[b]The Problem[/b]

During the holiday season, mall traffic is at its peak! Last minute shoppers, bargain hunters and tiny tots rushing to tell Santa their wish lists are filling up mall aisles everywhere! The Martins took their kids to the mall to see Santa. Mr. Martin had never seen a line so long! At 11:32 am he counted that there were 61 kids ahead of his and that Santa was spending 2 minutes 15 seconds, on average, with each child. If Santa maintained this rate, at what time would the Martin kids get to sit on Santa’s lap?

--------------------------------------------------------------------------------

More hours = More profit: On average, the Gadget Store profits $600 per hour it is open during regular weeks. Its regular hours are 9-5 every day of the week. Last week, due to the holidays, the store stayed open 3 hours longer per day and averaged 20% more profit per hour. How many more dollars of profit did The Gadget Store make during the holiday week than the average regular week?

--------------------------------------------------------------------------------

Does The Early Bird Get The Bargain?: Last week I witnessed the classic “percent discount” mistake at a local store. The store had everything on sale for 60% off, plus an additional 20% off you shopped before noon. The woman in front of me in line was shocked when the cashier gave her the final price for her purchase. She said, “That’s not 80% off the price on the tag!” The cashier then explained to her that she should only take 60% off of the original price, and then take 20% of that subtotal off. Because it sounds better, the store advertises “60% off, plus an additional 20% off”, but it’s really equal to a one-time discount of x%. What is the value of x?

How can you not know these problems, JEFF?!  :D

someone else can do 1, because i think jeff should know it  :roll:
[hide="2"]It was open for 8 hours a day on weekdays, and 11 hours during the holiday. It average 600*1.2=720 dollars an hour.

It made 33600 every week on normal weeks, during the holiday, it make 55440.
55440-33600=21840, assuming it was open on Sunday. [/hide]

[hide="3"]0.8*0.4=0.32, or 68% off. (0.8=20% off, 0.4=60% off). [/hide]

do u mean 61 children[b] including or not including [/b]the child sitting on Santa's lap?
also, did a child start sitting on Santa's lap the instant Mr. Martin looked at the clock? well, to mi understanding [hide="answer #1"]2 min and 15 secs times 61 is 122 min and 915 secs, which is 137 min 15 secs or 2 hours 17 min and 15 secs. 11:32 + 2:17:15 is 13:49:15 or 1:49:15 PM.[/hide]

1.  2.25 minutes per child and 61 children equals 120+15+2.25=137.25 minutes or
two hours and 17 minutes 15 seconds


Pete thinks of a number. He doubles it, adds 10, multiplies by 4,
and ends up with 120. What was his original number?

let $ x$ be the original number:
Our equation would then be $ 4(2x\plus{}10)\equal{}120$, dividing by 4 so our equation is now $ 2x\plus{}10\equal{}30  \Longrightarrow    2x\equal{}20 \Longrightarrow x\equal{}\boxed{10}$

Or, as an alternative, we can just work backwards!

Let's start with $ 120$. Pete multiplied by $ 4$, so now we divide.

$ \frac{120}{4} \equal{} 30$

He added $ 10$, so we subtract.

$ 30 \minus{} 10 \equal{} 20$

He doubled it so we "un-double" it.

$ \frac{20}{2} \equal{} 10$

Our answer is $ \boxed{10}$


a word has n characters. some of the characters are repeated. how do you find the number of permutations of r-letter words( n things taken r at a time) where r is smaller than n.

As far as I know, there is no way of working this out other than splitting it up into several cases. Maybe someone can say I'm wrong?

The only way that I know of where you can find permutations of a word with repeated letters is with ALL of the letters being used.

The formula is (number of elements!)/(number of each element!)

For example, is mississippi, there are 1 m's, 4 i's, 4 s's, and 2 p's, with 11 letters total. The number of permutations would be (11!)/(1!4!4!2!) = 34,650.

Hope this helps. Now, you might be able to break this down into different cases involving r letters.


real analysis
integration
real analysis unsolved

Show that if $ \parallel{} f_n  \parallel{}_1 \leq 2^{\minus{}n}$ for every $ n \geq 1$ then $ (f_n)_n$ converges to zero a.e.

This should follow from

If $ (f_n)_n$ is a sequence in $ L^1[0, 1]$ is such that $ \sum_{n\equal{}1}^{\infty}  \parallel{} f_n \parallel{}_1 < \infty$ then $ \sum_{n\equal{}1}^{\infty} | f_n(s) | < \infty$ for almost every $ s \in [0, 1]$.

However, I don't see how to show this.  I would appreciate some hints on how to proceed.

Suppose there is a set $ A\subset [0,1]$, with Lebesgue measure $ m(A) > 0$, such that $ \sum_{n \equal{} 1}^\infty |f_n(s)| > \sigma$ for all $ s\in A$.  Show that $ \sum_{n \equal{} 1}^\infty \Vert f_n \Vert_1 \equal{} \sum_{n \equal{} 1}^\infty \int_A|f_n(s)|ds \plus{} \sum_{n \equal{} 1}^\infty \int_{A^c}|f_n(t)|dt > \sigma m(A)$.  What happens as $ \sigma \to \infty$?

[hide="remark"]This proof seems a bit roundabout, as you're proving a rather general statement in order to then prove a simpler very specific one.  Here's what I think is a slightly more natural proof of the first statement:

Suppose the contrary, that there exists $ \epsilon > 0$ such that $ |f_n| > \epsilon$ for all $ n$ on a set $ A\subset[0,1]$ with $ m(A) > 0$.  Then $ \Vert f_n\Vert_1\geq \int_A |f_n(s)|ds\geq \epsilon m(A)$ for all $ n$, contradiction.[/hide]


Prove that all norms in $ \mathbb{R}^n$ are equivalent.

I'm not sure I understand, I mean, I always thought that there existed different norms because we want to prove that there are metric spaces with different norms.

Edit: Ah ok, thanks jmerry for the clarification on the terminology.

"Equivalent" doesn't mean that they're isometric- it means that they induce the same topology. Two norms $ \|\cdot\|_a$ and $ \|\cdot\|_b$ are equivalent if $ C_1\|x\|_a\le \|x\|_b\le C_2\|x\|_a$ for some constants $ C_1,C_2$ and all $ x$.

One can compare with the $ l_1$ norm of $ R^n$. For an arbitrary norm,
$ \parallel{}a_1e_1 \plus{} ... \plus{} a_ne_n\parallel{} \leq |a_1\parallel{}|e_1\parallel{} \plus{} ... \plus{} |a_n\parallel{}|e_n\parallel{} \leq \max_{1 \leq i \leq n} \parallel{}e_i\parallel{} (|a_1| \plus{} ... \plus{} |a_n|)$
On the other hand, we note that with respect to $ l_1$, $ \parallel{}.\parallel{}$ is continuous. since $ S^{n\minus{}1}$ is compact, $ \parallel{}.\parallel{}$ has a minimum on $ S^{n\minus{}1}$ (distance between $ l_1$), say $ \parallel{}x\parallel{} \ge c$ for any $ x \in S^{n\minus{}1}$. Then in general $ \parallel{}x\parallel{} \ge c|x|_1$ (in $ l_1$ norm).

This shows the equivalence of any norm with $ l_1$ norm, since equivalence is an equivalence relation we are done.


So I have a pretty decent grasp of the basic principles of chemistry - reaction mechanisms, bonding patterns, solubility rules, and a little bit of electro and thermochemistry.  My problems arise when I get questions which ask me about common substances not bonding because there exists some miniscule exception to standard bonding theories that the two substances exhibit.  Do any of you have any good recommendations for books / texts that I can use to reinforce these basic principles and exceptions?  A book at or below olympiad level would suit me best.

Thanks!

umm... This has been answered at least 10 times on this forum so far....

If you wanna get good at chemistry, u gotta do hard problems, and a lot of them. Whatever source can supply that is fine. Many people choose atkins' chem book. I liked zuhmdal myself. whatever floats your boat.

Thanks!!

yea.. sorry, I should've browsed the forums a little bit before I posted my question.. :blush:


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

:D

If $ \beta\equal{}(b_{1},b_{2},...,b_{n})'$

then $ \beta\beta'\equal{}$$ \begin{bmatrix}b_{1}b_{1}& b_{1}b_{2}&... & b_{1}b_{n}\\ b_{2}b_{1}& b_{2}b_{2}& ... & b_{2}b_{n}\\ ... & ... & ... & ...\\b_{n}b_{1}& b_{n}b_{2}& ... & b_{n}b_{n}\end{bmatrix}.$

[color=red]Find the EigenValues of the above matrix [/color]:lol:

OK,Assume $ b_{1} \not\equal{} 0$

we have:$ \begin{bmatrix}b_{1}b_{1}& b_{1}b_{2}&... & b_{1}b_{n}\\ b_{2}b_{1}& b_{2}b_{2}& ... & b_{2}b_{n}\\ ... & ... & ... & ...\\b_{n}b_{1}& b_{n}b_{2}& ... & b_{n}b_{n}\end{bmatrix}$ $ \simeq$ $ \begin{bmatrix}b_{1}b_{1}& 0&... & 0\\ 0& 0& ... & 0\\ ... & ... & ... & ...\\0& 0& ... & 0\end{bmatrix}$

OK, Find the EigenValues of its similar matrix is easy. But I can't Find the eigenvalues of $ \beta\beta'$. Is there some relationship between $ \beta\beta'$ and its similar matrix? :roll:

Be careful about the words you are using. It doesn't appear to me that you really mean similar (i.e. same linear transformation, different matrix representation using another basis). It seems that you really want to say row echelon form.

Anyway, to compute the characteristic polynomial for this matrix. Notice that
1. rank = 1 (unless all $ b_i \equal{} 0$, let's say this is not the case)
2. $ \beta$ is an eigenvector for this matrix. What is the corresponding eigenvalue $ \lambda$?
3. Do you know what the eigenvalues are and their multiplicities now?

$ \beta\beta',$ or as I'd prefer to write it, $ \beta\beta^T,$ is an $ n\times n$ matrix of rank $ 1.$ That means that $ 0$ is an eigenvalue of it of geometric multiplicity $ n-1.$

The algebraic multiplicity of an eigenvalue is always greater than or equal to the geometric multiplicity. So at least $ n-1$ of the eigenvalues (considered algebraically) is zero. But the sum of the eigenvalues is the trace of the matrix, and the trace of $ \beta\beta^T$ is $ b_1^2+b_2^2+\cdots+b{}^n{}^2.$

The eigenvalues are $ 0,0,\dots,0,b_1^2+b_2^2+\cdots+b{}^n{}^2.$

The action you called "similar" in your post was not, in fact, true matrix similarity.

$ B$ is similar to $ A$ if and only if there exists an invertible $ P$ such that $ B=P^{-1}AP.$

By contrast, $ B$ is row equivalent to $ A$ if and only if there exists an invertible $ P$ such that $ B=PA.$ What you used above wasn't even row equivalence (or column equivalence) but a combination of both: $ B=PAQ$ for invertible $ P$ and invertible $ Q.$

If two matrices are similar, then they have the same trace, the same determinant, the same characteristic polynomial, and the same eigenvalue.

Row equivalence preserves none of these things. Two row equivalent matrices will have different traces, different determinants, different eigenvalues. True, you can use row reduction to compute $ \det A,$ but that involves keeping track of the determinant of $ P.$


calculus
integration
function
trigonometry
calculus computations

\[ \int {\frac{{x^2 }}{{\left( {x\sin x \plus{} \cos x} \right)^2 }}} dx
\]

[quote="Xaenir"]
\[ \int {\frac {{x^2 }}{{\left( {x\sin x \plus{} \cos x} \right)^2 }}} dx
\]
[/quote]

look here please

http://www.mathlinks.ro/Forum/viewtopic.php?t=168932

and here :

http://www.mathlinks.ro/viewtopic.php?t=106333

hello, i have got $ \int\frac{x^2}{(x\sin(x)\plus{}\cos(x))^2}\,dx\equal{}\frac{\sin(x)\minus{}x\cos(x)}{\cos(x)\plus{}x\sin(x)}\plus{}C$.
Sonnhard.


limit integration geometry rectangle function logarithms calculus

It's a
[hide="cow."]Riemann sum in disguise on the interval $[0,1]$.[/hide]

Consider using rectangles of width $\frac{1}{n}$ on the interval $[0,1]$. What function would give you heights of

$\frac{1}{1+\frac{1}{n}}, \frac{1}{1+\frac{2}{n}}, \ldots, \frac{1}{1+\frac{n}{n}}$?

Alternatively, write the result in summation notation and try to match it to

$\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)$.


algebra
polynomial
superior algebra
superior algebra unsolved

Consider the finite field $ F = GF(2^h)$ with $ h$ a positive integer. A polynomial $ p(x)$ is called a normal permutation polynomial iff $ x\mapsto p(x)$ defines a permutation on the elements of $ F$ with $ p(0) = 0$ and $ p(1) = 1$.

If $ p(x)$ and $ \frac {p(x)}{x}$ are both normal permutation polynomials, show that $ f: F^*^2\to F^*: (x,y)\to \frac {1}{p(x)}\frac {1}{p(y)} + \frac {x}{p(x)}\frac {y}{p(y)}$ is not surjective.

[i][size=75]Remark: if f(x) is a normal permutation polynomial, then so is 1/f(x) with 1/0=0. Here F* denotes the nonzero elements of F.[/size][/i]

No one? Is it too hard or too ugly? :maybe:


A regular octagon with sides of length $ 1$ has $ 20$ diagonals. Compute the sum of the [u]squares[/u] of the lengths of these diagonals.

[hide="Answer"]

Using Pythagorean and counting diagonals got me $ 24 \plus{} 16\sqrt {2} \plus{} 16 \plus{} 8\sqrt {2} \plus{} 16 \plus{} 8\sqrt {2} \equal{} 56 \plus{} 32\sqrt {2}$[/hide]


There is a number, which when divided by $ \frac{2}{3}$ of $ \frac{4}{5}$ of $ 1\frac{1}{2}$ will produce 1. What is the square of the number? Express your answer as a common fraction.

