Im not going to write the restrictions because its hard to do it in latex.

1. How am i going to use the binomial theorm to show that
   $ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+... = 2^{n-1}$

2. Hard Combinatorial Proof. Use combinatorial proof to show that
  $ \binom{a+b}{c}=sum_{i=0}^{c}\binom{a}{i}\binom{b}{c-i}$

What is wrong with my summation symbol?

Please help me, first time to meet combinatorial proof. my exam is two days from now. This is for practice and analysis. Thanks.

you should write \ before sum...
\sum_{i=1}^n

[hide="hint"]
1. Expand
$ 1/2\cdot[(1+1)^{n}+(1-1)^{n}]$

2. The LHS counts number of ways of choosing c from a set of a+b elements. If you had 2 sets, one of $ a$ elements and one of $ b$ elements, what is the RHS counting?
[/hide]

for $ 1$ in the identity $ (x+y)^{n}=\sum_{i=0}^{n}\binom{n}{i}x^{i}y^{n-i}$ put $ x=1,y=-1$ we get that:
$ 0=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+...+(-1)^{n}\binom{n}{n}$
so we get that:
$ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+...=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+...$
but we know that $ LHS+RHS=2^{n}$ so we get that $ LHS=RHS=2^{n-1}$ as wanted...
------------------------------------
for $ 2$ as JavaMan said,we want tpo choose $ c$ elements among $ a+b$ elements,so we can do this in $ \binom{a+b}{c}$ ways
in the other hand we can do this in the following way:
devide the set into two subsets of $ a$ and $ b$ elements,and name them $ S_{a},S_{b}$ so we can do the job in the following ways:
$ i_{0})$we choose $ 0$ elements from $ S_{a}$ and $ c$ elements from $ S_{b}$ $ \Rightarrow \binom{a}{0}.\binom{b}{c}$.
$ i_{1})$we choose $ 1$ element from $ S_{a}$ and $ c-1$ elements from $ S_{b}$ $ \Rightarrow \binom{a}{1}.\binom{b}{c-1}$.
.
.
.
$ i_{c})$we choose $ c$ elements from $ S_{a}$ and $ 0$ lements from $ S_{b}$ $ \Rightarrow \binom{a}{c}.\binom{b}{0}$.
so we get that:
$ i_{0}+i_{1}+...+i_{c}=\sum_{i=0}^{n}\binom{a}{i}\binom{c-i}{b}$
so we get that:
$ \text{the number of ways to choose c elements among a+b elements}=\binom{a+b}{c}=\sum_{i=0}^{n}\binom{a}{i}\binom{c-i}{b}$
as wanted...

[hide="hint for 2"]
From the identity $ (1+x)^{a+b}=(1+x)^{a}(1+x)^{b}$ identify the coefficient of $ x^{c}$ from both members.
[/hide]

1. Use $ \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$. The sum becomes $ \sum^{n-1}_{i=0}\binom{n-1}{i}=2^{n-1}$.

2. (same as BaBeK Ghalebi) Think of it like this: You have $ a$ blue marbles and $ b$ red marbles. Suppose the marbles are distinguishable.
How many ways can you pick 0 blue marbles and $ c$ red marbles? $ \binom{a}{0}\binom{b}{c}$.
How many ways can you pick 1 blue marble and $ c-1$ red marbles? $ \binom{a}{1}\binom{b}{c-1}$.
Continue down the line and add them all up to get the number of ways to pick $ c$ marbles from the red and blue ones.
However, this must be equal to $ \binom{a+b}{c}$, or choosing $ c$ out of $ a+b$ marbles.


[quote="sharkman"]What is the area of 68' x 50' x 64' x 30'= sq.feet = acres?[/quote]

Huh? Are you asking for that value converted to acres??

My
box happens to be irregular i.e., 68' x 50' x 64' x 30'.  So how would
I calculate that? Should I figure the perimeter and then the area to
the closest sq.ft.?  I am trying to figure out the size of a piece of
land that I want to buy.  The land around here is sold in acres.  So I
need my answer to be converted to how many (partial) acres that would
be.

We don't have enough information; the lengths of the sides of a quadrilateral does not determine its area. (Think of a quadrilateral with all sides equal to 1 ... it could be a square for maximum area, or a really flat rhombus with very little area)


The standard reduction potential for H+(aq) is 0.00 V.
What is the reduction potential for a 1×10-3 M HCl
solution?

Answer:

[hide]-.178[/hide]

My question is how and why?

I think this is easier to see, if you consider the H+ reacting with something else.  Assume you have an imaginary reaction where the standard reduction potential is 0.00 V, with the H+ getting reduced.

The reduction equation is 2H+ + 2e- ---> H2

Then by Nernst equation, in the new conditions:
E=0-0.0257/2 x ln(1/(1x10^-3)^2)=-.178 V.
The new potential is the reduction potential for H+.  The imaginary substance getting oxidized is still at standard, so keep it at 1.


Let $ a,b,c$ be the roots of the equation $ x^3 \minus{} 2x^2 \plus{} x \plus{} 5 \equal{} 0$. Find the values of the following expressions.

$ (a)$ $ a^3 \plus{} b^3 \plus{} c^3$
$ (b)$ $ a^4 \plus{} b^4 \plus{} c^4$

Hello, tanhnam2902, your answer is correct. :lol:

Here is alternative solution:

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$
$ \Longleftrightarrow a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)\{(a \plus{} b \plus{} c)^2 \minus{} 3(ab \plus{} bc \plus{} ca)\}$

$ a^4 \plus{} b^4 \plus{} c^4 \equal{} (a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 2\{(ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(a \plus{} b \plus{} c)\}$

Maybe MathLinker would solve that by using Newton Sum.


A funny little problem :
Let $n \in \mathbb{N}$ .Find all $\mathbb{N}$ such that:
$n \mid 2^n -1$

I believe it's ben discussed many times before. Take $p$ to be the smalest prime which divides $n$, and you get $p|(2^{p-1}-1,2^n-1)=2^{(p-1),n}-1=1$, which is absurd, so there are no such $n$'s.

On the other hand, for every $a>2$ there are infinitely many $n$ s.t. $n|a^n-1$.

Exactly,But it is a little funny problem  :D


geometry
3D geometry
number theory
prime numbers

Prove that 7 is the only prime number that is 1 less than a perfect cube.

Let the perfect cube be $n^3$ for some natural number $n$.
Then $n^3-1=(n-1)(n^2+n+1)$ is a prime number.
Since prime numbers are only divisible by one and itself, one of $n-1$ and $n^2+n+1$ must be $1$, and the other must be the prime.
If $n-1=1$, then $n=2$ and $n^2+n+1=7$, so that works.
On the other hand if $n^2+n+1=1$, then $n=0$ or $n=-1$, which contradicts the definition of $n$.
Thus, $n^3-1=7$ is the only prime one less than a cube.

Corollary:
3 is the only prime that is one less than a perfect square, follows from:
$a^2-1^2=(a+1)(a-1)$


Prove without using the iran 96 famous inequality that if $a,b,c$ are sidelenghts of a triangle then: $bc(b-c)^{2}(2a^{2}-bc)+ca(c-a)^{2}(2b^{2}-ca)+ab(a-b)^{2}(2c^{2}-ab)\ge 0$


Find all primes $p$ such that $2^p + p^2$ is also prime. Prove there are no others.

Masoud Zargar

[hide]Notice that 2^x mod 6 is either 2 or 4.  And also notice that it can only be four when x is even.  The only even prime is two and 2^2+2^2=8 which is not prime.  So x must be odd and 2^x must be 2 mod 4.  Now, notice all odd primes are (1,3,5) mod 6.  But 3 mod 6 is only a prime when the number is 3 since 9,21,... are divisible by 3 and not primes.  Finally notice that 1^2=1 mod 6 and 5^2=25-24=1 mod 6 and when two is added to these, both equal 3 mod 6.  In other words, any prime that is 1 mod 6 or 5 mod 6 will result in a composite number divisible by 3.  When 3 is plugged in to the function, we get 3^2+2^3=9+8=17 which is a prime so only 3 works.[/hide]


I missed the last math jam that I really wanted to go to, so I went to check the Transcript, but it didn't seem that it all was there, like there should be a "next" or something to finish the discussion. Perhaps I'm just missing something, but any help would be greatly appreciated!

Thanks,
WindSlicer

I assume you're talking about the May 25 Math Jam -- that transcript is complete.  We sometimes delete the stuff like "Thank You" and "Good night" from the end of the chat, in the interest of saving space, so it may seem like it ends somewhat abruptly.

Ahh ok, yeah, that's probably it.

Well, thanks for the assistance!


trigonometry
limit
calculus
calculus computations

here is an easy and  fun question for you   :D

can you sum

$ \sin^{\minus{}1} \frac{1}{\sqrt 2} \plus{} \sin^{\minus{}1} \frac{\sqrt 2 \minus{}1}{\sqrt 6} \plus{} \sin^{\minus{}1} \frac{\sqrt 3\minus{}\sqrt 2 }{2\sqrt{3}} \plus{}  \sin^{\minus{}1} \frac{2 \minus{}\sqrt 3}{2\sqrt {5}}$

$ \plus{} \sin^{\minus{}1} \frac{\sqrt 5 \minus{} 2}{\sqrt {30}} \plus{}  \sin^{\minus{}1} \frac{\sqrt 6 \minus{}\sqrt 5}{\sqrt {42}}\plus{}  \sin^{\minus{}1} \frac{\sqrt 7 \minus{}\sqrt 6}{2\sqrt {14}} \plus{} \ldots\qquad \ldots$

The sum could be rewritten as
$ S \equal{}\sum_{n\equal{}1} ^{\infty} \sin^{\minus{}1} (\frac{\sqrt{n} \minus{} \sqrt{n\minus{}1}}{\sqrt{n(n\plus{}1)}})$

[quote="Carcul"]Misan, are we supposed to telescope this sum?[/quote]

yes.  :)

[hide="you wouldnt want to click here"]
First of all, \[ 1\minus{}\left(\frac{\sqrt{n}\minus{}\sqrt{n\minus{}1}}{\sqrt{n(n\plus{}1)}}\right)^2\equal{}\frac{n^2\plus{}n\minus{}(2n\minus{}1)\plus{}2\sqrt{n(n\minus{}1)}}{n(n\plus{}1)}\]\[ \equal{}\left(\frac{1\plus{}\sqrt{n(n\minus{}1)}}{\sqrt{n(n\plus{}1)}}\right)^2\]
The corresponding simpler identity used in the proof later (spot it yourself) is: $ \left(\frac{n}{\sqrt{n(n\plus{}1)}}\right)^2\plus{}\left(\frac{1}{\sqrt{n\plus{}1}}\right)^2\equal{}1$

(Good job coming up with that neat factorization, misan!! :D )

Now for the tedious part... suppose the $ n$-th partial sum is $ \sin^{\minus{}1}\sqrt{\frac{n\minus{}1}{n}}$.
Then \[ \sin(\text{n\plus{}1th partial sum})\equal{}\]\[ \frac{n\minus{}1}{\sqrt{n(n\minus{}1)}}\cdot \frac{1\plus{}\sqrt{n(n\minus{}1)}}{\sqrt{n(n\plus{}1)}} \plus{} \frac{1}{\sqrt{n}}\cdot \frac{\sqrt{n}\minus{}\sqrt{n\minus{}1}}{\sqrt{n(n\plus{}1)}}\]\[ \equal{}\frac{\sqrt{n}\plus{}(n\minus{}1)\sqrt{n}}{n\sqrt{n\plus{}1}}\equal{}\sqrt{\frac{n}{n\plus{}1}}\]
Just using sine angle sum.
[/hide]
In short, the boring answer is that the answer is $ \boxed{\sin^{\minus{}1}1}$

[hide]We have $ \arcsin x - \arcsin y = \arcsin(x\sqrt {1 - y^2} - y\sqrt {1 - x^2})$. If we consider $ x = \frac {1}{\sqrt {n}}$ and $ y = \frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}$ then

$ \arcsin (\frac {1}{\sqrt {n}}) - \arcsin (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}) = \arcsin (\frac {1}{\sqrt {n}}\sqrt {1 - (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}})^2} - \frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}\sqrt {1 - \frac {1}{n}})$.

But according to the expression derived above by Me@home,

$ 1 - (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}})^2 = (\frac {1 + \sqrt {n(n - 1)}}{\sqrt {n(n + 1)}})^2$,

and so

$ \arcsin (\frac {1}{\sqrt {n}}) - \arcsin (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}) = \arcsin(\frac {1}{\sqrt {n}} \frac {1 + \sqrt {n(n - 1)}}{\sqrt {n(n + 1)}} -$

$ - \frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}} \frac {\sqrt {n - 1}}{\sqrt {n}}) = \arcsin(\frac {1 + \sqrt {n(n - 1)} - \sqrt {n - 1}(\sqrt {n} - \sqrt {n - 1})}{n \sqrt {n + 1}}) =$

$ = \arcsin(\frac {1}{\sqrt {n + 1}})$

$ = > \arcsin (\frac {1}{\sqrt {n}}) - \arcsin (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}) = \arcsin(\frac {1}{\sqrt {n + 1}})$

$ = > \arcsin (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}) = \arcsin (\frac {1}{\sqrt {n}}) - \arcsin(\frac {1}{\sqrt {n + 1}})$.

Therefore, $ \sum_{n = 1}^{\infty}\arcsin (\frac {\sqrt {n} - \sqrt {n - 1}}{\sqrt {n(n + 1)}}) = \sum_{n = 1}^{\infty}(\arcsin (\frac {1}{\sqrt {n}}) - \arcsin(\frac {1}{\sqrt {n + 1}}))$

$ = \arcsin(1) - \lim_{n \to \infty}\arcsin(\frac {1}{\sqrt {n + 1}}) = \arcsin(1) - 0 = \arcsin(1)$.

