geometry
geometric transformation
rotation

How many ways can we arrange 5 boys and 5 girls in a circle if they must alternate (no two people of the same gender can stand next to each other)?

[quote="AstroPhys"][hide]
$ \frac {5! \cdot 5!}{10} \equal{} \boxed{1440 \ \text{ways}}$.
[/hide][/quote]

I am sorry to bother you, but how did you derive that out of the question?

Some how I get a different answer, but I'm pretty bad a combinatorics.... To form such a circle, we just "interlace" two circles of five boys and five girls. The number of ways to arrange each of the circles is $ \frac{5!}{5}\equal{}4!$, and the number of ways to "interlace" two given circles is $ 10$, so the answer is $ 4!\cdot 4!\cdot 10\equal{}5670$. Where AstroPhys derives his I think is the number of ways to arrange each of the diff. genders and then divide by the number of people in the circle to discount rotations.

Gah, I think this problem is supposed to be easy, but I'm getting confused...

Fix one boy at the top of the circle. Then we have 4! ways for the boys and 5! ways for the girls. So $ 4!\cdot 5!$

I think Temperal's argument is incorrect. It counts

$ B_{1}G_{1}B_{2}G_{2}B_{3}G_{3}B_{4}G_{4}B_{5}G_{5}$

and

$ B_{3}G_{2}B_{4}G_{3}B_{5}G_{4}B_{1}G_{5}B_{2}G_{1}$

as the same (I think...) So I think that Temperal undercounted...

But I think AstroPhys is correct, even though his answer is different from mine... Can someone clear this up?

AstroPhys and Temperal both used a $ 10$ where there should've been a $ 5$ (and that substitution makes both of their answers correct).  The issue is that there are $ 5$ different rotations of a given arrangement that keep boys in boys' seats and girls in girls' seats, not $ 10$ (equivalently, in Temperal's argument, there are $ 5$ ways to "interlace" two circles, not $ 10$ - the only thing that matters is the relative position of a particular boy and a particular girl.)

So both answers are correct?

I now kind of get it.

So the correct way to approch it is $ \frac{5!5!}{5}$?

Sorry, I'm kind of on the slow side.

According to what t0rajir0u said, which makes sense, vishalarul is correct.
[hide="So it is"]
$ 4! \cdot 5! \equal{} \boxed{2880 \ \text{ways}}$.
[/hide]

EDIT
$ \frac{5! \cdot 5!}{5} \equal{} 4! \cdot 5!$.

[quote="AstroPhys"]According to what t0rajir0u said, which makes sense, vishalarul is correct.
[hide="So it is"]
$ 4! \cdot 5! \equal{} \boxed{2880 \ \text{ways}}$.
[/hide]

EDIT
$ \frac {5! \cdot 5!}{5} \equal{} 4! \cdot 5!$.[/quote]
I see. But according to t0rajir0u, he said that both previous answers are correct. I am confused..

[quote="t0rajir0u"]AstroPhys and Temperal both used a $ 10$ where there should've been a $ 5$ (and that substitution makes both of their answers correct).[/quote]
 :wink:

For Temperal's, $ 4! \cdot 4! \cdot 5 \equal{} 4! \cdot 5!$.

Yes, I see now. I was considering any one boy between two girls, which makes little sense, now that I think of it.


calculus
integration
algebra
polynomial
real analysis
real analysis unsolved

I am trying to solve this integral to the nearest decimal place.

The equation is: 1 = integral of sinx/x from 0 to some x

Can i use the Riemann Sum (or a power series) with something such as:

x - x^3/3*3! + x^5/5*5! ..............

 I know i need to use something such as this because that equation cannot be expressed in terms of elementary functions.

I am unsure how that integral expressed in this format equal to 1 can be solved.  Maybe i have this whole thing wrong.  Any assistance would be great!

Thank you.

no, that doesn't seem the right way to me, since then you'll have a polynomial of huge degree, which you cannot solve.

I don't know what techniques in higher mathematics exist for this, but if you're looking specifically for the result it is x=1.064839726 :)


modular arithmetic
number theory unsolved
number theory

[quote="vnmathboy"]Find all prime numbers $p$ and $q$ satisfying:
$pq|2^p+2^q$[/quote]

This's an old prob.! Here's my nice sol. :D
We have
$p\mid 2^q+2$ and $q\mid 2^p+2$. If $p=q$ then $p^2\mid 2^{p+1}$ then $p=q=2$. If $p<q$ for exam., first if $p=2$ then $q=3$. Assume $p>2$ then
$p\mid 2^{q-1}+1$ and $q\mid 2^{p-1}+1$. Assume that $2^r|| p-1$ and $2^s|| q-1$
then $r,s\geq1$. From $p\mid 2^{q-1}+1$ it gives $p\equiv1\pmod{2^{s+1}}$ so $r\geq s+1$. From $q\mid 2^{p-1}+1$ it gives $q\equiv 1\pmod{2^{r+1}}$ so $s\geq r+1$. Both tell us it contraction! So $p=q=2$ or $p=2$ and $q=3$ and any permute left  ;)

[b]Note that[/b] if $p$ is prime which $p| 2^{2^tm}+1$ where $m$ is odd then $p\equiv1\pmod{2^{t+1}}$.

[quote="pluricomplex"]

[b]Note that[/b] if $ p$ is prime which $ p| 2^{2^tm} \plus{} 1$ where $ m$ is odd then $ p\equiv1\pmod{2^{t \plus{} 1}}$.[/quote].

Sorry but can you tell me why this is true?

A technique lemma: If $ a>1$, $ p$ is an odd prime and $ p\mid a^{2^n}\plus{}1$ then $ p\equiv1\pmod{2^{n\plus{}1}}$.
Proof. Let $ h$ be the $ ord_p(a)$ then by Fermat theorem $ h\mid p\minus{}1$. On the other hand, $ a^{2^{n\plus{}1}}\equiv1\pmod p$ so $ h\mid 2^{n\plus{}1}$. So we have $ h\equal{}2^k$ for some non-negative integer $ k$. We shall show that $ k\equal{}n\plus{}1$. Assume that $ k<n\plus{}1$ then $ 2^k\mid 2^n$ so $ p\mid a^h\minus{}1\equal{}a^{2^k} \minus{}1\mid a^{2^n}\minus{}1$. This implies $ p\mid \minus{}(a^{2^n}\minus{}1) \plus{}(a^{2^n}\plus{}1)\equal{}2$, contraction!.

So $ h\equal{}2^{n\plus{}1}$, but $ h\mid p\minus{}1$, this prove the lemma.

For your question, $ a\equal{}2^m$, any prime $ p\mid a^{2^t}\plus{}1$ must be odd. So we have the requirement.

1. $ p \equal{} q \implies p^2 \mid 2^{p \plus{} 1} \implies p \mid 4 \implies p \equal{} q \equal{} 2$.
2. $ p \equal{} 2 \implies q \mid 2^{q \minus{} 1} \plus{} 1 \implies 2(q \minus{} 1) \equal{} ord_q(2) \le \phi(q) \implies q \le 3$.
3. wlog $ 2 < p < q, pq \mid 2^p \plus{} 2^q \implies 2(q \minus{} 1) \equal{} ord_p(2) \mid \phi(p)$  and $ 2(p \minus{} 1) \equal{} ord_q(2) \mid \phi(q) \implies 4(q \minus{} 1) \le 2(p \minus{} 1) \le q \minus{} 1$.

So unique solutions are (2,2),(2,3),(3,2).

I don't see why
$ pq\mid 2^{p} \plus{} 2^{q}\implies 2(q \minus{} 1) \equal{} ord_{p}(2).$

I think it should be $ 2(q \minus{} 1) \equal{} ord_{p}(2) \cdot m$, where $ m$ is odd. Then $ 2(q \minus{} 1) \mid (p \minus{} 1)m$, and we end up using the same argument as pluricomplex.


Hi, how did you do on these three contests? I guess we can put them together.
Please post any comments.

My onnoficial score on the cayley is 134.

unofficial score on fermat: 138
i'm in western ontario region, any1 from there beat me? lol

I don't know yet, so how do you guys know?

I think I got 150 so far, but I don't have the question paper.

I got 126, the one mistake to end it all. :(
Now I work for Euclid.

[quote="blootzboy"]we know by like ya know checking over a bunch of times, comparing answers with people, stuff like that

btw wat region are u from and which contest did u write[/quote]

Michigan so don't worry.

Pascal.

150 unofficially here.

We have 1+1+1+1+1+1+1=7 unofficial perfect pascals in our state.

[quote="Inspired By Nature"]Wow, Pascal must have been too easy this year. My sister says it was, she got 132 (also unofficial). :)[/quote]

lol...cayley was pretty easy as well...the last question was sooo simple...a lot of ppl guessed and got it right...

The a's and b's of the cayley were easy to. I did them in 15 minutes.


linear algebra
matrix
search
algebra
polynomial
linear algebra unsolved

If $ A, B \in M_{n} (\Re)$, then when is the equality $ e^{A} e^{B} \equal{} e^{A\plus{}B}$ true?

... if $ AB\equal{}BA$. The inverse implication is not true in general.

Can you please elaborate a little, with a proof?

What did you try so far - and where did you get stuck after so little time? ;)

Check [url=http://limo.a-eskwadraat.nl/oudeopgaven/limo2008opg.pdf]this pdf[/url], problem 9 (which is in Dutch, but shouldn't be too hard to understand what they want from you) to see a more general expression for this thing, namely that

$ e^A e^B \equal{} e^{A \plus{} B \plus{} \frac12 [A,B]}$, where $ [A,B] : \equal{} AB \minus{} BA$.

Note, there are solutions to these problems provided, but where's the fun in that?

[quote="Tyl"]$ e^A e^B \equal{} e^{A \plus{} B \plus{} \frac12 [A,B]}$, where $ [A,B] : \equal{} AB \minus{} BA$.[/quote]... with the additional assumption that $ [A,[A,B]] \equal{} 0$ and $ [B,[B,A]] \equal{} 0$ as is requested in the referenced pdf. So we end up with $ [A,B]\equal{}0$ again such that the original relation holds.

See
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=364003777&t=191160
More precisely about the problem $ e^Ae^B\equal{}e^{A\plus{}B}$ with $ AB\not\equal{}BA$ over $ \mathbb{C}$ :
If $ n\equal{}2$ then OR $ e^A,e^B$ are scalar matrices
EITHER $ A,B$ are simultaneously triangularizable.
If $ n\equal{}3$ the result is unknown. It is a good subject of research.

[quote="loup blanc"]It is a good subject of research.[/quote]A surprising result: If $ A$ and $ B$ have algebraic integer entries, then $ [A,B]\equal{}0$ iff $ e^Ae^B\equal{}e^{A\plus{}B}$.

See Horn, Roger A.; Johnson, Charles R. (1991), Topics in Matrix Analysis, Cambridge University Press, ISBN 978-0-521-46713-1 which I found [url=http://en.wikipedia.org/wiki/Matrix_exponential#The_exponential_of_sums]here[/url].

[quote="Third Edition"]A surprising result: If $ A$ and $ B$ have algebraic integer entries, then $ [A,B] \equal{} 0$ iff $ e^Ae^B \equal{} e^{A \plus{} B}$.

See Horn, Roger A.; Johnson, Charles R. (1991), Topics in Matrix Analysis, Cambridge University Press, ISBN 978-0-521-46713-1 which I found [url=http://en.wikipedia.org/wiki/Matrix_exponential#The_exponential_of_sums]here[/url].[/quote]
Fascinating. Did you actually mean algebraic integers? The wikipedia entry says algebraic numbers, and I guess that makes a little bit more sense (you can always "clear denominators" by multiplying by a sufficiently large integer).

Hmm, certainly I should have written algebraic numbers. Thanks. Today I wanted to see the reference in google books, but unfortunately the pages did not seem to be made available. Just wondering how the assumption that the matrices have entries in a number field makes the not very intuitive implication possible. I have absolutely no idea ?!?!?

I probably have even less than absolutely no idea. This is really intriguing. Wikipedia says this is in pages 435-437 in the book, and I can actually see those pages in Google Books but they seem to be about a different topic altogether. Still looking...

[quote="randomgraph"]...pages 435-437 in the book, and I can actually see those pages in Google Books but they seem to be about a different topic altogether. Still looking...[/quote]
I don't know about google, but it is indeed mentioned in the book I have, at the bottom of page 436 and the top of page 437. There is no reference, no proof, which gives me hope that it might not be too difficult to prove :wink:

[quote="Third Edition"][quote="loup blanc"]It is a good subject of research.[/quote]A surprising result: If $ A$ and $ B$ have algebraic integer entries, then $ [A,B] \equal{} 0$ iff $ e^Ae^B \equal{} e^{A \plus{} B}$.

See Horn, Roger A.; Johnson, Charles R. (1991), Topics in Matrix Analysis, Cambridge University Press, ISBN 978-0-521-46713-1 which I found [url=http://en.wikipedia.org/wiki/Matrix_exponential#The_exponential_of_sums]here[/url].[/quote]
It's perhaps a typo. A true result is:
 if $ e^Ae^B\equal{}e^Be^A$ and if the entries of $ A,B$ are algebraic over $ \mathbb{Q}$ then $ AB\equal{}BA$.
Sketch of the proof:
1) (The entries of $ A$ are algebraic )    implies
(if $ u,v$ are $ A$-eigenvalues, then $ u\minus{}v\notin{2}i\pi\mathbb{Z}^*$ )     implies
($ A$ is a polynomial in $ e^A$).
2) $ e^Ae^B\equal{}e^Be^A$ implies $ AB\equal{}P(e^A)Q(e^B)\equal{}Q(e^B)P(e^A)\equal{}BA$.

There is not any typo in the Horn's book.
The mentioned result is $ e^Ae^B\equal{}e^Be^A$ iff .....


Let $ n\ge 2$ be a natural number and the complex numbers $ z,z_k, \ k\equal{}\overline{1,n}$ so that $ z\equal{}z_1z_2\plus{}z_2z_3\plus{}\dots \plus{}z_nz_1$ .

Prove that : $ |z_1|^2\plus{}|z_2|^2\plus{}\dots \plus{}|z_n|^2 \ge \max(\mathrm{Re} z,\mathrm{Im} z)$ .

we have the identity
$ \sum_{k=1}^n(|z_{k+1}|-|z_k|)^2 \ge 0$
$ \Longleftrightarrow \sum_{k=1}^n|z_k|^2 \ge \sum_{k=1}^n|z_k|.|z_{k+1}| \ge \left|\sum_{k=1}^n{z_k.z_{k+1} \right|=|z| \ge \max\{\text{Re}z,\text{Im}z\}}$


Let S be a set of sequences of length 15 formed by using the letters a and b such that every pair of sequences in S differ in at least 3 places. What is the maximum number of sequences in S?

