vector
linear algebra
matrix
algorithm
parameterization

Ahh .. Well, I'm having a bit of a problem with a couple of these questions .. so here's a screenshot from my assignment sheet .. any help will be much appreciated, guys ..
cheers!

[img]http://img.photobucket.com/albums/v505/Puj/questions.jpg[/img]

For question 8 note that a subset of a vector space is a subspace if and only if it is closed under addition and under multiplication by scalars.  So ask yourself whether a scalar multiple of a symmetric matrix is symmetric, and whether the sum of two symmetric matrices is symmetric.

#9 and #10 both rely on the technique of Gaussian elimination. That's THE algorithm that dominates this portion of a linear algebra course. I'm not going to do the row elimination for you, but I will say that both of these questions appear to have been artificially rigged to make computation easier. In particular, no fractions need appear.

As for question #10, here's what you do with the row echelon form you get as a result of row elimination. One check: the answer to (c) is that the rank is 3.

(a) The nonzero rows of the row echelon form can be used as a basis for the row space.

(b) Find the columns with pivots in the row echelon form. The columns of the original matrix corresponding to those locations form a basis for the column space.

(c) Count the number of pivots.

(d, e) This is the solution to the homogenous equation. Introduce a parameter for each column without a pivot. The number of such columns should be (# of columns of original matrix) - (rank). The vectors that multiply each such parameter form a basis of the null space.

(f) The number of parameters, and number of vectors, in your answers to (d), (e).

(g) The given particular solution plus the null space computed in (d).

---

Some authors make a distinction between "exercises" and "problems." Exercises accompany textbooks or lecture notes. They exist for the student to practice the techniques taught in that text or course, and to illustrate the concepts being taught. In preparing an exercise, the author starts from the particular concepts or techniques being taught, and designs a particular example to use those techniques or illustrate those concepts.

A "problem" is different. The problem itself is primary. Rather that predictably drawing from a clearly announced set of techniques, it simply says, "Here I am - use whatever you can to solve me."

Of course no such attempt to classify the world cleanly into two types can ever wholly succeed, and the edges will be gray and fuzzy. But the subject of this topic is very clearly a set of exercises. There's nothing wrong with that, and we wouldn't want to prohibit the posting of exercises - but the lifeblood of this site, the thing that attracts its regulars, is problems.


Michal Rolínek (GJK Praha)
Zbyněk Konečný (GKJB)
Miroslav Klimoš (GMK Bílovec)
Lenka Slavíková (G Mnichovo Hradiště)
Jiří Řihák (GKJB)
Hana Šormová (GKJB)

Can you post all problems from TST? Thanks!  :)

TST= Team Selection Test = výběrové soustředění v Kostelci (in Czech)

One problem from there:

$x,y,z$ positive reals satisfying x^2+y^2+z^2+2xyz=1.
Find minmal $M$ such that $x+y+z\leq M$.

PS: TST problems are hard, but not much interesting (often taken from other olympiads). National round problems are much more interesting:

https://cgi.math.muni.cz/~rvmo/ABC/59/A59iii.pdf
https://cgi.math.muni.cz/~rvmo/ABC/58/A58iii.pdf
https://cgi.math.muni.cz/~rvmo/ABC/57/A57iii.pdf

No time to translate these in English, sorry. You may try Google's translator.

The given restricted inequality yields immediately to Lagrange multipliers, leading to $M=3/2$ (reached for $x=y=z=1/2$).

Lagrange beats it, but we didn't know them at that time. Just of curiosity -- do you in Romania study Langrange in Olympiad level trainings?

The nice part is that one can say "let x,y,z be cosines of inner angles in acute triangle and we are done from Jensen, with equality in equilateral case". This trig solution and the Lagrange one are the only ones I know till today (didn't play with it much though).


Hi;
 I have squares in my plane and also circles.I want to know if for example a square(its interior) is fully covered by a circle.Maybe I will have examples as follow in my attached file.

Your help is appreciated

How can I  know if a square(its interior not its edges) is fully covered by a circle?


[quote="lhplhp"]Let $ (O_1)$ and $ (O_2)$ be two circle,$ (O_1)$ cuts $ (O_2)$ at $ P;Q$
$ C$ is a point on $ (O_1)$.$ CP,CQ$ cuts $ (O_2)$ at $ A;B$
Find locus of circumcenter of triangle $ ABC$[/quote]
you shouldn't post this problem (VLTT in VN) :mad:  :mad:  :mad:

to LHPLHP:ban hoc o LHP TPHCM dung o? toi rat that vong ve ban
     toi noi vay chac ban tu hieu


Help me please; I seem to have almost done it, but still I cannot do it. :(
$ 3(x\plus{}y\plus{}1)^2\plus{}1\geq 3xy\forall x, y\in\mathbf{R}$.
[hide]
From my intuition I know that I have to do it by converting the LHS-RHS into a sum of some squares; but I am feeling like "so near, yet so far!"[/hide]

[quote="Potla"]Help me please; I seem to have almost done it, but still I cannot do it. :(
$ 3(x \plus{} y \plus{} 1)^2 \plus{} 1\geq 3xy\forall x, y\in\mathbf{R}$.
[hide]
From my intuition I know that I have to do it by converting the LHS-RHS into a sum of some squares; but I am feeling like "so near, yet so far!"[/hide][/quote]
It is equivalent to: $ \frac {(3(x \plus{} y) \plus{} 4)^2}{4} \plus{} \frac {3}{4}(x \minus{} y)^2 \ge 0$
Which is obvious true.Equality holds when $ x \equal{} y \equal{} \frac { \minus{} 2}{3}$ :)

[quote="Potla"]Help me please; I seem to have almost done it, but still I cannot do it. :(
$ 3(x \plus{} y \plus{} 1)^2 \plus{} 1\geq 3xy\forall x, y\in\mathbf{R}$.
[hide]
From my intuition I know that I have to do it by converting the LHS-RHS into a sum of some squares; but I am feeling like "so near, yet so far!"[/hide][/quote]
Just look for the discriminant. It is $ \Delta \equal{} \minus{} 3(3y \plus{} 2)^2\le0$, and by the way this shows that equality holds iff $ x \equal{} y \equal{} \minus{} \frac23$ :)

This is Columbia 2001 . See [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1337586104&t=100716]here[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1337586104&t=63920]here[/url] , or [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1337586104&t=110841]here[/url]


function
limit
logarithms
ratio
real analysis
real analysis unsolved

Hi,can someone  help me with this proof ??
I have a implicit function $ \beta: R \rightarrow R$ define as number satisfying:
\[ \sum_{i\equal{}1}^m p_i^q r_i^{\beta(q)}\equal{}1\] where $ 0<p_i,r_i<1.$
We can note that \[ \lim_{q\rightarrow \minus{}\infty} \beta(q)\equal{}\infty\] and
\[ \lim_{q\rightarrow \infty} \beta(q)\equal{}\minus{}\infty.\]
Let \[ \alpha(q)\equal{}\frac{\sum_{i\equal{}1}^m p_i^q r_i^{\beta(q)} \log p_i}{\sum_{i\equal{}1}^m p_i^q r_i^{\beta(q)} \log r_i}.\]
So, I need to proof that, when $ q\rightarrow \minus{}\infty$
\[ \alpha_{\max}\equal{}\max_{1\leq i\leq m}\frac{\log p_i}{\log r_i}\] and
$ q\rightarrow \infty$
\[ \alpha_{\min}\equal{}\min_{1\leq i\leq m}\frac{\log p_i}{\log r_i}\]

I really apreciate any help.
Thanks

First of all, $ p_i^q r_i^{\beta(q)}\le 1$ for all $ q$. Let us assume for a moment that $ \frac{\log p_1}{\log r_1}\equal{}\alpha<\frac{\log p_i}{\log r_i}$ for $ i>1$. Then, for $ q>0$, we have $ \beta(q)\ge \minus{}\alpha q$ (otherwise $ p_i^q r_i^{\beta(q)}>1$. Now, for each $ i>1$, we get
\[ p_i^qr_i^{\beta(q)}\le p_i^q r_i^{\minus{}\alpha q}\equal{}(p_i r_i^{\minus{}\alpha})^q\to 0\]
as $ q\to\plus{}\infty$ because $ p_i r_i^{\minus{}\alpha}<1$ according to our assumption. Thus, we have the weighted sums of logarthms in our fraction with the weights about $ 0$ in all terms except the first and with the total sum of weights equal to $ 1$. Hence, the first term is the only one that really matters.

If you have more than one term with the same minimal ratio of logarithms, you still can observe that only these terms matter and the ratio for each of them is the same.

The other part is similar.


Let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3.$ Find all positive integers $ n$ such that the inequality\[ a^n\plus{}b^n\plus{}c^n \leq \frac{1}{a^n}\plus{}\frac{1}{b^n}\plus{}\frac{1}{c^n}\] holds true for all $ a,b,c$ with the above property.

First give an inequality between aythmetical average and harmonical average for $ \frac{1}{a^{n}}$ , $ \frac{1}{b^{n}}$ and $ \frac{1}{c^{n}}$
It'll go sth like that:
$ \frac{\frac{1}{a^{n}}\plus{}\frac{1}{b^{n}}\plus{}\frac{1}{c^{n}}}{3} \geq \frac{3}{a^{n}\plus{}b^{n}\plus{}c^{n}} \geq \frac{a^{n}\plus{}b^{n}\plus{}c^{n}}{3}$ The last inequality is to be proved:
$ \frac{3}{a^{n}\plus{}b^{n}\plus{}c^{n}} \geq \frac{a^{n}\plus{}b^{n}\plus{}c^{n}}{3}$
$ 9\geq (a^{n}\plus{}b^{n}\plus{}c^{n})^{2}$
$ a^{n}\plus{}b^{n}\plus{}c^{n} \leq 3$
$ \frac{a^{n}\plus{}b^{n}\plus{}c^{n}}{3}\leq 1$ and this is equivalent to
$ \sqrt[n]{\frac{a^{n}\plus{}b^{n}\plus{}c^{n}}{3}} \leq \sqrt[n]{1}\equal{}1\equal{}\frac{a\plus{}b\plus{}c}{3}$
The last inequality is proper for n=1 and is not true for any bigger positive integer given the generalised inequality between averages which states that $ \sqrt[p]{\frac{a_{1}^{p}\plus{}a_{2}^{p}\plus{}...\plus{}a_{n}^{p}}{n}} \geq \sqrt[q]{\frac{a_{1}^{q}\plus{}a_{2}^{q}\plus{}...\plus{}a_{q}^{p}}{n}}$ for $ p\geq q$ and $ n\in N$  :wink:

Well, the last inequality is not obvious, so I simply wrote it and then I proved it. Proving that inequality finishes our given inequality, because when you multiply by 3 the first line I wrote it is equivalent to our problem. :wink:

The second inequality is true (AM-HM), and we want to prove the first inequality. So if we prove the third one the problem will also hold true.

EDIT: Aaah, sorry, the inequality I gave is weaker then inequality given in the problem for $ n\geq 2$
Sorry for misunderstanding  :oops:

[quote="tchebytchev"][quote="polskimisiek"]
$ \frac {\frac {1}{a^{n}} \plus{} \frac {1}{b^{n}} \plus{} \frac {1}{c^{n}}}{3} \geq \frac {3}{a^{n} \plus{} b^{n} \plus{} c^{n}} \geq \frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{3}$ The last inequality is to be proved:
  :wink:[/quote]
why $ \frac {3}{a^{n} \plus{} b^{n} \plus{} c^{n}} \geq \frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{3}$?
the  problem is the  Generalization ,and posted before by perfect_radio here : http://www.mathlinks.ro/Forum/viewtopic.php?p=512704#512704[/quote]
(a^n+b^n+c^n)(1+1+1)^(n-1)>=(a+b+c)^n=3^n.
This can be easily proved by the improved Th. of Cauchy.
then a^n+b^n+c^n>=3.


Is there a way to prove that $ e^{ix}\equal{}i$sin$ x\plus{}$cos$ x$ without using calculus?

