calculus
integration
number theory
Diophantine equation
number theory unsolved

[quote="RobertuX"][quote="djimenez"][quote="Kapi"]Find $n$ such that $x^3-3xy^2+y^3=n$ has  integral solution[/quote]

What do you mean by an integral solution?[/quote]
The diophantine equation has solutions in integers.[/quote]
But in such case shouldn't it be trivial. To find a value of $n$, everything you have to do is to choose a pair $(x,y)\in\mathbb{Z}\times\mathbb{Z}$, plugg it into the equation and obtain a value of $n$, isn't it?

Now, it would be different if they ask "Find ALL values of $n$ such that $x^3-3xy^2+y^3=n$ has  integral solution".

Best,

[quote="djimenez"][/quote][quote="RobertuX"][quote="djimenez"][quote="Kapi"]Find $n$ such that $x^3-3xy^2+y^3=n$ has  integral solution[/quote]

What do you mean by an integral solution?[/quote]
The diophantine equation has solutions in integers.[/quote][quote="djimenez"]
But in such case shouldn't it be trivial. To find a value of $n$, everything you have to do is to choose a pair $(x,y)\in\mathbb{Z}\times\mathbb{Z}$, plugg it into the equation and obtain a value of $n$, isn't it?

Now, it would be different if they ask "Find ALL values of $n$ such that $x^3-3xy^2+y^3=n$ has  integral solution".

Best,[/quote]
U r right, sometimes people [b]forget  [/b] but I assume that if they dont use the term [b] any [/b]  then is find all.  ;)

My problem:
[quote="djimenez"]"Find ALL values of $n$ such that $x^3-3xy^2+y^3=n$ has  integral solution".

[/quote]


[quote="4865550150"]Let $ x_1,x_2,...,x_n$ be a set of positive numbers, where $ 0 < x_i \le 1$ and  $ \sum_{i \equal{} 1}^n x_i^2\ge \sum_{i \equal{} 1}^n x_i^3 > 1$.

Prove or disprove if $ \sum_{i \equal{} 1}^n x_i \equal{} n$ , $ n \ge 2$

$ \left[ \sqrt {\sum_{i \equal{} 1}^n x_i^3 \minus{} 1} \plus{} \sqrt {\sum_{i \equal{} 1}^n x_i^2 \minus{} 1} \right]\left[ \frac {1}{\sqrt {\sum_{i \equal{} 1}^n x_i^3 \minus{} 1}} \plus{} \sqrt {\sum_{i \equal{} 1}^n x_i^2 \minus{} 1} \right] \newline \ge 2n \minus{} 1 \plus{} \sqrt {\frac {\sum_{i \equal{} 1}^n x_i^3}{\sum_{i \equal{} 1}^n x_i^4}}$[/quote]
 :rotfl: by the conditions, $ x_i\equal{}1$ for all $ i$

[quote="spanferkel"] by the conditions, $ x_i \equal{} 1$ for all $ i$[/quote]

 :wallbash_red: i make very poor mistake.

In fact, i want to ask is:

Let $ x_1, x_2, ..., x_n$ be a set of positive numbers, where $ 0 < x_i \le 1$.

prove or dispove

$ \frac {\sum^n_{i \equal{} 1} x_i^2 \minus{} 1}{\sum^n_{i \equal{} 1} x_i^3 \minus{} 1} \ge \frac {\sum^n_{i \equal{} 1} x_i^3}{\sum^n_{i \equal{} 1} x_i^4}$.

(which is more easy easy easy....)


This is one of my examples from my textbook.. I don't think it is quite right, or at least, I don't see how they came up with what they have.

[hide="Quoted from book"]
[b]EXAMPLE 3:[/b] A box of 10 tubes of construction adhesive has two tubes with defective seals.  Three tubes are selected from the box by a customer.  If it is assumed the tubes are randomly arranged in the box, what is the probability that the customer will obtain no defective tubes?

[b]SOLUTION:[/b] Let the three events $ N_1, N_2,$ and $ N_3$ denote the outcomes that [i]"no defective is obtained on the 1st selection, "no defective is obtained on the second selection," [/i] and [i]no defective is obtained on the third selection,[/i] respectively.  Thus,

$ \displaystyle P(\text{No defective}) \equal{} P(N_1 \text{and} N_2 \text{and} N_3) \equal{} P(N_1) \cdot P(N_2 | N_1) \cdot P(N_3 | N_1 \text{and} N_2) \equal{} \frac {8}{10} \cdot \frac {6}{9} \cdot \frac {7}{24} \equal{} 0.29167$
[/hide]

Source: [i]Introduction to Statistics: A Calculus-Based Approach[/i] by Howard B. Christensen

I think that the book is wrong. For one thing, there is no way I end up with 7/24 when 7 is not even one of the factors of products. So, I think that the answer should be

$ \frac {8}{10} \cdot \frac {7}{9} \cdot \frac {6}{8} \equal{} \frac {7}{15}$

Can anyone confirm this? Or at least present me with solution that is actually correct?

I'm siding with you. The easy way to see it is as $ \binom{8}{3}/\binom{10}{3}$, which is your answer. Stat books are almost notorious for containing errors.

err... if you are doing it that way, then it is 7_C_2 / 10_ C_ 2

By way of explanation:  Bictor's expression for the answer counts the number of ways the customer can choose 3 tubes from among the 8 non-broken tubes, divided by the number of ways the customer can choose any 3 tubes.  Altheman's expression for the answer counts the number of ways we can choose 2 tubes to be broken from among the 7 the customer didn't choose, divided by the number of ways to choose any 2 of the 10 tubes to be broken.  Both answers (and Silverfalcon's, but not Silverfalcon's book's) are equal to $ \frac{7}{15}$.

Thanks for explanations!

I didn't know statistics book was notorious for its errors but anyway, no wonder why I couldn't understand it after staring it for 30 minutes!  :mad:

Another way to write the same thing is with a [[multinomial coefficient]], $ \frac {\binom{10}{3;\,2}}{\binom{10}{3} \cdot \binom{10}{2}}$, where $ \binom{10}{3;\,2} \equal{} \frac {10!}{3!2!5!}$ counts the number of ways to select one subset of size 3 and one disjoint subset of size 2 from a set of size 10.  (This is often written $ \binom{10}{3; \, 2; \, 5}$, too.)  Then Bictor's resolution is to note that $ \binom{10}{3;\,2} \equal{} \binom{10}{2} \cdot \binom{8}{3}$ while Althemans is to note that $ \binom{10}{3;\,2} \equal{} \binom{10}{3}\cdot \binom{7}{2}$ (and you can prove these with the same combinatorial arguments I suggested in my last proof, or with formulas and factorials, or even other methods).


[quote="mathStrugler"]In many books I stumbled accross ≡ symbol but I don't see how is this symbol different than good old equal symbol =.

Does anybody know any book which explains how to use mathematical language in detail?

Thank you[/quote]

$\equiv$ is used when dealing with equality in mods

$\equiv$ is the congruency sign. Say if you had the expression $12 \equiv 7 \pmod{5}$ you would read it as "12 is congruent to 7 modulo 5."

$\equiv$ means "is congruent" and is used in modular arithmetic.

The expresion $7\equiv2\pmod{5}$ is read "seven is congruent to 2 modulo 5." It means, more or less, that 7 has a remainder of 2 when divided by 5. For more information on modular arithmetic, you can look on AoPS Wiki or buy the new Intro to Number Theory book.

[quote="mathStrugler"]In many books I stumbled accross ≡ symbol but I don't see how is this symbol different than good old equal symbol =.
Does anybody know any book which explains how to use mathematical language in detail?
Thank you[/quote]
≡, is for congruences
=, is for equalities

[quote="t0rajir0u"][b]Definition:[/b]  $a \equiv b \bmod m$ if and only if $m | (b-a)$.
There are also some non-number-theoretic contexts where the $\equiv$ sign has a very specific meaning, but none come to mind at the moment.[/quote]
$\equiv$ can also be used for congruent triangles.

the 3 lined equal sign means "defined as" whille the regular = sign just represents an equality. Basically, they are the same thing except the three lined sign is just more significant and represents a stronger statement.

That's how I've seen it as well.  That is, $\equiv$ is the same as $: =$ in certain contexts.

= it's the equality sign and $\equiv$ it's the equivalent sign and it's used to:
1) for identities (e.g. $(a+b)^{2}\equiv a^{2}+2ab+b^{2}$ )
2) for congruences(e.g. $7\equiv2\pmod{5}$ )
3) in triangles equalities (e.g. $\triangle{ABC}\equiv \triangle{DEF}$ )

[quote="SplashD"]for congruences, you use $\cong$ not $\equiv$[/quote]
You can use either.

I have also come across it when doing series and stuff.  For example:

Prove that $\sum_{r=1}^{n}r(r+1) = \frac{1}{3}n(n+1)(n+2)$

The solution would start:
Let $u_{r}\equiv r(r+1)$    (1)

If I was saying (1), would I say"let $u_{r}$ be equivalent to $r(r+1)$"?

First post :D

That would be a usage corresponding to $: =$.  In other words, you are defining the meaning of $u_{r}$.

[quote="SplashD"]for congruences, you use $\cong$ not $\equiv$[/quote]

Well, it depends on what you mean by "congruences".

If you mean that triangle ABC is "congruent" to triangle DEF, you would use $\cong$.

However, if you mean that 842 is "congruent" to 42 modulo 100 (I have herad that terminology somewhere), you would use $\equiv$.

Also, as a notation for geometrical points that coincide with one another it's often used $A\equiv B$ rather than $A\cong B$.

We can also talk about generalized congruences in group, ring, and field theory, but that's a somewhat advanced topic  :D


hey....post our phsics marks in aits pt1. i think i could have got a good score..if i had the time to attempt integer tpe ques.
    i got- 80/160. :blush:

[quote="nuclear.parag"]its...all india test series of fiitjee!!...[/quote]
You should specifiy. There are sooooooo many aits here.


[quote="xxxyyyy"]How many individual digits are there in the numbers from $1$ to $10,000,000,000$?(Ten Billion)

And what is the sum of those individual digits?

No direct calculations please....[/quote]

10 billion*10/10*(1+2+3+4+5+6+7+8+9)+1=sum

so 10 billion*45+1 (because 10 billion has one 1)=sum of digits

you can see why by wrtting 1 as 0000000001 and so forth.

caculating individual digits is kind of boring:
9*1+9*10*2... +11 (because 10 billion has 11 digits) and so forth.

Or you can just do 10 billion*10+11-1*10-9*9-9*10*8 and so forth..


Prove that for any positive real numbers a, b this inequality is valid:
[img]http://content.imagesocket.com/images/Clipboard01325.jpg[/img]

the inequality on the right is equivalent to $ (a \minus{} b)^2\geq 0$.

for the one on the left, let $ a \equal{} x^2, b \equal{} y^2$. then the inequality becomes $ 5x^3y \plus{} 5xy^3\leq 2x^4 \plus{} 6x^2y^2 \plus{} 2y^4\Leftrightarrow 0\leq 2x^4 \minus{} 5x^3y \plus{} 6x^2y^2 \minus{} 5xy^3 \plus{} 2y^4\Leftrightarrow 0\leq (x \minus{} y)^4 \plus{} (x \minus{} y)(x^3 \minus{} y^3)$, the last of which is clearly true.

hello, alternatively squaring the inequality
$ \sqrt{ab}\le \frac{2(a^2\plus{}3ab\plus{}b^2}{5(a\plus{}b)}$ and simplifying and factoring we get
$ \frac{1}{25}\frac{(4a^2\plus{}7ab\plus{}4b^2)(a\minus{}b)^2}{(a\plus{}b)^2}\geq0$
Sonnhard.


Deer jumps over fence
With graceful leaping beauty.
Sine curve models height.

Six feet above ground,
Each leap requires $ 1$ second.
The equation, please?

When $ t\equal{}o$
The deer is over the fence
At its maximum height.

Find the sine or cosine curve modeling this.

[hide]$ y \equal{} |6\cos{\frac{\pi x}{2}}|$[/hide]


Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

[hide][u]4.5 hours.[/u] When the second cyclist starts, the first cyclist is 18 miles ahead. So, you solve for x in 10x = 6x + 18.[/hide]

(6 mph)(3 hours) = 18 miles, due to your topic title.


In connection with proof in geometry, indicate which one of the following statements is [i]incorrect[/i]:

$ \textbf{(A)}\ \text{Some statements are accepted without being proved.}$
$ \textbf{(B)}\ \text{In some instances there is more than one correct order in proving certain propositions.}$
$ \textbf{(C)}\ \text{Every term used in a proof must have been defined previously.}$
$ \textbf{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.}$
$ \textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.}$

[quote="AIME15"][hide="Click for solution"]
$ \textbf{(C)}$ must be incorrect because postulates must necessarily remain undefined.
[/hide][/quote]
[hide="Click for corrected explanation"]
$ \textbf{(C)}$ must be incorrect because some geometrical objects (like point and line) must necessarily remain undefined and postulates (or axioms) as truth.
[/hide]


We are attempting to put together a group of people who can answer questions about the major project competitions (Siemens Westinghouse and Intel).

Those of you who are interested in the Siemens Westinghouse and Intel programs (and considering the massive prizes, that should be many of you), please list here what questions you might have for them.

Just in general I want to know what type of process other winners have used in their research.  For example how they got their idea, what they did when they got a good idea, did they have help from a professional etc.

Also I would like to know what winners of the competition went on to do and if they made any more big contributions to science and/or mathematics.

I'm sure I can think of a lot more questions but this is all I have for now.