A more general solution :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cardinality+possible+largest&t=76633

Pierre.


This is the classic fly in a box problem, but I can't seem to get the right answer:

A rectangular room measures 30 feet in length and 12 feet in height, and the ends are 12 feet in width. A fly, with a broken wing, rests at a point one foot down from the ceiling at the middle of one end. A smudge of food is located one foot up from the floor at the middle of the other end. The fly has just enough energy to walk 40 feet. Show that there is a path along which the fly can walk that will enable it to get the food.

I can get ~40.7, but not 40.


Suppose there is a sequence of real numbers $ a_1,\,a_2,\,a_3,\,\ldots,\,a_n$. Prove that

$ (a_1 \minus{} a_2)^{2k \minus{} 1} \plus{} (a_2 \minus{} a_3)^{2k \minus{} 1} \plus{} \ldots \plus{} (a_{n} \minus{} a_1)^{2k \minus{} 1} \equal{} 0$

for all $ k \in \mathbb{N}$.

Sorry, it turns out that I assumed something that wasn't true when trying to prove this, so it is in fact false.  :blush: It only works for $ n \equal{} 2$.


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Suppose that $X$ is connected sum of two manifolds $M_{1}$ ,$M_{2}$obtained by cutting out a small open ball from each of  $M_{1}$ and $M_{2}$
($B_{1}$ from $M_{1}$ and $B_{2}$ from $M_{2}$ ) and gluing the resulting spaces together along their boundary spheres

Now show that :
Diffeomorphism type of $X$ is independent from choice of ball's ($B_{1}$ and $B_{2}$)

If any one know how to prove this  please help me even a sketch of proof would be a great help to me

$M_{1}$ and $M_{2}$ are assumed connected, of course.

I would try the following: (1) $M_{1}\setminus B_{1}$ is diffeomorphic to $M_{1}\setminus\{p\}$, where $p$ is some point of $B_{1}$. [Construct a diffeomorphism using a chart for $B_{1}$]

(2) $M_{1}\setminus\{p\}$ is diffeomorphic to $M_{1}\setminus\{q\}$ for any two points $p,q\in M_{1}$. [Hint: the set of all points $p'$ such that $M_{1}\setminus\{p'\}$ is diffeomorphic to $M_{1}\setminus\{p\}$ is open.]

(3) From (1) and (2) $M_{i}\setminus B_{i}$ is diffeomorphic to $M_{i}\setminus B_{i}'$ for any other ball $B_{i}'$. This is where I get stuck trying to glue two diffeomorphisms (for $i=1,2$) together.  :(

Don't shrink it to a point, just make it really small. Then use a path to link the centers of two holes together, then extend this path to a local flow near the path so that the flow maintains the diffeomorphism.  Actually it is almost the same as you did when shrinking the ball to a point.

But to explain more explicitly, I think it would be better adding a Riemannian metric on the manifold so that you can talk about the size, though not necessary.


inequalities
trigonometry
calculus
geometry unsolved
geometry

[hide="Here are two true jewelry !"] \[ \triangle ABC \] \[ \boxed {\ a\ge b,\ a\ge c\Longrightarrow \frac{b^2+c^2}{bc}+\frac{b+c}{a}\ge 2+\frac{1}{\sin \frac A2}\cdot \sqrt{1+\left(\frac{b-c}{2a}\right)^2}\ }\ge 2+\frac{1}{\sin \frac A2} \] \[ \boxed {\ A=90^{\circ}\Longrightarrow \frac{a^2}{bc}+\frac{b+c}{a}\ge 2+\sqrt 2\cdot \sqrt{1+\left(\frac{b-c}{2a}\right)^2}\ }\ge 2+\sqrt 2 \][/hide]

I have no doubt that you found such beautiful inequalities
But i tried to prove the second one 4 years ago! (unsuccessful attempt :oops: )
(It was a contest...)

Maybe a weaker inequality when $A=90^{\circ}$ (such it appears in Crux Mathematicorum), i.e.
$A=90\longrightarrow \frac{a^2}{bc}+\frac{b+c}{a}\ge 2+\sqrt2 \ !$

I apologize for not writing the details. After more or less trivial manipulations, the inequality becomes \[\left(4-\frac{(b+c)^{2}}{bc}+\frac{a^{2}}{bc}\right)\left(\frac{(b+c)^{2}}{bc}+\frac{b+c}{a}-4\right)^{2}\geq 4+\frac{(b+c)^{2}}{a^{2}}-4\frac{bc}{a^{2}}\eqno(*)\] Let $x=\frac{b+c}{a}$, $y=\frac{bc}{a^{2}}$ and $d=x^{2}-4y$. Most important is that $d \geq 0$ and $1-d \geq y$ (since $a \geq b,c$). Now, $(*)$ is equivalent to \[d\left((1-d)(d+2xy)-dx^{2}y^{2}-y^{3}+y^{2}\right) \geq 0\] and since $d \geq 0$, we can divide by $d$. Also, we can use $1-d \geq y$ and divide by $y$ to reduce it further to $d+2xy+y(1-y) \geq dx^{2}y$, which follows from trivial inequalities such as $0 \leq d,y \leq 1$ and  $1<x \leq 2$.

[quote="Virgil Nicula"]And the first inequality ?![/quote]
I think it's wrong! Try $ a\equal{}4.5,b\equal{}3.5,c\equal{}3 \Rightarrow LHS\minus{}RHS\equal{}\minus{}0.04929...<0$

[quote="Virgil Nicula"][size=130][color=darkred] $\boxed {\ a\ge b,\ a\ge c\Longrightarrow \frac {b^2 + c^2}{bc} + \frac {b + c}{a}\ge 2 + \frac {1}{\sin \frac A2}\cdot \sqrt {1 + (\frac {b - c}{2a})^2}}$ .[/color][/size][/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ \|\begin{array}{c} a\ge b \\\
\ a\ge c\end{array}\|$ $ \implies$ $ 2a\ge b + c$ $ \Longrightarrow$ $ \boxed {\ 2a(b + c)\ge 4bc\ }$ $ \implies$ $ 2a(b + c) + bc\ge bc\cdot\frac {bc - p(p - a)}{(p - b)(p - c)} + 4bc$ $ \implies$

$ 2a(b + c)(b - c)^2 + bc(b - c)^2\ge$ $ \frac {b^2c^2[a^2 - 4(p - b)(p - c)]}{(p - b)(p - c)} - \frac {bcp(p - a)(b - c)^2}{(p - b)(p - c)} + 4bc(b - c)^2 =$

$ \frac {a^2b^2c^2}{(p - b)(p - c)} - 4b^2c^2 - bc(b - c)^2\cdot\cot^2\frac A2 + 4bc(b - c)^2$ $ \implies$ $2a(b + c)(b - c)^2 + bc(b- c)^2\ge $

$\frac {a^2bc}{\sin^2\frac A2} + bc(b - c)^2(4 - \cot^2\frac A2)\ \boxed {: }$ $\ a^2bc$ $ \implies$ $ (\frac {b + c}{a})^2 + 2\cdot\frac {b + c}{a}\cdot \frac {(b - c)^2}{bc}\ge$ $ \frac {1}{\sin^2\frac A2} + (\frac {b - c}{a})^2\cdot\frac {4\sin^2\frac A2 - \cos\frac A2}{\sin^2\frac A2}$ $ \implies$

$ [\frac {(b - c)^2}{bc} + \frac {b + c}{a}]^2\ge [\frac {(b - c)^2}{bc}]^2 + \frac {1}{\sin^2\frac A2}\cdot[1 + (\frac {b - c}{a})^2\cdot\frac {3 - 5\cos A}{2}]$ . But $ \boxed {A\ge 60^{\circ}\implies\cos A\le \frac 12}$ $ \implies$

$ \frac {3 - 5\cos A}{2}\ge \frac 14$ and $ [\frac {(b - c)^2}{bc}]^2\ge 0$ . In conclusion, $ [\frac {(b - c)^2}{bc} + \frac {b + c}{a}]^2\ge\frac {1}{\sin^2\frac A2}\cdot[1 + (\frac {b - c}{2a})^2]$ $ \implies$

$ \frac {(b - c)^2}{bc} + \frac {b + c}{a}\ge\frac {1}{\sin\frac A2}\cdot\sqrt {1 + (\frac {b - c}{2a})^2}$ $ \implies$ $ \boxed {\frac {b^2 + c^2}{bc} + \frac {b + c}{a}\ge 2 + \frac {1}{\sin\frac A2}\cdot\sqrt {1 + (\frac {b - c}{2a})^2}}$

[b]Remark.[/b] I used the following well-known identities : $ \{\begin{array}{ccc}
p(p-a)+(p-b)(p-c)=bc & ; & (b-c)^2=a^2-4(p-b)(p-c)\\\\
2\sin^2\frac A2=1-\cos A & ; & 2\cos^2\frac A2=1+\cos A\\\\
\sin\frac A2=\sqrt {\frac {(p-b)(p-c)}{bc}} & ; & \tan\frac A2=\sqrt {\frac {(p-b)(p-c)}{p(p-a)}}\end{array}$ [/color]


graph theory
combinatorics proposed
combinatorics

Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$.

If $e(G_{n}) \geq\frac{n\sqrt{n}}{2}+\frac{n}{4}$ ,prove that $G_{n}$ contains $C_{4}$ .

We have a better result than desired
The maximum  numbers  edges of $G_{n}$ without cycle 4 be :
 $\frac{n+n. \sqrt{4.n-3}}{4}$.
You cand find solution of this result in "Extremal Combinatorics "


probability
function
calculus
integration
derivative
conditional probability
real analysis

guys maths is my weakest subject not coz i dont know but coz no 1 has been able to tell me how to

can some explain to me what is random variable and how does 1 solve a sum ...

what is random variable
Density Function - gaussian, uniform func ...blah blah

got an exam on wednesday and i have no clue what so ever also my integration & derivatives aint that good

i hope some 1 can help me :(

Welcome to this forum Manan. If you have some specific problem that you don't know how to solve, post here in the forum (after you have made a good try at solving it). There will be certainly someone willing to help you.

the problem is that i tried solving a few basic sums and was able to an extent ... but im unable to understand the concepts that have been told in the book

i want if possible some 1 could tell me the basic concepts used to solve these type of sums ...

i have questions like find marginal probability, conditional probability distribution of x iff y is given etc ...

What you're asking is simply too vague. We don't have time to help you with [b]everything[/b]- that's what classes and textbooks are for.


Let $ a,b,c\ge 0$, s.t. $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that:
$ x(1\minus{}y^2)(1\minus{}z^2)\plus{}y(1\minus{}x^2)(1\minus{}z^2)\plus{}z(1\minus{}x^2)(1\minus{}y^2)\leq \dfrac{4\sqrt{3}}{9}$

$ \sum{x(1\minus{}y^2)(1\minus{}z^2)}$
$ \equal{} \sum{x}\minus{}\sum_{sym}{xy^2}\plus{}xyz \sum{yz}$
$ \equal{}\sum{x}\minus{}(\sum{x}\sum{xy}\minus{}3xyz)\plus{}xyz\sum{yz}$
$ \equal{}4xyz$
$ \le\ 4 \sqrt{(\frac{xy\plus{}yz\plus{}zx}{3})^3}$
$ \equal{}\frac{4\sqrt{3}}{9}$


[quote="friendlist"]Solve equation:
$ 7^{15}+9^{17}-13^{x+1}=15^{7}$[/quote]

With some simple algebra, we get:

$ 7^{15}+9^{17}-15^{7}= 13^{x+1}$

The left hand side equals approximately $ 1.688 \cdot 10^{16}$. Now write that value to the base 13 by doing the following:

$ \log^{1.688 \cdot 10^{16}}_{13}= 14.5629$

$ 13^{14.5629}= 13^{x+1}$

Since the bases are the same, equate the exponents.

$ 14.5629 = x+1$

Thus $ x = 13.56290056\ldots$


Where can I take MOSPs' problems, I rarely see it in the network...

take a look here...

http://web.mit.edu/rwbarton/Public/mop/

[quote="Lovasz"]Thanks for the link.
There's many American here, but there's noone posted USA TSE some near year: 2004, ..., 2006. It's really bad. Can you?[/quote]

The [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=35&year=2004]2004 USA TST[/url] is already on AoPS.  The problems to the 2005 and 2006 USA TST's have not yet been released to the public.


inequalities
rearrangement inequality
inequalities proposed

Now I have a proof for your inequality:

You set a=x/y b= y/z c=z/x

Then you get

1/(x^2/(yz)+y^2/(xz)+z^2/(xy))+1/(yz/x^2+xz/y^2+yz/x^2)=<
1/(x/y+y/z+z/x)+1/(y/x+z/y+x/z).

From the rearrangement inequality for x^2,y^2,z^2 and 1/(xy),1/(xz),1/(yz) follows:

x^2/(yz)+y^2/(xz)+z^2/(xy)>=x^2/(xz)+y^2/(xy)+z^2/(yz)=x/z+y/x+z/y.

On the other hand it follows also:

x^2/(yz)+y^2/(xz)+z^2/(xy)>=x^2/(xy)+y^2/(zy)+z^2/(xz)=z/x+x/y+y/z

The same you have for the other term of the left side. So I've proven the following 4 inequalities:

1/(x^2/(yz)+y^2/(xz)+z^2/(xy))=<1/(x/y+y/z+z/x)
1/(x^2/(yz)+y^2/(xz)+z^2/(xy))=<1/(y/x+z/y+x/z)
1/(yz/x^2+xz/y^2+yz/x^2)=<1/(x/y+y/z+z/x)
1/(yz/x^2+xz/y^2+yz/x^2)=<1/(y/x+z/y+x/z)

The sum of these 4 inequalities gives us the required one.

Misha

[quote="xzlbq"][quote="ThAzN1"]For $a,b,c>0$ and $abc=1$, prove that

$\frac{1}{a^2b+b^2c+c^2a}+\frac{1}{ab^2+bc^2+ca^2} \leq \frac{1}{a+b+c}+\frac{1}{ab+bc+ca}$[/quote]

But holds:

$\frac{1}{a^2b+b^2c+c^2a}+\frac{1}{ab^2+bc^2+ca^2} \leq \frac{1}{a+b+c}+\frac{1}{a^2+b^2+c^2}$

Who does show noce?

BQ[/quote]

if easy?

