Show that for any quadrilateral inscribed in a circle of radius 1, the length of the shortest side is less than or equal to $ \sqrt{2}$.

Please give [b]hints only, no solutions.[/b] Thanks in advance!

(I realize that proof type problems are usually posted in pre-olympiad, but the earlier CMO problems are a lot easier than the current ones, so I posted it here. If this was incorrect, a mod could move this into the correct forum.)

[hide="hint"]
How is the length of a chord in a circle related to the arc it intercepts?
[/hide]

[hide="bigger hint"]
In a unit circle, any chord of length $ \sqrt{2}$ intercepts an arc of 90 degrees (because of the 1-1-$ \sqrt{2}$ triangle)
Also, any chord of length greater than $ \sqrt{2}$ intercepts an arc that's greater than 90 degrees.
[/hide]

[hide="really big hint"]
Use contradiction: What would happen if all of the sides of the quadrilateral were greater than $ \sqrt{2}$? (oh, and a circle has 360=4*90 degrees)[/hide]

1. What if the sides of the quadrilateral were [hide]equal[/hide]

2. What happens to length of one side if the length of another [hide]changes[/hide]?

[hide="my relatively big hint"]What's that theorem again? The one about the inscribed quadrilateral??? OH YEAH! GOOGLE![/hide]

Remember the pythagorean inequality thm: in a tri with length a and b, c<root(a squared plus b squared) if and only if angle C is less than 90 degrees; the converse/inverse holds too.  Also, involves pigeonhole principle.  This may be a slightly different solution than the one that the other users are suggesting, but it's the same conceptually and doesn't use contradiction; it uses pigeonhole.  Oh, and my solution involves the angle at the intersection of the quad's diagonals, not its arcs, but it's nearly the same.

By the way sorry: I don't know how to hide text with the blue underline!

@yugrey:

Press the hide button in the upper toolbar (under subject, subject description, message icons, whoa, there it is!)

Just don't forget to click it again before you submit your message.

No one's here; I'll jut post the solution.  OK,so by Pigeonhole, at least one angle at the center must be right or acute.  By the pythagorean inequality theorems, the length of the chord outside the acute triangle is less than root 2, while it is equal to root 2 for a right traingle.  Thus, there will always be one chord that is at most root 2 in length.

Q.E.D.


Euler
geometry
incenter
parallelogram
trigonometry
geometry proposed

[b]Problem 1.[/b] Given triangle $ ABC$ with three altitude $ AA',BB',CC'$ let $ I_a,I_b,I_c$ be incenter of triangle $ AB'C',BC'A',CA'B'$ prove that Euler line of $ I_aI_bI_c$ is connecting incenter of $ ABC$

[b]Problem 2.[/b] Given triangle $ ABC$ with three altitude $ AA',BB',CC'$ let $ I_a'$ be $ A$-excenter of triangle $ AB'C'$, $ I_b'$ be $ A'$-excenter of $ BC'A'$, $ I_c'$ be $ A'$-excenter of $ CA'B'$ prove that Euler line of $ I_a'I_b'I_c'$ is connecting $ A$-excenter of $ ABC$.

for problem （a）

△A’IbC’∽△A’IcC   =〉  △IbA’Ic∽△C’A’C
∠IIcIb=180-∠IbIcA’-∠A’IcC=180-(90-∠B)-(90+1/2∠B)=1/2∠B
∠IcIbIa=∠A’IbC’-∠A’IbIc-∠IaIbC’=90+1/2∠B-(90-∠C)-(90-∠A)=90-1/2∠B
So  IIc⊥IaIb
Analogously   IIa ⊥IbIc       IIb ⊥IaIc     =〉  I is the orthocenter of △IaIbIc
this completes the solution

problem  (b)  is the same to problem (a)  actually
A-excenter of △ABC is the orthocneter of Ia'Ib'Ic'   So the euler line pass through the point

Nice thanks  :lol: , here I mean the triangle $ I_aI_bI_c$ equal to contact triangle and actually they have the same Euler line, the similar result for excenter, we can prove $ I_bI_cB'C'$ is parallelogram when $ A'B'C'$ is contact triangle by $ I_bI_c \equal{} IA\sin A \equal{} B'C'$ and $ I_bI_c\perp IA$ as you shown.


limit
number theory
Diophantine equation

Given a 4-digit number with distinct digits.

If you take the largest possible four-digit number that can be made by rearranging the four digits, and you subtract from that the smallest possible four-digit number that can be made by rearranging the four digits, you get the number, but with the digits in reverse order.

Find the number.

[hide="answer"] Let $ a > b > c > d$, then we are looking for a number $ n$ composed of the digits $ a,b,c,d$ such that $ n \equal{} (1000a \plus{} 100b \plus{} 10c \plus{} d) \minus{} (1000d \minus{} 100c \minus{} 10b \minus{} a) \equal{} 999(a \minus{} d) \plus{} 90(b \minus{} c)$.  There are 24 possible permutations of $ (a,b,c,d)$, so in principle we can just look at the 24 equations arising from setting $ n$ equal to each of the 24 candidate numbers in reverse.  Many of these can be eliminated outright (for example we know $ a,d$ cannot be in the first or fourth position which immediately eliminates half the possibilities, we also know $ n$ is divisible by $ 9$ so $ a\plus{}b\plus{}c\plus{}d$ is a multiple of $ 9$, etc), and the rest of these can be handled case by case by showing the corresponding diophantine equation has no solutions with $ a,b,c,d\in [0,9]$; the only possible answer is $ 4716$. [/hide]

[hide="see this link"]
[url=http://en.wikipedia.org/wiki/D._R._Kaprekar]Kaprekar number[/url][/hide]

Ahh that's clever, the number in reverse must by definition be a fixed point of the iteration $ a_k \equal{} F(a_{k \minus{} 1}) \minus{} f(a_{k \minus{} 1})$ where $ F(x)$ gives the largest four digit number out of the digits of $ x$ and $ f(x)$ gives the smallest.  So just performing the iteration a few times with some seed value not a multiple of 1111 gives a cheap way to find an answer.  Of course once you've found an answer proving uniqueness is still equivalent to the original question, and it's not clear a priori that there are no cycles, but it's a quick and easy thing to try. :)

Yeah, for low digit numbers, a quick mathematica program will do it. In five digits, there is one possible cycle of length 2, 2 possible cycles of length 4, and the possibility that it degenerates to 0.

[b]Strange Conjecture:[/b] Let $ g(x) \equal{} F(x) \minus{} f(x)$, with F and f defined as it was in 2 posts above. Let $ D(x)$ denote the number of digits $ x$. Prove or disprove that exactly one of the following is true:

$ \blacktriangleright \qquad \forall n \in \mathbb{N} \; \;  D(g^n (x)) \equal{} D(x)$
$ \blacktriangleright \qquad \exists k \in \mathbb{N} \; : \; \forall n \in \mathbb{N}^{\geq k} , \; \; \lim_{n \to \infty} g^n (x) \equal{} 0$


I found a really interesting way to solve systems of congruences without having to combine them... it is an application of the Chinese Remainder Theorem...
I am attaching my proof, and the proof also shows the method to solve a system of congruences

The way to solve large linear congruences is to follow the methdo used to prove the theorem

I don't see how that's an application of the Chinese remainder theorem.  Most textbooks would call that the chinese remainder theorem, or at least a corollary to it (in which case the chinese remainder theorem would be that the system of $n$ congruences $x_i\equiv a_i\bmod b_i$ has a unique solution mod $\prod^n b_i$ if $(b_i,b_j)=1$ for all $i\not=j$)

In any event, that statement can be found on the mathworld website.

You're right... wrong word choice of application
But I did see the statement on the mathworld website.


What is the greatest possible product of a four-digit number and a three-digit number obtained from seven distinct digits?

From the numbers $ 9753$ and $ 864$ which equals $ \boxed{8426592}$.

how do you get the numbers...
how do you know your product is biggest...
$ (1000a\plus{}100b\plus{}10c\plus{}d)(100e\plus{}10f\plus{}g)$
expand and order coefficients from greatest to least, then assign numbers nine to 3 in order from greatest to least.


I tried out asymptote because people said it was good. How do I make dashed lines?

Nothing from metapost works :(

[quote="#H34N1"]I tried out asymptote because people said it was good. How do I make dashed lines?[/quote]It took me about 20 seconds to find the answer on the Asymptote web page:
[url]http://asymptote.sourceforge.net/doc/Pens.html#Pens[/url]

[quote]Nothing from metapost works :([/quote]Why should it?  Asymptote is "inspired by" MetaPost, but it's not MetaPost.   I haven't yet invested the time to learn Asymptote, but I'm assuming (based on everything I've heard) that once I do, it will be a lot less painful to use than MetaPost is.

[quote="DPatrick"][quote="#H34N1"]I tried out asymptote because people said it was good. How do I make dashed lines?[/quote]It took me about 20 seconds to find the answer on the Asymptote web page:
[url]http://asymptote.sourceforge.net/doc/Pens.html#Pens[/url][/quote]

Thank you. I scourged that website for about 10 minutes and couldn't find it :(

I imagine that website is rather bruised and battered and lacerated now, after you scourged it for so long. I can't imagine it was a pleasant experience for either of you.

Go look up scourged. http://www.onelook.com/?w=scourge

Ever read the Asymptote Manual thingy? http://asymptote.sourceforge.net/asymptote.pdf It has a very useful Table of Contents and Index.

It is bruised to the hard core :)

148 pages would take a pretty long time to find out how to make dashed lines though. :(

As I said, the Table of Contents and Index are quite useful to find a specific topic. Failing that, might I suggest the Find tool - Ctrl+F ?


algebra
polynomial
function
LaTeX
algebra proposed

$A$.prove that any polynomail $P(x)$ such that $P(x) \ge 0$ for all real $X$ can be written as the sum of the squares of two polynomails .

$B$(Hilbert`s problem ).

Prove that polynomail :

$x^2(x^2-y^2)(x^2-1)+y^2(y^2-x^2)(y^2-1)+(1-x^2)(1-y^2)$

is everywhere nonnegative ,but cannot be written as the sum of squares of any number of polynomails.

(One of the Hilbert`s problems ,solved by Artin and Schreier ,was to prove that such a polynomail can always be written as the sum of squares of rational functions )

P(x) >=0 So P(x) = Q(x)^2 H(x)
where H(x) = (x^2 + p1x +q1) ...
where pi^2 < 4qi

Now x^2 + p1x + q1 = (x+p1/2) ^2  + D^2
So Using the identity: (a^2+b^2)(c^2 +d^2) = (ad-bc)^2 +(ac+bd)^2
we are done.

A little explanation and use of latex would ,deliver what do you mean , ,Cause right now i just cant understant what ,you are trying to say.

a)
All real zeros occure an even number of times, so by factoring out we can assume that $P(x)$ has no real zeros, so it's of the type $P(x)= \prod (x-x_i)(x-\overline{x_i})$ with nonreal zeros $x_i$.
Now divde them up to $Q(x)=\prod (x-x_i)$a nd $\overline{Q}(x) = \prod (x-\overline{x_i})$.
Now write $Q(x)=R(x)+i \cdot I(x)$ such that $R$ has just real and $I$ just pure imaginary coefficients.
Then $P(x)=Q(x)\overline{Q}(x)=(R(x)+i \cdot I(x))(R(x)-i \cdot I(x)) = R(x)^2+I(x)^2$.

The same problem was posted some time ago.

I gave the standard solution for a) (very similar to ali's), but I remember I couldn't solve b).

so, how do you solve it?


function
ratio
geometric series
geometric sequence
calculus
calculus computations

[img]http://img25.imageshack.us/img25/2394/83113647.jpg[/img]

it looks very easy , but I traied to find a similar problem in my textbook and i didn't find any.

please help

$ \sum_{n\equal{}2}^{23}5^n\equal{}\sum_{m\equal{}1}^{22}5^2\cdot 5^{m\minus{}1}$ (i just used other inidces)
This is now a geometrical sequence $ b_m\equal{}b\cdot q^{m\minus{}1}$ with $ q\equal{}5$ and $ b\equal{}5^2$
So we can use the formula for geometrical series $ \sum_{m\equal{}1}^kb_m\equal{}b\frac{1\minus{}q^k}{1\minus{}q}$, in our case $ \sum_{m\equal{}1}^{22}b_m\equal{}5^2\frac{1\minus{}5^k}{1\minus{}5}$    (answer d)

[quote="valerie"]... in our case $ \sum_{m \equal{} 1}^{22}b_m \equal{} 5^2\frac {1 \minus{} 5^k}{1 \minus{} 5}$    (answer d)[/quote]

This should be $ \sum_{m \equal{} 1}^{22}b_m \equal{} 5^2\frac {1 \minus{} 5^{22}}{1 \minus{} 5}$

I teach my class to remember this verbally.

The sum of a finite geometric series is
\[ \frac{(\text{first term})\minus{}(\text{next term after the last})}{1\minus{}(\text{ratio})}\]
(The one exception is if the ratio equals $ 1.$ That's fairly easy to compute if you just think about it.)

In this particular case: the first term is $ 5^2,$ the next term after the last is $ 5^{24},$ the ratio is $ 5$, so the sum is $ \frac{5^2\minus{}5^{24}}{1\minus{}5}.$ From that form you have to do a bit of processing to get that it is the same as (d).

I do like this verbal organization - can you remember it that way?

[quote="Kent Merryfield"]

The sum of a finite geometric series is
\[ \frac {(\text{first term}) \minus{} (\text{next term after the last})}{1 \minus{} (\text{ratio})}\]
(The one exception is if the ratio equals $ 1.$ That's fairly easy to compute if you just think about it.)[/quote]

So do always I.

If $ a_i,\ a_{i \plus{} 1},\ \cdots a_{j \minus{} 1},\ a_{j}$ are a geometric progression, and the ratio is $ r\neq 1$, then these sum $ S(i,\ j)$ is :

By the definition of geometric progression, we have

$ S(i,\ j) \equal{} a_i \plus{} a_{i \plus{} 1} \plus{} \ \cdots \plus{} a_{j \minus{} 1} \plus{} a_{j}$

$ rS(i,\ j) \equal{} \ \ \plus{}a_{i \plus{} 1} \plus{} a_{i \plus{} 2} \plus{} \ \cdots \plus{} a_{j} \plus{} a_{j \plus{} 1}$

Subtracting both sides,

$ (1 \minus{} r)S(i,\ j) \equal{} a_i \minus{} a_{j \plus{} 1}$, yielding $ \boxed{\boxed{S(i,\ j) \equal{} \frac {a_i \minus{} a_{j \plus{} 1}}{1 \minus{} r}\ (r\neq 1)}}$

I always calculate the sum in imaging this scheme in my head.

