What do you say when you win a game?  Like, my friend beat me in chess and he said "I rocked your socks like a monkey" (that makes no sense).

When i win (in tennis) i say "PEANUT BUTTER JELLY!". (i don't know why).
So what do you say?

I have [i]won[/i]!

This is an inside joke from my history class.  We saw a video remake of the Mahabharata; in the scene where Shakuni beats Yudhisthira in dice repeatedly, Shakuni shouts "I have won!" every time he, well, wins.

[quote="Chocoboom"]i say

PWNED

XD[/quote]

Burn..............jk actually I say pwnded too :lol:

Today, in tennis, i hit an amazing shot, so i said "PEANUT BUTTER JELLY!".  THen the next point my opponent hit an amazing shot and said "PEANUT BUTTER JELLY" (to mock me).  But then i said "PLAGERISM!"...

There's no real point in saying anything : you just give an empathising
smile almost as if to say, " it was way out of your reach all the time!"

[quote="fireforce"]In playing computer games I some times say w00t :w00t:[/quote]

While I'm playing "Icy Tower" I say 'wow', 'amazing', 'yupi' etc :)

I say either 'Pwned' or 'Oh Snap'.
Dunno why though.

When I WIN  A GAME, I SAY CHECK OUT THE MASTERS INVITATIONAL MATH CONTEST! CLICK ON THE LINK ON MY SIGNATURE!!!!!!

One of my firends knew someone who would always say Good Call if he lost. It got so anyoying my friend let him win so his friend would not sound like a Miller Lite Commercial.  :P

How about "Cor lumme stone the crowss!", a Wodehouse classic?

A simple Yahoo! is often quite effective.


[quote="GameBot"]Hillary has eleven coins, all dimes and nickels. In total, the
coins are worth 75 cents. How many nickels does she have?[/quote]
[b]Method 1:[/b]
Solve the following system of equations, where:
d+n=11
10d+5n= 75
10d+5(11-d)=75

d=4, [b]n=7[/b]

[b]Method 2:[/b]
If you notice the sum is 7[b]5[/b], we realize there must be an odd number of nickels. We can slowly check 1, 3, 5, and ah-ha 7 nickels to get the answer.


complex analysis
complex analysis unsolved

prove that the series $\sum_{n=1}^{\infty}\frac{{z^{n}}}n$ converges at all pts of the unit circle except at $z=1$. Can anyone help? i tried to break it into real and imaginary parts and consider them seperately, but i'm getting stuck.

Use the Abel-Dirichlet test: the series $\sum_{k}a_{k}b_{k}$ converges if the sums $\sum_{k=1}^{n}a_{k}$ are bounded, $b_{k}\to 0$, and $\sum_{k}|b_{k}-b_{k+1}|$ converges.


matt recieved change totaling $\$$1.85 in nickels, dimes, and quarters. he recieved at least one coin of each type. what is the least # of coins he could've recieved?

didn't read so don't pay attention to this

[hide="Click to reveal hidden content"] Since Matt has to get ATLEAST  $\$$.40 change, since there has to be one nickel, one dime, one quarter, you have to find out the least number of coins needed to make $\$$ (1.85 - .40), or $\$$ 1.45. The least way to do that is 5 quarters and two dimes. So you get 3 (required number of coins) + 5 (5 quarters) + 2 (two dimes) = 10
So the least number of coins is 10. [/hide]

[hide]Same as making 185/5=37 cents from coins of 1 cent, 2 cent, and 5 cent worth. The smallest solutions is 1 nickel, 3 dimes, 6 pennies, or 10 coins total.[/hide]
This has nothing to do w/ prob.

[hide]
least coins needed to get $\$$1.45 and then add 3

7+3=10

guys, you really don't need to simplify something life this

this is like a #7 on a chapter sprint
[/hide]
jorian


Two circles are internally tangent. The smaller circle is also tangent to two perpendicular radii of the larger circle. What is the ratio of the area of the small circle to the area of the large circle? Express your answer as a decimal rounded to the nearest hundredth.

Actually its .17, but why is that the radius of the small circle?

I don't understand where your getting all these formulas or what the problem's asking. Could you explain better?

[geogebra]e7958dd061d50a3b44a0f3c79819f553ed3c0fd0[/geogebra]

It is easy to see from this diagram that $ R \equal{} r \plus{} r\sqrt2$.

This means that $ \frac rR\equal{}\frac1{1\plus{}\sqrt2}$.

We want the ratio of the two areas, which is $ \frac{\pi r^2}{\pi R^2}\equal{}\frac{r^2}{R^2}\equal{}\left(\frac rR\right)^2\equal{}\frac1{3\plus{}2\sqrt2}\cdot\frac{3\minus{}2\sqrt2}{3\minus{}2\sqrt2}\equal{}\frac{3\minus{}2\sqrt2}{9\minus{}8}\equal{}3\minus{}2\sqrt2\approx\boxed{0.17}$.

i agree with"the champ"
PS:what is your rating on ftw?
  :sleeping:  :sleep2:    :starwars:

BLAHHHH
[hide="No Algebra Since I Lost A Bet"]To find the ratio of areas, we need to know the ratio of radii.

We know that the ratio of areas is the square of the ratio of linear lengths.

First, we need to find a way to calculate each radius.
BIg Radius is distance between the [b]point of tangency[/b] and the center of the big circle.
Small Radius is distance between center of small circle and any point on the small circle.

We then consider the [b]center of smaller circle[/b], and its distance from the center of the large circle.
Assuming we can draw a square that has the two centers as opposite vertices, the distance between the two centers is $\sqrt2$ times the side length.

Now we consider the distance between the center of the smaller circle and the [b]point of tangency between the two circles[/b]
Since the point of tangency is on the smaller circle, we know that its distance from the center (small circle's) to the point of tangency is equal to the radius of the smaller circle.

By the given statements, we can prove that Side-Length-of-square is equal to the radius of the small circle. However, since this is a countdown question, you can either take it to be true or just make a non-rigorous observation that two sides of the square that share the center of the small circle as a center both are radii of the small circle.
So we have Large radius is equal to sum of Small Radius times $\sqrt2$ and Small Radius.
We can discover that the [b]ratio of radii[/b] is equal to the reciprocal of the sum of 1 and $\sqrt2$. By mutliplying both sides of the fraction by the denominator's conjugate, we can simplify it into the [b]positive difference of 1 and $\sqrt2$[/b]
And since we need the square of that ratio accurate to the hundredth's place, we should first evaluate it to accurate to the hundredth's and then square it.
$\sqrt2-1\approx 0.41$
Now squaring this in your head:
$0.41^2=(0.0001)(41\cdot40+41)=(0.0001)(1640+41)=(0.0001)(1681)\approx 0.17$
And that's our answer! 0.17
[/hide]

[hide="Extremely Non-Rigorous Substitution Evaluation"]
Let the larger radius be $1$ since that'll be on the denominator.
Now all we need to evaluate is the square of the smaller circle's radius ($r$).
We can find it too, by observing that the larger radius is $r\sqrt2+r=1$ (connect the centers, that's $r\sqrt2$)
so we have $r=\frac1{\sqrt2+1}$ from which follows $r^2=\frac1{1+2+2\sqrt2}=\frac1{3+2\sqrt2}$
[hide="scientific calculator"] [code]Press 2
Press Square Root (You should see 1.414...)
Press X
Press 2
Press + (You should see 2.828...)
Press 3
Press = (You should see 5.828...)
Press 1/x (You should see 0.171...)[/code]
Round the answer to hundredth's.[/hide]
And you get $0.17$ as your answer!
[/hide]

Never again...


If $F_n$ represents the nth Fibonacci number, were $F_0 = 0, F_1 = 1$, compute the sum  $\sum_{n=1}^{/infty}\frac{F_n}{F_{n+1}F_{n+2}}$.


I thought maybe I could use the product-to-sum identities but I couldn't get anything workable from it.

$\text{Evaluate } \cos12^\circ\cos24^\circ\cos48^\circ\cos96^\circ$.

Wow - pretty slick! Thanks shobber!

Just in case no one sees it, the sin12 at the numerator will cause a "chain reaction" of sine double angles to ultimately form sin192/16sin12, which reduces to -1/16.

An easy trick that might help

[hide]

Letting $P=\cos 12^o\cos 24^o\cos 48^o\cos 96^o$

Now multiply both side with $\sin 12^o$ you will get

$\sin 12^oP$
$=\sin 12^o\cos 12^o\cos 24^o\cos 48^o\cos 96^o$
$=\frac{1}{2}\sin 24^o\cos 24^o\cos 48^o\cos 96^o$
$=\frac{1}{4}\sin 48^o\cos 48^o\cos 96^o$
$=\frac{1}{8}\sin 96^o\cos 96^o$
$=\frac{1}{16}\sin 192^o$
$=-\frac{1}{16}\sin 12^o$

Now since $\sin 12^oP=-\frac{1}{16}\sin 12^o$ ..... what do you think $P=?$  :)
[/hide]

Opps ... shobber is faster than me  :P


You walk on a moving sidewalk. If you walk in the same direction as the belt, you can get from one end to the other in 30 sec. If you walk against it, it takes 120 sec. How long does it take to get to the other side if you just stand still on the belt?

Follow up: If you walk at 50 cm/s, how long is the belt?

Solution.

Let the speed of the belt be $ v_{b}$, let the walking speed of the person be $ v_{p}$. Let $ d$ be the length of the belt.

When the person is walking with the belt, the net speed is $ v_{p}\plus{}v_{b}$. When the person is walking against the belt, the net speed is $ v_{p}\minus{}v_{b}$. Since velocity is distance over time, we get:

$ v_{p}\plus{}v_{b}\equal{}\frac{d}{30}$

$ v_{p}\minus{}v_{b}\equal{}\frac{d}{120}$

Subtracting these two equations gives:

$ 2v_{b}\equal{} d\left(\frac{1}{30}\minus{}\frac{1}{120}\right) \equal{}\frac{d}{40}$

$ \Rightarrow v_{b}\equal{}\frac{d}{80}$

If you stand still, you will travel at the speed of the belt and the time taken will be:

$ t \equal{}\frac{d}{v_{b}}\equal{} 80\,s$

Substituting $ v_{b}\equal{}\frac{d}{80}$ and $ v_{m}\equal{} 0.5\, ms^{\minus{}1}$ into $ v_{m}\plus{}v_{b}\equal{}\frac{d}{30}$ gives:

$ 0.5\plus{}\frac{d}{80}\equal{}\frac{d}{30}$

$ \Rightarrow 120\plus{}3d \equal{} 8d$

$ \Rightarrow 5d \equal{} 120$

$ \Rightarrow d \equal{} 24\,m$

Shouldn't it be $ v_{b}\minus{}v_{p}\equal{}\frac{d}{120}$?

No...because that would imply that the speed of the belt is greater than the speed of the person. The person would never be able to walk to the other side.


geometry
probability
topology
search
Support
invariant
3D geometry

who is the best modern indian geometer. maybe dead.
and one who is present.
i dont see giant steps by india in geometry why is that?
maybe its the education system here?
also say who is the best mathman is india any field..
any potential riemann hypothesis solvers or 21st century ramanujan??
calcutta and south always has math talents, why is that?
eveybody give their view...

How can we decide who is the best 'mathman' in India? :maybe:

C'mon,let's conduct a lottery,
pick up names of some great Indian Mathmen and call any guy good at lottery probability :rotfl:

Akshanil? Gaurav? Some other young genius?  :huh:

Eh,speakin in real terms,they've yet got a long way to go.
May be older guys like Ramanujan,Mahalanobis,J Bose,S Bose.....................
One gets loads of them really
It's gonna be a tough lottery :(

M Prakash  :omighty:
[hide="."]chill just kidding  :rotfl: [/hide]

[hide]david shepherd jumping!!!! i completed 444 posts, a quadruple-nelson  [/hide]

lets have categories
*geometry and topology
*algebra, number theory ((i am sure ramanujan will get this))
*professor teaching , calculation and problem solving
*ancient
*young but future terence tao'sss
*all round dead mathematicians or present (potential Riemann hypothesis solvers or ones who could give some insights into unsolved problem but werent introduced to RH.)

also give names that had ability to solve greatest problems of HILBERT but werent even present or didnt know about the discovery!!!)

Gaurav Shukla is the one!!! And there's a friend of mine whom I know only by the name M. Chakraborti.

[quote="Rajiv"]Gaurav Shukla is the one!!![/quote]
 :ninja: Sounds familiar...

Yeah, he's the nut who creeps into your mirror when you sneak in front of it.

BTW, Abhishek Hemantkumar Dang (:omighty:) is a real master of Math!
[hide]And I am proud to say he studied in JPNV, a school started by RSS (and meant only for Brahmins), where my brother studies now! (Enter his name in the [url=http://www.imo-official.org/searchq.aspx]search[/url])[/hide]

His mom is d chem teacher in my school  :D

Psst... please don't tell her....
[hide]His cousin is a real nut, almost dead neck-up! She studied in my college[/hide]

How come u guys can support freaks like Bal Thackeray & co? :mad:
Dont worry,I 'll wipe out every RSS/BJP/SS supporter in the country beginning with Dang and Rajiv :diablo:

[quote="Scary math"]
Dont worry,I 'll wipe out every RSS/BJP/SS supporter in the country beginning with Dang and Rajiv :diablo:

[/quote]

Sagar Vaze too.

[url=http://en.wikipedia.org/wiki/Rashtriya_Swayamsevak_Sangh]RSS[/url] was established by a Brahmin, Dr. KB Hedgewar. By speaking ill about it thus, you, by no means, invite sympathy from me for your ignorance nor calm for your arrogance.