For $a,b,c>0$ and $abc=1$, prove that

$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)\geq (a^3+b^3+c^3)(a+b+c).$

BQ


vector
number theory
prime numbers
algebra unsolved
algebra

[1] Is it possible to find to find a finite set of reals, $ a_{1}, a_{2} ... a_{n}$, such that for every real number $ x$, we can find rational numbers $ q_{1}, q_{2} ... q_{n}$ (not necessarily non-zero), such that:
$ \sum_{1\leq i \leq n} a_{i} q_{i} \equal{} x$.

Also, perhaps a stronger problem:
[2] Is it possible to find to find a finite set of reals, $ a_{1}, a_{2} ... a_{n}$, such that for every algebraic number $ x$, we can find rational numbers $ q_{1}, q_{2} ... q_{n}$ (not necessarily non-zero), such that:
$ \sum_{1\leq i \leq n} a_{i} q_{i} \equal{} x$.

I think the answer to both questions is 'no'. Clearly, proving [2] has 'no' as the answer, would suffice to solve [1].

See http://mathworld.wolfram.com/HamelBasis.html .  The existence of such a basis (finite or otherwise) is based on the Axiom of Choice, and even if it exists it cannot be finite (since a finite extension of the rationals has finite transcendence degree, and the transcendence degree of $ \mathbb{R}$ over $ \mathbb{Q}$ is infinite.)

That answers [1] in the negative (alternatively, one can observe that there are only countably many such linear combinations).
[2] asks if the set of algebraic numbers is a finite dimensional vector space over $ \mathbb Q$. I'm pretty sure that square roots of prime numbers are linearly independent over $ \mathbb Q$, but can't think of a proof right now.


number theory
relatively prime
number theory proposed

[quote="yuyang"]Hi,

find all integers $ n$ such that  if $ n\vert ab\plus{}1$ then $ n\vert a\plus{}b$, where $ a$ and $ b$ are two positive integers.[/quote]

$ (\mathbb{N}\minus{}D(ab\plus{}1))\cup(D(ab\plus{}1)\cap D(a\plus{}b))$ where $ D(n)$ is the set of divisors of $ n$. :D

the answer is: all divisors of 24!
it is obvious that the condition is equivalent to saying $ x^{2}\equiv 1(\mod n)$ for all $ x$ relatively prime to $ n$


Well, to make this forum sort of active again, let's have a mafia game!  I'm the mod, and such.

Unless you're going to do something Connecticut-specific, I'm going to move this to Games or Fun Factory or wherever these things happen.

Okay, then it's postponed.  I first have to think of a CT idea. :P

sorry for souble posting but you could have the mafia poison hot dogs that townies eat?


Let $ \angle XYZ \equal{} 120$ and $ P$ a point on the bisector of $ \angle XYZ$ such as $ YP \equal{} YX \plus{} YZ$.Prove that $ \Delta PXZ$ is equilateral.

$ R(Y, 60^o)$, suppose $ X \to X'$, then triangle $ YXX'$ is equilateral. Since $ PY\equal{}YX\plus{}YZ\equal{}YX'\plus{}X'P$ and $ YX'\equal{}YX$, so $ YZ\equal{}X'P$. Since $ \angle{XYZ}\equal{}120^o \equal{} \angle{XX'P}$, so we have $ \triangle{XYZ} \cong \triangle{XX'P}$, which gives $ XZ\equal{}XP$. Similarly there is $ XZ\equal{}ZP$. Hence $ \triangle{PXZ}$ is equilateral.

[b]Proof 1:[/b]([color=red]geometrical)[/color]

[hide]Let $ E \in YP$ such as $ YE=XY$ and $ EP=YZ$.
$ \Delta XYE$ is equilateral
$ \{\begin{array}{cccc} XE=XY\\
\\
EP=YZ\\
\\
\angle XEP=\angle XYZ=120 \end{array}$ $ \implies$ $ \Delta XEP \equiv \Delta XYZ$ $ \implies XZ=XP$   ($ 1$).

$ \Delta AOP \equiv \Delta APB$ $ \implies$ $ XP=ZP$                   ($ 2$)
From ($ 1$) and ($ 2$) $ \implies$ $ XPZ$ equilateral.[/hide]

[color=red][b]Proof 2[/b][/color]
[hide]It is know the problem:
[color=darkblue][b]Let $ Y$ a point on the circumcircle of the equilateral triangle $ XZP$.Prove that $ XY+YZ=YP$[/b] [/color]      [color=red] [b](Schooten's theorem)[/b][/color]
Our problem could be considered as a reciprocal for this theorem[/hide]


induction
algorithm
inequalities
pigeonhole principle
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer,$ n\geq4$. $ n$ cards are arranged on a circle and the numbers $ 1$ and $ \minus{}1$ are written on each of the cards.In a $ question$ we may find the product of the numbers on any $ 3$ cards.What is the minimum number of $ questions$ needed to find out the product of all $ n$ numbers.

Wrong,there are $ 3$ cases,and I guess that the answers are the following:
For $ n\equal{}3k$ the answer is $ k$.
For $ n\equal{}3k\plus{}1$ the answer is $ k\plus{}2$.
For $ n\equal{}3k\plus{}2$ the answer is $ k\plus{}1$.
Anyway, I need a way to prove.Please help! :D

Sorry,big mistake:
For $ n\equal{}3k\plus{}1$ the answer is $ k\plus{}3$.
For $ n\equal{}3k\plus{}2$ the answer is $ k\plus{}2$.

I still need a way to prove. PLEASE HELP! :|

Sorry, but no one can solve? :(
Is this problem so hard?

The case $ n\equal{}3k$ is trivial, and the other two cases can be proved is easily by induction. :wink:

YEs induction does work for the rest 2 cases :lol:  :)

to find out the product of 3 more  cards you need one more question hence done
PLEASE try on the idea once before asking details(this is infact the solution :) )

@Johan Gunardi: ok. why does induction work for the other two cases?
in other words: which cards will guarantee you to know the remaining card or the remaining two cards knowing the fact that if we have $ N\equal{}3k$ the minimum number of questions is $ k$?

@gauravpatil: this proves that if we have $ N\equal{}3k$ the minimum number of questions is $ k$. How about $ 3k\plus{}1$ and $ 3k\plus{}2$

I found an algorithm that enables us to find the product in $ k\plus{}3$ questions fo $ n\equal{}3k\plus{}1, 3k\plus{}2$:

Pick a triple with product 1 and name as $ x$ the unknown card, and change each card by $ x$ (3 question more):

If the product changes three times, then $ x\equal{}\minus{}1$
If the product changes 2 times, then $ x\equal{}1$
If the product changes once, then $ x\equal{}\minus{}1$
If the product doesnt change, then $ x\equal{}1$

if $ n\equal{}3k\plus{}2$ then pick three cards with product -1 and substitute the three distinct subsets of two cards as the two unknowns $ (x,y)$ then(3 question more):

If the product doesn't change, $ xy\equal{}1$
If the product changes once, $ xy\equal{}\minus{}1$
If the product changes twice, $ xy\equal{}1$
If the product changes three times, $ xy\equal{}\minus{}1$

I think this is the minimum, but Inequalities Master: have you found any algorithm that made you conjecture these values? could you illustrate so?

if so, the case $ 3k\plus{}2$ would be solved, since the minimum is $ k\plus{}1$ operations (pigeonhole principle)

.In a qusetion we may find the product of the numbers on any 3 cards.
Does the three cards must be different?

Anyway, the answer to this question only infect the solution on n=4.
We can easily got that we need to use 2 questions(cards can be same) and 4 questions(cards must be different).
And when n=3k it is obviously.
when n=3k+1, where k is at least 2. We only need to ask one card for 3 times(as to ask card abc;ade&afg)
then make k-2 questions for the left cards. Totally k+1 questions.
We multiple all the answers and get the answer we need.
k questions are not enough beacuse you cannot ask for every card.
when n=3k+2, we need to ask two card for 3 times(as to ask card abc;abd&abe)
the make k-1 questions for the left cards. Totally k+2 questions.
k+1 questions are not enough beacuse you must ask for every card and then ask only one card for 2 times.
suppose we ask card x for 2 times, whether we ask card xxy(so strange) or ask xyz&xab, we cannot make sure the number of x(the former) and the product of xyzab(the latter) and the total product is unpredictable.


function
ceiling function
induction
modular arithmetic
algebra proposed
algebra

Given an integer $ m$, define the sequence $ \left\{a_{n}\right\}$ as follows:
\[ a_{1}\equal{}\frac{m}{2},\ a_{n\plus{}1}\equal{}a_{n}\left\lceil a_{n}\right\rceil,\textnormal{ if }n\geq 1\]
Find all values of $ m$ for which $ a_{2007}$ is the first integer appearing in the sequence.

Note: For a real number $ x$, $ \left\lceil x\right\rceil$ is defined as the smallest integer greater or equal to $ x$. For example, $ \left\lceil\pi\right\rceil\equal{}4$, $ \left\lceil 2007\right\rceil\equal{}2007$.

[quote="Rust"]$ m \equal{} 1\mod 2^{2007}$ work.[/quote]

Post your solution, please  :)

because $ a_{2}$ must not be an integer so $ a_{1}\equal{}\frac{m}{2}$ must'nt be an integer too because if it is an integer then $ \left\lceil a_{1}\right\rceil$ is an integer too and... this tells us that $ m\equiv1(\text{ mod }2)$ (or in other words $ m$ must be odd and for a number $ k_{1}\in\mathbb{Z}$, $ m \equal{} 2k_{1}\plus{}1$) then we have $ a_{2}\equal{}\frac{2k_{1}\plus{}1}{2}\times(k_{1}\plus{}1)$ now $ a_{3}$ must'nt be integer and similarly we must have $ (2k_{1}\plus{}1)(k_{1}\plus{}1) \equal{} 2k^{2}_{1}\plus{}3k_{1}\plus{}1\equiv1(\text{ mod }2)\Longrightarrow k_{1}\equiv0(\text{ mod }2)$ and this means there exists $ k_{2}\in\mathbb{Z}$ such that $ k_{1}\equal{} 2k_{2}$ and then as we know $ \left\lceil\frac{8k^{2}_{2}\plus{}6k_{2}\plus{}1}{2}\right\rceil \equal{} 4k^{2}_{2}\plus{}3k_{2}\plus{}1$ we have: $ a_{3}\equal{}\frac{(4k_{2}\plus{}1)(2k_{2}\plus{}1)(4k^{2}_{2}\plus{}3k_{2}\plus{}1)}{2}$  now $ a_{4}$ must'nt be an integer and similarly we must have $ (4k_{2}\plus{}1)(2k_{2}\plus{}1)(4k^{2}_{2}\plus{}3k_{2}\plus{}1)\equiv1(\text{ mod }2)\Longrightarrow k_{2}\equiv0(\text{ mod }2)$ and so there exists $ k_{3}\in\mathbb{Z}$ such that $ k_{2}\equal{} 2k_{3}$ and so on... :lol:

and atlast we can prove that if for $ n\in\mathbb{N}$, $ a_{n}$ is the first integer appearing in the sequence this means instead of $ a_{1},a_{2},\cdots,a_{n\minus{}2}$, the $ a_{n\minus{}1}$ must be noninteger too then there exists $ k_{n\minus{}2}\in\mathbb{Z}$  such that $ m \equal{} 2^{n\minus{}1}k_{n\minus{}2}\plus{}1$ but because $ a_{n}$ must be integer then $ k_{n\minus{}2}\equiv1(\text{ mod }2)$ and there exists $ k_{n\minus{}1}\in\mathbb{Z}$ such that $ k_{n\minus{}2}\equal{} 2k_{n\minus{}1}\plus{}1$ and $ m \equal{} 2^{n\minus{}1}\left(2k_{n\minus{}1}\plus{}1\right)\plus{}1$. So the answer is all numbers in the form $ \boxed{m \equal{} 2^{2006}\left(2s\plus{}1\right)\plus{}1}$ where $ s\in\mathbb{Z}$.Hope theres no miscalculations! :)

I get a slightly different solution, but maybe there is a typo  :wink:

Consider the map $ \Psi:\frac{1}{2}\mathbb{N}\rightarrow\frac{1}{2}\mathbb{N}\,,\,\Psi(x)\equal{}x\,\lceil x\rceil$ and define for $ k\geq 1$ the sets $ T_{k,\plus{}}\equal{}2^{k}\mathbb{N}\plus{}\frac{1}{2}\minus{}2^{k\minus{}1}$ and for $ k\geq 0$ similar sets $ T_{k,\minus{}}\equal{}2^{k}\mathbb{N}\plus{}\frac{1}{2}\minus{}2^{k}$.

We observe the following obvious properties:

(1) We have the disjoint decomposition $ \frac{1}{2}\mathbb{N}\equal{}\mathbb{N}\cup T_{0,\minus{}}$

(2) We have for all $ k\geq 0$ the disjoint decomposition $ T_{k,\minus{}}\equal{}T_{k\plus{}1,\plus{}}\cup T_{k\plus{}1,\minus{}}$ (hint: substitute $ x\equal{}2y$ resp $ x\equal{}2y\minus{}1$ in $ 2^{k}x\plus{}\frac{1}{2}\minus{}2^{k}$)

Application of (1),(2) yields the generalization (2') $ \forall k\geq0:\frac{1}{2}\mathbb{N}\equal{}\mathbb{N}\cup T_{k,\minus{}}\cup\bigcup_{v\equal{}1}^{k}T_{v,\plus{}}$ which allows to immediately verify:

(3) $ \Psi^{\minus{}1}(\mathbb{N})\equal{}\mathbb{N}\cup T_{1,\plus{}}$

(4) $ \Psi^{\minus{}1}(T_{k,\plus{}})\equal{}T_{k\plus{}1,\plus{}}$ (hint: simple induction and (2'))

Application of the previous properties requires $ a_{2006}\in T_{1,\plus{}}$ and pulling recursively back with (4) we get ultimately $ a_{1}\in T_{2600,\plus{}}$ which is equivalent to $ m\equiv 1\minus{}2^{2006}\bmod{2^{2007}}$

Let me try, put $ b_{n}\equal{} 2a_{n}$ then $ b_{n}\in\mathbb Z$ for all $ n\geq1$. $ a_{2007}$ is the first integer appearing in the sequence iff $ 2\mid b_{2007}$ and $ 2\nmid b_{2006}$ iff $ b_{2006}\equiv1\pmod 4$. Now we need some auxiliary steps:

Lemma 1. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv 1\pmod 4$ iff $ s\equiv\minus{}1\pmod 8$.

Lemma 2. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv\minus{}1\pmod 8$ iff $ s\equiv\minus{}5\pmod{16}$.

Lemma 3. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv\minus{}5\pmod{16}$ iff $ s\equiv3\pmod{32}$.

Lemma 4. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv 3\pmod{2^{n}}$ iff $ s\equiv3\pmod{2^{n\plus{}1}}$ for all $ n\geq3$.

