algebra
polynomial
logarithms
function
inequalities
complex numbers
complex analysis

I suppose you were going for something along the lines of:

$ i \pi \equal{} P(k) \minus{} 1 \iff e^{\pi i} \equal{} e^{P(k) \minus{}1} \equal{} e^{\minus{}1} \iff \minus{}1 \equal{} e^{\minus{}1}$

But as you can see, it doesn't really work out, so you may want to revise it (and clarify your wording... like, what $ a_n$ is)  :)

What?  Your post is nonsense.  Is $ k$ supposed to be the root of the polynomial $ a_n x^n \plus{} a_{n \minus{} 1} x^{n \minus{} 1} \plus{} ... \plus{} a_0$?  Then the given statements reduce to

$ \imath \pi \equal{} \minus{} 1$

which is absurd.  In either case, why add a bunch of terms that evaluate to $ 0$ by definition?

no sorry i forgot to write the $ ln$ i will edit.

[quote="binomial_4eva"]prove that:

$ i \equal{} \frac {1}{\pi}ln(a_nk^n \plus{} a_{n \minus{} 1}k^{n \minus{} 1} \plus{} \cdots \plus{} a_0 \minus{} 1)$
and
$ \pi \equal{} \frac {1}{i}ln(a_nk^n \plus{} a_{n \minus{} 1}k^{n \minus{} 1} \plus{} \cdots \plus{} a_0 \minus{} 1)$

where $ k$ is the root of the polynomial, $ a_j$ is the coefficients and $ i$ is the imaginary number.

....edited.....[/quote]
[hide]  We can prove both equations in one shot by showing that $ \pi i \equal{} \ln(a_nk^n \plus{} a_{n \minus{} 1}k^{n \minus{} 1} \plus{} \cdots \plus{} a_0 \minus{} 1)$. Call $ a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1}... \plus{} a_0 \equal{} P(x)$.  So, we have that the left hand side is $ P(k) \minus{} 1 \equal{} \minus{} 1$ since k is a root of $ P(x)$.  So, we have that $ \pi i \equal{} \ln (\minus{} 1)$.  Taking both sides to the $ e$th power, we have that $ e^{\pi i} \equal{} \minus{} 1$, which is obviously true. [/hide]

The complex logarithm is an infinitely-valued function, i.e. $ \log(\minus{}1) \equal{} \{\ldots, \minus{}3\pi, \minus{}\pi, \pi, 3\pi, \ldots\}$.  So the question, while not complete nonsense, is still wrong.  (The reason this happens is that $ e^{2\pi i} \equal{} 1$, so if $ e^a \equal{} b$ then also $ e^{a \plus{} 2\pi n i} \equal{} b$ for any $ n \in \bf Z$ -- thus, if there is one solution to this equation, there are infinitely many.  At most one of these will be real.)

[quote="ProtestanT"][quote="JBL"]The complex logarithm is an infinitely-valued function, i.e. $ \log( \minus{} 1) \equal{} \{\ldots, \minus{} 3\pi, \minus{} \pi, \pi, 3\pi, \ldots\}$.  So the question, while not complete nonsense, is still wrong.  (The reason this happens is that $ e^{2\pi i} \equal{} 1$, so if $ e^a \equal{} b$ then also $ e^{a \plus{} 2\pi n i} \equal{} b$ for any $ n \in \bf Z$ -- thus, if there is one solution to this equation, there are infinitely many.  At most one of these will be real.)[/quote]

it's asking to show that the equations are true, not asking for solutions to the equation. ($ k$, $ i$, $ a_j$, and $ \pi$ are already defined and $ n$ is given)
And like quatto showed, the equations are true. I don't think the question is wrong in that case.
even though $ \minus{} 1$ is equal to expressions other than $ e^{\pi i}$ (like $ i^2$, $ 1 \minus{} 2$, $ \frac {5}{ \minus{} 5}$, $ e^{7\pi i}$, etc), that doesn't mean that $ \minus{} 1 \equal{} e^{\pi i}$ is not true.[/quote]

The RHS evaluates to $ \ln (\minus{}1)$, which is not single-valued.  QuattoMaster's proof is incorrect:  it's not hard to see that he argued backwards.  Arguing forwards, we face the problem that JBL describes.

Allow me to expand on what t0r said.  False premises can lead to true conclusions.  (In fact, a false premise can lead via a valid chain of logic to [i]any[/i] conclusion.)  Quatto showed that if the given statement is true, then we can derive the equality $ e^{i \pi} \equal{} \minus{} 1$.  However, Quatto did not derive an equivalence between these two statements because one step (exponentiating) isn't reversible, precisely because the complex logarithm is multivalued.

[quote="t0rajir0u"][quote="ProtestanT"][quote="JBL"]The complex logarithm is an infinitely-valued function, i.e. $ \log( \minus{} 1) \equal{} \{\ldots, \minus{} 3\pi, \minus{} \pi, \pi, 3\pi, \ldots\}$.  So the question, while not complete nonsense, is still wrong.  (The reason this happens is that $ e^{2\pi i} \equal{} 1$, so if $ e^a \equal{} b$ then also $ e^{a \plus{} 2\pi n i} \equal{} b$ for any $ n \in \bf Z$ -- thus, if there is one solution to this equation, there are infinitely many.  At most one of these will be real.)[/quote]

it's asking to show that the equations are true, not asking for solutions to the equation. ($ k$, $ i$, $ a_j$, and $ \pi$ are already defined and $ n$ is given)
And like quatto showed, the equations are true. I don't think the question is wrong in that case.
even though $ \minus{} 1$ is equal to expressions other than $ e^{\pi i}$ (like $ i^2$, $ 1 \minus{} 2$, $ \frac {5}{ \minus{} 5}$, $ e^{7\pi i}$, etc), that doesn't mean that $ \minus{} 1 \equal{} e^{\pi i}$ is not true.[/quote]

The RHS evaluates to $ \ln ( \minus{} 1)$, which is not single-valued.  QuattoMaster's proof is incorrect:  it's not hard to see that he argued backwards.  Arguing forwards, we face the problem that JBL describes.[/quote]
does that mean that $ \ln( \minus{} 1)$ does not equal $ \pi i$?

$ i\pi$ is [b]one of[/b] the possible values for $ \ln(\minus{}1)$. It can be argued that it is the principal value, in which case we could define $ \ln$ to give the principal logarithm similarly to how we have $ \sqrt{\ }$ give the principal root. If we define the complex logarithm this way, then the equations above are true.

Wouldn't an equation be considered true if it doesn't have any false information?
Like saying $ x \equal{} 2$ is true if the previous statement is $ x^2 \equal{} 4$, but it doesn't contain all the possible answers. You can still prove that $ x \equal{} 2$ works for $ x^2 \equal{} 4$ even though $ x\equal{}2$ isn't all the possibilities.

$ x^2 \equal{} 4 \not\Rightarrow x \equal{} 2$, because $ x$ could be $ \minus{} 2$. What you actually showed is that $ x$ [b]could[/b] be $ 2$, which is different from $ x$ [b]is[/b] $ 2$.

[quote="binomial_4eva"]i think The QuattoMaster meant this:

this is how i came up with that: because $ e^{i\pi} + 1 = 0$ and $ a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 = 0$ when $ k$ is the root of the $ P(x)$

$ \therefore e^{i\pi} + 1 = a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0$
$ \implies a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1 = e^{i\pi}$
$ \implies \ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1}) = \ln{e^{i\pi}}$
$ \implies \ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1}) = i\pi$
$ \implies i = \frac {1}{\pi}\ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1})$[/quote]

You define $ k$ to be a root of $ P(x)$, so $ P(k) = 0$ by definition.  Why would you consider this a useful or meaningful addition to your problem?

[quote="ProtestanT"]Like saying $ x = 2$ is true if the previous statement is $ x^2 = 4$, but it doesn't contain all the possible answers. You can still prove that $ x = 2$ works for $ x^2 = 4$ even though $ x = 2$ isn't all the possibilities.[/quote]

That statement [b]does[/b] contain false information.  By saying that $ x$ must equal $ 2$ given that $ x^2 = 4$, you will be wrong if it turns out that $ x = -2$.

[b]Example:[/b]  Consider evaluating the quantity

$ x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... }}}$

It's not hard to see that

$ x^2 = 6 + x \Leftrightarrow$
$ x^2 - x - 6 = 0$

If I go on to conclude that $ x = -2$, this would be valid according to your logic, but this is [b]wrong.[/b]  $ x$ has only a single value in this case because the square root function is defined to return only the [b]positive[/b] root, so in fact $ x = 3$.

@ t0rajir0u - Technically you're correct, but I think ProtestanT simply means $ x\equal{}2$ is [i]one[/i] of the solutions, but not necessarily all of them.

Isn't your example different from ProtestanT's example? In his, the equation is $ x^2\equal{}4$, and no indication of $ x$ being only the principle root is made, so multiple answers are possible. Yours, however, has only one solution.

On the other hand, if we're talking in a 100% mathematically correct way, then saying $ x\equal{}2$ is incorrect.

[quote="i_like_pie"]@ t0rajir0u - Technically you're correct, but I think ProtestanT simply means $ x \equal{} 2$ is [i]one[/i] of the solutions, but not necessarily all of them.[/quote]

ProtestanT's question is in reference to the statement that $ \ln (\minus{}1) \equal{} \imath \pi$.  The validity of this statement depends on which branch cut of $ \ln z$ we are using.  Without any such information, the [b]implication[/b] (I am focusing solely on the nature of the logic behind the chain of implication involved, which is the same in both my example and ProtestanT's)

$ e^{\imath \pi} \equal{} \minus{}1 \implies \imath \pi \equal{} \ln (\minus{}1)$

is incorrect, as is the implication $ x^2 \equal{} 4 \implies x \equal{} 2$.  In both cases we are applying a multivalued function to both sides of an equation.

So it's more complicated than I thought... :ninja:

[quote="binomial_4eva"]$ a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1 = e^{i\pi}$
$ \implies \ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1}) = \ln{e^{i\pi}}$[/quote]

And this is the step that is incorrect.  It is [i]not[/i] true in the complex numbers that $ e^x = y$ implies $ x = \ln y$ (or, in other terms, $ e^x = e^y$ does not imply $ x = y$ in the complex numbers, even though it does in the real numbers).  This is confusing, but you should already be familiar with a similar example (and one that a couple of other posters have batted around for a few posts), namely that $ x^2 = y^2$ does not imply $ x = y$.  One can introduce terminology and other ways of dealing with this (again, some other posters have started batting this around: branches of the logarithm function and principle values and so on), but none of that is really necessary to see the flaw.
For what it's worth, I've seen a lot of smart people get tripped up on this (any students of Mr. Cocoros at Stuyvesant can ask him about it) -- it's a subtlety that is really easy to miss if you haven't had it pointed out to you before.  If you keep on taking math classes, the first time you'll examine this behavior in any kind of detail and rigor will be a complex analysis course in college.

Okay, from what i'm understanding from a few posts above, $ e^x \equal{} y$ and $ x \equal{} \ln y$ works one way but not the other. Which way(s) does the implying work and which way(s) does it not work?
(Does $ e^x \equal{} y \implies x \equal{} \ln y$ or $ x \equal{} \ln y \implies e^x \equal{} y$ work? Or is it neither?)

Also, could you provide an example of when it works for the hypothesis but not for the conclusion?

[This comment is now obsolete.]  It's bad practice to quote the entire post directly preceding yours: that is what everyone will assume you're responding too, anyhow, so all it does is clutter up the thread.

Ask and ye shall receive:

Good:
For any complex numbers $ x, y$, $ x \equal{} y \implies e^x \equal{} e^y$.  (Equivalently, $ x \equal{} \ln y \implies e^x \equal{} y$.)

Bad:
For complex numbers $ x, y$, $ e^x \equal{} e^y \not\implies x \equal{} y$.  (Equivalently, $ e^x \equal{} y \not\implies x \equal{} \ln y$.)

Way to remember this: $ e^0 \equal{} e^{2\pi i}$ but $ 0 \neq 2\pi i$.

An alternative way to say the same thing:  $ e^x$ is a proper function (so for every input it has only one output, and if two inputs are equal then the two outputs are equal) but $ \ln x$ is not (every input has multiple possible outputs (except 0, for which it is undefined)).

Yet another alternative way to say the same thing: for $ x \neq 0$, $ e^{\ln x} \equal{} x$ is an identity.  It's true no matter what branch of the logarithm we're considering (no matter how we chop up those many values).  However, there's no consistent way to make $ \ln$ a function so that $ \ln e^x \equal{} x$ is identically true: there will always be some $ x$ that is a counter-example.  (If $ \ln 1 \equal{} 0$ then $ x \equal{} 2\pi i$ is a counter-example.  If $ \ln 1 \neq 0$ then $ x \equal{} 0$ is a counter-example.)

[quote="t0rajir0u"][quote="binomial_4eva"]i think The QuattoMaster meant this:

this is how i came up with that: because $ e^{i\pi} + 1 = 0$ and $ a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 = 0$ when $ k$ is the root of the $ P(x)$

$ \therefore e^{i\pi} + 1 = a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0$
$ \implies a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1 = e^{i\pi}$
$ \implies \ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1}) = \ln{e^{i\pi}}$
$ \implies \ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1}) = i\pi$
$ \implies i = \frac {1}{\pi}\ln({a_nk^n + a_{n - 1}k^{n - 1} + \cdots + a_0 - 1})$[/quote]

You define $ k$ to be a root of $ P(x)$, so $ P(k) = 0$ by definition.  Why would you consider this a useful or meaningful addition to your problem?[/quote]

Never lose the holy curiousity "Albert Einstein"
because there is a relation. because then we might be able to show the roots of any polynomial using $ e$,$ \pi$,$ i$.

[quote="t0rajir0u"][quote="binomial_4eva"]because there is a relation. [/quote]

No, there isn't.  You [b]define[/b] $ k$ to be a root of $ P(x)$, which means that $ P(k) \equal{} 0$ [b]by definition.[/b]  That part of the problem is [b]totally unrelated[/b] to the complex logarithm part (which is wrong).