$ e^{ix}$ is a complex-valued function.  The techniques for extending the [i]real[/i] function $ e^z$ to complex values of $ z$ are not only calculus, they are complex analysis.  This function doesn't even [i]exist[/i] outside of calculus.


function inequalities quadratics conics parabola

Prove that:
[quote][b]1.[/b] $ x^2\plus{}y^2\plus{}z^2 \leq xyz\plus{}2$ , for any $ x,y,z \in [0,1]$[/quote]  [quote][b]2.[/b] $ a^2\plus{}b^2\plus{}c^2\leq a^2b\plus{}b^2c\plus{}c^2a\plus{}1$ , for any $ a,b,c \in [0,1]$[/quote]
[quote][b]3.[/b] $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca \leq (\alpha\minus{}\beta)^2$ for any $ a,b,c \in [\alpha,\beta]$[/quote]

1)
Let us consider a function $ f(x,y,z) \equal{} x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xyz$
Now, let us have z fixed and assume $ x \equal{} max\{x,y,z\}$
Now we consider pairs of numbers $ x,y$ and $ x \plus{} k,y \minus{} k$

Now, observe that $ (f(x,y,z)\leq f(x \plus{} k,y \minus{} k,z))\iff 2k(x \minus{} y) \plus{} 2k^{2} \plus{} kz(x \minus{} y) \plus{} k^{2}c\geq 0$
(of course $ (k\leq y)\wedge (k\leq 1\minus{}x)$)
Thus in order to maximize our function we can maximize x and minimize y. So we only have to consider our inequality in 2 cases:
1)one of the variables is equal to 0
2)one of the variables is equal to one

First case is obvious, because it is equivalent to $ x^{2} \plus{} z^{2}\leq 2$
The second case is also very easy, because:
$ (1 \plus{} y^{2} \plus{} z^{2}\leq yz \plus{} 2)$
and observe that $ ((y \minus{} 1)(z \minus{} 1)\geq 0)\iff yz \plus{} 1\geq y \plus{} z$, so:
$ yz \plus{} 1\geq y \plus{} z\geq y^{2} \plus{} z^{2}$ so the inequality holds :wink:

[hide="Solution to 1"]
$ x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xyz \minus{} 2\le 0$

Consider the left side as a quadratic in $ x$.  Its graph is a parabola that opens upward and has vertex at $ yz/2\le 1/2$.  We achieve the maximum when $ x$ is as far from the vertex as possible, which occurs at $ x \equal{} 1$.  Therefore, replacing $ x$ with 1 strictly increases the left side of our inequality.  Similarly, replacing $ y$ and $ z$ each with 1 also strictly increases the left side of our inequality.  Therefore

$ x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xyz \minus{} 2\le 1 \plus{} 1 \plus{} 1 \minus{} 1 \minus{} 2\le 0$ as desired.[/hide]
[hide="Solution to 2"]$ a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} a^{2}b \minus{} b^{2}c \minus{} c^{2}a \minus{} 1\le 0$
$ (1 \minus{} b)a^{2} \minus{} c^{2}a \plus{} b^{2} \plus{} c^{2} \minus{} b^{2}c \minus{} 1\le 0$

If all the variables are 1 then the inequality is clearly true, so assume $ b\ne 1$.  Consider this a quadratic in $ a$.  Its graph is a parabola that opens upward since $ 1 \minus{} b > 0$, so its maximum value occurs when $ a$ is 0 or 1.  Now we split the problem into cases:

Case 1: $ a \equal{} 0$.  We want to prove $ b^{2} \plus{} c^{2} \minus{} b^{2}c \minus{} 1\le 0$.  We have

$ b^{2} \plus{} c^{2} \minus{} b^{2}c \minus{} 1\le b^{2} \plus{} c \minus{} b^{2}c \minus{} 1 \equal{} (b^{2} \minus{} 1)(1 \minus{} c)$.  Since $ b^{2} \minus{} 1$ is negative and $ 1 \minus{} c$ is nonnegative, the desired inequality follows.

Case 2: $ a \equal{} 1$.  We want to prove $ b^{2} \minus{} b \minus{} b^{2}c\le 0$.  This is equivalent to $ b(b \minus{} 1 \minus{} bc)\le 0$.  But $ b$ is positive and $ b \minus{} 1 \minus{} bc\le b \plus{} c \minus{} 1 \minus{} bc \equal{} (b \minus{} 1)(1 \minus{} c)\le 0$, so this is true.

Therefore the original inequality holds.[/hide]
[hide="Solution to 3"]This is equivalent to $ (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}\le 2(\beta \minus{}\alpha)^{2}$.

Clearly we can add a constant to $ a,b,c,\alpha,\beta$ to obtain an equivalent inequality such that the condition $ a,b,c\in [\alpha,\beta]$ still holds.  Therefore assume WLOG $ \alpha\equal{}0$.  We have

$ (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}\le 2\beta^{2}$, where $ 0\le a,b,c \le\beta$.

Now substitute $ x\equal{}a\minus{}b$, $ y\equal{}b\minus{}c$, and $ z\equal{}c\minus{}a$.  We now want to prove

$ x^{2}\plus{}y^{2}\plus{}z^{2}\le 2\beta^{2}$, where $ \minus{}\beta\le x,y,z\le\beta$ and $ x\plus{}y\plus{}z\equal{}0$.

Assuming $ x,y,z$ are not all 0, then one or two of $ x,y,z$ must be non-negative.  If only one is non-negative, replace $ x,y,z$ with $ \minus{}x,\minus{}y,\minus{}z$, which still satisfy all our conditions and include two non-negative numbers.  WLOG $ x$ and $ y$ are non-negative.  Then

$ x^{2}\plus{}y^{2}\plus{}z^{2}\le x^{2}\plus{}2xy\plus{}y^{2}\plus{}z^{2}\le (x\plus{}y)^{2}\plus{}z^{2}\equal{}2z^{2}\le 2\beta^{2}$

as desired.[/hide]


It's a general polynomial.  What do you want us to do?


probability
Putnam
geometry
trigonometry
expected value
area of a triangle

If three random points are chosen on a unit circle and a triangle is formed and this process is repeated, what is the expected are of the overlap of the two triangles?

I think this is much too hard for this forum.

If I remember correctly, this problem is like an old Putnam problem that I really don't feel like looking for...only a lot harder.

If I had to venture a guess,
[hide="then I would say"]
let's remember that the area of a triangle is $\frac{1}{2}ab\sin{C}$.
The expected value of any angle in a triangle is 60 degrees. Therefore, the exval of any value of $\sin{C}$, given that C is in a triangle, is $\frac{\sqrt3}{2}$.
let's consider an equilat triangle in a unit circle. it's $ab$ value is $3$, maximizing $ab$.
the minimum $ab$ is zero. or really close to zero, whatever.
I'm kinda making stuff up here, but since we have an $ab=k$ relationship, the expected value of $ab$ in our case is $\frac{\sqrt2}{2}\times3=\frac{3\sqrt2}{2}$, and our expected $\frac{1}{2}ab\sin{C}=\frac{3\sqrt6}{4}$.
so we have two triangles of area $\frac{3\sqrt6}{4}$. the expected overlap is...uh...a geometric probability problem.
We are lucky since $(\frac{2}{\pi})\frac{3\sqrt6}{4}>1$.
The expected overlap is now $\frac{\frac{3\sqrt6}{4}+\frac{3\sqrt6}{2}-\pi}{2}=\frac{9\sqrt6}{8}-\frac{\pi}{2}$
[/hide]

N.B. I probably lied several times during that post, so don't assume this post is anywhere near correct.


function
algebra
functional equation
algebra proposed

f is periodic with period $ T \equal{} 2$, $ f(x \plus{} 2) \equal{} f(x)$.Is that the answer?

[hide]$ x\minus{}>x\plus{}1$ the functional equation gives: $ f(x\plus{}2)\equal{}\frac{5\minus{}f(x\plus{}1)}{1\plus{}f(x\plus{}1)}\equal{}\frac{6f(x)}{6}\equal{}f(x)$[/hide]

[i][u]solution[/u][/i]

Solving for $ f(x)$ the given equation becomes:

$ f(x)\equal{}\frac{5\minus{}f(x\plus{}1)}{1\plus{}f(x\plus{}1)}$. $ \boxed{1}$.

Now from $ f(x\plus{}1)\equal{}\frac{5\minus{}f(x)}{1\plus{}f(x)}$ setting $ x\minus{}\minus{}>x\minus{}1$ it becomes:
$ f(x)\equal{}\frac{5\minus{}f(x\minus{}1)}{1\plus{}f(x\minus{}1)}$ $ \boxed{2}$

From $ \boxed{1},\boxed{2}$ we have that:

$ \frac{5\minus{}f(x\plus{}1)}{1\plus{}f(x\plus{}1)}\equal{}\frac{5\minus{}f(x\minus{}1)}{1\plus{}f(x\plus{}1)} \Longleftrightarrow f(x\plus{}1)\equal{}f(x\minus{}1) \Longleftrightarrow f(x)\equal{}f(x\plus{}2)$

Done.

to generalize problems  $ f(x\plus{}c)\equal{}\frac{5\minus{}f(x)}{1\plus{}f(x)}$  where $ c \in\mathbb{R}$
 Azerbaijan   Land of the Fire

[i]solution to the generalization[/i]

Solving for $ f(x)$ we have that:
$ f(x)\equal{}\frac{5\minus{}f(x\plus{}c)}{1\plus{}f(x\plus{}c)}$. $ \boxed{1}$.

In $ f(x\plus{}c)\equal{}\frac{5\minus{}f(x)}{1\plus{}f(x)}$ setting $ x\minus{}\minus{}>x\minus{}c$ we take:

$ f(x)\equal{}\frac{5\minus{}f(x\minus{}c)}{1\plus{}f(x\minus{}c)}$. $ \boxed{2}$.

From $ \boxed{1},\boxed{2}$ we take that:

$ f(x\plus{}c)\equal{}f(x\minus{}c) \Longleftrightarrow f(x)\equal{}f(x\plus{}2c)$.

[quote="arqady"]Given function $ f: \mathbb R\rightarrow\mathbb R$ such that
\[ f(x + 1) = \frac {5 - f(x)}{1 + f(x)}\]
Prove that $ f$ is a periodic function.[/quote]
$ \iff f(x+1) + 1 = \frac{6}{1+f(x)}$
$ g(x) = f(x)+1: g(x)g(x+1) = 6$
$ g(x)g(x+1) = 6, g(x+1)g(x+2) = 6 \Rightarrow g(x) = g(x+2) \iff f(x) = f(x+2)$


[quote="FelixD"]Let $ M$ be the intersection of the diagonals of a convex quadrilateral $ ABCD$. Determine all such quadrilaterals for which there exists a line $ g$ that passes through $ M$ and intersects the side $ AB$ in $ P$ and the side $ CD$ in $ Q$, such that the four triangles $ APM$, $ BPM$, $ CQM$, $ DQM$ are similar.[/quote]
If the orientation of the triangles are according to the given data. (I mean if $ \triangle APM \sim \triangle BPM$ means, the angle at $ A$ equals the angle at $ B$ etc. ) The quadrilateral must be a square.

Sorry, I forgot to say that in this problem, $ \triangle APM \sim \triangle BPM$ does not necessarily mean that $ \angle A = \angle B$.  :wink:


Galois Theory
number theory proposed
number theory

Let $p$ be a prime. For any other prime $l$, denote by $\mathrm{ord}_{l}(p)$ the minimal integer $k$ such that $p^{k}\equiv 1\mod l$. Prove that for any $n$, there is a prime $l\neq p$ such that $p^{n}| \mathrm{ord}_{l}(p)$.

Of course, there is a standard proof for this, but I have a very funny "one-line" proof using infinite galois theory - try to find that one ;) .

Consider $ \overline{\mathbb{F}}_p$. Let $ \phi$ be the frobenius map $ x\mapsto x^p$. Then the fixed fields of $ \phi^{p^{n \minus{} 1}}$ and $ \phi^{p^n}$ are different. Take $ x$ such that $ x$ is in the fixed field of $ \phi^{p^n}$ but not in the fixed field of $ \phi^{p^{n \minus{} 1}}$. Let $ k$ be the order of $ x$.

Then $ k$ divides $ p^{p^n} \minus{} 1$, but not $ p^{p^{n \minus{} 1}} \minus{} 1$. Thus, the same is true for some prime power $ \ell^m$ dividing $ k$ and hence the order of $ p$ modulo $ \ell^m$ is $ p^n$. Let $ t$ be the order of $ p$ modulo $ \ell$. Then the order of $ p$ modulo $ \ell^m$ divides $ t\ell^{m \minus{} 1}$. As $ \ell\neq p$, we see that in fact $ t \equal{} p^n$.

(Well, there are variants that are more galois-theoretic and i suspect that my original proof was more like the following: Prove that $ \text{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p(\zeta_l|l\neq p))$ (here $ \zeta_l$ ranges over all primitive $ l$-th roots of unity, where $ l\neq p$ is a prime) is an inverse limit of groups with no $ p$-torsion. It suffices to see this for $ \text{Gal}(\mathbb{F}_p(\zeta_{l^k})/\mathbb{F}_p(\zeta_l))$ where it reduces to an argument similar to the above. Using $ \text{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p) \equal{} \prod_{l} \mathbb{Z}_l$ (over all primes $ l$, including $ l \equal{} p$), we see that $ \text{Gal}(\mathbb{F}_p(\zeta_l|l\neq p)/\mathbb{F}_p)$ contains a factor $ \mathbb{Z}_p$. Now think about where this may come from.)


The solution in the solution manual thing was "high-tech," using recursion, but I came up with a balls and urns like solution for this one. I changed the wording a little bit.

We have a sequence of numbers $a_{1},a_{2},\ldots,a_{10}$, such that:

$a_{i}=0$ or $1$ for all $i=1,2,\ldots,10$.
$a_{i}\cdot a_{i+1}=0$ for all $i=1,2,\ldots9$.
$a_{i}\cdot a_{i+2}=0$ for all $i=1,2,\ldots8$.

What's the question?  How many possible sequences?

This is almost identically AMC12 2007A #25.

I did a solution like that too.  You just have to realize that in any three consecutive $a_{i}$, 2 of them must be zero and then break it up into cases on how many 1s there are in the sequence.