So we have $ b_{2006}\equiv1\pmod 4$ iff $ b_{2005}\equiv\minus{}1\pmod 8$ iff $ b_{2004}\equiv\minus{}5\pmod{16}$  iff $ b_{2003}\equiv3\pmod{2^{5}}$ iff $ b_{2002}\equiv3\pmod{2^{6}}$ ... iff $ b_{1}\equiv3\pmod{2^{2007}}$ iff $ m\equiv 3\pmod{2^{2007}}$.

I solved this problem when I was very sleepy!

[quote="hjbrasch"]I get a slightly different solution, but maybe there is a typo  :wink:

Application of the previous properties requires $ a_{2006}\in T_{1,\plus{}}$ and pulling recursively back with (4) we get ultimately $ a_{1}\in T_{2600,\plus{}}$ which is equivalent to $ m\equiv 1\minus{}2^{2006}\bmod{2^{2007}}$[/quote]

well your answer is similar to what i've got! :)

$ m\equiv 1\minus{}2^{2006}\bmod{2^{2007}}$ means there exists $ s\in\mathbb{Z}$ such that $ m \equal{} 2^{2007}s\plus{}1\minus{}2^{2006}\equal{} 2^{2006}(2s\minus{}1)\plus{}1$

 i think the answer of both of us is correct :)

Oh I though the problem is for the integer part $ [x]$ . Sorry!  :blush:

Note that if $ a_{i}$ is not an integer it is of the form $ \frac{k}{2}$ where $ k$ is an integer.
Now suppose that $ a_{j}\equal{}\frac{2^{n}(2s_{j}\plus{}1)\plus{}1}{2}$ with $ n\ge 1$. We have that:

$ a_{j\plus{}1}\equal{}\frac{2^{n}(2s_{j}\plus{}1)\plus{}1}{2}\left(\frac{2^{n}(2s_{j}\plus{}1)\plus{}2}{2}\right)$
$ \implies 2a_{j\plus{}1}\equal{} 2^{2n\minus{}1}(2s_{j}\plus{}1)^{2}\plus{}2^{n}(2s_{j}\plus{}1)\plus{}2^{n\minus{}1}(2s_{j}\plus{}1)\plus{}1$

So we see that:

i) $ 2a_{j\plus{}1}\neq 1 (\bmod 2^{n})$
ii) $ 2a_{j\plus{}1}\equiv 1 (\bmod 2^{n})$

Or equivalently we have that there is an integer $ s_{j\plus{}1}$ such that $ a_{j\plus{}1}\equal{}\frac{2^{n\minus{}1}(2s_{j\plus{}1}\plus{}1)\plus{}1}{2}$.

We can right $ m \equal{} 2^{k}(2s\plus{}1)\plus{}1$ for some integers $ s$ and $ k\ge 0$.  The last result tells us that the first integer of the secuence is $ a_{k\plus{}1}$. So in order to have $ a_{2007}$ as the first integer of the secuence, we need to have $ k \equal{} 2006$. So we have that all the possible values of $ m$ are:

$ m \equal{} 2^{2006}(2s\plus{}1)\plus{}1$  for any integer $ s$.

I think the most natural way in which to look at the problem is base 2.

It's easy to see that m is odd in particular 4k+1 and that it cannot be only 1.

So we have something like m= .....1000000...0001 with some ammount of consecutive digits equal to zero. if we call k this ammount of zeros we can see that as a_1 has k zeros between the coma and the first one, and every time you iterate the function that ammount of zeros decreases exactly one time. And when we have no more zeros de ceiling will be even so the next term will be an integer.

So the ammount of zeros is 2005 which yields m=2^2006+1 mod 2^2007

sory about my english and no latex,

julian

it's identical with 2010 China Second Round,Test 2,problem 2.

[hide=Solution] Mine is pretty similar to sinajackson's, for starters is clear that $m$ can't be even and by plugging numbers it seems that numbers of the form $4k+1$ do the trick for $a_2$ (to not be an integer). So by setting $m=4k+1$ and evaluating, we get polynomials just like the ones appearing in sinajackson's solution, and in fact if arranged properly and keep setting the variable as an even number they all seem to have this form,
\[ a_{r}=\frac{8p(k_{r-1})^2+6p(k_{r-1})+1}{2}\]
where $p(k_{r-1})$ is a polynomial on $k_{r-1}$ with no constant term and whose value is even if and only if $k_{r-1}$ is even (provided that $k_{r-1}$ is an integer of course). So we're ready to do a proof by induction; The base cases are the little cases we did early to find the pattern, now suppose the above conditions follow for $a_r$, if we intend $a_{r+1}$ to not be an integer we need the numerator of $a_r$ to be of the form $4k+1$ or equivalently $p(k_{r-1})$ must be even, by the induction hypothesis that means that $k_{r-1}$ must be even, so let $k_{r-1}=2k_r$, evaluate the polynomial and factor a $2$ out of every term of the polynomial, we are left with
\[ a_{r}=\frac{4[8q(k_{r})^2+3q(k_{r})]+1}{2}\]\\
\[ \lceil{a_{r}} \rceil  =2[8q(k_{r})^2+3q(k_{r})]+1 \]\\
\[ a_{r+1}=a_{r} \lceil{a_{r}} \rceil = \frac{8[8q(k_{r})^2+3q(k_{r})]^2+6[8q(k_{r})^2+3q(k_{r})]+1}{2} \]

where $q(x)$ is just the result of doing $p(2x)$. Now set $g(k_r)=8q(k_{r})^2+3q(k_{r})$ getting  $a_{r+1}= \frac{8g(k_{r})^2+6g(k_{r})+1}{2}$ and where is not too difficult to see that $g(k_r)$ satisfies that it is a polynomial with no constant term and that it is even if and only if $k_r$ is even, and we're done!. In other words we have proved that if we choose $m=4k+1$ with $k$ "sufficiently even" we can keep the sequence from turning into an integer. In particular we need $k_{2005}$ to be the first odd $k_r$ so that $a_{2006}=\frac{4l+3}{2}$ and thus $a_{2007}$ is an integer, so $m=4\cdot2^{2004}k_{2005} +1$ where $k_{2005}$ is odd.

[/hide]


let a,b,c be posivetive real numbers such that
a n +b n +c n <(2 n +2)(a+b+c) n /4 n
prove that a,b,c are three sides of a triangle

assume $a\leq b\leq c$ and if $a+b\leq c$ if we can prove that
$a^{n}+b^{n}+c^{n}\geq \frac{2^{n}+2}{4^{n}}(a+b+c)^{n}$ we can have the contradiction.
it is the same as   $\frac{a^{n}+b^{n}+c^{n}}{(a+b+c)^{n}}\geq \frac{2^{n}+2}{4^{n}}$
let $\frac{a}{a+b+c}=x,\frac{b}{a+b+c}=y,\frac{c}{a+b+c}=z$
and we can get $x+y+z=1,z\geq\frac{1}{2},x,y\geq0$
we have to prove $x^{n}+y^{n}+z^{n}\geq\frac{2^{n}+2}{4^{n}}$
because  $x^{n}+y^{n}\geq\frac{(x+y)^{n}}{2^{n-1}}$
so we have to only to prove  $\frac{(1-z)^{n}}{2^{n-1}}+z^{n}\geq\frac{2^{n}+2}{4^{n}}$
for  $\frac{(1-z)^{n}}{2^{n-1}}+z^{n}$ is increase in $[\frac{1}{2},1]$

Nice solution, Zhaobin!!!
In my proof, i used the ineq.
m n-1 (x_1 n +...+x_m n ) >= (x_1+...+x_m) n


algebra
polynomial
inequalities unsolved
inequalities

find all polinomical $P(x)$ satisfying $\large P(x^2)=P(x).P(x+1)$ $\forall x \in R$

Plug x=0 and you get:
$P(0)=P(0)P(1)$, $P(0)(1-P(1))=0$, so either $P(0)=0$ or $P(1)=1$  (*)
a) if $P(0)=0$, then $P(x)=xW(x)$, where W(x) is some polynomial. By rewriting the original equation we have:
$x^{2}W(x^{2})=x(x+1)W(x)W(x+1)$, $xW(x^{2})=(x+1)W(x)W(x+1)$. By plugging x=0 we get $W(0)W(1)=0$, so either $W(0)=0$ or $W(1)=0$. If $W(0)=0$, then $W(x)=xW_{1}(x)$, where $W_{1}(x)$ is some polynomial - then we repeat step a). If $W(1)=0$, then $W(x)=(x-1)W_{1}(x)$. After rewriting the equation we have:
$(x^{2}-1)W_{1}(x^{2})=(x-1)xW_{1}(x)W_{1}(x+1)$, $(x+1)W_{1}(x^{2})=xW_{1}(x)W_{1}(x+1)$. By plugging x=1 we get $W_{1}(1)=0$, so we repeat the step above.
We proceed until we finally $P(x)=x^{k}(x-1)^{l}V(x)$, where $V(x)$ is such a polynomial that $V(0) \neq 0$ and $V(1) \neq 0$. Our equation takes form:
$x^{2k}(x^{2}-1)^{l}V(x^{2})=x^{k}(x-1)^{l}(x+1)^{k}x^{l}V(x)V(x+1)$.
I'll finish tomorrow (hope it isn't a dead-end  ;) )


let k, m, n be positive integers.
is it possible that 1/k, 1/m, 1/n form an arithmetic progression?
is it possible that 1/k + 1/m + 1/n = 1/10?

dunno how to bite it ;p

second part:

suppose $k\geq m\geq n$
$\Rightarrow\frac{3}{k}\geq\frac{1}{10}$
$\Rightarrow k\leq30$
From looking at the equation itself, we can see that $k>10$
So basically, all you have to do is check every single value for k from $(10,30]$
Next, do the same for m and n...
This will take a lot of work, but it's the main idea. When you put a smaller value instead of 10, it'll turn up WAY more easily...
Some are obvious though, like $(k,m,n)=(30,30,30)$ or $(20,40,40)$ etc.
I'm working on a full write-out... will take a couple of minutes though  ;)

[quote="JBL"]For the first part, there are many examples.  Take the largest of m, k, n to be 6l, the next to be 3l, and the last to be 2l for any integer l, for example.

As for the full write-out psycho promised, there are 116 solutions, with k, m, n falling between 11 and 12,210.  There is a solution with k = 12, m any number between 61 and 70 except 67.[/quote]

well yeah :)
it took me longer than expected, but i promised it  :lol:

no problem  :P
i was kinda bored anyway, but there were more answers than i forsaw  :blush:


draw 2 lines to cut the figure into 4 rectangles.
Then the area is 7*12=84
84+4*4+5*8=84+16+40=140


MATHCOUNTS
algebra
difference of squares
special factorizations

[quote="Ignite168"]Factor completely over the reals. $4r^{2}s^{2}-(r^{2}+s^{2}-t^{2})^{2}$[/quote]

[hide]Hint:
Use Difference of Squares as many times as possible and rearrange when needed.[/hide]

Ugly and advanced and out of MC this goes to Intermediate, where it might be too easy, but 1234567890 can decide that.

Not that I'm a mod or anything, but I think it's better in Basics.

How is this "ugly and advanced" ? It's simply difference of squares twice and associative property of addition once. Many more difficult problems come up on MathCounts using difference of squares.

I'm going to move this to Intermediate. Although algebra "work" may be bit over basic, the "concept" is not definitely intermediate. If it's "intermediate," it should require some sort of investigation but this problem is really straightforward.

Along with this whole debate about the problem level, the problem itself remained unsolved...


geometry
3D geometry
calculus
integration
sphere
analytic geometry
trigonometry

case 1 V denote the regoin in the cone x^2+y^2=z^2 (0<=z<=1)
case 2 V denote the region in x^2+y^2+z^2=z

when i use triple integration in sphere corrdinate in case 1 and 2
what should be the range of integration?( from where to where)

First of all, wrong section.  Post calculus problems in the College Playground section.

Do you want the volume inside the cone or below the cone?  Since $z$ is positive, you can just say $z=\sqrt{x^{2}+y^{2}}$ intersected with $z=1$.  Assuming it's the region inside the cone, you can say

\[ V=\pi-\int_{0}^{2\pi}\int_{0}^{1}r^{2}\,dr\,d\theta=\pi-\frac{2\pi}{3}=\frac{\pi}{3}  \]

Otherwise, if you are finding the volume under the cone, you would just take the second part $V-\pi$.  Notice that by doing it this way you only have to take the volume of a cylinder and subtract the volume under the cone.

If, on the other hand, you prefer spherical coordinates, you can say that $\rho=\sqrt{\rho^{2}\sin^{2}\phi+1}$ to get $\rho=\sec\phi$.  Here $\phi\in\left[0,\frac{\pi}{4}\right]$ and $\theta\in\left[0,2\pi\right]$.

And if it's easier for you to do, just let $\rho\in\left[0,\sqrt{2}\right]$ and subtract the "cap" of the sphere $x^{2}+y^{2}+z^{2}=2$ over the unit disk.

\[ \pi +\int_{0}^{\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\int_{0}^{2\pi}\rho^{2}\sin\phi\,d\theta\,d\phi\,d\rho-\int_{0}^{2\pi}\int_{0}^{1}r\sqrt{2-r^{2}}\,dr\,d\theta  \]

[quote="lingomaniac88"]If you don't mind me asking, what does something like $\left[0,\frac{\pi}{4}\right]$ mean?[/quote]

$x\in[a,b]\iff a\le x\le b;\qquad x\in(c,d)\iff c<x<d$

Oh yeah, sorry about that (I should have clarified).  There are many ways to do this problem; you could even use a single integral if you don't want to go through the steps I posted above:

\[ V=\pi\int_{0}^{1}y^{2}\,dy=\frac{\pi}{3},  \]

which is clearly the same answer as when done by means of a double / triple integral.  (You can see that this is the shortest way, but I'm assuming you wanted spherical coordinates so you could practice using them rather than simply to find the answer.)

As for problem #2, notice that

\[ x^{2}+y^{2}+z^{2}=z\implies x^{2}+y^{2}+\left(z-\frac{1}{2}\right)^{2}=\frac{1}{4}  \]

is just a sphere (moved upwards by $\frac{1}{2}$) of radius $\frac{1}{2}$.

