I wasn't an Alcumus taker yesterday, and POOF I am today. I don't mind being one, but was this intended? Is Alcumus open for everybody now?

Just to clear this up, I sent rrusczyk a PM to confirm this, and he said:
[quote="rrusczyk"]You are correct - was only open to classes, now open to everyone. Please share with friends! [/quote]

Ah.

I was really worried because I(and andersonw) am in WOOT.

Yeah, the two of you are pretty much past your Alcumus-appropriate days :(

Of course, if it goes well, I'll likely try to hire you both when you're in college to help out with it :)

when did alcumus come out? i just saw it but it looks like its been there for a while now. :?:  :!:

[quote="rrusczyk"]It has been non-public for a long time.  We've had students in our school, and some in a couple of the country's best MATHCOUNTS programs testing it for around a little under a year or so.[/quote]

Ha ha why was I testing it then? :)
Has alcumus extended from probability?

Probably not. If it will expand, I wanna help with Number Theory.

[quote="AIME15"]I know it's DEFINITELY not for everybody...[/quote]

Boy do I feel stupid now.

Hm...on the Landing page,it says x out of 557, but there's over 50000 registered users on the forum...!?

The x out of 557 is x out of the number of people who have played with Alcumus so far.

No reason to feel stupid -- until yesterday around midday, you were right.


geometry
circumcircle
trigonometry
complex numbers
geometry proposed

[quote="Ulanbek_Kyzylorda KTL"]P is the point on the circumcircle of the square ABCD between points C and D (on arc CD). Prove that PA^2-PB^2=PB*PD-PA*PC.[/quote]
Let $ a,b,c,d,p$ be the corresponding complex numbers to $ A,B,C,D,P$. Wlog assume $ a \equal{} 1, b \equal{} i, c \equal{} \minus{} 1, d \equal{} \minus{} i$, then $ p \equal{} e^{i\alpha} \equal{} \cos \alpha \plus{} i \sin \alpha$, where $ \alpha \in ( \pi; \frac {3\pi}{2} )$.
Then obviously $ |PA| \equal{} \sqrt {2 \minus{} 2 \cos \alpha}$, $ |PB| \equal{} \sqrt {2 \minus{} 2 \sin \alpha}$, $ |PC| \equal{} \sqrt {2 \plus{} 2 \cos \alpha}$, $ |PD| \equal{} \sqrt {2 \plus{} 2 \sin \alpha}$. Inserting gives:
$ |PA|^2 \minus{} |PB|^2 \equal{} |PB| \cdot |PC| \minus{} |PA| \cdot |PC| \iff$
$ \sin \alpha \minus{} \cos \alpha \equal{} \sqrt {1 \minus{} \sin^2 \alpha} \minus{} \sqrt {1 \minus{} \cos^2 \alpha}$.
But since $ \alpha \in ( \pi; \frac {3\pi}{2} )$: $ \sqrt {1 \minus{} \sin^2 \alpha} \equal{} \minus{} \cos \alpha$ and $ \sqrt {1 \minus{} \cos^2 \alpha} \equal{} \minus{} \sin \alpha$. Inserting yields the desired :) QED

[b]Lemma:[/b] $\triangle ABC$ is isosceles with $AB=AC=L$ and $ P$ is a point on the arc $ BC$ of its circumcircle that does not contain $A.$ Then $ PA^2 \minus{} L^2 \equal{} PB \cdot PC.$

Use the previous lemma for  the isosceles $ \triangle ABC$ and $\triangle  BAD$

$PB^2 \minus{} a^2 \equal{} PA \cdot PC \  , \  PA^2 \minus{} a^2 \equal{} PB \cdot PD$

$\Longrightarrow  PB^2 \minus{} PA^2 = PA \cdot PC \minus{} PB \cdot PD.$

Is that lemma easier to prove than proving the problem directly ? :)

Okay I have a [hide="solution by complex numbers"]
Let $ a,b,c,p$ be the complex numbers corresponding to $ A,B,C,P$. Wlog assume $ a \equal{} 1$, $ b \equal{} e^{i \alpha}$, $ c \equal{} e^{\minus{}i \alpha}$, $ p \equal{} e^{i \beta}$, for some $ \alpha, \beta \in [0;2 \pi )$.
Then $ PA^2 \equal{} 2 \minus{} 2 \cos \beta$, $ AB^2 \equal{} AC^2 \equal{} 2 \minus{} 2 \cos \alpha$. So $ PA^2 \minus{} L^2 \equal{} \cos \alpha \minus{} \cos \beta \ge 0$ (Because $ \alpha \le \beta \le 2\pi \minus{} \alpha$ by definition.
We easily see $ PB \equal{} \sqrt{2 \minus{} 2 \cos \alpha \cos \beta \minus{} 2 \sin \alpha \sin \beta }$, $ PC \equal{} \sqrt{2 \minus{} 2 \cos \alpha \cos \beta \plus{} 2 \sin \alpha \sin \beta }$. So $ PC \cdot PB \equal{} 2\sqrt{1 \minus{} 2 \cos \alpha \cos \beta \plus{} \cos^2 \alpha \cos^2 \beta \minus{} \sin^2 \alpha \sin^2 \beta}$. Since $ PA^2 \minus{} L^2 \ge 0$ we just have to prove:
$ 1 \minus{} 2 \cos \alpha \cos \beta \plus{} \cos^2 \alpha \cos^2 \beta \minus{} \sin^2 \alpha \sin^2 \beta \equal{} (\cos \alpha \minus{} \cos \beta)^2 \iff$
$ 1 \plus{} \cos^2 \alpha \cos^2 \beta \minus{} (1\minus{} \cos^2 \alpha)(1 \minus{} \cos^2 \beta) \equal{} \cos^2 \alpha \plus{} \cos^2 \beta$
Which is obvious :) (We also see from the proof that $ L^2 \minus{} AP^2 \equal{} PB \cdot PC$, when $ P$, lies on the other arc)
[/hide], but I would like to see your solution - which I guess is more classic? :)

[quote="Luis González"][b]Lemma:[/b] $\triangle ABC$ is isosceles with $AB=AC=L$ and $ P$ is a point on the arc $ BC$ of its circumcircle that does not contain $A.$ Then $ PA^2 \minus{} L^2 \equal{} PB \cdot PC.$
 [/quote]
See [url=http://www.artofproblemsolving.com/viewtopic.php?t=266756][color=red][b]here[/b][/color][/url] an usual and nice extension of this lemma.


Let $a$ be a non-zero real number. For each integer $n,$ we define $S_{n}=a^{n}+a^{-n}.$ Prove that if for some integer $k,$ the sums $S_{k}$ and $S_{k+1}$ are integers, then the sums $S_{n}$ are integers for all integer $n.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="rulin"]Let $ G$ be a finite group. If  $ G$ has a  $ \{p,q\}$-Hall subgroup for any  different prime divisors  $ \{p,q\}$ of $ |G|$, then whether or not $ G$ is solvable?[/quote]

This is called Hall's conjecture and was answered affirmatively in:

Arad, Zvi; Ward, Michael B. "New criteria for the solvability of finite groups."
J. Algebra 77 (1982), no. 1, 234-246.
http://www.ams.org/mathscinet-getitem?mr=665175
http://dx.doi.org/10.1016/0021-8693(82)90288-5

Thanks, JackSchmidt. have you a fulltext version? Could you post it online if you have?


What did you guys score on the AMC 10/12 a/b?

I scored 132 on amc 10b.

I second guessed myself on #17 and didn't answer it, even though I had the right answer- arghh.

sorry, I thought I made a separate line for "150", but obviously it didn't work like that.

oh well, but if anyone got a perfect 150- OMG- You're amazing and a Genius!! :)

Hmm official:
109.5 on 12A, 144 on 10B

hopefully 13 on AIME.

13 on aime?

thats like david benjamin-level...o.O

dude 13 on AIME?  pwnalicious.

I'm hoping for a 7.  (Though I phailed on 10B.)

[quote="cognos599"]i wanna go to USAMO no matter what the cutoff is.

so at least like 8 or 9.  :rotfl:[/quote]

It's not gonna happen.  Janglion might beat you.  HAHAHA jk

wow 109 on 10A and shooting for 13?
dang
hoping for a 11-ish here, maybe higher depending on test difficulty level

is this payback for me invading the texas forum? ne way, you got a 109 on 10A?

ra was referring to shawn's post above - shooting for a 13 on AIME with a 109.5 on the *12A*

he got a 144 on the 10 B though..

aiming for 9.. *must tie with richard* lol


$n$ is a positive integer, $S$ is a set consisting of $n^{2}+1$ positive integers. Suppose that for any $n+1$-element subset of S, there exist two different integers $x,y$, so that $y$ is a multiple of $x$. Prove that there exists a subset ${x_{1},x_{2},...,x_{n+1}}$ of $S$, such that $x_{i}|x_{i+1}$ for $i=1,2,...,n$.

Please continuous here http://www.mathlinks.ro/Forum/viewtopic.php?p=198301#198301


geometry
rectangle
3D geometry
function
videos
sphere
vector

When we design, create, build or arrange objects, the most basic, fundamental shape we use is the right angle and its progeny: the rectangle, square, cube and cuboid.

Here's the modern city of Johannesburg, beside the city of Kahun, near Giza in Egypt, built 5,000 years ago:

[img]http://skipall.com/3d.jpg[/img]

Both built on right angle grids.

Here's some game boards: chess, tic-tac-toe, sudoku, a crossword, scrabble:

[img]http://skipall.com/3e.jpg[/img]

all played on square grids.

If you're not commuting right now, chances are you're reading this inside a room... a room made from right angle corners, in a building built with right angle corners ... all made from right angle bricks.

The screen you're viewing this post on, all the windows it contains, the menus, the icon, character, button and picture spaces, and the pixels themselves ... are, from a visual standpoint, all designed with the right angle as the most rudimentary concept.

Here's various sports pitches: basketball, football, badminton, snooker, volleyball:

[img]http://skipall.com/3f.jpg[/img]

All rectangles.

Now pause for a moment and imagine how many times per day, across the world, a graph is shown:

[url]http://skipall.com/3g.jpg[/url]

and that graph is almost exclusively a right angle graph.

Why not draw the graph at 85 degrees? We could, of course, display the same information, with the same data points, on a graph at 85 degrees, and it would be perfectly valid:

[url]http://skipall.com/3h.jpg[/url]

But we don't. We show it at 90 degrees. Because 85 degrees feels wrong. And 90 degrees feels right.

Stand up from your desk now, right now, and look down at your body.
Now, which part of your body requires you to sleep on a rectangular bed?

90 degrees is perpendicular, orthogonal, square, perfect. Anything else, other than 180 and 270 degrees, is diagonal - imperfect.

Unless, of course, you're a bee. Just like humans are born onto a rectangular bed inside a rectangular room, so, bees are born into their own perfect, regular geometric shape: the hexagon.

[url]http://skipall.com/4m.jpg[/url]

Just like the human-made world is dominated by 90 degrees, so, the bee-created world, the honeycomb beehive, is dominated by the angle at the corner of a hexagon: 120 degrees.

To a bee, 90 degrees is diagonal.

Bees, if they could make graphs, would make them like this, with 120 degrees as the critical angle:

[url]http://skipall.com/4o.jpg[/url]

A bee created world would consist of hexagonal buildings, hexagonal screens, hexagonal books, hexagonal bedrooms, hexagonal games, graphs, doors and windows.

Consider the common staircase - which performs a simple function - the vertical transport of humans:

[url]http://skipall.com/3r.jpg[/url]

The horizontal portion, the tread, is horizontal for good reason - to avoid slipping. But the vertical portion, the riser, has no reason to be perfectly vertical.
The vertical riser could be slanted in either direction, like this:

[url]http://skipall.com/3s.jpg[/url]

The first case, the outward slant, is less safe than the standard right angle due to a smaller horizontal surface.
By that same logic, the second case, the inward slant, would be safer than the right angle version.
Yet across the world, standard staircase design remains mostly perpendicular.

Not because it's the best design. But because its the easiest design for humans to create and build.

At some point, way back in the evolution of our mind, something snapped into place, and 90 degrees became more important than all the other angles around it.