[quote="valerie"]$ \sum_{n \equal{} 2}^{23}5^n \equal{} \sum_{m \equal{} 1}^{22}5^2\cdot 5^{m \minus{} 1}$ (i just used other inidces)
This is now a geometrical sequence $ b_m \equal{} b\cdot q^{m \minus{} 1}$ with $ q \equal{} 5$ and $ b \equal{} 5^2$
So we can use the formula for geometrical series $ \sum_{m \equal{} 1}^kb_m \equal{} b\frac {1 \minus{} q^k}{1 \minus{} q}$, in our case $ \sum_{m \equal{} 1}^{22}b_m \equal{} 5^2\frac {1 \minus{} 5^k}{1 \minus{} 5}$    (answer d)[/quote]

yes, of course, it should be 22 and not k.  :blush:
I shouldn't copy and paste all the time (and forget to edit ^^)


What is the wavelength of a stream of electrons accelerated in an electron gun through a potential difference of 100 volts?

The strategy for solving the problem can be the following:

1 - Compute the velocity acquired by each electron.
2 - Use the De Broglie's relation to calculate the wavelenght of each electron.

As the electric force $ F_{e}$ is conservative, then the work done by the forces of the electric field applied by the gun is $ W(F_{e}) =-\Delta E_{p}$, where $ E_{p}$ is the potential energy. But the work-energy theorem also tell us that $ W(F_{e}) = \Delta E_{c}$, where $ E_{c}$ is the kinetic energy (we can neglect the work done by the weight, because the mass of an electron is so small). So we have:

$ -\Delta E_{p}= \Delta E_{c}\Rightarrow  qV = \frac{mv^{2}}{2}\Rightarrow  v = \sqrt{\frac{2qV}{m}}\Rightarrow  p = mv = \sqrt{2mqV}$,

where $ q$ is the elementary charge, $ m$ is the electron mass, and $ V$ is the potential difference. Now using De Broglie's relation

$ \lambda = \frac{h}{p}\Rightarrow  \lambda = \frac{h}{\sqrt{2mqV}}$.


geometry
3D geometry
topology
advanced fields
advanced fields unsolved

Prove that the unit cube is a topological manifold.

:huh: Naturally, an open subset of $\mathbb R^{n}$ is an $n$-manifold without boundary. Perhaps you assume that the cube is closed, in which case there is a little work involved in showing that the boundary points have neighborhoods with appropriate charts. Still, I don't see why this is a nice problem.


Sooo...who here laughed his/her but off while taking the AP Chem Exam today?  It was a total joke; no lie.

You're not allowed to discuss any of the specifics.  If you want to discuss the FR, wait 48 hours.

Haha, indeed, I'm aware of that, but thanks for the heads up.

If you want to discuss it (once you again), try collegeconfidential.com.  People will discuss it non-stop.


In a plane, 2002 segments are given, the sum of the lengths of which equals 1. Prove that there exists a line l in the given plane such that the sum of the lengths of the projections of the given segments on l does not exceed 2/3.

[i]Spain 2002[/i]

Don't care about the 2002, it's useless. Just suppose 'a lot of'.

Choose any point, and trough it you can draw a line, and can rotate that line anywhere you want. There you got all possible axisses for projection.

Now, suppose this was false, that all projections exceeded 2/3
that would mean that the total length would at least have to be \pi/3
But it is stated that the total length is only 1, so this is not possible. Proven.

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

Another way to do this is the reverse: To have optimal conditions, you need to fulfill these requirements: [list][*]the projection length should be about the same on every axis (d'uh)[*]each projection piece should be coming from only 1 segment[/list]
Consider the segments being in a circle (or a regular 2002con if you like), then the first requirement is ok, but the second isn't: all projections come from 2 pieces. The \emptyset with circumference 1 = 1/\pi, so doubled for the 2nd requirement we get a projection length 2/\pi

which is less than 2/3.


geometry
rectangle
analytic geometry
integration
inequalities
calculus
function

Let $ C$ denote the set of all rectangles in $ \mathbb{R}^2$ that contain the origin, and with sides parallel to the coordinate axis.

Consider the maximal operator associated to this family, namely

$ f^*_C$($ x$) = $ \sup_{R \in C}$ $ \frac {1}{m(R)}$ $ \int_{R}$ $ |{f(x \minus{} y)}|dy$.

(a) Then, $ f\to$ $ f^*_C$ does not satisfy the weak type inequality

    $ m$({$ x$: $ f^*_C (x)$ > $ \alpha$}) $ \le$ $ \frac {A}{\alpha}$ $ \|f\|_{L^1}$

    for all $ \alpha > 0$, all integrable $ f$, and some $ A > 0$.

(b) Using this, one can show that there exists $ f \in L^1(\mathbb{R})$ so that for $ R \in C$

$ \limsup_{diam(R) \to 0}$ $ \frac {1}{m(R)}$ $ \int_{R}$ $ f(x \minus{} y) dy$ = $ \infty$              for almost every $ x$.

Here $ diam(R)$ = $ \sup_{x,y \in R}$ |$ x \minus{} y$| equals the diameter of the rectangle.

I think the following book is  a good resource for this theory if someone really interested.
[b]Elias M. Stein[/b], [i]Singular Integrals and Differentiability properties of functions[/i], Princeton University Press, Princeton, New Jersey, 1970

That maximal operator - the so-called "strong maximal operator" - does send $ L\log L$ to weak $ L^1$ and as a consequence, we have

$ \lim_{\text{diam}(R)\to0}\frac1{m(R)}\int_Rf(x \minus{} y) \equal{} f(x)$ for almost every $ x,$

provided that $ |f|\log|f|$ is locally integrable.

Hence the example for part (b) above cannot be locally $ L\log L$ in any neighborhood.

The key paper is from the 1930's and is by Jessen, Marcinkiewicz, and Zygmund.

----

Stein's [i]Singular Integrals ...[/i] was the most important book I read during my graduate education. I second Myth's endorsement.


geometry
inequalities
trapezoid
inequalities unsolved

Show that for all point P inside to the triangle $ABC$,

$min(|PA|,|PB|,|PC|) + |PA| + |PB| + |PC| < |AB| + |BC| + |CA|$

Denote M,N,Q the midpoints respectively of AB, BC, CA. Noe the point P is in at least two of the trapezoid ABNP, BCMQ, ACMN, say ABNP and ACMN. so we have:
NP+BC/2+AC/2 >= PB+PA
AB/2+BC/2+MN >= PC+PA

and so AB+BC+CA >=2PA+PB+PC>= min(PB,PC,PA) + PA+PB+PC

euhmmmm... yes!

Just one thing, it's not $ABNP$ but $ABNQ$


function
induction
number theory unsolved
number theory

Find all fuctions $ f: \mathbb{N^{*}} \longrightarrow \mathbb{N^*}$ satisfying $f(1)\equal{}1$ and
\[f(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\equal{}f(a)^2\plus{}f(b)^2\plus{}f(c)^2\plus{}f(d)^2\]
for all $(a,b,c,d) \in \mathbb{N^*}^4$.

[quote="Abdek"]Find all fuctions $ f: \mathbb{N^{*}} \longrightarrow \mathbb{N^*}$ Satisfying:

$ \forall (a,b,c,d) \in \mathbb{N^*}^4$

$ f(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \equal{} f(a)^2 \plus{} f(b)^2 \plus{} f(c)^2 \plus{} f(d)^2$ and $ f(1) \equal{} 1$[/quote]

To simply our proof, denote $ P(x,y,z,t) \equal{} f^2(x) \plus{} f^2(y) \plus{} f^2(z) \plus{} f^2(t)$

From $ f(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \equal{} P(x,y,z,t)$ we obtain this following result:
If $ a_1,...,a_8$ be positive integers satisfying $ a_1^2 \plus{} ... \plus{} a_4^2 \equal{} a_5^2 \plus{} ... \plus{} a_8^2$ and $ f(a_i) \equal{} a_i \forall 1 \le i \le 7$, then $ f(a_8) \equal{} a_8$

Now we notice that $ P(2k \plus{} 1,k \minus{} 2,1,1) \equal{} P(2k \minus{} 1,k \plus{} 2,1,1)$ and $ P(2k,k \minus{} 5,1,1) \equal{} P(2k \minus{} 4,k \plus{} 3,1,1)$, so we will have $ f(n) \equal{} n \forall n \in N*$ if $ f(n) \equal{} n \forall 1 \le n \le 10$
- $ f(4) \equal{} 4f^2(1) \rightarrow f(4) \equal{} 4$
- $ P(5,2,1,1) \equal{} P(4,4,1,1)$ and $ P(5,1,1,1) \equal{} P(4,2,2,2)$, so $ f(2) \equal{} 2$ and $ f(5) \equal{} 5$
- $ P(3,3,3,1) \equal{} P(5,1,1,1) \rightarrow f(3) \equal{} 3$
- $ P(2,3,5,5) \equal{} P(1,1,5,6) \rightarrow f(6) \equal{} 6$
- $ f(7) \equal{} f^2(2) \plus{} 3f^2(1) \rightarrow f(7) \equal{} 7$
- $ P(8,1,1,1) \equal{} P(7,4,1,1) \rightarrow f(8) \equal{} 8$
- $ P(9,2,1,1) \equal{} P(7,6,1,1) \rightarrow f(9) \equal{} 9$
- $ P(10,10,3,1) \equal{} P(9,8,8,1) \rightarrow f(10) \equal{} 10$

And we conclude the only function is $ f(n) \equal{} n$

[quote="shoki"]we can easily get f(0)=0 and f(1)=1, f(2)=2,f(3)=3,f(4)=4
now we prove the rest by induction and we use the fact that every integer can be written as the sum of four squares:
we know that f(k)=k for all k=1,...,n-1 ; n>4
now we know that for some a,b,c,d we have n=a²+b²+c²+d² which implies f(n)=f(a)²+f(b)²+f(c)²+f(d)²
and since a,b,c,d < n so we get f(n)=a²+b²+c²+d²=n[/quote]

No. The only difficulty of this problem was that it involved $ \mathbb N^*$. So no $ f(0)$ and the equality is true only for elements $ >0$ and Lagrange theorem is true only for squares $ >\equal{}0$, so cant be used here.
Sorry.

sorry but i didn't understand what u said,(sorry for my poor english)
can u explain plz?
thx
(as i know $ N^*\equal{}N \cup \{0 \}$ ! plz tell me where i'm wrong)

[quote="shoki"]sorry but i didn't understand what u said,(sorry for my poor english)
can u explain plz?
thx
(as i know $ N^* \equal{} N \cup \{0 \}$ ! plz tell me where i'm wrong)[/quote]

$ \mathbb{N^*}\equal{}\mathbb{N}\minus{}\{0\}$

:!:   that's very amazing it seems that in iran $ N$ is different from the other countries! :huh:
but i'm sure that $ N^*\equal{}N \cup \{0 \}$ and it follows that $ 0\not \in N$ (at least here it is so!)
anyway,i'm really sorry ... :(

[quote="shoki"]:!:   that's very amazing it seems that in iran $ N$ is different from the other countries! :huh:
but i'm sure that $ N^* \equal{} N \cup \{0 \}$ and it follows that $ 0\not \in N$ (at least here it is so!)
anyway,i'm really sorry ... :([/quote]

According to me, $ \mathbb N^*$ is a notation used in countries where $ 0\in\mathbb N$ in order to describe $ \mathbb N \minus{} \{0\}$

Dont you use $ \mathbb R^*$ in order to describe $ \mathbb R\minus{}\{0\}$ ?

Nice solution ll931110   :)

In my solution I used the fact that

$ n^2 \plus{} n^2 \plus{} 3^2 \plus{} 1^2 \equal{} (n \plus{} 1)^2 \plus{} (n \minus{} 1)^2 \plus{} 2^2 \plus{} 2^2$

and it follows that $ f(n \plus{} 1)^2 \minus{} f(n)^2 \equal{} f(n)^2 \minus{} f(n \minus{} 1)^2 \plus{} f(3)^2 \plus{} f(1)^2 \minus{} 2f(2)^2$

we can easly get by replacing $ n$ by $ 3$ $ : f(3)^2 \minus{} f(2)^2 \equal{} 5 \implies f(3) \equal{} 3$ and $ f(2) \equal{} 2$ since $ 5$ is prime

and hence $ f(3)^2 \plus{} 1 \minus{} 2f(2)^2 > 0$ By the induction we can clealry see that $ f(n \plus{} 1)^2 \minus{} f(n)^2 > 0$ or $ f(n \plus{} 1)\ge f(n) \plus{} 1$

which guid us to $ f(n) \ge n$

we know that $ f(4^{2^n \minus{} 1}) \equal{} 4^{2^n \minus{} 1}$ (easy to prove by induction) hence if there exists some $ k < 4^{2^n \minus{} 1}$ such that $ f(k) > k \implies f(k \plus{} 1) > k \plus{} 1$ then $ f(4^{2^n \minus{} 1}) > 4^{2^n \minus{} 1}$ since $ f$ is increasing which is absurd

Mr pco Can you verify this solution ? please  :lol:

[quote="Abdek"]Nice solution ll931110   :)

In my solution I used the fact that

$ n^2 \plus{} n^2 \plus{} 3^2 \plus{} 1^2 \equal{} (n \plus{} 1)^2 \plus{} (n \minus{} 1)^2 \plus{} 2^2 \plus{} 2^2$

and it follows that $ f(n \plus{} 1)^2 \minus{} f(n)^2 \equal{} f(n)^2 \minus{} f(n \minus{} 1)^2 \plus{} f(3)^2 \plus{} f(1)^2 \minus{} 2f(2)^2$

we can easly get by replacing $ n$ by $ 3$ $ : f(3)^2 \minus{} f(2)^2 \equal{} 5 \implies f(3) \equal{} 3$ and $ f(2) \equal{} 2$ since $ 5$ is prime

and hence $ f(3)^2 \plus{} 1 \minus{} 2f(2)^2 > 0$ By the induction we can clealry see that $ f(n \plus{} 1)^2 \minus{} f(n)^2 > 0$ or $ f(n \plus{} 1)\ge f(n) \plus{} 1$

which guid us to $ f(n) \ge n$

we know that $ f(4^{2^n \minus{} 1}) \equal{} 4^{2^n \minus{} 1}$ (easy to prove by induction) hence if there exists some $ k < 4^{2^n \minus{} 1}$ such that $ f(k) > k \implies f(k \plus{} 1) > k \plus{} 1$ then $ f(4^{2^n \minus{} 1}) > 4^{2^n \minus{} 1}$ since $ f$ is increasing which is absurd [/quote]

We may make it simpler. Since $ n^2 \plus{} n^2 \plus{} 3^2 \plus{} 1^2 \equal{} (n \plus{} 1)^2 \plus{} (n \minus{} 1)^2 \plus{} 2^2 \plus{} 2^2$ implying $ f^2(n \plus{} 1) \equal{} 2f^2(n) \minus{} f^2(n \minus{} 1) \plus{} 2 \equal{} (n \plus{} 1)^2 \rightarrow f(n \plus{} 1) \equal{} n \plus{} 1$ (by induction)


Errm.. no.

