geometry
3D geometry
sphere
pyramid
dodecahedron
LaTeX
perimeter

1. Can someone tell me how to find the length of the diagonal of a cube.

2. Can anyone teach me any other Geometry stuff that could show up/be helpful for MathCounts.

Thanks,
Timmy

1) The internal diagonal is root3 times a side, proven by using the Pythagorean Theorem twice
2) Check Intro to Geometry

Thx, but i need any general help that ppl r willing to give instead of just a book.
Like...wat is the formula of a hexagon.

Greatly Thankful,
Tim

To start, you should know how to find the area of regular polygons. For an equilateral triangle, the formula is $ \frac{a^2 \sqrt{3}}{4}$, which is derived from using the pythagorean theorem. A square, obviously, is $ a^2$, pentagons...What is the formula for a pentagon? They don't seem to ask those kinds of questions often. Anyway, a hexagon is made up of 6 equilateral triangles, and octagons are made of 8 isosceles triangles.
You should know formulas for volumes and surface areas of polyhedra, especially common ones like cubes, spheres, cones, pyramids, etc.
Know that f+v-2=e (f=faces, v=vertices, e=edges) in any polyhedron.
There are probably several more that could be added, but I'm lazy and not gonna think of them all right now.
:D

Yes, but all you need to know is the formula for the area of an equilateral triangle and you can find the area of a regular hexagon. Why bother memorizing another formula?
There is no formula for isosceles triangles, seeing as there are all kinds of them.

Well then, how are we able to find the area of an octagon?

I highly doubt problems would just randomly ask you to find the area of an octagon with no additional info.

Instead of memorizing formulas for all this, why don't you just famliarize yourself with the general stuff?

Octagons are not that hard. It's just 7*(side length)^2, I think.
Decagons and dodecahedrons are different though.

[quote="returnofkiladi"][quote="mathcountsfreak"]To start, you should know how to find the area of regular polygons. For an equilateral triangle, the formula is $ \frac {a^2 \sqrt {3}}{4}$, which is derived from using the pythagorean theorem. A square, obviously, is $ a^2$, pentagons...What is the formula for a pentagon? They don't seem to ask those kinds of questions often. Anyway, a hexagon is made up of 6 equilateral triangles, and octagons are made of 8 isosceles triangles.
You should know formulas for volumes and surface areas of polyhedra, especially common ones like cubes, spheres, cones, pyramids, etc.
Know that f+v-2=e (f=faces, v=vertices, e=edges) in any polyhedron.
There are probably several more that could be added, but I'm lazy and not gonna think of them all right now.
:D[/quote]
Thx mathcountsfreak. Isn't there a more specific formula for area of a hexagon like $ \frac {6 * a^2 \sqrt {3}}{4}$ or something...
Anyways, thanks for the help. What is the formula for the area of an isosceles triangle because I NEED to no how to find area of octagon.
Thx,
Tim[/quote]

If the octagon is regular, than you could use 1/2 ap (sry i don't know latex) where a is the distance from the center to the midpoint of a side (like the radius of a circle) and p is the perimeter of the octagon.

there is actually a formula for a regular decagon based on the radius of it's circumcircle, but it's really pointless to memorize and I doubt mathcounts would ask it.  The dissection that can be used to produce the formula is also pretty hard to see.

I have one question: why can't you just look in the book?  It's not much to ask for, and they explain things much better than we, or at least I, can.

For any regular polygon, $ \text{Area}\equal{}as$, where $ a$ is the apothem and $ s$ is the semiperimeter.  This can be easily proven using $ A\equal{}\frac{bh}{2}$ for triangles.

[quote="returnofkiladi"]Thx for all ur help.
That's what I love about AoPS. [b]Everyone[/b]helps. Thank you all very much.

Is there any other general Geometry tips besides to find area/perimeter of regular polygons?[/quote]

2-dimensional geometry - many problems rely on your knowledge of triangles.  Therefore, knowing the pythagorean theorem, similar triangles, angle bisector theorem, areas, perimeters, and maybe some properties of quadrilaterals would be helpful

3-dimensional geometry - try to see if you can bring this down to a question about 2-dimensional geometry.

meh, google doesn't really help THAT much

I think that an excellent understanding of basic pythagorean triples will help in cd round...... :)

i have not read the rest of the topic so sorry if some of these have already been said:

1) 30-60-90 triangle side relations
2) 45-45-90 triangle side relations
3) formulae for areas and volumes of figures.
4) Euclid's theorem, the one that says f + v = e - 2
5) [b]POWER OF A POINT[/b] that is a big one
6) basic facts about equilateral triangles.

[quote="cognos599"]doesn't really belong in the mc forum but

Law of Cosines: if two sides of a Triangle are a and b and the included angle is $ \theta$ then the 3rd side, c =

${ a^2 + b^2 - 2ab\cos{\theta}}$. not usually used in mc alot but it is a really helpful tool in higher level trigonometry.[/quote]

sorry in the law of cosines $ c^2$ = that formula i gave not c = the formula.

What is power of a point? I never really bothered to learn it :P

like that whole set of theorems with angles and sides and tangents and secants. it is really helpful. it is hard to explain without diagrams plus there are like 5 theorems so here.

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

also look at the bottom with the intersecting chords theorems.

[quote="returnofkiladi"]What about a target that something like find the area of the triangle that has vertices at points (3,4),  (2,4), and (2006,2007) or something. Does neone remember that?[/quote]

I didnt get a significant response to this question, so I re-asked this one.

you could either use the shoelace method or make a giant rectangle around the quadrilateral (there was another point of (0,0)) and find the area of the rectangle and minus the two triangles.

I think remember a question like that from chapter last year. I just used shoelace, since the large numbers didn't matter since you had a calculator. Even without one, you can still probably do it in a minute or 2.

If you have a fear of rectangles or shoelaces, use heron's and the distance formulas, though only someone who has completely blanked on shoeleace and basic geometry would resort to that method.

[quote="returnofkiladi"][quote]I think remember a question like that from chapter last year. I just used shoelace, since the large numbers didn't matter since you had a calculator. Even without one, you can still probably do it in a minute or
[/quote]

How do you avoid large numbers with the calculator.
How do you do the shoelace formula.[/quote]

To use the shoelace, you need to know the all points. In this case, you do, so you may apply it.
Here is the mathcounts solution, and original question from last year using basic geometry : [url=http://www.mathcounts.org/webarticles/articlefiles/510-07_C_Solutions.pdf]Solution[/url]
[hide="shoelace method"]
The quadrilateral has (2006,2007) (1,1) (2,1) & (1,3) as vertices and we'll try to find the area using shoelace.

First, pick any one of the points - note that it may help for polygons with more vertices to draw a picture.
Next, write it down and go counterclockwise or clockwise and write down in order all the points below the first point, and end with the first point.
Write (I'm starting with (2006,2007)) :
2006  2007
1       3
1       1
2       1
2006  2007
Starting with the left side's uppermost number, multiply each value by the value on the right side diagonally down one number. For example, we will multiply 2006 by 3. We keep doing this until we get to 2*2007, the last product for the left side. Add all of these products up, and we get $ 2006*3 \plus{} 1 * 1 \plus{} 1*1 \plus{} 2*2007\equal{}10034$

Next start with the left side's lowermost number, and multiply diagonally one number up. So we start at 2006*1 and end with 1*2007; adding all products yields $ 1*2007 \plus{} 3*1 \plus{} 2*1 \plus{} 1*2006\equal{}4018$
Naming these two sums S1 and S2 respectively, the final step is:
Area = the absolute value of   $ \frac{S1\minus{}S2}{2}$
.      = 3008 [/hide]

Heron's formula states that, letting a, b, and c be the sides of a triangle, and s being their semi-perimeter ($ \frac{a\plus{}b\plus{}c}{2}$), the area of a triangle is:

$ \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$.

Power of a point states that for a point inside a circle, any two chords passing through it, split into 2 parts by the point, will always multiply to be the same.

For example, take a circle. Place Point P inside. Now take 4 points A,B,C,D such that A,P,C are collinear (Are on the same line) and such that B,P,D are collinear. Then, Power of a point tells us that $ BP \cdot PD \equal{} AP \cdot PC$.

[quote="BOGTRO"]
Power of a point states that for a point inside a circle, any two chords passing through it, split into 2 parts by the point, will always multiply to be the same.

For example, take a circle. Place Point P inside. Now take 4 points A,B,C,D such that A,P,C are collinear (Are on the same line) and such that B,P,D are collinear. Then, Power of a point tells us that $ BP \cdot PD \equal{} AP \cdot PC$.[/quote]

this is one of the things power of a point tells us.  there's also 3 others, which are for points outside the circle.


I think we need a new rank and star system.  As AoPS grow, many users are well above 2500 posts, and deserve anew rank for their long time commitment to AoPS (like Valentin, with over 5000 posts!!!).  I suggest a system in which the 5 star rank is 5000 posts.  If you think we need a new system of stars and ranks, please reply.

tons of more suggestions:
http://www.phpbb.com/phpBB/catdb.php?cat=21

[quote="Valentin Vornicu"][quote="mysmartmouth"]ike Valentin, with over 5000 posts!!![/quote]Thanks, but, as you can notice, I already have more than 5 stars :D

Anyway, the stars thing is just "for fun" and the system will most probably remain the same for now.[/quote]

I think that having stars is just for having fun, so if you put a limit at 5000 posts, is like thinking that having five stars is so far that you finally abandon it...

valentin said it would most likely remain the same.

[quote="José"][quote="Peter VDD"]There are people that post to contribute to the board. They will not post more due to stars.
There are people who post to raise their rank and post count. They will just spam more.
Not a good idea thus imo...[/quote]

Who are you talking to???[/quote]

He is talking about the members that posted before and after his reply.

[quote="orl"]
He is talking about the members that posted before and after his reply.[/quote]

So true...

[quote="Peter VDD"][quote="José"]sarcasm eh?[/quote]I think you often see sarcasm where there is none... there was none in my message for sure.[/quote]

I wasn't talking about your post, I was talking about orl's

[quote="José"][quote="Treething"]Either way, José's sarcasm detector is broken.[/quote]

Yes, you broke it[/quote]

That's compelling enough for me.  I'll have to decided if breaking somebody's sarcasm detector is a bannable offense...


logarithms inequalities function calculus derivative algebra domain

Prove that if $ x,y>1,$ then
$ \ln \frac{x\plus{}y}{2\sqrt{xy}} \ge \frac{x\plus{}y}{4xy}\minus{}\frac{1}{x\plus{}y}.$

Nice problem, canhang. Thanks for sharing  :)

[u][i][b]Solution[/b][/i][/u]

Without loss of generality, we may suppose $ x\ge y$ then let

$ t\equal{}\sqrt{\frac{x}{y}}$  with constraint $ t\ge 1$

Therefore, the original inequality can be written as follows

$ y\ln\left[\frac{1}{2}\left(t\plus{}\frac{1}{t}\right)\right] \ge \frac{t^2\plus{}1}{4t^2}\minus{}\frac{1}{t^2\plus{}1}$

Since $ y> 1$ so it suffices to prove that

$ \ln\left[\frac{1}{2}\left(t\plus{}\frac{1}{t}\right)\right] \ge \frac{t^2\plus{}1}{4t^2}\minus{}\frac{1}{t^2\plus{}1}$

We consider the following function with $ t$ is the variable.

$ f(t)\equal{}\ln\left[\frac{1}{2}\left(t\plus{}\frac{1}{t}\right)\right] \plus{}\frac{1}{t^2\plus{}1}\minus{}\frac{t^2\plus{}1}{4t^2}$

Taking the first derivative to investigate the behavior of $ f(t)$ in the domain $ t \in [1,\infty)$

We have

$ f'(t)\equal{} \frac{2t^6\minus{}3t^4\plus{}1}{2t^3(t^2\plus{}1)^2}$

However, using Cauchy inequality, we derive

$ 2t^6\plus{}1\equal{}t^6\plus{}t^6\plus{}1 \ge 3\sqrt[3]{t^12}\equal{}3t^4$

which infers that  $ 2t^6\minus{}3t^4\plus{}1 \ge 0$

Thus, $ f(t)$ is increasing on the interval $ [1,\infty)$

Hence, $ f(t) \ge f(1) \equal{}0$

We completed our proof.

[hide="Canhang2007"]Anh Can post solution cho bai 4 duoc khong anh? Em cam on[/hide]

[quote="can_hang2007"]Prove that if $ x,y > 1,$ then
$ \ln \frac {x \plus{} y}{2\sqrt {xy}} \ge \frac {x \plus{} y}{4xy} \minus{} \frac {1}{x \plus{} y}.$[/quote]

Set $ x \plus{} y \equal{} a,\ xy \equal{} b\ (x > 1,\ y > 1)$,

$ \Longleftrightarrow (x \minus{} 1) \plus{} (y \minus{} 1) > 0\ and\ (x \minus{} 1)(y \minus{} 1) > 0\ and\ x \minus{} 1,\ y \minus{} 1\in\mathbb{R}$

$ \Longleftrightarrow a > 2\ and\ a \minus{} 1 < b\leq\frac {1}{4}a^2$.

