I thought this one was short but fun enough.

Find three integers such that the sum of any two  is always divisible by the third.

Edit: Wrong word ><

One set could be (0,1,1)

2 divides 0
1 divides 1
1 divides 1

Any three integers whose sum equals $ 0$ does the trick.  :wink:

But, $ \left\{ \minus{}n \ , \ 0 \ , \ n \right\}$ makes $ n\minus{}n\equal{}0 \rightarrow \frac{0}{0}$.

So just pick the integers so that they are nonzero?

[quote="AstroPhys"]But, $ \left\{ \minus{} n \ , \ 0 \ , \ n \right\}$ makes $ n \minus{} n \equal{} 0 \rightarrow \frac {0}{0}$.[/quote]
We say that $ a | b$ iff there exists an integer $ c$ such that $ b \equal{} ac$.  It follows that $ 0 | 0$.  (In other words, divisibility can be defined without an explicit reference to the notion of a quotient.)  In fact, any integer divides zero.  On the other hand, zero does not divide any nonzero integer.

I suspect that mathgeniuse^lnx and nayel have not described the possible forms of all possible solutions.  Anyone care to find all of them (with proof)?  (Note that you can assume $ \gcd(a, b, c) \equal{} 1$.)

Oh my apologies I wrote it wrong. Find 3 natural numbers. That is if $ (a,b,c)$ is a solution than $ \min(a,b,c) > 0$

The sum of the larger two clearly can't divide the smallest unless it's zero.

Wrong word, I edited the question.
It should be the remaining number will divide into the sum of the other 2.

[hide="So far"]
The sets I found that work are
$ \left\{ n,n,n \right\}$
$ \left\{ 2n,n,n \right\}$.
That seems very far from the complete list.
[/hide]

Actually, that is the complete list.

[hide="Discussion"] It is still valid to make the assumption that $ \gcd(a, b, c) \equal{} 1$.  The conditions we are looking at can now be written

$ a, b, c | a \plus{} b \plus{} c$.

WLOG $ a \le b \le c$.  Since $ c$ is at least a third of $ a \plus{} b \plus{} c$, the quotient $ \frac{a \plus{} b \plus{} c}{c}$ is either $ 3$ or $ 2$.

[b]Case:[/b]  $ \frac{a \plus{} b \plus{} c}{c} \equal{} 2$.  Then $ a \plus{} b \equal{} c$ and we require that $ a, b | a \plus{} b$.  Because we made the assumption $ \gcd(a, b, c) \equal{} 1$, it follows that $ \gcd(a, b) \equal{} 1$ and therefore $ ab | a \plus{} b$, which (keeping with our assumption) holds only when $ (a, b) \equal{} (1, 1)$ and has no relatively prime solutions for $ a, b \ge 2$ because there $ ab \ge a \plus{} b$.  Our (general) solution is therefore $ \boxed{ (a, b, c) \equal{} (n, n, 2n) }$.

[b]Case:[/b]  $ \frac{a \plus{} b \plus{} c}{c} \equal{} 3$.  By assumption, this is possible only when $ a \equal{} b \equal{} c$, so our general solution is $ \boxed{ (a, b, c) \equal{} (n, n, n) }$. [/hide]

Hmm... What about $ \{1, 2, 3\}$?  Somewhere your discussion went wrong.  I think it's when you implicitly assume that $ a,b|2(a\plus{}b)$ means $ a,b|a\plus{}b$.  This is not the case if one of $ a, b$ is even and the other is odd.

Whoops, I messed up that case.  (I didn't add correctly at all.)

[hide="Fix"] The conditions in the first case are actually $ b | 2a \plus{} b, a | a \plus{} 2b \implies a, b | 2(a \plus{} b)$.  It is still valid to assume $ \gcd(a, b) \equal{} 1$, so $ ab | 2(a \plus{} b)$.  We get the extra solution $ (a, b) \equal{} (1, 2)$, but we still conclude that for $ a, b \ge 4$ there are no solutions and it's easy to check the small relatively prime solutions.  We therefore have $ \boxed{ (a, b, c) \equal{} (n, 2n, 3n) }$. [/hide]

Another way to solve it:

[hide="Beginning of a solution"]We have $ a,b,c|a\plus{}b\plus{}c$.  Let $ S\equal{}a\plus{}b\plus{}c$, and let $ a\equal{}S/x, b\equal{}S/y, c\equal{}S/z$ for integers $ x,y,z$.  Then we get $ S/x\plus{}S/y\plus{}S/z\equal{}S$, so $ 1/x\plus{}1/y\plus{}1/z\equal{}1$.  [/hide]

Yes those three are the only ones I got also >_<

$ (n,n,n) , (n,n,2n) , (n,2n,3n)$


joml88 is well-known as an MIT student because he always has perfect attendance. One day, he feels himself running late! He wants to be able to get to the campus on time, or else his attendance record is ruined. When he drives his car at a constant rate of 50 miles per hour, he arrives 5 minutes late. If he drives his car at 60 mph, he arrives 5 minutes early. At what speed does joml88 need to drive his car in order to arrive to MIT exactly on time?

does ANYONE EVER look in this forum at all?

Patrick and Ravi are playing the smiley game. The first person to post 4  :D s wins. Each person tries to be the first to post the  :D for a day, continuing until one person wins. Ravi is at his computer more often and has a 60% chance of winning the  :D for a certain day. What is the probability that Patrick wins the smiley game? Express your answer as a common fraction.

Oh, and guys, check out my blog for the craze about  :D !


The polynomial $ f(x) \equal{} x^{4} \plus{} ax^{3} \plus{} bx^{2} \plus{} cx \plus{} d$ has real coefficients, and $ f(2i) \equal{} f(2 \plus{} i) \equal{} 0.$ What is $ a \plus{} b \plus{} c \plus{} d?$

$ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 16$

I tried doing this problem with Vieta's (not the smartest approach :oops: ...) , but came up with a different answer than with factoring it and letting x=1.  :huh: I don't know what I did wrong, (arithmetic was checked with calculator.)

I'm assuming that $ \imath = \sqrt{-1}$ in this problem, correct?

So, you know two roots of the function, $ 2\imath$ and $ 2+\imath$.  Let the other two roots be $ x$ and $ y$.  Then, we know that

$ x+y+2\imath+2+\imath=-a$

$ 2\imath x+2\imath y+4\imath-2+2x+\imath x + 2y+\imath y+xy=b$

$ x(4\imath-2)+y(4\imath-2)+2\imath xy+(2+\imath)xy=-c$

$ xy(4\imath-2)=d$

So you just sum these up (well, flip a few signs, first) and see what happens?  It doesn't appear as though this is going to miraculously simplify without using any other tricks...

Fortunately, the coefficients are real, which means you can use [hide="powerful magic"] the Conjugate Root Theorem, which implies that $ \minus{} 2i, 2 \minus{} i$ are also roots. [/hide]
This allows you to calculate [hide="the average airspeed of an unladen swallow"] the sum of the coefficients directly; in particular,

$ f(x) \equal{} (x^2 \plus{} 4)(x^2 \minus{} 4x \plus{} 5)$

so

$ f(1) \equal{} 1 \plus{} (a \plus{} b \plus{} c \plus{} d) \equal{} (1 \plus{} 4)(1 \minus{} 4 \plus{} 5) \equal{} 10 \Leftrightarrow$
$ a \plus{} b \plus{} c \plus{} d \equal{} \boxed{9}$ [/hide]
If you used Vieta's to calculate the coefficients separately, it's most likely that you forgot to account for signs somewhere along the way.

[b]b-flat:[/b]  Vieta's won't help you calculate the other two roots without [hide="the powerful magic"] the Conjugate Root Theorem[/hide].  They're not unique without the extra condition.

:rotfl: Monty Python was a funny movie.
I knew that the other two roots must be $ \minus{} 2i$ and $ 2 \minus{} i$, but when I tried to use Vieta's to find the coefficients, it didn't work out. (Don't remind me that I was too stupid to do it the other ways). Perhaps there is something that I did wrong with Vieta's or a calculation mistake?

I found that $ a \equal{} \minus{} 1(2i \minus{} 2i \plus{} 2 \minus{} i \plus{} 2 \plus{} i) \equal{} \minus{} 4$,
$ b \equal{} (2i)( \minus{} 2i) \plus{} ( \minus{} 2i)(i \plus{} 2) \plus{} (i \plus{} 2)(i \minus{} 2) \equal{} 11 \minus{} 4i$
$ c \equal{} \minus{}(2i)( \minus{} 2i)(i \plus{} 2) \plus{} ( \minus{} 2i)(i \plus{} 2)(i \minus{} 2) \equal{} \minus{}8 \plus{} 6i$
$ d \equal{} (2i)( \minus{} 2i)(i \plus{} 2)(i \minus{} 2) \equal{} 20$

Well, for starters, $ b, c$ were given as real in the problem statement, so that can't possibly be right.

I only see three terms for $ b$ and two terms for $ c$.  The correct expression for $ b$ has six terms and the correct expression for $ c$ has four terms.  They should be

$ b \equal{} (2i)( \minus{} 2i) \plus{} (2i)(2 \plus{} i) \plus{} (2i)(2 \minus{} i) \plus{} ( \minus{} 2i)(2 \plus{} i) \plus{} ( \minus{} 2i)(2 \minus{} i) \plus{} (2 \plus{} i)(2 \minus{} i) \equal{} 9$
$ c \equal{} \minus{} \left( (2i)( \minus{} 2i)(2 \plus{} i) \plus{} (2i)( \minus{} 2i)(2 \minus{} i) \plus{} (2i)(2 \plus{} i)(2 \minus{} i) \plus{} ( \minus{} 2i)(2 \plus{} i)(2 \minus{} i) \right) \equal{} \minus{}16$

[quote="t0rajir0u"][b]b-flat:[/b]  Vieta's won't help you calculate the other two roots without [hide="the powerful magic"] the Conjugate Root Theorem[/hide].  They're not unique without the extra condition.[/quote]

I had thought that the author of the question was asking for a solution using Vieta's, not for the regular solution to the problem (he said that he already knew the other solution).


The product of all real roots of the equation $x^{\log_{10} x} = 10$ is

A. 1
B. -1
C. 10
D. $10^{-1}$
E. None of these

Silverfalcon, nice job! You posted all the 1984 problems. :)

[hide="Now for the problem"]$x^{logx}=10$
$logx^{logx}=log10$
$(logx)^2=1$
$logx=1~or~-1$
$x=10~or~\frac{1}{10}$
The answer is 1, or A[/hide]

Huh? If you click on the link to view all problems from the 1984 AMC12, we're still missing #16-19, 22, and 24.

[hide="Answer"]$x^\frac{\log x}=10\Rightarrow \log x^\frac{\log x}=\log x\cdot \log x=(\log x)^2=\log 10=1\Rightarrow \log x=\pm 1$
$x=10^{\pm 1}\Rightarrow 10^0=1\Rightarrow \boxed{A}$.[/hide]

[quote="Toby"]I know the answer.
[hide]The answer is C.[/hide][/quote]

Toby, I don't want you to be posting just a letter. That's NOT accepted and you should NOT do it.

Post explanation but your answer is wrong.  ;)

[hide="Sol"]
$10^a=x$
$x^{\frac{1}{a}}=10$
According to this equation,
$x^a=10$
So,
Equating the 2 equations,
$x^{a-\frac{1}{a}}=1$
$a2-1=0$
$a=1,-1$
So,
Product of the 2 roots,
$10^{-1}*10=1$
$\boxed{A}$
[/hide]

Has the 23rd one been done?I can't find it

It's in Intermediate. Click on the link given in the main post of this thread, then go down to 23, and click on the number. It'll take you to the post.


geometry
conics
ellipse
analytic geometry

There is an ellipse with foci K,L...
There are two points B,D on the ellipse such that KB and LD meet outside the ellipse at A and KD,LB meet inside the ellipse at C.

Prove that quad ABCD is a circumsciribed quad
(i.e. there exists a circle on which it is circumscribed)

[hide]If sb gonna try do it using coordinate geometry, the solution will be spat upon as a pure geometric solution exists... :rotfl:  :rotfl: [/hide]

[quote="SOUMYASHANT NAYAK"][quote="gauravpatil"]seems pretty obvious with poles and polars for an eclippse
sorry for the spelling
i cant remember it at the moment[/quote]

Would you elaborate more...
I know nothing more than the definitions of pole and polars for conic sections other than circles...

Can you just post some of the properties [b]without proof[/b]...
I guess it would be a nice exercise for me atleast...[/quote]

The definition is pretty much similar for other conics and so is the std equation.

I know the definitions as I remember mentioning...

What I asked for was [b]some properties [/b]like some preserving transformations....
But the link gaurav gave contains nothing of that sort ..just the polar equation of a conic....

