so the problem reads:

Find all triples (x,y,z) of positive integers such that x^3 + y^3 + z^3 = 2008.

Don't give me the answer because this is for a summer program admission problem set, but it's the last one I have to do and I'm having a lot of trouble with it.

If someone could give me a hint of where to go with this that would be amazing.

Well you have a limited amount of possible answers so listing them isn't too bad of an idea  :)

well ive listed what i believe is most, if not all of them... but i guess my real question was: how do i show that these are all of them?

(4,6,12) and all of the combinations of those numbers (6 altogether)
(10,10,2) and all of the combos of those numbers (3 altogether)

for a total of 9 solutions.

If you can prove that you have listed (and checked) them all, that qualifies as a proof (not a great one, but an airtight one, nevertheless).

Without loss of generality, if $ x \le y \le z$, then $ 3x^3 \le 2008 \implies x \le 8$.

[quote="niknar"]this is for a summer program admission problem set[/quote]

Er... asking for the answer here probably isn't a great idea, since it sort of defeats the purpose of asking a question for the purpose of admission.  Anyone else applying to the same program can see the answer here, too.

I suggest using modes to simplify the problem.

Notice:

(3n)^3 ≡ 0 (mod 9)
(3n+1)^3 ≡ 1 (mod 9)
(3n+2)^3 ≡ -1 (mod 9)

which you can easily check by expansion, and

2008 ≡ 1 (mod 9)

Therefore, your possible candidates are any arrangement of (0,0,1) or (-1,1,1)
and since x,y,z <13, you get

-1: 2,5,8,11
0: 3,6,9,12
1: 1,4,7,10

For a grand total of 80 triples, many of which you can eliminate due to being too small or too large.

[quote][quote]niknar wrote:
this is for a summer program admission problem set
[/quote]
t0rajir0u wrote:
Er... asking for the answer here probably isn't a great idea, since it sort of defeats the purpose of asking a question for the purpose of admission. Anyone else applying to the same program can see the answer here, too.

[/quote]
i agree. niknar, this is [b]definitely[/b] cheating!


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

for any map $\mathbb{R}^{n}\rightarrow \mathbb{S}^{n}$ for $n>1$ is uniformly homotopic to constant.

I don't want to worry about uniformity right now, so I'll just show that the map is homotopic to a constant map. In fact, the following generalization holds:

If $X$ is a contractible CW complex and $n\ge 2$ is a positive integer, then any map $f: X\to\mathbb S^{n}$ is homotopic to a constant map.

Proof

I'll use the following theorem (see, for example, Allen Hatcher, "Algebraic Topology", Chapter 4, Corollary 4.33):

$(*)$ If a map between simply connected CW complexes induces isomorphisms on all homology groups (with integer coefficients), then the map in question is a homotopy equivalence.

Let $X'$ be the wedge sum of $X$ and a sphere $\mathbb S^{n}$. In other words, we fix a point $x_{0}\in X$, and "glue" an $n$-sphere to $X$ in $x_{0}$. In fact, identify this new sphere with the one where $f$ takes its values, and the point $x_{0}$ with $f(x_{0})$. The map $f$ can be extended to $f: X'\to\mathbb S^{n}$ by setting $f$ to be the identity on $\mathbb S^{n}$. Both $X'$ and $\mathbb S^{n}$ are simply connected CW complexes, and $f$ defined as above certainly induces an isomorphism on homology. By $(*),\ f$ must be a homotopy equivalence. This means that we can find $g: \mathbb S^{n}\to X'$ such that $g\circ f$ is homotopic to the identity $1_{X'}$ of $X'$, and $f\circ g$ is homotopic to the identity $1_{\mathbb S^{n}}$ on $\mathbb S^{n}$.

Since $X$ is contractible and $f|\mathbb S_{n}=1_{\mathbb S_{n}}$, saying that $f\circ g$ is homotopic to the identity is the same as saying that $g$ is. Hence we may assume that $g=1_{\mathbb S_{n}}$, and $g\circ f$ being homotopic to $1_{X'}$ is now the same as $f: X'\to\mathbb S^{n}$ being homotopic to $1_{X'}$ (here, we view $\mathbb S^{n}$ as a subspace of $X'$). If $H: X'\times[0,1]\to X'$ is a homotopy from $f$ to $1_{X'}$ and $r: X'\to\mathbb S^{n}$ is the retraction of $X'$ onto $\mathbb S^{n}$ equal to the identity on $\mathbb S^{n}$ and sending $X$ to $x_{0}$, then $r\circ H$, restricted to $X\times [0,1]$, is a homotopy from $f: X\to\mathbb S^{n}$ to the constant map $X\to x_{0}$.


abstract algebra
vector
Functional Analysis
advanced fields
advanced fields unsolved

Let $E$ be a complex Banach space, and let $T$ be a bounded linear compact operator on $E$. Prove that:

(a) the kernel of $1-T$ is finite-dimensional;

(b) the range of $1-T$ is closed;

(c) the cokernel of $1-T$ (i.e. $E/(1-T)E$) is finite dimensional;

(d) the dimensions of $\ker(1-T)$ and $\mbox{coker}(1-T)$ are equal.

(a)
it suffices to show that the unit ball of $K=Ker(1-T)$ is compact. But $x$ belongs to $Ker(1-T) \cap B(0,1)_{E}$
iff $T(x)=x$ and $\parallel x\parallel \leq 1$. Hence the unit ball of $Ker(1-T)$ is included in the image $T(B_{E})$, and thus is compact.

(b)
$y_{n}=x_{n}-Tx_{n}$ a sequence of $Im(1-T)$ that converges towards $y \in E$. One has to prove that $y=X-TX$ for a certain $X \in E$.
To stay bounded and use compacity arguments, among all $x$ that satisfy $y_{n}=x-Tx$, $x_{n}$ can be choosen with minimal norm.
It is possible because $Ker(1-T)$ is finite dimensional. That means that one supposes that: $d(x_{n}, K)=\parallel x_{n}\parallel $.

Suppose that a subsequence of $x_{n}$ is bounded. Then, because $T$ is a compact operator, one can suppose (take an extraction) that $Tx_{n}$ converges
towards a vector $v$, and thus $x_{n}$ converges towards $y+v$. Then $X=y+v$ works : $y=X-TX$

So let's prove there is an extraction of $(x_{n})_{\mathbb{N}}$ that stay bounded. If it were not the case then $\parallel x_{n}\parallel \to \infty$.
Then $u_{n}=\frac{x_{n}}{\parallel x_{n}\parallel }$  would be a sequence of unit vectors such that: $u_{n}-Tu_{n}\to 0$. Because $T$ is compact
one can suppose (extraction) that $Tu_{n}$ converges and thus $u_{n}$ converges towards a unit vector $u \in K$.
But then: $d(u_{n},K)= \frac{d(x_{n},K)}{\parallel x_{n}\parallel }=\frac{\parallel x_{n}\parallel }{\parallel x_{n}\parallel }=1$. This is absurd because $d(u,K)=0$

(c)(d)
$Im(1-T)$ is closed thus $Im(1-T)=Ker((1-T)^{*})^{\bot}=Ker(1-T*)^{\bot}$
But it has already been proved on the forum that $T$ is compact iff $T^{*}$ is compact.
Thus $Ker(1-T^{*})$ is finite dimensional because of (1).

First of all, I think you're using some Hilbert space machinery (orthogonality and such). That can be repaired though. More importantly, is (d) proven here? :? You wrote (c)(d) before your last paragraph. What does that mean, exactly? :)


Can I get an explanation of the second sentence of the solution to #8 on Alegrbra test 2006:

http://web.mit.edu/hmmt/www/datafiles/solutions/2006/salg06.pdf

thanks.

Here's the original problem

$ (z^2 \plus{} 2)^2 \plus{} 2 \equal{} z$

Think of it this way

$ z' \equal{} z'^2 \plus{} 2$ where z' is some potential solution to the above quartic

substituting z = z', we see
$ (z'^2 \plus{} 2)^2 \plus{} 2 \equal{} z'$

$ (z')^2 \plus{} 2 \equal{} z'$

$ z' \equal{} z'$ which shows that values of z' are in fact solutions to the quartic

Write $ f(x) \equal{} z^2 \plus{} 2$.  The given equation is

$ f(f(x)) \equal{} x$

which is clearly satisfied by any $ z$ such that $ f(z) \equal{} z$.  This gives two of the roots; we can then divide out and find the other two.


trigonometry calculus integration derivative function logarithms algebra

I have seen people use Euler's Formula a lot but i still really don't understand its use and proof

please help me

thanks,
Dhaivat(uh huh that's my name)

Well, you could have been a little more specific: which Euler's formula is on your mind? I assume you meant $ e^{i\phi}=\textbf{cis} \phi= \cos \phi + i\sin \phi$. If this is what you wanted (there's also Euler's polyhedral formula, etc.) then I can make a few remarks.

The proof is pretty hard to understand without the basics in calculus: I've seen three/four proofs:

1. power series
2. complex integral
3. ODE
4. derivative of a complex function

Maybe the first one is the easiest for a beginner, so you can go through it (Wikipedia and MathWorld give it in brief).

The use is wide. First, you use it when multiplying and taking powers and/or roots of complex numbers, because it is the most convenient form for such actions. Later, it is useful when you need some kind of graphical representation in Argand plane: because you immediately know the polar angle and the length of the radius vector. It's easy to find use for the formula, once you get familiar with it.

Euler's formula: $ e^{i\theta} \equal{} \cos \theta \plus{} i \sin \theta$

Proof: By taylor's theorem we have are $ e^x \equal{} \sum_{k\equal{}0}^\infty \frac{x^k}{k!}$, $ \sin x \equal{} \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k x^{2k\plus{}1}}{(2k\plus{}1)!}$ and $ \cos x \equal{} \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k x^{2k}}{(2k)!}$.  If we plug $ x\equal{}i\theta$ into the series for $ e^x$ and simplify $ (i\theta)^k$, we can then collect the real part and the imaginary parts, which clearly gives $ \cos \theta \plus{} i \sin \theta$ (if you aren't convinced, try writing out a few terms)

The only issue with this proof is that you might not be familiar with Taylor's theorem or the calculus surrounding it.  Taylor's theorem: $ f(x) \equal{} \sum_{k\equal{}0}^\infty \frac{f^{(k)}(a)}{a!}(x\minus{}a)^k$ when the RHS converges and $ x$ is close enough to $ a$.  $ f^{(k)}(a)$ denotes the value of the $ k$th derivative at $ x\equal{}a$. ($ f^{(0)}$ is the original function).  In the cases of $ e^x$, $ \sin x$, and $ \cos x$, this series is equivalent to the functions everywhere.

Uses are almost endless.  Beyond complex powers of real numbers, we can also find logs of complex numbers: $ \ln(A\plus{}Bi) \equal{} 1/2\ln(A^2\plus{}B^2) \plus{} i\tan^{\minus{}1}(B/A)$ (Note that here the inverse tangent is used to mean any angle with that tangent which goes through the point $ (A,B)$).  So there are really and endless number of natural logs of each numbers, each differingby some multiple of $ 2\pi i$.

From there, logarithms with other bases can be found using the change of base formula, and complex numbers can be raised to complex powers.

Man i wish i knew calculus.....
Oooo well i am doing it next year....

Proving Euler's theorem using [i]real[/i] calculus is totally invalid since you aren't working with real functions - as I have repeatedly said, it can at best act as a motivation.  Euler's theorem is most straightforwardly viewed as a [i]definition[/i] of the expression $ e^{i \theta}$ (given that you define $ e^x$ in terms of the natural logarithm for real $ x$) compatible with the properties we want the exponential function to have.  Alternately, if you define $ e^x, \cos x, \sin x$ in terms of their power series alone, and for complex values of $ x$, this is a proof, but it assumes facts about the convergence of complex power series that don't immediately follow from the convergence of real power series.

Similarly, the proof from a differential equation assumes that you can treat $ i$ the same way you would treat a real number, which is again not a valid operation within the realm of real calculus.

only the proof requires calculus, not the use.  Anyway, the relevant derivatives are $ \frac{d}{dx}[ \cos x ] \equal{} \minus{}\sin x$, $ \frac{d}{dx}[\sin x] \equal{} \cos x$, and $ \frac{d}{dx}[e^x] \equal{} e^x$.  See [url]http://tutorial.math.lamar.edu/classes/calcI/calcI.aspx[/url] for general calc I material (you really just need to know enough to understand where the derivatives I gave came from and derivatives of polynomials) and [url]http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx[/url] for information on Taylor/Maclaurin series,

t0rajir0u, why does it matter if we are using results of real calculus on complex numbers?

Because the properties of real numbers do not necessarily carry over to the complex numbers.  For example, the conditions that a complex-valued function be differentiable are [i]much[/i] stricter than the conditions that a real-valued function be differentiable.  Without a proper definition of the complex derivative (which requires in turn a proper definition of what it means for a sequence of complex numbers to approach a limit) any proof is necessarily formal only.  In other words, we need the basics of [url=http://en.wikipedia.org/wiki/Complex_analysis]complex analysis[/url].

I can't remember but isn't there a nice geometric type proof alluded to in the Art and Craft of Problem Solving?

Isn't there also a proof involving matrices as well? Is that valid?

The most problematic part of any "proof" that doesn't explicitly invoke properties of complex-valued functions is what $ e^{ix}$ means for real values of $ x$.  One sensible way to define exponentiation on the real numbers is to define $ \ln x \equal{} \int_{1}^{x} \frac {dt}{t}$ and then to define $ e^x$ as its inverse function.  This definition doesn't carry over to the complex value $ ix$ unless we define a complex integral.

