MATHCOUNTS AMC AIME USA(J)MO USAMO FTW Putnam

Hello, I am beginning a new tournament. It is probably going to be the hardest tournament that I have ever created though, because this one is going to last long. Here is how it works. Everybody will sign up for the months of the year that they would like to compete in. You can not sign up for every month though, because, for some months of the year, there are requirements to compete for that month. You might have to be a Hodge Conjecture to compete in that round or something like that. Here is how it works;

Anybody can signup for the month of January. The top 3 for that month do not have to meet the requirements for the month of February. The champion of the month of January can compete in any month up to April, without having to meet the requirements. For February, you have to be a Hodge Conjecture or above to join. The top three for that month, and the champion of January, get to participate in March with the people that signed up and met the requirements. Anybody that is the champion of a month gets to compete in the following three months without having to meet the requirements. In December, the people competing will only have the first week to submit their answers. Then for the rest of that month, the champions of every month compete against each other to be the yearly champion. This competition will go on for years and years, until somebody locks it for some reason, or I stop running the competition. The yearly champion can compete in any of the following 6 months without having to meet the requirements. If the competition ends up running for a very long time, then I might have a competition between all the yearly champions.

I will be sending 40 questions to the competitors of every month. The first ten will be MOEMS level. The next ten will be Mathcounts level. The following ten problems will be AMC 12 level (not the easy ones). The next 6 questions are AIME level. The next three are USAMO level, and the last question will be IMO level.

Sign-ups start now, and you can only sign up for January, February, or March. The requirements are shown below;

January - Anybody can compete.
February - You must have been a Hodge Conjecture before.
March - Your highest skill level on FTW, must be higher than 1400.  (You must give me proof.)

Good Luck!!!

This can't be in Mathcounts then, if it has material more advanced then Mathcounts. Choose a better forum to put it in.

Also, you shouldn't have anything to do with FTW.

I sign up.

YEAH FINALLY A TOURNMANET WITH IMO LEVEL PROBLEMS!!!!!

BTW, Problem 8 on mine's is medium IMO level, so it might be hard for you to solve...

I sign up, Jan-Mar. My FTW rating is currently 1569.

I sign up jan-march also.  My FTW rating is 1881 (check high scores for proof).

can you send pm's when it starts?  Also, starting it in september, isn't that too early if the real tourney is starting in jan?

I sign up, but might drop out.
Highest FTW rating: 1426.

yikes i'm gonna get killed.......
can i still try the test anyway just for the heck of it tho?

Um, on the sub-title, it says there will be a champion for the year of '08.  Shoudln't it be '09?

Oops, yes I meant '09.

List;

January -
pythag011
lotsofmath
AIME15
Bugi
johnnybeard
boredtodeath

February -
pythag011
lotsofmath
AIME15
Bugi

March -
lotsofmath
AIME15
Bugi

[quote="nebula42"]I don't think it's a good idea to make a tournament with such varied level of difficulty because this is a middle school tournament forum, and so most of the people who visit this forum will be unable to make significant progress on USAMO or IMO problems. I think you need to cut the top level of the tournament down at least to AMC 12 or easier AIME questions which use only Mathcounts concepts (not trig). If you wish to, you can split this tournament up into two, with the harder tournament going into one of the high school forums.

Also, don't push the "Hodge Conjecture" requirement too far - soon people may start spamming just to have enough posts for the tournament.[/quote]

Okay, I won't base anything on rank any more, but what do you mean when you say that I split it up into 2 pieces?

The only reason why I have USAMO and IMO problems is because those are the ones that seperate the top 3 from everybody else.

There is nothing wrong with USAMO / IMO problems except that they are [i]far[/i] too hard for a middle school forum. Those in this forum who can make progress on these problems really should be visiting the high school forums instead.

If you still want to use such hard problems, I meant that you could split this tournament up into two tournaments, with the problems appropriate for middle school going in this forum, and the others (many AMC 12 and AIME questions and all USAMO / IMO questions) going to a high school forum.

I am going to pm the questions though, and then after the competitions, I am going to post the questions in their correct forum.

Just out of curiosity, where would I have to post the USAMO and IMO problems? In the olympiad section?

[quote="Wickedestjr"]
March - Your highest skill level on FTW, must be higher than 1400.  (You must give me proof.)[/quote]

Advice: Don't use FTW as a discerning factor, as (at the high ratings, say 1600s-1900s), it has basically turned into a memming/typing quickly contest instead of an actual problem solving contest.

I don't see this problem so much at the lower ratings (1200-1480 :D). (I only go on once every two months to see how good I REALLY am without memming, as I usually forget 90% of the questions during that time. It turns out I'm pretty okay.)

April;

pythag011
Yongyi781
lotsofmath
Bugi
AIME15

Graders;
Wickedestjr - Grades everybody's.
pythag011 - Grades everybody's but himself.
Yongyi781 - Grades everybody's but himself.
1=2 - Grades everybody.

Let me tell you now, that you must send me the grades that you have made, because I will submit them.

[quote="Wickedestjr"]April;

pythag011
Yongyi781
lotsofmath
Bugi
AIME15
[b][size=150][color=red]Poincare[/color][/size][/b]

Graders;
Wickedestjr - Grades everybody's.
pythag011 - Grades everybody's but himself.
Yongyi781 - Grades everybody's but himself.
1=2 - Grades everybody.

Let me tell you now, that you must send me the grades that you have made, because I will submit them.[/quote]

The people that grade the answers. Although, I will be checking everything at the end. There were be no more graders unless I must replace one of them for one of these reasons;

1. Quit
2. Wasn't active for one of the months.
3. Tried to grade themselves higher than they graded others if they were competing.
4. Was a grader for six months. (I am sure that there are others that would like to grade also.)

The consequences of doing number 2 would be, not being able to sign up for grading for 6 months.
The consequences of doing number 3 would be, could never be grader ever again, and couldn't play for six months.
The consequences of doing number 4 would be, won't be able to siqn up for 2 months. (They didn't do anything)

April;

pythag011
Yongyi781
lotsofmath
Bugi
AIME15
Poincare

Graders;
Wickedestjr - Grades everybody's.
pythag011 - Grades everybody's but himself.
Yongyi781 - Grades everybody's but himself.
1=2 - Grades everybody.

Also, at IMO, before the contest, people decide how many points to give out if the contestant showed something.

For example:

If he proves f(0) = 0, or f(x) is even or notices that f(x) = x is a solution, no points for him.
If he proves f(x) is injective, or f(x) is onto, 1 point.
If he proves f(x) is injective and f(x) is onto, 2 points.
If he has a correct solution, 7 points.
If he forgets to state the trivial case f(x) = 0 is not a solution, 6 points.

Also, typically:

0 points: Trivial work.
1 point: Small part solved.
2 point: Major progress.
3/4 points: 2-part problems only.
5 points: Complete solution, missing something easily filled in.
6 points: Complete solution, missing something trivial.
7 points: Complete solution, not missing anything.

I will join for April

If you need another grader then I am available.

Of course, pythag011 is right, that sort of grading is OK.
I would also like to be a grader if someone is replaced.

So...why do we need graders?

I thought that this would be locked if higher level problems were put in....

The higher level problems go in the harder sections of this forum.

April;

pythag011
Yongyi781
lotsofmath
Bugi
AIME15
Poincare
james4l

Graders;
Wickedestjr - Grades everybody's.
pythag011 - Grades everybody's but himself.
Yongyi781 - Grades everybody's but himself.
1=2 - Grades everybody.

Future Graders;
james4l
Bugi

Hmm, i guess this isn't very popular, but it might become more popular as months go by.

I don't think it's necessary to start signups so early.  We're still in November, and April signups are going on.

Well, I didn't want to forget about this idea, and I really wanted it to begin in January. I think I will post the first few non-highschool-and-above level questions on January 1st, at midnight. Or it will say midnight on this forum, but it will really be kind of early where I actually live.

Do you guys think that this would be a good qualification to participate in a round;

Ranked in the top half at least one month.

But I would do that in like July or August to give people a chance to do it.

We don't have to have a qualification, but you might like this one.

Wickedestjr, you don't have a lot of people so if you kick one-half every month...

That's true. I will consider the idea later when/if this tourney gets popular.
(which I don't think it will)


Can you divide all the positive 2-digit integers into two groups so that the product of the numbers in each group is the same?

[hide="hint"]
Think prime.