[quote="gauravpatil"]http://www.mecca.org/~halfacre/MATH/polarconics.htm
a basic one is pole polar duality
polar is  the chord of contact[/quote]

And how did that help you in solving the problem..??? :|


Is there anybody knows about Result of İBO and ICHO in 2006?

results of IChO 2006:

http://icho2006.kcsnet.or.kr/main/i_presskit/i_p_catalyzer.htm?qpage=i_f_catalyzer_en

Issue No.11 Monday 10 July 2006

Yeah there is:

[url=http://icho2006.kcsnet.or.kr/main/file_download.htm?filename=catalyzer_v11.pdf]PDF Version[/url]

Post non-math-but-science questions in http://www.mathlinks.ro/Forum/index.php?f=328 please.
And especially dont ask these questions in the middle of other threads.


function
inequalities
number theory proposed
number theory

Last week I could prove one of my easier conjectures and I'd like to know if someone else is able to find a much easier proof:

Define the function $A: \mathbb R \to \mathbb R$ by the following series:

$A(x) = \sum_{n=2}^\infty ~ \frac{1}{\phi(n)^{x}}$
where $\phi (n)$ denotes the Eulerian phi-Function.
Theorem: A(x) diverges for $x=1$ and converges for $x= 1+\epsilon$

I need at least 7 pages for a formal proof of this theorem using the basic identities of the phi-function, like the formula

$\phi (p) = p-1$
($p \in \mathbb P$)

I also need the multiplication-formula and the formula when n is a higher power of a prime number.
This way I get some information about the inequation $\phi (n)^{x}< n$ , that I'm able to construct a series I can compare with the Riemann-Zeta-function.
The divergence of the series for x=1 can be shown using a comparison by A(1) and the harmonic series of the prime numbers.

Is there a shorter proof using more elegant methods?

The inequality $\varphi(n) <n$ is trivial and implies divergence for $x=1$ because of minorization by the harmonic series.

The convergence for $x>1$ can be seen by majorization with $\zeta (\sqrt x)$:

We consider the inequality $\varphi(n)^{x}> c \cdot n$ with $x,c$ fixed.
Looking at prime powers first, we just want $(p-1)^{x}p^{x(k-1)}> c \cdot p^{k}$, being equivalent to $p^{k-1}> \sqrt[{x-1}]{\frac{c \cdot p}{(p-1)^{x}}}$. The left hand side grows with $k$ whereas the right hand side is constant. Especially, we get two observations:
a) fixing $p$, if the inequality is true for $k=1$, it's true for all $k$.
b) for any $p$, it is always true for all sufficient big $k$.

If $p > (2c)^\frac{1}{x-1}+1$, we get $(p-1)^{x-1}> 2c \geq c \cdot \frac{p}{p-1}$. This yields $(p-1)^{x}> c \cdot p$, giving our desired inequality in case of $n=p^{1}$. By observation a), it's also true for all $n=p^{k}$.
For every $p \leq (2c)^\frac{1}{x-1}+1$, we find $k$ such that the inequality is true for $n=p^{k}$ by observation b).

Let $P_{x,c}$ be the set of primes powers $p^{k}$ such that the inequality is wrong for $n=p^{k}$. We also set $Q_{x,c}$ to be the set of those $n$ that can be written as product of coprime elements of $P_{x,c}$. We've just seen that $P_{x,c}$ is finite, and thus $Q_{x,c}$ is, too.

We now look at the quotient $b(n) = \frac{\varphi(n)^{x}}{n}$. The function $b$ is multiplicative and $b(p^{k}) \geq 1$ for $p^{k}\not \in P_{x,1}$.
Thus $b(n)$ adopts some minimum, namely for some $n \in Q_{x,c}$.
Take $C_{x}$ with $\frac 1{C_{x}}= \mathrm{inf}_{n \in \mathbb{N}}\frac{\varphi(n)^{x}}{n}= \mathrm{min}_{n \in Q_{x,1}}\frac{\varphi(n)^{x}}{n}>0$.

Now for any $n$, we write $n=k \cdot m$ with $\gcd(k,m)=1$ and $k \in Q_{x,C_{x}}$.
If $n \not\in Q_{x,C}$, we get $m \not\in Q_{x,C}$ and  $\frac{\varphi(n)^{x}}{n}= \frac{\varphi(m)^{x}}{m}\cdot \frac{\varphi(k)^{x}}{k}> C \cdot \frac{1}{C}=1$ resulting in $\varphi(n)^{x}> n$ for all but finitely many $n$.

Setting $y= x^{2}$ (every $y>1$ can be expressed that way with some $x>1$), we have $\varphi(n)^{y}= \left(\varphi(n)^{x}\right)^{x}> n^{x}$.
Thus we can majorize the series over $\frac{1}{\varphi(n)^{y}}$ by that over $\frac 1{n^{x}}$, the latter being $\zeta(x)$ and convergent.


[quote="shinwoo"]find the remainder of

5^15 + 1 when it's divided by 3[/quote]

[hide="Solution"]  Without getting into mods, which I think is above the basics, and unnesecary for such a simple problem, we can use pattern recognition to aid us to the answer.

 Let's create a little table to demonstrate the remainders when the following numbers are divided by 3.

$ 5^1 \rightarrow 5 \rightarrow R 2$

$ 5^2 \rightarrow 25 \rightarrow R 1$

$ 5^3 \rightarrow 125 \rightarrow R 2$

$ 5^4 \rightarrow 625 \rightarrow R 1$

  As you can see:  $\frac{5^x}{3}$ leaves a remainder of 2 if $x$ is odd, and leaves a remainder of 1 if $x$ is even.  In our equation $\frac{5^{15}}{3}$  $x$ is odd meaning it will leave a remainder of $2$, however since the "plus 1" it will leave a remainder of $3$, meaning it will go in perfectly and leave no remainder. [/hide]

[quote="G-UNIT"][hide="Solution"]  Without getting into mods, which I think is above the basics, and unnesecary for such a simple problem, we can use pattern recognition to aid us to the answer.

 Let's create a little table to demonstrate the remainders when the following numbers are divided by 3.

$ 5^1 \rightarrow 5 \rightarrow R 2$

$ 5^2 \rightarrow 25 \rightarrow R 1$

$ 5^3 \rightarrow 125 \rightarrow R 2$

$ 5^4 \rightarrow 625 \rightarrow R 1$

  As you can see:  $\frac{5^x}{3}$ leaves a remainder of 2 if $x$ is odd, and leaves a remainder of 1 if $x$ is even.  In our equation $\frac{5^{15}}{3}$  $x$ is odd meaning it will leave a remainder of $\boxed{2}$. [/hide][/quote]
[hide="Umm no..."]
$5^{15}+1=(5+1)(5^{14}-5^{13}+5^{12}-...+5^2-5+1)$
I dont see remainder 2

[size=75][color=darkblue]<Moderator RCV's note:  ATTENTION middle schoolers.  You do NOT have to be able to perform the above factorization or even understand the above factorization.  I believe 10000th user was using a high-school technique to demonstrate that the number in question was divisible by 3.

10000th:  Please read the post titled "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=148301#p148301]Using this Forum[/url]".  Quoting Mr. Rusczyk:  "The Basics forum is for middle school students who are [b]just beginning[/b] with math contests.">[/color][/size][/hide]

[quote="10000th User"]Umm no...[/quote]
LOL :blush:  yea i forgot the "plus 1", ill go and edit it


Each degree in the Celsius temperature system is $ \frac{9}{5}$ of a degree in the Fahrenheit system, and $ 32^{\circ}\text{F} \equal{}  0^{\circ}\text{C}$.  What is the Fahrenheit temperature when the Celsius temperature is $ 45^{\circ}$?

for every 5 Celsius degrees, there are 9 Farenheit.

keeping this in mind

5x9=45

so replace the 5 with 9, 9x9=81.

but you have to add 32 because 0 C = 32 F

81 + 32 = [b]113[/b].

The general formula is:

$ \frac{9}{5}C \plus{} 32 \equal{} F$

($ C$ is Celsius, and $ F$ is Fahrenheit.)


in some geometry posts the term PTT and inversion is used. what is PTT.
its said that PTT is used in solving IMO1996 geometry problem

Maybe means Parallel tangents theorem (use the Search option)....a term coined by Shay Gueron in its American Mathematical Monthly paper.


The point $ I$ is the incenter of triangle $ ABC$. A circle centered at $ I$ meets $ BC$ at $ A_{1}$ and $ A_{2}$, $ CA$ at $ B_{1}$ and $ B_{2}$, and $ AB$ at $ C_{1}$ and $ C_{2}$, where the points occur around the circle in the order $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$, $ C_{2}$. Let $ A_{3}$, $ B_{3}$, $ C_{3}$ be the midpounts of the arcs $ A_{1}A_{2}$, $ B_{1}B_{2}$, $ C_{1}C_{2}$, respectively. The lines $ A_{2}A_{3}$ and $ B_{1}B_{3}$ meet at $ C_{4}$, $ B_{2}B_{3}$ and $ C_{1}C_{3}$ meet at $ A_{4}$, and $ C_{2}C_{3}$ and $ A_{1}A_{3}$ meet at $ B_{4}$. Prove that the lines $ A_{3}A_{4}$, $ B_{3}B_{4}$, $ C_{3}C_{4}$ are concurrent.

Easy to prove that $ C_{3},C_{1}$ are the symmetric points of $ B_{3},B_{2}$ wrt $ AI$,  so the bisector of angle $ \angle B_{3}A_{4}C_{3}$ is $ AI$. Similarly, the bisectors of the angles $ \angle C_{3}B_{4}A_{3}, \angle A_{3}C_{4}B_{3}$ are pass through $ I$. In the other hand, it's clear that the bisectors of the angles $ \angle A_{1}A_{3}A_{2}, \angle B_{1}B_{3}B_{2}, \angle C_{1}C_{3}C_{2}$ are $ A_{3}I,B_{3}I,C_{3}I$, respectively. So $ A_{3}B_{4}C_{3}A_{4}B_{3}C_{4}$ has a circle centered at $ I$ to incircle, and $ A_{3}A_{4},B_{3}B_{4},C_{3}C_{4}$ are concurrent from Brianchon's theorem


You have a 4 x 4 grid in a square shape with 16 squares. Just above this grid are attached two more squares, for a total of 18 little squares.

When you flick on a switch, 7 of the 18 squares will light up. The pattern is different and random each and every time you turn on the switch, but there will always only be 7 squares that light up. Let's say that you get to select 8 squares just before you turn the switch on. What is the probability that upon flicking on the switch that the 7 squares that light up will be a part of the 8 that you had selected just before turning on the switch?

The furthest I have gotten is that the denominator (number of combinations in universe) should be C(18,8) x C(18,7) and I can't even guarantee that this is correct. Goodluck!

hi, i see you have posted this in the intermediate forum.  Please don't double post.

[quote="blahblahblah"]You choose from $C^{18}_8$ combinations; $8$ of those combinations are good, so the answer is

$\frac {8}{C^{18}_8}$[/quote]

Very interesting insight.

But using your idea, how would you calculate the probability if you are now allowed to choose 9 squares instead of 8, or how about 10, 11?

The tricky one here is what if you were to choose all 18 squares... your probability should be 1, but I don't think it works out...

9: $\frac{C^{9}_7}{C^{18}_9}$=~0.07%

18: $\frac{C^{18}_7}{C^{18}_{18}}$=31824=318.24%

Let's say you have x total squares, then you at first choose m squares and then when you flick the switch, n squares light up ($m\gen$).  Then there are $\binom{m}{n}$ ways for the n lights to be a subset of the m squares, and $\binom{x}{n}$ ways for the n tiles to be chosen.  So we get $\frac{\binom{m}{n}}{\binom{x}{n}}=\frac{m!(x-n)!}{(m-n)!x!}$.  In this case, with $x=18$, $m=8$ and $n=7$, you get $\frac{8!11!}{1!18!}=\frac{1}{13\times17\times18}$.

the way that the squares are oriented does not matter at all...why does the problem have so much unnecessary info?


[quote="pcalin"]Let A= \sum yz/(y+z) 2  and B= \sum yz/[(x+y)(x+z)], where x,y,z are positive reals. Prove that:

a) A+3B \geq 3.
b) 4A+8B \geq 9.[/quote]
=============
The first one is equivalent to
 \sum x^3.y^3 + x^4.y.z >= \sum x^3.y^2.z
That holds by Murhead inequality
(3,3,0) > (3,2,1) and (4,1,1) > (3,2,1)

The second one seems to be false. Check for x = 3, y = 2, z = 1.

Namdung


I am practicing olympiad-style questions, and I would like to know if there is any way I could improve my solutions, since I don't want to lose marks due to presentation/lack of logical clarity. Here is my solution to one problem, so if anyone could point out holes/ ways that it could be improved I eould be very thankful.

Similarly, if the solution is fine, could someone please tell me what it is in the solution that the markers would be looking for, so that I can write up all my solutions in a similar way.

Many thanks in advance.

[hide="Problem"]A function $ f$ is defined over all the positive integers and satisfies:

$ f(1)\equal{}1996$

$ f(1)\plus{}f(2)\plus{}f(3)...\plus{}f(n)\equal{}n^2f(n)$

Calculate the exact value of $ f(1996)$[/hide]

[hide="Solution"][b]Lemma 1[/b]: The sum of the reciprocals of the first $ n$ triangular numbers is $ \frac{2n}{n\plus{}1}$.

[b]Proof[/b]: When $ n\equal{}1$ we have that the sum is $ \frac{2}{2}\equal{}1$ so the base case is done.

Now assume the result proven for $ k$ and prove it for $ k\plus{}1$.

We have that the sum is $ \frac{2k}{k\plus{}1}\plus{}\frac{1}{\frac{1}{2}(k\plus{}1)(k\plus{}2)}$.