The closest you can get by real methods is to prove that $ \text{cis}(\alpha \plus{} \beta) \equal{} \text{cis } \alpha \cdot \text{cis } \beta$ (which can be used to prove de Moivre's formula), which can be done in a variety of ways and probably provides a deeper understanding of the connection between complex numbers and rotation than Euler's formula itself.  Again, this is a [i]motivation[/i] of a [i]definition[/i] of $ e^{ix}$.


AMC
AIME
USA(J)MO
USAMO
analytic geometry
geometry
MATHCOUNTS

My older daughter is really good with math -- except arithmetic. Her first couple of years of schooling were more focused on language, thanks to a new dual language program she was enrolled in, so she didn't get lots of experience with basic add/sub/mult/div operations. Short of buying her a really nice calculator, what's the best way to bring her up to speed on the basics, in a way that she won't resent?

Thanks.

well, to put it in a nutshell, i believe that your daughter is not unskilled in basic arithmetic, but rather, i believe that her faults are simply careless errors. the reason i believe this is due to the fact that math is cumulative; everything builds up to the next level and the next. you cannot simply go straight into coordinate geometry without comprehension of simple geometry. Therefore, as BOGTRO said, the only way to get your daughter to improve is through rigorous training to solidify her skills. As i said before, it is not that your daughter cannot do basic operators; its just that she may need more time to calculate such problems, and the only way to correct that is by drilling her repeatedly.

To do it in a way that she doesn't resent, I would recommend giving her AMC or MATHCOUNTS style problems that have a lot of computations embedded in them. Doing lots of arithmetic computations feels less rote when you are doing them as part of a solution to a more involved problem.

i am really good at getting ideas to solve stuff, but my algebra is, well, lets put it this way-ok. (really needs to improve, is what my dad wood say. :))

i believe you should just give her basic problems, over and over and over.

that might help.

it probobably wood help me too....

maybe i should do that.

and let her do the Math is Cool.

Search it on google

it is great.

I remember when I was in fifth grade I had the same problem (except on a higher level, I had trouble with algebraic work). I understood how to solve problems, but I never got them right because I always screwed up during the manual work. The way I eventually solved the problem was by continued practice every day.
For your daughter, algebra (two, three-variable equations, quadratics) is probably a good way to improve her computational skills because they require computation at every step. As FibonacciFan said, straight basic operations may seem mundane to her, so find problems that are within her skill level (so she thinks she can do it) and require a lot of computation. Mathcounts, Math is Cool, AMC (but probably only AMC8 or first half of AMC10) are all good sources. So are the AoPS books and algebra/geometry textbooks.

I would also recommend the [url=http://zetamac.com/arithmetic/]Arithmetic Game[/url], especially if your daughter is a competitive person.  If not, it is still a good way to practice basic arithmetic, and with enough practice, arithmetic computations should become almost second-nature to her, which is very helpful on the road to success in MathCounts and other similar competitions.

Maybe her only problem is being in an environment that expects that she conforms with certain ways of doing things. Maybe she has a lot to learn, but there are better ways to teach it. She needs to know a number of words, symbols and definitions that people around her use, in order to communicate. She needs to be able to do calculations and find her own mistakes. Algorithms and formulas do not need to be memorized. There is no absolute need to conform with pre-ordained ways of solving problems, although iit may make life easier when in certain environments.

If the problem is that her school puts a lot of emphasis on "the basics," just try to survive until she gets beyond the level where they test for that. For my children, that was up to 8th grade. After that, they stopped giving them timed "computation skills" tests. All my life I had trouble remembering the times tables. I have to think for a moment to tell you how much is 8 + 5. When I add a string of numbers I count imaginary points in each digit: the three ends in the digit 3, four ends in a number 4, and so on. I don't think pencil and paper calculation speed matters. I doubt that anything in my life will depend on calculating quickly. It never held me back in a carrer involving science and math. What you need to know is how to get to the answer (even if you have to count on your fingers and toes) and how to find your own mistakes.

Calculators help, but we should not trust them farther than we can toss them. At work, I once had to calculate a result involving the factor
$ \frac{0.723\minus{}0.002}{0.716\minus{}0.002}$, which was written just that way on paper.
In a fanatical quest for speed, my boss turned to the calculator and entered the string of symbols
0.723-0.002/0.716-0.002,
confidently continuing from there. The wrong result was not questioned and quickly reported by phone a minute before the beginning of the business day, whil I was frantically waving and mouthing "that can't be right." Speed kills; sometimes you should slow down and think.
If someone does not realize that the result should be very close to $ 1$, and does not understand order of operations as well as the calculator does, the calculator may be asked the wrong question and wrong results will be reported.

Making children sit down to memorize "math facts" is popular, but insane. The same people who advocate that kind of memorization never sat down to memorize the phone numbers and addresses of a list of emergency contacts. That is information that they may need quickly, and could never deduce. They could count on their fingers to find how much is 8 + 5, but there is no way to calculate the unlisted phone number of aunt Sophie. I do remember a lot of contacts' phone numbers, but I never sat down to memorize them; they just got engraved in my memory from repeated use. For practice with "math facts," problems involving number theory and divisibility would engage the thinking part while naturally involving arithmetic.

Another popular idea is that formulas and algorithms need to be memorized. It is important to know what symbols stand for. It is important to understand why long division is done that way. Someone figured out a convenient way to do calculations, and we are free to use it after we understand the mechanism. It is not necessary to memorize the algorithm, because it can be re-deduced. "Carrying over the one" or "re-grouping" is not something you do following the magical rules handed over by a teacher. There is a reason why the nifty finger trick works for the nine times table. If that's the problem, help your daugter find the explanations she needs. It is insulting to a reasoning creature when such math "rules" and formulas are handed out without explanation.

If your child is mathematically challenged, then send her to a program called KUMON! KUMON helped me learn almost everything. It's an organization that you should have around your city. They give you worksheets and homework that helps you learn stuff from counting dogs, to BC College Level Math! They also have a reading level. KUMON is based on a 3A-O Level. For instance, Level G is the beginning of Algebra. It made me the top of my class! Most of my credit goes to KUMON! I am currently in Level K right now and I am studying Algebra 2! I highly reccomend this program. For more info, go to http://www.kumon.org

Sincerely,

J.C.

This has to be said.  I, too, resent simple arithmetic and rote memorization at any level of education, but we must all realize that the ability to memorize and calculate is the simplest mental task.  It is easier than doing USAMO problems, so think about it logically.  If A is easier than B, and you can't do A, then how in the world are you going to do B?  If you can't walk, you can't run.

The fact is, if you can't multiply and divide at lightning speed, you should probably learn how to do so.  Ability at basic arithmetic and ability at harder math are not as mutually exclusive as most people would like to think.  Arithmetic is a great indicator of overall intelligence and thinking speed and ability.  You've got to pound some stuff in.  Take any IMO competitor.  I have no doubt that they can do calculations much faster than any normal person can.

I believe what Z is saying is that he, who came in top 12 at nats, does not attribute any of his success to Kumon, whereas you say Kumon got you far, when you did not get as far as he did.

And MC has at most 5 logic problems every year, all of which are trivial.

[color=green][Moderator says: this post is responding to a now-deleted post.  Keep it civil, everyone, and cut it out with the both the flames and the pointless talking-oneself-up, please.][/color]

Sorry. :oops:  :blush:

If anyone is interested, some discussion about KuMon is here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=103239624&t=206882[/url]

I have to say that I agree strongly with KMST and disagree with many of the responses in this post. If you understand what you're doing from an intuitive and conceptual viewpoint, the calculations will follow. The former is what's more difficult. And in the end, both in pure and even in applied math, it's what is more important. The sense I get is that she understands basic arithmetic but isn't extremely fast at it. I was never very fast with arithmetic, yet I've been extremely successful. In fact, too much focus on brute calculations actually takes emphasis off intuitive understanding, which the students begin to lose sight of. This, in turn, becomes much more of a hiderance than a help.

I got my little brother to practice and learn his multiplication tables and basic calculations by introducing him to a really cool website:

http://www.freerice.com

For every problem you get right, 10 grains of rice are donated to those in need.  He loved watching the rice add up.  There's a good chance the math there will be too basic for what you're looking for, but they also do vocabulary, some chemical symbols, and foreign languages.  It's a really cool website!  Hope it helps!

[quote="1=2"][quote="byrdgyrl"]www.freerice.com[/quote]

This website is epic.

I learned from the McGraw hill textbooks when I was 7 or 8, but that was almost 6 or 7 years ago. I was seeing how fast I could get through the books (a month? three weeks? two weeks? a week?), and before I knew it I was ready for algebra. :)

I would suggest doing that, but the books must not have been published after 2003 if you are to use the exact same books I used.[/quote]

go.hrw.com is epic.
So is freerice, but that really doesn't teach a lot of advanced math.

Er, go.hrw.com is just lots of exercises, which isn't epic at all.

Speaking of calculus, I found the book http://www.amazon.com/Calculus-Easy-Way-Douglas-Downing/dp/0812091418 to be an amazing book for a first introduction to calculus. This is the book that I learned calculus from in 7th grade.

It's fairy tale style, where a group of characters is presented with a series of mathematical problems that lead to the concepts in calculus. You watch the characters' train of thought as they try one approach after another and develop the ideas of calculus in the same way that new mathematics is made. One particular benefit is that this book allows the reader to learn calculus with enormous motivation.

Sounds a little bit like me. Of course, for all problems (real problems) there are at least two solutions. Uf she is that creative, maybe she could get away with that and slowly work on that. This is the reason why I failed my last test, which was on Matrix operations.

what I did was do a lot of simple problems such as 1634/89 or find the surface area of a certain cylinder about 100 times a day

It helped but was not fun

[quote="Quickster94"]On the same note as Calc Rulz, you may want to check out the book The Number Devil by Enzensberger:
[url]http://www.amazon.com/Number-Devil-Mathematical-Adventure/dp/0805062998/ref=sr_1_1?ie=UTF8&s=books&qid=1242445064&sr=8-1[/url]

It introduces a lot of stuff in a way children can better understand (my sister got it, wish I had known about this book when I was a kid).  It does however relabel some math terms.
See here for summary: [url]http://en.wikipedia.org/wiki/The_number_devil[/url][/quote]

i love that book, i got it as a prize

666th post!!!


How do you prove that:
$2n+1 < n^{2}\qquad \forall n\geq 3$

Here is what I tried. The base case is simple and it obviously works. Now assume the above hypothesis. Then
$2n+1 < n^{2}$
$2n+1+2n+1 < n^{2}+2n+1$
$4n+2 < (n+1)^{2}$

But now I am stuck. Can someone please help me?

[quote="Farenhajt"]Your inductive step is $(n+1)^{2}>2(n+1)+1=2n+3$. Hence you just need to prove $4n+2>2n+3$.[/quote]I see. So
$2n+3<4n+2$
$\Rightarrow 2n+3 < 2n+2n+2$
$\Rightarrow 1 < 2n$
$\Rightarrow 1/2 < n$

This is obviously true because $n \geq 3$. QED!

And a direct (non-inductive) proof is also simple:

$n^{2}>2n+1\iff n^{2}-2n-1>0\iff n^{2}-2n+1>2\iff (n-1)^{2}>2$, and that's obviously true for $n\geqslant 3$

In general if $deg(P(x))<deg(Q(x))$ then for a large enough $k$, $Q(x)>P(x) \forall x>k$


quadratics function calculus integration

f(x) is a quadratic $ ax^2 \plus{} bx \plus{} c$ with nonegative integer coefficients $ f(0) \equal{} 5$ and $ f(1) \equal{} 7$. Find the sum of all possible values of f(x).

What if f(1)=9?

Um...all you can deduce is that $ f(x)\equal{}ax^2\plus{}(2\minus{}a)x\plus{}5$. And in any case, $ f(x)$ has an infinite number of values.

Yes but if I say find sum of all posible value of f(x)=x it is clearly 0. Thus if a>0 the answer is infinity, if a<0 the answer is negative infinity.

Hm... for your first one:

Since $ f(0) \equal{} 5$, $ c$ is clearly equal to $ 5$.  There are two possible ways for $ (a,b)$ for $ f(1)$ to equal $ 7$.  These are $ (2,0)$ and $ (1,1)$.

So $ f(x) \equal{} 2x^2 \plus{} 5$ or $ x^2 \plus{} x \plus{} 5$.

For your second one, we use the same logic:

However, $ (a,b)$ can be $ (4,0)$, $ (3,1)$, $ (2,2)$, or $ (1,3)$.

Thus, you have the following functions:

$ f(x) \equal{} 4x^2 \plus{} 5$

$ f(x) \equal{} 3x^2 \plus{} x \plus{} 5$

$ f(x) \equal{} 2x^2 \plus{} 2x \plus{} 5$

$ f(x) \equal{} x^2 \plus{} 3x \plus{} 5$

*Note that $ (0,b)$ cannot be used, because then, it wouldn't be a quadratic.

[quote="math154"]Did it ever say that the coefficients are integral? :P[/quote]

Meh... list the answers for us, abacadaea. ;)


logarithms
ratio
calculus
integration
search
geometric series
calculus computations

I have a problem...

the sum of n= 2 to Infinity

((ln(n))^n)/(n^2n)

Would you use comparison test here? compare it to (1/n^2)^n, or would this be considered a geo series? any ideas I would appreciate it..

