{F})$ on the intersection of the boxes $E$ and $F$. Therefore, the functions $T(\chi_{E})$ can be glued into one function $\phi\in L_{\rm loc}^{2}(\mathbb{R}^{n})$. Since $T$ is bounded, $\phi\in L^{\infty}(\mathbb{R}^{n})$ (consider a sequence of boxes shrinking to a point.) It remains to observe that the linear span of $\{\chi_{E}\}$ is dense in $L^{2}(\mathbb{R}^{n})$.


Let $p>0$, $q>0$ and $n\ge 0$ are integer numbers. How many integer numbers can be represented in the form $ip + jq$, where $i$ and $j$ are non-negative integers, and $i + j \leq n$.

Problem 5 is in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59015]here[/url]. So, i hope you can wait solution in 2 weeks  ;)

I cannot understand. You mean it is still being in contest?

Well, I am sorry them,.. i will wait :juggle:

I think it's a problem in CGMO 2004 but the problem has a condition that p and q is coprime.

It makes no difference if they are coprime...

I could have sworn this was in a Red MOP test. I even think I recall Tiankai taking a point off of my score for missing a case.

Can we answer it yet? Is it still in a competition?

Yes, I was also wondering about that..
I guess we should wait for MrMATH's answer (or another Vietnamese)  ;)

it is in VMEO
http://diendantoanhoc.net/forum/index.php?showtopic=6663


Prove that there are infinitely many sets of integers (a, b, c) such that gcd(a, b, c)=1 and a^2+b^2+c^2=a^3+b^3+c^3

It is rather difficult problem, nobody got a mark more then 2 (of 7 maximum) for it.


calculus
integration
function
logarithms
calculus computations

Let $ T \equal{} \int_0^{\ln2} \frac {2e^{3x} \plus{} e^{2x} \minus{} 1} {e^{3x} \plus{} e^{2x} \minus{} e^x \plus{} 1}dx$. Evaluate $ e^T$.

[hide="Solution"]
Let's look at the coefficients in the numerator:

The coefficient of $ e^{3x}$ is 2, the coefficient of $ e^{2x}$ is 1, the coefficient of $ e^{1x}$ is 0, the coefficient of $ e^{0x}$ is -1, a pattern, suggesting that we clear out a factor of $ e^x$ to get terms of the form $ ke^{kx}$.

We do so, and our fraction becomes $ \frac{2e^{2x}+e^x - e^{-x}}{e^{2x}+e^x + e^{-x}-1} = \frac{\frac{d}{dx} (e^{2x}+e^x - e^{-x}+1)}{e^{2x}+e^x + e^{-x}-1} = \frac{u'(x)}{u(x)}$

Integrating, we have:
$ T=\int_{0}^{\ln 2} \frac{u'(x)dx}{u(x)} = \int_{u(0)}^{u(\ln 2_} \frac{du}{u(x)} = \left( \ln |u(x)| \right)_{x=0}^{\ln 2}$

$ = \left( \ln |e^{2x}+e^x + e^{-x}-1| \right)_{x=0}^{\ln 2} = \ln \frac{11}{2} - \ln 2 = \ln \frac{11}{4} \implies \boxed{e^T = \frac{11}{4}}$[/hide]

I found this problem to be very contrived.


See http://www.mathlinks.ro/Forum/viewtopic.php?t=31545 .

  darij


For which $n$ does $ab \equiv -1 \mod n$ imply $a+b \equiv 0 \mod n$ ?


who can help me prove the AM-GM by Cauchy's Mean Theorem :?
THANKS :ninja:

sorry  :(  i think my internet has some mistakes


I was a member of the Florida Team for the 2008 HMMT, but not the 2009 HMMT.  Whom should I contact to become involved with the 2010 HMMT Team?

Email me: eliross2@aol.com

(I'm pretty sure you're already on my list; look for an email in the next two days.)

Were you not at the invite this weekend?

THERE IS NO 2010 HMMT FOR US :( :( :( :( :(  :( :( :( :( :(
*SOB* *SOB* *SOB* *SOB* *SOB* *SOB**SOB* *SOB* *SOB**SOB* *SOB* *SOB**SOB*
SEE HOW SAD I AM?