$ \frac{2k(k\plus{}2)}{(k\plus{}1)(k\plus{}2)}\plus{}\frac{2}{(k\plus{}1)(k\plus{}2)}$

$ \frac{2k^2\plus{}4k\plus{}2}{(k\plus{}1)(k\plus{}2)}$

$ \frac{(2k\plus{}2)(k\plus{}1)}{(k\plus{}1)(k\plus{}2)}$

$ \frac{2(k\plus{}1)}{(k\plus{}2)}$

$ \frac{2(k\plus{}1)}{(k\plus{}1)\plus{}1}$

So by the principle of mathematical induction, the result holds true for all $ n$.

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

We wish to show that $ f(n)\equal{}\frac{1}{\frac{1}{2}n(n\plus{}1)} \cdot f(1)$.

For $ n\equal{}1$ we have that $ f(1)\equal{}\frac{1}{\frac{1}{2}(1(2))} \cdot f(1) \equal{} \frac{1}{1} \cdot f(1) \equal{} f(1)$ so base case done.

Assume the result true for all integers up to and including $ k\minus{}1$, and prove the result for $ k$.

We know that $ k^2f(k)\equal{}f(1)\plus{}f(2)\plus{}...f(k)$

Subtract $ f(k)$ from both sides to obtain:

$ (k^2\minus{}1)f(k)\equal{}f(1)\plus{}f(2)\plus{}f(3)...\plus{}f(k\minus{}1)$

$ (k\plus{}1)(k\minus{}1)f(k)\equal{}(\frac{1}{1} \plus{} \frac{1}{3} \plus{} \frac{1}{6} ... \plus{} \frac{1}{\frac{1}{2}(k\minus{}1)(k)}) \cdot f(1)$

By Lemma 1, this becomes:

$ (k\plus{}1)(k\minus{}1)f(k)\equal{}\frac{2(k\minus{}1)}{k} \cdot f(1)$

$ f(k)\equal{}\frac{2}{k(k\plus{}1)} \cdot f(1)\equal{}\frac{1}{\frac{1}{2}k(k\plus{}1)} \cdot f(1)$

So, by the principle of mathematical induction, $ f(n)\equal{}\frac{1}{\frac{1}{2}n(n\plus{}1)}$ for all $ n$.

Using this, $ f(1996)\equal{}\frac{1}{\frac{1}{2} \cdot 1996 \cdot 1997} \cdot f(1) \equal{} \frac{2f(1)}{1996 \cdot 1997}$.

Substitute $ f(1)\equal{}1996$

$ f(1996)\equal{}\frac{2 \cdot 1996}{1996 \cdot 1997} \equal{} \frac{2}{1997}$[/hide]

The answer is correct.

Here is my solution:

For $ n\geq 2$, $ f(n) \equal{} n^2f(n) \minus{} (n \minus{} 1)^2f(n \minus{} 1)$

$ \Longleftrightarrow (n^2 \minus{} 1)f(n) \equal{} (n \minus{} 1)^2f(n \minus{} 1)$

$ \Longleftrightarrow (n \plus{} 1)f(n) \equal{} (n \minus{} 1)f(n \minus{} 1)\ \because n\geq 2$.

$ \Longleftrightarrow n(n \plus{} 1) f(n) \equal{} n(n \minus{} 1)f(n \minus{} 1)$

$ \Longleftrightarrow n(n \plus{} 1)f(n) \equal{} 2\cdot(2\minus{}1)f(1)$.

$ \therefore f(n) \equal{} \frac {2f(1)}{n(n \plus{} 1)}\ (n\geq 1)$, yielding $ f(1996) \equal{} \frac {2}{1997}$

I knew that the answer was correct, but I was asking if anyone could check the layout/presentation of my solution.

If there are gaps, or parts that are unclear, could you point them out.

Also, if the solution is fine, could you tell me what it is about my solution that makes it like that, i.e what the markers would be looking for so that I can write up all my solutions in a similar way.

The main reason I posted this topic was to minimise my chances of losing marks not due to an error in my mathematics, but in the way I present my solution.

As my opinion, your solution is long. Why do you solve the reccursive directly?

With a problem, I generally find it easier/quicker to continue with the first approach I find rather than try and find a simpler one. If it gets me nowhere, then I will search for another way.

Upon calculating $ f(n)$ for small $ n$, and putting them in terms of $ f(1)$, the triangular numbers leaped out to me as the obvious pattern. It seemed obvious for me to continue with that approach.

Well, this is my solution:

$ f(1)\plus{}f(2)\plus{}f(3)...\plus{}f(n)\equal{}n^{2}f(n)$

$ f(1)\plus{}f(2)\plus{}f(3)...\plus{}f(n)\plus{}f(n\plus{}1)\equal{}(n\plus{}1)^{2}f(n\plus{}1)$

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

$ f(n\plus{}1)\equal{}\frac{nf(n)}{n\plus{}2}$

Now we can substitute $ f(n)\equal{}\frac{(n\minus{}1)f(n\minus{}1)}{n\plus{}1}$ and continue substituting until we conclude that

$ f(n\plus{}1)\equal{}\frac{2n!f(1) }{(n\plus{}2)!} \implies f(n)\equal{}\frac{2f(1)}{n(n\plus{}1)}$

We can easily prove this by induction:

$ 1^{\circ}$ for $ n\equal{}1$ the statement is true

$ 2^{\circ}$ assume it's true for $ f(n)$

$ 3^{\circ} f(n\plus{}1)\equal{}\frac{nf(n)}{n\plus{}2}  \implies f(n\plus{}1)\equal{}\frac{n}{n\plus{}2} \cdot \frac{2f(1)}{n(n\plus{}1)} \implies \\ f(n\plus{}1)\equal{}\frac{2f(1)}{(n\plus{}1)(n\plus{}2)} \quad\blacksquare$

now we have that $ f(1996)\equal{}\boxed{\frac{2}{1997}}$

@kunny

In the 5th row of your solution you input $ n\equal{}2$ on RHS but you left LHS unchanged. Could you explain why it can be done?

[quote="Flame"]

@kunny

In the 5th row of your solution you input $ n \equal{} 2$ on RHS but you left LHS unchanged. Could you explain why it can be done?[/quote]

[quote="kunny"]The answer is correct.

Here is my solution:

For $ n\geq 2$, $ f(n) \equal{} n^2f(n) \minus{} (n \minus{} 1)^2f(n \minus{} 1)$

$ \Longleftrightarrow (n^2 \minus{} 1)f(n) \equal{} (n \minus{} 1)^2f(n \minus{} 1)$

$ \Longleftrightarrow (n \plus{} 1)f(n) \equal{} (n \minus{} 1)f(n \minus{} 1)\ \because n\geq 2$.

$ \Longleftrightarrow n(n \plus{} 1) f(n) \equal{} n(n \minus{} 1)f(n \minus{} 1)$[/quote]

Because $ \{n(n\plus{}1)f(n)\}$ is a constant sequense, so $ n(n\plus{}1)f(n)\equal{}2\cdot (2\minus{}1)f(2\minus{}1)\ \because n\geq 2.$

A friend of mine (nevfer) from another forum has authorized me to post his solution here as follows:
[url=http://img89.imageshack.us/i/nevfersol.gif/][img]http://img89.imageshack.us/img89/577/nevfersol.gif[/img][/url]

I like the solution. :lol:

Or from my post, use $ \frac{f(n)}{f(n\minus{}1)}\equal{}\frac{n\minus{}1}{n\plus{}1}$ repeatedly.

Or taking logarithm with base 10 of the both sides gives an answer.


The mean of six positive integers is 5, and the median is 6.
What is the largest the mode could be, given that the mode is
unique?


I've been arguing with my friend about lenses.
Suppose the curves y=f(x), y=g(x)   define a region in R^2, that we call a lens, and we say it's made out of water (or glass, or whatever specific thing you want to say).

And say we want to track one ray of light, say approaching the lens along the line y=mx+b.

Can't we just use snell's law, taking the "normal" to be perpendicular to the lens' tangent line?

you can. in fact, that is exactly how we get Lensmakers formula for a lens.


trigonometry
trig identities
Law of Cosines

I am waiting solution for weeks.Please find x angle using pure geometrical ways.(don't use trigonometry)

what's the point in solving a trig problem without using trig?

To solve the problem, substitute values for a and b first, and then use trig, such as the law of cosines and the law of sines.

and what are pure geometrical ways?

I mean you can use auxilary drawing or constructions.Perhaps you must extend any side of triangle or draw any triangle on the side of main triangle that congruent each other,like this.I mean "synthetic solution".

You can mabye use the fact that BD is an angle bisector.


Let $x^{*}=x-\lfloor{x}\rfloor$. How many numbers $b$ are there such that $(b^{*})^{2}-b^{2}$ is an integer and $1\le|b|<5$?

[i]Note: If this competition somehow hasn't ended yet (I checked the regulations and you were supposed to distribute the test from 7-9:30 yesterday morning or something like that) please remove this[/i]

Substitute $b=n+\frac{p}{q}$ so that $p<q, (p,q)=1$, and then just do casework. I got $1+3+5+7=\fbox{16}$ numbers total.

Unfortunately, this was a multiple choice question and the smallest answer of the multiple choices was 20. So you're missing some  :wink:

Don't forget the negative integers.

Oh yea, I forgot negatives. I'm just gonna blindly guess it's 32 then.

32 wasn't an answer choice either...

Brute-forcingly I'll list out the 20 positive numbers I got.

1, 1.5
2, 2.25, 2.5, 2.75
3+{0, 1/6, 2/6, ... 5/6}
4+{0, 1/8, 2/8, ... 7/8}

Oh I didn't even count integers

rofl

I'm too lazy fix my work

What I'm really interested in is how many negative numberswork. On the test I just assumed it was the same number as the number of positive numbers, so I put down 40 as the answer. However, someone else later pointed out that -1.25 works while 1.25 doesn't...

...except that led him to the answer 48, when the answer choices were

A) 20 B) 40 C) 49 D) 50 E) 58

I would have been trapped too.
This was my solution:

We must solve the problem for b=[b]+b* with b in (-5,-1]U[1,5).

Let's consider first the alternative problem with b in [-5,-1)U[1,5),
so that we can consider b with [b] in {-5,-4,-3,-2}U{1,2,3,4}.

b*^2-b^2 integer is equivalent to 2[b]b* integer,
so the solutions (if [b] isn't equal to 0) must have
b*=k/(2|[b]|) where k is integer in [0,2|[b]|),
so there are 2|[b]| solutions for every [b].

If we have [b] in {-5,-4,-3,-2}U{1,2,3,4}, then there must be 2(5+4+3+2+1+2+3+4)=48 solutions.

The only differences with our problem, b in (-5,-1]U[1,5),
is that we must exclude the solution b=-5 and we must include the solution b=-1,
so there should be 48 solutions as well.

I came up with a similar solution.

If $b = n+\frac{p}{q}$ as proposed earlier, then we want $-n^{2}-2n\frac{p}{q}$ to be an integer, as $n$ is the integer part of $b$, where $n \in{1,2,3,4,-2,-3,-4,-5}$ that allows for a full range of values such that $0 \le \frac{p}{q}< 1$. The only case missing is when $n =-1$, which yields $b =-1$. Thus we want to show that $q \mid 2np$.

$\frac{2np}{q}= k$, where $k$ is a non negative integer and $n$ is actually the magnitude of the value of $n$, thus $\frac{p}{q}= \frac{k}{2n}$, therefore $0 \le \frac{k}{2n}< 1$, for a given $n$ there are $2n$ solutions for $\frac{p}{q}$.

The total number of solutions is therefore $2(1+2+3+4+2+3+4+5)+1 = 48+1 = 49$

But it seems that you're including $b=-5$.


factorial
number theory
prime factorization

Find all solutions of $a^{2}+b^{2}=n!$ for positive integers $a,$ $b,$ $n$ with $a\le b$ and $n<14.$

[hide="Solution"] A number expressible as the sum of two squares must have the property that each prime of the form $4k+3$ in its prime factorization is of even degree.

We have the obvious $\boxed{a = 1, b = 1, n = 2}$.  We need $3$ to divide $n!$ an even number of times, which occurs at $n = 6$.  Indeed,

$6! = 720 = (12^{2})(1^{2}+2^{2})$

Which gives $\boxed{a = 12, b = 24, n = 6}$.  For $n > 6$ the prime $7$ can only divide $n$ once, so no other solutions are possible. [/hide]


$ N$ dwarfs of heights $ 1, 2, 3, ..., N$ are arranged in a circle. For each pair of neighbouring dwarfs the positive difference between the heights is calculated; the sum of these $ N$ differences is called the "total variance" $ V$ of the arrangement.

Find (with proof) the maximum and minimum possible values of $ V$.


Provide a synthesis of 2-methylcyclobutanone from 4-methylcyclohexanone.

(This problem features the general issue of making a [i]two carbon[/i] ring contraction of a cyclic ketone. My proposed method for this problem involves about 10 steps and 8 different reactions.)

Provide a synthesis of 2-methylcyclobutanone from 4-methylcyclohexanone.