Let $ f(b) \equal{} \ln \frac {a}{2\sqrt {b}} \plus{} \frac {1}{a} \minus{} \frac {a}{4b}\ (a \minus{} 1 < b\leq\frac {1}{4}a^2)$, we have $ f'(b) \equal{} \frac {a \minus{} 2b}{4b^2}$

Since $ a > 2\Longrightarrow \frac {a}{2} < a \minus{} 1 < \frac {1}{4}a^2$, so $ f'(b) < 0$, or $ f(b)$ is strictly decreasing fuction for $ a \minus{} 1 < b\leq\frac {1}{4}a^2$, yielding

$ f(b)\geq f\left(\frac {a^2}{4}\right) \equal{} 0$ which completes the proof.


If James starts at the lower left corner of an 8x8 chessboard and can only move one square up or one square right, in how many ways can he reach the upper right corner?

[hide]
I'm probably wrong. But anyways...
If it's an 8x8 board. Then the answer should be...

$ 2^{8}=256$

Answer: 256

[/hide]

Its only 14 steps (because we dont count first and last squares) so its $ \binom{14}{7}$

[quote="13375P34K43V312"]That's incorrect. By your logic, moving up 14 times in a row will get you to the desired square.

There must be 7 moves to the right and 7 moves up, so 14 moves total. Our answer is $ \dbinom{14}{7}$.[/quote]

yes i just fixed that (it didnt even tell me someone else had posted)

like right when you posted i was busy reading nexmus's post and it confused me

I ended updoing 14C0+14C1...+14C14 and got that 2^14 so yea..but yea 7 of them moves have to be up so any 7 in that 14 step pattern

It means 14 choose 7. It means combinatorics.

Meaning the number of ways we can choose 7 distinct objects from a pile of 14.
In this case, we choose 7 right turns out of 14 possible turns.

[quote="archimedes1"]Meaning the number of ways we can choose 7 distinct objects from a pile of 14.[/quote]

Or 7 steps to go up in a 14 step pattern in this case.

[quote="alexhhmun"]what does is the fourteen over 7 thing?  I've seen it often, never understood it, though.[/quote]

Wohoo! It's a wiki-moment:

Combinations, [url]http://www.artofproblemsolving.com/Wiki/index.php/Combinations[/url]

Not to be confused with permutations, [url]http://www.artofproblemsolving.com/Wiki/index.php/Permutations[/url]

start with the first position. you can move 1 block up or 1 block right. from those two points you can move 1 block up or 1 block right. this happens $ 14$ times so the answer is $ 2^{14}=$a big number.

we seen to all have different solutions.....


geometry
3D geometry
prism
pyramid
tetrahedron
Pythagorean Theorem

Find the volume of a regular triangular prism if all the edges and sides of the prism are $x$.  Find you answer in terms of $x$.

[hide](sqrt(3)/4)*x^3
Find the area of the base using pythagorean theorem and get the height, and then the area of the triangle. The volume of the prism is this multiplied by the height. [/hide]

[hide]
First you need to find the area of the base. Since it is an equilateral triangle, it is $\frac{x^2\sqrt{3}}{4}$.
Then you need to find the height.
It is still $x$, since all edges are the same.
Multiply to get $\frac{x^3\sqrt{3}}{4}$
[/hide]

didn't u just find the volume of a regular triangular pyramid, a tetrahedron, but the question asks for a triangular prism. so the later faces are squares with side lengths of x, right???

The area of an equilateral traingle is $\frac{\sqrt{3}}{4}x^2$, ehich is obviously true.  Now multiply by hieght/length $x$ to get $\frac{\sqrt{3}}{4}x^3$.  As a result, I believe that fireemblem13 is correct. :)

Masoud Zargar

[hide="Solution"] Because the area of an equilateral triangle is $\frac{x^2\sqrt{3}}{4}$ , we do base times height, height is x, so we have $\frac{x^3\sqrt{3}}{4}$[/hide]


Determine if you can complete the filling of the magic square with the natural numbers from 1 to 9 without repetition, so that the sum of any row is equal to any column or diagonal.

$ \begin{array}{ccc}
a & b & c\\
d & 9 & e\\
f & g & h\end{array}$

Portuguese question:
Determine se é possível completar o preenchimento do tabuleiro abaixo com os números naturais de 1 a 9, sem repetição, de modo que a soma de qualquer linha seja igual a de qualaquer coluna ou diagonal.

$ \sum_{i\equal{}1}^9 i \equal{} 45$.
So,
$ a\plus{}b\plus{}c\equal{}d\plus{}9\plus{}e\equal{}f\plus{}g\plus{}h\equal{}\frac{45}{3}\equal{}15$.
$ a\plus{}d\plus{}f\equal{}b\plus{}9\plus{}g\equal{}c\plus{}e\plus{}h\equal{}\frac{45}{3}\equal{}15$.

And $ a\plus{}9\plus{}h\equal{}c\plus{}9\plus{}f\equal{}15 \iff a\plus{}h\equal{}c\plus{}f\equal{}15\minus{}9\equal{}6$.
WLOG, let $ (a, h)\equal{}(1,5), (c, f)\equal{}(2,4)$.

Therefore $ a\plus{}b\plus{}c\equal{}1\plus{}b\plus{}2\equal{}15 \iff b\equal{}12 \not\in \{3,6,7,8\}$.

Thus, this magic square can [b]not[/b] complete.

Ah ok ;)
Good explanation. Thank you very much.

wait, can i do the same thing with a larger square? like a 4 by 4?

Yes ;)

But the sum of the numbers are 16.17/2 = 136 and 136/4 = 34 ;)


i'm pretty sure they're all from Disney, but can you guys confirm?

Drake and Josh is on Nickelodeon, according to the great wikipedia.

MIX IT ALL TOGETHER AND U NO THAT IT's THE BEST OF BOTH WORLDS...

take a guess.  :D

hmm...
hannah montana??? :D
omg, i can't think of any more lyrics.....

Wait, are we doing it for real?  Cuz I think all of these shows are so stupid...

EXCEPT FOR [size=200][color=violet]HANNAH MONTANA![/color][/size]

SUITE LIFE!!!!!

hannah montana's okay too =)

but....
i still like zack and cody :omighty:

Suite life is funny...hannah montana is..okay I guess

But zack and cody are better..=P

These shows are a wonderful way to turn you into a cheeto-eating mindless zombie.

That being said, I voted other. GO POWER RANGERS!!!

[color=blue]* ernie hates all of these shows, but he likes cheetos![/color]

I don't believe Power Rangers was ever on Disney, although I could be wrong.  It was FOX when I was younger.  Anyway, that was a great show.

i've never really like that show....
o well
look!
it's the twins!!! :yup:  :yup:

they don't have any emoticon with hair....

i would be posting something prases zack and cody said, but that may be against copyright law.....

no one cares if you just post it on a forum....

[size=59]i think, some restrictions apply look at website for more details  :D [/size]


function
algebra
polynomial
domain
rational function
calculus
calculus computations

I am learning to distinguish between rational and polynomial functions. A questions reads:

"State whether the function is a polynomial, a rational function(but
not a polynomial), or neither a polynomial nor a rational function. If
the function is a polynomial, give the degree."

g(x)= [x^2 - 2x - 8] / [x + 2]

I know that the top tier is a polynomial of degree 2 and the bottom tier is a polynomial of degree 1. By definition, a rational function is a polynomial divided by a polynomial where the lower tier cannot equal zero, which is what I seem to have here.

Since the function is rational, I do not have to give the degree, but can rational functions have a degree? Yes or no? If so, how would I calculate it.

Thanks,
Jack Daniels

The difference between the degrees of numerator and denominator.

So is this a rational function, or are you saying this is a polynomial?

It isn't a polynomial because a polynomial's domain is all real numbers, while the stated function is undefined at $ x \equal{} \minus{}2$.


It's a well known fact from Quantum Mechanics that electrons exist in fixed, quantised energy states. Also, when they move from one energy level to another, they emit or absorb the energy as required by $ \triangle E$ . Also, if a particular photon is absorbed , the apprpriate transition takes place. My question is regarding the second phenomenon. Suppose we consider a $H$ atom s.t. the energy difference between $1^{st} $ and $ 2^{nd} $ energy levels is $ 13.6 eV$ and theat between 2nd and 3rd is $ 3.4 eV$ . What if we give Energy =
(i} 13.7 eV
(ii)13.3 eV
(iii) 4 eV... etc..??
Do electrons jump and then come back if the energy difference is not appropraite or do they just stay in position and don't jump if the PRECISE Energy is not provided???

I think that electron can only absorb a photon with the right energy, so if you provide a 13.7 eV photon it won't be absorbed. I wonder how close to the right energy does the energy of photon has to be though... I mean - what is "exactly" in physics?

Well, I don't feel like competent to answer this, but even if the energy would have to be precisely the difference of states, a photon with larger energy could think of a lot of different ways how to dispose of the redundant energy, for example to give the electron a push (even though this has to comply with the momentum laws also (I think this can be observed in practice also - free atoms have slightly different energy states than those in a crystal lattice also because they take up more energy in the collision)) or emit another photon, or I don't know what. And I guess there is also some quantum uncertainity to this ... (?;-)

I was thinking on the same lines as kubus. It seems quite unrealistic that while there is a great deal of uncertainity in the Quantum world, the electrons shouls be so fastidious to accept the photon of the exact energy(or even approx. exact energy) to undergo transition. MAy be something else happens to the extra energy?!?!

Well, it can't be really of any energy, only of an energy close to the energy needed for the jump. That's how absorbtion lines are created - only photons with certain energies are absorbed by the <name your favorite gas> atoms and then reemitted in random direction so that people doing gas spectroscopy see blanks in the spectrum when they look throught this gas and can tell what gas it was. Photons of other energies are supposed to "pass right on by".
However, what I really think is that absorbtion of a photon with wrong energy is just much less probable than of that with energy close to $\triangle E$ and there is probably some way to get rid of / gain the extra energy other than the electron jump.


algebra
polynomial
quadratics
complex numbers
AMC

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

$ \textbf{(A)} \ \frac {1 \plus{} i \sqrt {11}}{2} \qquad \textbf{(B)} \ \frac {1 \plus{} i}{2} \qquad \textbf{(C)} \ \frac {1}{2} \plus{} i \qquad \textbf{(D)} \ 1 \plus{} \frac {i}{2} \qquad \textbf{(E)} \ \frac {1 \plus{} i \sqrt {13}}{2}$

Wow... this is a #23!!!
quick check with TI-89 gives [hide]$\boxed A$[/hide]

Show us instead of saying "I got this from calculator."

[hide]Since the polynomial has integer coefficients, and the other two roots are integer coefficients, than the quadratic formed by the two complex roots must also have integer coefficients to gaurantee the original polynomial has integer coefficients. Since they are integer coefficents, they are real and thus the two roots are complex conjugates. The sum of the roots (the coefficient of the x term) for any of the possible choices is always an integer, so we must look at the product of the roots. The product of A and the complex conjugate of A is an integer, so that is the one that we're looking for.[/hide]

[hide="solution"]Denote the polynomial $P(x)$.  If $P$ has exactly two roots, then by the FTA it must be of the form $P(x) = (x - a)^2(x-b)^2$.  Evidently $a$ and $b$ are a conjugate pair (as otherwise $P$ wouldn't have real coefficients), so $b = \overline{a}$.  In addition, if we want the coefficients to be integers, then $ab = a\overline{a}$ must be an integer.  Testing the complex numbers given, we see that only $(A)$ provides such a root.  Indeed, if we let $a = \frac{1 + i \sqrt{11}}{2}$, then $P(x) = (x - a)^2(x- \overline{a})^2 = x^4 - 2x^3 +7x^2 - 6x + 9$[/hide]

[hide]we know that the quadratic that gives the complex roots has integer coeffients, $x^2+bx+c$, $b$ is the negative sum of the roots, and all of the choices give an integer for b, and c is the product of the roots, and only $\boxed{A}$ gives an integer for c[/hide]

Of the given answer choices, only $\boxed{\textbf{(A)}}$ is an algebraic integer; indeed,
\[
\mathcal O_{\mathbb Q(\sqrt{-11})} = \mathbb Z\left[\frac{1+ \sqrt{-11}}2\right]
\]
since $-11\equiv 1\pmod 4$.  (The other choices fail because $\mathcal O_{\mathbb Q(i)} = \mathbb Z[i]$ and $\mathcal O_{\mathbb Q(\sqrt{-13})} = \mathbb Z[\sqrt{-13}]$.)


David ordered 4 black T-shirts and some white T-shirts. The price of one black T-shirt was twice that of each white T-shirt. When David received the order, it was found that the number of T-shirts of each color had been interchanged. This increased the bill by 50$ \%$. What is the ratio of the number of black T-shirts to the number of white T-shirts in the original, correct order David placed? Express your answer as a common fraction.

Let $ p$ be the price of a white T-shirt, and $ y$ the number of white T-shirts David ordered.
If there hadn't been any mistakes, David should have paid $ 8p\plus{}yp$. But there was a mistake so David paid $ 2yp\plus{}4p$.

So $ 2yp\plus{}4p \equal{} \frac{3}{2} (8p\plus{}yp)$, dividing by $ p$ yields $ 2y\plus{}4 \equal{} \frac{3}{2} (8\plus{}y)$, so $ 4y\plus{}8 \equal{} 3(8\plus{}y)$, $ 4y\plus{}8 \equal{} 24\plus{}3y$, $ y \equal{} 16$.

So the answer is $ \frac{4}{16} \equal{} \frac{1}{4}$.