[quote="binomial_4eva"]because then we might be able to show the roots of any polynomial using $ e$,$ \pi$,$ i$.[/quote]

We have told you repeatedly that this is not possible.  In fact, the first proof was published in 1824.  This is also completely irrelevant to the question; your proposed problem provides [b]no relationship[/b] between the polynomial expression and $ e, \pi, i$.

Please [b]study harder[/b] before you make your next post.[/quote]
I agree with everyone saying that I am wrong, sorry for my mistake :blush:

I know that everyone has already said this, but I want to clarify further:
I worked backwards, now let's work forwards.
\[ \ln (P(k)\minus{}1)\equal{}\ln (\minus{}1)\] $ \ln (\minus{}1)$ has more than one value.  In general, $ \ln (\minus{}1)$ can equal $ (2k\plus{}1)\pi i$ for an integer $ k$.  So, we have that \[ \ln (P(k)\minus{}1)\equal{}\ln (\minus{}1)\equal{}(2k\plus{}1)\pi i\] Thus giving us that $ \frac{\ln (P(k)\minus{}1)}{\pi}\equal{}\frac{(2k\plus{}1)\pi i}{\pi}\equal{}(2k\plus{}1)i$.  Setting $ k\equal{}0$, we obtain the inequality that you asked us to prove.  However, if $ k\ne 0$, then we obtain that your equation is false.

To binomial: The problem would be provable if we specified that $ k\equal{}0$, as we have just shown.

Two comments:
1) See the first comment of my previous post.  In this case, it's even more egregious because you aren't actually responding to the post you've quoted.

2) You've used $ k$ to mean two different things in the same equation, and that is very confusing.  Also, it makes the other 25 letters of the alphabet are jealous.


I was solving an inequality where I got myself stuck in the following inequality, may be true!! Can anybody tell me whether it is true or not:
\[ \boxed{a(b^2\minus{}c^2)\plus{}b(c^2\minus{}a^2)\plus{}c(a^2\minus{}b^2)\geq{0}}\]

[quote="Sunkern_sunflora"]I was solving an inequality where I got myself stuck in the following inequality, may be true!! Can anybody tell me whether it is true or not:
\[ \boxed{a(b^2 \minus{} c^2) \plus{} b(c^2 \minus{} a^2) \plus{} c(a^2 \minus{} b^2)\geq{0}}
\]
[/quote]
Any condition,if not $ a,b,c \in R$ this ineq is not right
Try it with$ c\equal{}0,a\equal{}2,b\equal{}1$

We can assume WLOG $ a \ge b \ge c$

Then
 $ a(b^2\minus{}c^2) \plus{} b(c^2\minus{}a^2) \plus{} c(a^2\minus{}b^2) \ge a(b^2\minus{}c^2) \plus{} a(a^2\minus{}b^2) \plus{} b(c^2\minus{}a^2)$
$ \equal{} a(a^2\minus{}c^2) \plus{} b(c^2\minus{}a^2) \equal{} (a\minus{}b)(a^2\minus{}c^2) \ge 0$

[quote="hsbhatt"]We can assume WLOG $ a \ge b \ge c$

Then
 $ a(b^2 \minus{} c^2) \plus{} b(c^2 \minus{} a^2) \plus{} c(a^2 \minus{} b^2) \ge a(b^2 \minus{} c^2) \plus{} a(a^2 \minus{} b^2) \plus{} b(c^2 \minus{} a^2)$
$ \equal{} a(a^2 \minus{} c^2) \plus{} b(c^2 \minus{} a^2) \equal{} (a \minus{} b)(a^2 \minus{} c^2) \ge 0$[/quote]
You are wrong,a basic mistake,you should check it carefully

You only can assume that $ a \geq b \geq c$ when a, b, c are symmetric.

I'm posting the actual problem with my approach of solution. That might help you guys a bit. You can also suggest alternative paths if my approach was wrong. Alright then, Here we go!
The actual problem was:
\[ \boxed{(\frac{a+2b}{a+2c})^3+(\frac{b+2c}{b+2a})^3+(\frac{c+2a}{c+2b})^3\geq{3}}\]

Where $ {a,b,c>0}$. My solution approach was:

$ \frac{\sum_{cyc}(\frac{a+2b}{a+2c})^3}{3}\geq{\prod(\frac{a+2b}{a+2c})}$ [From $ A.M.\geq{G.M.}$]

$ \Rightarrow(a+2b)(b+2c)(c+2a)\geq(b+2a)(c+2b)(a+2c)$

$ \Rightarrow{ab^2+bc^2+ca^2}\geq{a^2b+b^2c+c^2a}$

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

Right here I got stuck! :maybe:

Using AM-GM it remains to prove that
$ \sum \frac {a \plus{} 2b}{a \plus{} 2c}\geq 3$
from CS we got that
$ \sum \frac {(a \plus{} 2b)^2}{\sum (a \plus{} 2b)(a \plus{} 2c)} \geq \frac {9(a \plus{} b \plus{} c)^2}{\sum a^2 \plus{} 8\sum ab}$
so we have to prove that
$ 9(\sum a^2 \plus{} 2\sum ab)\geq 3\sum a^2 \plus{} 24\sum ab$
which is obviously...

another solution :
\[ \sum_{cyc}\frac {a \plus{} 2b}{a \plus{} 2c} \equal{} \sum_{cyc}\frac {a}{a \plus{} 2c} \plus{} \sum_{cyc}\frac {2b}{a \plus{} 2c}
\]

\[ \equal{} \sum_{cyc}\frac {a^2}{a^2 \plus{} 2ac} \plus{} 2\sum_{cyc}\frac {b^2}{ab \plus{} 2bc}
\]

\[ \geq \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)} \plus{} 2\sum_{cyc}\frac {(a \plus{} b \plus{} c)^2}{3(ab \plus{} bc \plus{} ca)}
\]

\[ \equal{} 1 \plus{} 2\sum_{cyc}\frac {(a \plus{} b \plus{} c)^2}{3(ab \plus{} bc \plus{} ca)} \geq 3
\]

\[ \therefore \sum_{cyc}\frac {a \plus{} 2b}{a \plus{} 2c} \geq 3
\]
By holder ;
\[ 9\sum_{cyc}(\frac {a \plus{} 2b}{a \plus{} 2c})^3 \geq (\sum_{cyc}\frac {a \plus{} 2b}{a \plus{} 2c})^3 \geq 27
\]

\[ \therefore \sum_{cyc}(\frac {a \plus{} 2b}{a \plus{} 2c})^3 \geq 3
\]

Thanks to both of you guys, for such nice solutions!


Let $m$ be a positive integer. Define the sequence $a_0, a_1, a_2, \cdots$ by $a_0 = 0,\; a_1 = m,$ and $a_{n+1} = m^2a_n - a_{n-1}$ for $n = 1,2,3,\cdots$.
Prove that an ordered pair $(a,b)$ of non-negative integers, with $a \leq b$, gives a solution to the equation
\[ {\displaystyle \frac{a^2 + b^2}{ab + 1} = m^2} \]
if and only if $(a,b)$ is of the form $(a_n,a_{n+1})$ for some $n \geq 0$.

We have $a^2+b^2-(ab+1)m^2=0,a>b$.
 $b^2+(m^2b-a)^2-m^2-m^2b(m^2b-a)=0$ Tefore if (a,b) is solution, then $(b,m^2b-a)$ is solution. If $b\gem$, then new solution satisfied $b>m^2b-a$. When a<m have not solutios. Therefore smallest solution is (m,0), and all other solutions can be find by this recurent relations.

New take on an old IMO problem (#6 from 1988).  The problem statement in that case was proving that if $ ab \plus{} 1$ divides $ a^2 \plus{} b^2$, then $ \frac {a^2 \plus{} b^2}{ab \plus{} 1} \equal{} d$ is a perfect square.  Possibly the most straightforward solution was by descent, and if I am not mistaken one of the ways to find the descent was to find a "reverse recursion" equivalent to the "forward recursion" given in this problem.  The conclusion followed the fact that the descend "stopped" iff $ d$ was a perfect square, otherwise it became infinite, absurd since $ a,b$ are positive integers...

Nice new spin on one of my all-time favorites, good choice Shobber and good job Rust!!!  :D


geometry
perimeter
inequalities
circumcircle
geometry proposed

Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.

When $\triangle ABC$ is right-angled, the inequality is trivial, thus we assume that $\triangle ABC$ is acute. Let $D,E,F$ denote the midpoints of $BC,CA,AB.$ Since $O$ becomes orthocenter of $\triangle DEF$ (also acute) then $O$ is inside of it $\Longrightarrow$ $O$ is inside the quadrilateral $BFEC,$ thus $BF+FE+EC>BO+OC$ $\Longrightarrow$ $a+b+c>4R$ $\Longrightarrow$ perimeter of $\triangle ABC$ is greater than two times its circumcircle diameter.


Vieta
algebra
polynomial
factorial
sum of roots

This can be done mentally, but I just want to know if there is an algebraic (or other) method to solve this problem.

Find the sum of the real solutions of $(x-2)(x-3)(x-4)(x-5)=360$.

Well, why not just go with the quickest way?

Isn't it Viete?

And could you elaborate, I forgot those.
What do you mean by the "easy way," nutz?

it is vieta not viete.
And vieta just relates roots to the coefs of the polynomial, and part of it says sum of roots of monic polynomial is the coef of the $x^{n-1}$ when the deg is n.
here, the adding/subtracting constant doesnt change the coef of the term we want, so the sum of roots stay the same(which is 2+3+4+5=14).

n!/(n-4)! = 360, where n=x-2.

Now, it is easy to find all possible solutions. I think this is alternative solution. If I made a mistake, someone tell me.

[quote="JavaMan"]You can use Vieta's formulas. It's simply $-\frac{a_{3}}{a_{4}}= 14$.[/quote]

How did this happen, sorry, I am not familiar with Vieta's Formula.

[quote="neelnal"]n!/(n-4)! = 360, where n=x-2.[/quote]

Now, it is easy to find all possible solutions. I think this is alternative solution. couple things
1. x is not necessarily an integer, so factorial is messed up
2. x is not necessarily a real number either. in fact, 2 solns for the equation are imaginary

[quote="lotrgreengrapes7926"][quote="junggi"]2. x is not necessarily a real number either. in fact, 2 solns for the equation are imaginary[/quote]
It asks for the sum of real solutions. Does Vieta take only real solutions?[/quote]
oh ok then nvm.
well factoring gives solns $x=-1,8$ so sum is $7$  :P
edit: vieta takes all solutions

Vieta's includes complex roots, so I think you have to do a bit more to find the sum of the real roots

If you group the expressions in the parentheses as $(x-3)(x-4)$ and $(x-2)(x-5)$, the equation becomes, \[(x^{2}-7x+12)(x^{2}-7x+10) = 360.\] Make the substition $a=x^{2}-7x+12$ and rewrite the equation as, \[a(a-2)=360.\] Solving yields $20$ and $-18$ as solutions. However, $a=18$ has complex solutions, so we don't count those. Setting $a=20$ yields the equation, \[x^{2}-7x-8=0,\] with both roots real. The sum of them is $7$.

EDIT: oops, got beat to it


topology
real analysis
real analysis unsolved

Let $(X,d)$ be a complete metric space. $A\subset X$ is a closed set. For any $\epsilon>0$, there exists a compact set $A_\epsilon$ such that $O(x,\epsilon)\cap A_\epsilon \neq \emptyset (\forall x\in A)$.
(where $O(x,d)$ means the open neighbour of point $x$ with radius of $d$)
Show that $A$ is compact.

You should make additional assumptions to make the statement true (for instance, to assume that $X$ is complete). Otherwise $X=A=(0,1)$, $A_\varepsilon=[\varepsilon, 1-\varepsilon]$ is a counterexample.

Closed subset of $X$, isn't it? And the entire space is always a closed subset of itself. Note that $X=(0,1)$ in my example, not $\mathbb R$. The fact that the metric in $X$ is induced by the metric in a wider space $\mathbb R$ and $A$ is not a closed subset of that wider space is completely irrelevant. $A=X$ is still closed in $X$.

Oh, yes, $X$ should be complete. Now I have modified it.

$A_{\frac\varepsilon 4}$ can be covered with finitely many bals of radius $\frac\varepsilon 4$. Call them $B_1(a_1),\ldots,B_n(a_n)$ ($a_i\in A_{\frac\varepsilon 4}$ are the centers). It's easy to see now that the balls of radius $\frac\varepsilon 2$ centered at $a_1,\ldots,a_n$ cover $A$, and if $x_i\in A$ is a point belonging to the ball of radius $\frac\varepsilon 2$ centered at $a_i$, then the balls of radius $\varepsilon$ centered at $x_i$ cover $A$.

That means that when regarded as a metric space (with the metric induced by $d$), $A$ is both complete and totally bounded, and thus compact.


trigonometry
AMC
AIME
geometry
quadratics
trig identities
Law of Sines

In $ \triangle ABC, \angle B \equal{} 3 \angle C$. If $ AB \equal{} 10$ and $ AC \equal{} 15$, find the length of $ BC$.

[hide="solution"] Consider point $ D$ on $ AC$ such that angles $ DBC$ and $ DCB$ are congruent. Note that angles $ ABD$ and $ ADB$ are also congruent.

This gives us a pair of isosceles triangles; we find $ AD\equal{}10$, $ BD\equal{}5$, and if we want, we can use Stewart's Theorem to find $ BC$.[/hide]

That is a pretty awesome (and more intuitive solution).

Does anyone have another using any form of trigonometry, like Law of Sines or Cosines? (I found this problem in a trigonometry handout)

[hide] We have, by the Law of Sines,

$ \frac {AC}{\sin B} \equal{} \frac {AB}{\sin C} \implies \frac {15}{\sin(3C)} \equal{} \frac {10}{\sin(C)}$.