Of course, this would be more simple if we start directly by "trying" our lucky with the calculation of

$ \arcsin (\frac {1}{\sqrt {n}}) - \arcsin(\frac {1}{\sqrt {n + 1}})$.[/hide]

[hide="I came up with my solution in the post box... but looking back there are always nicer/act of magic solutions (with a little bit of motivation...)"]
[hide="motiv"]For telescoping, we want to find some $ \{x_n\}$ that satisfies\[ \sin\left(\sin^{-1}x_{n+1}-\sin^{-1}x_n\right) \equiv \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\]
Basically, in order for anything to work out, $ x_n^2\equiv 1-y_n^2$ where $ y_n$ has a numerator in the field $ \mathbb{Z}[n]$.
This is just gibberish, the only sensible candidates for $ x_n$ will look like $ \sqrt{1-\frac{z_n^2}{z'_n}}$... leading us naturally to $ \frac{\sqrt{n-1}}{\sqrt{n}}$
[/hide]
We compute that \begin{eqnarray*}\sin\left(\sin^{-1}\frac{\sqrt{n}}{\sqrt{n+1}}-\sin^{-1}\frac{\sqrt{n-1}}{\sqrt{n}}\right)&=& \frac{\sqrt{n}}{\sqrt{n+1}}\cdot\sqrt{1-\left(\frac{\sqrt{n-1}}{\sqrt{n}}\right)^2} - \frac{\sqrt{n-1}}{\sqrt{n}}\cdot\sqrt{1-\left(\frac{\sqrt{n}}{\sqrt{n+1}}\right)^2}\\&=&\frac{1}{\sqrt{n+1}}-\frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\\&=& \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\\\sum \sin^{-1}\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}&=& \sin^{-1} 1 - \sin^{-1} \sqrt{\frac01} \ = \ \sin^{-1} 1\end{eqnarray*} :D Done
[/hide]


geometry
angle bisector
geometry proposed

In the square $ ABCD$, $ M \in (CD),N \in (AB)$ so that $ MN \| AD$.The inner bisectors of the angles $ \widehat{ANM}$ and $ \widehat{BNM}$ intersecting sides $ AD$ and $ BC$ in the points $ P$ and $ Q$, also on $ AM$ and $ BM$ in $ S$ and $ T$.Prove that the side $ ST$ contains center of gravity for triangle $ AMB$.

[b]Cristea's theorem:[/b] $ D,E,F$ are three points on the sides $ CB,BA,AC$ of $ \triangle ABC$ and $ M \in AD.$ Then the transversal $ EF$ goes through $ M$ if and only if
\[ \overline{DC} \cdot \frac {\overline{EB}}{\overline{EA}} \plus{} \overline{BD} \cdot \frac {\overline{FC}}{\overline{FA}} \equal{} \overline{BC} \cdot \frac {\overline{MD}}{\overline{MA}} \]
Particularly, $ EF$ passes through the centroid of $ \triangle ABC$ $ \Longleftrightarrow$ $ \frac {\overline{EB}}{\overline{AE}} \plus{} \frac {\overline{FC}}{\overline{AF}} \equal{} 1.$

$\bullet$ By angle bisector theorem in $ \triangle ANM$ and $ \triangle BNM,$ we have the relations

$ \frac {\overline{AS}}{\overline{SM}} \equal{} \frac {\overline{AN}}{\overline{MN}} \ \ , \ \ \frac {\overline{BT}}{\overline{TM}} \equal{} \frac {\overline{BN}}{\overline{MN}} \ \Longrightarrow$

$\frac {\overline{AS}}{\overline{SM}} \plus{} \frac {\overline{BT}}{\overline{TM}} \equal{} \frac {\overline{AN} \plus{} \overline{BN}}{\overline{MN}} \equal{} \frac {\overline{AB}}{\overline{MN}} \equal{} 1$

By Cristea's theorem we conclude that $ST$ goes through the centroid of  $ \triangle ABM.$

Can someoane to post solution for Cristea's theorem

Suppose that $ M \in EF.$ Let $ P \equiv EF \cap BC$ and apply Menelaus' theorem to the transversals $ \overline{PME}$ and $ \overline{PFM}$ in $ \triangle DAB$ and $ \triangle DAC$ to obtain the relations:

$ \overline{DC} \cdot \frac {\overline{EB}}{\overline{EA}} \equal{} \overline{DC} \cdot \frac {\overline{MD}}{\overline{MA}} \cdot \frac {\overline{PB}}{\overline{PD}} \ \ , \ \ \overline{BD} \cdot \frac {\overline{FC}}{\overline{FA}} \equal{} \overline{BD} \cdot \frac {\overline{PC}}{\overline{PD}} \cdot \frac {\overline{MD}}{\overline{MA}} \ \Longrightarrow$

$ \overline{DC} \cdot \frac {\overline{EB}}{\overline{EA}} \plus{} \overline{BD} \cdot \frac {\overline{FC}}{\overline{FA}} \equal{} \frac {\overline{MD}}{\overline{MA}} \cdot \frac { (\overline{DC} \cdot \overline{PB} \plus{} \overline{BD} \cdot \overline{PC})}{\overline{PD}}$

$ \equal{} \frac {\overline{MD}}{\overline{MA}} \cdot \frac {[(\overline{PD} \minus{} \overline{PC}) \cdot \overline{PB} \plus{} (\overline{PB} \minus{} \overline{PD}) \cdot \overline{PC}]}{\overline{PD}} \ \Longrightarrow$

$ \overline{DC} \cdot \frac {\overline{EB}}{\overline{EA}} \plus{} \overline{BD} \cdot \frac {\overline{FC}}{\overline{FA}} \equal{} \frac {\overline{MD}}{\overline{MA}} \cdot \frac {\overline{PD} \cdot (\overline{PB} \minus{} \overline{PC})}{\overline{PD}} \equal{} \overline{BC} \cdot \frac {\overline{MD}}{\overline{MA}}$

The converse is  proved analogously due to the converse of Menelaus' theorem.

Thank you for your kindness and promptness with which you answered my request!


geometry
parallelogram
analytic geometry
vector

AoPS 2, #158

Three vertices of parallelogram PQRS are P(-3, -2), Q(1, -5), R(9, 1), with P and Q diagonally opposite. What are the coordinates of S.

Ok, so does "diagonally opposite" mean that P and Q are adjacent to R and S, but not to one another?

[color=cyan]It sounds like it, although in that case the parallelogram should have been labeled PRQS . . . but I think you should assume they are non-adjacent.[/color]

Ok, but in the answer, it says that vector PQ = vector SR. How can this be if the are opposite? Shouldn't it then be vector PR = vector QS?

Yeah. Is that all you want, or do you want help with the problem?

Well, I had some trouble understand the solution in the text. I don't get why vector PQ = vector SR. if PQ and Sr are diagonals of the parallelogram.

er..yeah. It's a long story: I had seen your problem, planned to post my reply which you see above, started working on another problem, got back and posted the reply without realizing you guys had posted more, tried to delete my post, learned I could not delete posts that had been replied to....so here I am. Just forget all about me.

Hmm, i don't undestand y they are equal either...
perhaps dot-product?

The problem statement is just plain wrong. They mean to state that P and Q are adjacent.

O well, in that case, it makes sense then.

Yeah, everything makes sense now. Thanks.


complex numbers
geometry proposed
geometry

[quote="Megus"]Dissect a circle into $2n$ congruent parts by $n$ diameters. Choose an arbitrary point $M$ inside a circle. Prove that orthogonal projections of $M$ onto these diameters form a regular $n$-gon.

P.S. It is easy to prove it with complex numbers but is it possible to prove it without them?[/quote]

Let's mark this projections as $H_{1},..., H_{n}$ (Numeration is in the correct order -
the next projection lies on the next diameter)
Let the point O be the centre of the circle, and S is the middle of segment OM.
Then the projections lie on a circle with diamerter OM and centre S.
Then $\angle H_{i}SH_{i+1}=2\angle H_{i}OH_{i+1}=2\frac{2\pi}{2n}=\frac{2\pi}{n}$
So the polygon is regular.
V:)


Fie triunghiul $ABC$ si $O$ centrul cercului circumscris triunghiului $ABC$.Fie punctele $A',B',C'$ apartinand planului $(ABC)$ astfel incat $OA'BC$,$OB'CA$ si $OC'AB$ sa fie paralelograme.Demonstrati ca dreptele $AA'$,$BB'$,$CC'$ sunt concurente.

da asa e ..da e o pb f usoara.chiar si o solutie sintetica e f simpla.imi cer scuze pt aceasta pb

hm...
atunci eu cred ca paralelogramele sunt $OBA'C$, $OCB'A$, $OAC'B$. in cazul asta, $O$ poate sa fie orice punct si...
[hide="hint"]
segmentele respective se injumatatesc
[/hide]

[quote="Svejk"][color=darkred]([b][u]Enunt modificat[/u][/b]) Fie centrul $O$ al cercului circumscris triunghiului $ABC$ si punctele $A',B',C'$ apartinand planului $(ABC)$ astfel incat patrulaterele $OBA'C$,$OCB'A$, $OAC'B$ sa fie paralelograme. Demonstrati ca dreptele $AA'$,$BB'$,$CC'$ sunt concurente.[/color][/quote]

[color=darkblue][b]Metoda 1 (sintetica).[/b] Notam ortocentrul $H$ al triunghiului dat. Se observa ca cele trei paralelograme sunt romburi. Deoarece $OA'\parallel AH$ si $OA'=AH=2R|\cos A|$ etc rezulta ca patrulaterele $OA'HA$, $OB'HB$, $OC'HC$ sunt paralelograme cu diagonala comuna $OH$. In concluzie, dreptele $AA'$, $BB'$, $CC'$ sunt concurente in centrul $E$ al cercului lui Euler asociat triunghiului $ABC$ si punctul $E$ este mijlocul segmentelor $[AA']$, $[BB']$, $[CC']$.

[b]Metoda 2 (vectoriala).[/b] Alegem punctul $O$ ca centru al sistemului de vectori

si notam centrul de greutate $G$ al triunghiului $ABC$. Asadar,

$\left\{\begin{array}{c}A'=B+C\\\ B'=C+A\\\ C'=A+B\end{array}\right\|\mathrm{\ si\ }AA'\cap BB'\cap CC'\ne \emptyset\Longleftrightarrow$

exista $P\in AA'\cap BB'\cap CC'$ si $\{x,y,z\}\subset \mathrm R$ asfel incat

$P=xA'+(1-x)A=$ $yB'+(1-y)B=zC'+(1-z)C\Longleftrightarrow$

$P=(1-x)A+xB+xC=yA+(1-y)B+yC=zA+zB+(1-z)C\Longleftrightarrow$

$\left\{\begin{array}{c}P=(1-x)A+xB+xC\\\ 1-x=y=z\\\ x=1-y=z\\\ x=y=1-z\end{array}\right\|\Longleftrightarrow\left\{\begin{array}{c}P=\frac{1}{2}\cdot (A+B+C)\\\ x=y=z=\frac{1}{2}\end{array}{\right\|\Longleftrightarrow}$

$2\cdot P=3\cdot G\Longleftrightarrow 2\cdot \overline{OP}=3\cdot\overline{OG}\Longleftrightarrow 2\cdot\overline{OP}=\overline{OH}=2\cdot\overline{OE}\Longleftrightarrow P\equiv E$.[/color]


Hope the mods allow this topic here..

Let us use this space  to  reoprt some fun/unusual  items regarding college life:

Here is first post from me:
.  From Duke Chronicle ..

Now as a parent of two college students, it is reassruing that our money is put to good use: :? :?
[url=http://www.chronicle.duke.edu/vnews/display.v/ART/2005/03/04/42284d012585f]Rodents, insects plague GA dorm[/url]
[quote]In the past seven months, the East Campus dormitory has endured invasions of mice, cockroaches, ants and, most recently, rats...
...the girls watched in terror as a mouse ran under their desk, and they quickly alerted their residential adviser and neighbors of the problem. “We heard that mice run in packs of six, so we were really scared,?...[/quote]

Of  course my other son is at MIT where miceis so common that they don't even make news  : :(  :(

Well, that doesn't sound nice. I live in the older building of GATech Campus Housing, and there are no such problems. As the people from the housing department know that during the late fall and the winter, there may be some mice, they place mice poison, and I can tell you that I have seen just once a mouse on the building (in almost two years living here). Also, I have never seen cockroaches, even when I know that does not mean that there are none.

C'mon, if that almost doesn't happen here, and we are not Ivy Leage, why does it happen in Duke or MIT!

Best,

Yea I know.. as I have told my son, we did not have any mice problem either ..Not since  we had our own pet snake...:):)

You know, I like snakes a lot, but my wife would ask for the divorse if I carry one to home  :D .

Best,

[quote="Gyan"]... Of  course my other son is at MIT where miceis so common that they don't even make news  : :(  :([/quote] One strike against MIT when my son is ready for college  :D  Wait, wait, wait ... he says one bonus for, not one strike against  :blush:

Back when I was an undergraduate at Rice in the early 1970's, the campus trees were the winter roosting grounds for about half a million brown-headed cowbirds. There were certain sidewalks passing under trees that you didn't want to walk on. Also, on any given night, not all of a half million birds are going to live through the night. Those rumors about the food services going out and collecting them - rumors, of course.

Speaking of birds wasn't one famous hack by MIT  envolved  "training"   hundreds of birds to "land in unison"  at the sound of a whistle (from a striped shirt person) ... Imagine Football umpire's surprise  in the  Harvard-Yale game..:)

Sorry for x-post but I liked this video produced by MIT  Indian students a lot. Looks like they are not that homesick and learned to have fun too in MIT. Nice to see MIT is    home to so many foreign students who  are having fun.  Enjoy
(The title song means  "May we  sing in tune ...."  I liked it a lot. Enjoy... (See how many languages can you identify..?)

Get videos directly
Small (10 MB): http://www.mit.edu/~kapoor/mile_sur_small.wmv
Big(50 MB): http://web.mit.edu/anoop/www/msur/milesur50.mpg

Yesterday was pi day .. so MIT had a  wooden Pie on  its dome and a  huge banner with lots of pi digits...streched across .....

Any parent here whose son complained about LAN jack in student dorm? (It turns out that the what one thought was  LAN jack turned out  to be a mouse hole)

Yeah, but the postings are several days apart.