You got my attention :D [/hide]

primes have nothing to do with it, well at least in my solution,

heres my hint [hide] factorials[/hide]

[hide]
It's no and I did it with primes. I'd like to see your solution.[/hide]

Paladin's hint was right. It's all about the primes.
[hide]it doesn't work because of all the prime 2 digit numbers greater than 50[/hide]

[hide]Prime factorize the numbers. You'll only have 1 of the following (if you do it right): 53,59,61,67,71,73,79,83,89,93,97. You obviously can't divide those into two groups.[/hide]


If $ V \equal{} gt \plus{} V_0$ and $ S \equal{} \frac {1}{2}gt^2 \plus{} V_0t$, then $ t$ equals:

$ \textbf{(A)}\ \frac {2S}{V \plus{} V_0} \qquad\textbf{(B)}\ \frac {2S}{V \minus{} V_0} \qquad\textbf{(C)}\ \frac {2S}{V_0 \minus{} V} \qquad\textbf{(D)}\ \frac {2S}{V} \qquad\textbf{(E)}\ 2S \minus{} V$

[hide] Since $ V\equal{}gt\plus{}V_0$ solving for $ g$ we get $ g\equal{}\frac{V\minus{}V_0}{t}$. Substituting this with $ g$ in the second equation we get:

$ S\equal{}\frac{1}{2}\cdot \left (\frac{V\minus{}V_0}{t} \right )\cdot t^2 \plus{} V_0 \cdot t$
$ \Leftrightarrow S\equal{}\frac{t\cdot (V\minus{}V_0)\plus{}2V_0\cdot t}{2}$

$ \Leftrightarrow S\equal{}t \left (\frac{V\plus{}V_0}{2}\right )$
$ \Leftrightarrow t\equal{}\frac{2S}{V\plus{}V_0}$

The answer is A
[/hide]

[hide="Fast Solution"]
Solving for $ gt$ in the first equation, we obtain $ gt \equal{} V \minus{} V_0$. Substituting this into the second equation, we obtain:

$ S \equal{} \frac{(V \minus{} V_0)}{2}\cdot t \plus{} V_0\cdot t\\
S \equal{} \left(\frac{V \plus{} V_0}{2}\right)t\\
t \equal{} \boxed{\frac{2S}{V \plus{} V_0}\text{ (A)}}.$
[/hide]


Solve in the set of the real numbers the following system :

$ \{\begin{aligned} y^2+z^2-x^2&=\frac{4yz}{x}\\
z^2+x^2-y^2&=\frac{6zx}{y}\\
x^2+y^2-z^2&=\frac{8xy}{z}\end{aligned}$

hello, i have found
$ x\equal{}8,y\equal{}\frac {12}{5}\sqrt{10},z\equal{}\frac{8}{5}\sqrt{10}$
$ x\equal{}\minus{}8,y\equal{}\minus{}\frac {12}{5}\sqrt{10},z\equal{}\minus{}\frac{8}{5}\sqrt{10}$
Sonnhard.

[hide="Solution"]None of $ x,y,z$ can be zero. Multiply the equations by $ x^2, y^2, z^2$ respectively, to get

\begin{eqnarray}x^2y^2 + x^2z^2 - x^4 & = & 4xyz \\
y^2z^2 + x^2y^2 - y^4 & = & 6xyz \\
x^2z^2 + y^2z^2 - z^4 & = & 8xyz\end{eqnarray}

Now $ (1) - 2\cdot(2) + (3)\iff 2y^4 - y^2z^2 - x^2y^2 - x^4 + 2x^2z^2 - z^4 = 0$

That yields

$ y^2(2y^2 - x^2 - z^2) - (x^2 - z^2)^2 = 0$

$ y^2\left(2y^2 - \left(y^2 + {6xz\over y}\right)\right) - (x^2 - z^2)^2 = 0$

$ y^4 - 6xyz - (x^2 - z^2)^2 = 0$

$ (y^2 - x^2 + z^2)(y^2 + x^2 - z^2) - 6xyz = 0$

$ {4yz\over x}\cdot{8xy\over z} - 6xyz = 0$

$ 32y^2 - 6xyz = 0$

$ 16y - 3xz = 0$

Now plug $ xz = {16y\over 3}$ into the RHS of $ (2)$ and $ y = {3xz\over 16}$ into the RHS's of $ (1)$ and $ (3)$ to get

\begin{eqnarray}\setcounter{equation}{4} - x^2 + y^2 + {1\over 4}z^2 & = & 0 \\
x^2 - y^2 + z^2 & = & 32 \\
- {1\over 2}x^2 + y^2 - z^2 & = & 0\end{eqnarray}

This is a simple linear system in $ x^2,y^2,z^2$, whose solution is $ (x^2,y^2,z^2) = \left(64, {288\over 5}, {128\over 5}\right)$

From the original equations, we get that for the computed values, all of $ {yz\over x}, {zx\over y},{xy\over z}$ must be positive. Assume $ x$ is negative. Then from the first expression we get that $ y$ and $ z$ must be of the opposite sign, and then the expressions are positive. Similar discussion for $ y<0$ and $ z<0$.

Therefore the solutions to the original system are $ (x,y,z) \in\left\{\left(8, 12\sqrt {{2\over 5}},8\sqrt {{2\over 5}}\right),\left(8, -12\sqrt {{2\over 5}},-8\sqrt {{2\over 5}}\right),\left(-8, 12\sqrt {{2\over 5}},-8\sqrt {{2\over 5}}\right),\left(-8, -12\sqrt {{2\over 5}},8\sqrt {{2\over 5}}\right)\right\}$[/hide]


There are four apples on a plate, of weights $ 600,400,300,$ and $ 250$ grammes. Two pupils take an apple each (one by one) and start eating simultaneously. A pupil can take another apple only after he has eaten the previous apple. It is assumed that they equaly fast (measured in grammes per second). How should the first pupil do in order to eat as large amount as possible?

If the first pupil starts by eating the 600 g apple, then the second pupil can eat the 250 g apple and the 300 g apple and start eating the 400 g apple by the time the first pupil is done. The first pupil will only eat 600 g of apples.
If the first pupil starts by eating the 400 g apple, then the second pupil can eat at most one of the lighter apples and start to eat the 600 g apple by the time the first pupil is done. The second pupil will start with the 300 g apple. Then, by the time the second pupil is done withe 600 g apple, the first will have to eat the 250 g apple. The first pupil will have eaten 650 g of apples.
If the first pupil starts by eating the 300 g apple, then the second pupil can eat the 250 g apple and then start to eat the 600 g apple by the time the first pupil is done. The first pupil will then eat the 400 g apple, ending up with 700 g of apple eaten.
If the first pupil starts by eating the 250 g apple, then he can finish eating it before the second pupil finishes his apple. If the second pupil eats the 600 g apple, the first pupil can eat the 300 g and 400 g apples, ending up with 950 g. If the second pupil eats the 400 g or the 300 g apple, then the first pupil can start eating at most one of the apples left before the second pupil finishes. The first pupil will eat the 600 g apple, leaving the other to the second pupil. The first pupil will have eaten 850 g of apple. The second pupil will either eat the 400 g or 300 g apple because that will minimize the amount of apple the first pupil eats.
The first pupil will, under perfect play by the second pupil, eat 850 g of apple if he starts with the 250 g apple. If he starts with a different apple, the second pupil can make it so the first one eats less than 850 g. Thus the first pupil must eat the 250 g apple first. By perfect play, the second pupil will eat either the 300 g or the 400 g apple, so the first pupil must then eat the 600 g apple.
The first pupil will end up with 850 g, the maximum possible amount under perfect play by the second pupil.


How exactly was the Hardy Weinberg equation derived?

How did they know which factors contributed to microevolution?

Easy Exercise: In a generation, there are 2 individuals w2ith a homozygous recessive trait.  What are the Hardy Weinberg values?

Are you asking because you want to find out the answer yourself, or just to quiz other people? I'm very slightly confused by your post.

But regardless, I think I can answer some of what you're asking. About your second question, I don't really know what to say besides just careful thought and observation, but the Hardy-Weinberq Equation follows rather logically from basic classical genetics.

Consider a Pummett square: Each parent contributes one of two alleles, which each can either be an A allele, with a frequency p, or a B allele, with a frequency q. Hence, in a random offspring with random parents, the probability that a given parent has transferred to it  the A allele is p, and the B allele q. Since this must happen with both parents for any homozygous ofspring, the p^2 and q^2 terms in the Hardy-Weinberg Equation, the probability that any given offspring of random parents will be a homozygous A = The probabilty of event A and event A = p*p. The same logic can be applied to homozygous Bs.
Heterozygotes, on the other hand, require one allele of both types, so the probablility that one will occur is slightly different. For them, it is the probability that event A and event B will both occur, which can occur in one of two orders, and hence equals p*q + q*p = 2pq.

Thus, we have p^2 +2pq +q^2=1

And finally, of course, the simple p+q=1 is quite obvious, and doesn't really require elaboration. If there are two alleles in a population, the sum of their frequencies must equal one.

So there we have it. I hope that rather long-winded post helped somewhat. And sorry if the notation I used for the probabilities was a bit confusing. I don't usually actually write that stuff out and have by now long forgotten the proper way to do it.