Vieta
linear algebra
matrix
linear algebra theorems

There is such an identity

$\det(\lambda I_{n}-A)=\lambda^{n}-a_{n-1}\lambda^{n-1}+a_{n-2}\lambda^{n-2}+\ldots (-1)^{n}a_{0}$

for all $n$ by $n$ matrices $A$. It is easy to show that

$a_{0}=\lambda_{1}\lambda_{2}...\lambda_{n}$

$a_{1}=\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}+\ldots\lambda_{2}\lambda_{3}\cdots\lambda_{n}$

:

$a_{n-2}=\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\ldots \lambda_{n-1}\lambda_{n}$

$a_{n-1}=\lambda_{1}+\lambda_{2}+\ldots\lambda_{n}$

(where $\lambda_{1}\ldots\lambda_{n}$ are the eigenvalues of $A$) by Vieta's rule

How could we prove that

$a_{0}=\det(A)$

$a_{1}= \det\left[ \begin{array}{cccc}A_{1,1}&A_{1,2}&\ldots &A_{1,n-1}\\ A_{2,1}&A_{2,2}&\ldots&A_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots \\ A_{n-1,1}&A_{n-1,2}&\ldots&A_{n-1,n-1}\end{array}\right]+\ldots+\det\left[\begin{array}{cccc}A_{2,2}&A_{2,3}&\ldots &A_{2,n}\\ A_{3,2}&A_{3,3}&\ldots&A_{3,n}\\ \vdots&\vdots&\ddots&\vdots \\ A_{n,2}&A_{n,3}&\ldots&A_{n,n}\end{array}\right]$

:

$a_{n-2}=\det\left[ \begin{array}{cc}A_{1,1}&A_{1,2}\\ A_{2,1}&A_{2,2}\end{array}\right]+\ldots+\det\left[\begin{array}{cc}A_{1,1}&A_{1,n}\\ A_{n,1}&A_{n,n}\end{array}\right]+\ldots \det\left[\begin{array}{cc}A_{n-1,n-1}&A_{n-1,n}\\ A_{n,n-1}&A_{n,n}\end{array}\right]$

$a_{n-1}={\rm Tr}(A)$

 :?:

The proof is aldreay done by myself and is simple. I just want to see if anybody have a better solution

Note that

${\rm Tr}^{k}(A)=\det\left[\begin{array}{ccccc}A_{1,1}&A_{1,2}&\ldots&A_{1,k-1}&A_{1,k}\\ A_{2,1}&A_{2,2}&\ldots&A_{2,k-1}&A_{2,k}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ A_{k-1,1}&A_{k-1,2}&\ldots&A_{k-1,k-1}&A_{k-1,k}\\ A_{k,1}&A_{k,2}&\ldots&A_{k,k-1}&A_{k,k}\end{array}\right]+\ldots \det\left[\begin{array}{ccccc}A_{1,1}&A_{1,2}&\ldots&A_{1,k-1}&A_{1,k+1}\\ A_{2,1}&A_{2,2}&\ldots&A_{2,k-1}&A_{2,k+1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ A_{k-1,1}&A_{k-1,2}&\ldots&A_{k-1,k-1}&A_{k-1,k+1}\\ A_{k+1,1}&A_{k+1,2}&\ldots&A_{k+1,k-1}&A_{k+1,k+1}\end{array}\right]$

$+\det\left[\begin{array}{ccccc}A_{n-k+1,n-k+1}&A_{n-k+1,n-k+2}&\ldots&A_{n-k+1,n-1}&A_{n-k+1,n}\\ A_{n-k+2,n-k+1}&A_{n-k+2,n-k+2}&\ldots&A_{n-k+2,n-1}&A_{n-k+2,n}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ A_{n-1,n-k+1}&A_{n-1,n-k+2}&\ldots&A_{n-1,n-1}&A_{n-1,n}\\ A_{n,n-k+1}& A_{n,n-k+2}&\ldots&A_{n,n-1}&A_{n,n}\end{array}\right]$

is known as the "generalized $k^{th}$ trace" of the $n\times n$ matrix $A$    :D

Can any one help me with proving the

$ a_{0} \equal{} \det(A)$

$ a_{1} \equal{} \det\left[ \begin{array}{cccc}A_{1,1} & A_{1,2} & \ldots & A_{1,n \minus{} 1} \\
A_{2,1} & A_{2,2} & \ldots & A_{2,n \minus{} 1} \\
\vdots & \vdots & \ddots & \vdots \\
A_{n \minus{} 1,1} & A_{n \minus{} 1,2} & \ldots & A_{n \minus{} 1,n \minus{} 1}\end{array}\right] \plus{} \ldots \plus{} \det\left[\begin{array}{cccc}A_{2,2} & A_{2,3} & \ldots & A_{2,n} \\
A_{3,2} & A_{3,3} & \ldots & A_{3,n} \\
\vdots & \vdots & \ddots & \vdots \\
A_{n,2} & A_{n,3} & \ldots & A_{n,n}\end{array}\right]$

:

$ a_{n \minus{} 2} \equal{} \det\left[ \begin{array}{cc}A_{1,1} & A_{1,2} \\
A_{2,1} & A_{2,2}\end{array}\right] \plus{} \ldots \plus{} \det\left[\begin{array}{cc}A_{1,1} & A_{1,n} \\
A_{n,1} & A_{n,n}\end{array}\right] \plus{} \ldots \det\left[\begin{array}{cc}A_{n \minus{} 1,n \minus{} 1} & A_{n \minus{} 1,n} \\
A_{n,n \minus{} 1} & A_{n,n}\end{array}\right]$

$ a_{n \minus{} 1} \equal{} {\rm Tr}(A)$

identities?

Very interesting material. I just took an introductory course in linear algebra, and found it to be fascinating. We learned that a0 was the product of the eigenvalues and that a(n-1) was the sum of the eigenvalues, but never learned about the intermediate values. However, I deduced the values of a1, a2, a3, etc. simply by multiplication of binomials. Cool stuff, though.


[quote="pkerichang"]are there any advantages of sending solutions in early??? such as you get your score back earlier, etc...[/quote]
There are no advantages to sending in your solutions early, except that it might make things a little easier on the fax machine operators on the date of the postmark deadline. ;)

[quote="chess64"]oooh! Off-topic, but on the USAMTS website... "Round 1 Scores currently being uploaded."

YAY! :D[/quote]

Where?

[quote="towersfreak2006"]It says that on the site reached after clicking on the My USAMTS portion of the sidebar on usamts.org, if you are logged in. I don't think it's too meaningful at the moment, since that has been the same status since Friday. I recall DPatrick saying that scores can usually be entered on the same day, so we can probably expect results by tomorrow. (Or I am being too optimistic?)

I think the only advantage of turning solutions in early is a quick notification of having received your solutions. Depending on the type of person you are, that in itself may be worth it.[/quote]

They should be up today.

[quote="pkerichang"]are there any advantages of sending solutions in early??? such as you get your score back earlier, etc...[/quote]No.  We grade them all at once -- if you send it early it will just sit in our office (or in our email inbox) until we get ready to grade them.

My scores still aren't up... :(
How about you guys?

Yay! I just saw it! :) 18.... pretty good in my opinion :D

I got a 19. I was hoping I could use it to qualify for AIME. Gotta work hard if I a still to have a chance.  ;)

I got a commended solution for number 4 but my name wasn't on the list of commended people. I turned in the form, but maybe there was a mix-up or something.

22 here.  5 5 2 5 5.  I didn't expect to really get anything on 3..more expecting some critique, ie was i going in the right direction etc.  Instead, I got a score of 2, and the critique "Good try!".  Hmm...meh.
Well, having solved the others with confidence in the first few days, I worked on the other solutions to make them really really good.  Just wanted to make them very thorough and clear.
I've done that for round 2 now as well, but if I don't get scores back early from sending solutions in early, I'll just hold on to them for now, just in case I want to change something at some point before now and the deadline...


What is MAO?
I see this test frequently in AoPS Volume 1. :(  :(

This is $ \text{MA}\Theta$, or Mu Alpha Theta.  Look [url=http://www.artofproblemsolving.com/Wiki/index.php/Mu_Alpha_Theta]here[/url] and [url=http://www.mualphatheta.org/]here[/url].  I have never actually taken the test, so I can't tell you what it's like.

One huge In My Opinion before all of this:

I do not like the MAO style of testing. Without me trashing any individuals too much, there is not a lot of review going into MAO tests, so answers are often wrong (about 2-3 key changes per test) and even then the questions are more formula based and less interesting. They favor the student that can remember what certain formulas are and those that can add correctly over those that understand what they are doing.

As a Florida student, I am heavily into MAO seeing as I have MAO competitions 8 times a year :)
My opinions are nearly the same as monkeythyme, the tests in the past have been more formula based, but more recently, test writing has gone to the students who have recently graduated from high school and their tests are a lot more thinking/concepts than just formula. All in all, MAO is a fun competition that I highly enoy :)


You play the following game. You throw a six-sided fair die repeatedly. You may choose to stop after any throw, except that you must stop if you throw a 1. Your score is the number obtained on your last throw. Determine the strategy that you should adopt in order to maximize your expected score.

I surmise all the possible outcomes are equally likely (that is - getting a 1,2,3 .. or 6), so, if I do manage to carve out a strategy to maximize my expected value, and then if I roll the fair die, what if I got a 1 on the very first roll of the die?

How can I expect my strategy to work in a scenario like the one stated above?

That's irrelevant to maximizing your [b]expected[/b] score, which occurs over all possible games.

This is [url=http://www.admissionstests.cambridgeassessment.org.uk/adt/step]STEP[/url] 1999 II, Question 14

[hide]
Your tactic will essentially be "stop rolling if greater than a given number", this is because each result is independent on the previous ones, so you must take a decision based on what you have rolled. The probability of obtaining any of the scores in the "stop" range is the same (symmetry, relabelling, etc). Therefore, considering cases we have:

Stop after first:
$ \displaystyle \frac {1}{6}(1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6) \equal{} \frac {7}{2}$

Stop if larger than 2
$ \displaystyle \frac {1}{5}(1 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6) \equal{} \frac {19}{5}$

Stop if larger than 3
$ \displaystyle \frac {1}{4}(1 \plus{} 4 \plus{} 5 \plus{} 6) \equal{} 4$

Stop if larger than 4
$ \displaystyle \frac {1}{3}(1 \plus{} 5 \plus{} 6) \equal{} 4$

Stop if larger than 5
$ \displaystyle \frac {1}{2}(1 \plus{} 6) \equal{} \frac {7}{2}$

Therefore we should stop if we get greater than 3. (or 4, I guess we should consider variances)[/hide]

SimonM, how do you consider variances here?

By the way, are you writing STEP II and III?


group theory
superior algebra
superior algebra solved

It is well known that
 - the product of two infinite cylic groups is not a cyclic group;
 - the product of two cyclic groups of order $m$ and $n$ is cylic if and only if $(m,n)=1$.

What about the product of a finite cylic group and an infinite cylic group?

An infinite cyclic group contains no notrivial elements of finite order. The direct product of a finite group and an infinite group contains elements of finite order.

[quote="liyi"]What about the product of a finite cylic group and an infinite cylic group?[/quote]

It is for sure not cyclic!


what is the absolute value of $ 2\sqrt{34}\plus{}15i$?

[hide="Solution"]The absolute value of any complex number such that $ z \equal{} x \plus{} iy$ is also the value of its distance as a point $ (x,y)$ from the origin in the complex plane. A way to evaluate this is if the complex number $ z$ is given in the form $ z \equal{} x \plus{} iy$, then its absolute value is $ \sqrt {x^2 \plus{} y^2}$.
The absolute value of the complex number $ 2\sqrt {34} \plus{} 15i$ is $ \sqrt {(2\sqrt34)^2 \plus{} 15^2} \equal{} \sqrt {361}$, which simplifies to $ 19$.[/hide]


FTW
MATHCOUNTS
geometry
similar triangles

The only reason I'm posting this topic is because I want to know HOW FTW has helped you excel in Countdown.

For me, FTW has made me confident in Countdown and slightly faster. I am barely nervous in countdown and I now always feel excited that I'm going against people that are better than me.

Now about the bad things that FTW has influenced you on:

I personally think that FTW was an ingenious idea and it would attract a lot of people to AoPS. But here are some thoughts that I think that other people will think that is bad about FTW:

"I think there are too many multiusers!"
"IM LOSING TOO MUCH RATING!!!!"
"It's too hard! I don't want to go on FTW anymore..."

Levans stated that he wanted the people that lost a lot to keep playing. But the bad thing about this is that the people that lose a lot lose interest in FTW and just leaves AoPS forever. FTW is a great attraction. But how can we make the people that only go on FTW to get to post on the Forums more and stay on AoPS?

I've asked a lot but you don't need to answer all of them...  :wink:

hmm...

i don't know if this is a good example and i'm not trying to brag.

at chapter mathcounts, i got last place in countdown. after chapter i joined aops. then at state mathcounts, i got first place in countdown. go figure

For the people who lose a lot of rating, I would suggest that FTW is a learning tool... It isn't about who has the highest rating.

Instead, you should try to focus on finding the fast way to solve the problem. IF you do this then you can benefit from FTW.

This is exactly what I did when I first started playing. After each game I would review the problems and solve them on my own.