Next, we use the identity $ \sin 3\theta \equal{} 3\sin \theta \minus{} 4\sin^3 \theta$ (use $ \sin(\alpha \plus{} \beta) \equal{} \sin \alpha\cos \beta \plus{} \sin \beta\cos\alpha$ to derive it).

Thus,

$ \frac {15}{3\sin C \minus{} 4\sin^3 C} \equal{} \frac {10}{\sin C}$

$ \implies 40\sin^3 C \minus{} 15\sin C \equal{} 0 \implies sin C(40\sin^2 C \minus{} 15) \equal{} 0$.

Now, if $ \sin C \equal{} 0$, then $ C \equal{} 0^\circ$, which obviously can't happen, so

$ 40\sin^2 C \minus{} 15 \equal{} 0 \implies \sin^2 C \equal{} \frac {3}{8} \implies \sin C \equal{} \sqrt {\frac38}$ (we don't need to worry about the possibility of a negative value since $ 0^\circ < \angle A, \angle B, \angle C < 180^\circ$).

Then, using our expression for $ \sin B$ in terms of $ \sin C$, we get

$ \sin B \equal{} 3(\sqrt {\frac38}) \minus{} 4(\frac38)(\sqrt {\frac38})\implies \sin B \equal{} \frac32(\sqrt {\frac38})$.

Now, $ \angle A \equal{} 180^\circ \minus{} (\angle B \plus{} \angle C)$

So, utilizing the identities $ \sin \theta \equal{} \sin(180 \minus{} \theta)$,
$ \sin(\alpha \plus{} \beta) \equal{} \sin\alpha\cos\beta \plus{} \cos\alpha\sin\beta$,
$ \cos\theta \equal{} \sqrt {1 \minus{} \sin^2\theta}$

we get that $ \sin A \equal{} \sin(B \plus{} C)$ and

$ \sin(B \plus{} C) \equal{} \sin B\cos C \plus{} \sin C\cos B$

$ \equal{} \frac32(\sqrt\frac38)(\sqrt {\frac58} \plus{} \sqrt {\frac38}\sqrt {\frac {5}{32}}$

$ \equal{} \frac {4\sqrt {15}}{16} \equal{} \frac {\sqrt {15}}{4} \equal{} \sin A$.

Now, using the Law of Sines,

$ \frac {10}{\sin C} \equal{} \frac {BC}{\sin A}$

$ \implies BC \equal{} \frac {10(\sin A)}{\sin C}$

$ \implies BC \equal{} \frac {10(\frac {\sqrt {15}}{4})}{\sqrt {\frac38}}$

$ \implies BC \equal{} 5\sqrt {10}$.[/hide]

[quote="mathemagician1729"]Question: Is Stewart's theorem something you need to keep at the top of your head if you're trying to tackle AIME level geometry problems? Are there any other theorems which would also prove useful because I know that AoPS volume 2 treats geometry very little.[/quote]
Not really know off the top of your head but you should be able to prove it relatively easily with the Law of Cosines. Stewarts along with Ptolemy's, and Brahmagupta's are some that I know of that are relatively useful.

* I decided to relearn Stewart's Theorem but I am really confused now because I seem to get two different answers.

Using xpmath's idea and actually computing the answer after stewart's theorem, I get $ \frac{5\sqrt{10}}{2}$.

However, after doing it my way (basically plenty of algebra/trig manipulations) I ended up getting a quadratic which gave me

$ \frac{5\sqrt{10}}{2}$ and $ 5\sqrt{10}$.

I'm guessing the latter is incorrect but this worries me because that was gaussintraining's answer.

If you get a quadratic in a geometry problem with only one solution to the original problem, you can discard one of the solutions because it's nonsense (generally). Or perhaps both answers are correct (I don't know, I didn't do the problem).


As the title says, I am wondering if buying Mathematics of Choice is worth it. Some people tell me its an excellent book useful for counting techiniques that can come up on various math contests, and some tell me its too advanced and doesnt present the topics nicely. What are your opinions?

Without a doubt, Mathematics of Choice by Ivan Niven is worth buying. So is almost any book in that same New Mathematical Library series. We can get it from our public library here, which is the only reason I haven't already bought it.


algebra
polynomial
number theory
relatively prime
superior algebra
superior algebra unsolved

Show that if $ F$ is a field, $ p$ is a prime, and $ a\in F$, then the polynomial $ X^p\minus{}a\in F[X]$ is either irreducible or has a root in $ F$.

[color=red]Suppose $ \text{car}F\equal{}p$.[/color] Let $ K$ be the splitting field of $ X^p \minus{} \alpha$ over $ F$.
There is some element $ \beta\in K$ such that $ \beta^p \equal{} \alpha$.
Hence in $ K[X]$ we have $ X^p \minus{} \alpha \equal{} (X \minus{} \beta)^p$.

Let suppose $ X^p \minus{} \alpha$ be riducible in $ F[X]$.
Then it has a proper factor $ f\in F[X]$; in $ K[X]$ we have $ f(X) \equal{} (X \minus{} \beta)^n$ for some $ 0 < n < p$.
Conseguently, we have $ f(X) \equal{} X^n \minus{} \beta X^{n \minus{} 1} \plus{} \cdots$ hence $ \minus{} \beta\in F$.
We observe that $ n$ is relatively prime with $ p$, thus $ \beta\in F$ and $ X^p \minus{} \alpha$ has a root in $ F$.

[color=red]Edit: on remark by loup blanc.[/color]

Hi ficus,
$ X^{p}\minus{}\alpha\equal{}(X\minus{}\beta)^{p}$ ????
 A priori $ charac(K)\not\equal{}p$.

[quote="loup blanc"]Hi ficus,
$ X^{p} \minus{} \alpha \equal{} (X \minus{} \beta)^{p}$ ????
 A priori $ charac(K)\not \equal{} p$.[/quote]
I solved the case with $ \text{car}F \equal{} p$.
It remain the case $ \text{car}F\neq p$.

$ |F| \equal{} p$ is hardly a "case".  Anyway, somehow this problem comes up a lot, and I've yet to see anything better than the following:

Suppose that $ X^p\minus{}a$ factors as $ f\cdot g$, with $ k\equal{}\deg{f}$ strictly between $ 0$ and $ p$.  Then there exist integers $ m$ and $ n$ such that $ mk \plus{} np \equal{} 1$.

Let $ \beta$ be the product of the roots of $ f$.  We have $ \beta^p \equal{} a^k$.  But then $ a \equal{} a^{mk \plus{} np} \equal{} (\beta^m a^n)^p$, so $ a$ is a $ p$-th power in $ F$.


\[ \left\Vert \begin{array}{c}
x,y,z>0\\
a\equal{}x\plus{}\frac{1}{x}\\
b\equal{}y\plus{}\frac{1}{y}\\
c\equal{}xy\plus{}\frac{1}{xy}\end{array}\right\Vert \Longrightarrow abc\plus{}4\ge ab\plus{}bc\plus{}ca\]

Note that the given conditions imply that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} abc \plus{} 4$.
(You'll have to compute that yourself!)

Since $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4 \plus{} abc \ge ab \plus{} bc \plus{} ca$, we are done!

$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$ follows as it is equavalent to $ \frac {1}{2}[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \ge 0$ which is trivially true.


i was just going to ask if there were any peopoel that live in indiana that do not post here.  ill pm you about it.


calculus
integration
limit
conics
parabola
calculus computations

Problem: Prove that:

$ \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\frac {\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{2}}}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{{{\left( \sqrt {\frac {i}{n}} \minus{} \frac {i}{n} \right)}^{3}}}$
This problem is just begun from this excercise.
Example 3: Find the volume of the solid that lies under the paraboloid  $ z \equal{} {{x}^{2}}$ and above the region D in the xy-plane bounded by the line y = 2x and the parabola $ y \equal{} {{x}^{2}}$.
Solution 1: The parabol and the line intersect with equation   $ {{x}^{2}} \equal{} 2x\Leftrightarrow x \equal{} 0\,\,or\,\,x \equal{} 2\Rightarrow 0\le x\le 2$
$ \text{In}\,\,\text{interval}\,\,\left[ 0,2 \right]\,\,{{x}^{2}}\le 2x\Rightarrow {{x}^{2}}\le y\le 2x.\,\,So\,\,D \equal{} \left\{\left( x,y \right)|0\le x\le 2,\,\,{{x}^{2}}\le y\le 2x \right\}$

$ V \equal{} \iint\limits_{R}{\left( {{x}^{2}} \right).dA} \equal{} \int\limits_{0}^{2}{\int\limits_{{{x}^{2}}}^{2x}{\left( {{x}^{2}} \right).dydx}} \equal{} \int\limits_{0}^{2}{\left[ {{x}^{2}}.y \right]_{y \equal{} {{x}^{2}}}^{y \equal{} 2x}.dx} \equal{} \int\limits_{0}^{2}{\left( 2{{x}^{3}} \minus{} {{x}^{4}} \right).dx} \equal{} \left[ \frac {2{{x}^{4}}}{4} \minus{} \frac {{{x}^{5}}}{5} \right]_{0}^{2} \equal{} \frac {8}{5}.$

$ \text{Expressing}\,\,\iint\limits_{R}{{{x}^{2}}.dA}\,\,in\,\,double\,\,Riemann\,\,sum: x\in \left[ 0,2 \right]\Rightarrow \Delta x \equal{} \frac {2}{n},\,\,{{x}_{i}} \equal{} i.\Delta x \equal{} \frac {2i}{n}] \[{{x}^{2}}\le y\le 2x\Rightarrow \Delta y \equal{} \frac {2x \minus{} {{x}^{2}}}{n} \equal{} \frac {1}{n}.\left( \frac {4i}{n} \minus{} \frac {4{{i}^{2}}}{{{n}^{2}}} \right) \equal{} \frac {4\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{3}}},\,\,x_{i}^{2} \equal{} \frac {4{{i}^{2}}}{{{n}^{2}}}$

$ \Rightarrow \int\limits_{0}^{2}{\int\limits_{{{x}^{2}}}^{2x}{\left( {{x}^{2}} \right).dydx}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\sum\limits_{j \equal{} 1}^{n}{\left( \frac {4{{i}^{2}}}{{{n}^{2}}} \right).\frac {8\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{4}}}}}$

$ \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\sum\limits_{j \equal{} 1}^{n}{\frac {32{{i}^{2}}\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{6}}}}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\frac {32{{i}^{2}}\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{6}}}} \equal{} \frac {8}{5}.$
$ Solution\,\,2: \,\,y \equal{} 2x\Rightarrow x \equal{} \frac {y}{2},\,\,y \equal{} {{x}^{2}}\Rightarrow x \equal{} \sqrt {y}\,\,\left( because\,\,x > 0 \right)$

$ V \equal{} \iint\limits_{R}{{{x}^{2}}.dA} \equal{} \int\limits_{0}^{4}{\int\limits_{y/2}^{\sqrt {y}}{{{x}^{2}}.dxdy}} \equal{} \int\limits_{0}^{4}{\left[ \frac {{{x}^{3}}}{3} \right]_{x \equal{} y/2}^{x \equal{} \sqrt {y}}.dy} \equal{} \int\limits_{0}^{4}{\left( \frac {{{y}^{3/2}}}{3} \minus{} \frac {{{y}^{3}}}{24} \right).dy} \equal{} \left[ \frac {2{{y}^{5/2}}}{15} \minus{} \frac {{{y}^{4}}}{96} \right]_{0}^{4} \equal{} \frac {8}{5}.$

$ \text{Expressing}\,\,\iint\limits_{R}{{{x}^{2}}.dA}\,\,in\,\,double\,\,Riemann\,\,sum: y\in \left[ 0,4 \right]\Rightarrow \Delta y \equal{} \frac {4}{n},\,\,{{y}_{j}} \equal{} j.\Delta y \equal{} \frac {4j}{n},\,\,$

$ \frac {y}{2}\le x\le \sqrt {y}\Rightarrow \Delta x \equal{} \frac {2\sqrt {y} \minus{} y}{2n} \equal{} \frac {1}{2n}.\left( 4\sqrt {\frac {j}{n}} \minus{} \frac {4j}{n} \right) \equal{} \frac {2}{n}.\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right),\,\,$

$ {{x}_{i}} \equal{} i.\Delta x \equal{} \frac {2i}{n}.\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right),\,\,x_{i}^{2} \equal{} \frac {4{{i}^{2}}}{{{n}^{2}}}.{{\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right)}^{2}}$

$ \Rightarrow \int\limits_{0}^{4}{\int\limits_{y/2}^{\sqrt {y}}{{{x}^{2}}.dxdy}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\sum\limits_{j \equal{} 1}^{n}{\left( \frac {4{{i}^{2}}}{{{n}^{2}}} \right).{{\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right)}^{2}}.\frac {8}{{{n}^{2}}}.\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right)}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\sum\limits_{j \equal{} 1}^{n}{\frac {32{{i}^{2}}}{{{n}^{4}}}.{{\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right)}^{3}}}} \equal{} \frac {8}{5}.$

$ \Rightarrow \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\frac {32{{i}^{2}}\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{6}}}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\sum\limits_{j \equal{} 1}^{n}{\frac {32{{i}^{2}}}{{{n}^{4}}}.{{\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right)}^{3}}}} \equal{} \frac {8}{5}.$

$ \Rightarrow \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{\frac {\left( n.i \minus{} {{i}^{2}} \right)}{{{n}^{2}}}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{j \equal{} 1}^{n}{{{\left( \sqrt {\frac {j}{n}} \minus{} \frac {j}{n} \right)}^{3}}} \equal{} \underset{n\to \plus{} \infty }{\mathop{\lim }}\,\sum\limits_{i \equal{} 1}^{n}{{{\left( \sqrt {\frac {i}{n}} \minus{} \frac {i}{n} \right)}^{3}}}$

Explain what? :? (It seems that my English is not good enough to understand what the OP meant again. :( ).

Probably the OP wants clarification in his(er) results.