What types of projects win? Are you looking for something with practical applications or something that just shows creativity and ingenuity? Some people have said that certain contestants get especially impressive sounding projects merely by repeating or doing an experiment suggested by a mentor using a different substance or other object of study? Is this given the same weight as someone else who does their work independently through what they learn on their own? What resources are there available to students who cannot go to programs like RSI to do their research? Also, who are the judges? What is the experience like? Also, are there any benefits, e.g. internships, possibilities of doing more research, etc, after the competition? I am just tossing out some questions; I'll try to think up some more when I actually begin work on my project.

It seems to me that the top finishers in Olympiads and research contests are often disjoint. I guess that the skills and the types of motivation that these two sorts of contests require are not as similar as we'd imagine.

What's the difference between the Siemens-Westinghouse program and the Intel program?

How does one embark on a project? Do all winners tend to have professors as advisors? Do advisors usually suggest the subject to a student? Generally, who tends to be an advisor?

How does the [url=http://www.ditd.org/public/article.aspx?cid=1&mid=101&tp=2]Davidson Fellows[/url] program compare to the older science fair programs? Can we get Anders Kaseorg and Daniel Kane online here to answer questions about that program?

This is going to be a math jam, right? Or whatever it is, do you know around when the questions will be answered (beginning of summer, midsummer, in the fall, etc.)?

Out of curiousity, when will this take place?

Still not sure - we're rounding up people.  We hope for later in the summer or in the fall.

We have now scheduled a Siemens Westinghouse Math Jam.  Information can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=106071#106071]here[/url].

[quote="texas137"]I have an Intel question -

the instructions say that you can submit "up to 3" letters from teachers/advisors. In addition to general questions like a college app form would have (students strengths & weaknesses, work habits, blah, blah), the form asks "to what extent is the research the work of the student". So, if only one teacher knows anything about your research, should you just submit that one letter? Or should you also include a couple of letters from teachers just generally saying you're a great guy, but they don't know anything about your project?[/quote]

Please post this question at the link that I posted.  That is where the Siemens Westinghouse Competition people are compiling questions to answer.  Thank you.

[quote="MCrawford"]

Please post this question at the link that I posted.  That is where the Siemens Westinghouse Competition people are compiling questions to answer.  Thank you.[/quote]

Yes, but this is an Intel question, not a Siemen's Westinghouse question. It's clear from their instructions what Siemen's Westinghouse wants in the way of letters, but not for Intel. Should I still move?

Oops, sorry.  Don't post Intel questions there.

[quote]Also I would like to know what winners of the competition went on to do and if they made any more big contributions to science and/or mathematics.
[/quote]

I know I'm not even of Intel-age yet, but I know that Sheldon Glashow (Nobel Physics 1979) and Roald Hoffman (Nobel in Chemistry 1981, I believe) have both reached the finalist stage in Intel/Siemens Westinghouse.

[quote="texas137"]I have an Intel question -

the instructions say that you can submit "up to 3" letters from teachers/advisors. In addition to general questions like a college app form would have (students strengths & weaknesses, work habits, blah, blah), the form asks "to what extent is the research the work of the student". So, if only one teacher knows anything about your research, should you just submit that one letter? Or should you also include a couple of letters from teachers just generally saying you're a great guy, but they don't know anything about your project?[/quote]

Are they going to take letters of recommendation into consideration when they judge your project? If not, then what are they for?


abstract algebra
number theory
least common multiple
Ring Theory
group theory
modular arithmetic
superior algebra

1. Determine ALL the homomorphisms from $ (\mathbb Z_{2008}, \plus{} )$ to $ (\mathbb Z_{1230}, \plus{} )$.

2. Let $ f: \mathbb Z_{56}\to\mathbb Z_7\times\mathbb Z_8$ be given by $ f(x\mod56)\equal{}(x\mod7,x\mod8)$. Show that (i) $ f$ is bijective; (ii) $ f$ is isomorphism under addition.

1. First we prime factorize $ 2008$ and $ 1230$. We have $ 1230 \equal{} (2)(3)(5)(41)$, and $ 2008 \equal{} (2^3)(251)$. The homomorphism is completely determined by $ f(1)$ (with $ 1 \in \mathbb{Z}/2008$). The possible images of $ 1$ are $ \{x \in \mathbb{Z}/1230 | 2008x \equal{} 0\}$. The only such elements are $ 0$ and $ 615$, so there are only two homomorphisms.

2 is just the Chinese Remainder Theorem. In fact, it is a ring isomorphism, by the CRT for rings. If you'd like more explanation, please ask.

[quote="calc rulz"]The possible images of $ 1$ are $ \{x \in \mathbb{Z}/2008 | 2008x \equal{} 0\}$.[/quote]
Thank you very much. But the above set should be $ \{x \in \mathbb{Z}_{1230} | 2008x \equal{} 0\}$

[quote="calc rulz"]2 is just the Chinese Remainder Theorem. In fact, it is a ring isomorphism, by the CRT for rings. If you'd like more explanation, please ask.[/quote]
I am just learning group theory now. But if you wouldn't mind explaining to me in detail, I will be pleased to hear from you.

[quote="ifai"]1. Determine ALL the homomorphisms from $ (\mathbb Z_{2008}, \plus{} )$ to $ (\mathbb Z_{1230}, \plus{} )$.

2. Let $ f: \mathbb Z_{56}\to\mathbb Z_7\times\mathbb Z_8$ be given by $ f(x\mod56) \equal{} (x\mod7,x\mod8)$. Show that (i) $ f$ is bijective; (ii) $ f$ is isomorphism under addition.[/quote]

Good point. Note that $ \mathbb{Z}/2008$ is another notation for the integers mod 2008.

Let's say we have two integers $ a$ and $ b$. Then there is a homomorphism $ f: \mathbb{Z}_{lcm(a,b)} \to \mathbb{Z}_a \times \mathbb{Z}_b$ given in a similar way as your function. Let's say that $ f(x) \equal{} 0$, meaning that $ a|x$ and $ b|x$. Then a basic property of the least common multiple is that these two facts imply $ lcm(a,b) | x$, meaning that $ x \equal{} 0$ in $ \mathbb{Z}_{lcm(a,b)}$. This means that the kernel is $ 0$. You can check using basic properties of modular arithmetic that this is a homomorphism. So since the kernel is $ 0$, it must be injective. When is it bijective? Simply when $ lcm(a,b) \equal{} ab$, which is when $ a$ and $ b$ are relatively prime. This is true for $ a \equal{} 7$, $ b \equal{} 8$, right?

Think about what happens when $ a \equal{} 2$ and $ b \equal{} 4$. What elements of $ \mathbb{Z}_2 \times \mathbb{Z}_4$ are not in the image of $ f$? Why can't we define a homomorphism from $ \mathbb{Z}_8$ into $ \mathbb{Z}_2 \times \mathbb{Z}_4$? (i.e. why do we need to use the lcm?)


I like how AOPS advertises now: its on every AMC phamphlet and it is shown on google sponsered links. BTW how much does google receive for advertising.

https://adwords.google.com/select/ has all the details you'd ever want about the Google AdWords system.

A word to the wise: don't click on the google adwords thing if you already know about AoPS.  That just costs AoPS money.

Mathew came to a local math tournament and they handed out flyers with order forms one time. This is a good idea too. (Even though I realize that they can't come every time). I think a lot of AoPS has also spread by word of mouth, (which is how I learned about it..).

The mail also helps. (like AMC top scorers might be interested)

mathfanatic i'm not that stupid btw i never click on ad links unless google is really messed up (no good website) i use netscape instead of msn explorer because of the nice pop up add blocker ads make me so pissed of and i like how google has it on the right.


prove that for positive reals
\[a\ b\ c \]
we have the following inequality true:
\[\sqrt{\frac{ab}{(c+b)(c+a)}}+\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ca}{(b+c)(b+a)}}\le \frac{3}{2}. \]

By AM-GM $\sqrt{\frac{ab}{(c+b)(c+a)}}\leq\frac{1}{2}(\frac{a}{c+a}+\frac{b}{c+b})$, etc. Done :D

Note: $1=\frac{a}{a+b}+\frac{b}{a+b}$, etc.

[quote="anuj kumar"]oh! it became so easy.
nevertheless observe that its equvivalent to:
\[cosA+cosB+cosC\le3/2 \]
in any triangle.[/quote]
Why?  :maybe:

let $x=\sqrt{\frac{bc}{(a+b)(a+c)}}\ similarly\ for\ y\ and \ z$
then observe that
\[x^{2}+y^{2}+z^{2}+2xyz \]

\[=\frac{sum_{sym}(a^{2}b)+2abc}{(a+b)(b+c)(c+a)}=1.so\ we \ ca n\ substitute \ x=cosA,y=cosB,z=cosC apply jensen . \]
thanks.

Wow, thanks you very much, nice solution!  :wink:

[quote="anuj kumar"]let $x=\sqrt{\frac{bc}{(a+b)(a+c)}}$ and similarly for $y$ and $z$.
Then observe that
\[x^{2}+y^{2}+z^{2}+2xyz=\frac{\ds%Error. "fancy" is a bad command.
\big\large\sum_{sym}a^{2}b+2abc}{(a+b)(b+c)(c+a)}=1 \]
So we can substitute
\[x=\cos A, y=\cos B, z=\cos C \]
and apply jensen

thanks.[/quote]


limit
real analysis
real analysis unsolved

Let (X,p) be a metric space. For each pair of non-empty subsets A and B of X, we set d(A,B) = inf{p(x,y)|x in A, y in B}.

Two questions:

1. Suppose that both of the non-empty disjoint subsets A and B are compact. Prove that there exist points x_o in A and y_o in B with p(x_o,y_o) = d(A,B).

2. Suppose that (X,p) is a Euclidean space R^n, for some n > or = 1, endowed with the usual metric. Prove that if A and B are non-empty disjoint subsets of X, with A compact and B closed, then again there exist points x_o in A and y_o in B with p(x_o,y_o) = d(A,B).

$\rho(A,B)=\inf d(x,y)$ so there exists $x_{n},y_{n}$ such that $d(x_{n},y_{n}) \to \pho(A,B),n\to \infty$. A - is a compact, so there exist $x_{n_{k}}$ subsequence of $x_{n}$ such that $\exists \lim x_{n_{k}}= x \in A$. And as above exists $y_{n_{k_{l}}}$ subsequence of $y_{n_{k}}$ such that $\exists \lim y_{n_{k_{l}}}= y \in B$.
Now: d is continuous, so $d(x,y)=\lim d(x_{n_{k}},y_{n_{k_{l}}})=\rho(A,B)$

as above, let $r=\rho(A,B) = inf d(x,y) = \lim d(x_{n},y_{n}).$
exists $N$ such that $\forall n>N d(x_{n},y_{n})<2r.$ now $\exists x_{n_{k}}$ - subsequence of $x_{n}$ such that $x=\lim x_{n_{k}}$. So $\exists M : \forall k>M d(x,x_{n_{k}})<r.$ Thus for $k>max{M,N}$ we have ${d(y_{n_{k}},x})<d(y_{n_{k}},x_{n_{k}})+d(x_{n_{k}},x)<4r.$ So all $y_{n_{k}},k>max{N,M}$ are in $F=B\cap S(x,4r)$. The last set if bounded and closed so $F$ compact and we can use case 1)


function
trigonometry
calculus
calculus computations

Prove that

$ \phi_n(x) \equal{} \sin (n\pi x/L)$

has $ n\minus{}1$ zeros for $ 0 < x < L,$ where $ n \in N, L \in R^\plus{}$.


A box contains six cards. Three of the cards are black on both sides, one card is
black on one side and red on the other, and two of the cards are red on both sides. If
you pick a card at random from the box and see that the side facing you is red, what is
the probability that the other side is red? Express your answer as a common fraction.

So I set the cards as...
B B B R R R
B B B B R R

Since you pick a red card, it can only be the last 3 cards. Of the last three cards, the other side is red on only two of them. So, the answer appears to be 2/3. But the answer key says the answer is 4/5.

Help is much appreciated! :P

For the two cards that are red on both sides, you can pick up the card and have the side facing you be red in two ways (for each card): front or back.  So there are $ 2\cdot 2 \equal{} 4$ ways to pick up a red card with red on the other side.  For the card that is red on one side and black on the other, there is only one way: the side that is red.  So out of the total of $ 4 \plus{} 1 \equal{} 5$ ways you could choose the card, $ 4$ of the ways have red on the other side of the card.  So the probability is $ \boxed{\frac {4}{5}}$


[quote="hunter34"]The Youtube tags are not working. Instead of the video you get:

[youtube]vSKiKPXkM90[/youtube]

Can you fix this?[/quote]

You must have it set to disable BBCode.  Fix that (in your profile, or post by post) and it should work.


geometry
cyclic quadrilateral
geometry unsolved

Consider a cyclic quadrilateral ABCD. Let S be the intersection of diagonals AC and BD. Let M, N be the midpoints of the sides AD and BC, respectively. Let E, F be the feet of the perpendiculars of point S on the sides AB and CD, respectively. Prove that the lines AC, FM and EN are concurrent.