The human mind sees 90 degrees as the most perfect angle, because at it's root, spatially, the right angle is the simplest, smallest unit of mental processing.

All of this is explained further in this video:

[youtube]Kfkc6qu9Vvk[/youtube]

Please watch the full video set before replying here. Thanks.

Well A square/rectangle is the easiest design to work with and to do equations with. Curves are nearly impossible to make. On the other hand, straight lines are much easier. To find the area of a square, a cube, a rectangle; you just need lw or lwh. For a circle you need pir^2 and a sphere, 4/3 pi r^3

And how do you know that bees are based off of a 120º world? All the proof you have off of this is their hives.

Very simple reasons (but this is a likely crank anyways; by the way: deletion recommended ;) ):
- rectangular shapes can be stacked.
- mathematical orthogonality corresponds to this; this especially gives that rectangular axes are most natural.
- $ 90^\circ$ is half of straight $ 180^\circ$.

None of that given pictures can be optimised by using a non-rectangular shape (staircases, axes, etc.). Optimisation here happens for the total effects of use, cost, etc.). So it IS optimal as it is. That's more than enough reason.
For example, bee hives are optimal for what they do: building long round caverns for young bees. Human houses, each having it's own wall and streets (that should be straight for obvious reasons) between them, need to be rectangular.

But even more: there is no reason to assume that human brains have some special property regarding that.

To the original poster: please consult http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779 when making posts here.

[quote="ZetaX"](but this is a likely crank anyways; by the way: deletion recommended ;) )[/quote]  I think "likely" is understatement, but it seems harmless enough for now.  I'll keep an eye on it  :wink:

Haha go to about 6 minutes in the first video, he says the internal angle of a pentagon is 72º.

that's the exterior angle of a pentagon... >.<

Actually, the reason right angles are significant is their definition: they are the angles formed when two lines cut each other so that the adjacent angles are congruent.  In both Euclidean and general neutral geometry, there are many important theorems involving right angles.  (For example, if triangles have side-side-angle congruent, and if the angle is right, then the triangles are necessarily congruent; this can be proven without using the Pythagorean theorem or any other form of the parallel postulate.  If the known angle is anything else, then the triangles are not necessarily congruent.)  The choice of using right angles is not just choice of the number 90 -- degrees were created based on the characteristic that a full circle is 360 degrees, and it can be shown that a right angle is one-fourth of that (because when two lines intersect they divide the plane into four regions).

[quote]Now pause for a moment and imagine how many times per day, across the world, a graph is shown:

http://skipall.com/3g.jpg

and that graph is almost exclusively a right angle graph.

Why not draw the graph at 85 degrees? We could, of course, display the same information, with the same data points, on a graph at 85 degrees, and it would be perfectly valid:

http://skipall.com/3h.jpg

But we don't. We show it at 90 degrees. Because 85 degrees feels wrong. And 90 degrees feels right.[/quote]

Actually, the reason right angles seem so natural is that, by using perpendicular basis vectors, we can graph curves in $ \mathbb{R}^2$ in such a way that numerical symmetries correspond to geometric symmetries, e.g. even functions have graphs symmetrical about the y-axis.  And, in the two-dimensional plane, a parallelogram is the only shape with each of its sides parallel to an axis; the rectangle is an equiangular parallelogram.  If you had basis vectors that were not perpendicular, then the area of a parallelogram drawn with its sides parallel to the axes would not simply be its length times its width; instead it would depend on the angle between the axes.  In general, the "area" under a curve would not be $ \int f(x)\, dx$ (if you keep the same definition of "area").

Now I will attempt to give an example from physics.  Newtonian mechanics give rise to the use of a quantity called the moment of inertia, which proves to be the rotational analog of mass.  Suppose an object is rotated about an axis.  In calculating the moment of inertia ($ \iiint r^2 \rho(\mathbf{r})\,dV$), the distance $ r$ is taken perpendicular to the axis.  Why perpendicular?  Because that way the motion of the element of volume is coplanar with the center of its circle of rotation.  And inversely, this is rooted in the very nature of rotation -- the only intuitively sensible way to define what an "axis" of rotation is, is to call it the line perpendicular to the plane of rotation at its center; otherwise we don't know which line to choose because symmetry gives multiple options.

Right angles are more common in human creations than in nature, but that is because right angles are mathematically simple.  We can model the behavior of "irregular" objects, but it is easier to model that of "regular" objects -- so we build those.

What is the shortest path a particle can take from a given point to a given line?  The perpendicular.  This result does not require the parallel postulate to prove.

benzi455:
[quote]if triangles have side-side-angle congruent, and if the angle is right, then the triangles are necessarily congruent[/quote]
Correct.
[quote]If the known angle is anything else, then the triangles are not necessarily congruent.[/quote]
Close but incorrect. If the known angle is acute, the triangle has two possible solutions. But if the known angle is obtuse, the triangle is fixed. From wikipedia: "The side-side-angle condition proves congruence if the angle is obtuse or right."
Thus, in this particular argument, giving the all the angles between 90 and 180 the same significance as the right angle.
[quote]even functions have graphs symmetrical about the y-axis[/quote]

An even function drawn using an x axis of 0 degrees and a y axis at 60 degrees would also have symmetry through a line drawn at 60 degrees to the y axis

[quote]they are the angles formed when two lines cut each other so that the adjacent angles are congruent ... a full circle is 360 degrees, and it can be shown that a right angle is one-fourth of that[/quote]

Correct. However, dividing the circle in 4 to obtain a right angle, no matter how simple and beautiful it seems, remains an algorithm designed by humans, and thus not natural; nature performs no such division of the circle.

We could also use a similar algorithm to divide the circle in 6, 7 or 9 equal angles, creating the perfect geometric shapes the hexagon, septagon and nonagon, and label all of those natural:

[img]http://skipall.com/53.jpg[/img]

The hexagon appears in nature (the beehive); the septagon and nonagon do not (to the best of my knowledge)

[quote]this is rooted in the very nature of rotation -- the only intuitively sensible way to define what an "axis" of rotation is, is to call it the line perpendicular to the plane of rotation at its center; otherwise we don't know which line to choose because symmetry gives multiple options.[/quote]

For the earth, it's defined the other way around. The equator is defined as the intersection of a plane with the surface of the earth at right angles to the axis of rotation of the earth.
Regardless, unless the line is rendered in nature, any line drawn to represent the equator or the axis remains just that: an imaginary line drawn by humans.
Gravity / nature [i]does[/i] sometimes render such lines, at least, in the case of saturn, a circle of debris. The circle, being a curved line, cannot form part of a right angle.

Any natural rotating objects, such as a falling leaf, snowflake, raindrop or a galaxy, regardless of how humans describe their motions, have no actual line through their axis of rotation.

As well to define a diagonal plane of rotation at 60 degrees to the axis and call that natural.

The thread-opening post can also be found here:

http://www.thescienceforum.com/viewtopic.php?p=177346

Does AOPS have a rule against making exact copies of posts on other forums? IMO, it just pollutes the AOPS forums (and/or the other forums) and could be better accomplished by just posting a link to the post on the other forum and maybe giving a short summary.

EDIT: okay maybe that only applies to the round table or to really long posts; obviously that should not apply to contest problems

If you notice, those posts were made by the same author.

I can understand the concerns; if requested by a mod I will edit the post and link/ summarize, no problem.
The reason I do this is to get more opinion / criticism.

Argument as to why right angles make sense:

1. Pythagorean Theorem.
2. Trigonometry is much simpler when right angles are involved.
3. Perpendiculars are often used to find the area of shapes. Squares, rectangles, rhombi, and triangles are just a few.

However, the earth IS curved, leading to the conclusion that we should build things slightly curved. Which, of course, many things are (bridges, very long walls, etc.) but that isn't our immediate mindset.

[quote="spiderboy95"]we should build things slightly curved. Which, of course, many things are (bridges, very long walls, etc.)[/quote]  Random remark: the Verrazano Narrows Bridge (in New York City) was at one point the longest suspension bridge span in the world.  It has two towers, both perpendicular to the surface of the Earth.  But the tops of the towers are about an inch and a half further apart than the bases of the towers: http://www.mta.nyc.ny.us/bandt/html/veraz.htm

Right angles probably became so prominent in human life because we experience them every day even if we do not have any modern living conveniences. The upright human body is at right angles to the world around it (assuming the world is a plane). This led our ancestors to construct artificial objects perpendicular to the surface of the Earth. When this trend continued long enough, right angles became so prominent that they were accepted as a basis of human life. It's perfectly possible to graph Cartesian coordinates without perpendicular axes or use tetrahedrons as building blocks for structures. However, upright objects are the natural tendency of humans.
This probably resulted from the fact that, when faced with the existence of gravity, a 90 degree angle is the only angle that any object can balance at other than 0 or 180 degrees. Thus, any upright, free-standing structures in nature (ex. trees) had to be at right angles to the surface of Earth or very close. Humans copied the world around them to make their first buildings, then continued the trend through history.
Question-why is the 90 degree angle 90 degrees, and not perhaps 100 (following from the decimal system)? I assumed it was because it was 1/4 of a circle, and 360 has many positive factors. Is there any other reason?

Ihatepie:[quote]Well A square/rectangle is the easiest design to work with[/quote]
For humans. For bees, that shape is the hexagon.

[quote]... and to do equations with. Curves are nearly impossible to make. On the other hand, straight lines are much easier. To find the area of a square, a cube, a rectangle; you just need lw or lwh. For a circle you need pir^2 and a sphere, 4/3 pi r^3[/quote]
This is a classic example of living inside a system, and not being able to see it from the outside.
In the right angle world, a curve is difficult to make.
But in the perfectly valid polar coordinate system, the equation for the unit circle is simply r(θ)=1.
In the diagonal / rhombic world, with x and y axes at say 60º, straight lines are also simple eg. x = 3 or y = 7.

Ihatepie:[quote]And how do you know that bees are based off of a 120º world? All the proof you have off of this is their hives[/quote]
Ok I'm not actually suggesting that bees take out pencil and paper and protractors, it's really just an analogy to show the similarities between the bee fixation with 120º and the human fixation with 90º.

ZetaX:[quote]rectangular shapes can be stacked.[/quote]
Given a random array of rectangles or squares of different shapes and sizes, stackability is not a given.
Conrguent rectangles and squares can always be stacked.
There are 3 regular polygons that can be stacked: the equilateral triangle, the square, and the hexagon. Very few species on earth produce any kind of polygon. Bees mass produce another of those regular polygons, the hexagon.

[quote]this especially gives that rectangular axes are most natural.[/quote]
While I agree they appear natural to humans, rectangular axes do not appear in nature, other than in right angle crystals, and when multiple crystals form close together, they don't line up together, forming at arbitrary angles against one another.

[quote]None of that given pictures can be optimised by using a non-rectangular shape (staircases...[/quote]
As mentioned, the staircase can be optimised for safety with an inward sloping riser.

[quote]Optimisation here happens for the total effects of use, cost, etc.).[/quote]
In a world where everyone in the chain from design to manufacture to fitting uses the rectangle as the default shape, yes, the rectangle is the optimal shape.

[quote]Human houses, each having it's own wall and streets (that should be straight for obvious reasons) between them, need to be rectangular.[/quote]
Please state those obvious reasons. I hope they don't include the motor car, since as mentioned above, Kahun in ancient Egypt, like many other ancient cities, was also built on a right angle grid.

Even if you cite the efficient transit of the motor car through straight streets as a reason, the grid could be triangular or rhombic, and even then, all of square, triangular or rhombic grids require intersection controls / traffic lights which slow the traffic. Many cities not designed on a right angle grid work just fine.

Catal Hoyuk, a settlement in Anatolia dated to 7,000BC, used cuboid bricks and roughly rectangular buildings. Yet it had no streets at all. The buildings were stacked beside each other; the door was a hole in the roof. Even that long ago, the right angle building was within the very fabric of human society, before the invention of the street itself.