My question is to prove that:

$ \underset{n\to \plus{}\infty }{\mathop{\lim }}\,\sum\limits_{i\equal{}1}^{n}{\frac{\left( n.i\minus{}{{i}^{2}} \right)}{{{n}^{2}}}}\equal{}\underset{n\to \plus{}\infty }{\mathop{\lim }}\,\sum\limits_{i\equal{}1}^{n}{{{\left( \sqrt{\frac{i}{n}}\minus{}\frac{i}{n} \right)}^{3}}}$

:what?: Haven't you just presented the full proof yourself? I mean, it is, probably, not the shortest proof, but there is nothing wrong with it. :?

But I want to know exactly the result of limit.

I think you're a bit confused.  We have $ \sum_{i \equal{} 1}^n i \equal{} \frac {n(n \plus{} 1)}{2}$ and $ \sum_{i \equal{} 1}^n i^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$, so
\[ \sum_{i \equal{} 1}^n in \minus{} n^2 \equal{} \frac {n^2(n \plus{} 1)}{2} \minus{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \equal{} \frac {n^3 \minus{} n}{6}.\]
Thus the first limit is infinite (as is the second).  If you had another factor of $ \frac {1}{n}$, the one I've written would give $ \frac {1}{6}$ while the other would give (with a bit more work) $ \frac {1}{140}$.

Yeah, I understood it. If the limits are infinite, so they are equal at infinite point.


Increasing the radius of a cylinder by $ 6$ units increased the volume by $ y$ cubic units. Increasing the altitude of the cylinder by $ 6$ units also increases the volume by $ y$ cubic units. If the original altitude is $ 2$, then the original radius is:

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 6\pi \qquad\textbf{(E)}\ 8$

[hide="Click for solution"]
$ V\plus{}y\equal{}\pi h(r\plus{}6)^2\equal{}\pi r^2(h\plus{}6) \implies (r\plus{}6)^2 \cdot 2 \equal{} r^2 \cdot 8$ $ \implies 3r^2\minus{}12r\minus{}36\equal{}0 \implies r\equal{}6$ or $ \boxed{\textbf{(C)}}$.
[/hide]

Let the radius of the cylinder be $r$. We can build the equation:
$$\pi r^2 \cdot 2 - \pi (r+6)^2 \cdot 2 = \pi r^2 \cdot 2 - \pi r^2 \cdot 8 \Leftrightarrow r^2 \cdot 8 = (r+6)^2 \cdot 2 \Leftrightarrow 4r^2 = r^2 + 12r + 36 \Leftrightarrow r^2-4r-12 = 0 \Leftrightarrow r = \boxed{6}.$$


Find all the possible pairs of positive integers $(x,y)$, both with no prime factor greater than 5, such that $x^2 - y^2 = 2^k$ where $k$ is a non-negative integer.

[hide="My Q2 Working"]
I got some way with this proof, but didn't complete it (details, but not full proof, below).
I found (and proved) the general solutions $(2^a 3, 2^a)$ and $(2^a 5, 2^a 3)\ \ (\forall a \in \mathbb{N})$  without too much difficulty (although that probably means the proofs were dodgy...). There also existed however the possibility that a solution would be found if there exists a set of consecutive integers strictly increasing or decreasing thus $\{3^a, 2^b, 5^c\} (a,b,c \in \mathbb{N} )$. I could not prove that this was impossible, or find any such a sequence, so I just left it hanging... Anyone got any ideas?
[/hide]

That seems a problem to be solved by cases. Consider different prime factors. For example x = 2^a , y = 2^b, were a > b. It's easy to prove it's not possible. So you can try different combinations of those factors and find all answers.
Or maybe it's nice to immediately start with the expression x = (2^a1)*(3^a2)*(5^a3) and similar for y and then derive all solutions.

In either way it should be quite easy to solve it.

Here is my solution
 If $x^2-y^2=2^k$, then $x-y=2^m$ and $x+y=2^n$, where m,n are positive integers(since $x-y$ and $x+y$ is divisible by 2) and, evident, $m<n$. Hence $x=\frac{2^m+2^n}{2}$ and $y=\frac{2^n-2^m}{2}$, i.e. $x=2^s(2^t+1), y=2^s(2^t-1)$, where $s=m-1\geq 0, t=n-m\geq 1$. From condition follows, that $2^t-1$ and $2^t+1$ have only prime factors $3$ and $5$($2^t-1$ and $2^t+1$ is not divisible by $2$).
1) $2^t-1$ is divisible $15$. Hence $2^t+1=(2^t-1)+2$ is not divisible neither $3$ nor $5$ and $2^t+1=1$ :arrow: $2^t=0$ but it isn’t true.
2) $2^t-1$ is not divisible neither $3$ nor $5$. Hence $2^t-1=1$ :arrow: $2^t=2$ :arrow: $t=1$. It means $x=3*2^s, y=2^s$. It is, evident, solution of our equation.
3) $2^t-1=3^a$, $a$ is a positive integer. If $t$ is odd, then $2^t$ give the remainder $2$ modulo $3$ - contradiction. Hence $t=2u$, $u$ is a positive integer, and $2^u-1=3^c, 2^u+1=3^d$ :arrow: $3^d-3^c=2$ and it’s easy to see, that $c=0, d=1$ and $t=2(c+d)=2$. It means $x=5*2^s, y=3*2^s$. It is too, evident, solution of our equation.
4) $2^t-1=5^a$, $a$ is a positive integer. If $t$ is odd, then $2^t$ give the remainder $2$ or $3$ modulo $5$ - contradiction. Hence $t=2u$, $u$ is a positive integer, and $2^u-1=5^c, 2^u+1=5^d$ :arrow: $5^d-5^c=2$ but it, clearly, can’t be.
Also we get 2 types of an answer:
[b]1.[/b]  $x=3*2^s, y=2^s$. [b]2.[/b] $x=5*2^s, y=3*2^s$.


$ O$ is the circumcircle of $ \triangle ABC$.$ M$ is the midpoint of arc $ BC$($ A$ is not on this arc).circle$ (W)$ is internaly tangent to circle $ (O)$ at $ D$ and is tangent to $ BC$ too.from $ A$ we draw tangent $ AT$ to circle$ (W)$.$ P$ is on $ AB$ such that $ AP\equal{}AT$.($ B$,$ P$ are on the on side of $ A$).prove that $ \angle APD\equal{}90$

[quote="behdad.math.math"]$ O$ is the circumcircle of $ \triangle ABC$.$ M$ is the midpoint of arc $ BC$($ A$ is not on this arc).circle$ (W)$ is internaly tangent to circle $ (O)$ at $ D$ and is tangent to $ BC$ too.from $ A$ we draw tangent $ AT$ to circle$ (W)$.$ P$ is on $ AB$ such that $ AP \equal{} AT$.($ B$,$ P$ are on the on side of $ A$).prove that $ \angle APD \equal{} 90$[/quote]

My GeoGebra disagrees with this statement. Do you mean $ D\equal{}M$?

first i think $ M$ must be $ D$.so$ MB \equal{} MC$ now its well known that ($ M$ the midpoint of $ BC$ and $ D$ is the midpoint of arc) $ (AC)(BM) \plus{} (AB)(CM) \equal{} (BC)(AT)$ so $ AP \equal{} \frac {b \plus{} c}{2}$.the problem now is trivial for me because:
lemma.if $ R$ be the projection of $ D$ on $ AB$;then $ AR \equal{} \frac {b \plus{} c}{2}$  :)


[hide]int_{a}^{b} h(t) dt will mean the integral of h(t) from a to b.

int_{3}^{x^3} e^(t^2) dt = int_{1}^{x^3} e^(t^2) dt - int_{1}^{3} e^(t^2) dt = g(x^3) - g(3)[/hide]


Let $n$ be a positive integer. Is the greatest integer less than $(3+\sqrt{7})^{n}$ odd or even?

[hide="Solution"]Observe the sequence $b_{n}=(3+\sqrt{7})^{n}+(3-\sqrt{7})^{n}$. Its recursive equation can be found from the characteristic equation $(t-(3+\sqrt{7}))(t-(3-\sqrt{7}))=0\iff t^{2}-6t+2=0$, giving $b_{n+2}=6b_{n+1}-2b_{n}$, with $b_{1}=6, b_{2}=32$. Obviously, all $b_{n}$ are even, and $0<3-\sqrt{7}<1\implies 0<(3-\sqrt{7})^{n}<1$, which means that $\lfloor(3+\sqrt{7})^{n}\rfloor=\lfloor b_{n}-(3-\sqrt{7})^{n}\rfloor$ is odd.[/hide]

[quote="Farenhajt"]Its recursive equation can be found from the characteristic equation $(t-(3+\sqrt{7}))(t-(3-\sqrt{7}))=0\iff t^{2}-6t+2=0$. [/quote]

Can you say a little bit more about using such generating equations? I know the recursion can be generated by the identity

$x^{n}+y^{n}= (x+y)(x^{n}+y^{n})-xy(x^{n-1}+y^{n-1})$

but I haven't seen this before.

First, LHS of your identity should be $x^{n+1}+y^{n+1}$

Second, put all the terms on one side and observe the coefficients with $x^{n}+y^{n}$ and $x^{n-1}+y^{n-1}$. They are $-(x+y)$ and $xy$. Does that bring Vieta's formulas into mind? Therefore, if you write $(t-x)(t-y)=0$ and expand, you'll get $t^{2}=(x+y)t-xy$, i.e. the two coefficients appear immediately.


Let $a,b\in R$.Find all $f(x)$ such that:
$f(a-x)+f(x)=b, \forall x\in R.$

Take $f: ({a\over 2},\infty)\to\mathbb{R}$ arbitrary and extend it using $f(a-x)=b-f(x)$ for $x>{a\over 2}$ and $f({a\over 2})={b\over 2}$.


Let a, b, c be positive real numbers with $abc = 1$. Prove that
$\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2}\le{1}$

we know that $a^{2}+1\geq 2a$ so applying this to the inequality we get that:

$LHS\leq \frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}$

so we have to prove that $\frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}\leq 1$

[hide]so we have to prove that $\frac{a(2b+1)(2c+1)+b(2c+1)(2a+1)+c(2a+1)(2b+1)}{(2a+1)(2b+1)(2c+1)}\leq 1$

$\Rightarrow a(2b+1)(2c+1)+b(2c+1)(2a+1)+c(2a+1)(2b+1)\leq (2a+1)(2b+1)(2c+1)$

$\Rightarrow 12abc+a+b+c+4ab+4bc+4ca\leq 8abc+4ab+4bc+4ca+2a+2b+2c+1$

so using $abc=1$ we have to prove that $3\leq a+b+c$ which follows by $AM-GM$[/hide]

[quote="inom"]Let $a,\, b,\, c$ be positive real numbers with $abc = 1.$ Prove that
\[\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2}\le{1}\]
[/quote]
[b]Proof.[/b] We have:
\[\sum\frac{a}{a^{2}+2}\le 1\iff\sum\frac{a}{2a+1}\le 1\iff \sum\frac{1}{2a+1}\ge 1 \]
Let $a=\frac{x}{y},\,b=\frac{y}{z},\,c=\frac{z}{x},$ we have:
\[\sum\frac{1}{2a+1}=\sum\frac{y}{2x+y}\ge\frac{\left(\sum x\right)^{2}}{\sum\left(x^{2}+2zx\right)}=1 \]


factorial
function
calculus
integration
real analysis
combinatorics theorems
combinatorics

1!=1
Why 0!=1 was incorporated by Christian Kramp?

shall we exclude 0 & infinity from the number list?
or they will contradict.

We know $ n! \equal{} n(n\minus{}1)(n\minus{}2)\cdots 2 \cdot 1$. Or $ (n\plus{}1)! \equal{} (n\plus{}1)(n!)$ Choose $ n \equal{} 0$ so $ 1! \equal{} 1 \cdot 0!$ and $ 0! \equal{} 1! \Rightarrow 0! \equal{} 1$

All empty products are 1, and 0! is an empty product.  In addition, it's fully consistent with both the recursive definition of factorial and the definition of the gamma function and all other contexts in which factorials arise.  What this has to do with calculus and real analysis, though  :maybe:

In the same vein, we can interpret $ 0!$ as the number of permutations of the empty set, which is $ 1$.  (There is exactly one empty set.)

as t0rajir0u says;

there are only 1 way to pick 0 "things".

why [b]nC0[/b] = 1? because there's only 1 way to choose 0 objects from n objects, which is not to choose= 1 way.

similarly, 0!=1 because combinatorially, there is only 1 way to permute 0 object, which is not to permute= 1 way.


Who will win it all? I say UCONN! (Final Four: Uconn, Duke, Florida, Pitt). Texas might take out Duke, Nova might take out FLA, and UCLA for Pitt, but I still like my picks.

So I've never watched a game of colege basketball in my life, but I think George Washington, Gonzaga, Connecticut, and Ohio State will be in the FInal Four, with the championship game being between Gonzaga and Connecticut, and Gonzaga winning.

Im going with Villanova.

And where's Gonzaga on that poll?
i noticed only 1 and 2 seeds are on here.
maybe that's why

My final four is Duke, UCONN, Villanova, and Pittsburgh with Duke winning it all. Gonzaga isn't that good. They have to depend on Adam Morrison.

they do depend on adam morrison, but when he scored 14 points yesterday, they beat Indiana. :(  :mad: (i m a hoosier fan)

my final four choices are duke, gonzaga, u conn, and florida
and duke for the overall win

mine was
Duke-still alive and ready for LSU and possibly Texas

Gonzaga-still alive and waiting for UCLA and Memphis or Bradley

North Carolina-lost to George Mason :(

Villanova-still alive after a close call against Arizona in their home town

Finals: Villanova over Duke

I'm going with UConn.  That Kansas loss totally screwed up my bracket.

Bradley and Wichita State are only big surprise teams in the Sweet Sixteen.

GO MISSOURI VALLEY CONFERENCE!!!! :!:


geometry
geometric transformation
rotation
linear algebra
matrix
vector
analytic geometry

[b] 1. [/b] What constraints must the elements $ (R_{ij})$ of the three dimensional rotation matrix satisfy in order to preserve the length of $ \bold{A}$ (for all vectors $ \bold{A}$).

