I was just wondering how far people in here have got in the USACO, with both the training pages and competitions to be on the USA team for the IOI.

I'll start by saying that I just finished chapter 1 half an hour ago.  (But I've only done this for around a month.)

I just finished chapter 1 a couple weeks ago myself. I know almost nothing about programming so it's really slow going for me...

Nukular was invited to the training camp...

there have been people who have burned thru the entire training pages in a few months, but they were already good programmers.

I do not personally program, but my son and some of his friends have been very involved in USACO, so I know something about it. It's a great program!


Hello,

I know most of you have an experience with AMCs and other competition. So I want to ask you about the Geometry formula that you should know before entering such competitions it's impossible to know all of the formulas (at least for me  :oops: ).

Thanks in advance

the basic triangle identities, specially those ones who involve trigonometry and Quadrilateral formulas, there are more i mention some of them

i would  suggest  u have  a look  at few inequality problems  in  introductory  problems  section  of 103 trigonometric problems by titu andreescu


Hello, everyone. I have a problem with Latex that I am unable to resolve. I need to typeset simple arithmetic - things like 123 + 456 and 123 x 456. "Simple," you might say, but the problem is that I need to put the things vertically. For example, it should look like this:
     123
[u]x 456[/u]
56088

Except that the "123" is over the "456", and the "56088" is aligned under the "456." Now, I can't seem to find any way to typeset these sorts of things in Latex, and I don't know how to manually format it.

The reason I need to format such things is because I am writing a paper about increasing the speed of mental arithmetic calculations, and I need a vertical formatting in order to demonstrate my methods.

If someone could help me, I would be very grateful. Thank you!

It's ugly, but it can be done:

\[ \begin{array}{r*{4}{@{\hspace{0pt}}c}} &&1&2&3\\ \times&&4&5&6 \\ \hline 5&6&0&8&8 \end{array} \]

(Click on the expression to see the LaTeX code)

That's very ugly. But it gets the job done. Thanks for the assistance!


[size=150](1)[/size]
If $ a,b,c$ are the side of triangle, prove the inequality:
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\ge\sum_{cyc}ab(a\plus{}b)$
[size=150](2)[/size]
If $ A,B,C$ are the angle of triangle, prove that:
$ 4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\plus{}1\equal{}cosA\plus{}cosB\plus{}cosC$

[quote="Unidranix"][size=150](1)[/size]
If $ a,b,c$ are the side of triangle, prove the inequality:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\ge\sum_{cyc}ab(a \plus{} b)$
[/quote]
This is the Schur inequality for $ n\equal{}3$.
[hide="my solution"]WLOG we can assume that:$ a \ge b \ge c$
$ \sum{a^3\plus{}abc\minus{}a^2b\minus{}a^2c} \ge 0 \Leftrightarrow a(a\minus{}b)(a\minus{}c)\plus{}b(b\minus{}c)(b\minus{}a)\plus{}c(c\minus{}a)(c\minus{}b) \ge 0 \Leftrightarrow a(a\minus{}b)(a\minus{}c)\minus{}b(b\minus{}c)(a\minus{}b)\plus{}c(c\minus{}a)(c\minus{}b) \ge 0$
The first and the third term are positives..
So we have to prove that:
$ a(a\minus{}b)(a\minus{}c)\minus{}b(b\minus{}c)(a\minus{}b) \ge 0 \Leftrightarrow (a\minus{}b)(a^2\minus{}ac\minus{}b^2\plus{}bc) \ge 0 \Leftrightarrow (a\minus{}b)[(a\minus{}b)(a\plus{}b)\minus{}c(a\minus{}b)] \ge 0 \Leftrightarrow (a\minus{}b)^2(a\plus{}b\minus{}c) \ge 0$ , which is true......
Also the condition is useless[/hide]

[quote="Unidranix"]
[size=150](2)[/size]
If $ A,B,C$ are the angle of triangle, prove that:
$ 4sin\frac {A}{2}sin\frac {B}{2}sin\frac {C}{2} \plus{} 1 \equal{} cosA \plus{} cosB \plus{} cosC$[/quote]

$ L.H.S.\equal{}4\sin\frac {A}{2}\sin\frac {B}{2}\sin\frac {C}{2}\plus{}1$
$ \equal{}4\sin\frac {A}{2}\sin\frac {B}{2} \cos \frac {A\plus{}B}{2}\plus{}1$
$ \equal{}4\sin\frac {A}{2}\sin\frac {B}{2}(\cos \frac{A}{2} \cos \frac{B}{2}\minus{} \sin \frac{A}{2} \sin \frac{B}{2})\plus{}1$
$ \equal{}\sin A \sin B\minus{}(2\sin^2 \frac{A}{2})( 2\sin^2 \frac{B}{2})\plus{}1$
$ \equal{}\sin A \sin B\minus{}(1\minus{}\cos A)(1\minus{}\cos B)\plus{}1$
$ \equal{}\minus{}\cos (A\plus{}B)\plus{}\cos A\plus{}\cos B$
$ \equal{}\cos A\plus{}\cos B\plus{}\cos C$
$ \equal{}R.H.S.$

[quote="Dimitris X"][quote="Unidranix"][size=150](1)[/size]
If $ a,b,c$ are the side of triangle, prove the inequality:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\ge\sum_{cyc}ab(a \plus{} b)$
[/quote]
......$ \Leftrightarrow (a \minus{} b)[(a \minus{} b)(a \plus{} b) \minus{} c(a \minus{} b)] \ge 0 \Leftrightarrow (a \minus{} b)^2(a \plus{} b \minus{} c) \ge 0$ , which is true......
Also the condition is useless[/hide][/quote]
How will you say $ a\plus{}b\minus{}c>0$ if $ a,b,c$ are not the sides of a triangle? If $ a>b\plus{}c$ we have $ (a\minus{}b)^2(a\plus{}b\minus{}c)<0$
 :roll:

[quote="Potla"][quote="Dimitris X"][quote="Unidranix"][size=150](1)[/size]
If $ a,b,c$ are the side of triangle, prove the inequality:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\ge\sum_{cyc}ab(a \plus{} b)$
[/quote]
......$ \Leftrightarrow (a \minus{} b)[(a \minus{} b)(a \plus{} b) \minus{} c(a \minus{} b)] \ge 0 \Leftrightarrow (a \minus{} b)^2(a \plus{} b \minus{} c) \ge 0$ , which is true......
Also the condition is useless[/hide][/quote]
How will you say $ a \plus{} b \minus{} c > 0$ if $ a,b,c$ are not the sides of a triangle? If $ a > b \plus{} c$ we have $ (a \minus{} b)^2(a \plus{} b \minus{} c) < 0$
 :roll:[/quote]
Hi potla! he used that $ a\geq b\geq c$ see the beginning of the solution,

[quote="peine"][/quote][quote="Potla"][quote="Dimitris X"][quote="Unidranix"][size=150](1)[/size]
If $ a,b,c$ are the side of triangle, prove the inequality:
$ a^3 + b^3 + c^3 + 3abc\ge\sum_{cyc}ab(a + b)$
[/quote]
......$ \Leftrightarrow (a - b)[(a - b)(a + b) - c(a - b)] \ge 0 \Leftrightarrow (a - b)^2(a + b - c) \ge 0$ , which is true......
Also the condition is useless[/hide][/quote]
How will you say $ a + b - c > 0$ if $ a,b,c$ are not the sides of a triangle? If $ a > b + c$ we have $ (a - b)^2(a + b - c) < 0$
 :roll:[/quote][quote="peine"]
Hi potla! he used that $ a\geq b\geq c$ see the beginning of the solution,[/quote]
Oh!  :oops:  I must have not slept well yesterday...... :rotfl:
Sorry for the inconvenience,everyone! :(


abstract algebra
algebra
polynomial
group theory
superior algebra
superior algebra unsolved

i) Prove or disprove that a p-primary finite abelian group is generated by the elements that has largest order.

ii) Let $ F$ be a field and $ G$ the multiplicative group of the non zero elements of $ F$, prove using the Basis Theorem and the fact that a k degree polynomial has at most k roots that $ G$ is cyclic.

iii) Give an example of an abelian Group $ H\times K$ which contains a normal non trivial subgroup $ N$ such that $ N\cap (H\times \{ 1\} )\equal{}\{ 1\}$ and $ N\cap (\{ 1\}\times H  )\equal{}\{ (1,1\}$

Your post has several errors.  For example, is "elements that has" to be interpreted as "element that has" or "elements that have"?

And ii) is simply not true.

Steve


advanced fields
advanced fields unsolved

Let $A_{1},\dots,A_{n+1}$ be closed subsets of $S^{n+1}$ that cover $S^{n+1}$. Prove one of $A_{i}$ contains two antipodal points

A proof of this is contained in the second post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=72330]here[/url].


Let positive real numbers $ a,b$ satisfying $ a\plus{}b\equal{}a^4\plus{}b^4$. Prove that $ a^ab^b\le 1 \le a^{a^3}b^{b^3}$

This is a problem from Awesome Math Journal.
See [url]http://reflections.awesomemath.org/2008_5/MR_5_2008_solutions.pdf[/url]
[b]Page 29-31[/b]

There are two solutions in those pages.


Yes! And I don't see another way. Maybe someone see? Thank you.


Problem 5 is still missing, but we have the marking scheme :
So we have approx :
FRA1 18
FRA2 19
FRA3 8
FRA4 14
FRA5 17
FRA6 18

We think we will be around the 30th place...

CAN5 got 42/42 !!! And no Bulgarian will get 42...

what do you mean by "Problem 5 is still missing"? You mean that you don't have the marking scheme of Problem 5?

What about Chinese team?

Chinese team got 5 marks of 6p in problem 1 !!. They will almost surely receive all gold, but they will not be among the highest golds. One still has a chance of max. score.

[quote="Valentin Vornicu"]Chinese team got 5 marks of 6p in problem 1 !!.[/quote]
Is that low or ... maybe they should get 7p?

I don't know, but i've got a photo of him  :)
And these results are definitive.


induction
topology
real analysis
real analysis unsolved

Show that every open subset of $ \mathbb{R}$ is the union of a disjoint sequence of open intervals.

enumerate $ Q\equal{}\{r_1,r_2,... \}$

then define $ f(x)\equal{}\inf \{n\in N: r_n \in (\inf \{t\in R: (t,x]\subset O \}, \sup \{ t \in R: [x,t) \subset O \} ) \}$

$ f: O \to N$ then we know

$ O \equal{} \cup_{n \in N} f^{\minus{}1} (n)$ and it is a nice thing to notice that $ f^{\minus{}1} (n)$ is an open interval or empty

Open set $ O$;  $ \mathbb{Q} \equal{} \{r_1,r_2,\cdots\}$
Use the definition: An open interval is a (path)connected open subset of $ \mathbb{R}$.
Let $ A_1 \equal{} \cup\{(a,b) \subset O| r_1 \in (a,b)\}$
$ A_1$ is open and connected. (Connected, because if $ p,q \in A_1$, then there is a path from $ p$ to $ r_1$ and from $ r_1$ to $ q$).
(Induction) Let $ A_1,A_2,\cdots,A_{n\minus{}1}$ be defined.
Let $ r_{i_n}$ be the first element in  $ r_1,r_2,\cdots$ with $ r_{i_n}\notin A_k$ for $ k<n$.
Let $ A_n \equal{} \cup\{(a,b) \subset O| r_{i_n} \in (a,b)\}$.
$ A_n \cap A_k \equal{} \emptyset$ for all $ k<n$, because if $ x\in A_n \cap A_s$ for some $ s<n$, then
there is a path from every point of $ A_n$ to every point of $ A_s$ via $ x$, and so there is a path from $ r_{i_n}$ to $ r_{i_s}$ in an open interval $ A_s\cup A_n$, which has to be a subset of $ A_s$ (because $ A_s$ is the union of all intervals which contain $ r_{i_s}$). But $ r_{i_n} \notin A_s$.
Thus, every open set is a union of disjunct open intevals.

A minor English note: the word is "disjoint", not "disjunct". While the "disjunct" form appears as part of some other forms of the word, it isn't a word on its own.