You can tell by the wording that we are supposed to evaluate "$ \frac {2}3$ of $ \frac {4}5$ of $ 1\frac {1}{2}$" before dividing "a number" by that.  Note that the word "of" means "$ \times$".  Call "a number" $ x$.  Knowing that $ 1\frac {1}2 \equal{} \frac {3}{2}$, we have:
\[ \frac {x}{\cancel{\frac {2}{3}}\times\frac {4}{5}\times\cancel{\frac {3}{2}}} \equal{} 1\]
Simplifying:
\[ \frac {x}{\frac {4}{5}} \equal{} 1\implies 5x \equal{} 4\implies x \equal{} \frac {4}{5}\]
So our final answer is $ x^2 \equal{} \left(\frac {4}{5}\right)^2 \equal{} \frac {4^2}{5^2} \equal{} \boxed{\frac {16}{25}}$


number theory
greatest common divisor
relatively prime

Given that $ A \equal{} \{\frac{n}{24}\}$, $ n$ is a natural number, gcd$ (n,24) \equal{} 1$, and $ \frac{n}{24} < 2$, what is the sum of the elements of A?

24 = 2^3 * 3
That means it must be relatively prime if it isn't divisible by 2 and/or 3

Numbers less than 48 that satisfy this:
5,7,11,13,17,19,23,25,29,31,35,37,41,43,47

Sum: 383
Sum of members in set: 383/24

By definition, 1 is relatively prime to all whole numbers, so the answer is $ \frac{383\plus{}1}{24}\equal{}\boxed{16}$

just feel like saying you can pair the terms that sum to 2


calculus
integration
trigonometry
calculus computations

Could someone please help me with this
Int[0,2]Int[x,2] y^2sin(xy)dydx


Would you like to post a solution?  I remain perplexed  :huh:  If I can show the roots of the first equation are related to the $25^{th}$ roots of unity, I have a proof...

[quote="Altheman"][hide]yes...$y=x^25$, then say $r$ is a root of the first, then the roots of the second are

$x^25=r \implies (x-r)(x^24...)=0$, then since all roots of the first are also of the second, result follows[/hide][/quote]
Are you saying that r has to be a root of $x^{25} = r$? That is wrong. In fact, if $|r| \ne 1,0$ it cannot possibly be true, since the length of each root will be $|r|^{1/25}$.

It is obvious that we have no roots to $x^4 + x + 1 = 0$ for which $|r| = 0$.

Assume that we have a root $r$ for which $|r| = 1$. Then we have that $r^4 + r + 1 = 0, |r^4| = |r| = |1| = 1$. It is not hard to show that that for this to be true we MUST have one of the following:
Case 1: $1 = e^0, r = e^{\frac{2\pi}{3}i}, r^4 = e^{\frac{4\pi}{3}i}$.
Case 2: $1 = e^0, r^4 = e^{\frac{2\pi}{3}i}, r = e^{\frac{4\pi}{3}i}$.
The geometric interpretation of this is that if we have three complex numbers on the unit circle and the sum of them is 0, they must lie in a equilateral triangle.

You can easily check (just take the respective $r$ to the power four) that neither of these  two cases is possible. Therefore, there is no root $r_j$ to $x^4 + x + 1 = 0$ with length 1 or 0.

Since we have concluded that every solution to $x^{100} + x^{25} + 1 = 0$ must be on the form $x^{25} = r_j$, we've also concluded that the two polynomials does not have a single root in common (using the comment in the beginning). They therefore also cannot divide each other.

you are right... i don't know what i was doing with that factorization...i was thinking of $x^{25}-r^{25}=0$...but after my mistake, we see that they do not have common roots


function
trigonometry
limit
calculus
calculus computations

Does there exist a real-valued function $ f(x)$ on $ \mathbb{R }$ which is differentiable at every $ x\in \mathbb{R}$, but $ f '(x)$ is not continuous ?

$ f(x)\equal{}x^2\sin(\frac{1}{x})$ for $ x\neq 0$,
$ f(x)\equal{}0$ for $ x\equal{}0$.

[quote="feniks"]$ f(x) \equal{} x^2\sin(\frac {1}{x})$ for $ x\neq 0$,
$ f(x) \equal{} 0$ for $ x \equal{} 0$.[/quote]
$ \lim_{x \to 0} x \sin \left( \frac{1}{x}\right) \equal{} 0$
As, $ x \to 0, \; \frac{1}{x} \to \infty$ but $ f(\theta) \equal{} \sin \theta \in [\minus{}1, 1]$.
Hence, any number within $ [\minus{}1, 1] \times \lim_{x \to 0}x\equal{}0$.

[quote="Rajiv"][quote="feniks"]$ f(x) \equal{} x^2\sin(\frac {1}{x})$ for $ x\neq 0$,
$ f(x) \equal{} 0$ for $ x \equal{} 0$.[/quote]
$ \lim_{x \to 0} x \sin \left( \frac {1}{x}\right) \equal{} 0$
As, $ x \to 0, \; \frac {1}{x} \to \infty$ but $ f(\theta) \equal{} \sin \theta \in [ \minus{} 1, 1]$.
Hence, any number within $ [ \minus{} 1, 1] \times \lim_{x \to 0}x \equal{} 0$.[/quote]

Sorry,what means?

The value of the function is $ 0$, at $ x\equal{}0$, hence you need not add an extra "$ f(0)\equal{}0$, atleast I feel so.

[quote="Rajiv"]The value of the function is $ 0$, at $ x \equal{} 0$, hence you need not add an extra "$ f(0) \equal{} 0$, atleast I feel so.[/quote]

I think feniks' discription is conservatively careful, do you agree?

 :lol:  :D

[quote="Rajiv"]The value of the function is $ 0$, at $ x \equal{} 0$[/quote]
How can you know what $ f(0)$ is if it involves an expression $ \frac{1}{x}$?
The limit as x approaches 0 is 0, but $ \frac{1}{x}$ is still not defined at $ x\equal{}0$.

[quote="BlueVelvet"][quote="Rajiv"]The value of the function is $ 0$, at $ x \equal{} 0$[/quote]
How can you know what $ f(0)$ is if it involves an expression $ \frac {1}{x}$?
The limit as x approaches 0 is 0, but $ \frac {1}{x}$ is still not defined at $ x \equal{} 0$.[/quote]
Although $ \frac {1}{x} \to \infty$ as $ x \to 0$, we have a sine function which is bounded within $ [ \minus{} 1, 1]$. Hence $ \sin \left( \frac {1}{x}\right)$ is defined, finite. Hence you can multiply it with $ x$ to get a determined value, $ 0$.

See the problem is difference between [i][b]functional value[/b][/i] and [b][i] limiting value[/i][/b]. a function may have limiting value at a certain point but not be continuous. suppose we define $ f(0) \equal{} 5$. then our function is not continuous, but has limiting value at $ x \equal{} 0$.

[quote="BlueVelvet"][quote="Rajiv"]a sine function which is bounded within $ [ \minus{} 1, 1]$. Hence $ \sin \left( \frac {1}{x}\right)$ is defined, finite. [/quote]
This just doesn't make sense. Of course, whatever you put in the sine, the outcome will be between 1 and -1. But the problem is that we don't know [i]what [/i]to put in the sine. Forget about the $ x$ in front of the sine, and consider $ f(x) \equal{} \sin\left(\frac {1}{x}\right)$. This isn't defined at $ x \equal{} 0$. Multiplying by $ x$ does not change that. You seem to be confusing this with what [i]is "reasonable" to define, considering what value the function approaches near the undefined value[/i].

As another example, consider $ g(x) \equal{} \frac {x^2 \minus{} 1}{x \minus{} 1}$. In this way, $ g(1)$ is not defined because we would be dividing by zero. You might be tempted to say $ g(x) \equal{} \frac {x^2 \minus{} 1}{x \minus{} 1} \equal{} \frac {(x \minus{} 1)(x \plus{} 1)}{x \minus{} 1} \equal{} x \plus{} 1$, so that $ g(1) \equal{} 2$, but this is not contained in the formula. Or even better, the function $ h(x) \equal{} \frac {x}{x}$. It equals 1 everywhere, except at zero: there it's not defined.
To define a function properly, you just have to be explicit what value the function assigns to each point in its domain. No limits involved.

(sorry if it's offtopic)[/quote]
check post 2 carefully, function has been piecewise defined, [b]so it is reasonable[/b] :wink:  :)

[quote="BlueVelvet"][quote="Rajiv"]a sine function which is bounded within $ [ \minus{} 1, 1]$. Hence $ \sin \left( \frac {1}{x}\right)$ is defined, finite. [/quote]
This just doesn't make sense. Of course, whatever you put in the sine, the outcome will be between 1 and -1. But the problem is that we don't know [i]what [/i]to put in the sine. Forget about the $ x$ in front of the sine, and consider $ f(x) \equal{} \sin\left(\frac {1}{x}\right)$. This isn't defined at $ x \equal{} 0$. Multiplying by $ x$ does not change that. You seem to be confusing this with what [i]is "reasonable" to define, considering what value the function approaches near the undefined value[/i].

As another example, consider $ g(x) \equal{} \frac {x^2 \minus{} 1}{x \minus{} 1}$. In this way, $ g(1)$ is not defined because we would be dividing by zero. You might be tempted to say $ g(x) \equal{} \frac {x^2 \minus{} 1}{x \minus{} 1} \equal{} \frac {(x \minus{} 1)(x \plus{} 1)}{x \minus{} 1} \equal{} x \plus{} 1$, so that $ g(1) \equal{} 2$, but this is not contained in the formula. Or even better, the function $ h(x) \equal{} \frac {x}{x}$. It equals 1 everywhere, except at zero: there it's not defined.
To define a function properly, you just have to be explicit what value the function assigns to each point in its domain. No limits involved.

(sorry if it's offtopic)[/quote]

Thanks, I get it (as I already said before!!!)

[quote="ith_power"]check post 2 carefully, function has been piecewise defined, [b]so it is reasonable[/b] :wink:  :)[/quote]My message was still directed to the quote [quote="Rajiv"]The value of the function is $ 0$, at $ x \equal{} 0$, hence you need not add an extra "$ f(0) \equal{} 0$, atleast I feel so.[/quote] The original defintion in post 2 is of course correct.
[quote] (as I already said before!!!)[/quote]
Sorry ;) I thought my last post added something to the discussion.

(PS: I think it's a bit overdone to quote the whole message)


function
LaTeX
number theory
prime numbers
number theory unsolved

Let $p , q , r$ be distinct prime numbers and let $A$ be the set
  $A=\{ p^aq^br^c : 0\leq a , b,c \leq 5 \}$ .
 Find the smallest integer $n$ such that any $n$-element subset of $A$ contains two distinct elements $x,y$ such that $x$ divides $y$

Let $A=\{p_1 ^{a_1} \dots p_k^{a_k} ,a_i\leq m\}, \ and \ \deg(x) \ mean \ function \ A \to Z_+ \ \sum a_i .$ We have k=3,m=5. Let \[ B_n=\{y=p_1^{b_1} \dots p_k^{b_k}, \ b_i \leq m, \deg(y)=\sum b_i =n\} \]
Easyly to show, that for any x  exist y, that $\deg(y)=n \ \ y|x \ if \deg(x)>n, \ x|y \ if \ \deg(x)<n.$ My idea was chose $B_n \ (n=m),$ with capasity $C_{n+k-1} ^{k-1} , when \ n\leq m.$ I think, can be proved, that optimal subset is $B_n , when \ n=[km/2].$
In this case k=3, m=5 $\#B_5 =21, \ \#B_6=C_8 ^2 -3=25, \ \#B_7=28, \ \#B_8 =27$
Therefore, for case m=5,k=3 we have $N=1+\#B_7 =29.$

Sorry , I can't your solution . Please use Latex !


algorithm
MATHCOUNTS
algebra
binomial theorem

This is a formula for finding the square roots of numbers:

[hide]/b] y= 1/2(x+ z/x)

x is a guess at the square root of the original number z.

For example the square root of 3 with a guess of 2:

y= 1/2(2+ 3/2)

y= 1/2(3.5)

y= 1.75

Do the process again with why as the new guess:

y= 1/2(1.75+ 3/1.75)

y= 1/2(3.4642857......)

y= 1.732...

1.732 * 1.732= 2.999824[/hide]

The correct term is [i]algorithm[/i].  A formula is a single equation with the desired quantity on one side and a closed-form expression on the other; this is a rule for constructing a sequence that [i]approximates[/i] the desired quantity.  It is known as the [url=http://en.wikipedia.org/wiki/Babylonian_method#Babylonian_method]Babylonian method[/url].

There are other algorithms such as the [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Bakhshali_approximation]Bakhshali approximation[/url] and the [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit_by_digit_calculation]Digit by digit calculation[/url] the latter of which is my favorite as you won't have to go back and change any digits.

By the way, love the avatar t0rajir0u!

One method that is more "MATHCOUNTS-level" (that is, more easily explained) makes use of the generalized binomial theorem, which states that $ (x \plus{} y)^n \equal{} x^n \plus{} nx^{n \minus{} 1}y \plus{} \frac {n(n \minus{} 1)}{2!}x^{n \minus{} 2}y^2 \plus{} \cdots$ for all real $ n$ (compare this to the normal Binomial Theorem).

So if we wanted to find $ \sqrt {x^2\pm\epsilon}$ (where the smaller $ \epsilon$, the more accurate the calculation), we could write this as

$ (x^2 \plus{} \epsilon)^{\frac12} \equal{} (x^2)^{\frac12} \plus{} \frac12(x^2)^{ \minus{} \frac12}\epsilon \minus{} \frac18(x^2)^{ \minus{} \frac32}\epsilon^2 \plus{} \cdots \equal{} x \plus{} \frac\epsilon{2x} \minus{} \frac {\epsilon^2}{8x^3} \plus{} \cdots$, and similarly,

$ (x^2 \minus{} \epsilon)^{\frac12} \equal{} x \minus{} \frac\epsilon{2x} \minus{} \frac {\epsilon^2}{8x^3}\plus{}\cdots$.