$\therefore$ The volume is simply $\frac{4}{3}\pi\left(\frac{1}{8}\right)=\frac{\pi}{6}$.

\[ V=\int_{0}^{\frac{1}{2}}\int_{0}^{2\pi}\int_{0}^{\pi}\rho^{2}\sin\phi\,d\phi\,d\theta\,d\rho=\frac{\pi}{6}, \]

which is the same :).


algebra
polynomial
real analysis
real analysis unsolved

I knew it has been posted,but I can't find it,can you help me to find the link. :)
$f$ is defined $R \to R$,and $f \in C^{\infty}(R)$,and for every $x\in R$,there exist a $k$,such that $f^{(k)}(x)=0$,prove:$f$ is a polynomial

[quote="alekk"]small Hint: use Baire theorem ..[/quote]
Thank you,but I don't know what to do after use Baire's theorem. :(


search
number theory proposed
number theory

if we have polinome $P(k) = (x-1)(x-2)...(x-k) $
say we have $P(0)$ and we want to know how many roots does this
polinome $P(x)$ have we should solve $(-1)^k\cdot k!=P(0)$ trying to find k if i'm right.
now we can do a binary search here,constrainting more and more the interval
and we can find a solution this way very fast.

1. It is an inappropriate section for this problem.
2. Your question is unintelligible

the question will come later in this same topic about a problem i tried to solve
like this.


Prove that for any positive integers $a,b,n$,there exists a positive integer $x$ such that the number $\frac{ab+x^{n}}{a+b+x}$ is also an integer.


function
calculus
derivative
calculus open

Does anybody know (or heared) about Shilov's problem about family of functions  $f(x,y )$ on $[0,1]*[0,1]$ with two derivatives ?!?


Without calculators proved that
sin 54  -sin 18 =1\2

[quote="ashrafmod"]Without calculators proved that $\sin 54^{\circ}-\sin 18^{\circ}=\frac{1}{2}\ .$[/quote]
[color=darkblue]$\sin 54-\sin 18=\frac{1}{2}$ $\Longleftrightarrow$ $2\sin 36\sin 54-2\sin 36\sin 18=\sin 36$ $\Longleftrightarrow$

$(\cos 18-\cos 90)-(\cos 18-\cos 54)=\sin 36$ $\Longleftrightarrow$ $\cos 54=\sin 36$, what is truly.[/color]


induction
combinatorics proposed
combinatorics

In a rectangular array of nonnegative reals with $m$ rows and $n$ columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that $m=n$.

Easy induction. Just choose maximal number of
columns and rows with the same sum, delete them
 and apply hypothesis.

If $ n>m$, we can choose the sequence $ c_1, c_2, ... c_m$ of distinct naturals in $ 1,2,...,n$ such that the $ (i,c_i)$ entries are positive.

Hence, the sum of elements of the whole array = the sum of (sum of elements of each rows) = the sum of elements of the $ c_1, c_2, ..., c_m$ columns < the sum of (sum of elements of each columns) = sum of whole array. Contradiction!

Similar contradiction for $ n<m$. Hence, $ m\equal{}n$.

[quote="mathangel"]If $ n>m$, we can choose the sequence $ c_1, c_2, ... c_m$ of distinct naturals in $ 1,2,...,n$ such that the $ (i,c_i)$ entries are positive.[/quote]Counterexample: All positive elements are in row 1 and column 1.

[hide="my solution"]We proceed by induction on $m+n$. The result is obvious if $m+n=2$. Rearrange the rows such that the rows containing positive elements of column $c_1$ are $r_1,\ldots,r_{a_1}$. Next rearrange the columns such that the columns containing positive elements of rows $r_1,\ldots,r_{a_1}$ are $c_1,\ldots,c_{b_1}$. Clearly the sums of numbers in $r_1,\ldots,r_{a_1}$ and $c_1,\ldots,c_{b_1}$ are equal. Now we have a subboard $S_1$ which contains the intersections of rows $r_1,\ldots,r_{a_1}$ and columns $c_1,\ldots,c_{b_1}$. All elements in $r_1,\ldots,r_{a_1},c_1,\ldots,c_{b_1}$ outside $S_1$ are 0. So $S_1$ is a subboard such that all rows and columns have equal sums of numbers, call it $s$. Then $a_1s=b_1s$, which gives $a_1=b_1$. If we delete these $a_1$ rows and $b_1$ rows, we get a $(m-a_1)\times(n-b_1)$ board. By the hypothesis, we have $m-a_1=n-b_1$, therefore $m=n$.[/hide]

Consider the bipartite graph $G$ with partite sets $R,C$ with $|R|=m$ (representing the rows $r_1,r_2,\cdots ,r_m$) and $|C|=n$ (representing the columns $c_1,c_2,\cdots , c_n$). We make an edge from $r_i$ to $c_j$ if the entry at $(i,j)$ is positive.

Consider any set of $w$ rows, say $W = \{r_1,r_2,\cdots , r_w\}$, without loss of generality. Consider the neighbourhood $N(W)$ of these rows, and let there be $f(w)$ columns in this neighbourhood. Then the sum of the elements in each such column must be equal to the sum of elements in some row, and for each row in the $w$ rows, there must be some column in $N(W)$ whose sum of elements is equal to the sum of elements of this row. It follows easily that $f(w) \not < w$, that is, $|W| \le |N(W)|$.

By Hall's Marriage Theorem, there is a matching from $R$ to $C$, so $ m \leq n$. Similarly, $n \leq m$ (by considering the neighbourhoods of the subsets of $C$). Thus $m=n$.

Make a bipartite graph between rows and columns, and draw an edge if a row and column have a positive intersection. Weight the edge with that number. Assume the graph is connected (otherwise, decompose the graph into its connected components). Then each vertex has equal edge sum, so this forces $m=n$.

Let $R_i$ represent the $i$th row sum. Also, let $C_j$ denote the $j$th column sum.
To each non-zero element $a_{i,j},$ assign the weight $R_i^{-1}=C_{j}^{-1}$.
Hence,
$$n=\underbrace{1+\cdots+1}_{\text{n times}}=\sum_{j} \left( \sum_{i} \frac{a_{i,j}}{R_i} \right)=\sum_{i,j} \frac{a_{i,j}}{R_i}=\sum_{i,j} \frac{a_{i,j}}{C_j}=\sum_{i} \left( \sum_{j} \frac{a_{i,j}}{C_j} \right)=\underbrace{1+\cdots+1}_{\text{m times}}=m$$

Alternatively :

$\bullet$ construct a bipartite graph, with rows and columns as the vertex parts.

$\bullet$ Show that Hall's condition is satisfied.

$\bullet$ Deduce $m \le n.$

$\bullet$ Using symmetry deduce $n \le m.$

$\bullet$ Conclude $m=n.$

Really a pleasant problem.

[quote="adityaguharoy" ]$\bullet$ Show that Hall's condition is satisfied.[/quote]
[quote="biomathematics"]It follows easily that $f(w) \not < w$[/quote]
I have two concerns, hopefully I'm not misunderstanding the solutions
$\bullet$ Isn't proving Hall's condition stronger than the problem itself? Proving Hall's Condition for the case when the set of rows is the set of all rows already solves the problem.
$\bullet$ It's not obvious to prove either.
[quote="igrn"]
We proceed by induction on $m+n$. The result is obvious if $m+n=2$. Rearrange the rows such that the rows containing positive elements of column $c_1$ are $r_1,\ldots,r_{a_1}$. Next rearrange the columns such that the columns containing positive elements of rows $r_1,\ldots,r_{a_1}$ are $c_1,\ldots,c_{b_1}$. Clearly the sums of numbers in $r_1,\ldots,r_{a_1}$ and $c_1,\ldots,c_{b_1}$ are equal. Now we have a subboard $S_1$ which contains the intersections of rows $r_1,\ldots,r_{a_1}$ and columns $c_1,\ldots,c_{b_1}$. All elements in $r_1,\ldots,r_{a_1},c_1,\ldots,c_{b_1}$ outside $S_1$ are 0. So $S_1$ is a subboard such that all rows and columns have equal sums of numbers, call it $s$. Then $a_1s=b_1s$, which gives $a_1=b_1$. If we delete these $a_1$ rows and $b_1$ rows, we get a $(m-a_1)\times(n-b_1)$ board. By the hypothesis, we have $m-a_1=n-b_1$, therefore $m=n$.
[/quote]
I think this proof is wrong because we cannot assume after rearranging the rows and columns that we have a subboard such that all the rest of the elements not in that subboard but in the same rows or columns are all $0$s. Take, for instance,

\begin{tabular}{llll}
1 & 1 & 1 & 0 \\
1 & 0 & 0 & 2 \\
1 & 0 & 1 & 1 \\
0 & 2 & 1 & 0
\end{tabular}

Let the rectangular array be $A = (a_{ij}).$ Let $M_i$ be the row sum of $i$th row for $1 \leq i \leq m$ and $N_j$ be the column sum for $j$th row for $1\leq j \leq n.$ Furthermore, we have $M_i > 0$ and $N_j > 0.$ The problem statement says that whenever $a_{ij} >0$ for some $i, j$ then $M_i = N_j.$ We need to show $m=n.$
If $M_i = N_j$ whenever $a_{ij} >0,$ we have $\frac{a_{ij}}{M_i} = \frac{a_{ij}}{N_j}$ for all $i,j.$ Now, \begin{align*}
    \sum_{i, j} \frac{a_{ij}}{M_i} &= \sum_{i,j} \frac{a_{ij}}{N_j} \\ \implies \sum_{i=1}^m \left(\frac{1}{M_i} \sum_{j=1}^n a_{ij}\right) &= \sum_{j=1}^n \left(\frac{1}{N_j} \sum_{i=1}^m a_{ij}\right) \\ \implies \sum_{i=1}^m \left(\frac{1}{M_i} M_i\right) &=  \sum_{j=1}^n \left(\frac{1}{N_j} N_j\right) \\ \implies \sum_{i=1}^m 1 &= \sum_{j=1}^n 1 \\ \implies m &= n.
\end{align*}

I also think that the Hall solutions here are incomplete/faulty. In fact, when you try to check that Hall's condition holds, one arrives at the following solution:

Consider the bipartite graph with a row and column adjacent when they intersect at a positive entry. In each connected component all rows and columns must have the same sum $S$. But if there are $m_0$ rows and $n_0$ columns in a component, then calculating the total sum in two ways gives $m_0S = n_0S$, i.e. $m_0 = n_0$. Finally, summing through all components implies $m=n$.


LaTeX
number theory unsolved
number theory

Fins all ordered triples $ \left(a,b,c\right)$ of positive integers such that $ abc \plus{} ab \plus{} c \equal{} a^3$.

the ordered triplets (a,b,c) are :-
i)  (a,a-1,a)
ii) (a,1,a(a-1))
this was probably the easiest of all the postal problems. :)

Again, please use LaTeX, and don't just post answers but rather complete proofs.

[quote="Rushil"]Fins all ordered triples $(a,b,c)$ of positive integers such that $abc + ab+ c = a^3$[/quote]

Okey so let me try the complete sol though it seems bad way  :( , may be someone else like [b]venkata[/b] know  nicer and better solutions :
We show  all the ordered triplets $(a,b,c)$ are :
i) $(a,a-1,a)$
ii) $\Big(a,1,a(a-1)\Big)$

We have $a(a^2-b)=c(1+ab)$ cause $\gcd(a,ab+1)=1$ then $a\mid c$ we put $c=ak$ where $k\in Z_+$ and $a^2-b=(ab+1)k$ so $a^3+1=bka^2+ab+ka+1=(ka+1)(ab+1)$. Now use Four Number Theorem  we have $x,y,z,t$ which $(x,z)=1$ and
$a+1=xy,a^2-a+1=zt,ka+1=zy,ba+1=xt$. Because $k,b< a$ so $zy\leq a(a-1)+1=zt$ so $y\leq t$. Other hand $xt\leq (a-1)a+1=zt$ so $x\leq z$. If $z>x$ and $t>y$ then $k,b<a-1$ so $ka+1,ba+1\leq (a-1)^2=(xy-2)^2<(xy)^2$ so $xyzt<(xy)^2$ then $zt<xy$, contract!. So $z=x$ or $t=y$

1) IF $z=x$ then $b=a-1$ and $ka+1=zy=xy=a+1$ then $k=1$ so $c=a$ we have got $(a,b,c)=(a,a-1,a)$.
2) IF $t=y$ then $k=a-1$  so $c=a(a-1)$ and then $ba+1=xt=xy=a+1$ so $b=1$. We have got $(a,b,c)=\Big(a,1,a(a-1)\Big)$.

Maybe I do something wrong.

We have $a|c$. From equation we have $a^2>c\geq a>b$ then let $c=ad$ then we have $d<a$ then the equation comes to $abd+b+d=a^2$ then we have $a|b+d$

If $b+d>a$ contradiction!

Then we have $b+d=a$ the equation comes to $bd+1=a$. From this the solutions is $(a,b,c)=(xy+1,x,(xy+1)y)$

The equation is equivalent to $a^3+1=abc+ab+c+1=(ab+1)(c+1)$. Then
\[ \\ ab+1 \mid a^3+1 \\ \\ ab+1 \mid a^3-ab \\ \\ ab+1 \mid a^2-b \\ \\ ab+1 \mid a^2b-b^2=(ab)a-b^2 \\ \\ ab+1 \mid b^2+a \\ \\ab+1 \mid b^3+ab \\ \\ab+1 \mid b^3-1. \\ \]

Suppose $b \geq a^2$. Then, in the original equation, $ab+1 \geq a^3+1=RHS$, so we need $c+1 \leq 1, c \leq 0$ for equality, absurd because $c$ is positive. Then $b<a^2$, so $a^2-b>0$. Since $ab+1 \mid a^2-b$, we have $ab+1 \leq a^2-b$. If $b \geq a$, we have $a^2+1 \leq LHS \leq a^2-b$, absurd. So $b<a$. Since $ab+1 \mid b^2+a$, we have $ab+1 \leq b^2+a$. Since $b<a$, let $a=b+r$ with $r$ positive. Then $b^2+br+1 \leq b^2+b+r$, so $(b-1)(r-1)=br-b-r+1 \leq 0$. Then one of $b, r$ is at most $1$. If $b \leq 1$, since $b>0$ we have $b=1$, and in the original equation $(a+1)(c+1)=(a^3+1)$, so $c=a^2-a$ and this is a solution. If $r \leq 1$, we have $r=1$, $a-1=b$, so $ab+1=a^2-a+1$ and $c+1=a+1, c=a$. Solutions are $(k, 1, k^2-k)$ and $(k, k-1, k)$.