A $ 6$-inch and $ 18$-inch diameter pole are placed together and bound together with wire.  The length of the shortest wire that will go around them is:

$ \textbf{(A)}\ 12\sqrt{3}\plus{}16\pi \qquad
\textbf{(B)}\ 12\sqrt{3}\plus{}7\pi \qquad
\textbf{(C)}\ 12\sqrt{3}\plus{}14\pi \\
\textbf{(D)}\ 12\plus{}15\pi \qquad
\textbf{(E)}\ 24\pi$

[hide="solution"]
First calculate the length of wire that is not touching either circle.
Using the fact that tangents are perpendicular to the center, we can create a right angle triangle with sides $ 9\plus{}3,9\minus{}3, x$

$ \Rightarrow x \equal{} 6\sqrt{2^2\minus{}1} \equal{} 6\sqrt{3}$ There are two such lengths of this $ \Rightarrow 12\sqrt{3}$

Now If we draw the radius of each circle to their points of tangency then they both make an angle $ \theta$

The sum of both arcs which wire is touching is $ 3\theta \plus{} 9(2\pi \minus{} \theta) \equal{} 18\pi \minus{} 6\theta$

If we look at the triangle we drew we find that $ \cos\left(\frac{1}{2}\theta\right) \equal{} \frac{6}{12} \Longrightarrow \theta \equal{} \frac{2\pi}{3}$

$ \Rightarrow 18\pi \minus{} 6\frac{2\pi}{3} \equal{} 14\pi$

$ \Longrightarrow 12\sqrt{3} \plus{} 14\pi \longrightarrow \boxed{C}$
[/hide]


Can anybody teach me about balls and urns. AoPSWiki's article confused me. I think I learn better from you guys  :lol:

What specifically confused you? Tell us and then we can help you.  :)

So pretend you have red sticks yellow sticks blue sticks and pink sticks(4 total)
you can order them such that all the reds are on the left, then it's yellows, then blues, then pink. Then you put in 3 more white sticks, and put them in the places where the colors meet, separating them. so there are total 3+<number you need to choose>
now each one of these thingies are unique, so it's 3+<number you need to choose>  choose  3     , or $ \binom{n\plus{}r\minus{}1}{r\minus{}1}$  for $ n$ (multi)choose $ r$
I probably made you more confused :) but that's the point

[quote="AceOfDiamonds"]I probably made you more confused :) but that's the point[/quote]

bleh...I got this stuff until I read your post...

so ignoring your post (sorry  :lol: ), this is also known as stars and bars

so you have $ n$ kids to give $ k$ pieces of candy to...

you represent this with $ n\minus{}1$ bars because that gives you $ n$ slots between (or on the sides of) the (vertical) bars;
_ | _ | _ ...| _ | _

these slots represent the kids you can give the candy to...or the slots you can put the $ k$ stars (candy) in

you can put any star in any slot...the slots can contain no stars, one star, or multiple stars, e.g.
 _ | ** | _ | _ | * | _

...this is one way to distribute $ 3$ pieces of candy to $ 6$ kids...

so to find all the possible ways...there are $ n$ kids/slots, so $ n\minus{}1$ bars...there are $ k$ pieces of candy/stars...you are ordering the pieces of candy/stars between the bars...

so you're choosing $ k$ stars out of $ k\plus{}n\minus{}1$ stars and bars...

$ \binom{k\plus{}n\minus{}1}{k}$ ...which is equal to... $ \binom{k\plus{}n\minus{}1}{n\minus{}1}$


geometry
3D geometry
pyramid
area of a triangle
Heron\u0027s formula
AMC

A pyramid has a square base $ ABCD$ and vertex $ E$. The area of square $ ABCD$ is $ 196$, and the areas of $ \triangle{ABE}$ and $ \triangle{CDE}$ are $ 105$ and $ 91$, respectively. What is the volume of the pyramid?

$ \textbf{(A)}\ 392 \qquad
\textbf{(B)}\ 196\sqrt{6} \qquad
\textbf{(C)}\ 392\sqrt2 \qquad
\textbf{(D)}\ 392\sqrt3 \qquad
\textbf{(E)}\ 784$

I think some one has discussed this question before.
The question could be simplified as what is the height at the edge which is 14 units long in a triangle with edge length 13-14-15.

[hide]Since $ ABCD$ is a square, the area is equal to the length of the side squared. It follows that $ AB\equal{}14$. Note that \[ [ABE]\equal{}\frac{1}{2}\cdot 14\cdot d(E, AB)\equal{}105\implies d(E,AB)\equal{}15\] Similarly, $ d(E,CD)\equal{}13$. The locus of points that are of a constant distance from a line is a cylinder. The intersection of these two cylinders is two lines that are parallel and equidistant from the plane of $ ABCD$. We can take $ E$ on the plane that is perpendicular to $ ABCD$ through $ M$ and $ N$, the midpoints of $ AB$ and $ CD$, respectively. Since the plane $ EMN$ is perpendicular to the plane $ ABCD$, $ d(E,ABCD)\equal{}d(E, MN)\equal{}12$ which is the height to the $ 14$ side of a $ 13$, $ 14$, $ 15$ triangle. Then the total volume is \[ \frac{1}{3}\cdot 12\cdot 196\equal{}784\] so the answer is $ \boxed{E}$.[/hide]

[quote="Altheman"]Since $ ABCD$ is a square, the area is equal to the length of the side squared. It follows that $ AB \equal{} 14$. Note that
\[ [ABE] \equal{} \frac {1}{2}\cdot 14\cdot d(E, AB) \equal{} 105\implies d(E,AB) \equal{} 15
\]
Similarly, $ d(E,CD) \equal{} 13$. The locus of points that are of a constant distance from a line is a cylinder. The intersection of these two cylinders is two lines that are parallel and equidistant from the plane of $ ABCD$. We can take $ E$ on the plane that is perpendicular to $ ABCD$ through $ M$ and $ N$, the midpoints of $ AB$ and $ CD$, respectively. Since the plane $ EMN$ is perpendicular to the plane $ ABCD$, $ d(E,ABCD) \equal{} d(E, MN) \equal{} 12$ which is the height to the $ 14$ side of a $ 13$, $ 14$, $ 15$ triangle. Then the total volume is
\[ \frac {1}{3}\cdot 12\cdot 196 \equal{} 784
\]
so the answer is $ \boxed{E}$.[/quote]

What does $ d(E,AB)$ mean?  Thanks in advance!

$ d(E,AB)$ denotes the shortest distance from point $ E$ to line $ AB$; i.e., the length of the perpendicular dropped from $ E$ to $ AB$.

Can someone explain Altheman's solution when it talks about the locus of points that are a constant distance from a line? I don't understand which cylinders he's talking about nor about which lines he's forming the cylinders. And how does he find that $d(E, MN)=12$?

[hide]Given a point $B$ on a line and given a radius $r$, what is the set of points that are $r$ away from $B$ and more than $r$ away from any other point on the line?  If we have a point $E$ in this locus, we see that $EB$ must be perpendicular to the line; see the picture below.  $C$ is not in the locus because $C$ is closer to $A$ than $B$, but $D$ is in the locus.  It's an intuitive idea, so I won't prove it.  So the set of points we're looking for is just the set of points $E$ so that $BE\perp l$ and $BE=r$.  But the set of points $E$ with $BE\perp l$ is just a plane perpendicular to $l$ through $E$, and the points in that plane so that $BE=r$ is just a circle in that plane with radius $r$.  Thus the set of points that are $r$ away from $B$ and more than $r$ away from any other point on the line is just a circle around $B$ with radius $r$.  If we take the circles for all points on the line and join them up, we get a hollow cylinder; this cylinder is the set of points that are at least $r$ away from every point on the line, and exactly $r$ away from some point on the line.

So we draw cylinders of radius $15$ and $13$ around $AB$ and $CD$; we know that $E$ must lie on each of these.  The intersection of these is just two lines (imagine two rolled up pieces of paper, if they could pass though each other).  So we take $E$ to be the point described by Altheman, so that $EM=15$ and $EN=13$.  Also $MN=14$.  So we find the area of $\triangle EMN$ by Heron's formula to get $[EMN]=84$, and then also the area can be written as $\frac{MN\cdot d(E, MN)}{2}$ so that $d(E, MN)=12$.

Does this make it a bit clearer?  It's hard to explain 3D geometry without a readily available 3D geometry program.[/hide]


In the diagram, ΔABC is right-angled at C. Also, 2 sin B = 3 tan A. Determine
the measure of angle A.
Pic is in attachment!
Mind the fax thing......

This is quite simple if you just use a bunch of trig identities.

[hide]
We have $ \cos{A} \equal{} \sin{B}$ so we can rewrite the equation as

$ 2\cos{A} \equal{} \frac {3\sin{A}}{\cos{A}} \implies 2\cos^2{A} \equal{} 3\sin{A}.$

We can then apply the Pythagorean Identity involving sine and cosine.

$ 2\cos^2{A} \equal{} 3\sin{A} \implies 2\cos^2{A} \equal{} 3\sqrt {1 \minus{} \cos^2{A}} \implies 4\cos^4{A} \equal{} 9 \minus{} 9\cos^2{A} \implies 4\cos^4{A} \plus{} 9\cos^2{A} \minus{} 9 \equal{} 0 \implies (4\cos^2{A} \minus{} 3)(\cos^2{A} \plus{} 3) \equal{} 0.$

The 2nd case is obviously has no solutions, so we consider the first case.

$ 4\cos^2{A} \minus{} 3 \equal{} 0 \implies \cos^2{A} \equal{} \frac {3}{4} \implies \cos{A} \equal{} \pm \frac {\sqrt {3}}{2}$

We then have $ A \equal{} \boxed{30^\circ}$ (the other 3 angles don't work because they are either obtuse or reflex angles).
[/hide]

I wonder if there any short solutions to this. (Maybe with the application of more clever identites?)

Or, you could've just done $ \cos^2 A \equal{} 1 \minus{} \sin^2 A$.  :P

Bah I feel so dumb. But yea that would've avoided the fourth powers and the extra cases...


vector
parameterization
function
conics
parabola
advanced fields
advanced fields theorems

The question reads:

Consider $ u_x \equal{}0$ and the initial curve, C, $ y \equal{} x^2$

(a) Show that unless the initial data satisfies certain conditions, the initial value problem has no global solution.
(b) However, if P is any point of the intiial curve different from $ (0,0)$ Show that the initial value problem always has a solution in a sufficiently small neighborhood of P.
(c) Is this true for P = (0,0)?

The characteristic equation is $ 1 \xi_1 \equal{}0$
The normal to the curve is $ <2x>$

If $ <1 \xi_1> \cdot <2x> \equal{}0$
from which $ 2x \equal{} 0$ (eq. 1)

From a sketch of $ y\equal{}x^2$ and (eq. 1) I can conclude the following:

1)  at $ (x,y) \equal{} (0,0)$ the initial curve is characteristic w/r/t the operator.
2)  at $ (x,y) \equal{} (0,0)$ the normal vanishes

3)  at all other points the initial curve is not characteristic w/r/t the operator.
4)  If $ x \neq 0$ then the normal $ \neq 0$. Hence the curve is not characteristic at all points $ (x,y) \neq (0,0)$

In order for $ u(x,y)$ to be characteristic then $ \nabla u$ must vanish. $ \nabla u\equal{}0$ only vanishes at (0,0).
For (a) is the correct answer $ u(0,0) \equal{} (0,0)$?

I think any help or pointers answering (a) will help me answer (b) and (c)...

Reread the chapter about the method of characteristics for linear PDE in your textbook and start with answering the question [i]What are the characteristics of this equation?[/i].

[quote="fedja"][i]What are the characteristics of this equation?[/i].[/quote]

I note from the text:
1) A non-zero vector $ \xi \equal{} (\xi_1... \xi_n) \in R^n$ is called [i]characteristic[/i] w/r/t the operator if the characteristic equation of the principal part is equal to zero. I believe the text is essentially saying, $ \xi_1 \equal{} 0$ is the characteristic equation. This is an $ R^2$ problem so there should be a $ \xi_2$ present either as a variable or constant...? I believe $ \xi_2$ does not appear because either there isn't a second variable in the original equation or the second variable is constant.

2) A surface is characteristic at $ x^0$ w/r/t the operator if a vector normal to the surface at $ x^0$ defines a direction which is characteristic at every one of its points. In $ R^2$ the "surface" is called a "curve".

To get the normal to this curve, write the curve using a parameter $ x\equal{}f_1(t)$ and $ y\equal{}f_2(t)$
(noting, of course, the parameter function should be 1:1 :) )

$ \vec T \equal{} <\frac{dx} { dt}, \frac {dy} {dt}>$ and $ \vec N \equal{} <\frac{dy} { dt}, \frac {\minus{}dx} {dt}>$

Stopping at this point to allow you to take this discussion another direction?

Seems like either my memory betrays me or the Russian terminology is slightly different from the English one :( I mean, when you use the method of characteristics, you really reduce the linear PDE to an ODE that you solve along certain curves (they are curves, not surfaces, in every dimension!). How do you call these curves in English? Translated from Russian literally, they would be called "characteristic curves" or, simply, "characteristics". My question was "what are those curves for your particular equation?".