For example, by this, $ \sqrt {26} \equal{} \sqrt {5^2 \plus{} 1}\approx5.099$, which is very close to the actual value $ \sqrt {26}\approx5.099019514\ldots$.


induction
function
logarithms
inequalities
inequalities unsolved

Taking natural logarithm, we can rewrite as follows.

$\frac{\ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}<\frac{\ln \left(1+\frac{1}{n+1}\right)}{\frac{1}{n+1}}$

For $n=1,$ L.H.S.=$2,$ R.H.S.=$\frac{9}{4},$ yielding the claim is true.

Supposed that the given inequality for some $n\in \mathbb{N},$ we have

$\left(1+\frac{1}{n+1}\right)^{n+2}\geq \left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n}\right)^n,$ thus

$\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n+1}\right)^{n+1}$

$=\left(1+\frac{1}{n+1}\right)\left\{\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n+1}\right)\right\}$

$>\left(1+\frac{1}{n+1}\right)\left\{\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n}\right)\right\}>0$

$\Longleftrightarrow \left(1+\frac{1}{n+1}\right)^{n+2}\geq \left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n}\right)^n,$ yielding $\left(1+\frac{1}{n+1}\right)^{n+2}>\left(1+\frac{1}{n+1}\right)^{n+1}$

which completes the proof by induction.

[quote="Lovasz"]Why must we use induction? It would be better if using AM-GM or cansider the function $f(x)=\left(x+\frac1{x}\right)^x$. They're shorter.[/quote]
Because using induction is a part of our problem.

[quote="kunny"]For $n=1,$ L.H.S=$\frac{1}{2},$ R.H.S.=$\frac{9}{4},$ yielding the claim is true.[/quote]
$\displaystyle \left(1+\frac{1}{(1)}\right)^{(1)}=(1+1)^1=2$

Sorry for that I have just found mistakes in my solution. I'm confusing! :?:
But I found antoher proof without using induction or A.M.-G.M.

We have the following: \[1+\frac{1}{n+1}  = \frac{1  + n(1+\frac{1}{n})}{n+1} \geq \sqrt[n+1]{1\cdot (1+\frac1n)^n}\] which is what we want.

Alternative solution: you can study the monotony of the $f(x)=\frac{\ln (1+x)}{x}$ using the derivative.

[quote=hpasten]Show by induction that
$\displaystyle \left(1+\frac{1}{n}\right)^n\le \left(1+\frac{1}{n+1}\right)^{n+1}$
for every $n\in\mathbb{N}$

I know it`s too easy, for example using $AM\ge GM$, but here you must use induction.[/quote]

An alternate solution is as follows:
Let's define the sequence $\left(x_{n}:=\left(1+\frac{1}{n} \right)^{n}: n \in \mathbb{N} \right)$.Clearly it suffices to prove that $(x_{n})$ is nondecreasing.. But we already know that $(x_{n})$ is bounded  , nondecreasing sequence converging to $e$.

The monotony it is enought for prove the inequality.

[quote = Lukas8r20]
Is there an easy way to proof that it converges to $e$ when we deine $e$ as the factorial sum
[/quote]
See e.g. [url = https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function]here[/url] for the equivalence of the different definitions/characterizations for $e$.


Source: Furman University Wylie Math Tournament Senior Exam 2001 #21

Without using a calculator/tables, evaluate:
cos 87:^o:/sin 1:^o: - sin 87:^o:/cos 1:^o:.

(I omitted all the degree simbols since there are way to many of them.)

[hide]cos87/sin1-sin87/cos1

(cos1cos87-sin1sin87)/sin1cos1

(cos1sin3-sin1cos3)/sin1cos1

cos1sin3=1/2*(sin4-sin(-2))

sin1cos3=1/2*(sin4+sin(-2))

sin1cos1=1/2*(sin2+sin0)=1/2*sin2

(1/2*(sin4-sin(-2))-1/2*(sin4+sin(-2)))/(1/2*sin2)

(1/2*sin4-1/2*sin(-2)-1/2*sin4-1/2*sin(-2))/(1/2*sin2)

-sin(-2)/(1/2*sin2)

sin2/(1/2*sin2)

2[/hide]

Am I right?

Solution (in spoiler):

[hide]cos 87/sin 1 - sin 87/cos 1.

(cos 1 cos 87 - sin 87 sin 1) / sin 1 cos 1

Since sin x = cos (90-x) and cos x = sin (90-x),

(sin 89 cos 87 - sin 87 cos 89) / sin 1 cos 1

sin (89 - 87) / sin 1 cos 1

sin 2 / sin 1 cos 1

Since sin 2 = sin (2 * 1) = 2 sin 1 cos 1,

2 sin 1 cos 1 / sin 1 cos 1

2[/hide]

How did you get from

your solution wrote:[hide](sin 89 cos 87 - sin 87 cos 89) / sin 1 cos 1[/hide]
to
your solution wrote:[hide]sin (89 - 87) / sin 1 cos 1[/hide]

?

He got there using the identity sin (x-y) = sin x cos y - sin y cos x

I see...*sigh* I need more practice wth trigs...

Well there are people in this forum who are in colledge/university/etc. now...
For us, yes. [b]very[/b]

Nah. Trig is fun. More straightforward than algebra, if you ask me.


Η ακολουθία $ a_{0},a_{1},...,a_{n}$ ορίζεται ως εξής: $ a_{0}\equal{}\frac{1}{2}$ και $ a_{k\plus{}1}\equal{}a_{k}\plus{}\frac{1}{n}a^{2}_{k}$ για $ k\equal{}0,1,...,n\minus{}1$. Να αποδείξετε ότι $ 1\minus{}\frac{1}{n}<a_{n}<1$.

Χρονια πολλα! :)

Γιωργο αυτο το $ \frac{1}{n}$ στην αναδρομικη σχεση με μπερδευει πολυ. Μηπως εννοεις $ \frac{1}{k}$?
Γιατι αλλιως για καθε δοσμενο $ n$ παιρνουμε αλλη ακολουθια


integration
calculus
calculus computations

Find the volume of the solid generated by revolving about the region bounded by

$ x\equal{}y(4\minus{}y),x\minus{}2y\plus{}3\equal{}0,y\equal{}\minus{}1$ about the line $ y\equal{}\minus{}1$ .

I seem to get different answer each time i calculated .Maybe my approach is wrong.Can someone enlighten me?

It seems that there are some misprints in your question

[img]http://img232.imageshack.us/img232/4972/94271711pp0.jpg[/img]

 :?:

The region is bounded above by the line $ x\minus{}2y\plus{}3\equal{}0$ and bounded below by the curve $ x\equal{}y(4\minus{}y)$ . And we want to find the volume which revolve the region about $ y\equal{}\minus{}1$

If we set
\[ \begin{gathered} X: \equal{} x \plus{} 5 \hfill \\
Y: \equal{} y \plus{} 1 \hfill \\
\end{gathered}
\]
then the equations become
\[ \begin{gathered} X \minus{} 5 \equal{} \left( {Y \minus{} 1} \right)\left( {4 \minus{} \left( {Y \minus{} 1} \right)} \right) \hfill \\
\left( {X \minus{} 5} \right) \minus{} 2\left( {Y \minus{} 1} \right) \plus{} 3 \equal{} 0 \hfill \\
\end{gathered}
\]
or equivalently
\[ \begin{gathered} X \equal{} \minus{} Y^2 \plus{} 6Y \hfill \\
X \minus{} 2Y \equal{} 0. \hfill \\
\end{gathered}
\]

[img]http://img221.imageshack.us/img221/7083/clipboard01in0.jpg[/img]

Now the volume is equal to
\[ \pi \int_0^8 {\left( {\frac {X} {2} \minus{} \left( {3 \minus{} \sqrt {9 \minus{} X} } \right)} \right)^2 dX} \plus{} \pi \int_8^9 {\left( {\left( {3 \plus{} \sqrt {9 \minus{} X} } \right) \minus{} \left( {3 \minus{} \sqrt {9 \minus{} X} } \right)} \right)^2 dX} .
\]

To plot the graph as shown above, try the following Maple code

[code]> with(plots):
> E:=implicitplot(x=-y^2+6*y,x=-2..10,y=-7..10):
> F:=implicitplot(x=2*y,x=-2..10,y=-2..10):
> display({E, F});[/code]

 :D


Do you any good physics text book that could be comparable to the Art of Problem Solving series?

I'm looking for something that is beyond high school level, and focused more on contests and such.

Most physiscs books aren't aimed at contests.  To do well on a physics competition, you really have to know the physics.  The more problems you do, the better you'll do on competitions.

The Halliday and Resnick book uses calculus, but the problems are very good.

does UsaPho/ipho use calculus?  :D
and my brother said that halliday and resnick is too advanced for beginners is this true?

IPhO does not use calculus as far as I know.

Halliday and Resnick is the standard book in most freshman year physics classes so I as soon as you know calculus (basic differentiation and integrals), you can use that book.  The Halliday, Resnick, and Krane book is a bit tougher, but still good.


linear algebra
matrix
linear algebra unsolved

Prove or Disprove
a- A linear Transformation T :R^3 --> R^3 WITH EIGENVALUES  -7,2, 4  SUCH THAT R^3 IS NOT A THE Direct Sum of the eigenspaces  of T.

b- A linear Transformation T :R^3 --> R^3 WITH EIGENVALUES  -8, 4
Where 4 has multiplicity two
 SUCH THAT R^3 IS NOT have a basis of the eigenspaces  of T.

Thank u in Advance

For (b). we might as well give the matrix of the transformation in Jordan form:

$\begin{pmatrix}4&1&0\\0&4&0\\0&0&-8\end{pmatrix}$

The largest set of independent eigenvectors we can find has two elements and is thus not a basis for $\mathbb{R}^3.$

But for (a):False. A linear tranformation from $\mathbb{R}^3$ to $\mathbb{R}^3$ might as well be a $3\times 3$ matrix. As long as such a matrix has three distinct (real) eigenvalues, the three eigenvectors form an independent set, and thus span $\mathbb{R}^3.$ So the direct sum of the eigenspaces will be $\mathbb{R}^3$ and there will be no counterexample.


cac ban cho hoi LaTex la gi. Toi nghe noi no dung de go ki tu toan hoc. Cac ban co the bay toi cach su dung duoc khong

Ban co the vao Latex help o goc tren man hinh , hoac co the vao trang web toan TV http://www.diendantoanhoc.net

nhưng tôi nên download cái nào.Cảm ơn.(Tôi đã thử nhưng chua tìm ra )

Latex la 1 chuong trinh de go ky hieu Toan . Ban khong can phai dowload cai gi het , chi can hoc cach viet cau lenh de xuat hien cac ky hieu do . Ban co the click vao cac cong thuc toan (in dam) trong cac bai viet cua members de xem cach viet .


A pentagone ABCDE has AB=BC,  CD=DE, $ \widehat{ABC}\equal{}120$
y $ \widehat{CDE}\equal{}60$ y BD=2. Calculate the area of the pentagone.

See here a very similar problem...
it is solved in the same way as yours

Daniel

There's no link and this should be moved into geometry proposed&own problems.

:blush:  my mistake
Here is it
http://www.mathlinks.ro/Forum/viewtopic.php?p=863241#863241

Daniel


that's so correct that it's like...correct. ya

Syntax Error wrote:Find the smallest real value of x that satisfies the equation:

(x+5)(x:^2:-x-11)=x+5

some people like to put their reasoning you know, but since im an impatient moderator

Reasoning:[hide]sigh, so you want it to equal 0 preferably, so then (-5+5)(doesnt matter about this part)=-5+5

so 0=0

yah![/hide]

You really [b]are[/b] impatient. Not only was your last post doubled, the whole thread was doubled too.

Also I do agree AoPS is [b]unusually[/b] groggy (slow) currently...
and it stopped working for, like, 3 times already.

what the heck...some weird reasoning out there
Syntax, try this problem...

(x+5)(x:^2+5x-5)=(x+5). What's the smallest possible value of x?

-6. In this case the right side is x-5, not 0, so the answer is -5. (It won't work with the roots of the quadratic)

yes. that's right. however you would get the wrong answer if you used the same answer you did before.

Factoring is easier though

x :^2:  +5x -6 = (x+6)(x-1) so the smallest is x = -6

and didn't umm... Mystic's problem have x :^2: +5x - 5???

[quote="Syntax Error"]just take the roots of the second one:x:^2:+5x-6=0

to get: -5 :pm:  :sqrt: (25+24) all divided by 2

to get 1 or -6

pfft, i solved it before tare, but my internet connection is taking so long to submit stuff and stuff. pfft[/quote]

I know. It (AoPS server) can't handle all the posts. Mr. Rusczyk must be moving the data to the new server or something.

yeah, now all of a sudden, my thingie is fast again

OK, now that the problem is solved I'm going to change this to Normal, is that OK with you?

[hide]-5 satisfies the equality.  lets see if theres anything else.  set x doesn't equal -5 and divide that term out.

then x^2-x-12=0=(x-4)(x+3) --> x=-3 is another small one, but not quite as small.

answer: -5.[/hide]


Cramming for an exam tomorrow and looking through the past papers.

I'm not too sure how to approach the following problem. (There's something similar in every past paper)

Show that $ 1\plus{}i$ is a solution of the equation

$ x^6 \plus{} 2x^5 \plus{} 3x^4 \plus{} 2x^3 \plus{} 3x^2 \plus{} 2x \plus{} 2 \equal{} 0$

and find all other solutions (real or complex) of this equation.

Any help as to how to do this type of question would be appreciated. (Also, am I in the right forum?)

hello, if $ 1\plus{}i$ is a solution of your equation then $ 1\minus{}i$ is also a solution.
Sonnhard.

So if a complex number is a solution, it's conjugate is also a solution?

Only if the coefficients of [i]your[/i] polynomial happen to be real... ;)

Cool

So how do I go about finding the other solutions? Would I be right in thinking there be 4 more to find?

Am I dividing the expression by the factor $ x\minus{} (1 \plus{} i)$ perhaps?

That looks painful.