Let $ f: \mathbb{R}\to \mathbb{R},\: f(x)\equal{}\left\{\begin{array}{ll}0,\: \mbox{if}\:  x\in\mathbb{R\setminus Q}\\ \frac{1}{q^3}\: \mbox{if}\: x\equal{}\frac{p}{q},\: p,q\in\mathbb{Z},\: q>0,\: \frac{p}{q}\: \mbox{irreductible}\end{array}\right.$

Prove that $ f$ has derivate in every $ x_0\equal{}\sqrt{k},\: k\in\mathbb{N}$ where $ k$ is not a perfect square.

For any $ x \equal{} p/q$ as above we have $ |(x \minus{} x_0)(x \plus{} x_0)| \equal{} |x^2 \minus{} x_0^2| \equal{} \frac {|p^2 \minus{} kq^2|}{q^2}\ge \frac {1}{q^2}$ because the product on the left is not equal to zero. Hence $ |x \minus{} x_0|\ge \frac {1}{3x_0 q^2}$ if $ x$ is close to $ x_0$. Finally, $ \left|\frac {f(x) \minus{} f(x_0)}{x \minus{} x_0}\right| \equal{} \frac {1}{q^3|x \minus{} x_0|}\le \frac {3x_0}{q} \to 0$ as $ x\to x_0$.

By [url=http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem]Roth's theorem[/url] $ f$ is differentiable at all algebraic numbers. It's not necessarily differentiable at [url=http://en.wikipedia.org/wiki/Liouville_number]transcendental numbers[/url].

It's a very nice solution. I've got one too, but is more longer and difficult. Anyway congratulations.:D


inequalities
trigonometry
algebra unsolved
algebra

If the inequality $\sqrt{2}(2a+3)\cos\left(x-\frac{\pi}{4}\right)+\frac{6}{\sin x+\cos x}-2\sin 2x<3a+6$ holds for any real $x\in[0,\pi/2]$, find the value range of real number $a$ can be taken.

The answer is $a>3$.

Let $\sin x+\cos x=t$, by $(\sin x+\cos x)^2=1+\sin 2x$,we have $\sin 2x=t^2-1$.

Therefore we can rewrite the given inequality to $(2a+3)t+\frac{6}{t}-2(t^2-1)<3(a+2)t$.
Since $1\leqq t\leqq \sqrt{2}$ we obtain $2t^3-(2a+3)t^2+(3a+4)t-6>0\Longleftrightarrow (2t-3)(t^2-at+2)>0$,
where $1\leqq t\leqq \sqrt{2}$.Then $-1\leqq 2t-3\leqq 2\sqrt{2}-3<0$,so we should consider the condition for which $t^2-at+2>0$ at $1\leqq t\leqq \sqrt{2}$.
Rewriting this we obtain $t+\frac{2}{t}<a$ at $1\leqq t\leqq \sqrt{2}$.
In using the graph of $y=t+\frac{2}{t}\ (1\leqq t\leqq \sqrt{2})$,yielding $2\sqrt{2}\leqq t\leqq 3$.
Consequently the desired answer is $a>3$.


superior algebra
superior algebra unsolved

Here's the question: Let G be a group and a,b be elements of G such that ab is in Z(G) [aka. the center of G].  Prove ab=ba.

Here is MY question:  (let a inverse = a')
can I say a'b'(ab)a'b'=a'b'(ba)a'b'   so
(a'b'a)(ba'b')=(a'b'b)(aa'b')
can i just assume these are all communatative since they are in the center of a group which has the property ax-xa??
How do I prove ab=ba?  Please help!

That looks too complicated. If $ab$ is in the center, then $ab=aba^{-1}a=a^{-1}aba=ba$, where we use that $ab$ and $a^{-1}$ commute in the middle step.

Wait... the way I read the problem was that I'm trying to prove that its commutatvie.  So can I use the communtative property in trying to prove that its communative?

so ahendri2 i'm not sure that the reason given can be used, otherwise you could just say ab = ba because of communitivity which is a properity of the center of a group.

We know that $ab$ is in the center. We do not know that $a$ is in the center or that $b$ is in the center, and they don't have to be.

I'm a little confused by these posts... jmerry, can you give me any more specific help? Can anyone else help me too?

You can say that $ab$ commutes with everything, but you can't say that $a$ and $b$ commute with each other- you're trying to prove that. Looking closer, your firsat post is simply wrong- you can't do that.


Can someone help me with this:

What is the remainder when (100!/1+100!/2+100!/3+...+100!/99+100!/100) is divided by 101

I can try...
[hide]Note that $ 100!$
$ \equal{}1\cdot2\cdot3\cdot...\cdot98\cdot99\cdot100$
$ \equiv1\cdot2\cdot3\cdot...\cdot(\minus{}3)\cdot(\minus{}2)\cdot(\minus{}1)\pmod{101}$

So $ \frac{100!}{1}$ = $ \minus{}1\cdot2(\minus{}2)\cdot3(\minus{}3)\cdot4(\minus{}4)\cdot...\cdot50(\minus{}50)\pmod{101}$

and

$ \frac{100!}{100}$ = $ 1\cdot2(\minus{}2)\cdot3(\minus{}3)\cdot4(\minus{}4)\cdot...\cdot50(\minus{}50)\pmod{101}$

thus adding these two terms result in $ 0\pmod{101}$. The same goes for all the other fractions, they should have a corresponding pair that sums up to $ 0\pmod{101}$. Therefore the remainder should be $ \boxed{0}$.

Hope I made no mistakes
[/hide]

[hide="alternative solution"]
We know that 101 is prime.
By Wilson's theorem, $ 100!\equiv100\mod101$

Thus $ \sum^{100}_{k\equal{}1}\frac{100!}{k}\equiv\sum^{100}_{k\equal{}1}(\minus{}1)(k^{\minus{}1})\mod101$

Since 101 is prime, the set of inverses of all positive residues mod 101 is a complete set of positive residues mod 101. This means we have

$ \sum^{100}_{k\equal{}1}(\minus{}1)(k^{\minus{}1})\equiv\sum^{100}_{k\equal{}1}\minus{}k\equiv\minus{}\frac{100\times101}{2}\equiv\minus{}5050\equiv\fbox{0}\mod101$
[/hide]

thank you...but can someone explain mod...

$ a\equiv{b}\mod{c}$ simply means the remainder of $ a$ divided by $ c$ is $ b$.

For example
$ 10\equiv{1}\mod{3}$
$ 7\equiv{3}\mod{4}$

[quote="IrinaSharova"]$ a\equiv{b}\mod{c}$ simply means the remainder of $ a$ divided by $ c$ is $ b$.[/quote]

In more general usage, $ a\equiv b\bmod c$ means that $ c | a\minus{}b$.  Of course, both posters have used more complicated facts about modular arithmetic, including uniqueness of inverses $ \bmod$ a prime and Wilson's Theorem.  Neither are necessary.

[hide="A simple proof"] $ \frac{100!}{k}\plus{}\frac{100!}{101\minus{}k}\equal{}\frac{101!}{k(101\minus{}k)}$

When the appropriate fractions are paired up, we have

$ \sum_{k \equal{} 1}^{100}\frac{100!}{k}\equal{}\sum_{i \equal{} 1}^{50}\frac{101!}{i(101\minus{}i)}$

Since the denominators of the fractions only contain factors less than (and therefore not divisible by) $ 101$, and since we know that each term of the original sum (and therefore the entire sum) is an integer, and lastly since $ 101$ appears as a factor in the numerator, the conclusion follows. [/hide]


inequalities
geometry
inequalities unsolved

Given a triangle $ABC$ with sides $a,b,c$ and area $F$ prove that
for all $x,y,z>0$ we have
$xa^{4}+yb^{4}+zc^{4}\ge 16\sqrt{\frac{xy+yz+zx}{3}}F^{2}$
 :)

[quote="gemath"]Given a triangle $ABC$ with sides $a,b,c$ and area $F$ prove that
for all $x,y,z>0$ we have
$xa^{4}+yb^{4}+zc^{4}\ge 16\sqrt{\frac{xy+yz+zx}{3}}F^{2}$[/quote]

Unfortunately, this is wrong. Take a = 3, b = 3, c = 4, x = 16, y = 16, z = 2.

  Darij


Solve the Diophantine Equation:

$2^{x}+1=y^{2}$.

[hide="Big hint"] Subtract $1$ from both sides. [/hide]

[hide="my solution"]
$2^{x}=(y-1)(y+1)$
$2^{x}$ must factorize in two factors with difference $2$. Possible factors of $2^{x}$ are $\{1,2,4,8,16,...,2^{x}\}$. The only two factors differs by $2$ are $2$ and $4$, so $2^{x}=8 \Rightarrow x=3$ and $y=\pm 3$.
[/hide]

[hide]We can easily see that $2^{x}=(y+1)(y-1)$. Since the prime factorization is unique we deduce that $y+1=2^{k}$ and $y-1=2^{m}$ where $m>k>0$. $2^{k}-2^{m}=2$, hence $2^{k}(2^{m-k}-1)=2$ where $k=1$ and $m=2$. Therefore we get $x=y=3$.[/hide]

[hide="solution"]It is asking for integer solutions, so we subtract one from both sides: $2^{x}=(y+1)(y-1)$ That tells us y is odd. We employ inspection and see that 3,3 is a solution. We also see that 3,-3 is a solution. Those are the only two solutions.[/hide]


Does anyone know if the answers to the 1999 AP multiple choice are available online?

these are the answers... my teacher just gave them all to us, not sure if they are available online though
i kno it's kinda unclear, but i couldn't impove it. hope it's ok!