I still got the question wrong due to an arithmetic error.


algebra
polynomial
linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

Let $f(x)\in F_{p}[x]$ be a monic polynomial of degree d, with
distinct roots in $\overline{F_{p}}$. For each $d\geq j\geq 1$, let
$r_{j}(x) \in F_{p}[x]$ be the unique polynomial of dgree at most
$d-1$ such that

\[(x^{j-1})^{p}=r_{j}(x) \,\,modulo f(x)\]

and let \[r_{j}(x)=a_{1,j}+a_{2,j}x+...+a_{1,j}x^{d-1}\]

with $a_{1,j}\in F_{p}$. Let $N\in M_{(}dxd)(F_{p})$ be the $dxd$ matrix
$N=(a_{i,j})$. Let r be the dimension (over $F_{p}$) of $Ker (N-I)$.
Prove that $r$ is the number of irreducible factors of f(x).

the matrix $ N$ represents the frobenius endomorphism of $ A : \equal{}\mathbb{F}_{p}[x] / (f)$ with respect to the basis $ 1,x,...,x^{d\minus{}1}$ (but the dimension of the kernel does not depend on the basis). now if $ f$ is irreducible, $ A\cong\mathbb{F}_{p^{d}}$ and thus $ \ker(N\minus{}1)\cong\{a\in\mathbb{F}_{p^{d}}: a^{p}\minus{}a \equal{} 0\}\equal{}\mathbb{F}_{p}$, i.e. the dimension is $ 1$. for the general case, we may split $ f$ into coprime (since $ f$ is separable) irreducible factors $ f_{1},...,f_{r}$. by the chinese remainder theorem, we have $ A\cong\mathbb{F}_{p}[x]/(f_{1})\times ...\times\mathbb{F}_{p}[x]/(f_{r})$. the frobenius restricts to each factor. so we may arrange  that $ N \equal{} diag(N_{1},...,N_{r})$, where $ N_{i}$ are the matrices corresponding to $ f_{i}$. since $ \dim\ker(N_{i}\minus{}1) \equal{} 1$ for all $ i$, we deduce $ \dim\ker (N\minus{}1) \equal{} r$.


Ten positive integers are arranged around a circle. Each number is one more than the greatest common divisor of its two neighbors. What is the sum of the ten numbers?

[hide="constructive"] Clearly, no number can be $1$. So start with $2$ and $3$. Then build the sequence

$2, 3, 4, 3, 2, 2, 3, 4, 3, 2$.

As for a semi-proof/reasoning, if we have $a, b, c, d$ in a row we know $(a,c) = b-1$ so $c \ge b-1$. Also, $(b,d) = c-1$, so $b \ge c-1$. From this we get $b = c+1, c, c-1$, so any two numbers must be consecutive. Unless $a = c$, we know $(a,c) \le 2 \Rightarrow b \le 3$. Then just experiment. [/hide]


trigonometry
calculus
calculus computations

$ f(x) \equal{} e^{\sin x}$.

Find $ f'(x)$ and $ f''(x)$.

Hence find the first three terms of the Maclaurin series for $ f(x)$.

hello, we have $ f'(x)\equal{}e^{\sin(x)}\cos(x)$ and from here $ f^{''}(x)\equal{}e^{\sin(x)}(\cos(x))^2\minus{}e^{\sin(x)}\sin(x)$.
Sonnhard.

I make

$ f(0) \equal{} e^{0} \equal{} 1$

$ f'(0) \equal{} 1\times e^{0} \equal{} 1$

and

$ f''(0) \equal{} 1\times e^{0} \minus{} 0 \equal{} 1$

giving

$ f(x) \equal{} 1 \plus{} x \plus{} \frac {1}{2}x^2$

as the first three terms in the Maclaurin series.

Stimmt das?

Can someone check this in case I've made a silly mistake (I have a negative self-confidence coefficient).

Let's try it this way:

$ \sin x\equal{}x\minus{}\frac{x^3}6\plus{}O(x^5).$

$ e^x\equal{}1\plus{}x\plus{}\frac{x^2}2\plus{}\frac{x^3}6\plus{}O(x^4)$

$ e^{\sin x}\equal{}1\plus{}\sin x\plus{}\frac{\sin^2x}2\plus{}\frac{\sin^3x}{6}\plus{}O(x^4)$

$ \equal{}1\plus{}\left(x\minus{}\frac{x^3}6\right)\plus{}\frac{x^2}2\plus{}\frac{x^3}6\plus{}O(x^4)$

$ \equal{}1\plus{}x\plus{}\frac{x^2}2\plus{}0\cdot x^3\plus{}O(x^4).$

So we can read off $ f(0)\equal{}1,\ f'(0)\equal{}1,\ f''(0)\equal{}1,\ f'''(0)\equal{}0.$

Just for checking you can also try http://www58.wolframalpha.com/input/?i=Taylor+Series+for+e%5E%28sin%28x%29%29

But seeing it worked out as Dr. Merryfield did is definitely better for your understanding.

Thanks very much , gauss202, I don't have any problem understanding this: I've used Kent's technique before which I first saw in a SUMS (Springer Undergraduate Mathematics Series) text (University Mathematics, I think, or maybe Mathematical Analysis). But I haven't come across WolframAlpha before; I'm grateful to you for this. Thanks.


geometry
trigonometry
algebra
function
domain
cyclic quadrilateral
trig identities

Suppose you have 4 sticks on the ground with all different lengths.
Each successive stick differs from another in length by one unit.
However, your sticks might not necessarily be of integer length.  Or they could be of integer length.  Their lengths might be listed as x, x+1, x+2, x+3.  For this problem, you want to consider those sets of sticks with lengths that will form only convex-shaped quadrilaterals when the sticks are placed end to end
on level ground (that is, in one plane).

Question:
------------

If the desired result is to maximize the area out of from one of the potential quadrilateral configurations formed by the sticks with lengths of
x, x+1, x+2, and x+3, then is there a constant way of ordering of these sides around the quadrilateral that does it, OR is the answer "It depends on the specific lengths of the four sides?"

In order to maximize the area of a polygon (including a quadrilateral) with fixed side lengths, we inscribe that polygon in a circle. The order of the sides doesn't matter; this is easy to see once we have the circle. Convex/concave doesn't matter either; the concave shapes have smaller area anyway.

[hide="A proof of the quadrilateral case"]Let the quadrilateral be $ ABCD$, with sides $ AB=a,BC=b,CD=c,DA=d$. Let the angle at $ A$ be $ \theta$, and the angle at $ C$ be $ \phi$.
For $ ABCD$ to close up, $ BD$ must be the same length in triangles $ BCD$ and $ DAB$. By the Law of Cosines, this gives $ b^{2}+c^{2}-2bc\cos\phi=a^{2}+d^{2}-2ad\cos\theta$.
The area is then $ \frac12(bc\sin\phi+ad\sin\theta)$. We use Lagrange multipliers; the critical point comes when $ 2\lambda bc\sin\phi=\frac12 bc\cos\phi$ and $ -2\lambda ad\sin\theta=\frac12 ad\cos\theta$. Solving the system, $ \lambda=\frac14\cot\phi=-\frac14\cot\theta$, and $ \cot\phi=-\cot\theta$. Since $ \theta+\phi$ is less than a full circle, $ \theta$ and $ \phi$ must be complementary, making $ ABCD$ cyclic.
We should also check the extreme values; these occur when one angle of the quadrilateral is zero. If this occurs, the quadrilateral is not convex, and the area is obviously not maximal.[/hide]

[hide="A proof of the general case"]Lemma: If a polygon with five or more sides is not cyclic, there is another polygon with the same side lengths in the same order which has strictly greater area.
Proof: We assume that the polygon is convex; this is trivial if it isn't. Since the polygon is not cyclic, some four consecutive vertices don't lie on a circle; call these vertices $ A,B,C,D$, with $ AB,BC,CD$ sides of the polygon and $ AD$ a diagonal. Cut the polygon at diagonal $ AD$; the total area is the sum of the areas of the quadrilateral $ ABCD$ and the cutoff polygon. By the quadrilateral case, we can find a cyclic quadrilateral $ AB'C'D$ with the same sides and greater area than $ ABCD$ (there is a choice of orientation: make it stick out). Replace $ B$ with $ B'$ and $ C$ with $ C'$; the new polygon has greater area than the old.

This lemma assures us that a maximizing polygon must be cyclic. We can show that there is a maximizing polygon by parametrizing all possible polygons (with the given sides) by $ n-3$ of their angles, where $ n$ is the number of sides. The domain of possible angles is a compact set in $ \mathbb{R}^{n-3}$ and the area is a continuous function of these angles, so a maximum must exist.[/hide]


Let p and q be the radii of two circles through A, touching BC at B and C, respectively.
Show that pq = R^2.

This is a problem from Geometry Revisited 1.1
I can't visualize the picture, so can someone explain the problem to me?

I don't really get it. What's R? Is ABC a triangle?

Yeah and what does "through A" mean? Are they concentric circles or do both of the circle's circumferences touch point A?

Well, it sounds like you have two circles passing through a common point, each tangent to a segment (BC) at one of its endpoints.  In other words, two intersecting circles and their common external tangent.  However, there's obviously more information needed.  For example, we don't know what R is.  Perhaps there is a diagram there in an odd place?  Or information about this question somewhere nearby?

R is the radius of the circumcenter of Triangle ABC.
I am assuming that based on previous problems that the book has.

Like xirtamax says, R is the circumradius (this is a convention used in the book).
After thinking about this problem, I came to one conclusion: its wrong.  Clearly a circle through two given points can have any radius you want as long as it greater than than half the distance between the two points.  The two circles are completely independent of each other.  Thus pq can have an infinite number of values, whereas R, which is independent of p and q, can have only one value.


linear algebra
matrix
advanced fields
advanced fields solved

We have a linear space V, dimV=n, and a subspace W, dimW=k.We consider a bilinear symmetric form g defined on V.We call the set {x from v :g(x,u)=o for every u in W} the ortogonal of W.
Is it true or not that the ortogonal of the ortogonal of W is W?If not, what conditions must satisfy g for this to be true?

If g is not denerate then o(o(W))=W where o(W) is the orthogonal of W
g not degenerate means the matrix of the bilinear form g is invertible.


Putnam
calculus
MIT
college
complex numbers

Argh! I'm been inactive for way too long! It's been more than 2 years since my last (and first) IMO, and still stuck in the army. (You see, our country practices peace time conscription)

My problem now is that my problem solving skills are patchy. That is, I can do some stuff but I can't do others.

Does anyone have any advice on how I should best approach this? Has anyone else experienced a similar problem? I'm doing my best to do plenty of math, but I haven't quite been able to reach the standard that I was at before. I haven't been able to solve many IMO-standard questions lately, let alone putnam-standard questions. I guess it serves me right for slacking.

I guess I'll have plenty of ground to cover in college to even stand a chance in the putnam. Any advice on what I should do, what books would be good for preparing for the putnam?

I feel a bit like some hermit that's come down from his mountain to a big city. That's what the army does to you.

Well, math is a lot like riding a bicycle.  Maybe.

But since you're on this site, you've already got a good start: a source of problems.  Practice, and most likely it'll come back to you.

It may sound like "easy for me to say" .. but I  would say don't worry about it and enjoy.. scores in putnam etc would take care of themselves. Some colleges here (in US) have problem solving seminars type courses, ( specially tailored to Putnam type tests,  which  could   be fun)

Many great people have had these "dry spells" and self doubts, In his autobiagraphy  Feynman writes of one such spells where he  finally   said something like "Heck .. I will  work on whatever problem I like for fun" and ended  up doing   some really great work and he was "back in business".

Do your best to get into stuff, activities  and courses which you really enjoy and want to do.

I guess quick cures are too much to hope for! *grin*

I guess I'm just a little shell-shocked at my decline.

Does anyone know of any source of online courses for math and physics? I need to beef up my fundamentals. I think I've forgotten my calculus, matrices, complex numbers, uh, and just about everything, except the basic, easy stuff in all these areas. Would be nice if I could find notes on these online.

You could try MIT OCW [url]http://ocw.mit.edu/index.html[/url] .  The stuff on the site is mostly above me but I think that there is probably a lot of stuff that you would be interested in.

Uh oh. Clearly I am in trouble. I have to go borrow high school notes from my juniors to understand some of the stuff. How embarrassing.
*blames government for everything*

Hmmm... And I volunteered to train my juniors.
*desperately searches for quick fix*

I guess I will have to start from the topics I've always been good at, and throw plenty of obscure problems at them. One of my main problems seems to be that a few of my juniors look at any given list of problems we give them, and start saying, "seen 2 before *tick* seen 5 before *tick* seen 6 before *tick*", prompting us seniors to panic and dig out the backup problems. (If you're reading this - you know who you are! But it's great that you all are working so hard. Gold next year okay?)

Does anyone know of any place I can get really obscure problems that's suitable for a rather good team? Okay, I know it's not likely, but IF anyone knows of any place where I can get nice problems that few people know about (with solutions, I'm not very good yet), could you please PM me? Goodness, I wish I knew how the training's done in other countries.

To keep one's brain in top condition,
I recommend a daily cleaning with mental floss.


Here's a really lame riddle (the answer is so lame you're going to rate my post spam once you find it out).

There are 3 turtles racing on a non circular track.

The first turtle says that when he finished, he saw no turtles ahead of him.