It's not a geometric series, since neither $\ln n$ nor $n^{2}$ are constants. You want to do a comparison, but you'll need to estimate that logarithm first. The eventual comparison will be to a geometric series.

Comparisons to geometric series can be automated using the ratio test or the root test; have you tried one of those?

I tried the root test, and I think the limit i get is 0. using l'hopitals rule, the top part becomes 1/n and the bottom is 2n so according to a/b/c = a/(b*c) which is infintiy times infintiy i get 1/infinity which equals 0.

but i thought the root test was only for absolute convergence, the question is just asking for convergence or divergence not absolute.

Absolute convergence implies convergence. For a series with positive terms, they're the same thing.

The root test and ratio tests are comparison tests. All comparison tests work only on positive term series, and rely on the "monotone series alternative": an increasing sequence (for instance, the sequence of partial sums of a positive term series) either converges or tends to infinity.

All comparison tests work only for positive term series, but that then means that any comparison test - including the root and ratio tests - can be used as a test for absolute convergence.

The particular series is a positive term series. For a positive term series, convergence and absolute convergence are the same question. This particular series was set up to encourage the use of the root test, and indeed the root test shows that it converges.

The following tests are comparison tests, and hence tests for absolute convergence:

Simple (one-sided) comparison
Limit comparison
Integral test
Ratio test
Root test
Condensation test (not in most mainstream calculus books but often discussed on this site - you can search for it)

So what isn't a comparison test? There are two:

The "test for divergence": that is, if the terms don't go to zero, then the series diverges. It doesn't matter what sign the terms are.

The Alternating Series Test, which is capable of proving convergence even in the absence of absolute convergence (hence capable of proving conditional convergence.)

It is possible to improve on the Alternating series test: see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121138]here[/url] for one way.

Thanks for the clarification that helps. So my above example where I found the limit after using the root test to be 0. that would imply the series is convergent. Did I do the math correctly?

Yes, the limiting root should be 0, which is less than 1.


Find the real roots of the equation $\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}}=x$ [There are n radical signs]

$0$ and $3$ hold the equation. But I don't know all roots .

you have to solve the equoition $\sqrt{x+2\sqrt{3x}}=x$
and you will find all roots :oops:

There are not another solutions. Because if $0<x<3$ $g(x)=\sqrt{3x}>x$, and if $x>3$ $g(x)<x$.

$\sqrt{1+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}=x$ (m radicals)
Replace the final x of the left member by the entire expression for x:
$\sqrt{1+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}=x$ (2m radicals)
So:
$\sqrt{1+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}=x$ (3m radicals)
Thus:
$\lim_{K\to\infty}\sqrt{1+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}=x$
(K radicals)
Then:
$x=\sqrt{x+2[\sqrt{x+2\sqrt{2\sqrt{x+...}}}]}$, look that the expression between $[ ]=x$
$x=\sqrt{x+2x}$
$x(x-3)=0$
$x_{0}=0 ; x_{1}=3$

This is the same idea of this equation:
http://www.mathlinks.ro/Forum/viewtopic.php?t=127282


prove
$ 39pqr\equal{}4q^2r^2\plus{}27p^2$
where
$ p\equal{}abc
q\equal{}a\plus{}b\plus{}c
r\equal{}ab\plus{}bc\plus{}ca$

I can't see why you post it here, and it is disproved easily: take $ a\equal{}b\equal{}c\equal{}r\equal{}1,p\equal{}q\equal{}3$.


email
vector
calculus
calculus computations

I received this as an email message:

[quote]I have been reviewing some engineering books and came across this question that I couldn't solve. I would really appreciate your help in solving this problem.

Minimize the following objective function f(x1, x2) subject to the
constraints h(x1, x2) as shown below.

f(x1, x2) = (x1- 1.5)2 +( x2-1.5)2

h(x1, x2) = x1+ x2 -2=0

Use the method of Lagrange multipliers.

Thank you very much.
[/quote]

I'm going to guess that this is the correct statement:

Minimize

$f(x_{1},x_{2})=(x_{1}-1.5)^{2}+(x_{2}-1.5)^{2}$

subject to

$h(x_{1},x_{2})=x_{1}+x_{2}-2=0.$

If the constrained minimum of $f$ is attained at $(x_{1},x_{2})$, then the gradient $\nabla f(x_{1},x_{2})$ is parallel to $\nabla g(x_{1},x_{2})$. Two vectors are parallel if one is the scalar multiple of another, i.e. $\nabla f(x_{1},x_{2})=\lambda \nabla g(x_{1},x_{2})$. Here $\lambda$ is the Lagrange multiplier, so far unknown. Equating the components of two gradients, we get two equations involving $\lambda$, $x$, and $y$. The third equation is $g(x_{1},x_{2})=0$. Bring them together
\[\begin{cases}2(x_{1}-1.5)&=\lambda \\ 2(x_{2}-1.5)&=\lambda \\ x_{1}+x_{2}-2&=0\\ \end{cases}\]
and solve: $\lambda=-1$, $x_{1}=x_{2}=1$.

P.S. The neat thing about this example is that $\lambda$ (which we don't really want to know) easily decouples:
\[\begin{cases}2(x_{1}-1.5)&=2(x_{2}-1.5) \\ x_{1}+x_{2}-2&=0\\ \end{cases}\]
Too bad this does not happen more often.


In which of the following compounds does the bond between the central atom and chlorine have the greatest ionic character?

BCl3
CCl4
CaCl2
HCl
FeCl2

Ionic character is determined by a large electronegativity between two atoms. Electronegativity differences usually increase as atoms are farther apart on the periodic table. I think you can reason the answer based on that information

sorry, i did not see HCl was there. I think it would be HCl.

On the other hand, CaCl2 exits in the solid state as a stable net of ions, while HCl never does.

I still remember when a fellow student threw concentrated HCl into the trough, and my teacher had to tape up the smoke alarm to prevent it from sounding.  :P


limit
logarithms
calculus
integration
calculus computations

hello lovelies.  :D

can you find

$ \sqrt {\; \dfrac{2}{1}\; }\cdot \sqrt [3]{\; \dfrac{2^2}{1\cdot 3}\; }
\cdot \sqrt [4]{\; \dfrac{2^3\cdot 4}{1\cdot 3^3}\; }\cdot \sqrt [5]{\; \dfrac{2^4\cdot 4^4}{1\cdot 3^6\cdot 5}\; }\cdot \sqrt [6]{\; \dfrac{2^5\cdot 4^{10}\cdot 6}{1\cdot 3^{10}\cdot 5^5}\; }\cdot \sqrt [7]{\; \dfrac{2^6\cdot 4^{20}\cdot 6^{6}}{1\cdot 3^{15}\cdot 5^{15}\cdot 7}\; }\cdot \sqrt [8]{\; \dfrac{2^7\cdot 4^{35}\cdot 6^{21}\cdot 8}{1\cdot 3^{21}\cdot 5^{35}\cdot 7^7}\; }\cdot \sqrt [9]{\; \dfrac{2^8\cdot 4^{56}\cdot 6^{56}\cdot 8^8}{1\cdot 3^{28}\cdot 5^{70}\cdot 7^{28}\cdot 9} \; }
\cdot \sqrt [10]{\; \dfrac{2^9\cdot 4^{84}\cdot 6^{126}\cdot 8^{36}\cdot 10}{1\cdot 3^{36}\cdot 5^{126}\cdot 7^{84}\cdot 9^9} \; }\cdot \;\ldots\;\ldots\;\ldots$

answer below....  :oops:

[hide] did you try it out first ?  :| [hide] why not ?  :P [hide] that's called giving up ....  :lol:  it is not that hard...  :D  OK then, the answer is coming right up....  :rotfl:
$ \text{answer: }\quad\boxed{e^\gamma}$
[/hide][/hide][/hide]

Let our product will be $ \textbf{P}\equal{}\lim_{n\to\infty}{\textbf{P}_{n}}$, where $ \textbf{P}_{n}\equal{}\frac{\textbf{A}_{n}}{\textbf{B}_{n}}$. As I see for $ n\minus{}\textbf{th}$ term of it we have: $ \textbf{A}_{n}\equal{}\prod_{k} (2k)^{\sum_{m\equal{}2}^{n}\frac{1}{m}\cdot\textbf{C}_{m\minus{}1}^{2k\minus{}1}}$ and $ \textbf{B}_{n}\equal{}\prod_{k} (2k\minus{}1)^{\sum_{m\equal{}2}^{n}\frac{1}{m}\cdot\textbf{C}_{m\minus{}1}^{2k}}$.

Then we sholud write:
$ \ln\textbf{P}_{n}\equal{}\ln\textbf{A}_{n}\minus{}\ln\textbf{B}_{n}\equal{}$$ \sum_{k} \left(\ln (2k)\cdot\sum_{m\equal{}2}^{n}\frac{1}{m}\cdot\textbf{C}_{m\minus{}1}^{2k\minus{}1}\right)\minus{}\sum_{k} \left(\ln (2k\minus{}1)\cdot\sum_{m\equal{}2}^{n}\frac{1}{m}\cdot\textbf{C}_{m\minus{}1}^{2k}\right)$

I'll continue later...  :P

Misan, here are two products for you:

In the paper [url=http://arxiv.org/abs/math/0506319]Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent[/url] ([url=http://arxiv.org/PS_cache/math/pdf/0506/0506319v3.pdf]PDF version[/url]) Guillera & Sondow give the product (Theorem 5.6, equation (60) on pg. 20):

\[ e^{x} \equal{} \prod_{n \equal{} 1}^{\infty}\left(\prod_{k \equal{} 1}^{n} (1 \plus{} kx)^{{\scriptscriptstyle ( \minus{} 1)^{k \plus{} 1} }{\scriptstyle \binom{n}{k}}} \right) ^{\frac {1}{n}},
\]

for $ x\geq 0,$ to which I add

\[ \boxed{\frac {e^{u}\Gamma (u)}{\sqrt {2\pi e}} \equal{} \prod_{n \equal{} 0}^{\infty}\left(\prod_{k \equal{} 0}^{n} (k \plus{} u)^{{\scriptscriptstyle ( \minus{} 1)^{k} }{\scriptstyle \binom{n}{k} (k \plus{} u)}} \right) ^{\frac {1}{n \plus{} 1}},}
\]

for $ \mbox{Re} [u]\geq 0.$

I posted these products on [url=http://www.physicsforums.com/showthread.php?p=999960#post999960]physicsforums[/url] some time ago for a research paper I did at UCSB.

Note also that misan's product in the OP is given in the above cited paper as Example 5.8 on pg. 18.


Through the mid-point M on base BC of $ \triangle ABC$,a st.line DME is drawn so as to cut off equal parts AD and AE from AB and AC produced.Prove that BD=CE.
Thanks.

sine relation in $ \triangle BDM$ and $ \triangle CME$.

Can you please provide a geometric solution?My knowledge of trigonometry is very limited.

Thank you.

Take $ X\in AD$ such that $ BD\equal{}DX$. or $ D$ is mp. of $ BX$.
In $ \triangle BCX$ .
we have $ DM\parallel CX$. implies $ DE\parallel CX$.
since, $ \angle ADE \equal{} \angle AED$ .we get $ \angle AXC \equal{} \angle ACX$
$ AX\equal{}AC$
then,$ DX\equal{}CE$
or,   $ BD\equal{}CE$


inequalities
complex numbers
geometry unsolved
geometry

Let $M$ be the point inside triangle $ABC$. Prove that:
$\displaystyle \frac {MB.MC}{AC.AB}-\frac {MA.MC}{BC.AB}-\frac {MB.MA}{BC.AC} \leq 1$

Complex numbers can be usefull

$ \sum_{cyclic} \frac {(z-a)(z-b)}{(c-a)(c-b)} = 1 $

Also it can be solved using inversion.

Thank Nam Dung :D . I'll post my solution next week.

[quote="Namdung"]Complex numbers can be usefull
$ \sum_{cyclic} \frac {(z-a)(z-b)}{(c-a)(c-b)} = 1 $[/quote]
Great!
What else do we need?


$ x,y,z$ are [b]real numbers[/b] such that $ xyz\equal{}1$,prove that
\[ 4(1\plus{}x\plus{}y\plus{}z)(1\plus{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})\geq (1\plus{}x)^2(1\plus{}y)^2(1\plus{}z)^2\]
I'm sorry if someone have created it before.

[quote="Tourish"]$ x,y,z$ are [b]real numbers[/b] such that $ xyz \equal{} 1$,prove that
\[ 4(1 \plus{} x \plus{} y \plus{} z)(1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})\geq (1 \plus{} x)^2(1 \plus{} y)^2(1 \plus{} z)^2\]
[/quote]
I have proved that
\[ 4(1 \plus{} x \plus{} y \plus{} z)(1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})\leq (1 \plus{} x)^2(1 \plus{} y)^2(1 \plus{} z)^2\]
 :maybe:

[quote="arqady"]I have proved that
\[ 4(1 \plus{} x \plus{} y \plus{} z)(1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})\leq (1 \plus{} x)^2(1 \plus{} y)^2(1 \plus{} z)^2\]
:maybe:[/quote]
Of course ! Only AM-GM :
$ (1 \plus{} x)^2(1 \plus{} y)^2(1 \plus{} z)^2 \equal{} \left( (1\plus{}x \plus{} y \plus{} z) \plus{} (1\plus{}\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}) \right)^2 \geq 4(1 \plus{} x \plus{} y \plus{} z)(1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})$

Sorry for my typo, arqady and rachid.