[quote="shobber"]can you tell me where i can find the questions??[/quote]

all past AP exams are posted here:
http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/ChemTeamIndex2.html

However, the most recent one is the 1998... so it doesn't have the 1999 MC... sorry!... oh wait... it's already posted... so yeah... those are the other years' problems.

do you know where you can find other years' multiple choice questions?


One each tile in a set of dominoes, each integer 0 through 12 is paired exactly once with each integer 0 through 12. One number of the pair is written on one half of the domino, and the other number is written on the other half. If one domino is randomly selected from all dominoes in this set, what is the probability that the same number will be on both halves? Express your answer as a common fraction.

Shouldn't the numbers appearing on the domino be a combination of two numbers between 0 and [b]6[/b]?

You can have the numbers 0 to anything, really.
Anyway, there are a total of $ 13 \plus{} 12 \plus{} 11 \plus{} 10 \plus{}...\plus{} 1 \equal{} \frac{13\cdot14}{2} \equal{} 91$ domino combinations. Of these 91, 13 pairs exist. Thus, the probability of a pair is $ \frac{13}{91} \equal{} \boxed{\frac17}$.


floor function
inequalities unsolved
inequalities

Let $ m$ and $ n$ be positive integers with $ m > n \geq 2.$ Set $ S \equal{} \{1, 2, \ldots, m\},$ and $ T \equal{} \{a_l, a_2, \ldots, a_n\}$ is a subset of S such that every number in $ S$ is not divisible by any two distinct numbers in $ T.$ Prove that
\[ \sum^n_{i \equal{} 1} \frac {1}{a_i} < \frac {m \plus{} n}{m}.
\]

Here is my short solution.

Define $\mathbb{S}_k=\{x: a_k|x \text{ and } x\in\mathbb{S}\}$ for each natural number $k$ less than or equal to $n$.

It is easy to show that any two such sets are pairwise disjoint.

$|\mathbb{S}|=m$ so $|\mathbb{S}_k|=\left\lfloor\frac{m}{a_k}\right\rfloor$
So $\sum_{k=1}^{n}\left\lfloor\frac{m}{a_k}\right\rfloor=\sum_{k=1}^{n}|\mathbb{S}_k|\leq |\mathbb{S}|=m$
So $\sum_{k=1}^{n}\left(\left\lfloor\frac{m}{a_k}\right\rfloor+1\right)\leq m+n$
Now for each $k$, $\frac{m}{a_k}<\left\lfloor\frac{m}{a_k}\right\rfloor+1$
So $\sum_{k=1}^{n}\frac{m}{a_k}< m+n$
So $\sum_{k=1}^{n}\frac{1}{a_k}< \frac{m+n}{m}$ as required.


geometry
superior algebra
superior algebra unsolved

Is there an extension of $ \mathbb{Q}$ of degree $ 11$ unramified outside 43?

I don't understand the question. Does it mean that $ 43$ is the only prime ramified?

Yes.  The message is too small. Please make the message longer before submitting.

It's easy to show that it can't be a Galois extension. Does $ \mathbb{Q}(\sqrt[11]{43})$ work?

I think $ 11$ is inert since $ x^{11}\equiv43(mod$ $ 11)$ has a solution but $ x^{11}\equiv43(mod$ $ 11^2)$ does not.

I am not very familiar with this area, but I am interested in finding fields like this.

I think one of the previous answers was wrong, can someone verify this?

11 ramifies in Q(43^(1/11)).  If z^11 = 43, then (11,1+z)^11 = (11).

There are also Galois extensions where only 43 ramifies, for instance Q[X]/(X^3 +X^2 -14*X +8), but not of degree 11.

How rare are these fields?  Are there any of degree 11? degree 2?


for positive real numbers a,b,c,d denote $A=a^{3}+b^{3}+c^{3}+d^{3}$ and B=bcd+cda+dab+abc. Prove that
$(a+b+c+d)^{3}$=<$4A+24B$

It can be very quickly solved by a method posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20851]here[/url].


Euler
geometry
geometric transformation
reflection
circumcircle
trapezoid
incenter

In $ \triangle{ABC}$, points $ H$ and $ O$ are its orthocenter and circumcenter, respectively. Consider the parallel lines $ k,l,m$ passing through the vertices $ A,B,C$ and denote by $ k',l',m'$, respectively, the reflection of these lines in $ BC,CA,AB$.
 [b]a)[/b] Prove that $ k',l',m'$ are concurrent iff $ k,l,m$ are parallel to the [i]Euler line [/i]of the triangle.
 [b]b) [/b]If $ X$ is the point of concurrence(in (a)) prove that the circumcircle of $ \triangle{ABC}$ passes through the midpoint of $ [OX]$.

Let p be an arbitrary line through the orthocenter H, cutting BC, CA, AP at $ P_a,\ P_b,\ P_c.$ The reflections $ p_a,\ p_b,\ p_c$ of p in BC, CA, AB meet at the pole $ P \in (O)$ of the Simson line parallel to p. It is sufficient to show that $ p_a$ passes through P. The A-altitude AH meets (O) again at the reflection $ H_a$ of H in BC, the line p meets BC at $ P_a,$ and $ p_a \equiv H_aP_a.$ The parallel to p through A meet the circumcircle (O) again at A' and the normal to BC from A' meets (O) at P. Since $ AH_a \parallel A'P$ (both $ \perp BC$), $ AH_aA'P$ is a cyclic trapezoid, isosceles, $ \angle HH_aP \equiv \angle AH_aP \equal{} \angle H_aAA' \equal{} \angle H_aHP_a \equal{} \angle HH_aP_a,$ $ P \in p_a.$

Let q be an arbitrary line passing through the circumcenter O, cutting BC, CA, AB at $ Q_a,\ Q_b,\ Q_c.$ Let $ q_a,\ q_b,\ q_c$ be reflections of q in BC, CA, AB. If $ H \in q,$ q is identical with the Euler line OH, and the reflections $ q_a,\ q_b,\ q_c$ concur at the pole Q of the Simson line parallel to $ q \equiv OH.$ If $ H \notin q,$ let $ p \parallel q$ pass through the orthocenter H. We have $ \measuredangle (q_b,q_c) \mod \pi \equal{} 2 \measuredangle (CA, AB) \mod \pi \equal{} 2 \angle A \mod \pi$  and similarly, $ \measuredangle (q_c,q_a) \mod \pi \equal{} 2 \angle B \mod \pi,$ $ \measuredangle (q_a,q_b) \mod \pi \equal{} 2 \angle C \mod \pi.$ Since $ p_a,\ p_b,\ p_c$ concur in P and since the line pairs $ (p, q),\ (p_a, q_a),\ (p_b, q_b),\ (p_c, q_c)$ are pairwise equidistant, the lines $ q_a,\ q_b,\ q_c$ form a triangle $ \triangle UVW$ with incenter P. Because of their mutual angles, this triangle is similar to the circumcevian $ \triangle H_aH_bH_c$ WRT the orthocenter H.

Let the lines $ k \parallel l \parallel m \parallel q \equiv OH$ pass through the vertices A, B, C, cutting BC, CA, AB at K, L, M, and let $ k',\ l',\ m'$ be their reflections in BC, CA, AB. Denote $ A' \equiv k \cap q_a$ $ K' \equiv k' \cap q.$ $ Q_aA'KK'$ is a rhombus, $ A'K' \perp Q_aK \equiv BC \Longrightarrow$ $ A'K' \parallel AH,$ AHK'A' is a parallelogram, A'K' = AH. Let $ (O_a)$ be reflection of (O) in BC ($ O \in q \Longrightarrow O_a \in q_a$) and (N) the 9-point circle of the $ \triangle ABC.$ The circles $ (N) \sim (O_a)$ are centrally similar with center A and coefficient 1/2, N is the midpoint of $ AO_a$. But N is also the midpoint of OH $ \Longrightarrow$ $ AHO_aO$ is a parallelogram, its diagonals cutting each other in half, $ OO_a \equal{} AH \equal{} A'K'.$ Then $ OO_aK'A$ is a rectangle, $ Q_a$ is the midpoint od $ O_aA'.$ If OQ meet k' at X, $ q_a \equiv Q_aA' \equiv Q_aQ \parallel k' \equiv K'X$ is the midline of the $ \triangle OK'X,$ $ Q \in (O)$ is the midpoint of OX. Since the pole Q of the Simson line parallel to the Euler line OH, and the reflection X of O in Q are unique triangle points and $ X \in k',$ it follows that $ k',\ l',\ m'$ concur at X.