$ NE, MF$ cut $ AC$ at $ X, Y$ respectively. By Menelaus theorem for $ \triangle ABC, \triangle ADC$ cut by the lines $ NEX. MFY,$ $ \frac{\overline{XA}}{\overline{XC}} \div \frac{\overline{YA}}{\overline{YC}} \equal{} 1$ $ \Longrightarrow$ $ X \equiv Y.$


:first: :winner_first: 1=2
:winner_second: Dgreenb801
:winner_third: Boy Soprano II

Congratulations to the winners, though

[hide]
, *cough*, there wasn't exactly much competition.[/hide]

I won 2nd without even knowing it until now(April 4). Anyways, I like how not all the answers are filled in so you can do it yourself and feel proud of it :)


Let $ a,b,c>0 ,\alpha \ge 1, abc\equal{}1$. Prove that:
$ \sum \frac{a^{\alpha}}{(ab\plus{}a\plus{}1)^{\alpha\plus{}1}} \ge \frac{1}{a\plus{}b\plus{}c}$

[hide="A first step"]$ \sum_{cyc}\frac {a}{ab \plus{} a \plus{} 1} \equal{} 1$ (show this by the standard substitution) so the ineq. becomes $ (a \plus{} b \plus{} c)\sum_{cyc}\frac {x^{\alpha \plus{} 1}}{a}\ge 1$ with $ x \plus{} y \plus{} z \equal{} 1$. Maybe the inequality $ (a \plus{} b \plus{} c)\sum_{cyc}\frac {x^{\alpha \plus{} 1}}{a}\ge k\left(x^{\alpha} \plus{} y^{\alpha} \plus{} z^{\alpha}\right)$ can be proven for some suiting constant $ k$? :maybe:[/hide]

[quote="NguyenDungTN"]Let $ a,b,c > 0 ,\alpha \ge 1, abc \equal{} 1$. Prove that:
$ \sum \frac {a^{\alpha}}{(ab \plus{} a \plus{} 1)^{\alpha \plus{} 1}} \ge \frac {1}{a \plus{} b \plus{} c}$[/quote]

mmmm, I don't know may be the sign must be reversed, set $ a\equal{}b\equal{}c\equal{}1$:
$ \sum \frac {a^{\alpha}}{(ab \plus{} a \plus{} 1)^{\alpha \plus{} 1}}\equal{}\sum{\frac{1}{3^{\alpha\plus{}1}}}\equal{}\frac{1}{3^{\alpha}}\ge\frac {1}{a \plus{} b \plus{} c}\equal{}\frac{1}{3}\Leftrightarrow \alpha\le1$

So it may be:

$ \sum \frac {a^{\alpha}}{(ab \plus{} a \plus{} 1)^{\alpha \plus{} 1}} \le \frac {1}{a \plus{} b \plus{} c}$

 :)

Substituting $ a = \frac {y}{z},b = \frac {z}{x},c = \frac {x}{y}$ we obtain $ \frac {1}{ab + a + 1} = \frac {zx}{xy + yz + zx}\implies \sum_{cyc} \frac{1}{ab+a+1} = 1$. In addition the function $ f(x) = x^{\alpha + 1}$ is convex. Thus, by weighted Jensen we obtain
\begin{align*} \sum_{cyc}\frac {a^\alpha}{(ab + a + 1)^{\alpha + 1}} & = \sum_{cyc}\frac {1}{a}f\left(\frac {a}{ab + a + 1}\right) \\
& \ge\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right) f\left(\frac {\sum_{cyc}\frac {1}{ab + a + 1}}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}}\right) \\
& = \frac {1}{\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)^\alpha} \end{align*}
which is a pretty sharp inequality from which one can certainly derive the wanted one :)

[quote="bertram"]
$ \ge\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) f\left(\frac {\sum_{cyc}\frac {1}{ab \plus{} a \plus{} 1}}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\right) \equal{} \frac {1}{\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^\alpha}$
which is a pretty sharp inequality from which one can certainly derive the wanted one :)[/quote]
EDIT: oh i see...you used that clever substitution...ahaha

EDIT2: Okay but how do you get the desired inequality from there?  You obviously don't always have $ a\plus{}b\plus{}c \ge \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$...

Yeah,but as ElChapin has pointed out the inequality isn't true anyway. ;) An indication is that for $ \alpha\equal{}1$, equality holds, but not for $ \alpha>1$. I think it's a generalization of the inequality $ \sum_{cyc} \frac{a}{(ab\plus{}a\plus{}1)^2} \ge \frac{1}{a\plus{}b\plus{}c}$ (see the third post [url=http://www.mathlinks.ro/viewtopic.php?t=179622]here[/url]), I would generalize it in the following way:

Let $ a,b,c,\alpha>0$ with $ abc\equal{}1$. Prove that $ \sum_{cyc} \frac{a}{\left(ab\plus{}a\plus{}1\right)^{\alpha\plus{}1}} \ge \frac{1}{(a\plus{}b\plus{}c)^\alpha}$.


geometry
circumcircle
perpendicular bisector
geometry proposed

Let $ ABC$be a triangle with $ \angle BAC\neq 45^{\circ}$ and $ \angle AIO\equal{}90^{\circ}$, where $ (O)$ and $ (I)$ are its circumcircle and incircle, respectively. Choose a point $ D$ on the ray $ BC$ such that $ BD\equal{}AB\plus{}AC$. The tangent line through $ D$ touches $ (O)$ at $ E$. The tangent line through $ B$ of $ (O)$ meets $ DE$ at $ F$ and $ CF$ meets $ (O)$ at another point denoted by $ K$. Let $ G$ be the centroid of the triangle $ ABC$. Prove that $ IG$ is parallel to $ EK$.

Lemma: Let $ BC$ be arbitrary chord of a circle $ (O)$ and $ D$ point on the ray $ (BC,$ such that $ \overline{BD} \equal{} 2\ \overline{BC}.$ Tangent from $ D$ touches $ (O)$ at $ E$ (does not matter which one), tangent to $ (O)$ at $ B$ cuts $ DE$ at $ F,$ and $ CF$ cuts $ (O)$ at $ K.$ Then $ KE \parallel BC.$

Power of $ D$ to $ (O)$ is $ DE^2 \equal{} \overline {DB} \cdot \overline{DC} \equal{} 2\ CB^2.$ Tangent lengths $ FB \equal{} FE$ from $ F$ to $ (O)$ are equal. $ FC$ is the F-median of the $ \triangle BDF,$

$ FC^2 \equal{} \frac {_1}{^4}\left(2\ FB^2 \plus{} 2\ FD^2 \minus{} BD^2\right) \equal{} \frac {_1}{^4}\left(2\ FE^2 \plus{} 2\ (\overline {FE} \plus{} \overline {ED})^2 \minus{} 4\ BC^2\right) \equal{}$

$ \equal{} FE^2 \plus{} \overline {FE} \cdot \overline{ED} \equal{} \overline {FE} \cdot \overline{FD}$

Consequently, $ FC$ is tangent to the circumcircle of the $ \triangle CDE$ and the angles $ \angle FCE \equal{} \angle CDE$ are equal. Perpendicular bisector of the segment $ BC$ cuts the arc $ BC$ of $ (O)$ containing $ E$ at $ M.$ Since $ OM \perp CD$ and $ OE \perp ED$ it follows that the angles $ \angle MOE \equal{} \angle CDE$ are also equal. Combining, $ \angle KCE \equal{} \angle FCE \equal{} \angle MOE,$ $ \angle KOE \equal{} 2\ \angle KCE \equal{} 2\ \angle MOE,$ which means that $ OM$ is also the perpendicular bisector of the segment $ KE$ and $ KE \parallel BC.$

Problem: Condition $ \angle CAB \neq 45^\circ$ is completely unnecessary. $ AI \perp OI$ $ \Longleftrightarrow$ $ b \plus{} c \equal{} CA \plus{} AB \equal{} 2\ BC \equal{} 2a$ $ \Longleftrightarrow$ $ IG \parallel BC.$ See [url]http://www.mathlinks.ro/viewtopic.php?t=210342[/url]. By the problem conditions, $ b \plus{} c \equal{} \overline {BD} \equal{} 2\ \overline {BC} \equal{} 2a,$ hence by the lemma, $ KE \parallel BC.$ Again by the problem conditions, $ IG \parallel BC,$ hence $ KE \parallel IG.$ Nice obfuscation, though.

[quote="yetti"]...
[b]Lemma:[/b] Let $ BC$ be arbitrary chord of a circle $ (O)$ and $ D$ point on the ray $ (BC,$ such that $ \overline{BD} \equal{} 2\ \overline{BC}.$ Tangent from $ D$ touches $ (O)$ at $ E$ (does not matter which one), tangent to $ (O)$ at $ B$ cuts $ DE$ at $ F,$ and $ CF$ cuts $ (O)$ at $ K.$ Then $ KE \parallel BC.$
...[/quote]
[b]Proof.[/b] We have $ BKEC$ is a harmonic quadrilateral, it implies $ KE\cdot BC\equal{}KB\cdot EC$. On the other hand, $ BC\equal{}CD$, so we conclude that $ KE\cdot CD\equal{}KB\cdot EC$, i.e. $ \frac{KE}{KB}\equal{}\frac{CE}{CD}$. Combining with $ \measuredangle EKB\equal{}\measuredangle ECD$, we have $ \triangle EKB\sim\triangle ECD$, which follows $ \measuredangle EBK\equal{}\measuredangle EDC$. Therefore $ \measuredangle KEF\equal{}\measuredangle CDE$, hence $ KE\parallel CD\equiv BC$, as desired.


What do you believe is the current state of the US economy? Especially, since the effects of the hurricanes? As far as I can see, most economists surveyed believe the economic effect is only temporal, and they believe that we should be back to regular economic growth soon even though we can lose up to a bit less then a percentage of GDP for this year, I think. So they believe the effects are quite moderate. It seems also, that the high gas prices are just because of the great demand for oil, and I believe, as does the President, that we should look for new sources of energy before our oil is done and stop relying only on petroleum. So our economy I believe, even though not in a great boom, but in a fine economic expansion.

Your opinions?

The economy could be doing a lot better if so much of the tax pool was not in the hands of people who waste it.  This is always the case in a highly controlled economy.  This brings about the question: do we expect the level of waste to increase or decrease?

The advancement of technology means that economies are always increasing if they are not overly taxed by wasters who mitigate economic gains.

You also bring up the important issue of energy consumption.  Some historical economic gains have been what could be called "artificial technology" as those gains are not based purely on the ability of people to produce for themselves, but rather their ability to exploit existing resources.  It is unclear to me as to the exact difference this makes, though I'm sure we could back out a worst case scenerio calculations from the costs of costs of purely renewable resources.

However, that's a headache I'm not about to engage in.

[quote="hello"]So who do you think is being wasted in the tax pool?[/quote]

That's a very very long list.  Half of all money in the U.S. is spent by governments (federal, state, local), though my opinion is that far less than half of U.S. productivity is due to government services.

If government were efficient, much of it would be privatized.  Only the monopolization of force (military) really constitutes government.  The rest is just a political game based on all the money floating around.

Yes, I agree with you that government controlled companies are doomed to failure because they are monopolistic and have no competition, which means that whatever services they provide, there is no choice given to the individual. That is why I believe a free market system is superior to communist, because, there is choice (at least thats one of the things).

Now, I am taking public transportation to my CO-OP job and it takes me like 2-2.5 hours one way to get there. I take SEPTA (south eastern transportation authority) which is government owned, they also have a great possibility of striking now, which I think would be horrific, because I dont have any alternatives and I am not sure all the thousand of other Philadelphians and other Pennsylvanians do. Sorry for describing my situation.

[quote="hello"]Yes, I agree with you that government controlled companies are doomed to failure because they are monopolistic and have no competition, which means that whatever services they provide, there is no choice given to the individual. That is why I believe a free market system is superior to communist, because, there is choice (at least thats one of the things).

Cut[/quote]

As one extremely passionatr about business and one who has lived in Austria, Hungary, Taiwan, and the United States, I would like to point a few things out:

- It is merely impossible and improperable for one to pick a side and stick to it. However, I would say that the US economy is gradually building up due to low taxations.

- For those who are unaware of how the economy works, lower tax, yes, means less federal budget in a way, but more federal budget in another. Ironic? Yes, and no. Lower taxes mean more consumption, it's logical, and with more consumption, it means more federal budget. Therefore, the Bush administration isn't actually doing such a terrible job.

- One can also argue that the US economy, without looking at the stock market, as imploding at an exponential rate because the US government owes a couple of trillion dollars, and that will one day backfire when it can no longer maintain stabile interest payments. For instance, let's just say that the US federal governments owes one million one trillion dollars, which it will one day, the entire country will collapse and break like a pair of 1mm chopsticks.

- At this moment, higher and lower taxations don't really matter, because much of the budget is spent in a not-worth-fighting-for war. So, lower tax means that less fundings for the public. Higher tax means more weapons and more killing in Iraq. Let me leave the Americans to decide on this *avoids eye contact*

Well, as Bush has proven, the rate of taxation and the rate of federal spending don't have to be all that well correlated.

Yes, it appears that Bush is not a true fiscal conservative, I dont know what makes him a true Republican? Does he know that low taxes are a preequisite to low spending? He should be in a party that is called the democratic republicans, those who favor low taxes and high spending. Nice combination eh?

As both of the major parties have determined, what determines a true party member is a willingness to associate with the party as well as an ability to be elected.

Since the parties don't really have any independent power in the US, the most senior (in rank) member of the party ususally controls the party itself.  There was real problem with Kennedy when Clinton was President and ex-President.

Imagine for a moment if the Republicians go so mad at Bush that they kicked him out.  What would change other than making the man they have to work with angry.

That's why the parties are so alike, neither actually polices its members or representatives claiming affilation.

I think both parties have at least some affiliation with their ideas, thats why they often vote similarly, even though not always.