My method is rather long but is it right?  :idea:

1. Use Cl2 to chlorinate in the posn alpha to C=O
2. Use Favorskii rearrangement ( in presence of a base and then H+ ) to convert it into 3 methyl    cyclopentane 1 oic acid.
3. Decarboxylate to get methyl cyclopentane.
4. Use aq. NaOH and then dehydrate to get methyl cyclopentene
5. Use SeO2 to perform allylic hydroxylation and then reduce the double bond ( Use H2/Pt )
6. Oxidise the C-OH into C=O using PCC
7. Again perform $ \alpha$ chlorination using Cl2
8. Again perform Favorskii rearrangement and decarboxylate to get methyl cyclobutane
9. Chlorinate and then use alc. KOH to get methyl cyclobutene.
10. Use HBO and then PCC to get 2 methyl cyclobutanone.
 :)

1.alpha chlorination using Cl2
2.base OH- to get the cyclopropanone intermediate(favorskii)
3.cleave by OH- to get 3-methylcyclopentanecarboxylicacid
4.perform smidt reaction
5.NaNO2/HCl
6.Sarret reagent(PCC)
7.again one more favorskii reaction

[quote="Chemrock"]5.NaNO2/HCl[/quote]

Again? How many times does one have to say that alkyl diazonium salts are not stable?

I have used many of madness conversion....
i have made modifications...
have a look at it...

1. Use Cl2 to chlorinate in the posn alpha to C=O
2. Use Favorskii rearrangement ( in presence of a base and then H+ ) to convert it into 3 methyl cyclopentane 1 oic acid.
3. Use red P/Cl2 to get a-chlorination
4. Decarboxylate and use aq KOH .
5. perform opennaur oxidation.
6. Use Cl2 to get a-chlorination
7.Again perform Favorskii rearrangement and use red P/Cl2 to get a-chlorination.
8.decarboxylate and use aq.KOH.
9. Use PCC to get 2 methyl cyclobutanone.

[quote="Carcul"][quote="Madness"]4. Use aq. NaOH and then dehydrate to get methyl cyclopentene[/quote]

How can an elimination be performed on methyl cyclopentane?[/quote]

Please explain i still dont understand why not?

[quote]
madness ,in elimination reaction -OH is the nucleophile.....
now as the carbo cation is formed ...
Demjenov rearrangement takes place and the ring expands.
[/quote]
wait a min. am i reading this rite. first madness performs an elimination reaction on methyl cyclopentane(which is pretty amazing!). secondly, demjanov rearrangement takes place during the elimination( which is even more amazing!) :rotfl:  :rotfl:  :rotfl:

[quote="Valeriummaximum"]4. Decarboxylate and use aq KOH .
5. perform opennaur oxidation.
6. Use Cl2 to get a-chlorination[/quote]

Would steps 4. and 6. be selective?

[quote="Madness"]Please explain i still dont understand why not?[/quote]

Do we have any leaving group in methylcyclopentane?

yes i think steps 4 and 6 are selective...........

Is that so? In that case can you show with a mechanism why the elimination step is selective?


search
geometry
geometric transformation
algebra proposed
algebra

Solve: \[xy + yz + zx = 1,\] \[3\left(x + \frac{1}{x}\right) = 4\left(y + \frac{1}{y}\right) = 5\left(z + \frac{1}{z}\right).\]

Arne, as far as I remember this has been discussed some days ago, with the problems from Girls olympiad in China. Try to search please.

Sorry, I didn't know (and I can't find it either).

The topic is equation system posted by Zhaoli in algebra unsolved.

This is almost the same as Chinese Girl Mathematical Olympiad 2005 question 2.
Here is a link to the questions and solutions.
[url]http://gifted.hkedcity.net/Gifted/ActReview/2005CGMO/Solution.pdf[/url]
(Although it is in Chinese I think most of you can still understand or guess  :) some parts of it. Just feel free to ask for translation if necessary. ;) )


I have posted a proof here already
http://www.mathlinks.ro/Forum/viewtopic.php?p=268306#p268306


AMC
AIME
USA(J)MO
USAMO
AMC 10
AIME I
AIME II

How did everyone do on the AIME?  I am sad to report an expected score of 8 here, but on the plus side, I should qualify for USAMO which is all that matters.

11 for me... I thought I had done much better coming out of it though
I've spoken with Jane, John, and Damien, all of whom are probably taking the USAMO.  It seems John was our big winner with a 14  :D

[quote="jb05"]I have a first name? Anyway, 14. Makes up for my AMC scores.[/quote]

When did this happen?  Unless we are not talking about Berman...

[quote="jb05"]I have a first name? Anyway, 14. Makes up for my AMC scores.[/quote]

yeah I was seriously pretty confused as to who john was

and then I was like OH YEAH BERMAN

I got 9. That gives me a 210 index, which will probably be close to the USAMO cutoff.

Probably. It will be pushing it. I guess it's the difference between floor 7 and floor 8. I'm betting on floor 8, even with the unnecessarily large USAMO pool these last few years. By the way, the answers are up on the AMC page, but they are exactly the same as what we decided.

Hmm is anyone taking the B AIME?

I really doubt a 203 index will make USAMO  :(
Did Vivek take the AIME?

well a 191.5 made it last year... anyway I heard from teki that Vivek got a 6 or 7...  :(

anyway I will probably make USAMO with the exact same index as last year... (counting amc10 score)

EDIT: actually 197.5 was the index cutoff last year

The low index last year was due to a computation heavy (and thus high error rates) AIME, but this year, the AIME was much more standard and I would be a little surprised if 203 index makes it (though I hope it does for multiple reasons).  My index will be 224, if they use the AMC 10 again (well, they don't have a choice unless they want to look up my other score =D) for young people; otherwise, I'm screwed (189.5)...

I'm sure you'll be fine. This year's AIME seemed a lot easier than other recent ones, so I still think there will be high indices.

[quote="Teki-Teki"]8 for a 203 index.  FAIL.  I will probably make USAMO, though, so I am not too worried.

Actually, I am worried.  I cannot wait for AIME II to be over so that I know if I qualified for USAMO.  Darn.[/quote]

Woah

I failed too.. 2 stupid mistakes XD

I got an

[hide]
8
[/hide]

which gives me an index of

[hide]
80+138=[b]218[/b][/hide]


topology
group theory
abstract algebra
advanced fields
advanced fields unsolved

How to prove that a Möbius band is homotopy equivalent to $\mathbb{S}^{1}$? And how to show that the boundary of a Möbius band is not a retract of $M$?

point one: just see $M$ as the quotient of $[-1,1]\times[0,1]$ with respect to $(x,0) \tilde (-x,1)$, $S^{1}$ as $\{0\}\times [0,1]/\tilde \subset M$ and call $i$ the inclusion. of course, $F: M\times [0,1] \to M, F: ([(x,y)], t) \mapsto [(xt, y)]$ is an homotopy between $id_{M}$ and $i$.
so $S^{1}$ is a deformation retract.

point two: if $\partial M$ was a deformation retract of $M$, there'd be an isomorphism $i_{*}$ between the two fundamental groups $\pi_{1}(\partial M), \pi_{1}(M)$ given by inclusion. but take a generator $\alpha$ of $\pi_{1}(\partial M)$, and check that it generates $2\pi_{1}(M)$ (the group of squares).

Correction: $2\pi_{1}(M)$ is the (additive) group of even numbers, not of squares.

uhm.. it's the subgroup of even numbers in $(\mathbb{Z},+)$: i just meant the squares in $(\pi_{1}(M), *)$ (the operation is usually called product)...
anyway, this is just a matter of terminology :P
the substance is the same  :)


floor function
inequalities
function
algebra

Prove that, for any ${ n \in \mathbb N}$,
\[ \left\lfloor{n \over 3}\right\rfloor+\left\lfloor{n+2 \over 6}\right\rfloor+\left\lfloor{n+4 \over 6}\right\rfloor=\left\lfloor{n \over 2}\right\rfloor+\left\lfloor{n+3 \over 6}\right\rfloor. \]

Just an idea:
Consider several cases:
n = 6k;
n = 6k +1;
n = 6k +2;
n = 6k +3;
n = 6k +4;
n = 6k + 5.
No more other cases.
Sub. and compare both sides.

Modular arithmetic, as luimichael suggested, easily proves the equation. Is there a more elegant solution than just casework?

One more, prove the inequality,

$ \displaystyle \left\lfloor\left({3 \over 2}\right)^n\right\rfloor > \left({3 \over 2}\right)^n - 1 + {1 \over 2^{n + 1}.}$

[hide]Suppose $ \left\lfloor \left( \frac{3}{2} \right)^n \right\rfloor \equal{} \left( \frac{3}{2} \right)^n \minus{} x$ for $ 0 \leq x < 1$. $ \left( \frac{3}{2} \right)^n \equal{} \frac{3^n}{2^n}$, so the floor function can't take this down any more than $ \frac{2^n \minus{} 1}{2^n}$, giving an upper bound for $ x$. Thus \[ \left\lfloor \left( \frac{3}{2} \right)^n \right\rfloor \geq \left( \frac{3}{2} \right)^n \minus{} \frac{2^n \minus{} 1}{2^n} > \left( \frac{3}{2} \right)^n \minus{} 1 \plus{} \frac{1}{2^{n\plus{}1}}\][/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that every trasitive permutation representation $(G,X)$ is equivalent to a permutation representation of $G$ acting on the set of right (or left, as you want) cosets as $x\in G: Hg \mapsto Hgx$.

Hi Peter,

I hope I am not too late  :(

Let $C$ denote the set of cosets.
Equivalences are obtained by providing a bijection $\alpha$  between $X$ and $C$ ,and a group isomorphism between $G$ and $G$, such that

$\forall z\n X,\forall g\in G : (\alpha(z))^{\beta(g)}=\alpha(z^{g})$

We will (and that is probably also what we HAVE to do) assume $\beta$ is the identity on $G$

Choose a random fixed element $x\in X$.  Now consider the subgroup $H=G_{x}$  of $G$.
Now take any random element $y\in X$.  Consider a solution  $g$ of $x^{g}=y$ which exists as you assumed transitivity.
Let $\alpha(y)=H g \in C$
WARNING! :!:  You do have to prove that this map does not depend on the chosen solution $g$.
Finding the correspondence is, as usual, the hardest thing.  Injectivity of $\alpha$, its surjectivity and the
$\forall z\n X,\forall g\in G : (\alpha(z))^{\beta(g)}=\alpha(z^{g})$ property are all easy to check!

Enjoy!

[quote="fredbel6"]I hope I am not too late  :( [/quote]It's never to late to learn. ;) thanks!


Find all triples of $ (x, y, z)$ of positive intergers satisfying $ 1\plus{}{4}^{x}\plus{}{4}^{y}\equal{}z^2$.

let me try. Assume  $ x \ge y$. Then we get  $ 4^y (4^{x \minus{} y}  \plus{} 1) \equal{} (z \minus{} 1)(z \plus{} 1)$. Since  $ (4^y ,4^{x \minus{} y}  \plus{} 1) \equal{} 1$, then we have 2 conditions to check.
1-)  $ z \plus{} 1 \equal{} 4^y$  and  $ z \minus{} 1 \equal{} 4^{x \minus{} y}  \plus{} 1$  or  $ 2z \equal{} 4^y  \plus{} 4^{x \minus{} y}  \plus{} 1$. But this is becomes equality of even number and odd. So we got 2 conditions to check.
1-A-) $ y\equal{}0$ means $ z\equal{}0$ from  $ z \plus{} 1 \equal{} 4^y$. Contradiction.
1-B-) $ x\equal{}y$ means $ z\equal{}3$. From here we get solution $ (x,y,z)\equal{}(1,1,3)$.

2-) $ z \minus{} 1 \equal{} 4^y$  and  $ z \plus{} 1 \equal{} 4^{x \minus{} y}  \plus{} 1$. this condition giving contradiction since $ z$ can not be even and odd at same time.

So only answer is $ (1,1,3)$.

I hope it is true... :oops:

[quote="theSA"]let me try. Assume  $ x \ge y$. Then we get  $ 4^y (4^{x \minus{} y} \plus{} 1) \equal{} (z \minus{} 1)(z \plus{} 1)$. Since  $ (4^y ,4^{x \minus{} y} \plus{} 1) \equal{} 1$, then we have 2 conditions to check.
1-)  $ z \plus{} 1 \equal{} 4^y$  and  $ z \minus{} 1 \equal{} 4^{x \minus{} y} \plus{} 1$  or  $ 2z \equal{} 4^y \plus{} 4^{x \minus{} y} \plus{} 1$. But this is becomes equality of even number and odd. So we got 2 conditions to check.
1-A-) $ y \equal{} 0$ means $ z \equal{} 0$ from  $ z \plus{} 1 \equal{} 4^y$. Contradiction.
1-B-) $ x \equal{} y$ means $ z \equal{} 3$. From here we get solution $ (x,y,z) \equal{} (1,1,3)$.

2-) $ z \minus{} 1 \equal{} 4^y$  and  $ z \plus{} 1 \equal{} 4^{x \minus{} y} \plus{} 1$. this condition giving contradiction since $ z$ can not be even and odd at same time.

So only answer is $ (1,1,3)$.

I hope it is true... :oops:[/quote]

Sorry, but $ (k, 2k\minus{}1, 2^{2k\minus{}1}\plus{}1)$ is answer too! :(

Yes [b]Stephen[/b] you are right.
My solution is not complete.
Just the several cases.
indeed,it can be easily observed that all numbers are positive integers. There is also 1.
So  $ 1 \plus{} 2 \cdot 2^{2x \minus{} 1}  \plus{} 2^{2y}  \equal{} 1 \plus{} 4^x  \plus{} 4^y  \equal{} (1 \plus{} a)^2  \equal{} 1 \plus{} 2a \plus{} a^2$  which ecaxtly gives $ y\equal{}2x\minus{}1$.