Assume now that $ O \in q,\ H \notin q,$ $ p \parallel q,\ H \in p$ and let the lines $ k \parallel l \parallel m \parallel q$ and their reflections be defined similarly as before. Since $ k' \parallel q_a,\ l' \parallel q_b,\ m' \parallel q_c,$ the lines $ k',\ l',\ m'$ form a $ \triangle U'V'W'$ centrally similar to the $ \triangle UVW,$ possibly degenerated to a point. Since the line reflection pairs $ (p_a, q_a),\ (p_b, q_b),\ (p_c, q_c)$ are now different, define $ A' \equiv k \cap p_a,$ $ K' \equiv k' \cap p.$ Then we still have the rhombus $ A'P_aK'K,$ parallelogram AHK'A', and rectangle $ OO_aK'A'.$ Take a line $ u \parallel k',$ such that the line pairs $ (p, q),\ (p_a,q_a),\ (u, k')$ are pairwise equidistant, and define the lines $ v \parallel l',\ w \parallel m'$ similarly. The lines $ (p, q, p_a, q_a),\ (p, q, u, k')$ form congruent rhombi. Denoting their vertices $ A'' \equiv p_a \cap q,\ K'' \equiv u \cap q,$ A'' is the midpoint of OK''. If OP meets u at X, $ p_a \equiv P_aA'' \equiv P_aA' \equiv P_aP \parallel u \equiv K''X$ is the midline of the $ \triangle OK''X,$ $ P \in (O)$ is the midpoint of OX. Given the lines $ p \parallel q$ through H, O, respectively, the pole P of the Simson line parallel to $ p \parallel q$ and the reflection X of O in P are unique triangle points and $ X \in u,$ it follows that u, v, w concur at X. Since the line pairs $ (p, q),$ $ (p_a, q_a \equiv VW),$ $ (p_b,q_b \equiv WU),$ $ (p_c, q_c \equiv UV),$ $ (u, k' \equiv V'W'),$ $ (v, l' \equiv W'U'),$ $ (w, m' \equiv U'V')$ are pairwise equidistant, $ p_a,\ p_b,\ p_c$ concur at the incenter P of the $ \triangle UVW,$ and u, v, w concur at X, it follows that the $ \triangle UVW \cong \triangle U'V'W'$ are congruent, and X is incenter of the $ \triangle U'V'W'.$ Since the $ \triangle UVW$ degenerates to the point $ P \equiv Q$ iff $ p \equiv q \equiv OH$ is the Euler line, so does the $ \triangle U'V'W'.$

Related to Parry's reflection point, here is another nice result, which I gave on the Hyacinthos forum several weeks ago:

If $ A'B'C'$ is the tangential triangle of $ ABC$ (the triangle determined by the tangents at the vertices of $ ABC$ with respect to the circumcircle) and $ X$, $ Y$, $ Z$ are the reflections of $ A$, $ B$, $ C$ in the opposite sidelines, then the circumcircles of $ YZA'$, $ ZXB'$, $ XYC'$ are concurrent at Parry's reflection point.

The "dual" problem also holds, meaning, using the same notations, the circumcircles of $ XB'C'$, $ YC'A'$, $ ZA'B'$ are concurrent. (their perspector is not in the current edition of the ETC as Javier Francisco Garcia noticed.)

(Link to the first problem: [url]http://www.mathlinks.ro/viewtopic.php?t=177994[/url])


What is the smallest integer value $ n>1$ such that $ n!$ is a perfect square?

Yikes, why is this in High School Basics? The answer is there is no such value. Suppose $ n!$ is a square. A prime $ p$ greater than $ n/2$ and less than or equal to $ n$ exists by Bertrand's Postulate. This prime must have an exponent of $ 1$ in the prime factorization of $ n!$. Contradiction.

EDIT: To answer the below post, the fact that you almost need Bertrand's Postulate here is what made me question the placement. I wouldn't say that's an elementary and extremely well-known result at this level.

Nice!

I did not think this was too hard of a problem which is why I posted in High School Basics.


trigonometry
analytic geometry
graphing lines
slope

Find all solutions to the equation

$ \sec{x} \plus{} \tan{x} \equal{} 1$.

[hide="Funny story"]
No one in my pre-calc class including the teacher could solve this because they could only manipulate equations using direct trig identities. The teacher ended up removing the problem from the review homework and I realized just [i]after[/i] I left that I simply needed to square both sides and work from there >_>.
[/hide]

Can you show your steps?

EDIT: Fixed. I made a REALLY stupid mistake when factoring.

[hide="My Solution"]
$ \sec{x} \plus{} \tan{x} \equal{} 1 \\
\sec^2{x} \plus{} \tan^2{x} \plus{} 2\sec{x}\tan{x} \equal{} 1 \\
2\tan^2{x} \plus{} 2\sec{x}\tan{x} \equal{} 0 \\
2\tan{x}(\tan{x} \plus{} \sec{x}) \equal{} 0 \\
 \\
\tan{x} \plus{} \sec{x} \equal{} 0 \\
\frac {\sin{x} \plus{} 1}{\cos{x}} \equal{} 0 \\
\sin{x} \equal{} \minus{} 1 \\
x \equal{} \frac {3\pi}{2} \plus{} 2n\pi \\
\text{extraneous} \\
 \\
\tan{x} \equal{} 0 \\
x \equal{} \boxed{2n\pi}$
[/hide]

Your mistake occurs from line $ 3\rightarrow4$

Well

$ 1 \plus{} \sec{x} \equal{} 0 \\
\frac {1}{\cos{x}} \equal{} \minus{} 1 \\
\minus{} \cos{x} \equal{} 1 \\
\cos{x} \equal{} \minus{} 1$

Oh yea I see what you mean. It's only $ \pi \pm 2n\pi$.  :blush:

@ocha: I subtracted 1 from both sides and I replaced $ \sec^2{x} \minus{} 1$ with $ \tan^2{x}$ and then combined. I still don't see what I did wrong.

EDIT2: I see it.

Try $ x \equal{} \pi ,\ 3\pi .\ \cdots$.

In terms of identities, you're missing a pretty useful one.  $ \cos x \minus{} \sin x \equal{} \sqrt{2} \cos \left( x \minus{} \frac{3\pi}{4} \right)$, which gets you your answer in a very straightforward manner.

Ok thanks. Sorry for the trouble, I made a stupid factoring mistake >_>. Otherwise, I get the same solution as kunny.

@toraijirou: Interesting identity, I never saw that before.

[quote="t0rajir0u"]In terms of identities, you're missing a pretty useful one.  $ \cos x \minus{} \sin x \equal{} \sqrt {2} \cos \left( x \minus{} \frac {3\pi}{4} \right)$, which gets you your answer in a very straightforward manner.[/quote]

That's one of solution.

The problem can be solved only 10 seconds in our head. :wink:

Keyword:

[color=blue]Slope [/color][quote="kunny"]$ \frac {\sin x \minus{} ( \minus{} 1)}{\cos x \minus{} 0} \equal{} 1\ (\cos x\neq 0)$, yielding $ x \equal{} 2n\pi$ for $ n\in{\mathbb{Z}}$.[/quote]

[color=blue]Unit Circle [/color]

Solve the system of equation: $ X^2 \plus{} Y^2 \equal{} 1\ (X\neq 0)$ and $ Y \equal{} X \minus{} 1$.


[quote="carij"]I have ordered one book and it seems to me he is cheating.. just to get money and not send the book. Do you have any experience with him?[/quote]

He is [b]NOT[/b] reliable. I have sent him money to his address he wrote me and he didn't admit getting it.

[quote="carij"]He wrote me his post address, I sent him money, and about 10 days later he replied me:

[quote="pvthuan"]Hello friend:

I checked my account two days ago but I did not see the money.

I will wait for that.[/quote]

But he haven't wrote me number of his bank account, just the postal address... that is quite strange as well.

===

Anyway what can I do about it? I have got a post statement that I sent money to address he wrote me, and he is denying that he got it. What can I do?[/quote]

the only thing i can think of is call the police. did he give u his address?

[quote="mz94"]

the only thing i can think of is call the police. did he give u his address?[/quote]

Of course he did. It was the only contact he gave to me. I DO have a post statement that I have sent him 25 dollars, so I have warranty against his statement, that he didn't received anything.

Anyway of course, that if he will send me the book and admit getting my money I sent him, I will announce it here.

I only repeat what it was already said:

[quote="ZetaX"]Have you taken into account that the money may still be on the way before claiming such things here¿[/quote]

[quote="carij"]Anyway of course, that if he will send me the book and admit getting my money I sent him, I will announce it here.[/quote]

[quote="carij"]I have ordered one book and it seems to me he is cheating.. just to get money and not send the book. Do you have any experience with him?[/quote]

anyway...I don't have any experience but he is a writer of famous books and a very good problem solver and the moderator of the inequality section and I also can say many good things about him.(he was the chief guide of IMO as i know [url]http://www.mathlinks.ro/Forum/portal.php?t=159715[/url])

So i don't think that we u should worry for ur $ \$25$.I do not think that he will cheat u for just $ \$25$


induction
inequalities
algebra unsolved
algebra

Let $ (x_n)_{n \geq 0}$ be a real sequence given by the condition : $ x_0>0, x_1>0$ and $ x_{n\plus{}2}\equal{}\frac{2\plus{}x_{n\plus{}1}}{2\plus{}x_{n}}$, for all $ n\geq 0$. Show that $ (x_n)_{n \geq 0}$ is a convergent sequence.