Never write anything unless you know about it to the greatest possible depth.

No
I am shocked to see freaks who can be so biased on religion/caste basis.
How come ur so proud of what u got by birth and not by ur merits?Look beyond these narrow walls,let more light enter that little head of urs. :mad:

[quote="jaiawaj"]and they think iit's entrance exam is more difficult than MIT.
MIT doesent test the student's algebraic skill or hard work but also tests their right hemisphere of the brain.[/quote]
Really? Even IMO gold medalists are rejected from it.

@apratimgtr
Hello?
I meant in general,analytical geo is definetely more application based,and I never really meant to disregard what u and jai have to say.Yah,it is love and thirst for knowledge that helps in making ground-breaking discoveries,it's always not the discoverers themselves who invent applications.Infact dat's probably why Erdos never won fields medal,coz he was more of a problem-solver than theory developer
[hide]Anyway,let me confess,I am not exactly an IIT freak,so.....u know[/hide]
Anyway,I would luv to read dat book :)

nuclear_alchemist
              have u even tried to see a sample paper of mit entrance exam. and plz once try to see one and compare the two papers based on rigorous factors.

[hide="einstein"]"Any fool can make things bigger, more complex, and more violent. It takes a touch of genius-and a lot of courage-to move in the opposite
"[/hide]

also IMO gold medalits, they r not used to ur rigourous non-real abstract mathematics meant only for ghokantes, and they will never even learn them, even if they go through the routine of the gokhante iit guys. the IMO is a playground for mathematical geniuses not a playground of iit (full of brain stinking dumb guys who have succeded only in one think in life, i.e memorizing). if a platform like iit wouldnt be there then i am sure they(iitians) would be doctors.
also IMO gold medalists they have learnt the a,b,c of problem solving and are the real heros of future. just an education wont make u an engineer, u need creative art and like leonardo da vinci.

to  apratimgtr
              actually u misunderstood me. i know what is calculus and learnt it a long time back in a different way than taught here. but i think my algebra is not good in perspective of problem solving (any suggestion) .so how can i solve calculus problems?

What do you mean by rigourous non real abstact mathematics?IIT-jee has pre rigourous mathematics like any othe examination for high school students.And I think at IMO you need to be lot more rigour than IIT,as it is a long proof based examination.Abstract mathematics is real analysis,abstract algebra those you learn in college.I hope you know that modern geometry too is all about abstraction and deals with more general spaces. A

I think for algebra problems you can try Arthur Engel.It has good problems in polynomials,theory of equations,inequalities and sequences.For calculus problems maron is the best source in secondary level.You may also try ISI problems from 10+2,it has some interesting calculus problems.If you are looking for a rigourous development of real analysis you may try Rudin's Mathematical Analysis.

[quote="nuclear_alchemist"]I won't even bother to give an explanation to someone who thinks there is an entrance paper for MIT  :rotfl:  :rotfl:  :rotfl:[/quote]

Mate , .....  :rotfl:
Same here , I see many gaping holes in his/hers point of view , but i see no point in arguing , since it doesnt help neither me nor him/her . Id rather move on to the next problem in pss .  :P
Just one thing , answering ur question , why dont you use pure geometry in the IIT mathematics paper?? It is perfectly all right and you will crack problems in seconds. You seem more irritated @ the way others study and the nature of the jee papers , but thats not going to get you anywhere. It would do you more good if you do your pure mathematics and move on . Im not saying pure maths is bad , i myself appreciate it more that anything in the world , but you have to find a way around d jee paper if you want to crack it  :roll:

sagar is down with swine flu....please pray for him...
hope he recovers soon...

How is he now?
Even one of the campus students over here is probably havin it. :(

He'll recover anyways.... Swine flu is no worse than normal flu <_<

[quote="Sagar_Vaze"]

and i will tell wht they did to u at the swimming pool, and ur nickname and the funda behind it too  :rotfl: [/quote]

This looks interesting :D Do share with us also your 'experience' :rotfl:

@sumnath
Dont worry,I have got loads of worse nicknames,the best one means rotten in beng :rotfl:

Anyway,how's the swimming pool over there?The one over here is snake-infested and the source of all skin diseases in the campus.

[quote="Scary math"]
 The one over here is snake-infested and the source of all skin diseases in the campus.[/quote]

Are you serious? :o  :o

Yah,u get all sorts of snakes in dat water in monsoon.And a good quantity of bleaching powder as well. :P

[quote="Sagar_Vaze"]and i will tell wht they did to u at the swimming pool, and ur nickname and the funda behind it too  :rotfl:[/quote]

Tell us... we'd love to know. BTW, anyone heard of or participated in the Buzzer Round at IIT?

[hide="Registration"]You only have to go to IIT-K. The seniors take care of your registration...  :P  :P  :rotfl: [/hide]


Let $ {s_i}$ denote the sum of product of $ {a_i}$ taking it $ {i}$ at a time.
Then $ {s_1}=a_1+...a_n$.
Also the LHS in the original question is $ {1-s_1+s_2-...+...-...+s_n}$.
It is worthwile to know that the $ {nth}$ root of $ {n!}$ is approx. 1.
Also $ {s^k}>s_k$ because in each $ {s^k}$, the are $ {s_k}$ and $ {a_1^k,a_2^k,...,a_n^k}$.
So we are done.
If i am wrong i wish to know

Well, I wouldn't quite describe your post as "wrong," but rather as "lacking any identifiable chain of logic."  It does contain false statements, though.  For example, [quote="21 7 15"]It is worthwile to know that the $ {nth}$ root of $ {n!}$ is approx. 1.[/quote] is just false: $ \sqrt [n]{n!} \approx \frac {n}{e}$ is nowhere near 1 even for fairly small $ n$.  (For example, it's easy to see that $ 3^6 \equal{} 729 \approx 720 \equal{} 6!$ and so $ \sqrt[6]{6!} \approx 3$.  It grows from there.)

[hide="Proof"]$ (1 \plus{} x_{1})(1 \plus{} x_{2})\cdots (1 \plus{} x_{n}) \le \left(\frac {x_{1} \plus{} \cdots \plus{} x_{n} \plus{} n}{n}\right) \equal{} \left(\frac {S \plus{} n}{n}\right)^{n} \equal{} \left(1 \plus{} \frac {S}{n}\right)^{n} \equal{} \sum_{i \equal{} 0}^{n}\binom{n}{i}\frac {S^{n \minus{} i}}{n^{n \minus{} i}}$

where i used AM-GM.

It'll be enough to prove that $ \binom{n}{k}\frac {S^{n \minus{} k}}{n^{n \minus{} k}}\le \frac {S^{n \minus{} k}}{(n \minus{} k)!}$ for $ k \equal{}\overline{0,n}.$
Since, $ n! \le n^{n \minus{} k}\cdot k!$ for any $ n, k,$ we have:

$ n! \le n^{n \minus{} k}\cdot k! \Leftrightarrow \frac {n!}{k! (n \minus{} k)!}\cdot (n \minus{} k)! \le n^{n \minus{} k}$
$ \Leftrightarrow \frac {\binom{n}{k}}{n^{n \minus{} k}}\le \frac {1}{(n \minus{} k)!}$
$ \Leftrightarrow \binom{n}{k}\frac {S^{n \minus{} k}}{n^{n \minus{} k}}\le \frac {S^{n \minus{} k}}{(n \minus{} k)!},$

done.[/hide]


factorial induction algebra polynomial inequalities calculus integration

Problem:Find all  infinite sequence such that:
A_n=2 n *a+b is a square for every n which is a positive integer.
(a and b are  integers).
 :lol:

Just have a look here :
http://www.mathlinks.ro/viewtopic.php?t=4399&highlight=polish

You must have $a=0$ and then $b$ is a square. The converse is trivial.

Pierre.

Here's another way of solving this: Assume $a\ne 0$. Then we have a sequence of distinct numbers $k_n$ s.t. $k_1^2-2k_0^2=k_2^2-2k_1^2=\ldots=k_{n+1}^2-2k_n^2=\ldots=-b$. This means that $b$ has infinitely many decompositions in $\mathbb Z[\sqrt 2]$, which is absurd because it's well-known that $\mathbb Z[\sqrt 2]$ is an Euclidean and thus factorial ring.

Obviously, there are many solutions to this problem.  But the following is very elementary and easy to think of:

Suppose $x^2 = A_n=2^n a + b$ with $b>0$ and $n$ sufficiently large.  Then $(2x)^2 = 4A_n > 2^{n+2} a + b = A_{n+2} > (2x-1)^2$.
We derive a similar contradiction for $b<0$.

Well I have nice solution with induction I proved:
For each k $2^{k+1}$ divides a and $\displaystyle \frac a{2^k}+b$ is perfect square.

But nobody answered till now to an old question of my, posted long time ago: if p is a polynomial with integer coefficients and $ p(2^n)$ is a perfect square for all n, does it follow that p is the square of a polynomial? I think the answer is yes, but I have no idea about a proof.

harazi, here's a solution when p has even degree and leading coefficient 1. i do not use that it are powers of 2, i only use that we have an infinite subset of the integers. so suppose p has degree 2n. then p can be written as $p(x)=q(x)^2+r(x)$ where q has degree n and r has degree at most n-1. suppose $r\neq 0$. then $\sqrt{p(n)}$ is only finitely often equal to $q(n)$ since $r(n)$ has only a finite number of zeroes. thus we infinitely often have one of the cases $\sqrt{p(n)}\geq q(x)+1$ or $\sqrt{p(n)}\leq q(x)-1$, suppose we have the first. squaring the inequality gives(suppose we have $q(n)\geq 0$ infinitely often, else exchange q(x) by -q(x)) $p(n)\geq q(n)^2+2q(n)+1>q(n)^2+r(n)=p(n)$ for almost all n since $2q(n)+1$ has degree n whereas $r(n)$ has degree at most n-1. similarly for the other case, so we get a contradiction.
i don't know how to proceed in the case of odd degree, maybe there's a really simple way to exclude that case, i will try to find it.

Peter

[Edit: nonsense removed(I hope ;) )]

i just recognized that there is a little mistake in my last proof. actually we cannot garantuee q to have integral conefficients, but at least it has rational coefficients, so takes rational values with bounded denominator at integers, so our bounds for p must be chosen another way, but the general argument is the same.

Peter

my proof can be extended to cases with leading coefficient a square, but that are certainly not the interesting cases. sorry for posting. in the proof for harazi's statement it is important that we have powers of 2 and the proof must be in a completely other direction.

Peter

i found a solution using pell equation recurrence


Given log[x]+log{x}>=3 for positive real x. Prove that x>1001,998

[x]=n and {x}=x-n
so n(x-n)>=1000
x>=1000/n +n
the rest is exactly like before,...


trigonometry
geometry
cyclic quadrilateral
geometry proposed

[color=indigo][b][u]Problem 1.[/u][/b] Let be given a quadrilateral $ ABCD$ and $ I$ be its diagonals intersection. Let $ M,\,N$ be the midpoint of $ BC,\,CD,$ respectively. Assume that $ MI\perp AD$ and $ NI\perp AB.$ Prove that the quadrilateral $ ABCD$ is cyclic.

[b][u]Problem 2.[/u][/b] Let be given a [b]cyclic quadrilateral[/b] $ ABCD$ and $ I$ be its diagonals intersection. Let $ M,\,N$ be the midpoint of $ BC,\,CD,$ respectively. Prove that $ IM\perp AD\iff IN\perp AB$[/color]

[quote="April"][color=indigo][b][u]Problem 1.[/u][/b] Let be given a quadrilateral $ ABCD$ and $ I$ be its diagonals intersection. Let $ M,\,N$ be the midpoint of $ BC,\,CD,$ respectively. Assume that $ MI\perp AD$ and $ NI\perp AB.$ Prove that the quadrilateral $ ABCD$ is cyclic.
[/color][/quote]
[hide="Hint."] Denote $ P$ be the midpoint of $ AC.$ [/hide]

[size=134][color=darkblue]I edit again the your nice and easy problem - [b]a little jewel[/b] ![/size][/color]

[quote="Virgil Nicula"][size=117][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABCD$ be a convex quadrilateral for which denote $ \{\begin{array}{c}I\in AC\cap BD\\\ M\in (BC)\ ,\ MB=MC\\\ N\in (CD)\ ,\ NC=ND\end{array}$. Prove that

if two from the following four sentencies are truly, then and the another two sentencies are truly :[/size] $ \{\begin{array}{ccc}1 & \blacktriangleright & ABCD\--\ \mathrm{cyclically}\\\\ 2 & \blacktriangleright & IM\perp AD\\\\ 3 & \blacktriangleright & IN\perp AB\\\\ 4 & \blacktriangleright & AC\perp BD\end{array}$[/color][/quote]

Problem 1.
Denote $ P,M,N$ are midpoints of $ AC,BC,CD$ respectively.$ AC$ intersects $ BD$ at point $ I$.Since $ AB\bot NI$ and $ MP\|AB$ then $ MP\bot NI$ and $ NP\bot MI$.Since $ MN\|BD$and $ AC\bot MN$ then $ AC\bot BD$.Let $ MI$ intersects $ AD$ at $ K$ and $ NI$ intersects $ AB$ at $ L$.Since $ \angle CID=90$  then $ DN=NI$.So $ \angle BDC=\angle NDI=\angle NID=\angle LIA=\angle BAC$.
Problem2.
Let $ \angle IAK=\alpha$ and $ IDA=\beta$ then $ \angle MIC=(90-\alpha)$ and $ \angle MIB=(90-\beta)$ so $ \frac{MI}{\sin{\alpha}}=\frac{BM}{\sin{(90-\beta)}}$and $ \frac{MI}{\sin{\beta}}=\frac{BM}{\sin{(90-\alpha)}}$ $ \Rightarrow$ $ \sin2\beta=\sin2\alpha$.So if $ \alpha$ is not equal to $ \beta$ then $ \alpha+\beta=90$
$ \Rightarrow$  $ AC\bot BD$.