Let be $ a,b,c,d \in (0,1)$.Prove then
$ a) \sqrt{ab}\plus{}\sqrt{(1\minus{}a)(1\minus{}b)}\le 1$

$ b) \sqrt{abcd}\plus{}\sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)}<1$

[quote="soruz"]Let be $ a,b,c,d \in (0,1)$.Prove then
$ a) \sqrt {ab} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)}\le 1$

$ b) \sqrt {abcd} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)} < 1$[/quote]
a)
$ LHS\le \frac{a\plus{}b\plus{}1\minus{}a\plus{}1\minus{}b}{2}\equal{}1$
b)$ a,b,c,d \in (0,1)$ so  $ \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}<\sqrt[4]{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$
So  $ \sqrt {abcd} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}<\frac{a\plus{}b\plus{}c\plus{}d\plus{}1\minus{}a\plus{}1\minus{}b\plus{}1\minus{}c\plus{}1\minus{}d}{4}\equal{}1$
 :blush:


geometry
calculus
integration
rectangle
function
real analysis
puzzles

Hai folks, Can anyone tell me how to calculate the area of a circle as :pi*r*r where 'r' is the radius of the circle without using Definite Integration?

If anyone is interested, they can also provide the Definite Integrals method. :lol:

I would rather define it as---  The measurement of the surface within the boundary (circumference , in case of a circle). How would you define it, do you have any method as to prove that the area of a circle as: pi*r*r , without applying definite integration?  :maybe:

[quote="Gen8"]The measurement of the surface within the boundary (circumference , in case of a circle).[/quote]
What I mean is this:  from first principles, how do you define the area of anything that isn't a rectangle?  (You are surprisingly close with the word "measurement," but what does that word actually mean [i]mathematically?[/i])  If I gave you an arbitrary shape, how would you go about finding its area?

Hmm... isn't the formal defintion of area the definite integeral of the curve, if the function is integratable?

That is the intuitive point I'm attempting to drive at, although that is not the way mathematicians actually do things.  It turns out that to get the best results one [url=http://en.wikipedia.org/wiki/Lebesgue_measure]formally defines "length" and "area" first[/url] and then defines the integral in terms of lengths and areas.

In any case, what that means is more or less this:  the only more "fundamental" way to define the area of a circle than as an integral is in terms of the construction of something called the Lebesgue measure that you do [i]not[/i] want to deal with.

This is not to say there aren't ways you can gain intuition about why this relation should hold.  Consider the following argument:  the "average" circumference of a circle with radius $ a$ randomly chosen in the interval $ [0, r]$ should be the average of the maximum and minimum circumferences, which are $ 0$ and $ 2 \pi r$, which gives the average $ \pi r$.  It should also be $ \frac{A}{r}$.

I m posting the limit method
 Recall the definition of pi
c=2* pi *r
Divide the circle of radius r into 8 parts(I should say 8 equal sectors) by using 4 diagonals.
Put one part and then other part inverted.Thus it will take shape of parallelogram.
Just observe what happens if the no. of cuts are increased.
The figure starts resembling a rectangle as we increase the no. of cuts to the circle.
As we increase the no. of cuts indefinitely , or equivalently, we decrease the angle of each sectors decreases
The figure almost becomes a rectangle exactly.
area=(pi*r)*r
      =pi*r^2


Hi,

I clicked "who is online" and got the attached screenshot. Note that guests are supposedly viewing their profiles.

Is this a bug? Sorry if this has been reported; I searched and didn't find anything.

Because of the update delay, right when you sign out you become a guest, so if you were viewing your profile right before you signed out, that's what would happen.

Usually what happens is that we have bots and/or other people that are not logged in crawling the site. They cannot see profiles if they are not logged in, but they can type in the URL, and the page will say "Not Authorized". But they are still on that page, so that's what you see there.


vector
inequalities
function
real analysis
integration
probability and stats

Prove that for all reals x, y we have :

  F(x,y)<=sqrt(F(X)F(Y)).  where F is the distribution function.   im sorry for not using latex.

Does your distribution function F(x,y) have a density w.r.t. Lebesgue measure ?

If so we can say

$ F(x,y)\equal{}\int_{\minus{}\infty}^{y}\int_{\minus{}\infty}^{x}f(X,Y)dXdY$

$ F(x,y)\equal{}\int_{\minus{}\infty}^{y}\int_{\minus{}\infty}^{x}f(X,Y)^{1/2}f(X,Y)^{1/2}dXdY$

$ F(x,y)\leq (\int_{\minus{}\infty}^{y}\int_{\minus{}\infty}^{x}f(X,Y)dXdY)^{1/2}.(\int_{\minus{}\infty}^{y}\int_{\minus{}\infty}^{x}f(X,Y)dXdY)^{1/2}$ by cauchy schwartz and then by fubini

$ F(x,y)\leq (\int_{\minus{}\infty}^{y}(\int_{\minus{}\infty}^{x}f(X,Y)dX)dY)^{1/2}.(\int_{\minus{}\infty}^{x}(\int_{\minus{}\infty}^{y}f(X,Y)dY)dX)^{1/2}$

$ F(x,y)\leq (\int_{\minus{}\infty}^{y}f(Y)dY)^{1/2}.(\int_{\minus{}\infty}^{x}f(X)dX)^{1/2}\equal{}\sqrt{F(y).F(x)}$

If this is not the case I think it can still work as Cauchy Schwartz is a very general property.


I know the results come out the 14th, but when do the national finalists get called? I've heard that Marcy goes in the order 20th to 1st.

Supposedly there are ~125 points so an 80 cutoff is possible?

yeah, the finalists will get phone calls this week, and yes an 80 should be enough....SDS? haha

Well, SDS was only worth 4 points anyway. I was thinking about the sheer number of essay questions.
I simply counted ~100/150=1/3 for last year and therefore ~1/3*125=83.

I wouldn't be surprised if a high 70 can make it, though.

I nailed the genetic engineering question and due to partial credit got at least 2/3 of the rest of the part C, except the last few mathy genetics questions (which I got all of)...I just need to hope my part A score was at least a 30 because I'm pretty sure the part B was at least high 20s for me...

[quote="hillarryous"]I have not heard anything yet. We'll all know [b]for sure[/b] by the 14th.[/quote]

Hmm...going by last year, we will either find out today within the next hour or on Wednesday from 5  - 7 EST.

So, 5-7 EST on Tuesday and Wednesday? Wow, the calls are quite ceremonial. Marcy personally calls each finalist?

Why don't they post the results earlier if they have all the tests graded by then?

Well, nothing is etched in stone, but the reason this is done is to confirm that certain people will be able to attend the camp (it would look silly to go back and re-edit the list posted on the website).

Apparently, Marcy has started [url=http://talk.collegeconfidential.com/high-school-life/457638-usabo-2008-club-16.html]calling the finalists[/url] (read post 237).

According to that post, she goes from 1st to 20th, so if you haven't gotten a call yet, don't worry too much...

Does she finish calling to top 20 by the end of today. Or is it stretched over a few days?

If the former is true, I think it is safe for me to assume I didn't make it.

It looks like I didn't make it after all...oh well, time for usnco!

Does anyone know what the cutoff is?

Btw, my teacher said that CEE never posts it, so how do we know that last year's cutoff was 101? Do the finalists talk to each other at the camp?

Well, there were 21 finalists last year due to a tie, and we know that two students got 101, so we inferred that 101 was the cutoff. But as your teacher stated, the cutoff score is never announced, publicly or to the finalists.


[img]http://i5.photobucket.com/albums/y194/jelly119/05070501.jpg[/img]
ABCDEF is a hexagon with AB=BC=CD, AF=DE, $\angle{A}=\angle{B}=\angle{C}=\angle{D}=\angle{E}=\angle{F}$. Given that $\angle{FCE}=60^{\circ}$ and the area of $\triangle{FCE}$ is 60, find the area of ABCDEF.

[quote="Iversonfan2005"]I'm assuming those squares mean angles right? ;) Anyway, I'm solving this interesting problem.[/quote]

Ya...they are "angles". :oops:

By the way, I was stunned by this question since it's from some math contest for students in elementary school!

can anyone solve this. It is tooo hard. :?  :(  please real help needed.
Maybe yetti the great geometer could help :?  :(

[quote="mysmartmouth"]This is just a regular hexagon with 120 degree angles. [/quote]
How can this be a regular hexagon?????
In a regular hexagon [u]all[/u] sides are equal???? arent they???
[quote="mysmarmouth"]But from there I am confused.  A ratio of degrees to area gets you a total are of 120, but I don't think that's correct.[/quote]
It isnt.
How can you just find that.
Could you explain how, if it is possible?????

Maybe FE is parallel to BC, so it is a regular hexagon that has been "parallel-cut?"

$FE\parallel{BC}$ that is pretty much obvious but Ive tried everything but, it is not working.
I think it lacks information or its too hard. :?  :(

[quote="mysmartmouth"]This is just a regular hexagon with 120 degree angles.  But from there I am confused.  A ratio of degrees to area gets you a total are of 120, but I don't think that's correct.[/quote]

Uhh...you do know that this isn't a regular hexagon right? And no, it's not as simple as that....however, I do agree that each angle is $120$ degrees since all the angles in the hexagon are equal. However, a guess says that it is indeed $120$, don't know exactly how to deduce that though by mathematical reasoning, but just by switching around the shape.

[quote="frt"]I got an answer, though I'm not sure if it's correct.

Rotate $\triangle CDE$ around $C$ so that $CD$ overlaps $BC$. Let $D'$ be the new location of point $D$ after the rotation. Since $\angle D'BC=\angle BCD=120^\circ$, $D'B\parallel CD\parallel AF$.

Let $G$ be the point where $AB$ and $D'F$ intersect. Since $AF=D'B(=DE)$ and $AF\parallel D'B$, we see $\triangle AFG\cong\triangle BD'G$.

Now consider the area of the convex polygon $AFCD'B$.
$Area(AFCD'B)=Area(\triangle AFG)+Area(GFCD'B)=Area(\triangle BD'G)+Area(GFCD'B)=Area(\triangle D'FC)$

Since $D'C=CE$ and $\angle D'CF=\angle ECF=60^\circ$, we conclude $\triangle FCD'=\triangle FCE$. Therefore, the area of the hexagon is $2(60)=\boxed{120}$[/quote]

COOOOL solution, :10:  :10:  :coolspeak:  :omighty:  I think that is right.
I always hated areas :wallbash_red:  :wallbash:  :mad:  :?  :(

[quote="frt"]I got an answer, though I'm not sure if it's correct.

Rotate $\triangle CDE$ around $C$ so that $CD$ overlaps $BC$. Let $D'$ be the new location of point $D$ after the rotation. Since $\angle D'BC=\angle BCD=120^\circ$, $D'B\parallel CD\parallel AF$.

Let $G$ be the point where $AB$ and $D'F$ intersect. Since $AF=D'B(=DE)$ and $AF\parallel D'B$, we see $\triangle AFG\cong\triangle BD'G$.

Now consider the area of the convex polygon $AFCD'B$.
$Area(AFCD'B)=Area(\triangle AFG)+Area(GFCD'B)=Area(\triangle BD'G)+Area(GFCD'B)=Area(\triangle D'FC)$

Since $D'C=CE$ and $\angle D'CF=\angle ECF=60^\circ$, we conclude $\triangle FCD'=\triangle FCE$. Therefore, the area of the hexagon is $2(60)=\boxed{120}$[/quote]

That's right ! And it originally appeared in a contest from your country (Japan!!).

[quote="jelly119"]And it originally appeared in a contest from your country (Japan!!).[/quote]
Oh cool! :D
Is it from Japanese math olympiad for students in elementary school? If so, I'm so happy that I solved this problem because I've never solved any problem from this contest before (although it's for those who are much younger than me...)

[quote="riddler"]the chinese maths olympiad is supposed to be the hardest in the world, I think.[/quote]

Probably, considering they always win :P


inequalities
geometry
3D geometry
tetrahedron
inequalities proposed

Let $x,y,z,a,b,c$ six pozitive numbers. Give a [b]geometrical interpretation[/b] (in space) for the wellknown conditioned inequality (a particular case of the [u]C.B.S.'s inequality[/u]):

$\frac xa+\frac yb+\frac zc=1\Longrightarrow \frac{1}{x^2+y^2+z^2}\le \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}.$

Let $OABC$ be the tetrahedron with $O(0,0,0),A(a,0,0),B(0,b,0),C(0,0,c)\ (a>0,b>0,c>0),$ and $P(x,y,z)$ be the point which lie in the inside of the tetrahedron where $x>0,y>0,z>0.$
If $H$ be the foot of perpendicular to the plane $ABC: \ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,$  then by the formula of a point and plane,we have

$\overline{OP} \geq \overline{OH}\Longleftrightarrow \sqrt{x^2+y^2+z^2}\geq \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}},$ yielding $\frac{1}{x^2+y^2+z^2}\leq \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}.$


function
real analysis
real analysis unsolved

hi,
consider the functions $ f,g : \mathbb{R} \rightarrow \mathbb{R}$. If $ x_{0}$ is the unic fixpoint of $ f \circ g$ and $ g \circ f$ then $ f,g$ have a common unic fixpoint.

$ g(x_0)\equal{}g(f\circ g(x_0))\equal{}g\circ f(g(x_0))\Longrightarrow g(x_0)$ is fixed point of $ g\circ f\Longrightarrow g(x_0)\equal{}x_0$. Similarly, we conclude $ f(x_0)\equal{}x_0$. If $ f(x)\equal{}g(x)\equal{}x\Longrightarrow f\circ g(x)\equal{}x\equal{}g\circ f(x)\Longrightarrow x\equal{}x_0$ proving the uniqueness.


trigonometry
geometry
algebra unsolved
algebra

I'm wondering how to prove that
if x,y,z are angels of a triangle then tan(x/2)*tan(y/2) + tan(x/2)*tan(z/2) + tan(z/2)*tan(y/2) = 1.
I remember Darij Grinberg posting this, but I can't find his post. Could you help me??

Thanks in advance
Haile!!

if a+b+c=90 then tga = 1/tg(b+c)
tga(tgb + tgc)=1-tgb tgc
so...

Thank you very much. I made a geometrical proof using the fact that the point where the inscribed circle = P is tangent to BC, and the point where escribed is tangent to BC = Q are so that BP = CQ. From it, I got an equation
|BP|*|PC| = r*Ra where r is incircle radius and Ra is radius of escribed opposite to vertex A.
using that fact whole problem is easy cause tan(x/2)*tan(y/2) = r^2/|BP|*|PC|

Best regards

$A+B+C=\pi$
Then:
$\frac{A}{2}=\frac{\pi}{2}-\frac{B+C}{2}$

$\displaystyle\tan{\frac{A}{2}}=\frac{1}{\tan{\frac{B+C}{2}}}=\frac{1-\tan{\frac{B}{2}}\tan{\frac{C}{2}}}{\tan{\frac{B}{2}}+\tan{\frac{C}{2}}}$

Which implies the result.