[quote="rudyard"]Cool

So how do I go about finding the other solutions? Would I be right in thinking there be 4 more to find?

Am I dividing the expression by the factor $ x \minus{} (1 \plus{} i)$ perhaps?

That looks painful.[/quote]

$ (x\minus{}(1\plus{}i))(x\minus{}(1\minus{}i))\equal{}(x^2\minus{}2x\plus{}2)$
So we only need to divide the expression by the factor $ (x^2\minus{}2x\plus{}2)$

[quote="stephencheng"]

$ (x \minus{} (1 \plus{} i))(x \minus{} (1 \minus{} i)) \equal{} (x^2 \minus{} 2x \plus{} 2)$
So we only need to divide the expression by the factor $ (x^2 \minus{} 2x \plus{} 2)$

[/quote]

Thankyou very much! That rings a bell from a 9am lecture I dozily attended.

Is enough for me to crack on and get a few more marks anyway, I can do polynomial division without that pesky $ i$ happily enough.


In each blank below a single digit is inserted such that the following three-digit numbers, in this order, form and arithmetic sequence:

1 _ _,     _ _ 9,     2 _ 2,     _ 6 _,     2 _ _,     _ 3 _

What is the value of the next number in the sequence?

[hide]
I think the answer is $264$.

Between_ _ 9 and 2_2 there is an increase of a number whose ones digit is 3. With this you can fill in the numbers.

1_6, _ _ 9, 2_2, _65, 2_8, _31
From here it is easy to finish the problem
[/hide]

[hide] i am pretty sure that 364 is infact the correct answer.[/hide]

[quote="footballrocks41237"]Oh!  I know!

[hide]You know it increases by the number _$3$ because there is a difference of $3$ between $6$ and $9$...The first two ones digits...

You also know that _$65$ is $265$ because both of the ones that surround it are $2$__...  Given this, the last term has to be $331$, since it has to be greater than $265$.  Doing $331-265$ gets you $66$, and you can divide by $2$ to get a difference of $33$.  Therefore, the numbers must be $166, 199, 232, 265, 298, and 331$.  You can get this by adding $33$ to $265$ to get $298$, subtracting it over and over to get $232, 199, and 166$, and all of these numbers work with what they give you.  And then you add $33$ to $331$, to get $331+33=364$, the answer.[/hide][/quote]

I don't get it. Can you explain in more detail?

[hide="oops"]
I forgot I was carrying over into 300 :blush: .
[/hide]

[hide="solution"]We look at the units digits, and see that we can fnd the other units digits.

1-6,--9,2-2,-65,2-8,-31

And, looking at the tens and hundreds digits, we can determine some hundreds digits.

1-6,--9,2-2,265,2-8,331

Now we can find the common difference.

331-265=66

33=d

331+33=364[/hide]


limit
real analysis
real analysis theorems

if $\lim_{x\rightarrow a}f(x)=L_{1},\lim_{x\rightarrow a}g(x)=L_{2}$
$\lim_{x\rightarrow a}(f(x).g(x))=\lim_{x\rightarrow a}f(x).\lim_{x\rightarrow a}g(x)=L_{1}.L_{2}$
please give some proof for this

Those are all questions you can prove easily by using the $\epsilon$-definition of a limit. Have you tried ?

i have tried but i can't prove some parts!!can you send our proof?

This is standard textbook material - do you have a textbook that you are looking at?

i have 'silverman' but it doesn't have the proof of this theory!!

I'll not wright all proof, but I put an idea. First you have to pay attention that $f(x)$ and $g(x)$ are bounded, let say $|f(x)|<M$ and $|g(x)|<K$. And all you want to prove is $|f(x)g(x)-L_{1}L_{2}|<\epsilon$. So the trick is $|f(x)g(x)-L_{1}L_{2}|$=$|f(x)g(x)-L_{1}g(x)+L_{1}g(x)-L_{1}L_{2}|$<$|g(x)\parallel f(x)-L_{1}|+L_{1}|g(x)-L_{2}|$<$\epsilon(K+L_{1})$. Note thats not the proof it's only a way how to do it, you have to use epsilon delta definitions and consider some thing on you way.


Can someone show me how to factor $x^{4}-x^{3}-12x^{2}+7x-1$?  Thanks

[hide="Technique"] By RRT it has no rational roots, so we suppose it can be factored into two quadratics, like so:

$x^{4}-x^{3}-12x^{2}+7x-1 = (x^{2}+ax+b)(x^{2}+cx-\frac{1}{b})$

The calculation is rather annoying, but doable; expand and solve the resulting system. [/hide]

Thanks but is there an easier way of doing this?

I plugged it into my calculator and it gave me this:

$(x^{2}-4x+1)(x^{2}+3x-1)$

I'm not sure how you would do it without one though.

If RRT fails (as toraijrou pointed out) I think his method would be the quickest method.

[hide="stuck on t0rajir0u method"]$x^{4}-x^{3}-12x^{2}+7x-1 = (x^{2}+ax+b)(x^{2}+cx-\frac{1}{b})$

Expanding out the RHS, we get:
$x^{4}+(a+c)x^{3}+(b+a c-\frac{1}{b})x^{2}+(b c-\frac{a}{b})x-1$

So, our system is:

$a+c =-1$
$b^{2}+a b c-1 =-12 b$
$b^{2}c-a = 7 b$

since $b\neq 0$

From the first equation, $a =-1-c$. Plugging into the third equation,
$b^{2}c+1+c = 7 b$
$(b^{2}+1) c = 6 b-1$
${c = \frac{6 b-1}{b^{2}+1}}$

Re-arranging the second equation,
$b(b+ac+12) = 1$

Plugging in our value for $a$,
$b (b-(1+c)c+12) = 1$

Plugging in our value for $b$,
\[b \left(b-\frac{(b^{2}+6b)(6 b-1)}{(b^{2}+1)^{2}}+12 \right) = 1 \]
Now, I'm stuck.[/hide]

[hide="Suggestion"] $b$ is the root of two quadratics.  What value of $a$ causes them to have a common linear factor? [/hide]

[hide="this could resolve some nastiness"]
if we assume that we're factoring over $\mathbb{Z}$, then we can assume that $b=1$ because the only integers that multiply to -1 are 1 and -1

or we could just amend our original factoring

$x^{4}-x^{3}-12x^{2}+7x-1 = (x^{2}+ax+1)(x^{2}+cx-1)$

and work it from there
[/hide]


This is pretty well known ...
$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$.

So the polynomial you're looking for is
$f(n)=\frac{n^{2}}{2}+\frac{n}{2}$.

[quote="grizzland"]ok so in the format i needed it in above the would be
f(n)= [u]1/2[/u]n^2 +[u]1/2[/u]n +______

I think i went dumb because i cant figure out the last one[/quote]

0

[quote="grizzland"]Here is a small challenge for your entertainment and gratification. For any natural number n, let f(n) denote the sum of the number 1 to n. Thus f(1)=1, f(2)=1+2=2, f(3)=1+2+3=6, f(100)= 1+2+3+...+100=5050, etc

It turns out that f is a polynomial of degree 2 in n. figure out the coefficients of f:
f(n)=___n^2 +____n+____

There is a story about Carl Friedrich Gauss (1777-1855) who may have been the most outstanding mathematician in human history. According to the story, when Gauss was seven years old, his teacher at one stage was unhappy with the class and as a punishment he asked them to compute f(100).Gauss' class mates started writing the numbers from 1 to 100 on their paper, and adding those numbers. Gauss stared at the ceiling and then wrote the single number 5050 on the sheet and handed it in. You aren't Gauss, but you also aren't seven years old, so maybe you can figure out what he was thinking![/quote]

The way Gauss figured this out was by pairing the first and last numbers, then the second and second-to-last numbers, and etc, seeing that each pair gives a sum of $n+1$. Then there are half of the number of pairs as there are numbers, so you multiply $n+1$ by $\frac{n}{2}$ to obtain the formula. This technique is often called [i]Gaussian pairing[/i] in his honor.


Let $ a,b,c$ be positive real numbers. Prove that
\[ ab\plus{}bc\plus{}ca \ge \frac{4a^3(b\plus{}c\minus{}a)}{(b\plus{}c)^2}\plus{}\frac{4b^3(c\plus{}a\minus{}b)}{(c\plus{}a)^2}\plus{}\frac{4c^3(a\plus{}b\minus{}c)}{(a\plus{}b)^2}\]
:)

Hi Can_Hang. Thanks for the nice inequality. After hours of direct expansion  :blush: , i can't continue.
Can you please prove the following statement:
(all sums are cyclic)
$ A\equal{}\sum 4a^8 \plus{} 4a^7b\plus{}4a^7c\plus{}2a^4b^2c^2\plus{}2a^4b^4\plus{}14a^3b^3c\plus{}a^4b^3c\plus{}a^4c^3b$

$ B\equal{}\sum 4a^6b^2\plus{}4a^6c^2\plus{}3a^5b^3\plus{}3a^5c^3\plus{}9a^5b^2c\plus{}9a^5c^b$

Prove that, $ A\geq B$

Thank you ~

I think, i have found a solution. However, i use iran 96 inequality. Lets first define famous Iran 96, in a case some people may not know it.. (Actually, this inequality is first proposed by Ji-Chen to Crux in 1994)

For all positive real numbers $ a,b$ and $ c$,  the following inequality holds:

$ (ab \plus{} bc \plus{} ca)(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2})\geq \frac {9}{4}$

or equivalently:

$ ab \plus{} bc \plus{} ca \geq \frac {9[(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2]}{4[(a \plus{} b)(b \plus{} c) \plus{} (b \plus{} c)(c \plus{} a) \plus{} (c \plus{} a)(a \plus{} b)]}$

To make thing easier, lets make with sigma notation... Iran 96 implies:

$ \sum_{cyc}{ab}\geq \frac {9(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}{4\cdot \sum_{cyc}{(a \plus{} b)^2(b \plus{} c)^2}}$

Use the following substitution:
(note: this is not Ravi substitution, which is applied only for the sidelengths of triangle
$ a \plus{} b \equal{} x$
$ b \plus{} c \equal{} y$
$ c \plus{} a \equal{} z$
You can solve for each $ a, b$ and $ c$ , for example for the value of "$ a$":
$ a \equal{} \frac {x \plus{} z \minus{} y}{2}$ and similarly for $ b$ and $ c$...Observe that:
$ 4a^3 \equal{} \frac {(x \plus{} z \minus{} y)^3}{2}$. look at the RHS carefully:
$ \sum_{cyc}{4a^3(b \plus{} c \minus{} a)} \equal{} \sum_{cyc}{\frac {(x \plus{} z \minus{} y)^3(y \minus{} x \minus{} z)}{4}} \equal{} \sum_{cyc}{\frac { \minus{} (x \plus{} y \minus{} z)^4}{4}}$
Therefore LHS=negative value... We have
$ RHS \equal{} \frac {9x^2y^2z^2}{4\cdot \sum_{cyc}{x^2y^2}}$
RHS = positive value. So RHS-LHS=positve>zero...

Can, is my proof correct?

[quote="shaaam"]
$ \sum_{cyc}{4a^3(b \plus{} c \minus{} a)} \equal{} \sum_{cyc}{\frac {(x \plus{} z \minus{} y)^3(y \minus{} x \minus{} z)}{4}} \equal{} \sum_{cyc}{\frac { \minus{} (x \plus{} y \minus{} z)^4}{4}}$[/quote]
I read your proof, and I think you got mistake here. It must be
\[ \sum \frac{4a^3(b\plus{}c\minus{}a)}{(b\plus{}c)^2}\]
;) :)

[quote="can_hang2007"][quote="shaaam"]
$ \sum_{cyc}{4a^3(b \plus{} c \minus{} a)} \equal{} \sum_{cyc}{\frac {(x \plus{} z \minus{} y)^3(y \minus{} x \minus{} z)}{4}} \equal{} \sum_{cyc}{\frac { \minus{} (x \plus{} y \minus{} z)^4}{4}}$[/quote]
I read your proof, and I think you got mistake here. It must be
\[ \sum \frac {4a^3(b \plus{} c \minus{} a)}{(b \plus{} c)^2}
\]
;) :)[/quote]

well, i think it doesn't matter. because the denominator will be $ y^2$/ But the result will not change. Am i wrong?

[b]EDIT: [/b] I am really sorry for my mistake..

P.S: can you the write expression i showed in second post as sum of the squares?

[quote="shaaam"][quote="can_hang2007"][quote="shaaam"]
$ \sum_{cyc}{4a^3(b \plus{} c \minus{} a)} \equal{} \sum_{cyc}{\frac {(x \plus{} z \minus{} y)^3(y \minus{} x \minus{} z)}{4}} \equal{} \sum_{cyc}{\frac { \minus{} (x \plus{} y \minus{} z)^4}{4}}$[/quote]
I read your proof, and I think you got mistake here. It must be
\[ \sum \frac {4a^3(b \plus{} c \minus{} a)}{(b \plus{} c)^2}
\]
;) :)[/quote]

well, i think it doesn't matter. because the denominator will be $ y^2$/ But the result will not change. Am i wrong?[/quote]
I dont think so, since I think you got mistake in calculation, it must be
\[ \sum \frac{4a^3(b\plus{}c\minus{}a)}{(b\plus{}c)^2} \equal{}\sum \frac{(x\plus{}z\minus{}y)^3(3y\minus{}x\minus{}z)}{4y^2}\]
;):)


superior algebra
superior algebra unsolved

Prove that $ |\left<x,y|x^4\equal{}y^3\equal{}(xy)^3\equal{}1\right>|\leq 24$.

I believe this is false.

Let $ x \equal{} (1,2)(3,4,5,6)$ and $ y \equal{} (1,2,3)(4,6,5)$ be elements of $ S_6$ of orders $ 4$ and $ 3$ respectively. We have $ xy \equal{} (1,3,6)$ of order $ 3$. Thus $ x^4 \equal{} y^3 \equal{} (xy)^3 \equal{} 1$. However, the group $ \langle x,y \rangle$ is $ A_6$, a group of size $ 360$.