I used to never get any questions right on countdown. This year, I made it through the second round!

In 6th grade I won 0 cds at school and regionals. In 8th grade, the only cd I lost was to Bobby Shen.

[quote="akhil0422"]Though I'm on the forums a lot, I NEVER get on FTW or Alcumus. When I do get on FTW, it's only to chat.  :D

FTW is way too hard for me.  :([/quote]

But I said that that's what levans is trying to achieve! He wants the not so good players (no offense) to keep playing!

And that's not braggin yaofan. That happened to me. I'm oooookkkkk at written so I almost always get in the top 10. (nvm. always.) And at a mock competition, I was so scared I couldn't write because I was quivering so hard.

But then FTW and AoPS showed up. Hallelujah.

I think in general, just thinking about math for more time helps a lot. I first heard about MathCounts in January (and started doing math outside of school this year). This year at chapter in February, I didn't even know about similar triangles and the triangle area formula. Luckily, I qualified, even though I only had a 27. Next, at state, I got fifth place. Then, I found AoPS. Now I can do almost all the problems on MathCounts tests. I think that the most important thing is motivation. Getting one place away from qualifying in State really motivated me.

To answer the question "how has FTW helped?" I will say that it just makes you more experienced. (Even though I just discovered FTW a few weeks ago) I can say that the more problems you see, the more connections you can make between problems. It is like excersizing. If you exercise everyday and do the right things, you will become stronger and fitter. The same goes with math.

Well, FTW helped me solve problems faster and guess better.

It made me more interested to learn math because I have such a low rating

[quote="Mukherjee"]I am new to FTW(a regular loser) and I must say it is good.It makes me better.And I also agree with Midnight Moon.Losing makes you think .[/quote]

Yeah, and since I am indian everyone expects high of me!

[quote="professordad"]When I was in sixth grade and completely unaware of AoPS, let alone AoPS FTW, I got 63rd in chapter.  When I was in seventh grade and an experienced AoPS FTW user, I made it to the countdown round!  16th!  And 22nd in state.  A major improvement.  Now, with even more FTW use, I am not afraid of very talented people, and am eager to go head-to-head.  I can answer most questions within 10 seconds.  Thank you, For the Win.[/quote]

You got 63rd in chapter.  That's funny.  I got 63rd in state!(in 6th grade)

This year, as a seventh grader, I WILL DO BETTER.

After all, before AoPS, I still got first in chapter cd, but my chapter is kind of a joke so this is like what cd was like:

Moderator:  What is the value of 2,345+3,567
My Opponent:  uhh...........5,000?
Me(40 sec. later): 5,912

Moderator:  What is the area of a right triangle  triangle with sides of length 3,4, and 5?
My Opponent:  I know!  12
Maybach did not answer

Moderator:  What is the value of 1/2 + 1/3 + 1/5?
No one answered.  Maybach went on to the next round by getting 1 question right.  Whoohoo!

The championship round took 12 questions for me to get 3 questions right because me and my oppoonent were slow and careless.....but I still won 3-1.

And then I got something like 25/46 in chapter and it was still good enough for first written.

My goals for next year would be to get in the top 4 state so I can make nats.

And it seems like there are a bunch of AoPS'ers who get 5th at MC States.  And that really stinks.

But now, I can average 25 on MC Chapter Sprints, 6 on Chapter Targets, 23 on State Sprints, and 5 on State Targets.  Not perfect, but it's improving.

And even though I have a really fail rating(1160ish because of a bunch of 3-4 games)  I still get a kick out of playing and I can improve my reaction time. :D

hmm.. i just noticed this may help because FTW helps you remember questions(i do this unintentionally). I'ma going to try to get 1st in CD.

[sarcasm]Yes that is totally useful because you will be able to mem problems in cd[/sarcasm]

I have only seen one problem ever that I could have memmed (and that is probably >50 rl cds), and that is because the very nature of the problem made it such that it only worked with the numbers that it did.

Aha, my mind changed the meaning of memming formulas and problems.


algebra
difference of squares
special factorizations

well you see a pattern
when you subtract lets say $ 1^2\minus{}2^2\equal{}3$ and 3 is 1 more than 2
so all the way till $ 1998$ the sum is $ 1999$
but you still have to add $ 1999\plus{}1999^2$$ \equal{}1999$x$ 2000\equal{}3,998,000$

[quote="calculatorwiz"]well you see a pattern
when you subtract lets say $ 1^2 \minus{} 2^2 \equal{} 3$ and 3 is 1 more than 2
so all the way till $ 1998$ the sum is $ 1999$
but you still have to add $ 1999 \plus{} 1999^2$$ \equal{} 1999$x$ 2000 \equal{} 3,998,000$[/quote]That's incorrect... $ 1^2\minus{}2^2\equal{}\minus{}3$...

[quote="Stokes93"]I'm really bad with sigma notation (I'm assuming you need it for this problem) - so please explain all your steps!

Evaluate:

$ 1^2 \minus{} 2^2 \plus{} 3^2 \minus{} 4^2 \plus{} ... \minus{} 1998^2 \plus{} 1999^2$[/quote]

From difference of squares, it equals $ 1\plus{}2\plus{}3\plus{}\cdots\plus{}1999\equal{}1999\cdot2000\div2\equal{}1999000$

Don't need sum, but it is $ \sum_{k\equal{}1}^{n}k^2$


geometry
rectangle
inequalities open
inequalities

By  $ d(A,B) $  let us denote the euclidean distance , e.g. in  $ {\mathbb  R}^2. $  Suppose that in each village there is only  one triplet  (C,T,S), where :
      C= a church
      T= a townhall
      S= a school-house.
Let us fixed a triplet $ (p,q,r) $ of positive numbers  and identify the village as a  set in  $ {\mathbb R}^2. $
Someone intend to establish a restaurant (pub) in  a village.  In order to keep within the law,  the restaurant (=R) must satisfy conditions :
\[     d(R,C) > p  \; ,\; \; d(R,T) > q \; \;  \; \mbox{\rm and}\; \; \;  d(R,S)>r       .
\]
[b]Question:[/b] [i]Find the largest area of a village in      which it is not possible to  establish a restaurant.[/i]
Perhaps consider that the village is a  convex region. (  Try to generalize such a problem  .)
The problem is sugested by an old ,,open" problem proposed many years ago by Ion Ionescu in ,,Gazeta Matematica "-Problem 1297, time-period= $\approx $ 1910-1915.  He  asked for a solution when the village is an rectangle (in that case it's possible to give a solution ).

one should say that the village is necessarily convex: if it is not, just take a triangle CTS with sides r+q,p+r,p+q(maybe the ordering is wrong) and take the village to be union of circles around C,T,S with radius p,q,r respectively. that village cannot have a restaurant. if a village has larger area than $\pi (p^2+q^2+r^2)$, then there is some area which is none of the circles around C,T,S with radius p,q,r. somewhere in that area, we can build our restaurant.
if our village is necessarily convex, the problem actually seems hard.

Peter


Well, I was writing a text and then I started to use

[code]\begin{enumerate}
	\item
\end{enumerate}[/code]

the thing is that I can't move to the right the items, they're a bit sticked to the left.

Edited: I know what's happening, I did \item{text} but if I put \item{a)}, then the item is not so sticked to the left, any way, is the any solution if I want to use text inside the item?

I don't understand what you are doing. The enumerate environment is for numbered lists (so it makes no sense to use a)) and itemize is for bulleted lists where you can change the bullets to anything else.
See attached pdf for code and output. See also [[LaTeX:Layout#Lists]] and [url=http://www.tug.org/tutorials/tugindia/chap04-prn.pdf]Lists, etc[/url]


Let $ a_1, a_2, \cdots, a_n, b_1, b_2, \cdots, b_n$ be real numbers such that $ a_i \ge b_i$ for all $ 1\le i \le n$. Rearrange the sequence of numbers $ a_1, a_2, \cdots, a_n$ in monotonically increasing order. Rearrange the numbers $ b_1, b_2, \cdots, b_n$ in order too. Then prove that it still holds that $ a_i \ge b_i$ for all $ 1\le i \le n$.

I think this would more appropriate in Pre-Olympiad section.

WLOG $ a_{i}$ are arranged in increasing order (otherwise just switch them around until they are). Suppose the permutation of $ b_{i}$ where they are in increasing order, is : $ c_{1} \leq c_{2} \leq c_{3} ... \leq c_{n}$. Let $ b_{i} \equal{} c_{\sigma(i) }$. Then we know that $ a_{i} \geq c_{\sigma(i) }$, and we want to prove to prove $ a_{i} \geq c_{i}$. Since $ a_{i} \geq a_{j} \geq c_{\sigma(j) }$ for all $ j \leq i$, we have $ a_{i} \geq c_{\sigma(j)}$ for $ j \equal{} 1,2... i$. But for some $ j \equal{} t$, $ t \leq i$, $ c_{\sigma(t)} \geq c_{i}$ (since there are at most $ i \minus{} 1$ numbers $ c_{k}$ which are strictly smaller than $ c_{i}$, while we have $ i$ numbers here: $ c_{\sigma (j)}$ for $ j \equal{} 1,2... i$). This means $ a_{i} \geq c_{\sigma(t)} \geq c_{i}$, so $ a_{i} \geq c_{i}$, as required.

I give effectively the same solution with a slight algorithmic style how I proved it. Let $ a_i$ and $ b_j$ ($ i$ may or may not be equal to $ j$) be the least elements among $ \{a_n\}$ and $ \{b_n\}$ respectively. Obviously $ a_i < b_j$ else $ \exists a_k\neq a_i$ with $ a_k < b_j$ contradicting the minimality of $ a_i$.

Remove $ a_i$ and $ b_j$. We can go on removing  the pairs of least numbers from here on that takes finite number of steps.


Post here or PM me when you are ready to play. The initial rules are below, as well as the order of the players. This section will be updated whenever a rule is added.

math92 - -7
Hexaditidom - 11
k2c901_1 - 14
Doubtable - -4
paladin8 - 9
WindSlicer - 11
L_Li - 4
falconwing64 - 0

IMMUTABLE RULES

101. All players must always abide by all the rules then in effect, in the form in which they are then in effect.

102. A rule-change is any of the following: (1) the enactment, repeal, or amendment of a mutable rule; (2) the enactment, repeal, or amendment of an amendment of a mutable rule; or (3) the transmutation of an immutable rule into a mutable rule or vice versa.

103. All rule-changes proposed in the proper way shall be voted on. They will be adopted if and only if they receive the required number of votes.

104. Every player is an eligible voter. Every eligible voter must vote within 2 days of the proposal he/she is voting on.

105. No rule-change may take effect earlier than the moment of the completion of the vote that adopted it, even if its wording explicitly states otherwise. No rule-change may have retroactive application.

106. Each proposed rule-change shall be given a number for reference. The numbers shall begin with 301, and each rule-change proposed in the proper way shall receive the next successive integer, whether or not the proposal is adopted. If a rule is repealed and reenacted, it receives the number of the proposal to reenact it. If a rule is amended or transmuted, it receives the number of the proposal to amend or transmute it. If an amendment is amended or repealed, the entire rule of which it is a part of receives the number of the proposal to amend or repeal the amendment.

107. Rule-changes that transmute immutable rules into mutable rules may be adopted if and only if the vote is unanimous among the eligible voters.

108. In a conflict between a mutable and an immutable rule, the immutable rule takes precedence and the mutable rule shall be entirely void.

109. If a rule-change as proposed is unclear, ambiguous, paradoxical, or otherwise of questionable value, then the other players may suggest amendments or argue against the proposal before the vote. The proponent decides the final form in which the proposal is to be voted on and also decides the time to end debate and vote.

110. The state of affairs that constitutes winning may not be altered from achieving n points to any other state of affairs. Rules that establish a winner when play cannot continue may be enacted and (while they are mutable) be amended or repealed.

111. A player always has the option to forfeit the game rather than continue to play or incur a game penalty. No penalty worse than losing, in the judgment of the player to incur it, may be imposed.

112. There must always be at least one mutable rule. The adoption of rule-changes must never become completely impermissible.

113. Rule-changes that amend or repeal their own authority are permissible. No rule-change or type of move is impermissible solely on account of the self-reference or self-application of a rule.

114. Changing the rules is permitted only when a rule or set of rules explicitly or implicitly permits it.

MUTABLE RULES

201. Players shall alternate turns in a fixed order which will be randomly generated before the start of the game. Players shall take one whole turn apiece. Turns may not be skipped or passed, and parts of turns may not be omitted.

202. All players begin with zero points.

203. One turn consists of proposing one rule-change and having it voted on. At the end of the turn, $\lfloor{(P-291)\cdot f}$ points are added to the player's score, where $P$ is the proposal number and $f$ is the proportion of favorable votes received.

204. [Amended by rule 302]

205. Players who vote against winning proposals shall receive 10 points each.

206. An adopted rule-change takes full effect at the moment of the completion of the vote that adopted it.

207. When a proposed rule-change is defeated, the player who proposed it loses 10 points.

208. Each player has exactly one vote.

209. [Amended by rule 305]

210. At no time may there be more than 25 mutable rules.

211. Players may not conspire on the making of future rule-changes unless they are teammates.

212. If two or more mutable rules, or two or more immutable rules, conflict with each other, then the rule with the lowest ordinal number takes precedence. If at least one of the rules in conflict explicitly says that it defers to another rule or takes precedence over another rule, then such provisions shall supersede the numerical method for determining precedence. If two or more rules claim to take precedence over one another or to defer to one another, then the numerical method again governs.

