Gauss
polynomial
algebra
inequalities
trigonometry
Vieta
algebra proposed

prove that $x^n-x-1$ is irreducible over $\mathbb{Q}$ for all $n \geq 2$.

By Gauss lemma, it is enough to check the irreducibility in $\mathbb{Z}[X]$. Eisenstein criterion doesn't apply. This polynomial has no rational root (by the rational root test). I tried to prove the irreducibility in $\mathbb{Z}/2\mathbb{Z}$ by hand (which would be enough to conclude, since the polynomial is monic), but I couldn't find the trick (neither am I sure this works). Anyone ?

The wording of the problem is surely false. For instance if $n=5$, that polynomial is reducible over $\mathbb{Q}$.

[quote="Christian Hirsch"]Why is it reducible for $n=5$? I think for $n$ prime the problem is classic...[/quote]

how do you prove it when n is prime?

Hint: when $n$ is prime you first prove that it is irreducible in $\mathbb{Z}/n\mathbb{Z}$; as a result it will be irreducible in $\mathbb{Q}$.

It's enough to prove that it is irreducible $\mod n$. So lets work $\mod n$ from now.

Faktor $f(x)=x^n-x-1$ totally as product of polynomials. Since $f(x+1)=f(x)$, the shift $x \to x+1$ permutes the factors of $f$.
Take any factor $g(x)$.
When $g(x+k)=g(x)$ for $k \neq 0$ holds for all different $x$, then $g$ is constant or has at least degree $n$ (the polynomial $g(x)-g(0)$ has $n$ roots).
So every nontrivial factor fulfills $g(x+k) \neq g(x)$ for the $n-1$ different $k$, so all we can reach out of $g(x)$ by shifting are different. Since every factor has at least degree $1$, we get $n$ factors of degree $1$, thus all of them. But a factor of degree $1$ implies that there is a root, which is impossible by Fermat's Little Theorem.

This is a theorem of Selmer (the general case) and I find it an extremely beautiful and difficult way of showing that studying roots may always be a way to solve it. Here's the idea. Consider a factorization $X^{n}-X-1=f(X)g(X)$ for some integer polynomials $f,g$, nonconstant. Thus $f$ will have some roots $z_{1},z_{2},...,z_{s}$ of $X^{n}-X-1$, which are clearly pairwise distinct. Now,
the magical fact is the following:
   Lemma: for any root $z$ of $X^{n}-X-1$ one has $2 Re(z-1/z)>1/|z|^{2}-1$.
   Indeed, writing $ze^{it}$, the inequality comes down to $(1+2r\cos t)(r-1)>0$. However, $r^{2n}=|z|^{2n}=|z+1|^{2}=1+2r\cos t+r^{2}$, so what we need is $(r^{2n}-r^{2})(r^{2}-1)>0$, which is clear (modulo the easy exercise: show that $r$ is not $1$).
    This means that $2Re(z_{1}-1/z_{1})+...+2Re(z_{s}-1/z_{s})>1/|z_{1}|^{2}+...+1/|z_{s}|^{2}-s\geq 0$, because the product of $|z_{i}|$ is just $|f(0)|=1$. Thus $Re(z_{1}-1/z_{1}+...+z_{s}-1/z_{s})>0$. On the other hand, because $f$ is monic and has integer coefficients, $Re(z_{1}-1/z_{1}+...+z_{s}-1/z_{s})$ is an integer, so it is actually at least $1$. Working similarly with $g$, we deduce that $Re(z_{1}-1/z_{1}+...+z_{n}-1/z_{n})\geq 2$, where $z_{1},...?z_{n}$ are all roots of $X^{n}-X-1$. However, this is impossible, because by Vieta's formulae $z_{1}-1/z_{1}+...+z_{n}-1/z_{n}=1$. Done! By the way, Alekk, I see you posted this in the proposed section. Do you have another solution?

[quote="ZetaX"]When $g(x+k)=g(x)$ for $k \neq 0$ holds for all different $x$, then $g$ is constant or has at least degree $n$ (the polynomial $g(x)-g(0)$ has $n$ roots).[/quote]

I think this argument is valid only for prime $n$:
Take n=4. Then $g(x)=2x^{2}+2x$ has 4 roots but is not constant (as a polynomial).

he was talking about n-prime (should understand that from the fact that he mentioned Fermat's little theorem) :)

What about the polynomial $x^{k}-x+k?$ I know a proof about the irreducibility for $k=p^{n}.$


sum of x^2 is divergen starting at 1 going to infinity[/list][/u]


pigeonhole principle
modular arithmetic
number theory
relatively prime

Let $ n$ be a positive integer. Show that there always exists a positive integer $ m$, whose digits are all 0's or 3's, which is divisible by $ n$.

Pigeon hole, $ 3,33,333,3333,...,333...333$ where the last one has n+1 3's.

By pigeonhole, 2 of them are congruent mod n, subtract those 2 numbers to get your desired number.

[hide="Solution"]First consider $ n$ relatively prime to 10.  By Euler's Theorem we have $ 10^{\phi (n)}\equiv 1\pmod{n}$ and therefore $ 10^{k\phi (n)}\equiv 1\pmod{n}$ for all $ k$.  Therefore, $ 3\cdot 10^{\phi (n)}\plus{}3\cdot 10^{2\phi (n)}\plus{}\cdots \plus{}3\cdot 10^{n\phi (n)}\equiv 3n\equiv 0\pmod{n}$, and this number's digits are all 0's or 3's.

Now consider $ n$ not relatively prime to 10.  Let $ n\equal{}2^{a}5^{b}n'$, where $ n'$ is relatively prime to 10.  Then we can find $ m$ whose digits are all 0's or 3's that is divisible by $ n'$.  Adding sufficiently many 0's to the end of the decimal representation of $ m$ gives us a number whose digits are all 0's or 3's that is divisible by $ n$ as desired, and we're done.[/hide]


These are called Franel numbers. No closed form exists (however such a closed form exists for sums of first and second powers of binomials).
For more information: http://mathworld.wolfram.com/BinomialSums.html


calculus
integration
trigonometry
calculus computations

Let $ n\geq 1$ be a natural number. Calculate

$ \int\limits_{0}^{\frac{\pi}{2}}\frac{dx}{(\sqrt[n]{\sin x}\plus{}\sqrt[n]{\cos x})^{2n}}.$


number theory
greatest common divisor
algebra
polynomial
continued fraction
Rational Root Theorem

Could someone tell me what the easiest proof that shows$ \sqrt{2}$ is irrational?

thanks!

This is one of my favorites!

[hide="Proof"]
Assume that $ \sqrt{2}$ is rational. In this case, there must exist a fraction $ a/b$ that equals $ \sqrt{2}$. Now if $ a$ and $ b$ share any common factors, they simplify to $ p\q$, which is assumed to be the simplest fraction of  $ \sqrt{2}$. Either $ p$ and $ q$ must be odd, because if they were even, then they would not be the simplest form. Thus we have the following:

$ \frac{p}{q}=\sqrt{2}$

$ \frac{p^{2}}{q^{2}}= 2$

If we multiply both sides by $ q^{2}$ we get $ p^{2}= 2q^{2}$. From this, we deduce that $ p$ is even because anything times 2 would be even.

This is a contradiction, because $ p$ cannot be even and odd at the same time. Therefore, no such fraction exists to equal $ \sqrt{2}$ and therefore $ \sqrt{2}$ is irrational. QED.
[/hide]

[hide]

Don't know if this is the esiest...

Let $ \sqrt{2}\equal{}\frac{a}{b}$ with $ \gcd(a,b)\equal{}1$. Therefore $ 2b^{2}\equal{}a^{2}$

Since $ 2$ divides the Left Hand Side it also divides the Right Hand Side, implying that $ a$ should be even. So $ a\equal{}2m$:

$ 2b^{2}\equal{}4m^{2}\implies b^{2}\equal{}2m^{2}$. By the same argument $ b$ is even. Since they're both even they have a common factor, $ 2$, contradicting the definition.

[/hide]

well we prove this bu assuming that $ \sqrt{2}$ is rational,let $ \sqrt{2}\equal{}\frac{a}{b}$ where $ a$ and $ b$ are two positive integers such that $ \text{gcd}(a,b)\equal{}1$, then we must have $ 2\equal{}\frac{a^{2}}{b^{2}}$ or $ a^{2}\equal{}2b^{2}$. because the LHS is perfect square so $ 2b^{2}$ must be perfect square too but this is impossible because if we assume that $ b\equal{}2^{s}t$ where $ s$ is a nonnegative integer and $ t$ is odd then $ a^{2}\equal{}2^{2s\plus{}1}t$  therefore RHS is not perfect square and this is contradiction so $ \sqrt{2}$ cant be a rational number.

[quote="7h3.D3m0n.117"]
Both $ p$ and $ q$ must be odd, because if they were even, then they would not be the simplest form.
[/quote]
What about $ 2/5$ or $ 7/4$? I'm pretty sure those are in simplest form.

You need to slightly refine the argument. ;)

[quote="MellowMelon"]You need to slightly refine the argument. ;)[/quote]

Whoops  :blush:  fixed.

[quote="t0rajir0u"]Since three posts have given the same proof, I might as well list some common proofs including the one already given.

[hide="Proof 1"] Suppose $ \sqrt{2}\equal{}\frac{a}{b}$ with $ a, b$ integers.  Then $ a^{2}\equal{} 2b^{2}$.  The largest exponent of $ 2$ dividing the LHS is even, but the largest exponent of $ 2$ dividing the RHS is odd; contradiction. [/hide]
[hide="Proof 2 (my personal favorite)"] $ \sqrt{2}$ is a root of $ x^{2}\minus{}2 \equal{} 0$, and by the Rational Root Theorem this polynomial has no rational roots. [/hide]
[hide="Proof 3"] Suppose $ m$ is the smallest positive integer such that $ m\sqrt{2}$ is an integer, say $ n$.  Then

$ (n\minus{}m)\sqrt{2}\equal{} 2m\minus{}n$

Is also an integer.  But since $ n < 2m$ we have $ n\minus{}m$ is a positive integer less than $ m$ such that $ (n\minus{}m)\sqrt{2}$ is an integer.  Contradiction. [/hide]
[hide="Proof 4"] The continued fraction representation of $ \sqrt{2}$ is provably periodic, therefore infinite. [/hide][/quote]

number 2 is awesome.
Here is another proof (includse the root 2 proof)
http://en.wikipedia.org/wiki/Irrational_number


[quote="salt_n_vineger"]Hey, my name is Kendall and I'm in 11th grade.
I was wondering if anyone could help me with the problem my math extension teacher gave me.
It's like this: when a^2+(b^2-4b+1)a-(b^4-2b^3)=0, prove that b^2+a is square.[/quote]

I doubt that is true because when $b=1$ we have $a^2-2a+1=0$ which implies $a=1$ but $b^2+a=1+1=2$ is not a square ....

I agree with shyong, I think it's not true.


quadratics
LaTeX
Gauss
number theory unsolved
number theory

These aren't really problems, jsut little things that I was pondering about.

(1) Is the condition that 2 is a quadratic residue mod p, where p is prime;  equivalant to uhm, something simpler; like an easier condition p (e.g. perhaps p = 2 mod 3 or something, but that doesnt work). (i know it is the same as saying 2 ^ (p-1/2) == -1 (mod p), but i was wondering if you could simplify it further). note that quadratic reciprocity doesnt work.

(2) Prove that there are infinitely many primes of the form 2*3^s + 1.
Extnesion: prove there are infinitely many such primes that also satisfy the condition imposed in (1), i.e. there exist infinitely many s s.t. 2*3^s+1 is prime with 2*3^s + 1 | 2 ^ (3 ^ s) + 1.

Vintoh I had the impression that you know how to use $\LaTeX$. If so, please do it. If not, then please follow the link in my signature to quickly learn it ;)

about the first question:
[url=http://mathworld.wolfram.com/EulersCriterion.html]mathworld.wolfram.com/EulersCriterion.html[/url]
on your own - prove the Gauss Lemma [url=http://mathworld.wolfram.com/GausssLemma.html]mathworld.wolfram.com/GausssLemma.html[/url] and then you'll see how the quadratic reciprocity law works for $2$.

For the first one, $2$ is a quadratic residue modulo $p$ iff $\frac{p^2-1}8$ is even (or, in other words, $\left(\frac 2p\right)=(-1)^{\frac{p^2-1}8}$).

As for the second one, if I understand your non-$\LaTeX$-ed post (:)), you want to prove that there are infinitely many primes of the form $2\cdot 3^s+1$. I really think this is an unsolved problem.

hey,

thanks for that, i understand it now. yea, that is precisely what i meant with the prime thing in the second part. sorri about not using latex; well rite now i should probably be preparing for the imo, so i'll learn latex later, after the imo. sorry about that again.

Yeah, how do u learn Latex. I trued before but got nowhere. You get loads of webpages upon clicking signayure.

Bomb


[quote="A. S. K."]Please, can someone give me a link of some good PDF about algebra, with solutions? Thank you![/quote]
Hi, A.S.K
I think you should place your post at:[url]http://www.mathlinks.ro/Forum/index.php?f=121[/url]
or:[url]http://www.mathlinks.ro/Forum/index.php?f=71[/url]
and you can find some books in:[url]http://www.amazon.com[/url]
good luck!


geometry
parallelogram
ratio
perpendicular bisector
geometry unsolved

Let $ D$ the midpoint of side $ BC$ in a triangle $ ABC$.Let  $ E,Z$ be the projections of $ D$ on $ AB,AC$ resp/ly and $ T$ the intersection of the tangents at points $ E,Z$ to the circle with diameter AD . Prove that $ TB\equal{}TC$.