Solve the equation
x^2 - x - 1000=1000[(1+8000x)^1/2]

Please help!

hello, do you mean
$ x^2\minus{}x\minus{}1000\equal{}1000\sqrt{1\plus{}8000x}$?
Sonnhard.


[quote="SuperMath"]Let $\large x;y;z > 0$. Prove that:
$\large 5(x+y+z)^2.(x+y)(y+z)(z+x) \geq\ [3(x+y)(y+z)(z+x)+ xy(x+y)+yz(y+z)+zx(z+x)] [(x+y+z)^2+xy+yz+zx]$[/quote]
it is the same as:
$xy(x+y)(x-y)^2+yz(y+z)(y-z)^2+zx(z+x)(z-x)^2 \geq 0$


Solve this riddle:

I start with the letter E.
I end with the letter E.
I usually contain one letter.
But I am not the letter E.

What am I?

Good job! I actually couldn't figure out because I kept thinking it was I letter long :D .

HEY!!!

I GOT ANOTHER ANSWER!!!!!!!!!!!!!!

My answer is [hide="..."]EYE. It starts with the letter E. It ends with the letter E. It usually contains one letter. (remember that EYE is a homophone with I.) It is not the letter E.[/hide]

If you pronounce the word "EYE" and ask someone to write down what u said, they will probably write the word "I".

It still doesn't make much sense, I like my answer better :P

[quote="SplashD"]The answer is obviously
[hide="."]envelope[/hide][/quote]

Wow good job I never would have guessed!

I think the containing one letter was a big give-away

Alternate answer:

I am the [b]number[/b] e, at approximately 2.718281828. :roll:

That's what I first thought of when I read the title.

I have actually never seen this before...

I find the alternate answeres very original and amuzing :D .

[quote="mathgeniuse^ln(x)"][hide]Envelope is the answer right?[/hide][/quote]
yes. Next time please read all the posts above, because the answer has already been posted. :wink:

[hide="answer given above "]Envelope  is [/hide] obvious answer,  but I like the other answer given above too

Namely
[quote]...[/quote]

[hide]This is obviously one letter (S), it starts with 'e'  and ends with 'e'

If it is not clear,[hide]  The morse code for Letter S is "..." or three dots.

Morse code:   E = 'dot'
                     S = 'dot dot dot'
:)[/hide]
Think out of Box!
[/hide]

That solution is quite creative, Gyan. Now that I think about it, [hide="it can also be"] email message  :P [/hide] It's surprising that such a riddle can have that many solutions...

You can stop using your subjective words such as "obviously" and "obvious" as were used in this thread (as well as other threads).  Alternate answers from different users showed that what came natural for one person did not for another, and vice versa.

this is not as good but if you really think about it it is a [hide="AHHHH"]word or a noun[/hide]

:huh: How does that contain one letter and starts and ends with e?


inequalities algebra polynomial quadratics geometry trigonometry search

yes expand, get your hands dirty and use Muirheads inequality

I just  hate  Muirheads  inequality.... :?

It always work, and 7 marks are seven marks!

[quote="dante"]It seems like using [tex]\frac{x^2_1}{y_1} + ... +\frac{x^2_n}{y_n} \geq \frac{(x_1+...+x_n)^2}{y_1+...y_n} [/tex] would be a good idea, but afterwards, I end up with the sign switch on [tex]a^2 + b^2 + c^2 \geq ab + bc + ac[/tex]. Anyone have any ideas?[/quote]

this is because the ineq from Gm is much more stronger and Muirhead works just fine! :D

Well  there  must be  an elementery solution....I understand  Muirhead  works  but  you have  to  do  too  many  computations  which i  think  are  useless...

Indeed, but the elementary solution is too complicated ... ;)

I dont  have  the issue  with the  elementary solution....

The solution for this problem hasn't been posted yet from what I know, so far. But  the elementary solution is very hard, trust me! :D :D :D

[quote="dante"]Of course, we could do that to any symmetric inequality, but there must be a nicer way   :D .[/quote]

Actually, is it true that one can nuke every symmetric polynomial inequality with Muirhead?

And what about Hilbert(?)'s theorem that every non-negative polynomial can be written as a sum of quadratic forms? Can it be applied in such problems?

Finally, what about a^2 + ab + b^2 = (a^3 - b^3) / (a - b), maybe it could be of use? Or substituting a = (b' + c' - a') / 2 etc., with a', b', c' being the sides of a triangle, and then trigbashing the resulting expression?

Sorry if I'm posting bullshit, you obviously see I'm a beginner...

  darij

Darij for your first question i think it is not always true, for example, consider the inequlity given in mathlinks contest 3rd edition 5 round, after squaring and having a symetric thin in  :sqrt: a, :sqrt: b, :sqrt: c i couldnt kill it with muirheads because "i had a simetric sum that didnt majorize another one" in the wrong side, so I had to use WAM-GM. but one could also use Muirheads and later Schur or Am-Gm, and it works just fine.

Thanks, Pascual. I actually could guess that Muirhead is not everything ;-)

Now I think I've got a real idea. It may look strange but actually I find it quite promising. The inequality in question is equivalent to a rather simple-looking inequality from triangle geometry. Since I will use the letters a, b, c later in a more significant way, I rewrite the inequality from gazeta matematica using p, q, r instead of a, b, c:

If p, q, r > 0, then

  1 / (q^2 + qr + r^2) + 1 / (r^2 + rp + p^2) + 1 / (p^2 + pq + q^2) >= 9 / (p + q + r)^2.

Now, remember some well-known facts from triangle geometry:

Let ABC be a triangle. Erect equilateral triangles BA'C, CB'A, AC'B outwardly on the sides BC, CA, AB of triangle ABC. Then, the lines AA', BB', CC' concur at a point F. This point F is called the [b]Fermat point[/b] of the triangle ABC.

The point F lies in the interior of triangle ABC if and only if all angles of triangle ABC are < 120. Moreover, if this is the case, then we have the nice equations < BFC = < CFA = < AFB = 120 and AA' = BB' = CC' = AF + BF + CF. Finally, the Fermat point has its well-known minimal property, also holding only if all angles of triangle ABC are <= 120: If P is an arbitrary point in the plane of triangle ABC, then AP + BP + CP >= AF + BF + CF with equality only for P = F.

So why do I write this down? Well, consider the three positive numbers p, q, r. Take an arbitrary point F in the plane, and draw three half-lines through F forming 120 angles with each other. On one of these three half-lines, take a point A such that FA = p. On another one, take a point B such that FB = q. Finally, on the third one, take a point C such that FC = r. Then, the point F is the Fermat point of triangle ABC (since < BFC = < CFA = < AFB = 120). Since the point F lies in the interior of triangle ABC, all angles of triangle ABC are < 120.

The sides a = BC, b = CA, c = AB of triangle ABC can be found using the Cosine Law applied to triangles BFC, CFA, AFB, all of them having a 120 angle:

  a^2 = q^2 + qr + r^2;
  b^2 = r^2 + rp + p^2;
  c^2 = p^2 + pq + q^2.

And p + q + r = AF + BF + CF = d, where we use the letter d for the constant d = AA' = BB' = CC' = AF + BF + CF.

Hence, the inequality from gazeta matematica is equivalent to

  1 / a^2 + 1 / b^2 + 1 / c^2 >= 9 / d^2.

Therefore, it remains to prove this inequality for all triangles with angles < 120. Any idea? (Maybe it's in Bottema?)

  Darij

I remember to have posted long time ago that elementary solution. So, you better search in the solved section. I cannot search because my Internet-connection is really low low ... this summer.

Although the inequality $\displaystyle\sum_{cyc} \frac{(a-b)(a-c)}{b^2+bc+c^2} \ge 0$ has been fully discussed in the link Darij provided, I just want to post one more interesting proof which i found just now.

WLOG assume $a \ge b \ge c$.
For convenience, denote the denominator of each terms by $R,S,T$ respectively, it is obviously that $R,S,T>0$. Then the inequality now looks like
\[\frac{(a-b)(a-c)}{R}+\frac{(b-a)(b-c)}{S}+\frac{(c-b)(c-a)}{T} \ge 0 \Leftrightarrow\]
\[\frac{(a-b)T[(a-c)S-(b-c)R]+(c-b)(c-a)RS}{RST} \ge 0 \Leftrightarrow\]
\[\frac{(a-b)T(a^3-b^3)+(c-b)(c-a)RS}{RST} \ge 0\]
which is obviously true.  :)

I think it's Schur's inequal:[tex]\sum_{cyclic}\frac{1}{b^2+bc+c^2}(a-b)(a-c)[/tex] and we always have:[tex]\frac{1}{b^2+bc+c^2}-\frac{1}{a^2+ac+c^2}=\frac{(a-b)(a+b+c)}{(b^2+bc+c^2)(a^2+ac+c^2)}\geq 0[/tex] if [tex]a\geq b[/tex]

Well.... Muirhead is exactly what you used to find the AM-GM's...

This is stronger than Cauchy, right?
I wonder if Arqady can find a sum of squares for it?

[quote="nickolas"]If $a$, $b$, $c$ are positive reals, prove that $\frac 1{a^{2}+ab+b^{2}}+\frac 1{b^{2}+bc+c^{2}}+\frac1{c^{2}+ca+a^{2}}\geq \frac9{(a+b+c)^{2}}$.[/quote]

From this problem we have

Let triangle $ABC$ with $\max(A,B,C)<\frac{2\pi}{3}$ and $M$ in it such that $\widehat{AMB}=\widehat{BMC}=\widehat{CMA}=\frac{2\pi}{3}$. Prove that $MA+MB+MC\geq \frac{3.AB.BC.CA}{\sqrt{AB^{2}.BC^{2}+BC^{2}CA^{2}+CA^{2}.AB^{2}}}$. :D

[quote="nickolas"]If $a$, $b$, $c$ are positive reals, prove that $\frac 1{a^{2}+ab+b^{2}}+\frac 1{b^{2}+bc+c^{2}}+\frac1{c^{2}+ca+a^{2}}\geq \frac9{(a+b+c)^{2}}$.[/quote]

we trasform this problem to Problem(1940) Of Crux that well known to:Iran(96)
We aim to prove: {(a+b+c).^2/b.^2+b*c+c.^2)>={4*(a+b+c).^2/(b+c)>^2}. this implies that: (b.^2+c.^2-a*b-a*c).^2>=0

[quote="nickolas"]If $a$, $b$, $c$ are positive reals, prove that $\frac 1{a^{2}+ab+b^{2}}+\frac 1{b^{2}+bc+c^{2}}+\frac1{c^{2}+ca+a^{2}}\geq \frac9{(a+b+c)^{2}}$.[/quote]

It's Vasc's problem . You can find it in the page 91 of the book Old and New inequalities where a beautiful solution is proposed.

i have two diffrent solutions to it one is using  Vornicu schur and second mixing variable after assuming a+b+c=1;
but i dont have time to post even a single.

take a point P and

draw distances of a, b, c from it such that the three line segments make angle 120 degrees with the adjacent line segment.

let x, y, and z be the sides of the triangle so formed.
(P will be the fermat point of the triangle)

(a+b+c)^2 = 1/2(x^2 + y^2 + z^2 + 4 sqrt 3 * (area of the triangle))

a^2 + ab + b^2 = x^2 (by cosine rule)
similarly for the others

so the inequality reduces to

1/x^2  +  1/y^2  +  1/z^2  >=  18/(x^2 + y^2 + z^2 + 4 sqrt 3 * (area of the triangle))

cross multiply
and use (x/y + y/x >= 2)

we are left to prove

2 sqrt3 *(area of triangle)(1/x^2  +  1/y^2  +  1/z^2) >= 9

which can be easily proved by application of AM GM

(use x=2R sin A
and area of triangle = 2R^2 sin A *sin B*sin C)

QED

and of course

the equality holds if and only if

ABC is equilateral or x = y = z

or a = b = c

[quote="nickolas"]If $a$, $b$, $c$ are positive reals, prove that $\frac 1{a^{2}+ab+b^{2}}+\frac 1{b^{2}+bc+c^{2}}+\frac1{c^{2}+ca+a^{2}}\geq \frac9{(a+b+c)^{2}}$.[/quote]
I used to SOS method and Schur inequality to solve above inequality :P It is nice problem but not hard  :P

Just some information
This is an old inequality of VASC, appeared in "Old and new inequalites". And maybe it was posted here!

Please give more information about your proofs than just "by Vornicu-Schur" or "by AM-GM". These will hardly help anyone.

A list of topics where this inequality has been discussed:
http://www.mathlinks.ro/Forum/viewtopic.php?t=14082
http://www.mathlinks.ro/Forum/viewtopic.php?t=13280
http://www.mathlinks.ro/Forum/viewtopic.php?t=1738
http://www.mathlinks.ro/Forum/viewtopic.php?t=6364 (this one)

  darij


function
floor function
ceiling function
LaTeX

Knowing that $ \lfloor x \rfloor = \max\{n \in\mathbb{Z}\; | \; n \leq x\}$ and $ \lceil x \rceil = \min\{n \in\mathbb{Z}\; | \; n \geq x\}$, prove that:

1) $ x < \lfloor x \rfloor+1$

2) $ \lceil x \rceil < x+1$

3) $ \lfloor k+x \rfloor = k+\lfloor x \rfloor$

4) $ \lfloor x \rfloor =-\lceil-x \rceil$ as well as $ \lceil x \rceil =-\lfloor-x \rfloor$

Note: It might help greatly to state properties of a min and max function.

You need some spaces in you LaTeX: you've let the code "\le" touch the letter "b" which leads to the non-code "\leb".

[quote="krsattack"]Just let....[hide]$ x=a+b$, where $ a=\lfloor x\rfloor$ and $ 0\le b<1$
OR...
$ x=c-d$, where $ c=\lceil x\rceil$ and $ 0\le d<1$

We should be able to prove all of the properties using this idea:[hide="1)"]$ \lfloor x\rfloor+1=k+1>k+r$
[/hide]
[hide="2)"]$ \lceil x\rceil=x+d<x+1$[/hide]

The next two proofs are exactly the same, although they are somewhat trivial as well.[/hide][/quote]

I agree, but it remains to prove that $ 0 \leq x-\lfloor x \rfloor < 1 \;\; \forall x \in\mathbb{R^{+}}$ and $ -1 < x-\lceil x \rceil \leq 0 \;\; \forall x \in \mathbb{R^{-}}$.

Really.

If $ \lfloor x \rfloor$ is the greatest integer not exceeding $ x$.