You're right about the members of a party having similar ideas.  I think that is the primary reason for membership in a group who's idea's one can agree with.  However, if one looks back at the Blue Dogs in the Democratic Congress (Clinton and before), they had to be Democrats as the South was still voting the Civil War.  The majority would be better fits in the Republician party (many shifted when the Congress switched and they had a reason to tell their constituents -- more political power).  It just was that the party (either party) has no ability to kick someone out, even if it would do them some good.

[quote]Yes, I agree with you that government controlled companies are doomed to failure because they are monopolistic and have no competition, which means that whatever services they provide, there is no choice given to the individual. That is why I believe a free market system is superior to communist, because, there is choice (at least thats one of the things).
[/quote]
Yes and no.
Take care observing that several markets cannot work efficiently with the free market system.
If an activity has always decreasing marginal costs, it will always end in a monopoly.
And if a monopoly has to be, it would be much better if it is owned by Government.
Furthermore, it's not true that government companies are doomed to failure just because they have no competition.
Generally, their losses are caused by the undercost product sell, due to political targets.


Is it true?
suppose that d and d' are two parallel line and P is a fix point between them. and A and B are two points on d and d' so that angle APB=k which k is fix too.is it true that all the circles which pass from p and A and B,are coaxal?
Thank you.


Three numbers, a, b, and c, none zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Find b.

[quote="anirudh"]what is the answer? i am bewilderd. thanx[/quote]
[hide="This problem is easy if you know what are arithmetic and geometric progressions"]
$a,b,c$ arithmetic: $a+c=2b$ (1)
$a+1,b,c$ geometric: $(a+1)c=b^2$ (2)
$a,b,c+2$ geometric: $a(c+2)=b^2$ (3)

If you know a little algebra, you can do a little more with (2) and (3) and get $c=2a$. Substitute it into (1), you get $a+2a=2b$, $2b=3a$

Put both $c$ and $b$ into either (2) or (3), I choose (2) and get $(a+1)(2a)=(\frac32a)^2$

Solve (divide by $a$ because $a\neq 0$) and get $2a+2=\frac94a$, $a=8$, $b=12$, $c=16$[/hide]

I hope you learned, anirudh

Is this problem MATHCOUNTS level problem? This problem is like Getting Started...


Estw $ a,b$ pragmatikoi arithmoi tetoio wste:

                                        $ 9a^2\plus{}8ab\plus{}7b^2 \leq 6$

Apodeikste oti:
                                        $ 7a\plus{}5b\plus{}12ab \leq 9$

Υποθέτουμε ότι $ 7a\plus{}5b\plus{}12ab>9$ . Τότε θα έχουμε ότι $ 7a\plus{}5b\plus{}12ab\plus{}6>9\plus{}9a^2\plus{}7b^2\plus{}8ab$ ή ισοδύναμα

$ 7(a\minus{}\frac{1}{2})^2\plus{}5(b\minus{}\frac{1}{2})^2\plus{}2(a\minus{}b)^2<0$ που είναι άτοπο ! Άρα $ 7a\plus{}5b\plus{}12ab\leq 9$

Ωραία λύση sil!

[quote="r_boris"]Μπορούμε να δείξουμε οτι η πρώτη σχέση παριστάνει το εσωτερικό έλλειψης το οπόιο βρίσκεται κατω από την εφαπτομένη της στο σημείο (1/2,1/2) που είναι και εφαπτομένη της υπερβολής που παριστάνει  η 2η ισότητα. Το ότι είναι ανισότητα κανει τα σημεία του επιπέδου να βρίσκονται στο κατάλληλο χωρίο πάνω από την εφαπτομένη της και να ισχύει το ζητούμενο Η συμμετρία ολοκληρώνει την λύση[/quote]

Nice, very nice :)

Durandal 1707

[quote="Protonios"]arqady...hehe :rotfl:[/quote]
Οντως......
Τι κακος λυτης ανισοτητων....... :rotfl:  :rotfl:


modular arithmetic
number theory proposed
number theory

(a) Show that for every positive integer $ n$ there exists a set $ M_{n}$ consisting of $ n$ positive integers and having the following property:

(i) the arithmetic mean of elements in any nonempty susbset of $ M$ is an integer

(ii) the geometric mean of elements of any nonempty subset of M is an integer

(iii) both arithmetic mean and geometric mean of elements of any nonempty subset of $ M$ are integers.

(b) Is there an infinite set $ M$ of positive integers having property (1)?

(a)

(i) sure: just make sure all elements are divisible by $ n!$.

(ii) sure: just make sure all elements are $ n!$-powers of integers.

(iii) sure: just make sure all elements are both divisible by $ n!$ and are $ n!$-powers. For instance, let $ M_n \equal{} \{(n!)^{n!}, (2n!)^{n!}, (3n!)^{n!}, \ldots, (n\cdot n!)^{n!} \}$. For $ 1 \leq k \leq n$, all elements are divisible by $ k$ and are $ k$th powers. So the sum of $ k$ elements is divisible by $ k$, and the product of $ k$ elements has a $ k$th root which is an integer.

(b) No. Let $ a<b$ be the smallest members of $ M$, and let $ p$ be a prime greater than $ b$. Thus $ a \neq b \pmod{p}$. Take any $ p\minus{}1$ memebers of $ M$ other than $ a,b$, say $ x_1,\ldots,x_{p\minus{}1}$. The sum $ S \equal{} a\plus{}x_1\plus{}x_2\plus{}\ldots\plus{}x_{p\minus{}1}$ must be divisible by $ p$, but then the sum $ S' \equal{} b\plus{}x_1\plus{}x_2\plus{} \ldots \plus{} x_{p\minus{}1} \equal{} S\plus{}(b\minus{}a)$ cannot be divisible by $ p$.


geometry
analytic geometry
geometry proposed

[quote="minhtuan_lequydon_dn"][quote="darij grinberg"][quote="minhtuan_lequydon_dn"]Let delta ABC and M is the point in delta ABC . Let A' , B' , C' on BC , CA , AB with A'B' // AM , B'C' // BM , C'A' // CM. Prove that : 3.S(A'B'C')   =< S(ABC)  :P  :P  :P[/quote]

http://www.mathlinks.ro/viewtopic.php?t=32873
Do you have a proof without affine transformations or computations?

  darij[/quote]

Oh, Your proof for the lines AQ, BQ, CQ are the altitudes of triangle ABC ?? but they don't ...[/quote]

As I said, you can reduce the general case to the case when AQ, BQ, CQ are the altitudes of triangle ABC with the help of affine transformations. (Please read my post, not just glance over it.)

It's not what I consider a really beautiful proof, though.

  darij

[quote="darij grinberg"][/quote][quote="minhtuan_lequydon_dn"][quote="darij grinberg"][quote="minhtuan_lequydon_dn"]Let delta ABC and M is the point in delta ABC . Let A' , B' , C' on BC , CA , AB with A'B' // AM , B'C' // BM , C'A' // CM. Prove that : 3.S(A'B'C')   =< S(ABC)  :P  :P  :P[/quote]

http://www.mathlinks.ro/viewtopic.php?t=32873
Do you have a proof without affine transformations or computations?

  darij[/quote]

Oh, Your proof for the lines AQ, BQ, CQ are the altitudes of triangle ABC ?? but they don't ...[/quote][quote="darij grinberg"]

As I said, you can reduce the general case to the case when AQ, BQ, CQ are the altitudes of triangle ABC with the help of affine transformations. (Please read my post, not just glance over it.)

It's not what I consider a really beautiful proof, though.

  darij[/quote]
I don't study affine transformations . Are you have a another proof for " For every triangle ABC and every point N inside triangle ABC, there exists an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point N to a point N' such that the point N' is the orthocenter of triangle A'B'C'. "  :(  :(  :(

Not really. You should read jmerry's proof if you don't like mine - it is more elementary. Although jmerry mentions barycentric coordinates, he doesn't really use them - just read
"For the proof, set $ m\equal{}\frac{CM}{BC}$, $ n\equal{}\frac{AM}{CA}$, $ p\equal{}\frac{BM}{AB}$, $ a\equal{}\frac{\left|BCQ\right|}{\left|ABC\right|}$, $ b\equal{}\frac{\left|CAQ\right|}{\left|ABC\right|}$, and $ c\equal{}\frac{\left|ABQ\right|}{\left|ABC\right|}$."
instead of
"For the proof, use barycentric coordinates: $ M \equal{} (0,m,1 \minus{} m),\, N \equal{} (1 \minus{} n,0,n),\, P \equal{} (p,0,1 \minus{} p),\, Q \equal{} (a,b,c)$."

  darij


A solution of phosphoric acid is titrated with a solution of sodium hydroxide. If 50.00 ml of the phosphoric acid is titrated with 0.125 M NaOH, it requires 45.76 ml of the sodium hydroxide to reach the equivalence point.What is the equivalence point? How about its end point?
My answer: I think equivalence point is simply like: 50+45.76=95.76. The end point should be the same as equivalence point. I'm not sure on both answers though. Anybody wants to help me?

First ask yourself what is the equivalence point? What is the end point?

[hide=" Alternatively"] Which one is the one you can "see" and which one is theoretical?[/hide]

[hide=" Answer"] Equivalence Point: the point in the titration at which the amount of added standard reagent is exactly equivalent to the amount of analyte.

End Point: the point in a titration when a physical change occurs that is associated with the condition of chemical equivalence.

Titration Error = $ E_t \equal{} V_{ep} \minus{} V_{eq}$.[/hide]


vector
linear algebra
matrix
number theory unsolved
number theory

http://www.mathlinks.ro/viewtopic.php?p=356507#356507

How is this ques related to the eigen vector or eigen value of a matrix??


function algebra polynomial logarithms domain calculus derivative

consider any function f(x) .

is there any mathematical procedure where we can find whether the function f(x) is derivable
in any given interval?

what i mean is if we consider a polynomial function we generally say that a polynomial function is derivable
anywhere in {R},but isnt there any mathematical procedure by taylor or any other scientist
where we can ell a given function is derivable in any given interval (a,b),not  by just  verifying by taking every point
and imagining?

consider a function log(((x^2)+ab)/((x)(a+b))) in (a,b) where a>0,b>0  so is this function derivable in (a,b)

prove mathematcally not just taking and verifying some points

please reply!!!

In symbols... Take $ f$ and $ g$ differentiable in an interval $ I$. Then $ h$ is differentiable in $ I$ if:

a) $ h(x) \equal{} f(x) \plus{} g(x)$ (and $ h'(x) \equal{} f'(x) \plus{} g'(x)$).
b) $ h(x) \equal{} f(x).g(x)$ (and $ h'(x) \equal{} f'(x)g(x) \plus{} f(x)g'(x)$)

If $ f$ is a differentiable function defined in an interval $ I$ and the range of $ f$ is contained in the domain of a differentiable function $ g$, to conclude $ h$ is differentiable it suffices to have

c) $ h(x) \equal{} g(f(x))$ (and $ h'(x) \equal{} f'(g(x)).g'(x)$)

From this, we conclude, assuming $ f$ and $ g$ are differentiable in an interval $ I$, that $ h$ is differentiable if

d) $ h(x) \equal{} f(x) \minus{} g(x)$ (to prove this, it suffices to prove $ \minus{} g$ is differentiable, because $ h \equal{} f \plus{} ( \minus{} g)$. And $ \minus{} g$ is the product of $ \minus{} 1$ and $ g$, both differentiable).
e) $ h(x) \equal{} f(x)^{g(x)}$ and $ Im f \subset R_ \plus{}$ (to prove this, we see that $ f^g \equal{} e^{glnf}$. $ lnf$ is the composition of $ f$ and $ ln$, hence differentiable (see [i]c[/i]). $ gln f$ is the product of $ g$ and $ ln f$, hence differentiable. Thus, $ f^g$ is the composition of $ e^x$ and other differentiable function).
f) $ h(x) \equal{} f(x)^t$, if $ f(x) > 0$ and t is a constant function (particular case of [i]e[/i]).
g) $ h(x) \equal{} \frac {f(x)}{g(x)}$ and $ g(x) > 0$ (to prove this, notice this is the product of $ f$ and $ \frac {1}{g} \equal{} g^{ \minus{} 1}$. The later is differentiable because of [i]f[/i]). The case $ g(x) < 0$ is obtained just by changing a sign.
h) $ h(x) \equal{} log_{f(x)}g(x)$ if $ f(x), g(x) > 0$ (to prove this, remember that $ log_f g \equal{} \frac {log f}{log g}$. We have a fraction of differentiable functions, then applying $ g$ suffices).

Polynomials are differentiable functions because we can train them by adding and multiplying the indentidy with constant functions. For example, the product of $ f(x) \equal{} x$ and $ g(x) \equal{} x$ is $ x^2$, thus $ x^2$ is differentiable. Multiplying this by the constant function $ a$, we see $ ax^2$ is differentiable. If we prove similarly that $ bx$ and $ c$ are differentiable, we conclude $ ax^2 \plus{} bx \plus{} c$ is differentiable.

[quote="§outh§tar"][quote="pascal12"][quote="t0rajir0u"]Functions built from elementary functions should be differentiable almost everywhere because we can trust the product and chain rules.  The only exceptions are points at which certain functions aren't defined and/or have sharp kinks (for example, when a denominator becomes zero, the argument of a logarithm becomes non-positive, or the argument of an absolute value is zero), so you just need to check for those possibilities.