I am guessing that it must equal the identity matrix?

[b] 2. [/b] How do the components of a vector transform under a translation of coordinates ($ \bar{x} \equal{} x, \ \ \bar{y} \equal{} y\minus{}a, \  \ \bar{z} \equal{} z$)? The components dont change at all (e.g. because vectors are independent of coordinate system)?

[b] 3. [/b] How do components of a vector transform under an inversion of coordinates ($ \bar{x} \equal{} \minus{}x, \ \ \ \bar{y} \equal{} \minus{} y, \ \ \ \bar{z} \equal{} \minus{}z$)? Guessing that the components change in sign, but stay the same in magnitude?

[b] 4. [/b] How does the cross product of two vectors transform under an inversion. Is the cross product of two pseudovectors, a vector or a pseudovector?

Is it $ \bold{A} \times \bold{B} \equal{} \minus{} \bold{B} \times \bold{A}$? The cross product of two pseudovectors is a pseudovector? Two examples of pseudovectors in classical mechanics are torque and angular momentum?

[b] 5. [/b] How does the scalar triple product of three vectors transform under inversions (e.g. pseudoscalar).

So $ \bold{A} \cdot (\bold{B} \times \bold{C}) \equal{} \minus{}(\bold{B} \times \bold{C}) \cdot \bold{A}$?

For #1, the well-known answer is that the columns of $ R$ should be an orthonormal set - that is, the dot product of any two columns is zero, while the dot product of any column with itself is $ 1.$ There are many, many more examples of that than merely the identity matrix (although the identity matrix does count.)

This can also be written as $ R^TR\equal{}I.$

The proof involves appropriate use of inner products (dot product). Write our vectors as columns. Then the dot product can be written as matrix multiplication: $ x\cdot y\equal{} x^Ty.$

Now, we prove that for any vectors $ x$ and $ y,$ we have that $ (Rx)\cdot(Ry)\equal{}x\cdot y.$ Just set it up as matrix multiplication and follow your nose:

$ (Rx)\cdot(Ry)\equal{}(Rx)^T(Ry)\equal{}(y^TR^T)(Ry)\equal{}x^T(R^TR)y\equal{}x^TIy\equal{}x^Ty\equal{} x\cdot y.$

In particular, $ \|Rx\|^2\equal{}(Rx)\cdot(Rx)\equal{}x\cdot x\equal{}\|x\|^2,$ so $ R$ preserves lengths. And since we can find the angle between vectors from dot products and lengths, then $ R$ preserves angles as well.

We can show that either $ \det(R)\equal{}1$ or $ \det(R)\equal{}\minus{}1.$ If the determinant is $ 1,$ then $ R$ preserves orientation (right-handedness versus left-handedness) and can be called a rotation; if $ \det(R)\equal{}\minus{}1,$ then $ R$ reverses orientation and includes a reflection.

Key words: orthogonal matrix, special orthogonal matrix, $ O(3),$ $ SO(3).$


FTW
MATHCOUNTS
AMC
AIME
ceiling function
Gauss
probability

This is...[size=150][b]AIME15's Tournament...V2![/b][/size]

There will be $ 5$ rounds, in increasing difficulty, plus a countdown round (not FTW) to [b]help[/b] determine the final four spots.  In the earlier rounds, the difficulty will range from MATHCOUNTS Chapter Round level to State Round level.  Toward the later rounds, the difficulty will range from MATHCOUNTS State Round level to MATHCOUNTS National level.  Each round will have $ 15$ problems, posted here in the forum.  A full solution will later be provided to each problem.  Calculators will be allowed, but please do not use any internet aids.  You cannot communicate with anyone else about this test.  You are on your honor.  You will be allotted $ 15$ minutes per round.  The problems in each round are in no particular order.  Scoring will be $ 4$ points for a correct answer and $ 1$ point for no answer.

Additionally, there will be $ 3$ Master's Round, one in the first, third, and fifth rounds.  Solutions will be required, and each will be worth $ 20$ points.

Your score WILL carry on from the last round.  Half your TOTAL score from the previous round will be added to your score from the current round to find the TOTAL score from the current round.

There will be no limit in sign-ups, and you can sign up through PM or posting.   Reminder PM's will go out at the beginning of each round.  Signups will end on February 14, 2009.

[b][u]SIGNUPS:[/u][/b]
1. athunder
2. nikeballa96
3. ReseesPieces
4. Twin Prime Conjecture
5. 007math
6. permutations
7. Bugi
8. isabella2296
9. vamathletes
10. ArtofMath
11. MiddleSchooler
12. dragon96
13. ernie
14. AIME_is_hard
15. Discrete_Math
16. FantasyLover
17. THECHAMP
18. dafjbomb
19. mathemagician1729
20. jonathanchou711
21. HydraPyros

NIKE JOINS.

[hide="."] argh, too short [/hide]

I will sign up to fial......but i gues ill sign up...

Finnaly... I join.
Hopefully no. 1 this time.

I guess i'll join too, even though i PHAIL.

put me in... even though i fail at cd...

[color=white]I join.  Get better eyes.[/color]

I have sent solutions exactly 2 weeks ago (and 8 minutes past  :lol: ) to pythag, no response.

[quote="AIME15"]Everyone, do pythag's Bonus Round.  That will be round 2.

Send solutions to pythag011, not me.  If they are not graded within 3 days, send to me.[/quote]
He graded my solutions... and PM me my result! I got a first perfect score! 30 points for me!

If activity doesn't pick up within 12.894384573 hours, I will declare this over.

I sent my solutions... like a month ago?

[quote="BOGTRO"][quote="AIME15"]Time to PM pythag.[/quote]

I mean to pythag like a month ago.[/quote]

Blah... I think I got a LOT of pms and now most of my old messages disappeared.

AIME15, Can I do a new bonus round? This one will be on a much more advanced theorem.

Evidently so.

I am on the verge of canceling it.

The tournament was canceled 7 months ago.

I find it strange that some one would say that they join a tournament without reading the previous post which says that the tournament is canceled in big bold letters....
unless they are trying to be funny and failed
(no offense to anyone)

Indeed, mathdriven (or as we call him at school, Shi-man :D) is attempting to be funny.

Can I join now?

jk :rotfl:

But seriously, the thread should be locked........


Bobby is numbering the pages of a book starting at 1, what number will he be on when he has written the 2008th "2"?

wow...ok...there are like trillions of these problems...are you trying to challenge us or do you need help???

If you need help--
[hide="hint"]find the number of 2's between 1 and 100 and use that info to find the 2's between all other sets of hundreds except 200-299 where there is atleast one 2 in every number[/hide]

Every 100, there are 20 2's.
For 200's, there are 20 2's for the units and tens, and 100 2's from the hundreds digit.
Therefore, every thousand there are $ 10\cdot20 \plus{}100\equal{}300$ 2's
Now, in 2000's, theres 1000 2's from the thousands place, and the 300 2's from the units, tens, and hundreds.
Now we get as close to 2008 as possible in thousands.
$ 300\plus{}300\plus{}1300$
Meaning at $ 3000$, you have used $ 1900$ 2's.
Now you need to get to 2008.
$ 1900\plus{}20\equal{}1920$
Now you're at 3199.
Now, you need to get $ 88$ twos.
Now, we'll get to 50.
From 3199 to 3250, you've used $ 51\plus{}5\plus{}10\equal{}66$ twos.
So now we need 20 more.
Go to 3262.
12 twos from the hundreds.
1 two from the 62.
So that's 13.
Now we need to get 7 more.
$ \boxed{3269}$
Meh, I probably messed up.

I was waiting so long to say this:
Good job!!! That's correct :D

[quote="Yongyi781"]By the way, Peijin, "Failed AIME" is too generic. It could mean you got a score of between 0 (0%) to 9 (60%) out of 15.  :D[/quote]
Technically, though, contest scores for hard contests generally are considered "failing" only if they're like 30%.

[quote="Yongyi781"]Yes, you messed up :P

You were correct up to the point where you said 1920 2's is at $ 3199$.

Actually, it's at $ 3099$. Did you skip the whole 3000's? At $ 3199$, you're at 1940 2's. At $ 3299$, you're at 2060 2's, which is too much, so you know the number is between $ 3200$ and $ 3299$.

$ 3209$: 1951 2's
$ 3219$: 1962 2's
$ 3229$: 1983 2's (You count the 2's in the hundreds as well as the tens!)
$ 3239$: 1994 2's
$ 3249$: 2005 2's

Count 3 more, namely, $ 3250$, $ 3251$, and $ 3252$. And then, you're there. $ \boxed{3252}$. He is in the middle of writing $ 3252$ when he hits the $ 2008^{th}$ 2.

By the way, Peijin, "Failed AIME" is too generic. It could mean you got a score of between $ 0$ (0%) to $ 9$ (60%) out of 15. :D[/quote]
Oh....
Dang.
I always forget the 0's.
:(


the answr to ur question is $ \frac {12}{13}$.... is it right madness??

my method is find the general term of the series and then multiply and divide by its conjugate ... and on furhter siplification we get the deries as $ \frac {1}{n^1/2*(n\plus{}1) \plus{} (n \plus{} 1)^1/2*n}$...

post full soln plz  :D

and what is the genl term? mazhupaada da, fulla sollu ennal this is a magazine sum  :rotfl:

come on mate.. i have already told u the solution over the phone...

i don mind posting it either...

now as i have menitioned the general term in my earlier post..

multiply and divide by its conjugate......then simplify it and it transforms to a telescopic summation......

then the summation becomes....

$ \frac {1}{n^1/2} \minus{} \frac {1}{(n \plus{} 1)^1/2}$... so the answer is $ {1} \minus{} \frac {1}{13}$ and by this method i got this answer...

The crux of this question is to find the general term. One first appearence, it seems to be highly misleading but simple observations can lead to the general term and then this becomes straight forward.

yes the genl term can be confusing to find  :D after that its child's play


If x > y > 0 and xy = 2, what is the smallest possible value of (x^2 + y^2)/(x - y)?

Let $x-y = a$. Therefore, we have:

$\frac{a^2+4}a$. By AM-GM, we get:

$\frac{a^2+4}a = a+\frac{4}a \ge 2 \sqrt {4} = \boxed{4}$

:P
EPIC REVIVE


function
trigonometry
geometry
angle bisector

A projectile $ P$ is launched from point $ A$ in a perfect semicircle $ O$ of radius $ 1$ mile. After traveling at a constant velocity for $ 1$ minute, $ P$ returns to the ground at point $ B$. Find a function $ f(t)$ that describes $ d$, where $ t$ is time in seconds.
[asy]size(250);
pointpen=black;
real deg=135;
pair O=origin,A=(-1,0),B=(1,0),P=dir(deg);
D(B--A--arc(O,1,180,0),linewidth(0.7),MidArrow);
D(A--P--O,dashed+linewidth(0.7));
MA("\theta",11,black,P,O,A,0.2,black+linewidth(0.7));
MP("A",D(A),SSW);
MP("B",D(B),S);
MP("O",D(O),S);
MP("P",D(P),NW);
MP("d",midpoint(A--P),ESE);[/asy]

[hide="Solution."]After $ t$ seconds have elapsed, the projectile has covered $ \frac{t}{60}$ of the total distance and $ \theta \equal{} \frac{\pi t}{60}$.  You can draw $ \angle ABP$ which intercepts the same arc as $ \angle AOP$ and whose measure is $ \frac{\theta}{2}$, or $ \frac{\pi t }{120}$.  Since $ \angle APB$ is a right angle, the length of $ d$ is the sine of $ \angle ABP$ multiplied by the diameter of the semicircle, or $ 2$.  Thus, $ f(t) \equal{} 2sin(\frac{\pi t }{120})$.[/hide]

I don't know...a little different

[hide]Use law of cosines...

$ f(t) \equal{} \sqrt {1^2 \plus{} 1^2 \minus{} 2(1)(1)\cos\theta}$

Since $ \theta \equal{} \left(\frac {t}{60}\right)180 \equal{} 3t$,

$ f(t) \equal{} \sqrt {2 \minus{} 2\cos3t}$, in terms of degrees...

Obviously, it can be simplified, but I'm too lazy to[/hide]

The two answers given are indeed both correct, which can be confirmed by graphing them. :)

[hide="My Solution"]Drawing the angle bisector of $ \theta$, we see that $ \sin\frac {\theta}{2} \equal{} \frac {d}{2}$. Thus, $ d \equal{} 2\sin\frac {\theta}{2}$.

Since the projectile will arrive at $ B$ in $ 60$ seconds and the total distance is $ \pi$, we can express $ \theta$ as $ \frac {\pi t}{60}$.

Replacing $ \theta$ with $ \frac {\pi t}{60}$ in $ 2\sin\frac {\theta}{2}$, we conclude that $ f(t) \equal{} 2\sin\frac {\pi t}{120}$.[/hide]


for all $a,b,c\in \mathbb{R}$ such that $abc=1$ suppose that
$\frac{1}{a^3(b+c)} +\frac{1}{b^3(c+a)(a+b)} +\frac{1}{c^3(a+b)(b+c)} \ge \frac{3}{2}$

[quote="silouan"]Shortlist ? which one ?[/quote]

A question also by me... :)

Besides this topic should be in Inequalities unsolved Problems
or in Inequalities proposed and own Problems...

[quote="pooyan_seva"]it was in one the shortlists betwen the years 95_2001
i'm not sure!
thanks for your replies but i want some other inequalities with nice changings . because i know the solution.[/quote]

Why don't yu post it?? :P

[quote="zhaoli"]It is IMO 1995-2, by the way, a, b, c should be positive real numbers.[/quote]
It is not just the problem.see [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=IMO+1995&t=60434]here[/url]
and I think the ineq which pooyan_seva wrote is :
$\frac{1}{a^3(b+c)(c+a)} +\frac{1}{b^3(c+a)(a+b)} +\frac{1}{c^3(a+b)(b+c)} \ge \frac{3}{4}$
and anybody have a solution?