The second turtle says that when he finished, he saw 1 turtle ahead of him.

However, the third turtle also says that when he finished, he saw 1 turtle ahead of him.

How could the be?

Because turtle two and three finished at the same time.  :rotfl:  Due to their extremely poor vision they can't see the turtle running along side them.

[hide]
The second place turtle blocks the view of the thiird place turtle, so the third place turtle can't see the first place turtle.
[/hide]

The first turtle was very far ahead of the 2nd, and the 2nd was slightly far ahead of the third. The first one finished, and started crawling toward the hallway for the prize. By the time the 2nd finished, the 1st one almost reached the hallway to the other room for his prize. When the 3rd turtle finished, the 1st turtle already was in the hallway for the 1st place prize, and can't be seen from the racetrack. He can only see the 2nd turtle crawling in that direction.

In other words, the 1st one started leaving the track when he finished, was still in sight when the 2nd one finished, but went too far away by the time the 3rd turtle finished.

You have introduced a "hallway" and a "prize" out of nowhere in your story
for your lameness.  Instead, I would have wanted the answer to have hinged
upon 'bad/poor' puns to enjoy/suffer through for the lameness.

All of these are wrong  :P
The answer is way lamer than any of these  :wink:

When the third turtle reached the finish line the first turtle had already gone to the bar for a drink.

[quote="Fraenkel"]The turtles 2 and 3 are conjoined twins.[/quote]

 :rotfl:  Yeah, also they are probably the ones where they have two conjoined head.

[hide="the lamest thing I can think of"]
One turtle randomly got sucked into a black hole.
[/hide]

[hide="Well, this seems like a lame answer"]A few of the turtles have second sight.[/hide]
[hide="Lamest answer possible"]THE SOUND OF ONE HAND CLAPPING![/hide]

[hide]The third turtle started from way behind.  When the first and second turtles finished, the first kept going, and the second one stopped.  Therefore when the third one finished, it only saw the second turtle.[/hide]

EDIT: Never mind.  I saw that ProtestanT had this idea before me.

[quote="DonkeyKong"][quote="Temperal"]No, I've heard this riddle before.

The answer:

[hide="It really is lame"]
The third turtle lied.[/hide][/quote]
Phewahahahahaha!!!! :rotfl:[/quote]

 :rotfl:

Yeah I pretty much died of laughter when I saw this answer...

[quote="Temperal"]No, I've heard this riddle before.

The answer:

[hide="It really is lame"]
The third turtle lied.[/hide][/quote]

My head just exploded.

[quote="Nerd_of_the_Ages"]the answer is so lame you're going to rate my post spam once you find it out[/quote]

And it looks like that is turning out to be the case.

You know, I think what we SHOULD be worrying about is not lying turtles, but talking turtles in general.

I would have freaked out right after the first turtle said there was no one in front of him.

The version I heard was about the order of a hare, tortoise, and a kangaroo.

He can't have lied if he only saw one, and in all competitions with racing and turtles, theres hallways and prizes.

But anyways the answer was pretty funny :D

[quote="Temperal"][quote="ProtestanT"]He can't have lied if he only saw one, and in all competitions with racing and turtles, theres hallways and prizes.[/quote]

Uh, what? Why can't he have lied if he only saw one?[/quote]
because then he said the truth, and saying the truth isn't lying, therefore he can't have lied in that situation since the problem forces him to say the truth, which is only seeing one.

edit-By That situation i meant the situation where he only saw one because the first one went to another hallway/exited the stadium. It's the most likely situation because theres almost always prizes for races with racers being interviewed.

Oh, you meant if the first turtle went to the "hallway" for the "prize" and "interview". Got it. That makes sense.

But you were warned to give the lamest answer!

The pea-brained turtle definitely don't know how to lie :P
[hide="What i am thinking is"]
the one who tell this lame riddle lies :D
[/hide]

:huh:
Hi

Why would I lie. It's a lame joke, turtles can lie in jokes.  :P

since its NON circular the 3rd turtle shouldnt be able to see the first turtle because the 2nd turtle is blocking the way.


Do we know how many M7m's he's giving to each kid?
If its one each wouldn't it be 5 choose 3.

[hide]5 choose 3 = $\frac{5\cdot4\cdot3}{3!}=10$[/hide]

[quote="anirudh"][quote="math92"]In how many ways can Horatio give M&m's to three kids? (m&m's come in red, yellow, green, brown, and tan.)[/quote]

5c3[/quote]
can you give me a number?

[quote="mathmanman"]10 possibilties then. :)[/quote]

Isn't it a number?  :P

i was talking about anirudh because he wrote "5c3" :)

[hide]
there are three kids and 5 m&m's. So the answer has to be 5C3 = 10[/hide]

there are a certain number of ways of dividing 5 M&M's among 3 kids.
Let's say that we have child A, child B and child C.

First, how many different ways would you divide them up if you allowed only one M&M per child? In other words, how many different combinations of colors would you have using only 3 M&M's at a time?

Second, how many different ways would you divide them up if you allowed 2 for one child and one each for the other two. In other words, how many combinations would you have using 4 M&M's at a time?

Third, how many different ways would you divide them up if you allowed 2 for two children and only one for the third child, using all 5 M&M's?

Will be working on a solution.

The solution to the first part : if 3 M&M's are handed out

[hide]There are five different colors with three different children.

The answer would be a permutation, not a combination.

[tex] _5  P  _3 = (5)(4)(3)  =  60[/tex]

So just handing out 3 M&M's alone would create 60 different ways.

solution to the second part will follow in another post[/hide]

solution for the second part: if 4 M&M's are handed out

[hide]
let's say child A has 2 M&M's and child B and C each have one.

now supposing that A retains those two M&M's while B and C are given diferent combinations from the remaining three.

now suppose that B is given one and retains that one leaving only 2 for C to choose from.

we then have [tex] _2P_1 = (2)(1) =2[/tex]

every time we give B a different M&M of the remaining three, C will have a different combination of colors to choose from.

so then we have [tex]3(_2P_1)[/tex] or 6 different combinations between B & C.

so for each combination of 2 that A has, there are 6 different combinations between B & C.

The next thing we need to find out is how many combinations is possible for A.

That would be [tex]_5C_2 = \frac {(5)(4)}{(1)(2)} = 10[/tex]

If it is possible for A to have 10 different combinations of 2 then we would have a total of [tex] (10)(6) = 60[/tex] combinations among the three children.

Now repeat this process when B has two M&M's and C has two M&M's

there will then be a grand total of [tex](60)(3) = 180[/tex] or 180 combinations
[/hide]

the solution to the third part will follow in another post

solution for the third part: if 5 M&M's are handed out

[hide]
let's say that child A and Child B are both given 2 M&M's and child C is given one

now supposing that A retains those two M&M's while B and C are given different
combinations from the remaining three

now suppose that B is given two and retains those two leaving only 1 for C to choose from. This would obviously give us only one combination.

Next, we need to find how many possible combinations child B can obtain from the remaining three M&M's.

that would give us: [tex]_3C_2 = \frac {(3)(2)}{(1)(2)} = 3[/tex]

so with A retaining 2 M&M's there are a total of 3 combinations between childs B & C

as in the second part child A was able to obtain 10 different combinations

therefore, there will be a total of [tex] (10)(3) = 30[/tex] or 30 combinations among the three children.

then taking into account the situations where A & C have 2 M&M's and B & C have two M&M's you will have a grand total of [tex](30)(3) = 90[/tex] or 90 combinations

adding the total of all three parts, we have [tex]60 + 180 + 90 = 330[/tex] or 330 combinations.
[/hide]

These are only cases where all the children will have at least one or two M&M's

You might want to check me out on my answers

If you think about it, the problem is undefined.  Although, math92 specified the types of M&Ms we are dealing with, he did not tell us the quantity of M&M's of each colour, thus; making it impossible to say how many was we can distribute an undefined amount to three children.

LOL, guess what, another ordered partitions problem.

Assuming the three boys are distinct :D (unless there clones)

 Heres my solution:

The number of ways we can distribute 5 distinct MM"S is the same as finding how we can distribute the partitions of 5, in regard to the distinct candy.

5,0,0=3 ways

4,1,0=5 * 3 * 2 *1=30 ways

3,2,0=10*3*2*1=60 ways

3,1,1=10*2*1*3*2*1=120 ways

2,2,1=10*3*1*3*2*1=180 ways

  For a total sum of 393 ways.


I like finding basic electromagnetic problems to solve. I don't know where to post this - but it does involve calculus.

Consider a copper wire that is a continuous circle of radius r. Let the area of the circle be 1 square meter.
 Let the diameter of the wire be very much smaller than r. Let the wire carry a steady current of 1 ampere.  What is the total flux in Webers that passes through the circle?

I assume that the flux density is greatest near the wire and is smallest at the axis of the circle. Evidently the flux density is proportional to 1/r^2

I have a solution for this but don't know if it is correct

So I will try to give an answer to this.  Expect several postings on this.

 The cirlce of copper wire is on a plane surface. The area of the plane encirled by the wire is one square meter. There is a continuous current of 1 ampere flowing in the wire. The area of the plane encompassed by the circle is divided into many small washers of equal width $\Delta r$.  Each washer has an area of $\pi 2r\Delta r$

Then the flux is calulated for each washer and the approximate flux is the sum of the flux for every washer where the sum of all washers are equal the area of the circle.

If the washers are made of smaller widths then eventually we will have to take the sum of an infinite (number) of washers. This of course leads to an integration

I will post the rest of this later.  comments?

how about using amperes law, whisch states-

$\int B\,{\vec{\{dl}}$= $\mu_{0}$/ $i_{net}$ = Flux thru an amperean surface . As of now I have not been able to identify any such surface but still......

Yes I realize that Amperes law is needed here. Also the names Biot and Savart come up when you investigate Amperes law.  Let me state that I have not picked an easy problem to solve so it seems.  Anyone out there interested in electromagnetics??

We live in an electromagnetic world. Our eyes can see a portion of the electromagnetic spectrum.

Come on people open your eyes and ears and get yourself away from war craft games and find out WHY for a change instead of letting the geniuses of the world do it for you.

Anyway it seems that no one is interested in the most fundamental laws of electromagnetics so I will now end this thread forever.


AMC
AMC 12
AIME
geometry
USA(J)MO
USAMO
email

If I study with AoPS Volume 1 and 2, can I get at least 100 on AMC12?
By the way, will AoPS Volume 2 help me do well on AIME??

Yes, those two books should get you a pretty decent score on AMC12 and AIME, along with practicing old tests.

Art of Problem Solving 1 and 2 alone will give you the skills needed for problem solving. But, to really do well on an AMC12, you should also do their specialized books, like the intermediate combinatorics and intermediate algebra. I hear an intermediate geometry is coming out soon also.

You should also do practice tests, as these will help you manage your time during the real thing.

If you understand everything very well in AoPS 1 and AoPS 2, you should be able to make USAMO.

By "understand everything very well" I mean being able to solve almost all of the problems.

Thanks guys!
I have one more question.  How does award system work for AMC 10 and 12?
If you qualify for AIME, then do you get a medal?

no offense, but I would suggest concentrating more on the mathematics part than the prizes. but, the stuff is on the AMC FAQ.

Yes, the prizes come and go, but the overall prize is the development of the mathematical mind.

(And making AIME/USAMO) :P

The skills on an AMC12 can only be learned through a combination of practice and talent. A book cannot assure you a certain score; AoPS volumes 1 and 2 pretty much give all the actual mathematical tools you need.

do you think AoPS 1 is enough to get 100+ (yes i know, a low goal, but nonetheless, a goal), or should I get the 2nd book?

I qualified to AIME with the minimum score on AMC 12 last year without knowing a lot of the topics in Volume 1.  If you study it thoroughly and understand how to do most/all of the problems in Volume 1, you should be able to qualify for AIME.

[quote="stanford1463"]Ok thanks...but you know any specific chapters from Volume I that I should know? And specific formulas/methods/techniques?[/quote]
It's stressed in the book that you shouldn't learn formulas or methods but learn where they come from. If you know how to derive a formula, you won't need to memorize it.  :)

DUDE DO NOT JUST SKIM THROUGH THE BOOK OR YOU WON'T EVEN GET A 50 ON THE AMC 12 DO ALL THE CHAPTERS NO MATTER HOW USELESS YOU THINK THEY ARE

That's not necessarily true. If you're just aiming to make AIME, I don't think you need to do all the chapters. Infact, some chapters such as chapter 8 (limits) aren't even needed for AIME. You should still read over each chapter sometime though.

Also, I think that if you even [i]have[/i] Volume 2, you should be able to easily get over 50 on the AMC 12. Volume 1 contains enough to give you the tools necessary to get 100+.

^ ok thanks, that's the answer I was looking for. So, I will comprehensively study the Volume I book..and btw I got a high 80/low 90 last year already...I can get 10 questions right, but I'm aiming to get 16 questions right, and leave the rest blank this year, so I'll guarantee a 100+

Note that you only need to get 3 questions right and leave the rest blank to get over 50.

Yeah.  The scoring system works like this:

6 points for correct answer.
1.5 points for blank (used to be 2.5, but changed now)
0 points for incorrect.