Arbitrary functions, on the other hand, have to be investigated.  Differentiability is a more basic notion than many of the other properties that we impose on functions.[/quote]

Agreed. And along the same train of though, any function which is continuous and smooth is differentiable.[/quote]

Isn't that what smooth means?[/quote]

Nope. Continuous just means that it's "connected" at every point. Techincally, that the limit of a function to point a equals the function of a at all a. Smooth means that it doesn't have any kinks, which is not implied by continuous.

Nope.  "Smooth" means "has derivatives of all orders."

How would you explain the use of the phrase "a function which is sufficiently smooth"?

Also, [url=http://mathworld.wolfram.com/SmoothFunction.html]mathworld[/url] seems to lean to what I thought 'smooth' meant as well, that is, "has 'a lot of' derivatives".

Anyways, calling a function smooth means it has at least a derivative of order one, so well obviously a function which is smooth is also differentiable (and thus continuous etc).

Mathematicians use "smooth" in both a rigorous and a loose sense.  The rigorous meaning is $ C^{\infty}$ (at least as I have seen it; see [url=http://en.wikipedia.org/wiki/Smooth_function]Wikipedia[/url]), while the loose sense depends on context (e.g. "has a Fourier transform"), although I think this is sloppy and "well-behaved" works better here.  I have also seen "smooth" used to mean $ C^1$.  In any case, I think we are in agreement that "smooth" is at least as strong as $ C^1$.


Apparently we forgot to post these results when they were available last month...

Check out the official results for the 2005 Four-by-Four Competition at http://www.natassessment.com and take the time to congratulate your peers.  This contest was harder than we intended; we will try to make next year's problems a bit more accessible.  I don't know if the difficulty had anything to do with this, but Vestavia Hills swept the first-place awards!

School:  Vestavia Hills High School, Vestavia Hills, AL

"Rock" (younger) division:
     Vestavia Hills High School, Vestavia Hills, AL
          Evan Zhao
          Yuehan Huang
          Seva Tchernov
          Zhenya Kaliberova

"Paper" (middle) division:
     Vestavia Hills High School, Vestavia Hills, AL
          Rajan Ananthan
          Laura Sims
          Sara K. Klay
          Jeanette Dooley

"Scissors" (older) division:
     Vestavia Hills High School, Vestavia Hills, AL
          Ellie Barr
          Jonathan Waite
          David Harris
          Ivy Wu

[quote="tom_clymer"] This contest was harder than we intended; we will try to make next year's problems a bit more accessible.  I don't know if the difficulty had anything to do with this, but Vestavia Hills swept the first-place awards![/quote]

I thought they were good problems, but a lot that I tried would have been doable for me with 10-30 more seconds each (though apparently our other groups didn't have as much of a problem with that...). I guess you can't change anything but the problems without ruining the 4/4/4 thing.

notice that in the paper division, Bellevue had three guys... their grade only add up to 32. Shouldnt they be in Rock division???

p.s. no offense to Vestavia(good job on the sweep), but if Bellevue were to go to Rock, then they would break the Sweep!!!

RedDevil,

     The divisions are based upon the sum of the grades of the team members, but to encourage schools to reach out beyond core math team members, we assume that any unfilled slot on a team is filled with a freshman.  This makes the sum of the grades for this team 41, which puts them in the paper division.  Nonetheless, Bellevue's award-winning performance with just three people is pretty impressive!


[color=cyan]Hey, I definitely didn't make this question up.  Although it is pretty funny.[/color]

[color=cyan]I doubt highly that this particular version originated with mathcounts.  It is the approximate difficulty of a tough mathcounts question, or a medium AMC 10 or medium-easy AMC 12 question.[/color]

darkquantum wrote:During a bike marathon, Hubert was already 3/8 of the way across a narrow bridge. A train traveling at 60 miles per hour was approaching the bridge from behind him. Hubert, using his crazy math skills, calculated that if he were to continue biking forwards at x miles per hour, he and the train would meet at the farther end of the bridge. Hubert also figured out that if he were to bike backwards at x miles per hour, he and the train would meet at the closer end of the bridge. In both cases, Hubert gets hit by the train just as he thinks hes escaped. Compute x.

Hmm...I'm pretty sure getting hit just as you think you've escaped means that you get hit at the very beginning (or end) of the bridge.  Thus:

[hide]Let the length of the bridge equal y.

In the time it takes Hubert to move y/4 miles (since the difference between one distance and another is 5y/8 - 3y/8 = y/4), the train has moved y miles (since it went from one end of the bridge to the other.  So Hubert is moving 1/4 times the speed, and his speed is 15 mph.

[/hide]


inequalities
induction
inequalities unsolved

Let $ n$ be positive integer. Prove the following inequality:

$ e<(\frac{(n\plus{}1)^{2n\plus{}1}}{(n!)^{2}})^{\frac{1}{2n}}<e^{1\plus{}\frac{1}{12(n\plus{}1)}}$

[quote]from Stirling number aproximation.[/quote]

Sorry, but I don't know what is it Can you give some links etc. and please can you explain it.

[quote="canada"]Let $ n$ be positive integer. Prove the following inequality:

$ e < (\frac {(n + 1)^{2n + 1}}{(n!)^{2}})^{\frac {1}{2n}}$[/quote]

This part is equivalent to:   $ e^{2n} < \frac {(n + 1)^{2n + 1}}{(n!)^{2}}$

From here will proceed by induction:

☺$ n=1$:
Just need to verify that $ e^{2}<\frac {(1 + 1)^{2 + 1}}{(1!)^{2}}=8$

☺Lets suppose it is true for $ n=k\ge1$, we have:
${ \frac {(k + 2)^{2k+ 3}}{((k+1)!)^{2}}=
\frac {(n + 1)^{2n + 1}}{(n!)^{2}}\frac{(k+2)^{2k+3}}{(k+1)^{2k+3}}>
e^{2k}(\frac{k+2}{k+1}})^{2k+3}>e^{2k+2} \Leftrightarrow$
$ (\frac{k+2}{k+1})^{2k+3}>e^2$

Write the LHS as $ ((1+\frac{1}{k+1})^{k+1})^2(1+\frac{1}{k+1})$ and verify this is greater than $ e^2$ for $ k=1$.
Now just note that for integers $ k\ge1$ is a decreasing sequence wich tends to $ e^2$.

 :)


logarithms
number theory proposed
number theory

Prove that there exists a number
\[ N\in\{2^{3i}\;|\;i\in\mathbb{N}\}\]
such that $ N=\overline{2007a_{0}a_{1}\cdots a_{n}}$ in decimal expansion.

OK. This is just a draft of the solution.
The question asks to find natural numbers x, i such that
$ 2007. 10^{x}< 8^{i}< 2008 .10^{x}$
$ 2007 < \frac{8^{i}}{10^{x}}< 2008$
$ 2007 <2^{3i-x}5^{-x}<2008$
$ \ln{2007}< (3i-x)\ln{2}-x\ln{5}< \ln{2008}$
Let $ \ln 2 = u$, $ \ln 5 = v$. Since u and v are "lineary independent":doesn't exist
$ h \in R : \frac{\ln 2}{h}\in N \& \frac{\ln 5}{h}\in N$ thus every real can be approximated with number of the form
$ au+bv$, for a, b integers.


I can't seem to figure out where the archive is. Can someone help me?

http://www.artofproblemsolving.com/Forum/index.php?f=152

Of course, you'll only be able to view classes that you have taken in the past.


vector
function
algebra
system of equations
linear algebra

What is linear independence? How can you determine whether two equations are linear independent in respect to one another?

Let's say we have a set of numbers $ S\equal{}\{x_{1},\ldots,x_{n}\}$.  Now let's say we have a set of scalars $ \{a_{1},\ldots,a_{n}\}$.  We say that $ S$ is linearly independent if the ONLY solution to
$ a_{1}x_{1}\plus{}\cdots\plus{}a_{n}x_{n}\equal{}0$
is $ a_{1}\equal{}\cdots\equal{}a_{n}\equal{}0$.  If there is some other set of scalars, not all zero, that make the equation work, then $ S$ is linearly dependent.

When we talk about linear independence of equations, our "numbers" (which are [url=http://mathworld.wolfram.com/Vector.html]vectors[/url]) are expressions $ f(x, y, z, ...) \equal{} 0$.  If a system of equations is linearly dependent, at least one equation can be removed from the system and it will still have the same solution set.

For example, if we have functions $ f_{1}$, $ f_{2}$,..., $ f_{n\minus{}1}$, $ f_{n}$, then
$ \text{span}(f_{1},\ldots,f_{n})\equal{}\text{span}(f_{1},\ldots,f_{n\minus{}1})$
is linearly dependent.

What you meant to say was that that condition means that $ f_1, f_2, ... f_n$ are linearly dependent.


Let $ ABC$ be an equilateral triangle. Let $ D,E, F,M,N,$ and $ P$ be the mid-points of $ BC, CA, AB, FD, FB,$ and $ DC$ respectively.

[b](a)[/b] Show that the line segments $ AM,EN,$ and $ FP$ are concurrent.
[b](b)[/b] Let $ O$ be the point of intersection of $ AM,EN,$ and $ FP.$ Find $ OM : OF : ON : OE : OP : OA.$

If ABCD is the quadrangle, O ond I its centers of excircle and incircle respectively, R the intersection of AC and BD, E and F the intersections of AB and CD, respectively BC and AD, then it's easy to show that OR and IR are perpendicular to EF.

Best regards,
sunken rock


inequalities
n-variable inequality
Germany
TST

Let $ n$ be a positive integer such that $ n\geq 3$. Let $ a_1$, $ a_2$, ..., $ a_n$ and $ b_1$, $ b_2$, ..., $ b_n$ be $ 2n$ positive real numbers satisfying the equations
\[ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 1, \quad \text{and} \quad b_1^2 \plus{} b_2^2 \plus{} ... \plus{} b_n^2 \equal{} 1.\]
Prove the inequality
\[a_1\left(b_1 \plus{} a_2\right) \plus{} a_2\left(b_2 \plus{} a_3\right) \plus{} ... \plus{} a_{n \minus{} 1}\left(b_{n \minus{} 1} \plus{} a_n\right) \plus{} a_n\left(b_n \plus{} a_1\right) < 1.\]

I have an idea. [url=http://www.mathlinks.ro/viewtopic.php?p=190966#190966]I can prove[/url] if $ a_1 \plus{} a_2 \plus{} .... \plus{} a_n \equal{} 1$ then the following holds:

          $ a_1a_2 \plus{} a_2a_3 \plus{} .... \plus{} a_{n \minus{} 1}a_n \leq 1/4.$

Why do we have $\sum a_i b_i \leq (\sum a_i^2)( \sum b_i^2)$ ??

[quote="dashmiz"]it's chushy shwartz inequality[/quote]Wrong, Cauchy-Schwartz inequality is $ \left ( \sum a_ib_i \right)^2 \leq \left( \sum a_i^2 \right) \left( \sum b_i^2 \right)$ ;) But the inequality used in the proof is correct (it has a square root :) ).

sorry, but why is $\sqrt{\sum a_{i}^2 }+ \sum a_{i}a_{i+1}<1$ ? and what inequality is used for $\displaystyle \sqrt{U} + S < \sqrt{U+2S} < 1$ ?

solution
$S=\sum_{i=1}^n a_ib_i+\sum_{i=1}^{n-1} a_ia_{i+1}+a_na_1$
objective is to prove S<1

Consider,$(\sum_{i=1}^n a_i)^2=1 $
$\implies \sum_{i=1}^n a_i^2+2\sum_{i\neq j}^n a_ia_j=1$
$\implies \sum_{i=1}^n a_ia_{i+1} +a_na_1<1/2-\sum_{i=1}^n a_i^2/2$
Therefore, $S<\sum_{i=1}^n a_ib_i+1/2-\sum_{i=1}^n a_i^2/2$
Now $\sum_{i=1}^n (b_i-a_i)^2>0$
$\implies \sum_{i=1}^n a_i^2+\sum_{i=1}^n b_i^2-2\sum_{i=1}^n a_ib_i>0$
$\sum_{i=1}^n a_ib_i<1/2+\sum_{i=1}^n a_i^2/2$
plugging the inequality in the inequality derived before
$S<1/2+\sum_{i=1}^n a_i^2/2+1/2-\sum_{i=1}^n a_i^2/2$
$\implies S<1$

Inequality would be more interesting if numbers were nonnegative so equality would hold when $a_i=b_i=1$ for some $i$ and $a_j=b_j=0$ for all $j\neq i$.

This way, using

$\sum a_i^2+2\sum a_ia_{i+1} <\left(\sum a_i \right)^2=1$

and am-gm on two numbers: $2ab\leq a^2+b^2$, we have:

$2\sum a_i(a_{i+1}+b_i)=2\sum a_ib_i+2\sum a_ia_{i+1} \leq \sum a_i^2+\sum b_i^2+2\sum a_ia_{i+1} <2$

[quote="prowler"]$\sum a_ib_i + \sum a_{i}a_{i+1}\leq \sqrt{(\sum a_{i}^2)(\sum b_{i}^2)}+ \sum a_{i}a_{i+1}
= \sqrt{\sum a_{i}^2} + \sum a_{i}a_{i+1} < 1 $

Note $U = \sum a_{i}^2$, and $S = \sum a_{i}a_{i+1}$ and  $1 > U + 2S$

We have to prove $ \sqrt{U} + S < 1$
[/quote]
Since $S<\frac{1-U}{2}$, it remains to prove that $\sqrt{U}+\frac{1-U}{2}<1$, which is $\left(\sqrt U-1\right)^2>0$.


function
calculus
real analysis
real analysis unsolved

if $ g$ is continuous and non-decreasing on [0,1], satisfying $ g'(x)\equal{}0$ for all but a countable set, prove that $ g$ is a constant function.