[quote="April"][color=indigo][b][u]Problem 2.[/u][/b] Let be given a [b]cyclic quadrilateral[/b] $ ABCD$ and $ I$ be its diagonals intersection. Let $ M,\,N$ be the midpoint of $ BC,\,CD,$ respectively. Prove that $ IM\perp AD\iff IN\perp AB$[/color][/quote]
[b]Proof.[/b]
$ \bullet$ We only have to prove that: [b]If $ IN\perp AB$ then $ IM\perp AD.$[/b]

Denote by $ P,\,Q$ the midpoints of $ AC,\,IC,$ respectively. And denote by $ H,\,K$ the orthogonal projections of $ I$ on $ AB,\,CD,$ respectively.

We have $ \angle HBI=\angle KCI\implies\angle BIH=\angle DIN=\angle KIC\quad (1)$
On the other hand, $ QN\parallel ID,$ this implies that $ \angle DIN=\angle INQ\quad (2)$
And $ \triangle IKC$ is right angled at $ K$ and $ Q$ is the midpoint of $ IC,$ so $ \angle QKI=\angle KIC\quad (3)$

From $ (1),\,(2)$ and $ (3),$ we have $ \angle INQ=\angle IKQ,$ which implies that the quadrilateral $ IKNQ$ is cyclic.
Thus $ \angle IQN=90^\circ\implies IQ\perp MN\quad (*)$
Notice that $ PM\parallel AB$ and $ IN\perp AB,$ so $ PM\perp IN\quad (**)$

From $ (*)$ and $ (**)$ we have $ P$ is the orthocentre of triangle $ \triangle IMN.$
Hence $ MI\perp NP,$ i.e. $ IM\perp AD$

[quote="April"][quote="April"][color=indigo][b][u]Problem 2.[/u][/b] Let be given a [b]cyclic quadrilateral[/b] $ ABCD$ and $ I$ be its diagonals intersection. Let $ M,\,N$ be the midpoint of $ BC,\,CD,$ respectively. Prove that $ IM\perp AD\iff IN\perp AB$[/color][/quote]
[b]Proof.[/b]
$ \bullet$ We only have to prove that: [b]If $ IN\perp AB$ then $ IM\perp AD.$[/b]

Denote by $ P,\,Q$ the midpoints of $ AC,\,IC,$ respectively. And denote by $ H,\,K$ the orthogonal projections of $ I$ on $ AB,\,CD,$ respectively.

We have $ \angle HBI=\angle KCI\implies\angle BIH=\angle DIN=\angle KIC\quad (1)$
On the other hand, $ QN\parallel ID,$ this implies that $ \angle DIN=\angle INQ\quad (2)$
And $ \triangle IKC$ is right angled at $ K$ and $ Q$ is the midpoint of $ IC,$ so $ \angle QKI=\angle KIC\quad (3)$

From $ (1),\,(2)$ and $ (3),$ we have $ \angle INQ=\angle IKQ,$ wich implies that the quadrilateral $ IKNQ$ is cyclic.
Thus $ \angle IQN =90^\circ\implies IQ\perp MN\quad (*)$
And notice that $ PM\parallel AB$ and $ IN\perp AB,$ so $ PM\perp IN\quad (**)$

From $ (*)$ and $ (**)$ we have $ P$ is the orthocentre of triangle $ \triangle IMN.$
Hence $ MI\perp NP,$ i.e. $ IM\perp AD$[/quote]

 To this [b]nice[/b] solution I have added the image(and corrected some typos).


[hide="Hint that practically solves the problem"]$ \cos 2x \equal{} \cos^2 x \minus{} \sin^2 x$[/hide]

$ \cos 2x\equal{}5\sin ^ 2 x\minus{}\cos ^ 2 x$
$ \Longleftrightarrow \cos ^ 2x\minus{}\sin ^ 2 x\equal{}5\sin ^ 2 x\minus{}\cos ^ 2 x$
$ \Longleftrightarrow \tan x\equal{}\pm \frac{1}{\sqrt{3}}$


algebra
polynomial
number theory
relatively prime
combinatorics proposed
combinatorics

We call a finite sequence $\{a_k\}_{1\leq k\leq n}$, $p$-balanced if the sums
\[ s(k,p) = a_k + a_{k+p}+ a_{k+2p} + \cdots  \] are all equal among themselves for $k=1,2,\ldots,p$.

Prove that if a sequence of 50 reals is 3, 5, 7, 11, 13 and 17-balanced, then all its terms are equal with 0.

[hide="what I managed until now"]
We can make each $p$-balancing condition into a system of $p-1$ linear homogenous equations:
\[ \begin{array}{ccc} a_1 + a_{p+1} + \cdots - a_{2}-a_{2+p} - \cdots & = & 0  \\
a_1 + a_{p+1} + \cdots - a_{3}-a_{3+p} - \cdots & = & 0  \\
\vdots & \cdots & \vdots \\
a_1 + a_{p+1} + \cdots - a_{p-1}-a_{2p-1} - \cdots & = & 0  \\
\end{array}
\]
Now this means that the statement of the problem is translated into the following: we are given a system of linear homogenous of 3-1+5-1+7-1+11-1+13-1+17-1 = 50 (!) equations, and 50 unknowns, show that the only solution is the trivial solution. By Cramer all we have to prove now is that the determinant of the system is non-zero.

However computing that determinant is anything but an easy task. So any other ideas? :)
[/hide]

Nevermind this, I solved it myself after all. At first I thought this was divisible with $1+x+\cdots +x^{p-1}$, for $p$ prime, because of a mistake in computation, and I just corrected my computations, and it is divisible with that after all :) Problem solved. :D

We can now create an auxiliary problem, prove that the determinant above (only with 1, 0 and -1) is non-zero :lol:

Suppose a sequence $\{ a_{k} \}, k=1,\ldots, n $ is $p$-balanced.
Take a complex number $ \omega $, such that $ \omega^{p}=1, \omega
\neq 1 $ and consider
\[ S_{\omega} = \sum_{k=1}^{n} a_{k} \omega^{k} \].
Then we have \[ S_{\omega} = \sum_{k=1}^{p} s(k,p)\omega^{k} \]
and this is equal to zero, because $ s(1,p) = s(2,p) = \ldots =
s(p,p) $ and \[ \sum_{k=1}^{p} \omega^{k} = 0 \]. So we get that
if $\{ a_{k} \}, k=1,\ldots, n $ is $p$-balanced, then for every $
\omega $, such that $ \omega^{p}=1, \omega \neq 1 $, we have
\[ S_{\omega} = \sum_{k=1}^{n} a_{k} \omega^{k} = 0 \].
Now if we have our sequence $ \{ a_{k} \}, k=1,\ldots, 50 $ of
length 50, which is 3,5,7,11,13,17 balanced, then there are
exactly 2+4+6+10+12+16=50 roots of unity ( not counting $1$ ) of
orders 3,5,7,11,13,17. Denote them by $
\omega_{1},\omega_{2},\ldots ,\omega_{50} $ , and we have for each
$ \omega_{i} $ that \[ \sum_{k=1}^{50} a_{k} \omega_{i}^{k} =0 \].
Now if we denote by \[ P(z) = \sum_{k=1}^{50} a_{k} z^{k-1} \] a
polynomial of $z$ then it is of degree $49$ and at least has $50$
different roots $ \omega_{1},\omega_{2},\ldots ,\omega_{50} $. It
means that it is identically zero, so \[ a_{1} = a_{2} = \ldots =
a_{50} = 0 \]

Leva1980, I give the [b]OFFICIAL[/b] solution which is basically the same as yours.

Let $z$ be the complex $p-th$ root of unity that is not equal to 1. Since

\[0 = z^p - 1 = (z-1)(z^{p-1} + z^{p-2} + \ldots + z + 1),\]

we have

\[z^{p-1} + z^{p-2} + \ldots + z + 1 = 0.\]

Now we associate with any $p-$balanced sequence $\left( a_k \right)$ a polynomial

\[f(x) = \sum^n_{k=1} a_k \cdot x^{k-1}\]

Note that

\[a_1 + a_{p+1} \cdot z^{p} + a_{2 \cdot p+1} \cdot z^{2 \cdot p} + \ldots = a_{1} + a_{p+1} + a_{2 \cdot p+1} + \ldots,\]

\[a_2 + a_{p+2} \cdot z^{p} + a_{2 \cdot p+2} \cdot z^{2 \cdot p} + \ldots = a_{2} + a_{p+2} + a_{2 \cdot p+2} + \ldots,\]

and so on. Denoting the sum

\[a_{1} + a_{p+1} + a_{2 \cdot p+1} + \ldots\]

by $S$, we obtain that

\[f(z) = S \cdot \left( 1 + z + \ldots + z^{p-1} \right) = 0.\]

All sums are the same of the form

\[a_{k} + a_{p+k} + a_{2 \cdot p+k} + \ldots\]

are the same for any $k$ by the property of the sequence $\left( a_i \right)$ There are exactly $p$ $p-$th roots of unity for
any $p$ and exactly $p-1$ of them are not equal to 1. Because it is proved that each of these $p-1$ number is the root
of a polynomial formed by a $p-$ balanced sequence, we conclude that the polynomial $f(x)$ formed by the given sequence
has at least

\[2 + 4 + 6 + 10 + 12 + 16 = 50\]

roots. Note that any two numbers $p,q$ from the set

\[\{3,5,7,11,13,17\}\]

are relatively prime and therefore the the $p-$th root of unity can conincide with the $q-$th root of unity if and only if
they are equal to 1. Thus, we see that a certain polynomial $f(x)$ of the $49-$ degree has no less than 50 distinct
roots, which implies

\[f(x) \equiv 0,\]

that is,

\[a_{1} = a_{2} = \ldots = a_{50} = 0.\]

[quote="Pascual2005"]jaja, is this the only solution?[/quote]You could try computing that $50\times 50$ determinant I reached trying that other method :)  (see my hidden note in my first post).

jaja, no thanks! i will let that to a computer!


geometry
circumcircle
projective geometry
geometry proposed

Let $P$ be a interior point of the triangle $ABC$. I denote:
$\blacktriangleright$ the second intersections $B'$, $C'$ of the lines $BP$, $CP$ respectively with the circumcircle $w$;
$\blacktriangleright$ the points $M\in AB$, $N\in AC$ so that $P\in MN$; $A'\in C'M\cap B'N$;
$\blacktriangleright$ the circumcenters $O_1$, $O_2$ of the triangles $BA'N$, $CA'M$ respectively.
Prove that $\widehat {O_1OO_2}\equiv \widehat {BAC}\ .$

[b]Remark.[/b] If the point $P$ is the circumcenter of the triangle $ABC$, then we obtain the proposed problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=16567

Consider the hexagon AC'BA'CB' with A, C', B, C, B' concyclic on the circumcircle w. Since the intersections $P \equiv BB' \cap CC',\ M \equiv AB \cap C'A',\ N \equiv CA \cap A'B'$ are collinear, by Pascal's theorem, A' also lies on the circumcircle w, on the arc BC opposite to A. Hence $\angle BA'C = 180^\circ-\angle BAC.$ But $O_{1}O,\ O_{2}O$ are perpendicular bisectors of the segments BA', CA', hence $\angle O_{1}OO_{2}= 180^\circ-\angle BA'C = \angle BAC.$


Does anyone have the answers to the Sprint and Target rounds for Nationals 1990-1991? Any help wold be appreciated

I have them somewhere. I'll try to find them...


A 5-inch by 7-inch picture is placed on a rectangular piece of red paper such that a 0.5-inch wide red border is seen along each side of the picture. What is the area of the visible red border, in square inches?

[hide]The perimeter of the picture is $ 5\plus{}5\plus{}7\plus{}7 \equal{}24$, so multiplying that by $ 0.5$ we have our primary estimate of $ 12$.  However, we have not accounted for the four squares of side length $ 0.5$ on the corners, so we must add those in to get our final answer, $ 13$.[/hide]

How does finding the perimeter give us the area?

I just found the area of the red border minus the area of the picture.

7+.5+.5=8  and 5+.5+.5=6  These are the lengths of the red paper.
8*6-7*5=48-35=13              Total area of red paper shown


algorithm
geometry
3D geometry
analytic geometry
parameterization
topology
function

I am trying to do a computer visualization of a surface in complex dimensions 2. I choose simple quintic equation:
$z^{5}_{1}+z^{5}_{2}= 1$
which has some an interesting meaning for me, but its not important here.

I also implemented algorithm for producing plots of 3D surfaces that are defined with algebraic equations. It is called Marching Cubes and it simply checks how surface intersects a grid of cubes, so it produces a triangles to approximate the surface. The same technique could be used for 4D, where instead of cubes, it uses hypercubes and extracts tetrahedras to approximate surface. I tested algorithm for hyperspheres and it works perfectly, however this equation in 4D (mentioned above) gives strange results.

Generally, in algorithm, it should iterate through all 4 coordinates (x,y,z,w) to produce plot, like 0.0 < x < 1.0,  0.0 < y < 1.0, 0.0 < z < 1.0, 0.0 < w < 1.0. The problem is, that I have two complex dimensions and I try to map them to 4 real dimension to produce a surface.

How can I do this and what methods could be used? Or could it be simply accomplished by:
$z_{1}= x+iy$
$z_{2}= z+iw$?