[quote]But even more: there is no reason to assume that human brains have some special property regarding that.[/quote]
On the contrary, there is every reason to believe it. I will give the anthropological evidence in a post below.

benzi455:[quote]then the area of a parallelogram drawn with its sides parallel to the axes would not simply be its length times its width; instead it would depend on the angle between the axes...
(if you keep the same definition of "area").[/quote]

You've hit the nail on the head. Defining 1 unit of area requires an ideological leap from one to two dimensions, and is not restricted to the square. A sector of a circle could be considered a unit of area. Likewise, any closed 2d geometric shape could be considered a unit of area, and there are an infinite variety of non-regular polygons that tile with no gaps.
If you desire to use, say, 60º x and y axes, simply define 1 unit of diamond shaped area = 1 x 1 diamond length.
[img]http://skipall.com/d2.jpg[/img]

As you correctly mention, the area will not be the same if expressed in square units.
But when expressed in diamond units, if you place 3 diamonds up and 4 across, you have the same answer: 3 x 4 = 12 diamond-shaped units of area.
[img]http://skipall.com/d3.jpg[/img]

I agree, it looks weird, skewed and wrong. That's because you're human. To a bee, the square is skewed and wrong.

A similar ideological leap is required when going to 3 dimensions, and again it's a simple matter to define a diamondoid unit of volume, so that eg 3 x 4 x 5 = 60 units of diamondoid volume.
[img]http://skipall.com/d4.jpg[/img]

[quote]Right angles are more common in human creations than in nature, but that is because right angles are mathematically simple.[/quote]
If using cartesian coordinates. In polar coordinates, the circle is simple. The right angle is complex.

Even non-congruent rectangular forms can be stacked. But stacking non-congruent triangles will cause chaos. For the stairs, you ignored the higher difficulti tu build rhomboid stairs.
I gave reasons why rectangular axes are most natural that weren't objected by you (instead, you put one of my conclusions out of context). For example, that it's half the angle of a straight line, so all angles in a 2-dimensional coordinate system are equal.

Circular forms are more effective for houses (converning the amount of stone needed). This excludes triangular forms. Now hexagonal forms are impractical as streets couldn't be built as straight lines (yes, this was also relevant more than 2000 years ago) and as they need more knowledge than rectangular ones. So rectangular forms are optimal for that.
There are exceptions, and as you may know, there were not always only rectangular shapes: A lot of buildings are circular, octogonal, or sometimes even other forms.

In fact I've no clue how you come even to think that there is a non-natural reason (by the way, you didn't state that reason). Humans and nature just try to optimise. So if it's not a reason of thought, it's a reason of evolution. Nothing more. You always just considered one effect of (non)rectangular shape at once, not the average effect concerning the sum of everything. You can't just deny something because it's not the perfect solution to one out of many problems this has to solve.
Also, there are tons of rectangulr situations like walls being perpendicular to the ground (something you seem to have no problem with).

[quote="evilhamster"]The upright human body is at right angles to the world around it (assuming the world is a plane). This led our ancestors to construct artificial objects perpendicular to the surface of the Earth... Thus, any upright, free-standing structures in nature (ex. trees) had to be at right angles to the surface of Earth or very close...right angles became so prominent that they were accepted as a basis of human life. ...However, upright objects are the natural tendency of humans.[/quote]
The human race has covered its world in perfectly flat, horizontal surfaces. But natural surfaces are often haphazard, curved, slanted. On these surfaces trees grow at diagonal angles to the ground.
The number of times a tree naturally makes a true right angle with the ground on earth is exactly zero.
Similarly, the right angle grid, so favored by humans, also exists naturally in exactly zero places in the entire known universe.

Long before mathematics, physics, or construction with bricks, the human mind had an unnatural fixation with the right angle, and the right angle grid.
And here is the unquestionable proof of that:

[img]http://skipall.com/ub.jpg[/img]

This is a grid of near right angles, drawn on the wall of a cave in Altamira in spain, in the Old Stone Age, ~16,000 years ago. It predates writing, agriculture, the wheel and the invention of metal tools.
Many of the angles are rough and oblique, but overall there is no question it points to a right angle grid.
Such a grid does not exist in nature, therefore it is a pure human creation, and came straight out of the mind of our ancestors.

How we now interpret this grid, or what we decide it meant to our ancestors, is, for the purpose of this discussion, irrelevant.
What remains on the wall is the evidence, and this evidence shows that the human mind had a fixation with rectangular and perpendicular forms, long before any systems of planning or building gridded cities.

This fixation translates itself into many areas of modern life.

Here is another example:
Here's 24 bottles of beer, packed in a standard crate the normal way.

[img]http://skipall.com/8y.jpg[/img]

Billions and billions of bottles of beer make their way around the world every day of the week.

Now surely, if there is a constant need for optimization and efficiency, and if land prices for storage come at a premium, surely we would by now have found the best way to pack a few bottles of beer into a box.

For simplicity let's call the diameter of a bottle 1 unit. So the area of the above box is 6 x 4 = 24 units.

Now, increase the length of the box by ~3%, also increasing the area by 3%, and pack the bottles in a 4-3 hexagonal formation.

You can now place the mythical 25th bottle of beer into the box:

[img]http://skipall.com/8z.jpg[/img]

giving you an extra 4% beer for a space saving of ~1%.

Ok, 1%, no big deal.

Now do it the other way. Increase the width of the box by ~11.6%, and pack in a 6-5 hexagonal formation.

http://skipall.com/90.jpg

Now you can place an extra 4 bottles of beer into the box, giving you an extra ~17% beer for a space saving of ~5%.

If you use the (less common) 8 x 6 = 48 square packing, you can actually place 2 extra bottles in without even changing the shape of the box:

http://skipall.com/94.jpg

Go ahead and count them.
Similarly, a 5 x 8 = 40 square packed box will yield 41 bottles with hexagonal packing.

Here's the top 10 most efficient hex-in-rectangle packing formations for 48 or less bottles:

http://skipall.com/uc.jpg

If you want smaller boxes, here's the best formations for boxes of 20 or less:

http://skipall.com/ud.jpg

Now multiply that by every crate of beer in the world and it adds up. Fuel costs for transport would not be reduced much, as the weight of one bottle of beer is the same, but the land costs for storage and in retail would make a difference.
Further, hex-packing reduces material used in each box (usually cardboard, known as secondary packaging) by a similar percentage. This would result in ~5% savings in all associated energy and resources used in packaging manufacture.

Many cylindical products come packed with individual plastic or cardboard wrap for 4 or 6 items. The superior efficiency of hexagonal packing begins at 11 items (for a 4-3-4 hex formation). Any square-packed box of 12 or more is always beaten by a similar or nearby number of hex-packed objects.

This applies to literally every cylindical product the world over, packed in a box of 12 or more - a staggering number of products by any account.

Packing machinery would have to be reconfigured to pack alternate rows offset from each other, however this doesn't seem to be a problem for the cigarette industry, which has long packed 20 cigarettes into a box with a 7-6-7 hexagonal formation, for a space and packaging saving of 4.58%.

The efficiency of hexagonal packing has been known for thousands of years.
But we don't use hexagonal packing. We use square packing.
Not only is there no rational or logical reason to use a square formation, there is a very very good reason not to use it.

No law, equation or formula in mathematics or physics can explain this.

Square packing of cylindrical objects is used because it fits into the [i]mind[/i] of every person in the chain from manufacturing to retail.

The human mind loves to see objects arranged in a square formation. It [i]feels[/i] right, even when it goes against simple mathematical efficiency.

From the construction standpoint, a triangular house (including equilateral triangles) is just too cramped. Acute angled-corners do not support efficient use of space. A quadrilateral is the next polygon with a higher number of sides---and we use a regular quadrilateral both for consistency and for ease of tessellation (just as bees build regular hexagonal hives).

On a related note, Google the "honeycomb house", designed by architect Frank Lloyd Wright.

As posters before me have said, the simplest definition of a right angle is an angle that bisects a straight angle. You can't get much simpler than that.

As for your example of the cave drawing, it is unclear what that is a drawing of. That could be a drawing of some sort of bowl, woven from grass or something. The curved edges on the bottom especially invalidate that this was some sort of rectangular grid drawn by right angle-obsessed cave-painters.

Hexagonal packing of cylindrical objects is most natural in a hexagonal package. However, since the rectangular container has already become the established container (for convenience reasons only, one being splitting material into four pieces instead of six), and it is intuitive to pack objects rectangularly in a rectangular container.

As ZeTaX said, the right angle is so widely used because of its simplicity, and, while it may not be perfect in all situations, it simply evolved from practicality, not because of some human obsession with right angles.

Wow. Well I might as well go point by point.

The human race did not cover its world it perfectly flat horizontal surfaces. How about you stand on a boat in the middle of a windless ocean. Additionally, it's impossible to create any true angle. If angles could be measured with infinite precision then yes we could create a perfect any-angle. But that does not exist.

Your picture is moot, that does not look like a right angled grid to me at all.Maybe if all the rows or columns were constructed similarly to the column in the middle, but that looks like it serves a different purpose.

There is a very big reason why things are packed in rows and columns. It is much more secure that way. Objects will not wiggle in transport if they are packed that way. If they are packed hexagonally, then the package needs to be very tight around the objects otherwise the objects will wiggle and possibly break.

Finally, right-angles actually do occur in nature at the molecular level. Look up VSEPR models and also Covalent-Network solids.


number theory
least common multiple
greatest common divisor
logarithms
relatively prime
real analysis
real analysis unsolved

Here is a very very beautiful problem that I had during an oral examination: look at the sequence $x_n=\sum_{k=1}^n{d^2(k)}$ where $d(n)$ is the number of divisors of $n$. Prove that there exist positive constants $a,b$ such that $an (ln n)^3\leq x_n\leq b n (ln n)^3$.
   I find it very difficult and the solution that the examinator showed to me is absolutely genial.

One way to approach the problem is to start with the identity $\sum_{k=1}^n d(k)^2=\sum_{a,b\ge 1}\left[\frac n{\text{LCM}(a,b)}\right]$ where $\text{LCM}$ stands for the least common multiple and $[\cdot]$ stands for the entire part. The estimate from above is then very straightforward: all one needs to note is that the last sum can be rewritten as $\sum_{a,b,d\ge 1; \text{gcd}(a,b)=1}\left[\frac n{abd}\right]$ where $\text{gcd}$ stands for the greatest common divisor. In the last sum non-zero terms may arise only if $a,b,d\le n$. Restricting them to this range and omitting the entire part and the condition that $a$ and $b$ are relatively prime, we get about $n(\ln n)^3$.
To get the estimate from below, note that the entire part can be omitted if $a,b,d\le\sqrt[3]n$ (the factor of $2$ may be lost but we do not care). Thus, the sum in question is at least $\frac 12 n\Bigl(\sum_{1\le d\le\sqrt[3]n}\frac 1d\Bigr)\Bigl(\sum_{1\le a,b\le \sqrt[3]n; \text{gcd}(a,b)=1}\frac1{ab}\Bigr)\approx \frac n2\frac{\ln n}3\sigma(\sqrt[3]n)$  where $\sigma(x)=\sum_{1\le a,b\le x; \text{gcd}(a,b)=1}\frac1{ab}$.
To estimate $\sigma(x)$ from below, just note that $(\ln x)^2\approx \sum_{1\le a,b\le x}\frac 1{ab}=\sum_{d\ge 1}\frac 1{d^2}\sigma(x/d)\le \frac{\pi^2}{6}\sigma(x)$. The rest should be clear.


algebra
polynomial
quadratics
quadratic formula

Note: I do want to solve this so, please don't give me the answer!

Here's the problem:
[img]http://jqsoftware.com/images/Polynomial.gif[/img]

My problem is that I have never come across a polynomial of this type.
(BTW: I learned how to solve Polynomials with VideoText Algebra)

[quote="CPUFreak91"]$3x^{4a} + 2x^{2a}y^b -y^{2b}$[/quote]

I suppose that you mean $3x^{4a} + 2x^{2a}y^b -y^{2b}=0$

(or simply you want to factor this)

You may notice that you have $x^{4a}$ and $x^{2a}$, that is the first exponent ($4a$) is twice the other ($2a$)

The same happens for $y$

How could we obtain $x^{4a}$ starting from $x^{2a}$ ???