$ \implies x \not < \lfloor x \rfloor \implies x \geq \lfloor x \rfloor \implies x-\lfloor x \rfloor \geq 0$

Also, $ \lfloor x \rfloor+1 \; (> \lfloor x \rfloor)$ is an integer that exceeds $ x$, $ \implies \lfloor x \rfloor+1 > x \implies 1 > x-\lfloor x \rfloor$

I'm not getting why $ \lfloor x \rfloor =-\lceil-x \rceil$ though I think I have a rough idea how to do it. Basically, I got:

$ \lfloor x \rfloor \leq x$ and $ \lceil x \rceil \geq x$

$ \lfloor-x \rfloor \leq-x$ (questionable step)

$ -\lfloor-x \rfloor \geq-(-x)$

$ -\lfloor-x \rfloor \geq x$

Since $ \lfloor-x \rfloor \geq x$ and $ \lceil x \rceil \geq x$, it would seem that I'm close to a conclusion. However, making $ x = \lfloor x \rfloor$ or something similar doesn't work because in that case, $ \lfloor x \rfloor = \lceil x \rceil$ automatically.

:(

[b]i.[/b] If $ x$ is an integer, it's trivially true.
[b]ii.[/b] If $ x$ is not an integer, let $ x \equal{}\lfloor x\rfloor\plus{}\{ x\}$
$ \implies\minus{}x \equal{}\minus{}\lfloor x\rfloor\minus{}\{ x\}\equal{}\minus{}(\lfloor x\rfloor\plus{}1)\plus{}(1\minus{}\{ x\})$
$ 0 <\{ x\}< 1\implies 0 < 1\minus{}\{ x\}< 1$

So, $ \minus{}\lfloor\minus{}x\rfloor \equal{}\minus{}\lfloor\minus{}(\lfloor x\rfloor\plus{}1)\plus{}(1\minus{}\{ x\})\rfloor \equal{}\lfloor x\rfloor\plus{}1 \equal{}\lceil x\rceil$

Alternatively, you can imagine flipping $ \lfloor x\rfloor$ across the horizontal and vertical axes to get $ \lceil x\rceil$.


geometry
geometric transformation
reflection
circumcircle
geometry unsolved

If $ A'B'C'$ is the tangential triangle of $ ABC$ (the triangle determined by the tangents at the vertices of $ ABC$ with respect to the circumcircle) and $ X$, $ Y$, $ Z$ are the reflections of $ A$, $ B$, $ C$ in the opposite sidelines, then the circumcircles of $ YZA'$, $ ZXB'$, $ XYC'$ are concurrent at Parry's reflection point.

Any 'non-barycentric' solutions are more than welcomed.

Let $ u \parallel v \parallel w$ be 3 parallel lines with arbitrary direction, passing through A, B, C. Their reflections u', v', w' in BC, CA, AB pass through the reflections X, Y, Z of A, B, C in BC, CA, AB. They intersect at points $ U \equiv v' \cap w',\ V \equiv w' \cap u',\ W \equiv u' \cap v'.$ Regardless of the direction of $ u \parallel v \parallel w$, they form constant angles $ \measuredangle YUZ \mod \pi \equal{} \measuredangle (v', w') \mod \pi \equal{} 2 \measuredangle (CB, BA) \mod \pi \equal{} 2 \angle BAC \mod \pi,$ $ \measuredangle ZVX \mod \pi \equal{} \measuredangle (w', u') \mod \pi \equal{} 2 \angle CBA \mod \pi,$ $ \measuredangle XWY \mod \pi \equal{} \measuredangle (u', v') \mod \pi \equal{} 2 \angle ACB \mod \pi.$ For the tangential $ \triangle A'B'C',$ we also have $ \measuredangle C'A'B' \mod \pi \equal{} 2 \measuredangle BAC \mod \pi,$ $ \measuredangle A'B'C' \mod \pi \equal{} 2 \measuredangle CBA \mod \pi,$ $ \measuredangle B'C'A \mod \pi \equal{} 2 \measuredangle ACB \mod \pi.$

AB' = CB', BZ = BC = CY, $ \measuredangle A'BZ \equal{} \measuredangle ACB \plus{} \pi \minus{} \angle CBA \equal{} 2 \measuredangle ACB \plus{} \measuredangle BAC,$ $ \measuredangle A'CY \equal{} 2 \measuredangle ACB \plus{} \measuredangle BAC.$ Consequently, the $ \triangle A'BZ \cong \triangle A'CY$ are directly congruent by SAS, $ \measuredangle BA'Z \equal{} \measuredangle CA'Y \Longrightarrow$ $ \measuredangle YA'Z \equal{} \measuredangle C'A'B' \Longrightarrow$ $ U \in \odot (YZA')$ and similarly, $ V \in \odot (ZXB'),$ $ W \in \odot (XYC').$ Since u', v', w' concur in Parry's reflection point R, i.e., $ U \equiv V \equiv W \equiv R,$ iff $ u \parallel v \parallel w \parallel OH$ and since their pairwise intersection U, V, W are on the circumcircles $ \odot (YZA'), \odot (ZXB'), \odot (XYC'),$ these circumcircles also concur in R.


:(  I have always been confused with basic operations concerning computers.

Could someone notify me about the procedures in which I add a graphic signature?
If the answer is copy and paste, it didn't work on my computer.

[quote="Hikaru79"]He means an image in your signiature. No, it can't be done on AoPS, that's disabled. The admins probably figured it would be way too distracting if every person had a picture on every post they made (besides their avatar, I mean).[/quote]You can use one image in your signature using the [img ] tag. Read the BBCode FAQ for how to actually use the img tag.

But first you must upload the image to a web hosting site like [url=http://xs.to/index.php]XS[/url].  That's because you can't upload a signature picture directly from your computer.  Don't ask me why. :P


inequalities
integration
inequalities theorems

I can't find the proof of the following version of Almansi-Wirtinger inequality:

$ f: [0,2\pi]\to\mathbb{R}$ differentiable and $ f'$ continuous, $ f(0) \equal{} f(2\pi) \equal{} 0$

prove that $ \int_0^{2\pi}f(x)^2dx\le 4\int_0^{2\pi}f'(x)^2dx$

edited - coefficient "4" was missing

Edit -
it works fine, there is "4" missing in above post
thanks


Determine all triples of nonnegative integers $ (a,b,c)$ such that: $ 2^a\plus{}3^b\equal{}4^c$.

Clearly $ c \geq 1$, so $ 4^c$ is even.

If $ a \geq 1$, we get $ 3^b$ is even, a contradiction.

So $ a\equal{}0$. We must solve $ 3^b \plus{} 1 \equal{} 4^c$. But this is $ 3^b \equal{} 4^c \minus{} 1 \equal{} (2^c \plus{} 1)(2^c \minus{} 1)$, so both $ 2^c\plus{}1$ and $ 2^c\minus{}1$ are powers of 3, so they must be $ 1$ and $ 3$, which yields $ c\equal{}1$ and $ b\equal{}1$.

The only solution is $ (0,1,1)$.

[quote="moldovan"]Determine all triples of nonnegative integers $ (a,b,c)$ such that: $ 2^a \plus{} 3^b \equal{} 4^c$.[/quote]
Clearly $ a\equal{}0$.The equation becomes $ 3^b\plus{}1\equal{}4^c$.If $ b\equal{}1$ then $ c\equal{}1$.Otherwise,we get $ c > 1$,hence $ 8|3^b\plus{}1$ but this is a contradiction since $ 3^b\plus{}1 \equiv 2,4 (mod 8)$.Therefore,the only solution is $ (0,1,1)$.


analytic geometry
geometry
parallelogram

Let B and D be two fixed points. Aand C are two variable points opposite to the line BD such that ABCD is a parallelogram. Two circles S1 and S2 are inscribed in triangle ABD and CBD respectively touching the line at P and Q respectively. If PQ is 2 and BD = 4 then show that angle AED is either less than 60° or greater than 120° where E is the mid point of PQ.

Um...in your second sentence, did you mean "A and C are variable points on opposite sides of BD", or in other words, AC and BD intersect?

Yes. They are actually the diagonals of the parallelogram ABCD.


guys, i dont know how you ppl remember all the reagents and their uses but i think its a real pain in the ass. besides there are too many reagents and there are no proper sources to search for them. so why dont we start posting all the reagents we know( at least the important ones) over here. we could organise it later. i know it is boring to type all the reagents but it would be useful for everyone especially for those trying for jee n the olympiads

:spam: DONT POST TWICE :spam:

1. Lemeiux Reagent
2. Schiffs Reagent
3. Nessler's reagent
4. Millon's reagent
5. Adam's Catalyst
6. Lindlar's Catalyst
7. Fieser's soln

Want more?

can u tell me their uses, not the mechanism, that would take too long. only their uses and what it consists of( at least the last one).

Fieser's soln is the last problem of nsec 2006 thats why i included it  :wink:

8.Collins reagent
9.Grignard Reagents
10.Lawesson reagent
11.PCC
12.Wittig's reagent

I WILL POST MORE OF THESE IN MY NEXT POST :wink: hey!!!!!!!!!!!!!!!

The "Wittig reagent" you mentioned is seldom know that way: the correct name is "phosphorus ylide".


Find all $ x;b \in Z$ such that
$ b|a\plus{}1$ and $ a|b^2\plus{}2$

Hi,How do you mean "find $ x;b$.Do you mean find $ a$ and $ b$ ?
I think this can be solved using Euler's function.(just first idea)

xxx001: If you don't answer enndb0x' objection, you shouldn't expect any answers  :wink:

Kamyar1991: You may want to at least sketch a proof. You missed some solutions  :wink:


geometry
reflection
incenter
circumcircle
radical axis

[quote="rohitsingh0812"]Let $ABC$ be a triangle, and let $D$, $E$, $F$ be the points of contact of its incircle $\omega$ with its sides $BC$, $CA$, $AB$, respectively. Let $K$ be the point of intersection of the line $AD$ with the incircle $\omega$ different from $D$, and let $M$ be the point of intersection of the line $EF$ with the line perpendicular to $AD$ passing through $K$. Prove that $AM$ is parallel to $BC$.[/quote]

[color=darkblue][b]Nice and easy problem ![/b]

[b]Proof.[/b] Construct the line $l$ for which $A\in l$ and $l\parallel BC\ .$ Denote : the reflection $N$ of the point $D$ w.r.t. the incenter $I$ ;
$S\in EF\cap AI$ ; $L\in l\cap DI$ ; $M_{1}\in l\cap KN$ ; $M_{2}\in l\cap EF\ .$
$\blacktriangleright$ The points $K$ and $L$ belong to the circle with the diameter $DM_{1}\Longrightarrow$ $AK\cdot AD=AL\cdot AM_{1}\ \ (1)\ .$
$\blacktriangleright$ the points $S$ and $L$ belong to the circle with the diameter $IM_{2}\Longrightarrow$ $AS\cdot AI=AL\cdot AM_{2}\ \ (2)\ .$
$AK\cdot AD=AF^{2}=AS\cdot AI$ $\Longrightarrow$ $AK\cdot AD=AS\cdot AI\ .$ From the previous relations $(1)$ and $(2)$
obtain $AL\cdot AM_{1}=AL\cdot AM_{2}\ ,$ i.e. $M_{1}\equiv M_{2}\ (\equiv M)\ .$ In conclusion, $AM\parallel BC\ .$

[b]Remark 1.[/b] The vertex $A$ is the radical center for the circumcircles of the quadrilaterals $KDML\ ,\ KSID\ ,\ SIML$ and the incircle of the triangle $ABC\ .$

[b]Remark 2.[/b] Can use (for another proof) the following remarkable property :
[u]The intersection between the lines $AN$ and $BC$ belongs to the $A$- exincircle of the triangle $ABC\ .$[/u][/color]

Dear Mathlinkers,
A scheme of my synthetical proof.
Let 1 the incircle, 2 the circumcircle of AEF, 3 the circumcircle of AKD' where D' is the antipode of D wrt 1 and X the second intersection of 2 and 3.
1. D' is on 1 and on 2.
Let K' and D" the second point of intersection of AK, DD' with 2.
2. KD' // K'D"
3. Accordling to Reim's theorem applied to 2 and 3, D", D' and X are collinear.
4. D, D', D" and X are collinear.
5. Accordling to Reim's theorem applied to 1 and 3, AX // BC.
6. Accordling to Monge's theorem applied to 1, 2 and 3,	AX, KD' and EF are concurrent at M…
Sincerely
Jean-Louis

Let $Q$ the second intersection of $MP$ and the incircle it s obvious that is the antipode of $D$ .Let $R=AQ\cap \odot(I)$ .we have $DR\cap PQ$ is on the polar of $A$ i.e $PQ\cap DR =M$ noticing that $Q$ is the orthocenter of $DAM$ leads $\angle AMD=\angle RPD= \angle RDC$ which means that $BC \parallel AM$
[b]RH HAS[/b]

1. Draw a line $l$ parallel to $BC$ through $A$. Let $DI$ intersect $l$ in $L$. Also, let $N$ be the point diametrically opposite to $D$ on the incircle ($\Gamma_1$).
2. There is a circle $\Gamma_2$ passing through $K, N, A$ and $L$ ($AK \perp KD$ and $AL \perp LN$).
3. Also, consider $\Gamma_3$ - the circle passing through $A, I, E$ and $F$.

Now,
4. $KN$ ($\perp AD$) is the radical axis of $\Gamma_1$ and $\Gamma_2$.
5. $l$ is the radical axis of $\Gamma_2$ and $\Gamma_3$.
6. $EF$ is the radical axis of $\Gamma_1$ and $\Gamma_3$.

Thus, $KN$, $l$ and $EF$ are concurrent at $M$, which implies $AM \parallel BC$.

Let $D'=\overline{KM}\cap\omega,\ N=\overline{DD'}\cap\overline{EF},\ L= \overline{AN}\cap\overline{BC}$. It's a known fact that $L$ is the midpoint of $ BC$. By Pascal's theorem on $FFKD'EE$, $A,\ \overline{FK}\cap\overline{ED'},\ \overline{EK}\cap\overline{FD'}$ are collinear; therefore, invoking Brokard's theorem, $AN $ is the polar of $M$ wrt. $\omega$. Hence, $(F,E;N,M)=-1$. Projecting from $A$, we get $(B, C; L, \overline{AM}\cap\overline{BC}=-1)$. Since $L$ is the midpoint of $BC$, we conclude that $P_{\infty}=\overline{AM}\cap\overline{BC}$, thus $AM \parallel BC$.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/01.%201.%20Parallele%20a%20un%20cote%20du%20triangle%20passant%20par%20un%20sommet%20du%20triangle.pdf

p. 11-12.

Sincerely
Jean-Louis


King School is planning a field trip.  Every student must be accompanied by a guide, and no guide can accompany more than twelve students.  What is the minimum number of guides required to accompany 147 students?

[hide]
147/12=12.25 so that rounds up to $ \boxed{13} $[/hide]