Your solution is very nice ,I didn't notice that $ (1\plus{}x)(1\plus{}y)(1\plus{}z)\equal{}....$.I can prove the following one easily now.

\[ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2}\geq\sqrt{(abc\plus{}ab^2\plus{}bc^2\plus{}ca^2)(abc\plus{}a^2b\plus{}b^2c\plus{}c^2a)}\]


Στο δεύτερο μέλος αυτά που έχουν 2 μπροστά είναι τα διπλάσια γινόμενα των αριθμών του πρώτου μέλους οπότε αρκεί να αποδείξουμε ότι

$ (\frac{k^3m}{l}\plus{}\frac{l^3k}{m}\plus{}\frac{m^3l}{k})^2\geq 3(k^5m\plus{}m^5l\plus{}l^5k)$

Η τελευταία προκύπτει από την ανισότητα $ (a^2\plus{}b^2\plus{}c^2)^2\geq 3(a^3b\plus{}b^3c\plus{}c^3a)$  (της οποίας η απόδειξη έχει παρουσιαστεί στο φόρουμ , αν θέλετε θα δώσω το λινκ) για τους αριθμούς $ a\equal{}\sqrt{\frac{k^3m}{l}}$  κτλ ..

[quote="silouan"]
$ (\frac {k^3m}{l} \plus{} \frac {l^3k}{m} \plus{} \frac {m^3l}{k})^2\geq 3(k^5m \plus{} m^5l \plus{} l^5k)$
[/quote]

Πιο απλά διαίρεσε με klm και Tschebytsev.

Θες να πεις δηλαδή ότι  η ανισότητα του Vasc (δλδ η $ (a^2\plus{}b^2\plus{}c^2)^2\geq 3(a^3b\plus{}b^3c\plus{}c^3a)$ ) που είναι ισοδύναμη με την παραπάνω , αποδεικνύεται με Chebychev ?  :P  Θα ήθελα πολύ να το δώ αυτό γιατί εκτός από την ισότητα $ a\equal{}b\equal{}c$ η παραπάνω έχει ισότητα και για
$ (\sin^2\frac{4\pi}{7}, \sin^2\frac{2\pi}{7}, \sin^2\frac{\pi}{7})$  (αυτή πώς θα στη δώσει ο Chebychev ?  :P )

όχι εντελώς απευθείας:

$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$

άρα
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \frac {ab \plus{} bc \plus{} ca}{3}$

από Tscebytsev τώρα αυτό

$ \ge \frac {a^3b \plus{} b^3c \plus{} c^3a}{3}$ q.e.d.

Δεν είναι σωστό?

Νομίζω πως όχι ... Εκτός και αν έχω ξεχάσει εντελώς και τα λίγα μαθηματικά που ήξερα
Αν θυμάμαι καλά  όταν $ a^2\geq b^2\geq c^2$  τότε $ ab\geq ca\geq bc$

:blush: Δίκιο έχεις! Λάθος μου...

Sil παντα γρηγορο πιστολι

[quote="silouan"]Εκτός και αν έχω ξεχάσει εντελώς και τα [b]λίγα[/b] μαθηματικά που ήξερα[/quote]

Ha. Ha. Mrrrrfffffffhahahahahahaahahahahaaha :rotfl: :rotfl:

[quote="Durandal"][quote="silouan"]Εκτός και αν έχω ξεχάσει εντελώς και τα [b]λίγα[/b] μαθηματικά που ήξερα[/quote]

Ha. Ha. Mrrrrfffffffhahahahahahaahahahahaaha [/quote]

Συμφωνώ και επαυξάνω!!!!!!!!!!!!!!!

 :rotfl:  :rotfl:  :rotfl:

Αυτή πάντως η ανισότητα του Vasc είναι απίστευτα sharp παρά το ότι ανάγεται σε SOS. Μερικές ανισότητες έχουν τρομερή μαγεία.

Α ρε Sil έχει πάντως πλάκα το statement σου γιατί ξέρω άλλα άτομα που νομίζουν ότι ξέρουν μαθηματικά ενώ βρίσκονται παγιδευμένοι στο στενό τετραγωνάκι της άγνοιάς τους... :rotfl:

[quote="silouan"]Στο δεύτερο μέλος αυτά που έχουν 2 μπροστά είναι τα διπλάσια γινόμενα των αριθμών του πρώτου μέλους οπότε αρκεί να αποδείξουμε ότι

$ (\frac {k^3m}{l} \plus{} \frac {l^3k}{m} \plus{} \frac {m^3l}{k})^2\geq 3(k^5m \plus{} m^5l \plus{} l^5k)$

Η τελευταία προκύπτει από την ανισότητα $ (a^2 \plus{} b^2 \plus{} c^2)^2\geq 3(a^3b \plus{} b^3c \plus{} c^3a)$  (της οποίας η απόδειξη έχει παρουσιαστεί στο φόρουμ , αν θέλετε θα δώσω το λινκ) για τους αριθμούς $ a \equal{} \sqrt {\frac {k^3m}{l}}$  κτλ ..[/quote]

Δωσε μια λινκ αν μπορεις :P


algebra
difference of squares
special factorizations

[hide]Noticing that this expression is in the form: $(a-b)^n(a+b)^n$, we can apply the difference of squares and properties of exponents to express it in the form: $(a^2-b^2)^n$.  Doing this to our problem, we have:

$\displaystyle \left(\displaystyle \left(\frac{i\sqrt3}{2}\right)^2-\displaystyle \left(\frac{1}{2}\right)^2\right)^{2005}$
$\displaystyle \left(\displaystyle \left(\frac{-3}{4}\right)-\displaystyle \left(\frac{1}{4}\right)\right)^{2005}$
$(-1)^{2005}=\boxed{-1}$[/hide]


Sorry, I wrote something wrong and the code wasn't appear. I corrected.

[quote="Mateescu Constantin"]If  $ 0\ < \ a\ < \ b$, prove that

$ \sqrt {ab}\ < \ \left(\frac {b \minus{} a}{n(\sqrt [n]{b} \minus{} \sqrt [n]{a})}\right)^\frac {n}{n \minus{} 1}\ < \ \frac {a \plus{} b}{2},\ \forall\ n\in\mathbb{N},\ n\ge 2$.[/quote]

 Using AM-GM

  $ \left(\frac {b \minus{} a}{n(\sqrt [n]{b} \minus{} \sqrt [n]{a})}\right)^\frac {n}{n \minus{} 1} \equal{} (\frac {\sqrt [n]{b}^{n \minus{} 1} \plus{} ... \plus{} \sqrt [n]{a}^{n \minus{} 1}}{n})^{\frac {n}{n \minus{} 1}}\ge \sqrt {ab}$

 $ \left(\frac {b \minus{} a}{n(\sqrt [n]{b} \minus{} \sqrt [n]{a})}\right)^\frac {n}{n \minus{} 1}\equal{} (\frac {\sqrt [n]{b}^{n \minus{} 1} \plus{} ... \plus{} \sqrt [n]{a}^{n \minus{} 1}}{n})^{\frac {n}{n \minus{} 1}} \le \left(\frac{\sqrt[n]{a}^{n\minus{}1}\plus{}\sqrt[n]{b}^{n\minus{}1}}{2}\right)^{\frac{n}{n\minus{}1}}\le\frac{\sqrt[n]{a}^n\plus{}\sqrt[n]{b}^n}{2}\equal{}\frac{a\plus{}b}{2}$


geometry
rectangle
trapezoid
geometry proposed

Circles $ C_1$ and $ C_2$ with centers at $ O_1$ and $ O_2$ respectively meet at points $ A$ and $ B$. The radii $ O_1B$ and $ O_2B$ meet $ C_1$ and $ C_2$ at $ F$ and$ E$. The line through $ B$ parallel to $ EF$ intersects $ C_1$ again at $ M$ and $ C_2$ again at $ N$.  Prove that $ MN \equal{} AE \plus{} AF$.

$\angle BAE = 90^o = \angle BAF$.

$AB$ is in common between $\Delta BAF$ and $\Delta BAE$, so $A \in FE$.

Now $\angle AEB = \angle EBN$ and $\angle BNE = \angle BAE = 90^o$.

For ASA $\Delta BAE \equiv \Delta BNE$ $\Longrightarrow$ $AE = BN$.

Also $\Delta BAF \equiv \Delta BMF$ $\Longrightarrow$ $AF = BM$.

Adding this $MN = AF + AE$.

Since $ \widehat{BAF}\equal{}\widehat{BAE}\equal{}\widehat{ABM}\equal{}\widehat{ABN}$ we obtain that $ ABMF$ and $ ABNE$ are rectangles, i.e., $ AF\equal{}MB, AE\equal{}NB$. Thus: $ MN\equal{}MB\plus{}BN\equal{}AE\plus{}AF$ and we are done.

I think the problem has been stated wrong. It should have been $ O_1B \cap (O_2) \equal{} \{B,E\}$ and $ O_2B \cap (O_1) \equal{} \{B,F\}$

Please someone post a solution to this problem (to the correct version). Thanks.

Isosceles triangles $\triangle BFO_1$ and $\triangle BEO_2$ with apices $O_1,O_2$ are clearly similar $\Longrightarrow$ $\angle FO_1E=\angle EO_2F$ $\Longrightarrow$ $EFO_1O_2$ is cyclic $\Longrightarrow$ $\angle BFE= \angle BO_1O_2.$ But $\angle BO_1O_2=\frac{_1}{^2}\angle BO_1A=\angle BMA,$ thus $\angle BMA=\angle BFE$ $\Longrightarrow$ $AM \parallel BF,$ i.e. $ABFM$ is an isosceles trapezoid with congruent diagonals $AF=BM.$ Similarly, $AE=BN$ and the conclusion follows.


floor function
induction
algebra unsolved
algebra

Let $d>0 \in \mathbb{R}$. Consider the set

$\displaystyle S_n(d)=\{s=\frac1{x_1}+\cdots+\frac1{x_n} \vert x_i \in \mathbb{N}, s < d\}$.

Prove that $S_n(d)$ has a maximum element.

We can easily get a lower bound for the maximum, by  considering $x_i=\left\lfloor\frac{n}{d}\right\rfloor+1$.  We show by induction that there are only finitely many $n$-tuples satisfying $d>\sum_{i=1}^n\frac{1}{x_i}\ge\frac{n}{\left\lfloor\frac{n}{d}\right\rfloor+1}$.  The result is trivial for $n=1$, so we assume that it is true for $n=k-1$.

It is easy to see that if the sum of $k$ terms is at least $km$, then at least one of the terms is at least $m$. Without loss of generality, we assume that $x_i\le x_j$ for $i<j$, so we get $\frac{1}{x_1}\ge\frac{1}{\left\lfloor\frac{k}{d}\right\rfloor+1}$, or $x_1\le\left\lfloor\frac{k}{d}\right\rfloor+1$.

This allows only finitely many values for $x_1$, and for each one there are only finitely many $k-1$-tuples satisfying $d-\frac{1}{x_1}>\sum_{i=2}^k\frac{1}{x_i}\ge\frac{k}{\left\lfloor\frac{k}{d}\right\rfloor+1}-\frac{1}{x_1}$, which completes the inductive step.

Since there are only finitely many elements in this range, the set must have a maximum element in this range which, by construction, must be the global maximum for the set.


1. Two tangents to a circle are parallel and intersect the circle at A and B. A tangent to the circle at X (where X is not A nor B) intersects the first tangent at Y and the second tangent at Z. Prove that AY*BX is independent of the position of Z.

2. Tom and Bill live on the same street and has to deliver a package to each other. They simultaneously start walking from their homes and meet 600 yards from Bill's house. Then they continue and meet 700 yards from Tom's house. How far apart are their houses?

[hide="1"]
is not possible because you can prove that X=B, or I'm an idiot
[/hide]

[hide="2"]
Two lines on the time - distance from Bill's house plane should not intersect at more than 1 point.
[/hide]


A convex quadrilateral $ ABCD$ is inscribed in a circle. The tangents to the circle through $ A$ and $ C$ intersect at a point $ P$, such that this point $ P$ does not lie on $ BD$, and such that $ PA^{2}=PB\cdot PD$. Prove that the line $ BD$ passes through the midpoint of $ AC$.

Darij we are waiting for you ... Please prove this ...

Davron

Come on guys, this problem was already posted 3 times on the forum.
See, for example: http://www.mathlinks.ro/Forum/viewtopic.php?t=77685

Excuse me because I am not Darij, but I'll try to prove this

Let $E$ be second intersection point of line $BP$ and circle. Then
$PB\cdot PE=PA^2=PB\cdot PD$ so $PE=PD$.$F$ is intersection point of
line $BD$ and circle. Now, $\triangle PEF \cong \triangle PBD$, so
quadrilateral $BFED$ is a trapezoid.

$AC$ is a polar of point $P$ wrt circle so lines $BD,FE$ and $AC$
are concurrent. Line passing through P and midpoints of $BF$ and
$ED$ (remember $PE=PD$ and $PB=PF$) is also passing through center
of circle (because $BFDE$ is a trapezoid), so it's passing through
midpoint of $AC$ and through interection point of diagonals $BD$ and
$EF$. So intersection point of lines $AC,BD$ and $EF$ is midpoint of
$AC$. The end

If you don't understand just ask :)

bye


modular arithmetic
algebra proposed
algebra

$p$ is a prime
$a_{i}$ integers
How many solutions has the equation $a_{1}^{2}+.......+a_{n}^{2}=1$ in $Z/pZ$


Anyone here participate in the Canadian Computing Competition? Junior or Senior?