Babis

[hide="hint or ..."]
it may be posted before ,
it sufficies to prove that $ T$ lies on the radical axe of the circles with diameter $ BD$ and $ CD$
for this we draw these circles and then by denoting the point of their intersections with $ ET$ and $ ZT$ by $ E'$ and $ Z'$ respectively we use sinuses law to prove that $ EE'\equal{}ZZ'$ and since $ ET\equal{}ZT$ we are done!
[/hide]

Let $ M,N$ be the midpoints of $ AC,AB$ and $ K \equiv MN \cap EF.$ In the parallelogram $ AMDN,$ we have $ \angle NDE \equal{} \angle MDF,$ due to $ \angle BND \equal{} \angle CMD.$ Let $ L$ be the projection of $ D$ on $ MN.$  From the quadrilaterals $ LDEN$ and $ LDFM$ inscribed in the circles with diameters $ND,MD,$ we have $ \angle NLE \equal{} \angle NDE \equal{} \angle MDF \equal{} \angle MLF$ $ \Longrightarrow$ $ LD,MN$ bisects $ \angle ELF$ internally and externally. If $ J \equiv LD \cap EF,$ then cross ratio $ (E,F,J,K)$ is harmonic $\Longrightarrow$ $K$ is the pole of the perpendicular bisector $ LD$ of $ BC$ WRT the circle $ \omega$ with diameter $ AD.$ Since $ T$ is the pole of $ EF$ WRT $ \omega$  $\Longrightarrow$ $T \in LD,$ i.e. $ TB \equal{} TC,$ as desired.


Sam decides to start a rumor. Sam tells the rumor to her three friends. Each of Sam’s three friends then tells the rumor to three friends who have not heard the rumor. This continues for five total cycles. Sam telling her three friends was the first cycle. How many people, not including Sam, will have heard the rumor when the fifth cycle is complete?

Keep multiplying by four to count the three new people, as well as the old one-
3 other than Sam
12 after2
48
192
768 after five

$ 768$

120
Each cycle, every person tells 3 more people so you would multiply your current number by 3 each cycle.
so the pattern is: 1,3,9,27,81.
add them up and you get 120 (not including sam).
(I believe)

Umm...I think that the people who have told the rumor once don't tell it anymore...
I'm not sure though, the question does not specify it.

Well, first of all, after the first cycle, there are 3 people, not 1 person who knows the secret.

And, my solution is if, assuming, each friend is able to tell as many times as they want, which now that I think about it, is probably incorrect. So,

after 1-3 total, 3 new
after 2- 12 total, 9 new
after 3- 39 total, 27 new
after 4- 120 total, 81 new
after 5- 363 total, 243 new.

Add by powers of 3, I think...

Oh oops, I did one less cycle than there should be.

I think your mistake, guys, is that you forgot to add the powers together:
3^1 + 3^2 + 3^3 + 3^4 + 3^5 = 3 + 9 + 27 + 81 + 243 =  [b]363[/b].


vector
symmetry
invariant
real analysis
real analysis unsolved

Consider the system of differential equations $ X'(t)\equal{}A(t)X(t)$ where $ X: [0,1]\to \mathbb C^n$ (sic!) and $ A(t)$ is a continuous real symmetric $ n\times n$ matrix. Suppose that $ x_0,x_1$ are two [i]generic[/i] vectors in $ \mathbb C^n$. Is it always possible to choose $ A$ so that the solution starting at $ X(0)\equal{}x_0$ ends at $ X(1)\equal{}x_1$? This is obviously false when $ n\equal{}1$ and a bit less obviously false when $ n\equal{}2$, but you can assume $ n$ to be as large as you wish. By "generic" I mean that you can exclude any set of measure $ 0$, say.

We can represent $ X$ as a pair of vectors in $ \mathbb R^n$. Let's consider the set $ L$ of all ordered pairs $ (u,v)$ of linearly independent vectors in $ \mathbb R^n$. When $ n > 2$, $ L$ is connected. We fix a "starting point" $ (u_0,v_0)$ in $ L$ and let $ F$ be the set of all possible "finish points" $ (u_1,v_1)$. It seems to me that $ F$ is open in $ L$ (and, by symmetry, closed).

My reason to think that $ F$ is open is that it is invariant under any "unidirectional stretch", where distances along a chosen line are multiplied by a positive factor. I haven't made the calculations, but this should suffice to move a pair of vectors slightly in any direction.


LaTeX
real analysis
real analysis unsolved

problem17: x1>1, x_(n+1)x_n-2x_n+1=0, n natural.find lim(n->infinity)(x_n)^n.

For the love of all that is legible, use Latex!
[quote="martin1985"]$x_{1}>1$, $x_{n+1}x_{n}-2x_{n}+1=0$, $n$ natural.  Find $\lim_{n\to \infty}x_{n}^{n}$.[/quote]

[hide="A beginning"]I claim that $x_{n}= \frac{nx_{1}-(n-1)}{(n-1)x_{1}-(n-2)}$.  This is clear for $n = 1$, and solving for $x_{n+1}$ in the recursion relation gives the desired result by induction.  Note that for $n > 1$,
\[\frac{nx_{1}-(n-1)}{(n-1)x_{1}-(n-2)}= \frac{n}{n-1}-\frac{1}{(x-1)(n-1)^{2}+(n-1)}\]
and so
\[\frac{n}{n-1}-\frac{1}{(x-1)(n-1)^{2}}<x_{n}< \frac{n}{n-1}. \]
[/hide]


calculus
integration
trigonometry
function
calculus computations

Find
\[ \int_0^1 (1 - x^{2/3})^{3/2}dx  \].

How is this supposed to be done? Letting Maple find the primitive doesn't even give you a real result.

Chase substitutions.
Let $u=1-x^{\frac23}$, $x=(1-u)^{\frac32}$, so $dx=-\frac32\sqrt{1-u}$. The integral becomes $\frac32\cdot\int_0^1 u^{\frac32}\sqrt{1-u}\,du=\frac32\cdot\int_0^1 u\sqrt{u-u^2}\,du$. This can be evaluated by conventional means; writing that outer $u$ as $u-\frac12+\frac12$ would help.

And $\int_0^1u^{\frac32}(1-u)^{\frac12}\,du$ is a specific case of the beta integral:

$\int_0^1x^{a-1}(1-x)^{b-1}\,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$

In this particular case,

$\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)}{\Gamma(4)}= \frac{\frac32\cdot\frac12\cdot\sqrt{\pi}\cdot\frac12\cdot\sqrt{\pi}}{3!}=\frac{\pi}{16}.$

[added in edit]

Note from jmerry's post that this integral is multiplied by $\frac32.$ So the original integral is $\frac32\cdot\frac{\pi}{16}=\frac{3\pi}{32}.$ Hence, jmerry, kunny, and I are all in agreement.

Thanks. Is it necessary to complete the square and make trig substitutions and stuff, or is there a faster way (even using your nifty trick I still end up having to use trigonometric substitutions)?

Yes! This is relatred to asteroid, $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1\ (0\leq x\leq 1),$so you can set $x=\cos ^ 3 \theta, y=\sin ^ 3 \theta\ \left(0\leq \theta \leq\frac{\pi}{2}\right).$

We have $\int_0^1 \left(1-x^{\frac{2}{3}}\right)^{\frac{3}{2}} dx=\int_{\frac{\pi}{2}}^0 \sin ^ 3 \theta\cdot 3\cos ^ 2 \theta(-\sin \theta)d\theta=3\int_0^{\frac{\pi}{2}}(\sin ^ 4 \theta-\sin ^ 6 \theta)d\theta.$

Now let $I_n=\int_0^{\frac{\pi}{2}} \sin ^ n d\theta\ (n=0,1,2,\cdots)$, we have $I_n=\frac{n-1}{n}I_{n-2}\ (n=2,3,\cdots),$

Our definite integral is $3(I_4-I_6)=3\left(I_4-\frac{5}{6}I_4\right)=\frac{1}{2}I_4=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{1}{2}I_0=\frac{3}{16}\cdot \frac{\pi}{2}=\frac{3}{32}\pi.$

That is a nice solution. Easier than the other one (although coming up with the substitution seems difficult).

I think this is a difficult problem. I wouldn't have solved it :( Maybe I need more experience. Thanks everyone.

Here's a variation on the beta integral:

$\int_0^{\frac{\pi}2}\cos^{\alpha}\theta\sin^{\beta}\theta\,d\theta = \frac{\Gamma\left(\frac{\alpha+1}2\right)\Gamma\left(\frac{\beta +1}2\right)} {2\Gamma\left(\frac{\alpha+\beta+2}2\right)}$

The most interesting particular case of this is

$\int_0^{\frac{\pi}2}\cos^{2j}\theta\sin^{2k}\theta\,d\theta =\frac{1\cdot3\cdots(2j-1)\cdot1 \cdot 3\cdots (2k-1)\pi}{2^{j+k+1}(j+k)!}.$

kunny computed the $j=1,k=2$ case:

$\int_0^{\frac{\pi}2}\cos^2\theta\sin^4\theta\,d\theta= \frac{1\cdot1\cdot3\cdot\pi}{2^43!} =\frac{\pi}{32}.$

Although I do appreciate it, I don't want to use methods that are beyond what this course is about. I barely know what the gamma function is! Some perspective is always nice though.

I thought I'd use this thread for a different question... is there a way to generally integrate (a + x^2)^n for positive integer n WITHOUT expanding it, which can be supertedious? We went through how to integrate 1/(x^2 + a)^n today (positive integer n), and I thought it was very complicated, so I expect the answer to be no.

The trig substitution will get you into trigonometric power integrals.

If you make the substitution $x=a\tan\theta,$ then

$\int\frac{dx}{a^2+x^2}$ becomes $\frac1{a^{2n-1}}\int\cos^{2n-2}\theta\,d\theta$

$\int(a^2_x{}^2){}^n\,dx$ becomes $a^{2n+1}\int\sec^{2n+2}\theta\,d\theta.$

The usually-recommended way to integrate an even power of the secant would simply undo the substitution and take you back to the original polynomial. If you don't want to do that, you can use an integration-by-parts based reduction formula method.

Of course, you could do the reduction without the trig substitution:

$\int(a^2+x{}^2){}^n\,dx=\int a^2(a^2+x{}^2){}^{n-1}\,dx+\int x^2(a^2+x{}^2){}^{n-1}\,dx$ where

$\int x^2(a^2+x{}^2){}^{n-1}\,dx=\frac x{2n}(a^2+x{}^2){}^n-\frac1{2n}\int(a^2+x{}^2){}^n\,dx.$

Now we do some algebra to solve for $\int(a^2+x{}^2){}^n\,dx$ in terms of $\int(a^2+x{}^2){}^{n-1}\,dx$

I'll let you finish off the details.

$\int x^2(a^2+x^2)^{n-1}\,dx=\frac x{2n}(a^2+x^2)^n-\frac1{2n}\int(a^2+x^2)^n\,dx.$
How did you get this? While it is right, I don't see it being integration-by-parts at all! :)

edit: ahh, I see now that it's partial integration of $\frac1{2n}\int(a^2+x^2)^n\,dx.$. Nevermind. Tricky indeed.


function
logarithms
calculus
calculus computations

Assume that the rate of change in the number of miles $ s$ of road cleared per hour by a snowplow is inversely proportional to the height $ h$ of the snow.

(a)  Write and solve the differential equation to find $ s$ as a function of $ h$.

(b)  Find the particular solution if $ s \equal{} 25$ miles when $ h \equal{} 2$ inches and $ s \equal{} 12$ miles when $ h \equal{} 10$ inches.

$ \frac {dS}{dh} \equal{} \frac {k}{h}$ for some constant k

This is a separable differential equation, which can be solved with a simple integration. Do that and plug in your given values to find your constant of porportionality.

I get $ S \equal{} 30.6 \minus{} 8.07 \ln h$


A vessel initially contains $ V_0$ L of water with $ x_0$kg salt dissolved in it. [b]Pure water[/b] is added to it from the top at the rate of $ m$ L/s and the solution is allowed to flow out from an outlet at the bottom at the rate of $ n$L/s such that $ m > n$ and salt concentration is assumed to be uniformly distributed throughout the mixture all the time.

Therefore, find amount of salt (in kg) left in the vessel (in dissolved state) after time $ t$.

Agreed! :lol:
Thank You. :)

-$ Rajiv$ $ Patki$


Find all integers $n$, such that the following number is a
perfect square \[N= n^4 + 6n^3 + 11n^2 +3n+31. \]

This is quite an easy one for a chinese olimpiad, isn't it so?

I edited the problem to more complicated one

Yes, bounding always does kill problems like this.


Να αποδείξετε ότι η ακολουθία

$ a_{n} \equal{} \frac {1}{\sqrt {2}}[a_{0}\sqrt {1 \plus{} a_{n \minus{} 1}} \minus{} \sqrt {1 \minus{} a_{0}^{2}}\sqrt {1 \minus{} a_{n \minus{} 1}}]$

με $ a_{0}\in[0,1]$ συγκλίνει...

Πρέπει να παραδεχθώ ότι δεν έχω καταφέρει και πολλά.... :blush:

(Πρόλαβα και το άλλαξα....)