Thank you.

[quote="Alteran"]$z_{1}= x+iy$, $z_{2}= z+iw$[/quote]
That's the right way to make two complex dimensions into four real dimensions. Of course, handling four dimensions graphically is nontrivial.

You should be looking for triangles, not tetrahedra; your surface here is a 2-dimensional object in 4-dimensional space, not a 3-d object.

Thank you. Yes, I though it to be the right way.. however when I use parameterization for this equation, I get compact curled manifold, which consists of 25 patches glued together (I used complex superquadrics). Picture is quite nice and it is possible to deduce some interesting properties of the manifold. However it should be equally satisfied with algebraic representation and when I do a plot in 4D, I get noncompact wave like surface, far from that I have seen with parameterization (of course, I set w to constant and do not interate through all values).
So, I just want to understand, what is wrong. By the way, why my surface is 2D in 4D, but not 3D?

Roughly speaking, a system of $m$ equations over real numbers with $n$ variables should have a $(n-m)$ dimensional set of solutions. One equation over complex numbers is two equations over real numbers (because both real and imaginary parts must match). This is why your surface has $4-2=2$ dimensions.

One way to represent the fourth coordinate on a graph is by color: $(1,2,3,4)$ could be a green dot at $(1,2,3)$, while $(4,3,2,1)$ is a blue dot at $(4,3,2)$.

Got it.
So basically, software need to solve these equations?
I tried to find all intersections of this equation
$z^{5}_{1}+z^{5}_{2}-1 = 0$
with hypercubes in 4D, points where equation is $\geq 0$

However, for 3D, algorithm extracts set of triangles, because when surface intersects the cube (there are 255 ways to intersect a cube), it is possible to triangulate whole slice, there could be 1 triangle, 2 or 4 triangles for one slice. In 4D, algorithm extracts a set of tetrahedras. I tried to display only one side of it and still it is far from parametric version. If I use equation of real numbers, like hypersurface $x^{2}+y^{2}+z^{2}+w^{2}-1 = 0$, then I see whole sphere.

But, you told me, that I have actually 2D surface, I wonder what I did wrong. May be here its not possible to just fix w coordinate. Basically, I calculate it as:
$0 < x < 1$
$0 < y < 1$
$0 < z < 1$
$w = 0$
$z_{1}= x+iy$
$z_{2}= z+iw$
$(z_{1})^{5}+(z_{2})^{5}-1 = 0$

That complex equation is really a system of two real equations; is your algorithm set up to handle that?

Actually no, its just put values x,y,z,w there and calculates the equation. Algorithm only need to know what is the value of the equation in some particular points in 4D space (x,y,z,w), which are vertices of hypercubes (I divided 4D space to small hypercubes). Then, what is the reason to solve a system, it has 4 variables and z1 is independent from z2, as I can see. Can you give me more hints?

OK, seems like I did it. Solution was more-less trivial :blush:
We have complex function $f(x)=c$, and if we set $c=0$ then we have two constraints $real(f)=0$ and $imag(f)=0$. It is possible to extract tetrahedras according first equation and then reduce dimensionality with second equation (simple contouring).


I dunno where I should post this so any moderator people can move it to its proper location...

How many ways can one throw  n  dice each with  f  number of faces?

i guess i should have been more specific. I meant how many combinations can you throw n dice each with f faces. Like where order doesnt matter, for example, when having 2 six-sided dice, rolling a 2 then a 1 would be counted as basically the same thing as rolling a 1 then a 2. So in this scenario, have 2 six-sided dice there would be 21 different combinations of throwing 2 dice each with f faces... i hope this clears it up some

[hide]There are a certain number of 1s, 2s, 3s,...,fs. All that matters is how many there are, not in what order they are. Denote these numbers by a1,a2,a3,..,af respectively. Then we are finding all solutions to the equation a1+a2+a3+...+af=n. In this case, a1,...,af are [b]nonnegative[/b]. We change the equation so that a1+a2+a3+...+af=n+f. Note that the answers have a one-to-one correspondence in these two equations but in this one, a1,...,af are [b]positive[/b]. Break up the n+f into lines:
||||||||||...
And insert f-1 addition symbols between them. The first number of lines represents a1, then a2, a3, ..., af. Therefore, the answer is [b]n+f-1 choose f-1[/b].[/hide]


inequalities
function
inequalities unsolved

Let $ x,y,z$ be real numbers greater than or equal to $ 1.$ Prove that
\[ \prod(x^{2} \minus{} 2x \plus{} 2)\le (xyz)^{2} \minus{} 2xyz \plus{} 2.\]

Set $ x\equal{}a\plus{}1$ , $ y\equal{}b\plus{}1$ and $ z\equal{}c\plus{}1$ , so $ a,b,c\ge 0$

then the inequality turns to

$ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\le ((a\plus{}1)(b\plus{}1)(c\plus{}1))^2\minus{}2(a\plus{}1)(b\plus{}1)(c\plus{}1)\plus{}2$

$ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\le \left(2(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}2\plus{}(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\plus{}2abc(ab\plus{}bc\plus{}ca\plus{}2a\plus{}2b\plus{}2c)\right)\minus{}2(a\plus{}1)(b\plus{}1)(c\plus{}1)\plus{}2$

$ 0\le 2abc(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)$ obviously true

so equality holds if and only if at least one of $ a,b,c$ equal to $ 0$ so at least one of $ x,y,z$ equal to $ 1$

[quote="matrix41"]Set $ x \equal{} a \plus{} 1$ , $ y \equal{} b \plus{} 1$ and $ z \equal{} c \plus{} 1$ , so $ a,b,c\ge 0$

then the inequality turns to

$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\le ((a \plus{} 1)(b \plus{} 1)(c \plus{} 1))^2 \minus{} 2(a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \plus{} 2$

$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\le \left(2(a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \minus{} 2 \plus{} (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \plus{} 2abc(ab \plus{} bc \plus{} ca \plus{} 2a \plus{} 2b \plus{} 2c)\right) \minus{} 2(a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \plus{} 2$

$ 0\le 2abc(a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca)$ obviously true

so equality holds if and only if at least one of $ a,b,c$ equal to $ 0$ so at least one of $ x,y,z$ equal to $ 1$[/quote]

we have the same solution... but I got $ 0\le 4abc \sum a \plus{} 2 \sum ab \plus{} 2abc \sum ab \plus{} 2 \sum ab \sum a,$ equality holds when at least two of $ a,b,c$ equals $ 0$ :D

Note that we only have to prove it for two variables, that is, $ (x^{2}\minus{}2x\plus{}2)(y^{2}\minus{}2y\plus{}2)\le (xy)^{2}\minus{}2xy\plus{}2$. Then we can repeat with $ (x,y)\rightarrow (xy,z)$. However expanding (unless I have made an error) turns this into $ (x\minus{}1)(y\minus{}1)(x\plus{}y\minus{}1)\ge0$, which is obviously true.

indeet  CatalystofNostalgia, we can use your ineq to prove the general one , or we can use  theorems with convex/concave functions to the function $ f(e^x)$ where $ f(x)\equal{} ln( x^2\minus{}2x\plus{}1)$


See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284260]here[/url]. The image just shouldn't be that big!  :o

Let's try here too:
[asy]size(50);
draw((-7,-7)--(7,7));[/asy]

It seems to be so. I will take a look at it.

unitsize doesn't work quite right sometimes too; generally, though, I'm able to use one of the two to get the right size.

Bug is fixed. Size now works as it worked before. Additionally while debugging I've noticed that the comments inside the code weren't working properly, and that is now fixed as well

[asy]size(75);
// we draw a line
draw(origin--(2,0));
/* then we draw the points */
dot(origin); dot((2,0));[/asy]

Currently 150 is the default size (square picture.)

[asy]size(2);
draw((-7,-7)--(7,7));[/asy]

size two is the same as size 50

It's cached image, from when the size rendering didn't work :)


Can you simplify
$\frac{\cos{x}\sec{x}+\cos{x}-\sin{x}\sec{x}\tan{x}}{(\sec{x}+1)^2}$ anymore?

I think that the maximum we can do is:
$\frac{1+\cos x-\frac{\sin^2x}{\cos^2x}}{\left(\frac{1+\cos x}{\cos x}\right)^2} = \frac{\frac{\cos^2x+\cos^3x-1+\cos^2x}{\cos^2x}}{\frac{\left(1+\cos x\right)^2}{\cos^2x}} = \frac{\cos^3x+2\cos^2x-1}{\left(1+\cos x\right)^2} = \frac{\cos^2x+\cos x-1}{\cos x+1}$.

He cancels off the same factor from numerator and denominator .

Notice here $cos^3 x+2cos^2 x -1 = (cos x +1)(cos^2 x+cos x -1)$

but I duno whether it is right or not cause i come out with something like

$\frac 12 -\frac 32 \cdot tan^2 \frac{x}{2}$  :?

[quote="shobber"][quote="236factorial"]My books says something else, this was intended to be an easy calculus problem, but I just needed help with this last step. Maybe i started the problem wrong.[/quote]
Can you post the calculus solution? Never heard of the solving this kind of problem with calculus.[/quote]

You must've misunderstood (i didn't make it clear), it was the final step to a calculus problem (but I can't post that here, it'll get split). there's no calculus solution to this problem, I don't think...

I think it isn't correct $\frac{1}{2} - \frac{3}{2} \cdot \tan^2\frac{\alpha}{2}$.

[quote="joml88"]Looks like you were trying to do a differentiation problem...[/quote]

Yes, the quotient rule, I must've screwed up. In the answer, it had $\tan$ inside it.
shyong isn't correct however.


A rectangular box has sides with squares 84, 70 and 30 squareunits. Whats the volume of the box?

[hide="Hint 1"] Let $a, b, c$ be the length, width, and height, respectively.  Then $ab = 84, bc = 70, ca = 30$ and we want to find $abc$. [/hide]
[hide="Hint 2"] $abc = \sqrt{ ab \cdot bc \cdot ca}$ [/hide]

[hide="answer"]
I just happened to notice the dimensions to be $5$ x $6$ x $14$. The volume would therefore be $420 units^{3}$.

t0rajir0u's way is very useful, particularly when the numbers are not so small. However, in this case, the numbers are manageable and we just need to look for common factors.[/hide]

[hide="Big Hint"]Another possibility is just to use the substitution method to solve the three equations ab=84, bc=70, and ca=30.  Once you find all three variables, just multiply them together to get the volume.  Of course, t0rajir0u's solution is much quicker.[/hide]

[hide="Answer"]a, b, and c are the sides. You are given ab, bc, and ac. to find abc, multiply ab, bc, and ac, and take the square root of that. The answer becomes [b]420[/b].[/hide]


Let $ a,$ $ b$ and $ c$ are side lengths of triangle. Prove that:
\[ (ab\minus{}c^2)\sqrt{a\plus{}b\minus{}c}\plus{}(ac\minus{}b^2)\sqrt{a\plus{}c\minus{}b}\plus{}(bc\minus{}a^2)\sqrt{b\plus{}c\minus{}a}\geq0\]

[color=darkblue]Substitute $ \|\begin{array}{cc} a = x + y \\
b = y + z \\
c = z + x \end{array}$

Then we have to prove:
$ \begin{array}{cc} \displaystyle \sqrt {2} \sum ((x + y)(y + z) - (x + z)^2)\sqrt {y} \ge 0 \Leftrightarrow \\
\Leftrightarrow \displaystyle \sqrt {2}(\sum x - \sum\sqrt {xy})(\sum \sqrt {x^3} + \sqrt {xyz}) \end{array}$
Which is always true by $ Am - Gm$.[/color]


group theory
abstract algebra
superior algebra
superior algebra unsolved

let G be a group with subgroups H and K s.t.HK=G.show H intersection K= identity.

Don't they have to be [i]proper[/i] subgroups?

@who: This is obviously wrong. Please check missing conditions.


What percent discount is equivalent to a $ 40\%$ discount followed by a $ 10\%$ discount?

and 40 percent discount means taking 40 percent away from 100 percent. Therefore we have 60 percent left. Similarly, taking 10 percent off is the same as 90 percent left. So .6*.9=.54. Since the problem is asking for the discount it is
 1-.54=.46= 46 percent discount

[hide=Easy]Let's say the total amount is $100 dollars.$
$40$ percent away means subtracting 40 dollars from 100 dollars.
$100-40=60$
Then we take $10$ percent away from $60$, or subtract $6$ dollars.
$60-6=54.$
Since the new amount is 54 dollars, the discount is $100-54$ or $46$ dollars.
Answer: $\boxed{46}$[/hide]
ThX!

[hide]This is pretty easy. 40 percent means 40 out of 100. So out of that, we have 60 percent left. 10 percent of 60 percent is 4. 60-4 = 54 Then, subtract 54 from 100 to get 46[/hide]


Let $ a, b, c$ positive real numbers. Prove or disprove that

$ \sum(\frac{2a^2 \plus{} b^2c^2 \plus{} 2bc \plus{}1}{b^2 \plus{} bc \plus{} c^2})\ge \frac{22}{3}$.

This inequality seems wrong.
Try $ a\equal{}b\equal{}c\equal{}1$.
The LHS $ \equal{} 6 < \frac{22}{3}\equal{}$ RHS.


Does a.e conv. of a seq. of funct. => measure conv. and what about the converse?

Could you state explicitly what is mean by "measure convergence"?

f_n converges in measure to f
if for every d positive,
the measure of
...x: Abs(f_n(x) - f(x)) greater or equal d...
tends to 0
as n tends to infinity.