[quote="zhaobin"]
and I think the ineq which pooyan_seva wrote is :
$\frac{1}{a^3(b+c)(c+a)} +\frac{1}{b^3(c+a)(a+b)} +\frac{1}{c^3(a+b)(b+c)} \ge \frac{3}{4}$
and anybody have a solution?[/quote]

I'm assuming $a, b, c$ are positive, since otherwise the inequality does not follow.
Change $a$ to $\frac{1}{a}$, $b$ to $\frac{1}{b}$, $c$ to $\frac{1}{c}$.
You get $\frac{a^3c}{(b+c)(c+a)} + \frac{b^3a}{(c+a)(a+b)} + \frac{c^3b}{(a+b)(b+c)} \ge \frac{3}{4}$
Now I expanded it, :blush: it becomes:
\[ 4(a^4c+a^2+b^4a+b^2+c^4b+c^2) \ge 3(a+b)(b+c)(c+a) \]
By AM-GM:
\[ a^2 + c^2 + a^4c \ge 3a^2c \]
\[ a^2 + b^4a + a^4c \ge 3a^2b \]
and similarly for $3b^2a$, $3b^2c$, $3c^2a$, and $3c^2b$
Also \[ a^4c+b^2+b^4a+b^2+c^3b+c^2 \ge 6abc \]
The result follows.

$a,b,c>0$  and $abc=1$   I think. thanks

[quote="cezar lupu"]Zhaobin, in what circumstances does your inequality from post 7 hold? :)[/quote]
sorry,I haven't quote the condition...sorry


The function $f(x) = |1-2x|$ is defined on the interval [0, 1].  How many solutions does the equation $f(f(f(x))) = \frac{x}{2}$ have?

This problem look's nice.  :?
[hide] By calculating i found 8 solutions between [0,1]. The solutions are $x\in( \frac{2}{15},\frac{6}{15},\frac{10}{15},\frac{14}{15},\frac{2}{17},\frac{6}{17},\frac{10}{17},\frac{14}{17} )$[/hide]

I hope it's right. I have a classes now, and i don't have time to post the entire solution. I'll post it later.

Is there other ways which avoid calculating the results??
The problem says "how many solutions", not tells us to actually work out the solutions......

The calculs were not so hard. I think this was a exercise to rezistence  :)


I currently have a post rating of 1.8 :blush:(you'd think some of my threads in the game forum could drum that up) and it was once down to 1.7.  I wanted to know how bad some other people got it because some people rate posts as spam for no reason..................Answer the poll too please.

Where do you see what your post rating is?

Right below your avatar, location, and post count.  If your new to AoPS you might get a "Not Rated Yet".

It was 4.3 before I entered the Missouri forum. It is now a 3.3.

I am currently "not rated yet". I assume that's a zero?

$ 1.7$ w00t

@maybach: [url]http://www.artofproblemsolving.com/Forum/index.php?f=244[/url]

My lowest post rating was 1.8. Since ratings below 2 imply decreased privileges, I tried to improve my post rating. It's 2.8 now :)

I currently have a post rating of 1.8.  What kind of decreased priviliges will I have to go through?

Also, please rate my posts here 3,4,5, or 6 cubes.  For FF standards, they're not spam at all, and people have been unfairly rated my posts lately.

If you wish to, you could always contact an administrator or moderator for help on improving your post quality level, and you could always make a new thread with a poll, "How good are my posts?" for quick results.

2.7 was my lowest.

I just realized that I could rate bad posts badly to get revenge on my haters! :P
Seriously don't abuse it.
And FF ratings don't count.

Mine is 3.  What is a FF rating, and how can you see which of your posts have been rated?

Current Rating: 2.7
Lowest Rating: 2.3
Highest Rating: 4.3

mine was you have not been rated yet or something like that
now it is 1 since everyone thinks i spam:(

i hate how many people dont take my posts seriously. they just treat it as spam. its like telling somebody something, and they take it like a joke.

\me is more grumpy.

Highest-2.4
Lowest-2
Current-2

No post rating yet.  :(   I want a rating...  Preferably higher than 2...

2.6 i think  :(  that means all my posts are total spam

mine is 2.7.....
how do you see which posts got rated?

I think everybody should worry more about the quality of their posts than the ratings they get!

My current rating is $ 4.2$ after having reached $ 4.5$, and it has nothing to do with the quality of the math I write (it went down, as I was actually posting more good math, imho). I invite anybody skeptical to take a look at, say, my last $ 50$ posts.

Do I care? Nope. If rating was a consistent and coherent effort by peers to evaluate my work, I would. But it is too random an action to be of any consequence, so I don't.

So my advice to all is be your own rater!

That was the best post mavropnevma, I give a 6 :D
That's what I keep saying in QS&A when someone makes his 1000th thread about his rating. They aren't important. At all. My is 3.4 and I don't care. The rating system on this site is good because it is introduced, but bad because people rate bad posts 90% So just do your best, and don't care about the value. If you know that you are posting well, it's ok.

I hardly talk out of online class forums. People give me low post on PURPOSE and then I'm going to be muted on FTW for a whole month again. :(

Mine's 4.7. Although everyone will probably rate this post 1 star, and make it drop back down again.

Mine's 4.5.
It's been like that for a long time...

Did they remove this feature or something?

(aha 11 year bump)


We call $ Lucky$ a 6-digit number $ abcdef$ such that:

a) His 6 digits are distinct.
b) $ a\plus{}b\plus{}c\equal{}d\plus{}e\plus{}f$.

Prove that the if sum of all Lucky Numbers is $ S$, then $ 2002\mid S$.

well, if $ A\equal{}abcdef$ is lucky, so is its associate $ A'\equal{}defabc$, obviously. denoting the three digit numbers $ abc$ and $ def$ by $ x$ and $ y$, respectively, we have $ A\plus{}A'\equal{}1000x\plus{}y\plus{}1000y\plus{}x\equal{}1001(x\plus{}y)$. we can pair each lucky number with its associate except if it's its own associate. but such a number looks like $ abcabc$, which is clearly divisible by $ 1001$. so clearly $ 1001|S$. it only remains to show that $ 2|S$. but that's obvious!!


FTW
MATHCOUNTS
AMC
AMC 12
AIME
USA(J)MO
USAMO

Aww cmon Jason, stop rubbing it my face.
:(

I still think I'm dreaming....

My score was definitely not something I can talk about it anyway what so ever  :lol: (that was an edit!!!)

As expected, I didn't make nats, since i didn't do MC :P

ehh...did some states already have their state competitions already!!!gosh mine is in like 3 weeks!!! im really nervous!!!!!any tips

@mathcrazed, you didnt have your state competition yet or your not going to nats???

[quote="unimpossible"][quote="mathcrazed"]Nah. I failed[/quote]

good job for getting first  :D

i hope i can make nats.  My last chance to =)[/quote]

what state are you in??  on ftw u are pretty good, i think ull make it

Yes  :lol: I got 1st countdown, and 2nd written round. Our team got 5th by tiny bit :(

I wish I got 1st, but it is my first and last year, so it's ok

Nope. Failure. Not completely unexpected, either...:embarrassed:

1st countdown (unofficial), but something like 12th or 13th written.

I power phail.

EDIT: 12th I think, but i have poor short term memory (and poor long term memory for that matter).

[quote="mathcrazed"]Nah. I failed[/quote]

No you didn't.

To the world: See mathcrazed's blog.

I uber power phail, if that's possible.

I hate buzzer systems.

EDIT: mathcountsfreak, you know what place you got.

Not even remotely close.  I got 23/46 and got 63rd place(however, I did better than most of my team anyway).  Better luck(I hope) for 7th grade.

[quote="Yongyi781"]I never study for anything. I just don't have the time for it :D

... except possibly MATHCOUNTS![/quote]

You're genious!  :rotfl:

Shouldn't this topic just be deleted, as it is not applicable anymore?

I obviously failed in my attempt in state so ya :D

Well, that doesn't seem to be happening, hmm?

[quote="l Bob l"]Just a straight-forward question. Did you make NATS?  You may elaborate on your answer if you want.

I made nats, although I did not think I would. I got a X (where |X-23| < 24 ) Placed second and my team Clarke got 1st.[/quote]

does that mean that your score could've been a 46.5?

It's a joke and a slight phail on Bob's part. No, that score isn't achievable.

My score = Bob's Score/2 (no that's not a typo-I got 63rd place at Indiana and feel kinda depressed :blush: )

GUYS.
look at the dates of posts!
come on, this is total and complete spam.
a mod needs to lock this. NOW.

Please don't post in threads that have been inactive for a long time.


Four whole numbers, when added three at a time, give the sums $180$, $197$, $208$, and $222$. What is the largest of the four numbers?

$\text{(A)} \ 77 \qquad \text{(B)} \ 83 \qquad \text{(C)} \ 89 \qquad \text{(D)} \ 95 \qquad \text{(E)} \ \text{cannot be determined}$

[hide]Let the numbers be $a,b,c,d$. We're given the values of $a+b+c$, $b+c+d$, $a+c+d$, and $a+b+d$. If we add these all together, we get $3(a+b+c+d)=807\implies a+b+c+d=269$.

Subtracting the smallest triple gives the largest number, so our answer is $269-180=89$.[/hide]

[hide]
Each number is added three times to get the results 180, 197, 208, 222. So add those numbers up and divide by three giving you the sum of all four numbers, $269$. You want the largest so take the sum of the smallest three numbers, $180$, and do $269 - 180$ giving you $89$.[/hide]


Hello,

I am hoping someone can help explain these terms to me, i cant find a clear and simple explanation anywhere

epipolar line - cant find this anywhere, well nothing i understand, a one paragraph   explanation (if possible) would be perfect, I dont need huge detail just a high level understanding.

least squares matching - I have done alot of reading on this, maybe if you could point me towards some good resourses on this topic.

Help with one or both definitions would be great.

Thanks
J.

Epipolar line: Google immediately gives you [url=http://homepages.inf.ed.ac.uk/rbf/CVonline/LOCAL_COPIES/OWENS/LECT10/node3.html]this page[/url] from a set of lecture notes on computer vision at the University of Edinburgh.


Is it possible to take $ 10$ red and $ 10$ green point on a line so, that the following condition is satisfied: We can number both red and green points with natural numbers from $ 1$ to $ 10$ so, that for any $ i$, there is exactly $ i$ points from these $ 20$ points between number $ i$ red point and number $ i$ green point?

I think it's true:
( we don't use the colors .) Suppose that ,it's possible to do this. If one of the numbers $ a$ be between two $ b$ ,we said $ b$ [i]covered[/i] $ a$,($ b...a...b$). It's possible that ,$ b$ [i]covered[/i] both $ a$($ b...a...a...b$) or $ a$ [i]covered[/i] both $ b$ ($ a...b...b...a$), or one $ a$ [i]covered[/i]  one $ b$ and one $ b$ [i]covered[/i] one $ a$ ($ a...b...a...b$). If we choose any two numbers, this two numbers create two [i]covered[/i] numbers. Then the number of [i]covered[/i] numbers are even. On the other hand, the numbers $ 1$ [i]covered[/i] one number and the numbers $ 2$ [i]covered[/i] two numbers and so on.Then the number of [i]covered[/i] numbers $ \equal{} 1\plus{}2\plus{}...\plus{}10\equal{}55$ and it's contradiction.

Nice solution! I solved this problem the same but I used the other words. :)


For a given $ a\in\mathbb{R}\,$: $ \forall x,y\in\mathbb{R}\,\,f(x\plus{}y)\equal{}f(x)\cdot f(a\minus{}y)\plus{}f(y)\cdot f(a\minus{}x)$ assuming $ f$ is a real-valued function and $ f(0)\equal{}\frac 12$.
$ f(2008)\equal{}?$

-) $ x\equal{}y\equal{}0: f(0)\equal{}2f(0)f(a)\Leftrightarrow f(a)\equal{}\frac12$
-) $ y\equal{}0: f(x)\equal{}f(x)f(a)\plus{}f(0)f(a\minus{}x)\Leftrightarrow f(x)\equal{}f(a\minus{}x)$
$ \Rightarrow f(x\plus{}y)\equal{}2f(x)f(y)$
-) $ y\equal{}a: f(x\plus{}a)\equal{}2f(x)f(a)\equal{}f(x)\Rightarrow f(a\minus{}x)\equal{}f(x)\equal{}f(x\minus{}a)$
$ \Rightarrow f(x)\equal{}f(\minus{}x)$
Then $ f(x\plus{}y)\equal{}2f(x)f(y)\equal{}2f(x)f(\minus{}y)\equal{}f(x\minus{}y)$ for all $ x,y$, it mean $ f(x)\equal{}c\equal{}\frac12$ $ \forall x$


I have a 50g tank (not including substrate and animals). I want to do daily water changes so that at the end of 6 months I will have replaced 95% of the water present at the beginning of the water change regimen. I want to replace the same volume every day.

1. What volume should I replace each day?
2. What is a general formula I could use for larger or smaller tanks? Or for changing the amount of time over which I replace the 95%. (So I could use it if I decide to set up a X
gallon tank and have my water 'turn over' period every Y months.)
3. What is a general formula I could use that contains the variables from (2) but also allows me to change the periodicity (say I get lazy and only want to do weekly water changes.)

It is not even calculus, just algebra.

Assume that the time period between water changes is $t$ and the portion of water you change each time is $\alpha$. Then, after time $T$, you've done $T/t$ changes and the portion of old water that is still in the tank is $(1-\alpha)^{T/t}$ (I assume that the old and the new water mix up completely between the changes). You want this portion to be, say, $\beta$, so you get the equation
\[(1-\alpha)^{T/t}=\beta\,.\]
This allows you to find any variable if you know all other ones. For your original question, you have $t=1$, $T=183$ (in days), $\beta=0.05$. So, you get
\[(1-\alpha)^{183/1}=0.05\]
or
\[\alpha=1-0.05^{1/183}\,.\]
Now you should really take your calculator to compute that :). Since I have none nearby at the moment, let me just estimate this number as $0.016$. This is the portion of water you want to change daily. If you want the amount in gallons, you just multiply it by the tank volume: $0.016\cdot 50=0.8$ gal.
With weekly changes the story will be the same except you'll use $t=7$. The result will be slightly below $7$ times the result for daily changes. Again, without a calculator I can only make an educated guess and say something like $5.5$ gal.