Kent, mlok, or somebody else, can you refresh my memory about the corresponding terminology? :)

I think that it is the right terminology: http://en.wikipedia.org/wiki/Method_of_characteristics
and this is also the same name in french :-)

At this point I'll recognize a couple of typo's in my original post... $ 2x$ is not normal but tangent...

[quote="fedja"]I mean, when you use the method of characteristics, you really reduce the linear PDE to an ODE that you solve along certain curves (they are curves, not surfaces, in every dimension!). [/quote]

Yes!! the text (and the link in previous post) confirms your comment. I see the next step is forming an ODE on the curve using the normal to the curve. The characteristics curves are the solutions to this ODE.  The tangent to the curve is $ \frac {dx} {dt}, \frac {dy} {dt}$.  The normal to the curve is then $ \frac {dy} {dt}, \frac { \minus{} dx} {dt}$

I see (in the text) the ODE equation as the dot product of the coefficients of the characteristic equation and the normal to the curve set equal to zero... in this problem the characteristic equation is $ \xi_1 \equal{} 0$ and the normal to the curve is $ \frac {dy} {dt}, \frac { \minus{} dx_1} {dt}$ thus $ < a_1, a_2 > \cdot < dy, \minus{} dx > \equal{} 0$ hence $ 1 dy \equal{} 0$. which implies $ y \equal{} constant$

Therefore, I believe the answer to your question "what are the characteristic curves to the equation?" is that the characteristic curves are the lines, $ y \equal{} c$ (translated back into Russian :))

Great. Now think of this: it is enough to know the value at one point on the characteristic curve to recover all values on that curve from the PDE. At how many points does each horizontal line intersect the parabola?

Horizontal lines typically intersect the parabola at 2 points of intersection; just 1 point at point of tangency.

Now, once we typically have two intersection points and one is enough to determine the values on the entire line, there should be a relation between the values at these 2 points for a global solution to exist. What is this relation? Once you answer this question, you finish (a).

The only relation between the 2 points, as they relate to the line, is that they satisfy the equation of the line. In this case a horizontal line with $ \frac{dy}{dx} \equal{}0$

I'm asking not about the relation between [b]points[/b] but about the relation between the [b]values[/b] of the solution at those points. :roll:

The values of the solution at those points will be equal?

My God...  Who thought this up...

I think I understand, Solution values along a characteristic curve are equal.

The initial data must be a point on a characteristic curve in order for a global solution to exist? (or should I go get my Russian dictionary again?)

Seems like you got the idea, except your last sentence doesn't make sense. What you should say is [i]"in order for a global solution to exist, the initial data must have equal values at each 2 points of the parabola symmetric with respect to the $ y$ axis"[/i] (or something equivalent to it).

Excellent fedja... thank you...

Working part b and c....

For part b)
The characteristic curves are $ y\equal{}c$, Hence $ y\equal{}x^2$ is not characteristic except at one point, the vertex of the parabola. By definition of characteristic, the characteristic equation is not equal to zero, hence $ \xi_1 \neq 0$ and so, $ <1> \cdot <dy> \neq 0$ or $ dy \neq 0$.  The "Existence Theorem" (previous chapter) asserts that if $ P(x_0, y_0,z_0) \frac{dy_0(t_0)}{dt} \minus{} Q(x_0, y_0,z_0) \frac{dx_0(t_0)}{dt} \neq 0$ then in a neighborhood of U of $ (x_0, y_0)$ there exists a unique solution of $ P(x_0, y_0,z_0) z_x \minus{} Q(x_0, y_0,z_0)z_y \equal{} R(x_0, y_0,z_0)$ with initial condition $ (x_0, y_0)$ at every point of C contained in U. And so, in the neighborhood of P a unique solution exists. This should satisfy for part b)?

For part c)
At the point $ (0,0)$ the curve $ y\equal{}x^2$ is characteristic w/r/t to operator hence,  $ \xi_1\equal{} 0$. As such, $ <1> \cdot <dy> \equal{} 0$ hence, $ dy\equal{}0$ and the condition of the existence theorem is not satisfied. The theorem is not written "if and only if".  I'm not sure what is implied here?  No existence? No uniqueness?


calculus
integration
algebra
function
domain
algorithm
Ring Theory

salut!
daca are cineva timp va rog sa faceti o lista cu numele asociate structurilor algebrice in romana si in engleza
mi-ar folosi f mult :)

Mpai avand in vedere forumul in care ai postat, cred ca lista ar suna cam asa:

          Monoid -> Semi-Group
          Grup -> Group
          Inel   -> Ring
          Domeniu de Integritate, comutativ -> Integral Domain
          Inel cu algoritm de divizibilitate -> Euclidean Domain
          Corp -> Skew-Field
          Corp Comutativ -> Field
          Extensie minima in care un polinom f e perfect reductibil -> Splitting Field of f (just kidding...mai urla daca am ratat ceva...)

din ce carte ai luat toate denumirile astea ?
inele de fractii m-ar mai interesa
inele de fractii peste domenii de integritate (probabil ring over integrity domain)
homomorfism
epimorfism
monomorfism
etc *morfism
si stiu ca pentru *morfism sunt aici mai multe denumiri care inseamna
acelasi lucru
multumesc mult :)

HMm...pai

  inel de fractii -> ring of fractions
  daca inelul tau original e domeniu de integritate se cheama field of fractions (aici iarasi vezi ca la ei field si integral domain sunt automat comutative)
   homo, izo, si endo morfismele se traduc prin, evident, homo, iso si endo morphisms,
  la epi si mono nu stiu la ce te referi...

Aici apar vreo 8 feluri de morfisme: http://en.wikipedia.org/wiki/Morphism .

mmm...Dar ce probleme faci de te-ai lovit de astea? :D

[quote="Stgmihnea"]

          Monoid -> Semi-Group
[/quote]
semigrup$\ne$monoid
semigrup -> semigroup
monoid -> monoid(=semigrup cu element unitate)
corp se traduce mai des - "division ring"

Nu si in cartea mea....but anyway, nu conteaza cum le spui ca doar n-o sa-i spui football unui inel Euclidian...


Attention duelists, my hair says that super special awesome Yugioh: The Abridged Series will henceforth commence in America.

Hm, episode was pretty awesome. The first Bakura joke was ok, the second was kinda lame, the third was hilarious. OHOHHAOAHAOHAAH BUGGARATION.

Btw, SNAPE KILLS D

Something is with Yugioh and gay clowns.

Hopefully there aren't any Dark Magician Girl fanboys on AoPS. Those people are rather annoying.

Mokuba is getting better and better (Charmander, Char! Shut up Charmander!; EPIC MANEUVER; Now, let's go and get you a pet tarantula. What? But I wanted a pet rabbit. YOU WILL GET A TARANTULA AND YOU WILL LIKE IT. Later... Ok, now let's go get you a pet scorpion. But I thought we were going to get a tarantula! It's always about what you want, isn't it, Mokuba...)

There are too many good jokes, lol.

For those of you who don't know, I'm referring to this parody of the Yugioh show:

http://www.yugiohtheabridgedseries.com


The needle of the scale in the bulk food section of a supermarket is 26 cm long.  Find the distance the tip of the needle travels when it rotates 64 degrees.


Three distinct points $ P{(3{u}^2, 2{u}^3)}, Q{(3{v}^2, 2{v}^3), R{(3{w}^2, 2{w}^3)}}$ lie on the curve $ 27{y}^2=4{x}^3$. Show that if $ P, Q, R$ are in a line then $ uv+vw+wu=0$.

hello, let the line be in the form $ y\equal{}mx\plus{}n$, from here we have the system
$ 2u^3\equal{}3u^2m\plus{}n$
$ 2v^3\equal{}3mv^2\plus{}n$
we calculate the line
$ y\equal{}\frac{2}{3}\frac{u^2\plus{}uv\plus{}v^2}{u\plus{}v}x\minus{}\frac{2u^2v^2}{u\plus{}v}$
Inserting the coordinates of our last point we have
$ w^3\minus{}\frac{(u^2\plus{}uv\plus{}v^2)w^2}{u\plus{}v}\plus{}\frac{u^2v^2}{u\plus{}v}\equal{}0$
From here we get after a simple calculation $ uv\plus{}uw\plus{}wv\equal{}0$.
Sonnhard.


integration
inequalities
real analysis
real analysis unsolved

Suppose $ 0< a< b $ and let $ f:[a,b]\to {\mathbb R} , f(x)=\arctan{x}. $  According to the Lagrange mean-value theorem, there  exists at least a point $\xi \in (a,b)  $ such that   $   f(b) - f(a) =(b-a)f'(\xi)\;  . $
[i]  Prove that  $\xi $ is unique, and moreover that [/i] $ \;   \sqrt{ab} < \xi < \sqrt{ \frac{a^2+b^2}{2} }\;  . $

The uniqueness of $\xi$ is easy to see, just because that [tex]f'(x) = \frac{1}{1+x^2}[/tex] is monotonic.

and we know that $\arctan x < x\; (x>0)$.

[tex]\arctan b - \arctan a  = \arctan \left(\frac{b-a}{1+ab}\right) < \frac{b-a}{1+ab} = (b-a) f'(\sqrt{ab})[/tex]

this implies that $f'(\xi) < f'(\sqrt{ab})$, that is, $\xi > \sqrt{ab}$.


Susan's family inherited $30,000 from her parents. They spent 1/2 of the inheritance for a car, gave 1/4 of the inheritance to charity, and spent 1/5 of the inheritance for a trip. How much money is left from the original inheritance?

For full credit, you must do the following:

1. show or describe each step of your work, even if you did it in your head or used a calculator

2. write an explanation stating the mathematical reason(s) why you chose each of your steps.

when we answer the question, do we have to do the number 2 about why you chose that step? that's like one of the math tournaments I've been to, it's call Group Problem Solving, you get 3 questions, 4 people, and 45 minutes. THe problems are not hard but you have to explain every tiny detail, and I hated it.

PSSA usually looks for your ability to explain your solutions

so,

since it's from one of the PSSA practice question, I suggest you to write them down

[quote]Susan's family inherited $30,000 from her parents. They spent 1/2 of the inheritance for a car, gave 1/4 of the inheritance to charity, and spent 1/5 of the inheritance for a trip. How much money is left from the original inheritage[/quote]

 [hide]if you want to find out how much is left from the original inheritage, you have to find the sum of the money they spent on the car, charity, and the trip, then subtract it from 30000

to find the amount of money spent on the car, you have to find 1/2 of 30,000, and it's the same thing as 30000/2, so they spent 15000 on the car

they spent 1/4 of the money on charity, so 1/4 times 30000 is equal to 30000/4, which is 7500

they spent 1/5 of it on a trip so 1/5 * 30000 = 6000

then you add them up and subtract them from 30000. So 15000+7500+6000 = 13500 + 15000 = 28500

and you subtract that from 30000, 30000 - 28500 = 1500. so they have 1500 left from the inheritage[/hide]

this is a good answer, but I would rather make the denominators of the car charity and trip and just subtract it from the total

either ways are good

I just don't like arithmetic


Let $\{a_{n}\}$ be an arithmetical sequence with common difference being a prime. Suppose that some perfect $j$-th power and some perfect $k$-th power belong to this sequence, where $j,k$ are coprime positive integers. Prove that perfect $jk$-th power belong to $\{a_{n}\}$.

Probably I'm idiot, but this is completely trivial: so you need to show that if $x$ is a $j$-th power in $Z/pZ$ and also a $k$-th power, then it's a $jk$-th power. Take $a$ and $b$ roots of order $j,k$ for $x$. By Bezout theorem there are $uj+vk=1$. Then $a^{u}b^{v}$ is a root of order $jk$ for $x$.  :maybe:

You're right Harazi. Honestly speaking I haven't thought about it, just seen it for a sec today and it seemed nice to me (reminded me of this known problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=658). I think we can erase condition on primality of common difference and the thesis still hold, doesn't it? (this probably would be more interesting)
Sorry for posting such a triviality  :(

Yes, the result holds without the primality condition od the common difference. This follows from an appropriate use of the chinese remainder theorem and generalize the other problem (in the given link, which was from the ISL 97).

Pierre.

Some time ago I was given to solve the generalised version of the ISL problem proving that a $[j,k]$ th power will be in the sequence (the proof is almost the same with the original)


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:


Consider the circle $ (O)$ and three collinear points $ X$, $ Y$, $ H$ that are not on this circle such that $ \overline{HX}\cdot\overline{HY}\neq\mathcal{P}_{H/(O)}$. A straight line $ d$ through $ H$ meets $ (O)$ at two points $ M$ and $ N$. $ MX$ and $ NY$ intersect with $ (O)$ again at $ P$ and $ Q$ respectively. Show that as the line $ d$ through $ H$ varies, the line connecting $ P$ and $ Q$ always passes through a fixed point


let (C)be a circle and AB a chordofit.N is a mobile pointon AB>draw two circles that each one gothrough point N and
one of them is tangent to (C) in point A.and the other is tangent to (C) in point B.
findthe locus of point M.wich is the second inter section of the two circle.