Thought this might be the right time to tell those California AoPSers heading east for college …

Light years ago I came from the Orient for college in the east coast. One freezing, windy January day I was walking back to the dorm from class when I came to a large parking lot. I could either walk along two nicely plowed sidewalks, or I could walk diagonally through the parking lot. The parking lot was covered with the beautiful white stuff. I decided to save a few wind-chilled steps. Good math decision on a fine day, but that was not a fine day.

By the time I got to the middle of the parking lot, snow was up to my neck! The best I could say then was that my body was no longer under the attack of the howling wind. Should I trace my way back, or risked drowning in snow? Not really sure that I could trace my way back correctly, I went forward (perhaps still wanting to save those few steps).

I lived to tell the story, and you don’t have to wonder why I am NOT moving back to the east coast.

[quote="Gyan"]... MIT ...  MIT ... MIT ...[/quote] This one's for you, Gyan  :D

http://www.caltechvsmit.com/

Humane traps that release rats! At least people are finally civil to rats!

[quote="MathChauffeur"]Thought this might be the right time to tell those California AoPSers heading east for college …

Light years ago I came from the Orient for college in the east coast. One freezing, windy January day I was walking back .....[/quote]

From today's [url=http://www.boston.com/news/science/articles/2006/11/05/neuroscientist_backs_recruiter/]news[/url]:
[b]WINTER 101: [/b]
[quote]We did a double take last week when we saw this on Harvard Law School's calendar: "What to Wear in Winter Climates." The seminar, requested by a student from Southern California, drew 62 participants, including students from Iran and India. Hot cider, hot chocolate, and snowflake cookies were served. It seems that Dean Elena Kagan's administration is taking customer service to new heights.[/quote]
Look at the end  of this item: [url=http://www.boston.com/news/science/articles/2006/11/05/neuroscientist_backs_recruiter/]Link[/url]

[quote="Gyan"]... Harvard Law School's calendar: "What to Wear in Winter Climates." The seminar, requested by a student from Southern California, drew 62 ... administration is taking customer service to new heights.[/quote]
Good piece!  :lol:

I wonder, for those students who have gone to sunny California for college from cold weather, what do they think about going back after college?

Educational institute or big business? That is the question!  :D

[quote="Gyan"][quote="MathChauffeur"] This one's for you, Gyan  :D
http://www.caltechvsmit.com/[/quote]
Did you also see: http://www.mitcannon.com/
Old news, but I am sure, you have read this
[url]http://www.boston.com/news/local/articles/2006/04/07/cannon_is_fodder_in_mit_caltech_rivalry/[/url]
:) :)[/quote] Yes, indeed  :D . And I am glad that Harvey Mudd (which to me is the 3rd Tech) is in the news because of it.

Seems like the page http://www.caltechvsmit.com is no longer valid. For those who are curious about one of CalTech's more creative prank sequence recently, read http://slashdot.org/articles/05/04/11/0010245.shtml
and see the attached picture!


Prove that the positive roots of equation \[x(x+1)(x+2)...(x+1981)=1\]is less than $\frac{1}{1981!}$

Are you talking about each single root or the sum or product of them?

[hide]
So pretend we were to multiply out the LHS.  We would get:

\[x^{1982}+\cdots+1981!x=1\]

with all the terms on the left being positive.  Now, if we plug $\frac{1}{1981!}$ we see that we have some positive number plus $1981!(\frac{1}{1981!}.$  So the LHS is greater than one when $x=\frac{1}{1981!}$ and when $x$ increases so does the LHS so the roots must be less than $\frac{1}{1981!}$. [/hide]


simplify the expression
71. cube root of 2/3
77. 1/squareroot of a + squareroot of a.

If you could please explain how you arrived at the answers it would be a great help.

[hide="77"]
I will let $ A\equal{}\sqrt{A}$ for simplicity.

$ \frac{1}{\sqrt{a}} \plus{} \sqrt{a}$

$ \frac{1}{A} \plus{} A$

$ \frac{A^2\plus{}1}{A}$

$ \frac{a\plus{}1}{\sqrt{a}}$

Or if you want to rationalize the denominator:

$ \frac{\sqrt{a^3}\plus{}\sqrt{a}}{a}$[/hide]

this is sounds strangly home work-like... :wacko:

(cube root of 18)/3

(squareroot of a)+((squareroot of a)/a)

this should be explained in almost every high school textbook...

Cube root of 2/3

First of all, you can simplify it to cube root of 2 over cube root of three.  Multiply both sides by cube root of three*cube root of three.  Now, you have cube root of 2*cube root of 3*cube root of three/three, which comes to cube root of 18/3:
looks like:

(cbrt. 18)/3

1 over square root of a plus square root of A

this is hard to explain... 1/sqrt a * sqrt a/sqrt a = sqrt a/a...

answer: sqrt a/a + sqrt a

easy as 3.1415926...
 :whistling:


The "obvious" roots are $2 \pi n$ for $n \in \bf Z$.  However, there are also roots $2\pi n \pm d$ where $d$ is a real number approximately equal to 0.514703759....  I can't figure out a "nice" form for this number, unfortunately.  It is possible to do away with the trig altogether by making the substitution $x = \arcsin y$, but the result doesn't seem any easier to handle than the original.

It seems that a good first manipulation of the left hand side would be

\[ (\frac{1}{2})^{2\sin^2{x}}+\frac{1}{2}=\frac{1}{2}((\frac{1}{2})^{2\sin^2{x}-1}+1)=\frac{1}{2}(2^{\cos{2x}}+1)  \]

It would be helpful to be able to express $4\cos^3{x}-1$ in terms of $\cos{2x}$.

[quote="K81o7"]
It would be helpful to be able to express $4\cos^3{x}-1$ in terms of $\cos{2x}$.[/quote]

Sorry, maybe, but complex.

Not "impossible," just "unattractive."  The path you are going down is not substantively different than my suggestion of taking $x \to \arcsin y$.


Find the expectations and variance of the products produced between two repairs of a machine. if the probability of defects is p and the machine is repaired if defects is k.

The question is very unclear.
Although I suspect that the number of interest is sum of $ k$ independent geometrically distributed (with an obvious answer in this case).


lim of expression have to be or not to be?
$\lim_{x\to 0}\frac{2xy}{x^{2}+y^{2}}$
$_{y\to 0}$

 :(

Strictly written definition of limit $\lim_{(a,b)\to (x,y)}f(a,b)$:
[color=blue][i]If for any sequence of points ${(a_{n},b_{n})}_{n=1}^{\infty}$ satisfying \[\lim_{n\to\infty}(a_{n},b_{n})=(x,y)\] the limit \[\lim_{n\to \infty}f(a_{n},b_{n})\] has the same value $g$, then we define \[\lim_{(a,b)\to (x,y)}f(a,b)=g.\] [/i][/color]
If $x_{n}=0$ and $y_{n}=\frac{1}{n}$, we have \[\frac{2x_{n}y_{n}}{x_{n}^{2}+y_{n}^{2}}=0,\] and for $x_{n}=y_{n}=\frac{1}{n}$: \[\frac{2x_{n}y_{n}}{x_{n}^{2}+y_{n}^{2}}= \frac{\frac{2}{n^{2}}}{\frac{2}{n^{2}}}=1,\] so the limit [b]doesn't exist[/b].

[quote="RohanSingh"]Substituting y=mx, and then taking limit x tends to zero will get you
$\frac{m}{1+m^{2}}$. Which is not uniqe. Hence limit doesn't exists.[/quote]
If I didn't miscalculated, it should be $\frac{2m}{1+m^{2}}$ instead of $\frac{m}{1+m^{2}}$ (of course coefficient $2$ isn't important and existing the limit doesn't depend on it). For $y=mx$, $m=\mathrm{const}$ we get \[\frac{2xy}{x^{2}+y^{2}}=\frac{2mx^{2}}{x^{2}+m^{2}x^{2}}= \boxed{\frac{2m}{1+m^{2}}},\] so you don't have to take the limit $x\to 0$; the expression is constant for constant $\frac{y}{x}$.

When (x,y) approach (0,0) you might try converting to polar coordinates.

x^2 + y^2 = r^2

x = r cos(theta)
y = r sin(theta)

so your limit becomes

lim (r->0)   [2(rcos(theta))(rsin(theta))/r^2]
which is equal to 2cos(theta)sin(theta), which depeds on our choice of theta.  So it doesn't exist.

[quote="Rzeszut"][quote="RohanSingh"]Sorry, I missed 2 in the expression.
Thanks Rez[/quote]
If you call me "Rez", use $\LaTeX$: $\Re\left( z \right)$ or $\mathrm{Re}\left( z \right)$.[/quote]

You are right Rzeszut. Anyway You can't be imaginary.


Given a positive prime $p$ and two positive integers $a,b$, prove that

$\binom{pa}{pb} \equiv \binom{a}{b} \mod{p^3}$.

The Olympiad Number Theory section is more appropriate for this.

Oh. OK. I didn't know how hard it was... I'll go post it over there.


Πολυωνυμο π(χ) εχει την ιδιοτητα π(χ)=π(χ-1) για καθε χ.
Δειξτε οτι ειναι σταθερο.

 :)

Αν θέσουμε στην δεδομένη σχέση όπου χ το χ-1 παίρνουμε
$p(x-1)=p(x-2)$ .Ara $p(x)=p(x-1)=p(x-2)$ .Όμοια μπορούμε να δείξουμε ότι
$p(x)=p(x-1)=.....=p(x-k)$  για κάθε κ πραγματικό .Άρα για δύο τυχαίους m,n πραγματικούς έχουμε $p(m)=p(n)$  άρα p σταθερό

Σιλουανε δεν μπορω να καταλαβω πως απο τους ακεραιους 1,2,3...και τα χ-1,χ-2,χ-3,..μεταβαινεις στο χ-κ οπου κ ειναι πραγματικος....
Socrates,μηπως πρεπει να δωσεις και αλλα στοιχεια οπως που ανηκει το χ,δηλαδη για καθε χ εΖ η για καθε χεR?Και επισης οι συντελεστες του Π(χ) ειναι πραγματικοι?Γιατι ετσι που το βλεπω μου φαινεται οτι το π(καθει πραγματικου αριθμου)=π(α) οπου α ειναι ενας πραγματικος αριθμος που ανηκει σε ενα διαστημα πλατους 1,δηλαδη π(1,7)=π(0,7)και  π(1345,7)=.......=π(1,7)=π(0,7) και επισης π(567,35)=....=π(0,35) και ετσι καθε το π(Χ) αναγεται παντα σε ενα π(α)οπου α ε(0,1),
ομως δεν μπορεις να αποδειξεις οτι ολα αυτο τα π(χ)ειναι ισα,δηλαδη οτι π(0,7)=π(0,35)=π(α) για καθε α ε(0,1).Δεν ξερω αν καταλαβαινετε αυτο που θελω να πω,αλλα νομιζω οτι κατι λειπει.Για παραδειγμα π(χ)={χ}(το κλασματικο μερος του χ) εχουμε π(χ)=π(χ-1) αλλα το π(χ) δεν ειναι σταθερο.

ΡΑΛΛΗΣ :D

έχεις δίκιο .Λάθος :blush:

Με πραγματικους συντελεστες και για καθε χ στο R!

 :)

Σωστες οι παρατηρησεις του Ραλλη. αλλα αν παρατηρησουμε οτι το π(χ) ως πολυονυμο ειναι μια συνεχεις συναρτηση και αφου η διαφορα π(χ)-π(χ-1) ειναι ο συντελεστης διευθυνσης του π(χ) (θεωρω ως αριθμητη το χ-(χ-1)) ο  οποιος απο την υποθεση ειναι ισος με 0.αρα αν παρεις 3οποιαδηποτε σημεια της συνεχους αυτης καμπυλης ειναι προφανες οτι το γραφημα της ειναι ευθεια γραμμη.δηλ π(χ)=σταθ. αλλα αυτο δεν ειναι και πολυ μαθηματικη αποδειξη :lol:

Παραθετω τη λυση του σιλουανου

[quote]Θεωρουμε το πολυωνυμο $Q(x)=P(x-1)-P(0)$ και παρατηρουμε οτι εχει βαθμο ν
αλλα ν+1 ριζες. οποτε ειναι σταθερο. Αμεση συνεπεια το Ρ ειναι και αυτο σταθερο! [/quote]

 :)

:coolspeak:  νομιζω οτι το ιδιο προκυπτει αν αντι για χ γραψεις αχ+β α,β Ε R. με την ιδια εξυπνη λυση.


If you have a globe, and you cut off the top and bottom of the globe at the two tropics (at 23.5 degrees north and south) what is the surface area not counting the flat circle on the top and bottom.

I was thinking about somehow unfolding the "earth" and calculate the area between two curves. I don't know if it works. But, if it gets too calculus-like, we can start a new thread in the calc fforum.

Thanks a lot in advance.

The formula for the surface area of  a spherical cap of height $ h$ for a sphere with radius $ r$ is given by $ S \equal{} 2\pi rh$. If you'd like to see the derivation (calculus), you can go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193544]here[/url].

You can take it from there.

edit: As for unfolding the earth, that would be a bit complicated as 3D surfaces don't really like being turned into 2D surfaces. This has a little bit to do with (so far as I know) cartographic projections and the different ways you can make a map. The above formula is derived using a 'surface of revolution formula' and thus contains no such error, albeit the earth is an oblate spheroid; not quite a sphere.

Nonetheless, an interesting idea and one that might very well be valid.


In how many ways can a pair of students be selected from a group of five students?


If a,b,c>0 and a+b+c=12.Find the max of $ ab^2c^3 $


integration calculus derivative function geometry rectangle conics

Find the max value for $ \int_0^1 f(x) dx$

#1
f'' < 0 on [0,1]
f'(0)=1 and f'(1) = 0
f(0)=1 and f(1)= 0

#2
f' < 0 on [0,1]
f(0)=1 and f(1) = 0

#3
f'' < 0 on [0,1]
f(0)=1 and f(1) = 1

I made the up... and I've been trying to think of the answer... what type of problem are these?