I mean, you  have to express $x^{4a}$ in terms of $x^{2a}$

When you'll do this, you'll have a polynomial of 2nd degree with two variables $w=x^{2a}$ and $z=y^b$

Let us know about what you've done  :)

sorry, yes I want to factor this.
Thanks for the tip. I'm working on it now.

[quote="riddler"][hide=" another hint"]
$3p^2=2p^2+p^2$
[/hide][/quote]nice  ;)

[quote="pontios"](or simply you want to factor this)[/quote]

I should write "factorize". Excuse my english ...

[hide]$x^{2a} = m, y^b = n$ so you have $3m^2+2mn-n^2.$  Using basic factoring, (or the quadratic formula), you can get the factorization $(3m-n)(m+n) = (3x^{2a}-y^b)(x^{2a}+y^b)$.[/hide]


This seems really obvious, but why isn't 1 a prime number? What would it be classified as then???

1 is not considered a prime by most mathematicians, because a prime is defined to have two factors - itself and 1, while 1 only has one factor - i.e., itself.

[quote="hsiljak"]Wikipedia:

[quote]One was formerly considered prime by some mathematicians, using the definition that a prime is divisible only by one and itself. However, this complicates the fundamental theorem of arithmetic, so modern definitions exclude units. The last professional mathematician to publicly label 1 a prime number was Henri Lebesgue in 1899.[/quote]

In other words, i.e. 12=2x3x4, and 12=1x2x3x4, which would mean (not really,but...) that it has two prime factorizations. To avoid such 'nonsense', one isn't a prime (it's just one of the reasons, but first I remembered right now).[/quote]

2x3x4=24

I agree with u nick42.....

1 is a unique number....

[quote="mewto55555"]

2x3x4=24[/quote]

Lapsus mentis. 2x2x3.  :lol:

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[quote="stevenmeow"]1 is neither prime nor composite, by the way[/quote]
It's nothing...?

[quote="FantasyLover"]prime=2 factors
composite=3 or more factors

1=only has 1 factor

so neither prime nor composite[/quote]
i get it...thanks a ton!
1 is so lonely...having only one factor =[

I believe that I read somewhere that 0 and 1 are their own categories.

Some mathematicians actually do refer to 1 as "lonely" rather than prime or composite.


The second step, $(a^{2}+b^{2}+c^{2}) \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge 27$, is just Holder.

[b]arqady[/b] may I ask how you find such counterexamples?  :o

I try to prove and if it is not true, the counterexample turns out.

[quote="arqady"][quote="Gibbenergy"] I have proved that:
For all $a,b,c> 0$  such that $a^{2}+b^{2}+c^{2}=3$ then
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq a^{3}+b^{3}+c^{3}$
[/quote]
I think it is wrong. Try $a=b=\frac{4}{3\sqrt3}$ and $c=\frac{7}{3\sqrt3}.$ :wink:[/quote]

  Your counterexample is good, Arqady. Surely, there exist fault in my proof. And I have checked it for an hour but unable to find the mistake. I will spend more time to figure out this one.
  Thank you

[quote="silouan"]I found an one line solution for the initial one :
It is equivalent to
$\frac{(a-1)^{2}(1+a)^{2}(2+a)}{2(a^{2}+a)(a+1)}\geq 0$

This solution shows that the inequality holds for $a,b,c\geq-2$  :)[/quote]

That actually holds if $a, b, c \le-2$ or $a, b, c \ge 0$ because the $a^{2}+a$ simplifies to $a(a+1)$, so it really just depends on the $a$ and $a+2$...

Can't you use partial fraction decomposition?
$\frac{1}{a^{2}+a}$
$=\frac{1}{(a)(a+1)}$
$=\frac{1}{a}-\frac{1}{a+1}$

Applying this repeatedly, it suffices to show:
$=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{b}-\frac{1}{b+1}+\frac{1}{c}-\frac{1}{c+1}\geq \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$

It suffices to show:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3$

The root mean square is greater than the harmonic mean, so:

$\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$1\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3$

Thanks, K81o7.

[quote="vishalarul"]Can't you use partial fraction decomposition?
$\frac{1}{a^{2}+a}$
$=\frac{1}{(a)(a+1)}$
$=\frac{1}{a}-\frac{1}{a+1}$

Applying this repeatedly, it suffices to show:
$=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{b}-\frac{1}{b+1}+\frac{1}{c}-\frac{1}{c+1}\geq \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$

It suffices to show:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3$

I'm not sure how to do this, so any help would be appreciated.[/quote]

That's actually what I did.  Think harmonic mean...

The original inequality can be proved by Chebyshev's easily.


$0<m,n\le 1,\ m+n\le 1\Longrightarrow \frac{(m-1)^2+n(2m-n)}{(n-1)^2+m(2n-m)}\ge \min \left \{ \frac mn,\frac nm\right \}$.


What is your favorite classic book? Please include title and author.

whats a classic book, do you mean something that ilustrated classics made a version of, something thats victorian or romantic as opposed to modern(ie present not modern literature) or perhaps just something an english major/teacher considers good literature?

By classic, I mean what english major would consider good literature. Examples will be like Moby Dick, The Old Man and the Sea, Tales of Two Cities, and so on.

If I define a "classic" book as a book more than 100 years old that people still think is a good book, then I would recommend, among many others

A Tale of Two Cities, by Charles Dickens (mentioned first only because you mentioned it)

a collection of stories, essays, and poems by Edgar Allen Poe

a collection of short stories by Leo Tolstoy

Huckleberry Finn, by Mark Twain

A Connecticut Yankee in King Arthur's Court, by Mark Twain (I like this one better than Huckleberry Finn)

Looking Backward, by Edward Bellamy (this one is more debatable as a "classic," but it is still available for reading today, long after it was published)

I've just been paging through a list I kept when I was in high school of books I read up to that time. I read too much science fiction, and not enough "classic" literature (by the definition above) when I was a kid.

A really good book written during my lifetime, which I highly recommend, is [url=http://www.amazon.com/gp/product/0449213447/]The Chosen[/url] by Chaim Potok. It's not old, but it's as good as any classic.

Try

[url=http://www.ala.org/ala/yalsa/booklistsawards/outstandingbooks/outstandingbooks.htm]Outstanding Books for the College-Bound[/url]

or

[url=http://www.collegeboard.com/student/plan/boost-your-skills/23628.html]101 Great Books[/url]

for more suggestions.

Enjoy your reading.

I am what non math people may describe as  a boring culture barbarian : I read next to nothing in fiction (I spend most of my time reading interviews and articles...)

There is only one, but a very good one, I can recommend : 1984 by George Orwell

[quote="fredbel6"]There is only one, but a very good one, I can recommend : 1984 by George Orwell[/quote]

I would have to agree; that is probably one of my favorite books of all time.  Others include [u]The Scarlet Letter[/u] (Nathaniel Hawthorne,) and, of course, [u]A Tale of Two Cities.[/u]

We just read Pride and Prejudice and I LOVED it. I was really surprised, I've avoided Jane Austen for most of my life because I thought her stuff would be boring but I actually stayed up at night reading it (I usually only do that with Harry Potter and Douglas Adams). Siddhartha was really good, has anyone read that one? Also... Le Morte de Arthur. Another one that LOOKED seriously boring, but ended up being wonderful.

1984 isnt 100 years old

For classic litterature (ie more than 100 years old) I like Cyrano de Bergerac and Beowulf

A book I never finished but really enjoyed what I read (the only copy I was able to find was missing pages in the middle) was Water Margin which is definately classic chinese litterature (they are reprinting it in June)

For almost classic litterature (ie early 20th century) I enjoyed The trial by kafka and we by zamayatin

personally I prefer more modern classics (1920-1970) including Kurt Vonnegut, Pale Fire by Nabokov, As I Lay Dying, and Old Man and the Sea

The Leopard, by Giovanni Lampedusa, is one of my favorites.

I also echo Tokenadult's affinity for Tolstoy, and love many of his shorter novellas, including The Death of Ivan Ilych and The Kreutzer Sonata.

The Davinci Code and it's sequel Angles and Demons :laugh:  :yup:

[quote="Iversonfan2005"]...Angles...[/quote]
Public service/pet peeve announcement: if it has wings and a halo, it's an "angel."  If it's acute, obtuse or right, it's an "angle."

I'll second the nomination of A Conneticut Yankee in King Arthur's Court.  I feel like 50 years is probably long enough ago to make something classic, and with that frame I'd toss in Orwell's Down and Out in Paris and London, which I found to be more engrossing than 1984.  (Actually, I was travelling through Europe by train when I was reading it, and at one point had to remind myself that I wasn't living day-to-day and could afford a restaurant dinner that night.)

The Stranger by Camus, though it is part of the modern era of literature( modern but not contemporary).

Julius Caesar by William Shakespeare although it is a play.

UUUGH. how could you like [i]As I Lay Dying[/i]? i thought that book was TERRIBLE.

Why was it terrible? I'd agree that the plot wasn't exciting but I found the way the book was narrated quite interesting.

My favourite classic by far is The Bell Jar by Sylvia Plath. I love that book to death.

Conversely, I'd rather chew carpet than read Exodus, Bury My Heart at Wounded Knee or The Fountainhead again. I would use those books when I needed to fall asleep.

I am surprised nobody has said The Art of Problem Solving!

Dune is a good one.  How about the Romance of the Three Kingdoms?  It's like the most amazing classic chinese text ever.

Mine would be Great Expectations-Charles Dickens

[quote="Sunny"][quote="shobber"]It is hard for me to say what my favourite is but I can tell you which book I dislike most: Pride and Prejudice![/quote]
I loved that book!  But I wasn't forced to read it.  I find books I read in school...I don't like as much as when I read for fun.  But Pride and Prejudice...I thought was an awesome book.[/quote]
Maybe this is because you are a girl. I haven't met a single boy who claims he love the book Pride and Prejudice.

A Connecticut Yankee in King Arthur's Court by Mark Twain was a good read, and you can find a copy free online easily.

The Three Musketeers by Alexander Dumas was pretty good also

For books in general, I plan to read these 4 books over the summer:
I'm currently reading The Cider House Rules by John Irving
Next, I will read Presumed Innocent by Scott Turow,
Either The Blessing Way, Angela's Ashes, Huckleberry Finn, or The Color Purple next, and lastly,
The Catcher in the Rye by J.D. Salinger

I have finished reading 1984 and Animal Farm by George Owell. It is indeed a great book!

[quote="shobber"]
Maybe this is because you are a girl. I haven't met a single boy who claims he love the book Pride and Prejudice.[/quote]

Hahahaha ok, point taken.  Definitely a girly kinda book.   :D

LotR were my favorite classic books.  i hated oliver twist. it took me a month to read, compared to me reading harry potter 5 in about 8 hours.  it was really boring.

[quote="Djole"]For 'classic' meaning a book at least 100 years old, I'd go for (almost) any russian realist and Dickens' 'David Copperfield' ('Oliver Twist' and 'Great Expectations' were good, too).

If 'classic' means simply a widely recognised literary work (without limiting to pre-20th century literature), my definite favorite is 'A Hundred Years of Solitude' by Gabriel Garcia Marquez (and most of his other books except 'Autumn of the Patriarch'). Eco's 'Foucault's Pendulum' is also a very good book, though some found it tiresome; 'The Catcher in the Rye' and 'The Little Prince' were also really enjoyable.

A bit off-topic: Some 20th century books which I also found to be 'classics' of their own genres would include Orwell's '1984' and Huxsley's 'Brave New World', 'The Big Four' by A. Christie, 'The Hitch-hiker's Guide to the Galaxy' (of course) and everything else by Douglas Adams, and a lot of books by Terry Pratchett. LotR was quite likeable, too.[/quote]

I was going to pick up Brave New World but then my friend was saying something along the lines of "Don't read that book, it's the worst ever. One kid is all emoey because he's not perfect and wants to kill everything." He also said something similar, although to a lesser extent about Catcher in the Rye though  :huh: And Brave New World wouldn't give me credit for Summer Reading as it's a lower difficulty level book so I put it off for now.

I recommend Animal Farm by George Well in addition to my 2 earlier recommendations by the way. Cider House Rules was pretty good in my opinion, but it's pretty controversial and it's definitely only for people who can stomach things like Steven King.