Just a simple observation.

A related problem: http://www.mathlinks.ro/viewtopic.php?t=245391

[hide="my solution"]WLOG $ x\ge y$.

$ 1\plus{}4^x\plus{}4^y>(2^x)^2\implies 1\plus{}4^x\plus{}4^y\ge(2^x\plus{}1)^2\equal{}1\plus{}4^x\plus{}2^{x\plus{}1}\implies 4^y\ge2^{x\plus{}1}\implies 2y\ge x\plus{}1\implies x\le2y\minus{}1$ (1)

$ z$ is odd, let $ z\equal{}2k\plus{}1$
$ 1\plus{}4^x\plus{}4^y\equal{}1\plus{}4k\plus{}4k^2\implies 4^{x\minus{}1}\plus{}4^{y\minus{}1}\equal{}k(k\plus{}1)\implies4^{y\minus{}1}(4^{x\minus{}y}\plus{}1)\equal{}k(k\plus{}1)$
One of $ k$ or $ k\plus{}1$ is divisible by $ 4^{y\minus{}1}$ $ \implies k\plus{}1\ge4^{y\minus{}1}$.

If $ k\plus{}1\equal{}4^{y\minus{}1}$, $ 4^{x\minus{}y}\plus{}1\equal{}4^{y\minus{}1}\minus{}1$, clearly impossible.

$ k\ge4^{y\minus{}1}$ $ \implies$ $ 4^{y\minus{}1}(4^{x\minus{}y}\plus{}1)\equal{}k(k\plus{}1)\ge4^{y\minus{}1}(4^{y\minus{}1}\plus{}1)\implies x\minus{}y\ge y\minus{}1\implies x\ge2y\minus{}1$ (2)

Combining (1) and (2), we get $ x\equal{}2y\minus{}1$.

Solution: $ (a,2a\minus{}1,2^{2a\minus{}1}\plus{}1)$ and $ (2a\minus{}1,a,2^{2a\minus{}1}\plus{}1)$.[/hide]

:D


let
$x_{1},x_{2},....,x_{2006}$be positive real number satsfying the condition
$\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+...+\frac{1}{1+x_{2006}}=1$
find the minimal posible value of the exepression
$x_{1}x_{2}....x_{2006}$

Well, very recently posted in inequalities :wink:
[hide="solution"]
Let $y_{i}=\frac{1}{1+x_{1}}$ so $\sum y_{i}=1$
We claim that $\prod x_{i}\geq 2005\iff \prod \frac{1-y_{i}}{y_{i}}\geq 2005$
Note that: $\frac{1-y_{i}}{y_{i}}=\frac{\sum_{j\not =i}y_{j}}{y_{i}}\geq \frac{2005}{y_{i}}\sqrt[2005]{\prod_{j\not =i}y_{i}}$
Thus, $\prod x_{i}\geq 2005$ when it all cancels out (sorry hard to type)
[/hide]


I'm doing an independent study with one of my high school's math teachers and a few other students next year, and we're looking for good textbooks to use.  We're planning on covering multivariable calculus, linear algebra, and (basic) differential equations in a yearlong course, so we'll need textbook(s) that cover all of these topics.

It seems like Stewart's Multivariable Calculus is the standard multivar book, but are there any other good (or better) ones?  Also, I have no experience with linear algebra and differential equations textbooks, so can anybody give me good recommendations for those as well?

I've found Multivariable Calculus by George Cain & James Herod a pretty good beginners course text.

A friend of mine was using
Differential Equations and Linear Algebra First Edition. by Jerry Farlow James E. Hall
for his math course a couple of years back.
 Also Stanley I. Grossmann has authored some very good books on all your criteria

Gilbert Strang has two really nice books out for linear algebra and multivariable calculus, although Stewart's is good as well.

When you get to the vector calculus part of multivariable calculus, try to get a hold of [i]Vector Calculus[/i] by Marsden and Tromba.

Div, Grad, Curl, and All That is a good book for vector Calculus.

Boyce and DiPrima is a really good book for differential equations.

I agree, Strang is an excellent math author.


You have 5 red shirts, 5 green shirts, 6 pairs of pants, 8 green hats and 8 red hats, all of which are distinct. How many outfits can you make consisting of one shirt, one pair of pants, and one hat without having the same color of shirts and hats?

You have Shirt- 5R 5G
             Pants-    6
             Hat-   8R 8G

If you choose one of the red shirts, then you have six choices for pants, and 8 choices for a hat, since you can't wear the same color. You have the same numbers if you choose a green shirt. So, our answer is 5*6*8*2=480.[/i]


If we let A be a point on the circle, B must be within 1/6 of the circumference on either side of A.  So unless I'm mistaken, it's just 1/3.

I know this sounds odd, but it is ture:
That is not necessarily the correct answer, even though there are no flaws in your logic or math.

I think it depends on the random process you use to draw the chord.

Hello, towbomb!

This is a classic paradox in the study of Probability.
grobber is absolutely correct!
There are at least three different approachs and answers.

[quote]What is the probability that the length of a chord randomly drawn in a circle
 will be smaller than the circle's radius?[/quote]

Inscribe a regular hexagon is a circle of radius R.
Start at the 6:00 position and label the vertices counterclockwise: A-B-C-D-E-F.

[b][1] Method #1[/b] (Suggested by ThAzNi)
Draw a tangent to the circle at A, call it $T_1AT_2$.

Let $P$ be any point on the circumference of the circle.
If $AP < R$, then $P$ must lie on arc $AB$ or on arc $AF$.

Since $\angle BAT_2 = \angle FAT1 = 30^o$, the probability is: $\frac{60^o}{180^o} \;= \;\frac{1}{3}$

[b][2]  Method #2[/b]
In the above diagram, draw diameter AD.
We will consider all chords perpendicular to this diameter.

Draw chord BF, intersecting diameter AD at G.
Draw chord CE, intersecting diameter AD at H.
Note that BF = CE = R.

A chord will be shorter than R if its midpoint lies on GA or HD.
Hence, the probability is: $\displaystyle{\frac{GA + HD}{AD}}$

It can be shown that $GA = HD = \left(1 - \frac{\sqrt{3}}{2}\right)R $

The probability is: $\displaystyle{\frac{2(1 - \frac{\sqrt{3}}{2})R}{2R} \;= \;1 - \frac{\sqrt{3}}{2} \;\approx \;0.134 }$

[b]Method #3[/b]
In the diagram, inscirbe a circle in hexagon ABCDEF.
This circle has radius $\frac{\sqrt{3}}{2}R$ and area $\frac{3}{4}\pi R^2$

Consider the midpoints of all the possible chords.
If the chord is to be shorter than R, its midpoint must lie in the outer ring.

Then the probability is: $\displaystyle{\frac{\text{Area of ring}}{\text{Area of circle}} \;= \;\frac{\frac{1}{4}\pi R^2}{\pi R^2} \;= \;\frac{1}{4}}$


If (tan A) + (tan [i]m[/i]A) + (tan [i]n[/i]A) = (tan A)(tan [i]m[/i]A)(tan [i]n[/i]A), then show that A = (3.1416[i] r[/i])/([i]m[/i] + [i]n[/i] + 1), where [i]r [/i]is any integer.

If $ \tan A\plus{}\tan mA\plus{}\tan nA\equal{}(\tan A)(\tan mA)(\tan nA)$,

then $ A\plus{}mA\plus{}nA\equal{}r\pi$ where 'r' is an integer

$ A\equal{}\frac{r \pi}{1\plus{}m\plus{}n}$

[b]picture74[/b], you are right. Knowing this common identity tan (A) + tan (B) + tan (C) =  (tan A)(tan B)(tan C), if A + B + C = 180 degrees; your solution becomes mechanical in nature. . . . nice conclusion though.

Here's a hint for proving that

tan(a) + tan(b) + tan(c) = tan(a) tan(b) tan(c)

if and only if a + b + c is a multiple of π as discussed in this blog post.

Use the identity

tan(α + β) = (tan(α) + tan(β)) / (1 - tan(α) tan(β))

first with α = a + b and β = c. Then apply it again with α = a and β = b. Do some algebra and show that the identity you're trying to prove holds if and only if tan(a+b+c) = 0, i.e. if and only if a + b + c is a multiple of π.
 [url=http://simulationcredit1.com][color=#F0F0F0][u]voiture simulation de credit immobilier en ligne[/u][/color][/url][color=#F0F0F0] - Simulation credit immobilier en ligne. L’Internet a rendu encore plus facile d’obtenir de l’aide pour trouver le credit immobilier.[/color]


induction
topology
Ring Theory
superior algebra
superior algebra unsolved

Let $A$ be a ring with $2^n+1$ elements, where $n$ is a positive integer and let
\[ M = \{ k \in\mathbb{Z} \mid k \geq 2, \ x^k =x , \ \forall \ x\in A \} . \]
Prove that the following statements are equivalent:

a) $A$ is a field;
b) $M$ is not empty and the smallest element in $M$ is $2^n+1$.

[i]Marian Andronache[/i]

[quote="harazi"]It seems that Marian Andronache's beautiful and hard problems are the highlight of every national olympiad and this one is really the most beautiful of them all. Did anyone find a solution for this problem?[/quote]Of course. Andrei Stefanescu :P which also received a special prize for this. But there was another solution from Banda Narcisa, which also received a special prize.

I dont' know if it's very fair to post a solution, because I participated at the olympiad, but.. here goes the solution I found
for a) => b) I used Catalan's lemma and .. of course... because it wasn't permitted to use it.. i got nothing on that
for b)=> a)
Suppose A is not a field. It is obvious that there is no $x\in A$ which is not inversable s.t. $x,x^2,\ldots, x^{2^n}$ are all different (else we would have that $2^n+1=|A|\ge |\{0,1\}|+|\{x,x^2,...,x^{2^n}\}|=2^n+2$ contradiction)
So we must have some $1\le i< j\le 2^n$ so that $x^i=x^j$. From this follows that $x^{2^n-j+1}\cdot x^i=x^{2^n+1}=x$. So there is a $k$, $2\le k\le 2^n$ s.t. $x^k=x$. Consider the least positive integer k for which $x^k=x$. Then by induction we get that $x^{k+m(k-1)}=x$. Now it is obvious that we can find a $m$ s.t. $2^n+1=k+m(k-1)$ (else we can find a $m'$ s.t. $d=2^n+1-k-m'(k-1)<k-1$ and then $x=x^{2^n+1}=x^{d+k+m'(k-1)}=x^d\cdot x^{k+m'(k-1)}=x^d\cdot x=x^{d+1}$, so $x=x^{d+1}$ where $d+1<k$ contradiction).
So there is a positive  integer $m$ s.t. $k+m(k-1)=2^n+1\Rightarrow k=\frac{2^n}{m+1}+1$, so $k$ is of the form $2^r+1, r<n$. (1)
Now because $A$ is not a field it follows that if $a\in U(A)$ then its order is of the form $2^u$ with $u<n$.(2)
From (1) and (2) we deduce that we can find for any element  $a\in A$ a number of the form $2^{t_a}+1$ with $t_a<n$ s.t. $a^{2^{t_a}+1}=a$. Considering $t$ the maximum from the set $\{t_a|a\in A\}$ and using the fact that if $a^u=a$ then $a^{u+m(u-1)}=a$ for any positive integer $m$ we deduce that $a^{2^t}+1=a$ for any $a\in A$ and since $2^t+1<2^n+1$ we get that the least element of $M$ is not $2^n+1$, contradiction. So $A$ is a field.

[quote="Valentin Vornicu"] Of course. Andrei Stefanescu :P which also received a special prize for this. But there was another solution from Banda Narcisa, which also received a special prize.[/quote]

For Andrei Stefanescu and Banda Narcisa

Could you post your nice solution about this nice problem ?

If $A$ is a field, because it is finite, we have that the group $(A^*,\cdot)$ is cyclic, and it has $2^n$ elements. This means $x^{2^n}=1, \forall x\in A^*\Rightarrow x^{2^n+1}=x,\forall x\in A\Rightarrow 2^n+1\in M$. Also, because $A^*$ is cyclic, we have an element $\lambda\in A^*$ with $ord(\lambda)=2^n$, so $2^n+1$ is the smallest element in $M$.

well, I think u used it, when you said $A=L\cup \{0\}$, because $|L|=2^n$.

Kuba, can you post Catalan's lemma?? I'm really interrested. :)

Cezar, look here http://en.wikipedia.org/wiki/Catalan's_conjecture . I think you knew it.

PS: Because of the ' sign you can't follow the link just by a click. You have to copy-paste it in the address bar. Any remedy to this problem?

Sorry, I realised the mistake in my problem. Deleted that part.

Any further thoughts on this (extremely hard) problem?

The official solution is a monster... it uses lots of ideas.

Well, I can give a solution that maybe gives some insight in how the problem was created. It however uses some more advanced techniques - you might want to read Atiyah-MacDonald: Introduction to Commutative Algebra(the first chapter with exercises should be enough).

!!!Attention: A ring is always commutative with unity in my solution!!!

Lemma 1. Let $\text{R}$ be a ring. Assume that for each $x\in \text{R}$, there exists an $n>1$ such that $x^{n}=x$. Then every prime ideal of $\text{R}$ is maximal.