Any chunk of information is appreciated. Thanks.

One possibility would be to subtract $ \delta x$ and a small non-decreasing function that has a positive jump at every point of that exceptional countable set and to get a new function $ f$. Then, for every point  $ x$, we'll have $ f(y)\le f(x)$ for all $ y>x$ sufficiently close to $ x$. ;)

Looks like homework. Is it? What are the rules of this forum regarding that?

To give hints rather than full solutions and to ask to show what the poster has done himself first if the problem is routine (this one is not).

Considering the Cantor-Lesbege function on [0,1],it is non-decreasing,and the derive is 0 a.e.But it is not a constant.
I think you mean the function could be derived everywhere not "except a countable set",and the value of the derive is 0 a.e..
However,if it is so,then we could use the Newton-Lebniz Formula,by Lesbege Calculous,it is done.

How do we a priori know that $ g$ doesn't have a singular part?

[quote="ficsur"]if $ g$ is continuous and non-decreasing on [0,1], satisfying $ g'(x) \equal{} 0$ for all but a countable set, prove that $ g$ is a constant function.

Once you prove this you will have a proof that the interval [0,1] is not countable.  No more need for the "controversial" diagonal argument.

Yes, but the nested interval lemma (which is in the heart of all this analytic business anyway) gives you such proof much faster ;).

[quote="sofeshue"]The example by zhongym8903 is not a good one since the Cantor ternary set is uncountable.[/quote]
Oops..
Sorry,I didn't consider the problem carefully...


geometry
circumcircle
geometric transformation
reflection
incenter
geometry proposed

Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$. The $ A \minus{}$ altitude of $ \triangle ABC$ intersects $ (O)$ again at $ A_0$, the same definition holds for $ B_0$ and $ C_0$. The tangents at $ B_0,$ $ C_0$ intersect at $ A_1$. Let $ A_2$ be the antipole of $ A_0$ with respect to $ (O)$. Denote $ A_3$ by the second intersection of $ A_1A_2$ and $ (O)$. Let $ A_4$ be the intersection of $ A_0A_3$ and the tangent at $ A$ of the circumcircle $ (O)$ of $ \triangle ABC$. Define similarly for $ B_4$ and $ C_4$. Prove that $ A_4,$ $ B_4,$ $ C_4$ are collinear.

[i]Best regard,[/i]
[b]mathVNpro[/b]

[quote]Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$. The $ A \minus{}$ altitude of $ \triangle ABC$ intersects $ (O)$ again at $ A_0$, the same definition holds for $ B_0$ and $ C_0$. The tangents at $ B_0,$ $ C_0$ intersect at $ A_1$. Let $ A_2$ be the antipole of $ A_0$ with respect to $ (O)$. Denote $ A_3$ by the second intersection of $ A_1A_2$ and $ (O)$. Let $ A_4$ be the intersection of $ A_0A_3$ and the tangent at $ A$ of the circumcircle $ (O)$ of $ \triangle ABC$. Define similarly for $ B_4$ and $ C_4$. Prove that $ A_4,$ $ B_4,$ $ C_4$ are collinear.[/quote]
Since no one has joined in solving this problem  :lol:  , so I will offer a hint: http://xtec.net/~qcastell/ttw/ttweng/definicions/d_mig_arc2_p.html

Thanks dear Minh for that big hint.

Let $ B_1$ be the intersection of the tangent of $ (O)$ at $ A_0,C_0$ and $ C_1$ the intersection of the tangents of $ (O)$ at $ A_0,B_0.$ Reflections of the orthocenter $ H$ of $ \triangle ABC$ about $ BC,CA,AB$ are precisely $ A_0,B_0,C_0.$  Then it follows that $ \triangle A_0B_0C_0$ and $ \triangle ABC$ are centrally similar with center $ H$ and factor $ 2.$ Since $ H$ is incenter or of the orthic triangle, then $ H$ is incenter of $ \triangle A_0B_0C_0$ $ \Longrightarrow$ $ A$ the is midpoint of the arc $ B_0C_0$ and similarly vertices $ B,C$ are the midpoints of the arcs $ C_0A_0$ and $ A_0B_0$ as well. From the fact that $ A_2$ is the antipode of $ A_0$ WRT $ (O),$ it  follows that $ A_1A_2$ is the $ A_1$-Nagel ray of $ \triangle A_1B_1C_1$ meeting $ B_1C_1$ at the reflection $ N_1$ of $ A_0$ about its midpoint $ M_1.$  Circle with diameter $ A_0N_1$ passing through $ A_3$  is then orthogonal to $ (O)$ $ \Longrightarrow$ Tangent of $ (O)$ at $ A_3$ passes through $ M_1,$ therefore $ A_4$ is the pole of $ AM_1$ WRT $ (O).$ The  line $ AM_1$ passes through the second mid-arc point $ J$ of $ \triangle A_1B_1C_1$ as long as $ \triangle A_0B_0C_0$ is acute $ \Longrightarrow$ $ A_4$ lies on the polar of $ J$ WRT $ (O)$ $ \Longrightarrow$ $ A_4,B_4,C_4$ lie on the polar of $ J$ WRT $ (O).$


function
geometry
geometric transformation
reflection
combinatorics proposed
combinatorics

[size=150]A cricket team of 11 players wearing jersey numbers from 1 to 11 (obviously none of them same) have to be seated around a circular table. How many ways are there to seat them if the difference between the jersey numbers of two adjacent players should be either 1 or 2?

The answer might surprize you. :rotfl:  [/size]

[quote="JBL"]http://www.research.att.com/~njas/sequences/A003274

No nice closed form, although it has a rational generating function and therefore a relatively simple recurrence relation.[/quote]

A very nice closed form exists, beleive me. Firstly because its a contest problem and secondly because i've constructed a nearly perfect solution. Will post it in a few days. Hint: What's special about jersey no.1 and no.11? Hmmm....

You're right, although for bad reasons (I missed the word "circular" in the question statement).  With a circular table, the solution is straightforward: for $ n \equal{} 1$ or $ n \equal{} 2$ there is only one arrangement and it is good.  For larger $ n$, 1 must lie between 2 and 3; then adjacent to 2 must be 4, adjacent to 3 must be 5, etc., and these chains eventually meet on the opposite side with $ n$ adjacent to $ n \minus{} 1$ and $ n \minus{} 2$.  The only choice we have is the orientation of the triple 2-1-3, so there are only 2 such arrangements.

Great, that's the correct argument.
The answer is 2 (taking care of orientations).

Seems like you can help me with this one too:

A "m X n" chessboard has to be covered with "1 X 2" dominoes. In how many ways can this be done, if at least one of m and n is even?

Adding to JBL's proof, for all even $ n \geq 6$ we have just one arrangement and for all odd $ n\geq5$ we have two arrangements.

That's not true: for all $ n \geq 3$ we have two arrangements.  (They are reflections of each other.)


What is the ratio of the area of triangle $ BDC$ to the area of triangle $ ADC$? Express your answer as a common fraction.

[asy]draw((0,0)--(10sqrt(3),0)--(0,10)--(0,0)--cycle,linewidth(2));
draw(5sqrt(3)*dir(60)--(0,0),linewidth(2));

draw((1,0)--(1,1)--(0,1),linewidth(1));
draw(5sqrt(3)*dir(60)+1*dir(-120)--(5sqrt(3)-1)*dir(60)+1*dir(-30)--5sqrt(3)*dir(60)+1*dir(-30),linewidth(1));

label("A",(10sqrt(3),0),E);
label("B",(0,10),W);
label("C",(0,0),SW);
label("D",5sqrt(3)*dir(60),NE);
label("$30^{\circ}$",(12,0),N);[/asy]

If you fill in all the angles, you will see both triangles are 30-60-90 triangles. In triangle ABC, BC is half of the hypotenuse, so the side lengths of BDC are half of ABC. Thus, the area ratio of BDC to ABC is $ 1: 4$

great solution
the only problem, is that the question asks what the ratio is of the area of triangle $ BDC$ to [b]$ ADC$[/b] not $ ABC$

Oh right...

Both triangles are similar. The ratio of BC to AC is $ \sqrt{3}: 3$, so the ratio of the area is $ 3: 9$ or $ \frac{1}{3}$

How do you get[b] sqrt3/3 [/b]as the ratio. Could you  please explain?

Thanks!


function
algebra
system of equations
algebra proposed

Find all functions $f,g,h: \mathbb R \to \mathbb R$ such that
\[ f(x+y^3)+g(x^3+y)=h(xy)\]
for all $x,y \in \mathbb R$.

Find all functions $ f,g,h: R\minus{}>R$ such that:
$ f(x\plus{}y^3)\plus{}g(x^3\plus{}y)\equal{}h(xy)$,for all $ x,y$ in $ R$

[quote="Adriana N."]Find all functions $ f,g,h: R \minus{} > R$ such that:
$ f(x \plus{} y^3) \plus{} g(x^3 \plus{} y) \equal{} h(xy)$,for all $ x,y$ in $ R$[/quote]

Here is a rather ugly solution. I hope someone will propose a better one :

1) $ f(x)\equal{}g(x)\plus{}c$
Swapping $ x$ and $ y$, we get $ f(x\plus{}y^3)\plus{}g(x^3\plus{}y)\equal{}f(y\plus{}x^3)\plus{}g(y^3\plus{}x)$.
Let $ k(x)\equal{}f(x)\minus{}g(x)$. We get $ k(x\plus{}y^3)\equal{}k(x^3\plus{}y)$
The system of equations : $ x\plus{}y^3\equal{}u$ and $ x^3\plus{}y\equal{}v$ is equivalent to :
$ y\equal{}v\minus{}x^3$ and $ x\plus{}(v\minus{}x^3)^3\equal{}u$
Since the degree of the second equation is odd ($ 9$), this system always has real solution and then $ k(u)\equal{}k(v)$ $ \forall u,v$
So $ k(x)\equal{}c$ constant
Q.E.D.

2) $ f(x)\equal{}$constant
The new equation is now $ f(x\plus{}y^3)\plus{}f(x^3\plus{}y)\equal{}h(xy)\minus{}c$
And so, replacing $ x$ with $ \frac xz$ and $ y$ with $ yz$ : $ f(\frac xz\plus{}y^3z^3)\plus{}f(\frac{x^3}{z^3}\plus{}yz)\equal{}h(xy)\minus{}c$

And so : $ f(x\plus{}y^3)\plus{}f(x^3\plus{}y)\equal{}$ $ f(\frac xz\plus{}y^3z^3)\plus{}f(\frac{x^3}{z^3}\plus{}yz)$ $ \forall x,y$, $ \forall z\neq 0$

Let now $ y\equal{}\frac{x^3(z^2\plus{}z\plus{}1)}{z^3}$. This value implies $ x^3\plus{}y\equal{}\frac{x^3}{z^3}\plus{}yz$ and the above equality becomes :

$ f(x\plus{}y^3)\equal{}f(\frac xz\plus{}y^3z^3)$ $ \forall x$ $ \forall z\neq 0$ and with $ y\equal{}\frac{x^3(z^2\plus{}z\plus{}1)}{z^3}$

Then we'll try to see if the following system has real solutions :
$ y\equal{}\frac{x^3(z^2\plus{}z\plus{}1)}{z^3}$
$ x\plus{}y^3\equal{}u$
$ \frac xz\plus{}y^3z^3\equal{}v$

This system is equivalent to :
$ y\equal{}\frac{x^3(z^2\plus{}z\plus{}1)}{z^3}$
$ x\plus{}y^3\equal{}u$
$ \frac xz\plus{}y^3z^3\equal{}v$ $ \iff$ $ \frac xz\plus{}(u\minus{}x)z^3\equal{}v$ $ \iff$ $ x\equal{}z\frac{v\minus{}uz^3}{1\minus{}z^4}$ if we consider now $ z\notin\{\minus{}1,0,\plus{}1\}$

So :
$ x\equal{}z\frac{v\minus{}uz^3}{1\minus{}z^4}$
$ y\equal{}\frac{x^3(z^2\plus{}z\plus{}1)}{z^3}$
$ z\frac{v\minus{}uz^3}{1\minus{}z^4}\plus{}\left(\frac{(z\frac{v\minus{}uz^3}{1\minus{}z^4})^3(z^2\plus{}z\plus{}1)}{z^3}\right)^3\equal{}u$

This last equation may be written :

$ P(z)\equal{}z(v\minus{}uz^3)(1\minus{}z^4)^8\plus{}(v\minus{}uz^3)^9(z^2\plus{}z\plus{}1)^3\minus{}u(1\minus{}z^4)^9\equal{}0$
The highest power term of $ P(z)$ is $ (v\minus{}u^9)z^{33}$ and so this equation always has a solution (odd degree) whenever $ v\neq u^9$
If we want that this solution $ \notin\{\minus{}1,0,\plus{}1\}$, we also need to add some constraints :
$ P(0)\neq 0$ $ \iff$ $ v^9\minus{}u\neq 0$
$ P(1)\neq 0$ $ \iff$ $ (v\minus{}u)^9\neq 0$
$ P(\minus{}1)\neq 0$ $ \iff$ $ (v\plus{}u)^9\neq 0$

So we have $ f(u)\equal{}f(v)$ $ \forall u,v$ such that $ v\neq u^9$ and $ u\neq v^9$ and $ u\neq \minus{}v$

From this, it's immediate to conclude $ f(u)\equal{}f(v)$ $ \forall u,v$
Q.E.D.