[quote="conan_naruto236"]Let $ (x_n)_{n \geq 0}$ be a real sequence given by the condition : $ x_0 > 0, x_1 > 0$ and $ x_{n \plus{} 2} \equal{} \frac {2 \plus{} x_{n \plus{} 1}}{2 \plus{} x_{n}}$, for all $ n\geq 0$. Show that $ (x_n)_{n \geq 0}$ is a convergent sequence.[/quote]
Please help me ??? nobody help me.  :(

Let $ A$ be such that $ A\equal{} max\{2,x_1\}$ . It's easy to see, by induction, that the sequence is bounded : $ 0< x_n, x_{n\plus{}1} \leq A \Longrightarrow  0 < x_{n\plus{}2}  < \frac {2\plus{}A}2 \leq A$.
Now I denote $ L\equal{}$ the set of limit_points of the sequence $ x_n$ .  Limit_point = limit of a convergent sub_sequence of $ x_n$ .
Because of $ x_n$ is bounded, $ L$ is bounded and  non empty  , and there is $ M\equal{}sup(L) \geq \ m\equal{}inf(L)  \geq 0$ (superior/inferior bounds)
From $ x_{n \plus{} 2} \equal{} \frac {2 \plus{} x_{n \plus{} 1}}{2 \plus{} x_{n}}$ we obtain $ \forall_{\lambda \in L}, \frac{2\plus{}m}{2\plus{}M} \leq \lambda \leq \frac{2\plus{}M}{2\plus{}m}$,  so :
$ \frac{2\plus{}m}{2\plus{}M} \leq m \Longleftrightarrow 2\minus{}m \leq m.M$ , and $ M\leq \frac{2\plus{}M}{2\plus{}m}  \Longleftrightarrow m.M  \leq 2\minus{}M$ , so $ m \geq M$
But $ M\geq m$, so finally $ m\equal{}M\equal{}1$ , so $ x_n$ is convergent and $ x_n \rightarrow 1$ ..  :cool:

[quote="conan_naruto236"]why do you write ??? :?: Please explain me. :(
[quote="Diogene"]
F$ \forall_{\lambda \in L}, \frac {2 \plus{} m}{2 \plus{} M} \leq \lambda \leq \frac {2 \plus{} M}{2 \plus{} m}$,  so :
$ \frac {2 \plus{} m}{2 \plus{} M} \leq m \Longleftrightarrow 2 \minus{} m \leq m.M$ , and $ M\leq \frac {2 \plus{} M}{2 \plus{} m} \Longleftrightarrow m.M \leq 2 \minus{} M$ , so $ m \geq M$[/quote][/quote]
Let $ \lambda$ be an element of $ L$(set of limit points), so there is a sub-sequence $ x_{p(n)\plus{}2}$ convergent to $ \lambda$ .
The sequence $ x_{p(n)\plus{}1}$ is bounded , so there a convergent sub-sequence $ x_{p(q(n))\plus{}1}$, and let $ a$ be its limit.
The sequence $ x_{p(q(n))}$ is bounded , so there a convergent sub-sequence $ x_{p(q(r(n)))\plus{}1}$, and let $ b$ be its limit : $ x_{p(q(r(n)))}\rightarrow b$
$ x_{p(q(r(n)))\plus{}2}$ is a sub-sequence of $ x_{p(n)\plus{}2}$ so $ x_{p(q(r(n)))\plus{}2}\rightarrow\lambda$ ..;.. $ x_{p(q(r(n)))\plus{}1}$ is a sub-sequence of $ x_{p(q(n))\plus{}1}$ so $ x_{p(q(r(n)))\plus{}1}\rightarrow a$ .
From $ x_{p(q(r(n))) \plus{} 2} \equal{} \frac {2 \plus{} x_{p(q(r(n))) \plus{} 1}}{2 \plus{} x_{p(q(r(n)))}}$ we obtain $ \frac{2\plus{}a}{2\plus{}b} \equal{} \lambda$. But $ 0\leq m \leq a , b \leq M$ , so $ \frac{2\plus{}m}{2\plus{}M}\leq  \frac{2\plus{}a}{2\plus{}b}\equal{}\lambda \leq  \frac{2\plus{}M}{2\plus{}m}$
From  $ \forall_{\lambda \in L}, \frac{2\plus{}m}{2\plus{}M} \leq \lambda$ we obtain $ \frac{2\plus{}m}{2\plus{}M} \leq inf(L)\equal{}m$ so $ 2\minus{}m\leq m.M$
In the same way we obtain $ M\leq \frac{2\plus{}M}{2\plus{}m}$ so $ m.M\leq 2\minus{}M$ .. and so on ..
Do you follow me ? :cool:

[quote="conan_naruto236"]why do you write ??? :?: Please explain me. :(
[quote="Diogene"]
F$ \forall_{\lambda \in L}, \frac {2 \plus{} m}{2 \plus{} M} \leq \lambda \leq \frac {2 \plus{} M}{2 \plus{} m}$[/quote][/quote]

Alternatively to Diogene's proof take some arbitrary $ \varepsilon > 0$, then for all but finitely many x_n we have $ m \minus{} \varepsilon\le x_n\le M \plus{} \varepsilon$ implying
\[ \frac {2 \plus{} m \minus{} \varepsilon}{2 \plus{} M \plus{} \varepsilon}\le \frac {2 \plus{} x_{n \plus{} 1}}{2 \plus{} x_n} \equal{} x_{n \plus{} 2}\le \frac {2 \plus{} M \plus{} \varepsilon}{2 \plus{} m \minus{} \varepsilon}.
\]
Now pick some limit point $ \lambda$ and approximate it by a sub-sequence of (x_n). Plugging this into the inequality yields
\[ \frac {2 \plus{} m \minus{} \varepsilon}{2 \plus{} M \plus{} \varepsilon}\le \lambda\le \frac {2 \plus{} M \plus{} \varepsilon}{2 \plus{} m \minus{} \varepsilon}.
\]
The limit $ \varepsilon\to 0$ yields the desired result.


[url]http://www.thesmokinggun.com/archive/1013043mackris12.html[/url]

It's sort of disgusting that I'm so partisan that this makes me happy.  I don't care, I'm still happy!  Serves you right, you evil, evil man.

Ummmmm, this probably shouldn't be viewed by anyone under the age of 15.[/url]

the http://www.fark.com headline for this was...

"Bill O'Reilly being asked to pay up $60 mil for trying to show off his factor"

I must say, thesmokinggun.com sometimes scares me with the things they dig up...

wut exactly is this? someone give me a summary of wuts happening...

Do all of you read a newspaper every day?

If you don't read a newspaper, you could at least read the headlines on [url=http://news.google.com/news]Google News[/url] and then follow links to the major stories that interest you. That's how I learned about what the issue is around Bill O'Reilly.

is this law suit the reason the oreilly factor hasnt been on for a while?

I hope this lawsuit is why his show won't be on for 10-20 years. :)

According to some Al Franken book (I think it's [i]Lies[/i]), there are a bunch of sex scenes in [i]Those Who Trespass[/i], a Bill O'Reilly book.

i dont think bill oreilly would do those things...

But ...but .. from Boston  globe ...[img]http://cache.boston.com/bonzai-fba/Globe_Graphic/2004/10/16/1097983731_2324.gif[/img]

[quote="Rep123max"]Oh, I forgot the most ironic part of all of this, he just came out with a book for kids![/quote]

And even more ironic... the charges were dropped. Haha, isn't that funny when people jump to conclusions all over something and it gets dismissed? :)

[quote] the charges were dropped.[/quote]
It was settled out of court .  It just means that the case will not go to court.

Just as a terminology issue, "charges" isn't usually what one uses to refer to the allegations of the plaintiff in a civil suit -- it's reserved for criminal cases, at least in common usage.  Similarly, one doesn't say "the charges were dropped" when a case was settled -- usually, when a case is settled, especially when there's a non-disclosure agreement, it means the defendant was trying to cut his or her losses.

[quote="gameworld7"]whats the "haha! yes!" part of this? isn't this a bad thing?[/quote]

I know!  Bill O'reilly is great.


function
quadratics
linear algebra
matrix
calculus
calculus computations

Find the maximum and minimum values of the function $ f(x,y) \equal{} x^2 \plus{} y^2 \minus{}xy \plus{} x \plus{} y$ in the region $ R \equal{} \{ (x,y) \in \mathbb{R} ^2: x \leq 0 \wedge y \leq 0 \wedge x \plus{} y \geq \minus{}3 \}$.