Does it seem to anyone else like having to wait over 6 months for these individuals (not me) to be recognized appears to be an inordinately long time? I guess I'm just a bit impatient .. my apologies to the AMC organization for my anxiousness to see this list.  :o   Does anyone know if the names of those individuals who scored perfect scores are listed anywhere else?

Well they have the list of AMC 10 perfect scores, and AMC 12 perfect scores, so you could create your own list.

I only know of one double perfect score offhand.

well, if you mean what i think your intending (took 10a&12b or 12a&10b and perfect scored both), from the amc's list, there's only one individual who did that this year (and you can find that on the list yourself :p)

or do you mean for every year?

although only a couple of people responded to the question, a whopping 117 + Aopser's read the post titled "AMC 10 and 12 Perfect Scorers 2009" WITHIN ONLY ABOUT 24 HOURS OF BEING POSTED so obviously there is much interest in knowing (either by people in general who are curious who these scorers were or from the scorers themselves (huh, were there 117 of them?? I have no idea ...)   :P

AMC - Please post the list soon!! Will it at least be posted prior to the start of the next school year in Sept.?   Strangely enough, the person I know who did score a perfect score was asked for their picture many, many months ago .. but it never got posted - at least not as yet.

Did you look on the [url=http://www.unl.edu/amc/]AMC site[/url]?  There's a link to the list of perfect scorers on the front page.

Oh, actually I was clicking on the list to the left that says AMC 10 and then it opens to a page where it highlights perfect scorers in pink but only shows up to 2008 -- I was waiting 'till they posted the names and pictures in that nicely formatted page so you can scroll over the states and pictures and see the names at a glance, but I had not seen the chart you pointed out, so thank you very much! (though I still hope they haven't abandoned the page with the pictures altogether)


[hide="1"]The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius of the circle?[/hide]

The below problem is a slight variation of the above AMC 10 problem.

[hide="2"]The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its [b]inscribed[/b] circle. What is the radius of the circle?[/hide]

1.
[hide]$ 3s\equal{}\pi r^2$, and $ \dfrac{s^2\sqrt{3}}{4}\equal{}\dfrac{s^3}{4r}$.
$ s\equal{}r\sqrt 3$
$ r\equal{}\dfrac{3\sqrt{3}}{\pi}$.[/hide]

2.
[hide]$ 3s\equal{}\pi r^2$, and $ \dfrac{s^2\sqrt{3}}{4}\equal{}\dfrac{3rs}{2}$
$ s\equal{}2r\sqrt3$
$ r\equal{}\dfrac{6\sqrt 3}{\pi}$[/hide]


Howś can I solve it
X² -5x -36 is congruent to 0 mod103

[hide]$(x+4)(x-9) \equiv 0 \bmod 103$. Since $103$ is prime, $x+4 \equiv 0 \vee x-9 \equiv 0$.[/hide]


calculus
integration
trigonometry
calculus computations

Find the antiderivatives:

$ \int{\frac{1}{\cos{x}-1}dx}$

$ \int{\frac{4}{4x^2+4x+65}dx}$


algebra
polynomial
quadratics
symmetry
geometry
3D geometry
parameterization

Find all solutions for $ \sqrt[3]{xyz} \equal{} \frac{x\plus{}y\plus{}z}{3}$ for $ x,y,z$ distinct integers.

I found a class of infinitely many solutions, but I can't prove that these are all of the solutions.

$ x \plus{} y \plus{} z \minus{} 3 \sqrt [3]{xyz} \equal{} \frac {1}{2} (\sqrt [3]{x} \plus{} \sqrt [3]{y} \plus{} \sqrt [3]{z})( (\sqrt [3]{x} \minus{} \sqrt [3]{y})^2 \plus{} (\sqrt [3]{y} \minus{} \sqrt [3]{z})^2 \plus{} (\sqrt [3]{z} \minus{} \sqrt [3]{x})^2)$

When the first factor is equal to $ 0$, we have the solution set $ x \equal{} a^3, y \equal{} b^3, z \equal{} \minus{} (a \plus{} b)^3$.  When the second factor is equal to $ 0$, we have the solution set $ x \equal{} y \equal{} z$.

for the second part, they dont have to be cubes. Im fairly sure this applies in the first case too.

Hmm.  You're right about the second case, but most integer choices of $ x, y$ won't produce an integer value of $ z \equal{} \minus{} (\sqrt [3]{x} \plus{} \sqrt [3]{y})^3 \equal{} \minus{} x \minus{} y \minus{} 3 \sqrt [3]{xy} ( \sqrt[3]{x} \plus{} \sqrt[3]{y})$.

But trying to show that just reduces back to the original problem! I have gotten it down to trying to solve $ 2d^2 \plus{} 2c^2 \minus{} 5dc \equal{} 0$ in $ \mathbb{N}$ With some other conditions on $ d,c$ but im a bit tired so i might give it a go tomorrow

If $ z$ is an integer then $ \sqrt [3]{x^2 y} \plus{} \sqrt [3]{xy^2} \equal{} r$ is rational, and

$ \sqrt [3]{xy^2} \equal{} \frac {1}{x} \left( \sqrt [3]{x^2 y} \right)^2$.

Assume by way of contradiction that $ \sqrt [3]{x^2 y}$ is irrational; then the minimal polynomial of $ \sqrt [3]{x^2 y}$ is a quadratic

$ t^2 \plus{} xt \equal{} rx$.

But if $ \sqrt [3]{x^2 y}$ is not itself rational then its minimal polynomial must be

$ t^3 \minus{} x^2 y \equal{} 0$.

Contradiction.  Hence $ \sqrt [3]{x^2 y}$ (and $ \sqrt [3]{xy^2}$ by symmetry) are both themselves rational (hence, integers by the Rational Root Theorem).  Assuming WLOG that $ \gcd(x, y) \equal{} 1$, we conclude that $ x, y$ are both cubes as desired.

Edit:  I'm not sure how the quadratic you gave plays into this, but it factors as $ (2c \minus{} d)(c \minus{} 2d) \equal{} 0$.

Now thats an elegent argument! Applause! :clap2:

Thanks all. This is my solution:

$ (x\plus{}y\plus{}z)^3 \equal{} 27xyz$

Assume $ d \equal{} \gcd(x,y,z)$. WLOG, $ d\equal{}1$ because if not, we can always divide each side by $ d^3$.

Now, let $ e$ be a prime number that divides both $ x$ and $ y$. If $ e | z$ then $ e | d$, contradiction.
So $ e$ does not divide $ z$. But then $ e^2 | (x\plus{}y\plus{}z)^3$ that means $ e | x\plus{}y\plus{}z$ thus $ e | z$. Contradiction.
So $ \gcd(x,y) \equal{} 1$.
Similarly, $ \gcd(y,z) \equal{} \gcd(z,x) \equal{} 1$.
Therefore, they all have to be cubic.

From here, it's easy to see that the solution has to be in the form of $ (x,y,z) \equal{} (a^3,b^3,\minus{}(a\plus{}b)^3)$ and of its permutations and multiples.

The parameterization you gave leaves out $ (x, y, z) \equal{} (1, 1, 1)$ and multiples.  (The most appropriate replacement for the last step is the factorization I gave.)


i have been curious to know if there is a specific method by which we can solve a rubicks cube.can we use programming?

[quote="Poincare"]there's lots. For a beginner (such as me), I would say to use the cross method or just brute force the crap out of it (the latter is highly unrecommended as it can take years)[/quote]

I usually use the 2*2*2 box method instead of the cross method...Either way it's definitely programmable.

But if you want to get really spiffy, you could try to program it to recognize certain series of moves as redundant or reducible to other smaller series of moves.

can you please give links regarding these methods

This is the 2*2*2 box method, also known as the Petrus method,

http://www.lar5.com/cube/


analytic geometry
geometry
perpendicular bisector

This if from the New Jersey forum, since it has not been answered in months I thought I would post it here.

[quote="tiredepartment"]In the coordinate plane, the endpoints of a chord of a circle have coordinates $(-1-2\sqrt{3},4)$ and $(-1+2\sqrt{3},4)$.  If the degree measure of the minor arc determined by this chord is $120^{o}$, determine the corodinates of all possible locations for the center of this circle.[/quote]

[hide="outline of solution"]Draw radii from the endpoints of the chord, and form some 30-60-90's. There are two possible centers.[/hide]

[quote="AstroPhys"]This if from the New Jersey forum, since it has not been answered in months I thought I would post it here.

[quote="tiredepartment"]In the coordinate plane, the endpoints of a chord of a circle have coordinates $(-1-2\sqrt{3},4)$ and $(-1+2\sqrt{3},4)$.  If the degree measure of the minor arc determined by this chord is $120^{o}$, determine the coordinates of all possible locations for the center of this circle.[/quote][/quote]
Solution:
[hide]Well call the center of the circle $(h,k)$.  We can figure out $h$ because the $y$ coordinates of the chord are the same.  Simply average the $x$ coordinates:

$h=\frac{[(-1-2\sqrt{3})+(-1+2\sqrt{3})]}{2}$
$h=\frac{-2}{2}=-1$

So now we know our center is going to occur at $(-1,k)$.  Note that the distance between the two points is $|-1-2\sqrt{3}+(-1+2\sqrt{3})|=2$

The radius will go from the center of the circle to each of the endpoints of the curve so we will have an isosceles triangle with sides $r,r,2$ and degree angles $120, 30, 30$.  We can draw a segment from the center such that it is a perpendicular bisector of the chord, producing two equal $30-60-90$ triangles.  Call this perpendicular bisector $d$.

Setting up a comparison:

$\frac{1}{\sqrt{3}}= \frac{d}{1}$
$d=\frac{\sqrt{3}}{3}$

$d$ is the distance from the center of the chord to the center thus we have two solutions:

$\boxed{(-1,4 \pm \frac{\sqrt{3}}{3})}$[/hide]

You know that you can draw a diameter that is perpendicular to and bisects the chord because you can do that with any chord. You also know this diameter is straight up vertical which means your x coordinate will not change but your y will. You know the center of that chord is -1,4 because that is the average of the two end points. We know that intersecting chords always multiply to get the same product.

so 2root3 squared gets you 12. We now know the two parts of the chord will multiply to be 12. Also since the minor arc formed is 120 this is the perfect opportunity to draw a equilateral triangle using the original chord as one of its side. This will allow the height of the triangle using the original chord as the base to be equal to one part of the chord that will multiply out to be 12. Using the Pythagoras theorem we get that height to be 6 so the other piece must be 2. 2 is the essential number as that is how far the center is away from the midpoint of the chord. Since the x coordinate is going to stay the same (-1) only the y will change. You can either get 6 or 2 because you can add or subtract 2.

Hence the answers for the center of the circle are

(-1,2) or (-1,6)

[hide="Correct answers"]
$ (\minus{}1,2),(\minus{}1,6)$.
[/hide]
This topic is too old to be revived, though the correct answer was never posted.


symmetry
geometry
3D geometry
tetrahedron
group theory
abstract algebra
superior algebra

Find, with proof,  the symmetry group of a 4 dimensional tetrahedron.

I assume you mean the 4-dimensional regular simplex, which has 5 vertices ("tetrahedron" seems like the wrong word).

The symmetries are completely determined by their action on the vertices, so the symmetry group is a subgroup of the permutation group $S_5$. Since every pair of vertices is connected by an edge of equal length, every permutation determines a symmetry and the group is $S_5$.


I've come across a question that asks me to find an elementary matrix E such that EA=B where matrix A and matrix B are given.  I'm not sure how to go about this, can anybody help with the steps I should follow to solve a question of this nature.

Do I take matrix B and multiply it by the Inverse of matrix A? :huh:

You are probably looking for an elementary matrix that corresponds to swapping two rows, adding one to another, or multiplying one row by a scalar.  Without knowing what A or B are, however, it's impossible to tell you any more.

Here is a general form:
The following equation $AX + XB = C$ is called Sylvester equation. Using some theory u can get from this to the following equation : $SY+YT=D$, where $S=U^tAU,T=V^tBV,Y=U^tXV,D=U^tCV$ are in Schur-form. This equation has sollution iff $\lambda_i (A) + \lambda_j (B) \neq 0$.
Taking $A=0$ u get your equation $XB=C$. So this has sollution iff $\lambda_j (B) \neq 0$

This is called as SCHUR-SCHUR method. There are more metods like Hessenberg-Schur, Metod of eigenvalues and eigenvectors,etc..

[quote="jmerry"]Multiplication on the left by an elementary matrix is a row operation. Look at the two matrices, find the changed row or pair of rows, and figure out what row operation got you there. At that point, it's easy to translate the row operation into an elementary matrix.[/quote]

A handy way of finding the matrix corresponding to a sequence of row operations is to write down an identity matrix next to the matrix you're working on and do the row operations on both at the same time.

davidbrw: Listen to jmerry and rgep. They have what you need. What jianovidiu posted was out of place on this particular thread; don't worry about it if you're having trouble understanding it.


Find $ a,b,c \in \mathbb{R}$ such that $ f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous on $ \mathbb{R}$, where $ f(x)\equal{}\displaystyle\lim_{n\to\infty}\frac{ax^2\plus{}bx\plus{}c\plus{}e^{nx}}{1\plus{}c \cdot e^{nx}}$.

If $ x>0$ then $ f(x)\equal{}\frac{1}{c}$.

If $ x\equal{}0$ then $ f(x)\equal{}1$.

If $ x<0$ then $ f(x)\equal{}ax^2\plus{}bx\plus{}c$

$ f$ is continios, so $ \frac{1}{c}\equal{}1$, this gives $ c\equal{}1$.

Answer $ a,b\in\mathbb{R}$ and $ c\equal{}1$.