Hope this helps :).

Thanks Fedja, I'll work on it. I thought that Calculus would be involved.

Thanks again for the feedback.


inequalities
trigonometry
inequalities unsolved

Let $ \theta_1, \theta_2,\ldots , \theta_{2008}$ be real numbers. Find the maximum value of

$ \sin\theta_1\cos\theta_2 \plus{} \sin\theta_2\cos\theta_3 \plus{} \ldots \plus{} \sin\theta_{2007}\cos\theta_{2008} \plus{} \sin\theta_{2008}\cos\theta_1$

$ \sin\theta_1\cos\theta_2 \plus{} \sin\theta_2\cos\theta_3 \plus{} \ldots \plus{} \sin\theta_{2007}\cos\theta_{2008} \plus{} \sin\theta_{2008}\cos\theta_1\leq\frac{1}{2} \sum (\sin^2\theta_i\plus{}\cos^2\theta_{i\plus{}1}) \equal{} 1004$
Equality holds when $ \theta_i$ is constant

[quote="brianchung11"]$ \sin\theta_1\cos\theta_2 \plus{} \sin\theta_2\cos\theta_3 \plus{} \ldots \plus{} \sin\theta_{2007}\cos\theta_{2008} \plus{} \sin\theta_{2008}\cos\theta_1\leq\frac {1}{2} \sum (\sin^2\theta_i \plus{} \cos^2\theta_{i \plus{} 1}) \equal{} 1004$
Equality holds when $ \theta_i$ is constant[/quote]
The constant is $ \frac{\pi}{4}$  :roll:


function
calculus
derivative
calculus computations

How do I use a function f(x) to graph f'(x)?  Is there any special trick or something?  I did look on Google, but I did not find anything.  Thanks for your help.

I've found [url=http://mathdl.maa.org/mathDL/4/?pa=content&sa=viewDocument&nodeId=409]this[/url] derivative plotter while browsing [url=http://mathdl.maa.org/]MathDL[/url] long time ago :)
You can also find many other useful applications from this good site.

The thing is, I  am not given an actual equation for the graph of f(x).  I am just shown a graph of a function, then which of the following graphs is f'(x)

Just remember the geometric meaning of the derivative, as the instantaneous slope. If $f$ is increasing on an interval, then $f'$ is positive there; if $f$ is decreasing, then $f'$ is negative. Maximum and minumum points should correspond to zeros of the derivative, but sharp corners or cusps in the graph of $f$ correspond to discontinuities in the graph of $f'$. Reading in the other direction, a region in which $f'$ is increasing is a region in which the graph of $f$ is concave upward, with the corresponding opposite conclusion where appropriate.


calculus
integration
function
calculus computations

(1) Let $f(x)$ be the differentiable and increasing function such that $f(0)=0.$Prove that $\int_{0}^{1}f(x)f'(x)dx\geq \frac{1}{2}\left(\int_{0}^{1}f(x)dx\right)^{2}.$

(2) $g_{n}(x)=x^{2n+1}+a_{n}x+b_{n}\ (n=1,\ 2,\ 3,\ \cdots)$ satisfies $\int_{-1}^{1}(px+q)g_{n}(x)dx=0$ for all linear equations $px+q.$
Find $a_{n},\ b_{n}.$

[quote="kunny"](1) Let $f(x)$ be the differentiable and increasing function such that $f(0)=0.$Prove that $\int_{0}^{1}f(x)f'(x)dx\geq \frac{1}{2}\left(\int_{0}^{1}f(x)dx\right)^{2}.$[/quote]

Since $f$ is an increasing function, and $f(0)=0,$ it follows that if $0\leq x\leq 1,$ then $0\leq f(x)\leq f(1)$ so that $0\leq \int_{0}^{1}f(x) dx\leq f(1)$ and hence that

\[0\leq \left(\int_{0}^{1}f(x) dx\right)^{2}\leq f^{2}(1)\qquad\qquad\qquad (1).\]

Also note that

\[\int_{0}^{1}f(x)f'(x)dx = \int_{0}^{f(1)}udu = \frac1{2}\left[u^{2}\right]_{u=0}^{f(1)}= \frac1{2}f^{2}(1)\qquad\qquad\qquad (2).\]

Combining (1) and (2) gives the required result.

[quote="kunny"](2) $g_{n}(x)=x^{2n+1}+a_{n}x+b_{n}\ (n=1,\ 2,\ 3,\ \cdots)$ satisfies $\int_{-1}^{1}(px+q)g_{n}(x)dx=0$ for all linear equations $px+q.$
Find $a_{n},\ b_{n}.$[/quote]

$(px+q)g_{n}(x) = px^{2n+2}+qx^{2n+1}+a_{n}px^{2}+(a_{n}q+b_{n}p)x+b_{n}q$, and integrating this for $x$ in $[-1, 1]$ gives $0 = \frac{2p}{2n+3}+\frac{2a_{n}p}{3}+2b_{n}q$ for all $p, q$ from which $a_{n}=-\frac{3}{2n+3}$ and $b_{n}= 0$.

I suppose alternatively, one could have looked at the cases $p = 0, q = 1$ to show $b_{n}= 0$ and $p = 1, q = 0$ to get the value of $a_{n}$, but it's basically the same thing.

[quote="benorin"][quote="kunny"](1) Let $f(x)$ be the differentiable and increasing function such that $f(0)=0.$Prove that $\int_{0}^{1}f(x)f'(x)dx\geq \frac{1}{2}\left(\int_{0}^{1}f(x)dx\right)^{2}.$[/quote]

Since $f$ is an increasing function, and $f(0)=0,$ it follows that if $0\leq x\leq 1,$ then $0\leq f(x)\leq f(1)$ so that $0\leq \int_{0}^{1}f(x) dx\leq f(1)$ and hence that
\[0\leq \left(\int_{0}^{1}f(x) dx\right)^{2}\leq f^{2}(1)\qquad\qquad\qquad (1). \]
Also note that
\[\int_{0}^{1}f(x)f'(x)dx = \int_{0}^{f(1)}udu = \frac1{2}\left[u^{2}\right]_{u=0}^{f(1)}= \frac1{2}f^{2}(1)\qquad\qquad\qquad (2). \]
Combining (1) and (2) gives the required result.[/quote]

how did you get your equivalent expression for $\int_{0}^{1}f(x)f'(x)dx$?

Substitute $u=f(x)\Rightarrow du=f^\prime (x)dx$ so that $0\leq x\leq 1\Rightarrow f(0)\leq f(x)\leq f(1)\Leftrightarrow 0\leq u\leq f(1)$ and the integral becomes $\int_{0}^{1}f(x)f'(x)dx = \int_{0}^{f(1)}udu$


M is a set of positive real numbers with the property that their decimal representation digits are 0 or 7.
For example, 700/99=7.0707... and 77.707 are in M.
What is the smallest number of elements of M such that their sum is 1?

We can choose the following numbers:

$ x_1 \equal{} 0.777777...$
$ x_2 \equal{} 0.077777...$
$ x_3 \equal{} 0.070777...$
$ x_4 \equal{} 0.070777...$
$ x_5 \equal{} 0.000777...$
$ x_6 \equal{} 0.000707...$
$ x_7 \equal{} 0.000707...$
$ x_8 \equal{} 0.000700...$

The numbers are rational and their periods are are the six digits after the point. It is not difficult to verify their sum is $ 0.999999... \equal{} 1$.

Now it suffices to prove that there are no seven numbers satisfying the desired condition. Suppose by absurd there are such numbers. Clearly, exactly one of them has $ 7$ as its first digit (after the point), as in the example with eight numbers. If three or less of them have $ 7$ as second digit, the sum is at most $ 0.7 \plus{} 3\times0.07 \plus{} 7\times0.007777... < 1$. If five or more of them have, the sum is gonna be $ \ge 1.05$. Then exactly four of them have $ 7$ as the second digits. Now, let´s see the third digit of these numbers. If three of them are $ 7$´s, our sum is $ \ge 0.7 \plus{} 4\times0.07 \plus{} 3\times0.007 \equal{} 1.001$, wich is impossible. Else, however, the maximum attainable value is $ 0.7 \plus{} 4\times0.07 \plus{} 2\times0.007 \plus{} 7\times0.000777777... \equal{} 0.999444444... < 1$. So there is no way of getting $ 1$.

Beautiful problem!


I was reading http://www.mathlinks.ro/Forum/viewtopic.php?t=837 (just for curiosity, the oldest topics in the forum =P )

It come to that, although how to do elementary operations with complex numbers, I have no idea what are complex numbers O.O

In fact, this was bothering me for a while, I've just some random proof while lurking on the internet (looking on euler's archive, i think) and it was used quite a bit on them.

So any suggestions? I am much more interested in the "philosophical" concept of complex numbers, but I don't to delve too deep on it for now... So something interesting, readable, not lacking i depth but not too deep either.

(Probably I`am buying AoPS's intermediate algebra, but I feel it is not exactly what I am looking for)

Complex numbers are basically numbers of the form $ x\plus{}iy$ where $ i\equal{}\sqrt{\minus{}1}$.

Here's some more info - http://mathworld.wolfram.com/ComplexNumber.html

AoPS V2 has complex numbers, but I'm not sure if it's the best book.

http://www.amazon.com/Complex-Numbers-Z-Titu-Andreescu/dp/0817643265

Might be a little too deep though.


[hide]The get the biggest perimeter, you want the least number of tiles in the inside.  THat means, we would just put one tile next to another tile, get the average edges touched per tile at the highest it can be.

That means that two sides would be 1 foot, and the other sides are 30 feet.  30+30+1+1 = 62 feet.[/hide]

Oops, I accidentally found the smallest perimeter.

[hide]

Infinity, if you can cut the squares into infinitely small pieces.

Assuming they can't be broken into pieces, 62.

[/hide]

I don't think infinity can possibly be a correct answer. It's sort of like that Zeno's paradox about movement. It's an infinite sum that always adds up to a constant, 62 (I think).

Well, if you were capable of hacking the tiles into pieces that, put together, would make a circle, then the circumference would be the least possible perimeter. But in the context of the problem, the least possible perimeter would be [hide]22, since you want the two dimensions to be as close to each other as possible, 5x6 is the closest they can get under the circumstances, and 2(5+6)=22.[/hide]

[quote]What is the greatest number of feet in the perimeter of a rectangular floor that can be covered by 30 whole 1' x 1' square tiles?[/quote]

[hide]to get the greatest perimeter, it has dimentions 1X30

the perimeter is 62[/hide]

[hide]to get greatest perimeter we want one side to be the shortest and the other to be the longest.... so we get: a 1x30 rectangle which gives a perimeter of [b]62[/b][/hide]

[quote="LynnelleYe"][quote="eryaman"]Actually yif, 4*sqrt(30) would be the smallest permineter.  But if only integers are allowed Lynelle would be right.[/quote]
Um. I got [hide="this"]$2\sqrt{30\pi}$[/hide] for the smallest perimeter. Here's how I calculated it...
[hide]The area is 30. We can solve for the radius of the circle with said area.
$\pi r^2=30$

$r^2=\frac{30}{\pi}$

$r=\frac{\sqrt{30}}{\sqrt{\pi}}$
$2\pi r$ is the circumference of the circle, so
$p=2\pi\cdot\frac{\sqrt{30}}{\sqrt{\pi}}$

$=\frac{2\pi\sqrt{30\pi}}{\pi}$

$=2\sqrt{30\pi}$
[/hide]
Is there a mistake?[/quote]

It looks right to me.  I think that what eryman meant was that $4\sqrt{30}$ was the smallest perimeter for a rectangle.

The problem seems to have morphed from rectangular tiles to any figure with any dimensions.  I supposed it's good to get a taste of isoperimetric problems...

[quote="LynnelleYe"][quote="eryaman"]Actually yif, 4*sqrt(30) would be the smallest permineter.  But if only integers are allowed Lynelle would be right.[/quote]
Um. I got [hide="this"]$2\sqrt{30\pi}$[/hide] for the smallest perimeter. Here's how I calculated it...
[hide]The area is 30. We can solve for the radius of the circle with said area.
$\pi r^2=30$

$r^2=\frac{30}{\pi}$

$r=\frac{\sqrt{30}}{\sqrt{\pi}}$
$2\pi r$ is the circumference of the circle, so
$p=2\pi\cdot\frac{\sqrt{30}}{\sqrt{\pi}}$

$=\frac{2\pi\sqrt{30\pi}}{\pi}$

$=2\sqrt{30\pi}$
[/hide]
Is there a mistake?[/quote]

Yes, that's right Lynelle.

Eryaman- I was actually just giving examples for perimeters that would be smaller than 22 given, but good job picking that up.

[quote="LynnelleYe"]Yes, by 'in the context of the problem' I meant integers-only.[/quote]

Gotcha.

What if the tiles don't have a negligible thickness and you start stacking them up to find surface are?

It would depend on the thickness, obviously...

[hide]In the problem, they stated "whole tiles".
I'm assuming that means that you can't break the tile in half.  Therefore, the longest perimeter is a 30 by 1 figure with a perimeter of 62.
However, I think that if you could break the tiles into fractional piece squares, you will not get infinity as the ultimate perimeter.  The long sides will always be the same.  For example, 1/4 squares there's going to be 120 of them and we know that a rectangle that has longest possible and shortest possible sides will have the greatest perimeter. 120*1/4=30 which is multiplied twice for both long sides.  However the shortest only measure 1/4th each so the perimeter is 60 and 1/2.  Ultimately, I think 60 will be the ultimate perimeter[/hide].

[quote=Stormstar--]\bump
\8char[/quote]

... Was that necessary?

[quote=Stormstar--]of course[/quote]

Why did you revive the thread then?