Is that just basically axiomate physics? Everything is a model, they can just get more accurate

Your question is content-free without definitions of the terms "everything," "world," and most importantly "mathematical formula."

[quote="binomial_4eva"]Prove or disprove that everything in the world has a mathematical formula.[/quote]

Quantum physics. Gödel. No.

Perhaps move this to the [url=http://www.mathlinks.ro/index.php?f=131]Other Problem Solving Topics Forum[/url] or [url=http://www.mathlinks.ro/index.php?f=138]The Round Table Forum[/url]...

[quote="binomial_4eva"]Prove or disprove that everything in the world has a mathematical formula.[/quote]

are you a mathematical formula  :D ?

lol no i am not a mathematical formula but somethings in me contains mathematical expressions or numbers or may be a formula who knows lol.

binomial__4eva=one person$ \equal{}b\equal{}p$  isnt this an expression?

[quote="binomial_4eva"]somethings in me contains mathematical expressions or numbers or may be a formula who knows[/quote]
No.  Numbers are abstract concepts -- they don't have physical reality.

[quote="binomial_4eva"]binomial__4eva=one person$ \equal{} b \equal{} p$  isnt this an expression?[/quote]
Not a meaningful one.

Even [url=http://colin.shoddybattle.com/suicide/theory.htm]suicide[/url] has a formula now...

The question I think you are [b]trying[/b] to ask is whether all physical phenomena can be described in terms of mathematical models.  (You are wording this question very badly.)

Unfortunately, this is a question outside the scope of mathematics, so it cannot be answered with mathematics.  It's even outside the realm of science.  Perhaps you should ask about this in the Round Table.

Congratulations for discovering one of the most famous open problems in existence. :P

You might considering reading some books on logic and philosophy. Be forewarned that the stuff can get bizarre and esoteric, which is why I personally seldom read such books.


induction
ratio
algebra
polynomial
binomial theorem
geometric series

$Prove: \left(x^{n}-y^{n}\right) = \left(x-y\right)\left(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\right)$

I think it's pretty simple, but I can't figure it out. I'm supposed to prove it with induction, but I am having serious problems. Hints are needed more than solutions  :lol:

Thanks.

EDIT: Got it fixed.

[hide="Hint without induction"]Find the sum of the geometric series $x^{n-1}, x^{n-2}y, x^{n-3}y^{2}, \cdots, y^{n-1}$.[/hide]

I know that

$\sum_{i=0}^{n-1}\left(x^{n}\right)$
is given by the equation $\frac{x^{n}-1}{x-1}$, is that the equation I should use to find the sum?

It seems that the series would be something like

$\sum_{i=0}^{n}\left(x^{n-i}y^{i}\right)$. Is this correct?

[hide="Hint for inductive proof"]$x^{n+1}-y^{n+1}=x^{n+1}-x^{n}y+x^{n}y-y^{n+1}=\dots$ (factorize and rearrange)[/hide]

Since $x^{n-1-i}y^{i}$ forms a geometrical sequence with common ratio $\frac{y}{x}$, so:

$\sum_{i=0}^{n-1}x^{n-1-i}y^{i}=\frac{x^{n-1}\left( 1-\frac{y^{n}}{x^{n}}\right) }{1-\frac{y}{x}}=\frac{x^{n}-y^{n}}{x-y}$

[quote="TylerB"]Ok. Now I'm even more confused. I can't see any way to factor that in such a way that would give me the answer I was looking for.

Could someone give me the solution so I can read through it and then say to myself "Whoa! why didn't you think of that?"[/quote]

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

Now substitute $x^{n}-y^{n}$ with the inductive presumption, extract the common factor $x-y$ and you'll get what you need. If you're not sure what "mathematical induction" is, get informed in advance.

I'd use polynomial division
that's how I proved the binomial theorem (that is what this is called, right?)
at a camp this summer

Well, it's a binomial, but the only times where I would use binomial theorem are when two variables are being added together.. ie

$(x+y)^{n}$. Not as a difference of ns

[quote="Gyan"]Sometimes direct way is the best! (Just try out the multiplication!)

For convenience ...

Let  $z =x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\dots xy^{n-2}+y^{n-1}$
Then see what is $xz$  and   $yz$
subtract one from other ..  (after shifting one to the right etc...)

You will get $xz-yz = x^{n}-y^{n}$  (All other terms cancels out!)

Or $(x-y) z = x^{n}-y^{n}$ (Which is what you want to prove)

(This is nothing but Geometric series sum .. as pointed out above)[/quote]
Wow. I like that proof the best :P. Thanks.


conics
parabola
LaTeX
analytic geometry
graphing lines
slope
geometry unsolved

I'm new on the forum and therefore maybe i have to introduce myself before asking for a geometric problem, but i need the solution of this soon and so i'm jumping some netiquette's rules. I'm sorry for this, for my bad English and beacause i don't use the LaTex yet.  ;)

**Anyway**
Given the equations of two parabolas (|:y=-x^2+4x and |':y=-x^2+6x), let be P a variable point with coordinates (0,t^2) with t>0. Determine the two contact points T and T' between the tangents that passing for P and the two parabolas, contained in the first quadrant.
Note: Between the two points pass a straight line parallel to the ass of ordinates

Let the tangency points $T_1, T_2$ of the line through P with the 2 given parabolas have the coordinates $T = (x_1, y_1)$ and $T_2 = (x_2, y_2)$. The slopes of the tangents $t_1,\ t_2$ are

$m_1 = \frac{\text d(-x^2 + 4x)}{\text dx}|_{x = x_1} = -2x_1 + 4$

$m_2 = \frac{\text d(-x^2 + 6x)}{\text dx}|_{x = x_2} = -2x_2 + 6$

The equations of the tangents $t_1,\ t_2$ passing through the point $P = (0, t^2)$ and with the slopes $m_1, m_2$, respectively, are

$t_1: \ \ y - t^2 = m_1 x = (-2x_1 + 4)x$

$t_2: \ \ y - t^2 = m_2 x = (-2x_2 + 6)x$

Since these 2 lines pass throught the points $T_1 = (x_1, y_1) = (x_1, -x_1^2 + 4x_1)$ and $T_2 = (x_2, y_2) = (x_2, -x_2^2 + 4x_2)$, we get equations for $x_1, x_2$ in terms of t:

$T_1 \in t_1: \ \ y_1 - t^2 = -x_1^2 + 4x_1 - t^2 = (-2x_1 + 4)x_1\ \Longrightarrow x_1 = t$

$T_2 \in t_2: \ \ y_2 - t^2 = -x_1^2 + 6x_1 - t^2 = (-2x_2 + 6)x_2\ \Longrightarrow x_2 = t$

It follows that $T_1 = (t, -t^2 + 4t),\ T_2 = (t, -t^2 + 6t)$.

Well done, it was easy (middle school's problem), but I wanted be sure of my solution. Thank you.  :roll:


algebra
polynomial
calculus
integration
system of equations

Any ideas for solving this one?

Two of the roots of a polynomial equation with integral coefficients are
sqrt(5+sqrt 3) and sqrt(7-sqrt 2).  If the equation is
209(x^4) + (x^a) - 24(x^k) - p(x^2) + m = 0, find the value of a+k+p+m.

[hide]Other two roots are the "conjugates"  5-  :sqrt: 3 and 7 +  :sqrt: 2. [/hide]

[quote="polymorphic"]Other two roots are the "conjugates" 5- :sqrt: 3 and 7 + :sqrt: 2.[/quote]

That doesn't work; I think you missed two radicals.  The given roots are $\sqrt{5-\sqrt3}$ and $\sqrt{7+\sqrt2}$.

To solve the problem, you could just plug in each given root to get a system of equations in terms of a, k, p, and m.

Find a way to express those roots outside of the radical.

[hide]sorry...you're right towtomb.  I now realize that there was a problem; even if the problem were the way i read it, i'd get 209x^2 instead of 209x^4.  So instead of doing that, do something pretty similar: the polynomial factors as (x + sqrt(5 + sqrt 3))(x - sqrt(5 + sqrt3))(x  + sqrt(5 - sqrt3))(x - sqrt(5-sqrt3))(x - sqrt(7 - sqrt2))(x + sqrt(7 - sqrt2))(x - sqrt(7 + sqrt2))(x + sqrt(7 + sqrt2)).  So now this polynomial has degree 8 and we'll have the 209x^4 term that we need  :D [/hide]


[quote="moldovan"]There are given $ 100$ integers $ a_1,a_2,...,a_{100}.$ Prove that among these numbers there are some of them whose sum is divisible by $ 100$.[/quote]
[hide="Hint"]
[color=blue]
Define $ \ b_r \equal{} \sum_{i \equal{} 1}^{r}a_i$ for $ \ r \equal{} 1,\ 2,\ 3,\cdots,\ 100 \ .$  Now consider the complete residue class of $ 100 ,\ \{0,\ 1,\ 2,\ 3\ \cdots,\ 99\}$
[/color]
[/hide]

[hide="Complete Solution"]
[color=blue]
There are $ \ 100$ types of   $ \ b_r .$ If one of them is divisible by $ \ 100$, we are done with the problem. Let none of them is divisible by $ \ 100\ .$ But in this case atleast two $ \ b_r$ will belong to same residue class of $ \ 100$ (as their can $ \ 99$ residue classes apart from $ \ 0$),their difference (which will be sum of some of the $ \ a_{i^s}$) will be divisible by $ \ 100$.
Hence among $ \ 100$ numbers there are some of them whose sum is divisible by $ 100$.
[/color]
[/hide]


Hi. I had linear algebra a while back, and this proof, which seems like it should be a piece of cake, is giving me all kinds of trouble.

[i]Let A and B be nxn diagonalizable matrices. Show that AB is diagonalizable.[/i]

I've proved a special case (where none of the eigenvalues of A or B are equal to zero) by a simple argument with inverses, but can't make the jump to a qualification-free conclusion. Any help? Thank you in advance.

Thank you...I suppose I should read through the vaults before posting, eh?

Or even the most recently posted topics - I put that example up just yesterday. And you might notice that none of the eigenvalues in my example are zero, so you might want to reexamine your argument.


Hi, could someone help me with the following question please:

3. (a) Find constants A and B so that
((1)/(2 − x)(4 − x)) = (A/(2-x)) + (B(4-x))

Mutiply both sides by $(2-x)(4-x)$ to get $1=A(4-x)+B(2-x)$ .This must be true for any $x$. For $x=4$ we get $1=-2*B$ or $B=-\frac 12$ and for $x=2$ we get $A= \frac 12$

[hide="Solution"]$\frac{1}{(2-x)(4-x)}=\frac{a}{2-x}+\frac{b}{4-x}$
$\frac{1}{(2-x)(4-x)}=\frac{a(4-x)}{(2-x)(4-x)}+\frac{b(2-x)}{(2-x)(4-x)}$

$1=a(4-x)+b(2-x)$
This equality have to be true $\forall x\in{\mathbb{R}- [2,4]}$, so let's take $x=0$ and $x=1$, we obtain system of two equation:

$1=4a+2b$
$1=3a+b$

Now we have to solve it

we multiply the second equation by $-2$ and we get: $-2=-6a-2b$ now we add this to the first one and we obtain: $-1=-2a-2b+2b \iff -1=-2a \iff a=\frac{1}{2}$ We put $a$ to the first equation and compute : $1=2+2b \iff b=-\frac{1}{2}$

The answer is:
$\boxed {a=\frac{1}{2} \quad b=-\frac{1}{2}}$[/hide]


ABC has incenter I and line l is tangent to the incircle of ABC. A', B' and C' are points on l such that AIA', BIB' and CIC' are right angles. Show that AA', BB', CC' are concurrent.

http://www.mathlinks.ro/Forum/viewtopic.php?t=18907

  darij


Alexis read a book with more than 100 and fewer than 200 pages.
The sum of the three digits in the number of pages is 10. The
second digit is twice the last digit. How many pages did Alexis'
book have?

Since it's more than 100 pages and less than 200, we know the hundreds digit is a 1. The sum of the digits is 10, and since one of them is 1, the other digits have to add up to 9. Here are our possibilities:

118

127

136

145

154

163

172

181

We know the tens digit is twice the units digit, so we can see that our answer is $ \boxed{163}$.

Ernie! What did I JUST tell you? People know the answers!!!!!!!!

What they want is the HOW! C'mon Ernie, you used to explain things just fine!  :(

gah!

[color=blue]* ernie is lazy.[/color]

This problem is self-explanatory.

Maybe to some people, but if they asked it on the forum, they obviously want to know how and why.

(And anyways, if you're "too lazy", how come you're doing this in the first place? Hmmm?   :bruce: )

[hide] Since the book is more than 100 pages and less than 200, we know the hundreds digit has to be a 1. The sum of the digits is 10. Since one digit is taken care of (1), the other digits MUST add up to nine.
Possibilities:

118
127
136
145
154
163
172
181

the tens digit is twice the units digit, so we see that our answer is $ \boxed{163}$ pages! [/hide]

[quote="ernie"]OMG, this is like the third post I saw with like two people answering with the same solution and method.

No offense, but its kinda spamming.[/quote]
well sometimes i like to try out the problems and stuff...

You can always try them out on a piece of paper, if you really wanna.


Bari   or   Txerq:
Inch  ka  chka?????? :lol:

[quote="Tigranhak"]Bari   or   Txerq:
Inch  ka  chka?????? :lol:[/quote]

English please


How many square residues are there mod 2^n?

All residues mod $ 2^{n}$ of quadratics are perfect square numbers.

Prove:
All integers can be expressed in on of the forms
$ 2^{n}, 2^{n}+1,...,2^{n}+2^{n}-1$

Let's look at the quadratics:
$ (2^{n})^{2}=2^{n+1}$
$ (2^{n}+1)^{2}= 2^{n+1}+2\cdot 2^{n}+1$
...
$ (2^{n}+2^{n}-1)^{2}= 2^{n+1}+2\cdot 2^{n}\cdot (2^{n}-1)+(2^{n}-1)^{2}$

As you see, in every column there is every factor without the last one $ \equiv 0$ mod $ 2^{n}$. So we just have to see at the quadratics, so all of the integer numbers $ x$ with $ x^{2}\geq 2^{n}$ there is aquadratic residue.