EDIT: I found those particular elements using GAP, but I suspected the claim was unlikely since triangle groups $ \langle x,y | x^l \equal{} y^m \equal{} (xy)^n \rangle$ with $ \frac {1}{l} \plus{} \frac {1}{m} \plus{} \frac {1}{n} < 1$ are hyperbolic and "large". You should expect finite groups when $ \frac {1}{l} \plus{} \frac {1}{m} \plus{} \frac {1}{n} > 1$ (the "spherical" case).

Now that I think about it, the group $ \langle x,y | x^4 \equal{} y^3 \equal{} (xy)^2 \equal{} 1\rangle$ is spherical, and according to GAP indeed has size $ 24$. Is this what you meant?


EF is parallel to BC
BD=DC
Prove that EK=KF

Refer to picture below.

Bye bye, save proofs for getting started, not the basics.  :D

Similar triangles, as before:

AKE and ABD are similar so BD=EK*k
AKF and ACD are similar so CD=KF*k

Giving

BD=CD
EK*k=KF*k
EK=KF


Support
induction
symmetry
strong induction

I think I know the answer, but I'm looking for an easy proof to this problem:

Twenty points are marked on the circumference of a circle.  Two players play the following game.  On each turn, one connects two of the 20 points with a segment, according to the following rules:
  1. a segment can appear only once     during a game;
  2. no two segments can intersect,     except on the endpoints;
  3. the player who has no choice     loses.
Assuming both players use their best strategy, which one (first or second) is guaranteed to win?

Well...

[hide]At the end of the game, there will be 18 triangles formed. There will be 37 segments in total. Therefore, player 1 will make the last move.[/hide] - edited because I can't count

Did I understand the question right?

Yes, except there are 37 segments drawn.

Wheeps, I subtracted 1 instead. Yeah, it should be 37.

TripleM wrote:Or another way:
[hide]
Join point 1 to point 11 (counting them in order). Then play symmetrically.
[/hide]

[hide]Actually, it doesn't matter how they play.  If a chimp went first, he would always win.  The question was to *prove* that player 1 will always win.[/hide]

Does anyone have a proof to support their answer?

It was to prove it using the best strategies -- it just happens that any strategy is the best strategy.  So TripleM was just as correct as the others. (I'm not sure that you were saying he was wrong, but it sounded like it.)

[quote="rcv"][quote="JBL"]It was to prove it using the best strategies -- it just happens that any strategy is the best strategy.  So TripleM was just as correct as the others.[/quote]

You're right, JBL.  TripleM, your strategy works and the proof that player 1 always wins is implied by your strategy.  Sorry if it sounded like I was claiming otherwise.

But I think smiley wants a proof that *every* strategy provides a win for Player 1.[/quote]

Well, the question did say "Assuming both players use their best strategy". In which case all you need to do is find a strategy for the winning player. But I suppose if someone *did* want to show that the first player always won, then they'd be happy :)

Well, to prove that it always ended in a first player victory I suppose you would have to prove a few things:

1) In the finishing position, the original 20-gon is disected into triangles. Put a different way, if the polygon is not yet triangluated, then it is possible to draw a segment somewhere under the given rules.

2) The polygon is triangulated if and only if there are 17 segments drawn.

From those 2 propositions you should be able to conclude that player 1 will be the person who puts in the 17 (and final) segment.

Those were the points of my first post - however shouldn't there be 37 segments in total, since the original 20-gon was not drawn?

You can do that easily with [hide]strong induction -- assume that it's true for any n-gon up to (and including) a k-gon.  First player draws any diagonal.  We then have 2 less-than-or-equal-to-k-gons, so apply the inductive hypothesis and we're done.[/hide]

Also, it hardly matters in this case whether the outside segments are drawn since they always can be drawn, no matter what the state of the game.  So Osiris is right, but it doesn't affect the outcome at all, or even the game-play.

Here is a (probably much messier) problem: what if the segments can't intersect anywhere, including endpoints?  (This does not mean they have to be parallel -- only the segments must not intersect each other, not the lines they are part of.)

Ya, ya'll are right... i forgot about the edges of the 20-gon.

JBL wrote:Here is a (probably much messier) problem: what if the segments can't intersect anywhere, including endpoints?  (This does not mean they have to be parallel -- only the segments must not intersect each other, not the lines they are part of.)

Oh, I know this one! The wonderful [hide]catalan numbers[/hide]!

A few variations on this problem and its proofs have been bugging me for a few days - then I couldn't find the problem!

1.  Does N even versus N odd matter?
2.  Does including vs. excluding the N edges of the N-gon matter?

This yields four cases.  Here's my analysis of
1.  N even, edges included.  P1 wins
   Both TripleM's symmetry argument and Joel's induction work (but see note below).  Note that the dissection segment becomes the first move in both new games.

2.  N odd, edges included. P1 wins
    Though there is no 'middle' diagonal, I believe that the symmetry argument be modified (how?)  Induction proof is valid.  (Note that even if N even, the induction proof required that P1 wins for both odd and even N upon dissection.)

3.  N odd, edges not included.  P2 wins
    Since N-3 is even, Player 2 always wins.
    Symmetry argument works: Player 2 matches each move of P1.
    Induction is more 'interesting' since the dissection is into an odd game and an even game.

4.  N even, edges not included.  P1 wins
   Original symmetry argument is valid.
   Induction again more involved since the first dissection can be into odd,odd games or even,even games.

Conclusion:  With minor modifications, I believe  that the symmetry argument is valid throughout.  Induction requires care.  We don't have a response to Zabelman/s challenge of two items to prove.  (Basically, the 'simplest' argument for all 4 cases relies on "exactly N-3 non-edges can always be drawn" - but why?)

Note:  The induction proof proves that P1 wins for all N when edges included.  But the proof does not seem to mention or rely on 'edges' issue.  Hence why wouldn't same proof 'prove' that P1 wins for all N when edges not included?  Is there a gap in the proof?


Weird question. Just take out your graphic calculator? Or think about it a bit. Goes to $ 1$ when $ x$ goes to zero. Then, it takes off sky-rocket with increasing values of $ x$, reaching for example 10 billion at $ x \equal{} 10$.

Haven't got a graphical calculator.

It seems clear that for x > 0, x^x is always real and that the graph is continuous, going down from (0,1) to (1,1) and then up quickly.

However, for x < 0, x^x may not exist.

For x = -m/n, x^x > 0 when m is even. When m and n are both odd, x^x < 0 (I think) and when m is odd and n is even, x^x seems to be imaginary. Therefore to the left of the y-axis the graph has two branches which are not continuous but are everywhere dense.

Is this correct?

Any thoughts on my post about cubic curves?

If you want something for $ x<0,$ the interesting graph would be $ y\equal{}|x|^x.$ Or, more fully,
\[ y\equal{}f(x)\equal{}\begin{cases}|x|^x&x\ne0\\1&x\equal{}0\end{cases}\]
This function would have a local minimum at $ x\equal{}e^{\minus{}1},$ a local maximum at $ x\equal{}\minus{}e^{\minus{}1},$ and a vertical tangent line at $ x\equal{}0.$


function
Putnam
calculus
algebra unsolved
algebra

The function $f: R\to R$ satisfies $f'''(x)$ is a continuos function in $R$.
Prove that $\exists a\in R$ such that
$f(a)f'(a)f''(a)f'''(a)\geq 0$

[quote="Diogene"][quote="phamduyhiep"]The funtion $f: R\to R$ satisfies $f'''(x)$ is a continuos function in $R$.
Prove that $\exists a\in R$ such that
$f(a)f'(a)f''(a)f'''(a)\geq 0$[/quote]

What about $f(x)=e^{-x}$ ?

 :cool:[/quote]

 :huh:

What about it ???  :)

I didn't see the contracdiction here!
With $f(x)=e^{-x}$,we have $f(x)f'(x)f''(x)f'''(x)=(e^{-x})^4\geq 0\forall x\in R$

[quote="zhaobin"]if not,we have for $\any x \in R$,then  $f(x)f'(x)f\"(x)f\"'(x) <0$
so $f\"'(x) \neq 0$because $f\"'(x)$ is a continuos function,
so f(x),f'(x),f"(x),f"'(x) must have the only one kind sign when $x \in R$
Case $f\"'(x) >0$,so $f\"(x)$ is increasing about $x$.
case $f\"(0)>0$,so $f'(x) \ge f'(0)+xf\"(0)>0$,when x is large enough.
so $f'(x)>0,f\"(x)>0$ when  $x \in R$.
same reason we can prove:$f(x)>0$ when  $x \in R$.
contradictions,

We can also find the contradictions when
$f\"'(x) <0$[/quote]
 well done!zhao bin!now i am interested in alberage.

This is [url=http://www.kalva.demon.co.uk/putnam/psoln/psol983.html]Putnam 1998/A3[/url]

This probably belongs in the "College Playground" section - either in the Calculus-Analysis problems, or in the Putnam section.


Dear group members,

Does anyone have more info about the death of Serge Lang ? -announced
at the following link :( -

http://www.math.yale.edu/public_html/index.html

Pray for the souls of the dead!

Serge Lang was the first person whose book in written English encountered with me.

kunny

RIP.

I attended a lecture given by him. He put me on the spot (just pointed to me at random) and asked me to come up with a differentiable real function which has an inverse that is not differentiable. I said $sgn(x)x^2$, but then he called on someone else.  :roll: :roll: (the other guy said $x^3$). Maybe he thought I said $sin(x)x^2$ instead. Maybe he meant $C^\infty$ instead of differentiable. I guess we'll never know now.

Anyway, it amazes me how one guy could know so much (and write so many books)! A great loss to mathematics.


I think I got 2
[hide](it just had to be something as beutiful as a 3-4-5, didn't it? A + C = 90, so B was a right angle)[/hide]
and 4
[hide](by showing that x = a^5 - a^2 - 1, y = ax gave a solution for all intger a >= 2)[/hide]
plus a load of random workings on the other questions (and a brief sorry to the markers - I didn't think to remove my working on the problems I solved, so there's a lot of paper under "rough work" ;).

How did people find it?

They are at:

http://www.bmoc.maths.org.uk/home/bmo2-2008.pdf

Did you ever do the Intermediate Maths Olympiad?

Hmmm for #1 [hide]$ \left( \sum x^2 \right)^3 \ge (x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{}zx)^2 \equal{} 1$[/hide]

[quote="probability1.01"]Hmmm for #1 [hide]$ \left( \sum x^2 \right)^3 \ge (x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)^2 \equal{} 1$[/hide][/quote]
How do you get any of this?

@dgreenb801,
$ (\sum a^2)^3 \equal{} (\sum a^2)(\sum a^2)^2$
But we know from Power MEan inequality that $ \sum x^2 \ge \frac {(\sum x)^2}{3}$.
Also we know that $ \sum x^2 \ge \sum xy$.
Thus we get probability 1.01's lemma.

I just did #1 really fast
[hide]Let $ S\equal{}x^2\plus{}y^2\plus{}z^2$ and $ w\equal{}x\plus{}y\plus{}z$
You can factor the equation they give as a condition, and then write it in terms of $ S$ and $ w$:
$ w(S\minus{}\dfrac{w^2\minus{}S}{2})\equal{}1$.
Solving for $ S$, you get
$ S\equal{}\dfrac{w^2\plus{}\frac{2}{w} }{3} \ge 1$ by AM-GM.[/hide]


Hi,arqady,can you post the solution now? :blush:


I saw a lab tool that I can't remember the name of.

Info on the tool:
- used to grind up tissues that were contained in small test tubes (for the one I saw, it was for grinding up mouse lung tissue that was treated earlier with radiation)
- has a disposable tip that was used to grind up tissue

What's this tool called?

i think it must be a centrifuge attachment.... :maybe:

If it was called that, it had another name...

It did not look like it was attached to the centrifuge. You could stick part of the tool into a small tube to mash, not really grind, up small bits of tissue.

Once the tissue was mashed up, you could extract the dna and rna using a column in another tube to filter and a centrifuge.

If for DNA/RNA extraction, as was mentioned in the third post, I believe the proper tool is a sonicator.

"In biological applications, sonication may be sufficient to disrupt or deactivate a biological material. For example, sonication is often used to disrupt cell membranes and release cellular contents."
-[url]http://en.wikipedia.org/wiki/Sonicator[/url]

oh right yeah and it vibrates and rotates and the actual vibrating and rotating touching the cell mashes it up.

so I mean its more of a mechanical thing... Just to mash up soft tissue in a tube.

think of sticking your finger into a tube and rotating and wiggling your finger to mash up something in the tube.... yeah so its does that.


modular arithmetic
number theory unsolved
number theory

Suppose that $ \{a_1,a_2,...,a_n\}$ and $ \{b_1,b_2,...,b_k\}$ are two sets of positive integers with the same sum of elements, and that sum is less than $ nk$. Prove that it's possible to omit some (but not all) summands from both sides of the equality $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}b_1\plus{}b_2\plus{}...\plus{}b_k$ in such a way that the equality remains valid.

an idea  is to use induction...if you want a full solution ask me and I will post it ( the solution is from a text ,is NOT mine)

Daniel

[hide="how about this?"]We are searching for some set $ A$ of $ a_i$ and some set $ B$ of $ b_i$ such that $ \displaystyle\sum_{a_i\in A}a_i\minus{}\displaystyle\sum_{b_i\in B}b_i\equal{}0$, with not all of $ a_i$ in $ A$ or vice versa.  So, we prove that such sets exist.