The quadratic form generated by the matrix $ \begin{bmatrix}2 & \minus{} 1 \\
\minus{} 1 & 2\end{bmatrix}$ is positive definite.

Among other things, that means that the values of $ f(x,y)$ restricted to any straight line are concave upwards. Hence the maximum should occur at some vertex of the triangle. I'll leave it to someone else to determine whether the minimum lies in the interior or along an edge.


can we state that Y is a triangulare matrix?

[quote="Vialli"]$U(M_{3}(R))$[/quote]

Could someone explain this notation. I am new to linear algebra   :!:

I am not sure but I think this is how you can get Y without guessing:

the egenvalue of matrixes $A,B$ have to be the same and they are indeed the same ($0,(\pm\sqrt{105}+9)/2$)

Because the matrix's $A,B$ have three different eigenvalues each than they must have the decomposotions:

$A = M*D*M^{-1}$
$B = V*D*V^{-1}$

Now let $Y = M*V^{-1}$ and so $A=Y*B*Y^{-1}$

I think that $U(M_{3}(R))$ is the set of $3 \times 3$ matrices with coefficients from $\mathbb{R}$ that have determinant $1$.

We are sure that  $det(Y) = 1$???   :?

[quote="Ramanujan"]

We are sure that  $det(Y) = 1$???   :?[/quote]

Yes am I sure.

[quote="Vialli"]$A \in U(M_{3}(R))$ means that $det A \neq 0$.[/quote]
The usual notation for this is $A\in GL_3(\mathbb{R})$. The GL stands for "general linear group". There are also some variations on where to put the 3.


real analysis
function
limit
real analysis unsolved

Suppose that $f: [0,1]\to R$ is a bounded Lebesgue measurable function such that for all $x$, $f$ coincides almost everywhere with a function that has a limit at $x$. Then $f$ coincides almost everywhere with a continuous function.

Let $f_{x},\tilde f_{x}$ be two functions with limits in $x$ and differing from $f$ on a nullset. We can then find $x_{n}\to x$ such that $\forall n, f_{x}(x_{n})=\tilde f_{x}(x_{n})=f(x_{n})$, which means that the two limits in $x$ coincide, and their common value depends only on $x$. Denote this value by $\ell_{x}$. It is now clear that there is only one reasonable candidate for a continuous function $g$ differing from $f$ on a nullset: $g(x)=\ell_{x}$. We have to prove that this definition does indeed work, i.e. that $g$ defined as above differs from $f$ on a nullset and that it is continuous.

Proof of the continuity of $g$

Fix $x\in[0,1]$ and $\varepsilon>0$. We can find a neighborhood $U$ of $x$ almost all of whose points differ from $\ell_{x}$ by less than $\frac\varepsilon 2$. Now pick an arbitrary $y\in U$. There is a neighborhood $V\subset U$ of $y$ almost all of whose points differ from $\ell_{y}$ by less than $\frac\varepsilon 2$. This means that almost all points in $V$ differ from both $\ell_{x},\ell_{y}$ by less than $\frac\varepsilon 2$ and hence, in particular, $|\ell_{x}-\ell_{y}|<\varepsilon$. This happens for all $y\in U$, so we’re done: $g(x)=\ell_{x},\ \forall x$ is indeed continuous.

Proof of the fact that $f=g$ almost everywhere

It suffices to fix an arbitrary $\varepsilon>0$ and show that $\mu(S)=0$, where $S =\{x\ |\ g(x)-f(x)>\varepsilon\}$. Assuming the contrary, $S$ contains a point $x$ where it has density $1$ (that is, $\lim_{h\to 0}\frac{\mu((x-h,x+h)\cap S)}{2h}=1$), by Lebesgue's density theorem. This implies that we can find $x_{n}\to x,\ x_{n}\in S$ such that $f(x_{n})\to\ell_{x}$. However, by the continuity of $g$, we also have $g(x_{n})\to g(x)=\ell_{x}$, and since $g(x_{n})-f(x_{n})>\varepsilon,\ \forall n$, we have our contradiction.


Prove that a+b+c+1/(abc)>=3+9^(1/3) where sum(a^3)=1 and a,b,c>0.

bye bye!!

[b]Generalization:[/b]

we also have that:

a1+a2+..+an+1/(a1*a2*..*an)>=sqrt(n)+sqrt(n^n) with sum(a1^2)=1 and a1,..an>=0

which is done like the other one!!! :D :D

cheers :D :D


function
limit
integration
inequalities
real analysis
real analysis unsolved

I am trying to learn math, and working on the following problem, PLEASE give me your comments for my "solution" .  Thanks so much .

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

[b]Let $ (X, M, \mu)$ be a positive measure space with $ \mu(X) < \plus{} \infty$

show that $ f : X\rightarrow [0, \plus{} \infty)$ is integrable iff the series $ \sum_{n\geq 0}\mu(\{ x: f(x)\geq n\})$ converges[/b]

Here is my proof :

for each $ n\geq 0$, let $ A_{n} \equal{} \{ x : f(x)\geq n\}$ then $ A_{n}\subset X$ therefore $ \mu(A_{n})\leq \mu(X) < \infty$. Moreover, I have $ A_{n \plus{} 1}\subset A_{n}$

Since $ f$ is integrable, I have $ \{ x : f(x) \equal{} \infty\}$ has zero measure
and $ \bigcap_{n\geq 0} A_{n} \equal{} \{ x : f(x) \equal{} \infty\}$ then I have $ \lim_{n\rightarrow \infty}(A_{n}) \equal{} \mu(\{ x : f(x) \equal{} \infty\}) \equal{} 0$

so given $ \epsilon > 0$ , there exists $ N$ st : $ \mu(A_{n}) < \frac {\epsilon}{2^{n}}$ for all $ n > N$

So I have
$ \sum_{n\geq 0}\mu(\{ x: f(x)\geq n\}) \equal{} \sum_{n \equal{} 0}^{N}\mu(\{ x: f(x)\geq n\}) \plus{} \sum_{n\geq N}\mu(\{ x: f(x)\geq n\})\leq \sum_{n \equal{} 0}^{N}\mu(\{ x: f(x)\geq n\}) \plus{} \sum_{n \geq 0}\frac {\epsilon}{2^{n}} < \infty$

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

For other direction, assume that the series $ \sum_{n\geq 0}\mu(\{ x: f(x)\geq n\})$ converges

then I have $ \lim_{n\rightarrow\infty}\mu(\{ x: f(x)\geq n\}) \equal{} 0$ then I have $ 0\leq f\leq M$ is bounded a.e
so I have $ \int_{X}fd\mu\leq \int_{X}Md\mu \equal{} M\mu(X) < \infty$, so $ f$ is integralbe.

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

Am I right ?

Thanks so much

[quote="1234567a"]Am I right ? [/quote]
No.

Let's look at part (b):

[quote]For other direction, assume that the series $ \sum_{n\geq 0}\mu(\{ x: f(x)\geq n\})$ converges

then I have $ \lim_{n\rightarrow\infty}\mu(\{ x: f(x)\geq n\}) \equal{} 0$ then I have $ 0\leq f\leq M$ is bounded a.e [/quote]

You cannot get that $ f$ is bounded from that. Let's look at a particular case. Let $ f(x)\equal{}\frac1{\sqrt{x}}$ on $ [0,1].$

Then $ \{x: f(x)\ge n\}\equal{}\left(0,\frac1{n^2}\right)$ and $ \mu(\{ x: f(x)\geq n\}) \equal{}\frac1{n^2}.$

This satisfies the conditions you want - that is the terms of a convergent series - and $ f$ is most certainly not bounded (or essentially bounded, which is what really matters).

Your proof of the first part is also dubious. The biggest problem is that you never actually used the hypothesis that $ f$ is integrable. OK, you did use the fact that $ f$ is finite almost everywhere, but let $ f(x)\equal{}\frac1x$ on $ [0,1].$ That's finite almost everywhere but not integrable; if your proof worked at all it should work for that function, and it doesn't. After all
\[ \sum_{n\equal{}0}^{\infty}\mu(\{x: f(x)\ge n\})\equal{}1\plus{}\sum_{n\equal{}1}^{\infty}\frac1n\equal{}\infty.\]

Thanks Prof. Kent for your comments

I will keep working on this problem and may need your help then

Consider this inequality:
\[ \sum_{n\equal{}0}^{\infty}n\left(\mathbf{1}_{A_n}\minus{}\mathbf{1}_{A_{n\plus{}1}}\right)\le f\le \sum_{n\equal{}0}^{\infty}(n\plus{}1)\left(\mathbf{1}_{A_n}\minus{}\mathbf{1}_{A_{n\plus{}1}}\right)\]
Here I'm using the notation $ \mathbf{1}_A$ for the function that is $ 1$ on $ A$ and zero otherwise.

Note that $ \int\mathbf{1}_{A_n}\equal{}\mu(A_n),$ and consider the uses of summation by parts.