This is quite a pointless game. In this game, there will be 5 rounds. In each round, I will post "Start!". After this, please post here as soon as you can. The first one to post will recieve 10 points, 2nd 7, 3rd 5, 4th-6th 2, 7th-10th 1 points. Whoever has the most points at the end is the champion. After 24 hours, from when I make the post to start a round, all other competitiors who have not posted will recieve 0 points. The time between rounds varies, but is usually reasonably soon after the last round ended. Hope you play! Sign up, there's ten ppl who can play. I will post the points standings often.

Locked for being really, as you say, a "pointless" topic.


http://www.deviantart.com

My page is [url=http://hexaditidom.deviantart.com]here[/url], anyone else have one?

Sweet...so we got artists here as well...

[url=http://mutant-knight.deviantart.com]My Gallery[/url]


A bicycle is rolling down a circular hill whose radius is 9.00 m.  As the drawing illustrates, the angular displacement of the bicycle is 0.960.  The radius of each wheel is 0.400 m.  What is the angle (in radians) through which each tire rotates?

[quote="phucnv87"][color=blue]Read this first [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=84055]Rules for the Introductory Physics Forum[/url]

[quote]Kindly note the following rules for posting in this forum.

1) This Forum is for High School (non-AP) , Non- Calculus problems and simple College level discussion. Homework problems at College level can be discussed here. For more advanced theoretical, Calculus-based as well as Olympiad problems, kindly use the Advanced Physics Forum.

2) Problems and exercises of simple nature ( 'cookbook' problems) mainly requiring application of formulae and homework problems can be posted here. Recreational and 'fun' problems should also be posted here. [b]However note that no problem shall be fully solved for you and you should try to explain what you didnt understand or were unable to do while solving the problem.
[/b]
3) Clarification of doubts on specific High school or Freshman College level Physics Theory are welcome here. You can also seek help on course material. Advanced theoretical discussions should be done at the Advanced Physics Forum.

For more information, kindly contact the Moderator. Suggestions are welcome.[/quote][/color][/quote] Thank you kindly for your reminder.  I am not asking for problems to be fully solved, just need foot in the right direction.  You could have also sent me a pm whch would have been the adult thing to do.  God bless!

ok
we have the angle X=
the radii of the hill=9m
s=the length of the arc infront of $X$
$s=rX$
so we can find $s$
the radii of the wheels:R=
so we have the circumference of them =$2(3.14)R$=P
$s/P$*$2(3.14)$ gives us the angle in radiant


Find x if the angle of incidence is 40 degrees.

NOTE:

A picture is given showing angle ADC = 70 degree

Angle CDE = x degrees

Angle EDB = 70 degrees

All on a straight line = 180 degrees

Do I add 70 + 70 + x = 180 and solve for x?

what is the angle of incidence for? is there a mirror?
u can solve for x that way provided that B, Dand C lie on a straight line


A deaf old man always had the same schedule: he woke at 7, ate at 8, and took a walk across the nearby railroad tracks at 9:30. He knew he'd be safe, because the first train always came by at 10:30, and he was always far past when it came. However, one day, despite leaving at the right time, and following his schedule exactly,  he was hit by the train. Why was he hit by the train?

[hide="HAHAHAHHAHAHAHA got this in the first post"]DAYLIGHT SAVING TIME[/hide]

haha no one cares about this one anymore cuz i got it HAHAHAHAHHA soxfan

crazy, in 2008 nobody actually talked about video games

stooooooooooooooooooooop pls why are you bumping useless posts


Given an $ nxn$ table with nonnegative entries such that the sums of entries in each column and row are equal, a player plays the following game: The steps of the game consist of choosing $ n$ cells no two of which share a column or a row, and subtracting the same number from each of the entries of the $ n$ cells, provided that the resulting table still has nonnegative entries. Prove that the player can change all the entries to $ 0$.

you want to marry each row off to a different column ;). it's been posted before:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173634


geometry
geometric transformation
rotation
geometry proposed

(Kiev olympiad, 8--9) Reconstruct the square $ ABCD$, given its vertex $ A$ and distances of vertices $ B$ and $ D$ from a fixed point $ O$ in the plane.

There are usually four such squares (or two, or one, or none).  Suppose $ x \equal{} OB$ and $ y \equal{} OD$.  Let $ O'$ be the image of $ O$ under a rotation about $ A$ by $ 90$ degrees (you have a freedom to choose the clockwise or the counterclockwise direction, and it doesn't matter).  Let $ D_1$ and $ D_2$ be the points in which the circle at $ O'$ with radius $ x$ meets the circle at $ O$ with radius $ y$.  Also, let $ B_3$ and $ B_4$ be the points of intersection between the circle at $ O'$ with radius $ y$ and the circle at $ O$ with radius $ x$.  The points $ D_1$ and $ D_2$ determine $ \left(B_1,C_1\right)$ and $ \left(B_2,C_2\right)$.  Likewise, $ B_3$ and $ B_4$ determine $ \left(C_3,D_3\right)$ and $ \left(C_4,D_4\right)$.  Now, we've got the desired $ AB_iC_iD_i$ for $ i \equal{} 1,2,3,4$.


(1) If f(1-x)+2f(x) = sin^2(x).    means sin square x.
Then prove that sin(f(x))<1/sqrt(2.) means square root of 2.


It's October 4th, but is it at midnight October 4th or a bit later? Is the e-mail checked during SUnday? I'm thinking of finally submitting my solutions on SUnday (Unfortunately, I'm a procrastinator and will need to take advantage of the coming Saturday).

If you are faxing or submitting by email, your submission is due on midnight on the night of Monday, October 4.  (This means that 11:59 pm on October 4 is OK, but 12:01 am on October 5 is not OK.)  But don't wait until the last minute!  We will not be responsible if, at 11:55 pm on October 4th, your fax machine jams, your email server crashes, etc.

If you are submitting by mail, your submission must be postmarked no later than October 4.  There is no need to submit by FedEx, Express Mail, or any other such express service: as long as it is postmarked October 4 or earlier, it will be accepted (even if it arrives several days after October 4).


group theory
abstract algebra
calculus
integration
algebra
function
domain

Let $(A,+,\cdot)$ be a finite unit ring, with $n\geq 3$ elements in which there exist [b]exactly[/b] $\dfrac {n+1}2$ perfect squares (e.g. a number $b\in A$ is called a perfect square if and only if there exists an $a\in A$ such that $b=a^2$). Prove that

a) $1+1$ is invertible;

b) $(A,+,\cdot)$ is a field.

[i]Proposed by Marian Andronache[/i]

In a finite ring, an element being invertible is equivalent to it not being a divisor of $0$. If $2$ were a divisor of $0$ (to the left or to the right), then we would have an $a\ne 0$ for which $a+a=0$, so $\{0,a\}$ would be a $2$-element subgroup of $(A,+)$. This is impossible, since $A$ has $n$ elements, and $n$ is odd.

Let's look at the squares of non-zero elements. If $x^2=a$, then $(-x)^2=a$, and $x\ne -x$ for all non-zero $x$. This means that the only way we can have $\frac {n-1}2$ non-zero squares is if $x^2=y^2$ implies $x=y$ or $x=-y$.

Assume now there are two elements $a,b$ s.t. $ab=0$. Then $(ba)^2=baba=0$, meaning that $ba=0$ as well. Then we have $(a+b)^2=(a-b)^2=a^2+b^2$. This means that either $a+b=a-b$ or $a+b=b-a$, which, in turn, means that one of $a,b$ must be $0$. We have thus shown that no non-zero element can be a divisor of $0$, so all non-zero elements are invertible, Q.E.D.

Grobber, I think that you would have been the only one with max. points at this olympiad :D

integral domains are not necessarily fields, unless they are finite.


modular arithmetic
number theory unsolved
number theory

[quote="irantst"]hi
question:
assume that "p" is a prime number and p>3
prove that
A=(7^p)-(6^p)-1 is devisible by 43
tnx :maybe:[/quote]

Let $ A_p \equal{} 7^p \minus{} 6^p \minus{} 1$

$ 7^pA_p \equal{} 49^p \minus{} 42^p \minus{} 7^p$ and so $ 7^pA_p \equal{} 6^p \minus{} ( \minus{} 1)^p \minus{} 7^p\pmod{43}$

So, if $ p$ is odd, $ 7^pA_p \equal{} 6^p \plus{} 1 \minus{} 7^p \equal{} \minus{} A_p\pmod{43}$ and so $ A_p(7^p \plus{} 1) \equal{} 0\pmod{43}$ $ \forall p$  odd.

Now, it's easy to see that $ 7^p \plus{} 1 \equal{} 0\pmod{43}$ $ \iff$ $ p \equal{} 3\pmod 6$

And so $ \boxed{A_p \equal{} 0\pmod{43}\text{ }\forall\text{ odd positive integer }p\ne 3\pmod 6}$

Hence the result.

I think that It was for Iran,stage 2.
yes??


Let [tex]M[/tex] be the set of terms of a geometric progression.

What is the largest possible number of elements in the intersection of the set [tex]M[/tex] and the set:
a) [tex]\{2^n-1 | n\in\mathbb{Z}\}[/tex] ; b) [tex]\{2^n+1 | n\in\mathbb{Z}\}[/tex]

Very large ressemblance with a beautiful problem of Laurentiu Panaitopol from TST 1994 in Romania: the sequence $3^n-2^n$ does not contain three-terms geometric progressions. The proof I knew was based on some ugly inequalities.

If the proof is ugly, why do you call a problem beautiful?

Because the result it states is nice. Probably I'm wrong, but there are many results so nice to state and so ugly proofs.

Is the answer to (b) two?

Because it seems to me that $(2^{a}+1)/(2^{b}+1)=(2^{b}+1)/(2^{c}+1)$ has no solutions a,b,c in the natural numbers (cross multiply and look at the base two expansion).


Shanille O'Keal shoots freethrows on a basketball court. She hits the first and misses the second, and thereafter the probability that she hits the next shot is equal to the proportion of shots she has hit so far. What is the probability that she hits exactly $ 50$ of her first $ 100$ shots.

It is readily checked out that at any throw the (combined) probability of a hit is $ \frac {1} {2}$. Therefore all $ 2^{98}$ continuations are equi-probable, so the probability of hitting exactly $ 50$ out of the first $ 100$ is $ \frac {1} {2^{98}} \genfrac {(} {)} {0pt} {} {98} {49}$.

Apologies - I am guilty of what they call d'Alembert syndrome! The first part was OK, but the conclusion was wrong by miscounting.

If we start graphing all the possible outcomes, we can notice that after $ n$ shots each of the numbers from the interval $ [1,n\minus{}1]$ will appear $ n\minus{}2$ times, so all the possible number of hits are equally likely to happen. We can prove this by induction.

We know that after $ 2$ shots Shanille hit exactly once, so there are equal chances of hit or miss after the third shot, therefore $ 1$ and $ 2$ will appear once each. This is our base of induction.

Let's assume this will be true after $ n$ shots.

Let $ m$ be the number of hits after $ n$ shots, $ m\in [1,n\minus{}1]$. Probability that she will hit is $ \frac{m}{n}$.
Let's write down all the possible $ m$'s and count the possible outcomes.

$ m\equal{}1$

$ 1$ will appear $ n\minus{}1$ times in the graph
$ 2$ will apear $ 1$ time

$ m\equal{}2$

$ 2$ will appear $ n\minus{}2$ times
$ 3$ will appear $ 2$ times

$ m\equal{}3$

$ 3$ will appear $ n\minus{}3$ times
$ 4$ will appear $ 3$ times

$ \cdots$

$ m\equal{}n\minus{}2$

$ n\minus{}2$ will appear $ 2$ times
$ n\minus{}1$ will appear $ n\minus{}2$ times

$ m\equal{}n\minus{}1$

$ n\minus{}1$ will appear $ 1$ time
$ n$ will appear $ n\minus{}1$ times

Summing we get that each $ m$ will appear exactly $ n\minus{}1$ times. $ QED$

After $ 100$ shots total number of hits can be any number in interval $ [1,99]$ (each being equally probable), therefore probability that she will hit exactly $ 50$ shots is $ \boxed{\frac{1}{99}}$.

[quote="keyree10"]Shanille O'Keal shoots freethrows on a basketball court. She hits the first and misses the second, and thereafter the [b]probability that she hits the next shot is equal to the proportion of shots she has hit so far.[/b] What is the probability that she hits exactly $ 50$ of her first $ 100$ shots.[/quote]

mavra, you missed the bold

[hide="Solution"]
Let 1 denote a hit, 0 denote a miss. Then we're looking for sequences starting with 10 which contain fifty 0s and fifty 1s of which there are $ \binom{98}{49}$.

All of these are equi-probable since it is easy to check that replacing "01" in the sequence with "10" doesn't change the probability of the sequence occurring, so we can continue doing this until we have the configuration 10 0000...49 of these...000 1111...49 of these...111 which has a probability of occurring of $ \frac{49!^2}{99!}$. Multiplying this with the number of possible sequences gives $ \boxed{\frac{1}{99}}$.
[/hide]


calculus
integration
trigonometry
calculus computations

I have another integral for you:
$ I\equal{}$ $ \int_{0}^{\frac {\pi}{4}} \frac {dx}{1\plus{}sin^2x}$

this is easy man....

write the $ {sin^2x}$ as $ \frac{1}{cosec^2x}$ and then write the $ {cosec^2x}$ in the denominator as $ {1\plus{}cot^2x}$...and then substitute $ {cotx}{\equal{}}{t}$ and then alter the limits and solve...

 i thyink u can handle it from here..

[hide]$ \int\frac{1}{1 \plus{} \sin^2x}\,dx \equal{} \int \frac{1}{2\minus{} \cos^2x}\,dx \equal{} \int \frac{\sec^2x}{2 \sec^2x\minus{}1}\,dx$

$ \equal{} \int \frac{\sec^2x}{1 \plus{} 2 \tan^2x}\,dx \equal{} \frac{1}{\sqrt{2}}\int \frac{\sqrt{2} \sec^2x}{1 \plus{} (\sqrt{2} \tan x)^2}\,dx \equal{} \frac{1}{\sqrt{2}} \arctan(\sqrt{2} \tan x) \plus{} C$[/hide]

that was almost the same method i used Carcul....


I've been curious about how to solve decimals as rational fractions, as i find that i encounter those kind of questions alot, but i can never solve them... any help? Any generic example would be helpful. :)

Thanks in advance,
the origamiguy

Could you be more specific? For example, do you mean repeating decimals?

well, repeating decimals mostly, but i think that non-repeating decimals are impossible...??

Non-repeating decimals are irrational, right?

Anyway, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=282644]here's a link[/url] that explains the basics of repeating decimals.