[quote=Stormstar--]to show you how old things are[/quote]

?

[quote=Sullymath]this may be the oldest thread on aops.[/quote]

It isn't.
https://artofproblemsolving.com/community/c6h2

There is no need for this revive since the OP has long been answered.

Kinda weird to think that this thread was made before I was born though. Brings back so many memories and so much nostalgia.

well now since it's revived...

[quote=yif man12][quote="LynnelleYe"]Well, the problem statement says 'whole' tiles, so in this context infinity wouldn't be allowed. But if you were allowed to split the 1x30 rectangle in half the long way and put the two pieces together end to end, and then do that again, and again, and again, the perimeter could definitely get much bigger than 62 and would probably approach infinity (I don't know much in those areas so I don't know for sure).[/quote]

Bingo.

What if the tiles were divided into .5x.5 times? What's the greatest perimeter?

How about $\frac{1}{3}$ by $\frac{1}{3}$ tiles? What's the greatest perimeter then?

Another interesting question is about the smallest possible perimeter. What's that for both of those?[/quote]

[hide=i got...]
1x1=62

0.5x0.5=121 (there are 120 tiles now)

1/3x1/3=180 2/3 (there are 270 tiles now)
[/hide]

[Hide=Answer]62. You want the figure to be as spread out as possible. ( Make it as close to a line you can make it)[/Hide]


geometry
trapezoid
ratio
3D geometry
tetrahedron
similar triangles

In trapezoid $ABCD$, base $AB$ is half as long as base $CD$. Diagonals $AC$ and $BD$ intersect at $F$, and point $E$ is chosen on line segment $BC$ such that $BE:EC=1:3$. If $EF$ is extended to intersect $AD$ at $G$, find the ratio $EF:FG$.

would the answer happen to be $\boxed{7:8}$?
i used mass points, and the similar triangles $\triangle ABF \sim \triangle CDF$ and the ratio $\frac{1}{2}$
mass points are confusing though, i did not learn them well, but it seems to work

Yes, $7:8$ is correct. Indeed, mass points do work in this problem. So post your solution :)

well, the mass points make somewhat of a contradiction but i just ignored it
since we get $BE:EC = 1:3$ let $BE = y$ and E$C = 3y$
(I am assuming that you understand how mass points work)
using the idea of the fulcrum and balancing (w is for weight) $W_B * y = W_C * 3y$
balancing that gives $W_B = 3$ and $W_C = 1$
(note that any masses would have worked, just so that they are in the correct ratio)
therefore, the middle $W_E = 3 + 1 = 4$
by similar triangles, $BF:FD = 1:2$ and $AF:FC = 1:2$
so applying the balancing procedure, $W_A = 2$ and $W_D = 1.5$
then the middle $W_G = 2 + 1.5 = 3.5$
then by balancing of mass points,
$W_G* FG = EF * W_E$
$\frac{EF}{FG} = \frac{3.5}{4} =\frac{7}{8}$
The only problem is that using the principle that the mass of the center equals the mass of the 2 ends of the fulcrum would give you 3 masses for F, and i don't know why that is, maybe a flaw? or i might not know the procedure correctly

I was aware of that problem, or "flaw". Actually I don't know why it occurs either. Can anyone explain? :|

But interestingly, mass points give the correct answer anyway :)
(You can also get the same answer by brute forcing with a lot of area ratios :D )

i suppose that the only thing that you should do is not place any masses that contradict any of your other ideas, and does anybody have a reason for why this works, i have seen the procedure, but no justification showing that it works, also another thing that is odd is that it works in 3d as well, as in real life, the centroid is the center of mass, so if you have a triangle, with masses on each of the verticies, you can label the centroid as the sum of those masses, in my geo book, this is the proof that they gave for the way the centroid of a tetrahedron separates the centroid of a face and the opposite side (the ratio is 1:3)

Hmm..even though I use mass points obsessively, I never knew [i]why[/i] it worked.  Anyone know and would like to share?


algebra polynomial function limit LaTeX MATHCOUNTS

There exists a function $ f(x)$ which can be expressed as a polynomial with degree 3, which satisfies
$ f(4)\equal{}16$
$ f(3)\equal{}8$
$ f(2)\equal{}4$
$ f(1)\equal{}2$

a) Will the output of this cubic function ever be $ 0$?  (If so, for which value(s) of $ x$?)
b) Find the function.

[hide]i don't know, i didn't actually get a polynomial with degree 3. i got a much simpler function that satisfies all the given value. $ 2^{x}$. test it out, it fits. im guessing ur looking for a different function. otherwise, its obvious, the output of my function can never be 0[/hide]

You can get an infinite number of functions that satisfy these four points. :wink:
However, you are givne that it's a cubic functions.  A function of degree $ n$ is uniquely defined by $ n\plus{}1$ points.  (For example, 2 points determine a line (deg. 1))

Using Cubic Regression, I get (with $ R^{2}\equal{} 1$):

$ y \equal{} 0.333...x^{3}\minus{}0.999...x^{2}\plus{}2.666...x\plus{}0$

I supposed this can be cleaned up to:

$ y \equal{}\frac{1}{3}x^{3}\minus{}x^{2}\plus{}\frac{8}{3}x$

Well, there you have it. The polynomial equals 0 at x=0

The only way I know to solve it algebraically is with simultaneous equations:

$ 2 \equal{} a\plus{}b\plus{}c\plus{}d$
$ 4 \equal{} (2^{3})a\plus{}(2^{2})b\plus{}2c\plus{}d$
$ 8 \equal{} (3^{3})a\plus{}(3^{2})b\plus{}3c\plus{}d$
$ 16 \equal{}(4^{3})a\plus{}(4^{2})b\plus{}4c\plus{}d$

Good luck with that!

[quote="archimedes1"]You can get an infinite number of functions that satisfy these four points. :wink:
quote]

really? i never knew that. thanks. u learn something new everyday.

Yes.

$ \lim_{x\to\minus{}\infty}a^{x}\equal{}0,a\not\equal{} 0$

[quote="Kevin K."]I have a question...

Does 2^x = 0 when x = -infinity?

Sorry, i don't know how to do the sideways [size=200]8[/size].[/quote]
its called latex make sure you put the dollar signs arround whatever you want. heres the code
[code]$\infty$[/code]

[quote="Identity"]Using Cubic Regression, I get (with $ R^{2}\equal{} 1$):

$ y \equal{} 0.333...x^{3}\minus{}0.999...x^{2}\plus{}2.666...x\plus{}0$

I supposed this can be cleaned up to:

$ y \equal{}\frac{1}{3}x^{3}\minus{}x^{2}\plus{}\frac{8}{3}x$

Well, there you have it. The polynomial equals 0 at x=0

The only way I know to solve it algebraically is with simultaneous equations:

$ 2 \equal{} a\plus{}b\plus{}c\plus{}d$
$ 4 \equal{} (2^{3})a\plus{}(2^{2})b\plus{}2c\plus{}d$
$ 8 \equal{} (3^{3})a\plus{}(3^{2})b\plus{}3c\plus{}d$
$ 16 \equal{} (4^{3})a\plus{}(4^{2})b\plus{}4c\plus{}d$

Good luck with that![/quote]
You are correct.
However, that system is not as hard as it looks.  I don't think there is a super-elegant way to solve it, but you can solve it reasonably fast by adding/subtracting equations from each other and setting values equal.

[quote="abacadaea"]still, it a lot easier just to say its 2^x.[/quote]

but you can also say

OH I FOUND THE FUNCTION DEFINE THE FUNCTION $ \alpha\beta\gamma(x)$ THE FUNCTION DEFINED FOR $ x\equal{}1,2,3,4$ THAT BRINGS $ 1,2,3,4$ to $ 2,4,8,16$ SO THE FUNCTION GIVES $ \alpha\beta\gamma(x)$ EXACTLY QED

but that's not what the question's asking, is it?

[quote="the future"][quote="archimedes1"][quote="abacadaea"]still, it a lot easier just to say its 2^x.[/quote]
You can't!  That's the point!  It's given as a cubic function, not an exponential function.[/quote]

I dont think this is mathcounts lvl and it defiently isnt a mathcounts type of question.[/quote]
Actually I'm pretty sure this is on par with some of the harder stuff in mathcounts.
Also, what does def[b]ient[/b]ly mean?
[hide="interesting solution"]
This function greatly resembles the function $ 2^{n}$ (as pointed out by numerous people).
Now consider the function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}...\plus{}\binom{n}{k}$. We know that if k>n (by our definition of the choose function) the value of $ \binom{n}{k}$ is 0.
So if we pick an n<k the value of this is $ 2^{n}$, since $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}...\plus{}\binom{n}{n}\equal{}2^{n}$.
So we have that the function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n}{3}\plus{}\binom{n}{4}$ would work. Unfortunately, this is a quartic.
The function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n}{3}$ comes close, but when n=4, the value is 1 less than what we desire. So we wish to find a new function for which f(1,2,3)=0, but f(4)=1. Since 1, 2, and 3 are roots of this new polynomial, we can just scale $ (x\minus{}1)(x\minus{}2)(x\minus{}3)$ by $ \frac{1}{6}$ to get the desired function (note that this is equal to $ \binom{n\minus{}1}{3}$).
So we have the function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n\minus{}1}{3}\plus{}\binom{n}{3}\equal{} 1\plus{}n\plus{}\frac{n^{2}\minus{}n}{2}\plus{}\frac{(n\minus{}1)(n\minus{}2)(n\minus{}3)}{6}\plus{}\frac{n(n\minus{}1)(n\minus{}2)}{6}$ works.
Simplifying, we get $ y \equal{}\frac{1}{3}x^{3}\minus{}x^{2}\plus{}\frac{8}{3}x$.
[/hide]

I don't know if this has been posted yet, but I saw this in some other topic...

It's called Lagrange Interpolation, and I first saw it here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141069[/url]

Anyway, that makes problems like this extremely easy.

[hide="Solution"]By the method of Lagrange Interpolation, we get that the equation is:

\[ f(x) \equal{}\minus{}\frac{1}{3}\cdot (x\minus{}2)(x\minus{}3)(x\minus{}4)\minus{}2 (x\minus{}1)(x\minus{}3)(x\minus{}4)\minus{}4 (x\minus{}1)(x\minus{}2)(x\minus{}4)\plus{}\frac{8}{3}\cdot (x\minus{}1)(x\minus{}2)(x\minus{}3)\]

Just plug in $ 1,2,3,4$ and you should get the values $ 2,4,8,16$. Since this is a cubic (as required), this must be the desired function.

Since it's a cubic, it MUST cross the y-axis somewhere (if you're skeptical, try finding a cubic that doesn't cross the y-axis). So it must be $ 0$ somewhere. (I don't think the solutions are particularly nice, though). [/hide]

[quote="vishalarul"]
$ f(x) \equal{}\minus{}\frac{1}{3}\cdot (x\minus{}2)(x\minus{}3)(x\minus{}4)\minus{}2 (x\minus{}1)(x\minus{}3)(x\minus{}4)\minus{}4 (x\minus{}1)(x\minus{}2)(x\minus{}4)\plus{}\frac{8}{3}\cdot (x\minus{}1)(x\minus{}2)(x\minus{}3)\]$
Just plug in $ 1,2,3,4$ and you should get the values $ 2,4,8,16$. Since this is a cubic (as required), this must be the desired function.

Since it's a cubic, it MUST cross the y-axis somewhere (if you're skeptical, try finding a cubic that doesn't cross the y-axis). So it must be $ 0$ somewhere. [b](I don't think the solutions are particularly nice, though). [/b][/hide][/quote]
See, that's the bad thing about this method which apparently make the problem so easy.  It's a great clever method, but who wants to go through all the simplification to get the polynomial in standard form?  I wouldn't want to expand and simplify that mess.  If you use th other methods, you can easily see that $ x$ is a factor of the polynomial and so $ f(0) \equal{} 0$.

[quote="archimedes1"][quote="diophantient"]
[hide="interesting solution"]
This function greatly resembles the function $ 2^{n}$ (as pointed out by numerous people).
Now consider the function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}...\plus{}\binom{n}{k}$. We know that if k>n (by our definition of the choose function) the value of $ \binom{n}{k}$ is 0.
So if we pick an n<k the value of this is $ 2^{n}$, since $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}...\plus{}\binom{n}{n}\equal{} 2^{n}$.
So we have that the function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n}{3}\plus{}\binom{n}{4}$ would work. Unfortunately, this is a quartic.
The function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n}{3}$ comes close, but when n=4, the value is 1 less than what we desire. So we wish to find a new function for which f(1,2,3)=0, but f(4)=1. Since 1, 2, and 3 are roots of this new polynomial, we can just scale $ (x\minus{}1)(x\minus{}2)(x\minus{}3)$ by $ \frac{1}{6}$ to get the desired function (note that this is equal to $ \binom{n\minus{}1}{3}$).
So we have the function $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n\minus{}1}{3}\plus{}\binom{n}{3}\equal{} 1\plus{}n\plus{}\frac{n^{2}\minus{}n}{2}\plus{}\frac{(n\minus{}1)(n\minus{}2)(n\minus{}3)}{6}\plus{}\frac{n(n\minus{}1)(n\minus{}2)}{6}$ works.
Simplifying, we get $ y \equal{}\frac{1}{3}x^{3}\minus{}x^{2}\plus{}\frac{8}{3}x$.
[/hide][/quote]
That's awesome!  So, (please correct me if I'm wrong) you basically found a function that looks like what we need, but didn't fit the criteria, so we had to make it cubic.  But then it  didn't work for 4, so you had to find another function to add to it which wouldn't alter the values for 1,2,3 but increased the value for 4 by 1.  You wrote a poly with 1,2,3 as roots and then made it so that 4 gives 1.  Then you just added that to the original function to make it work?[/quote]
Yeah, that's my solution. Although you don't have to do the whole function with 1,2,3 being roots, if you think about it for a minute you can probably come up with $ \binom{n\minus{}1}{3}$.


y% of x is x% more than x% of 50, while x% of y is y% more than y% of 25. Find x.