213. If players disagree about the legality of a move or interpretation of a rule, then the player preceding the one moving is to be the Judge and decide the question. This process is called invoking Judgment. When Judgment has been invoked, the next player may not begin his or her turn without the consent of a majority of the other players. The Judge's Judgment may be overruled only by a unanimous vote of the other players taken before the next turn is begun, or by the Moderator (jmadsen). If a Judge's Judgment is overruled, then the player preceding the Judge in the playing order becomes the new Judge, and so on, except that no player is to be Judge during his or her own turn or during the turn of a teammate. Unless a Judge is overruled, one Judge settles all questions arising from the game until the next turn is begun, including questions as to his or her own legitimacy and jurisdiction as Judge.

214. If the rules are changed so that further play is impossible, or if the legality of a move cannot be determined with finality, or if by the Judge's best reasoning, not overruled, a move appears equally legal and illegal, then the first player unable to complete a turn is the winner. This rule takes precedence over every other rule determining the winner.

302. A rule-change is adopted if and only if the vote is supported by a simple majority of voters who voted before the 2-day deadline.

305. The winner is the first player to achieve 100 points.

306. Any player to vote for a winning proposal receives 5 points.

Umm...i would play too...but the rules spin my head around... :P

i am ready to play....so confused...  :stink:

jli and falconwing64 have been added, and chess64 said he couldn't play. I am still waiting for GoBraves and baseballfreak2. Since math92 is the first player, the game will start when he makes a proposal.

EVERYONE MUST VOTE BY PM, NOT BY POSTING.

[quote="Hexaditidom"]Sure, ready.

Wondering- Why is it called Nomic?[/quote]

it's in one of hofstadter's books, mimicking an (actual) system of law, I think

[quote="math92"]okay, here goes something:
Players who vote against winning proposals shall receive 10 points each.[/quote]

I'm dumb, someone help me out here. I see that this is the same as rule 205. What are we trying to do here-- repeal it, amend it, what?

Proposal 305:
Amend Rule 209 to:

"The winner is the first player to achieve 100 points."

This game will take too long... as you can see we only have one person with a positive score right now.

Hmmm. According to Rule 205, 2 of us must receive 10 points each.

Proposal 306:

"Every player to vote for a winning proposal gains 5 points."

Proposal 306
Accepted (3-1)
Score: +11
Also, k2c901_1 gets 10 points for voting nay.
Those who voted aye do not get 5 points because no retroactive rules.

proposal 307:

modify mutable rule 208 to:
Each player has one vote for each proposal, except for the person who made the proposal, who has zero votes for that proposal.

(to balance out the growing pts possible per proposal + 5 bonus from that last rule proposed)

[i]Edit:

Let's see how clear I can make this. If there is another post by you with more than a hundred of the same smileys and nothing else, I'll ask for you to be banned. :) Okee...enjoy your stay Mr. Potato.[/i]

Proposal 307
Defeated 1-3
+4 -10
L_Li gets -6 points.

falconwing, if you do not post a proposal by tomorrow your turn will be skipped.

math92, you may make a proposal.
falconwing64, you may still make a proposal if you do so before math92.

EDIT: It looks like people have lost interest. I understand, as the game was going very slowly. I have never done this before, so I didn't know how it would work out. It seems like it would be better in real time.

Sorry about the triple post.

a pity, it started getting interesting what proposals where coming up

It seemed like a good idea, but it's way over my head at this moment.

Proposal 308: A person who submits a proposal that is passed gains 15 points.

If this is illegal, please don't blame me.  I really don't get this game.


Source: ACoPS
Find a formula for the sum $r+2r^2+3r^3+...+nr^n$, and generalize.

[hide="my approach"]
I first let the sum equals to $S$.
Then
$S=r+2r^2+3r^3+...+nr^n$
$\frac{S}{r}=1+2r+3r^2+...+nr^{n-1}$.
After that, I found $\frac{S}{r}-S$, and so on..
Is it correct? or is there any other way??
Thanks,
[/hide]

Wait, what are you asking?  I believe that way will work out, because you get a geometric series.

this general problem has been posted before...that is pretty much the way that you solve it, i suppose another way that is more work is to write it

$r$
$r^2+r^2$
$r^3+r^3+r^3$
$r^4+r^4+r^4+r^4$
...

then add the columns using geometric series, to get

$r*\frac{1-r^{n} }{1-r}+r^2*\frac{1-r^{n-1} }{1-r}+...+r^n*\frac{1-r}{1-r}$

$\frac{1}{1-r}*\left( r-r^{n+1}+r^2-r^{n+1}+r^3-r^{n+1}+...+r^n-r^{n+1}\right)$

$\frac{1}{1-r}*\left( r*\frac{1-r^n}{1-r}-n*r^{n+1}\right)$

$\frac{1}{(1-r)^2}*(r-r^{n+1}-n*r^{n+1}+n*r^{n+2})$

$\frac{r-(n+1)*r^{n+1}+n*r^{n+2}}{(1-r)^2}$

[hide="another way"]
$S=r+2r^2+3r^3+...+nr^n$
$rS=r^2+2r^3+...+(n-1)r^n+nr^{n+1}$

Subtract the two equations to get
$S-rS=r+r^2+r^3+...+r^n-nr^{n+1}$

Solving for $S$ gives $S=\frac{r(1-r^n)-n(1-r)r^{n+1}}{(1-r)^2}$, which could possibly be simplified more.  (not sure?)[/hide]


calculus
derivative
superior algebra
superior algebra unsolved

Let $ F$ a field with $ char(F) \equal{} p>0$. Let $ f(X)\equal{}X^p\minus{}a \in F[X]$

Suppose it doesn't split on F. So $ \not \exists \alpha \in F\ s.t.\ \alpha^p \equal{} a$. But we can say $ \alpha \in K,\ \exists K \geq F$.

So in K it is $ \alpha^p\equal{}a$. We can write in K[X]: $ f(X) \equal{} (X\minus{}\alpha)^p$ which is clearly to have only one root.

But right here I have a problem: if  I compute the formal derivative it is $ f'(X) \equal{} \minus{}a$ which has no zeros at all so the roots of f must be simple!
why is that?

$ f^{\prime}\left(X\right)$ is $ 0$, not $ \minus{}a$. ;)

  darij


limit
logarithms
function
algebra unsolved
algebra

For positive real $a$, we can use l'Hôpital's rule and
$\lim_{x\to 0}\frac{a^{x}-1}{x}=\lim_{x\to 0}\frac{a^{x}\ln a}{1}=\ln a$.
We can also use Taylor's series:
$\lim_{x\to 0}\frac{a^{x}-1}{x}=\lim_{x\to 0}\frac{\frac{1}{0!}+\frac{(x\ln a)}{1!}+\frac{(x\ln a)^{2}}{2!}+\dots-1}{x}=\ln a$

[quote="Amir.S"]Can anyone prove it without "[b]l'Hôpital's rule [/b]" and "[b]Taylor's series[/b]"?[/quote]
It can be prove without "Hopital's rule", but necessary Taylor's series for define $a^{x}, \ ln(a)$.

[quote="lordWings"]For positive real $a$, we can use l'Hôpital's rule and
$\lim_{x\to 0}\frac{a^{x}-1}{x}=\lim_{x\to 0}\frac{a^{x}\ln a}{1}=\ln a$.
...[/quote]

can any one send a proof for:
$(a^{x}-1)'=a^{x}\ln a$

[u]to rust[/u] :
can we define $\lim_{n\rightarrow \infty}(1+\frac{1}{n})^{n}=e$
then prove the problem without Taylor's series?

[quote="Amir.S"]can any one send a proof for:
$(a^{x}-1)'=a^{x}\ln a$[/quote]
How about this?
$(a^{x}-1)'=(a^{x})'=(e^{x\ln a})'=(e^{x\ln a})\ln a=a^{x}\ln a$

now , what's the proof for $(e^{u})'=e^{u}u'$? :wink:

[quote="Amir.S"]find:

$\lim_{x\rightarrow 0}\frac{a^{x}-1}{x}$[/quote]

If i am not wrong , such limit was found in this way :

Well naturally THEY  wanted to differentiate the function $f(x)=a^{x}$  which is \[\lim_{h \rightarrow 0}\frac{a^{x+h}-a^{x}}{h}= a^{x}\lim_{h \rightarrow 0}\frac{a^{h}-1}{h}\] Now for different $a$, this $\lim_{h \rightarrow 0}\frac{a^{h}-1}{h}$was turned out to be some weird constant (now we know its $ln(a)$)
So the natural question was when this $\lim$ will be one.indeed for $a=e$ ,we will have one for the limit , but to have an idea where this comes from Assume for $a=b$ the limit would be one, then \[\lim_{h \rightarrow 0}\frac{b^{h}-1}{h}=1\] so[i] Vaguely[/i] for a very small $h$ \[\frac{b^{h}-1}{h}\approx 1 \longrightarrow b \approx (1+h)^{\frac{1}{h}}\] Hence it was natural to think that \[b=\lim_{h \rightarrow 0}(1+h)^{\frac{1}{h}}\] Indeed this is true and we can prove it in a rigorous way.

Now if we accept that $\lim_{h \rightarrow 0 }\frac{e^{h}-1}{h}=1$ then \[\lim_{h \rightarrow 0}\frac{a^{h}-1}{h}= \lim_{h \rightarrow 0}\frac{e^{ln(a)*h}-1}{h}=\lim_{h \rightarrow 0}\frac{e^{ln(a)*h}-1}{ln(a)*h}*ln(a) = 1*ln(a)=ln(a)\] ( We just replaced $a= e^{log_{e}^{a}}=e^{ln(a)}$)

$\{\begin{array}{c}\lim_{x\to\infty}(1+\frac{1}{x})^{x}=e\\\\ \lim_{x\to e}\ln x=1\end{array}$ $\implies$ $\lim_{x\to\infty}\ln(1+\frac{1}{x})^{x}=1$ $\implies$ $\lim_{x\to\infty}x\ln(1+\frac{1}{x})=1$ $\implies$ $\boxed{\ \lim_{x\searrow 0}\frac{\ln (1+x)}{x}=1\ }\ .$ Analogously for $x\nearrow 0\ .$

$\{\begin{array}{c}\lim_{x\searrow 0}e^{x}=1\\\\ \lim_{x\searrow 0}\frac{\ln (1+x)}{x}=1\end{array}$ $\implies$ $\lim_{x\searrow 0}\frac{\ln [1+(e^{x}-1)]}{(e^{x}-1)}=1$ $\implies$ $\boxed{\ \lim_{x\searrow 0}\frac{e^{x}-1}{x}=1\ }\ .$ Analogously for $x\nearrow 0\ .$

$\lim_{x\to 0}\frac{a^{x}-1}{x}=\lim_{x\to 0}\frac{e^{x\ln a}-1}{x\ln a}\cdot\ln a=\ln a$ $\implies$ $\boxed{\ \lim_{x\to 0}\frac{a^{x}-1}{x}=\ln a\ }\ .$

Thus, must prove only $\lim_{x\to e}\ln x=1$ and $\lim_{x\to 0}e^{x}=1\ !$

[quote] \[x \nearrow 0\] [/quote]

 :D   and what do you mean by $\nearrow$,
you mean $x \rightarrow 0^{+}$

[quote="lomos_lupin"][quote]What do you mean by $\nearrow$,
you mean $x \rightarrow 0^{+}$[/quote][/quote]
" $x\rightarrow 0^{+}$ " means "$x$ tends from right hand to $0$ " ,  i.e. " $x\rightarrow 0\mathrm{\ and\ }x>0$ ".
" $x\searrow 0$ " means "$x$ tends decreasing to $0$ ", i.e. $x$ is an arbitrary decreasing sequence $x_{n}\searrow 0\ .$
Prove easily these notions are equivalently in this case, i.e. if $\lim_{x\searrow 0}f(x)$ and $\lim_{x\rightarrow 0^{+}}f(x)$ exist, then they are equally.


inequalities ARML geometry algebra polynomial function rectangle

When do you recognize when to apply the inequality $ \frac{a}{b}\plus{}\frac{b}{a}>\equal{}2$ in problems?

I'm wondering because I've seen the inequality used in two arml geometry problems dealing with maximizing a length or area but I didn't see any relationship between the two problems or how one would have thought to use that inequality.

Also in general, what methods are there to maximizes certain variables in geometry problems? How do maximization or minimization in geometry differ from other maximization or inequality problems?

Let P(x) be a polynomial with non-negative coefficients.

Find the minimum of P(x)+P(1/x) where x ranges through the positive reals, as a function of the coefficients of P(x).

That is actually a very specific case of the [[AM-GM]] Inequality, which is in turn a special case of the [[Power Mean Inequality]]* -- to obtain your inequality, apply AM-GM to the list {a/b, b/a}.