Situation 1 isn't possible (wherever the first derivative hits 0, the value of the function will be positive).  For number 2, the supremum is 1 (every curve is under the rectangle, and we can get arbitrarily close).  For 3, we can get arbitrarily large values (take, for example, the downward-facing parabolas passing through (0, 1) and (1, 1).

[quote="thefuturebird"]
#1
f'' < 0 on [0,1]
f'(0)=1 and f'(1) = 0
f(0)=1 and f(1)= 0

[/quote]

if $ f(0) \equal{} 1, f(1) \equal{} 0$ then $ \exists \xi \in [0,1]: \ f'(\xi) \equal{} \frac {f(1) \minus{} f(0)}{1 \minus{} 0} \equal{} \minus{} 1 < 0$. So $ \exists t \in [\xi,1]: \ f''(t) \equal{} \frac {f'(1) \minus{} f'(\xi)}{1 \minus{} \xi} \equal{} \frac {1}{1 \minus{} \xi} > 0$... contradiction


Let $ a,b$ and $ c$ be side lengths of a triangle , prove that $ \sum_{cyc}\frac {b \plus{} c}{a^2} \geq \frac {3}{2(a\plus{}b\plus{}c)}$

[quote="enndb0x"]Let $ a,b$ and $ c$ be side lengths of a triangle , prove that $ \sum_{cyc}\frac {b \plus{} c}{a^2} \geq \frac {3}{2}$[/quote]
It's wrong. Try $ a\equal{}b\equal{}c\rightarrow\plus{}\infty.$ :wink:

sorry , it is

$ \sum_{cyc}\frac{b\plus{}c}{a^2} >\frac{3}{2(a\plus{}b\plus{}c)}$

[quote="enndb0x"]Let $ a,b$ and $ c$ be side lengths of a triangle , prove that $ \sum_{cyc}\frac {b \plus{} c}{a^2} \geq \frac {3}{2(a \plus{} b \plus{} c)}$[/quote]
For all positive $ a,$ $ b$ and $ c$ we obtain:
$ \sum_{cyc}\frac {b \plus{} c}{a^2}\geq\frac{18}{a\plus{}b\plus{}c}>\frac {3}{2(a \plus{} b \plus{} c)}.$ :wink:

Well looks that was too weak arqady,can you post your full proof for the inequality.

my solution was :

$ \sum_{cyc}\frac{b\plus{}c}{a^2}  >\sum_{cyc}\frac{a}{(b\plus{}c)^2} \geq \frac{1}{3}(a\plus{}b\plus{}c)\sum_{cyc}\frac{1}{(a\plus{}b)^2 }>\frac{3}{2(a\plus{}b\plus{}c)}$

[quote="arqady"]
For all positive $ a,$ $ b$ and $ c$ we obtain:
$ \sum_{cyc}\frac {b \plus{} c}{a^2}\geq\frac {18}{a \plus{} b \plus{} c} .$ :wink:[/quote]

 Result from AM-GM

@enndb0x: You can just expand and use muirhead :) Or you can use cauchy-schwartz:
$ (\sum_{cyc}(b\plus{}c))(\sum_{cyc}\frac {b \plus{} c}{a^2}) \geq (\sum_{cyc}\frac {b \plus{} c}{a})^2 \ge 6^2 \equal{} 36$, which is equivalent to the desired.


Are given $ A \equal{} (x|x \equal{} 6n \plus{} 2)$ and $ B \equal{} (x|x \equal{} 6n \plus{} 3)$  n is natural .

a) Prove that  there is not any perfect square in A.
b) Prove tha there are infinitely many perfect squares in B.

[hide="trivial"]
a)2 is not a square mod 6
b)3 is a square mod six
[/hide]

what he means is that
i)when squares are divided by 6, the remainder is never 2
ii)when squares are divided by 6, the remainder can be 3

i.e. $ (3^2)(mod6)\equal{}3$

As no number gives remainder 2, there is no squares for i)
As there is infinite numbers that gives remainder 3, there is infinite number of squares for ii)

[quote="max1546"]what he means is that
i)when squares are divided by 6, the remainder is never 2
ii)when squares are divided by 6, the remainder can be 3

i.e. $ (3^2)(mod6) \equal{} 3$

As no number gives remainder 2, there is no squares for i)
As there is infinite numbers that gives remainder 3, there is infinite number of squares for ii)[/quote]
[hide="this may be another solution"]
For any $ x\in \mathbb {N} , x\equiv \left\{ 0,1,2,3,4,5\right\} \pmod 6$

$ \therefore x^2\equiv \left\{ 0,1,4,3\right\} \pmod 6$

$ A\cap \left\{ 0,1,4,3\right\}_{\pmod 6 } \equal{} \phi , B\subset  \left\{ 0,1,4,3\right\}_{\pmod 6}$

[/hide]


algebra
polynomial
calculus
ARML
calculus computations

How do you find a double tangent of a polynomial whose degree is 4 or more? That is, a line that is tangent in two places (sorry if that sounds contradictory).

It doesn't have to be calculus at all - this can be attacked with polynomial algebra.

You want $ P(x)\minus{}ax\minus{}b\equal{}(x\minus{}x_1)^2(x\minus{}x_2)^2Q(x).$

Sample problem: (ARML 2003, Individual #8):

The graph of $ f(x)\equal{}x^4\plus{}4x^3\minus{}16x^2\plus{}6x\minus{}5$ has a common tangent line at $ x\equal{}p$ and $ x\equal{}q.$ Compute the product $ pq.$


geometry
circumcircle
power of a point
radical axis
geometry proposed

Given triangle $ ABC$ and its circumcenter $ O$. Denote $ P$ a point such that the pedal triangle $ A'B'C'$ of $ P$ wrt $ \Delta ABC$ is the cevian triangle of point $ Q$ wrt $ \Delta ABC$. Prove that $ (APA'),(BPB')$ and $ (CPC')$ concur at a point which lies on $ OP$.

We only need to understand that the pedal triangle of $ P$ is also a cevian triangle of $ \Delta ABC$.

I'm changing the notations according to my solution.

[quote]Given triangle $\triangle ABC$ and its circumcenter $ O.$ Denote $ P$ a point such that the pedal triangle $\triangle XYZ$ of $ P$ WRT $ \triangle ABC$ is the cevian triangle of point $ Q$ WRT $ \triangle ABC.$ Prove that the circles $\odot (APX),$ $\odot(BPY)$ and $\odot(CPZ)$ concur at a point which lies on $ OP.$[/quote]
Let $ \triangle A'B'C'$ be the antipedal triangle of $ P$ with respect to $ \triangle ABC,$ $ A',B',C'$ againts $ A,B,C.$ I'ts clear that $ \triangle PAC' \sim \triangle PZB$ and $ \triangle PAB' \sim \triangle PYC,$ because of $ \angle PC'A \equal{} \angle PBA$ and $ \angle PB'A \equal{} \angle PCA.$  Thus

$ \frac {\overline{PA}}{\overline{PZ}} \equal{} \frac {\overline{AC'}}{\overline{BZ}} \ , \ \frac {\overline{PA}}{\overline{PY}} \equal{} \frac {\overline{AB'}}{\overline{CY}}  \Longrightarrow  \frac {\overline{AB'}}{\overline{AC'}} \equal{} \frac {\overline{PZ}}{\overline{PY}} \cdot \frac {\overline{CY}}{\overline{BZ}} \ \ (1)$

By cyclic exchange we have the following relations:

$ \frac {\overline{BC'}}{\overline{BA'}} \equal{} \frac {\overline{PX}}{\overline{PZ}} \cdot \frac {\overline{AZ}}{\overline{CX}} \ \ (2) \ \ \frac {\overline{CA'}}{\overline{CB'}} \equal{} \frac {\overline{PY}}{\overline{PX}} \cdot \frac {\overline{BX}}{\overline{AY}} \ \ (3)$

Multiplying $ (1),(2)$ and $ (3)$ together yields by Ceva's theorem that $ AA',BB',CC'$ concur.

Hence $ \triangle ABC$ and $ \triangle A'B'C'$ are perspective $\Longrightarrow$ $ A_0 \equiv BC \cap B'C',$ $B_0 \equiv CA \cap C'A'$ and $ C_0 \equiv AB \cap A'B'$ are collinear on their perspective axis $ \ell.$ Since the centers $ O_a,O_b,O_c$ of the circles $ \odot(PAX),$ $\odot(PBY),$ $\odot(PCZ)$ are the midpoints of $ PA_0,PB_0,PC_0,$ respectively, it follows that $ O_a,O_b,O_c$ are also collinear. Thus $ (O_a),(O_b),(O_c)$ are coaxal having the straight line $ \ell'$ orthogonal to $ \ell$ through $ P$ as their common radical axis.

On the other hand, note that $ \triangle ABC$ and $ \triangle A'B'C'$ are orthologic having $ P$ and the intersection $ P'$ of the perpendiculars from $ A',B',C'$ to $ BC,CA,AB$ as their two orthology center. $ P'$ is obviously the isogonal conjugate of $ P$ WRT $ \triangle A'B'C'.$ Then $ (O)$ becomes the pedal circle of $ (P,P')$ WRT $ \triangle A'B'C',$ whose center is the midpoint of $\overline{ PP'}$ $ \Longrightarrow$ $ O \in PP'.$ By Sondat's theorem the straight line $ PP'$ connecting both orthology centers is orthogonal to their perspective axis $ A_0B_0C_0.$ Hence $ PP'$ is identical to $ \ell'$ $ \Longrightarrow$ radical axis $ \ell'$ of $ (O_a),(O_b),(O_c)$ goes trhough $ O.$
_____________________________________________________________

Big remark from this configuration:

Let $ P_0$ be the second intersection of the circles $ (O_a),(O_b),(O_c)$ lying on $ \ell.$ Let the isogonal ray of $ AX$ WRT $ \angle BAC$ cut $ \ell'$ at $ Q'.$ Angle $ \omega$ between $ AO$ and $ AQ'$ equals the angle between their isogonal rays in $ \angle BAC,$ i.e. angle between the A-altitude $ AD$ and the ray $ AX$ equals $ \omega.$  Since $ AD \parallel PX$ and $ XPAA_0$ is cyclic, we get

$ \angle PXA \equal{} \angle PP_0A \equal{} \angle OAQ' \equal{} \omega \Longrightarrow \odot(AQ'P_0)$ is tangent to $ OA$ through $ A$

$ \Longrightarrow$ $ OA^2 \equal{} R^2 \equal{} OP_0\cdot OQ'.$ Thus, isogonals of $ BY,CZ$ cut $ \ell'$ at the same $ Q',$ the isogonal conjugate of $ Q$ WRT $ \triangle ABC$ and hence the inverse of $ P_0$ in $ (O)$ $ \Longrightarrow$ $ Q' \in \ell'.$
_____________________________________________________________
This is the proof of the supporting lemma that we needed to finish [url=http://www.artofproblemsolving.com/viewtopic.php?t=297530]this problem[/url]

Thanks you dear [b]Luis[/b] for your very nice solution  :lol:. I only have the solution which uses the lemma above and I will post it later.

Dear Linh, Luis and Mathlinkers,
revisiting your nice problem in order to make something clear in my mind, I observe that P is the pivot of ABC wrt A', B', C'.
It follows the equivalence which can be prove synthetically without any calculation :
1. AA', BB', CC' concur at Q
2. the circle AA'P, BB'C and CC'P are concurrent (at R).
Now, the top of your problem is to prove this equivalence
1. the pivot P is a PC-point
2. O is on PR
Sincerely
Jean-Louis

Dear Linh, Luis and Mathlinkers,
I have say in my last message that,
if, AA', BB', CC' are concurrent at Q
then, the circles AA'P, BB'C and CC'P are concurrent (at R).
To day with the interesting Luis's ideas, I have proved that
the isogonal of Q, the centre O of of ABC and R are collinear.
So the Linh's problem can be seen as a particular case...
I think that there is much more to say...
Sincerely
Jean-Louis

Dear Jean Louis, moreover R is the inverse of the isogonal conjugate Q' of Q in the circumcircle ABC (see my remark between bold lines) and yes, I agree with you, seems like there's more behind this configuration. Indeed, Darboux/Lucas cubics have many properties that may be studied synthetically, but could you explain why Linh's problem can be seen as a particular case?.

Dear Luis and Mathlinkers,
when Q becomes a PC-point...
Yes also for the inverse that I have seing in one of your message.
Sincerely
Jean-Louis

Dear Mathlinkers,
for a synthetic proof you can see
http://perso.orange.fr/jl.ayme vol. 6 Ménélaüs d'Alexandrie et La Marquis Giovani Ceva ; Approfonodissement d'un problème du Monthly de 1917, p. 23, 27.
Sincerely
Jean-Louis


The integers $ a_1,...,a_n$ give at least $ k\plus{}1$ different remainders when they are divided by $ n\plus{}k$. Prove that some of them add up to an integer divisible by $ n\plus{}k$.

[hide]
Assume to the contrary that no sum is divisible by $ n\plus{}k$.

Note that if given two different sums of sets of a elements such that one is a subset of the other, then they two sums must have distinct remainders. Otherwise, we could subtract them to get another subset whose sum is divisible by $ n\plus{}k$.

Lemma. Under this assumption, given $ k\plus{}1$ distinct residues mod $ n\plus{}k$, there are at least $ 2k\plus{}1$ distinct resides among the different possible sums of these $ m$ elements.

Proof. Let the $ k\plus{}1$ distinct residues be $ a_1,a_2,...,a_{k\plus{}1}$, and let $ S$ be the sum of all of them. Then consider the $ 2k\plus{}3$ sums

$ a_1$
$ a_2$
...
$ a_{k\plus{}1}$

and

$ a_2\plus{}a_3\plus{}\cdots \plus{} a_{k\plus{}1}\equal{}S\minus{}a_1$
$ a_1\plus{}a_3\plus{}\cdots \plus{} a_{k\plus{}1}\equal{}S\minus{}a_2$
...
$ a_1\plus{}a_2\plus{}\cdots \plus{} a_{k}\equal{}S\minus{}a_{k\plus{}1}$

and finally

$ a_1\plus{}a_2\plus{}\cdots \plus{} a_{k\plus{}1}\equal{}S$.