That's a good article, but AIME15, you should probably consider adding in the proofs.

Here is the proof:

Let's have the decimal, $ 0.\overline{x},$ where $ x$ is a digit. Let $ y\equal{}0.\overline{x}.$ Therefore, $ 10y \equal{} x.\overline{x}.$ $ 10y\minus{}y\equal{}x.\overline{x} \minus{} 0.\overline{x},$ which gives us $ 9y\equal{}x,$ thus, $ y\equal{}0.\overline{x}\equal{}\dfrac{x}{9}.$

Let's prove it for a more general case. Let's have a repeating block of $ i$ digits, where $ i$ is an integer, and also $ i$ is greater than or equal to $ 1,$ but less than or equal to $ 9,$ as that is the maximum number of digits we can have repeated. So, our decimal is of the form $ 0.\overline{x_1x_2x_3\dots x_i}.$ Again, we let this equal $ y.$ So, we have $ 0.\overline{x_1x_2x_3\dots x_i} \equal{} y.$ Thus, $ 10^iy \equal{} x_1x_2x_3\dots x_i.\overline{x_1x_2x_3\dots x_i}.$ Subtracting $ y$ from this cancels out the numbers after the decimal point, giving us $ 10^iy \minus{} y \equal{} x_1x_2x_3\dots x_i.\overline{x_1x_2x_3\dots x_i} \minus{} 0.\overline{x_1x_2x_3\dots x_i} \equal{} x_1x_2x_3\dots x_i.$
$ 10^iy\minus{}y \equal{} (10^i \minus{} 1)y \equal{} (\underbrace{999\dots9}_{\text{i times}})y.$ Therefore, $ (\underbrace{999\dots9}_{\text{i times}})y \equal{} x_1x_2x_3\dots x_i,$  which gives us $ y\equal{}\dfrac{x_1x_2x_3\dots x_i}{\underbrace{999\dots9}_{\text{i times}}}.$

oh, thanks very much also to limac for for the proof; that really helped me understand why that method works... :)


Hi,
Could anybody tell me whether purposive sampling affect the results of OLS regression estimation? If yes why? If no, why?
Tnks
arijit

hello everyone,

actually i have a very important work to do using stata 9.but unfortunately i'm new to that software.i was just wondering,if i could find anyone to help me out.

Please reply,it's very important.

With Thanks


Write out Judith's interpretation of 11111, you'll see a pattern.  The answer is [2187/12] (ceiling function) 182, A.


algebra
polynomial
number theory
prime numbers
algebra open

[quote="hxy09"]Let $ f(x)$ be a non-constant integer-coefficient polynomial
Prove or disprove:The set of prime numbers dividing at least one number between $ f(p)$($ p$ is a prime) is infinite.[/quote]
I don't understand the question very much. Prime number p and prime number in f(p) is the same one, isn't it? If that, of course, the answer is no. If not, this set is infinite. One way to solve this problem is to consider the sum:
$ S_1 \equal{} \sum_{p\in P} \frac {1}{\sqrt [n]{|P(p)|}}$ where n is the degree of this polynomial
It is not hard to show that this sum tends to finite if S (set of all prime divisors of P(p) ) has only finite elements. However, we already know that this sum tends to infinite. That gives the contradiction.

Thank you so much!This is just what I want :)
By the way,merry Christmas my dear friend :lol:


we have triangle ABC with AB<AC and there are heights from B and C to the sides AC and AB an intersects at points B' and C'.The line B'C' intersects with line CB at point P and AP intersects with line MH at K, where M is midpoint of BC and H is orthocenter. Prove that KM is bisector of angle B'KB.

Consider the inversion through pole $ H$, power $ \overline {HB}.\overline {HB'} \equal{} \overline {HC}.\overline {HC'} \equal{} k^2$. It is easy to notice that $ I(H,k^2): M\mapsto K\Longrightarrow K\in (AB'C')$. Hence, $ \angle HKB' \equal{} \angle HAC \equal{} \angle HBC$. But it is a very well-known fact that $ MB',MC'$ are the tangent of $ M$ wrt $ AB'C'$. Thus, $ MB'^2 \equal{} MC'^2 \equal{} \overline {MH}.\overline {MK}$.
But $ MC' \equal{} MB\Longrightarrow MB^2 \equal{} \overline {MH}.\overline {MK}$, which implies $ MB$ is the tangent of $ (KHB)\Longrightarrow \angle HBC \equal{} \angle HKB$. Therefore, $ \angle HKB \equal{} \angle HKB'$, which implies that $ KH$ or $ KM$ is the bisector of $ \angle BKB'$
Our proof is completed. :lol:

Line $AP$ is the polar of $H$ with respect to the circle $\odot (B'C'BC)$ whose center is $ M.$ Therefore, $ KH \perp AP$ $\Longrightarrow$ pencil of lines $A(B,C,H,P)$ is harmonic $\Longrightarrow$ $ B,B',H$ and the projection of $ B'$ onto $ PA$ from $ B$ are harmonically separeted. Since $ KH \perp  AP,$ it follows that lines $KM$ and $AP$ bisects $\angle BKB'$ internally and externally.

[quote="luisgeometria"]Line $AP$ is the polar of $H$ with respect to the circle $\odot (B'C'BC)$ whose center is $ M$. Therefore, $ KH \perp AP$ $\Longrightarrow$ Line pencil $A(B,C,H,P)$ is harmonic $\Longrightarrow$ $ B,B',H$ and the projection of $ B'$ onto $ PA$ from $ B$ are harmonically separeted. Since $ KH \perp  AP$, it follows that lines $KM$ and $AP$ bisects $\angle BKB'$ internally and externally.[/quote]
Nice my friend :lol:

Hello! Here's my solution not using any non-olympiad methods or facts:
   By the well known lemma, the lines $ PH$ and $ AM$ are perpendicular (Too easy to prove eg by counting $ PM$, $ PA$, $ HM$ and
$ HA$). Consider the triangle $ APM$: It's obvious that $ AH$ and $ PH$ are altitudes in this triangle. So, as the altitudes of any triangle concur, $ MH$ is perpendicular to line $ AP$, implying that $ \angle{MKA}$ is right angle. Define by $ A'$ the intersection point of lines $ AH$ and $ BC$. Than, the quadrilaterals $ AKA'M$ and $ AC'A'C$ are both cyclic and $ KH \cdot HM \equal{} AH \cdot A'H \equal{} BH \cdot B'H$, implying  $ KH \cdot HM \equal{} BH \cdot B'H$. The last equality means that $ KBMB'$ is also cyclic, but on the other hand, we have $ B'M \equal{} BM \equal{} CM$. So, in circle, circumscribed on quadrilateral $ KBMB'$, the chords $ B'M$ and $ BM$ are equal, giving $ \angle{BKM} \equal{} \angle{B'KM}$, Q.E.D. Lasha Lakirbaia. ;)


Solve the natural equation: $ (x\plus{}y)^{x}\equal{}x^{y}$

If $ p|x$ $ yord_{p}(x)\equal{}xord_{p}(x\plus{}y)$, therefore exist $ z,a<b$, suth that $ x\equal{}z^{a},x\plus{}y\equal{}z^{b}\to y\equal{}z^{a}(z^{b\minus{}a}\minus{}1)$.
It give $ bx\equal{}ay\to x|y,\ a|y$, therefore we can take $ z\equal{}x, a\equal{}1.$
Your equation equavalent to $ x^{b\minus{}1}\equal{}1\plus{}b$. $ b\equal{}2$ give solution $ x\equal{}3,y\equal{}6$, $ b\equal{}3$ give solution $ x\equal{}2,y\equal{}6$.
If $ b\ge 4$, then $ x^{b\minus{}1}\ge 2^{b\minus{}1}>1\plus{}b$. Therefore there are not another solutions.

Are the $ ord_{p}(x)$ is the $ n$ min satisfy $ x^{n}\equiv 1(mod p)$?
And why $ bx\equal{}ay\to x|y,\ a|y$?

$ ord_{p}(x)$ is maximal m, suth that $ p^{m}|x$.

@nktp , i think you from Vietnam, and in Vietnam, have some book about $ ord$, Ex "Các bài toán cơ bản về số học - Phan Huy Khải".You can find some things about $ ord$, it is very nice. :)

Ok! Thanks quangpbc! Yes, I am from viet nam! I have solved this problem by myself!


hallo,
maybe easy:
Solve the qeuation in $Z$:
$5x^{2}-4y^{2}=17$.  :)

Have not solution, because $4y^{2}+17=2 \ or \ 1 \ or \ 3(mod \ 5)$.


During a drive to Chicago, Rob asks his dad, “Are we there yet?”  Professor Dad says, “No, but we are 3/8 of the way there.”  Just 24 miles later, Rob asks again, “Are we there yet?”  Professor Dad responds, “Not yet, but now we are 5/8 of the way there.”  How many miles are they now  from Chicago?

A.16  B.30   C.36   D.60   E.96

[hide="Solution"]$ 24$ miles is a quarter of the way. Scale both quantites by $ 4$, so $ 96$ miles is the whole way. Thus, the answer is $ \boxed{E}$.

[/hide]

[hide=" click to reveal hidden content"]

$ 24$ miles is $ \frac{5}{8}\minus{} \frac{3}{8} \equal{} \frac{1}{4}$ of the way. So $ \frac{1}{8}$ of the way is $ 12$ miles. They still have $ \frac{3}{8}$ to go $ \equal{} 36$ miles.

[/hide]


algebra
system of equations
algebra unsolved

Slove the system of equations
$ \sqrt{3\minus{}2x^2y\minus{}x^4y^2}\plus{}x^2(1\minus{}2x^2)\equal{}y^2$
$ 1\plus{}\sqrt{1\plus{}(x\plus{}y)^2}\equal{}y^2(2x^3\minus{}y^2\plus{}1)\minus{}x^2(1\plus{}x^2)^2$

The second equation has no real solution.

[quote="chien than"]Slove the system of equations
$ \sqrt {3 \minus{} 2x^2y \minus{} x^4y^2} \plus{} x^2(1 \minus{} 2x^2) \equal{} y^2$
$ 1 \plus{} \sqrt {1 \plus{} (x \plus{} y)^2} \equal{} y^2(2x^3 \minus{} y^2 \plus{} 1) \minus{} x^2(1 \plus{} x^2)^2$[/quote]

Do we solve in complex numbers?


limit
probability
number theory
greatest common divisor
number theory unsolved

I read in a book that it has been proved there are infinitely many numbers n for which $ n^2+1 $ is square-free. I thought about this problem, but couldn't solve it. Does anyone know a solution? And another problem: there are infinitely many n for which both n and n+1 are square-free. And if I remember corectly, Laurentiu Panaitopol even proved that there are infinitely many n for which n, n+1 and n+2 are square-free.

I read in the book "O introducere in aritmetica si teoria numerelor" by L. Panaitopol and Al. Gica that if $Q(n)$ is the number of square-free numbers $\leq\ n$, then $Q(x)=\frac 6{\pi^2} x+O(\sqrt x)$. This shows that $\frac {Q(x)}x\rightarrow \frac 6{\pi^2}$ when $x\rightarrow \infty$, so this takes care of the second problem, because assuming that for a sufficiently large $n$ at most one of $n,\ n+1$ is square free says that for sufficiently large $n$ we have $\frac {Q(n)}n<\frac 12+\epsilon$, with $\epsilon>0$ as small as we want. On the other hand, $\frac 6{\pi^2}>\frac 12$, and all of this yields a contradiction.

Here's a way to show that $\displaystyle \lim_{x\to \infty}\frac {Q(x)}x=\frac 6{\pi^2}$: The limit of $\frac {Q(x)}x$ must be the probability to pick a square-free number when we choose a random positive integer. At the same time, this probability must be $P=(1-\frac 1{p_1^2})(1-\frac 1{p_2^2})\cdots$. We have $\frac {\pi^2}6=1+\frac 1{2^2}+\cdots=\frac 1{1-\frac 1{p_1^2}}\cdot \frac 1{1-\frac 1{p_2^2}}\cdot \cdots=\frac 1P$ and we're done.

Why not?

Anyway, I expect that the third problem (the one solved by Laurentiu Panaitopol) has to be solved by completely different methods, since $\frac 23>\frac 6{\pi^2}$ and the same reasoning as above doesn't work. Where did you read about $n^2+1$, Harazi? I've searched the Net but could find nothing.

[I've modified this; it contained some nonsense :D]

I read this in a book of Leo Moser, which appeared in 1959: Introduction to number theory.

Here's something interesting:

[url]http://mathworld.wolfram.com/Squarefree.html[/url]

What's really cool is that we can find something on that page about the asymptotic desnity of the square-free Gaussian integers, which might be what we need for the problem about $n^2+1=(n+i)(n-i)$. I have no idea how to use that info exactly, but it might be helpful.

We could make the assumption that for sufficiently large $n$ $n^2+1$ isn't square-free, and then for sufficiently large $n$ no multiples of $n+i$ are square-free, and maybe this gives an asymptotic density smaller than the one cited on Mathworld, thus yielding a contradiction. I'll have to say again that maybe all of this is bogus :).

I think I have a approach on the $ n^2 + 1$ problem using the density of a sequence.
Suppose that for all sufficently large n ,  $ n^2 + 1 $ is not square-free.
It means that for all sufficently large n ,  $ n^2 + 1 $ is divisible by $ p_k ^2 $, for some prime $ p_k $ of the form 4t + 1.
Now, for each $ p_k $ of the form 4t+1 there are exactly 2 numbers $ r_k , s_k \in $ { 1; 2 ; ... ; $p_k ^2 - 1 $} such that $ r_k ^ 2 + 1 $ and  $ s_k ^ 2 + 1 $ are both divisible by $ p_k ^2 $, and $ r_k + s_k = p_k ^ 2 $ ( because if $a^ 2 + 1 $ and $b^ 2 + 1 $ are both divisible by $ p_k ^2 $ then (a-b)(a+b) is divisble by $p_k ^ 2 $, but gcd (a ; $p_k $) =1,
so we can't have both (a-b) and (a+b) divisible by $ p_k $, so either (a-b) or (a+b) is divisble by $ p_k^2 $ ).
Thus, for each sufficently large n, there exist a prime $ p_k $ of the form 4t+1 for which n is congruent to $ r_k $ or $ s_k $ mod $ p_k^2 $.
So, the union of the all arithmetic progressions of the form {$ r_k + m p_k ^ 2 $} and {$ s_k $ + $ m p_k ^ 2 $} with m, k  $ \in $ N ( and $ p_k $ is the k th  prime which is congruent to 1 mod 4 ), this means the density ( which clearly exist ) of this union must be 1. However, this density is exactly $ \alpha $ = 2 ( $ \sum $ $ \frac {1} {p ^ 2 } $) <1 (where the sum sums over all primes of the form 4t+1), a contradiction and we are done
( Let N be a sufficently large number. Then there are at least N-c integers x in [1;N] for which $ x^ 2 + 1 $ is not square-free ( c is an absolute constant ). Thus all those x must lie in an arithmetic progressions of the form {$ r_k + m p_k ^ 2 $} or {$ s_k + m p_k ^ 2 $} , where the prime $ p_k $ must also lie in [1;N] and =1 mod 4 ( for if $x^2 + 1 $ is divisible by $ p_k ^ 2 $  and x $\leq $ N then $ p_k \leq $ N ).
So we have N - c $ \leq  \sum  \frac { N + r_k + N + s_k } { p_k ^ 2 }
\leq \alpha N + M ( N ) $ ( * ) , where M(N) is the number of primes $ p_k $ less than N and congruent to 1 mod 4. We know that $ \frac { M( N ) } { N } $ tends to 0 as N tends to infinity , so after dividing N from ( * ), and let N tend to infinity we will have 1 $\leq \alpha $ < 1, a contradiction. This replaces the use of density ).