Proof. Let $\mathfrak{p}$ be a prime ideal and consider $\text{R}/\mathfrak{p}$. we have $x(x^{n-1}-1)=0$. As $\mathfrak{p}$ is prime, it follows that either $x=0$ or $x^{n-1}=1$ in $\text{R}/\mathfrak{p}$. therefore every nonzero element $x$ has an inverse in $\text{R}/\mathfrak{p}$(namely $x^{n-2}$). therefore $\text{R}/\mathfrak{p}$ is a field and this means that $\mathfrak{p}$ is maximal.

For our ring $\text{R}$, it follows that $\text{Spec}\text{R}$, consisting of a finite number of points each of which is closed, is totally disconnected.

Lemma 2. Suppose $\text{R}$ is a ring such that $\text{Spec}\text{R}$ is disconnected. Then $\text{R}=\text{R}_{1}\times \text{R}_{2}$ for two nonzero rings $\text{R}_{1},\text{R}_{2}$.

Proof. It is an exercise in Atiyah-MacDonald: Introduction to Commutative Algebra. In our situation, $\text{R}$ contains no nonzero nilpotent elements, this makes the proof easier:
Consider two components $V(\mathfrak{a})$, $V(\mathfrak{b})$ of $\text{Spec}\text{R}$ (here, $V(E)$ means the set of prime ideals of $\text{R}$ that contain $E$; we can assume that $E$ is an ideal. these sets are the closed sets of a topology on $\text{Spec}\text{R}$, the set of prime ideals of $\text{R}$, called the Zariski topology). As their intersection is empty, it follows that $\mathfrak{a}+\mathfrak{b}=\text{R}$, so there is $a\in \mathfrak{a}$ and $b\in \mathfrak{b}$ such that $a+b=1$. Trivial inclusions show that the components are also equal to $V(a)$ and $V(b)$. as their union is everything, it follows that $ab$ is in every prime ideal. this means that $ab$ is nilpotent. by our assumption, $ab=0$. now $b=1-a$, so we have proven that $a$ is idempotent. now, an idempotent $a$ induces a splitting $\text{R}=a\text{R}\times (1-a)\text{R}$, as is easily shown.

Using Lemma 2, we can write $\text{R}= \prod_{i=1}^{k}\text{R}_{i}$ where all $\text{R}_{i}$'s are rings with exactly one prime ideal. As $\text{R}$ contains no nonzero nilpotents, the same is true for each $\text{R}_{i}$.

Lemma 3. Let $\text{R}$ be a ring with exactly one prime ideal and without nonzero nilpotents. then $\text{R}$ is a field.

Proof. The intersection of all prime ideals is the set of nilpotent elements. therefore $(0)$ is the only prime ideal, meaning that $\text{R}$ is a field.

All in all, we have proved that $\text{R}$ is a product of fields $\text{R}_{i}$. note that yet, we didnt use the cardinality of $\text{R}$ nor the fact, that $2^{n}+1$ is minimal in $M$.

Let $q_{i}= |\text{R}_{i}|$. Then it is known that $x^{q_{i}}=x$ in $\text{R}_{i}$ and $q_{i}$ is minimal with this property. As we also have $x^{2^{n}+1}=x$, it follows that $q_{i}-1|2^{n}$; therefore $q_{i}=2^{n_{i}}+1$ with $n_{i}\leq n$. now, if $\text{R}$ wasnt a field itself, we would have $n_{i}<n$, therefore $x^{2^{n-1}+1}=x$ is true in each $\text{R}_{i}$, thus also in $\text{R}$, contradiction.

In fact, it is easy to prove that we can replace $2^{n}+1$ by any natural number $k>1$ in the problem, if we assume that rings are commutative with unity.

Replace that $2^{n}+1$ with any positive integer and the statement still holds (with no commutativity assumptions). More precisely:

Let $A$ be a finite ring with unit, and define $M$ as before to be the set of natural numbers $k\ge 2$ such that $x^{k}=x,\ \forall x\in A$. Then, the following are equivalent:

(a) $A$ is a field;
(b) $M$ is non-empty and the smallest element of $M$ is $|A|$, the number of elements of $A$.

I don't know if its suitable for an olympiad in this form though :).

Edit:

Oops, I hadn't seen Peter's last remark when I posted. Anyway, as I said, commutativity is not a necessary assumption (it follows from everything else).


Prove that there do not exist integers $a, b, c, d$ (not all zero) such that

$a^{2}+5b^{2}-2c^{2}-2cd-3d^{2}= 0$

[hide="2click or not 2click = ?"]Double the equation and have
(1) $2a^{2}-c^{2}=5b^{2}-10d^{2}$, where $c$ was redenoted $c: =2c+d$.
WLOG we may consider that there is a variable which is not devided by 5.
So $5|2a^{2}-c^{2}$. Considering the remainders we find only one possibility to satisfy: $a\equiv c \equiv 0 (mod 5)$
Hence from (1) we have
(2) $5|2d^{2}-b^{2}$ and again only one possibility.
Contradiction with the assumption.[/hide]


complex numbers
algebra unsolved
algebra

find  (x;y):

      x²-y²=3x-4y
      2xy= 4x+3y

Multiply the second equation by $ i$ and add to the first:

$ x^2 \plus{} 2xyi \minus{} y^2 \equal{} 3x \minus{} 4y \plus{} 4ix \plus{} 3iy$

$ (x \plus{} yi)^2 \equal{} x(3 \plus{} 4i) \plus{} iy(3 \plus{} 4i)$

$ (x \plus{} yi)\left[(x \plus{} yi) \minus{} (3 \plus{} 4i)\right] \equal{} 0$

Hence the solutions are $ (0,0)$ and $ (3,4)$

There are complex solutions too...

$ (x,y)\equal{}\left(\frac{3}{2}\plus{}2i,2\minus{}\frac{3}{2}i\right),\left(\frac{3}{2}\minus{}2i,2\plus{}\frac{3}{2}i\right)$

Note that we must verify that the two solutions produced by Farenhajt satisfy the original system.

I don't think we do, because equality of complex numbers (where we consider $ x,y$ to be real) actually compresses (the) two equations into one, hence the equivalences are preserved throughout.

[quote="Farenhajt"]I don't think we do, because equality of complex numbers (where we consider $ x,y$ to be real) actually compresses (the) two equations into one.[/quote]

Oh right...as long as $ x,y$ are real. [which is what the problem statement probably should include...]

[quote="greatestmaths"]find  (x;y):

      x²-y²=3x-4y
      2xy= 4x+3y[/quote]
Note that if $ x \equal{} 0$ then $ y \equal{} 0$
Easy to prove $ x \neq \frac {3}{2}$
If $ x \neq 0$
From $ 2xy \equal{} 4x \plus{} 3y$ we have $ y \equal{} \frac {4x}{2x \minus{} 3}$
From $ x^2 \minus{} y^2 \equal{} 3x \minus{} 4y$ we have $ x^2 \minus{} \frac {16x^2}{4x^2 \minus{} 12x \plus{} 9} \equal{} 3x \minus{} \frac {16x}{2x \minus{} 3}$
=>$ 8x^4 \minus{} 60x^3 \plus{} 194x^2 \minus{} 333x \plus{} 225 \equal{} 0$
=>$ (x \minus{} 3)(x \minus{} \frac {3}{2})(8x^2 \minus{} 24x \plus{} 50) \equal{} 0$
=>$ x \equal{} 3$=>$ y \equal{} 4$
Conclusion $ (x;y) \equal{} (3;4);(0;0)$

[quote="greatestmaths"][color=darkred]Solve over $ \mathcal R$ the system [/color]$ \{\begin{array}{ccc} x^2 - y^2 & = & 3x - 4y\ . \\
 \\
2xy & = & 4x + 3y\ .\end{array}$[/quote]
[color=darkblue]This old system is well-known. Very nice the [b]Farenhajt[/b]'s proof in which he profiteer by the particularity of the system. I"ll apply the [b]general method[/b] for such systems.

[b]Proof.[/b] Observe that $ x = 0\Longleftrightarrow y = 0$, i.e.  $ (0,0)$ is a solution

of the given system. Suppose $ xy\ne 0$ and denote $ \frac xt = \frac y1\ .$

$ \|\begin{array}{ccc} x^2 - y^2 & = & 3x - 4y \\
 \\
2xy & = & 4x + 3y\end{array}\|$ $ \implies$ $ \frac {x^2 - y^2}{2xy} = \frac {3x - 4y}{4x + 3y}$ $ \implies$ $ \frac {t^2 - 1}{2t} = \frac {3t - 4}{4t + 3}$ $ \implies$

$ 4t^3 - 3t^2 + 4t - 3 = 0$ $ \implies$ $ (4t - 3)(t^2 + 1) = 0$ $ \implies$ $ t = \frac 34$ because $ t\in\mathcal R\ .$ Thus,

$ 4x = 3y$, i.e. $ y = \frac 43\cdot x$ and the second equation of the given system becomes

$ 2x\cdot\frac {4x}{3} = 4x + 4x$, i.e. $ x^2 = 3x$ from where ($ x\ne 0$) obtain the second solution $ (3,4)\ .$[/color]

[quote="Altheman"]There are complex solutions too...

$ (x,y) \equal{} \left(\frac {3}{2} \plus{} 2i,2 \minus{} \frac {3}{2}i\right),\left(\frac {3}{2} \minus{} 2i,2 \plus{} \frac {3}{2}i\right)$
[/quote]

How did you find them  :roll: ?

You can start from my form of the system

$ (x\plus{}yi)[(x\plus{}yi)\minus{}(3\plus{}4i)]\equal{}0$

and then take $ x\plus{}yi\equal{}0\iff y\equal{}ix$ or $ x\plus{}yi\equal{}3\plus{}4i\iff y\equal{}4\plus{}i(x\minus{}3)$

[quote="Farenhajt"]You can start from my form of the system

$ (x \plus{} yi)[(x \plus{} yi) \minus{} (3 \plus{} 4i)] \equal{} 0$

and then take $ x \plus{} yi \equal{} 0\iff y \equal{} ix$ or $ x \plus{} yi \equal{} 3 \plus{} 4i\iff y \equal{} 4 \plus{} i(x \minus{} 3)$[/quote]

OK, and then we must set: $ x\equal{}a\plus{}ib$ where $ a,b$ are some reals I think.

No, you just plug those into one equation or the other:

$ 2x\cdot xi\equal{}4x\plus{}3ix\iff x(2xi\minus{}(4\plus{}3i))\equal{}0\iff x\equal{}0\lor x\equal{}{3\over 2}\minus{}2i$, and so forth

[quote="Farenhajt"]No, you just plug those into one equation or the other:

$ 2x\cdot xi \equal{} 4x \plus{} 3ix\iff x(2xi \minus{} (4 \plus{} 3i)) \equal{} 0\iff x \equal{} 0\lor x \equal{} {3\over 2} \minus{} 2i$, and so forth[/quote]

Now I see
Thanks  :wink:


function
number theory
number theory open

[quote="wpolly"]A positive integer is called Fermat-Lucas-Carmichael Number iff:
1) it's odd, squarefree, and have at least three distinct prime divisors;
2) for every prime $p|n$, we have $p+1|n+1$ and $p-1|n-1$.

construct an example of Fermat-Lucas-Carmichael number or give a proof that such numbers don't exist.

Note: those numbers are both Fermat-pseudoprime to any base and Lucas-pseudoprime to any Lucas Sequence.[/quote]
Of course, is an interesting problem, is really to probe that if exists an exmaple of course must veriify

0. $n$ odd,
1. $\mu(n)\ne 0$, Moebious function,
2. $n=p(1+\fbox{p-1}(1+p))$, $n+1=(1+\fbox{p-1}p)(p+1)$ and $n-1=(1+\fbox{p+1}p)(p-1)$, where $\fbox{p+1}$ and $\fbox{p-1}$ are the no commun factors o $p+1$ and $p-1$  respectively.

Good luck to find some example.

B.R.
RX  TCM

This is discussed in "Unsolved Problems in Number Theory" (3rd edition) by Richard Guy (Springer Verlag, 2004) in section A13.  I believe it is still an open problem.


quadratics
geometry
3D geometry
algebra
polynomial
AMC

Well, it would probably be easiest just to solve for x and y and then raise each to the sixth power on this one, instead of looking for some slick technique.

Hmm...how about doing factoring, such as in case x:^2:+y:^2:=(x+y)(x-y) then using the first two givens.
...although I know Mr. Crawford went over this in Math Jam before...

[quote="Fierytycoon"]Then just have x, y be roots of a quadratic and use newton's sums.[/quote]
If you could kindly tell me what "newton's sums" is, then that'll be great...stupid me for trying to solve with brute force...I think you're right, tetrahedr0n. "Anyone want to second that?" (a la mi profesor de matematicas :wink: )

i too lazy to check his answer for calculations, but thats the way i would have done it. and thats the way in the aops book.

Tare, Newton's Sums are described in AoPS 2. Here is a previous thread about it:

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=575&highlight=newtons+sums

I used a strange method, and its only an approximate I don't know why...

x^6+y^6=(x^3+y^3)(x^3-y^3)
(x^3+y^3)(x^3-y^3)=(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)
(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)=4*(16-6)*(x-y)(16-2)

This is where it gets tricky  :roll:

x-y=sqrt((x+y)^2-4xy)=sqrt(8)

Plug everything in and we get 1583.92....

ooo, now if only we could find x-y without actually calculating x and y, that method would be the nicest one yet.