Well, 'Animal Farm' has pretty much the same moral as '1984', but I somehow liked '1984' better... As for 'Brave New World', I considered it to be a very good book, especially the ending (the last several chapters, when the entire philosophy of the new society is analysed). And since its rather short, it shouldn't present a reading problem (I finished it in a day).

The Stranger by Albert Camus

Not a classic in the sense of year when it was written but a classic in the sense of classicness.

[b]the stranger by camus[/b]
...[b]werther and faust by goethe[/b]
father goriot by balzac
all by jules verne
cronicles of dune by herbert (frank, not brian)

Has anyone here read [u]The Grapes of Wrath[/u]?  I just started, it's my summer reading for English next year.


trigonometry
ratio
calculus
geometric sequence
geometric series
absolute value

I don't know if this belongs here, but I don't know where it belongs.

How come 1/2 + 1/3 + 1/4 + 1/5 + ... doesn't converge? But 1/2 + 1/4 + 1/8 + 1/16 + ... does?

the way i learned it from my trigonometry class is this

$\frac12\geq\frac12$
$\frac13+\frac14>\frac14+\frac14=\frac12$
$\frac15+\frac16+\frac17+\frac18>\frac18+\frac18+\frac18+\frac18=\frac12$
and so on... then you finally get
$\frac12+\frac13+\frac14+\cdots>\frac12+\frac12+\frac12+\cdots$
which is greater than any number

for the second you can think of a geometric sequence with common ratio $\frac12$ and the sum is equal to 1

$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots$

$(\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7})+\cdots$

Either group in paranthesis is $\geq \frac{1}{2}$ so the sequence diverges.

EDIT: Palytoxin beat me  :)

Notice that if we try to find the sum of this series, there is no common ratio. Do you know how to program? Make a program on your calculator to find the sum, and show the sum after each addition. You'll see it bigger and bigger.

[quote]Notice that if we try to find the sum of this series, there is no common ratio. Do you know how to program? Make a program on your calculator to find the sum, and show the sum after each addition. You'll see it bigger and bigger.[/quote]

Yes, but when you add positives to positives, the sum is [i]supposed[/i] to get bigger and bigger.  :P Thanks for the explanations, though.

Yes, there is no common ratio. I just didn't know why the sequence doesn't converge if there isn't a common ratio.

And yes, I know how to program.

It should be pretty easy to see why $\frac 12 +\frac 13 + \frac 14+\cdots$ diverges, right (just look at the proof provided above)?  To see why $\frac 12 +\frac 14 +\frac 18+\cdots$ converges let us first look at finite sums.  Consider the sum, $S_n$ where

\[S_n=\frac 12+\frac 14+\cdots+\frac{1}{2^n}.\]

Using the formula for the sum of a finite geometric series we see that $S_n=\frac 12 \frac{1-(\frac 12)^{n+1}}{1-\frac 12}=(1-(\frac 12)^{n+1}).$

Now we are getting in to limits.  Our original sum, $\frac 12+\frac 14+\cdots$, is actually the limit as $n$ goes to infinity of $S_n$.  So we look at $S_n$, if the limit as $n$ goes to infinity exists then the original series converges.  Looking at the formula for $S_n$, i.e. $1-(\frac 12)^{n+1}$, we consider what happens when $n$ gets really, really big.  When this happens $(\frac 12)^{n+1}$ gets really, really small.  So, when we take the limit as $n$ goes to infinity we consider the $(\frac 12)^{n+1}$ to be zero.  Thus the sum equals 1.

A common ratio has nothing to do with the sum of a sequence converging.  Consider the sum $1+2+4+8+16+\cdots$.  It has a common ratio but obviously diverges.

Convergence/divergence is a vast topic.  In calculus you will learn how to deal with it in a whole lot more detail and rigour.

Well, of course when the ratio is greater than 1 it's not going to converge. But what about when the ratio is less than 1? What is an example of that kind of sequence that doesn't converge?

All geometric sequences with the absolute value of their ratio $r$ less than 1 converge.  What exactly do you mean by your other question?  An example of a sequence that does not converge is $\frac 12 + \frac 13+\frac 14+\cdots$

[quote="joml88 wrote:"]
A common ratio has nothing to do with the sum of a sequence converging.  Consider the sum $1+2+4+8+16+\cdots$.  It has a common ratio but obviously diverges.
[/quote]

My comment was in response to chess64: "Notice that if we try to find a sum of this series, there is no common ratio. Do you know how to program? ......"

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

Next topic. I understand that a sequence converges if it approaches a limit as more and more terms are added. What is meant by diverging? For example, does the well-known sequence 1, 2, 3, 4, 5, ... diverge or not?

[quote="mathnerd314"][quote="joml88 wrote:"]
A common ratio has nothing to do with the sum of a sequence converging.  Consider the sum $1+2+4+8+16+\cdots$.  It has a common ratio but obviously diverges.
[/quote]

My comment was in response to chess64: "Notice that if we try to find a sum of this series, there is no common ratio. Do you know how to program? ......"[/quote]

Jeez!! :roll: Sorry, I made a mistake!! I just meant that it would be hard to find the sum anyways!!

[quote="Holier Than Thou"][quote="joml88"]All geometric sequences with ratio $r$ less than 1 converge. [/quote]

Actually, to be more specific/politically correct/whatever, all geometric sequences with $|r|< 1$ converge.

[quote="mathnerd314"]What is meant by diverging? For example, does the well-known sequence 1, 2, 3, 4, 5, ... diverge or not?[/quote]

Diverging is the sum either approaching $+\infty$ or approaching $-\infty$ (so yes, $1, 2, 3, 4, 5 \cdots$ diverges), or bouncing back and forth (ex: $2, -4, 8, -16, \cdots$).[/quote]

Yes, I meant $|r|<1$.  Thanks for catching that, I've edited it now.

You said
[quote="joml88"]
A common ratio has nothing to do with the sum of a sequence converging.  [/quote]

And I said (I'll reword it to make it more readable) Yeah, it's obvious that the sum you posted (1+2+4..) doesn't converge because it's ratio is greater than 1.  Can you give of an example where $0<r<1$ and the sequence diverges?

Ok, the ratio does have to do with convergence but what I meant, and should have said, is that just because you have a common ratio between terms, it doesn't mean that the series converges.

Also, I already said that every geometric series with the absolute value of their ratio less than 1 converges, i.e. $|r|<1$.  Thus every geometric series with $-1<r<1$ converges.

Ah ok I didn't see where you said that before(I see it now), but thanks for the clarification.

I wasn't sure if there were some cases that even though $-1<r<1$ the sequence diverges.

Anything that doesn't converge, diverges.  Things can diverge by approaching infinity or by bouncing around a lot.  For example, the series 1 - 1 + 1 - 1 + 1 - 1 + ... (with common ratio -1) diverges, although clearly parts of it exhibit some convergent behavior.

Completely random thought:

Have you ever heard of Zeno's Paradoxes?
A particular one states that before you travel some distance (which we shall scale down to $1$), you must travel half the distance first, $\frac{1}{2}$. Of the remaining distance, you must travel half of that, $\frac{1}{2}*\frac{1}{2}$ before the other. (And so on...)  Although it appears that even this total distance of $1$ cannot be traveled, eventually the shortening time it takes to travel each step compensates for this.

Okay, so that wasn't an effective argument that $\sum_{n=1}^\infty{\left(\frac{1}{2}\right)^n}=1$, but it does provide a logical "reason" that the summation converges.

Anyone read [u]Infinity: The Quest to Think the Unthinkable[/u] by Brian Clegg? He talks about Zeno's paradoxes.

Also, I think the one you just mentioned towersfreak he actually looked at the other way around: before you can move $\frac12$ the distance across the room you have to move $\frac14$ of it, before $\frac14$, $\frac18$... He was trying to make a point about motion, arguing that it doesn't really exist, because you can't move an infinitesimal distance, or, therefore, any distance.


There were so many warnings given to the US prior to the attack on the US port.

1: The US decoded a transmission that stated that an attack was going to occur (they didn't know where, though, so I sympathize...but shouldn't the President at least start mobilizing?)
2: A neutral cruise liner intercepted and decoded a transmission between two Japanese aircraft carriers that were going to the raid...and reported it to the US.
3: Prior to the attack, Pearl Harbor detected an incoming wave of hostile aircraft and reported it...but got the reply that the harbor was expected a large incoming fleet of airplanes from California - which was in the opposite direction.

The US were given so many warnings. Roosevelt refused to respond, so could it be that he wanted to end the Great Depression very desperately, and he wanted the US to get into war, knowing it would end the economic breakdown???

If that is so, then did Roosevelt really want the death of that many people and ships?

[quote="JBL"]Geez, the eye-roll quotient on this forum really has gone up over the past week or so.[/quote]Uh, I probably sound stupid for asking this, but what is the eye-roll quotient?

I doubt Roosevelt would have wanted something like Pearl Harbor to happen. Even if he were trying to get the Japanese to attack to give him a reason to go to war (and I doubt that's true), just detecting that that incoming wave of aircraft was Japanese would have been enough reason. There wouldn't have been any need to let them actually do serious damage.

I assume that roosevelt just didn't add up the clues. people are stupid that way. :wink:

why wouldn't he want pearl harbor? if it was destroyed, the american positions in the pacific would have to be withdrawn. otherwise they would have been isolated and destroyed. the war would have gone on much longer.

I'm not sure that he was being stupid by not realizing that the attack would be at Pearl Harbor - most people were expecting any attack to take place in the Phillippines, a location obviously closer to Japan.


Let $ a$ and $ b$ be positive real numbers such that $ a \plus{} b \equal{} 2$. Prove that

$ a^\frac 2{a} \plus{} b^\frac 2{b} \le 2$.

Can you generalize this inequality to $ n$ numbers ?

[quote="Vasc"]Let $ a$ and $ b$ be positive real numbers such that $ a + b = 2$. Prove that

$ a^\frac 2{a} + b^\frac 2{b} \le 2$.

Can you generalize this inequality to $ n$ numbers ?[/quote]
By Bernoulli's Inequality, we have
\[ \left( \frac{1}{a}\right)^{\frac{2}{a}} \ge 1+\frac{2}{a}\left( \frac{1}{a} -1\right)  =\frac{a^2-2a+2}{a^2}\]
Hence
\[ a^{\frac{2}{a}} \le \frac{a^2}{a^2-2a+2}\]
Similarly
\[ b^2^{\frac{2}{b}} \le \frac{b^2}{b^2-2b+2} =\frac{(a-2)^2}{a^2-2a+2}\]
Thus
\[ a^\frac 2{a} + b^\frac 2{b} \le \frac{a^2}{a^2-2a+2}+\frac{(a-2)^2}{a^2-2a+2} =2\]
We are done. :)

I think that the following inequality holds for $ n$ variables (but I haven't checked it yet)
Let $ a_1,a_2,\ldots, a_n$ $ (n \ge 2)$ be positive real numbers such that $ a_1+a_2+\cdots +a_n =n$. Then
\[ \sum^{n}_{i=1} a^{\frac{n}{(n-1)a_i}}_{i} \le n\]
:)

[quote="can_hang2007"][quote="Vasc"]Let $ a$ and $ b$ be positive real numbers such that $ a + b = 2$. Prove that

$ a^\frac 2{a} + b^\frac 2{b} \le 2$.