There's no dirth of AM-GM problems on AoPS -- searching returns pages and pages of results  :wink:

As for your particular inequality, a direct application would be -- [i]"a rectangle has area 1; find the minimum possible perimeter of the rectangle"[/i]

[i]* where the power-mean at an exponent of 0 is equivalent to the geometric mean[/i]


calculus
integration
trigonometry
real analysis
real analysis unsolved

Can you prove

$ \int_0^1 \int_0^1 s^{\alpha \minus{} 1} \, (1 \minus{} s)^{\beta \minus{} 1}\, t^{\alpha \minus{} 1}\, (1 \minus{} t)^{\beta \minus{} 1}\, |s \minus{} t|^{2\gamma}\, ds\, dt$

$ \equal{} \frac {\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha \plus{} \beta \plus{} \gamma)}\,\frac {\Gamma(\alpha \plus{} \gamma)\Gamma(\beta \plus{} \gamma)\Gamma(1 \plus{} 2\gamma)}{\Gamma(\alpha \plus{} \beta \plus{} 2\gamma)\Gamma(1 \plus{} \gamma)}$

for $ \text{Re}(\alpha),\text{Re} (\beta),\text{Re}(\gamma) > 0$

This is just  what I' thinking Now, not the solution ..
what about substitution $ s \equal{} \sin^{2} x$ and $ t \equal{} \sin^{2} y$?
I have not checked but I think there will be something more nice.
 for example  you will get something like this :
$ (\sin x \sin y )^{2\alpha \minus{} 1} (\cos y \cos x)^{2\beta \minus{} 1} |\sin (x \plus{} y) \sin (x \minus{} y)|^{2\gamma}$
Note that : $ \sin x \sin y \equal{} (\cos(x \minus{} y) \minus{} \cos (x \plus{} y))/2$  and $ \cos x \cos y \equal{} (\cos (x \minus{} y) \plus{} \cos (x \plus{} y))/2$
after substitution all  of this and also substitution $ x \minus{} y \equal{} p$ and $ x \plus{} y \equal{} q$ you will get Jacobian equal to $ 1/2$ So I think there will be something more nice than this one.

My first instinct was to introduce two new variables and have a quadruple integral involving $ x^{\alpha\minus{}1}e^{\minus{}x}y^{\beta\minus{}1}e^{\minus{}y}$ and $ z^{\alpha\minus{}1}e^{\minus{}z}w^{\beta\minus{}1}e^{\minus{}w}.$ That is, $ x\equal{}su,y\equal{}(1\minus{}s)u,z\equal{}tv,w\equal{}(1\minus{}t)v.$ I still need to figure out what to do with $ |s\minus{}t|$ and I need to ask why I'd do that at all - it needs plenty of work that I haven't done.


geometry
circumcircle
trigonometry
geometry unsolved

Let $ABC$ be an acute-angled triangle. The tangents to its circumcircle at
$A, B, C$ form a triangle $PQR$ with $C \in PQ$ and $B \in PR$. Let $C_{1}$ be the foot of the altitude from $C$ in $\Delta ABC$ . Prove that $CC_{1}$ bisects $\widehat{QC_{1}P}$ .

Hint: use trigonometric Ceva for a triangle $C_{1}BC$, point $P$ and for a triangle $C_{1}AC$, point Q. You'll get that $\tan \angle CC_{1}P =\tan \angle CC_{1}Q$. QED

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55918 or http://www.mathlinks.ro/Forum/viewtopic.php?t=59664 .

  darij


logarithms
calculus
integration
calculus computations

Ascertain the nature of the following series:

$ S = \sum_{n=2}^{+\infty}\frac{1}{(\ln n)^{\sqrt{\ln n}}}$.

For sufficiently large $ n,$ $ (\ln n)^{\sqrt{\ln n}}<n.$ Hence, this series diverges by comparison to the harmonic series.

We're trying to show that $ (\ln n)^{\sqrt{\ln n}}<n.$ Take the logarithm of both sides:

$ \sqrt{\ln n}\ln\ln n\stackrel{?}{<}\ln n$

$ \ln\ln n\stackrel{?}{<}\sqrt{\ln n}$

Letting $ u=\ln n,$ this is $ \ln u\stackrel{?}{<}\sqrt{u}$ which is true for sufficiently large $ u.$

Thanks Professor Merryfield. When I first solved this problem, I did something similar. First:

$ (\ln n)^{\sqrt{\ln n}}= e^{\sqrt{\ln n}\cdot \ln ln n}= n^{\frac{\ln \ln n}{\sqrt{\ln n}}}$.

But $ \frac{\ln \ln n}{\sqrt{\ln n}}< 1$ for $ n \ge 2$, because

$ \ln \ln n < \sqrt{\ln n}$

(let $ \ln n = t^{2}$)

$ \ln(t^{2}) < t$

$ \Rightarrow \ln t < \frac{t}{2}$, which is true for $ n \ge 2$. So, $ n^{\frac{\ln \ln n}{\sqrt{\ln n}}}< n$, and so the series diverges by comparision with the harmonic series. We could also use the Integral Test and the Cauchy Condensation Test.


superior algebra
superior algebra solved

How many elements that have an inverse are in $Z(\sqrt{2}, *)$ ?

infinitely many
consider $(3+2\sqrt{2})^n$ and $(3-2\sqrt{2})^n$


In a convex hexagon $ABCDEF$ triangles $ABC , CDE , EFA$ are similar. Find conditions on these triangles under which triangle $ACE$ is equilateral if and only if so is $BDF.$

See this :
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=138536[/url]

Hi guys!

The problem is not as easy as you may have thought. I solved it by using the new method of triangle super imposing of triangles ABC and CDE, then they must be equaled. I'm advised to not put a link here, and I don't know how to add graphics to this site. So I can not post the details.

Vo Duc Dien


i have what may seem to be a fundamental question.
i was doing a particular problem in aops vol. 2 and when reading the solution i saw that they went from
[b]$ (\dfrac{4}{3})^x\equal{}\dfrac{8\cdot\sqrt{3}}{9}$ from which we know that $ x\equal{}\dfrac{3}{2}$[/b] how did they just know that? it's not an obvious exponent like x^2 or x^3. how did they deduce that?

[quote="mihail911"]i have what may seem to be a fundamental question.
i was doing a particular problem in aops vol. 2 and when reading the solution i saw that they went from
[b]$ (\dfrac{4}{3})^x \equal{} \dfrac{8\cdot\sqrt {3}}{9}$ from which we know that $ x \equal{} \dfrac{3}{2}$[/b] how did they just know that? it's not an obvious exponent like x^2 or x^3. how did they deduce that?[/quote] divide both sides by $ \sqrt{3}$, and 8 is $ 4^\frac{3}{2}$, and $ 3\sqrt{3}\equal{}3^{3/2}$

I would write both sides in terms of powers of $ 2$ and $ 3$.  Specifically,

$ \left( \frac {4}{3} \right)^x \equal{} 2^{2x} \cdot 3^{ \minus{} x}$

and

$ \frac {8 \sqrt {3} }{9} \equal{} 2^3 \cdot 3^{ \minus{} \frac {3}{2} }$

from which we conclude that $ x \equal{} \frac {3}{2}$.  (To use this rigorously, square both sides and cite prime factorization.)

Edit:  A more general approach is to take the log of both sides and use some logarithm properties to simplify both sides, but in this specific case that's overkill.


topology
combinatorics proposed
combinatorics

Some cities of the country are connected by air flights in both directions. It is known that it is possible to reach every city from any another (probably, with changes) and there are exactly 100 flights from each city. Some $ m$ flights have been canceled because of bad weather conditions. For which maximum $ m$ it is still possible to travel between each two cities (probably, with changes)?

It is not clear what is "maximum $ m$" since such $ m$ may depend on the topology of the graph (formed by cities-vertices and flights-edges).
Is it a minimax problem, i.e., to find "maximum $ m$" that will work for [i]any[/i] 100-regular graph?
Or is it a maximax problem, i.e., to find the absolutely "maximum $ m$" that works for [i]some[/i] 100-regular graph?

[quote="maxal"]
Is it a minimax problem, i.e., to find "maximum $ m$" that will work for [i]any[/i] 100-regular graph?
[/quote]
Yes.


parameterization
function
algebra proposed
algebra

Lets $k$ be a real number,find all functions $f$ such that:
$f$:$R\to R$
$f(x)+f(y)^2=kf(x+y^2)$

[quote="lordWings"]With $x=y=0$ we get $f(0)+f(0)^2=kf(0)$, which yields two possibilities: either $f(0)=0$, or $f(0)=k-1$.
With $y=0$ we get $f(x)+f(0)^2=kf(x)$, then $f(x)=\frac{f(0)^2}{k-1}$, which is a constant.
Then, there are two solutions: $f(x)=0$ and $f(x)=k+1$.[/quote]

$f(x)=k-1$! :)

Thanks, you are right, there was an obvious typography error.
I meant: There are two solutions: $f(x)=0$ and $f(x)=k-1$.
At least, this one was easy.


Under preferences, there is a hide forums tab.
You can then tell it which ones you want closed/opened.
:)

There is no hide function for the Valley Math Circle (under National & Local Communities) nor on the India section. Can this be fixed?

[quote="tjhance"]Usually, this website has always known my preferences of what sections of the site I want to be opened. For instance, it would keep the Middle School forums closed but the Olympiad Section open. However, a few days ago I cleared my cookies, due to problems with my browser. Now, the forum index reverts to default settings every time I visit it. Now, I have to open up the Olympiad Section to view it every time I visit the forum index. I would like to be able to view all my favorite sections automatically. How do I get the site to know my preferences?[/quote]Wait are you saying that it doesn't remember the ones you click in/out?

Edit: you are right, the settings are not saved anymore for some reason. Moving to Support.

Bug solved. The site should remember your layout now. You need to press F5 once on the index page so that the cache for the js file regenerates (from your browser).

Yes, this is working now.
However, the Hide Forums panel does not have drop-downs for a few of the forums, such as India.

[quote="alkjash"]However, the Hide Forums panel does not have drop-downs for a few of the forums, such as India.[/quote]

Yeah, this was put under the "Wishlist" for new features of the forum.


n>1 and a positive numbers
$ 2^{n}\minus{}1$ this not divided by n

Can you please rephrase the question, I do not understand it. Are you asking for which positive integer n is there 2^n-1 which cannot be divided to create an integer. If that was the question, then the answer is 2.

I prove it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1610384#1610384]here[/url].


[url]http://www.cemc.uwaterloo.ca/english/contests/past_contest/2003/cayley.pdf[/url]

Here's a contest of Gr.10 Canadian Math Contest.

I don't really understand 23,24, and 25 even after I have read the solutions.  Can someone give me a hint on how to solve them or post their solution?

Thanks.

Typically, the official solutions are usually short. Let me help you a bit with #23
I'll post others as I get the time since it's late at night now.

#23) Draw a rough pyramid with square base on it. So there are 4 triangular faces. Let's do some labelling so that you can keep track of the work: label the top vertex as E and the square base with A,B,C,D, and then label the center of the square base as F. Label the midpoint of side AB as M. Work with one triangular face, choose face EAB. This is where algebra comes in:

Let h=height of pyramid=EF
Let x=length of side of square base=AB
Let t=altitude of triangle EAB=EM
Let S=area of square base
Let T=area of triangular face

Consider square ABCD.  S = x^2

Consider triangle EAB.  T = (t * x) / 2.  So t = 2T / x

Consider triangle EMF.  By Pythagoras theorem, EM^2 = EF^2 + MF^2  or  t^2 = h^2 + (x / 2)^2
Solve for h: h^2 = t^2 - (x^2) / 4 = (4T^2) / x^2 - S / 4 = 4T^2 / S  -  S / 4
We are given T = 840 and S = 1440,
h^2 = (4 * 840^2) / 1440  -  1440 / 4 = (4 * 840 * 840) / (12 * 12 * 10)  -  360
h^2 = 4 * 70 * 7  -  360 = 10 * (14^2 - 36) = 10 * (196 - 36)  = 1600
h = 40 because h>0
Answer is 40 or [b]B[/b]

Tomorrow I (or anyone else) can post #24 and #25. There are a lot of ways of doing #24 (one way is brute force and really count through, not recommended)(another way is actually counting combinations and using the fact that there partitioned classes according to congruence mod 3). #25 is easy if you can actually think the "big picture" of the successive angles. Sorry I gtg.


could someone tell what this means? (translation to english please!!! )

介紹一個超好玩o既江湖網俾你啊，呢個係全香港最大型最有人氣o既江湖網站
，通常都有300幾人在線。o係果度可以同人結婚啦，生仔啦，起屋啦，打交啦，殺人啦，仲有寵物啦，國戰啦，好多野玩啊。快ｄ玩啦~~~網址係

It's normal for you not understand it, because it is written in Cantonese... :D
介紹一個超好玩o既江湖網俾你啊，呢個係全香港最大型最有人氣o既江湖網站
For this sentence, I personally cannot come up with any English word to replace 江湖. Anyway, it rougly means this:
Introduce an extremely playable 江湖 website to me! This is the hugest and most popular 江湖 website in Hong Kong!
通常都有300幾人在線。o係果度可以同人結婚啦，生仔啦，起屋啦，打交啦，殺人啦，仲有寵物啦，國戰啦，好多野玩啊。快ｄ玩啦~~~網址係
and this means:
usually there are more than 300 people online. Here you can marry to other people, give births to babies, construct houses, fights, kills, and also keep animals and "national war", many things to play! Come here to play quickly. The website is

Sorry for poor English translation :blush:  :blush:

如果没有英文翻译,这段文字对我还真有点困难......

btw, shinwoo, where did you find this "advertisement"?