$a,b,c$be the lenght  of triangle .prove that :
$a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq 0$

[url=http://diendantoanhoc.net/forum/index.php?showtopic=11885]see here[/url]
  and [url=http://toantuoitho.nxbgd.com.vn]math for student have age  $\geq$ 12[/url]

[quote="tranthanhnam"][url=http://diendantoanhoc.net/forum/index.php?showtopic=11885]see here[/url]
  and [url=http://toantuoitho.nxbgd.com.vn]math for student have little age[/url][/quote]
   i think it is easy  , it isn't hard   .
  i want to have this complete solution


calculus
integration
algebra
function
domain
inequalities
calculus computations

Give the other 5 iterated integrals of: $\int_{0}^{2}\int_{0}^{y^{3}}\int_{0}^{y^{2}}f(x,y,z)dzdxdy$
my answer so far:
=$\int\int\int_{E} f(x,y,z)dV$ with $E=\left\{(x,y,z)|0\leq z\leq y^{2}, 0\leq x\leq y^{3}, 0\leq y\leq2\right\}$

xy projection: $D_{1}=\left\{(x,y)|\leq x\leq , \leq y\leq\right\}$
yz projection: $D_{2}=\left\{(y,z)|\leq z\leq ,\leq y\leq\right\}$
xz projection: $D_{3}=\left\{(x,z)|\leq z\leq ,\leq x\leq\right\}$

$\int_{0}^{8}\int_{\sqrt[3]{z}}^{2}\int_{0}^{y^{3}}f(x,y,z)dzdydx$=
$\int_{}^{}\int_{}^{}\int_{}^{}f(x,y,z)dydzdx$=
$\int_{}^{}\int_{}^{}\int_{}^{}f(x,y,z)dydxdz$=
$\int_{0}^{4}\int_{\sqrt{z}}^{2}\int_{0}^{y^{3}}f(x,y,z)dxdydz$=
$\int_{0}^{2}\int_{0}^{y^{2}}\int_{0}^{y^{3}}f(x,y,z)dxdzdy$

Okay, I admit it I am lost at this point. I was trying to draw them but not getting far.
Thanks fedja will try that in a little while.

Let's do the $dx\,dy\,dz$ one. The outer integration is over $dz$. Since $0\le y\le 2$ and $0\le z\le y^2$, the range for $z$ is $0\le z\le 4$. Once you decide that, the range for $y$ with fixed $z$ becomes $\sqrt z\le y\le 2$ and the range for $x$ with fixed $y$ and $z$ is $0\le x\le y^3$. Thus, the integral is $\int_0^4\int_{\sqrt z}^2\int_0^{y^3}f(x,y,z)\,dx\,dy\,dz$. Try to do the other ones yourself. The idea is the same for each integral: first determine the range of the variable over which the outer integration is taken, then, with this variable fixed, determine the range of the second variable and then, with two variables fixed, determine the range of the last variable.

Okay so doing $dzdydx$ The outer is $dx$. $0\leq x\leq y^{3}, 0\leq y\leq2$ So x range becomes: $0\leq x\leq8$ therefore y range: $\sqrt[3]{x}\leq y\leq 2$ so then: x & y fixed $0\leq z\leq y^{2}$
$\int_{0}^{8}\int_{\sqrt[3]{x}}^{2}\int_{0}^{y^{2}}f(x,y,z)dzdydx$

Is this right? And also my book mentions using projections on the 3 different planes as a method. ergo why I put $D_{1},D_{2},D_{3}$ Is that a usable method or is this better?

[quote="lajekahr"]
Is this right? And also my book mentions using projections on the 3 different planes as a method. ergo why I put $D_{1},D_{2},D_{3}$ Is that a usable method or is this better?[/quote]
Yes, this is correct. As to projections, you'll get them automatically this way from the limits for 2 outer integrals (or, if you somehow can see them immediately, you can  get the limits for 2 outer integrals from the projections).

okay the last 2 have my baffled: $dydxdz$ and $dydzdx$
Cause I get:
$0\leq z\leq y^{2}, 0\leq x\leq y^{3}$ or
$0\leq z\leq 4, 0\leq x\leq y^{3}$ or
$0\leq z\leq y^{2}, 0\leq x\leq 8$
How do I get past the y?

[quote="fedja"]Here is $dydxdz$. The range of $z$ is $[0,4]$; the range of $x$ for fixed $z$ is still $[0,8]$. Now, for fixed $x$ and $z$,
we must have both $x<y^3$ and $z<y^2$, so $y>\max\{\sqrt[3]x,\sqrt y\}$. This gives $\int_0^4\int_0^8\int_{\max\{\sqrt[3]x,\sqrt z\}}^2 f(x,y,z)\,dy\,dx\,dz$.[/quote]

$\max\{\sqrt[3]x,\sqrt z\}$???

How on earth can you integrate that? Say if $f(x,y,z)=y$
$\int_0^4\int_0^8\int_{\max\{\sqrt[3]x,\sqrt z\}}^2 y\,dy\,dx\,dz\Rightarrow$
$\int_0^4\int_0^8 \frac{1}{2}y^2]_{\max\{\sqrt[3]x,\sqrt z\}}^{2}\,dx\,dz=$
$\int_0^4\int_0^8 \frac{1}{2}(4-\max\{\sqrt[3]x,\sqrt z\}^{2})\,dx\,dz$???

Well, if you want to actually integrate something this way, you'll have the split the domain of integration into 2 parts: one where $\sqrt[3]x>\sqrt z$ and one where the inverse inequality holds. So, your triple integral becomes
\[ \int_0^4\int_0^{z^{3/2}}\int_{\sqrt z}^2 f(x,y,z)\,dy\,dx\,dz+\int_0^4\int_{z^{3/2}}^8\int_{\sqrt[3]x}^2 f(x,y,z)\,dy\,dx\,dz\,.  \]
Does it make sense now?

So this part cannot be solved with a single triple integral then?

That is a single triple integral- it's just not a form we would use to calculate the value unless we absolutely had to. Splitting it is a matter of practicality, not necessity.


[quote="broniran"]Let $ a_{i} > 0$ and $ a_{1}a_{2}...a_{n} \equal{} 1$. Prove that:
  $ (n \minus{} 1 \plus{} a_{1})(n \minus{} 1 \plus{} a_{2})...(n \minus{} 1 \plus{} a_{n})\ge n^n$[/quote]
With AM-GM we have:

$ a_{1} \plus{} n \minus{} 1 \equal{} a_{1} \plus{} 1 \plus{} 1 \plus{} ... \plus{} 1 \geq n\sqrt [n]{a_{1}}$ $ ( 1$ is repeted $ n \minus{} 1$ times$ ),$

similarly to $ a_{2},a_{3},...,a_{n},$then their product gives:

$ (n \minus{} 1 \plus{} a_{1})(n \minus{} 1 \plus{} a_{2})...(n \minus{} 1 \plus{} a_{n})\ge n^n\sqrt [n]{a_{1}a_{2}...a_{n}} \equal{} n^n.$


linear algebra
matrix
linear algebra unsolved

Can you solve it detail?
let $A$ be matrix $m$x$n$ have range be 1
prove that exist two matrix $B_{mx1}$ and$C_{1xn}$ such that :
   $A=BC$

I assume the condition is that $A$ has [b]rank[/b] equal to $1.$

That means that the columns of $A$ span a space of dimension $1.$ That means that the nonzero columns are all scalar multiples of each other. Assume there is at least one nonzero column (else the problem is trivial.) Call this column $v_{k}.$ Then $\{v_{k}\}$ is a basis for the column space of $A$ and there exist scalars $c_{j}$ such that $v_{j}=c_{j}v_{k}.$ (Clearly, this includes $c_{k}=1.$) Now, let $B=v_{k}$ and $C=\begin{bmatrix}c_{1}&c_{2}&\cdots&c_{n}\end{bmatrix}.$ We have that $BC=A.$


Amy gained 5 pounds a week to end at 480 pounds at the end of today. How much did she weigh a year ago? (assume a year is 52 weeks)

amy weighed 220 pounds a year ago. 5*52 is 260 pounds, so, a year ago, she weighed 480-260, or 220 pounds.

rotfl...

[hide="WeightGain Solution"]

$52 \times 5=260 \text{ lbs.}$

$480 - 260 = 220 \text{ lbs.}$

$\text{She gained 220 lbs.}$

:rotfl::rotfl::rotfl::rotfl::rotfl::rotfl:[/hide]

hwenterprise, she didn't gain that many pounds.. Correct anyway.

[quote="h78reg"]interesting problem........i wonder if it's based on a true story. :rotfl:[/quote]

Hopefully not  :rotfl:

I'm sure that has happened to some before. The round table would be a good place to talk about this  :D

i wonder how many visitors that post is gonna get..... :D

You should see for yourself  :D . Any post about that would be split. Let's not spam  ;)


Find all continuous functions f from R to R such that
$f(xy)=xf(y)+yf(x)$

Put $g(x)=\frac{f(x)}{x}(x\not =0)$ then $g(xy)=g(x)+g(y)\forall x,y\not = 0$. This equation is well know.  :wink:


Let BM be the median relative to side AC of triangle ABC. Let P be a point on MC. If we consider the parallel to BM passing trough P and we denote with N and Q the point where it touches the sides BC and  AB prove that:

$ PN \plus{} PQ \equal{} 2BM$

[hide="Solution"]$ \left\|\begin{array}{cc}
\triangle CPN\sim\triangle CMB & \implies  \frac{PN}{BM}\equal{}\frac{CP}{CM} \\
 \\
\triangle ABM\sim\triangle AQP & \implies  \frac{PQ}{BM}\equal{}\frac{AP}{AM}\end{array}\right| \bigoplus \implies$ $ \frac{PN\plus{}PQ}{BM}\equal{}\frac{CP}{CM}\plus{}\frac{AP}{AM}\equal{}\frac{AC}{AM}\equal{}2$ .[/hide]


Hello,
Just wondering when you guys think you will be get Round 4 Results. I'd only like to know so I can stop checking the USAMTS site 5 times/day till the appointed hour(s) roll around :).

Thanks,
sran

In some other topic, I believe the week of April 11th was mentioned.  Can't wait![/quote]

[quote="DPatrick"]We hope to release the scores sometime during the week of April 11.[/quote]And you don't have to keep checking the USAMTS site -- we'll post an announcement here when the scores are released.


I'm not sure has it ever been proposed yet.   :)

Let $ a_{1} \geq a_{2} \geq ... \geq a_{n} \geq 0$, $ b_{1} \geq b_{2} \geq ... \geq b_{n} \geq 0$ and $ \left(\pi\left(1\right),\pi\left(2\right),...,\pi\left(n\right)\right)$ be any permutation of $ \left(1,2,...,n\right)$.
Prove that
\[ \sqrt{a_{1}\plus{}b_{1}}\plus{}\sqrt{a_{2}\plus{}b_{2}}\plus{}...\plus{}\sqrt{a_{n}\plus{}b_{n}} \leq \sqrt{a_{1}\plus{}b_{\pi\left(1\right)}}\plus{}\sqrt{a_{2}\plus{}b_{\pi\left(2\right)}}\plus{}...\plus{}\sqrt{a_{n}\plus{}b_{\pi\left(n\right)}}\]
 :D

[quote="epsilonist"]I'm not sure has it ever been proposed yet.   :)

Let $ a_{1} \geq a_{2} \geq ... \geq a_{n} \geq 0$, $ b_{1} \geq b_{2} \geq ... \geq b_{n} \geq 0$ and $ \left(\pi\left(1\right),\pi\left(2\right),...,\pi\left(n\right)\right)$ be any permutation of $ \left(1,2,...,n\right)$.
Prove that
\[ \sqrt {a_{1} \plus{} b_{1}} \plus{} \sqrt {a_{2} \plus{} b_{2}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b_{n}} \leq \sqrt {a_{1} \plus{} b_{\pi\left(1\right)}} \plus{} \sqrt {a_{2} \plus{} b_{\pi\left(2\right)}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b_{\pi\left(n\right)}}
\]
:D[/quote]

We can prove $ a_1\ge a_2\ge0$, $ b_1\ge b_2\ge0$
$ \Rightarrow \sqrt {a_1 \plus{} b_1} \plus{} \sqrt {a_2 \plus{} b_2}\le \sqrt {a_1 \plus{} b_2} \plus{} \sqrt {a_2 \plus{} b_1}$
(Squaring and expanding), and using induction.

and I think it is also true:
Let $ c_i\equal{}b_{n\minus{}i\plus{}1}$
$ \Rightarrow \sqrt {a_{1} \plus{} b_{\pi\left(1\right)}} \plus{} \sqrt {a_{2} \plus{} b_{\pi\left(2\right)}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b_{\pi\left(n\right)}}\le \sqrt{a_1\plus{}c_1}\plus{}\sqrt{a_2\plus{}c_2}\plus{}\cdots\plus{}\sqrt{a_n\plus{}c_n}$.