Ελπίζω να μην έκανα καμιά κοτσάνα αλλά σωστό μου φαίνεται :maybe:
[hide]Καταρχήν θα δείξουμε ότι η ακολουθία είναι καλά ορισμένη $ n \in N$. Το $ a_0$ είναι καλώς ορισμένο και μάλιστα $ a_0 \in [0,1] \subset [ \minus{} 1,1]$.Υποθέτουμε ότι το $ a_n$ είναι καλώς ορισμένο δηλαδή $ a_{n \minus{} 1} \in [ \minus{} 1,1]$.Θα δείξουμε ότι και $ a_{n \plus{} 1}$ είναι καλά ορισμένο δηλαδή $ a_n \in [ \minus{} 1,1]$.
Πραγματικά είναι
$ a_n \equal{} {\frac {1}{\sqrt {2}} \left[ a_0 \sqrt {1 \plus{} a_{n \minus{} 1}} \minus{} \sqrt {1 \minus{} {a_0}^2} \sqrt {1 \minus{} a_{n \minus{} 1}} \right]} \le { \frac {1}{ \sqrt {2}} [a_0 \sqrt {1 \plus{} a_{n \minus{} 1}}]} \le {\frac {1}{\sqrt {2}} [a_0 \sqrt {1 \plus{} 1}] \equal{} a_0} \le {1}$.
Επίσης $ a_{n \minus{} 1} \in [ \minus{} 1,1] \Rightarrow \sqrt {1 \minus{} {a_{n \minus{} 1}}^2} \in [ \minus{} 1,1] \Rightarrow \minus{} \sqrt {1 \minus{} {a_{n \minus{} 1}}^2} \ge \minus{} 1$ και ακόμη ισχύει $ \sqrt {1 \minus{} a_{n \minus{} 1}} \le \sqrt {2} \Rightarrow \minus{} \sqrt {1 \minus{} a_{n \minus{} 1}} \ge \minus{} \sqrt {2}$.
Άρα τελικά έχουμε
$ a_n \equal{} {\frac {1}{\sqrt {2}} \left[ a_0 \sqrt {1 \plus{} a_{n \minus{} 1}} \minus{} \sqrt {1 \minus{} {a_0}^2} \sqrt {1 \minus{} a_{n \minus{} 1}} \right]} \geq {\frac {1}{\sqrt {2}} \left[ \minus{} \sqrt {1 \minus{} {a_0}^2} \sqrt {1 \minus{} a_0} \right]} \geq {\frac {1}{\sqrt {2}} \left[ \minus{} \sqrt {2} \sqrt {1 \minus{} {a_0}^2} \right]} \equal{} \minus{} \sqrt {1 \minus{} {a_0}^2} \ge \minus{} 1$.
Άρα κάθε στοιχείο της ακολουθίας $ a_n$ είναι καλώς ορισμένο και ανήκει στο διάστημα $ [ \minus{} 1,1]$.Επίσης από τον τρόπο απόδειξης του ότι  $ a_n \le 1$ προκύπτει όει $ a_1 \le a_0$.Υποθέτουμε τώρα ότι $ a_n \le a_{n \minus{} 1} , n \in N$ και έχουμε ότι:
\[ a_{n \plus{} 1} \minus{} a_n \equal{} {\frac {1}{\sqrt {2}} \left[ a_0 \sqrt {1 \plus{} a_{n}} \minus{} \sqrt {1 \minus{} {a_0}^2} \sqrt {1 \minus{} a_n} \minus{} a_0 \sqrt {1 \plus{} a_{n \minus{} 1}} \minus{} \sqrt {1 \minus{} {a_0}^2} \sqrt {1 \minus{} a_{n \minus{} 1}} \right]} \equal{} {\frac {1}{\sqrt {2}} \left[ a_0 \left( \sqrt {1 \plus{} a_{n}} \minus{} \sqrt {1 \plus{} a_{n \minus{} 1}} \right) \plus{} \sqrt {1 \minus{} {a_0}^2} \left( \minus{} \sqrt {1 \minus{} a_n} \plus{} \sqrt {1 \minus{} a_{n \minus{} 1}} \right) \right]} \le 0\] γιατί  $ \left( \sqrt {1 \plus{} a_{n}} \minus{} \sqrt {1 \plus{} a_{n \minus{} 1}} \right) \le 0$ και $ \left( \minus{} \sqrt {1 \minus{} a_n} \plus{} \sqrt {1 \minus{} a_{n \minus{} 1}} \right) \le 0$.Ως φθίνουσα και φραγμένη λοιπόν η ακολουθία συγκλίνει :).[/hide]

EDITED[hide]
Έστω $ \phi\in \left[0,\frac {\pi}{2}\right]$ ώστε $ a_0 \equal{} \cos \phi$. Θέτω $ \theta_0 \equal{} \frac {\phi}{2} \Rightarrow a_0 \equal{} \cos 2\theta_0$ όπου $ \theta_0 \in\left[0,\frac {\pi}{4}\right]$

Ας υποθέσουμε ότι "τυχαίνει" για κάποιο $ k$ να είναι $ a_k \equal{} \cos 2\theta_k$, για κάποιο $ \theta_k \in\left[0,\frac {\pi}{2}\right]$

$ a_{k \plus{} 1} \equal{} \frac {1}{\sqrt {2}}\left[a_{0}\sqrt {1 \plus{} a_{k}} \minus{} \sqrt {1 \minus{} a_{0}^{2}}\sqrt {1 \minus{} a_{k}}\right] \Rightarrow$

$ a_{k \plus{} 1} \equal{} \frac {1}{\sqrt {2}}\left[\cos \phi\sqrt {1 \plus{} \cos 2\theta_k} \minus{} |\sin \phi|\sqrt {1 \minus{} \cos 2\theta_k}\right] \Rightarrow$

$ a_{k \plus{} 1} \equal{} \frac {1}{\sqrt {2}}\left[\cos \phi\sqrt {2}|\cos \theta_k| \minus{} \sin \phi\sqrt {2}|\sin \theta_k|\right] \Rightarrow$

$ a_{k \plus{} 1} \equal{} \cos \phi\cos \theta_k \minus{} \sin \phi\sin \theta_k \Rightarrow$

$ a_{k \plus{} 1} \equal{} \cos (\phi \plus{} \theta_k)$

Αρα θα υπάρχει $ \theta_{k \plus{} 1} \equal{} \frac {\phi \plus{} \theta_k}{2}$ με την ιδιότητα $ a_{k \plus{} 1} \equal{} \cos(2\theta_{k \plus{} 1})$

Το $ \theta_{k \plus{} 1}$ ως μέσος όρος των $ \phi,\theta_k$, θα βρίσκεται κι αυτό μέσα στο $ \left[0,\frac {\pi}{2}\right]$

Συνοψίζοντας έχουμε:

$ \exists \theta_0 \in\left[0,\frac {\pi}{4}\right]$ τέτοιος ώστε $ a_0 \equal{} \cos 2\theta_0$

Αν $ \exists \theta_k \in\left[0,\frac {\pi}{2}\right]: a_k \equal{} \cos 2\theta_k$ τότε $ a_{k \plus{} 1} \equal{} \cos(2\theta_{k \plus{} 1})$, όπου $ \theta_{k \plus{} 1} \equal{} \frac {\phi \plus{} \theta_k}{2}$

Επομένως ισχύει για όλα τα $ a_k$ και αποδεικνύουμε εύκολα $ \theta_k \equal{} \phi \minus{} \phi\cdot 2^{ \minus{} (k \plus{} 1)}$

Είναι $ \theta_k\to\phi$ και επομένως $ a_k\to\cos 2\phi \equal{} 2a_0^2 \minus{} 1$[/hide]

Ωραία προσέγγιση pontios :)

Durandal 1707


Take all those 9-digit multiples of 1125  in which every digit is different.

How many of them have even digits in the odd places

(1st,  3rd,  5th, 7th, and 9th digits)?

[quote="Treething"]None, they all have 5 in the units' place?[/quote]
Why can't they have a 0 in the units place?

[hide="Solution"] Well, it might be nice to know that

$1125 = 3^{2}\cdot 5^{3}$

A multiple of $9$ needs digits that sum to a multiple of $9$, which is easy enough.  Since we are using nine digits out of $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$, which sum to $45$, we need to chuck out $9$.

A multiple of $125$ only needs to have last three digits

$125, 250, 375, 500, 625, 750, 875$

We want even digits in the $7^{th}$ and $9^{th}$ places, which is satisfied only by the ending $250$.

We now have $6$ more digits to determine, the $1^{st}, 3^{rd}$, and $5^{th}$ of which must be even.  Since every digit is different, we can only pick from the digits

$1, 3, 4, 6, 7, 8$

We have three slots for even digits and three for odd digits.   That gives us $3!^{2}= \boxed{36}$ permutations. [/hide]


My AP teacher says wording is very important in the essays, and that the graders shouldn't have to infer anything from your writing. Would the phrase "form equals function" be sufficiant enough in an essay or should I explain further (i.e. temperature breaks hydrogen bonds so secondary structures of proteins are affected which would determine how the tertiary structures form which etc.)?

Thanks  :) !

Most people would normally be able to decipher what you mean.  However, I would explain it.  My AP Bio teacher gave me some very good advice, which helped me to get a 5 quite easily.  Write everything that you can think of and don't stop writing.  If you think that something might need a little bit more explaining, create the extra sentence to do so.

If you're worried about the time pressure of the essays, consider this.  I am an extremely slow writer.  In fact, I did not answer one whole part of one of the free response questions of last year's exam.  While most people finished quickly, I took my time to write every little detail I could think of.  Each part of each question is only worth a few points on whatever raw score they use to calculate their AP grade.  In general, just make sure your clear.  If I was grading an exam, I would want to see somebody explain what they mean instead of spit out a phrase that their teacher gave them (not saying this is necessarily true, but "form equals functions" sounds iffy).


function
complex analysis
complex analysis unsolved

Let f(z) is an entire function and |f(z)| $ \le e^x$  throughout the plane. What we can say about f(z)

Hint: what properties does $ g(z)\equal{}e^{\minus{}z}f(z)$ have?


A circle of radius 1 is inscribed in a pentagon (as shown in the figure below). The ratio of the area of the pentagon to its perimeter is

[img]http://i204.photobucket.com/albums/bb97/hasan4444/Problem1.jpg[/img]

$ \textbf{(A)}\ \frac {1}{2} \qquad \textbf{(B)}\ \frac {1}{3} \qquad \textbf{(C)}\ \frac {1}{4} \qquad \textbf{(D)}\ \frac {1}{5} \qquad \textbf{(E)}\ \frac {1}{6}$

Let $ O$ be the center of the circle, $ M\in AB, \ N\in BC, \ P\in CD, \ Q\in DE, \ R\in EA$ the points where the circle touch the pentagon.

Then,
$ AM\equal{}AR\equal{}x, \ BM\equal{}BN\equal{}y, \ CN\equal{}CP\equal{}z, \ DP\equal{}DQ\equal{}u, \ EQ\equal{}ER\equal{}v$

The perimeter is $ p\equal{}2(x\plus{}y\plus{}z\plus{}u\plus{}v)$

The area is $ S\equal{}area(AOB)\plus{}area(BOC)\plus{}area(COD)\plus{}area(DOE)\plus{}area(EOA)\equal{}$

$ \equal{}\frac{x\plus{}y}{2}\plus{}\frac{y\plus{}z}{2}\plus{}\frac{z\plus{}u}{2}\plus{}\frac{u\plus{}v}{2}\plus{}\frac{v\plus{}x}{2}\equal{}x\plus{}y\plus{}x\plus{}u\plus{}v$

Then $ \frac{S}{p}\equal{}\frac{1}{2}$


[quote="now a ranger"]ok theres a right triangle
 ABCD
 where C
 is a right angle
 and AB is the hypotenuse
between A and B is point D
 between point A and C is E
then AD = 4
 BD=8
 EC=2

DE is perpendiclar to AB.

Find the distance from D to BC.[/quote]
[hide]So we have two similiar triangles ADE and ACB. Make AE equal to x. So now we have an equation to solve. $\frac{4}{x+2}=\frac{x}{12}$. Solving for x we get that x=6. Then draw a perpendicular line from point D to CB. Call that point F. So now we have another similiar triangles. ACB to DFB. So call DF x. We have another equation to solve. $\frac{8}{x}=\frac{12}{8}$. Solving for x which is our answer is $x=\frac{16}{3}$.[/hide]


abstract algebra
group theory
superior algebra
abelian group

Let G a group, $n \in N, n \geq 2$ s.t.:
$x^ny^n = y^nx^n$
$x^{n+1}y^{n+1} = y^{n+1}x^{n+1}$
$x^{n+2}y^{n+2} = y^{n+2}x^{n+2}$
$\forall x, y \in G$
Then G is abelian.

Nice one!
I try to solve it but I only solve it when n is an odd.
Has anyone solved for n=2?

@behemoth: Yes. ;) I know this exercise from another forum, so I won't post "my" solution. But the idea is to show $x^{n-1} y^{n-1} = y^{n-1} x^{n-1}$, i.e. the same equations for $n-1$, from which you can derive them for $n=1$.

Can you explain more clealy ! In my solution ,the hypothesis n is an odd is very important .And what's the forum you said about? :lol:

Oh, I mixed it up with another problem (where the equations are replaced by $(xy)^n = x^n y^n$ , $(xy)^{n+1} = x^{n+1} y^{n+1}$, $(xy)^{n+2} = x^{n+2} y^{n+2}$).  :blush: I'll try to put it down to that.

@mareleG: Are you sure, that this holds in general? Maybe something is missing? ;)

@behemoth:

This automatic prover
http://www-unix.mcs.anl.gov/~mccune/prover9/
can give a proof for n = 2.
(I used Windows version LADR805A-win.zip, the version LADR206A-win.zip seems not to know all the options I use.)

The input file I used is this:
[code]% Some options, see the sparse documentation of the prover.
set(auto).
assign(max_seconds, 300).
set(r_proofs).
set(rx_proofs).
clear(standard_proofs).
% This option yields the shortest proof I could find.
assign(max_weight, 39).

clauses(sos).
% These clauses are all-quantified formulas, where x,y,z,u are the variables.

% Our group axioms.
% e is a constant, * is binary, ' is unary
e * x = x # label(neutral_element).
x' * x = e # label(inverse_element).
(x * y) * z = x * (y * z) # label(associativity).
% This formula can be derived, but giving it makes the proof 2 lines shorter. :)
(x')' = x # label(inversion_involution).

x * (x * (y * y)) = y * (y * (x * x)) # label(squares_commute).
x * (x * (x * (y * (y * y)))) = y * (y * (y * (x * (x * x)))) # label(cubes_commute).
% Commutativity of fourth powers follows from the squares.