[quote="anuj kumar"]If a,b,c are positive reals and a+b+c=3
   Prove that        a/(ab+1)+b/(bc+1)+c/(ca+1)>=3/2.
   Anuj.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689


How do I change the font size so that the labels aren't so small?

EDIT: Never mind, I got it. defaultpen(fontsize(50));

You can also choose different color/font size for each labelling command. Just put [i]color+fontsize(number)[/i] to the pen argument. For instance,
label("$ A$",(0,0),S,magenta+fontsize(40));
will produce a big letter $ A$ in magenta color.

Since typing "magenta+fontsize(40)" several times may be too tiresome, it is usually a good idea to define all pens you are going to use in the beginning. So, writing
pen m40=magenta+fontsize(40);
in the beginning and using m40 every time you need that big colorful letter may spare some typing time.

Of course, resetting the defaultpen also works (and is all you need if you want all your labels the same size and color).


Which is the best/easiest way to prove the following problem?

XYZ is a triangle with angle X = 90
XW is an altitude of triangle XYZ.
Show that triangles XYZ, XWZ and XYW are all similar.

Is it by showing the triangles are equiangular or by showing the ratio of the sides are equal?

I think I’ve done it by showing the triangles are equiangular.
angle X = angle ZWX = angle YWX = 90 (from definition of altitude of a triangle)

angle X + angle Y + angle Z = 180 (angles of a triangle = 180)

(Not sure if I can do the following, I've chosen the angles that look similar from a drawing of the problem, or have I assumed too much?)
angle Z = 90 – angle Y = angle ZWX – angle ZXW = 90 - angle ZXW
so angle ZXW = angle Y
angle WYX is angle Y
therefore angle Y = angle WYX = angle ZXW

angle Y = 90 – angle Z = angle YWX – angle YXW = 90 – angle YXW
so angle YXW = angle Z
angle WZX is angle Z
therefore angle Z = angle WZX = angle YXW

therefore triangles are equiangular which implies triangles are similar

Any hints answers would be greatly appreciated.

I just give you the following hints:
-We only know something about angles.
-$ XW\perp AB\Rightarrow \angle XWA\equal{}\angle BWX \equal{} 90^\circ \equal{} \angle YXZ.$

So your way (proving equingularity) seems to be motivated quite well, and I just think you overdid it a little because you could have written down the solution in two lines. :D


absolute value
combinatorics unsolved
combinatorics

Let $ n$ be a natural number, and we consider the sequence $ a_1, a_2 \ldots , a_{2n}$ where  $ a_i \in (\minus{}1,0,1)$
If we make the sum of consecutive members of the sequence, starting from one with an odd index and finishing in one with and even index, the result is $ \le 2$ and $ \ge \minus{}2$
How many sequence are there satisfying this conditions?

I didn't have time to make all the calculations but here's some thoughts:
Since I couldn't find any nice combinatorial argument I thought of a "brute force approach". We see that each pair has a sum in $ S\equal{}\{\minus{}2,\minus{}1,0,1,2\}$ but we have three types of $ 0$, two types of $ \pm 1$ and one type of $ \pm 2$.
Let $ F(n)$ be the number of sequences we are looking at. Let $ f(n)$ denote the number of sequences of length $ n$ with entries in $ S'\equal{}S/\{0\}$ (taking different types into account) so that all consecutive sums have absolute value $ \le 2$, so that we have
\[ F(n)\equal{}\sum_{k\equal{}0}^n3^{n\minus{}k}f(k)\binom{n}{k}\]
Now let $ g(n),h(n),g'(n),h'(n)$ be respectively the number of sequences of length $ n$ with entries in $ S'$ so that the absolute value of the sum of consecutive terms doesn't exceed 2 and so that the sums of the first $ r$ terms ($ 1\le r \le n$) lies in $ \{\minus{}2,\minus{}1,0\},\{0,1,2\},\{\minus{}2,\minus{}1,0,1\},\{\minus{}1,0,1,2\}$ respectively.
Then we have
\[ f(n)\equal{}g(n\minus{}1)\plus{}h(n\minus{}1)\plus{}2g'(n\minus{}1)\plus{}2h'(n\minus{}1)\]
\[ g(n)\equal{}h(n\minus{}1)\plus{}2g'(n\minus{}1),h(n)\equal{}g(n\minus{}1)\plus{}2h'(n\minus{}1)\]
\[ g'(n)\equal{}2g(n\minus{}1)\plus{}2h'(n\minus{}1)\plus{}h(n\minus{}1),h'(n)\equal{}2h(n\minus{}1)\plus{}2g'(n\minus{}1)\plus{}g(n\minus{}1)\]
If we let $ a(n)\equal{}g(n)\plus{}h(n)$ and $ b(n)\equal{}g'(n)\plus{}h'(n)$ we have
\[ f(n)\equal{}a(n\minus{}1)\plus{}2b(n\minus{}1)\]
\[ a(n)\equal{}a(n\minus{}1)\plus{}2b(n\minus{}1)\qquad ,\qquad b(n)\equal{}3a(n\minus{}1)\plus{}2b(n\minus{}1)\]
Now everything can be computed recursively and with some algebra we should be able to get the answer.


function
search
number theory unsolved
number theory

Let $a$, $b$, $n$ be positive integers with $a,b > 1$ and $\gcd(a,b) = 1$. Prove that $n$ divides $\phi\left(a^{n}+b^{n}\right)$.

Search function! Posted very recently! (well, $-$ instead of $+$, but both are proven there)


geometry
analytic geometry
trigonometry
\/closed

What are, in short, the topics are covered in the Introduction to Geometry course thats not included in its book, if there are any?

i believe the course is based off the book but it goes a little more indepth...anywayz, go to the online school section that has the list of classes and a little bit about each, there's a course outline about what is done each of the 24 weeks. that should help you.

Coordinate geometry and intro to trigonometry are the only two I see on sight.
I haven't taken the course, but I do have the book. It also seems like the class goes more in depth with 3-D curved surfaces.

[quote="Anaxerzia"]Oh, thats it? My coordinate geometry and trig are pretty good though, so I'll probably just go through the book. Thanks :-)[/quote]
Those are the only one's I recognize on sight, it's been a good few months since I last used my copy of Intro to Geo.


$ \dfrac{1}{10}\plus{}\dfrac{9}{100}\plus{}\dfrac{9}{1000}\plus{}\dfrac{7}{10000}\equal{}$
\\ \\ Express your answer as a common fraction.

Our first step is to find the LCD - Least Common Denominator.

We can quickly find that the LCD is equal to $ 10000$. Now we change all the fractions so that they have the same denominator.

$ \frac{1}{10} \times \frac{1000}{1000} = \frac{1000}{10000}$

$ \frac{9}{100} \times \frac{100}{100} = \frac{900}{10000}$

$ \frac{9}{1000} \times \frac{10}{10} = \frac{90}{10000}$

$ \frac{7}{10000}$ already has our LCD.

Now we simply add.

$ \frac{1000}{10000} + \frac{900}{10000} + \frac{90}{10000} + \frac{7}{10000} \implies
\boxed {\frac{1997}{10000}}$


Mr. and Mrs. Lopez have two children.  When they get into their family car, two people sit in the front, and the other two sit in the back.  Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat.  How many seating arrangements are possible?

$ \textbf{(A) } 4 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 48$

[hide]First choose one of Mr. or Mrs. Lopez to drive.  This can be done in 2 ways.  The remaining 3 people can be seated in 6 ways.  So the answer is 2*6=12. B.[/hide]


Mr. Jones sold two pipes at $ \$ 1.20$ each. Based on the cost, his profit one was $ 20 \%$ and his loss on the other was $ 20 \%$. On the sale of the pipes, he:

$ \textbf{(A)}\ \text{broke even} \qquad\textbf{(B)}\ \text{lost } 4\text{ cents} \qquad\textbf{(C)}\ \text{gained } 4\text{ cents} \qquad\textbf{(D)}\ \text{lost } 10 \text{ cents} \qquad\textbf{(E)}\ \text{gained } 10 \text{ cents}$

So $ 120\equal{}1.2X$ for one pipe, and $ 120\equal{}.8Y$.  Solving for the costs of the pipes, $ X\equal{}100$ and $ Y\equal{}150$ (in cents).  Hence the total value is $ X\plus{}Y\equal{}250$, and he sold the pipes for $ 240$, so he lost 10 cents.


How do you solve abs(x^2+2x-6)<3x without a calculator?

General strategy:  in solving an inequality $ f(x) < g(x)$, find the points at which $ f(x) \equal{} g(x)$ and find out what happens in between those points.


Qn: Construct a 9 x 9 square with { 1, 2, 3,.... 80, 81} with the sum of numbers in every row, column and 3 x 3 square equal.

If it means anything to you, this is from Russia 1995 :)
Bye


$ABCD$ is a convex quadrilateral. How do you construct a point $E$ such that $\triangle{ABE}$ is similar to $\triangle{CDE}$? Please help!

We know the ratio of similarity |AB|:|CD|=r, so we can just draw the two Apollonius circles that are
1) locus of points X such that |AX|:|CX|=r
2) locus of points X such that |BX|:|DX|=r
Intersections of these circles are the points E.

I suppose this is really stupid, but how do you construct these Apollonius circles :?  :oops: ?? (I've never even heard of Apollonius circles before now, my knowledge of geometry isn't very good)

http://mathworld.wolfram.com/ApolloniusCircle.html
http://planetmath.org/encyclopedia/ApolloniusCircle.html

:)


This might be old but i came across it a few days back and its AWESOME

A  teacher has got some $blue$  and  $red$  stickers.He  has  atleast  one  of  each.He has a certain number of students.He  sticks  one  sticker  on  everyone's  back.A  student  cannot  see what  colour  sticker  is on  his  back.Other  student's  can  see which  sticker  is on  another's  back but cannot  tell them  that.Now  the  students are extraordinarily  intelligent.The  teacher  goes out of  the class  and  after  every hour,he  comes  back in  the class and announces "$\text{Who ever knows he has a red sticker on his back,please come out}$",and  some  students come out.He  does  this  evry  hour.Now tell  me  how many  students  will come  out  after  each hour(like how many afetr the 1st hour,then the second hour,etc.They do not have to be same)
 :lol:

how one student can know about the sticker on his/her back without seeing or asking.

If only one student has a red sticker, then he will see that everyone else has black and will conclude that he must have red (so in this case he'll leave after the first hour). If 2 (call them A and B) students have red stickers, then after the first hour no one will leave (as the above doesn't apply, and they have no other way of knowing). But, A will know that (because B) didn't leave, B must have seen someone with a red sticker. Hence, A realises that he himself has a red sticker (as all of the other students have black stickers). B will use the same logic to deduce that he too has a red sticker, and they will both leave after the second hour.

If we add another student with a red sticker, then we can use the same logic to show that all three will leave after the third hour, etc. So hence the answer is, if n students have red stickers, then n students will leave after the $n^{th}$ hour. Am I right?

I don't know.Anyone else can confirm or something?


algebra
polynomial
number theory unsolved
number theory

Prove that if the number $ ax^2\plus{}bx\plus{}c$ is divisible by $ 7$ and $ c$ is not divisible by $ 7$, then there is a number $ y$ so that the number $ cy^2\plus{}by\plus{}a$ is divisible by $ 7$ $ (a,b,c,x,y \in \mathbb{N}).$

For a generalization: if $ Q(x) \equal{} a_{k}x^{k} \plus{} .... \plus{} a_1x \plus{} a_0$ and $ R(x) \equal{} a_0x^{k} \plus{} a_1x^{k \minus{} 1} \plus{} ... \plus{} a_{k}$ are polynomials with integer coefficients and $ p$ a prime such that $ a_0$ isn't divisable by $ p$ then for any integer $ m$ with $ p|Q(m)$
there exist an integer $ n$ with $ p|R(n)$.
Prove $ \gcd(p,m) \equal{} 1$ and choose $ n$ such that $ mn\equiv1\mod{p}$


We denote $\ I$ the incenter of the triangle $\ ABC$ and $\ M,N,P$ the tangential points between the incirqle and the sides of $\ ABC$, whith $\ M,N,P$ on $\ (BC),(CA),(AB)$.Show that $\ MI,NP$ and the median from $\ A$ are concurrent.
[hide="I tried"]I tried to solve this problem, but if I show that we have the relation $\frac{BC}{AB+AC-BC}=\frac{RS}{SA}$, where$\ R$ is the midpoint of $\ (BC)$ and $\ S$ the intersection between $\ AR$ and $\ NP$, then the problem is finished.But I don't know...[/hide]

This is Theorem 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4, written in a slightly different way.

  darij

[color=darkblue][b]Enunciation.[/b] Denote $I$ the incenter of the triangle $\ ABC$ and $\ M\ ,\ N\ ,\ P$ the tangential points between the incircle and the sides of $\ ABC$,
whith $\ M\ ,\ N\ ,\ P$ on $(BC)\ ,\ (CA)\ ,\ (AB)$ respectively. .Show that $MI\ ,\ PN$ and the median from $A$ are concurrently.