Can you generalize this inequality to $ n$ numbers ?[/quote]
By Bernoulli's Inequality, we have
\[ \left( \frac {1}{a}\right)^{\frac {2}{a}} \ge 1 + \frac {2}{a}\left( \frac {1}{a} - 1\right) = \frac {a^2 - 2a + 2}{a^2}
\]
Hence
\[ a^{\frac {2}{a}} \le \frac {a^2}{a^2 - 2a + 2}
\]
Similarly
\[ b^2^{\frac {2}{b}} \le \frac {b^2}{b^2 - 2b + 2} = \frac {(a - 2)^2}{a^2 - 2a + 2}
\]
Thus
\[ a^\frac 2{a} + b^\frac 2{b} \le \frac {a^2}{a^2 - 2a + 2} + \frac {(a - 2)^2}{a^2 - 2a + 2} = 2
\]
We are done.[/quote]
Nice. I proved it in the same way. :wink:

[quote="can_hang2007"] I think that the following inequality holds for $ n$ variables (but I haven't checked it yet)
Let $ a_1,a_2,\ldots, a_n$ $ (n \ge 2)$ be positive real numbers such that $ a_1 + a_2 + \cdots + a_n = n$. Then
\[ \sum^{n}_{i = 1} a^{\frac {n}{(n - 1)a_i}}_{i} \le n
\]
:)[/quote]
Yes, it is. :wink:


real analysis
function
combinatorics proposed
combinatorics

[quote="Lovasz"]Prove that if $0<m\leq k \leq n$, then:
\[{m \choose 0}{n \choose k}+{m \choose 1}{n \choose k-1}+\cdots+{m \choose m}{n \choose k-m}={m+n \choose k}. \]
[/quote]
The coifficents at the $x^{k}$ must be equal in both sides of the following expration
$(x+1)^{m+n}=(x+1)^{m}*(x+1)^{n}$
But,in fact, they are equal respectively
${m+n \choose k}$ and ${m \choose 0}{n \choose k}+{m \choose 1}{n \choose k-1}+\cdots+{m \choose m}{n \choose k-m}$. :wink:

This is simply the well-known Vandermonde's convolution formula.

Let us consider $m$ blue and $n$ red balls. The right side is the number of ways in which we can choose $k$ balls among these $m+n$ balls. The left side is just the same...the $i-th$ sumnand is the number of ways in which we can choose $i$ blue and $k-i$ red balls.

I posted a proof of this identity in the Pre-Olympiad Forum last night.  There are two proofs that I know of.  One is the above using a combinatorial argument, and the other is using generating functions (Tik's solution).  Are there any other proofs?

You can also give a "geometric" proof of the identity. The interpretation goes as follows. Consider a grid of $m \times n$. Suppose you want to go from the lower left corner $A$ to the upper right corner $B$, but you're allowed to move only to the north or to the east. Then it's easily proved you can travel from $A$ to $B$ in $\dbinom{m+n}{n}$ different ways. Using this idea you can give a proof of the Vandermonde's identitiy and many others :) (sorry for my bad English :blush:).

Let's give a "geometrical" proof of an equivalent form of the Vandermonde's identity. We're going to show that \[\sum_{j=0}^{r}\binom{n+k-r}{n-j}\binom{r}{j}=\binom{n+k}{n}\] Consider a grid of $n \times k$. As usual, we want to travel from the left lower corner $A(0,\, 0)$ of the grid, to the upper right corner $B(n,\, k)$, and we are allowed to move to the north or to the east. We know there are $\dbinom{n+k}{k}$ different was to go from $A$ to $B$. Now, in order to classify all the paths, we draw the line $\ell$ through the points $(n,\, k-r)$ and $(n-r,\, k)$. It's clear that every path from $A$ to $B$ must pass through an unique point of $\ell$. We classify the paths according to this fact. It's easy to see that the points of the grid that belongs to $\ell$ are of the form $(n-j,\, k-r+j)$, $0\leq j \leq r$. Suppose that we want to count the number of paths passing through $P(n-j,\, k-r+j)$. From $A$, we can reach $P$ in $\dbinom{n-j+k-r+j}{n-j}=\dbinom{n+k-r}{n-j}$ ways. Then we can reach $B$ from $P$ in $\dbinom{r}{j}$ diferent ways. Hence we can go from $A$ to $B$ passing throught $P$ in $\dbinom{n+k-r}{n-j}\dbinom{r}{j}$ ways. Summing over $j$ we get the result. :wink:


geometry
rectangle
geometric transformation
rotation

In square ABCD, there is an interior point P.  If PA=10, PB=6, PC=14, find the area of the square.

no sry..
The message is too small. Please make the message longer before submitting.

oh, yeah. i see what i did wrong. my figure implied that P was on the diagonal, so i got carried away doing the wrong thing. i'll try again

[quote="daermon"][hide]
first, draw two lines through P that are perpendicular to each of the sides.  it will divide the square into four rectangles.  in the rectangle with P and B, let r be the vertical side and s be the horizontal.  if we let x be the side length, the sides of the rectangle with P and A are r and x-s.  similarly, the one with P and C s and x-r.  now, we plug each of these into pythag:
$r^{2}+s^{2}=36\\ r^{+}s^{2}+x^{2}-2xr=144\\ r^{2}+s^{2}+x^{2}-2xs=196$.  Now, we subtract the top equation from each of the lower ones:
$x^{2}-2xr=108\\ x^{2}-2xs=160$.  Now subtract the upper from the lower:
$x(r-s)=26$.
hmm
i remember doing this problem before but for some reason i forget how to continue... :maybe: [/hide][/quote]
[hide] Let's start out the same way, drawing perpindiculars PX and PY to AB and BC respectively such that X is on AB and Y is on BC.  Say that $PX=x$ and say that $PY=y$.  Now, also say that $AB=BC=CD=AD=s\Rightarrow [ABCD]=s^{2}$.  Now, we can say that \[XB=YP=y\Rightarrow PB^{2}=XB^{2}+PX^{2}=x^{2}+y^{2}\Rightarrow x^{2}+y^{2}=36\]  Now, also note that \[YC=s-PX=s-x\Rightarrow PC^{2}=YC^{2}+PY^{2}=y^{2}+(s-x)^{2}\Rightarrow x^{2}+s^{2}-2sx+y^{2}=196\]  Finally, we can note that \[AX=s-PY=s-y\Rightarrow PA^{2}=XP^{2}+AX^{2}=x^{2}+(s-y)^{2}\Rightarrow 100=x^{2}+y^{2}-2ys+s^{2}\]  Now, plugging in $x^{2}+y^{2}=36$, we get that \[100=x^{2}+y^{2}-2ys+s^{2}=36-2ys+s^{2}\Rightarrow 2ys=s^{2}-64\Rightarrow y=\frac{s^{2}-64}{2s}\]  We can also say that \[196=x^{2}+s^{2}-2sx+y^{2}=36+s^{2}-2sx\Rightarrow s^{2}-160=2sx\Rightarrow x=\frac{s^{2}-160}{2s}\]  Now, plugging in our values for $x$ and $y$ in terms of $s$, we get that \[x^{2}+y^{2}=36\Rightarrow (\frac{s^{2}-160}{2s})^{2}+(\frac{s^{2}-64}{2s})^{2}=36\Rightarrow \frac{s^{4}-320s^{2}+25600+s^{4}-128s+4096}{4s^{2}}=36\Rightarrow \frac{2(s^{4}-224s^{2}+14848)}{4s^{2}}=36\Rightarrow s^{4}-224s^{2}+14848=72s^{2}\Rightarrow s^{4}-296s^{2}+14848=0\] From here, we can say that $w=s^{2}$, so \[w^{2}-296w+14848=0\Rightarrow w=64, w=232\]  Logically, 64 is too small, so $\boxed{232}$ is the answer.  [/hide]

Ya thats right..I was thinking of that method but it was too much computation for me.

[quote="Altheman"]use a rotation[/quote]
how? :maybe:


geometry
inequalities
trigonometry
geometry proposed

Let $ABC$ be a right angled triangle with hypotenuse $BC$.Point $D$ is chosen on the hypotenuse such that $BD=BA$,$E$ is on $AC$ such that angle $ADE$=angle $ABC$.Prove that:$CE>\frac{1}{3}CA$

Let $\angle ABC=\alpha$ by law of sin we get
$\frac{CE}{\sin(CDE)}=\frac{CD}{\sin(CED)}$
and $\angle CDE=\pi-(\angle EDA+\angle ADB)=\pi-(\alpha+\frac{\pi-\alpha}{2})=\frac{\pi-\alpha}{2}$
$\angle CED=\angle EDA+\angle EAD=\alpha+\frac{\pi}{2}-\frac{\pi-\alpha}{2}=\frac{3\alpha}{2}$
Thus we get
\[CE=\frac{CD}{\sin(CED)}\sin(CDE)=(BC-AB)\frac{\cos\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}}=(\frac{CA}{\sin\alpha}-\frac{CA}{\tan\alpha})\frac{\cos\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}}=CA\frac{1-\cos\alpha}{\sin\alpha}\frac{\cos\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}}=CA\frac{\sin\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}}\]
we now prove
\[\frac{\sin\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}}>\frac{1}{3},\ 0\le\alpha\le\frac{\pi}{2}\]
ideed
\[\frac{\sin\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}}=\frac{\sin\frac{\alpha}{2}}{3\sin\frac{\alpha}{2}-4\sin^{3}\frac{\alpha}{2}}=\frac{1}{3-4\sin^{2}\frac{\alpha}{2}}>\frac{1}{3}\]


function
real analysis
real analysis unsolved

Show that for any real number $ c>0$ and any functions $ \mathrm{f},\mathrm{g}: [0,2c]\to[0,2c]$, there exist $ x,y\in[0,2c]$ such that $ |xy\minus{}\mathrm{f}(x)\plus{}\mathrm{g}(y)|\geq c^2$.

[hide="What I’ve done so far."]If $ c\geq\frac{2}{3}$, then $ 4c^2\minus{}2c\geq c^2$; then taking $ x\equal{}y\equal{}2c$ will satisfy the inequality. Thus it suffices to consider the case $ c<\frac{2}{3}$. Unfortunately this is where I’ve been stuck for ages. [img]http://i188.photobucket.com/albums/z198/fairjane/Blind.gif[/img][/hide]

Forget the red herring on the range of $ f$ and $ g$; they might as well map into $ \mathbb{R}$.

It suffices to consider the case $ x\in \{0,2c\},y\in \{0,2c\}$- the corners of the square. Can you find an equality case?

Ok,I now just complete the jmerry's idea.
Let $ h(x,y) \equal{} xy \minus{} f(x) \plus{} g(y)$
Then,
$ h(0,0) \plus{} h(2c,2c) \minus{} h(0,2c) \minus{} h(2c,0) \equal{} 4c^2$
which means,
$ |h(0,0)| \plus{} |h(2c,2c)| \plus{} |h(0,2c)| \plus{} |h(2c,0)| \ge 4c^2$
The rest seems to be not hard :)

And the equality case may happen when $ f,g$ are linear functions such that.
$ f(0)\equal{}2c^2,f(2c)\equal{}0,g(0)\equal{}c^2,g(2c)\equal{}3c^2$


Congrats =]

It doesn't matter if it is 35 points lower than everyone else (which I highly doubt anyway.. at least.. definitely not 35 points lower than what I could get...  :? ).. as long as you've improved.. it's always good. =)

[quote="samus"]I scored a 115 while doing the 2000 AMC 12 in contest-like situations. Its a full 9.5 points higher than my last one: my highest score yet! :cool: (though its probably 35 points down from the rest of you guys :( )[/quote]

115 is VERY respectable :-D be proud of yourself

not even 5% of the people registered here have scored a perfect on the AMC 12 so no worries...

The averages were in the 60s, the average winning score was 92.26, 5% qualified when normally only 2% get above 100, 115 is bigger than the national honor roll cutoff, 8 perfect scores, when only 6 got perfect in 2001, and 37 in 2002,  some how it doesn't seem like you got a bad score.

I think my brother got 165, but I'm not sure. Is it possible?

No 165 is impossible for the AMC. 150 is the highest.  He might have gotten that for his USAMO index which is AMC+10*AIME.


In this post, http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1589224#1589224 a web address enclosed in [url] tags dropped the trailing right parenthesis, resulting in a broken link.  This seems like the sort of thing that [url] tags shouldn't do.

Fixed (might have broken something else meanwhile :) )


a)find all positive integers n for which the equation    \[(a^{a})^{n}= b^{b}\] has a solution in positive integers a,b greater then 1.
b)find all positive integers a,b satisfying  \[(a^{a})^{5}= b^{b}\]

a) It is easy to show, that $b=a^{k}$, therefore $n=ka^{k-1},k\in N$.
b)Because for integer a>1 equation $5=ka^{k-1}$ had not solution we get $a=1=b$.