介紹一個超好玩o既江湖網俾你啊，呢個係全香港最大型最有人氣o既江湖網站
，通常都有300幾人在線。o係果度可以同人結婚啦，生仔啦，起屋啦，打交啦，殺人啦，仲有寵物啦，國戰啦，好多野玩啊。快ｄ玩啦~~~網址係

Let me introduce an extremely worth-playing Jiang-Hu(you can treat it as a society) site to you! This is the largest and the most popular Jiang-Hu site in Hong Kong. Usually there are more than 300 people online. There you can marry others, give birth to babies, build houses, fight with others, kill others. There are also pets, wars between countries. To conclude there are many things to play. Come now~~~The URL is

[quote="bugzpodder"]April Fools?  :D[/quote]

jack, why do you think so?
This kind of sites really exists in hong kong..
you know, many ppl enjoy staying in the virtual world rather than the real one..

[quote="siuhochung"]
you know, many ppl enjoy staying in the virtual world rather than the real one..[/quote]
Yes, one of them is me :P

[quote="bugzpodder"]because it is or close to april 1st when he posted that msg?

can you give me some good virtual world links?  i would like to check them out..   :D[/quote]

I don't really know, because I do not go to this kind of forum.
But I can tell you that in Hong Kong, many people use leoboard(hope i dun hv spelling mistake) (a cgi forum) or discuz!(a php forum). Both forums stress on modifications and both have a complete system of marriage, fight, those things. Maybe you can find some forums using these 2 forum softwares.


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

let ABCD is a cyclic quadrilateral that : (AB,CD)=E and (BD,AC)=F .consider, the circumcircle of AFD  cross the circumcircle of BFC at H and F.  prove that :\[ <EHF\equal{}90\]


Suppose that $a>1,b>1,c>1$.Prove the following inequality for positive integer$ n $.

\[(a-1)(b-1)(c-1)n^3+\left[(a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)\right]n^2\]
\[+(a+b+c-3)n+1\leqq(abc)^n\]

Isn't it  pretty easy? I mean you just put $ a=1+x, b=1+y, c=1+z$ and you develop everything. It is clear that the terms in the developping also appear in the left-hand side. Probably I misunderstood something.

I hope harazi will write solution instead of unnecessary remarks next time.

Let $x=(a-1)(b-1)(c-1)$; $y=(a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)$, $z=(a-1)+(b-1)+(c-1)$.
By induction. For $n=1$ we have identity.
Suppose we proved that $(x+y+z+1)^n\geq n^3x+n^2y+nz+1$.
Consider $(x+y+z+1)^{n+1}\geq (n^3x+n^2y+nz+1)(x+y+z+1)=S$.
We have $S\geq (n^3x+n^2yz+nyz+x)+(n^2y+nz^2+1)+(nz+z)+1$. Note that $yz\geq 3x$; $z^2\geq 2y$ $\Rightarrow$ $S\geq (n+1)^3x+(n+1)^2y+(n+1)z+1$.

To Myth, I see. This problem is based on Bernoulli's inequality,or tangent line.

kunny

Thank you for your kind words, Myth! Kind as usual... My solution, of course much more idiot than yours, based on some "unnecessary remarks" is:
   For n=2 observe that in the development of $ (1+x+y+z+xy+yz+zx+xyz)^2$ we have $ 1+(x+y+z)^2+2(x+y+z)+2(xy+yz+zx)+2(x+y+z)(xy+yz+zx)$ and since $ (x+y+z)(xy+yz+zx)$ is at least $ 9xyz$ we are done. For n at least 3, in the same development we have $ 1+n(x+y+z)+\frac{n(n-1)(x+y+z)^2}{2}+\frac{n(n-1)(n-2)(x+y+z)^3}{6}$ and a small AM-GM and $ (x+y+z)^2>= 3(xy+yz+zx)$ solve the problem.

For $a>1,b>1,c>1$,we have $a^n\geqq n(a-1)+1,b^n\geqq n(b-1)+1,c^n\geqq n(c-1)+1$.

so by multiplying both sides of each inequality,we obtain$(abc)^n\geqq\left[n(a-1)+1\right]\left[n(b-1)+1\right]\left[n(c-1)+1\right]$.

Therefore we expand the right hand equation,we are done.

kunny

Please, write my name correctly: Myth

Your solution is super, especially in comparison with my and harazi's ones.


PPP President Benazir Bhutto is dead  :mad:   :(

Oh my goodness! My condolences to everyone who supported her, and her family.

This is a great loss for Pakistan.
 :mad:  :mad:  :(

hey didn't have a look at it !!!!
will watch now now on the TV
 :ninja:


Dear Hungkhtn and the others, I would like to see your nice solution. :)

Very nice, Can. I solve it by the same manner I did post here (in my file)

http://www.mathlinks.ro/Forum/viewtopic.php?t=145431&sid=78d17d618e0e6211efb17efb693139b3

Rewrite the inequality to
\[(a-b)^{2}\left(\frac{(a+b)^{2}}{a^{2}b^{2}}-\frac{9}{a^{2}+b^{2}+c^{2}}\right)+(c-a)(c-b)\left(\frac{(c+a)(c+b)}{a^{2}c^{2}}-\frac{9}{a^{2}+b^{2}+c^{2}}\right). \]
Let
\[M=\frac{(a+b)^{2}}{a^{2}b^{2}}-\frac{9}{a^{2}+b^{2}+c^{2}}\]

\[N=\frac{(c+a)(c+b)}{a^{2}c^{2}}-\frac{9}{a^{2}+b^{2}+c^{2}}. \]
We may assume that $a\ge b\ge c$, then $N\ge 0$ (it is very interesting). If $a-b\le b-c$ then
\[(c-a)(c-b) \ge 2(a-b)^{2}\]
and the inequality is proved if we can prove that
\[M+2N \ge \frac{(c+a)(c+b)}{a^{2}c^{2}}-\frac{10}{a^{2}+b^{2}+c^{2}}\ge 0. \]
Indeed, this inequality is equivalent to
\[(c+a)(c+b)(a^{2}+b^{2}+c^{2}) \ge 10a^{2}c^{2}, \]
Since  $b\ge \frac{a+c}{2}$, we refer
\[(c+a)(c+b)(a^{2}+b^{2}+c^{2}) \ge (c+a)\left(c+\frac{a+c}{2}\right)\left(a^{2}+c^{2}+\frac{1}{4}(a+c)^{2}\right) \]

\[\ge 2\sqrt{ac}\cdot\sqrt{3ac}\cdot 3ac>10ac. \]
So it remains to consider the case $a-b\ge b-c$. In this case, we will prove that $M\ge 0$ or
\[(a+b)^{2}(a+b+c)^{2}\ge 9a^{2}b^{2}. \]
If $a\ge 2b$ then this inequality is true immediately because $a^{2}+b^{2}\ge \frac{5}{2}ab$. So we may assume that $a\le 2b$. Because $c\ge 2b-a$, we will prove next that
\[(a+b)^{2}\left(a^{2}+b^{2}+(2b-a)^{2}\right)^{2}\ge 9a^{2}b^{2}. \]
Let $x=\frac{a}{b}$, then $1\le x\le 2$. The inequality becomes
\[(x+1)^{2}(2x^{2}-4x+5) \ge 9x^{2}\]
or
\[2x^{4}-10x^{2}+6x+5 \ge 0. \]
This can be check easily. We have done.

Remark. The inequality $N\ge 0$ can be proved due to (sinc $c\le b$)
\[(c+a)(2c)(a^{2}+2c^{2}) \ge 9c^{2}a^{2}\]
or
\[2a^{3}-7a^{2}c+4c^{2}a+4c^{3}\ge 0 \]
or
\[(a-c)^{2}(2a+c) \ge 0. \]
What is your solution, Can?

[quote="tanpham90"]Use the following inequality ( easy with SOS of hungkhtn  :lol:  ) :
$a,b,c >0$
$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\geq\ 9\frac{\frac{5}{2}(a^{2}+b^{2}+c^{2})-\frac{3}{2}(ab+bc+ca)}{(a+b+c)^{2}}$

[hide]And the harder :

$a,b,c >0$
$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\geq\ 9\frac{4(a^{2}+b^{2}+c^{2})-3(ab+bc+ca)}{(a+b+c)^{2}}$[/hide]

Finally put $x=a^{2}+b^{2}+c^{2}$ and $y=ab+bc+ca$ we are done ! :lol:[/quote]
The second inequality is wrong, for example with $a=2.5,b=1.5,c=1$. :wink:
The first can't be use to solve my inequality since the inequality
\[9\frac{\frac{5}{2}(a^{2}+b^{2}+c^{2})-\frac{3}{2}(ab+bc+ca)}{(a+b+c)^{2}}+\frac{9(ab+bc+ca)}{(a+b+c)^{2}}\ge 12 \]
is false for $a=b=1,c=0.9$ :wink:

Oh , I see ............  :blush:
By the way , both of them can be kill with computer !
Assume $a=min(a,b,c)$ so that we have : $b=a+x$ and $c=a+y$
After expanding , we are done , but I'm too busy right now , I'll post the solution later !

Dear Toanhocmuonmau, did you solve them by SOS. I tried the first without SOS but I cannot  :(   Can you please help me ? Thank you very much. I find these inequalities very nice but hard  :)

\[\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\ge \frac{3}{2}\sum\frac{a+b}{c}-6 \]
and
\[\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\ge \frac{10(a^{4}+b^{4}+c^{4})-a^{2}b^{2}-b^{2}c^{2}-c^{2}a^{2}}{(a^{2}+b^{2}+c^{2})^{2}}\]

The second solution of mine for the first is used the following
\[x^{2}+y^{2}+z^{2}+\frac{15}{2}\ge \frac{11}{4}(x+y+z)+\frac{3}{4}(xy+yz+zx) \]
where $x,y,z$ be postive real numbers such that $xyz=1$.
:)

[quote="toanhocmuonmau"][quote="tanpham90"]This is the sharper , but I can't ........................  :!:  :blush:
$a,b,c >0$
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\ 2 \sqrt[3]{\frac{9(a^{3}+b^{3}+c^{3})}{(a+b)(b+c)(c+a)}}$[/quote]
I don't think it's sharper than my inequality, Tan
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3\left(\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)^{2/3}\]
For example with $a=b=1,c=0.5$, we have
\[3\left(\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)^{2/3}\ge 2 \sqrt[3]{\frac{9(a^{3}+b^{3}+c^{3})}{(a+b)(b+c)(c+a)}}\]
And for $a=b=1,c=1.1$, we have
\[3\left(\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)^{2/3}\le 2 \sqrt[3]{\frac{9(a^{3}+b^{3}+c^{3})}{(a+b)(b+c)(c+a)}}\]
:wink:[/quote]
Tan, your inequality can be prove easily by the same method with my problem. :wink:

[quote="toanhocmuonmau"]
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3\left(\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)^{2/3}\]
[/quote]
Can anyone have a nice solution for it? :)

[quote="toanhocmuonmau"]Prove that for any nonnegative real numbers $a,b,c$ we have
\[\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}+\frac{9(ab+bc+ca)}{a^{2}+b^{2}+c^{2}}\ge 12 \]
:)[/quote]
  A stronger inequality:

 $\sum{cyc}\frac{a^{2}}{b^{2}}+9,5\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\geq\ 12,5$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=8803[/img]
i like that be true :oops:

But it was wrong, certainly. Since
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3\]
\[\frac{9(ab+bc+Ca)}{a^{2}+b^{2}+c^{2}}\le 9\]