Please dont start discussions until tomorrow after 3 because some schools in Waterloo did not get a chance to write the cotnest as the server was down you can PM each other if you want

i guess we can start discussing about the problems now..ya they were easier then last year but somehow i managed to screw up lol..i'm not confident with some of my solutions

Anyway so how did everyone else do?

Yeah, I did the junior, and I think that I did pretty well... I know that I got the first 4 problems, but may have missed on some of the trickier test cases....
For the fifth one, I hardcoded some values.... lol :blush:  :D

I hope that even one test case will be right, though for the fifth problem, I think that you needed DP...??

- Nawahd Werdna

Does anyone know what the stage 2 cutoff will be this year? It's most likely that it will increase from last year  :wink:

Ah.... in the 60s? That's pretty darn high. LOL

yes but a lot of ppl were easily able to get the first four (like i'm sure its more then 20)..

I could be wrong..but I'm sure it will be high this year..

Christ, I forgot to DP in problem 4, so now my solution is going to go overtime on the big test data (which they'll probably have >.<). And problem 5? Gah, I talked it over with a few 2,3,4th year compsci guys and finally figured out the excessively simple answer. In fact, I almost achieved that answer by miscoding my poorly written DP solution.

Darn it, I'm not going to go to stage 2  :noo:

There are quite a lot DPs in Senior, but far easier than last year's Game of Life.

I honestly found Game of Life to be *easier* than this year's crop -- I didn't officially participate so maybe it was the stress :P

The solution I found is very efficient but does not actually DP. It stores partial results in a set, then checks each previous iteration. Anyway, I found a nice way (using the given constraints) to reduce the search space from 2^20 to something like 2^8 which is very easily manageable.

so does anybody have an idea for the cutoff i thought it would be high but i guess not maybe its like mid 50's or something...(btw two years ago it was like 57)

What do you guys program in? I was forced to program in V Basic, which totally sucked. :blush:

No! VB is the best programming language! (for beginners like me)
I actually used Pascal, which is extremely old-fashioned.

VB's error handling is a lot better than some of the other languages out there, ie. Turing

Two of my school mates got screwed over because of Turing.

I used VB and I got 57 on senior.

My computer teacher handed in the marking late, so CCC did not accept it! I could have gone to stage 2 too!!

 :cursing:


[quote="fizsle"]Thanks Everyone that was very helpful! :lol:[/quote]
Thats very nice of you, but you can also post this in the topic you recieved help from. :)


What's your favorite Linux distro? Considering there's a topic for Windows version and Mac version, it's only just that I add one for Linux :)...

I don't use Linux myself but almost everyone I know uses the ubuntu distribution.

ubuntu is nice....and I recommend it for all first-time linux users. However, I use Gentoo. Why? Well, first, I used ubuntu 6.06 Dapper Drake. It was nice and all, everything worked, but then I got a little annoyed that it was kind of slow. So I switched to Feisty Fawn. BUT!! suddenly ndiswrapper doesn't work on feisty fawn, so I have to use the bcm-fwcutter blah blah blah for my wireless card. Well this makes my wireless card like 4x slower. Also, the network manager kept randomly disconnecting me and stuff, so that annoyed me. So I heard that Gentoo was a nice distribution and I was planning to make the switch at some point anyway, so I downloaded a live CD and booted it. First thing I noticed was that the wireless card didn't work :(. However, it just said firmware not found, so I put the firmware onto a flash drive, booted off the live CD again and set up the networking. The I found out that Gentoo's network managers are MUCH better than ubuntu's (meaning that dhcpcd is simply godly). So I happily made the switch. Well, not entirely happily. It took me four days to get it installed properly. But no problem, I wanted it to be my distro, so I got it done. I also found out that ndiswrapper works with my wireless card on Gentoo! Yay! Well, I spent a few days happily installing X11, Xorg-server, gdm, gnome, beryl, kde, fluxbox, openoffice, firefox, links, blah blah blah. Then I really noticed something. Everything was faster than feisty fawn, not to mention dapper! That just made me really happy. So, Gentoo has been my distro for probably 6-9 months. Ubuntu lasted 3-6. (Yeah, I started linux just a year ago). I've been extremely happy with it.

Yeah yeah I have a broadcom wireless card. It makes me sad too :(


AMC
USA(J)MO
USAMO
Olimpiada de matematicas

Un hexágono convexo $ABCDEF$ está inscrito en una circunferencia de tal manera que $AB = CD = EF$ y las diagonales $AD$,$BE$y$CF$ concurren en un punto. Sea $P$ el punto de intersección de $AD$ y $CE$. Demuestra que
$CP/PE = (AC/CE)^{2}$

El problema es de la nacional de Estados Unidos si la memoria no me falla, y se resuelve por proporcionalidad. :D

Aqui hay algunas soluciones de este problema 3 de la USAMO de 1994.
http://www.mathlinks.ro/Forum/viewtopic.php?t=21738
 :lol:


Q1. Is HDD is really a random access memory device?

Q2. Data is stored only on one surface of the extreme plates in HDD. Is it true in modern technologies?

Q1: I think so.

Q2: http://en.wikipedia.org/wiki/Hard_disk_drive It looks like it.  :)


Hi, could someone please explain how to solve this problem (provide answer and solution)?:

For how many three-element sets of positive integers {a,b,c} is it true that a x b x c = 2310? [/b]

I am unsure of whether this is a subset problem or some sort of arrangement problem. [/b]

First, 2310 = 2*3*5*7*11. There are 5 ways to split up 5 things (prime divisors) into a set of 3:

{0,0,5}
{0,1,4}
{0,2,3}
{1,1,3}
{1,2,2}

If the primes are split up into {0,0,5}, then there is only [b]1[/b] way to do it: put all of them into the third position.

If it is {0,1,4}, there are C(5,4)=5 ways to choose 4 of them, and the fifth merely falls into place. so there are [b]5[/b] in this case.

Continue going through the cases like this, and you'll have the answer.

HINT: [hide]abc=2310

abc=2*3*5*7*11

You have 5 factors, arranged so that there are a, b, c.  So you have to figure out how many diff ways you can arrange them.[/hide]

I would guess:

[hide]

For a, it can be composed of 1, 2, or 3 factors.

*1 factor, 5 choices

*2 factors, 5C2=10 choices

*3 factors, 5C3=1o choices

Now for b you have to take into account what you took for a:

*1 factor, 4 choices

*2 factors, 3C2=3 choices

*3 factors, 2C3=0 choices (impossible)

For c, you have to take into account a and b:

*1 factor, 3 choices

*2 factors, 1 choice

*3 factors, impossible

But you can make any combination of using 1 factor, then 2, then 3, etc. so I am a little confused.  Probably just writing them out would be the easiest.

abc =2*3*5*7*11

(a)*(b)*(c)

(2)*(3)*(5*7*11) <--or others in this form

5 options * 4 options * 1 option

(2)*(3*5*7)*(11) <--new form

5 options * 4C3=4 options * 1 option

(2*3*5)*(7)*(11)

5C3=10 options * 2 options * 1 option

(2*3)*(5*7)*(11)

5C2 * 3C2 *1

(2*3)*(5)*(7*11)

5C2 * 3 * 1

(2)*(3*5)*(7*11)

5 * 4C2 * 1

So I think you have 20+20+20+30+30+30 = 150 ways

[/hide]

Don't trust me on this though!

zabelman, your explanation makes sense to me:

{0,0,5} = 5C5=1 way.
{0,1,4} = 5C4=5 ways.
{0,2,3} = 5C3-> Should this just be simply 5C3 ways? = 10 ways
{1,1,3} = 5C1*4C1*3C3=20 ways?
{1,2,2} = 5C2*5C2*1C1 = 30 ways (not sure about this part).

Adding these possibilities up: 1+5+10+20+30=66 ways. However, I think there may be repeats in the way the prime factors are grouped. Also, I think this problem talks about the combination of the factors, not their respective permutations. A set of {2 X 1155 X 1} would be identical to {1155 X 2 X 1}. The answer to this problem (from the answer sheet) is [b]40[/b] such sets (they have no explanation), but I can't see how this is.

JS1527, adding up your numbers: 5+10+10+4+3+3+1 yields 36, which is close to 40. Perhaps a few combinations were not added or something...0

I don't think we should consider the partition {0, 0, 5}.  Back in the original problem, that would lead to the set {1, 1, 2310}, but that is a two-element set, not a three-element set.  In other words, a, b, and c should be different.

ah, good point! then 40 is the answer, since all of the other sets have distinct elements.

Whoa I was way off...  Sorry about that.

So the solution of the problem lies in cleverly applying a combanatorics concept in what seems like a set problem. However, can you explain why division by two is necessary? (in other words, how do you know to divide exactly by 2?)

im guessing that when you have {1,2,2} the 2's can be switched and it will be the same thing, like:
{2,15,77} and {2,77,15} are the same, but both are counted using the 5C2 * 3C2 formula. Thus, you divide by 2 to get rid of the extras.

Julie, you double-counted quite a few cases.  Notive where you have counted (2 factors)(1 factor)(2 factors) and (2 factors)(2 factors)(1 factor).  Because you used chooses, these are really the same case.  Also, you missed the cases where one of the factors is 1.

Aren't combinatorics problems fun?  There are always many ways to solve them.

Let's put three empty buckets on the floor, one red, one white, and one blue.

Toss each of the 5 distinct prime factors into a bucket.  Each factor can be tossed into one of 3 buckets, so there are 3^5=243 ways to toss the factors into buckets.

Ravi convinces me that if two buckets are empty, we can't count that as a legitimate set of three.  There are three ways this can happen.  243-3=240.

These buckets can be arranged 3! different ways, so the number of distinct sets is just 240/6=40.


geometry calculus trigonometry integration conics ellipse function

Im currently solving some calculus problems in preparation for our upcoming exam.
Im having a hard time with problems regarding area of polar curves.

Problem 1: Given equations $r = \frac{4}{2-\cos\theta}$ and $r = 4\cos3\theta$ setup definite integral equal to the area outside the rose and inside the conic

Problem 2:given a) $r = 1$ and b) $r = 1-3\sin\theta$ find the area of the region outside equation a but inside equation b.

What i have for Problem 1 is an ellipse with directrix $x = 4$ and vertices $(\frac{4}{3},\frac{\pi}{2})$ and $(4,0)$. I have also sketched the 3 petaled rose and with this i get a pt of intersection $(4,0)$. I am lost after this part.

For Problem 2: I have solved that their pts of intersection occur at $\theta = 0,\pi$ I dont know what to do next.

Hope somebody could teach me how to solve these problems. Thanks

When you have a function given in cartesian coordinates, y = f(x), to calculate areas you use something like $\int^{b}_{a}f(x)\,dx$.
When you want to compute areas for "functions" given in polar coordinates, $r = f(\theta)$, what do we use?

The formula for getting the area of polar curves is

$A = \frac{1}{2}\int^{b}_{a}[f(\theta)]^{2}d\theta$ with $a \leq \theta \leq b$ and $r = f(\theta)$

Um I know this is a little bit impatience on my part but I hope someone could help me what to do next with the problems. I am still very much lost at this point. Thanks.

These are the curves.
It becomes easier if you divide your integral into parts like for example,
considering only the first quadrant the curves meet at $\theta_{1}$ (not at the axis) which one can find by solving
a trig eq.

so integrate the first curve $f_{1}(\theta)=4\cos{3\theta}$
\[\frac12\int_{0}^{\theta_{1}}f_{1}(\theta)^{2}\]
Subtract this from integral over blue curve over same interval
\[\frac12\int_{0}^{\theta_{1}}f_{2}(\theta)^{2}-\frac12\int_{0}^{\theta_{1}}f_{1}(\theta)^{2}\]
then add the rest namely
\[\frac12\int_{0}^{\theta_{1}}f_{2}(\theta)^{2}-\frac12\int_{0}^{\theta_{1}}f_{1}(\theta)^{2}+\frac12\int_{\theta_{2}}^{\theta_{2}}f_{2}(\theta)^{2}\]
There may be an easier way but this should work.

[color=green]Se also[/color]
[url]http://tutorial.math.lamar.edu/AllBrowsers/2414/PolarArea.asp[/url]


superior algebra
superior algebra unsolved

Let $ g$ be an element of a group $ G$,and suppose that $ |G|\equal{}mn$ where $ m$ and $ n$ are coprime.Show that there are unique elements $ x$ and $ y$ of $ G$ such that $ xy\equal{}g\equal{}yx$ and $ x^m\equal{}1\equal{}y^n$

I can only show the uniqueness of $ x$ and $ y$,how to show the existence of $ x$ and $ y$?

Write $ 1\equal{}am\plus{}bn$ (by Bezout). Then take $ x\equal{}g^{bn}, y\equal{}g^{am}$.


How do I make labelled block-walking diagrams on LaTex look like this:

[img]http://www.artofproblemsolving.com/Classes/AIME/Images/2005aime1_13second.gif[/img] and [img]http://www.artofproblemsolving.com/Classes/AIME/Images/2005aime1_13paths.gif[/img]?

(These are the pictures that Mathew posted for last year's AIME Math Jam.)

I'm pretty certain that these diagrams were not done in LaTeX -- they look like they were done in Fireworks.

It is possible to do diagrams like this directly in LaTeX, but not easy.  The much easier approach is to use a graphics program to create a .png or .jpg file, and then import this file into LaTeX using the \includegraphics command.  See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]this page[/url] for details.


This should be in the Games and Fun forum.

anyway:

Battle of Thermopolyae (sp?) i dunno it was all i can think of at the time...=)

hmm
maybe interview anne frank before she died?  all we have is stuff before she left the annex

Why would you want to know whether a God exists or not?