Now, all of these sums are subsets of the final one, so they are all distinct from it. Further, since all the $ a_i$ are distinct, we also have that any two of $ S\minus{}a_i$ and $ S\minus{}a_j$ are distinct if $ i$ and $ j$ are distinct.

This just leaves the comparison between $ a_i$ and $ S\minus{}a_j$. If $ a_i\ne a_j$, then $ a_i$ is a subset of $ S\minus{}a_j$, so they are distinct. Finally, consider the two sums $ a_i$ and $ S\minus{}a_i$. Suppose they have the same remainder when divided by $ n\plus{}k$. Then

$ a_i\equiv S\minus{}a_i\mod n\plus{}k$
$ 2a_i\equiv S\mod n\plus{}k$

But this modular equation has at most two solutions for $ a_i$ for a given $ S$ (and only one if $ n\plus{}k$ is odd, and none if $ n\plus{}k$ is even and $ S$ is odd). So, given the $ 2k\plus{}3$ sums listed, there are at most two cases where two of these sums are the same. Therefore, there are at least $ 2k\plus{}1$ distinct residues.

This completes the lemma.

Now, consider our sequence of $ n$ elements. Divide it into two sequences: $ a_1,a_2,...,a_{k\plus{}1}$ are distinct elements and $ b_1,b_2,...,b_{n\minus{}k\minus{}1}$ are the remaining elements. Now, take the $ 2k\plus{}1$ distinct sums from $ a_1,a_2,...,a_{k\plus{}1}$ that we know to be distinct, and then $ n\minus{}k\minus{}1$ more sums:

$ (a_1\plus{}a_2\plus{}...\plus{}a_{k\plus{}1})\plus{}b_1$
$ (a_1\plus{}a_2\plus{}...\plus{}a_{k\plus{}1})\plus{}b_1\plus{}b_2$
...
$ (a_1\plus{}a_2\plus{}...\plus{}a_{k\plus{}1})\plus{}b_1\plus{}b_2\plus{}...\plus{}b_{n\minus{}k\minus{}1}$

Now, this gives us $ (2k\plus{}1)\plus{}(n\minus{}k\minus{}1)\equal{}n\plus{}k$ total sums. We know that any two of the first $ 2k\plus{}1$ sums are distinct. If given two of the $ n\minus{}k\minus{}1$ sums, one is a subset of the other, so those subsets are distinct. The same applies if you have one of the $ 2k\plus{}1$ sums and one of the $ n\minus{}k\minus{}1$ sums. Thus, considering these $ n\plus{}k$ sums, they all have distinct residues, so one must be divisible by $ n\plus{}k$.[/hide]


linear algebra
matrix
induction
algebra
polynomial

(a) Show that the Cayley-Hamilton theorem is true for all $ 2 \times 2$ matrices.

(b) Let
\[ \mathsf{A} \equal{} \begin{pmatrix}3 & 4 \\ \minus{}1 & \minus{}1 \end{pmatrix}.\]Find the characteristic equation for $ \mathsf{A}$. Verify that $ \mathsf{A}^2 \equal{} 2\mathsf{A} \minus{} \mathsf{I}$. ([b]I can do this much of part (b)[/b])

Is $ \mathsf{A}$ diagonalisable?

Show by induction that $ \mathsf{A}^n$ lies in the two-dimensional subspace (of the space of $ 2 \times 2$ real matrices) spanned by $ \mathsf{A}$ and $ \mathsf{I}$, i.e. show that there exists real numbers $ \alpha_n$ and $ \beta_n$ such that
\[ \mathsf{A}^n \equal{} \alpha_n \mathsf{A} \plus{} \beta_n \mathsf{I}.\]Find a recurrence relation (i.e. a difference equation) for $ \alpha_n$ and $ \beta_n$, and hence find an explicit formula for $ \mathsf{A}^n$

A is not diagonalizable, since both eigenvalues are equal. In fact, if $ A \in M_{2 \times 2} (\mathbb{R}), A \neq \alpha I_2, \alpha \in \mathbb{R}$, has both eigenvalues equal, then A is not diagonalizable. Proof: Suppose there exists invertible $ P$ such that $ A \equal{} PDP^{ \minus{} 1}$, with $ D$ diagonal. But on this case, we would have $ D \equal{} \alpha I_2$, and $ A \equal{} P(\alpha I_2)P^{ \minus{} 1} \equal{} \alpha I_2$, contrary to the hypothesis. $ \Diamond$

Since A is of order 2, its characteristic polynomial will look like $ c(x) \equal{} x^2\minus{}ux \plus{} v$. Now let p(x) be an arbitrary polynomial, with degree $ \geq 2$. If q(x), r(x) are the quotient and remainder on division of p(x) by c(x), then p(x) = q(x) c(x) + r(x), with degree[r(x)] $ \leq 1$. Therefore, Cayley-Hamilton theorem implies that $ p(A) \equal{} r(A)$, and of course that we have $ r(x) \equal{} \alpha x \plus{} \beta$. In your case, $ p(x) \equal{} x^n$. I will leave the rest to you.

I understand the Cayley-Hamilton theorem to be "every square matrix satisfies its own characteristic equation" but I don't know how to apply it.

[quote="Carcul"]Now let p(x) be an arbitrary polynomial, with degree $ \geq 2$. If q(x), r(x) are the quotient and remainder on division of p(x) by c(x), then p(x) = q(x) c(x) + r(x), with degree[r(x)] $ \leq 1$.[/quote]
Why do we want to come up with this arbitrary polynomial $ p(x)$ and what is the connection to $ r(x)$?


Euler
inequalities unsolved
inequalities

Let ABC be a triangle inscribed in a circle of radius R

 and a, b, c be respectly sides of the triangle ABC.

 Prove that:

 1/ab + 1/bc + 1/ca >= 1/R^2.

 (HINT: Euler Ineq.   R>=2r
   r: inradio of the triangle ABC)

It's already on the forum: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15798[/url].

It's even twice on the forum: http://www.mathlinks.ro/Forum/viewtopic.php?t=15805


AMC
USA(J)MO
USAMO
summer program
PROMYS
LaTeX
geometry

[I know that there are a few forums about this, but I wasn't sure if I was supposed to piggyback on one of those and post this or not..so I posted a new topic]
So I'm thinking of applying to RSI this year, except I'm not too sure if I'm "qualified" enough. any opinions would be great!
I did absolutely horrible on the PSAT (209, ouch). Do I have to send that score in? Or could I just send my ACT (I got a 35 composite)?
Also, my gpa is not awesome, probably somewhere around a 4.2-4.3? How much is that going to hurt me?
On the good side, I have done research during the summer at Northwestern University, gone to math camp, and am very active in math team and science clubs at my school.

The only way to make sure you don't get in is by not applying.

I believe you are required to put your PSAT score on the application, but with your ACT score there also I'm sure they won't hold it against you. GPA is all over the place on different scales; if you don't have strong grades, that obviously doesn't help. Having rigorous classes on your transcript (i.e. AP instead of regular) is important, as are all of the other things you mentioned that you have done.

Indeed, you should apply, regardless of whether you think you will get in or not. If you make it, great, but even if you don't it's a pretty good precursor to college applications (if you really want to find some other "worthwhile" reason...).

[quote="juju1414"]I have done research[/quote]

Great!  You should apply :)

If you apply, you have a 6% chance of getting in.  If you don't apply, you have a 0% chance of getting in.

Which is better, you think? :)

[quote="iPod Nano"]If you apply, you have a 6% chance of getting in.[/quote]
That is, if the admissions process is completely random - which I hope is not the case.
Anyways, juju, I think your extra-curriculars will help a lot. Good luck!

Yeah, I'm still working on my application now.
..we'll see how things turn out.
Thanks for the encouragement guys!

I know this is one year old, but I am applying this year with limited/no (we'll say no) research experience.

Will this adversely affect my chances and possibly bring it down to like 0?

Thanks

Not that it wouldn't help, but that's certainly not a prerequisite.  Most of my friends in RSI '07 (myself included) were there with no prior research experience.

How does a regular aopser (say USAMO qualifier + various math awards but not MOP) and who takes BC calc in 10th grade and will take a year of university classess before RSI do any real research? I mean what could I think of with knowing just linear alg + NT (promys) + multivar that a real mathematician could'nt think of? I am a fairly accomplished chess and squash player (nationally ranked in both) as well as poet. Would this help my chances of getting in? Could I do anything real?

That pretty much describes my qualifications before RSI '07 (minus multivariable!).  You don't need a huge amount of background to do research in certain fields (with the possible exception of the field I attempted); a willingness to learn and explore interesting ideas is much more important.

If you're not familiar with the details, RSI assigns you a mentor who will play a pivotal role in your research.  I'm not suggesting that your mentor will do your project for you, but having a mentor to guide your ideas is invaluable.

Well im still a tenth grader ; and i just want to do PROMYS this summer. I know NT decently; ive taken intermediate NT and done V2. Also im taking NT at PuMaC div A in a month. Any advice for getting in to PROMYS? Thanks alot youv'e given me some hope  :)

[quote="t0rajir0u"]$ \text{\LaTeX}$ the problems and turn them in on time?  Also, do them correctly.  :)  Those were my three biggest mistakes the first time.

If you haven't gone yet, you may not be aware of how rigorous the program is there.  I strongly advise you to either take a second-year class or do a project and to attend the counselor's mini-lectures at night to get the most out of the experience; it'll prepare you for a future in mathematics better than you could think.  (That last bit of advice is mostly because in my year, Dustin Clausen gave a mini-lecture on algebraic geometry that would've helped me on my RSI project considerably had I attended :oops:)[/quote]

Why $ \text{\LaTeX}$ them? u went more than once?

The first time I applied to PROMYS I wrote the answers to the application test on paper (with pretty terrible handwriting), didn't do the problems very well, and turned them in a few days before the deadline, so one or more of those things got me rejected.  And the second time I did none of those things :)

so anyways back to RSI..any suggestions for looking at material for sub-field? Cause I picked a subfield, now I don't have any previous info on it..should I just look it up online and talk about some problems? Would this work? I really want to get in..but I know I will get rejected with the 4% acceptance rate and all..and also I don't have any previous research/USAMO/nationally ranked stuff..

thats what i did with the subfields, although i did not go into detail..i dont think you were (besides most people arent going to be super familar with their topics yet, since thats what RSI is for)

Also couple more questions.

I think announcement for chem olympiad is sometime in march.  we have to turn in rsi forms by mid-jan, but rsi doesnt return them until late march.  lets say if we make the chem olympiad, can we tell them is through email or something?

same for lets say getting perfect on amc 12 or something?

Also for teacher recommendations, do you usually give them a brag sheet or something?
Because for one of the questions, its like "how has the student pursued independent learning"...i dont really think my teachers would know what i have done outside of school...so am i suppose to tell them this so they can write it in their essays?

Thanks

^hey stanford, dont worry, alot of people are like you.  infact, i didnt even make the qualifications for the 220 psat recommendation (dont know if they let me sub in sat score though)...and the only major award i have is usamo, with no research.
your heart cant be measured with numbers...

It wouldn't hurt, but I don't know if it'll affect the decision process.

That's not what your teachers are there for; the recommendations, like the short answer questions, are to evaluate your ability to do [i]future[/i] research.  Personally, I would advise those of you applying to worry more about the short answers than about scores.  A lot of maturity and intellectual depth can show through there.

Just a quick question for all those who have applied to the RSI in the past or are planning to apply to this year's 2009 program:

Do you think we should list out our math awards in a resume format or would it hurt just to mention them as a collective whole (for instance, consistently scoring in the top ten individual places)? 3 pages is a pretty short page limit for all that we need to talk about..

Regional awards aren't really worth talking about except generally and you shouldn't have that many national-level awards (although if you do, feel free to put them all down :P ).

haha thanks t0rajir0u, that makes sense.

Now to figure out what I've done that can possibly transcend that 5% acceptance rate...  :|

this is so sad that i'm actually laughing at my own situation.


calculus
integration
absolute value
real analysis
real analysis unsolved

[b]Hello guys :

find the following integral

$ \int\limits_0^\lambda {\left[ {2 \minus{} \frac {{2(x \minus{} m)}}{{\left| {x \minus{} m} \right| \plus{} m}}} \right]} dx{\rm{ ; 0}}\langle {\rm{m}} \le \lambda$[/b]

note that [   ] means greatest integer number . and |    | absolute value fn
Good luck

Let $ I \equal{} \int_{0}^{\lambda}{\left[{2 \minus{} \frac {2(x \minus{} m)}{\left|{x \minus{} m}\right| \plus{} m}}\right]dx}$
$ I \equal{} \int_{0}^{m}{\left[{2 \minus{} \frac {2(x \minus{} m)}{2m \minus{} x}}\right]dx} \plus{} \int_{m}^{\lambda}{\left[{2 \minus{} \frac {2(x \minus{} m)}{x}}\right]dx}$
$ I \equal{} \int_{0}^{m}{\left[{2 \plus{} \frac {2m \minus{} x \minus{} x}{2m \minus{} x}}\right]dx} \plus{} \int_{m}^{\lambda}{\left[{\frac {2m}{x}}\right]dx}$
$ I \equal{} \int_{0}^{m}{\left[{4 \minus{} \frac {2m}{2m \minus{} x}}\right]dx} \plus{} \int_{m}^{\lambda}{\left[{\frac {2m}{x}}\right]dx}$
Now put $ x \equal{} m\cdot a$. We will have:
$ dx \equal{} m\cdot da$
$ I \equal{} \int_{0}^{1}{\left[{4 \minus{} \frac {2m}{2m \minus{} m\cdot a}}\right]m\cdot da} \plus{} \int_{1}^{\frac {\lambda}{m}}{\left[{\frac {2m}{m\cdot a}}\right]m\cdot da}$
$ I \equal{} m\cdot (\int_{0}^{1}{\left[{4 \minus{} \frac {2}{2 \minus{} a}}\right]da} \plus{} \int_{1}^{\frac {\lambda}{m}}{\left[{\frac {2}{a}}\right]da})$
$ \int_{0}^{1}{\left[{4 \minus{} \frac {2}{2 \minus{} a}}\right]da} \equal{} 2$
$ \int_{1}^{\frac {\lambda}{m}}{\left[{\frac {2}{a}}\right]da} \equal{} 1$, if $ \frac {\lambda}{m}\geq 2$ and $ \frac {\lambda}{m}\minus{}1$ if $ \frac {\lambda}{m} < 2$
So the answer is:
For $ m\leq \frac {\lambda}{2}$, $ I \equal{} 3m$.
For $ m > \frac {\lambda}{2}$, $ I \equal{} m \plus{} \lambda$.