[b]Proof[/b] Suppose w.l.o.g. that $b>c$. Denote : the middlepoint $D$ of the side $[BC]$ ; the projection $E$ of the point $A$ to the side-line $BC$ ; $L\in NP\cap BC$ ; $X\in MI\cap AD$ ; $Y\in LN\cap AD$. Prove easily that $DM=\frac{b-c}{2}$, $ME=\frac{(b-c)(p-a)}{a}$, $\frac{LM}{p-b}=\frac{LC}{p-c}=\frac{a}{b-c}$, $LD=\frac{a^{2}}{2(b-c)}$, $LC=\frac{a(p-c)}{b-c}$.
$\blacktriangleright XM\parallel AE\Longleftrightarrow$ $\frac{XA}{XD}=\frac{ME}{MD}$ $\Longleftrightarrow$ $\frac{XA}{XD}=\frac{(b-c)(p-a)}{a}\cdot\frac{2}{b-c}$ $\Longleftrightarrow$ $\boxed{\ \frac{XA}{XD}=\frac{b+c-a}{a}\ }\ .$
$\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline{LYN}$ for the triangle $ADC$ : $\frac{LD}{LC}\cdot\frac{NC}{NA}\cdot\frac{YA}{YD}=1\Longleftrightarrow$ $\frac{a^{2}}{2(b-c)}\cdot\frac{b-c}{a(p-c)}\cdot \frac{p-c}{p-a}\cdot\frac{YA}{YD}=1$ $\Longleftrightarrow$ $\boxed{\ \frac{YA}{YD}=\frac{b+c-a}{a}\ }\ .$ Therefore, $\boxed{\ \frac{XA}{XD}=\frac{YA}{YD}\ }=\frac{b+c-a}{a}\Longleftrightarrow X\equiv Y$, i.e. the lines $AD$, $NP$, $MI$ are concurrently.

[b]Remark.[/b] A similar problem may be put for the $A$- exincircle of the triangle $ABC$. [b]"Denote the middlepoint $D$ of the side $[BC]$ and the points $M'\in BC$, $N'\in CA$, $P'\in AB$ where the $A$- exincircle touches the sides of the triangle $ABC$. Then the lines $AD$, $N'P'$, $I_{a}M'$ are concurrently."[/b] [/color]


find the sum of

tan inverse of 3/(n^2+n-1) whre n goes from 1 to infinity

sorry if the problem is unclear, i dont know latex :oops:
thanks in advance.


Solve $ \frac{\sin x}{\cos3x}\plus{}\frac{\sin3x}{\cos9x}\plus{}\frac{\sin9x}{\cos27x}\equal{}0$

Since I'm fairly new to trig, can someone solve this as an example for me?

Thanks

Do you just use the trignometric formulas like triple angle formula because it gets really messy from there. you get cosx^3 and stuff. So I bet there is a shortcut.
 :o

well, [b]x=0[/b] is one obvious solution   :alien:

I calculated by Mathematica.
I was able to know the answer with $ x\equal{}\frac{i}{26}\pi$ for $ \minus{}25\le i \le25$ without $ i\equal{}\pm 13$.

[hide="Hint"]Use this formula:
$ \tan3a \minus{} \tan a \equal{} \frac {2\sin a}{\cos3a}$[/hide]

By doing what Kyoshiro said, you will get 3 equations, you will add them and get 1/2(tg27x-tgx)=0.

From there I got solution x=(k*PI)/26 (turn that into sin26x=0).

Sorry for not using Latex.

amazing factorization [b]SONNHARD![/b]But how to do it is hard for me, any further insight on this?

[b]Solution[/b]

Denote the expression by $ M$
$ M \equal{} \frac {\sin x\cos x}{\cos 3x\cos x} \plus{} \frac {\sin 3x\cos 3x}{\cos 9x\cos 3x} \plus{} \frac {\sin 9x\cos 9x}{\cos 27x\cos 9x}$

$ 2M \equal{} \frac {\sin(3x \minus{} x) }{\cos 3x\cos x} \plus{} \frac {\sin(9x \minus{} 3x) }{\cos 9x\cos 3x} \plus{} \frac {\sin(27x \minus{} 9x) }{\cos 27x\cos 9x}$

$ \equal{} (\tan 3x \minus{} \tan x) \plus{} (\tan 9x \minus{} \tan3x) \plus{} (\tan 27x \minus{} \tan 9x)$
which gives $ \tan27x \minus{} \tan x \equal{} 0$ so $ 27x \equal{} n\pi \plus{} x$ which gives $ \boxed{x \equal{} \frac {n\pi}{26}}$ where $ n$ ia an integer.

Guys do you like it??

 :showoff:

[quote="aadil"]i think we can  multiply maths mechanic's expression by 2 and try to use product or sum formula[/quote]

But i have already arrived at the answer!!
So why are u saying like that?? :(


Please see the system announcement shown [url=http://www.artofproblemsolving.com/Alcumus/Landing.php]here[/url].

I saw that all it says is "This is a test" just like everything else (Bug report validatied, response to bug report, etc.)

[quote="levans"]Wow, I can't believe no one has reported that bug previously.

Go ahead and check it now.[/quote]

I've reported it twice.

[quote]I've reported it twice.[/quote]
You have? How did you report it? Here in the forums? I don't recall seeing anything, but it's quite possible I missed it.


algebra
function
domain
interval notation

Find the domain of the function
\[ f(x)\equal{}\frac{\sqrt{2\plus{}x}}{1\minus{}x}\]

in interval notation.

[hide="Solution"]From the denominator, $ x\ne 1$.
From the numerator, $ 2\plus{}x\geq 0\implies x\geq \minus{}2$.

Then $ \boxed{x\in[\minus{}2,1)\cup(1,\infty)}$.[/hide]

What do those signs mean?

I got x≥-2 and x can't be 1

That's exactly what he wrote in interval notation.
In interval notation, a set of numbers is bounded by its outer limits, which are specified by the intervals.  For example, $ [a,b)$ describes the set of all real numbers between $ a$ and $ b$, where $ a$ is inclusive but $ b$ is not.  The curved bracket means that the given element is not in the set, while the square bracket indicates that the given element is in the set.  Hence $ [1,2]$ is all the real numbers from $ 1$ to $ 2$ including both of them, while $ (1,2)$ is all the real numbers from $ 1$ to $ 2$ excluding both of them.  Oh, and the $ \cup$ sign means that you are taking all the elements from both sets and merging them (such that the overlap of the sets only appear once in their union).


A company has five directors. The regulations of the company require that any majority (three or more) of the directors should be able to open its strongroom, but any minority (two or less) should not be able to do so. The strongroom is equipped with ten locks, so that it can only be opened when keys to all ten locks are available. Find all positive integers $n$ such that it is possible to give each of the directors a set of keys to $n$ different locks, according to the requirements and regulations of the company.

If n>4 (such as 5,6,7...) then two or lesser than two directors will be able to open the strongroom. And if n<4(such as 3,2,1,...)than three or more than three directors will not be able to open it.

But if n=4 then two or less than two directors will not be able to open it and three or more then three directors will be able to open it .
So n=4

[Please infom me if I am right or not]

Assume the company has $ 2n\plus{}1$ executives.
a) What is the minimum number $ l$ of locks necessary such that each majority ($ n\plus{}1$ or more) may open the vault, but no minority ($ n$ or less) may open the vault?
b) What are the possible numbers $ k$ of keys that each executive should be given in this case? (each executive should be given the same number of keys, otherwise those with less keys would be extremely jealous!!!   :P )

Solution (hidden) below for part (a):

[hide]
a) For each one of the $ \binom{2n\plus{}1}{n}$ minorities of exactly $ n$ executives, there is at least one lock that this minority cannot open.  Assume that there are two distinct minorities of exactly $ n$ executives that cannot open the same lock.  Thus, the majority formed by their union (since they are distinct minorities, they differ at least by one executive, so their union contains at least $ n\plus{}1$ executives) cannot open that lock.  So the number of locks is at least equal to the number of minorities with exactly $ n$ executives, i.e., $ l\geq\binom{2n\plus{}1}{n}$.  With exactly $ l \equal{}\binom{2n\plus{}1}{n}$, let us number both the minorities of exactly $ n$ executives, and the locks, from 1 to $ \binom{2n\plus{}1}{n}$, and let us give an executive a key to the $ i$-th lock iff he does not belong to the $ i$-th minority, so this lock (and hence the vault) will not be opened by the $ i$-th minority of $ n$ executives (and thus by any minorities contained therein).  On the other hand, consider any majority of at least $ n\plus{}1$ executives that cannot open the vault, or there is at least one lock, the $ j$-th, that they cannot open.  The majority is then contained in the minority of exactly $ n$ executives to which the $ j$-th key was not given, which is absurd.  So the minimum number of locks is $ l \equal{}\binom{2n\plus{}1}{n}\equal{}\binom{2n\plus{}1}{n\plus{}1}$.
[/hide]

Solution (hidden) below for part (b):

[hide]
b) So $ l \equal{}\binom{2n\plus{}1}{n}\equal{}\binom{2n\plus{}1}{n\plus{}1}$, which in the case $ n \equal{} 2$ yields $ l \equal{} 10$, as in the original problem.  Assume that there is a key to which less than $ n\plus{}1$ copies are given to the executives.  Therefore, there are at least $ n\plus{}1$ executives that do not have this key, and their majority cannot open the vault.  So the minimum (total) number of keys is $ (n\plus{}1)l \equal{} (2n\plus{}1)\binom{2n}{n}$, and since each executive receives the same number and there are $ 2n\plus{}1$ executives, each one receives at least $ \binom{2n}{n}$ keys.  Assume that each one of the $ 2n\plus{}1$ executives is given at least $ k \equal{}\binom{2n}{n}\plus{}1$ keys.  The total number of keys divided by the number of locks is then $ \frac{(2n\plus{}1)k}{l}>\frac{(2n\plus{}1)\binom{2n}{n}}{\binom{2n\plus{}1}{n}}\equal{} n\plus{}1$, so there is at least one key given to at least $ n\plus{}2$ executives.  This means that there is one key that any minority of $ n$ executives may open, since there are at most $ n\minus{}1$ executives that lack the key to it.  Or, the number of locks that may not be opened by at least one of the $ l$ minorities of $ n$ executives is at most $ l\minus{}1$, and by Dirichlet's pigeonhole principle at least one lock is not opened by at least two distinct such minorities, or as above it is not opened by at least one majority.  Contradiction, so the number of keys given to each executive is exactly $ \binom{2n}{n}$.  In our case, each executive receives exactly $ k \equal{}\binom{4}{2}\equal{} 6$ keys.
[/hide]

This is a nice solution. I myself just did the brute force solution (considered all 10 possibilities of numbers of locks to give to each director and just went eliminating them one by one by considering the unions of the subsets of keys)

Directors A and B cannot open all the locks together, so there
is a lock (say number 1), which neither can open, but all other directors
must have key number 1. The same reasoning applies to all 10
unordered pairs of directors; each pair will be missing a particular key,
and all other directors will own that key. This tells us how the 10 keys
must be distributed to have the desired effect (the only choice being the
particular labelling of the keys). Each director is missing a different key
for each of the 4 directors with whom he or she can form a pair, so n =6.

Just pasted it :p


Let $M$ be a point inside a triangle $ABC$ and let $p_{a},p_{b},p_{c}$ be the distances from $M$ to $BC,CA,AB.$ Then for all $k\geq 1\in\mathbb{R}$ we have \[\sum MA^{k}\geq2\sum p_{a}^{k}.\]

:D ,use lemma used to prove erdos mordell ineq.  :P


geometry
3D geometry
sphere
function
algebra
polynomial
vector

[size=150][color=darkblue]17-th Iranian Mathematical Olympiad 1999/2000[/color][/size]

[b]Second Round  [/b]

1.Does there exist a natural number $N$ which is a power of$2$, such that one
can permute its decimal digits to obtain a different power of $2$?

2..Three non-coplanar circles are given in the Euclidean space so that they
are pairwise tangent. Prove that there exists a sphere that passes through
all the three circles.

3.In a deck of $n > 1$ cards, some digits from $1$ to$8$are written on each card.
A digit may occur more than once, but at most once on a certain card.
On each card at least one digit is written, and no two cards are denoted
by the same set of digits. Suppose that for every $k=1,2,\dots,7$ digits, the
number of cards that contain at least one of them is even. Find $n$.

 4.A sequence of natural numbers $c_1, c_2,\dots$ is called PERFECT if every natural
number $m$ with $1\lem\le c_1 +\dots+ c_n$ can be represented as
$m =\frac{c_1}{a_1}+\frac{c_2}{a_2}+\dots+\frac{c_n}{a_n}$
Given $n$, find the maximum possible value of $c_n$ in a perfect sequence $(c_i)$.

5.Circles $C_1$ and $C_2$ with centers at $O_1$ and $O_2$ respectively meet at points
$A$ and $B$. The radii $O_1B$ and $O_2B$ meet $C_1$ and $C_2$ at $F$ and$E$. The line
through$B$parallel to $EF$intersects $C_1$again at $M$ and $C_2$ again at $N$.
Prove that $MN = AE + AF$.

6.Two triangles $ABC$and $A'B'C'$ are positioned in the space such that the
length of every side of $\triangle ABC$ is not less than $a$, and the length of every
side of $\triangle A'B'C'$ is not less than $a'$. Prove that one can select a vertex
of $\triangle ABC$ and a vertex of $\triangle A'B'C'$ so that the distance between the two
selected vertices is not less than $\sqrt{\frac{a^2 + a'^2}{3}}$

[size=150][color=darkblue]17-th Iranian Mathematical Olympiad 1999/2000[/color][/size]

[b]Third Round[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5433]1.[/url] Two circles intersect at points $A$ and $B$. A line $\ell$ through $A$ meets these
circles again at$C$ and$D$. Let$M$ and$N$be the midpoints of the arcs $BC$
and$BD$ not containing$A$, and $K$ be the midpoint of $CD$. Show that
$\angle MKN = 90^{\circ}$.