I don't see any importance to Anne Frank and why she is so special.

Anyway, I would go back to 9/11 cause those people purporting those conspiracy theories like bubka are seriously pissing me off and it would be nice to prove that the government had no involvement in 9/11

[quote="junggi"]how Kurt Cobain died.
man that dude was a genius...[/quote]
"oh whoops this shotgun is pointed at my head i wonder what happens if i pull the trigger"

I would go back and tell Doc he gets shot in the future.   :D

[quote="Ignite168"]
Anyway, I would go back to 9/11 cause those people purporting those conspiracy theories like bubka are seriously pissing me off and it would be nice to prove that the government had no involvement in 9/11[/quote]
Alternatively, just put a few bullets in their head.

Actually that's probably what I'd do.

this should not be in the fun and games section because it involves history and sciences and opinion. (lots of opinion) I think a good topick for the round table to disguss. Oh by the way, you can also do things like to study the behavior of the dodo or what colors exactly were the dinosaurs.

$\textrm{Opinion}\neq \textrm{Interesting discussion}$.  In particular, threads that involve everyone listing their own answer to a question (rather than people responding to each other) don't normally belong in RT.

To see if Pocahontis really WAS a real person.

pocohauntis probably was. She was documented to go to europe (I think twice) had a few children and died of smal pox

[quote="ckck"]to see what really killed the dinosaurs. =D[/quote]
OMG LOOK A BIG ________(insert what killed the dinosaurs) AAAAAAAAAAHHHHHHHHHHHHHH

kaboom

[size=59]silence[/size]

Hmm maybe the start of the Big Bang. It's kinda hard in my mind to grasp that NOTHING could exist at one point.

to see how troy was actually sacked (prove or disprove homer) :ninja:  :lol:

sorry to post here again...

To see who ACTUALLY shot JR.

[quote="ashpotter"]If you could go back in time to see how somthing exactly happened, what would you see. An example would be that to see if  a religious figure accually existed (this would most likely be chosen by an atheist) Another example would be to see who really shot JFK or how elvis or amelia earhart died[/quote]

there were many events in my own life that i once wanted to revisit

but then having lived the future of those events, i realised that for every event in time there is a reason

life is like a big puzzle, and everything was done because of something

and so its better to live the past as the present only once ;)

uh-oh, I'd better prepare for this, by making a machine that gives a completely random (quantumly random) decision, and observing the results a couple times.

When I go back in time, I'll check whether I observe the same events (from a differentt angle) as the first time around.

[quote="alkjash"]If I could go back in time, I would go back to the beginning of mankind and tell them that the Big Bang made the universe and evolution created the humans, so they'd never invent all those religions.[/quote]
HEAR HEAR

but then history would be altered and we'd never exist...
aww shucks :ninja:

And you probably also tell little kids there's no Santa Claus.


trigonometry
geometry
3D geometry
trig identities
Law of Sines
geometry unsolved

In a triangle ABC, if SinA*SinB*SinC=1/2, then prove that it is a right-angled isosceles triangle

In a right isosceles triangle, the $ \angle A\equal{}45^{\circ}$, $ \angle B\equal{}45^{\circ}$, and $ \angle C\equal{}90^{\circ}$.  Then $ \sin{A}\sin{B}\sin{C}\equal{}\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}\cdot 1\equal{}\frac{1}{2}$.

I don't think this is correct. By the Law of sines,
$ \frac {\sin a}{a} \equal{} \frac {\sin b}{b} \equal{} \frac {\sin c}{c} \equal{} \frac {1}{2R}$.
The product of the first three equals the cube of the last, hence
$ \frac {\sin a \sin b \sin c}{abc} \equal{} \frac {1}{8R^3}$
$ \sin a \sin b \sin c \equal{} \frac {abc}{4R}\cdot\frac {1}{2R^2}$
$ \sin a \sin b \sin c \equal{} K/2R^2$
for any triangle, where K is the area. So
$ K/2R^2 \equal{} 1/2$
$ K \equal{} R^2$.
But in a circle of radius 1, more than a right isoceles triangle would have an area of 1, since the largest possible area would be an inscribed equilateral triangle, of area $ 3\sqrt {3}/4>1$.


inequalities trigonometry limit function calculus derivative algebra

$ x(t)$ solves $ y'\equal{}x^2 \plus{} y^2, \,\, y(0)\equal{}1$.

Show that $ \displaystyle \frac{1}{1\minus{}x}\leq y(x)\leq \tan{(x\plus{}\frac{\pi}{4})}$

and $ \displaystyle \lim\limits_{x\rightarrow x_1}\equal{}\infty$ for some $ x_1\in[\frac{\pi}{4},1]$

The inequality holds for the largest interval of nonnegative positive $ x$ such that $ y$ is defined, right?
[hide]
First, note that $ y^\prime$ is nonnegative so $ y\geq 1$ for positive $ x$ (in particular, $ y$ is nonzero). We also get $ y^\prime \geq y^2$. As a result, the function $ f(x)\equal{}x\plus{}\frac{1}{y}$ has derivative $ 1\minus{}\frac{y^\prime}{y^2}$ so is decreasing. But $ f(0)\equal{}1$ so
\[ x\plus{}\frac{1}{y}\leq 1\Rightarrow y\geq \frac{1}{1\minus{}x}\]
Also, $ x\plus{}\frac{1}{y}\leq 1$ and $ y$ positive implies $ x$ must be less than 1. Then $ y^\prime <y^2\plus{}1$, so $ g(x)\equal{}\arctan y\minus{}x$ is decreasing. As $ g(0)\equal{}\frac{\pi}{4}$, we get
\[ \arctan y\minus{}x\leq \frac{\pi}{4}\Rightarrow y\leq \tan \left( x\plus{}\frac{\pi}{4}\right)\]
Now as $ y$ is increasing and cannot be defined for $ x\equal{}1$, we must have $ y$ unbounded from above on $ [0,x_1)$, where $ x_1$ is the least upper bound of the domain of $ y$ and is less than or equal to 1. We must then have $ \lim_{x\rightarrow x_1}\equal{}\infty$. As $ y\leq \tan \left( x\plus{}\frac{\pi}{4}\right)$, $ y$ is bounded from above for $ x<\frac{\pi}{4}$, and so $ x_1\geq \frac{\pi}{4}$.
[/hide]

The estimate $ x_1\le 1$ is very good: numerically $ x_1\approx 0.96981$. Although this ODE can be solved in terms of Gamma and Bessel functions, it's one of those cases when the explicit solution is too explicit to be shown in public.


AMC
AIME
function
LaTeX
Gauss
arithmetic series

A function $f$ defined on the set of integers satisfies

\[f(x)+f(x+3) = x^2\]

for any integer $x$. If $f(19) = 94$, find $f(94)$.

I think this problem's level is AIME but I'm curious to know if you can solve without some calculations I had to do.

Try this problem (It's fun!) and post your solution too. Also, please tell me what do you think about my solution

[hide="My answer"]

In our current equation, we can't do anything directly to find $f(94)$ since it will involve $f(\text{other number})$. But we do have one $f(x)$ that is smaller than 94 and since it's smaller, I decide to choose the function using $f(x+3)$ since setting as $f(x)$ will give $f(94) = 94^2-f(97)$. which involves $x$ greater than 94. Therefore,  I put $f(x)$ into right side and got $f(x+3) = x^2-f(x)$.

I decide to plug in some values in starting with $x=19$.

$f(22) = f(19+3) = 19^2-f(19)\\
=19^2-94$

$f(25) = f(22+3) = 22^2-f(22)\\
=22^2-(19^2-94)\\
=22^2-19^2+94$

$f(28) = f(25+3) = 25^2-f(25)\\
=25^2-(22^2-19^2+94)\\
=25^2-22^2+19^2-94$

From these, I see that $f(x) = (x-3)^2 - (x-6)^2 + (x-9)^2 - (x-12)^2 \cdots$ till it reach $19^2$ and 94 at the end.

So, using this idea, we can find $f(94)$.

$f(94) = 91^2-88^2+85^2-82^2+79^2-76^2+73^2-70^2+67^2-64^2+61^2-58^2+55^2-52^2+49^2-46^2+43^2-40^2+37^2-34^2+31^2-28^2$

Or:

$f(94) = 91^2-88^2+85^2-82^2+79^2-76^2+73^2-70^2+67^2-64^2+61^2-58^2+55^2-52^2+49^2-46^2+43^2-40^2+37^2-34^2+31^2-(25^2-22^2+19^2-94)$

And in this case, I used calculator to compute the value but because for instance, $91^2-88^2 = (91+88)(91-88)$ we can use this idea to write out each sum and then calculate the whole sum.

When we do that, $f(94) = 4303 \blacksquare$. [/hide]

i used a similar approach to silverfalcon.
[hide]using $f(x)=x^2-f(x+3)$, i plugged a couple values in and realized that what would be created was:
$19^2-22^2+25^2-....+85^2-88^2+91^2-f(94)=94$
using lots of difference of two squares, it became
$-3852+91^2-f(94)=94$
$f(94)=4335$
however this is not the same answer as silverfalcon. i couldn't easily find a mistake tho a simple one might have been made. i'm too tired to check anymore. good night.[/hide]

Yeah, see that's the problem. When you get into tons of squares or multiplication of many large or semi-large numbers, you can make a mistake and not notice it and just waste 30 mins or more on a problem that you will eventually get wrong.

There's got to be an easier way to do this.. I hope.

And if I were to actually solve it, I would have used SilverFalcon's method, at least that's what first popped into my mind when I read it, but then a flashing light came on in my head that screamed BIG NUMBERS BIG NUMBERS BIG NUMBERS, heh.

I basically did the same thing as everyone else to start it.
$f(94)=91^2-88^2+85^2-82^2+79^2-...-22^2+19^2-f(19)$
Then I grouped each pair of squares in two and factored it (except the last two parts because there is an uneven number):
$f(94)=(91-88)(91+88)+(85-82)(85+82)+...+19^2-f(19)$
Then you can write it as an arithmetic series, with a little extra:
$f(94)=(3)(91+88+85+82+79+76+...+22)+19^2-f(19)$
$f(94)=\frac{(3)(91+22)(24)}{2}+19^2-f(19)$
$f(94)=\frac{(3)(91+22)(24)}{2}+19^2-f(19)$
$f(94)=4068+361-94$
$f(94)=4335$

I don't know where the difference in answers is coming from.  The difference between them is 32, if that helps.

[quote="Wumbate"]
$f(94)=4335$
I don't know where the difference in answers is coming from.  The difference between them is 32, if that helps.[/quote]
hey, that's my answer. (wonder which answer is right? :? )

I agree. Which one is the correct answer then?

Hm, Wumbate, how did you go from step 3 to step 4 in your calculations? Just count your first latex line as step 1, next latex line as step 2 etc. etc.

It really simplifies the problem and I can't see where it's coming from.

Thanks!

EDIT: Oh did you use the sum of an arithmetic series formula?

Yes, it is the arithmetic series formula.

I figured out what Gauss did:
His number 4711 is equal to this:
$91^2-88^2+85^2-82^2+...+31^2$
Then he left out $28^2$

But that wouldn't give an answer as close as it was if he just left that out.
He also subtracted $25^2-22^2+19^2-94$ where it should have been added.

I know that - as I said, I used both methods to calculate it.

[quote="blahblahblah"]Silverfalcon and Gauss (although I suspect he just looked at the sum) are wrong.  They deleted the term $28^2$ from consideration.  I also get $4335$.  Gauss's method yields that result as well if you correct the bounds of the sums.[/quote]

Agree. I confused $28^2$ and $f(28)$.  :blush:  :)


When do we get these?

Do they address things like food, laundry, etc?

You should have already gotten a large envelope, including, among other things, an AMY packet (mine was induction, although I don't know whether everyone got the same one).

I'm assuming that if I make any further comments on it, zacacox will edit them out :wink: .

Well... i didn't get mine yet, but it may still be in the mail,  since US - Canada mail takes forever...  :lol:

- Nawahd Werdna

You send them your check, and when they get it, they send you the pre-camp materials, I'm pretty sure. At least, that's what my e-mail said.

I mailed my money and things, but all I have received is my student info forms to fill out.

[quote="Altheman"]
Do they address things like food, laundry, etc?[/quote]

Speaking purely from last year's experience, there are laundry machines in each apartment and are free so you don't need to bring any change for them. They also have laundry detergent available. All meals are covered. Even if you go somewhere for an outing and it is during a meal, they will give you money for food. However, I suggest you bring some of your own snacks and drinks just in case you get the munchies late at night  :P
Oh, also bring your own water bottles just in case. Those always seem to run out pretty quickly during a frisbee game.

Like I said, this is based on last year. Don't sue me if I turn out to be wrong.

[quote="Max300"]Like I said, this is based on last year. Don't sue me if I turn out to be wrong.[/quote]
Doggone it, I could have used that extra money :yankchain:  :roll: !

ok so i got an induction packet + a bunch of forms

...

anybody starting to work on the packet...?

I've started working on the packet, but it is HARD!  I think the difficulty level starts at the admissions test and goes up from there ... and that is ignoring the advanced section. :wacko:

Uhhh.... I didn't get anything yet... :maybe:  :maybe:  :|

Hmm, I got my packet a long time ago (I took the A), is it required that we write up our solutions and bring the packet to camp? Because I think I lost mine (I did solve like all the problems in the easy section and around ten in the medium section)

I think you were supposed to turn in your solutions after three weeks.