Let $ a_1,a_2,...,a_n$ be positive real numbers, and $ n\in N, n\geq2$, such that $ a_1\cdot a_2\cdot...\cdot a_n \equal{} 1$. For $ k\in N, k\geq2,$ prove that:
$ \frac {1}{(k \minus{} 1)a^k_1\plus{}1} \plus{} \frac {1}{(k \minus{} 1)a^k_2\plus{}1} \plus{} \cdots \plus{} \frac {1}{(k \minus{} 1)a^k_n\plus{}1} \plus{} \frac {k \minus{} 1}{k}\cdot(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)\geq n$

This is the right form for my inequality. It was a typo in my first post. Sorry!

[quote="marin.bancos"]Let $ a_1,a_2,...,a_n$ be positive real numbers, and $ n\in N, n\geq2$, such that $ a_1\cdot a_2\cdot...\cdot a_n \equal{} 1$. For $ k\in N, k\geq2,$ prove that:
$ \frac {1}{(k \minus{} 1)a^k_1} \plus{} \frac {1}{(k \minus{} 1)a^k_2} \plus{} \cdots \plus{} \frac {1}{(k \minus{} 1)a^k_n} \plus{} \frac {k \minus{} 1}{k}\cdot(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)\geq n$[/quote]
By AM-GM, we have the stronger but still trivial
\[ \sum_{i\equal{}1}^n \frac {1}{ka^k_i}\plus{}\frac {k \minus{} 1}{k}\cdot(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)\geq \frac {1}{k}\cdot n\plus{}\frac {k \minus{} 1}{k}\cdot n\equal{}n\]  :rotfl:

[quote="marin.bancos"]Let $ a_1,a_2,...,a_n$ be positive real numbers, and $ n\in N, n\geq2$, such that $ a_1\cdot a_2\cdot...\cdot a_n \equal{} 1$. For $ k\in N, k\geq2,$ prove that:
$ \frac {1}{(k \minus{} 1)a^k_1} \plus{} \frac {1}{(k \minus{} 1)a^k_2} \plus{} \cdots \plus{} \frac {1}{(k \minus{} 1)a^k_n} \plus{} \frac {k \minus{} 1}{k}\cdot(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)\geq n$[/quote]
$ \sum_{i\equal{}1}^n \frac{1}{(k\minus{}1)(a_1)^k} \equal{} \frac{1}{k\minus{}1}\sum_{i\equal{}1}^n \frac{1}{(a_1)^k} \ge \frac{n}{k\minus{}1}$ (AM-GM)
$ \frac{k\minus{}1}{k} \sum_{i\equal{}1}^n \ge \frac{k\minus{}1}{k} n$ (AM-GM)
Summing up we get:
$ LHS \ge n \left ( \frac{1}{k\minus{}1} \plus{} \frac{k\minus{}1}{k} \right ) > n \left ( \frac{1}{k} \plus{} \frac{k\minus{}1}{k} \right ) \equal{} n$
We never archeive equality

I've made the correction on the initial form of my inequality. There was a typo. Sorry, [b]Mathias_DK[/b]! Sorry, [b]spanferkel[/b]!

[quote="marin.bancos"]Let $ a_1,a_2,...,a_n$ be positive real numbers, and $ n\in N, n\geq2$, such that $ a_1\cdot a_2\cdot...\cdot a_n \equal{} 1$. For $ k\in N, k\geq2,$ prove that:
$ \frac {1}{(k \minus{} 1)a^k_1 \plus{} 1} \plus{} \frac {1}{(k \minus{} 1)a^k_2 \plus{} 1} \plus{} \cdots \plus{} \frac {1}{(k \minus{} 1)a^k_n \plus{} 1} \plus{} \frac {k \minus{} 1}{k}\cdot(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)\geq n$

This is the right form for my inequality. It was a typo in my first post. Sorry![/quote]

Just use :

$ \frac {1}{(k \minus{} 1)a^k_i \plus{} 1} \plus{} \frac {k \minus{} 1}{k}a_i \ge 1$  :P

But I think that the condition $ \prod_{i\equal{}1}^n a_i \equal{} 1$ is not neccesery to solve this problem  :)

Dear [b]Abdek[/b],
You're absolutelly right! This constraint is not necessary! The proof works without this additional condition!
Maybe a better statement for the problem is the following:

Let $ a_1,a_2,...,a_n$ be positive real numbers, and $ n\in N, n\geq 2$, such that $ a_1\plus{}a_2\plus{}\cdots\plus{} a_n\equal{}n$. Prove that