2.Let $A$ and $B$ be arbitrary finite sets and let $f: A\longrightarrow B$ and $g: B\longrightarrow A$
be functions such that $g$ is not onto. Prove that there is a subset $S$ of $A$ such that
$\frac{A}{S}=g(\frac{B}{f(S)})$.

3.Suppose $f : \mathbb{N} \longrightarrow \mathbb{N}$ is a function that satisfies $f(1) = 1$ and
$f(n + 1) =\{\begin{array}{cc} f(n)+2&\mbox{if}\ n=f(f(n)-n+1),f(n)+1& \mbox{Otherwise}\end {array}$
$(a)$ Prove that $f(f(n)-n + 1)$ is either $n$ or $n + 1$.
$(b)$ Determine$f$.

4.Let us denote $\prod = \{(x, y) | y > 0\}$. We call a semicircle in $\prod$ with
center on the $x-\text{axis}$ a semi-line. Two intersecting semi-lines determine
four semi-angles. A bisector of a semi-angle is a semi-line that bisects
the semi-angle. Prove that in every semi-triangle (determined by three
semi-lines) the bisectors are concurrent.

5.Determine all functions $f : \mathbb{N} \longrightarrow\longrightarrow{N}$ such that:
$(i)\ \ f(m) = 1 \text{if and only if} m = 1$
$(ii)\ \ \text{If}\ d=(m,n)\ ,\ \text{then} f(mn) =\frac{f(m)f(n)}{f(d)}$
$(iii)\ \ f^{2000}(m) = m\ \text{for every} m \in \mathbb{N}$.

6.Let$n$ points be given on a circle and let $nk + 1$ chords between these
points be drawn, where $2k+1 < n$. Show that it is possible to select $k+1$
of the chords so that no two of them intersect.

[size=150][color=darkblue]17-th Iranian Mathematical Olympiad 1999/2000[/color][/size]

[b]Fourth Round[/b]
 1.In a tennis tournament where $n$ players $A_1,A_2,\dots,A_n$ take part, any two
players play at most one match, and $k\le n(n-1)$
$2$ matches are played. The winner of a match gets $1$ point while the loser gets $0$. Prove that a sequence
$d_1,d_2,\dots,d_n$ of nonnegative integers can be the sequence of scores of the
players ($d_i$ being the score of$A_i$) if and only if
$(i)\ \ d_1 + d_2 +\dots+ d_n = k$, and
$(ii)\ \text{for any} X\subset\{A_1,\dots,A_n\}$, the number of matches between the players in $X$ is at most $\sum_{A_j\in X}d_j$

2.Isosceles triangles $A_3A_1O_2$ and $A_1A_2O_3$ are constructed on the sides of
a triangle $A_1A_2A_3$ as the bases, outside the triangle. Let $O_1$ be a point
outside $\Delta A_1A_2A_3$ such that
$\angle O_1A_3A_2 =\frac 12\angle A_1O_3A_2$ and $\angle O_1A_2A_3 =\frac 12\angle A_1O_2A_3$.
Prove that $A_1O_1\perp O_2O_3$, and if $T$ is the projection of $O_1$ onto $A_2A_3$,
then $\frac{A_1O_1}{O_2O_3} = 2\frac{O_1T}{A_2A_3}$.

3.A circle$\Gamma$ with radius $R$ and center $\omega$, and a line $d$ are drawn on a plane,
such that the distance of $\omega$ from $d$ is greater than $R$. Two points $M$ and
$N$ vary on $d$ so that the circle with diameter $MN$ is tangent to $\Gamma$. Prove
that there is a point $P$ in the plane from which all the segments $MN$ are
visible at a constant angle.

4.Let $n$ be a positive integer. Suppose $S$ is a set of ordered $n-\mbox{tuples}$ of
nonnegative integers such that, whenever $(a_1,\dots,an)\in S$ and $b_i$ are nonnegative integers with$b_i\le a_i$, the $n-\text{tuple}$  $(b_1,\dots,b_n)$ is also in $S$. If $h_m$
is the number of elements of $S$ with the sum of components equal to$m$,
prove that $h_m$ is a polynomial in $m$ for all sufficiently large$m$.

5.Suppose that $a, b, c$ are real numbers such that for all positive numbers
$x_1,x_2,\dots,x_n$ we have

$(\frac{1}{n}\sum_{i=1}^nx_i)^a(\frac{1}{n}\sum_{i=1}^nx_i^2)^b(\frac{1}{n}\sum_{i=1}^nx_i^3)^c\ge 1$

Prove that vector $(a, b, c)$ is a nonnegative linear combination of vectors
$(-2, 1, 0)$ and $(-1, 2,-1)$.

6.Prove that for every natural number $n$ there exists a polynomial $p(x)$ with
integer coefficients such that $p(1),p(2),\dots,p(n)$ are distinct powers of $2$.

Has anybody posted those problems already, if so what are the URLs?

Amir.S can you post the problems missing at
http://imocompendium.com/index.php?options=oth|Irn|Ne|Ne&ttn=Iran&p=0
and mathlinks.ro.
I tried some Iranian problems in the past and I think they are excellent.

[b]17-th Iranian Mathematical Olympiad 1999/2000

Second Round[/b]
1 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65852
2 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65857
3 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65865
4 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65853
5 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65850
6 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65851

[b]Third Round[/b]
1 http://www.mathlinks.ro/Forum/viewtopic.php?t=5433
2 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65859
3 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65860
4 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65858
5 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65863
6 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65867

[b]Fourth Round[/b]
1 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65868
2 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65855
3 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65854
4 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65864
5 http://www.mathlinks.ro/viewtopic.php?search_id=779290791&t=65866
6 http://www.mathlinks.ro/Forum/viewtopic.php?p=1365189#1365189


My school was only registered for the 12A, but another high school nearby is taking the 12B.  I completely blew the 12A so I wanna take the B test.  If I want to take the 12B at that high school, do I need to fill out any forms so that the AMC knows I am not actually apart of that school?

Thanks

Well, on the answer sheet, you would probably want to put your school's address for where they send the score report...

You would need to get permission from the school to take the AMC 12, and to be able to take the AIME at your school, you would have to pay a 15 dollar fee or something like that.


If $ x,y\in\left[0,\frac{\pi}{2}\right]$ prove that:

\[ 8\leq\left(\sin(x)\plus{}\sin(y)\plus{}\cos(x)\plus{}\cos(y)\right)\left(\frac{2}{\sin(x)\plus{}\cos(x)}\plus{}\frac{2}{\sin(y)\plus{}\cos(y)}\right)\leq 4\plus{}3\sqrt{2}\]

[hide]Let $ \sin{x}\plus{}\cos{x}\equal{}a$, $ \sin{y}\plus{}\cos{y}\equal{}b$.  Then the left-hand inequality reduces to $ \frac{a}{b}\plus{}\frac{b}{a}\ge 2$, which is AM-GM.  The right-hand inequality reduces to $ \frac{a}{b}\plus{}\frac{b}{a}\le\frac{3}{\sqrt{2}}$, which follows from the fact that $ a, b\in [1,\sqrt{2}]$.[/hide]

[quote="Svejk"]$ \{x,y\}\subset[0,\frac{\pi}{2}]$ $ \implies$ $ 4\le (\sin x+\sin y+\cos x+\cos y)\cdot(\frac{1}{\sin x+\cos x}+\frac{1}{\sin y+\cos y})\le\frac{4+3\sqrt 2}{2}$.[/quote]
[color=darkblue]From the well-known property  $ 0\le t\le\frac{\pi}{2}\implies\sin t+\cos t\in[1,\sqrt 2]$ and using the substitution $ \{\begin{array}{ccc}a & = &\sin x+\cos x\\ \ b & = &\sin y+\cos y\end{array}$

the upper inequality becomes $ \{a,b\}\subset[1,\sqrt 2]$ $ \implies$ $ 4\le (a+b)(\frac{1}{a}+\frac{1}{b})\le\frac{4+3\sqrt 2}{2}$.[/color]
[quote][color=darkred][size=134][u][b]Schweitzer[/b]'s inequaliy.[/u][/size]$ \{\begin{array}{c}n\in\mathbb N^{*}\ ,\ k\in\overline{1,n}\\ \\ a_{k}\in [A,B]\subset\mathbb R^{*}_{+}\end{array}\|$ $ \implies$ $ n^{2}\le\boxed{\sum_{k = 1}^{n}a_{k}\cdot\sum_{k = 1}^{n}\frac{1}{a_{k}}\le\frac{(A+B)^{2}}{4AB}\cdot n^{2}}\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] For any $ k\in\overline{1,n}$, $ a_{k}\in [A,B]$ $ \Longleftrightarrow$ $ (a_{k}-A)(a_{k}-B)\le 0$ $ \Longleftrightarrow$ $ a_{k}+\frac{AB}{a_{k}}\le A+B$. Thus, $ \sum_{k = 1}^{n}a_{k}+AB\sum_{k = 1}^{n}\frac{1}{a_{k}}\le n(A+B)\ \ (*)$.
But $ 4\cdot\sum_{k = 1}^{n}a_{k}\cdot AB\sum_{k = 1}^{n}\frac{1}{a_{k}}\le(\sum_{k = 1}^{n}a_{k}+AB\sum_{k = 1}^{n}\frac{1}{a_{k}})^{2}\le n^{2}(A+B )^{2}$ $ \implies$ ${ \sum_{k = 1}^{n}a_{k}\cdot\sum_{k = 1}^{n}\frac{1}{a_{k}}\le\frac{(A+B)^{2}}{4AB}\cdot n^{2}}\ .$

[b]Particular case.[/b] For $ \{a,b\}\subset[1,\sqrt 2]$ obtain $ 4\le (a+b)(\frac{1}{a}+\frac{1}{b})\le\frac{4+3\sqrt 2}{2}$.

[b]Remark.[/b] Prove similarly the lower easy extension of the [b]Schweitzer[/b]'s inequality :[/color]

[quote][color=darkred][size=134][u][b]Kantorovic[/b]'s inequality[/u].[/size] $ \{\begin{array}{c}n\in\mathbb N^{*}\ ,\ k\in\overline{1,n}\\ \\ a_{k}\in [A,B]\subset\mathbb R^{*}_{+}\ , t_{k}> 0\end{array}\|$ $ \implies$ $ (\sum_{k = 1}^{n}t_{k})^{2}\le\boxed{\sum_{k = 1}^{n}t_{k}a_{k}\cdot\sum_{k = 1}^{n}\frac{t_{k}}{a_{k}}\le\frac{(A+B)^{2}}{4AB}\cdot(\sum_{k = 1}^{n}t_{k})^{2}}\ .$[/color][/quote]

[color=darkblue]The LHS of the both remarkable inequalities are particular cases of [b]C.B.S[/b].'s inquality.[/color]

We can find a proof for Schweitzer's inequality also here
http://www.mathlinks.ro/viewtopic.php?t=154015


If $m,n\in{N}$ , find the minimum value for $|12^m-5^n|$ .

I know the answer is (i.e minimal value) [tex]7[/tex]. I'll try to prove that.

This has been discussed a very very long time ago...  :P
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=289[/url]


Determine all positive integers $ n$ for which there is a coloring of all points in space so that each of the following conditions is satisfied:

(i) Each point is painted in exactly one color.
(ii) Exactly $ n$ colors are used.
(iii) Each line is painted in at most two different colors.

You see at this link: http://www.mathlinks.ro/viewtopic.php?t=15624


algebra
polynomial
function
domain
advanced fields
advanced fields theorems

I'm reading a text about 'affine algebraic sets'. However, I don't feel comfortable when they introduce the notation $k[V]$ which is isomorphic to $k[\mathbb{A}^{n}]/\mathcal{I}(V)$. This is my current point of view.

Let $\mathbb{A}^{n}$ be an affine $n$-space and $k$ a field. Then...

1) We pick some polynomials out of $k[x_{1},...,x_{n}]$.
2) We consider them as $k$-valued functions, i.e. $f: (a_{1},...,a_{n}) \mapsto f(a_{1},...,a_{n}) \in k$.
3) These $k$-valued polynomial-functions form a ring: $k[\mathbb{A}^{n}]$.
4) Let $V$ be an affine algebraic set, then $\mathcal{I}(V)$ consists out of all those functions which vanish on $V$, i.e. $f(v_{1},...,v_{n})=0$ for all $(v_{1},...,v_{n}) \in V$. This is an ideal in $k[\mathbb{A}^{n}]$.
5) Now we are going to form the quotient. This would mean:

"all $k$-valued polynomial-functions with domain $\mathbb{A}^{n}$" modulo "all $k$-valued functions which vanish on $V$" = "all $k$-valued polynomial-functions with domain $V$".

This last interpretation doesn't make sense to me. Can someone tell me where I went wrong? If only the polynomial-functions with domain in $V$ survived when factoring, where would the functions with domain not $V$ go?

You said that $k[V]$ was "all $k$-valued polynomials with domain $V$."  The second phrase should be thought of as "all *restrictions* of $k$-valued polynomials to the domain $V$."

Two functions on the affine plane *restrict* to the same function on $V$ if and only if their difference is zero on $V$.  As far as $V$ is concerned, two such functions are the same function.  Therefore, algebraically, this is the same as saying that any function that is zero on $V$ should be identified with the zero function.  This is what it means to mod out by $I(V)$.

To put it slightly differently: You are interested in the ring of "good" functions on $V$, where "good" means that the function is induced by a polynomial function on $\mathbb A^{n}$. Some polynomials correspond to the same "good" function on $V$ (precisely if their difference lies in $\mathcal I(V)$).

BTW: To make you really feel uncomfortable, it should be pointed out that there are authors who use $k[V]$ even for non-affine varieties to denote $\Gamma(V,\mathcal O_{V})$. owk