% A and B are constants.
A * B != B * A # label(denial_of_commutativity).

end_of_list.
[/code]

This produces a proof of 60 lines (including assumptions), but the inference of the formulas is quite difficult, as always with these "low level" theorems. The more detailed proof, containing intermediate results, is 420 lines long, and only combines one or two previous formulas at a time.

LiMa

@LiMa: Ouh ... isn't it possible to find an easy proof?

For $n$ odd, it follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=89832]this problem[/url]. I guess ZetaX was inspired by this topic.

[quote="-oo-"]Ouh ... isn't it possible to find an easy proof?[/quote]

I am working on the "proof" the program gave me, and will post it as soon as I'm done converting it into a human readably form.

LiMa

Here is my rewrite of the proof for n = 2.

We have a group with these additional axioms:

S. x x y y = y y x x.
C. x x x y y y = y y y x x x.

In the following, only the usage of these axioms and the previous lemmas are mentioned.
Transformations involving usual group theory are sometimes hinted at by brackets.

L1. x' y y x = x y y x'.
x' y y x =
x' y y x x x' = [S]
x' x x y y x' =
x y y x'

L2. y' x x y = x' y x x y' x.
y' x x y =
(y' x'') x (x x') y =
(x' y)' x x (x' y) = [L1]
(x' y) x x (x' y)' =
x' y x x y' x

L3. x y y x' y = y x y y x'.
x y y x' y =
y (y' x y y x' y) = [L2]
y x' y y x = [L1]
y x y y x'

L4. x' y x = x y x y' x' x' y.
x' y x =
x' (x' y')' =
x x' x' (x' y')' (x' y')' (x' y') = [S]
x (x' y')' (x' y')' x' x' (x' y') =
x (x' y')' (x' y')' x' (x' x' y' y') y = [S]
x y'' x'' (x' y')' x' y' y' x' x' y =
x y x ((x' y')' x' y') y' x' x' y =
x y x y' x' x' y

L5. x' x' x' y x x x = y y y x y' y' x'.
y' y' y' x y y y =
y' y' y' (x' x' x x) x y y y =
(y' y' y' x' x') x (x x y y y) =
(x x y y y)' x (x x y y y) = [L4]
(x x y y y) x (x x y y y) x' (x x y y y)' (x x y y y)' x =
x x y y y (x x x y y y) x' y' y' y' x' x' y' y' y' x' x' x = [C]
x x y y y y y y x x (x x') y' y' y' x' x' y' y' y' x' (x' x) =
x x (y y y y y y x x) y' y' y' x' x' y' y' y' x' = [S]
x x x x y y y (y y y y' y' y') x' x' y' y' y' x' =
x x x x y (y y x' x') y' y' y' x' = [S]
x x x x y x' x' (y y y' y') y' x' =
x x x x y x' x' y' x' = [L3]
x x x x x' y x' x' y' =
x x x y x' x' y'

L6. y x y x x y' = x x y x.
y x y x x y' =
(x' y')' (x' y')' x (x x') y' = [S]
x x (x' y')' (x' y')' x' y' =
x x (x' y')' y x x' y' =
x x y x

L7. x' y y x' y y x = y y x' y y.
y' x x y' x x y =
y' x x y' x x x x' y =
(y' x) x (y' x) x x (x y')' = [L6]
x x (y' x) x =
x x y' x x

L8. x x y x x = x x x x y.
x x y x x =
x x y' y y x x = [S]
x x y' x x y y =
((x x y')' (x x y')) x x y' x x (y' y) y y =
(x x y')' (x x y') (x x y') (x x y') y y y = [C]
(x x y')' y y y (x x y') (x x y') (x x y') =
(x x y')' y (y y (x x y') (x x y') x x) y' = [S]
(x x y')' y ((x x y') (x x y') x x y (y y')) =
(y x' x') y x x (y' x x y' x x y) = [L7]
y x' x' y x x (x x y' x x) =
y x' x' (y x x y') (y x x y') x x = [S]
y x' x' x x (y x x y') (y x x y') =
y y x x x x y' = [S]
x x x x y

L9. y x x = x x y.
y x x =
x' x' x x y x x = [L8]
x' x' x x x x y =
x x y

Thm. x y = y x.
y x =
x x' y x =
x x' x' x' x x y x = [L9]
x x' x' x' y x x x = [L5]
x y y y x y' y' x' = [L9]
x y y y y' y' x x' =
x y

I don't think that this proof can be transformed into a proof for other values of n.

LiMa

Thank you Lima but frankly,I cant follow your solution. :D Can you explain your step and I will try to do the step myself?

The proof is purely syntactic and all the (implicit) quantifiers range only over the elements of the group. All formulas having free variables are meant to be all-quantified, so the axiom
  S. x x y y = y y x x.
means
  Axiom S. forall x forall y (x^2 y^2 = y^2 x^2).

The first lemma
[code]L1. x' y y x = x y y x'.
x' y y x =
x' y y x x x' = [S]
x' x x y y x' =
x y y x
[/code]
contains a typo. The last line should read
  x y y x'

This reads as:

Lemma 1. $\forall x \forall y: x^{-1} y^2 x = x y^2 x^{-1}$.
Proof:
$x^{-1} y^2 x =$
$x^{-1} y^2 x^2 x^{-1} =$ [using axiom S]
$x^{-1} x^2 y^2 x^{-1} =$
$x y^2 x^{-1}$
q.e.d.

One thing:
x'' = (x')' = (x^-1)^-1 = x

I hope you can decipher the proofs now. :)
If you can't follow them, let me know where.

LiMa

Generalizing Prover9's proof for n = 4, I found a proof of this proposition:

Let $G$ be a group, $n \geq 2$ s.th.
$\forall x \in G\ \forall y \in G\ (x^n y^n = y^n x^n)$
$\forall x \in G\ \forall y \in G\ (x^{n+1} y^{n+1} = y^{n+1} x^{n+1})$
Then $G$ is abelian.

The proof is still quite technical, but doesn't use the hypothesis for n+2.

LiMa

My apologies for reviving this topic, but I fail to see why the following solution is invalid (it seems to be okay  :wink:):

[u] Solution.[/u] Let $ k\geq 2$.  Let $ (xy)^k\equal{}x^ky^k,..., (xy)^{k\plus{}2}\equal{}x^{k\plus{}2}y^{k\plus{}2}$.  From the second equation, after cancelling the first $ x$ and the last $ y$ from both sides, we have $ (yx)^k\equal{}x^ky^k$.  Using the first equation, we have $ (yx)^k\equal{}(xy)^k\iff (yx)^{\minus{}k}\equal{}(xy)^{\minus{}k}$.  Similarly, using the second and third equations, we have $ (yx)^{k\plus{}1}\equal{}(xy)^{k\plus{}1}$.  Multiplying the two new equations, we have $ xy\equal{}yx$, that is, $ G$ is an abelian group. $ \Box$

[quote="boxedexe"][u] Solution.[/u] Let $ k\geq 2$.  Let $ (xy)^k \equal{} x^ky^k,..., (xy)^{k \plus{} 2} \equal{} x^{k \plus{} 2}y^{k \plus{} 2}$.[/quote]
It's a different problem, right?  -oo- mentioned the same thing.

No wonder it seemed to be extremely easy.  :blush:

conjecture:
G is a group.
$\exists m,n\in\mathbb{Z}^+ \forall x,y\in G$
gcd(m,n)=1
$x^m y^m =y^m x^m$
$x^n y^n =y^n x^n$
then G is abelian.

related:
https://artofproblemsolving.com/community/c7h559672
https://artofproblemsolving.com/community/c7h556009
https://artofproblemsolving.com/community/c7h1900517

It's true and appear several times in this community.


[hide="Here is what I did"]
Choose a random number. Do the process described in the problem and note the remaining number. If you were lucky and the number remaining was $ 11$, then your answer would be the number you chose. Otherwise, it will be less than or greater than $ 11$ by a certain amount. However much it is off, subtract that from the number you guessed for the answer.

For example, I chose $ 3$ as my random number. The number remaining was $ 5$, and $ 11 \equal{} 5 \plus{} 6$, so my answer was $ 3 \plus{} 6 \equal{} \boxed{9}$. If you want, you can check by erasing 9 first and seeing what you have left.[/hide]


geometry
3D geometry
algebra
difference of squares
special factorizations
factorization
difference of cubes

[quote="Ponnamperuma"]I'd call them expressions rather than equations!...  :wink: [/quote]

I see. :)

[quote="Ponnamperuma"]

(1) Just difference of squares
(3) Just difference of cubes
(4) Just sum of cubes

(2) $ 9x^{2}-12xy+4y^{2}-16=(3x-2y)^{2}-16$
$ =(3x-2y+4)(3x-2y-4)$
(5) $ 3x^{2}+7xy-6y^{2}=3x^{2}+9xy-2xy-6y^{2}=3x(x+3y)-2y(x+3y)$
$ =(3x-2y)(x+3y)$
(7) $ x^{3}-8y^{3}+x^{2}+2xy+4y^{2}$
$ =(x-2y)(x^{2}+2xy+4y^{2})+(x^{2}+2xy+4y^{2})$
$ =(x-2y+1)(x^{2}+2xy+4y^{2})$[/quote]

Your answer is correct. :)

[quote="Ponnamperuma"]I don't have time for the others... sorry!...   :)[/quote]

No problem. :lol:

[quote="LideR_M"](6) $ 3\left(x-\frac{2y-5}{3}\right)\left(x+3y-2\right)$ (by $ ax^{2}+bx=c=a(x-x_{1})(x-x_{2})$)
And I havent got an idea on (8).[/quote]

Your answer is correct. :) You can also write the answer as $ (3x-2y+5)(x+3y-2)$.

For $ (8)$, you can use the formula $ \boxed{a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)}$.

[quote="DiscreetFourierTransform"]8)
$ (x+1-2y)(4y^{2}+2xy+2y+x^{2}-x+1)$[/quote]
$ =(x-2y+1)(x^{2}+2xy+4y^{2}-x+2y+1)$.
Your answer is correct. :)


quadratics
algebra
function
domain
calculus
integration
Ring Theory

I'm working on a project on power sums in the quadratic field ${\bf QF}(m)=\{r+s\sqrt{m}\},\ r,s\in\mathbb{Q}$, and it would be very convenient if for each element $\alpha\in{\bf QD}(m)=\{a+b\sqrt{m}\},\ a,b\in\mathbb{Z}$ not divisible by some prime element $p\in{\bf QD}(m)$ there exists some $\alpha^{-1}$ such that $p\mid\alpha\alpha^{-1}-1$.  Is that the case in general?

Are you sure that the $\bf QD(m)$ is really what you want since it's not always a dedekind domain¿

But even then we still have your property (and in all other number fields etc., too).
There are $N(p)$ (the norm) different residues $\mod p$. This number is especially finite.
${\bf QD}(m) \mod p$ is an integral domain (since $p$ is prime) with a finite number of elements. Thus a simple counting argument (if $a \mod p$ is nonzero, then $ax$ runs through all different residues if $x$ does since no elements can occure twice) gives that this domain is already a field, thus every residue $\neq 0$ is invertible, which is your required property.


The graphs of the equaitons $y=x^{2}-2x-8$ and $y=2x-3$ intersect at point(s) with coordinates $(a,b)$. What is the sum of all possible values of b?

Would you use systems on this?

[hide] Plugging a and b in for x and y respectively, we get $a^{2}-2a-8=b=2a-3$.  Thus, $a^{2}-2a-8=2a-3\Rightarrow a^{2}-4a-5=0\Rightarrow a=5, a=-1$.  Plugging in these values in the second given equation, we get that b=2(5)-3=7 or b=2(-1)-3=-5.  Adding the two possible values of b, we get 7+(-5)=$2$ [/hide]


1. In a Slime Volleyball game, the first player to score 4 points wins.  If Nirav's probability of winning each point is 20%, what is the probability to two significant digits that nirav wins the game?

2. A magician has three "quarters": one is a standard, fair quarter; the other two quarters are fair, but each has two heads and no tails.  You randomly select one of the three quarters and flip it twice.  If you get "heads" twice, what is the probability that you selected the standard quarter?

umm...the second one seems way to easy

[hide]If you choose a coin ony two of the three coins have the ability to yeild two heads in a row.  One of these is the standard and one is the fake.  so the probablility of the one you picked to be the standard is just $ \boxed{\frac{1}{2}}$

Like i said it seeems way too easy!!![/hide]

Well, maybe the fact that the answer is [hide]$ \frac{1}{9}$[/hide] will change your mind about its difficulty. ;)

yes it definately changed my mind but i cant see how you would get that answer.

Its conditional probability

$ \frac{\frac{1}{3}\times\frac{1}{4}}{\frac{2}{3}\plus{}\frac{1}{3}\times\frac{1}{4}}\equal{}\frac{1}{9}$

So... the probability of that coin getting two heads divided by the probability of all coins getting two heads?  Now I feel stupid.

I got the first one, myself... eventually.


trigonometry
inequalities
geometry proposed
geometry

Hi to all mathlinkers.
I believe this time I have something interesting to show.