[quote="xpmath"]That's not necessarily true. If you're just aiming to make AIME, I don't think you need to do all the chapters. Infact, some chapters such as chapter 8 (limits) aren't even needed for AIME. You should still read over each chapter sometime though.[/quote]

Off the top of my head, I can remember that invariance/sequence problem [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_14]AIME 2007 I #14[/url] and that highway problem [url=http://www.artofproblemsolving.com/Wiki/index.php/2008_AIME_I_Problems/Problem_12]AIME 2008 I #12[/url] that needed at least an intuitive sense of limits.

But anyway, I agree with AIME15: I managed to scrape a 103.5 on last year's AMC 12, qualifying for AIME, after only skimming AoPS 1/2 for some 2 weeks before, doing maybe a problem or two from every other chapter of Vol 1, not even opening Vol 2, and I hadn't done any contest math since AMC/MATHCOUNTS 3 years before then. If you have just random math "goodness" swimming around in your head, sometimes it just takes a good book like AoPS 1/2 to get you straightened out.

I solved number 12 on aime this year.  i remembered trying out different values for the speed, and dividing, and i noticed that it went up, then down.  I knew nothing about limits.

[quote="rOadToAIME"]What are the perfect scores for AMC12, AIME, and USAMO?[/quote]
150, 15, and 42, respectively

A perfect score on the AMC 12 is a 150 (with 25 questions), on the AIME it is a 15 (with 15 questions) and on the USAMO it is 42 (with 6 questions).

I looked up the official AMC site, but I'm still confused since it's my first time taking this test. :)
I found out that my school does not offer AMC. However, there's a university near where I live that offers the test.  How do I register for AMC12 if I'm going to take the test at the university not at my high school?  What's the deadline and cost for the registration?
By the way, I'm the only one who's going to take AMC.  My school doesn't have a math club.

You can just talk to the college and take it there.  I did that for AMC 12 last year at my high school (was in 6th grade).

That question's been answered...3 times now?  You don't need to revive a thread, especially to answer an already-answered question.

this should help you, not necessarily for aops 1/2, but for books that could help on amc-usamo in general:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1789292352&t=242439

So what are the chapters in aops that usually show up in AMC 12? What chapter is really important?

SilverFalcon's AMC guide answers that.  Email him for information.


floor function
LaTeX
advanced fields
advanced fields unsolved

prop1:  given n non-colinear points s.t. they form a convex polygonal region, then the number of convex quadrilaterals is $ \binom{n}{3}*(n \minus{} 3)/4$.

prop2:  add 1 point inside those n points as in prop 1, to form total n+1 points, then the number of convex quadrilaterals is $ \binom{n}{3}*(n \minus{} 3)/4 \plus{} C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3)$.

prop3, add 2 random points inside those n points to form n+2 points, then the number of convex quadrilaterals is at least $ \binom{n}{3}*(n \minus{} 3)/4 \plus{} 2*C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3) \plus{} \lfloor_{n/2}\rfloor*n \minus{} \sum_{t \equal{} 1}^{\lfloor_{n/2}\rfloor}{2t}$

prop4, add 3 random points inside, total n+3 points, then the number is at least $ \binom{n}{3}*(n \minus{} 3)/4 \plus{} 3*C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3) \plus{} 3*\lfloor_{n/2}\rfloor*n \minus{} 3*\sum_{t \equal{} 1}^{\lfloor_{n/2}\rfloor}{2t}$

prop5, add 4 random points, total n+4 points, then the number is at least $ \binom{n}{3}*(n \minus{} 3)/4 \plus{} 4*C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3) \plus{} 4*\lfloor_{n/2}\rfloor*n \minus{} 4*\sum_{t \equal{} 1}^{\lfloor_{n/2}\rfloor}{2t} \plus{} 1$

theorem 1:  given total n+k points, s.t. n points form a convex polygonal region, and k points are inside the convex polygonal region, then the number of xonvex quadrilaterals is at least $ \binom{n}{3}*(n \minus{} 3)/4 \plus{} k*C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3) \plus{} \binom{k}{2}*\lfloor_{n/2}\rfloor*n \minus{} \binom{k}{2}*\sum_{t \equal{} 1}^{\lfloor_{n/2}\rfloor}{2t} \plus{} \lfloor_{k/4}\rfloor$

Draft of theorem 2: given $ n \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_k \plus{} s$ ($ s$ is less than 5) points such that $ a_1$ points form a convex polygonal region and such that all other points are inside this convex polygonal region; $ a_2$ points form another convex polygonal region such that all points except $ a_1$ points are in this convex polygonal region, and so on; and $ a_k$ points, which is inside all other convex polygonal regions, also form a convex polygonal region s.t. only $ s$ points are inside it, then using theorem 1 we can derive a lower bound for the number of convex quadrilaterals formed, not necessarily the least upper bound.

Note, $ C_{q}$ is the $ q$th Catalan number.

The proofs are very simple and elementary, so I omit it here, also partly because it's so difficult to type.  I think anyone who has spent some time on this problem can see how I derived those equations.  I welcome anyone and everyone to correct me if I am wrong.

A few quick LaTeX suggestions to make this more readable:

\cdot instead of * is a more attractive multiplication dot; \frac{}{} makes a nice fraction; and I don't know why you're using subscripts in the \lfloor n/4 \rfloor construction, but it makes it very difficult to read.  (If you're going to use both the \lfloor and \frac commands, you can also use \left\lfoor \frac{}{}\right\rfloor to resize.)

Also, for everyone else, this seems to have come from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316679


geometry
trigonometry
circumcircle
LaTeX
angle bisector
perpendicular bisector

hi!

I'm new here and i'm very happy yo see that there is such a group!

i want to share with you something that crossed my mind.

is it me or in any triangle a bisector of an angle is between the height and the
median? am i right? i've tried to prove it but withoust any success.

so, if there is a proof (only in geomtry, because i'm don't know trigonometry yet) i'd like to see it!

thanks!!!

Put AD, AX, AM as the altitude, angle bisector, and median from A.

Let $AC \ge AB$.

Now the perpendicular bisector of BC meets the circumcircle in the arc BC not containing A at K, and in the arc BC containing A at K'.  But AXK is collinear and A is in the minor arc BK', so X is necessarily to the left of M.

Also, A in minor arc BK' guarantees that AD is to the left of AX, since AX necessarily goes to the right (to K which is to the right of A) while AD goes straight down.

hi

thank you but i didn't understand the proof. can you draw a sketch or explain the proof in another way?

thank

Now you draw a circle, with inscribed triangle ABC. Let $AC> AB$ (since if $AC = AB$, it is isoseles triangle, and altitue, angle bisector and median are the same).

Draw altitue, angle bisector and median $AD$, $AX$ and $AM$ respectively.

From $M,$draw the perpendicular line with $BC$, cuts the circle at two points named $K$and $K'$ ($K$ is in the arc $BC$ without $A$ and $K'$ is in the arc $BC$ with $A$). Now you can see that $AXK$ must be collinear because $MK$ is perpendicular to $BC$ => $K$ is the mid point of $BC$. Besides, $AX$ is the bisector of angle $A$,so $AX$ must pass the mid point of arc $BC$, which is $K$. So $AXK$ is collinear.

And as sheketli explained as above:

A is in the minor arc BK', so X is necessarily to the left of M.

Also, A in minor arc BK' guarantees that AD is to the left of AX, since AX necessarily goes to the right (to K which is to the right of A) while AD goes straight down.

(Just sketch it, and you can see right away. Sorry I don't know how to use Latex to sketch that for you)


Al walks down to the bottom of an escalator that is moving up and he counts 150 steps.  Bob walks up to the top of the escalator and counts 75 steps.  If Al's speed is 3 times bob's speed, how many stps are visible on the escalator at any given time?

The only solution given when I searched it I didnt quite understand and I want to know various ways of doing these types of problems.

Thanks.

Assuming you read 4everwise's two line solution, the [url=http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_10]AoPSWiki[/url] might yelp you. *advertises*  :)


ceiling function
algebra unsolved
algebra

A=(2+Sqroot 3)^6
B>A
B is positive integer

What minumum of B is ?

[quote="telemat"]Can you explain a little
and what is result?[/quote]

What Rust said is :

Let $ A_n\equal{}(2\plus{}\sqrt{3})^n$
Let $ C_n\equal{}(2\minus{}\sqrt{3})^n$ :

1) $ A_n\plus{}C_n$ is a positive integer $ \forall n>0$
2) $ 0<2\minus{}\sqrt{3}<1$ and so $ 0<C_n<1$ $ \forall n>0$

So $ A_n\plus{}C_n\minus{}1<A_n<A_n\plus{}C_n$ $ \forall n>0$

So $ \lceil A_n\rceil\equal{}A_n\plus{}C_n$

And since you are looking for $ \lceil A_6\rceil$, the number requested is
$ A_6\plus{}C_6\equal{}(2\plus{}\sqrt{3})^6\plus{}(2\minus{}\sqrt{3})^6$ $ \equal{}2(2^6\plus{}15\times 2^4\times 3\plus{}15\times 2^2\times 3^2\plus{}3^3)$ $ \equal{}2702$

And Rust gave you a general method for $ \lceil(2\plus{}\sqrt{3})^n\rceil$.


calculus
derivative
ratio
advanced fields
advanced fields theorems

V and T are vectors.

If $ \vec V$ is parallel to $ \; \nabla u$ and
If $ \vec T$ is parallel to $ \; \nabla u$ this implies that
$ \vec T \equal{} \mu \vec V$ where is $ \mu$ is a real number.

Does this imply that the directional derivatives, $ D_t u \equal{} D_v u$, are equal as well?

Even if you demand normalization in your directional derivative, the answer is no- because $ \mu$ could be negative.
If you know $ \mu$, the answer is predicatable; it differs by a factor of $ \mu$ or $ \text{sgn}(\mu)$, depending on your definitions.

Thank you for the quick reply.  Just adding more detail and a follow up question:

I'm sure you already know, but $ \mu$ is coming from the ratio:

$ \frac{\frac {dx_0 (t_0)}{dt}} {a(x_0,y_0)} = \frac{\frac {dy_0 (t_0)}{dt}} {b(x_0,y_0)} = \mu$

which comes from:

\begin{align} a u_x + b u_y +c u = f \end{align}

I will be able to determine $ \mu$ from eq (1)? Either numerically or in equation form?


Determine the limit of the following sequence:
a_n=sum(k=1 to n) from (k!)/(x_1+1)(x_2+2)...(x_k+k) where x_n is a sequence of positive numbers, x_n is convergent to 1.

If x_n <1 for n>=n0 we have that a_n>1/n+1 so sum of a_n is divergent.
But what if x_n>1 for n>=n0 or x_n has an infinity of terms lower tha 1 and an infinity of terms greater than 1?


linear algebra
matrix
search
algebra
polynomial
geometry
calculus

Hi every body!
I'm finding some documents about companion matrix. I'm very pleased when you send me some files or post them on forum about it, so that I can get some thing to do my topic. Thanks very much

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127175]This problem[/url] that I posted would be one place to start.

The argument by jmerry [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28984]here[/url] is another thing well worth looking at. You can also search for other mentions of rational canonical form.

Any single $n$-th order constant coefficient linear differential equation (we're talking about ODE here) can be rewritten as a first-order linear system of differential equations $y'=Ay,$ where $A$ is an $n\times n$ matrix that is a companion matrix.

A matrix is similar to a companion matrix if and only if its minimal polynomial is equal to its characteristic polynomial- this excludes notable matrices such as the identity. Look up "rational canonical form" for more information.

[quote="dhdung"]can an any matrix be brought back to a companion metrix? And what are the role of companion matrix in algebra?[/quote]

Role of matrices with a simple structure - to make some facts obvious, so diagonal, row- and column- simplified matrices, triangle matrices, etc. - object of such observations and just collections.

Companion matrix is used not in matrix algebra - only, but in different areas ODE including: controllability, observability of Kalman, his lecture, SPb last summer http://www.artspb.com/indexkalman.html

http://mathworld.wolfram.com/CompanionMatrix.html

Due the simple structure, it's the most popular matrix together with diagonal form.

Just if matrix has no a simple form, it could be block-conversed, to some similar or derivative form, if reasons are enough to do this operation. Matrix algebra is'nt a frozen area (as itseems, on the first look), where all forms determined and collected, like in "Wolfram Book", it's living field for researches. All depends on your aims and tasks. Due simple matrices you get orientation and find some common language, it's a role too.

[quote="dhdung"]Prove that : A companion matrix is always similar to a matrix that has rational canonical form.[/quote]

Rational canonical form is form, that consists companion matrices as blocks, by determination.

http://mathworld.wolfram.com/RationalCanonicalForm.html

You have no needs to prove it.

Let's take matrix that has rational canonical form, block-structure follows by some eigen-values are equal eath-other (see link), so it's enough to make these numbers different, and you will have only one block, i.e. "pure" companion matrix. Those matrices (original and "pure" companion) are similar in this sense.

Sorry I had a mistake.
Prove that : A square matrix always similar to a matrix that has rational canonical from.

[quote="dhdung"]Please help me to solve it.[/quote]

Similar matrices are those, who share eigen values (if do not observe a special case of equal to eath-other eigen values, where appears those blocks and need some additional structure signs). The special case leads to blocks, so we have some clouds on the horizon, not very serious - but a field for algebraizations (to life of people learning algebra was not so boring, obviously).