Let $ A_i\equal{}\displaystyle\sum_{j\equal{}1}^ia_i$ and let $ B_i\equal{}\displaystyle\sum_{j\equal{}1}^ib_i$ and let $ A_n\equal{}B_k\equal{}s$.  Consider the set $ X\equal{}\{A_i\plus{}B_j\}$ which has cardinality $ nk$.  First, we prove that modulo $ s$, this set has two equal elements.  There are $ nk$ elements and $ s$ remainder classes.  Since $ s\le nk\minus{}1$, by pigeon hole principle, there are two equal elements modulo $ s$.  So, we let $ A_i\plus{}B_j\equiv A_p\plus{}B_q$.  Obviously, we cannot have $ i\equal{}p$ since this implies $ p\equal{}q$ since $ 0<B_j,B_q\le s$ which contradicts the fact that these two elements are distinct.  So, we assume without loss of generality that $ i>p$.  We have two cases:

Case 1: $ j>q$

We have

$ f\equal{}\displaystyle\sum_{d\equal{}p\plus{}1}^i a_d\plus{}\displaystyle\sum_{d\equal{}q\plus{}1}^j b_d\equiv0\pmod{s}$

Since $ 0\le f<A_i\plus{}B_j\le A_n\plus{}B_k\equal{}2s$ we have that either $ f\equal{}0$ or $ f\equal{}s$.  Obviously, $ f\ne0$ since all $ a_d,b_d$ are positive.  Thus, $ f\equal{}s$.  So,

$ 0\equal{}f\minus{}s\equal{}\displaystyle\sum_{d\equal{}p\plus{}1}^i a_d\minus{}\displaystyle\sum_{0<d\le q\text{ or }j<d\le k} b_d$

and we're done.

Case 2: $ q>j$

We have

$ f\equal{}\displaystyle\sum_{d\equal{}p\plus{}1}^i a_d\minus{}\displaystyle\sum_{d\equal{}j\plus{}1}^q b_d\equiv0\pmod{s}$

Since $ f<A_i\le A_n\equal{}s$, and $ f>\minus{}B_q\ge B_k\equal{}\minus{}s$ we have that $ f\equal{}0$ and we're done.
[/hide]


SOrry edited....

i didnt read the question properly...

now as per doing by my method  i.e put x=0... so a0=1....

differentiate and then put x=0 and we get a1=n*(n+1)/2...........

so by applying this method we get Ar=(n+r-1)Cr........

i am lost after  this step...

Answer is Wrong...

[hide]It is $ \binom{2n}{n}$. Method is right. Please $ \text{\LaTeX}$ your post above and write proper steps.... I'll post a more elegant method after that. [/hide]

ok answer is $ \binom{2n}{n}$

made a small mistake in the last term..

$ a_r\equal{}\binom{n\plus{}r\minus{}1}{r}$

then use the rule $ \binom{n}{r\minus{}1} \plus{}\binom{n}{r} \equal{} \binom{n\plus{}1}{r}$
so it evaluvates to ur answer shreyas :D


floor function
algebra
polynomial
quadratics
function
trigonometry
AMC

Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005$. Find $\lfloor P\rfloor$.

[hide="Answer"]$x^4-4x^3+6x^2-4x=0\Rightarrow x^4-4x^3+6x^2-4x+1=1=(x-1)^4$, so we have $(x-1)^4-1=0$. $y=x+1\Rightarrow y^4-1=0=(y^2-1)(y^2+1)=(y-1)(y+1)(y^2+1)$, so the complex roots are $y=\pm i\Rightarrow x-1=\pm i\Rightarrow x=1\pm i$, and the product is $(1+i)(1-i)=\boxed{2}$.[/hide]

Err... I'm confused. Did you leave something out of the original equation? Where did the $2005$ come from? And what did you do in your second solution? :?

[quote="JesusFreak197"]Err... I'm confused. Did you leave something out of the original equation? Where did the $2005$ come from? And what did you do in your second solution? :?[/quote]

I'm sorry, I posted the problem incorrectly, hehe. :fool:

Our two solutions would have been correct if I hadn't misread the problem.  :)

Okay, I figured there was some problem. I was a little curious as to why the greatest integer function was necessary. :roll:

[hide="Answer"]$x^4-4x^3+6x^2-4x=2005\Rightarrow x^4-4x^3+6x^2-4x+1=2006=(x-1)^4$
$y=x-1\Rightarrow y^4-2006=0=(y^2-\sqrt{2006})(y^2+\sqrt{2006})$
The two roots of the first part are both real, so we focus on the second part: $y=\pm i\sqrt[4]{2006}=x-1\Rightarrow x=1\pm i\sqrt[4]{2006}$, so the product of the roots is $(1+i\sqrt[4]{2006})(1-i\sqrt[4]{2006})=1+\sqrt{2006}=P\Rightarrow \lfloor P\rfloor=\boxed{45}$.[/hide]

[hide]Adding 1 to both sides, we get $(x-1)^4 = 2005.$ The roots of 2005 are $2006^{1/4} \text {cis}\ 0$, $2006^{1/4} \text {cis}\ \pi/2$, $2006^{1/4} \text {cis}\ \pi$, and $2006^{1/4} \text {cis}\ 3\pi/2$. The second and fourth ones are complex. Rewriting in rectangular form, we have $-2006^{1/4}i+1$ and $2006^{1/4}i+1$. Their product is a difference of squares, $P=1^2-(2006^{1/4}i)^2 = 1+\sqrt{2006}$. Thus $\lfloor P \rfloor = 1+44=45$.[/hide]

(Go interm trig class :D)

[quote="JesusFreak197"]Uhh... that's hugely unnecessarily complicated. :?[/quote]

It's not really complicated :huh: It's just using trigonometry instead of the normal way. Unless you don't know DeMoivre's theorem, it shouldn't be too confusing. ;)

It was a pretty simple application of Demoivre's theorem, I agree. That seemed like the simplest solution to me at the time.  :D

Was it just me, or were there alot of fast solutions to alot of the 2005 AIME A?

For one, I don't know DeMoivre's Theorem (I don't think I do... at least, not by that name), and also, the other method is much simpler than going into the whole $\text{cis}$ thing...

[quote="4everwise"][hide="Solution"]We start off by noticing that the polynomial looks suspiciously like $(x-1)^4$.

Adding 1 to both sides, we get $x^4-4x^3+6x^2-4x+1=(x-1)^4=2006$
Thus, $(x-1)^2=-\sqrt{2006}=x^2-2x+1$, since we only want the nonreal roots.

Th sum of the roots is thus $1+ \sqrt{2006}$, since the product of the roots of a quadratic is the constant term divided by the leading coefficient.

When we apply the greatest integer function, our answer is $45$.[/hide][/quote]
how did u think of that? :D

The coefficients of the given polynomial look like the fourth row of Pascal's triangle, i.e. the coefficients of $(x-1)^4$, so we add $1$ and factor to get $(x-1)^4=2006$. Then $x$ can be $1\pm i\sqrt[4]{2006}$ or $1\pm\sqrt[4]{2006}$. So $$P=(1-i\sqrt[4]{2006})(1+i\sqrt[4]{2006})=1+\sqrt{2006}.$$ After some computation, we find that $\lfloor \sqrt{2006}\rfloor=44$, so the answer is $\boxed{45}$.

[b]Solution:[/b] Exploit the symmetry. Notice that $x^4-4x^3+6x^2-4x$ looks oddly like $x^4-4x^3+6x^2-4x+1 = (x-1)^4$. So making that adjustment, we have that $(x-1)^4 = 2006 \Rightarrow x = 1 \pm \sqrt[4]{2006}$. Clearly, these will be the only real roots. So using Vieta's, we have that $P = \frac{2005}{\sqrt{2006} - 1} \Rightarrow \lfloor{P}\rfloor = \boxed{045}$.


How many non-negative integers are less than $ 10\pi$?

That's just 31.4159...

so ans is 32, because we include 0.


geometry
geometric transformation
homothety
rotation
geometry unsolved

P. point of the triangle abc, so measuring the inside corner PAC equal measure angle PBC.     D is the middle of [AB].     L is projected on the vertical told P. (BC), M is projected on the vertical told P. (AC).     See chart below.     The DM=DL.

[quote="ashrafmod"]Given are $\triangle ABC$ and a interior point $P\ :$ $\begin{array}{c}\widehat{PAC}\equiv\widehat{PBC}\\ D\in AB\ ,\ DA=DB\\ L\in CB\ ,\ PL\perp CB\\ M\in CA\ ,\ PM\perp CA\end{array}$ $\implies DM=DL\ .$[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=61873 applied to the triangle $APB\ .$

let's do some of darij's work. :lol:

first I restate the problem.

[b]Problem[/b].[color=blue]in cyclic quadrilareal $ABCD$ let $P$ be intersection of $AC$ and $BD$ and let $H_{1},H_{2}$ be projection of $P$ on $BC$ and $AD$.
prove that $MH_{1}=MH_{2}$ where $M$ is midpoint of $CD$[/color]

[i]proof.[/i]let $\angle BCA=\angle BDA=\alpha$
and let $\chi_{\alpha}^{k}(O)$ denote a spiral homothecy around center $O$ angle $\alpha$ and $k$ be the ratio.
let $f(\mathcal A)=(\chi_{\alpha}^{\frac{DH_{2}}{DP}}\circ \chi_{\alpha}^{\frac{CP}{CH_{1}}})(O) \left(\mathcal A \right)$

hence we have $f(H_{1})=H_{2}$ and $f(M)=M$ so it means $M$ is a constant point under this function.

and cause $\frac{CP}{CH_{1}}.\frac{CH_{2}}{DP}=1$ hence $f(\mathcal{A})$ is a rotation  around center $M$.

hence $MH_{1}=MH_{2}$


geometry
superior algebra
superior algebra unsolved

Look at Liu: Algebraic Geometry and Arithmetic Curves pages 28-29.


Can somebody please post some programs that you have made on a TI-83 plus calculator? :)

Well, I suppose I can share a very simple one, that divides a number into primes.
I'm recalling this one out of my memory, so probably it will have a few bugs.

Disp "Which number:"
Input A
B=2
Lbl Q
If A=1
Quit (I am not sure this commando exists, but some equivalent exists for sure)
If frac(Q/B)=0
Goto W
B+1-->B
Goto Q

Lbl W
Disp B
Q/B-->Q
Goto Q

What kinds do you want?  I'll post some later for sample (prime factorization, messup your calculator, radical simplifier...).

Here is a program that gives the prime factorization of a number:
NOTE:  In LAB, the L means the list thingie.
[hide="PRIMEFAC"]
:1->Y
: “Input Number? “,A
:0->dim(LAB)
:A->B
:2->X
:Goto 3
:Lbl 5
:1+X X
:If A=1:Goto 2
:If B/A>A:Goto 2
:Lbl 3
:Lbl 6
:If X=1: Goto 2
:If fPart(A/X)=0
:Then
:1+dim(LAB)->dim(LAB)
:X->LAB(Y)
:Y+1->Y
:A/X A
:If fPart(A/X)≠ 0:Goto 5
:Goto 6
:End
:Goto 5
:Lbl 2
:Disp LAB
:Delvar LAB
[/hide]

Here is a harmless program that changes the numbers you put in to incorrect answers.  Press ON to exit at any time.

[hide="MESSUP"]
:ClrHome
:While 1
:Input ““,A
:randInt(1,4) X
:randInt(-10,10)  B
:If X=1:Disp AB
:If X=2:Disp B+A
:If X=3:Disp A-B
:If X=4:Disp A/B
:End
[/hide]

Tell me some other types of programs you might want.

I have some games that I programmed, but they are too big to post it here(10814 bytes are considered to be big, right?). :D

Sorry for this double post. Below is a program that draws a fractal tree. It asks you to input O, which needs to be a positive integer. below 5 is not that interesting, and above 10 takes really long time to run it.

[hide="PROGRAM: TREE"]
: ClrHome
: Output(2,3,”FRACTAL
: Output(3,3,”TREE
: Output(4,3,”GENERATOR
: Output(7,2,”BY DAVID RHEE
: Output(8,2,”[PRESS ENTER]
: Pause
: ClrHome
: Prompt O
: Degree
: GridOff
: AxesOff
: ClrDraw
: FnOff
: PlotsOff
: Zstandard
: Zsquare
: 0→X
: -10→Y
: 4.5→Z
: 90→θ
: Prgm BRANCH
[/hide]

[hide="PROGRAM: BRANCH"]
: Line(X,Y,X+Zcos(θ),Y+Zsin(θ
: If O=0
: Return
: O-1→O
: X→L₃(dim(L₃)+1
: Y→L₄(dim(L₄)+1
: X+Zcos(θ)→X
: Y+Zsin(θ)→Y
: Z→L₁(dim(L₁)+1
: θ→L₂(dim(L₂)+1
: Z(.7+.2rand)→Z
: θ+20rand+20→θ
: Prgm BRANCH
: L₁(dim(L₁)+1)→Z
: L₂(dim(L₂)+1)→θ
: Z(.7+.2rand)→Z
: θ-20rand-20→θ
: Prgm BRANCH
: L₁(dim(L₁))→Z
: L₂(dim(L₂))→Z
: L₃(dim(L₃))→Z
: L₄(dim(L₄))→Z
: dim(L₁)-1→dim(L₁
: dim(L₂)-1→dim(L₂
: dim(L₃)-1→dim(L₃
: dim(L₄)-1→dim(L₄
: O+1→O
[/hide]

_____________________________________________________
EDITED: Sorry. It was missing about 5 lines of the program. :blush:

I posted one in my previous post that gives the prime factorization of any number.  I edited it to make the lists work better.  If I have some spare time, I could post my own version of Hobo Hunt 5, but because it is about 10,000 bytes, it would be a huge post.

How do you get into assembly language?  I see commands in the catalog that relate to it, but nothing else.

[quote="1234567890"]How about a prime factorization thingie? I always make errors in factoring.[/quote]

 The one I made uses a picture of primes(?). I generated a picture so that the p th pixel is off iff p is prime or 1. This one picture stores primes up to 5985, which causes the program to run faster.