[quote="NextPeace"]
and I think it is also true:
Let $ c_i \equal{} b_{n \minus{} i \plus{} 1}$
$ \Rightarrow \sqrt {a_{1} \plus{} b_{\pi\left(1\right)}} \plus{} \sqrt {a_{2} \plus{} b_{\pi\left(2\right)}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b_{\pi\left(n\right)}}\le \sqrt {a_1 \plus{} c_1} \plus{} \sqrt {a_2 \plus{} c_2} \plus{} \cdots \plus{} \sqrt {a_n \plus{} c_n}$.[/quote]
Yes, it is!
By the way, Now I'm thinking about another one,
Let $ b\equal{} \sqrt[n]{b_{1}b_{2}...b_{n}}$, then
\[ \sqrt{a_{1}\plus{}b_{1}} \plus{} \sqrt{a_{2}\plus{}b_{2}} \plus{} ... \plus{} \sqrt{a_{n}\plus{}b_{n}} \geq \sqrt{a_{1}\plus{}b}\plus{} \sqrt{a_{2}\plus{}b}\plus{} ... \plus{} \sqrt{a_{n} \plus{} b}\]

[quote="epsilonist"]
Let $ b \equal{} \sqrt [n]{b_{1}b_{2}...b_{n}}$, then
\[ \sqrt {a_{1} \plus{} b_{1}} \plus{} \sqrt {a_{2} \plus{} b_{2}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b_{n}} \geq \sqrt {a_{1} \plus{} b} \plus{} \sqrt {a_{2} \plus{} b} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b}
\]
[/quote]
Could anyone help me?  :)
I'm wondering that this statement is true or not?  :?:

[quote="epsilonist"]
Let $ b \equal{} \sqrt [n]{b_{1}b_{2}...b_{n}}$, then
\[ \sqrt {a_{1} \plus{} b_{1}} \plus{} \sqrt {a_{2} \plus{} b_{2}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b_{n}} \geq \sqrt {a_{1} \plus{} b} \plus{} \sqrt {a_{2} \plus{} b} \plus{} ... \plus{} \sqrt {a_{n} \plus{} b}
\]
[/quote]

try $ n \equal{} 2,a_1 \equal{} 100,a_2 \equal{} 1,b_1 \equal{} 4,b_2 \equal{} 1$.
$ \Rightarrow \sqrt {104} \plus{} \sqrt {2}\ge\sqrt {102} \plus{} \sqrt {3}$, but it is not true. :wink:

But, I think next inequalities are true :

1.Let $ a \equal{} \sqrt [n]{a_1a_2...a_n},b \equal{} \sqrt [n]{b_1b_2...b_n}$
$ \Rightarrow \sum \sqrt {a_i \plus{} b_i}\ge n\sqrt {a \plus{} b}$
(Oh, sorry. It is weaker than Minkowski's Inequality.)

2.Let $ a_{i,j}\ge a_{i,k}\ge0$ $ \forall j\le k$, and $ \{a_{i,\pi(i,j)}\}$ be any permutation of $ \{a_{i,j}\}$, and positive p,

1)If $ 0 < p < 1$
$ \Rightarrow \sum (a_{1,1} \plus{} a_{2,1} \plus{} ... \plus{} a_{n,1})^p\le \sum(a_{1,\pi(1,1)} \plus{} a_{2,\pi(2,1)} \plus{} ... \plus{} a_{n,\pi(n,1)})^p$

2)If $ p > 1$
$ \Rightarrow \sum (a_{1,1} \plus{} a_{2,1} \plus{} ... \plus{} a_{n,1})^p\ge \sum(a_{1,\pi(1,1)} \plus{} a_{2,\pi(2,1)} \plus{} ... \plus{} a_{n,\pi(n,1)})^p$

3.$ \sum_1^k a_i\ge \sum_1^k b_i\forall 1\ge k\ge n \minus{} 1$ and $ \sum_1^n a_i \equal{} \sum_1^n b_i ,c_i\ge0$

1)If $ 0 < p < 1$
$ \Rightarrow \sum(a_i \plus{} c_i)^p\le \sum(b_i \plus{} c_i)^p$

2)If $ p > 1$
$ \Rightarrow \sum(a_i \plus{} c_i)^p\ge \sum(b_i \plus{} c_i)^p$

[quote="NextPeace"]
2.Let $ a_{i,j}\ge a_{i,k}\ge0$ $ \forall j\le k$, and $ \{a_{i,\pi(i,j)}\}$ be any permutation of $ \{a_{i,j}\}$, and positive p,

1)If $ 0 < p < 1$
$ \Rightarrow \sum (a_{1,1} \plus{} a_{2,1} \plus{} ... \plus{} a_{n,1})^p\le \sum(a_{1,\pi(1,1)} \plus{} a_{2,\pi(2,1)} \plus{} ... \plus{} a_{n,\pi(n,1)})^p$

2)If $ p > 1$
$ \Rightarrow \sum (a_{1,1} \plus{} a_{2,1} \plus{} ... \plus{} a_{n,1})^p\ge \sum(a_{1,\pi(1,1)} \plus{} a_{2,\pi(2,1)} \plus{} ... \plus{} a_{n,\pi(n,1)})^p$
[/quote]
This is straightforward. For each $ j$ let $ x_j: \equal{}a_{1,j} \plus{} a_{2,j} \plus{} ... \plus{} a_{n,j}$ and $ y_j : \equal{} a_{1,\pi(1,j)} \plus{} a_{2,\pi(2,j)} \plus{} ... \plus{} a_{n,\pi(n,j)}$ Then trivially for all $ k$
$ \sum_{j\equal{}1}^kx_j\ge \sum_{j\equal{}1}^ky_j$ (and equality for $ k\equal{}n$). So $ x\succ y$ and the rest is Karamata.

[quote="NextPeace"]
3.$ \sum_1^k a_i\ge \sum_1^k b_i\forall 1\ge k\ge n \minus{} 1$ and $ \sum_1^n a_i \equal{} \sum_1^n b_i ,c_i\ge0$

1)If $ 0 < p < 1$
$ \Rightarrow \sum(a_i \plus{} c_i)^p\le \sum(b_i \plus{} c_i)^p$

2)If $ p > 1$
$ \Rightarrow \sum(a_i \plus{} c_i)^p\ge \sum(b_i \plus{} c_i)^p$[/quote]
This one is wrong. We can have $ max(b_i\plus{}c_i)> max(a_i\plus{}c_i)$
 e.g. $ n\equal{}2, a\equal{}(2,0), b\equal{}(1,1), c\equal{}(0,2)$. Then $ a\succ b$ but $ a\plus{}c \prec b\plus{}c$, just the reverse!  :(


Find all pairs of positive integers $(a,n)$ for which any prime divisor of $a^{\phi(n)}-1$ is a divisor of $n$. Here we work with Euler's function.

well if $(a,n)=1$ then this means that $a^{\varphi(n)}-1 | n$, but we know that also $n | a^{\varphi(n)}-1$ . As a consequence we must have $a^{\varphi(n)}= n+1$ !!!!!!

We don't have $a^{\varphi(n)}-1|n$...remember, it's only PRIME factors.


What is the remainder when 5^207 is divided by 7? I've seen a lot of these problems, and i always miss them. Is there some sort of formula, because if there is, Please tell me.

hello, we have $ 5\equiv\minus{}2 \mod 7$, $ 5^2\equiv4 \mod 7$, $ 5^3\equiv\minus{}8\equiv\minus{}1 \mod 7$ and hence $ 5^{207}\equal{}(5^3)^{207}\equiv(\minus{}1)^{69}\equiv\minus{}1 \mod 7$.
Sonnhard.

Why do we start with -2? Could it not be any other number?


inequalities
search
inequalities proposed

Let $ a;b;c$ be positive real numbers
Prove that
$ abc(a\plus{}b\plus{}c)^3 \geq \sum_{cyc}[a(8c^3\plus{}1)]$

[quote="friendlist"]Let $ a;b;c$ be positive real numbers
Prove that
$ abc(a \plus{} b \plus{} c)^3 \geq \sum_{cyc}[a(8c^3 \plus{} 1)]$[/quote]
Are you sure?

It's obviously not true if b is extremely small and a and c are extremely large.

[quote="friendlist"]Let $ a;b;c$ be positive real numbers
Prove that
$ abc(a \plus{} b \plus{} c)^3 \geq \sum_{cyc}[a(8c^3 \plus{} 1)]$[/quote]
Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that
$ (a \plus{} b \plus{} c)^3 \geq \sum_{cyc}[a(8c^3 \plus{} 1)]$
I'm sorry!

[quote="friendlist"][quote="friendlist"]Let $ a;b;c$ be positive real numbers
Prove that
$ abc(a \plus{} b \plus{} c)^3 \geq \sum_{cyc}[a(8c^3 \plus{} 1)]$[/quote]
Let $ a;b;c$ be positive real numbers satisfying $ abc \equal{} 1$
Prove that
$ (a \plus{} b \plus{} c)^3 \geq \sum_{cyc}[a(8c^3 \plus{} 1)]$
I'm sorry![/quote]

Are you solving the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=168836&start=20]T10/275[/url] ?

I think you shouldn't post the problems in MYM on this.  :wink:

http://www.mathlinks.ro/Forum/search.php?search_author=friendlist


That was an easy one.  Hydro had my answer.

Do you need to prove that 1, 3, 7, and 9 can be used?

If so, just look at the numbers 13 and 79.

[hide]If a prime contained any even digit, a permutation would be divisible by $2$. If If contained a $5$, a permutaion would be divisible by $5$. In either case, this happens when the last digit is even or $5$. This leaves $1$, $3$, $7$, and $9$ as the only valid numbers.
$\mathcal{QED}$[/hide]

[quote="drunner2007"]An absolute prime is a prime such that every permutation of its digits yields a prime. Show that an absolute prime with two or more digits may only contain the digits $1$, $3$, $7$, and $9$.[/quote]

[hide]If it has the number $5$, one permutation must by divisible by $5$ (one which last number is $5$). Then,if you have the numbers $0,2,4,6,8$, a permutation has to be even (when the last digit is $0,2,4,6$ or $8$). Then,  you can only use numbers $1,3,7,9$[/hide]

[quote="hydro"][hide]
clearly it can't contain 0 or an even digit, cause in this case at least one of its permutations would be an even number, so not prime. For the same reason it can't contains 5 cause one of its permutations would be divisible by 5...[/hide][/quote]

That is the same reasoning that I used.  And it seems to be correct.

The only numbers that can end a prime are 1, 3, 7, and 9, since a 2 or more digit number is divisible by 2 if the units digit is even, and divisible by 5 if the units digit is zero or five.

$PWNED!$ *cough* i mean $QED$

[quote="jclarkemathL314159"]$PWNED!$ [/quote]

Fanatic of Darij????

[quote="drunner2007"]hahaha

Follow up: A [i]cyclic prime[/i] is a prime such that every cyclic permutation of its digits yields a prime. For example, 79 and 97 are cyclic primes. Find all three digit cyclic primes that are not absolute primes.[/quote]

What are absolute primes and cyclic permutations?

[quote="b-flat"]I'm not sure what a cyclic permutation is, but an absolute prime is a prime such that every permutation of its digits yields a prime.  That was directly from the first post  ;) where the actual question was.

Maybe a cyclic permutation is when the digits stay in the same order (like in a circle), but different positions, like 123 and 231.[/quote]

Yeah, that's right. As another example, the cyclic permutations of abcd would be abcd, bcda, cdab, and dabc.

[quote="drunner2007"]hahaha

Follow up: A [i]cyclic prime[/i] is a prime such that every cyclic permutation of its digits yields a prime. For example, 79 and 97 are cyclic primes. Find all three digit cyclic primes that are not absolute primes.[/quote]
[hide]
The only digits you can use are 1, 3, 7, and 9. That leaves 137, 731, 139, 931, 179, 971, 973, and 379. 9+3-1 = 11, so 913 and 319 don't work. This leaves the cyclic permutations of 137, 731, 179, 971, 973, and 379. 937, 973, 791, and 371 are divisible by 7 leaving 731 and 971. 731 is divislbe by 17 while all cyclic permutations of 971 are prime. Since the cyclic permutations of 179 are the rest of the permutations and we already found out at least one of those is not prime. The answer is 971, 719, and 197.[/hide]


geometry
rectangle
similar triangles
analytic geometry
angle bisector

Let $ M$ and $ N$ be the midpoints of sides $ AD$ and $ BC$ of rectangle $ ABCD$, respectively.  Let $ P$ be a point on ray $ CD$ but not on side $ CD$.  Let point $ Q$ be the intersection of $ AC$ and $ PM$.  Prove that $ \angle MNQ\equal{}\angle MNP$.

[hide="ugly and probably wrong"]
call the intersection of MN and AC, F.  Draw the midpoint of DC and call that G.
then drop the perpendicular from Q to MN, call that H.
by similar triangles $ \frac{QJ}{NC}\equal{}\frac{JF}{FN}\equal{}\frac{JF}{HC}$
also by similar triangles $ \frac{QJ}{NC}\equal{}\frac{QM}{PM}\equal{}\frac{MF}{PC\minus{}MF}\equal{}\frac{MF}{PC\minus{}CH}\equal{}\frac{MF}{PH}\equal{}\frac{NF}{PH}$
so $ \frac{QJ}{NC}\equal{}\frac{JF\plus{}NF}{HC\plus{}PH}\equal{}\frac{JN}{PC}$
therefore triangles QJN and NCP are similar
so $ \angle NCP\equal{}\angle QNJ\equal{}\angle MNP$

yeah probably wrong
[/hide]

[img]http://i227.photobucket.com/albums/dd72/TPFC/OK.png[/img]
Apply Menelaus theorem for triangle $ ACD$  we have:
$ \frac{QA}{QC}\times\frac{PC}{PD}\times\frac{MD}{MA}\equal{}1$
Contruct a line over $ Q$ and parallel $ AD$ cut $ BD$ at $ R$, we have:
$ \frac{QA}{QC}\equal{}\frac{RD}{RB},\frac{MD}{MA}\equal{}1\equal{}\frac{NB}{NC}$ then
$ \frac{RD}{RB}\times\frac{PC}{PD}\times\frac{NB}{NC}\equal{}1$
It mean $ P,R,N$ are collinear! (Apply Menelaus theorem for triangle $ BCD$) Then OK!

Obviously, I havent done this...

But could you prove it also by using analytic geometry and prove using distance formulas/intersection points that NQ/QM=NP/PM which means that MN is the angle bisector of NQP.

[quote="CircleSquared"]@tdl: I don't see how your result leads to what the problem asked.[/quote]
Because $ P,R,N$ are collinear then $ \angle{MNQ}\equal{}\angle{MNR}\equal{}\angle{MNP}$


Prove that:
$\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}}\geq \frac{2}{(1+\sqrt{xy})^{2}}$
where x;y are positive real numbers satisfying $xy \geq \frac{1}{4}$

it has been posted before. just expand everything and group the squares. I'm lazy to do it right now.


function
trigonometry
real analysis
real analysis unsolved

Hi everybody,

For our next exam we  -unfortunately- have to approximate function by other (integrable) function at critical points. Those other function are well known to be integrable and then by theorems in our course, one can conclude that the function which is bounded by the approximation is also integrable.