[quote="bui_haiha2000"]Given a triangle $ ABC$. Let $ X$, $ Y$, $ Z$ be points on the segments $ \left[BC\right]$, $ \left[CA\right]$, $ \left[AB\right]$, respectively, such that $ AZ \equal{} BX \equal{} CY$ and $ \widehat{BAX} \equal{} \widehat{CBY} \equal{} \widehat{ACZ}$.
Prove that triangle $ ABC$ is equilateral.[/quote]

It is a problem proposed from me for the Bulgarian Math Olympiad.
It is a problem - invented from me as a statement at my own and solved by two russian guys at lib.mexmat.ru's forum.
I would like to thank them.
It is a long time discussed and proved to be a hard problem.
(you may see the discussions below)

http://www.mathlinks.ro/viewtopic.php?t=203616
http://www.mathlinks.ro/viewtopic.php?t=99674
http://www.mathlinks.ro/viewtopic.php?t=163712

Here is the solution I have (in the file attached). I have no time to translate it but if you have questions you may ask.
What is your opinion about this problem ... may it be generalized? May you invent some similar problems?
Have you ever seen the problem ... if not it may be called Bobi's theorem by my name (by analogy with Moorley's theorem) ...  :)

Could you tell me please when did you find this problem? Because as we can see, [b]bui_haiha2000[/b] has posted this problem a long time ago. And my teacher, [b]Nguyen Minh Ha[/b] also solved this problem about two years ago when I've just passed in the tenth grade (I'm sorry, I don't remember his solution  :( ). So I think it can't be called as "Bobi's theorem"

:( I didn't know this problem before... I started random drawings and I wondered if the statement is true. I was surprised to see the problem on mathlinks.

The result is really nice, but I can't actually say that I like this kind of problems, mostly because of the impossibility of seeing a neat synthetic proof. Here's another similar (and somehow related?) theorem, which was proposed as the IBM Research Group August 1998 problem.

[color=darkblue][b]Theorem[/b][/color]. Let $ X$, $ Y$, $ Z$ be points on the sides $ BC$, $ CA$, $ AB$ of triangle $ ABC$, such that $ AX \equal{} BY \equal{} CZ$. If the triangle $ XYZ$ is equilateral, then $ ABC$ is also equilateral.

I didn't think too much on it, but I doubt that there is a clear synthetic proof here as well.

vittasko has done great work trying to find syntetic proof of the problem proposed by me.
Probably he is genious but I cannot understand all the details ot its solution.
If it is possible he may post sistematized solution here.
About the solution I posted - I think it is really
beautiful and shows that solution using trigonometry may be very attractive. Try to understand all its details and you will see.
Your problem is probably well known and nice. It is far easier than mine I had success in solving it on another forum some time ago.
http://www.mathlinks.ro/viewtopic.php?p=558585#558585 is a similar problem.
I think that the problems of such kind may be generalized...
If on my drawing we choose three pairs of equal symmetric elements of kinds:
1) angle11=angle12=angle13 and angle21=angle22=angle23
2) segment11=segment12=segment13 and angle11=angle12=angle13
3) segment11=segment12=segment13 and segment21=segment22=segment23
i suppose we may say that ABC is equaliteral but I cannot prove it.

Example - try to solve the problem below:

[b]It is given triangle ABC with points M, N, P on its sides (internally). U, V, W are intersection points of AN and CM, BP and AN, CM and BP. Triangle UVW is equaliteral. If AM=BN=CP is it true that ABC is equaliteral?[/b]

I think it is also another very hard and open problem.

[quote="bui_haiha2000"]$ \|\begin{array}{c} \triangle\ ABC \\
= = = = = = = = = = = = = = = = = = = \\
\ X\in [BC]\ ,\ Y\in [CA]\ ,\ Z\in [AB] \\
 \\
AZ = BX = CY = x \\
 \\
m(\widehat{ACZ}) = m(\widehat{BAX}) = m(\widehat{CBY}) = \phi\end{array}\|\ \implies\ AB = BC = CA$ .[/quote]

[b][u]Proof.[/u][/b] Apply the [b]Sinus' theorem[/b] in the triangles with the common side $ [BC]$ :

$ \|\begin{array}{cccc} BCY\ : & \frac {a}{x} = \frac {\sin (C + \phi)}{\sin\phi} & \implies & \tan\phi = \frac {x\sin C}{a - x\cos C} \\
 \\
BCZ\ : & \frac {a}{c - x} = \frac {\sin (A + \phi )}{\sin (C - \phi )} & \implies & \tan\phi = \frac {x\sin A}{b - x\cos A}\end{array}\|\implies$ $ \frac {\sin C}{a - x\cos C} =$ $ \frac {\sin A}{b - x\cos A}\ \implies$

$ x = \frac {a\sin A - b\sin C}{\sin (A - C)} =$ $ \frac {a^2 - bc}{a\cos C - c\cos A} =$ $ \frac {2b(a^2 - bc)}{(a^2 + b^2 - c^2) - (b^2 + c^2 - a^2)}$ $ \implies\ x = \frac {b(a^2 - bc)}{a^2 - c^2}$ .

Otherwise, if denote $ E\in BY\cap AC$ and $ F\in BC$ so that $ EF\perp BC$ , then $ \tan \phi =\tan\widehat {CBY}=\frac {EF}{AF}=\frac {x\sin C}{a-x\cos C}$ .

Apply analogously the [b]Sinus' theorem[/b] to the pairs of thiangles : $ \|\begin{array}{ccc} (\ CAZ\ ,\ CAX\ )\ : & \implies & x = \frac {c(b^2 - ca)}{b^2 - a^2} \\
 \\
(\ ABX\ ,\ ABY\ )\ : & \implies & x = \frac {a(c^2 - ab)}{c^2 - b^2}\end{array}$ .

Thus, $ \boxed {\ x = \frac {a(c^2 - ab)}{c^2 - b^2} = \frac {b(a^2 - bc)}{a^2 - c^2} = \frac {c(b^2 - ca)}{b^2 - a^2} = \frac {abc[(a^2 + b^2 + c^2) - (ab + bc + ca)]}{(a - b)(b - c)(c - a)(a + b + c)}\ }$ a.s.o.

[quote="borislav_mirchev"]Virgil Nicula, I cannot understand your idea.
What is the intention of the last chain of equalities? (The last row.)[/quote]
Here is the true problem : how from the last chain of equalities you can obtain $ a \equal{} b \equal{} c\ !$ See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=205887][color=red][b]here[/b][/color][/url] .

To Virgil Nicula:
Yes it seems to be an interesting problem after following the hyperlink.
A man in Bulgarian Math Forum and I tried to solve my initial problem using
similar approach but unfortunately we both failed.

Do you know the solution of the problem proposed by you.
If you can solve this problem, please share your solution with us.
What do you think about the solution in .doc file I posted?

http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv22n3/body/PDF/page107-116.pdf?file=page107-116

In page 108 - the following problem was proposed. I'm wondering - how it is supposed to be solved.

I invented this problem at my own independently...

This problem was posted here......."It is given triangle ABC with points M, N, P on its sides (internally). U, V, W are intersection points of AN and CM, BP and AN, CM and BP. Triangle UVW is equaliteral. If AM=BN=CP is it true that ABC is equaliteral?"

A similar looking problem(more easy one) appeared in the Indian TST 2008 on 16/05/2008...

It is given triangle ABC with points D,E,F on sides BC,CA,AB such that AD=BE=CF.
Suppose the line segments AD,BE,CF are not concurrent and enclose an equilateral triangle.Is ABC necessarly equilateral?

Can you post hints or solution to the problem from Indian TST?
It seems Menelaus and inequalities should be used to prove the triangle is equaliteral.


$ f(x)\equal{}ax^2\plus{}bx\plus{}c$. $ |f(x)| \leq 1$ for all $ x\in [0,1]$. prove that $ f'(0) \leq 8$


[color=darkblue]Consider the following sequence which is defined recurrently thus:

$x_{1}\in (0,1)$ and for any $n\in N^{*}$, $x_{n+1}=x_{n}\cdot \left(2-2^{x_{n}}\right)$.

Prove that $\lim_{x\to\infty}x_{n}=0$ and ascertain $\lim_{n\to\infty}\left(1+x_{n}\right)^{n}$.[/color]

[hide]By induction on $n$ , we have $0<x_{n}<1\forall n$. From hypothesis $\frac{x_{n+1}}{x_{n}}=2-2^{x_{n}}<1$, thefore exist $\lim_{n\to\infty}x_{n}=L$, also $L=L(2-2^{L})\rightarrow L=0$ because $L<1$.[/hide]


function
limit
geometry
complex analysis
complex analysis unsolved

Take for example $ f(z)=\text{Csc}(1/z)$.  This function has isolated poles at $ z=\pm \frac{1}{k\pi}$. So in between these, we can construct a Laurent series convergent in the annulus $ \frac{1}{(k+1)\pi}<|z|<\frac{1}{k\pi}$ with:
\begin{align*}
a_n&=\frac{1}{2\pi i}\mathop\oint\limits_{C_k} \frac{f(z)}{z^{n+1}}dz;
\quad C_k=r_ke^{it},\; r_k=\frac{2}{(2k+1)\pi}\\
b_n&=\frac{1}{2\pi i}\mathop\oint\limits_{C_k} f(z)z^{n-1}dz
\end{align*}
(1)  How can the coefficients be determined more easily?
(2) What can be said about the sets $ \{a_n\}$ and $ \{b_n\}$ as functions of $ k$?  Do the terms eventually become zero for any $ k$?
(3) Will the properties of this set be similar to those of any other non-isolated singularity?

I don't know.  Books should talk more about non-isolated singularities I think . . .

The series you're thinking about behaves just like any other Laurent series with simple poles on the inner circle and the outer circle. The singularity at zero is irrelevant. If $ f(z)\equal{}\sum_{n\text{ is odd}} a_nz^n$ on this annulus, $ \lim_{\substack{n\to\infty\\ n\text{ is odd}}}a_n\left(\frac1{k\pi}\right)^n\equal{}a$ and $ \lim_{\substack{n\to\minus{}\infty\\ n\text{ is odd}}}a_n\left(\frac1{k\pi}\right)^n\equal{}b$ where $ a$ and $ b$ are some nonzero complex numbers.
The even terms are always zero, since this is an odd function. The odd terms are never zero.

Hi JMerry.  Thanks!  Apparently Laurent series for annular regions have non-terminating singular parts?  Guess I need to work on that although I have calculated Laurent series for example:
\[ f(z) \equal{} \frac {A}{z \minus{} z_1} \plus{} \frac {B}{z \minus{} z_2}
\]
valid in the annular region between $ |z_1|$ and $ |z_2]$ the answer of which is:
\[ f(z) \equal{} \minus{} \sum_{n \equal{} 0}^{\infty}\frac {B}{z_2^{n \plus{} 1}} z^n \plus{} \sum_{n \equal{} 1}^{\infty}Az_1^{n \minus{} 1}\frac {1}{z^n}
\]
I do not understand the limits you posted.  Might you explain that please . . . I'll try and figure it out though.

Also, might I challenge the relevance of the singularity at the origin?  You mind?  It's non-isolated geometry forces us to construct Laurent series (near the singularity) in terms of annular regions.  These series have non-terminating singular parts (I think).  I can begin to see why we still classify these types of singularities as "non-isolated essential singularities":  Any Laurent series for a non-isolated singularity (near the singularity) has a non-terminating singular part, and the singularity still meets the criteria for Casorati-Weierstrass.

The exact same limits would hold if our function were $ \frac{z}{z^2\minus{}\frac1{k^2\pi^2}}\plus{}\frac1{z^2\minus{}\frac1{(k\plus{}1)^2\pi^2}}$- an odd function with relevant poles in the same locations. In fact, $ a$ and $ b$ depend only on the residues at those poles. It's simply the radius of convergence, inner and outer. That's why I say the singularity at zero doesn't matter- it's the singularities further out that control the series' behavior.

Now, you might try to build a series that applied arbitrarily close to zero- but that would be an asymptotic series, not a Laurent series. Also, it would diverge everywhere.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be any group with no proper, nontrivial subgroups, and assume that |G| >1. Prove that G must be isomorphic to Zp for some prime p.

That G has no nontrivial proper subgroups means that every element not equal to the identity must have order |G|; in other words, G is a cyclic group generated by any element $x \in G$. For every element to have the same order, |G| must be prime, for if $|G| = mn$ where $m \le n$ then the element $x^m$ would have order $n \ne |G|$. Define the isomorphism $\varphi$ sending $x^a \rightarrow a \in \mathbb {Z}_p$.


calculus
integration
logarithms
trigonometry
calculus computations

More generally: Let $ a>0$ and let

$ I(a)=\int_{0}^{\pi/2}\frac{\ln(1+a\sin^{2}x)}{\sin^{2}x}dx$. Then

$ I'(a)=\int_{0}^{\pi/2}\frac{dx}{1+a\sin^{2}x}$. Using the substitution $ \tan x=t$ we get after some simple calculations that

$ I'(a)=\int_{0}^{\infty}\frac{dt}{1+(a+1)t^{2}}=\frac{\pi}{2\sqrt{a+1}}$.

Hence

$ I(a)=\pi\sqrt{a+1}+C$. We get , since  $ I(0)=0$ , that $ C=-\pi$ and hence

$ I(a)=\pi(\sqrt{a+1}-1)$.


so I'm working on this problem about windshield wipers....yeah I know...lame. But it involves finding the area of two different windshield wipers and then creating a new one that has a larger area being cleaned.