[b]Yeeeeeeeess :clap2:  :10:

You have a nice mind in Math.

Nice work dude.

:D[/b]

:surf:  :D  Thanks, great problem  :wink:

[quote="Yuriy Solovyov"]$ \int_{0}^{1}{\left[{4 \minus{} \frac {2}{2 \minus{} a}}\right]da} \equal{} 2$
$ \int_{1}^{\frac {\lambda}{m}}{\left[{\frac {2}{a}}\right]da} \equal{} 1$, if $ \frac {\lambda}{m}\geq 2$ and $ \frac {\lambda}{m}$ if $ \frac {\lambda}{m} < 2$
So the answer is:
For $ m\leq \frac {\lambda}{2}$, $ I \equal{} 3m$.
For $ m > \frac {\lambda}{2}$, $ I \equal{} 2m \plus{} \lambda$.[/quote]
Can someone explain why
$ \int_{0}^{1}{\left[{4 \minus{} \frac {2}{2 \minus{} a}}\right]da} \equal{} 2$
$ \int_{1}^{\frac {\lambda}{m}}{\left[{\frac {2}{a}}\right]da} \equal{} 1$, if $ \frac {\lambda}{m}\geq 2$ and $ \frac {\lambda}{m}$ if $ \frac {\lambda}{m} < 2$
Thanks a lot!!!

[quote="soruz"]Nice problem!!!
Can someone explain why
$ \int_{0}^{1}{\left[{4 \minus{} \frac {2}{2 \minus{} a}}\right]da} \equal{} 2$
and $ \int_{1}^{\frac {\lambda}{m}}{\left[{\frac {2}{a}}\right]da} \equal{} 1$.
Thanks a lot!!![/quote]

Let $ f(a) \equal{} 4 \minus{} \frac {2}{2 \minus{} a}$.

$ 2 < f(a) < 3$ if $ 0 < a < 1$ and $ f(0) \equal{} 2$, $ f(1) \equal{} 3$ and only in this points.

That`s why $ \left[f(a)\right] \equal{} 2$ if $ 0 < a < 1$.

$ \int_{0}^{1}{\left[f(a)\right]da \equal{} 2\cdot (1 \minus{} 0) \equal{} 2}$.

The second: $ g(a) \equal{} \frac {2}{a}$.

If $ \frac {\lambda}{m} < 2$ we have:

$ \left[g(a)\right] \equal{} 1$ ($ 1 < \frac {2}{a} < 2$) and $ \int_{1}^{\frac {\lambda}{m}}{\left[g(a)\right]}da \equal{} 1\cdot (\frac {\lambda}{m} \minus{} 1)$

and

If $ \frac {\lambda}{m} \geq 2$ we have $ \left[g(a)\right] \equal{} 0$ for $ a > 2$.

$ \int_{1}^{\frac {\lambda}{m}}{\left[g(a)\right]}da \equal{} \int_{1}^{2}{\left[g(a)\right]}da \plus{} \int_{2}^{\frac {\lambda}{m}}{\left[g(a)\right]}da \equal{} 1\cdot (2\minus{}1) \plus{} 0\cdot (\frac{\lambda}{m}\minus{}2) \equal{} 1$


In a triangle ABC,  <BAC = 120. Let D, E, F be the intersection of the bisectors with the opposite sides.
Show that the circle with diameter EF lies in D (in others words prove that <EDF = 90)

A cant solve it

maybe you can help

I like this problem very much. Please see here: http://www.mathlinks.ro/Forum/viewtopic.php?t=114998  :lol:


can anyone give me some basic trigonometry formulas?

Sure :
http://www.sosmath.com/trig/prodform/prodform.html
http://mathforum.org/dr.math/faq/formulas/faq.trig.html
http://mathworld.wolfram.com/topics/TrigonometricIdentities.html

 :)

$sin= \frac{opposite}{hypotenuse}$
$cos= \frac{adjacent}{hypotenuse}$
$tan= \frac{opposite}{adjacent}$
$sec= \frac{hypotenuse}{opposite}$
$csc= \frac{hypotenuse}{adjacent}$
$cot= \frac{adjacent}{opposite}$

I'm only here cuz it was on the mc forum.

sine is $\frac{opposite}{hypotenuse}$
cosine is $\frac{adjacent}{hypotenuse}$
tangent is $\frac{opposite}{adjacent}$
cotangent is $\frac{adjacent}{opposite}$
secant is $\frac{hypotenuse}{adjacent}$
cosecant is $\frac{hypotenuse}{opposite}$

easy way to remember:

sine vs. cosine
both over hypotenuse
s closer to o and c closer to a

tan vs. cot
which on top
same, t closer to o, c closer to a

secant vs. cosecant
look at sines and cosines
there is an s-c combination

:spam:
I just posted the correct list 5 minutes ago.

On the lines of trig:
What is the function for things like sine, cosine, tangent, etc? For example, how could you figure out, lets say, sin(83) without a calculator or trig table?

Learn calculus and sum an infinite series by hand.

Good way to memorize sin, cos, and tan:

[b]O[/b]scar [b]h[/b]ad [b]a[/b] [b]h[/b]eap [b]o[/b]f [b]a[/b]pples.
(o=opposite, h=hypotenuse, a=adjacent). It is in the order of sin, cos, tan.

co[i]tangent[/i] (cot) is very easy to memorize; it is the reciprocal of tangent (tan).

csc and sec... I don't know any good tricks. I'm not sure whether X_wing1redleader is right.

Wait a second... there's a disagreement on sec and csc between the first and second replies.

It's easy to memorize - just as tangent and cotangent are mutually reciprocal and ONE of them begins in "co", in the same manner [b][u]co[/u][/b]secant is the reciprocal of [u]sine[/u] and [u]secant[/u] is the reciprocal of [b][u]co[/u][/b]sine.

[quote="SnowStorm"]Yea, ud have to sum something like:
$83 - \frac{83^3}{3!} + \frac{83^5}{5!} - \frac{83^7}{7!} + \dots$

You could get a pretty good approximation though using the first 3 terms.[/quote]

now that does not look right does it...you have to use radians when using the taylor series

[quote="Apprentice"]Just think SohCahToa
It's the name of a Native American tribe chief   ;)[/quote]

Yes!!!!! SOH CAH TOA

SOH=Some Old Heavy
CAH=Cows Are Heavier
TOA=Than Others Are

Some Old Heavy Cows Are Heavier Than Others Are.

I always remember it this way. ;)

Can't you evaluate the trig functions using the half and double and sum and difference formulas for sin and cos?

4(sinx)^3 - 3sinx + sin(3x)=0
I think that is a correct formula. You can find the sine of 1 degree with it.

[quote="0714446459923"]If you had way too much time on your hands, you could find the sin of 83 degrees using the half angle formula and the angle addition formula.[/quote]
I don't think you can... You can find $\sin{84^\circ}$ and $\cos{84^\circ}$ though.


geometry
integration
function
algebra
domain
calculus
calculus computations

Find the area between the curves $ \frac{4x^{2}}{\pi^{2}}$ and $ \sin(x)$ from the origin to the first intersection point of the two functions.

I'm posting this question mostly to see how one would go about finding the area between two curves before it's taught in class.

[hide=" my attempt at a solution"]

Well finding the intersection between $ \frac{4x^{2}}{\pi^{2}}$ and $ \sin(x)$, we get $ \frac{\pi}{2}$ (kind of set that up for convenience). Thus we find:

$ \left| \int_{0}^{\frac{\pi}{2}} \frac{4x^{2}}{\pi^{2}} \minus{} \sin(x) \right|$
$ \equal{} \frac{4}{\pi^{2}}\left(\frac{x^{3}}{3}\right) \plus{} \cos(x) \; \Bigg|_{0}^{\frac{\pi}{2}}$
$ \equal{} \left( \frac{4}{\pi^{2}}\left(\frac{\frac{\pi^{3}}{8}}{3}\right) \minus{} 0 \right) \plus{} (0 \minus{} 1)$
$ \equal{} \left|\frac{\pi}{6} \minus{} 1\right|$
$ \equal{} \boxed{1 \minus{} \frac{\pi}{6}}$
[/hide]

Is this a suitable method (assuming the answer's even right?) and are there other ways of finding areas between multiple functions?

[quote="Prime factorization"]...are there other ways of finding areas between multiple functions?[/quote]

Yeah. Assuming you want to find area of a domain bounded by $ y \equal{} f(x)$, $ y \equal{} g(x)$ and $ x \equal{} a$, $ x \equal{} b$ ($ f(x) < g(x)$ for all $ a\leq x \leq b$). The answer is
\[ \iint\limits_{\begin{subarray}{l} a \leqslant x \leqslant b \\
f\left( x \right) \leqslant y \leqslant g\left( x \right) \end{subarray} }{dxdy} \equal{} \int_a^b {\left( {\int_{f\left( x \right)}^{g\left( x \right)} {dy} } \right)dx} .
\]


function
integration
combinatorial geometry
advanced fields
advanced fields unsolved

How about this for a change :)

$K=$ compact convex substset of a Hasdorff locally convex TVS.
$S(K)$ continuous convex functions on $K$.

Then $S(K)-S(K)=\{f-g: f,g\in S(K)\}$ is dense in $C(K)$.

That's obviously false (try $K=[0,1]$). What do you really mean?

sory about that, now it should be correct.

Nice problem :). I'll replace $(K)$ with $_K$ everywhere, because it's easier to type.

Obviously, $V_K=S_K-S_K$ is a subspace of $C_K$. Proving that it's dense in $C_K$ is equivalent to proving that every bounded linear functional $L$ of $C_K$ which vanishes on $V_K$ vanishes identically.

Let $L$ be such a functional. By the Riesz representation theorem, there is a (unique) Borel regular real measure $\mu$ on $K$ such that $L(f)=\int_Kf\ \mbox d\mu,\ \forall f\in C_K$. By this I mean that there is a real measurable function $h$ with $|h|=1$ and a positive regular Borel measure $|\mu|$, the total variation of $\mu$, such that for all $f\in C_k$, we have $L(f)=\int_Kfh\ \mbox d|\mu|\ (*)$.

For every convex compact subset $E$ of $K$, there is a convex continuous function $f_E: K\to[0,1]$ with $f_E^{-1}(0)=E$. We can take, for instance, $\widetilde E$ to be the closed convex hull of $(E,0)\cup(K,1)$ in the locally convex TVS $X\times R$ ($X$ is the space we were working in), and then define $f_E$ by $f_E(x)=\inf\{y\in[0,1]\ |\ (x,y)\in\widetilde E\}$. The claimed properties are easy to check (I hope I didn’t make any mistakes :)). Given such a function $f_E$, define $f_E^n$ to be $f_E$ whenever $f_E$ is $\ge\frac 1n$, and $\frac 1n$ whenever $f_E<\frac 1n$. Finally, define $\varphi_E^n=n\left(f_E^n-f_E\right)$. $\varphi_E^n$ belongs to $V_K$, and as $n\to\infty,\ \varphi_E^n$ goes to $\chi_E$, the characteristic function of $E$, in the sense of $L^1(|\mu|)$. Since $L(\varphi_E^n)=0,\ \forall n$ we get, by $(*),\ L(\chi_E)=0$. In other words, $\mu(E)=0$ for all convex compact subsets $E$ of $K$.

For every convex open subset $G$ of $K$ and any $x\in G,\ G$ is the union of the closures of $G_n=x+\left(1-\frac 1n\right)(G-x)$. By applying the Dominated Convergence Theorem for $\chi_{G_n}h\to\chi_Gh$ and $\int_K\chi_{G_n}h\ \mbox d|\mu|=0,\ \forall n$, we get $\mu(G)=0$ for open convex subsets $G$ of $K$ as well.

Finally, we have $\mu(F)=0$ for all closed subsets $F$ of $K$, since every $F$ is compact so $\chi_F$ can be approximated arbitrarily well by characteristic functions of finite unions of open convex sets, after which we apply the Dominated Convergence Theorem again.

The above shows that $\mu$ is identically zero, so we’re done.


Given a triangle $ABC$ for which $C=90$ degrees, prove that given $n$ points inside it, we can name them $P_1, P_2 , \ldots , P_n$ in some way such that:

$\sum^{n-1}_{k=1} \left( P_K P_{k+1} \right)^2 \leq AB^2$ (the sum is over the consecutive square of the segments from $1$ up to $n-1$).

[i]Edited by orl.[/i]

I've seen this problem before in Donald J. Newman's "A Problem Seminar". Besides, no "assumption on the minimal value of $n$" can make the problem as it was posted true, since the points can be as close to each other as you want.

Let for a broken line $ M$ - $ M_{1},M_{2}, \ldots, M_{k}$, $ f(M)$ be the sum sum of the consequtive squares of its segments.

It can be shown by induction on $ n$ that there exists broken line $ A,P_{1}, \ldots, P_{n}, B$ such that $ f(A,P_{1}, \ldots, P_{n}, B)\leq AB^{2}$.
If $ n=0$ or $ n=1$ it is obvious!
Suppose we proved this statement for $ n=0,1,...,k$. Lets examine the case $ n=k+1$.
Draw the altitude $ CH$.

If there are points in either triangle $ AHC$ and $ BHC$ then using the induction hypothesis
$ f(A,P_{1}, \ldots, P_{t}, C)\leq AC^{2}$ and
$ f(C,P_{t+1}, \ldots, P_{k+1}, B)\leq CB^{2}$.
Using that $ P_{t}C^{2}+CP_{t+1}^{2}\geq P_{t}P_{t+1}^{2}$ (because $ \angle{P_{t}CP_{t+1}}$ is non-obtuse) and $ AC^{2}+BC^{2}=AB^{2}$ we prove the step $ n=k+1$ !!!!