However, I'm really bad at approximating such functions and it's going to cost me precious points (even I understand the theorem in the first place). I found already some approximations myself (listed underneath), most of them derived with the mean value theorem. Are there any other -very common- function that behave bad at crtical points (i.e. at 0 or $\infty$), but still have a nice approximation.

$\boxed{e^{x}> x^{n}\ with \ x > 0}\\ \boxed{\exp(-y) < y^{-m}\ and\ \exp(-1/x^{2}) < x^{2m}}\\ \boxed{1-\frac{1}{x}< \ln(x) < x-1 \ with \ x>1}\\ \boxed{|1-\cos(x)| \leq |x| \ with\ x \in \mathbb{R}}\\ \boxed{|\tan(x)| \geq |x| \ with \ x \in \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[}\\ \boxed{Bgtan(x) < x \ with \ x>0}\\ \boxed{Bgtan(x) > \frac{x}{1+x^{2}}\ with \ x>0}$

Thanks in advance

Some ones I like to use commonly:

$e^{x}\ge 1+x \forall x \in \mathbb{R}$
$\sin x < x \forall x > 0$


im no mood to solve stuff rite now but ill post the method all the same
find the no of children of that dude(dont remember the name and too lazy to look again) and pramila
the total no of fights btw children of diff parents as children of same parents dont fight(talk abt favoritism :P ) = total no of fights that can take place - [the number of fights among the children of same parents]
so 24c2-[xc2 + (x+1)c2 + (23-2x)c2] the max value of this shud give the answer

oops  :oops: then how come i cant see 203 newhewre
saw it now its 204 :wallbash:
and wth i cant have more than 3 smilies again priveleges of the mod

god i must be really slow at typing, pardesi beat me by 2 min. :rotfl:


Let T and U be orthogonal projections in V with inner product <,> , such that TU=UT=0. Prove that
   the orthogonal complement of Im(U) is equal to  Im(T).

 P.d. I  have that Im(T) is contained in the orthogonal complement of Im(U) , im missing the second contention.

The statement is false. Counterexample: $ V\equal{}{\mathbb R}^3$, $ T$ is an orthogonal projection on $ \langle e_1\rangle \equiv \{\lambda e_1: \lambda \in \mathbb R\}$, $ U$ is an orthogonal projection on  $ \langle e_2\rangle \equiv \{\lambda e_2: \lambda \in \mathbb R\}$. Then $ \mathrm{Im}(T)$ is a line, $ \mathrm{Im}(U)^\perp$ is a plane.

So true. I wish my T.A. would check his homework for some typos or something before handing it out to us. I really should have found that counterexample.  :oops: . Anyway thanks ver much.


Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{8}{a+b+c}\ge \frac{6}{\sqrt{ab+bc+ca}}.\]

To see an original contest, visit the Vietnam Inequality Forum at http://www.bdt.ch06k.com.

See you there,

Is it too hard? Paying attention to the case of equality  :oops:

[hide]
WLOG let $ab+bc+ca=1$. Then the right hand side is 6, and we have
\[\frac{1}{a+b}=\frac{ab+bc+ca}{a+b}=c+\frac{ab}{a+b}\geq c+\frac{ab}{a+b+c}\]
Similar inequalities hold for the second and third terms, so we get
\[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{8}{a+b+c}\geq a+b+c+\frac{9}{a+b+c}\]
Finally, AM-GM on $a+b+c$ and $\frac{9}{a+b+c}$ finishes the problem off. As for the equality case...
\[a=\frac{3+\sqrt{5}}{2},\ b=\frac{3-\sqrt{5}}{2},\ c=0\]
 :D
[/hide]

[quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{8}{a+b+c}\ge \frac{6}{\sqrt{ab+bc+ca}}. \]
To see an original contest, visit the Vietnam Inequality Forum at http://www.bdt.ch06k.com.

See you there,[/quote]
Setting $p=a+b+c,q=ab+bc+ca$ then the inequality becomes
\[\frac{p^{2}+q}{pq-r}+\frac{8}{p}\ge \frac{9}{\sqrt{q}}\]
We have
\[\frac{p^{2}+q}{pq-r}+\frac{8}{p}\ge \frac{p^{2}+q}{pq}+\frac{8}{p}=\frac{p}{q}+\frac{9}{p}\ge 2\sqrt{\frac{p}{q}\cdot \frac{9}{p}}=\frac{6}{\sqrt{q}}\]
:)
\]

Thank you very much for your very nice proof, THMM and scorpius119. A member of VI Forum solves this problem here

http://www.ch06k.com/forum/showthread.php?t=43

by a quite similar one as THMM.

[quote="A3K08"][quote="scorpius119"][hide]
Finally, AM-GM on $a+b+c$ and $\frac{9}{a+b+c}$ finishes the problem off.
[/hide][/quote]
By AM-GM, we have $(a+b+c)+(\frac{9}{a+b+c})\geq \sqrt{\frac{9(a+b+c)}{a+b+c}}=\sqrt{9}=3\neq 6$ => Wrong![/quote]
No, you're wrong!
$(a+b+c)+(\frac{9}{a+b+c})\geq 2\sqrt{\frac{9(a+b+c)}{a+b+c}}=2\sqrt{9}=6$

[quote="evarist"][color=purple]
   Can you prove or diprove the following inequality for me, hungkhtn :
 $\sqrt{xy+yz+zx}(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})\le \frac{3\sqrt{3}}{2}$
  Thank alot  :D
[/color][/quote]

Consider $y=z=1,\ x\rightarrow \infty$.  :wink:


quadratics
trigonometry
geometry
3D geometry
algebra
complex numbers
polynomial

some good questions you guys can practice with. Please hide soln. if you want to answer

[color=red]1.[/color]Solve the following quadratic equation, expressing the roots in the form $ a\plus{}ib$, where a,b are real.
$ x^{2}\cos^{2}\theta\plus{}x\sin2\theta\plus{}1\equal{}0$

[color=red]2.[/color]Find the complex factors of $ x^{2}\plus{}2ax\plus{}a^{2}\plus{}b^{2}$

[color=red]3.[/color]solve for $ w_{1},w_{2}$ $ \equal{}\equal{}\equal{}\Rightarrow $ $ 2w_{1}\plus{}3iw_{2}\equal{}0$

.......................................$ (1\minus{}i)w_{1}\plus{}2w_{2}\equal{}i\minus{}7$

[color=red]4.[/color]If $ z\equal{}\cos\theta\plus{}i\sin\theta$, prove that $ \frac{2}{1\plus{}z}\equal{}{1\plus{}\cos\theta\minus{}isin\theta}{1\plus{}cos\theta}$=$ 1\minus{}i\tan\frac{\theta}{2}$

[color=red]5.[/color]Express $ (3\plus{}2i)(5\plus{}4i)$ and $ (3\minus{}2i)(5\minus{}4i)$ in the form $ a\plus{}ib$. Hence find the prime factors of $ 7^{2}\plus{}22^{2}$

[color=red]6.[/color]$ 1\minus{}2i$ is one root of the equation $ x^{2}\plus{}(1\plus{}i)x\plus{}k\equal{}0$. Find the other root and the value of $ k$.

[color=red]7.[/color]Find the modulus and argument of each of the complex numbers $ z_{1}\equal{}2i$ and $ z_{2}\equal{}1\plus{}\sqrt3i$. Mark on an Argand diagram the points P,Q,R and S representing $ z_{1},z_{2},z_{1}\plus{}z_{2}$ and $ z_{1}\minus{}z_{2}$ respectively. Deduce the exact values of $ arg(z_{1}\plus{}z_{2})$ and $ arg(z_{1}\minus{}z_{2})$.

[color=red]8.[/color]If $ |z|\equal{}r$ and $ arg z \equal{}\theta$, show that $ \frac{z}{z^{2}\plus{}r^{2}}$ is real and give its value.

[color=red]9.[/color]$ 1,\omega$ and $ \omega^{2}$ are the cube roots of unity. State the values of $ \omega^{3}$ and $ 1\plus{}\omega\plus{}\omega^{2}$. Hence show that $ (1\plus{}\omega^{2})^{12}\equal{}1$ and $ (1\minus{}\omega)(1\minus{}\omega^{2})(1\minus{}\omega^{4})(1\minus{}\omega^{5})(1\minus{}\omega^{7})(1\minus{}\omega^{8}) \equal{} 27$

[color=red]10.[/color]Show that the roots of $ z^{6}\plus{}z^{3}\plus{}1\equal{}0$ are among the roots of $ z^{9}\minus{}1\equal{}0$. Hence find the roots of $ z^{6}\plus{}z^{3}\plus{}1\equal{}0$ in mod-arg form.

[color=red]11.[/color]The point P represents the complex number $ z$ on an Argand diagram. Describe the locus of P in each of the following cases
(a)$ |z|\equal{}|z\minus{}2|$ (b)$ arg(z\minus{}2)$=$ arg(z\plus{}2)\plus{}\frac{\pi}{2}$

[hide="Answer for 1"]The solutions are $ \minus{}\tan\theta\pm i$.[/hide]

[hide="6"]

$ \text{Sum Of Roots}\equal{}\minus{}1\minus{}i\Rightarrow\minus{}2\plus{}i$ is the other root.

$ k \equal{} (1\minus{}2i)(\minus{}2\plus{}i) \equal{} 5i$.[/hide]

[hide="Solution for 4"] Assume WLOG $ \theta\in [0, 2\pi)$. We have $ \frac{2}{1\plus{}z}\equal{}\frac{2(1\plus{}\overline{z})}{|1\plus{}z|^{2}}\equal{}\frac{2(\cos\theta\plus{}1\minus{}i\sin\theta)}{2(1\plus{}\cos\theta)}\equal{}\frac{1\plus{}\cos\theta\minus{}i\sin\theta}{1\plus{}\cos\theta}$ (which I supposed you meant - there is no fraction in your problem).
For the last part, we have $ \frac{1\plus{}\cos\theta\minus{}i\sin\theta}{1\plus{}\cos\theta}\equal{} 1\minus{}i\frac{\sin\theta}{1\plus{}\cos\theta}$, which is equal to $ 1\minus{}i\tan\frac{\theta}{2}$ iff $ \frac{\sin\theta}{1\plus{}\cos\theta}\equal{}\tan\frac{\theta}{2}$, equivalent to $ \frac{\sin\theta}{1\plus{}\cos\theta}\equal{}\pm\sqrt{\frac{1\minus{}\cos\theta}{1\plus{}\cos\theta}}$, or $ \sin\theta \equal{}\pm\sqrt{1\minus{}\cos^{2}\theta}$, which is true (the choice of $ \pm$ depends on $ \theta$). The expression is valid for $ \theta\neq\pi$.
[/hide]
[hide="Solution for 10"] First, we want to show that every solution to $ z^{6}\plus{}z^{3}\plus{}1 \equal{} 0$ satisfies $ z^{9}\minus{}1 \equal{} 0$. However, $ z^{6}\plus{}z^{3}\plus{}1 \equal{} 0$ implies $ (z^{3}\minus{}1)(z^{6}\plus{}z^{3}\plus{}1) \equal{} 0$, or equivalently $ z^{9}\minus{}1 \equal{} 0$, which was to be proved.
To find the solutions, we see that setting $ \omega \equal{} z^{3}$ yields $ \omega^{2}\plus{}\omega\plus{}1 \equal{} 0$, and hence $ (\omega\minus{}1)(\omega^{2}\plus{}\omega\plus{}1) \equal{} 0$, which is $ \omega^{3}\equal{} 1$, and $ \omega\neq 1$. So $ \omega\in\left\{\cos\frac{2\pi}{3}\plus{}i\sin\frac{2\pi}{3},\cos\frac{4\pi}{3}\plus{}i\sin\frac{4\pi}{3}\right\}$, and $ z$ is a third root of $ \omega$. Finally, we get $ z\in\left\{\cos\frac{2\pi}{9}\plus{}i\sin\frac{2\pi}{9},\cos\frac{8\pi}{9}\plus{}i\sin\frac{8\pi}{9},\cos\frac{14\pi}{9}\plus{}i\sin\frac{14\pi}{9},\right.$
$ \left.\cos\frac{4\pi}{9}\plus{}i\sin\frac{4\pi}{9},\cos\frac{10\pi}{9}\plus{}i\sin\frac{10\pi}{9},\cos\frac{16\pi}{9}\plus{}i\sin\frac{16\pi}{9}\right\}$, which verify the given equation.
[/hide]

[quote="bos1234"][color=red]2.[/color]Find the complex factors of $ x^{2}\plus{}2ax\plus{}a^{2}\plus{}b^{2}$[/quote]
[hide="2"]$ x^{2}\plus{}2ax\plus{}a^{2}\plus{}b^{2}\equal{}(x\minus{}r_{1})(x\minus{}r_{2})$, where $ r_{1},r_{2}$ are the solutions of $ x^{2}\plus{}2ax\plus{}a^{2}\plus{}b^{2}\equal{}0$.

$ x\equal{}\frac{\minus{}2a\pm\sqrt{4a^{2}\minus{}4a^{2}\minus{}4b^{2}}}{2}\equal{}\frac{\minus{}2a\pm2bi}{2}\equal{}\minus{}a\pm bi$

$ x^{2}\plus{}2ax\plus{}a^{2}\plus{}b^{2}\equal{}\boxed{(x\plus{}a\plus{}bi)(x\plus{}a\minus{}bi)}$[/hide]

[quote="bos1234"]
[color=red]2.[/color]Find the complex factors of $ x^{2}+2ax+a^{2}+b^{2}$[/quote]
[hide="Straightforward"]
\begin{eqnarray*}x^{2}+2ax+a^{2}+b^{2}&=& 0\\ x^{2}+2ax+a^{2}&=&-b^{2}\\ \left({x+a}\right)^{2}&=&-b^{2}\\ x+a &=&\pm bi\\ x &=&\pm bi-a\end{eqnarray*}
[/hide]


[color=cyan]Hubert flips through a standard deck of 52 cards.  After flipping over how many cards from the top is the probability greater than  that he has already flipped over all 13 hearts?[/color]

[hide] The probability that he's flipped through all 13 hearts is equal to the probability that there are no hearts remaining in the stack. Starting from the bottom of the deck, the last card has a 3/4 chance of not being a heart. If the last card is not a heart, the second to last card has a 38/51 probability of not being a heart. The probability of the bottom two cards not being hearts is 19/34, which is still larger than 1/2. However, multiply this by 37/50 and the probability of the bottom three cards not being hearts is less than 1/2 -> there's a greater chance than not that there's a heart remaining in the last 3 cards -> there's a greater chance that he hasn't flipped through all 13 hearts. So he would have to flip over 52-2= 50 cards. [/hide]