Line of coefficients of companion matrix (or rational canonical form, in common) depend on its eigen values  known way (look properties), these eigen values free in choise.

It means - there is no problems to find for 'any' square matrix - a companion matrix (or rational canonical form, in common) choosing the same eigen values for both of them. Hope, it solves doubts a simpliest way.


inequalities
function
induction
inequalities unsolved

Oh this inequality is not as hard as I thought.  :)

Without loss of generality we can assume $x_{1} \ge x_{2} \ge ... \ge x_{n}$. Then \[\displaystyle\sum_{i=1}^n \frac{x_{i}}{1+\frac{x_{1}x_{2}x_{3}...x_{n}}{x_{i}}} \le \displaystyle\sum_{i=1}^n \frac{x_{i}}{1+x_{2}x_{3}...x_{n}} = \frac{x_{1}+...+x_{n}}{1+x_{2}x_{3}...x_{n}}.\]
I am going to prove that \[\frac{x_{1}+...+x_{n}}{1+x_{2}x_{3}...x_{n}} \le n-1,\]or \[x_{1}+...+x_{n} \le n-1 + (n-1)x_{2}x_{3}...x_{n}.\]
If $x_{2}...x_{n-1} \ge \frac{1}{n-1}$, it is obviously true, so we just consider the case $x_{2}...x_{n-1} < \frac{1}{n-1}$. Before writing the proof, i firstly introduce two lemmas.

lemma 1: $x_{2}+...+x_{n-1} \le x_{2}...x_{n-1}+n-3$ for $n \ge 3$.
It can be easily proved by induction.

lemma 2: $n^n \ge (n+1)^{n-1}$ for positive integer $n$.
It can be proved by AM-GM.

By lemma 2, $x_{n} < \frac{n-2}{n-1}$, otherwise $x_{2}...x_{n-1} \ge x_{n}^{n-2} \ge (\frac{n-2}{n-1})^{n-2} \ge \frac{1}{n-1}$, which contradicts to our assumption. Hence by it and lemma 1,
\[x_{1}+...+x_{n} \le 1+x_{2}+...+x_{n-1}+\frac{n-2}{n-1} \le 1 + n-3 + x_{2}...x_{n-1}+\frac{n-2}{n-1} \le n-1 \le n-1+(n-1)x_{2}...x_{n}.\]


modular arithmetic
number theory unsolved
number theory

Hi ,

Prove that if p>3 is a prime then

$6(p-4)! \equiv 1 mod p$

Looks like its related to Wilson Theorem ??


The numbers w and z are complex conjugates.  The sum of the squares of w and z is 7.  The sum of the cubes of w and z is 10.  What is the largest value that w+z can have?

Here is my solution:
Lets define $ w\equal{}a\plus{}bi$ and $ z\equal{}a\minus{}bi$
Sum of $ w\plus{}z\equal{}2a$

$ w^2\plus{}z^2\equal{}(a\plus{}bi)^2\plus{}(a\minus{}bi)^2\equal{}$
$ \equal{}(a^2\plus{}2abi\minus{}b^2)\plus{}(a^2\minus{}2abi\minus{}b^2)\equal{}2(a^2\minus{}b^2)\equal{}7 \Rightarrow a^2\minus{}b^2\equal{}\frac{7}{2}$

And

$ w^3\plus{}z^3\equal{}(a\plus{}bi)^3\plus{}(a\minus{}bi)^3\equal{}(a^3\plus{}3a^2bi\minus{}3ab^2\minus{}b^3i)\plus{}(a^3\minus{}a^2bi\minus{}ab^2\plus{}b^3i)\equal{}$
$ \equal{}2a^3\minus{}6ab^2\equal{}2a(a^2\minus{}b^2)\equal{}10$
$ 2a\equal{}\frac{10}{a^2\minus{}b^2}\equal{}\frac{20}{7}$
So the answer is $ \frac{20}{7}$

[quote="NapoleonXIV"]
$ \equal{} 2a^3 \minus{} 6ab^2 \equal{} 2a(a^2 \minus{} b^2) \equal{} 10$
[/quote]
You sure about that?

[hide="hint"]
Factor $ w^3\plus{}z^3$, then do some manipulation to get it in terms of $ w\plus{}z$ and $ w^2\plus{}z^2$ only.
[/hide]

scorpius119's hint is ultimately equivalent to NapoleonXIV's equations:
$ 7\equal{}2a^2\minus{}2b^2$,
$ 10\equal{}2a^3\minus{}6ab^2$.
[hide="Solution"]I hope I don't mess up.
From the first equation,
$ 6b^2\equal{}6a^2\minus{}21$ ($ b$ is real, so $ 6a^2>21$, or $ (2a)^2>14$),
and then the second equation is equivalent to
$ (2a)^2\minus{}21(2a)\plus{}20\equal{}0$,
with three solutions for $ 2a\equal{}w\plus{}z$: $ \minus{}4$, $ 1$, $ 5$
(two, once eliminated $ 1$ because we've stated before that we must have $ (2a)^2>14$).
The largest value is $ 5$.[/hide]

Let $ s$ be the sum of the complex numbers and $ p$ be the product of those numbers. Then $ s^2\minus{}2p\equal{}7$ and $ s^3\minus{}3sp\equal{}10$.Then

$ 2s^3\minus{}3s(s^2\minus{}7)\equal{}20$

$ s\equal{}4,1,\minus{}5$

you could check the contest archives on aops. im pretty sure this problem is somewhere in 1983 AIME.  :wink:


symmetry
inequalities
LaTeX
inequalities proposed

Given positive integers $ a,b,c$. Prove that,
\[ \frac {a}{\sqrt {a^2 \plus{} 2bc}} \plus{} \frac {b}{\sqrt{b^2 \plus{} 2ca}} \plus{} \frac {c}{\sqrt{c^2 \plus{} 2ab}} \leq \frac {a \plus{} b \plus{} c}{\sqrt {ab \plus{} bc \plus{} ca}}
\]

a stronger inequality sigmacyclic$ (a/(a^2\plus{}2bc))<(a\plus{}b\plus{}c)/(ab\plus{}bc\plus{}ca)$holds. (using cauchy you can obtain the inequality above)
Vornicu Schur makes quick work of it.

[quote="ywl"]a stronger inequality sigmacyclic(a/(a^2+2bc))<(a+b+c)/(ab+bc+ca) holds. (using cauchy you can obtain the inequality above)
Vornicu Schur makes quick work of it.[/quote]

If you can do it then post your solution.

ok. but I don't know latex so I am probably not the best person to do it.
ETS that $ sigma(a(1/(ab \plus{} bc \plus{} ca) \minus{} 1/(a^2 \plus{} 2bc) > 0$
which is equivalent to the following $ sigma(a\minus{}b)*(a\minus{}c)*(a/(a^2\plus{}2bc))$>0
but in $ sigma(a\minus{}b)(a\minus{}c)x>0$ where x,y,z>0 and a>b>c
It is enough to show ax>by (see the PDF file by Darij Grinberg(It is in the Second page of major announcements))
Which is trivial.

To use LaTeX just put dollar signs and between them write formula,that you need.For more information click on the bottom $ Latex help$ on the top your screen. :wink:

[quote="ywl"]ok. but I don't know latex so I am probably not the best person to do it.
ETS that sigma(a(1/(ab+bc+ca)-1/(a^2+2bc)>0
which is equivalent to the following sigma(a-b)*(a-c)*(a/(a^2+2bc))>0
but in sigmacyclic (a-b)(a-c)x>0 where x,y,z>0 and a>b>c
It is enough to show ax>by (see the PDF file by Darij Grinberg(It is in the Second page of major announcements))
Which is trivial.[/quote]

This is not solution for my problem.

You seem not have read my posts clearly. :(
I have proved a stronger inequality.
use my inequality $ (a\plus{}b\plus{}c)^2/(ab\plus{}bc\plus{}ca)>(a\plus{}b\plus{}c)(sigma(a/(a^2\plus{}2bc))$
and then $ (a\plus{}b\plus{}c)(sigma (a/(a^2\plus{}2bc))>sigma(a/root(a^2\plus{}2bc))^2$

[quote="zaya_yc"] $ \frac {a}{a + b + c}\sqrt {\frac {1}{a^{2} + 2bc}} + \frac {b}{a + b + c}\sqrt {\frac {1}{b^{2} + 2ac}} + \frac {c}{a + b + c} \sqrt {\frac {1}{c^{2} + 2ab}}\leq\sqrt {\frac {1}{\frac {a(a^{2} + 2bc)}{a + b + c} + \frac {b(b^{2} + 2ac)}{a + b + c} + \frac {c(c^{2} + 2ab)}{a + b + c}}}$

[/quote]

I think the sign is reverse because $ f(x)=x^{-1/2}$ is convex.


Let x, y, z such that x> y> z> 0, 3 real numbers. Find x, y, z such that
      x-z + 1 / [3 (x-y) (y-z)] = 3 :?:

by substituting
x-y= a
y-z= b
x-z= a+b
a,b is in R
then it gives $ 3a^2b\plus{}3b^2a\plus{}1\equal{}9ab$
seems to me there are no restrictions in a,b , therefore infinite amount of x y z

wrote same thing (less the last phrase) at a little contest and got a big 0.
the solution is something about inequality of the means and gives the solution something like
x=y+1/3        y=z+1/3  (sure,not those solutions).... but how is the problem solved?..:(  i've just heard others of
getting something like that...but how can anyone explain?(yes,i know the inequality of means but i dont find how can that help me,especialy because of that 3 of denomitor)

Through a little bit of manipulation
$ \frac1{3(x\minus{}y)(y\minus{}z)}\equal{}3\minus{}x\plus{}z$
$ 1\equal{}3(3\minus{}x\minus{}z)(x\minus{}y)(y\minus{}z)$

$ \frac{(3\minus{}x\plus{}z)\plus{}(x\minus{}y)\plus{}(x\plus{}z)}3\ge((3\minus{}x\plus{}z)(x\minus{}y)(y\minus{}z))^{1/3}$
$ 1\ge((3\minus{}x\plus{}z)(x\minus{}y)(y\minus{}z))^{1/3}$
$ 1\ge((3\minus{}x\plus{}z)(x\minus{}y)(y\minus{}z))$
$ 3\ge(3(3\minus{}x\plus{}z)(x\minus{}y)(y\minus{}z))$
so if $ x\minus{}y \equal{} y\minus{}z \equal{} 3\minus{}x\plus{}z$

something happens... I'm not sure what but I think I'm on the right track
Maybe now we'll use the  a and b?

[quote="RobRoobiks"]Through a little bit of manipulation
$ \frac1{3(x \minus{} y)(y \minus{} z)} \equal{} 3 \minus{} x \plus{} z$
$ 1 \equal{} 3(3 \minus{} x \minus{} z)(x \minus{} y)(y \minus{} z)$

$ \frac {(3 \minus{} x \plus{} z) \plus{} (x \minus{} y) \plus{} (x \plus{} z)}3\ge((3 \minus{} x \plus{} z)(x \minus{} y)(y \minus{} z))^{1/3}$
$ 1\ge((3 \minus{} x \plus{} z)(x \minus{} y)(y \minus{} z))^{1/3}$
$ 1\ge((3 \minus{} x \plus{} z)(x \minus{} y)(y \minus{} z))$
$ 3\ge(3(3 \minus{} x \plus{} z)(x \minus{} y)(y \minus{} z))$
so if $ x \minus{} y \equal{} y \minus{} z \equal{} 3 \minus{} x \plus{} z$

something happens... I'm not sure what but I think I'm on the right track
Maybe now we'll use the  a and b?[/quote]
Very nice what u did there :) ,im sure that you are on the right track. should get the equality if a=b=3-a-b ,that means a=b=1 ,but 2+ 1/3 =7/3 ,dont understand why it dosn't work.Your post still helped me alot,thanks :D

[quote="Balint Florin"]

x=y+1/3 y=z+1/3 (sure,not those solutions).... but how is the problem solved?.. i've just heard others of
getting something like that

[/quote]
[quote="RobRoobiks"]
so if $ x \minus{} y \equal{} y \minus{} z \equal{} 3 \minus{} x \plus{} z$
something happens... I'm not sure what but I think I'm on the right track
Maybe now we'll use the  a and b?[/quote]
I don't have it at this time.

Some work:

x - y = 3 - x + z becomes
2x = y + z + 3 . . . (*)

y - z = 3 - x + z becomes
x = -y + 2z + 3 . . . (**)

Add (*) to (**):

3x = 3z + 6 becomes
x = z + 2 (***)

Substitute (***) into the equation y - z = 3 - x + z,
and solve for y in terms of z:

y = z + 1

And notice that z + 2 > z + 1 > z.

Let z = t, t a real number greater than zero.

Solutions:

$ (t \plus{} 2, t \plus{} 1, t)$, but these don't satisfy the original equation.

Balint Florin had

x-z + 1 / [3 (x-y) (y-z)] = 3

Now, if the original equation had instead been

(x - z + 1)/[3(x - y)(x - z)] = 3, . . . (****)

then the values would work in that, but you would expect to get different
solutions working with (****) from the outset.