The standard wiper consists of a 12-inch blade attached rigedly to a 12-inch arm. The middle of the blade is at the end of the arm, and the arm rotates back and forth, making a 45 degree angle with the horizontal at each end of its motion. The new model also has a 12-inch blade attached at the midpoint to a 12-inch arm. But in this case, the blade pivots so that it's always vertical. The arm rotates 90 degrees as in the standard model.
Find which wiper is better in terms of area being swept, and design a new wiper that still uses 12-inch blades and rotates 90 degrees.

original: [img]http://i36.photobucket.com/albums/e16/RedSOX84/pow4pic.jpg[/img]

new model: [img]http://i36.photobucket.com/albums/e16/RedSOX84/wiper.jpg[/img]

anybody have any ideas or suggestions to help???[/img]

[hide="Answer."]Calculate the area wiped by each. Let's call the areas $A_{f}$ and $A_{v}$ the areas of the first one and the vertical wipers, respectively. The areas wiped by the first one is $72\pi$. The vertical one, if you turn the arms into legs of a right triangle, has a with of $12\sqrt{2}$ across the wiping area. Also, if you look at the picture, the area under the bottom curve is equal to the area of the curve on the top. So take a length of $12$ for this area and multiply it like a rectangle, and you get $144\sqrt{2}$. To find which one is larger let's assume a random area is larger. Let's take $72\pi>144\sqrt{2}\\\pi>2\sqrt{2}\\3.14159265483\dots>2(1.414)\\3.14159265483\dots>2.828$ which is correct. So the original model wipes a larger area. :P [/hide]


I'm not quite sure whether this belongs to Getting Started or Intermediate, so I'm just going to post it here.

in finding all solutions to 15x-22y=330, we add 22y to both sides to get 15x=330+22y

My question is whether we should consider this equation (mod 2) or (mod 22).  Does it make a difference?

Try dividing both sides by 15, and see how that helps.

So we have x=22+(22/15)y.  I guess we could use (mod 22).  Could we also use (mod 2) though?

Diophantine Equations are equations with more than one variable whose solutions are integers

I see how we get (mod 15), but can we also use (mod 22) or (mod 2)?


[quote="phuong"]Let $x,y,z>0$. Prove that
$(x+y)(y+z)(z+x)+xyz \leq\sqrt{(x+y)(y+z)(z+x)}.(\sqrt{xy(x+y)}+\sqrt{yz(y+z)}+\sqrt{zx(z+x)})$[/quote]
Maybe it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=466390&highlight=#466390 ? ;)


[quote="xzlbq"][size=150]
Let $ x,y,z \geq 0$,
$ \frac {1}{x^t}(\frac {1}{x \plus{} y} \minus{} \frac {1}{y \plus{} z}) \plus{} \frac {1}{y^t}(\frac {1}{y \plus{} z} \minus{} \frac {1}{z \plus{} x}) \plus{} \frac {1}{z^t}(\frac {1}{z \plus{} x} \minus{} \frac {1}{x \plus{} y}) \geq 0(t \geq 0)$

$ \frac {1}{x^t}(\frac {1}{x \plus{} y} \minus{} \frac {1}{y \plus{} z}) \plus{} \frac {1}{y^t}(\frac {1}{y \plus{} z} \minus{} \frac {1}{z \plus{} x}) \plus{} \frac {1}{z^t}(\frac {1}{z \plus{} x} \minus{} \frac {1}{x \plus{} y}) \le 0(t < 0)$
[/size][/quote]

$ \Longleftrightarrow x^{t \plus{} 2}z^t \plus{} y^{t \plus{} 2}x^t \plus{} z^{t \plus{} 2}y^t \minus{} y^tz^tx^2 \minus{} x^tz^ty^2 \minus{} x^ty^tz^2\geq 0$

 :lol:

Let $ x_1,x_2,...,x_n \geq 0,n \geq 3,\sum_{i \equal{} 1}^n{x_i} \equal{} S,x_0 \equal{} x_n$,prove that:

$ \sum{\frac {1}{x_i^t}(\frac {1}{S \minus{} x_{i \minus{} 1}} \minus{} \frac {1}{S \minus{} x_i})} \geq0 (t \geq 0);$

$ \sum{\frac {1}{x_i^t}(\frac {1}{S \minus{} x_{i \minus{} 1}} \minus{} \frac {1}{S \minus{} x_i})} \le0 (t \le 0);$

BQ


MATHCOUNTS
geometry
probability
calculus
AoPS Books
\/closed

Well, through i want to discuss a issue regarding AOPS Books and may be others too. This morning I read posters like:
"Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk. "
"two titles for math olympiads for elementary and middle schools"
"MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick. "
and more ....
dont u think this is a sort of training for olypiads.this is just training them and restricting the natural talent. for instance, in a country there is very aspiration to get into a institute and there are coaching institutes that have come up. its adverse effect is that the institute is not getting the right students. the students they get although pass exams but arenot good.

similarly, gere, i think students shud be let to prepare by themselves and training for a specific type of material is not good

What?  The AoPS books (if you've ever read one) encourage people to think outside the box and think creatively.  If anything, it does the exact opposite than restrict you.

I had a very hard time understanding the post, but, it seems u are saying that AoPS books restrict talent and make us narrow-minded and study to hard?  I don't think ull find many ppl here, if any, who agree with that...

I'm going to attempt to give an explanation instead of just an objection.

Consider two ways of teaching math:

[b](1)[/b] One way involves teaching math as a set of procedures. You'll learn how to crank out these computations, and the more comfortable you get with these procedures, the faster you can do such computations, or you can do more complicated computations. Many tests, like certain entrance exams, or the AP test, etc, are designed to test procedural knowledge. And you're right in that for these kinds of tests, teaching to the test is horrible.

[b](2)[/b]Another way of teaching math, though, portrays it as a web of interconnected concepts. It shows how mathematical thinking flows from one idea to another, how different branches of mathematics are related or are parallel, and in general tries to give the reader a panoramic view of the subject.

[i]This [/i]is how the Art of Problem Solving books approach math. When they say that these books are good for preparing for the AMC's, they aren't saying that it's some list of tips and tricks for optimizing testing performance. What they are saying is that it will help the reader develop his/her mathematical creativity and scope, while introducing the reader to helpful concepts in mathematics. This combination will help the reader to become a better problem solver.

Here, "preparing for the AMC's" is a means of advertising the book rather than a statement of the book's objective. Anyone who reads the book will see that it will improve their performance in pretty much any mathematics competition, not just the AMC's. So it's really not teaching to the test. I would argue that the creative thinking skills that these texts help build are relevant beyond mathematics, too.

I hope this clears your misunderstanding :)

[quote="analytic"]
dont u think this is a sort of training for olypiads.this is just training them and restricting the natural talent. for instance, in a country there is very aspiration to get into a institute and there are coaching institutes that have come up. its adverse effect is that the institute is not getting the right students. the students they get although pass exams but arenot good.

[/quote]

The aops books do the opposite of this.

Is the OP proposing that math be taught similar to in an AP course?  Calculus in an AP course is based purely on knowing a few facts and knowing how to use a few methods.  AoPS books focus on starting from the ground up-there are basics, and then we can assemble them into more powerful results.  With this technique, even if you can't memorize every formula, you'll understand the material so you can just re-derive it.

[quote="program4"]the introduction books are too hard to be called 'introduction'
they should call it 'intermediate' and 'advanced'
or it is that im just bad at math(i'm in 5th grade)
Also, are they going to make an Advanced series?(Advanced Geometry would be IMO level :wink: )[/quote]

Erm they the level lower than introduction would be called 'School Method' ... cough cough we're talking about AoPSers here, the future stars of the world.

um no you are not bad... you wouldn't be here.

I think they are. I read somewhere after they finish the Intermediate series, they will continue on to college level or something.

Hmm, the Introductions aren't a bad level at all, in my opinion. I started reading Intro to C & P in March, last year in 6th grade. If you go at a pace that suits you, and do [i]all[/i] problems, then it should be okay. Just go slowly and go back to review if you have to. The Interms are so much harder than Intros, which are essential.

i think thats why
sometimes i try to do 1 chapter in a day :wink:

Remember: this is the beauty of independent study; don't worry about about speed or how far you get. Just worry about what you understand. No one will know if you need to go back and review something that you think should be easy. This is purely for you, and only you. :)


Solve the following for $ (a,b,c,d)$

$ a \plus{} 2b \plus{} 2c \plus{} 2d \equal{} 31\\
3a \plus{} 2b \plus{} 3c \plus{} 3d \equal{} 29\\
5a \plus{} 5b \plus{} 4c \plus{} 5d \equal{} 17\\
7a \plus{} 7b \plus{} 7c \plus{} 6d \equal{} 3$.

[hide]Add them all up and divide by 16 to get a+b+c+d=5. Multiply that equation by 2, 3, 5, and 7 respectively to get values of a,b,c,d.[/hide]

[hide]If you add all of them up, you get:
$ 16a\plus{}16b\plus{}16c\plus{}16d\equal{}80 \longmapsto a\plus{}b\plus{}c\plus{}d\equal{}5$
(call this equation $ E$).
Now, multiply $ E$ by $ 2$ and subtract the first equation
to find $ a$.  Then, plug in $ a$ to the second equation, subtract
it to the side with the other constant, and multiply $ E$ by $ 3$.
Once again, subtract $ 3E$ from the second equation to find
$ b$...and so on and so forth.[/hide]

hello, i have found $ a\equal{}\minus{}21,b\equal{}\minus{}14,c\equal{}8,d\equal{}32$.
Sonnhard.


This  year  IMO  6  problems  is  a   hot   topic  in  the   forum  these days!
the  questions(pdf )  can  be   found     the  following  websites:

http://imo-official.org/problems.aspx

http://www.mathlinks.ro/resources.php?c=1&cid=16&year=2009

I  want  to  know   how   to   write   the   code   with   Miktex+texniccenter    to   get   the   pdf?

Have   anyone    the   tex  file    for   the   IMO  problems?

Many  thanks!

Let   us  take  a  look   at    the   first   page,
How  to  insert  the  following  content  into  the  pdf:

50th  International          ..............................................Language: English
       Mathematical           .............................................
        Olympiad
Bremen  Germany    2009   ............................................    Day:1

The  length   of   the  horizontal  line   is   longer   than  textwidth

How  to  get  that?
thanks!


Determinati toate solutiile complexe ale ecuatiei \[z^{4}-7z^{3}+24z^{2}-41z+35=0\] Stiu ca trebuie sa desfaci membrul stang al ecuatiei in produs de cateva paranteze, dar nu mi-am dat seama.

$\, \Longleftrightarrow \, \, \left( z^{2}-3z+5 \right) \left( z^{2}-4z+7 \right) = 0$.

Metoda de rezolvare: Fie $\left( z^{2}+az+b \right) \left( z^{2}+cx+d \right) = z^{4}-7z^{3}+24z^{2}-41 z+35$. Apoi rezolvi sistemul (vezi ca $bd=35$ si presupui ca $b=5$, $d=7$, dupa care $a,c$ ies destul de usor).

Cred ca e $\ z^{2}+cz+d$, nu?Ai pus $\ x$ in loc de $\ z$.


Just use x^3/(x^2+xy+y^2) >= (2x-y)/3    n times and we get the result.

Namdung

It would be interesting to see when we have
  sum (x_i)^3/((x_i)^2-x_i*x_(i+1)+(x_i+1)^2)>=sum x_i. I already proved it for n=3 in C 59 post. What about other values of n?

Sorry about this, but I have renamed all my posts with ineqs (so far just half, but I will modify them all also retyped them in LaTEX), so C 59 is now E2.40


It's rather easy to find out that if $n$ is even , then \[ a_n=3a_{n-1}-a_{n-2}  \] and if $n$ is odd, then \[ a_n=2a_{n-1}-a_{n-2}  \] we can easily prove this by induction. With this result ,as $a_1=a_2=a_3=1$, our statement is rather straightforward :)


I m just wondering, our Fairbanks AK Chapter had like 110 people :D

Ours had 60+ students and like 12 teachers/coaches/people who corrected the stuff.
So around 72+ people in all.
Just an estimate though.

yea... ours in sandiego  was [size=150]AT LEAST[/size] 100 ppl

At least 50 kids (11 schools)
Plus 17 staff or so...
Our chapter is a small one
(Ours is Seacoast, NH)

dang, seems like my chapter was crowded... Lol

anyway i got into state :D

wow, we had like very little people, two teams had 8, and 5 teams had 4, and some individuals... so around 40

About 30. 6 teams and 8 other individuals.

I don't remember how many people there were but there wasn't that much, considering it's southern New Mexico

Wow, 9 for 173rd/300!  We had 163 people, and I thought that was a lot.

[quote="Fowi Hunter"]Wow do you guys live in really small states or something?  We had at least 300+ people.  There was one person on our other team who got 9 points!  But some people must have done real bad because he got 173rd place.[/quote]

So 122 people got less than 9?? I had 163 people at the Santa Clara chapter.

did u guys had danish and orange juice? lol we had them

Our school was an hour and a half late to the chapter competition site, and there were no more than 25 people at the competition, which was very few.

My chapter was medium sized about 120 students maybe more.
Went home with a shiny trophy, shiny medal, (not shiny) registration form, and my dad had the "Coach of First Place Individual" plaque. Count down round was funny because my opponent had the same first name so only about half the people there knew who won.

We also had like 300-400 people, but that's b/c we have over 30 schools and most schools send four individuals and some send alternates (we usually have two to four of the alternates actually show up).. so that makes sense why like the top 50 people and anyone on a top 8 team makes states...

Well we live in CT... and our chapter is like half of CT so ya 50 does make sense...

HA!!!  :D  we're in i think the biggest chapter in utah and only 5 schools came and all are going to state our team of 6th graders beat the junior high that we're all gonna go to  :D

Haha, my chapter was 2 chapters COMBINED and there was only like 3 or 4 teams. Everyone qualified for state. One girl on my team got a 10 total, and she was in 8th place. :D I'd say there was about 20 competitors, tops.

[quote="kidwithshirt"]top 50?

thats messed up[/quote]

Anybody in the top 50 makes it to states, regardless of what their team does. It's that way because Connecticut is a fairly dense state.

we had like a kajillion. jk. we had maybe 170 people total. but it was a really good round. probably the 16th place person had like a 36. but we texans are smart. :) anyway. yeah. ill leave now

[color=orange][size=75]<Edited by moderator solafidefarms>[/size][/color]

whoa. our chapter contest only had 32 people. i think 20 people made state.

I believe about 200 something amount of people.