The following theorem answers your locus question:

[color=blue][b]Theorem 1.[/b] Let k be a circle and A and B two points on it. Let N be a point on the line AB. Let u be the circle through the points A and N touching the circle k at A, and let v be the circle through the points B and N touching the circle k at B. Let M be the point of intersection of the circles u and v apart from N. Then, the point M lies on the circumcircle of triangle AOB, where O is the center of the circle k.[/color]

[i]Proof.[/i] Since the circles u and k touch each other at the point A, they have a common tangent at this point A. Similarly, the circles v and k have a common tangent at the point B. Let these two common tangents intersect at a point R.

Since the line AR is the tangent to the circle k at the point A, while O is the center of the circle k, we have $AR\perp OA$. Thus, < (AR; OA) = 90°. Similarly, < (BR; OB) = 90°. Thus,

  < ARB = < (AR; BR) = < (AR; OA) + < (OA; BR) = 90° + < (OA; BR) = < (BR; OB) + < (OA; BR) = < (OA; OB) = < AOB.

Since the line AR is the tangent to the circle u at the point A, the tangent-chordal angle theorem yields < AMN = < (AR; AN). Similarly, < BMN = < (BR; BN). Thus,

  < AMB = < AMN - < BMN = < (AR; AN) - < (BR; BN) = < (AR; AB) - < (BR; AB) = < (AR; BR) = < ARB = < AOB.

Hence, the point M lies on the circumcircle of triangle AOB, and Theorem 1 is proven.

  darij


Solve equation:
$ 9x^6  \plus{} (x^4  \plus{} x^2  \plus{} 1)^2  \equal{} 18x^5$

[hide]The LHS is positive, so $ x$ must be positive.
By AM-GM $ x^4 \plus{} x^2 \plus{} 1 \ge 3x^2$.
Squaring both sides
$ (x^4 \plus{} x^2 \plus{} 1)^2\ge 9x^4$.
$ 9x^6 \plus{} (x^4 \plus{} x^2 \plus{} 1)^2\ge 9x^6 \plus{} 9x^4 \ge 18 x^5$ by AM-GM.
The equality case is when $ x^4\equal{}x^2\equal{}1$ and $ 9x^6 \equal{} 9x^4$, so $ x \equal{} 1$ is the only solution.[/hide]

hello, for $ x>0$ we have
$ 9x^6\plus{}(x^4\plus{}x^2\plus{}1)^2\minus{}18x^5\equal{}(x\minus{}1)^2(x^6\plus{}2x^5\plus{}14x^4\plus{}8x^3\plus{}5x^2\plus{}2x\plus{}1)$
from here we get $ x\equal{}1$ as the only real solution.
Sonnhard.


I know this has a trivial solution using results involving poles and polars, but my meagre notes say it can be done with inversion.  Any thoughts?

Let quadrilateral $ABCD$ be inscribed in a circle.  Suppose lines $AB$ and $DC$ intersect at $P$ and lines $AD$ and $BC$ intersect at $Q$.  From $Q$, construct the two tangents $QE$ and $QF$ to the circle where $E$ and $F$ are the points of tangency.  Prove that $P$, $E$, and $F$ are collinear.

[url=http://olimpiadi.ing.unipi.it/modules.php?op=modload&name=Forums&file=viewtopic&topic=2374&forum=5&start=20]here[/url]

If you have some little familiarity with italian, see message "Inserito 04-06-2004 alle ore 11:44" from sprmnt21, you can find  some tipical projective theorems proved using only classical euclidean arguments. Combining those results you can prove the claim.

I'm afraid the only Italian I know are some bad words, but thanks for the reference.

Anyway, I think I have worked it out, at least partially.


I'm new to Asymptote, so I have a few questions:

1) Do you have to install anything to do asymptote?
2) In the beginning of an Asy. code, do you use <asy> or [asy]?

You need to install the Asymptote package. You can google it, but it kinda depends on what OS you have. On linux it's a simple matter of copy/pasting some files. You might get binaries with Windows or Mac OS.

It's <asy> in the AoPSwiki, [asy] on the Forums, and \begin{asy}..\end{asy} in a $ \LaTeX$ file (btw the \end{asy} has to be in its own line, with NOTHING else).

So can I use asymptote on the forums without the package?

ah. yes you can use Asymptote on the forums without installing anything. You only have to install them if you're going to produce Asymptote images on your own computer. :)


[quote="delegat"]If $ x$, $ y$ and $ z$ are positive real numbers prove that:

$ \sum _{cyclic}\frac {2}{x^{2} \plus{} xy \plus{} xz \minus{} yz} \geq \sum_{cyclic}\frac {1}{xy}$[/quote]

$ P\equal{}x\plus{}y\plus{}z$
$ \sum _{cyclic}\frac{2}{x^{2}\plus{}xy\plus{}xz\minus{}yz}\equal{}\sum_{cyclic}\frac{2}{xP\minus{}yz}\equal{}\sum_{cyclic}(\frac{1}{xP\minus{}yz}\plus{}\frac{1}{yP\minus{}xz})\equal{}\sum_{cyclic}\frac{(x\plus{}y)^{2}}{(xP\minus{}yz)(yP\minus{}xz)}\equal{}\sum\frac{(x\plus{}y)^{2}}{xyP^{2}\minus{}(x^{2}\plus{}y^{2})zP\plus{}xyz^{2}}\geq\sum_{cyclic}\frac{(x\plus{}y)^{2}}{xyP^{2}\minus{}(2xy)zP\plus{}xyz^{2}}\equal{}\sum_{cyclic}\frac{(x\plus{}y)^{2}}{xy(P\minus{}z)^{2}}\equal{}\sum_{cyclic}\frac{1}{xy}$

$ x\equal{}y\equal{}z$

[quote="delegat"]If $ x$, $ y$ and $ z$ are positive real numbers prove that:

$ \sum _{cyclic}\frac {2}{x^{2} \plus{} xy \plus{} xz \minus{} yz} \geq \sum_{cyclic}\frac {1}{xy}$[/quote]
It's wrong.
Try $ x \equal{} 0.41$ and $ y \equal{} z \equal{} 1.$ :wink:

I can not agree with you [b] arqady[/b] Actually what is wrong in [b]zaya_yc[/b]'s proof :?:

[quote="zaya_yc"][quote="delegat"]If $ x$, $ y$ and $ z$ are positive real numbers prove that:

$ \sum _{cyclic}\frac {2}{x^{2} \plus{} xy \plus{} xz \minus{} yz} \geq \sum_{cyclic}\frac {1}{xy}$[/quote]

$ P \equal{} x \plus{} y \plus{} z$
$ \sum _{cyclic}\frac {2}{x^{2} \plus{} xy \plus{} xz \minus{} yz} \equal{} \sum_{cyclic}\frac {2}{xP \minus{} yz} \equal{} \sum_{cyclic}(\frac {1}{xP \minus{} yz} \plus{} \frac {1}{yP \minus{} xz}) \equal{} \sum_{cyclic}\frac {(x \plus{} y)^{2}}{(xP \minus{} yz)(yP \minus{} xz)} \equal{} \sum\frac {(x \plus{} y)^{2}}{xyP^{2} \minus{} (x^{2} \plus{} y^{2})zP \plus{} xyz^{2}}\geq\sum_{cyclic}\frac {(x \plus{} y)^{2}}{xyP^{2} \minus{} (2xy)zP \plus{} xyz^{2}} \equal{} \sum_{cyclic}\frac {(x \plus{} y)^{2}}{xy(P \minus{} z)^{2}} \equal{} \sum_{cyclic}\frac {1}{xy}$

$ x \equal{} y \equal{} z$[/quote]

$ \sum\frac {(x \plus{} y)^{2}}{xyP^{2} \minus{} (x^{2} \plus{} y^{2})zP \plus{} xyz^{2}}\geq\sum_{cyclic}\frac {(x \plus{} y)^{2}}{xyP^{2} \minus{} (2xy)zP \plus{} xyz^{2}}$ is not true because the denominators is not necessary positives

[quote="delegat"]If $ x$, $ y$ and $ z$ are positive real numbers prove that:

$ \sum _{cyclic}\frac {2}{x^{2} \plus{} xy \plus{} xz \minus{} yz} \geq \sum_{cyclic}\frac {1}{xy}$[/quote]

It is a hard problem and the sign is reversed

$ L \minus{} R \equal{} \frac { \minus{} (x \plus{} y \plus{} z)\left((x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \plus{} 4\sum_{cyc}xy(xy \minus{} yz)(xy \minus{} zx)\right)}{\Pi (x^2 \plus{} xy \plus{} xz \minus{} yz)xy}$

[quote="delegat"]I can not agree with you [b] arqady[/b][/quote]
Why? What about my counterexample? Does it not work?  :wink:

[quote]Why? What about my counterexample? Does it not work? Wink[/quote]

I am sorry You were right [b]arqady[/b] I have not noticed that denominator is not necessary positive.
Actually I think that inequality holds if $ xy$, $ yz$ and $ zx$ are triangle sidelengths.


Prove that t^(n−1)+t^(1−n) < t^n+t^−n when t is not equal to 1, t >0 and n is a positive integer.

Here's what I have:
[hide]
$ t^{n-1}+t^{1-n}< t^{n}+t^{-n}$
We know $ t^{n}$ is positive, so multiply the inequality by that:
$ t^{2n-1}+t < t^{2n}+1$
$ t^{2n-1}-t^{2n}< 1-t$
$ t^{2n-1}(1-t) < (1-t)$
If $ (1-t)$ is positive, then $ t<1$ and $ t^{2n-1}<1$, so the inequality holds.

If $ (1-t)$ is negative, then $ t>1$ and the inequality becomes:
$ t^{2n-1}> 1$
which is true.[/hide]

The general principle at hand:

For $ w>x \geq y >z>0, \; GM[wz]=GM[xy]$, it must be true that $ AM[wz]>AM[xy]$

[b]Proof:[/b]

Let $ w/x=c>1$

$ x \geq y$
$ x > y/c$
$ x(c-1)>y(c-1)/c=y(1-1/c)$
$ xc-x>y-y/c$
$ w-x>y-z$
$ w+z>x+y$


number theory
greatest common divisor
relatively prime

gcd (p,q) =1
if m is greater than or equal to pq. Show that m can be expressed in : m=px+qy, where both x and y are nonnegative.

We can strengthen the claim; we only require $m > pq-p-q$.

[hide]

$px+qy=m$
$qy\equiv m \mod p$
It is possible to find some positive integer $y<p$ such that this equation works (I'm not sure how to prove that...)

Now if $y<p$ then $qy<qp$. Since $m>qp$, $m>qy$, and then $m-qy>0$.

Solving for x,

$px+qy=m$
$px=m-qy$
$x=\frac{m-qy}{p}$

Since $m\equiv qy \mod p$, $p$ divides $m-qy$, so x is an integer. We can see that it is also positive because the numerator and denominator are both postive.

[/hide]

edit: i think i found a proof for the part that was missing

[hide]

I need to prove that for relatively prime p and q and an integer m, there exists an integer $y<p$ such that
$qy\equiv m \mod p$

I can get rid of the $<p$ requirement, because if I find any integer $y$ that works, I can add or subtract p's to get it within the correct interval.

By Euler's Totient Theorem,
$q^{\phi(p)}=1 \mod p$
$q\bullet q^{\phi(p)-1}\equiv 1 \mod p$
$q\bullet q^{\phi(p)-1}m\equiv m \mod p$

Thus $y=q^{\phi(p)-1}m$ minus an appropriate amount of p to put it within the correct interval.

[/hide]


Okay, so you know that there are $ 25 \plus{} 2$ people in both classes.  You simply have a system of equations:

$ e \plus{} f \equal{} 27$

$ e \equal{} 2f$

Solving reveals $ e \equal{} 18$.  Since there are $ 2$ people taking both English and French, our answer is $ 18\minus{}2$ or $ \boxed{16}$.

[quote="GameBot"]One day Max says to Liz, ``Out of the 25 people taking either English or French, you and I are the only two taking both.'' Liz, being mathematically inclined, responds by pointing out that there are exactly twice as many people in the English class as there are in the French class. How many people are taking English but not French?[/quote]

This problem needs reformulation. The 'exclusive or', that is 'either...or", should be replaced by inclusive or, that is "or". Otherwise we should assume that the 25 count is without Max and Liz, and this is not what is meant (judging by the answer).. -mc

[quote=mathsolver]Excuse me, when is 16 two times more than 7??[/quote]

$16$ is the number of kids taking English ONLY. So $18$ is the total number of kids taking English. And $7$ is the number of kids taking French ONLY, thus $9$ is the number of kids taking French. They are saying there are twice as many people TOTAL in English class than in French class, not ONLY.