If the equation *is* supposed to be the given, then using

y = z + a and x = z + a + b, for positive real numbers a and b, gives

$ a \plus{} b \plus{} \frac{1}{3ab} \equal{} 3$.

If $ b$ is set to $ \frac13$, for instance, then $ a$ = (the positive real values) $ \frac {4 \pm \sqrt{7}}{3}$.

I don't know if there is an error in the original equation.

[quote]
I don't know if there is an error in the original equation.[/quote]
I'll punch those who made the problem if there is one :P ,i think its actualy the brother of my class teacher (lol) :lol:


Hello,

I am having some trouble with the following two problems. Any help will be appreciated. Thank you!

---

1. A 15 kg child slides, from rest, down a 4.0 m playground slide. The slide makes a 40° angle with the ground. The child's speed at the bottom is 3.2 m/s. What was the force of friction that the slide was exerting on the child?

2. A worker on the roof of a building dropped a 0.125 kg wrench over the edge. Another workman on the 8th floor saw the wrench pass and determined that its speed was 33.1 m/s from the 8th floor. The first floor of the building is 12.0 m high and each successive floor is 8.00 m high. Neglecting air friction, how many floors does the building have?

[u]My Work:[/u]

1. At the top of the slide,
$ E_p \equal{} mgh \equal{} 588.6 J \equal{} E_t$

At the bottom of the slide,
$ E_k \equal{} \frac {mv^2}{2} \equal{} 76.8 J$

Therefore, potential energy at the end of the slide is $ 588.6 J \minus{} 76.8 J$ = $ 511.8 J$.

How would I use this information to find the $ F_f$?

2. Based on the information, I have no trouble finding:

$ E_k \equal{} 68.5 J$
$ E_g \equal{} 83.4 J$
$ E_t \equal{} E_k \plus{} E_g \equal{} 151.9 J$

However, I am not sure how I would use this information to find the number of floors. I tried to do this:

Number of floors = $ \frac{151.9 J \minus{} 12 m}{8 m}$ but I would get 17.5 floors. The correct answer is
 [hide]15 floors[/hide].

[quote="catalan"]
1. At the top of the slide,
$ E_p \equal{} mgh \equal{} 588.6 J \equal{} E_t$[/quote]

The height is not equal to $ 4.0\ m$, because that is the length of the slide. Instead, the height is $ 4.0\cdot \sin 40^o$. This will yield a different initial potential energy of $ \approx 378\ J$.

[quote="catalan"]
At the bottom of the slide,
$ E_k \equal{} \frac {mv^2}{2} \equal{} 76.8 J$
[...]
How would I use this information to find the $ F_f$?[/quote]

First, note that the potential energy lost to friction is instead $ \approx 301.2\ J$. By the work-energy principle, this energy is equal to the work done by friction over a distance of $ 4\ m$. Thus $ F\equal{}\frac{W}{d}\equal{}\frac{301.2}{4}\ N\approx \boxed{75.3\ N}$.

Thank you very much for your reply, jvenezuela716. I got the answer.

Also, I apologise for forgetting to post my work on question #2. It has been posted.

[quote="catalan"]
2. A worker on the roof of a building dropped a 0.125 kg wrench over the edge. Another workman on the 8th floor saw the wrench pass and determined that its speed was 33.1 m/s from the 8th floor. The first floor of the building is 12.0 m high and each successive floor is 8.00 m high. Neglecting air friction, how many floors does the building have[/quote]

[hide="Solution."]
In falling from the roof to the 8th floor, the wrench traveled some distance $ x$ and lost gravitational potential energy $ mgx$. The wrench also gained an equal amount of kinetic energy $ \frac{1}{2}mv^2$. Neglecting air resistance, these values must be equal:

$ mgx\equal{}\frac{1}{2}mv^2 \implies x\equal{}\frac{v^2}{2g}\equal{}\frac{33.1^2}{2\cdot9.8}\ m\approx 55.9\ m$.

This corresponds to having fallen $ \frac{55.9}{8}\approx 6.99\equal{}7$ floors. Add this to the remaining $ 8$ floors the wrench has to fall, we find the building to have $ 7\plus{}8\equal{}\boxed{15}$ floors.

It's interesting to note that this is irrespective of the mass of the wrench.
[/hide]


A circle with radius 1 is inscribed in an equilateral triangle.  Progressively smaller circles are inscribed at each of the vertices.  These smaller circles are successively congruent and are tangent to two of the sides of the triangle. The question is....what is the sum of the circumference of all the circles?

The incircle has a circumference of $2\pi$.  The next three circles will be tangent to the larger circle and two sides of the triangle, because it's an equilateral triangle, we can construct 3 smaller equilateral triangles at the verticies of the larger triangle, such that their incircles will be the desired circles.  These ratio of the side of the original triangle to the length of the side of one of these new triangles is 3, and thus the sum of these circuferences will be $3\left(\frac{2\pi}{3}\right) = 2\pi$.  Therefore the next set of circles would have a sum of circumferences of $\frac{2}{3}\pi$.  So we need to evaluate $2\pi + \sum_{n=0}^\infty 2\pi/3^n = 2\pi+2\pi\cdot\frac{3}{2} = 5\pi$.

okay, despite sounding a little newbish, but, how do you evaluate
$\sum_{n=0}^\infty 2\pi/3^n$
Is there a certain way to look at it?
Thanks a lot

It's jsut $2\pi$ times a geometric series. The sum for an infinite geometric series is $\frac{a_1}{1-r}$ where $a_1$ is the first term.

Here's a quick proof:

$S=a_1+a_1r+a_1r^2+...$

$S-a_1 = r(a_1+a_1r+a_1r^2+...) = rS$

Solving for $S$, we have
\[ S=\frac{a_1}{1-r} \]

Note that $|r|<1$ for the series to converge.

$\displaystyle \sum_{n=0}^\infty \frac{2\pi}{3^n} = 2\pi \sum_{n=0}^\infty \frac{1}{3^n}$

Notice that this is an infinite geometric series.  Each term is a third of the previous term; that ratio is the common ratio.  The sum of this series is then:

$\frac{a_0}{1-r}$

where $a_0$ is the first term of the series and $r$ is the common ratio.  Evaluate:

$\displaystyle 2\pi \sum_{n=0}^\infty \frac{1}{3^n} = 2\pi \cdot \frac{1}{1-\frac{1}{3}} = 2\pi \cdot \frac{3}{2} = 3\pi$

edit: beat me to it, lol

I have one question though, shouldn't the answer be $5\pi$ since he asks for all the circles? ;)

I know it is.  I was just explaining the infinite sum.

[quote="pkothari13"]How do you know "The ratio of the side of the original triangle to the length of the side of one of these new triangles is 3?"[/quote]

Draw in the lines, and let's say the three new triangles formed are also equilateral, so each angle is $60$, that would mean that each angle of the hexagon inside would be $120$ so its a regular hexagon, so each of those triangles are also equilateral, and since the sidelength of each of those triangles are $\frac{1}{3}$ of the bigger triangle, that means that the radii of the circles are also in the same ratio.

And plus since the original triangle is equilateral and the segments that you can draw to make the new triangles can only go through one point and be tangent to the circle making a $90$ degree angle and making the bases of the triangles parallel, so the triangles will be similar


logarithms
calculus
calculus computations

[quote="lambern314"]Hello Everyone, I'm trying to compute the value of the series:

S=1 +1/2 - 1/3 - 1/4 + 1/5 +1/6 - 1/7 -1/8.....

to infinity, accurate to at least 10 decimal places; unfortunately my program can only compute 7 decimal places before it dies;

Can anyone help me reach the ten decimal goal?

Thanks

so far its: 1.1319717[/quote]

i don't understand , ummm...... why don't you just evaluate the series  :(  ?

$ S\equal{}\left(1\minus{}\dfrac{1}{3}\plus{}\dfrac{1}{5}\minus{}\dfrac{1}{7}\plus{}\dfrac{1}{9}\minus{}\;\plus{}\;\ldots\right) \plus{} \dfrac{1}{2}\left(1\minus{}\dfrac{1}{2}\plus{}\dfrac{1}{3}\minus{}\dfrac{1}{4}\plus{}\dfrac{1}{5}\minus{}\;\plus{}\;\ldots\right)$
$ \equal{}\boxed{\dfrac{\pi}{4}\plus{}\dfrac{\ln\,2}{2}}$

Thanks, i wasn't sure exacly how to evaluate the ln(2)/2 part, I was trying something else, Thanks

The thing can be computed in closed form, but I won't spoil your fun and go there just now.

For consistent patterns, let's just look at $ S_4,S_8,S_{12},\dots.$ Using a spreadsheet, I ran that out to $ S_{1024}.$

This gives the following partial sums:

$ S_4 \equal{} 0.916666667$
$ S_8 \equal{} 1.01547619$
$ S_{16} \equal{} 1.071529859$
$ S_{32} \equal{} 1.101224274$
$ S_{64} \equal{} 1.11647067$
$ S_{128} \equal{} 1.124190006$
$ S_{256} \equal{} 1.128073163$
$ S_{512} \equal{} 1.13002054$
$ S_{1024} \equal{} 1.130995668$

(In fact, those numbers are stored in the spreadsheet to standard machine accuracy, which is 15+ places.)

Why the power of 2 pattern? Because I'm going to use those numbers as input for a Richardson extrapolation table. Consult some standard numerical analysis textbook for just what that is. I'm assuming errors of all orders - namely, I'm assuming that if the sum is $ S,$ then I can write
\[ S_n \equal{} S \plus{} \frac {\alpha}{n} \plus{} \frac {\beta}{n^2} \plus{} \frac {\gamma}{n^3} \plus{} \cdots
\]
In the far corner of that Richardson table, I get, to 14 places, $ 1.13197175367744.$ I'm pretty confident in the first 12 of those places.

As Kent did, we assume that we don't know a closed form for the result. It's an opportunity to speak about numerical analysis. We sum blocks of length $ 4$. Then
$ S\equal{}\sum_{p\equal{}1}^{\infty}a_p$ with $ a_p\equal{}\dfrac{32p^2\minus{}24p\plus{}3}{4(4p\minus{}3)(2p\minus{}1)(4p\minus{}1)p}$; $ S\equal{}S_n\plus{}R_{n\plus{}1}$; with the maple command "taylor" you can obtain $ a_p\equal{}1/(4p^2)\plus{}3/(16p^3)\plus{}17/(128p^4)\plus{}\cdots$. With the help of the Maple command "eulermac" you can obtain: $ R_n\equal{}U_n\plus{}O(1/n^6)$  with $ U_n\equal{}1/(4n)\plus{}7/(32n^2)\plus{}23/(128n^3)\plus{}35/(256n^4)\plus{}197/(2048n^5)$. Then $ \sum_{p\equal{}1}^{20000}a_p\plus{}U_{20001}$ gives $ S$ with $ 27$ or $ 28$ digits.


Why is there so much less activity here for C testers?

Most people applied on B, and more on A than on C. The people who applied test C are likely hopeful MOPpers who got pretty much annihilated by the USAMO, myself included.

I was a sort of hopeful (not really) MOPper and I took Test B...but like you, got owned on the USAMO. Do they like accept the same amount of people (around 40) from each test? Since like the release results a week after each test.

Well I based my previous claim on the fact that if I made MOP, I wouldn't go to AMP, and if I applied test B, then I would have had to have paid before knowing my MOP decision...

I'd bet that Dr. A has the tests made so that he knows how many people will score a certain score on the Admit test, so when they apply doesn't matter because he's going to take the best.

[quote="FightTheTide"]Well I based my previous claim on the fact that if I made MOP, I wouldn't go to AMP, and if I applied test B, then I would have had to have paid before knowing my MOP decision...[/quote]
they give you a full refund if you make mop and decide not to go

Really? If I'd known that I'd have applied Test A and got $ \$$329.50 off. Crap.


Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ be positive integers sastisfying $ abc \equal{} def$. Prove that the integer $ a\left(b^2 \plus{} c^2\right) \plus{} d\left(e^2 \plus{} f^2\right)$ is composite.

Setting the product equal to 60 and trying the 40 combinations that give odd sums, all of them are composite. I think it's true.

I sort of remember proving this before on the forum.

We have $a(b^2+c^2)+d(e^2+f^2)=a(b+c)^2+d(e-f)^2=a(b-c)^2+d(e+f)^2$.

From here it should be easy, since it's well-known that a prime $p$ has at most one decomposition of the form $p=ax^2+by^2$.

Hmm.Yes.Yes,........Grobber You are right


Just wondering how it is determined. Family income? Siblings in other colleges? What?

It's usually based on FAFSA. In a very general sense, yes, it is based on their estimates of your family's income, possessions, and expenses. It encompasses most of the relevant information you could think of, since it's based on your parents' tax returns.

It varies from university to university.  In general, elite schools only make you pay what your family can afford which they use the FAFSA to help determine.  A lot of elite schools don't offer merit scholarships.  On the other hand, in state schools usually have tons of merit based scholarships (at least for instate students).  Other than that, I don't know any specifics.  You should look at the websites of the colleges you're interested in.

The College Board publication Primer on Economics for Financial Aid Professionals

http://www.collegeboard.com/prod_downloads/highered/fa/Economics-Primer-2004.pdf

will tell you more than you want to know about the methodology underlying financial aid awards. For the applicant, the best strategy is to apply to various colleges, and compare offers.