This is the program to generate such picture. It takes about 10 minutes to run it. After running this program once, it can be deleted.
[hide="PROGRAM: PRIME"]
: GridOff
: AxesOff
: ClrDraw
: FnOff
: PlotsOff
: {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71}→L₁
: For(A,1,20,1
: For(B, L₁(A),iPart(5985/ L₁(A)),1
: Pxl-On(iPart((L₁(A)B-1)/95),round(95fPart((L₁(A)B-1)/95
: End
: End
: UnArchive Pic1
: StorePic Pic1
: Archive Pic1
[/hide]

Here is my program to factor a positive integer using the picture.
[hide="PROGRAM: FACTOR"]
: Clrdraw
: GridOff
: AxesOff
: PlotsOff
: FnOff
: UnArchive Pic1
: RecallPic Pic1
: Archive Pic1
: Lbl 0
: ClrHome
: Prompt A
: If A<1 or 0≠fPart(A
: Goto 0
: ClrHome
: Output(1,1,A
: If A=1
: Then
: Output(2,1,1
: Pause
: Goto 0
: End
: If A≤5985
: Then
: If not(pxl-Test(iPart((A-1)/95),round(95fPart((A-1)/95
: Then
: Ouput(2,1,A
: Pause
: Goto 0
: End
: End
: 1→Y
: 16→X
: 0→C
: While 0=fPart(A/2
: C+1→C
: A/2→A
: End
: If C
: Then
: Output(1,16,”*2
: X+2→X
: prgm L
: If C≠1
: Then
: Output(Y,X,”^
: X+1→X
: prgm L
: Output(Y,X,C
: X+iPart(log(C))+1→X
: prgm L
: End
: End
: 3→B
: Lbl 1
: 0→C
: While 0=fPart(A/B
: C+1→C
; A/B→A
: End
: If C
: Then
: Output(Y,X,”*
: X+1→X
: prgm L
: Output(Y,X,B
: X+iPart(log(B))+1→X
: prgm L
: If C≠1
: Then
: Output(Y,X,”^
: X+1→X
: prgm L
: Output(Y,X,C
: X+iPart(log(C))+1→X
: prgm L
: End
: End
: B+2→B
: If B>5985
: Goto 3
: While pxl-Test(iPart((B-1)/95),round(95fPart((B-1)/95
: B+2→B
: If B>5985
: Goto 3
: End
: Lbl 3
: If B²>A
: Goto 2
: Output (8,1,B
: Goto 1
: Lbl 2
: If A≠1
: Then
: Output(Y,X,”*
: X+1→X
: prgm L
: Output(Y,X,A
: End
: Output(1,16,” “
: Pause
: Goto 0
[/hide]

Below is a program that prgm FACTOR needs. This program changes lines(this task is also used in a lot of other programs, so I made it into a program.)
[hide="PROGRAM: L"]
While X>16
X-16→X
Y+1→Y
End
[/hide]


AMC
AMC 12
AMC 12 A
AIME
search
MATHCOUNTS
Putnam

What did everyone get?
I got 108.
Coudn't do any of the last four.

I got a 103.5 (15 right, 9 skipped)
Not bad for an 8th grader :D

My reaction to the last 4 problems was :stretcher:

Yea I didn't even look at 2 of them.
Can't believe I didn't realize you could add the three equations on 17!!!

Well...I'd rather not say...let's just say failed...At least I won't have to go eat dinner during the actual AMC12...

I failed very badly, with like 5 stupid mistakes.
I'm in 7th grade though, so I still have time to get better.

[quote="firecricket91"]oh ok..

but technically you couldve made this like 120 minutes instead of 75 minutes.

Because if the point of it was for people to learn how to do knew problems, then time shouldnt be such a constraint.[/quote]

True. But I wanted people to at least attempt as if they were taking real AMC.  :D

Well, thanks for everything, SilverFalcon.
It was great practice.

I agree. Hopefully the real AMC will be a bit easier  :P

Very well done... though very difficult.

[quote="firecricket91"]Well no duh you guys failed.

Lets see here...

#12 is the exact problem from 1990 AIME #9
#16 is the exact problem from 1984 AIME #11
#19 was an old AIME problem (it was also one of the 10-15 ones)
#20 is the exact problem from 1988 AIME #12

I know for a fact that #21 was the 1991 AHSME #30
Also I that #14 was the 1990 AHSME #30
#17 was also an old AHSME #30 (forgot which year-just search it)

#9 was an old AHSME problem (like 20-30 problem range)
#7 was an old MATHCOUNTS national problem
Even #1-I would consider that like a #10 on a regular AMC 12

I dont know where #22-25 came from but if the ones before that were considered to be very hard AIME problem, then obviously this test is not AMC level.  Infact its problaby harder than an AIME test (the last 15 questions can easily be used to make an AIME)

Thats why I didnt do this AMC because I couldve just easily done some old AIME problems and old ASHME/AMC 12 #25-30s.

Also if you got 120+ on this test, you probably can get over 8 on AIME easily considering this was only 75 mins.[/quote]
#1 was like #23 on 1993 National Mathcounts (might be wrong).

[quote="Silverfalcon"]firecricket91, what you say is true but let me tell you this.  This exam was hard, yes, and I did use a lot of AIME problem and one question (one contestant recognized it!) was from Putnam exam. So technically, this exam fails as a mock AMC exam.

But I made it like this so that people can learn from taking this exam.  This was NEVER meant to be an exam where you can get your approximate scores.  Those were my past mock exams (you can check some of them out on AoPS wiki). This one, on the other hand, meant to provide challenges, and if you broke 100, you probably can get at least 120 on real AMC.

Knowing this, let's not say that we "failed" this test since it was meant to be very difficult. I'm going to post solutions to all problems eventually (hopefully before real AMC) so make sure you read them! They'll be good review as I tried to be pretty thorough.[/quote]

yeah, thanks silverfalcon. now we know to be ready for anything (like a mock amc that's more like a mock aime). it was cool though, with a great variety of problems.

Great Problems... I worked on them today and timed out 25 minutes.  Here's my breakdown:

18 Correct
6 Blank
None Wrong
13 Was a Freebie

Assuming 13 was 6 points, I got a 123.  After I stopped timing myself though, I started to try the last three problems, which I didn't even get to look at.  I was able to do the last one pretty easily.  I haven't started 23 or 24 yet...

Anyway, again, great problems.  I haven't seen most of these questions (I think I counted 2 of them that I recognized... I haven't worked through too many former AIME problems).

If 13 was a freebie I got 114, if its just worth 1.5 as a blank answer I got 109.5
(17-0-8 or 16-0-9)...If I got like 1 more hour I think I mightve gotten like 3 more correct but I am really slow and I have to work on my time (good thing AIME is 3 hours)

Yea this was really hard AMC especially the last 3 or so geometry problems which I didnt even look at, but now you say that they are putnam and AIME probelms I feel better.


analytic geometry
graphing lines
slope
AMC
AIME
calculus
conics

A car travels due east at $\frac 23$ mile per minute on a long, straight road.  At the same time, a circular storm, whose radius is 51 miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute.  At time $t=0,$ the center of the storm is 110 miles due north of the car.  At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle.  Find $\frac 12(t_1+t_2).$

[hide]
Just for convenience, I will solve using the center of the storm as the origin of cartesian coordinate system (in the storm's reference point)
The horizontal distance between car and storm = x
The vertical distance between car and storm = y
$x_0 = 0 \text{ mi}$, $y_0=110 \text{ mi}$
assuming that the storm moves $45^o$ below the x-axis,
the horizontal component of storm's speed = $\frac{1}{2}$
the vertical component of storm's speed = $\frac{1}{2}$

$\frac{dy}{dt}=\frac{1}{2}-0$
$\text{(rate of change of y)}=\text{(difference between vertical components of velocities)}$
$\frac{dx}{dt}=\frac{2}{3}-\frac{1}{2}$
$\text{(rate of change of x)}=\text{(difference between horizontal components of velocities)}$
$\frac{dy}{dx}=3$
$\text{(slope of path of car relative to the storm)}$

Since the car is 110 mi below the storm originally, the equations to solve are:
$x^2+y^2=51^2$
$y=3x-110$
(solving these equations for x will give us the horizontal distances at which the path traveled by the car intersects the storm's circumference)

Substituting, we get $10x^2-660x+7499=0$
Luckily we are not asked for the exact times, because that would be ugly..
Instead, we find the sum of the distances by adding the two roots together.
$x=\frac{-(-660)\pm\sqrt{(-660)^2-4(10)(7499)}}{2(10)}$
$\frac{x_1+x_2}{2}=33\text{ mi}$
The time that it takes to travel this amount of horizontal distance is (using d=rt)
$\frac{x_1+x_2}{2}=33=\frac{1}{6}\frac{t_1+t_2}{2}$
$\frac{t_1+t_2}{2}=\boxed{198}\text{ min}$

...I hope I did not make any mistakes this time...Well, at least it is a 3 digit integer, since it's AIME  :|

[/hide]

this can be solved completly without using any calculus (i guess u could just say $\Delta$ instead of $d$), like every problem in this forum.

i used the starting point of the car as the orgin and east/ south etc. normally configured.

that way the point $C$ at which the car is at the time $t$ is $C(\frac23 t|0)$ and the equation of the circle is $(x-\frac12t)^2+(y-(51-\frac12t))^2 = 51^2$.
by plugging in $x$ and $y$ one gets:
\[ 10t^2-3960t+341964=10(t-198)^2-50076=0 \]

therefore the center of that parabola is at $t=198$ ...

[hide="a simpler solution"]
Using coordinates, we see the location of the car is $(\frac 2 3 t, 0)$ and the location of the storm is $(\frac 1 2 t, 110 - \frac 1 2 t)$ at time $t$.  Thus, by distance formula, when the car is in the storm, we have $\left( \frac 1 6 t \right)^2 + \left( 110 - \frac 1 2 t \right) ^2 \le 51^2$.  Since we want the times where the car enters and leaves, we just want the roots of $t$.  And the answer wants the sum of the roots, which by expanding and using vieta, gives $t_1 + t_2 = 396 \implies \frac{t_1 + t_2}{2} = \boxed{198}$
Interestingly enough, this value would be the same for any radius of the storm.
[/hide]

I let the storm be a circle of radius $51$ centered at the origin, and let the car's path be a line starting at $(0, 110)$, with slope 3.

A car travels due east at $ 2/3$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $ 51$ miles, moves southeast at $ 1/2 \sqrt2$ mile per minute. At time $ t \equal{} 0$, the center of the storm is $ 110$ miles due north of the car. At time $ t \equal{} tsub1$ minutes, the car enters the storm circle, and at time $ t \equal{} tsub2$ minutes, the car leaves the storm circle. Find $ 1/2(tsub1 \plus{} tsub2)$

Please bear with my inability to use Latex, or something. I'm still fairly new to this site.

All right then, so what I thought of doing was creating a diagram in which the storm is $ 110$ miles directly above the car. Therefore, the edge of the storm is only $ 59$ miles above the car at this point. Because its moving $ 1/2 \sqrt2$ mph, it must be moving $ 1/2$ miles southward every minute.

Okay, so if its $ 59$ miles above, then it should take $ 118$ minutes to reach the car right?

In that time, the storm will have also moved $ 59$ miles east, and the car will have moved $ 78 \plus{} 2/3$ miles east. So, then I just calculated how much time it would take for the car to move atleast $ 51$ miles east of the storm so that it cannot be inside it (because the radius of the storm is 51).

Well, the overtake speed of the car is $ 1/6$ miles per minute ($ 2/3 \minus{} 1/2$). When the car enters the storm, the car will have already progressed $ 78 \plus{} 2/3 \minus{} 59 \equal{} 19 1/3$ miles east of the storm's center. It needs only to go $ 51 \minus{} 19 \plus{} 1/3 \equal{} 31 \plus{} 1/3$ miles more. At the overtake speed, this will take $ 188$ minutes. Therefore, if the car stays in the storm for $ 188$ minutes, I find that the answer is $ 94$

Well, if I did anything wrong or assumed something (and I'm sure I did), I hope someone will tell me. Thanks for the help in advance

lots of mistakes...
[quote]
Okay, so if its 59 miles above, then it should take 118 minutes to reach the car right?[/quote]

No.  Keep in mind that the storm is circular, not square.  The southern edge of the storm reaches the x-axis after 118 minutes.  But the car has a larger x-velocity than the storm, and thus does not remain directly below the storm.

Looking at the rest of your math, it seems you assumed a square storm.  The storm is a circle - as it moves south, the distance along the line y=0 varies.  The car does not have to end up 51 miles east of the center (i.e., all 51 miles in the x-direction); rather, the car just has to be 51 miles from the center of the storm - a component of this 51 will be in the y direction.

btw, quick latex tutorial:
a_b produces $ a_b$
a^b produces $ a^b$
\frac{a}{b} produces $ \frac{a}{b}$

one last example:  to write a sub 22 we would do a_{22} which produces $ a_{22}$, not a_22 which produces $ a_22$ - latex doesn't recognize anything past the first character after a _ or ^ as in the exponent/subscript unless you use the braces to group them in.


n th term of an AP series is 1/2 n(n+1). find sum of n terms of the series.

[hide]The $n$th term is $n\triangle$, so the sum of $n$ terms would be $\frac{n(n+1)(n+2)}{6}$.[/hide]


geometry
3D geometry
sphere
Putnam
linear algebra
matrix
induction

This problem is inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244052]Putnam 2008 B3.[/url]

For $ 1\le m\le n$ suppose that $ A$ is an $ n\times m$ real matrix such that $ A^TA \equal{} r^2I$ for some real $ r$ and also each diagonal entry of $ AA^T$ is less than or equal to $ 1.$

a) Prove that $ r^2\le\frac {n}{m}.$

b) Conjecture: is it true that for all $ m$ and $ n,$ there exists such a matrix with $ r^2 \equal{} \frac {n}{m}\,?$

(The problem asked on the Putnam is equivalent to this for $ n\equal{}4,m\equal{}2.$ We're convinced that the conjecture is true for $ m\equal{}2$ and all $ n,$ so the question is now really about $ m\ge 3.$)

I'll throw in the proof of part a), which is quite short.

Since $ A^TA \equal{} r^2I$ and $ A^TA$ is $ m\times m,$ then $ \text{trace}\,A^TA \equal{} mr^2.$

If every diagonal element of $ AA^T$ is smaller than $ 1$ and $ AA^T$ is $ n\times n,$ then $ \text{trace}\,AA^T\le n.$

$ \text{trace}\,A^TA \equal{} \text{trace}\,AA^T$ so $ mr^2\le n$ or $ r^2\le\frac {n}{m}.$

This is something I found online from Noam Elkies:

[quote]
What's the largest circle that will fit into a unit cube? A tough question, suggested by the recent Putnam problem that asked the same question in a hypercube. It turns out that there's a nice answer even for the general question of the largest k-dimensional ball that will fit into an n-dimensional hypercube of unit side --- in each case the maximal diameter is the square root of n/k --- but the proof is beyond our scope.
[/quote]

So your conjecture does indeed appear to be correct, although that doesn't help at all with the proof.