3) Synthesis of solutions :
$ f(x)\equal{}a$
$ g(x)\equal{}b$
$ h(x)\equal{}a\plus{}b$


What we must prove is x+y+z+xyz(xy+yz+zx)<=4*3^(1/2)/9+xy(x+y)+yz(y+z)+zx(z+x). By adding 3xyz to both sides we get x+y+z+4xyz (we can eliminate xy+yz+zx since it?s equal to 1) <= 4*3^(1/2)/9 + x+y+z. By subtracting x+y+z from both sides and dividing by 4 we are left with xyz<=3^(1/2)/9, which is true by AM-GM applied to xy, yz, zx.


Here's a fun problem.

A customer bought 4 items at a 7-11 store on the 11th of July. Checking out, the price was found to be $7.11. The customer asked how the cashier got the price because of the coicidential number involved. The cashier responded that he multiplied the 4 prices together. Outraged, the customer said that the prices should be added instead of multiplied. The cashier told the customer not to worry for the sum of the prices was also exactly $7.11. The prices are in perfect dollars and cents. What are the 4 prices?

I'm curious for other people's thoughts, ideas, approahes, etc... to this problem.

Oops.... i forgot that the dollar signs would convert it to latex. Let me rewrite the problem.

Here's a fun problem.

A customer bought 4 items at a 7-11 store on the 11th of July. Checking out, the price was found to be 7 dollars and 11 cents. The customer Inquired about the "peculiarity" of the price and was told by the cashier that he multiplied the costs together. The customer protested that the prices should be a sum instead of a product but the cashier reassured the customer that everything was find because the sum of the 4 prices was also 7 dollars and 11 cents. The prices are in perfect dollars and cents. What are the 4 prices?

I'm curious on other people's approaches, ideas, solutions, etc... to this problem.

3.16,
1.20,
1.25,
1.5

i posted this problem a while ago, with a solution, basically, you multiply the prices by 100 so they are integers and factor, then use even/odd things to find the answer

look here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=711&t=44345[/url]

O wow thx. I prime factorized 711000000 and knew that one of the prices had to be a multiple of 79 and then I was left with the other factors and was too lazy to continue. haha. But yea, your solution and approach was what I speculated to be.


Find an analytic map from $ \mathbb{C}\setminus\{z\in\mathbb{R}: \minus{}1\le z\le 1\}$ to the open unit disk.

It has something to do with $ z\plus{}\frac{1}{z}$.

infact the required function is the inverse of the Joukowski function $ w\equal{}\frac{1}{2}.(z\plus{}\frac{1}{z})$ with the proper branch chosen.


A wholesaler purchased a pair of jeans from the manufacturer  for $ \$40$. The wholesaler then tried to sell the jeans for $ 140\%$ of the purchase price. When customers wouldn't pay that price, the jeans were placed on sale at a $ 20\%$ discount off the selling price. At the end of the season, a final reduction of $ 50\%$ off the lowered selling price was given. How many dollars was the final price of the jeans? Express your answer to the nearest cent.

the pair of jeans was bought for $ \$40$

the jeans then were put up for sale at $ 140\%$ the purchase price
which means that the jeans were on sale for $ \$40\times1.4\equal{}\$56$

later, a $ 20\%$ discout was put on it
that means, the jeans were $ \$56\times0.8\equal{}\$44.8$

finally, a $ 50\%$ reduction was placed on the jeans
so, the final cost of the jeans were $ \$44.8\times0.5\equal{}\$22.4$

the final price of the jeans was $ \boxed{\$22.40}$

[quote="SuperNerd123"]$ 40\times1.4 \equal{} 56$

$ 56\times0.8 \equal{} 44.8$

$ 44.8\times0.5 \equal{} \boxed{22.4}$
 :mad:
*EDIT* Beaten[/quote]

FYI: Don't forget the 0. :wink: It's mandatory in an actual Mathcounts competition, or else the graders mark it wrong. Same goes for FTW.

I've typed quite a few answers without an unnecessarily zero, and they got marked as correct.  :wink:

Erm, ok. Strange then, seeing how on a real Mathcounts test those kinds of forms of answers would be marked wrong...


inequalities
limit
calculus
calculus computations

prove that

lim (n tends to inf)  L_n / Log n  = 1/ pi

where L_n = 1 + (1/2)^2 + [(1/2)(3/4)]^2 + ... ... + [(1/2)(3/4)...((2n-1)/2n)]^2

the only thing I know is that the it may be related to wallis formula,
and 1,(1/2), (1/2)(3/4)... are all the coefficients of the binomial series for  1/(sqrt(1-x))

Anybody can help me.
Thanks a lots.

Yes, it is. If $a_{n}=\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}$, you need to show that $\lim_{n\to\infty}\frac{a_{n}^{2}}{\frac1n}=\frac1{\pi}$.

After that, it's an application of the Cesaro principle: if $b_{n}$ are positive with $\sum b_{n}$ divergent and $\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=c$, $\lim_{n\to\infty}\frac{a_{1}+a_{2}+\cdots+a_{n}}{b_{1}+b_{2}+\cdots+b_{n}}=c$.


Suppose $ S_{1}, ..., S_{k}$ are distinct subsets of $ \{ 1, 2, \ldots, n\}$ and that for any two of them $ |S_{i}\cup S_{j}| \le n-4$ holds. Find the maximal possible value of $ k$.

$ k\geq  2^{(n-4)}$
because $ S_{i}$ are all the subsets of $ \{1,2,\dots,n-4\}$

Sure, this is clear. How about an upper bound?


superior algebra
superior algebra solved

Let $A$ be a ring where the following implication holds: $x^2=0\Rightarrow x=0$
Let $M=\{a\in A\mid a^2=a\}$
a) Prove that if $a,b\in M$ then $a-2ab+b\in M$
b) Prove that if $M$ is finite then the cardinal of $M$ is a power of $2$.

For all $a,b\in M$, compute $(ab-aba)^2$ and $(ba-aba)^2$. They're both $0$, which means that $ab=aba=ba$, so each two elements in $M$ commute. From here, $(a)$ follows easily. As for $(b)$, it's easy (just computations) to show that $a \star b:=a+b-2ab$ defines an operation which turns $(M,\star)$ into a group in which every element has order $2$. If it's finite, then it must have an order equal to a power of $2$.


A group of 12 resistors is arranged along the edges of a cube as shown in the diagram below.  The vertices of the cube are labeled a-h.

[Diagram of cube $abcdefgh$, with edges $ab$, $ac$, $db$, $dc$, $ef$, $eg$, $hf$, $hg$, $ae$, $bf$, $dh$, $cg$, with resistors in each edge]

a. (13 pts) The resistance between each pair of vertices is as follows:
$R_{ab}=R_{ac}=R_{ae}=3.0 \Omega$
$R_{cg}=R_{ef}=R_{bd}=8.0 \Omega$
$R_{cd}=R_{bf}=R_{eg}=12.0 \Omega$
$R_{dh}=R_{fh}=R_{gh}=1.0 \Omega$
What is the equivalent resistance between points $a$ and $h$?

b. (12 pts) The three $12.0\Omega$ resistors are replaced by identical capacitors.  $C_{cd}=C_{bf}=C_{eg}=15\mu F$.  A $12.0V$ battery is attached across points $a$ and $h$ and the circuit is allowed to run for a long period of time.  What is the charge ($Q_{cd}$, $Q_{bf}$, $Q_{eg}$) on each capacitor after this long period of time?

I don't have a scanned copy of the diagram; I just included a description.

I can't see any symmetries :maybe:. Am I wrong, or is this pure brute force?

i brute-forced it.  kinda disappointing to see a blah problem like this on the usapho.

could you explain what brute force means?

In this case, brute force means writing up Kirchoff's rules (or equivalently, the circuit node equations) and solving the dozen or so equations  :mad: Dull and slow, but it works if you don't see an elegant solution.

So... What's the solution? (Of course I haven't done it, it'd take an hour to get right :))

I'm trying to find some symmetry, but not with much luck :( .
The kirchoff solution doesnt need any thinking, thats what I hate in the national olympiad problems, they are just calculations infinitely, and no real physics is required. :mad:

There's a lot of symmetry, only two different paths are possible from the opposite corners. Set the voltage drop equal in the two paths, some junction laws, and you're done.

Part a)

There is pretty obvious symmetry:

Three $3 \Omega$ resistors, each branching into a $12 \Omega$ and a $8 \Omega$ resistor.

Each $8 \Omega$ resistor then joins with a $12 \Omega$ resistor (but not the same one that it branched off of the $3 \Omega$ resistor with) and after this junction there is a $1 \Omega$ resistor in each case.

So, $I_{3}= I_{8}+I_{12}$, and $12 I_{12}= 8 I_{8}$.

Then, $I_{8}= \frac{3}{2}I_{12}$.

Therefore, $I_{3}= \frac{5}{2}I_{12}$.

So, $[3(\frac{5}{2})+12+1(\frac{5}{2})] I_{12}= V$.

Thus, $I_{12}= \frac{V}{22}$.

Therefore, $I_{3}= \frac{5V}{44}$.

So, $I = 3 I_{3}= \frac{15V}{44}$.

$R_{eff}= \frac{V}{I}$, so $R_{eff}= \frac{44}{15}$.

For part b)

Now, the resistance of each path is $12 \Omega$.

Thus, the current through each of the 3 paths is $1 A$.

Therefore, $\frac{q}{C}= 1 A * 8 \Omega = 8 V$.

So, $q = 8 * 15 = 120 \mu C$.

Yeah... I somehow put $360 \mu C$ when I took it.  :-/

That's not obvious to me :( Ok, thanks, I think I got it... Well, can we see other USAPhO problems?

oh sh§t I drew a wrong diagram switched up the sides. :mad:
now I see the symmetry. Nice solution PTyanan. :D


If you get any of the codes and not neccessarily (sp?) the last one, you can PM it to me and I'll say if you got it right or not.

i believe code 1 is incorrect.
(the last t shouldnt be there)

Well, whether it's incorrect or not, many people have already finished Code 3, so I don't think it's a problem.


How many real numbers $x$ satisfy the equation $3^{2x+2}-3^{x+3}-3^{x}+3=0$?

$\textbf {(A) } 0 \qquad \textbf {(B) } 1 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 3 \qquad \textbf {(E) } 4$

[hide="Solution"]Notice that we have a nice quadratic here. $9(3^x)^2-28(3^x)+3=0$. :D

Solving the quadratic $3^x=3, -\frac19$, so the only solutions are $x=1,-2$. [/hide]

[hide="The Answer"]
$3^{2x+2}-3^{x+3}-3^{x}+3=3^{2x} \cdot 9-3^x \cdot 27-3^x+3$, by letting $3^x=y$, we form the quadratic $9y^2-28y+3=0$ and since it's a quadratic, it has $\boxed{2}$ solutions.[/hide]

[hide="alternative"]
you can also factor it:

$3^x(3^{x+2}-1)-3(3^{x+2}-1)=0 \rightarrow (3^x-3)(3^{x+2}-1)=0$

and the two solutions follow from there.[/hide]


How do you solve this thing ??

Let AB and CD be distinct parallel chords of a circle and let P be a point on their common perpendicular bisector. The lines AP and CP meet the circle again at Q and R respectively. If S is the intersection of the lines BR and QD, prove that PS is parallel to AB

Thanks a real lot in advance.........

Argh... The problem is a trivial corollary from the Pascal theorem. The point P can be an arbitrary point in the plane, it doesn't need to lie on the common perpendicular bisector of the chords AB and CD. And the solution of the problem is as follows: Since the hexagon ABRCDQ is cyclic, by the Pascal theorem, the points $AB\cap CD$, $BR\cap DQ$ and $RC\cap QA$ are collinear. But $BR\cap DQ=S$ and $RC\cap QA=P$. Thus, the points $AB\cap CD$, S and P are collinear. In other words, the line PS passes through the point $AB\cap CD$. But since the lines AB and CD are parallel, their point of intersection $AB\cap CD$ is an infinite point, and the line PS, passing through this infinite point, must also be parallel to these lines AB and CD. Thus, the problem is solved.

  Darij

a problem from our university? :D WOW!

(besides I think those links may vanish in a while, now it's cage.ugent.be/...)


supose that $f(n)$ denotes to  the sum of all  (positive)divisors of $n$.now is the below statement true?:
for every natural number like $k$,there exist $n$ such that  $f(n)=nk$

I think that this problem is in the book of number teory which is written by mirzakhani and zavareh.

i wonder if you are familiar with that book. is this book translated in english?
[b]but however i didnt find the problem there.
if you found it ,let me know.
be aware that was of my own,and i have proved that for$n=1,2,3,4$
if you know more let me know. :?: [/b]


Αν στην $ \sum_{k\equal{}0}^na_kx^ky^{(k)}\equal{}0(1)$ θέσουμε $ x\equal{}e^t$ τότε αυτή παίρνει τη μορφή $ \sum_{k\equal{}0}^nA_ky^{(k)}\equal{}0(2)$.Υπάρχει γενικός τύπος που να δίνει τους συντελεστές της $ (2)$ συναρτήσει των συντελεστών της $ (1)$ και αν ναι πώς βγαίνει;

Μπορείς να καταλήξεις σε έναν γενικό τύπο, χρησιμοποιώντας τον κανόνα του Leibniz για παραγώγους υψηλότερης τάξης:
\[ (f\cdot g)^{(n)} \equal{} \sum_{k \equal{} 0}^n \binom{n}{k} f^{(n \minus{} k)} g^{(k)}\]
Στην πράξη πάντως, απλώς παίρνεις τη διαφορική που έχεις και κάνεις το μετ/μο με το χέρι.