[quote=mathsolver]A post get edited or something, because I can't find my reference of 16 is 2 times more than 7.[/quote]

use a quote next time.

Also, this problem looks more like algebra than C+P to me

[hide=Solution]
Let x be the number of students taking only French then:
$x+2$ = Total French student
Likewise $2(x+2)$ = Total English Student
Then:
$$ (x+2)+2(x+2) -2 = 25  $$
$$ 3x+4 = 25 $$
$$x = 7 $$
Then since Total French student is $x+2 = 7+2 = 9 $
Total English student is $9*2= 18 $
Since we need students taking English only we subtract Max and Liz:
Which gives us 16.
[/hide]


who do you think will win the NL Championships and go to the world series? I Vote for Cardnals

No idea. I haven't payed very much attention.

Seen your avatar that doesn't mean much.

Cards or braves. The NL west is just plain pathetic, but the Stros could do some damage in the playoffs if they win the wild card. If anyone else wins the wild card, they won't make it past the first round.

I think that the Braves are a strong choice, they always finish the year strong.  The Cardnals have a better record, and a better team.  The Padres have high potential to be the worst team to make the playoffs ever(Actually i think that is clinched) :P .  The Astros are another strong team, and the phillies and Nationals are giving them a hard time in the Wild Card chase. This is going to be an exciting year

Go Braves, i hope they win the nl.  I want the yankees to win it all this year though.

rootin for the stros but cardinals are gonna win their too strong

Cardinals.

Carpenter - Mulder are pitching really well, solid bullpen, ...
the only problem is hitting tho.. Rolen's out, Walker's not at the level what cards want him to be yet.

if Pujols can lead the team with some hits, they'll win the world series.

[quote="GoBraves"]This has to be the Braves year ;)[/quote]
they have strong team this year.
Smlotz and Hudson will be scary to face during the playoff and andrew jones is having a great season.

but their bullpen..
yanks could have won so many times in the past because theyve had Mariano Rivera.
I'm wondering how much farnsworth can help.

i'm a diamondbacks fan so i'll be rooting for them. who knows what will happen in the worst division  :lol:
2nd choice is the braves.

heh, lots of people are voting for the D-Backs.  If they manage to make it they get to replace the Miracle Mets(They pretty much clinched a worse record than them even if they won the rest of their games.) They could be the "dynamic d-backs".


algebra
polynomial
function
quadratics
system of equations

OK...[hide]Find a monique polynomial with the greatest degree is 4. With the roots x,y,z and t with the property[/hide]

here's a start

[hide]
Note that the 4th initial equation is equivalent to $ (xt-yz)^{2}=0 \implies xt=yz$.

Consider a function $ f(x)=c_{4}x^{4}+c_{3}x^{3}+c_{2}x^{2}+c_{1}x+c_{0}$ whose roots are $ x,y,z,t$. Let $ s_{k}=x^{k}+y^{k}+z^{k}+t^{k}$. By Newton's Sums, we have
$ \begin{cases}12c_{4}+c_{3}=0 \\ 50c_{4}+12c_{3}+2c_{2}=0 \\ 252c_{4}+50c_{3}+12c_{2}+3c_{1}=0 \end{cases}$
Simplifying these expressions should give $ \frac{c_{1}}{c_{3}}= \frac{c_{1}/c_{4}}{c_{3}/c_{4}}=\frac{8}{3}$. Now we just wade through all the algebra using symmetric sums and whatnot to find $ x,y,z,t$. it's a lot of work though. is there an easier way maybe :maybe: ?

[/hide]

let $ a=x+t$, $ A=xt$, $ b=y+z$, $ B=yz$. The third equation gives $ A=B$. The other equations are:
$ a+b=12$, (1)
$ a^{2}-2A+b^{2}-2B=50$, (2)
$ a^{3}-3aA+b^{3}-3bB=252$. (3)

(2) becomes: $ a^{2}-4A+(12-a)^{2}=50\implies A=\frac{1}{2}*(a^{2}-12a+47)$.
(3) becomes $ a^{3}-3a(A)+(12-a)^{3}-3(12-a)(A)=252\implies A=a^{2}-12a+41$.

Solving, we get $ a=5,7$ [wlog $ 5$]. Then $ b=7$. $ A=B=6$. We obtain quadratics to get that:

$ (x,t)=(2,3)$, $ (y,z)=(1,6)$. We can switch within the ordered pair or switch one ordered pair with another to obtain all solutions.


Let P(n) is the sum of the first n prime numbers. Prove that there exists at least 1 perfect square between P(n+1) and P(n) with n>=1

It suffices to prove that $P_{n+1} - P_n > 1 + 2 \sqrt {P_n}.$

But, for $n \geq 5$, we have $P_n = 2 + 3 + \cdots + p_n < 1+3+5+7+9+\cdots + p_n$, the sum of all the odd integers from $1$ to $p_n$, which is well-known to be equal to $\frac {(1+p_n)^2} 4.$
Thus, $\sqrt {P_n} < \frac 1 2 (1+p_n)$ so that $P_{n+1}-P_n = p_{n+1} \geq 2 + p_n > 1 + 2\sqrt {P_n}$, for $n \geq 5.$

It is straightforward to verify that the desired result also holds for $n \leq 5$, andweare done.

Pierre.

Let $0< x<y$ be real.
Let $m$ be an integer. Then $x<m^2<y$ iff $\sqrt x < m < \sqrt y$.
Thus, if $\sqrt y - \sqrt x > 1$ there exists a integer $m$ such that $x<m^2<y$.
But $\sqrt y - \sqrt x > 1$ iff $y-x > 1 + 2 \sqrt x.$

Now, put $x = P_n,y=P_{n+1}$ and you have the above starting point.

Pierre.


AMC
USA(J)MO
USAMO
geometry
geometric transformation
reflection
MATHCOUNTS

I don't know if any companies still do this, but in the 1980s, AT&T or Bell Labs or whatever it was called made too much money (they were regulated since they were a monopoly), so they had to give some of it away. So what they did was to hire mathematicians to do research for them. I think Ron Graham and Andrew Odlyzko worked there for a while.

Yeah, I believe one of the main researchers of quantum computing works for Bell Labs.

Simon--Just out of curiosity, what are you planning on doing? I'm guessing you're going to say that you'll just keep learning more... :)

I really don't know what I'll end up doing. Probably math professor would be a good career for me, but I am not a good speaker (I get really nervous when I have to speak in public), so that is a burden that I will need to overcome before I can get a job as a professor. But in the meantime, I'll just be learning more, as you said. I'll be an undergraduate for a while, and then I'll go on to graduate school and stay there until I get a PhD. At least, that's the plan.

There are plenty of math folks on Wall Street and in various hedge funds.  My old company (D.E. Shaw) used to (and probably still does) hire top math types pretty aggressively - they found me by mailing to past USAMO winners.  (It does help to learn a little computer programming if you're interested in taking your math/physics acumen to a trading career.)

[quote="ComplexZeta"]AT&T or Bell Labs or whatever it was called made too much money (they were regulated since they were a monopoly), so they had to give some of it away. So what they did was to hire mathematicians to do research for them. I think Ron Graham and Andrew Odlyzko worked there for a while.[/quote]

Yes, Graham and Odlyzko both were at Bell Labs for quite a while. Odlyzko is currently a professor at the U of Minnesota, I'm pretty sure.

[quote="MagnusRex"]Ok, I really like math but I don't know all that much about what jobs you can do with math.[/quote]

Now I'll tell you a little, which is most of what I know  :)  about my local friends who are a couple both with Ph.D.s in math. The husband in that couple works as a consultant for one of the major high-tech consulting firms. I don't know EXACTLY what he does, and probably I shouldn't, to preserve client confidentiality, but basically what he does is applied mathematical problem-solving. Someone has some real-world problem that they want to solve in an efficient manner, and the consulting firm suggests things to do, and the part of the problem that pertains to mathematics gets referred to the guy I know. There is great flexibility in this kind of work--the man I know lives FAR from the headquarters city of his firm, and as long as he comes up with a good answer in a reasonable amount of time, he can set his own hours. He travels occasionally to meet clients, but gets to have significant family involvement, which is something a lawyer with the same income level would probably not have, for instance. His wife tells me that sometimes he has to go to a quiet room and just THINK for hours to come up with his problem solutions.

Interestingly, neither member of this couple has any involvement in "competition culture" of the kind we discuss here on this forum. They actively encourage their children's interest in math and science, but actively DIScourage them from getting involved in math competitions. They think competition problems are all based on MUCH too short time limits and don't really reflect the real work of most mathematicians (or, at least, most U. of Chicago Ph.D.s that they know). I respect their opinion, but nonetheless have gradually got my oldest son more and more involved in competition culture. I would agree that math contests of the AMC and Mathcounts kind are neither strictly necessary nor sufficient for promoting mathematical development of young people, but they do help my son get more involved and to think more flexibly about math.

I know another local example, a woman who majored in math as an undergraduate and then went to medical school and eventually became a neurosurgeon. She tells me, with a smile, that her undergraduate advisor, a woman math professor, told her that she would be wasting her life by going to medical school. I've heard of other instances of math majors becoming medical doctors, which is pretty easy for the undergraduates who remember to take the usual pre-med courses as electives while studying for their math major.

Several university Web sites I have seen suggest that all people in today's generation who desire to study math should be sure to learn a lot of computer programming skills. My local friend, when asked by another friend what he did in his job, said, "Mostly I write software." That's true of a lot of mathematicians these days.

Some people become "merely" engineers or "only" computer scientists after pursuing an interest in math, but at the age of most participants in this forum, one can hardly get too much "pure" math.

Hope this helps!

[quote="MagnusRex"]Ok, I really like math but I don't know all that much about what jobs you can do with math....so could I have some ideas? (i don't need the obvious like accountant, teacher....like something interesting...)[/quote]

besides math professor, you can do mathematical modeling for a financial company, the U.S. Department of Defense needs math whizes to develop new technology, or be a math teacher

[quote="failure"]The NSA (which sponsors USAMTS) claims to employ a lot of mathematicians.[/quote]

Indeed, the claim is that NSA (National Security Agency of United States federal government) is the largest single employer of mathematicians in the world. That's quite a plausible claim too.

I interviewed with NSA when I was an undergraduate (about 1980, as I remember) because it is also one of the largest employers of Chinese language majors in the world. I declined to go farther through the job-seeking process when I learned that most NSA employees, for very understandable security reasons, are restricted in their foreign travel. Instead, after graduation I went to Taiwan to further my language studies and met the girl who is now my wife of twenty years. But for the person who is willing to give up some individual freedom (to travel, and to publish) to protect the freedom that all Americans enjoy, NSA might be a very good career fit.

Private companies beat public work any day!

Actually, this won't be for long because stock options will be over soon... I'll either need to work in another country or hope there's an exemption for start-ups, bah!

Why should you need to be more obsessed?

[quote="MagnusRex"]That's cool. There's an idea...math professor...although I would probably have to be more obsessed with math than I already am ;)[/quote]


A die is tossed 8 times. Each toss is recorded, and is formed into an 8 digit number. What is the probability that the number formed is a multiple of 8?

How do you do this quick?

[hide="Hint"]All you really need to know is that the last three digits form a multiple of $8$. Ignore all the digits in front of it. [/hide]

Yeah, but that still takes a long time. This was on a sprint.

It doesn't take that long.

Possible last 3 digits are starting with a 1 or 2 are:
112
136
144
152
216
224
232
256
264

All the rest repeat. For last 3 digits starting with 1, 3, or 5, there are 4 possibilities. For last 3 digits starting with 2, 4, or 6, there are 5 possibilities, for a total of 27 possibilities that work. There are 6^3 total possibilities, so the answer is $\frac{1}{8}$.

What test was this from?

Oh. It kinda helps to see that the odd and even hundreds digits have the same number of multiples.

This was from Nat 2003 Sprint.

$\frac{1}{8}$ is an intuitive answer as well, so if you are rushing and have 1 minute left you could guess it right.

[quote="Ubemaya"]I don't think it's really intuitive in this case because you have a weird restriction...

Actually I'm kinda surprised that that's the answer.[/quote]
Well one in every eight numbers is divisible by 8. That isn't guaranteed to be right, though the answer makes sense.


Im not sure if this question belongs here:

What is the maximum volume of a  box can you create from a piece of cardboard with area of 2700 cm^2.  The box have an overlapping cover on the front, left and right side.  The area of the overlap should be the same as the size of the overlapped sides.