[quote="Rust"]a) It is easy to show, that $b=a^{k}$, therefore $n=ka^{k-1},k\in N$.[/quote]

It's easy but I'm affraid it's wrong (see my demo below)

[quote="Rust"]b)Because for integer a>1 equation $5=ka^{k-1}$ had not solution we get $a=1=b$.[/quote]

Indeed, there is :  $a=2^{8}$ and $b=2^{10}$ :
$(a^{a})^{n}=(2^{8*2^{8}})^{5}=2^{5*2^{11}}$
$b^{b}=2^{10*2^{10}}=2^{5*2^{11}}$

[quote="kimnimalar"]a)find all positive integers n for which the equation
\[(a^{a})^{n}= b^{b}\] has a solution in positive integers a,b greater then 1.[/quote]

The answer is : every $n >0$ different from 2.
Demo:

Let $a=\prod p_{i}^{a_{i}}$ and $b=\prod p_{i}^{b_{i}}$ with the same set of $p_{i}$ (and maybe some $a_{i}$ or $b_{i}$ equal to zero)

We have : $a_{i}an = b_{i}b$ and so $b_{i}=\frac{an}{b}a_{i}$ and so $b=a^{r}$ for some $r\in\mathbb{Q}$ (This is the error of RUST : he said $k\in\mathbb{N}$)

So $r=\frac{p}{q}$ (p and q coprime), $b=a^{\frac{p}{q}}$, and $\frac{p}{q}=\frac{an}{b}=\frac{an}{a^{\frac{p}{q}}}=\frac{n}{a^{\frac{p}{q}-1}}$, and so $n=\frac{p}{q}a^{\frac{p}{q}-1}$

For $n>2$, $n=\frac{p}{q}a^{\frac{p}{q}-1}$ always have at least one solution : $\frac{p}{q}=\frac{n}{n-1}$ and $a=(n-1)^{n-1}$
This gives $a=(n-1)^{n-1}$ and $b=(n-1)^{n}$

For $n=2$, $2=\frac{p}{q}a^{\frac{p}{q}-1}$ implies $p>q$ and $p=2$ (else any prime $x$ dividing $p$ would divide 2) and so $q=1$ and $a=1$, and so no solution.

For $n=1$, a lot of solutions exist ($a=b$)

Hence the answer (every $n >0$ different from 2).

[quote="kimnimalar"]b)find all positive integers a,b satisfying
\[(a^{a})^{5}= b^{b}\][/quote]

For $n=5$, $5=\frac{p}{q}a^{\frac{p}{q}-1}$ implies $p>q$ and $p=5$. So $q=1, 2, 3$ or $4$

And the only solution is $q=4$ and so :
$a=2^{8}$ and $b=2^{10}$ :
$(a^{a})^{n}=(2^{8*2^{8}})^{5}=2^{5*2^{11}}$
$b^{b}=2^{10*2^{10}}=2^{5*2^{11}}$

--
Patrick


conics parabola quadratics geometry ellipse

i learned from algebra that a parabola is a graph shaped like the one for $y=x^2$

However you should note that the U can be upside down and the roots can be complex.

If you want the other definition, it is the locus of points equidistant from its focus and its directix.

Palytoxin's explanation seems to be perfectly adequate to answer the question.  All parabolas take the general shape of the curve $y=x^2.$

Generally in middle school, the parabolas studied are of the "smiling" or "frowning" type.  Yes, the curve may narrower or wider, and it may be shifted up or down, left or right.  But the general U-shape remains the same.

4everwise / Hamster1800:  Please be careful not to exceed the level of this forum!  This discussion could easily extend to Algebra 2 or above, which would be a High School topic.  Thank you.

g8sk8r, didn't you take geometry this school year?  I thought they teach parabolas in Geometry  :?.  I learned parabolas in Algebra 1.

Yeah, I was trying to give a general idea without using algebra 2 terms.

By the way, how do you pronounce this word.
do you stress the second or the third syllable?
I hear everyone stress the third syllable (and it sounds better that way), but the dictionary tells about stressing the second syllable.

although 4everwise wasn't correct in saying that each quadratic has 2 roots. Most have 2 roots. Half have real roots and half don't (i've been told not to go there). The other few are perfect squares and only have one root. Examples are:

$y = x^2$
$y = x^2 + 2x + 1$
$y = x^2 + 6x + 9$
$y = x^2 - 20x + 100$

236factorial, which word are you refering to? If you mean parabola, I've always heard people stress the second syllable

[quote="Hamster1800"] If you mean parabola, I've always heard people stress the second syllable[/quote]

yeah, par-AB-ola.

Hmm, the geometrical defininion (non-coodinate) Is (I belive) the set of all points that are eqidistant from a point called the focus and the line called the directrix. I think this is too high level but I hope that it is clear. Try [url=http://www.google.com]Google[/url] for a nice definition.

you pronouce it: perabala
that's how my teacher pronounces it and that's the best i can write it.

[quote="236factorial"]By the way, how do you pronounce this word.[/quote]
Follow this link and click the speaker:  [url=http://www.m-w.com/cgi-bin/dictionary?book=Dictionary&va=parabola]parabola[/url]

I think I tend to pronounce it stressing the 3rd syllable because there is a vowel at the end of the word, and in the spanish language, that usually means the syllable before the last syllable is stressed (I'm not spanish)

End it on the pronunciation. Let's get back on topic.

i might have learned what the "U" shaped figures were, i just forgot the term for them.

[quote="4everwise"]A parabola is the graph for any quadratic function. It always has two roots, and when graphed, is in the shape of a U.[/quote]

Actually, the shape can resemble a U or an upside down U.

[quote="IntrepidMath"]Hmm, the geometrical defininion (non-coodinate) Is (I belive) the set of all points that have a constant sum from a point called the focus and the line called the directrix. I think this is too high level but I hope that it is clear. Try [url=http://www.google.com]Google[/url] for a nice definition.[/quote]

That's not correct. It seems you're mixing the definitions of the ellipse and parabola together. The geometrical definition of a parabola is the set of all points equidistant from a line (directrix) and a point not on the line (focus).


Let a,,b,c be positive numbers such that $ abc\equal{}1$.Prove that

$ \frac{a\plus{}b\plus{}c}{3} \ge (\frac{a^2\plus{}b^2\plus{}c^2}{3})^{\frac{1}{5}}$

[quote="skywalkerJ.L."]Let a,,b,c be positive numbers such that $ abc \equal{} 1$.Prove that

$ \frac {a \plus{} b \plus{} c}{3} \ge (\frac {a^2 \plus{} b^2 \plus{} c^2}{3})^{\frac {1}{5}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919

[quote="skywalkerJ.L."]Let $ x,y,z$ be positive numbers such that $ xyz \equal{} 1$. Prove that

$ \frac {x \plus{} y \plus{} z}{3}\geq \sqrt [5]{\frac {x^2 \plus{} y^2 \plus{} z^2}{3}}.$[/quote]
[hide="Proof."]$ (x \plus{} y \plus{} z)^5 \minus{} 81xyz(x^2 \plus{} y^2 \plus{} z^2)\equiv F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} s \plus{} t)$

$ \equal{} 27(s^2 \plus{} st \plus{} t^2)x^3 \plus{} 9(2s \plus{} t)^3x^2 \plus{} 3(26s^4 \plus{} 52s^3t \plus{} 39s^2t^2 \plus{} 13st^3 \plus{} 5t^4)x$

$ \plus{} (2s \plus{} t)^5\geq 0,$

which is clearly true for $ x\leq y\leq z$.[/hide]
A nice proof see here : http://www.mathlinks.ro/viewtopic.php?t=214684

A stronger inequality see here : http://www.mathlinks.ro/viewtopic.php?t=62923


integration
trigonometry
calculus
calculus computations

Find $ \int \frac{1}{1 \minus{} \sin x \plus{} \sin^2x}\,dx$.

The only one way I think is making the universal trigonometrical substitution/


APMO 2003.
Let a,b,c be the sides of a triangle, with a+b+c=1, and let n>=2 be an integer. Show that:
(an +bn)1/n +(bn +cn)1/n + (cn +an)1/n <1+ 21/n/2

The problem is actually fairly simple once you realize how we can get the equality. the equality can be reached when two of the sides approach 0.5 and the other side approaches zero. keeping this in mind, let f(a,b,c) be the l.h.s expression. then it is easy to verify that if a>b>c
then increasing a and decreasing c has the effect of incresing the l.h.s.we have to keep in mind that at each point we should have b+c>a. Now notice that increasing b and deacreasing a again increase the l.h.s, and also note that the latter change always keeps a,b,c triangular.
Consider S=f(a+d,b,c-d)-f(a,b,c)
then (a+d)^n +(c-d)^n >=a^n+c^n  (*)
since f is convex with repect to each variable, therefore
f'(x) is deacreasing and therefore toghther with(*) we have verified the above argument. The result follows almost immidiately.(you need to still check a number of conditions, because you have to maintain the triangular property of the numbers a,b and c)


You would like to withdraw some money from your bank account using a nearby ATM. However, since it's the Christmas time, the ATM is almost empty - it only contains $a_1$ bills of denomination $d_1$, $a_2$ bills of denomination $d_2$, ... $a_k$ bills of denomination $d_k$. Is it possible to withdraw exactly $N$ dollars? If not, what is the greatest sum less than $N$ than can be withdrawn? $N<100000$, $a_i<1000$, $d_i<1000$, $k<10$.

In the general case the problem is NP-complete.

Let C[i][j] be true if you can obtain sum j with first i bills. You can build C[i][j] from C[i-1][j-k*D[i]] k <= A[i]. To get an O(N*K) solution, you know that at each j you compute you need to see if there is a true value C[i-1][x] , x havind the same remainder as j, mod D[i]. So, before you compute C[i], take the line C[i-1] and put the true values in lists acording to their remainder mod D[i].  You can use a heap or a deque now, such that when you compute C[i][j] you look at values from C[i-1] that are at least j-A[i]*D[i]. I hope I was clear enough. :)

Wow, I like the idea with the remainders, however, I don't see how to do the final heap / queue part efficiently. You may have as many as 1000 different reminders so having this many queues is not a good idea and building a heap is... well, I'll have to think about it :).

If you feel like thinking about this problem a little more, I'll tell you that there's a better way to calculate the C[i] array. Try to exploit the fact that unlike in common knapsack problems, this time you only need to calculate a true / false value.

It takes O(N) to build the deques / O(N*lg N) for heaps. Although there can be 1000 heaps/deques, the total size used is always N. It should be obvious how to do this with heaps.. when computing C[i][j], remove the top of the heap until it's index is >= j-A[i]*D[i], so that you can use it. After that, see if the top is < j, and if so set C[i][j] to true. Update the heap if A[i-1][j] was true. The whole thing can be done with deques in O(1) amortized time.

You said i should think about the fact that you only compute true / false values. Do you mean optimizing it using bit operations? The brute approach O(N*K*A[i]) can be made 32 times faster...


The width of a rectangle is 2/3 of its length and its perimeter is 4 units.  Express as a common fraction the number of square units in the area of the rectangle.

[hide]5413[/hide]

Did you guess or did you actual have some thinking involved?  Solution!!!!!!

[hide]The width of a rectangle is 2/3 of its length and its perimeter is 4 units. Express as a common fraction the number of square units in the area of the rectangle.

That is:2x/3+2x/3+x+x=4.

10x/3=4

10x=12

x=6/5.

6/5*2/3.

The width is 4/5.

The height is 6/5.