I think the AMY packet was just extra material we could work on if we had time... at least, that's what it appeared to be.

Does anyone know when we can talk about the problems in the packets?

[quote="junggi"][quote="mustafa"]In my email it said that we should try to solve at least ten problems, and that we would go over them at camp.[/quote]
really?
I didn't know about the 2nd part...but that doesn't make any sense, some of us got different packets (some got pigeonhole some got induction)[/quote]

Maybe we will split up :wink:
[quote="Titu Andreescu in my e-mail"]
The fourth one is an AwesomeMath Year-round (AMY) segment for you to work on. Please choose a level (introductory or advanced) you are more comfortable with and try to solve at least 10 problems. We will go over these problems at the camp. Please bring them with you.
[/quote]

Yea...I got the induction packet. The easy section...is not that easy...I guess I have seen many of those problems though...The hard section is like a wall of text....I am going to have to do some work with those problems...

In the advanced section (Induction), I only recognized about half of the theorems they presented, and that was mainly because of AoPS 2.  To be honest, I had to struggle just to solve ten of the problems.

blah...I am getting stuck on some of the induction problems... AIM, anybody? [since the woot room is closed... :( ]

sn: altheman1

[quote="b-flat"]I was under the impression that the packets were extra material to get you ready for camp, not something you had to send in solutions to.

I got one on the Pidgeonhole Principle and the easy section is rather easy  :rotfl:[/quote]

That's because they decided to include a medium section, which is pretty difficult (although a lot of the problems have become well-known). The general difficulty level seems to be

Easy: meh
Medium: IMO problems
Hard: easier than medium

the easy section still has olympiad problems...this is good...i know i am in for a challenge this summer  :)


calculus integration symmetry inequalities derivative probability

http://www.iitd.ac.in/cgi-bin/nph-p/http/10.80.10.8/jee/Mmaths.htm

Let's see if we can combine to solve all of them.  Spoiler your answers, of course.  I'll start.

1.  [hide]Write |z-a| = k|z-b| --> (x-a1)^2 + (y-a2)^2 = k^2[(x-b1)^2+(y-b2)^2], collect terms, and complete the square.[/hide]

2.  Not entirely sure yet, but use the fact that a,b,c,d are distinct, and that, using anticommutativity of the cross product, (a-d) X (b X c) = 0.  So either a-d and bxc are parallel, or b,c are parallel.

3.  [hide]Integrate with C=0, check that both endpoints of the interval give P(x)=0, Rolle's Theorem yields the desired result.[/hide]

4.  Haven't calculated it, but looks easy.

5.  (None)

6.  (None)

7.  Haven't worked on it yet, but M^T is necessarily the inverse of M.

8.  [hide]Take the cos x out, differentiate using the chain rule and product rule, the term with -sin x and the ugly integral drops out (because sin pi=0), the rest is easily calculated.[/hide]

9.  (None)

10. Equivalent to proving (x+y)(1-x)(1-y) <= x(1-y) + y(1-x), x,y < 1, x+y<1, not immediately sure how do to this.

11. [hide]It's a linear homogeneous differential equation with solution y = x^2 + (c+1)x + c + 3.  Given that (2,0) is on the curve, c = -3 and y = x^2 - 2x.  The rest is simple.[/hide]

12. Symmetry gets you halfway there, not sure about the rest.

13. (None)

14. (None)

15. [hide]The inequality clearly follows if we can prove that (1+x)^7/(7^(7/3)*x^4) > 1 for x >= 0.

Differentiating, we get that it is decreasing over 0 < x < 4/3, and increasing everywhere else.  Checking the critical points x=0 and x=4/3 show that it is greater than 1 at both points.  The inequality is valid for any x, so we can choose a,b,c E R+ to get the desired result.[/hide]

16.  [hide]It's easier than it looks.  Use the Taylor MVT to get sin x + 2x >= x - x^3/6 , then multiply by 3/pi (which makes the inequality strict), then note that 9x - x^3/2 - 3x^2 - 3x >=0 in 0 <= x <= pi/2.  The result follows.[/hide]

17. (None)

18. See JBL's later post

19. See JBL's later post

20. (None)

the link isn't working for me, would it be possible to post the problems?

The link's not working for me anymore, either.  :(

Thanks, Mathkid.

18. [hide]To be differentiable, it must be continuous.  Taking the limit as  from above, we see a/2 = 1/2, so a = 1.  Once we know a, we can find the limit of the derivative from above at 0 and then set the limit of the derivative from below at 0 equal to it and solve in order to get the given relationship between b and c.[/hide]

19. [hide]Straight-forward.  Take 3 cases, Simply compute the number of ways the given events can occur, multiply by the probability that their respective conditions held, and divide by total possible outcomes.  Their answer at the bottom is essentially a full solution.[/hide]

My solutions to two of the questions are incomplete:

12.  Using symmetry, the integral is equivalent to twice the integral of pi/(2-cos(x+pi/3)) from 0 to pi/3.  Now how the heck do we evaluate that?

15.  I AM-GM-ed terms wrong, haven't bothered to try and see if the AM-GM approach actually works for the right terms.

Correction for 15.

The inequality clearly follows if we can prove that (1+x)^7/(7^(7/3)*x^4) > 1 for x >= 0.

Differentiating, we get that it is decreasing over 0 < x < 4/3, and increasing everywhere else.  Checking the critical points x=0 and x=4/3 show that it is greater than 1 at both points.  The inequality is valid for any x, so we can choose a,b,c E R+ to get the desired result.

Oh, and by the way, the best possible constant we can have for a >= in the inequality would be, instead of 7^7, approximately 1691405 which is slightly greater than 2*7^7.

Hi Blahblahblah..
It gives me great pleaure to see IIT entrnace problems here on this forum..

As a matter of fact I happen to be an IITian...had cleared the IIT-JEE (Indian Institute of Technology-Joiint Entrance Examination) in 1997..

u get some real challengers not only in Maths but also in the other two papers viz. Physics and Chemistry..

I still love to recall those days of solving I.E. Irodov's problems in general physics, while preparing for this exams..

I'm saying so much of stuff because, for every Indian student [quote]seriously[/quote] aspiring for a career in Engineering , getting into one of the seven IITs is a dream.

This is because out of 1-1.5 lakh (1 lakh = 0.1 million) students that take this exam, some 3000 finally manage to get through..

that was a brief background of IIT-JEE..

I find Problem 6 cool... And though there is a combinatorial way to solve it, which was probably taken by most of the students, I wanted to note that it can be also shown by a simple telescoping product:

[hide]$\frac{\left( n^{2}\right) !}{\left( n!\right) ^{n}}=\binom{n^{2}}{\left( n-1\right) n}\cdot \binom{\left( n-1\right) n}{\left( n-2\right) n}\cdot ...\cdot \binom{3n}{2n}\cdot \binom{2n}{n}\cdot \binom{n}{0}$.[/hide]

  Darij


:blush:
last prob was easy yall got it right, rotflol.
Here's a bit harder one.

pi+po=20
pe+pp=35
pa=pi+3
a=sqrt(36)

Q: What is the value of ((E/P)(O+I)/O)+A???

[quote="Ihatepie"]impossible to solve, you have 4 unknown variables and 3 equations.[/quote]

In the other thread he added a $ p\equal{}5$  :wink: .

[quote="Ihatepie"]impossible to solve, you have 4 unknown variables and 3 equations.[/quote]

That isn't necessarily a valid reason; the problem asks for an expression in terms of the variables, not the variables themselves.

For example , suppose we had:

A+B = 4

A+C = 2

Find (A+B)(A+C); this problem can be solved but it has 3 variables and 2 equations.  I'm not saying the given expression is necessarily doable, just maybe try to give a better reason.

This problem isn't necessarily impossible to solve, if you define "impossible."  It's actually dependent on the value of p more than the other variables.  For example, the previous problem is the EXACT SAME THING as this problem EXCEPT that it has the fact p=5.  What does that mean?  It means that if we had p=6, the answer to the problem would be different than if we had p=5 (do the math).  Therefore, the answer isn't any one answer, just as saying y=-x+2 doesn't have exactly one answer for x and y.

If you really want me to, I can write out the whole thing as a function of p, but that would take quite a while...


I want to make the greater than sign.

Since \ngtr works, I tried \gtr. But it doesn't work. Why?

You can use $\ge$ (\ge) or $\le$ (\le).  Otherwise, just use $>$ or $<$, which is just > and < with dollar signs.

To negate something, you can use:

$\not>$ (\not>)
$\not<$ (\not<)
$\not\ge$ (\not\ge)
$\not\le$ (\not\le)

Hope that helps!

Also:

$\ngtr$ (\ngtr)


Ένα γράμμα βρίσκεται μέσα σε ένα από τα 5 συρτάρια με πιθανότητα [b]p[/b].Έστω ότι ψάξαμε στα τέσσερα από αυτά χωρίς αποτέλεσμα.Ποιά η πιθανότητα να βρίσκεται στο πέμπτο; Ποιά η αντίστοιχη αριθμητική της τιμή, στην περίπτωση που κάθε συρτάρι έχει την ίδια πιθανότητα να περιέχει το γράμμα και [b]p[/b]=0,5;

Παρακάλω όποιος μπορεί να υποδείξει την λύση της άσκησης.

THMMY?? δε βοηθάμε ΤΗΜΜΥτες...
εννοείς πως έχει πιθανότητα p να υπάρχει σε κάποιο συρτάρι?
τότε θα έλεγα πως είναι p αλλά δεν είμαι σίγουρος... ή μπορεί να θέλω να σαμποτάρω το απσου:-p

Να μην είσαι καθόλου σίγουρος....Όσον αφορά το ειρωνικό ύφος δεν νομίζω ότι το ΣΗΜΜΥ έχει υπεροχή του ΤΗΜΜΥ,αν και από του χρόνου μπορεί να μην είμαι πλέον ΤΗΜΜΥτης αφου έχω unconditional offer από Imperial και UCL(department of electrical and electronic engineer).Ευχαριστώ πάντως για την βοήθεια ρε μεγάλε...........Έυχομαι καλή χρονιά με πολλές πολλές καταληψεις(μιας και που έχετε την πρωτιά σε αυτό τον τομέα)!!!!

Υ.Γ Το 2008 θα αρχίσετε 3ο εξάμηνο????

τώρα έχουμε ένα νέο φρούτο, κλειστή σχολή;-p
εμείς είμαστε αγωνιστές καθότι έχουμε πολύπλευρη μόρφωση και όχι στείρα αναπαραγωγή πεπαλαιωμένων γνώσεων που οδηγούν στην εξειδίκευση και στην κατασκευή επιστημόνων ρομπότι στην υπηρεσία του συστήματος:-p
πω πω τι έγραψα πάλι, πρέπει να ασχοληθώ επαγγελματικά μου φαίνεται :rotfl:
lolol, δεν έχω κάτι με τους ΤΗΜΜΥτες, απλώς έχει πλάκα:)
και ούτε θέλω κατάληψη
το Φράγκι τον ξέρεις?

[quote="thatin"]τώρα έχουμε ένα νέο φρούτο, κλειστή σχολή;-p
εμείς είμαστε αγωνιστές καθότι έχουμε πολύπλευρη μόρφωση και όχι στείρα αναπαραγωγή πεπαλαιωμένων γνώσεων που οδηγούν στην εξειδίκευση και στην κατασκευή επιστημόνων ρομπότι στην υπηρεσία του συστήματος:-p
[/quote]

Ρε συ Θάνο, τι παίρνεις;  :rotfl:  :P

αν πας σε κάποιες συνελεύσεις προς το τέλος για να ψηφίσεις ακούς κατα λάθος κάτι τέτοια και ζαλίζεσαι:-p
φαντάσου τι παθαίνουν αυτοί που πάνε σε κάθε συνέλευση και την παρακολουθούν όλη!!


function
inequalities
calculus
calculus computations

Find all the functions defined on $(-\infty ;+\infty)$ satisfying the inequality
 $f(tx+(1-t)y)\leq  tf(x)+(1-t)f(y)$ for all x,y,t.

A geometric interpretation of this inequality: the graph of $f$ lies below any of its secant lines. This forces $f$ to be an affine function. Indeed, suppose that the graph of $f$ contains three non-collinear points -- each vertex of this triangle must lie below the extension of the opposite side, which is clearly impossible.


How is your USAMO index calculated if you qualify for and participate in the AIME by virtue of your USAMTS score?

The USAMTS acts as if you scored a 100 on the AMC-12; if you can take the AMC-10/12, that's the optimal way to qualify, of course.

So a score of 75 after three rounds would be no different than 74? All qualifying scores on th USAMTS contribute 100 points to your USAMO index, right? Thank you.


geometry
analytic geometry
area of a triangle

[quote="13375P34K43V312"]A triangle has two vertices at (0,0) and (2,0). It's third point is to be chosen on the circle with center (3,2) and radius 1. What is the maximum area of this triangle?[/quote]

Hint: The area of the triangle is given by $A=\frac1{2}xy$ and it must be that $(x-3)^{2}+(y-2)^{2}=1$.