I can't find a better way to explain this, but I think the diagram will help. Same letters mean same area. (The box is open with its lid lifted.)

do different letters mean different area

Not really.  There is no restriction on the areas as long as you can get the maximum volume. I think this can be solved not using calculus, i just cant figure it out.

well i think its gotta be a cube because i think a cube has the largest volume, so then 2700/9 gives us 300 then $300^\frac{3}{2}$ or u can do  sqrt 300 then that to the power of 3 and i think thats the max im not sure


What is the smallest natural number greater than $ 1\plus{}\frac{1}{3} \plus{} \frac{1}{6} \plus{} \frac{1}{10} \plus{} \frac{1}{15} \plus{} \frac{1}{21}$?

hello, we get
$ 1\plus{}\frac{1}{3}\plus{}\frac{1}{6}\plus{}\frac{1}{10}\plus{}\frac{1}{15}\plus{}\frac{1}{21}\equal{}\frac{12}{7}$, so the smallest natural number which is greater than $ \frac{12}{7}$ is $ 2$.
Sonnhard.

on the other hand, you can see that $ \frac13\plus{}\frac16\plus{}\frac{1}{10}\plus{}\frac{1}{15}\plus{}\frac{1}{21}$ is pretty small, and just guess that the answer is 2.

$ \frac{1}{3}\plus{}\frac{1}{6}$ is $ \frac{1}{2}$, and because the other three fractions are each clearly less than $ \frac{1}{6}$, they will sum to less than $ \frac{1}{2}$, so the whole will sum to less than $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{2}\equal{}\boxed{2}$.

Recall that $ \frac{1}{1(2)}\plus{}\frac{1}{2(3)}\plus{}\frac{1}{3(4)}\dots\plus{}\frac{1}{n(n\plus{}1)}\equal{}\frac{n}{n\plus{}1}$, which is easily shown by induction. Now, the current sum, generalized, is

$ \frac{1}{\frac{1(2)}{2}}\plus{}\frac{1}{\frac{2(3)}{2}}\plus{}\frac{1}{\frac{3(4)}{2}}\dots\plus{}\frac{1}{\frac{n(n\plus{}1)}{2}}$

$ \implies \frac{2}{1(2)}\plus{}\frac{2}{2(3)}\plus{}\frac{2}{3(4)}\dots\plus{}\frac{2}{n(n\plus{}1)}$

$ \implies 2(\frac{1}{1(2)}\plus{}\frac{1}{2(3)}\plus{}\frac{1}{3(4)}\dots\plus{}\frac{1}{n(n\plus{}1)})$.

$ \implies \frac{2n}{n\plus{}1}$.

Plugging in the specific case of $ n\equal{}6$ (since $ \frac{6(7)}{2}\equal{}21$), we have

$ \frac{2(6)}{7}\equal{}\frac{12}{7}<\boxed{2}$.

This was a troll question on FTW. Recall that the numerators on ALL the fractions is 1, + 1(from the beginning),so the sum will surely not get anywhere more than 2. Ans. 2


AMC AIME ARML FTW USA(J)MO USAMO MATHCOUNTS

I go through such an enormous amount of scratch paper that I can't keep it around. I use random recycled paper my dad gives me. I currently have a 3-inch stack in front of me (mostly used). I write pretty large.

Sometimes it's organized line by line - depends.

My scratch paper in contests usually follows one of two formats:

1. Big diagram - diagram spanning the width of the page, with some extra margin above the diagram. Main column of work starts centered underneath the diagram, continued to the right. Side-work written diagonally on the left of the main column, or if necessary, above the diagram

2. Small/no diagram - diagram in upper left. Space directly under diagram left as margin. Remaining space used for main work, in columns, as appropriate. Side-work written diagonally in the margin under the diagram

Because of how I format my scratchwork, I like to use plain white paper with no lines. I also use large writing. The point of my scratchwork is to document my train of thought. I usually keep my scratchwork after the AMC, AIME, or ARML, but not after any other contest. Sometimes I go back through them and revisit my train of thought. It's quite nice.

Outside of contests, my scratchwork is much less "nice", mainly because I use used paper instead of nice paper, and have to work around whats already there.

In general, I like reading my own scratch paper much more than other people's scratch paper -- I just feel more connected with my own work. When I read my own work, its as if I'm explaining my solution to myself. When I want to see someone else's solution, I'd rather they explain it to me.

Actually, my scratch work is about the same for contests and outside of contests.

I usually try to write in the same direction, but a lot of my work looks messed up, and sometimes it goes down and sometimes it goes left...

There's usually a bunch of small diagrams everywhere, and there's usually a bunch of stuff that's crossed out.

There's also a bunch of calculations (usually multiplication) everywhere..

My scratch paper is terribly messy. On FTW, I just write wherever my pencil comes down on the paper. There are usually equations facing in every direction, and sometimes I can't read my own writing. :P In MC, all of the equations are usually right side-up and readable, but still unorganised and messy.

My scratch paper is sometimes organized line-by-line (when I'm doing proofs or other longer AoPS problems), but mostly it's just a random jumble of mathematics and poorly drawn diagrams. Is it advantageous for you to organize your scrap paper?

haha this is what i do:

i pull out a sheet from the printer and then right where ever i can find room

and if there is no more paper... i just right on the table... seriously (pencil of course)

and for problems like this: n*1/2*1/3*1/7= 1/4*1/3*1/7 (or something like that)

i actually write on the computer monitor crossing off fractions :rotfl:

hm... don't do what i do. im trying to keep myself organized as welll =]

During state testing it's all over the place, lined with doodles and drawings, circles, dots, triangles, numbers, an occasional penguin, and eraser shavings. Not to mention stuff crossed out.

Yeah my work is usually a mess. On a contest like the usamo or aime, I am very organized, but for casual/practice problem solving, my work is all over the place. I will sometimes have work on the same problem on about 3 pieces of paper because I run out of room on one, due to there already being 80% of a page's worth of work on it before beginning. So my papers get all over the place, and my room becomes a mess, and i finally have to go through occasionally and clean it all up...

my scratch work is always a mess.  I have trouble going back and following my train of thought, one thing I should improve on...

[quote="joml88"][quote="splatyango"]i think rruczyk wrote an aritcle (i forget where) about the advantages of organizing scratch paper..[/quote]

It's right [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]here[/url].[/quote]

hmmmm i should try that sometime my paper looks like the one that he put for "unorganized"  :rotfl: i seriously just write  on anything and anywhere... it turns out [size=200]VERY[/size] messy :blush:

but that does make sense... this kid that scored highest at my school uses one piece of paper for each problem.......................and writes organizedly

Usually depends for me. If the problems are Speed based (MATHCOUNTS ARML) my equations are sprawled everywhere.

But if it is olympiad math it is very neat.It sorta looks like:

Diagram (If needed) 1/4 to half a page.

Equations.

Well for geometry I usually draw pretty small diagrams by hand like 4 in by 4 in or smaller. I can usually guess things from a diagram that size. I also usually draw two or three different diagrams. I also have small diagrams that I draw on the side if I'm trying to add elements like lines to facilitate the use of a common theorem. Then I write my computations or whatever on a separate sheet of paper. I have two columns: 'know' and 'want' and I try and connect them. In the know column, I note major steps with like an asterisk or something and other stuff is less organized. Whenever I solve a problem, I always take the time to write a coherent solution.

[quote="archimedes1"]Just curious, as I've never really gotten a chance to peek at another problem solver's work...

What does your rough work usually look like?  Is it organized line by line, or is it scattered all over the place?
What do you usually use for scratch paper?  Do you just write on whatever is close at hand?

I find it amusing to discover and browse through my old scratch work when it is lying about.  Does anyone else do this?[/quote]
80% of the time it's scattered

for some wierd reason, i try to squeeze all of my work in the teeniest, tiniest little corner. i usually cant read it, but i do it anyways. (what? i'm "conserving paper"!)

sometimes i do that too: i write really tiny because the work i need to work out i should do in my head...but i don't!

Most of the time, I just write wherever I can, so there are writings everywhere (including numbers, scribbles, doodles, a few stick figures, and some other stuff) on what ever I can get my pen(cil) on. Sometimes, when I'm solving an algebra problem on paper, I do most of the steps in my head.

I do MY math upside down!

But yeah, my writing skills are pretty bad. Here's my scratch work process:

1.Look at paper. 2. Find a blank spot. 3. Scribble numbers down. 4. Look back at paper. 5. look back at scrap work and forget why I wrote them down.

That pretty much summarizes what I do too!  :D


search
geometry
parallelogram
number theory
prime factorization
relatively prime
AMC

For these, you definitely need to show some work....

Difficulty: Intermediate

i)  Source: Wisconsin Mathematics, Science, and Engineering Talent Search, Oct. 1994

If m is a positive integer, can m(m+1) be the 7th power of an integer?  Why or why not?

ii) Source: Wisconsin Mathematics, Science, and Engineering Talent Search, Oct. 1994

Suppose that diagonals AC and BD each divide quadrilateral ABCD into two triangles of equal area.  Prove that ABCD is a parallelogram.

iii) Source: Wisconsin Mathematics, Science, and Engineering Talent Search, Oct. 1995

Suppose that S is a set consisting of three positive integers and that the sum of any two members of S is a perfect square.  For instance, S could be {5, 20, 44} or {10, 54, 90}.  Prove that S contains at most one odd number.

iv) Source: Wisconsin Mathematics, Science, and Engineering Talent Search, Oct. 1997

Prove that n! can never be a multiple of 2^n for any positive integer n.  [Recall that for each positive integer n, n! denotes the product (1)(2)(3)(...)(n)]

This is a problem inspired by number iii:
Either prove or disprove the existence of a set of 4 integers such that the sum of any two members of the set are perfect squares.  That is, show that there either do or do not exist [b]distinct [/b]integers {a, b, c, d} such that (a + b), (a + c), (a + d), (b + c), (b + d), and (c + d) are all perfect squares.

iv)[hide] if you factor n! into a prime factorization, you get a factor of 2^x, where x = floor function.(n/2) + floor function.(n/4) + floor function.(n/8) ..., which wil always be smaller than n [/hide]

That wasn't really a proof.  Although your last statement is true, you didn't justify it.

For iv. I think the what dark meant is:

[hide]exponent of 2 in n! = [n/2]+[n/2^2]+... < n/2+n/2^2+.. = n. I think everything here is clear.[/hide]

For iii:

[hide]Let's assume there are >= 2 odd numbers. Let these numbers be a and b and let the other number be c. Then perfect squares c+a and c+b have the same parity but different velues (mod 4), which is impossible, since a perfect square = 0 or 1 (mod 4), so any perfect square can choose between an odd remainder and an even remainder (mod 4).[/hide]

ml2: I'm afraid that method isn't going to work very well.. theres actually a much easier method that would get you the answer in about 2 lines.

i) [hide]m and (m + 1) are always relatively prime; that means they have no common divisors other than 1. Then, both m and (m + 1) have to be the 7th powers of some integers; but that is impossible, so m(m + 1) isn't a 7th power of some integer.[/hide]

ii) [hide]Let the diagonals intersect at O. Construct perpendiculars AE and CF from A and C to BD, respectively. Then AE = CF. Thus triangle AEO is congruent to triangle CFO => AO = OC. Similarly BO = OD. Only in a parallelogram does the diagonals bisect each other, thus ABCD is a parallelogram.[/hide]

What is the explanation to iii? I am a bit confused by the one given.

iii) Clarification of grobber's solution

[hide]Let the two odd numbers be a and b; let the even one be c. In order for a + b to be a perfect square, one of them has to be 1 (mod 4) and the other 3 (mod 4); but then either (c + a) or (c + b) will be 2 (mod 4) and thus not a square.[/hide]

Osiris, that's not quite right.
Mod 4, every perfect square is 0 or 1.  If we have 2 of our numbers, a and b, both odd, they cannot both be congruent to 1 or both congruent to 3, because then their sum would be congruent to 2 and thus not a perfect square.  So one of them is congruent to 1, the other to 3.  No matter what the third number is, mod 4, a + c and b + c will differ by 2 mod 4.  However, then they cannot both be perfect squares, since all perfect squares are congruent to 0 or 1 mod 4, and thus the difference between two perfect squares can only be 0, 1, or 3 mod 4.