Andrea flips a coin repeatedly, continuing until she either flips two heads in a row (the sequence $ HH$), or flips tails followed by heads (the sequence $ TH$).  What is the probability that she will stop after flipping $ HH$?

Source: $ HMMT$

I love this problem.

[hide]Assume Andrea flips tails. Then, it is impossible to get two heads consecutively because the first such head would necessarily follow a tail, yielding the sequence $ TH$.

Therefore, the sequence must either consist of two heads, or either zero heads or one head followed by an arbitrary number of tails, followed by a head. The only sequence that ends in $ HH$ first is the very first sequence mentioned, which occurs $ \displaystyle\frac{1}{4}$ of the time. The probability is therefore $ \displaystyle\frac{1}{4}$[/hide]

Yup.
I think it's a cool problem because at first, you want to just put down $ \frac{1}{2}$, but that's not the right answer.


[quote="ml2"]Will the AIME Problem Seminar be moved to archives?[/quote]

There will not be archives for the AIME Problem Seminar.


problem: You and your friend are in two adjacent boggies(compartments) of a train with infinite number of boggies.Both of you donot know in which boggie the other person is.Both of you move towards front or back boggie with the intention of meeting the other.A move consists of a person moving from one boggie to other and assume that both make moves simultaneously.What is the average(expected) number of moves after which both of you meet ?

Sorry, I am so stupid, but you mean [i]meet[/i] as in "[i]your paths cross[/i] right? (i.e. since the two people can never be in the same boggie at same time (parity))


Show that the equation $ x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}t^{3}\equal{}1999$ has infinitely many solutions in quadruples $ (x,y,z,t)$ of integers.

[quote="manjil"]Show that the equation $ x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} t^{3} \equal{} 1999$ has infinitely many solutions in quadruples $ (x,y,z,t)$ of integers.[/quote]
$ (10\plus{}60n^{3})^{3}\plus{}(10\minus{}60n^{3})^{3}\plus{}(\minus{}60n^{2})^{3}\plus{}(\minus{}1)^{3}\equal{}1999$

[quote="Moonmathpi496"][quote="Fermat's Little Turtle"]Why post just a solution? ><[/quote]
When we are trying to show that there are $ \infty$-ly many solutions , it is enough to construct a series of solution like [b]Allnames[/b] did.[/quote]
Yes,i guessed that root,and with $ n\in \mathbb Z$ we have the proof
Who can post a naturaler solution

[quote="manjil"]Can we replace $ 1999$ with any prime $ p$?[/quote]
I dont think so,because it is very hard

[quote="manjil"]Can we replace $ 1999$ with any prime $ p$?[/quote]
In allnames solution, he doesn't use that $ 1999$ is a prime. He uses that $ 1999 \equal{} 2 \cdot 10^3 \minus{} 1^3 \equal{} 2 k^3 \minus{} m^3$
In general $ (k \plus{} 6kn^3)^3 \plus{} (k \minus{} 6kn^3)^3 \plus{} (\minus{}6kn^2)^3 \plus{} (\minus{}m)^3 \equal{} 2k^3 \minus{} m^3$

Thanks a lot Mathias_DK. That was something I was looking for.


In a sports contest a total of m medals were awarded over n days. On the first day one medal and 1/7 of the remaining medals were awarded. On the second day two medals and 1/7 of the remaining medals were awarded, and so on. On the last day, the remaining n medals were awarded. How many medals were awarded, and over how many days?

I read Kalva's solution, but the answer seems to jump too quickly for me... I was wondering if anyone could explain it to me.

Hi -  you can start in the reverse and looks like the answer will come out  (7-1)^2. medals in (7-1) days.

Test it out for smaller divison say  (instead of  1/7th - use 1 /4th) the answer would  be 9.

First day                      1 +  (8/4)  = 3 medals
Second day                  2+ ( 4/4)  = 3 medals
Last day                                         3 medals.

Now  16 medals would work if the problem asked for  (1/5) etc... And for the given problem, the answer is 36.
 just write   down total number of medals etc...   and you get a expression in 7^(something)/6^(something_else) and that something_else has to be zero  to give  integer solution. which will give that number of days = 6.

One should not put IMO problems in the Pre-Olympiad section.


An angle is given (<180 degree of course) with a vertex X. Points A and B lie on different arms of this angle and PA=PB. Point Y should be placed inside this angle, and we have to say when the sum AY+BY is the smallest.

I cannot find any answer......

And sorry for my english.

I don't know why you even mentioned the point P but...

Think triangle inequality.

This should not be in the olympiad section...

[u]Sorry, this is the right question i think:
[/u]
[b]An angle is given (<180 degree) with a vertex X and a point P placed inside this angle. Points A and B lie on different arms of this angle, and XA=XB and the sum PA+PB is the smallest. I ask: when this sum is the smallest?
[/b]
It is not so easy, that's why i placed it here. Can anybody help me?

This is problem 2 from 1st series of 1st round of 59th Polish Mathematical Olympiad. The 1st series of the competition ends at 8.10.2007. Moderators, please lock the topic.

[quote="Rand0m"]NO. This is not the problem from polish mathematical olympiad. Read the question there.[/quote]
You are partially right. But the answer is: when $ \measuredangle APX\equal{}\measuredangle BPX$ and the competition problem asks to prove it.

[quote="Rand0m"]Yes. I am not asking for it. Man what do you want from me?

Ok. Can anybody help me?[/quote]
I want you to stop posting problems from current Polish MO (or more general versions of them), because it spoils the competition even if you don't intend that.

IT IS NOT A QUESTION FROM POL MATH OLYMPIAD. MODERATOR PLZ DO SOMETHING.... Who is this guy Rzeszut?

Can anybody help me? This is not internet competition, don't be afraid.........

this is the right question i think:

An angle is given (<180 degree) with a vertex X and a point P placed inside this angle. Points A and B lie on different arms of this angle, and XA=XB and the sum PA+PB is the smallest. I ask: when this sum is the smallest?

It is not so easy, that's why i placed it here. Can anybody help me?

I saw the same problem as number 834 of Geometriagon at [url]http://www.polarprof.org/geometriagon[/url] . There nobody solved it so far, and I not too. It is classified level 4, so it must not be so easy. If anyone wish to solve it, there is no problem for the olympiad, because the solutions are visible only to solvers: the only public information is that someone solved it and when. So, good luck

Mario Belluomo


ratio
probability
absolute value
geometric series

What is the formula for an infinite geometric sequence???

Here's the formula

$\frac{1st\ term}{1\  -\  common \ product}$

For this example

$3\ +\ 2\ + \frac{4}{3} + \frac{8}{9}+..........$

Since you are multiplying by $\frac{2}{3}$ each time, use this formula :

$\frac{3}{1 - \frac{2}{3}} = \frac{3}{\frac{1}{3}}=\ 9$

[quote]Wait... are you all refering to the sum of the terms in an infinite geometric sequence?[/quote]

Yes, frost13's reply gives the formula for the sum of an infinite geometric series.

Of course, the "common product" in frost13's formula (usually called the "common ratio", btw) must have absolute value less than 1, otherwise, the series doesn't converge (has no value).

Yakko

Kevin...I went over that with you the day we were given that problem... *Rollseyes*

[quote]This is for the sum right?[/quote]

Yes, for the sum of all terms of an infinite geometric series.

For example, what is 1 + 1/3 + 1/9 + 1/27 + ... ?  The first term is 1, the common ratio is 1/3, so the sum is

1/(1 - (1/3)) = 3/2.

Yakko

[quote="white_horse_king88"]Now for one which isn't infinite. It it:

$\frac{a - ar^{n-1}}{1-r}$, where $a$ is the first term, $ar^{n-1}$ is the last term, and $r$ is the constant ratio. Where $r$ does not equal 1.[/quote]

Question: The notation is a bit confusing, so can you show an example of the application of this formula? Thanks...

[quote]
Question: The notation is a bit confusing, so can you show an example of the application of this formula? Thanks...[/quote]

Example: What is the sum of the first 10 terms of the geometric series 2,6,18,... ?

The formula for the sum of a finite geometric series is (quoting from white_horse_king):

$\frac{a - ar^{n-1}}{1-r}$, where $a$ is the first term, $ar^{n-1}$ is the last term, and $r$ is the constant ratio. Where $r$ does not equal 1.

In this problem, $a=2$ (the first term), $r=3$ (the common ratio), and $n=10$ (the number of terms).  The answer is 19682.

Yakko

[quote="hockeyadi23"][quote="lightspeed"][quote="hockeyadi23"]Here's a problem that applies this formula: Source=Some National Sprint...
I don't exactly remember the problem, but I remember the important stuff:
John, Clyde, and Bill play a coin-flipping game, where to win you need to flip a head. If one flips a tail, then the next person goes. If the three go in that order, then what's the probability that Bill wins? Express your answer as a common fraction.

I'll post the actual solution tomorrow so that you all have enough time to try this problem.[/quote]

Does that problem have revelvance to infinite Geometric sequences?[/quote]

Yes in fact it does.[/quote]

Well, wouldn't it only be an infinite sequence if they all like flipped tails for a long time?

[quote="lightspeed"][quote="hockeyadi23"]Just do the problem, and you'll see. It was number thirty on some National. Since it's a probability question, it does involve an infinite geometric sequence.[/quote]

Lol, I'm not good with probablility! :(
I need some lessons on probability.[/quote]

Yes, typically, and for me especially, combinatorics, including counting and probability is the hardest.. that shows why they loaded this chapter with combinatorics problems... back to my problem, does anyone have a solution... or do I have to post it?

Here's the problem again if you don't want to bother seeing it on the last page:
[quote]John, Clyde, and Bill play a coin-flipping game, where to win you need to flip a head. If one flips a tail, then the next person goes. If the three go in that order, then what's the probability that Bill wins? Express your answer as a common fraction. [/quote]

lol
solution:
[hide]
Obviously, each person only goes every third time, and therefore the common ratio is (1/2)^3. The probability he wins the first time is 1/8. This is the first term in the geometric sequence. 1/8 / 1 - 1/8 = 1/7
[/hide][/hide]


there is a branch of geometry that says that two parallel lines intersect
at $+ \infty$.how in gods name is this possible ?

I think not intersects=intersects at $+\infty$.

:rotfl: this is really mad stuff ??!??!?!
how can this be
it surpasses my logic,i really cannot grasp this

See at a very far thing (for example, the moon) off your eyes and will save oneself by the strabism (anorthopia) !.


Show that $\forall a,b,c>0$ we have $ \sum \frac{a^3}{bc} \geq \sum \frac{a^2+b^2}{2c} \geq \sum a$

cheers! :D :D :cool:

By rearrangement :

a^4 + b^4 + c^4 >= a^3*b + b^3*c + c^3*a
and
a^4 + b^4 + c^4 >= a^3*c + b^3*a + c^3*b

Adding the two and dividing by abc gives the first one.

Again by rearrangement :

a^3*b + b^3*c + c^3*a >= a^2*b*c + b^2*c*a + c^2*a*c
a^3*c + b^3*a + c^3*b >= a^2*b*c + b^2*c*a + c^2*a*c

Again adding the two and dividing by abc gives the othe ineq. (equality iff a=b=c)


hi
how does one pronounce the word "fourier" in English? is it like "4EA"?
thanks

Close enough. (With the accent on the first syllable.) We try to make at least vague accommodations to the pronunciation of names that originated in other languages, and this name is French. That isn't to say that we try that hard; we don't really give a fully correct French pronunciation to such names as "Cauchy" or "Legendre."

In the same vein, "Riemann" should be pronounced REE - mahn. And the initial consonant in "Weierstrass" should be a "V" sound in English, not that everyone does it that way.


My solution:

First, it's known that
If a conic meet the sides $BC$, $CA$, $AB$ of a triangle in three pairs of points $A_{1}$, $A_{2}$; $B_{1}$, $B_{2}$; $C_{1}$, $C_{2}$
then $\frac{A_{1}B}{A_{1}C}\cdot\frac{A_{2}B}{A_{2}C}\cdot \frac{B_{1}C}{B_{1}A}\cdot\frac{B_{2}C}{B_{2}A}\cdot
\frac{C_{1}A}{C_{1}B}\cdot\frac{C_{2}A}{C_{2}B}=1$.

Problem*:
Let $ABC$ be a triangle, $M\in AB$, $N\in AC$, $X\in BC$ and $\left\{Y\right\}=AX\cap MN$.
Prove that $\frac{YM}{YN}=\frac{AM}{AN}\cdot\frac{XB}{XC}\cdot\frac{AC}{AB}$.

Solution:

$\frac{YM}{YN}=\frac{AM\cdot sin\left(\angle YAM\right)}{AN\cdot sin\left(\angle YAN\right)}=\frac{AM}{AN}\cdot\frac{sin\left(\angle XAB\right)}{sin\left(\angle XAC\right)}=\frac{AM}{AN}\cdot\frac{XB}{AB}\cdot\frac{AC}{XC}\Rightarrow$
$\Rightarrow\frac{YM}{YN}=\frac{AM}{AN}\cdot\frac{XB}{XC}\cdot\frac{AC}{AB}$.