If all the points are inside one of triangles $ AHC$ and $ BHC$ then do the same for it.  But this may happen only finite number of times because the triangle-s area tends to 0 while the c.o. of the points $ P_{1}, \ldots, P_{n}$ is constant.

$ READY!$

Why am I wrong? I don't understand.

Can you explain what you mean please?

[quote="ttd"]you are wrong.[/quote]

Bump up. Why is his solution wrong, except that the argument with the triangles' areas tending to 0 needs a little clarification (in fact, it is also important that these triangles are all similar, so their diameters also tend to 0) ?

  Darij

[quote="ttd"]you are wrong.[/quote]
Why is it wrong @ttd ? I think it is a right solution.  :huh:


2.303 log10 (k2/k1) = Ea/R (T2-T1/T1 T2).

Estimate an activation energy for the reaction, calculated from the Arrhenius equation. (Assume that the rate of a reaction is ten times faster at 32celsius, than it is at 22degrees Celsius.)

R is the gas constant, 8.31 j mol-1 K-1.

Can you please tell me where is the difficulty? You have the relation between the rate constants (k2 = 10k1) and the two temperatures, so you just have to substitute. Or am I missing something?


[color=darkblue]
[u][b]Problem 1[/b][/u]
Let $ (u_n)$ is a number sequence, let $ u_1 \equal{} a$, $ u_{n \plus{} 1} \equal{} u_n \plus{} (n \plus{} 1)^3$. Find formula of $ u_n$?

[u][b]Problem 2[/b][/u]
Let $ (u_n)$ is a number sequence, let $ u_1 \equal{} a$, $ u_{n \plus{} 1} \equal{} u_n \plus{} (n \plus{} 1)^4$. Find formula of $ u_n$?

[u][b]Problem 3[/b][/u]
Let $ (u_n)$ is a number sequence, let $ u_1 \equal{} a$, $ u_{n \plus{} 1} \equal{} u_n \plus{} (n \plus{} 1)^5$. Find formula of $ u_n$?
[/color]

Add $ a\minus{}1$ to each of (24), (25), and (26) from [url=http://mathworld.wolfram.com/PowerSum.html]here[/url].

Ever heard of telescopic sums?
$ u_{n\plus{}1}\minus{}u_n\equal{}\cdots$


What is $\frac{1}{2} \%$ of 100?  Express your answer as a decimal.

[hide="Solution"]

When I approach these types of questions, I often turn the percentage into a fraction by dividing it by 100.

So,

\begin{align*}
\frac{1}{2}\% &= \frac{\frac{1}{2}}{100}\\
                                &= \frac{1}{200}
\end{align*}

Therefore,

\[
\frac{1}{200} \cdot 100 = 0.5
\]

[/hide]

[hide="Answer"]0.5[/hide]
[hide="Solution"]Someone else probably already wrote it[/hide]


probability
parameterization
function
calculus
calculus computations

Firstly, its not a homework question.. I am trying to come up with the optimum probability for a cryptographic protocol ..(the value $x$ should be the probability that a biased coin tosses 1).. I need to find the optimum value for $x$ such that in the end of the following game, event $E_1$
occurs.

$Begin Game$

$Initiailize$: Inpute three parameters $n, m,k$ such that $n > m$.

$Phase~1$: Generate $n$ random bits $b=(b_1, b_2,\ldots b_n)$ such that $\Pr[b_j =1]=x~\forall~j~1\leq j\leq n$ (independent of any other bit $b_i$)

$Phase~2$: Repeat $k$ times: For each run $j$, choose upto $m$ bits from the $n$ bits $b$ without looking at  their values. If none of chosen bits are $1$, abort game, else continue till all $k$ tries are over.

$Phase~3$: finally, choose upto $m$ bits again one last time. If any of the chosen bits are 1, abort game.

$End Game$: Call it event $E_1$ if game did not abort in $phase~2$ and in $phase~3$.

Question: What is optimum value of $x$ (given $m, n,k$) such that $\Pr[E_1]$ is maximum?

Rgds
Amit

I think I would approach it as follows: at first, don't worry about $x$.  Suppose that $l$ of the $b_i$ are 1s.  For each $l$, compute the probability $p_l$ that $E_1$ occurs.  Then, take the product $P(\textrm{exactly l of the }b_i\textrm{ equal 1}) \cdot p_l$ and sum over all $l$.  This gives you the function you have to maximize, and some experimentation, calculus or numerical appropximation should yield the best $x$.


A triangle has sides of length $ a,b,c$ and its circumcircle has radius $ R$. Prove that the triangle is right-angled if and only if $ a^2\plus{}b^2\plus{}c^2\equal{}8R^2$.

[hide]It is easy to see that if $ a^2\plus{}b^2\equal{}c^2$ then we have $ \frac{a^2\plus{}b^2\plus{}c^2}{8R^2}\equal{}\frac{c^2}{4R^2}\equal{}\sin^2C\equal{}1$ and similarly if $ a^2\plus{}c^2\equal{}b^2$ or $ b^2\plus{}c^2\equal{}a^2$.

As for the second part in the proof, we have

$ a^2\plus{}b^2\plus{}c^2\equal{}8R^2$

$ \Rightarrow a^2\plus{}b^2\minus{}2ab\cos C\plus{}c^2\equal{}8R^2\minus{}2ab\cos C$

$ \Rightarrow 2c^2\equal{}8R^2\minus{}2ab\cos C$

$ \Rightarrow \frac{c^2}{4R^2}\equal{}1\minus{}\frac{ab}{4R^2}\cos C$

$ \Rightarrow \sin^2C\equal{}1\minus{}\sin A\sin B\cos C$

$ \Rightarrow 1\minus{}\sin^2C\equal{}\sin A\sin B\cos C$

$ \Rightarrow \cos^2C\equal{}\sin A\sin B\cos C$

This gives either $ \cos C \equal{}0$ which gives a right triangle or $ \sin A\sin B\equal{}\cos C$.  Since $ C\equal{}\pi\minus{}(A\plus{}B)$ this becomes

$ \sin A\sin B\equal{}\minus{}\cos (A\plus{}B)$

$ \Rightarrow\sin A\sin B\equal{}\sin A\sin B\minus{}\cos A\cos B$

$ \Rightarrow\cos A\cos B\equal{}0$

which gives one of $ \cos A, \cos B$ is $ 0$ and thus the triangle is right angled.[/hide]

$ \sum a^2=8R^2\iff \sum 4R^2\sin^2 A=2\iff \sum \sin^2A=2$

$ \sin^2(\pi-B-C)+1-\cos^2B+1-\cos^2C=2\iff \sin^2(B+C)=\cos^2B+\cos^2C$

$ \sin^2B\cos^2C+\sin^2C\cos^2B+2\sin B\sin C\cos B\cos C=\cos^2B+\cos^2C$

$ (\sin^2B-1)\cos^2C+(\sin^2C-1)\cos^2B+2\sin B\sin C\cos B\cos C=0$

$ 2\cos^2B\cos^2C=2\sin B\sin C\cos B\cos C$

$ 2\cos B\cos C(\cosh\cosh-\sinh\sinh)=0$

$ 2\cos B\cos C\cos (B+C)=0$
$ -2\cos B\cos C\cos A=0$

And the result follows.


quadratics
modular arithmetic
number theory
prime numbers

Prove that all numbers $ q\equal{}p^{2}\minus{}1$ where p>3 is a prime number
can always be divided by 24

I proved that $ p^{2} \equiv 1 mod(8)$
but next I don't know what to do
can someone help me out

thanks

as p>3, and p2-1 p2 and p2+1 are consecutive. 1 of them has to be divisible by 3. since it is not p2 so one of the other 2 must be.

how do I know that $ p^{2}\minus{}1$ can be divided by 3? and not $ p^{2}\plus{}1$ ?

[quote="Abelois"]how do I know that $ p^{2} \minus{} 1$ can be divided by 3? and not $ p^{2} \plus{} 1$ ?[/quote]
I don't think m00li use the right method. But this is really a easy one (even in Intermediate Topics).
First method: apply Fermat's little theorem directly, we get $ p^{3\minus{}1}\equal{}1(mod \;\; 3)$ for $ gcd(p,3)\equal{}1$

Second method: only two cases, so
Case 1: $ p\equal{}1(mod 3)$, then $ p^2\equal{}1^2\equal{}1(mod \;\; 3)$
Case 2: $ p\equal{}2(mod 3)$, then $ p^2\equal{}2^2\equal{}1(mod \;\; 3)$

oke thanks

I didn't know fermats little theorem

sorry for that absurd reply. i meant p-1, p and p+1 instead of p2-1...

Hello, one can even solve this question by making use of Quadratic Reciprocity theroem witht he knowledge of legendre's symbols.
Thanks.
Aravind.

My trial

We know that for prime numbers $ >3$, can be expressed in the form of $ 6k\pm 1$

So $ p^2\minus{}1\equiv 0 \pmod{24}$
$ 36k^2\pm12k\plus{}1\minus{}1\equiv 36k^2\pm12k\pmod {24}$

We have to provve that $ 24|36k^2\plus{}12k\equal{}12k(3k\plus{}1)$
If $ k$ is an even number, $ 24|12k$
If $ k$ is an odd number, $ 3k\plus{}1$ will be even, so $ 24|12k(3k\plus{}1)$

$ QED$

I hope my solution is correct :)

m00li,
at least use "^" for exponentiation.  Use more spacing, commas, and type out
"one" and "two" when used as adjectives:

Redo:
-------

[quote="Arrange your tan"]as p > 3, and p^2 - 1, p^2, and p^2 + 1 are consecutive.  One of them
has to be divisible by 3.  Since it is not p^2, so one of the other two must be.[/quote]

$ q\equal{}(p\minus{}1)(p\plus{}1)$ So since $ p$ is odd, we have two consecutive even numbers, so one is a multiple of four giving that $ q$ has a factor of $ 8$. Now note that $ p\minus{}1,p,p\plus{}1$ are consecutive, so $ 3$ divides one of them. Since $ 3$ does not divide $ p$, $ 3$ divides either $ p\minus{}1$ of $ p\plus{}1$ giving a factor of $ 3$. Since we have a factor of $ 8$, and a factor of $ 3$ we are done.


search
combinatorics unsolved
combinatorics

A polygon with 2n+1 vertices is given.Show that it is possible to assign 1,2,....,4n+2 to vertices and midpoint of the sides of polygon so

that for each side the sum of the three numbers assigned to it is the same.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=78458059&t=30749

Please use the search function.  (Words I searched for in this case: "polygon" and "assign."  If you had included the source, it would have been even easier.)


geometry
rectangle
linear algebra
matrix
vector
combinatorics open
combinatorics

Here is a problem I've been thinking about for some time with no success

Is it possible to arrange the numbers 1, 2, ..., 6n, in a 3 by 2n matrix
so that the sum of the entries in each row and column is a multiple of
(6n+1)? Of course, here n is a positive integer and each of the numbers
has to be used exactly once.

In general, the following is known to be true:

Given positive integers m\ge 5 and n\ge 2, m odd and n even, it is possible
to construct an m by n matrix with elements using the first mn positive integers
such that every row sum and every column sum is a multiple of (mn+1).

For some reason, the case m=3 still eludes me.

thanks,

Dan

It seems to be possible, though my construction is fairly ugly and involves a lot of case analysis.  I'll try and walk through how I arrived at it here.

My thought was to first find a configuration so that the columns all add up to a multiple of $ 6n\plus{}1$, then permute the elements in each column to make the rows work too.  The idea here is that we have so few constraints (only 3 rows to worry about), and so many options (many columns to switch around in) that hopefully something will just work out.  So first, let's look at the columns.

We know the sum of the elements in the whole grid is $ 3n(6n\plus{}1)$, and there are $ 2n$ columns, so an average column will sum to $ \frac{3}{2}(6n\plus{}1)$.  This suggests making half of the columns sum to $ 6n\plus{}1$ and half to $ 12n\plus{}2$.  So let's start by placing $ 6n$ in a column.  The column will have to sum to $ 12n\plus{}2$, and a logical choice for elements summing to $ 6n\plus{}2$ are $ 3n\plus{}2$ and $ 3n$.

We start the next column with $ 6n\minus{}1$, and need to pick two elements summing to $ 6n\plus{}1$.  Again, we pick from the center, choosing $ 3n\plus{}4$ and $ 3n\minus{}2$.  We continue in this fashion, each time decreasing two rows by $ 1$ and increasing the third by $ 2$ (to keep the column sums constant).  After $ n$ columns have been filled we have.

[code]6n    6n-1  6n-2  6n-3  ...  5n+1
3n+2  3n+4  3n+6  3n+8  ...  5n
3n    3n-1  3n-2  3n-3  ...  2n+1 [/code]

Now we're stuck, both because we can't decrease the first row any further (we've used $ 5n$), and because we've used up our $ 12n\plus{}2$ columns already.  The remaining numbers to use are $ 1, 2, \dots 2n$ and $ 3n\plus{}1, \dots 5n\minus{}1$.

We'll stick the $ n$ large numbers in our first row, and we're left trying to find pairs of numbers between $ 1$ and $ 2n$ that add up to $ n\plus{}2$ (to match $ 5n\minus{}1$), $ n\plus{}4$, etc.  Looks like $ (n\plus{}1, 1), (n\plus{}2, 2), \dots$ will work nicely.  This leaves us with

[code]6n    6n-1  6n-2  6n-3  ...  5n+1  ||  5n-1  5n-3  5n-5  ...  3n+1
3n+2  3n+4  3n+6  3n+8  ...  5n    ||  n+1   n+2   n+3   ...  2n
3n    3n-1  3n-2  3n-3  ...  2n+1  ||  1     2     3     ...  n[/code]

We're still left with the row sums.  Right now we in a sense have 6 different blocks, adding up to

[code] 0.5n(11n+1) || 4n^2
    n(4n+1)  || 0.5n(3n+1)
 0.5n(5n+1)  || 0.5n(n+1)[/code]

Our goal will be to make each row add up to $ n(6n\plus{}1)$.  As it turns out, two of our blocks already do this.  Moving the bottom right block to the top and sliding the other two blocks downwards gives us

[code]6n    6n-1  6n-2  6n-3  ...  5n+1  ||  1     2     3     ...  n
3n+2  3n+4  3n+6  3n+8  ...  5n    ||  5n-1  5n-3  5n-5  ...  3n+1
3n    3n-1  3n-2  3n-3  ...  2n+1  ||  n+1   n+2   n+3   ...  2n,[/code]
whose first row has the desired sum, but whose second row adds up to $ n(8n\plus{}1)$ instead of $ n(6n\plus{}1)$.  So now we're looking to keep the first row and all the columns fixed, but move $ 2n^2$ from the middle row to the bottom row by swapping elements there with the bottom row elements in the same column.