The total is 24/25

[/hide]

oops sorry.
[color=white]if height is x, then width is 2/3 of x. the perimeter is 2x+4x/3 which is 10x/3. 10x/3=4. x=4*3/10, which is 6/5. That makes the height 6/5 and the width 4/5. 6/5*4/5=24/25[/color]


I've seen Pick's Formula mentioned in this forum before.  Can anyone please explain it?  Oh, and I've also see the Shoelace Theorem (?) in the forum.  Could you smart peoples out there explain that too?  Thanks!   :)

If you wanted the area of this figure:
[asy]import graph;
for (int i=0; i<9; ++i)
{
 for (int j=0; j<9; ++j)
 {
   dot((i,j),green);
  }
}
draw((0,0)--(8,0),Arrow);
draw((0,0)--(0,8),Arrow);
pair a=(7,3);
pair b=(3,2);
pair c=(1,7);
dot(a,red);
dot(b,red);
dot(c,red);
dot((4,5),red);
dot((3,3),blue);
dot((3,4),blue);
dot((3,5),blue);
dot((2,5),blue);
dot((2,6),blue);
dot((4,3),blue);
dot((4,4),blue);
dot((5,3),blue);
dot((5,4),blue);
dot((6,3),blue);
draw(a--b--c--cycle,blue);[/asy]

We can use the formula:
$ i \plus{} \frac {b}{2} \minus{} 1 \equal{} 10 \plus{} \frac {4}{2} \minus{} 1 \equal{} \boxed{11}$
$ i$ is the number of dots inside the figure and $ b$ is the number on the edges
I think that is faster than using the distance formula or shoelace

[url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]Wiki is your friend[/url]


hey
solve this!
abc=1 a,b,c are real different from one prove that
2<1/abs[a(1-b)]+1/abs[b(1-c)]+1/abs[c(1-a)]

this is a problem I have created
i thought to share it with you here at ml
maybe its not that difficult or challenging...
 :P

Or maybe it'd be better if it was easily readable :)

[quote="Albanian Eagle"]
$ abc \equal{} 1$
$ a,b,c \in \mathbb{R} \minus{} \{1\}$
prove that
$ 2 < \frac {1}{\left|a\left(1 \minus{} b\right)\right|} \plus{} \frac {1}{\left|b\left(1 \minus{} c\right)\right|} \plus{} \frac {1}{\left|c\left(1 \minus{} a\right)\right|}$[/quote]

that's true but i dont have internet home and i write from an internet cafe
i didnt have time so i wrote it like that.
sorry
 :(

Bah, you don't have to be sorry for that ;)

I have an idea! :?
First we may suppose a,b,c are all positive, or WLOG a<0, b<0 and denote the right side of  the inequality by F(a,b,c). We have F(a,b,c)>F(-a,-b,c) when a and b are both negative.
then let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ which x,y,z are all positive.
We just need to prove $\sum_{cyc}{|\frac{yz}{xz-xy}|}>2$
Again let $A=xy, B=yz, C=zx$ WLOG $A \ge B \ge C>0$
We get $\sum_{cyc}{|\frac{A}{B-C}|}>2$
that is $\frac{A}{B-C}+\frac{C}{A-B}+\frac{B}{A-C}>2$
Suppose $A-B=P>0, B-C=Q>0$
we have $\frac{C}{P}+\frac{C+P}{Q}+\frac{C+Q}{P+Q}>1$
since$\frac{p}{q}>\frac{p}{p+q}$ so
$\frac{p}{q}+\frac{q}{p+q}>1$  and therefore
$\frac{C}{P}+\frac{C+P}{Q}+\frac{C+Q}{P+Q}>\frac{p}{q}+\frac{q}{p+q}>1$

why is F(a,b,c)>F(-a,-b,c)

and just for a challenge
Show that 2 can not be replaced by a bigger real number.


Two numbers, $ a$ and $ b$, are randomly selected without replacement from the set $ \{2, 3, 4, 5, 6\}$. What is the probability that the fraction $ \frac{a}{b}$ is less than 1 and can be expressed as a terminating decimal? Express your answer as a common fraction.


counting
distinguishability
floor function

[hide="I am not so good at counting"]

but I will try anyway
if the boxes are distinguishable but the balls are not...
let's start by putting 1 red ball and 1 white ball in each box
we now have r-n red and w-n white balls left.
for each of the r-n reds remaining, we have n choices for the box to put it in.
this gives n^(r-n) ways to distribute the remaining reds.
same goes for the whites.
but to cover the case when n>r or n>w (making [n^(r-n)]<1 or [n^(w-n)]<1), the answer should be
$\left\lfloor{n^{r-n}}\right\rfloor\times\left\lfloor{n^{w-n}}\right\rfloor$
[/hide]

[quote="Phelpedo"]In how many ways can we place $r$ red balls and $w$ white balls in $n$ boxes so that each box contains at least one ball of each color.[/quote]

[hide]Well, we know that we've used $n$ balls for both red and white.

So, we know have $r+w-2n$ balls.

Now, we need to put these $r+w-2n$ balls into the n boxes.

By Ballz and Urnz, we know that this is just $\binom{r+w-n-1}{n-1}$.[/hide]

I don't think that's quite right.

[hide] I did the same thing up to the point where we want to put $r-n$ red balls in $n$ boxes and $w-n$ white balls in $n$ boxes. The former can be done in $\binom{r-1}{r-n}$ ways and the latter can be done in $\binom{w-1}{w-n}$ ways. [/hide]

EDIT: mathgeniuse^lnx, ball and urn can only be applied to indistinguishable balls (if they are red and white, they are distinguishable).

[hide]Oh, so we consider each case separately.

There are r-n red balls into n boxes, and w-n white balls into n boxes.

So the answer is

$\binom{r-1}{n-1}*\binom{w-1}{n-1}$.[/hide]


For a convex n-gon $P_n$, we say that a convex quadrangle $Q$ is a diagonal-quadrangle of $P_n$, if its vertices are vertices of $P_n$ and it's sides are diagonales of $P_n$.
Let $d_n$ be the number of diagonal-quadrangles of a convex n-gon. Determine $d_n$ for all $n\geq 8$.

You are right Kondr. The answer in general (instead of quadriteral we consider $k$-gon with $k\leq \frac{n}{2}$) is $\frac{n}{k} \binom{n-k+1}{k}$.

The problem equals to:
Choose k points among n points on a circle, such that any two points are not consecutive.

So the answer should be $\binom{n-k}{k}$.

Since for n=8 and k=2 the answer should be 2, the formula may be
$\frac n k \binom {n-k-1} {k-1}$.


It's possible it could be factored using .. oh my goodness, I forgot what it was called for a minute. Synthetic division. Just goes to show I haven't had enough time to do math.

This is how far i've gotten:

$ax^3+bx^2+cx+d=0$

$x^3+\frac bax^2 + \frac cax + \frac da = 0$

$y = x + \frac b{3a}$

$y^3+\frac{ac-b^2}{a^2}y + \frac{4b^3-9bc+27ad}{27a^3} = 0$

^ that's the deflated cubic which *supposedly* can be turned into a quadratic^

[quote="4everwise"]Everybody knows the formula for quadratics: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
If anybody is interested in starting a chain to "attempt" to find a formula for cubic equations, I think it would be interesting.

We could always start with $ax^3+bx^2+cx+d$. However, "a" can be factored out from the beginning. A re-ordering of variables gives $stuff(x^3+ax^2+bx+c)$

If anybody has any good ideas of where to go next, feel free to post.[/quote]
a topic already exists in [quote="236factorial"]Someone put the formula in this thread
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41829[/url]

It's very nasty.[/quote]
i dont understand why you have to make another topic about it
i read carefully your post and it looks like you are looking for the steps on how to obtain that formula

start with $ax^3+bx^2+cx+d=0$ and divide thru by $a$ since $a\neq 0$

and get
\[x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}=0\]
now let $x=y-\frac{b}{3a}$ and substitute and simplify to get
\[y^3+\frac{3ac-b^2}{3a^2}y+\frac{27a^2d+2b^3-9abc}{27a^3}=0\]
let $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{27a^2d+2b^3-9abc}{27a^3}$
so new equation looks like this \[y^3+py+q=0\]
now note the resemblance between above equation and this:
\[(u-v)^3+3uv(u-v)=u^3-v^3\]
so if we let the solution to be $y=u-v$ and $p=3uv$ we have
\[(u-v)^3+p(u-v)+q=0\]
\[u^3-v^3+3uv^2-3u^2v+3uv(u-v)+q=u^3-v^3+q=0\]
but we know $3uv=p$ so we get rid of $v$ and get
\[u^3-\frac{p^3}{27u^3}+q=0\]
solve for $u^3$ in $u^6+qu^3-\frac{p^3}{27}=0$ and get
\[u^3=-\frac{q}{2}+\frac12\sqrt{\Delta}\]
where the discriminant $\Delta=q^2+\frac{p^3}{27}$
$v^3$ can be found easily too:
\[v^3=u^3+q=\frac{q}{2}+\frac12\sqrt{\Delta}\]
ok now lets start to put all the formulas together by back-substituting all those big expressions
\[y=u-v=\sqrt[3]{-\frac{q}{2}+\frac12\sqrt{\Delta}}-\sqrt[3]{\frac{q}{2}+\frac12\sqrt{\Delta}}\]

\[=\displaystyle \sqrt[3]{-\frac{q}{2}+\frac12\sqrt{q^2+\frac{p^3}{27}}}-\sqrt[3]{\frac{q}{2}+\frac12\sqrt{q^2+\frac{p^3}{27}}}\]

\[=\displaystyle \quad\quad\sqrt[3]{-\frac{27a^2d+2b^3-9abc}{54a^3}+\frac12\sqrt{\left(\frac{27a^2d+2b^3-9abc}{27a^3}\right)^2+\left(\frac{3ac-b^2}{9a^2}\right)^3}}-\sqrt[3]{\frac{27a^2d+2b^3-9abc}{54a^3}+\frac12\sqrt{\left(\frac{27a^2d+2b^3-9abc}{27a^3}\right)^2+\left(\frac{3ac-b^2}{9a^2}\right)^3}}\]

and so

\[x=y-\frac{b}{3a}\]
\[=\displaystyle -\frac{b}{3a}+\sqrt[3]{-\frac{27a^2d+2b^3-9abc}{54a^3}+\frac12\sqrt{\left(\frac{27a^2d+2b^3-9abc}{27a^3}\right)^2+\left(\frac{3ac-b^2}{9a^2}\right)^3}}-\sqrt[3]{\frac{27a^2d+2b^3-9abc}{54a^3}+\frac12\sqrt{\left(\frac{27a^2d+2b^3-9abc}{27a^3}\right)^2+\left(\frac{3ac-b^2}{9a^2}\right)^3}}\]

phew, this is insane and please do not ask this question again... :surrender:

[quote="Palytoxin"]\[y^3+\frac{3ac-b^2}{3a^2}y+\frac{27a^2d+2b^3-9abc}{27a^3}=0\]
let $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{27a^2d+2b^3-9abc}{27a^3}$
so new equation looks like this \[y^3+px+q=0\]
now note the resemblance between above equation and this:
\[(u-v)^3+3uv(u-v)=u^3-v^3\]
[/quote]

I don't exactly see the resemblence. I understand before and after that but this step confuses me a TON

resemblance as in $a^2+abc$ and $(m+n)^2+k(m+n)$ if you let $a=m+n$ and $k=bc$... substitute and you will get from one equation to another


[quote="kurt.math"]Nicole and Maria play against each other, in the third round of the Australian Open. In this tournament, the first player to win 2 sets wins the match. The probability that Nicole wins any set is 70%.

(i) Find the probability that the game will last two sets only.

(ii) Find the probability that Nicole wins the match.

(iii) Find the probability that Maria wins the match.

Thankyou =)[/quote]

[hide="(i)"]
if the game lasts two sets, either maria swept, or nicole swept. the probability that nicole sweeps is $ .7\times.7\equal{}.49$. the probability that maria sweeps is $ .3\times.3\equal{}.09$. the sum of these is $ .58$ or $ \boxed{58\%}$ [/hide]

[hide="(ii)"]
nicole can win in three ways: W-W, W-L-W, or L-W-W. the probability of the first one, as we calculated in part (i), is $ .49$. the probability of the second (and the third, they will be equal. do you see why?) is $ .7\times.3\times.7$, which equals $ .147$. the third will be the same, so $ .147$ again. summing these, we get $ .49\plus{}.147\plus{}.147\equal{}.784\equal{}\boxed{78.4\%}$ [/hide]

[hide="(iii)"]
because a tie is impossible, either maria will win, or nicole will win. thus, the probability that maria wins is simply $ 1\minus{}(\text{the probability that Nicole wins})$. thus, our answer is $ 1\minus{}.784\equal{}.216\equal{}\boxed{21.6\%}$ [/hide]

Hope that helped! :)