My chapter only had about 50 ppl. 6 people are going to state.

HOW DOES EVERYONE IN YOUR CHAPTER MAKE STATES!!! I thought CT sent a lot with 50 since we are wicked dense in terms of population.. but there are chapters that upper 30s wouldn't even crack the top 10! I should move to Las Vegas (as I heard earlier, the nats qualifying score is like a 25 out of 46) (you could get that with the spring alone).

Any one in any chapter in Ohio? Just curious...

We had around 300 students at our chapter, but the competition was fierce.


[quote="gnosis"]You have that $\lim_{x\to3}\frac{x^2+x+2}{x^2-2x-3}=\frac{14}{0_{-}}=-\infty$ for $x<3$
              and $\lim_{x\to3}\frac{x^2+x+2}{x^2-2x-3}=\frac{14}{0_{+}}=+\infty$ for $x>3$
so your limit does not exist.[/quote


function
continued fraction
algebra unsolved
algebra

Find all real-valued functions $f$ on the reals such that $f(-x) = -f(x)$, $f(x+1) = f(x) + 1$ for all $x$, and $f\left(\dfrac{1}{x}\right) = \dfrac{f(x)}{x^2}$ for $x \not = 0$.

I did not use condition (1)... dunno if it is correct...

From (3) we get: $x \cdot f(\dfrac{1}{x})=f(x) \cdot \dfrac{1}{x}$. Let $g(x)=x \cdot f(\dfrac{1}{x})$ hence: $g(x)=g(\dfrac{1}{x})$. Hence $g(x)=k$. Hence: $f(x)=kx$. Put it in (2) we get: $k=1$......

 :?

[quote="shobber"] $g(x)=g(\dfrac{1}{x})$. Hence $g(x)=k$. [/quote]

I don't think it's correct - I mean $g(x)=g(\dfrac{1}{x})$ doesn't imply  $g(x)=k$ as it is. You need some more arguments.

i proved these:
$f(x)=x,x \in Z$ and $f(\frac{1}{x})=\frac{1}{x},x \in Z$and i will think about it more. :D

[quote="Rushil"]Find all real-valued functions $ f$ on the reals such that $ f( \minus{} x) \equal{} \minus{} f(x)$, $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ for all $ x$, and $ f\left(\dfrac{1}{x}\right) \equal{} \dfrac{f(x)}{x^2}$ for $ x \not \equal{} 0$.[/quote]
It is trivial that $ f(0)\equal{}0$ for $ f(0)\equal{}f(\minus{}0)\equal{}\minus{}f(0)$, hence $ f(1)\equal{}f(0\plus{}1)\equal{}f(0)\plus{}1\equal{}1$.
Now, for $ x \neq 0,1$, one has
$ \begin{cases}f(x) \equal{} x^{2}f(\dfrac{1}{x})\equal{}x^{2}(1\minus{}f(\dfrac{x\minus{}1}{x}))
\\ f(\dfrac{x\minus{}1}{x}) \equal{} \dfrac{(x\minus{}1)^{2}f(\dfrac{x}{x\minus{}1})}{x^{2}} \equal{} \dfrac{(x\minus{}1)^{2}(1\minus{}f(\dfrac{1}{1\minus{}x}))}{x^{2}}
\\f(\dfrac{1}{1\minus{}x}) \equal{} \dfrac{f(1\minus{}x)}{(1\minus{}x)^{2}} \equal{} \dfrac{1\minus{}f(x)}{(1\minus{}x)^{2}}
\end{cases}$
Solving the system gives $ f(x)\equal{}x$ for $ x \neq 0,1$, combine with above gives $ f(x)\equal{}x$ for all $ x$.

[quote="Rushil"]Find all real-valued functions $ f$ on the reals such that
$ \boxed{1} \rightarrow f( \minus{} x) \equal{} \minus{} f(x)$
$ \boxed{2} \rightarrow f(x \plus{} 1) \equal{} f(x) \plus{} 1  \ \forall  \ x \in \mathbb{R}$
$ \boxed{3} \rightarrow f(\frac{1}{x}) \equal{} \frac{f(x)}{x^2}$ for $ x \not \equal{} 0$.[/quote]

I have quite an easy and achievable proof for this...
Putting $ x \equal{} 0$ in $ \boxed{1}$ gives $ f(0) \equal{} 0$.

Now let $ g(x) \equal{} f(x) \minus{} x$
Therefore,
$ \boxed{1} \rightarrow f( \minus{} x) \equal{} \minus{} f(x) \Rightarrow g(\minus{}x) \equal{} \minus{}g(x)$
$ \boxed{2} \rightarrow f(x \plus{} 1) \equal{} f(x) \plus{} 1 \ \forall  \ x \in \mathbb{R} \Rightarrow g(x\plus{}1) \equal{} g(x)$
$ \boxed{3} \rightarrow f\left(\dfrac{1}{x}\right) \equal{} \dfrac{f(x)}{x^2}$ for $ x \not \equal{} 0 \Rightarrow g\left(\dfrac{1}{x}\right) \equal{} \dfrac{g(x)}{x^2}$

Now, Let these be our new $ \boxed{1},\boxed{2},\boxed{3}$ conditions.

\[ g(\frac{1}{x\plus{}1}) \equal{} \frac{g(x\plus{}1)}{(x\plus{}1)^2} \ (\text{From} \  \boxed{3}) \equal{} \frac{g(x)}{(x\plus{}1)^2} \ (\text{From} \ \boxed{2})\]
Now, since $ g(1\minus{}x) \equal{} \minus{}g(x)$ can be almost trivially deduced from $ \boxed{2}$ and $ \boxed{1}$
\[ \Longrightarrow g(\frac{x}{x\plus{}1})\equal{}g(1\minus{}\frac{1}{x\plus{}1}) \equal{} \minus{}g(\frac{1}{x\plus{}1}) \equal{} \frac{\minus{}g(x)}{(x\plus{}1)^2}\]
Applying $ \boxed{3}$ on this, we get
\[ g(\frac{x\plus{}1}{x})\equal{} \frac{\minus{}g(x)}{x^2}\]
But $ g(\frac{x\plus{}1}{x}) \equal{} g(1 \plus{} \frac{1}{x}) \equal{} g(\frac{1}{x}) \  (\text{by}  \ \boxed{2}) \equal{} \frac{g(x)}{x^2}$
So, on equating..we get
\[ \frac{\minus{}g(x)}{x^2} \equal{} \frac{g(x)}{x^2} \Rightarrow g(x) \equal{} 0  \ \text{If}  \ x \not \equal{} 0\]
$ \Rightarrow f(x) \equal{} x  \ \forall \ x \in \mathbb{R}$ since the case $ x\equal{}0$ has already been proved...
We Are Done... :lol:
$ Q.E.D.$

I think there is an intuitive way to look at this equation.
You can deduce that f(x)=x for all integers, 1/integers and integers + 1/integers.
Now, let us say you start with any rational number, c, between 0 and 1, at which you want to find the value of f.
You can infer f(c) from f(1/c), but since 1/c > 1, you can drop the integer part. So, you basically get {1/c}, where {} denotes fractional part. Now, the denominator of this is going to be the numerator of c, which is less than the denominator of c. Continuing this, you are ultimately going to hit a point when your numerator is 1. Now you know f at that value. You can retrace all the steps and conclude that f(c)=c.
Now, you have established that $ f(x)\equal{}x$ for all $ \mathbf{rational}$ x.

I am not sure if this can somehow be extended to irrational numbers. If f can be proved to be continuous, then you can conclude that f(x)=x for all x.

[quote="mahajan.raghu"]I think there is an intuitive way to look at this equation.
You can deduce that $ f(x) \equal{} x$ for all integers, $ 1/integers \ and \ integers \plus{} 1/integers$.
Now, let us say you start with any number, $ c$, between $ 0$ and $ 1$, at which you want to find the value of $ f$.
You can infer $ f(c)$ from $ f(1/c)$, but since $ 1/c > 1$, you can drop the integer part. So, you basically get {$ 1/c$}, where {$ .$} denotes fractional part. Now, the denominator of this is going to be the numerator of $ c$, which is less than the denominator of $ c$. Continuing this, you are ultimately going to hit a point when your numerator is $ 1$. Now you know $ f$ at that value. You can retrace all the steps and conclude that $ f(c) \equal{} c$.
Now, you have established that $ f(x) \equal{} x$ for all $ x$.
[/quote]

So, this is something like making the continued fraction of an irrational no. and saying that it converges to it ?? Please reply fast bhaiya

Also $f(x) = f(x) + n$ by induction:

Base case $n=1$
This is given.

Assume $n=k$ holds for our equation. Then $n=k+1$ should hold:
\[f(x+k+1)=f(x+k) + 1=f(x) + (k+1)\]
So $f(x + n) = f(x) + n$

f(x)+f(-x)=0
Putting X=0 we get f(0)=0
Suppose you want to find f(k) for some k
f(x+1)=f(x)+1
f(x+2)=f(x+1)+1
.
.
.
.
.
f(x+k)=f(x+k-1)+1
Adding all these we get f(x+k)=f(x)+k
Now put x=0
And we get f(k)=k


Post here if you're going to Furman!

I'm Billy the Georgian :D

um...I think everyone on this forum is going (the single digit number of people that come to this forum anyway :P )

[quote="furious"]summerville will be going  :)[/quote]

Yep, that includes me, of course :P

I think the track team might kill me for missing two Saturday meets in a row...o well.

[quote="joml88"][quote="winluo2"]Hmm, we went to Furman last year but decided to go to CofC instead this year. What's the general consensus on which is better?[/quote]

I dunno.  I've never been to Furman.  The competition looks tougher at Furman but judging solely by contest, I think C of C offers more.[/quote]

well, i'm sure it's just another contest you can add to your win list :P

last years furman seemed harder than the ones from previous years...but i was a lot worse at math a year ago, so who knows.

some kid from illinois won, i think. not entirely sure though.

Woohoo! Go Illinois! haha.... i still have no idea why we randomly went to Furman last year and CofC this year...

[quote="arkmastermind"]last years furman seemed harder than the ones from previous years[/quote]

Agreed :D

dont worry i will furman just to let you know.  if i was going, that is.


[quote="fallinlovewithmaths"]In a triangle ABC, point P is taken on the side BC (P lies between B and C). m(BAP)=30 , m(ABC)=24. AB=PC , then what's the angle m(ACP) = ?[/quote]
I think its not corect :maybe:

[quote="dima ukraine"]I think its not corect :maybe:[/quote]
Are you sure? The problem defines the triangle according to angles completely (it's constructable with ruler and compass given a side).

[hide="My solution:"][geogebra]4f378dadedbd72459edd51095ba24a7fbd9aa9ee[/geogebra]

By the sine rule $ \frac {AB}{\sin \angle APB} \equal{} \frac {BP}{\sin \angle BAP}$

$ \implies \frac {AB}{\sin \frac {7\pi}{10}} \equal{} \frac {BP}{\sin \frac {\pi}{6}}\implies \frac {BP}{AB} \equal{} \frac {\sin \frac {\pi}{6}}{\sin \frac {7\pi}{10}} \equal{} \frac {\frac {1}{2}}{\frac {\sqrt {5} \plus{} 1}{4}} \equal{} \frac {2}{\sqrt {5} \plus{} 1} \equal{}$
$ \frac {2(\sqrt {5} \minus{} 1)}{5 \minus{} 1} \equal{} \frac {\sqrt {5} \minus{} 1}{2} \equal{} \frac {\sqrt {5} \plus{} 1}{2} \minus{} 1 \equal{} \frac {AB}{BP} \minus{} 1$

$ \implies \frac {BP}{AB} \plus{} 1 \equal{} \frac {AB}{BP}\implies \frac {BP \plus{} AB \equal{} BP \plus{} PC \equal{} BC}{AB} \equal{} \frac {AB}{BP}$

The triangles $ ABC$ and $ PAB$ have a common angle $ \angle ABC$ and also $ \frac {BC}{AB} \equal{} \frac {AB}{BP}$

Hence they are similar.

Thus $ \angle ACB \equal{} \angle PAB \equal{} \boxed{\frac {\pi}{6}}$

P.S: If you want proof of $ \sin \frac {7\pi}{10} \equal{} \frac {\sqrt {5} \plus{} 1}{4}$, ask me. It has a nice geometric proof. It's the main thing in the solution.[/hide]


function
geometry
combinatorics unsolved
combinatorics

$n-omino$ is polyomino with $n$ sqaure .so the problem is: How many $n-omino$ can we make?

This problem is _quite_ unsolved; I believe there is a 'formula' (in the form of a nasty generating function) for convex polyominoes by area, and there may or may not be one for general polyominoes by area...

Regardless, if there is, it's not clean, and its in the form of a generating function at best.


how do you use latex?
how would you write 3x-2=4x
in latex code

This [code]$3x-2=4x$[/code] looks like $ 3x\minus{}2\equal{}4x$. You put dollar signs around what you want in math form.

Going through [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]this[/url] will help a lot.


Se da numarul $\sqrt{11...1}$,cu $2n$ cifre.Aflati partea intreaga si primele $n$ zecimale ale numarului dat.

Vreau sa vad daca sunt mai multe solutii pentru problema.Eu am folosit(si demonstrat) inegalitatea $33...3$(cu $n$ cifre) $<$$\sqrt{11...1}$(cu $2n$ cifre) $<$ $333...3+1$($n$ cifre) deci partea intreaga e nr $33..3$ cu $n$ cifre,iar primele $n$ zecimale sunt $3$ de $n$ dupa o alta inegaliate demonstrata.

arata cum ai demonstrat inegalitatea aceea pls!