Each swap we do decreases the middle row, with the decreases being (by column)

[code]2  5  8  11  ...  3n-1 ||  4n-2  4n-5  4n-8  ...  n+1[/code]

We need to find a subset of this "decrease vector" summing to $ 2n^2$.  If $ n$ is even, we're done already, since we can just swap exactly half of the pairs $ (2, 4n\minus{}2), (5, 4n\minus{}5) \dots$ over.  So let's assume that $ n\equal{}2k\plus{}1$.  If we swap any $ k$ of the pairs, we're almost there, but still have to find $ 2n$ to swap over.  We'll do this by cases.

If $ n\equal{}3j\plus{}1$, then $ 2n$ appears in the first block of differences.  We'll just swap $ 2n$, along with $ k$ pairs $ (q, 4n\minus{}q)$  (making sure not to choose $ q\equal{}n$).

If $ n\equal{}3j$, then $ 2$ appears in the first block of differences, and $ 2n\minus{}2$ in the second block.  We swap them, along with $ k$ pairs $ (q, 4n\minus{}q)$ where $ q$ is neither $ 2$ nor $ 2n\plus{}2$.

If $ n\equal{}3j\plus{}2$, then both $ 2n\minus{}2$ and $ 2$ appear in the first block, and we swap both over, along with $ k$ pairs $ (q, 4n\minus{}q)$, where $ q$ is neither $ 2$ nor $ 2n\minus{}2$.


linear algebra
matrix
superior algebra
superior algebra solved

$e^M=I$ if and only if $M$ is diagonalizable and every eigenvalue of $M$ is an integer multiple of $2\pi i.$

Every $M$ is similar to a matrix in Jordan canonical form. But by explicit calculation, the exponential of a nontrivial Jordan block would contain nonzero entries above the main diagonal. So the condition $e^M=I$ forces $M$ to be diagonalizable.

So we assume $M$ is similar to $\text{diag}\{\lambda_1,\lambda_2,\dots, \lambda_n\}.$ Then $e^M$ is similar to $\text{diag}\{e^{\lambda_1}, e^{\lambda_2},\dots,e^{\lambda_n}\}.$ That is the identity matrix iff each $\lambda_k$ is an integer multiple of $2\pi i.$

A nice example: $M=\begin{bmatrix} 0 & -2\pi \\ 2\pi & 0 \end{bmatrix}.$

Nice  :)
there is another solution with the decomposition $M=N+D$ where $N$ is nilpotent ,$D$ diagonalizable, $ND=DN$.
Then it is easy to compute $exp(M)$ and conclude..


two events are 1 km apart and simultaneous in one inertial frame. In another frame, the two events are 2km apart - what is the time difference between the events in this second inertial frame?

Im having a bit of a shocker with relativity at the moment, but I fumbled around and eventually got a number out, 5.77*10^-9 seconds. could someone tell me if this is correct, and even if it is could you show me the way it is meant to be done using the lorentz transformations. thanks/

All you need to know is that the square of the space-time distance $c^2t^2 - x^2$ is invariant under Lorentz transformations, which can be easily verified by substituting the Lorentz transformation $x'(x, t),\ t'(x, t)$ with an arbitrary v < c into $c^2t'^2 - x'^2$. Choosing the space-time origins in both systems at the event 1 (i.e., $x_1 = x_1' = 0$, $t_1 = t_1' = 0$), then the square of the invariant space-time distance of the event 2 from the common origin is

$c^2t_2^2 - x_2^2 = c^2t_2'^2 - x_2'^2$

$c = 3 \times 10^8\ \text{m/sec}$ is the speed of light, $x_2 = 1\ \text{km} = 1000\ \text m$ and $t_2 = 0$, because the events 1,2 are simultaneous in one system, while $x_2' = 2\ \text{km} = 2000\ \text m$ and $t_2'$ is to be calculated in the other system:

$t_2' = \frac{\sqrt{x_2'^2 - x_2^2}}{c} = \frac{10^3 \sqrt 3}{3 \times 10^8} \doteq 5.77 \times 10^{-6}\ \text{sec} = 5.77\ \mu\text{sec}$

You forgot to convert kilometers into meters.


Doritoes:20

cheetos:20

Lays:20

Pringles:20

Fritos:20

funyuns:20

gardettos:20

Er... cheetos aren't chips, but idc.

[color=green][b]Doritoes: 21[/b][/color]
cheetos: 20
Lays: 20
Pringles: 20
Fritos: 20
funyuns: 20
[b][color=red]gardettos: 19[/color][/b]

[color=red][b]Doritoes: 20[/b][/color]
cheetos: 20
[color=green]Lays: 21[/color]
Pringles: 20
Fritos: 20
funyuns: 20
gardettos: 19

[color=green][b]Doritoes: 21[/b][/color]
cheetos: 20
[color=red][b]Lays: 20[/b][/color]
Pringles: 20
Fritos: 20
funyuns: 20
gardettos: 19[/quote]

Doritoes: 20
cheetos: 21
Lays: 20
Pringles: 20
Fritos: 20
funyuns: 20
gardettos: 19[/quote][/quote]

Doritoes: 20
cheetos: 20
Lays: 20
Pringles: 20
Fritos: 20
funyuns: 21
gardettos: 19

Doritoes: 19
cheetos: 20
Lays: 21
Pringles: 20
Fritos: 20
funyuns: 21
gardettos: 19

Doritoes: 19
[color=green]cheetos: 21[/color]
Lays: 21
Pringles: 20
Fritos: 20
funyuns: 21
[color=red]gardettos: 18[/color]

what are gardettos?

Doritoes: 19
cheetos: 21
Lays: 21
Pringles: 21 (delicious!!! but expensive :()
Fritos: 20
funyuns: 20 (no idea what it is)
gardettos: 18

Doritoes: 20
cheetos: 21
Lays: 21
Pringles: 20
Fritos: 20
funyuns: 21
gardettos: 17

+doritoes
-gardettoes

[b][color=green]Doritoes: 21[/color][/b]
cheetos: 21
funyuns: 21 <--- these are???
[b][color=red]Lays: 20[/color][/b]
Pringles: 20
Fritos: 20
gardettos: 17

http://en.wikipedia.org/wiki/Funyuns

http://www.generalmills.com/corporate/brands/brand.aspx?catID=59

wikipedia is your friend.

kill funyuns please.

Doritoes: 21
cheetos: 21
[color=red]funyuns: 20[/color]
[color=green]Lays: 21[/color]
Pringles: 20
Fritos: 20
gardettos: 17
 DOWN WITH FUNYUNS!!!!! GO LAYS!!!!!

Doritos: 20
cheetos: 21
Funyuns: 20
Lays: 21
Pringles: 21
Fritos: 20
Gardettos: 17

you've been spelling doritos wrong this whole time.. -_-

Doritos: 21
cheetos: 21
Funyuns: 20
Lays: 20
Pringles: 21
Fritos: 20
Gardettos: 17

Doritos: 21
cheetos: 22 (delicious!)
Funyuns: 20
Lays: 20
Pringles: 21
Fritos: 20
Gardettos: 16 (everyone are killing them, so...)


Let $ n$ be a positive integer and let $ a_n$ be the number of ways to write $ n$ as a sum of positive integers, such that any $ 2$ summands differ by at least $ 2$.
Also, let $ b_n$ be the number of ways to write $ n$ as a sum of positive integers of the form $ 5k\plus{}1$ or $ 5k\minus{}1$($ k \in Z$).
Prove that $ \frac{a_n}{b_n}$ is a constant $ \forall n \in N^*$

I assume you're talking about integer partitions here: as written, it's false, since $ b_3 \equal{} 1$ (just 1 + 1 + 1) and $ b_4 \equal{} 2$ (4 and 1 + 1 + 1 + 1) while $ a_3 \equal{} 1$ (just 3) while $ a_4 \equal{} 3$ (1 + 3, 3 + 1 and 4).  (Also, that's a rather contrived way to say that the two numbers are the same.)

It looks similar to the fact that partitions into distinct parts are in bijection with partitions into odd parts, but some related results are quite tricky.  (For example, partitions into parts congruent to $ \pm 1 \mod 6$ are in bijection with parts that differ by 3 and such that any two parts divisible by 3 differ by at least 6.)  The definition of $ b_n$ suggests that one of the Rogers-Ramanujan identities could conceivably show up in a proof (though perhaps not, since they don't have a nice combinatorial proof, as far as I know).


vector
advanced fields
advanced fields unsolved

Let $ \Sigma$ be a local section of the ODE $ \dot{x}\equal{}v(x)$, where $ v$ is a locally lipschitz function.  Let $ \phi$ be the solution passing through $ \xi$ which exists and is unique since $ v$ is locally lipschitz.   Suppose that the interior of $ \Sigma$ contains points of the omega limit set of $ \xi$.   Prove there is an increasing nonbounded sequence of times $ t_n$ such that $ \phi(t_n) \in \Sigma$ for all $ n$

What is "a local section"? A smooth surface that is transversal to the vector field? Then it is next to obvious: you come arbitrarily close to some inner point $ p$ of $ \Sigma$ at arbitrarily large time $ t$. Then for the times $ t\plus{}\delta$ and $ t\minus{}\delta$ you are still in a small neighborhood of $ p$ (if $ \delta>0$ is small enough) and on different sides of $ \Sigma$ (if your distance to $ \Sigma$ at the time $ t$ is much less than $ \delta$).

why can't the orbit of $ \xi$ approach $ \Sigma$ by just one side?

I have the picture in my mind of this case so if you were not able to get what I was trying to say let me know and if you were please help me again  :oops:

DANKEN!

Because the vector field is [b]transversal[/b] to $ \Sigma$, so you cannot come close and just turn around. The picture you have in mind is, probably, of the line $ \Sigma$ tangent to the unit circle and the trajectory that is a slowly unwinding spiral having the circle as its limit or something similar. Note that in this case the field is tangent to $ \Sigma$ at $ p$.


1. Fine the ecuation of the plane which intersects $A(1,2,3)$,$B(4,5,6)$,$C(1,1,1)$. (only knowing that teh ec. of a plane is $ax+by+cz+d=0$ because that's all that i've learned).

2. Find the afix of the 3rd point of an equilateral triangle if the other ones are $z_{1}=1$ and $z_{2}=2+i$.

1.
[hide]
We can let a=1, because even if it isn't, we can always multiply both sides of the equation of the plane by 1/a.

.:x+by+cz+d=0
Subbing in points

1+2b+3c+d=0
4+5b+6c+d=0
1+b+c+d=0

Three equations, three unknowns... you can solve it from here.
I'll finish off the question when I have time.[/hide]

[quote="dk2"]1.
[hide]
 We can let a=1, because even if it isn't, we can always multiply both sides of the equation of the plane by 1/a.

.:x+by+cz+d=0
Subbing in points

1+2b+3c+d=0
4+5b+6c+d=0
1+b+c+d=0

Three equations, three unknowns... you can solve it from here.
I'll finish off the question when I have time.
[/hide][/quote]

Solving the trilinear system that dk2 proposed yields $b=-2$, $c=1$, and $d=0$... i hope  :P

Well, i did number 2, i'll post my work if its right...
[hide] i found two points that'll make equilateral triangle:

$z_{3}=\frac{3-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}i$

and $z_3=\frac{3+\sqrt{3}}{2}+\frac{1-\sqrt{3}}{2}i$[/hide]

hmm i have some time now so if my solution above was wrong... please correct it. [hide]Since it was given that $z_1=1$ and $z_2=2+i$, i just let that be expressed as $(1,0)$ and $(2,1)$. The finding the distance between those two points i obtained $\sqrt{2}$. To find a third point making an equilateral triangle, i made two circles with centers at $(1,0)$ and $(2,1)$, both circles having a radius of $\sqrt{2}$.
The equations for the circles were:
$(x-1)^2+y^2=2$ and $(x-2)^2+(y-1)^2=2$
Setting them equal and simplifying gives a linear equation: $y=-x+2$. It should be observed that this line goes through both points of interection of the previous two circles. Then, to find the points of interection, one can set $(-x+2)^2=2-(x-1)^2$. Simplifying the equation yields the quadratic $2x^2-6x+3=0$
Solving the quadratic yields $x=\frac{3-\sqrt{3}}{2}$ and $x=\frac{3+\sqrt{3}}{2}$. Plugging both roots into $y=-x+2$ yields the following two points of interection:
$z_{3}=\frac{3-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}i$ and $z_3=\frac{3+\sqrt{3}}{2}+\frac{1-\sqrt{3}}{2}i$ when one changes the point from the form $(a,b)$ to $a+bi$[/hide]


[code]! Missing delimiter (. inserted).[/code]That is telling you that you forgot the [b].[/b] after \right and LaTeX has put it in for you, which is why the result is fine.
\left and \right have to be followed by something; using [b].[/b] tells LaTeX you don't want anything printed.

Centred in what way? I presume you don't just mean centred horizontally on the page which just uses \ [ and \ ] instead of dollar signs?

No, the effect he attains with array{c}. Mine is equivalent with an array{l}, so how do i make it look like array{c}?

Cases is deliberately defined to be left-aligned columns. I suspect that behind the scenes it uses \newenvironment which will put together \left, \right and aligned or gathered enviroments. So what you would need to do (in a document, not on the forum) is to use array or whatever to do what you want then use it to define your own environment pvddd say. From then on you will be able to use \begin{pvdd}...\end{pvdd} instead of cases.