That would be the nicest....if
x^6+y^6=(x^3+y^3)(x^3-y^3) is correct.

Take a look at it again.

Hint, this expression is FALSE!!!!

Oh, heh, was reading his solution quickly, missed that.

So now I like my solutions with Newton's Sums better.  :D

LoL... I see now.. Why is my answer so close?

:lol: , entertaining. What'll Newton say when he sees this... :lol:

[color=cyan]If you check out the two roots for x and y, they are 2+:rt2: and 2-:rt2:.  Now, this second root is just a tad over 1/2, which means its sixth power is just a tad over 1/64, which is not a very big number at all.  So, the difference between subtracting and adding it is relatively insignificant.

You might also try factoring as a sum of two cubes -- it might be a slightly quicker method, although no promises.[/color]

i think the newton sums would have been a lot faster!

Who wants to show their work for the Newton's Sums... I'm too lazy to bring out the AoPS book.  :lol:

usually its not a good idea to use Newton's sums, since there's usually a more clever way of solving the problem.  This is cause it takes a lot of time usually and you're very prone to making arithmetic errors when doing this method.  But since this is only a quad. equation, it is a little faster than if it were a polynomial of higher degree.  Its still not much faster, if it all, than any of the other solutions posted.

wow
Newton's sums
such complicated mathematical terminology
i like just plain old algebra


A spring has a force constant of 419.1 N/m.
Find the potential energy stored in the spring when the spring is
a)stretched 4.35 cm from equilibrium.
b)compressed 2.27 cm from equilibrium.
c)unstreched.

I need help with this problem.
I know that elastic potential energy = .5*spring constant*(distance compressed or stretched)^2.
So, for A, would I simply multiply .5 by 419.1 and (4.35)^2 to get 3965.209875 J?

For B, muliply .5 by 419.1 and (2.27)^2 to get 1079.490195 J?

For C, would the answer simply be 0 since the distnace compressed or stretched is 0?

Did I do this work correctly?  When it asked for the potential energy, do I find the elastic PE and gravitational PE and add them together?  I have a feeling that I need to find the gravitational PE for this problem.  Please help.

Yes, the potential energy of the spring is really $E = \frac{1}{2}ky^2$, where $k$ is the spring constant and $y$ is the displacement off the equilibrium position. ((This is because the force is $ky$ opposing the displacement - summing (or better, integrating) the work done against it, that is, $W = F.\Delta y$ over a larger stretch yields the above result for energy))

However, the results you got are not exactly correct, because you ommited the dimensions of the quantities. You yourself said that the values are $k=419.1\ N/m$ and $y=4.35\ cm$, so when you plug this into the equation, you get

$E = \frac{1}{2}(419.1 N/m)(4.35 cm)^2 = \frac{1}{2}(419.1 N/m)(0.0435 m)^2 \approx 0.397\ J$
or $\frac{1}{2}(419.1 N/m)(4.35 cm)^2 \approx 3970\ N.cm^2/m \neq 3970 J$

Another thing is, that you shouldn't simply write out all the digits you calculator gives you as a result. Difference between 3965.209875 J and 3965.209876 J is really unmeasurable in most circumstances, and since the values you were given had around 3 significant digits, your answer shouldn't really contain more.

Concerning the gravitational potential energy, it is equal to $E = mgh$ if you want just the formula, but the problem statement really asked for the "potential energy stored in the spring". And if you wanted to find the potential energy, then I'd ask - potential energy of what? The spring? Or of something attached to it? How is the spring oriented in space? Does it lie on the table? Or is something hanging on it? What is its mass?


Hi, I am applying for the AMSP, and I don't know how well I need to do on the admission test to get in. Right now, I answered 6.5 questions out of 10, and I'm wondering how many did you guys answer when you got in?

That's plenty. I procrastinated writing the solutions, and only turned in 6. I was fine. I think most people do even less than that. I think the test is just there to see that you tried and put effort into solving math problems.

cool. :D i finished 7.5 problems. ill see you at the camp then. :D

YAY! i'm up to 8.5 problems now!! i feel smart. :)

I don't think the admission test is the most important aspect in your admission into AMSP. Just for your info, I did like around 6 problems last year? I didn't do any this year because the test was waived for returning students!?!!!!  :D ..

Yea. 6.5 did you say? or 6? That is REALLY plenty. Last year, people did like 4 and got in. Really. As long as you can get stuff on the paper for some questions, you'll be fine.

If you're coming to Dallas, see you there!

aw, i applied to Santa Cruz. :(

was there people who did all of them?

Oh. Santa Cruz, I have never been to so I can't say, but Dallas was awesome last year, and I'm guessing Santa Cruz is too? The experiences are like a little bit different.

There are certainly a lot of EXCEPTIONALLY, and like VERY EXCEPTIONALLY smart people who come to AMSP. I'm sure a couple did all 10, but it's not like it's expected of an average applicant. That's why it says something like "You aren't expected to solve all of these problems fully" or something like that I think.. It said something like that last year.

Have fun. Enjoy your time at camp because AMSP can get pretty fun. :D

When I went a few years ago, basically everyone that applied got in. It was the time of my life.


geometry
circumcircle
symmetry
trigonometry
cyclic quadrilateral
geometry solved

This is a good problem!  :lol:
I don't have a complete answer to this problem but, in this moment, I try to implement Elementary Projective Geometry.And I try to pass this solution in English :D .In this article you can have more info and really many ideas to finish your problem!
But this article is in Portuguese  :(

http://www.obm.org.br/eureka/artigos/geometria.doc

for some reason i think O, H, E and F form an orthocentric quadrangle (i.e. H is the orthocentre of OEF, which implies E is the orthocentre of OHF, etc), and not just HO iz perpendicular to FE.
i dunno if diz will work well, but maybe we could start wif O,A, E and F - and then somehow construct ABCD given H, E and F (i dunt noe how 2 do that) - and then prove OA=OB=OC=OD.

I drew a picture and found that H isn't the orthocentre of OEF.

Well, H is the orthocenter of OEF, since O is the orthocenter of EFH. The latter is because triangle EFH is autopolar with respect to the circumcircle of the quadrilateral ABCD, and the orthocenter of a triangle autopolar with respect to a circle always coincides with the center of the circle.

  Darij

A triangle is autopolar wrt a circle if every side of the triangle is the polar of the opposite vertex wrt the given circle.

wrt=with respect to

Yes, precisely. And here is a proof why the triangle EFH is autopolar with respect to the circumcircle of the quadrilateral ABCD:

If we regard not the cyclic quadrilateral ABCD, but the points A, B, C, D lying on one circle k, without mentioning their order, then the points E, F and H are "equal in rights": The point E is the meet of AB and CD, the point F is the meet of BC and DA, and the point H is the meet of AC and BD. Now, a well-known theorem (Theorem 1 of the first posting in [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6]this CTK discussion thread[/url]) states that F lies on the polar of E with respect to k. Similarly, H lies on the polar of E with respect to k. Hence, the line FH is the polar of E with respect to k. Analogous reasoning shows that the line HE is the polar of F with respect to k, and the line EF is the polar of H with respect to k. Hence, any sideline of the triangle EFH is the polar of the corresponding opposite vertex with respect to k, and thus, triangle EFH is autopolar with respect to k. Qed..

  Darij

Hi.

How would you prove the orthocenter of a triangle autopolar with respect to a circle is the centre of the circle?

Vinoth

The polar p of a point P wrt a circle is a line. For symmetry reasons p is perpendicular to OP and we're done. From here we derive OA perpendicular to BC, OB perpendicular to CA Q.E.D.

hi.

just interested, are there any other useful theoroms/results associated with autopolar triangles? And is it true that for any triangle ABC, there exists a circle w s.t. ABC is autopolar wrt to w? And with a fixed circle w, are all triangles autopolar wrt w given by that construction (taking 4 arbitrary points and then the 3 meets of the corresponding pairs of lines)?

[quote="vinoth_90_2004"]is it true that for any triangle ABC, there exists a circle w s.t. ABC is autopolar wrt to w?[/quote]

If you restrict yourself to real circles (circles whose radii are real numbers, i. e. circles we are used to in Euclidean geometry), then you can prove that for a triangle ABC, there exists a real circle w such that ABC is autopolar with respect to w if and only if the triangle ABC is obtuse-angled. In this case, this circle w is called the [i]polar circle[/i] of the obtuse-angled triangle ABC, and it has many properties indeed: The center of w is the orthocenter of triangle ABC. The inversion with respect to w maps the points A, B, C to the feet of the atitudes of triangle ABC. Hence, this inversion with respect to w maps the circumcircle of triangle ABC to the nine-point circle of triangle. The radius of w is $2R\sqrt{-\cos A\cos B\cos C}$, where R is the circumradius of triangle ABC.

But if our triangle ABC is acute-angled, then there is only an imaginary circle (i. e. a circle with imaginary radius $2R\sqrt{-\cos A\cos B\cos C}$) with respect to which the triangle ABC is autopolar (what, of course, makes sense only if you work in the complex plane).

[quote="vinoth_90_2004"]And with a fixed circle w, are all triangles autopolar wrt w given by that construction (taking 4 arbitrary points and then the 3 meets of the corresponding pairs of lines)?[/quote]

Yes. If you have a triangle EFH autopolar with respect to w, then take an arbitrary line through E which meets w at two points A and B, then let the line FB meet w at the point C (apart from B), and let the line EC meet w at the point D (apart from C). You can easily see that the line DA passes through F, too, and that H is the meet of the lines AC and BD.

  Darij

[quote="liyi"]ABCD is a quadrangle which is inscribed in a circle with center O.
AB and CD meet at E.
AD and BC meet at F.
AC and BD meet at H.
Show that OH is perpendicular to EF.[/quote]

For a synthetic proof, see my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i], Theorem 9 [b]h)[/b].

  darij

Also you can see at:  http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

Kostas Vittas.


Go ahead and post the AMC 10 and 12A answers here first.


Find all the integers $a, b, c$ with $1<a<b<c$ such that $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$

[quote="vulgarfraction"]$(a-1)(b-1)(c-1) | abc-1 \Leftrightarrow abc-(a+b+c)+(ab+bc+ca)-1 \leq abc-1 \Leftrightarrow ab+bc+ca-a-b-c \leq 0 \Leftrightarrow a(b-1)+b(c-1)+c(a-1) \leq 0$

Contradiction, so no such cases.[/quote]

$(a-1)(b-1)(c-1) \neq abc-(a+b+c)+(ab+bc+ca)-1$.

Ergh. Sorry.

I thought I had the answer to this but it was the case with two variables, which is easy.

For this one.. I've got $a-1|bc+1$ etc. but I don't think that's much use.


suppse there is a spherical uncharged conductor with a cavity of any arbitrary size inside with a charge $q$ inside the cavity.then how does one show that the charge $-q$ that is accumulated on the surface of the cavity is such that it negates the electric field due to $q$ alone?

The induced charges will cancel out only the field inside the conductor (i.e. inside the metal), not necessarily in the cavity itself; are you asking for the proof of this?

no i am not atlking about the field inside the cavity.what i am telling is it was written in the book that the charge $-q$ that distributes over the cavity is such that it negates the effect of the electric field due to $q$ anywhere[b] outside[/b] the cavity.how does one prove this

Ok, now your question makes sense ;) But I'm not sure I can prove the claim, although I know about it. First of all, we can agree that the charges induced on the inner surface of the conductor add up to $-q$, which follows from Gauss' theorem. But then, by conservation of charge, a total charge of $q$ should be induced on the outer surface of the conductor. Applying Gauss' law again to the surface encompassing the entire conductor we get that the total flux is not zero  :maybe: Is this where you're stuck, too? In fact, I think that this statement is not true, in general. However, for a grounded conductor, the statement is true, since then we don't have to worry about conservation of charge and it's fairly easy to prove the claim.

well that's not exactly what it says ..the theory tells that if you ahve a cavity inside some conductor and put a charge $q$ inside the cavity thena charge $-q$ accumulates on the outer/inner suface of cavity in such a way that the electric field at any external point other than the cavity due to the charge inside the cavity is exactly equal and opposite to taht due to $-$ ...

it doesn't speak of the other cahrge distibution on the outer surface of the conductor at all!!

That can only hold for grounded conductors, though. You can't have a charge $-q$ just 'appearing' anywhere on the conductor, without $q$ appearing elsewhere on the same conductor, right?

yes the $q$ does appear but taht is what doesn't matter what the theory says is only about the[b] cahrge $q$ inside the cavity[/b] and the charge $-q$ onthe surface of cavity and how it redistributes

But not having anything on the outer surface is crucial for the proof, so you've got to suppose that. The proof, as far as I know, utilizes the fact that the curl of the el. field within a closed contour is equal to its circulation along the contour (Stokes' theorem) and that this quantity is always zero in electrostatics (a Maxwell's equation). But I'm not really an expert on this, so I'll stop here  :(

let me try ...but i am very unsure of what exactly i am trying to do...thsi theory as such  seems too general :)

Well, it is, and that's what makes it so beautiful. But there's not much sense trying to do this if you're not very familiar with multivariable calculus, so... Good luck!

no just a bit of multivariable learnt it a yaer or so ago....well let's hope atleast to see a proof :D

well it's too late but for any other future references let me have a picture of the situation(i did thsi yesterday before posting but couldn't find where it was...)