Is anyone going to give my problem a try?


calculus
integration
trigonometry
logarithms
function
calculus computations

What would the definate intergral from 0 to pi/2 of tan(x) be?[/code]

Your calc textbook [i]probably[/i] has a table of integrals inside the front cover.  If it doesn't,
(edited: right...)
$\int \tan x \, dx =-\ln | \cos x |$

t0rajir0u, you forgot $|\ \ |.$ and .... :(

$\int tanx.dx =\int \frac{-d(cosx)}{cosx}=-lncosx+C$ ( C=const)

In any case, $\int_{0}^{\frac{\pi}2}\tan x\,dx=\infty.$ The integral diverges.

you can also write -ln (cos x) as ln (1/cos x) = ln(sec x). That is how the antiderivative of the tangent function is defined in my text.


calculus
integration
real analysis
LaTeX

I have a rather nasty integral from doing Lebesgue product measure:

$ \int_{(X \times Y) \setminus (X' \times Y')} \abs{f(x,y)}\ d(\mu \times \nu)$

Attempting to type several of these on the same line is leading to problems.  Is there an easy way to lower the limit so that it is centered below the integral?

Edit: Alternatively, I would settle for just having the $ \abs{f(x,y)}$ appear directly above the limit.

Use \limits and move f(x,y) to the left with \hspace
$ \int\limits_{(X \times Y) \setminus (X' \times Y')}\hspace{\minus{}2.5em} |f(x,y)|\ d(\mu \times \nu)$
What is \abs? This gives an undefined control sequence error so use | |.

Thanks, exactly what I wanted.  \abs is my own command:

[code]\providecommand{\abs}[1]{\left\lvert#1\right\rvert}[/code]

Forgot to take it out.

While it does work, it still feels forced to me.  If I change up my integral limits to something smaller or larger, then I'm going to have to change the -2.5em to other values.  In a homework assignment, this is fine.  But if I was typing up something longer, I wouldn't want to spend an entire afternoon playing around with em values to make things look nice.

It feels like there should be some way to have LaTeX automatically format it for me.  I know just because there should doesn't mean there is, but still...

There are always cases where you have to intervene since LaTeX cannot automatically deal with every case; often, for example, you have to add extra vertical spaces where the lines are too close together. If you want something more automatic then you can get LaTeX to measure the width of the lower limit automatically as in:
[code]\newlength{\integrand}
\newcommand{\myintegral}[1]{\settowidth{\integrand}{$#1$}\ensuremath{\int\limits_{#1} \hspace{-0.3\integrand}}}
$\myintegral{(X \times Y) \setminus (X' \times Y')} |f(x,y)|\ d(\mu \times \nu)$[/code]

This is exactly the kind of thing I was looking for, thanks.


Dear Mam/Sir:

I need to create a formula for scheduling several sports games on several fields.  The games themselves will involve two teams and each game will have varying durations, and each field can have different availability.  For instance, say we have the following:

Games
1) Team A vs Team B (Duration = 40 minutes, can be played on field 1 or 2)
2) Team C vs Team D (Duration = 30 minutes, can be played on field 2 or 3)
3) Team E vs Team F (Duration = 35 minutes, can be played on field 1 or 3)

Fields:
1) Field 1 (Availability = 2 hours)
2) Field 2 (Availability = 3 hours)
3) Field 3 (Availability = 1 hour)

Based on the games and fields provided above, can you please assist me with creating a formula that will schedule the games on the correct field based on duration and availability?

I was thinking about doing something like this:
1) For each field, have a priority # = duration of games scheduled on field/availability of field
2) For each game, get the overall field availability = sum of field priorities/number of fields
3) Schedule the game with the highest overall field availability
4) Recalculate field priority # and overall field availability
5) Schedule the next game
6) Repeat steps 4 and 5 until all games are scheduled

Would this work?

What you want to do is schedule the hardest games to fit first. A game is hard to fit if the number of possible spots is low. So use a heap, insert the games and make it sorted by avaliability. Then extract the root, schedule it, rebuild the heap. The heap will be effective as the priorities won't change much.[/code]


function
logarithms
limit
calculus
integration
calculus computations

There isn't a linear function that bounds all primes so it's like a p-series p > 1...

No this is not a p-series with p>1. This is a p-series with p=1.
$ \sum_{prime} \frac{1}{q} \equal{} \sum_{prime} \frac{1}{q^1}.$

As to "why", I am not sure if that is even possible to answer. Sum of the twin primes converges, but primes do not. Same with Harmonic series. How would you explain "why" the harmonic series is divergent? Is an explanation synonymous with proof to you? If so,
[url=http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2004;task=show_msg;msg=0760.0001]Proof[/url]

Is there a linear function which bounds all integers? So why does the harmonic series diverge?

The [url=http://en.wikipedia.org/wiki/Prime_number_theorem]prime number theorem[/url] provides estimates for the distribution of primes. A crude but effective approximation for the $ k$th prime $ p_k$ is that it is about $ k\ln k$ - at least in the sense that $ \lim_{k\to\infty}\frac{p_k}{k\ln k}\equal{}1.$

That means that the series $ \sum_k\frac{1}{p_k}$ is limit comparable to $ \sum_k\frac1{k\ln k}.$

That series can be shown to diverge by either the integral test or the condensation test.

Of course, Euler's classic proof is the divergence of $ \sum_k\frac1{p_k}$ is both easier (because it doesn't require the massive technicality that goes into a proof of the prime number theorem) and more satisfying. Do you know that proof?


probability
geometry
geometric transformation
rotation

Three numbers, a_1, a_2, a_3 , are drawn randomly and without replacement from the set {1,2,3 ... 1000}. Three other numbers, b_1, b_2, b_3 , are then drawn randomly and without replacement from the remaining set of 997 numbers. Let P be the probability that, after suitable rotation, a brick of dimensions  b_1*b_2*b_3 can be enclosed in a box of dimension a_1*a_2*a_3, with the sides of the brick parallel to the sides of the box. If P is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Assume, without a loss of generality, that $0 \le a_1<a_2<a_3\le1000$ and $1\le b_1<b_2<b_3 \le 1000$.  Therefore, for the brick to go in the box, $a_1>b_1$, $a_2>b_2$, and $a_3>b_3$.  I don't know how to calculate the probability of this, but it's a start.

Yeah, that's what I did too.  I don't know how to calculate this probability since the chances a_1 is larger than b_1 isn't just 1/2.

Then again, maybe P(a_1 is larger than b_1) is just 1/2 because I really don't see how it could be anything else...

The probability that $a_1>b_1$ is the same that $b_1>a_1$, so
$P(a_1>b_1) = \frac 12$


logarithms
function
inequalities
inequalities proposed

Here's a bit  :)  complicated version of the problem from one Polish Olympiad:

Let $f_i(x)=x^i + \sqrt{p}x + r_{i-2} = 0$ for $i=3,4,...,n$ and $\frac{p}{r_k}=e^k$. Prove:
$\frac{n+1}{2}ln x + ln 4 \leq \frac{n+1}{2}$

$x^pr_p \geq 0$ for $p=1,2,...,n$

LOL... you should think about proposing problems for the German (former GDR) olympiad... most of their problems resemble this one very much in beauty... but PLEASE, start proposing only when I've finished school...  :D  :D

And, BTW, should x satisfy all the equations $f_i(x)=x^i + \sqrt{p}x + r_{i-2} = 0$ for all i = 3, 4, ..., n simultaneously? I bet finding such an x would be a bit difficult in the general case...  :D

  dg

Definitely, if Megus uses $\ln x$ and $\sqrt p$ then $x,p>0$, but in this case $x^i+\sqrt px+r_{i-1}>0$.

Darij: yeah, x satisfy simultaneously all the equations.

Myth: I don't see why it must be greater than zero  :?  - $p$ could possibly be zero, couldn't it ?

Ok - as I thought - I forgot about one important condition - $x^nr_n \geq 0$ for any $n \in N$ Shame on me  :blush:

Because no progress was made so far here is a hint:

[color=yellow] first - use method like in this problem for every function n times  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18942[/url] summarize all the inequalities; use Cauchy's inequality... [/color]

I hope this will help a bit


algebra
polynomial
inequalities
inequalities proposed

Let $ P(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$, be a polynomial which satisfy the following condition:

$ |P(x)|\leq 1$ for every $ x$ with $ |x|\leq 1$.

Prove that $ |a|\plus{}|b|\plus{}|c|\plus{}|d|\leq 7$

Alexandros

$ P(0),\ P(\pm 1)\Longrightarrow Triangle Inequality$.

Could you please give more details about the solution using triangle inequality?

Alexandros

Sorry for my delaying reply, cretanman.

$ P( \minus{} 1) \equal{} \minus{} a \plus{} b \minus{} c \plus{} d\ \cdots [1]$
$ P(1) \equal{} a \plus{} b \plus{} c \plus{} d\ \ \ \ \ \ \cdots [2]$
$ P(0) \equal{} d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots [3]$

$ \therefore b \equal{} \frac {P( \minus{} 1) \plus{} P(1) \minus{} 2P(0)}{2}, c \equal{} \frac {P(1) \minus{} P( \minus{} 1)}{2} \minus{} a$, thus

$ |a| \plus{} |b| \plus{} |c| \plus{} |d| \equal{} |a| \plus{} \left|\frac {P( \minus{} 1) \plus{} P(1) \minus{} 2P(0)}{2}\right| \plus{} \left|\frac {P(1) \minus{} P( \minus{} 1)}{2} \minus{} a\right| \plus{} |P(0)|$

$ \leq |a| \plus{} \frac {1}{2}(|P( \minus{} 1)| \plus{} |P(1)| \plus{} 2|P(0)|) \plus{} \left|\frac {P(1) \minus{} P( \minus{} 1)}{2} \minus{} a\right| \plus{} |P(0)|$

$ \leq |a| \plus{} \left|\frac {P(1) \minus{} P( \minus{} 1)}{2} \minus{} a\right| \plus{} 3$.

kunny I think that you don't get the desired result using this technique.

For example the polynomial $ P(x)\equal{}4x^3\minus{}3x$ (Tchebychev polynomial for $ n\equal{}3$) satisfies the condition of the problem (so you cannot improve the upper bound)  but obviously you don't get that $ |a| \plus{} \left|\frac {P(1) \minus{} P( \minus{} 1)}{2} \minus{} a\right| \plus{} 3 \leq 7$

Any other ideas for this problem?

Maybe we need to use $ P(\pm 1/2)$

Alexandros


[quote="amirhtlusa"]source: 1996 IRMO, first round.
this is a very easy problem:
how many natural numbers, less than $100000$ exist, that it's equal to it's reverce?
(for example the reverce of $1375$, would be $5731$)[/quote]

wouldnt we just call that "palindrome"???

[quote="Rodolphe2005"]choose one digit for the first and fifth digits
choose one digit for the second and fourth digits
choose one digit for the third digit
There is $9\times 9\times 9=729$ reverse numbers[/quote]
your answer is incorrect :D

[quote="K81o7"]Number of such 1 digit numbers=9
Number of such 2 digit numbers=9
Nmber of such 3 digit numbers=90
Number of such 4 digit numbers=90
Number of such 5 digit numbers=900
So it's 1098..  For 1 digit numbers, every one works.  For 2 digit numbers, it's just the number of 1 digit numbers (because you can repeat each.  For 3, the middle digit can be any of the 10 digits, and the outside two can be any but 0.  So it's 9*10=90.  For 4, it's just the number of 2 digit numbers, which is 90.  For 5, the middle digit can be any digit, the thousands and tens digits can be any digit, and the two outside ones can be any digit but 0.  So it's 9*10*10=900.  Simple perm problem.[/quote]
$1098$, is correct. :lol:

oups I have just took numbers of 5 digits and forget that we can take $0$


Simplify and express your answer as a common fraction:
$ \displaystyle\frac{\frac{2}{3!}\plus{}1}{\frac{1}{2!}\plus{}4}$.


[quote="shobber"]I have a very stupid question: Does $\sqrt{n}$ have a limit when $n \to \infty$? Prove it.[/quote]

take any $\alpha \in \mathbb{R}$.  for any $n \geq \alpha^2$ we have $\sqrt{n} \geq \alpha$. the sequence is increasing, so $\lim_{n \to \infty} \sqrt{n} = \infty$


MATHCOUNTS has notified me that it is now OK to discuss the 2007 MATHCOUNTS Chapter contest.

School contests may also be discussed.

Copies of the 2007 School and Chapter contests can be downloaded from [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=54]mathcounts.org[/url]


superior algebra
superior algebra unsolved

In any $ S_n$ show that every element is the product of at most two cycles.

Proof by example:  $ (87654321) \times (12)(3)(45678) \equal{} (283) \Leftrightarrow (12)(3)(45678) \equal{} (12345678) \times (283)$.

(I'm being a little tongue-in-cheek here, but I hope you can see the sketch of an argument I'm hinting at.  Given a cycle decomposition, we take the first cycle to be the permutation we obtain by dropping the parentheses in the cycle decomposition, and the second cycle is built from the last element of each cycle.)