(This is from near the bottom of http://www.math.harvard.edu/~elkies/FS24i.08/)

I have no proof, but found an equivalent question that might be easier to solve. I don't know if it helps, but I guess won't hurt either.

Does there exist an orthonormal basis $ u_1,...,u_m$ of an m-dimensional subspace of $ \mathbb{R}^n$ such that $ \sum_{k \equal{} 1}^{n}(u_k)_i^2 \equal{} \frac {m}{n}$ for all $ i \equal{} 1,...,n$?
(where $ (u_k)_i$ denotes the $ i$th component of $ u_k$)

Kent Merryfield, would you mind posting the general example you found for m=2?

In their [url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2008s.pdf]solutions document[/url], Kedlaya and Ng present an argument due to Dani Kane.  See the remark after the solution to B3.

[quote="Peter"]Kent Merryfield, would you mind posting the general example you found for m=2?[/quote]
For $ m\equal{}2,n\equal{}2,$ let $ A\equal{}I.$ This works for all $ m\equal{}n$ cases.

For $ m\equal{}2,n\equal{}3,$ let $ A\equal{}\begin{bmatrix}\frac{\sqrt{3}}2&\frac12\\ \minus{}\frac{\sqrt{3}}2&\frac12\\ 0&\minus{}1\end{bmatrix}.$

Then $ A^TA\equal{}\begin{bmatrix}\frac32&0\\ 0&\frac32\end{bmatrix}$ and $ AA^T\equal{}\begin{bmatrix}1&\minus{}\frac12&\minus{}\frac12\\ \minus{}\frac12&1&\minus{}\frac12\\ \minus{}\frac12&\minus{}\frac12&1\end{bmatrix}.$

Geometrically, this corresponds to placing a circle in a cube in the plane orthogonal to $ (1,1,1)$ through the center. If you slice the cube with that plane, you get a regular hexagon, and a circle inscribed in that regular hexagon has radius $ \sqrt{\frac32}.$

Now, suppose we have an $ A$ that is maximal for this theorem for some $ n$ and $ m\equal{}2.$ That is, $ A$ is $ n\times 2,$ $ A^TA\equal{}\frac n2\,I,$ and every diagonal entry fo $ AA^T$ equals $ 1.$

Let $ B$ be the $ (n\plus{}2)\times 2$ matrix $ B\equal{}\begin{pmatrix}A\\I\end{pmatrix}.$

Then $ B^TB\equal{}A^TA\plus{}I\equal{}\frac{n\plus{}2}2\,I$ and $ BB^T\equal{}\begin{pmatrix}AA^T&A\\ A^T&I\end{pmatrix},$ which satisfies our conditions.

The sets up an induction in which we eventually come down to the $ 2\times 2$ identity or the $ 3\times 2$ case.

This was all laid out in the exchange between me and Albanian Eagle on the original topic, although not this explicitly.


geometry
incenter
circumcircle
trigonometry
geometric transformation
reflection
geometry proposed

Let $ ABC$ be triangle. Let incenter,circumcenter of triangle$ ABC$ be $ I$,$ O$. line $ l$ is perpendicular to $ IO$ and pass$ I$. $ l$ meet external bisect of angle$ A$ at $ P$ and $ BC$ at $ Q$. Prove that $ PI: IQ=2: 1$

[quote="Leonhard Euler"][color=darkred]Let $ I,O$ be incenter and circumcenter of $ \triangle BC$. The line $ d$ for which $ d\perp IO$ , $ I\in d$ meet external bisector of angle $ A$ at $ P$ and $ BC$ at $ Q$.
Prove that $ IP=2\cdot IQ$.[/color][/quote]
[color=darkred][size=134][b]Nice problem ![/b][/size][/color]

[color=darkblue] [b]Proof I[/b] ([u]metric[/u]). Suppose w.l.o.g. $ b>c$. Denote $ \{\begin{array}{c}D\in (BC)\ ,\ ID\perp BC\\\ M\in (BC)\ ,\ MB=MC\\\ A'\in AI\cap OM\ ,\ m(\widehat{AIP})=\phi\end{array}$. Observe that $ \{\begin{array}{c}IA\cdot\sin\frac{A}{2}=r\\\\ 2\cdot DM=b-c\end{array}$.

$ 1\blacktriangleright$ Apply the Sinus' theorem in $ \triangle AIO$ : $ \frac{AO}{\sin\widehat{AIO}}=\frac{IO}{\sin\widehat{IAO}}$ $ \implies$ $ \frac{R}{\sin (90^{\circ}\pm \phi )}=\frac{IO}{\sin\frac{B-C}{2}}=\frac{2\cdot IO\cdot\cos\frac{B+C}{2}}{\sin B-\sin C}=\frac{2\cdot IO\cdot\sin\frac{A}{2}}{b-c}\cdot 2R$ $ \implies$
$ \cos\phi =\frac{b-c}{4\cdot IO\cdot \sin\frac{A}{2}}$. Observe that $ IP=\frac{IA}{\cos\phi}$. Thus, $ IP=\frac{4\cdot IA\cdot IO\cdot \sin\frac{A}{2}}{b-c}$ $ \implies$ $ \implies$ $ \boxed{IP=\frac{4r}{b-c}\cdot IO}\ \ (1)$.

$ 2\blacktriangleright$ Observe that $ \tan\widehat{DMI}=\frac{ID}{DM}=\frac{2r}{b-c}$ and the quadrilateral $ IOMQ$ is cyclically, i.e. $ \widehat{QOI}\equiv\widehat{QMI}\equiv\ (180^{\circ}-)\widehat{DMI}$. Thus,
from $ \triangle QOI$ ($ IO\perp IQ$) obtain $ IQ=IO\cdot\tan |\widehat{QOI}|=IO\cdot\frac{2r}{b-c}$, i.e. $ \boxed{IQ=\frac{2r}{b-c}\cdot IO}\ \ (2)$. From the relations $ (1)$ and $ (2)$ obtain $ IP=2\cdot IQ$.

[b]Proof II[/b] ([u]synthetic - [b]April's idea[/b][/u]). Denote the second intersections $ A'$, $ B'$, $ C'$ of the circumcircle with the lines $ IA$, $ IB$, $ IC$ respectively.

Is well-known or prove easily that the point $ A$ is the reflection of the point $ I$ w.r.t. the line $ B'C'$  ($ B'I=B'A$, $ C'I=C'A$ and $ B'C'\perp AI$).

Thus, $ B'C'\parallel AP$ and $ RP=RI$. Apply the [b]theorem of the butterfly[/b] : $ \{\begin{array}{c}d\perp IO\ ,\ I\in d\\\ I\in BB'\cap CC'\\\ R\in B'C'\cap d\ ,\ Q\in BC\cap d\end{array}\|$ $ \implies$ $ IR=IQ$.

Therefore, $ PR=RI=IQ$, i.e. $ \boxed{IP=2\cdot IQ}$.[/color]

[hide="Hint."] Use Butterfly Theorem  :wink: [/hide]


In the plane there is a close broken line with $ 2006$ vertices(It is a continuous line with $ 2006$ vertices which ends at the same point it starts), no $ 3$ collinear and no $ 3$ sides concurrent; find the maximum number of intersection points of the line.

can you post a solution, please(because i cant see your archive). I think the answer is $ \frac {2(2003) \plus{} 2004(2002)}{2}$ because i found an example with that but i fail in prove that is the maximum

I feel I do not understand the question completely.Also I am not aware of Erdos's result relating to this.I would like to clarify whether my understanding of the problem is correct. Does [i]a[/i] broken line with 4 vertices is this?(pic)
http://www.mathlinks.ro/viewtopic.php?mode=attach&id=11451

In case my understanding is correct, I have the solution.
Observe that a broken line with 4 vertices can have atmost one intersection. A broken line with $ n$ vertices can also be viewed as end to end join of $ n\minus{}1$ line segments. If I add another vertex so as to maximize the number of intersection points, the line segment joining $ n$th point and the $ n\plus{}1$th point cuts through all the previous line segments except the line joining $ n\minus{}1$th point and $ n$th point, thus generating $ n\minus{}2$ points.

$ f(n)$ denotes the max number of intersections of a broken lines with n vertices.
$ f(n\plus{}1)$=$ f(n)$+$ (n\minus{}2)$   with $ f(3)\equal{}0$

Clearly, $ f(n)$= $ \sum (n\minus{}3)$
$ f(2006)\equal{}2003.2004/2$

srikanth, the example you gave is correct(is a broken line, but no a closed broken line) but the problem say a closed broken line, in you example if you join v4 and v1, that is a close broken line

Thank you for the clarification Cuenca. Here is the solution.

Continuing the argument from my previous post
Let $ f(n)$ be the maximum number of intersections of a open broken line with n vertices.
Let $ g(n)$ be the maximum number of intersections of a closed broken line with n vertices.

Then we have the relation,
$ g(n)$ = $ f(n)$ + $ n\minus{}3$
When $ n\plus{}1$ th point is introduced (in a open broken line of n vertices) , it coincides with point $ 1$. It cuts all the line segments except $ 1\minus{}\minus{}2$ , $ (n\minus{}1)\minus{}\minus{}n$.

$ g(n)$= $ (n\minus{}3)(n\minus{}2)/2$ + $ n\minus{}3$

(qed?)


While working on something I stumbled across a certain problem dealing with Combinatorics. I have been unable to solve it myself.
It is the following:

Find the general solution of the following recursion:

$ F(n,k) \equal{}\begin{cases} &F(n\minus{}1,k\minus{}1) \plus{} k \cdot F(n\minus{}1,k)  \, \, \qquad \text{if } 2 \leq k \leq n\plus{}1\\
&0 \quad \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{if } k > n\plus{}1
\end{cases}$
where:
$ F(n,1) \equal{} 1$,
$ F(n,n\plus{}1) \equal{} 1$.

Could anyone help me out with this?

These are the [url=http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Recurrence_relation]Stirling numbers of the second kind[/url].

Interesting. I sort of know those, but not that they satisfied this recursion.

Then we have:

Let $ 0 < p < 1$ be given.
If $ U_n : \equal{} \sum\limits_{k \equal{} 0}^{\infty} k^n p^k$, then

$ U_{n\plus{}1} \equal{} \sum_{k \equal{} 1}^{n \plus{} 1} S(n,k) \frac {k! p^k}{(1 \minus{} p)^{k \plus{} 1}}$,

if I am not mistaken with indices.

Correct.  This identity is equivalent to the identity $ k^n \equal{} \sum_{r \equal{} 0}^{k} S(n, r) {k \choose r} r!$ together with the identity

$ \sum_{k\equal{}0}^{\infty} {k \choose r} p^k \equal{} \frac{p^r}{(1 \minus{} p)^{r\plus{}1}}$.


modular arithmetic
number theory
relatively prime
Diophantine Equations

Find a pair of relatively prime four digit natural numbers $A$ and $B$ such that for all natural numbers $m$ and $n$, $\vert A^m -B^n \vert \ge 400$.

We can also use $ A\equal{}2001,B\equal{}1000$ since $ A^m\minus{}B^n\equiv 1001\pmod{3000}$.


[quote="Rijul saini"]Let $ P(x)$ be a polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x)$ and $ R(x)$, again with integer coefficients, such that
[b](i)[/b] $ P(x) \cdot Q(x)$ is a polynomial in $ x^2$ , and
[b](ii)[/b] $ P(x) \cdot R(x)$ is a polynomial in $ x^3$.[/quote]

Let $ P(x)\equal{}\sum a_mx^m$

[b](i)[/b] Let $ Q(x)\equal{}\sum a_{2j}x^{2j} \minus{} \sum a_{2k\plus{}1}x^{2k\plus{}1}$
Then $ P(x) \cdot Q(x)\equal{}(\sum a_{2j}x^{2j})^2 \minus{} (\sum a_{2k\plus{}1}x^{2k\plus{}1})^2$ is a polynomial in $ x^2$

[b](ii)[/b] We have $ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$
Now put $ a\equal{}\sum a_{3i}x^{3i}, b\equal{}\sum a_{3j\plus{}1}x^{3j\plus{}1}, c\equal{}\sum a_{3k\plus{}2}x^{3k\plus{}2}$ we have $ P(x)\equal{}a\plus{}b\plus{}c$
and $ a^3\plus{}b^3\plus{}c^3\minus{}3abc$ is a polynomial in $ x^3$. So put $ R(x)\equal{}a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca$ and we are done.


[quote="IndianaMan"]I just signed up for an account after I learned about this contest from a couple of my friends.  However, I did not even do rounds 1,2, or 3.  Will there be any benefit for doing round 4?  I know that it will definently help me and is going to be fun, but is it possible to win any prizes?

Thanks[/quote]
http://www.usamts.org/Rules/U_RulesPrizes.php

You need 40 points to get an Honorable Mention, so one round isn't enough.

[quote="IndianaMan"]I just signed up for an account after I learned about this contest from a couple of my friends.  However, I did not even do rounds 1,2, or 3.  Will there be any benefit for doing round 4?  I know that it will definently help me and is going to be fun, but is it possible to win any prizes?

Thanks[/quote]

Unfortunately, no. The minimum score to recieve a prize is 40 points, which cannot be achieved in one round, which has a maximum of 25 points.

But do the problems anyway! It will be a great experience and practice for next year.

Doing the problems just for fun would definitely be a good experience for you. It'll help you get in the good mood for doing well next year, simply to familiarize yourself with USAMTS, how to write rigorous solutions, and how to be creative in writing them. It's really a good contest, in my opinion, because it like stresses creative thinking and problem-solving strategies. Even if you don't win, you'd still enjoy the experience of at least doing the problems, because most importantly, they are fun! :)