$ \frac{1}{(n\minus{}1)a^{n}_{1}\plus{}1}\plus{}\frac{1}{(n\minus{}1)a^{n}_{2}\plus{}1}\plus{}\cdots\plus{}\frac{1}{(n\minus{}1)a^{n}_{n}\plus{}1}\geq 1$


algebra
polynomial
function
absolute value
algebra unsolved

hey ppl help me with this one
Let f be a polynomial with real coeffcients such that for each positive integer n the equation f(x) = n has at least one rational solution. Find f.
i hope i'm posting in the rite forum
best regards

suppose f(x)=adx^d+ad-1x^(d-1) +...+a0
Notice that for large enough values of n, f(x)=n has at most two solutions.
Now we will assume that  d>=2. (handling d=1 is trivial)
then f'(x) will be a function of x and therefore rate of increase/decrease of function will grow faster and faster and at some point its absolute value is going to be bigger than k, where k is a huge number. let y1n,...,ytn be the sequence of rationals such that f(yj)= n
then define xn to be max(abs(y1n),...,abs(ytn))
clearly as stated above the sequences y1n,...,ytn will consist of at most two terms for all n>M where M is a constant.
we will first show that d=even.
suppose d=odd. then there exisits N such that if m>N. then f(x)=m has at most one solution. But clearly since d>1 abs(f'(x)) will be bigger than 1 at some point, and therefore if f(xj)=n then since f'(xj)>1 , equation f(y)=n+1 will not have any solutions. hence n should be even.
Now let n to be even. clearly ad should be positive.
again there will be a number N such that beyond N: f'(x)>2
hence if f(xj) =n for some huge n and negative xj, then f(xj) misses the point n+1, and n+2 for all x<xj but we can't have f(x)=n+1 and n+2 both on the other side( that is for positive x) hence d=1.

and so

It is about 1am here, but I am sure ali's post is an absurd! :huh:

Suppose $\deg f=d$. Let $q_{i}$ be a rational root of $f(x)=i$. Then equalities $f(q_{1})=1$, ..., $f(q_{d+1})=d+1$ gives us $d+1$ linear equations on all $d+1$ coefficients of $f$. It is clear that such a system with [b]rational[/b] coefficients has only rational roots. Therefore, coefficients of $f$ are rational numbers. Thus $f(x)=\frac{1}{m}P(x)$, where $P\in Z[x]$ and $m$ is a natural number.

Suppose $d\geq 2$. After some linear rational transformations we can get $P$ is monic (it will also change $m$). We have $P(x)-mn=0$ has a rational root $x_{n}$ for all natural $n$. But since $P$ is monic, this root is integer. We see $m=P(x_{n+1})-P({x_{n}})$. Hence $x_{n+1}-x_{n}\mid m$, i.e. $1\leq |x_{n+1}-x_{n}|\leq m$. But $x_{n}$ takes arbitrary large values as $n$ grows. Therefore, $|P(x_{n+1})-P({x_{n}})|$ is unbounded. Contradiction.


A satellite circles the earth in 48 hours.  A second satellite, travelling in the opposite direction and at the same altitude, passes the first satellite every 16 hours.  How many hours does it take the second satellite to circle the earth?

not exactly...if the second satellite were going that fast, they would meet [i]more often[/i] than 1/3 of an hour.

To see why, consider the first satellite as standing still (it pretty much is, compared to how fast you've got the second one going). Now, the second one will pass it every 1/3 of an hour, right? When the first satellite starts going in the opposite direction, they'll meet each other even more often. Be careful; this was a tricky problem (for me).

oh boy. SHOWDOWN! WILL THE MYSTIC EMERGE VICTORIOUS AGAIN? AGAINST AN ANSWER KEY??! *collective gasp from audience*

errr, mystic, i see a flaw in ur reasoning

the 1/3 in Chinaboy's work refers to the circumference of the earth, meaning 1/3 of the distance traveled, not 1/3 of an hour

whoops when I highlighted chinaboy's spoiler, I only highlighted the first line for some reason and thought he gave 1/3 as the answer. *sigh* now you see why I dislike spoiler

OOHHHHHHH THE MYSTIC HAS BEEN SHOT DOWN. BUT ONLY TO RISE AGAIN! STRONGER THAN EVER BEFORE....

satellite one's movement rate is 1/48.  satellite two's is R.

hold satellite one constant and pretend satellite two moves at rate (R+1/48).

Then by d=vt, (R+1/48)16=1 --> R=1/24