Find all nonnegative integers $m$ such that:
$a_m=(2^{2m+1})^2+1$ is divisible by at most two different primes.

[quote="rgep"]Aurefeuillian[/quote]

From where does this word come?
Anyway, in this particular case, this is Sophie Germain identity.

Pierre.

Well, the quick answer is "from the preface to [i]Factorisations of $b^n\pm1$[/i] by Brillhart et al, available online as [url=http://homes.cerias.purdue.edu/~ssw/cun/third/index.html]The Third Edition of the Cunningham Book[/url] at Sam Wagstaff's site".  The name is given in Dickson's [i]History[/i], volume I, attributed via Cunningham to Aurefeuille.


geometry
parallelogram
trigonometry
rhombus
geometry unsolved

Let $ ABCD$ be parallelogram and points $ P$ and $ Q$ are in segments $ AB$ and $ BC$ respectively satisfying $ AP \equal{} CQ$. Let $ R$ be the intesection of $ AQ$ and $ CP$. Prove that $ DR$ is the bisector of the angle $ ADC$.
Note: I would like to see a "pure" geometry solution. I have solved this problem using the sine theorem several times and proving that sin(ADR)=sin(CDR).

Just curious...would you consider Ceva to be a "pure geometry" solution?  I wouldn't say it's pure geometry, but neither is it trig bash...

Ceva is a direct consequence of area ratios...so yes.

I didn't look at the link, but I suppose that the proof is something along the lines of: let $ E\in DR\cap BC$, then prove $ DE\equal{}DC$.

[quote="Altheman"]Ceva is a direct consequence of area ratios...so yes.
[/quote]

Do keep in mind, though, that most of trigonometry is a direct consequence of similar triangles.  :wink:

Dear Mathlinkers,
I present my pure geometric solution.
Let Dp the parallel to BC passing through P, Dq the parallel to AB passing through Q, U the point of intersection of Dq with DA, V the point of intersection of Dp with CD, and X the point of intersection of Dp and Dq.
Accordling to Pappus's theorem, D, X and R are collinear. (The hexagone considered is degenerated)
DUXV being a rhombus (losange), DX is the bissector of <ADC.
Sincerely
Jean-Louis

Call M the intersection of CP and AD, N the one of AQ and CD, T the one of DR and AC. As AQ, CP and DT pass through R, apply Ceva to the triangle ADC and the 3 cevians from its vertices and get AT/TC=AD/CD.

Best regards,
sunken rock


calculus
real analysis
real analysis theorems

HI
could one of you please post the proof of the widely used L'Hospital's rule....haven't come across it yet...
Shyam

L'Hopital's theorem has more than one forms and can be found just in any advanced calculus textbook. So I recommend that you get yourself an analysis book and read the proof on your own, but if you then have a question you may ask.

Here is a nice and easy one...

http://en.wikipedia.org/wiki/L%27Hospital%27s_rule#Proofs_of_l.27H.C3.B4pital.27s_rule


function
limit
trigonometry
inequalities
real analysis
real analysis solved

I'm afraid this is going to be too easy..:) It's kind of cute, though.

Find a derivable $f:\mathbb R\to\mathbb R$ s.t. $\lim_{x\to\infty}\frac{f(x)}{x^2}=1$ but $\frac{f'(x)}{2x}$ does not tend to $1$ as $x\to\infty$. Can such a function be convex?

A simple example is $f(x)=x^2+x\sin x$. The first limit is 1; the second limit doesn't exist, since $\frac{f'(x)}{2x} = 1 + \frac{\cos x}2 + \frac{\sin x}{2x}$.

I haven't decided on the last question, but this example certainly fails to be convex.

There are no examples for convex functions.

To see why, let $\epsilon$ and $x$ be positive numbers.  By convexity,
\[f'(x) \leq \frac{f((1+\epsilon)x) - f(x)}{\epsilon x}.\]
Dividing by $2x$ gives
\[\frac{f'(x)}{2x}
  \leq \frac{f((1+\epsilon)x)}{2\epsilon x^2}
    - \frac{f(x)}{2 \epsilon x^2}.\]
Taking the limit as $x$ approaches infinity gives
\[\limsup_{x \rightarrow \infty} \frac{f'(x)}{2x}
 \leq \frac{(1+\epsilon)^2}{2\epsilon} - \frac{1}{2 \epsilon}
  = 1 + \frac{\epsilon}{2}.\]
Because $\epsilon$ is arbitrary, we have
\[\limsup_{x \rightarrow \infty} \frac{f'(x)}{2x} \leq 1.\]
That's half of what we want.

For the other half, we can use the inequality
\[f'(x) \geq \frac{f(x) - f((1 - \epsilon)x)}{\epsilon x}.\]

Thanks! I did something very similar, but in more "words" :).


The sequence $a_1, a_2,..., a_{2000}$ of real numbers satisfies the condition
\[a_1^3+a_2^3+...+a_n^3=(a_1+a_2+...+a_n)^2\]
for all $n$, $1\leq n \leq 2000$. Prove that every element of the sequence is an integer.

Obviously the general case also holds.
We may assume $a_i\neq0$ and $a_i+a_{i+1}\neq0$.
For n=1, we have $a_1=1$.
Suppose $a_n=n$ for all n. Then\[1^3+...+n^3+a_{n+1}^3=(1+...+n+a_{n+1})^2\]Solving this gives $a_{n+1}=n+1$.

I see most of the problems where taken from  Russian olympiad!

I dont know, I have them in a book: "Problems and solutions from arround the world" i think they are from 2000-2001 . About a web site with them as i understand you can only find them in Russian, but i am not very sure!

Yes,I'm pretty sure I've already seen this problem

The problem was proposed in 1978 at the county level of RMO (I was a competitor at that time...and I solved it :)) under the more restrictive condition all $a_i$'s are positive reals (proposed by L. Panaitopol).
However, if this condition is dropped, there are infinitely many sequences satisfying the given equality.
[quote="mecrazywong"]
We may assume $a_i\neq0$ and $a_i+a_{i+1}\neq0$.
[/quote]
We cannot assume this WLOG, otherwise, we miss solutions like $1,-1,1,-1,\ldots$ or like $1,2,-2,2,3,-3,\ldots$.
The problem, as stated at ARO 2000 can be solved by induction, using the following (strenghted) induction hypothesis: for each $n$, the numbers $a_1,a_2,\ldots ,a_n$ are integers and their sum is a triangular number, that is, for each $n$ there exists $k_n$ such that $a_1+a_2+\ldots a_n=\frac {k_n(k_n+1)}{2}$.

very nice problem i have seen this problem in "problems from national and regional olympiad 2000-01"(maa)
and it was the problem of our imo selection test2002


one thousand nine hundred and ninety-eight airports - each connected in both directions to three other ones (exactly 3). Any one of these can be reached from any other one, by just a sequence of flights.

How you close down 2 hundred of these (airports) in such a way that any remaining airportts can be reached by a sequence of flights from any of the other remaining airports. Any two of those closed are also not connected by a single flight.

I'd appreciate any help (preferably soon  :)  [V.close 2 deadline :( ]), so far no-one knows the solution :? .

THANK YOU!!

What deadline????
Hmmm....I think we will wait a little bit....

Pierre.

Hi Pierre,
             Yes, there exists a deadline. I am posting this question, fully conscious of the fact that it is a question with a deadline. But, you may not be aware that I am allowed help with the solution (both from outside and inside) if I acknowledge the source. The only reason why you don't see any of my attempts is because they have all failed.
The fact of the matter is that my own teachers have not been able to provide a satisfactory attempt at providing an avenue of exploration that might possibly yield a solution to the problem. If you wish me to post my attempts I have made to provide solution, I will gladly do so. As for my teacher's attempts, they have not come up with any. After all, it is strange that you should not notice that it was [b]I[/b] who gave access to the knowledge that there is a deadline to this question. It [i]is[/i] unfortunate my teachers have not been able to locate the source of the question, maybe that would have helped. I hope I have not inconvenienced the forum with my rather lengthy post, I have done so in the hope that we have cleared up any possible misunderstandings. Thank you.  :)

Quddusaliquddus

So, my question is : Why is there a deadline if you are trying to solve these problems with your teachers?

Pierre.

We have a saying over here, I will give you the under-18 version:     "Assumptions are the mother of all muck-ups"

You see, I have now only to regurgitate my erstwhile attempt at making the atmosphere more fruitful and therefore I have come to a position where I am desecrating the purpose of the forum.

Instead of getting into a possible rant, please read the following:

"I am allowed help with the solution (both from outside and inside) if I acknowledge the source."

"The fact of the matter is that my own teachers have not been able to provide a satisfactory attempt at providing an avenue of exploration that might possibly yield a solution to the problem."

"As for my teacher's attempts, they have not come up with any."

"After all, it is strange that you should not notice that it was I who gave access to the knowledge that there is a deadline to this question."

I will make it even more uncomplicated for you to understand (if that's possible!):

[b]I[/b] am trying to solve these problems.
Teachers have a list of problems that they hand out.
Teachers forget, are ignorant of, or have lost the solution to the problem. I am assessed at my attempt at the solution.
I am allowed help because the questions are problems to be solved - without presence of guidance there can be no hope of learning.
This is not a book-based question if you didnt realise. Rather it is open-ended!

My question to you is, why shouldn't there be a deadline to the solution? They are not hoping for me to come back to them in 10 years time in order to hand the solution in.
Annoying little facts like our nonexistence of our immortality and financial considerations get in the way of that happening at the present moment (lets cross our fingers, maybe in the future we can these).
I dont know your present location, but we here in Great Britain have just finished our educational year. Therefore this might also have something to do with having a deadline, or might it not? ..... you tell me

Could you please stop being so incredulous? It seems to me that I am wasting my breath because either you can't help me and you're still questioning my motives, or you [i]can [/i]help me but you're questioning my motives instead. Do you get browny points for obstructing the way for people in getting answers to their problems? Please, try and hold your horns before diving into such matters.

:D  :D  :D you are cute when you are angry...

I cannot believe that your teachers are not able to solve these so easy problems...There is no problem : anyone can solve them for you if he is willing to.
But, please, be more honest with your homework...

Pierre.

Don't you see how rediculous it sounds to say I havn't been honest? It is my honesty that got me into trouble at the first place. I'm the one who told you of the deadline at the first place!! Now, if I hadn't told you- then I would have been 'honest' - but only in your eyes - but not in reality. I give up. You were only questioning me for fun. If you were not, then why don't you accpet the (rather lengthy) explanation i have posted all over the forum? Why would you question me if you were not going to belive me? You had already made your mind up hadn't you? I give up. Thank you for your time.

This problem is not equivalent to the russian one (which is a hard one indeed)  as you omit to say that the airports which has to be closed are never joined by a single flight.
This ommission leads to a trivial problem, as I mentioned above : Let's consider the simple undirected graph G whose vertices are the airports, two of them joined by an edge if and only if they can be joined by a single flight.

From the statements of the problem, the graph G is connected. It is well-known that G has a spanning tree. So closing airports leaves by leaves gives the desired result, and you can close any number of airports....

Pierre.

Pierre! See problem's text: we can't close airports which connected by single flight!

Should I remind that someone must complete his proof...?  ;)


Hello!

Let $ ABC$ be a triangle, $ M_{a}$, $ M_{b}$, $ M_{c}$ are the midpoints of the sides $ BC$, $ CA$, $ AB$ and $ I_{a}$, $ I_{b}$, $ I_{c}$ are the excenters of the triangle $ ABC$.
Prove that the lines $ I_{a}M_{a}$, $ I_{b}M_{b}$, $ I_{c}M_{c}$ are concurrent.

Best regards, Petrisor Neagoe  :)

The incenter $ I$ of the reference $ \triangle ABC$ is orthocenter of the excentral $ \triangle I_aI_bI_c$ $ \Longrightarrow$ $ I_bI_cBC$ is cyclic $ \Longrightarrow$ $ BC$ is antiparallel of $ I_bI_c$ WRT $ \angle I_cI_aI_b$ $ \Longrightarrow$ $ I_aM_a$ is the $ I_a$-symmedian of the $ \triangle I_aI_bI_c.$ Similarly, $ I_bM_b, I_cM_c$ are its $ I_b$-, $ I_c$-symmedians, concurrent at its symmedian point.

The point of concurrency is called the Mittenpunkt. :wink:
If $ H$ is the orthocenter and $ V$ is the Bevan point then prove that the Mittenpunkt lies on the line $ VH$.

Dear Mathlinkers,
for more news about the Mittenpunckt see for example
http://perso.orange.fr/jl.ayme
vol. 3  Cinq théorèmes de Christian von Nagel p. 12.
Sincerely
Jean-Louis


When rolling two fair, six-sided die, what is the probability of rolling a sum of 7 or 11? Express as a common fraction.

The answer is 2/9, but how do you get it? I got 5/36, but it's wrong.

rolling a 7--6 ways
rolling an 11--2 ways

(6+2)/36=2/9


Give a, b, c>0.Prove that:
$\large \frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ca}}+\frac{1}{\sqrt{4c^2+ab}}> \frac{4}{a+b+c}$

This problem is quite easy.The method was used by me is mixing varible :D  :D

It is not always safe to describe an inequality as easy. Here, that is a hard one proposed by hungkhtn. It was unsolved for months you know.