[hide] SInce the area of a triangle is half the base times the height, and you know the base is 2, you just need to maximize the y coord. It is clearly 3 so the area is clearly 3.[/hide]

[hide] you see that the triangle has a base of 2, because if you count, the number of unit squares from (0,0) to (2,0) is 2. You know the base, and you know the formula for area of the triangle,
$\frac{1}{2}bh$. this becomes:
$\frac{2}{2}h$ when you substitute the base in and then multiply.
this also simplifies to:
$h$. So now, you know that the area of the triangle will be equal to the height, which is the $y$ coordinate value for the point that completes the triangle. Then, you can take a look at the restriction for the point; it must be on the circle with radius 1 and center at the point(3,2). You want the northernmost point, which is easily gotten by adding one to the y-coordinate of the center. So, the coordinate that you choose becomes (3,3). you take the $y$ coordinate, 3, and since you know from the previous equation that the area of the triangle equals the height, the area must be:
$\boxed{3}$
[/hide]


symmetry
number theory unsolved
number theory

[quote="April"]How many pairs $ (m,n)$ of positive integers with $ m < n$ fulfill the equation $ \frac {3}{2008} \equal{} \frac 1m \plus{} \frac 1n$?[/quote]
It's equivalent to: $ 9mn \equal{} 3*2008(m\plus{}n) \iff (3m\minus{}2008)(3n\minus{}2008) \equal{} 2008^2$. Thus $ 3m\minus{}2008 \equiv 3n\minus{}2008 \equiv \minus{}1 \bmod 3$. So if we get numbers $ ab \equal{} 2008$ such that $ 3m\minus{}2008 \equal{} a$ and $ 3n\minus{}2008 \equal{} b$ we want $ a \equiv b \equiv \minus{}1 \bmod$. Which happens iff $ a \equiv \minus{}1 \bmod 3$. There are $ 10$ positive numbers $ a$ such that $ a \equiv \minus{}1 \bmod 3$. Because of symmetry $ 5$ of them will have $ a > b$ so the answer is $ \boxed{5}$

We can say that some $d$ is the greatest common divisor of $m$ and $n$, so we can write $m=dx$ and $n=dy$. We will now put it in the equation:
$\frac {3}{2008} = \frac {1}{dx}+ \frac {1}{dy}$ or $3dxy=2008(x+y)$. Now, $2008=2 \cdot 2 \cdot 2 \cdot 251$, so we have few cases that can satisfy this equation, because $2008$ is not divisible by $3$, that means that $x+y$ has to be divisible, so these are pairs of $x,y$ that will satisfy this condition: $(x,y)=(1,2);(1,8);(1,251);(1,1004);(4,251)$
We can find $d= \frac {2008(x+y)}{3xy}$, so $d=(1004);(753);(672);(670);(170)$
Finally $(m,n)=(1004,2008);(753,6024);(672,672 \cdot 251);(670,670 \cdot 1004);(680,170 \cdot 251)$
So, there are $5$ solutions for this equation :)


function
floor function
algebra
algebra proposed

Prove that $ m! \ge (n!)^{[\frac{m}{n}]}$ ,for all positive integers $ m$ and $ n$

[quote="Victory.US"]Prove that $ m! \ge (n!)^{[\frac {m}{n}]}$ ,for all positive integers $ m$ and $ n$[/quote]

Is $ [\frac {m}{n}]$ floor function?

If so , then for $ m \ge n$, $ m! \ge (n \cdot \lfloor \frac {m}{n} \rfloor )! \ge (n!)^{\lfloor \frac{m}{n} \rfloor}$ ,

for $ m<n$, $ m! \ge (n!)^{\lfloor \frac {m}{n} \rfloor}\equal{}(n!)^0\equal{}1$


In this: http://www.osinfofrom.us/solprob1/solution_problem1.html

on the line that starts with: "In triangle ABC", does the author of the solution just notice this or is it a theorem or a result of one?

It would help to note that $ \triangle BDC\sim \triangle ABC$ because $ \angle BAC\equal{}\angle DBC$ and $ \angle C\equal{}\angle C$.  Thus $ \angle ABC\equal{}135^{\circ}$ and the rest is just from law of sines.

300th post!


Chinese Southeast Mathematical Olympiad 2006 has just finished and the problems are posted in the national olympiad forum. Valentin could you please post them in the Resources as well?

Valentin has not come here for quite a long time. We must wait for other moderators to do it.


The centers of all faces of a cube are joined to form an octahedron. The centers of all the faces of this octahedron are now joined to form a smaller cube. What is the ratio of an edge of the smaller cube to an edge of the original cube?

This is somewhat intuitive...

Say the side length of the large cube is $ s.$ Note that the side length of the octahedron is  \[ \sqrt{(\frac{s}{2})^2+(\frac{s}{2})^2}=\frac{s\sqrt{2}}{2}.\]
Now think about the center of a cube that is $ \frac{1}{8}$ of the large cube (which means its side length is $ \frac{s}{2}),$ and shares one vertex and 3 edges (in other words, it's placed in the corner of the large cube). This is the exact center of the triangular face of the octahedron. The distance between them is $ 2(\frac{\frac{s}{2}}{2})=\frac{s}{2},$ thus the ratio is $ \boxed{\frac{1}{2}}.$


modular arithmetic
number theory unsolved
number theory

1.If N= (3^a)(5^b)(7^a) ,then

(i) find total number of divisors of N of the form of (4k-1).

(ii)find total number of divisors of N of the form of (4k+1).

(iii)find total number of divisors of N which are divisible by 225.

(iv)find total number of divisors of N  which are divisible by 75.

2.Let f(n)= ((1000)^n)/(n!) , then for which of the  following four values

(n=997,998,999,1000)

 is f(n)  greatest

1)Consider $ f(n)$ as follow
$ f(n)\equal{}(\minus{}1)^{n\minus{}1}$ if n odd
$ f(n)\equal{}0$ if n even.
Then $ f(n)$ is multiple.
So you can find out .
2)Quite easy.

thanx mszew.
hello,TTsphn what is f(n). I do not understand.Also i have edited problem 1.

$ f(n) \equal{} 1, n \equal{} 4k\plus{}1$
$ f(n) \equal{}\minus{}1 , n \equal{} 4k\plus{}3$
$ f(n) \equal{} 0 , n \equal{} 2k$
I have a problem from my teacher
$ f(n) \equal{}\sum_{d|n,d odd}(\minus{}1)^{d\minus{}1}$
Prove that $ f(n)$ different from 0 then $ n \equal{} a^{2}\plus{}b^{2}$

[quote="pankajsinha"]1.If N= (3^a)(5^b)(7^a) ,then

(i) find total number of divisors of N of the form of (4k-1).

(ii)find total number of divisors of N of the form of (4k+1).

(iii)find total number of divisors of N which are divisible by 225.

(iv)find total number of divisors of N  which are divisible by 75.

[/quote]

$ N\equal{}3^{a}\cdot5^{b}\cdot7^{a}\equal{}21^{a}\cdot5^{b}$

i) and ii) Since $ 21\equiv1\pmod{4}$, $ 21^{a}\equiv1^{a}\equiv1\pmod{4}$, and since $ 5\equiv1\pmod{4}$, $ 5^{b}\equiv1^{b}\equiv1\pmod{4}$. Thus $ 21^{a}\cdot5^{b}\equiv{}1\pmod{4}$. So no divisors of $ N$ are in the form 4k-1 and all are in the form 4k+1.

You haven't supplied any information about $ a$ or $ b$, so I cannot answer (iii) or (iv).

Hello Anaxerzia, if a=4,then is 27 which is a divisor of N  not of the form 4k-1.
Also,TTsphn i still do not understand what you are trying to say by defining f(n).

If we choose odd number of 3's and even number of 7's and vice versa and multiply by any

 number of 5's then we get a divisor which is of the form 4k-1. So how should i count the cases.

Is there adirect way to arrive at the answer.

[quote="pankajsinha"]1.If N= (3^a)(5^b)(7^a) ,then

(i) find total number of divisors of N of the form of (4k-1).

(ii)find total number of divisors of N of the form of (4k+1).

(iii)find total number of divisors of N which are divisible by 225.

(iv)find total number of divisors of N  which are divisible by 75.

[/quote]
If a prime divisor of N is 1 mod 4 then all its powers are 1 mod 4.
If a prime divisor of N is 3 mod 4, its even powers are 1 mod 4 and odd powers are 3 mod 4.
Considering only positive divisors, No. of positive divisors of form 4k+1=([a/2]+1)^2*(b+1).
No. of positive divisors of N of form 4k-1 = (a-[a/2])^2*(b+1), where [x]: step function.
If x is a divisor of N, then -x is a divisor of N. If x=1 mod 4 then -x = -1 mod 4.
There will be equal no. of divisors of form 4k+1 and 4k-1, if we take into consideration negative divisors also.
So, No. of divisors of form 4k+1 or 4k-1 = Total No. of divisors of N/2 = (a+1)^2*(b+1).

No. of divisors of N divisible by x = No. of divisors of N/x
iii) If x=225 then ans = 2*(a^2-1)*(b-1) if both a,b>1. else 0
iv) If x=75 then ans = 2*a*(a+1)*(b-1) if a>0 and b>1. else 0


calculus
integration
combinatorics unsolved
combinatorics

[quote="MNrule"]Each side of a triangle is an integral number in centimetre, no side exceeding $ c$ centimetres. Prove that the number of different triangles which can be formed is $ \frac {1}{24}c(c \plus{} 2)(2c \plus{} 5)$ or $ \frac {1}{24}(c \plus{} 1)(c \plus{} 3)(2c \plus{} 1)$, according as $ c$ is even or odd. Also, show that the number of isoceles or equilateral triangles is $ \frac {3}{4}c^2$ or $ \frac {1}{4}(3c^2 \plus{} 1)$, according as $ c$ is even or odd.[/quote]

From you post here : [url]http://www.mathlinks.ro/viewtopic.php?p=1509215#1509215[/url]

[quote="MNrule"]The sides of a triangle are a, b and c where a, b and c are integers and $ a\leq b\leq c$. If c is given, show that the number of different triangles is $ \frac {1}{4}c(c \plus{} 2)$ or $ \frac {1}{4}(c \plus{} 1)^2$ , according as c is even or odd. Also, show that the number of isoceles or equilateral triangle is $ \frac {1}{2}(3c \minus{} 2)$ or $ \frac {1}{2}(3c \minus{} 1)$ , according as c is even or odd.[/quote]

When $ c$ is even, number of different triangles formed
$ \equal{}\frac {1}{4}(1 \plus{} 1)^2\plus{}\frac {1}{4}(2)(2 \plus{} 2)\plus{}\frac {1}{4}(3 \plus{} 1)^2\plus{}\frac {1}{4}(4)(4 \plus{} 2)\plus{}...\plus{}\frac {1}{4}((c\minus{}1) \plus{} 1)^2\plus{}\frac {1}{4}c(c \plus{} 2)$

$ \equal{}\frac{1}{2}(1\plus{}1)(1\plus{}2)\plus{}\frac{1}{2}(3\plus{}1)(3\plus{}2)\plus{}.....\plus{}\frac{1}{2}c(c\plus{}1)$
$ \equal{}\frac{1}{2}[(2^2\plus{}4^2\plus{}6^2\plus{}...\plus{}c^2)\plus{}(2\plus{}4\plus{}6\plus{}...\plus{}c)]$
$ \equal{}2[1^2\plus{}2^2\plus{}...\plus{}(\frac{c}{2})^2]\plus{}(1\plus{}2\plus{}3\plus{}...\plus{}\frac{c}{2})$
$ \equal{}2 \times \frac{\frac{c}{2}(\frac{c}{2}\plus{}1)(2 \times \frac{c}{2}\plus{}1)}{6}\plus{}\frac{1}{8}c(c\plus{}2)$
$ \equal{}\frac {1}{24}c(c \plus{} 2)(2c \plus{} 5)$

When $ c$ is odd, number of different triangles formed
$ \equal{}\frac {1}{24}(c\plus{}1)((c\plus{}1) \plus{} 2)(2(c\plus{}1) \plus{} 5)\minus{}\frac{1}{4}(c\plus{}1)((c\plus{}1)\plus{}2)$
$ \equal{}\frac{1}{24}(c\plus{}1)(c\plus{}3)[(2c\plus{}7)\minus{}6]$
$ \equal{}\frac {1}{24}(c \plus{} 1)(c \plus{} 3)(2c \plus{} 1)$

When $ c$ is even , number of isoceles or equilateral triangles
$ \equal{}[\frac{1}{2}(3(1)\minus{}1)\plus{}\frac{1}{2}(3(3)\minus{}1)\plus{}\frac{1}{2}(3(5)\minus{}1)\plus{}...\plus{}\frac{1}{2}(3(c\minus{}1)\minus{}1)]$
$ \plus{}[\frac{1}{2}(3(2)\minus{}2)\plus{}\frac{1}{2}(3(4)\minus{}2)\plus{}\frac{1}{2}(3(6)\minus{}2)\plus{}....\plus{}\frac{1}{2}(3c\minus{}2)]$
$ \equal{}\frac{3}{2}(1\plus{}2\plus{}3\plus{}...\plus{}c)\minus{}\frac{1}{2} \times \frac{c}{2}\minus{}\frac{c}{2}$
$ \equal{}\frac{3}{4}c(c\plus{}1)\minus{}\frac{3}{4}c$
$ \equal{}\frac{3}{4}c^2$

When $ c$ is odd , number of isoceles or equilateral triangles
$ \equal{}\frac{3}{4}(c\plus{}1)^2\minus{}\frac{1}{2}(3(c\plus{}1)\minus{}2)$
$ \equal{}\frac{3c^2\plus{}6c\plus{}3\minus{}(6c\plus{}2)}{4}$
$ \equal{}\frac{1}{4}(3c^2\plus{}1)$


You serious? WTF! :cursing:

I really want to see a game at InBev stadium...Or will someone else buy naming rights?