Cheerio,

Durandal 1707

Εντάξει στην πράξη εννοείται ότι το κάνεις με το χέρι για τάξης 2-3. :lol: Απλά είπα αν υπάρχει γενικός τύπος γιατί όταν τις κάναμε δεν είπαμε τίποτα τέτοιο.
Ο τύπος του Leibnitz είναι για παραγώγιση γινομένου.Εδώ όμως είναι σύνθεση και δεν βλέπω πώς μπορεί να εφαρμοστεί (για να βρεις τα $ \frac{d^ky}{dt^k}$ εκτός αν εννοείς κάτι άλλο). :|

[quote="giannis18"]Ο τύπος του Leibnitz είναι για παραγώγιση γινομένου.Εδώ όμως είναι σύνθεση και δεν βλέπω πώς μπορεί να εφαρμοστεί (για να βρεις τα $ \frac {d^ky}{dt^k}$ εκτός αν εννοείς κάτι άλλο). :|[/quote]

Το γινόμενο εμφανίζεται εξαιτίας του κανόνα της αλυσίδας επειδή αυτό που σε νοιάζει τελικά είναι να υπολογίσεις τον τελεστή $ \left(e^{ \minus{} t}\frac {d}{dt}\right)^n$. Δεν είναι ακριβώς το ίδιο (sorry αν σε μπέρδεψα, sloppiness on my part :( ) αλλά η λογική είναι παρόμοια.

Αν θέλεις το αποτέλεσμα κατευθείαν, υπάρχει ο τύπος του di Bruno για τη σύνθεση συναρτήσεων (I had to look it up though :D) που σου δίνει για την $ Y(t(x)) \equiv y(x)$:
$ \frac {d^n}{dx^n} Y(t(x)) \equal{} \sum \frac {n!}{m_1 !m_2 ! \cdots m_n !} \frac {d^{m}Y}{dt^m} \prod_{k \equal{} 1}^n\left(\frac {( \minus{} 1)^{k \plus{} 1}}{k x^k}\right)^{m_k}$

όπου $ m \equal{} m_1 \plus{} \cdots \plus{} m_n$ και το άθροισμα είναι πάνω στα partitioning $ m_1,\ldots m_n\geq 0$ που ικανοποιούν τη σχέση: $ m_1 \plus{} 2m_2 \plus{} nm_n \equal{} n$.

(Και ναι, 100% θα μου έχουν φύγει $ n\to\infty$ πράξεις και typos)

Hope this helps. Cheerio,

Durandal 1707

Ευχαριστώ πολύ :)


geometry
parallelogram
rectangle
analytic geometry

Determine the area in square units of parallelogram $ ABCD$
with vertices $ A (\minus{}6, 2)$, $ B (12, 2)$, $ C (15, 8)$ and $ D (\minus{}3, 8)$.

it is a rectangle with dimensions 18 by 6, so $ \boxed{108}$

Notice that since points $ A$ and $ B$ have the same $ y$-coordinate, it is parallel to the $ x$-axis. Thus, a difference in $ y$ will be the height, since it is perpendicular to the base (since they are parallel to the $ y$ and $ x$-axes, respectively). Thus, the height is $ 8 \minus{} 2 \equal{} 6$. The base is $ 12 \minus{} ( \minus{} 6) \equal{} 15 \minus{} ( \minus{} 3) \equal{} 18$. We plug these values into the formula for the area of a parallelogram, $ Bh$, to find that the area is $ 6(18) \equal{} \fbox{108}$.

EDIT: WHOOPS! Beaten.


[quote="Virgil Nicula"]$\{a,b,c\}\subset \mathrm R$ $\Longrightarrow$ $\left(3abc+\sum a^{3}\right)\cdot\sum a\ge 2\cdot\sum a^{2}\cdot\sum bc$.[/quote]

Doesn't it reduce to Schur after you expand?

Sorry to bump but this problem shouldn't stay without an answer....

First of all, I'm heuristically assuming that Nicula's $\ds\sum$ refers to $\ds\sum_{cyc}$ NOT $\ds\sum_{sym}$

Then \[\iff 3\sum a^{2}bc+\sum a^{4}+\sum_{sym}a^{3}b \geq  2 \sum a^{2}bc+2\sum_{sym}a^{3}b \]
\[\iff \sum a^{2}bc+\sum a^{4}\geq \sum_{sym}a^{3}b \]
Or, in Muirhead notation \[[2,1,1]+[4,0,0]\geq 2[3,1,1] \]
This certainly isn't normal schurs, but it might be what some people call the fourth degree schurs... i'm not quite "schur" about that one :P .
Maybe it's schur for $r=2$: \[\sum a^{2}(a-b)(a-c)\geq 0 \iff \sum a^{4}+a^{2}bc \geq  \sum_{sym}a^{3}b\]

well I guess it is just schur's for $r=2$ :)


Euler
geometry
circumcircle
geometry proposed

B,C are two points liyng on the circle $ (\omega)$. A point $ A$ moves on $ (\omega)$ such that triangle ABC isn't isosceles. D,E,F are orthogonal projections of A,B,C on BC,CA,AB respectively. $ M\equal{}DF\cap AC,N\equal{}DE\cap AB$. The 9-point circle of triangle ABC has center K. Prove that the perpendicular line from K to MN passes through a fixed point when A moves on $ (\omega)$

Is know that the Orthic Axis (the perspectrix between orthic triangle and reference triangle) is perpendicular to the Euler line, so the perpendicular line from K to MN is the euler line of ABC and it pass trought the circumcenter of ABC that is fixed.


Two touching circles $S$ and $T$ have a common tangent which meets $S$ at $A$ and $T$ at $B$. Let $AP$ be a diameter of $S$ and let the tangent from $P$ to $T$ touch it at $Q$. Show that $AP=PQ$.

I don't get the problem. What does the tanget from P to T mean?

The line that is tangent to both circles.  Teh tangent meets both cirlces at two different points.

[hide="another solution"]$AP \perp AB$ because $AB$ is tangent to $S$ at $A$. $AP \perp PQ$ because $PQ$ is tangent to $S$ at $P$.  This implies that $AB \parallel PQ$.  Since both circles are tangent to parallel lines, they have the same diameter.  Therefore, $PQ=ST$, which is sum of the radii, which are equal, meaning that $PQ$ has the same length as the diameters of the circles.  Thus, $AP=PQ$.[/hide]

That's wrong. $PQ$ is not necessarily tangent to $S$.

[quote="tim1234133"]Two touching circles $S$ and $T$ have a common tangent which meets $S$ at $A$ and $T$ at $B$. Let $AP$ be a diameter of $S$ and [b]let the tangent from $P$ to $T$ touch it at $Q$.[/b] Show that $AP=PQ$.[/quote]

That's stated in the problem...

The problem implies that $PQ$ is tangent to $T$, but this does not necessarily mean that $PQ$ is also tangent to $S$.  Arne says that you used the fact that $PQ$ is tangent to $S$ which is clearly not true in all cases.

Ok.  Sorry about that, then.  I just misread the problem...

Btw, can someone tell me what angle chasing is?

[i]Angle Chasing[/i] is simply using the conditions given in the problem to chase angles and get some results in solving the problem. I can't quite explain what it is.

Basically, it's [i]algebraic[/i] manipulation with known angle measures to obtain unknown angle measures, using well-established angle properties and relations, for example the relation between inscribed angles, between the central and the inscribed angles, between external angle in a triangle and the non-adjacent internal angles, the relation between opposite angles, between the angles with perpendicular and/or parallel sides, and so on and so on.

For a pure example of angle chasing, see the solution to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109079]this[/url]

[quote="Arne"]This is nice. Simple angle chasing shows that $P$, $K$ (the tangency point of the circles) and $B$ are collinear. Hence $PQ^{2}= PS \cdot PB$ (power of a point). But $PA^{2}= PS \cdot PB$ by similar triangles, so we're done.[/quote]
An intensely larger problem I solved today had your conclusive step as its conclusive step, also! (I had never seen this method before, and now twice in a day! Needless to say I have little geom. experience... but I can solve many at olympiad level)

There is a very nice solution, which (after I was told it existed) I managed to find, but not in the exam.

Invert the figure with respect to the circle centre P through A. Clearly the line AB and the circle S are mutually inverse, which implies that T (being tangent to both at specific points) is its own inverse, which implies that Q (being the only point of T on line PQ) is its own inverse, which implies that Q lies on the circle centre P through A, and we're done.


trigonometry
calculus
calculus computations

\[ \begin{gathered}
  whats{\text{ }}f'\left( x \right)? \hfill \\
  \sqrt[3]{{\left( {\sin x} \right)^2 }} \hfill \\
\end{gathered}\]

I put it in maple and I got two results  :|

You can do it by hand and get one result...

Use the chain rule.  This expression is equivalent to $ \sin^{2/3}{x}$, so we get $ \frac{2}{3}\sin^{\minus{}1/3}{x}\cos{x}\equal{}\frac{2\cos{x}}{3\sqrt[3]{\sin{x}}}$.

[quote="JRav"]You can do it by hand and get one result...

Use the chain rule.  This expression is equivalent to $ \sin^{2/3}{x}$, so we get $ \frac {2}{3}\sin^{ \minus{} 1/3}{x}\cos{x} \equal{} \frac {2\cos{x}}{3\sqrt [3]{\sin{x}}}$.[/quote]

I did it lol
But I get confused when I got the another result in maple.
Thanks for your time :D

What did maple spit out?  Try to avoid using computer algebra software when the problem can be easily done by hand.  I'll admit, I checked my answer in Mathematica, but I already did it by hand first and Mathematica was purely to check it.


ratio
geometry
circumcircle
trigonometry
geometry unsolved

Consider $ ABC$ is a triangle with acutes angles and $ D$ is a point inside this triangle that $ \widehat{ADB}$ $ \equal{}$ $ \widehat{ACB}$ $ \plus{}$ $ 90$ and $ AC$ $ \times$ $ BD$ $ \equal{}$ $ AD$ $ \times$ $ BC$.
find the ratio of $ \frac {AB . CD}{AC . BD}$

It it is a problem similar to IMO 1994 - Turkey - problem. It may be solved using trigonometry.

The solution I have is in Bulgarian so I should translate it ... it is for any angle not only for 90 degrees. I saw your post too late I must catch the train. But I'll sketch the idea ... draw circumcircle of ABC continue some lines and make some additional drawings then use sine law and algebra ... you also may try to find IMO Compendium book - it may be solved there. If for a week no one post a solution I'll translate mine.

You mean 1993:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14096[/url]

LET $ CK$ be perpendicular $ AC$ and $ AC \equal{} CK$
therefore, as $ AC*BD \equal{} BC*AD$ and $ \angle ADB \equal{} \angle KCB$
therefore $ \Delta ADB$ is similar to $ \Delta KCB$
$ \Longrightarrow \angle DBA \equal{} \angle CBK$
$ \Longrightarrow \angle KBA \equal{} \angle CBD$
also $ \frac {AB}{KB} \equal{} \frac {BD}{BC}$ (from similarity of triangles)
therefore, $ \Delta BDC$ is similar to $ \Delta BAK$
$ \Longrightarrow \frac {CD}{AK} \equal{} \frac {BD}{AB} \Longrightarrow \frac {CD*AB}{AK*BD} \equal{} 1$ and $ BK \equal{} \sqrt {2} AC$
therefore, the given ratio is $ \sqrt {2}$


For much simpler calculation, use the substitution $ k = x + 3$ to get

$ \begin{*eqnarray}(x + 2)^2(x + 3)^3(x + 4)^4 - 2 \\
= (k - 1)^2k^3(k + 1)^4 - 2 \\
= k^3(k + 1)^2(k^2 - 1)^2 - 2 \\
= k^3(k^2 + 2k + 1)(k^4 - 2k^2 + 1) - 2 \\
= k^3(k^6 + 2k^5 - k^4 - 4k^3 - k^2 + 2k + 1) - 2 \\
= k^9 + 2k^8 - k^7 - 4k^6 - k^5 + 2k^4 + k^3 - 2 \\
= 0\end{*eqnarray}$

which only has one real solution $ k\approx1.215157646\implies x\approx\boxed{ - 1.784842354}$.

Does anyone have an idea how to manually solve the last equation?

Edit: Fixed an error.

Um? I don't think it has an elementary solution. If it does, my computer can't find it.


Find the minimum value of the sum:
\[ \frac{sin\alpha\cdot sin\beta}{sin^{2}\frac{\gamma}{2}}+\frac{sin\beta\cdot sin\gamma}{sin^{2}\frac{\alpha}{2}}+\frac{sin\gamma\cdot sin\alpha}{sin^{2}\frac{\beta}{2}},\]
where $ \alpha,\beta,\gamma-$ angles of the triange.

Please try to prove that $ f(\alpha,\beta,\gamma)$ $ \geq$ $ f(\alpha,$\frac{(\beta+\gamma})-2$ ,$\frac{(\beta+\gamma)}-2$ )$. With assume that $ \alpha=min{\alpha,\beta,\gamma}$ or $ \alpha=max{\alpha,\beta,\gamma}$


[quote="GameBot"]Charlie rides her bike on a route that is 22 miles long. It takes her 165 minutes to complete the ride. What is her resulting average speed, in miles per hour?[/quote]

Isn't Charlie a boy's name?

Since the route is 22 miles long and it takes her 165 minutes, she takes $ 165 \div 22 \equal{} 7.5$ minutes to cover one mile. We then have $ 60 \div 7.5 \equal{} 8$ miles per hour.