Hi
I searched about this topic in Mathlinks but did not find anything. I decided to search it in Google and i find a topic about one of the ITT Problem in Mathlink witch solved by Mr.Omid Hatami (My fellow citizen). I think if we have a part about this tournament,  for describe problems, solve it and speak around this tournament is little for you (and too for me) interesting.

Thank  :roll:

Oh sorry you wrote about this contest here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=26017


Putnam
email
MIT
college
college contests

Aren't the Putnam 2005 results supposed to be out by now? Anyone has any idea?  My school reopens on Monday. Is that when I could know my score? :?:

Thanx in advance.

As of Friday, March 17, I hadn't received them yet. They do get mailed to me.

Is there a website where all of the statistics for a given year are posted?

Eventually they will get placed here: [url]http://www.unl.edu/amc/a-activities/a7-problems/putnam/[/url]

Does anybody know when we will ge the results for the putnam exam 2005? Thanks.

Last year they came out about this time.  I am assuming it will be this or next week (I didn't take it, but I'm interested to see how the schools rank).

last year I got an email from my math department on the 30th...so i'd expect results to come in sometime next week.

This past weekend, I talked to Byron Walden from Santa Clara U., which administers the exam. (Bonus challenge: figure out where this conversation happened. It has nothing to do with the Putnam, or even with mathematics at all.) He reported that results had been mailed sometime last week.

The AMC web site that David mentioned will be updated as soon as the mailing arrives at MIT and is passed on from Hartley Rogers to myself; I'm hoping that will happen sometime this week.

[quote] (Bonus challenge: figure out where this conversation happened. It has nothing to do with the Putnam, or even with mathematics at all.) [/quote]

American Crossword Puzzle Solving Championship?
http://www.stamfordadvocate.com/news/local/scn-sa-crossword1mar27,0,7411564.story?coll=stam-news-local-headlines

Dr. Kedlaya, way to go! Congrats on your splendid performance at the 29th annual American Crossword Puzzle Tournament!
 :)

I just got the results in the mail today. I'll start a new topic for it.

I picked up the list of winners this morning and have transcribed it to the AMC web site. I haven't gotten the stats chart yet, but that's merely because Hartley Rogers didn't leave me a copy; I should have it either today or Monday, and it'll go up soon thereafter.

Oh, and Marc982 guessed correctly: Byron Walden and I were both at the American Crossword Puzzle Tournament, Byron as a constructor (his was the most difficult puzzle of the event), and I as a solver.

just wondering, how do MIT students get notified of their scores? does an email get sent out to everyone who took the Putnam? or is it available at the UMA? Is monday too early for finding out?
Thank you


SPOJ Problem Set
153. Balancing the Stone
Problem code: SCALES

You are given scales for weighing loads. On the left side lies a single stone of known weight W<2^N. You own a set of N different weights, weighing 1, 2, 4, ..., 2^(N-1) units of mass respectively. Determine how many possible ways there are of placing some weights on the sides of the scales, so as to balance them (put them in a state of equilibrium). Output this value modulo a small integer D.

Input
The input begins with the integer t, the number of test cases. Then t test cases follow.

For each test case, the first line contains three integers: N L D, where N denotes the number of weights at your disposal, L is the length of the binary representation of number W, and D is the modulus (1<= L<= N<= 1000000, 2<= D<=100). The second line contains the value of W, encoded in the binary system as a sequence of exactly L characters 0 or 1 without separating spaces.

Output
For each test case, output a single line containing one integer - the calculated number of possible weight placements, modulo D.

well i have a recurrence relation worked out.assume W  to be given in decimal form.
let's call the number of ways to balance a weight W using N different weights given be ways(W,N);
then if W=1 return N;
if(W%2==0)return ways(W/2,N-1)
else return ways(W/2+1,N-1)+ways(W/2-1,N-1)

can be easily rephrased in terms of binary representation of W.
but how can i use dynamic programming to solve such a problem.(or is there an other way)
coz the limits on N and L (length of the representation of W) are too huge,arent they??
please help

Very, very nice problem.
No, dynamic programming won't get you anywhere; although the formulae you find are useful if you notice something.

Heres my solution and how I got there:
Key observation:
[hide]If you fix W, then ways(W,N) is [b]linear[/b] in N. Obviously true for W = 1, and any higher W is found by just a sum of lower W in your formula.[/hide]
Rest of solution:
[hide]Now consider any W, any let it be of length (in binary) n.

Lets first calculate ways(W,n+1). If you don't use the 2^n weight, there is precisely 1 way (writing W in binary). If you do, we need to recursively calculate the answer for 2^n-W, with n+1 weights. Lets call the value we get there x. Then ways(W,n+1) = x+1.

Now we calculate ways(W,n+2). If we don't use the 2^(n+1) weight, there are x+1 ways as above. If we do, then we must subtract the 2^n weight, leaving us with x ways again. Thus ways(W,n+2) = 2x+1.

That means ways(W,N) = aN + b, where a = x and b = 1 - xn, ie ways(W,N) = xN + 1 - xn.

So the final method will look something like:

[code]
ans(W, N):
  n = length of W in binary
  x = ans(2^n - W, n+1)
  return (x*N + 1 - x*n) mod D
[/code]

Its then relatively easy to work backwards and turn that into something iterative. I'll leave that to you, for now.
[/hide]

i think there is some minor thing wrong in the explanation u gave triplem..
u say that if wieght W has n bits in its binary representation then
way(W,n)=1;
but this is wront ,,look at W=5;
it has 3 bits and 2 ways namely 5=4+1 and 5=4+2-1
and also x=way(2^n-W,n) not n+1..in the pseudocode
so i guess it makes the iterative version a bit difficult...
i have got quite a few TLE'S on spoj...trying to implement a map-based memoized version :) ..sumthing like
map< pair<string,int> ,int > ...which i know is too slow...
anymore ideas
best regards
vasu

Hm, you're right. It seemed to work perfectly fine when I last did it (a while ago, I've lost my code now since I changed computers), so maybe I did it right but typed it up wrong. I'm thinking about it again now.

ok cool i got an AC with the iterative version in 0.20 secs
beautiful problem:)
here's the code
[code]
#include<iostream>
using namespace std;
char s[1000001];
int main()
{
    int ncases;
    scanf("%d",&ncases);
    while(ncases--)
    {
                   int n,l,d,i,t1=1,t2=1;
                   scanf("%d %d %d",&n,&l,&d);
                   scanf("%s",&s);
                   bool flag=false;
                   for(i=l-1;i>=0;i--)
                   {
                                      if(flag)
                                      {
                                              if(s[i]=='1')
                                              {
                                                           t2+=t1;
                                                           t2%=d;
                                              }
                                              else
                                              {
                                                  t1+=t2;
                                                  t1%=d;
                                              }
                                      }
                                      if(!flag &&s[i]=='1')flag=true;
                   }
			t1+=((n-l)*t2);
                   printf("%d\n",(t1%d));
    }
    return 0;
}
[/code]
i dont think there's anything to explain;)
best regards

Why shouldn't the brute force recursion work? Either you do nothing in which case the bits remain the same, or you subtract a power of two, in which case you can reverse the sign of the result (as your goal is zero), and you end up with the complement of the initial bitstring. So at each position in the string there are only two states: the original string and its complement. Thus if you do a memoized recursion (or I guess you could DP it if you are tricky enough), there are only $2N$ states, and each step in the recursion is $O(1)$, so this solution would be $O(N)$.

Is this correct, or did I miss something?

yes brute recursion will be fine as well:)
just that i was doing something wayward earlier...storing the two's complement of the original bitstring everytime in a different array
so it was timing out


probability
number theory
prime factorization

[hide="Outline for 1"] Have two cases, one where you find the probability of drawing a red from the first and red from the second, and one where you find the probability of drawing a blue from the first and red from the second.[/hide]
[hide="hint for 2"] A fraction of the form $ \frac{a}{b}$ is terminating if and only if the prime factorization of b consists of only 2s and 5s.[/hide]

Well, you've counted both cases for #1, so you should add now.

Your #2 looks right on a first glance.

for #1 you only find the probability of the red one which would be 7/40. you dont need to find the probability of the blue.

edit: this was for nick42.


number theory
relatively prime
number theory unsolved

Let $ k$ be an integer and let $ n$ be a positive integer. Show that the interval $ [kn,(k\plus{}1)n]$ contains an integer that is relatively prime to every element of the set $ \{1,2,\ldots,n\}$.

This result seems to be a recent one. I don't know if a proof is available online, and I haven't yet found one myself.

I'm sorry, people, something is wrong.
Between 13 x 168 and 13 x 169, every integer is divisible by a prime <13.
I suppose I shouldn't believe everything I read on the internet. :oops: Neither should you! :D


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $A\in\mathbb{R}^{n\times n}$ such that $3A^3=A^2+A+I$. Show that the sequence $A^k$ converges to an idempotent matrix. (idempotent: $B^2=B$)

[quote="Arne"]How is convergence of a sequence of matrices defined? :huh:[/quote]Convergence of each of its elements? :? it doesn't even matter here...

[quote="Arne"]How is convergence of a sequence of matrices defined? :huh:[/quote]

Take any norm, since they are all equivalent  :)

$A$ is diagonalisable since its minimal polynomial is splitted.

$A^k$ is similar to $diag ({a_1}^k,...,,{a_n}^k)$ with $a_i \in \{ 1,-1/3+i\sqrt{2}/3, -1/3+i\sqrt{2}/3\}$

So $A^k$ converge toward a matrix $diag (1,1, ..., 0)$ which is indeed idempotent.

Hope it's correct.


absolute value
combinatorics unsolved
combinatorics

There are $k$ positive integers written along a circle. Between every two consecutive numbers, we write the absolute value of their difference and erase the original numbers. We can repeat this procedure over and over again. Find all $k$ such that all numbers along the circle will be eventually zero, regardless of the choice of the original numbers.

You might find [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=difference&t=1073]this[/url] useful. I hope it's correct. Maybe you could tell me your opinion, since nobody bothered to read that post :).


algorithm
ceiling function
arithmetic sequence
algebra unsolved
algebra

You are in a building with n storeies ( n floors ) , you have two plates , you know that there is a a smalest k in [1,n] so that if you throw a plate from i whit i>=K then your plate will be broken , what is the minimum of steps using your two plates to find the storey k , and what is your strategy ? ( of cours if you throw a plate from a storey i<k then it will not be broken and you can use it again  :P )

I hope that the problem is clear ! any ideas ? maybe that have relation whit sorting algorithm but I have no idea about that !

I think the idea is to do something like this, for $ n \equal{} 26$:

Throw a plate in the $ 4^{\text{th}}$ floor.
If it is not broken, throw the plate in the $ 10^{\text{th}}$ floor.
If it is not broken, throw the plate in the $ 15^{\text{th}}$ floor.
If it is not broken, throw the plate in the $ 19^{\text{th}}$ floor.
If it is not broken, throw the plate in the $ 22^{\text{nd}}$ floor.
If it is not broken, throw the plate in the $ 24^{\text{th}}$ floor.
If it is not broken, throw the plate in the $ 25^{\text{th}}$ floor.

(observe that $ 25 \minus{} 24, 24 \minus{} 22, 22 \minus{} 19, 19 \minus{} 15, 15 \minus{} 10$ and $ 10 \minus{} 4$ are an arithmetic progression)

If the plate didn't break, $ k \equal{} 26$. If the plate broke when you threw it, say, from the $ 19^{\text{th}}$ floor, try the floors $ 16$, $ 17$ and $ 18$, in this order. So $ 7$ steps suffice. In general, (I think) our number of trials $ x$ must satisfy $ 1 \plus{} 2 \plus{} ... \plus{} (x \minus{} 1) \plus{} x \ge n \minus{} 1$, i.e., its minimum is $ x \equal{} \lceil \frac {\minus{} 1 \plus{} \sqrt {8n \minus{} 7}}{2} \rceil$.

But the argument must be very boring, so I don't want to discover it.

thanks , I think the x is satisfing x(x+1)/2>=n-1>=x(x-1)/2 to have the minimum of trials, actually , the solution could be : the minum existe ( finite nembre of possibilities ) let k be that one, the strategy consisting of , try the floor number k-1, if it doesnt broke jump to the floor k-1+(k-2), and so intil to jump to k-1+k-2+...+1 ( and of cours we should show that k-1+k-2+...+1<=n-1 , that easy because if not we can take the smallest k verifing that and with the same stratgey we find k'<k so that in k' steps we find de floor in question ) , after the k-1+k-2+...+1=k(k-1)/2 we should have at most 2 floors to reach the n th floor ( because until now our strategy allow as to find the floor in question in just k-1 steps ( simple calculation as explane feliz at the end of his post ) , and so at most k steps let us know the floor in question and so k is verifying k(k-1)/2<=n-1<k(k+1)/2