[quote="toanhocmuonmau"]
How about this one?
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge 3\left(\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{ab \plus{} bc \plus{} ca}\right)^{2/3}
\]
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ ( where $ v > 0$ ), $ abc \equal{} w^3$ and $ \sqrt [3]{\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} ac \plus{} bc}} \equal{} p.$
Then $ p\geq1,$  $ u^2 \equal{} \frac {p^3 \plus{} 2}{3}v^2$ and
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge 3\left(\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{ab \plus{} bc \plus{} ca}\right)^{2/3}\Leftrightarrow\sum_{cyc}a^2c\geq 3p^2w^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2b \plus{} a^2c)\geq6p^2w^3 \plus{} \sum_{cyc}(a^2b \minus{} a^2c)\Leftrightarrow$
$ \Leftrightarrow9uv^2 \minus{} 3w^3\geq6p^2w^3 \plus{} (a \minus{} b)(a \minus{} c)(b \minus{} c).$
But $ 9uv^2 \minus{} 3w^3\geq6p^2w^3$ is true because $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0\Leftrightarrow w^6 \minus{} 2(3uv^2 \minus{} 2u^3)w^3 \plus{} 4v^6 \minus{} 3u^2v^4\leq0,$
which gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, we need to prove here that $ 3uv^2\geq(1 \plus{} 2p^2)(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}).$
But $ 3uv^2\geq(1 \plus{} 2p^2)(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3})\Leftrightarrow$
$ \Leftrightarrow3\geq(1 \plus{} 2p^2)\left(3 \minus{} \frac {2(p^3 \plus{} 2)}{3} \plus{} 2\sqrt {\frac {(p^3 \minus{} 1)^3}{9(p^3 \plus{} 2)}}\right)\Leftrightarrow$
$ \Leftrightarrow(p \minus{} 1)^2(2p^6 \plus{} p^5 \plus{} p^4 \minus{} 3p^3 \minus{} p^2 \plus{} 2p \plus{} 1)\geq0,$ which is obvious.
Hence, enough to prove that $ 9uv^2 \minus{} 3w^3\geq6p^2w^3 \plus{} \sqrt {(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}.$
But $ 9uv^2 \minus{} 3w^3\geq6p^2w^3 \plus{} \sqrt {(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}\Leftrightarrow$
$ \Leftrightarrow(3uv^2 \minus{} (1 \plus{} 2p^2)w^3)^2\geq3( \minus{} w^6 \plus{} 2(3uv^2 \minus{} 2u^3)w^3 \minus{} 4v^6 \plus{} 3u^2v^4)\Leftrightarrow$
$ \Leftrightarrow(1 \plus{} p^2 \plus{} p^4)w^6 \minus{} 3(2uv^2 \minus{} u^3 \plus{} uv^2p^2)w^3 \plus{} 3v^6\geq0\Leftrightarrow$
$ \Leftrightarrow(1 \plus{} p^2 \plus{} p^4)w^6 \minus{} 3(2 \minus{} \frac {p^3 \plus{} 2}{3} \plus{} p^2)\sqrt {\frac {p^3 \plus{} 2}{3}}v^3w^3 \plus{} 3v^6\geq0\Leftrightarrow$
$ \Leftrightarrow(1 \plus{} p^2 \plus{} p^4)w^6 \minus{} (4 \plus{} 3p^2 \minus{} p^3)\sqrt {\frac {p^3 \plus{} 2}{3}}v^3w^3 \plus{} 3v^6\geq0.$
For $ t\geq1$ we obtain: $ 4 \plus{} 3p^2 \minus{} p^3\leq0\Leftrightarrow p\geq3.356...$
Id est, for $ p\geq3.357$ the original inequality is proved.
Let $ 1\leq p < 3.357.$
Thus, it remains to prove that
$ \left((4 \plus{} 3p^2 \minus{} p^3)\sqrt {\frac {p^3 \plus{} 2}{3}}\right)^2 \minus{} 12(1 \plus{} p^2 \plus{} p^4)\leq0,$ which is equivalent to
$ (p \minus{} 1)^2(p^7 \minus{} 4p^6 \minus{} 2p^4 \plus{} 8p^3 \minus{} 8p \minus{} 4)\leq0,$ which is true for all $ 1\leq p < 3.357.$ :)

thanks for your interest, arqady. I like this solution :)
By the way, my solution using Cauchy Schwarz Inequality :)

[quote="can_hang2007"]thanks for your interest, arqady. I like this solution :)[/quote]
Thank you!
[quote="can_hang2007"]By the way, my solution using Cauchy Schwarz Inequality :)[/quote]
Cool! :ninja:

Here is my proof:

Using the inequality:

\[(x+y+z)^3\ge \frac{27}{4}(x^2y+y^2z+z^2x+xyz)\]

for $x=\displaystyle\frac{a}{b}$, $y=\displaystyle\frac{b}{c}$ and $z=\displaystyle\frac{c}{a}$ we get:

\[\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^3\ge \frac{27}{4}\left(\frac{a^3+b^3+c^3}{abc}+1\right)\]

Now we'll prove the following inequality:

\[\frac{a^3+b^3+c^3}{4abc}+\frac{1}{4}\ge \frac{(a^2+b^2+c^2)^2}{(ab+bc+ca)^2}\]

Observe that:

\[LHS-RHS=\]

\[=(a^2+b^2+c^2-ab-bc-ca)\left(\frac{a+b+c}{2abc}-\frac{(a+b)^2+(b+c)^2+(c+a)^2}{(ab+bc+ca)^2}\right)\]

and because:

\[(a+b+c)(ab+bc+ca)^2-2abc((a+b)^2+(b+c)^2+(c+a)^2)=\]

\[c^3(a-b)^2+a^3(b-c)^2+b^3(c-a)^2\ge 0\]

the additional inequality is proved. Then it follows that:

\[\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^3\ge 27\frac{(a^2+b^2+c^2)^2}{(ab+bc+ca)^2}\]

Or:

\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3\sqrt[3]{\frac{(a^2+b^2+c^2)^2}{(ab+bc+ca)^2}}\]

Done.


Slove this equation: $ cos^3x.cos3x\plus{}sin^3x.sin3x\equal{}\frac{\sqrt{2}}{4}$

[hide]
$ \sin 3x \equal{} 3 \sin x\minus{} 4 \sin^3x$ and $ \cos 3x \equal{} 4 \cos^3x \minus{} 3 \cos x$

$ \Rightarrow 4\cos^6x\minus{}4\sin^6x\plus{}3\sin^4x\minus{}3\cos^4x\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow 4(\cos^3x\minus{}\sin^3x)(\cos^3x\plus{}\sin^3x)\plus{}3(\sin^2x\minus{}\cos^2x)(\sin^2x\plus{}\cos^2x)\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow 4(\cos^2x\minus{}\sin^2x)(\cos^4x\plus{}\sin^4x\plus{}\sin^2x\cos^2x)\plus{}3(\sin^2x\minus{}\cos^2x)(\sin^2x\plus{}\cos^2x)\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow 4\cos 2x(1\minus{}\sin^2x\cos^2x)\minus{}3\cos 2x\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow 4\cos 2x(1\minus{}(\frac{1\minus{}\cos 2x}{2})(\frac{1\plus{}\cos 2x}{2}))\minus{}3\cos 2x\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow 4\cos 2x(1\minus{}\frac{1\minus{}\cos^2 2x}{4})\minus{}3\cos 2x\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow \cos 2x(3\plus{}\cos^2 2x)\minus{}3\cos 2x\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow \cos^3 2x\equal{}\frac{\sqrt{2}}{4}$

$ \Rightarrow \cos 2x\equal{}\frac{\sqrt{2}}{2}$

$ x\equal{}k\pi\pm\frac{\pi}{8}$ for $ k$ integer.
[/hide]

:P THANK YOU VERY MUCH. IT'S A NICE SOLUTION.


function
calculus
derivative
trigonometry
real analysis
real analysis unsolved

Hello, I have an exam soon, can someone please help me on this question ?

Suppose that f(x) is a twice differentiable function which satisfies f′′(x) = −f(x) for all x. We show that f(x) = Acos x + B sin x for certain constants A and B as follows:

(a) Show that f(x) cos x − f′(x) sin x is constant.
(b) Show that f(x) sin x + f′(x) cos x is constant.
(c) Use the first two parts to complete the solution.

Take the derivative of the expression in (a) using the product rule: you will get zero. Therefore, $f(x)\cos x-f'(x)\sin x=C$, where $C$ is a constant. Similarly, $f(x)\sin x+f'(x)\cos x=D$ where $D$ is a constant. Now solve for $f(x)$ from these two linear equations.


Extremal combinatorics
modular arithmetic
pigeonhole principle
number theory
combinatorics
IMO
IMO 1985

Given a set $M$ of $1985$ distinct positive integers, none of which has a prime divisor greater than $23$, prove that $M$ contains a subset of $4$ elements whose product is the $4$th power of an integer.

Given a number with prime factors $\le 23$ of the form $n=2^{a_1}\cdot 3^{a_2}\cdot\ldots\cdot 23^{a_9}$, we can identify it with the vector $v=(a_1\pmod 4,\ldots,a_9\pmod 4)\subset\mathbb Z_4^9$, where $\mathbb Z_4$ is the cyclic group of order $4$. We must thus prove that given any $1985$ elements of $\mathbb Z_4^9$, we can find $4$ which sum up to $0$. In fact, that $1985$ is there only because the contest was held in $1985$ :). $2^9\cdot 4-(2^9-1)=1537$ suffices.

Consider the following assertion, which we denote by $P_k(n)$, where $n$ is a positive integer $\ge 2$ and $k$ is a positive integer:

Given $2^kn-(2^k-1)$ elements of $\mathbb Z_n^k$, we can find $n$ which sum up to $0$.

We'll prove that $(*)\ P_k(a),P_k(b)\Rightarrow P_k(ab)$. Then, since $P_k(2)$ is a simple pigeonhole, $P_k(4)$ follows, and since what we want is $P_9(4)$, we're done.

Assume that $P_k(a),P_k(b)$ hold, and consider $2^kab-(2^k-1)$ positive natural numbers. Take a group of $2^ka-(2^k-1)$ and select $a$ with sum divisible by $a$. Then, repeat this procedure with the remaining $2^ka-(2^k-1)-a$ numbers, until we have $2^kb-(2^k-1)$ groups of $a$ elements, each one with sum divisible by $a$. Now divide these sums by $a$, and from the collection (multiset) of $2^kb-(2^k-1)$ elements that we get, extract $b$ with sum divisible by $b$. Multiply $a$ back, and take the $ab$ elements which enter these $b$ sums. This is our collection of $ab$ elements with sum divisible by $b$.

P.S.

$P_1(n)$ is the well-known Erdos-Ginzburg-Ziv Theorem, discussed several times on the forum. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14797]Here[/url]'s one of the older links for example. In that topic, Charlydif wrote that $P_2(n)$ is no longer a conjecture, due to C. Reiher.

For a moment let's pretend we are asked for 2 numbers multiplying to a square. We have $9$ primes, so we need $2^9+1$ numbers to find two which's exponents mod 2 are congruent (so they multiply to square). Doing this many times we can find $(1985-(2^9+1))/2 \ge 2^9+1$ pairs of numbers which multiply to squares. So we have $2^9+1$ square numbers, and by the same logic two of them will multiply to a fourth power. We're done.

I find this one very nice.

As @JuanOrtiz did, notice that $2^9+1$ numbers are enough to find among them 2 nbs which product gives a square.

So we start with $2^9+1$ nbs, we know we have a square already. We put the 2 nbs aside, add 2 other nbs and repeat the process. Since each of the 9 exponents is either 0 or 2 mod 4, then there are also $2^9$ choices for a square.

This forces us to find $2^9+1$ squares so we need to add 2 $2^9$ times so we need at least $$2^9+1+2 (2^9)=1537.$$
We are done.


geometry
integration
trigonometry
calculus
calculus computations

Find the area under the curve (2a-x)y^2 = x^3 and x=2a.

can anyone please gv  me a detailed solution?
thank you...

I'm assuming you're looking for the area enclosed by the two curves.  I'm also going to assume $ a > 0$ (if $ a < 0$ it's not hard to see the answer is the same).  Solving for $ y$ gives $ y = \pm \sqrt {\frac {x^3}{2a - x}}$, so the region is symmetric about the x-axis, and hence the area we're looking for is twice the area enclosed by the x-axis, the line $ x = 2a$, and the upper half of the first curve.
\[ A = 2\int_0^{2a}{\sqrt {\frac {x^3}{2a - x}}dx}
\]
Let $ x = 2a\sin^2 u$, then formally $ dx = 4a\sin u \cos u du$ so the integral becomes
\[ A = 16a^2\int_0^{\frac {\pi}{2}}{\sqrt {\frac {2a \sin ^8 u \cos ^2 u}{2a(1 - \sin ^2 u)}}du = 16a^2\int_0^{\frac {\pi}{2}}\sin ^4 u du = \boxed{3 \pi a^2}
}\]
the integral $ \int_0^{\frac {\pi}{2}} \sin ^4 x dx = \frac {3\pi}{16}$ is easily evaluated by applying the identities $ \sin ^2 a = \frac {1}{2}\left(1 - \cos 2a\right)$ and $ \cos ^2 a = \frac {1}{2}\left(1 + \cos 2a\right)$ to find $ \sin ^4 x = \frac {3}{8} - \frac {1}{2}\cos 2x + \frac {1}{8} \cos 4x$.


A sequence of integers is obtained by starting with three digits $d_1$, $d_2$, and $d_3$.  The fourth term of the sequence is the units digit of $d_1 + d_2 + d_3$.  Each succeeding term is the units digit of the sum of the three previous terms.  What numbers belong in the first, second, and third positions respectively to complete the sequence?
\[ \underline{\ \ }, \underline{\ \ }, \underline{\ \ }, 1, 1, 1 \]

[hide] Also, another solution is 1,1,9 (remember, they said units' digit).[/hide]

C'mon you guys, -1 isn't a digit! Read the problem again.

[hide]The third blank is 9 because 9+1+1=11, and you can't get a number to reach 21 or any subsequent number with units digit one. Therefore, using the same logic, the other two blanks are both 1. 1 1 9.[/hide]

[hide]First tried 1,0,0 but it didn't work.  I was stuck until I reread the problem and realized 1,1,9 works.[/hide]
I'm just wondering, does any part of this series repeat?

Try and see, I'm sure it does eventually.


Culegere  Ganga, clasa a xi-a, pag.284, probl. 9.

Se da  $f: R\to (0,\infty )$  astfel incat  $f(x+1)-f(x)>0$  pentru orice  $x\in R.$
Sa se arate ca  $f(x)\to \infty$  cand  $x\to \infty$ .

E gresita. Daca $f(x) = \frac{\pi}2+\arctan x$, atunci $f(x+1)-f(x) > 0, \, \forall x \in \mathbb R$, dar $\lim_{x \to \infty}f(x) = \pi$. ([i]Editat usor.[/i])

totusi daca in loc de 0 la inegalitate ar aparea orice constanta
strict pozitiva cred ca problema ar fi corecta