Inegalitatea este echivalenta,prin ridicare la patrat,cu
$33..3$(cu $n$ cifre)$^2<11...1$(cu $2n$ cifre)$< (33..3 +1)^2$.
Iar inegalitatea asta se poate demonstra folosind faptul ca
$(33...3)\cdot (33...3 +1)=11..122..2$    $(1)$,
cu $n$ de $3$ fiecare factor din membru stang si cu $n$ de $1$ si $n$ de $2$ in membrul drept.Ori din relatia $(1)$ se poate scoate cu cat este egal $11..1$(cu $2n$ cifre) si de acolo este doar o treaba de verificare!
Sper ca te-am ajutat. :)

multumesc,da ai gresit semnul inegalitatii :rotfl:  :harhar:

33...3(cu n cifre) la patrat e mai mic decat 11....1(cu 2n cifre) :rotfl:

:oops_sign:   yep.. era mai mic.Celelalte subiecte de la concurs au fost crima..mai putin una cu o ecuatie simetrica.

poti sa le scri pls!ce ai facut la concurs :?:ca de la mine de la scoala n-a fost nimeni

[quote="andyciup"]O sa le scriu pe seara,ca acum am de facut niste probleme pana la 6.Am facut binisor,zic eu,locul 2. Dar subiectele mi s-au parut grele.Chiar sunt curios ce rezolvare este la una din probleme(le pun pe seara).Eu am vazut si baremul si imi este mai simplu sa zic rezolvarea,dar la concurs mi-a fost mai greu.Per total eu zic ca este bine!Uite pun acum una din probleme..[/quote]
    Mai stii si alte concursuri la care sunt invitate judetele Olt sau Valcea? :roll:


Give a set [tex] M [/tex] which have at least four elements. If [tex] a,b,c \in M [/tex] then [tex] a^2+bc \in Q [/tex].
i)Prove that there exists [tex] k \in N [/tex] satisfying [tex] \sqrt{k}a \in Q, \forall a \in M [/tex].
ii) Is the conclusion right if [tex] M  [/tex] just have three elements. :)

No,we can't. :) And they are [tex] \neq 0 [/tex]  too.


#1. ToBe (graphic)

[img]http://artofproblemsolving.com/Forum/files/phear_179.jpg[/img]

#2. DarkKnight (graphic)

[img]http://img358.imageshack.us/img358/3337/fear1rd.jpg[/img]

#3. ccy (graphic)

[img]http://img.photobucket.com/albums/v57/ccy111/prob/entry.png[/img]

#4. -Terminator- (graphic)

[img]http://artofproblemsolving.com/Forum/files/phear_174.jpg[/img]

You have one week to vote. Next Tuesday we will find out the winner! I will vote in case of a tie.

NOTE: The competitors CAN vote for themselves.

Umm... Isn't it the end of the poll now?

Yes, sorry, didn't have time to close it. Will close it now.


I am an undergraduate in mathematics, who might be stuck at a jr college for a while :(. I pushed my self back a few semesters to fill in a few gaps I felt I still had after high school. The reason I have so many gaps is I was taken out of school at the second grade to be home schooled, but my parents didn't have the resources. Eventually, I came here to live with my grandmother at 15, and she found a special ed school that help me some :( with filling in that major gap.
   The one main problem I face is the fact that I can learn techniques, but have a hard time gaining a natural understanding for things around me [learning content vs skill (which explains the reason why I have trouble with some problem solving)]. My best subject is still math, but that seems to be dieing down also. When, I was a child, I kept my self occupied by learning math, even though, I didn't understand what it was I was learning (like hermann grassman's theory of extension) and (matrices).
  As you can see, math in my past horrific life was a way of keeping my sanity.
The only problem now is all the roadblocks I keep running into when traveling down the math major path. The number one roadblock was slowing my self to the point I might never catch up in time, because I am on undergraduate scholarship that only pays for 4 years of schooling, and my skills are not good enough even now to gain mathematical competitive scholarships.
  Being a math major with gaps, and a learning disability has made things quite difficult, but I don't really know what other paths to go on. Since, my life has always been in the math books or on math sites.
  I need some advice ASAP.

So just to clarify, you are now enrolled full-time in a junior college? Have you applied for undergraduate admission anywhere?

Well, I start second semester next tuesday, and i am full time, but because learning is hard for me i can't take more then 12 hours at a time plus i can't work and do school at the same time. That is why i am stuck within scholarship time limits.

  As for undergraduate admissions, I have yet to talk to the math dept at the university of memphis, which I plan on doing soon. Your options are pretty limited with a Jr college. First of all they barely break the ice in the subject matter of math. So, even if i keep close to a 4.0 gpa, which right now I have a 3.5 in college. I sill couldn't get into the perfect schools for mathematics.

  It is interesting what I have gotten my self into.  Everytime i look into the fueture, there is duanting obsticles in the way of being an excelent math major.


three points are chosen from the interval [0,1] .What’s the possibility of chosing these points such that the longest distance between two of them is =< 1/2

[quote="cosinerburc"]...possibility...[/quote]
The term for it is [b]probability[/b]. :D


modular arithmetic
number theory unsolved
number theory

1. Let $S$ be the set of all $(a,b,c,d,e,f)$, where $a,b,c,d,e ,f$ are integers such that $a^2+b^2+c^2+d^2+e^2=f^2$. Find the largest $k$ which divides $abcdef$  for all members of $S$.
2. Show that the number of 5 tuples $(a,b,c,d,e)$ such that $abcde =5(bcde+acde+abce+abcd)$ is odd.

[i]Edited by Myth[/i]

For problem 2
Should it be "abcde = 5(bcde + acde + abde + abce + abcd)" or
"abcde = 5(bcde + acde + abce + abcd)"?

first one:
$(a,b,c,d,e,f)=(1,1,1,2,3,4)$
is in our Set
so $k|24$
on the other hand
if exist $(a,b,c,d,e,f)$ such $abcdef\neq 0 \pmod 3$
then:
$a^2+b^2+c^2+d^2+e^2=2\pmod 3$
$ f^2=1\pmod 3$
contradiction
if exist $(a,b,c,d,e,f)$ such $abcdef\neq 0 \pmod 2$
then
$a^2+b^2+c^2+d^2+e^2=5\pmod 8$
$ f^2=1\pmod 8$
now if we have $2||a$
and $bcdef\neq 0 \pmod 2$
then
$a^2+b^2+c^2+d^2+e^2=0 \pmod 8$
$f^2=1 \pmod 8$
contradiction
...
we can prove in this way that $8|abcdef$
and we have $3|abcdef$
so $k=24$


geometry
similar triangles
combinatorics unsolved
combinatorics

a.) Prove that for $n = 4$ or $n \geq 6$ each triangle $ABC$ can be decomposed in $n$ similar (not necessarily congruent) triangles.

b.) Show: An equilateral triangle can neither be composed in 3 nor 5 triangles.

c.) Is there a triangle $ABC$ which can be decomposed in 3 and 5 triangles, analogously to a.). Either give an example or prove that there is not such a triangle.

For the first one, let $ABC$ be our triangle. Divide its sides in half to get four triangles similar to the large one. This also shows that if we can divide it in $n$ triangles, thn we can also divide it in $n+3$ triangles, by dividing one of the triangles in $4$.

For $7$, divide the large triangle in $4$, and then divide one of the small resulting triangles in $4$ again.

For $6$, take $M\in AB,N\in BC$ s.t. $\frac{AB}{MB}=\frac{AC}{NC}=3$, take $R$ to be the midpoint of $MN$, and $P,Q\in BC$ s.t. $BP=PQ=QC$. The triangles $AMN,MBP,PRM,RPQ,QNR,NQC$ are the ones we're looking for.

For $8$ we can do something very similar to the construction for $6$, excapt that this time we divide $BC$ in four equal parts, $MN$ in three equal parts, and we take $M,N$ as above s.t. $\frac{AB}{MB}=\frac{AC}{NC}=4$.

As for the third one, a right-angled isosceles triangle can clearly be divided in as many similar triangles as we want (for any $n\ge 2$).

There is something that I do not understand in b) :
Divide an equilateral triangle $T=ABC$ into the three triangle $ABG,BCG,CAG$ where $G$ is the center of $T$.
Thus these three triangles are similar so...

Pierre.

The result for b) can be generalized as follows :

Let $T$ be a triangle whith angles $\alpha , \beta , \gamma$ in degree.
We will say that $T$ is free when for all integers $a,b,c$, if $a \alpha + b \beta + c \gamma = 0$ then $a=b=c=0$.

Then, we have the following result :
We cannot divide a free triangle into $2,3$ or $5$ similar triangles.

Pierre.


superior algebra
superior algebra unsolved

1)Prove that the field of quotients of Z[i] is all of Q[i].

2) Define

Q( sqrt(d) ) = {a+b*sqrt(d) | a,b E Q }

for any square free integer d. C

 i)Prove that Q( sqrt(a) , sqrt(b) ) = Q ( sqrt(a) + sqrt (b) for any square-free integers a and b.
ii) show for distinct integers a and b, that Q ( sqrt(a) ) and Q (sqrt(b)) are distinct.

(2 problems I couldn't do).

1) Well, use the definitions  :wink:

2)
i) The only thing that has to be shown is: $ \sqrt a, \sqrt b$ can be expressed (using $ \plus{},\minus{},\cdot,/$) by rational numbers and $ \sqrt a \plus{} \sqrt b$. Start with considering $ \frac 1{\sqrt a \plus{} \sqrt b}$.
ii) Show that $ a$ isn't a square in $ \mathbb Q[b]$.


Let $ p,q$ be primes such that $ p>2q$ ,and a sequence $ \{a_n\}$ is difined as follow:

$ a_n\equal{}p^n\minus{}4(\minus{}q)^n$　$ (n\equal{}1,2,3,\cdots )$.

Find the smallest value of $ pq$  such that all $ a_n (n\equal{}1,2,3,\cdots )$ are multiple of $ 3$ .

[quote="Bovy11"]Let $ p,q$ be primes such that $ p > 2q$ ,and a sequence $ \{a_n\}$ is difined as follow:

$ a_n = p^n - 4( - q)^n$　$ (n = 1,2,3,\cdots )$.

Find the smallest value of $ pq$  such that all $ a_n (n = 1,2,3,\cdots )$ are multiple of $ 3$ .[/quote]

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$ a_n$ satisfies the recursion \begin{align*}a_n = p^n - 4(-q)^n &= [p^{n-1} - 4(-q)^{n-1}](p-q) + pq[p^{n-2} - 4q^{n-2}] \\ &= (p-q) a_{n-1} + (pq) a_{n-2}\end{align*} Thus, $ 3 | a_i$ for all $ i$ iff $ 3 | a_1, a_2 = p + 4q, p^2 - 4q^2$.

If $ 3 | p$, then $ 3 | q$, contradiction. So we have $ p + 4q \equiv p + q \equiv 0 \pmod{3}$, and if $ 3 | p+q$, then $ 3|p^2 - 4q^2 = [(p + q) - 3q](p + 2q)$ also. The smallest such primes are $ (p,q) = (7,2)$, and $ pq = \boxed{14}$.
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function
real analysis
real analysis unsolved

i hope i haven't asked this before... :D

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$ \sum_{n=0}^\infty\;\frac{1}{1+n^{2}+n^{4}}\;=\;\boxed{\frac{1}{2}+\frac{\pi}{2\;\sqrt 3}\cdot\tanh\left(\frac{\pi\;\sqrt 3}{2}\right)}$

I will sketch a solution that uses complex variables, but it is really messy.  I will just have to wait and see if anyone solves this without complex variables.

Create the function:  $ f(z)=\frac{ \pi \cot \pi z}{z^{4}+z^{2}+1}$ and consider the square contour with vertices at:  $ (n+\frac{1}{2}) (\pm 1 \pm i)$.

Now in addition to the integers, $ f(z)$ will also have residues at $ -(-1)^{\frac{1}{3}},(-1)^{\frac{1}{3}},-(-1)^{\frac{2}{3}},(-1)^{\frac{2}{3}}$.

The residues at those points are respectively:

$ \frac{\pi \cot( (-1)^{\frac{1}{3}}\pi)}{2 (-2+(-1)^{\frac{1}{3}})}, \hspace{2 mm}\frac{\pi \cot( (-1)^{\frac{1}{3}}\pi)}{2 (-2+(-1)^{\frac{1}{3}})}, \hspace{2 mm}\frac{\pi \cot( (-1)^{\frac{2}{3}}\pi)}{2 (2+(-1)^{\frac{2}{3}})}, \hspace{2 mm}\frac{\pi \cot( (-1)^{\frac{2}{3}}\pi)}{2 (2+(-1)^{\frac{2}{3}})}$

It can be verified that the sum of those four residues can be simplified to the following expression:  $ -\frac{\pi \tanh( \frac{\pi \sqrt 3}{2})}{\sqrt 3}$   :)

Therefore we have (suppressing the usual estimates):

$ \oint_{\gamma_{n}}f(z) dz = 2 \pi i \left( \sum_{-n \leq k \leq n}\frac{1}{k^{4}+k^{2}+1}-\frac{\pi \tanh( \frac{\pi \sqrt 3}{2})}{\sqrt 3}\right)$

Taking the limit gives:

$ \sum_{k=-\infty}^{\infty}\frac{1}{k^{4}+k^{2}+1}= \frac{\pi \tanh( \frac{\pi \sqrt 3}{2})}{\sqrt 3}$

Since the sum is even, we get:

$ \sum_{k=1}^{\infty}\frac{1}{k^{4}+k^{2}+1}= \frac{\pi \tanh( \frac{\pi \sqrt 3}{2})}{2 \sqrt 3}-\frac{1}{2}$

Now just adjust the summation to start at $ 0$ to get:

$ \sum_{k=0}^{\infty}\frac{1}{k^{4}+k^{2}+1}= \frac{\pi \tanh( \frac{\pi \sqrt 3}{2})}{2 \sqrt 3}+\frac{1}{2}$