Thanks prof. Kent   :) , I got the problem !


If A ={a,b,c,d,)
then a relation R ={(a,b) (b,a ) (a,a)} is given to be symmetric and transitive only.

but when i choose (b,a) and then (a,b) then we should have (b,b) element.(according to definition of transitive).which is not in R .IS the answer wrong.it should not be transitive .is the relation must be true for every element.
pl explain the fallacy

[quote="jmerry"]That's not transitive. Wherever you got this from has a mistake.[/quote]
thanks sir


How many total days were there in the years 2001 through 2004?

That is 4 years. So it should be $ 365*4\equal{}1460$. But, 2004 was a leap year, so add 1 for a final answer of $ 1461$ days.


[b]How is Precalculus different than College Algebra and Trigonometry combined?[/b]

The College Board is closing out the option of taking trigonometry and replacing it with [url=http://www.collegeboard.com/student/testing/clep/ex_pcal.html]precalculus[/url].
I wanted to "CLEP-out" of college algebra and then test out of trigonometry.
However, the College Board his disabled the latter from being taken after June 30th of 2006.

I will now have to study precalculus instead of college algebra and then trignometry.

[b]Because of this I have a few questions:
[/b]
Why would studying precalculus be different than studying college algebra and trignometry?

Why would it be different than picking up a college algebra and trigonometry book and studying them?

What type of material would be covered that is different than both combined?

How is it all different?

It isn't really different.  I think that's why they won't offer both of them.

At most colleges, a course name like "College Algebra" just means "Precalculus." Precalculus is simply the current name for a course that covers advanced algebra and trigonometry. I have seen a very specific example: James Stewart has a textbook titled College Algebra, which I picked up at a used book sale after buying his textbook titled Precalculus: Mathematics for Calculus. When I compared the contents of the two books, I found out that they are almost identical. The two books have some difference in the order of chapters, and some difference in the length of a few chapters, but a professor could teach essentially the same course out of either book.

Sometimes College Algebra and Trig, combined (let's call that CATc) is the same as PreCalc and sometimes it's different.  CATc rarely includes a study of rates of change, but a lot of PreCalc courses include rates of change, especially where PreCalc is taught as a non-terminal course.

The syllabus the College Board is covering looks like traditional CATc content with a 'reform' flavor added (that's the rule-of-four and modelling stuff).  If you want to self-study this syllabus, the Hughes-Hallet et. al. precalculus book might be a good choice (I think it's called something like Functions Modelling Change).

[quote="tokenadult"]James Stewart has a textbook titled College Algebra, which I picked up at a used book sale after buying his textbook titled Precalculus: Mathematics for Calculus.[/quote]
For the most part, those books were not written by James Stewart. I know both of the primary authors, Saleem Watson and Lothar Redlin, quite well. They first wrote the precalculus book; after that, the publisher approached them about modifying it to be a college algebra book.

Yes, I was simply referring to James Stewart as the "publicity" author for those titles, as it is noted in the books themselves that the other authors had much to do with the content of the books.

[url=http://www.amazon.com/gp/product/0534434215/]Precalculus by James Stewart and others[/url]

[url=http://www.amazon.com/gp/product/0534405991/]College Algebra by James Stewart and others[/url]

I was able to find both of these books (newer editions of each than the ones I mentioned above) by searching on "Stewart" and each book's title  on Amazon.com, which I'm sure is what the publisher wants to happen. They appear to be pretty good books, which is a credit to Stewart's co-authors, although my favorite precalculus book is still David Cohen's [url=http://www.amazon.com/gp/product/0534402305/]Precalculus with Unit-Circle Trigonometry[/url].

Thanks for setting the record straight.

This is a good question. I have always been wondering this myself, just out of curiosity. The college i will be going to in the fall offers college algebra, trigonometry, and precalculus all as seperate courses. Since the school is on the quarter system, i'm assuming each class is a quarter long. So if you prepare for calculus by taking college algebra and trig, you'll spend 2 quarters, while you can complete prealculus in one quarter. I am pretty sure that precalculus is meant mainly for students who won't end their mathematics education at this level. There were alot of people in my "Trigonometry/Math Analysis" class in high school that had no intentions on going to AP calculus (if they weren't a senior) or taking more math in college. Most of the students wouldn't stop complaning about how hard the class was and how all the stuff we were learning wasn't relevant to the real world.  I think college algebra and trigonometry are meant more so for students who don't plan on taking more advanced math courses. Alot of people take college algebra just to meet the general education math requirement. I think because of these reasons, college algebra and trigonometry is a watered down version of precalculus which stresses more on manipulation and computational skills, rather than more conceptual things like how functions behave and whatnot. My brother went to the same high school as me and took up to Trig'math analysis as well. Then he took college algebra as a freshman in college because he had no intentions on going to study further math. He said that college algebra was really easy and that it was basically a mix between my school's Algebra 2 and Trig/Math analysis classes.

So the bottom line is:
Precalculus is a bit more rigorous and demanding in order to prepare you for calculus and higher mathematics

College Algebra/Trig is meant more as a terminal course and most of the students in this class don't plan on studying much math, so some topics deemed important for calculus might be de-emphasized.

or at least this is what i think...


geometry
geometric transformation
reflection
search
articles
angle bisector

"Prove that if $ P$ is a pedal-cevian point of $ \Delta ABC$ then so its isogonal conjugate."

NOTE: P is a pedal-cevian point if the feet of the perpendiculars dropped from P to the sides of $ \Delta ABC$ when joined form a cevian triangle.

In other words if $ D,E,F$ are the feet of the perpendiculars from $ P$ then
$ \frac {AF}{FB}. \frac {BD}{DC}. \frac {CE}{EA} \equal{} 1$
we have to prove that if $ P'$ is the isogonal conjugate of P and $ D', E', F'$ are the feet of the perpendiculars from $ P'$, then
$ \frac {AF'}{F'B}. \frac {BD'}{D'C}. \frac {CE'}{E'A} \equal{} 1$

[quote="1=2"]Define "isogonal conjugate".[/quote]

If line $ l$ is the reflection of line $ m$, along the angle bisector of an angle then lines $ l$ and $ m$ are said to be isogonal conjugates of each other with respect to the given angle.

In the given problem,$ \angle{PAB}\equal{}\angle{P'AC}$.

(Similarly prove the statement wrt angles $ B$ and $ C$.)

[quote="1=2"]Define "isogonal conjugate".[/quote]
You know, instead of being lazy, you could've spent one minute on google, or used mathworld or wikipedia. Since you are lazy, I'll spare you the minute.
http://mathworld.wolfram.com/IsogonalConjugate.html
http://en.wikipedia.org/wiki/Isogonal_conjugate
http://www.google.com/search?hl=en&q=%22isogonal+conjugate%22&btnG=Google+Search

This post is lying unanswered for a long time....
nobody wants to try???

Dear Mathlinkers,
an article (last of three) concerning the first synthetic proof of the PC-lines and the de Longchamps point has been put on my website
http://perso.orange.fr/jl.ayme vol. 6 Pedal-cevian lines go through the de Longchamps's point, p. 7
Sincerely
Jean-Louis


inequalities
SFFT
special factorizations

I guessed these right(it was logical), but I want to see how to do it algebrarically.

20. If $ x$,$ y$, and $ z$ are positive integers such that $ x^2\plus{}y^2\plus{}z^2\equal{}174$, what is the greatest possible value of $ x\plus{}y\plus{}z$?

23. If $ x$ and $ y$ are positive numbers less than 20 for which $ x\plus{}y\plus{}xy\equal{}76$, what is $ x\plus{}y$?

[hide="Hint for 20?"]A variation of the AM-GM inequality can be used here, if I'm not mistaken.[/hide]

I'm not sure about 20, but for 23:
[hide="solution"]using simon's favorite factoring trick and adding 1 to both sides of the equation, you get
$ xy \plus{} x \plus{} y \plus{} 1 \equal{} 77$
$ x(y\plus{}1) \plus{} (y\plus{}1) \equal{} 77$
$ (y\plus{}1)(x\plus{}1) \equal{} 77$
From here, its obvious that y is 10 and x is 6 (or vice versa, but it doesn't matter), x+y is [b]16[/b][/hide]

[hide="For twenty maybe"]
RMS-AM instead?
[/hide]

[quote="mathcountsfreak"]I'm not sure about 20, but for 23:
[hide="solution"]using simon's favorite factoring trick and adding 1 to both sides of the equation, you get
$ xy \plus{} x \plus{} y \plus{} 1 \equal{} 77$
$ x(y \plus{} 1) \plus{} (y \plus{} 1) \equal{} 77$
$ (y \plus{} 1)(x \plus{} 1) \equal{} 77$
From here, its obvious that y is 10 and x is 6 (or vice versa, but it doesn't matter), x+y is [b]16[/b][/hide][/quote]
I see. Thank you.

@7h3.D3m0n.117 & AstroPhys: I am not familiar with that method. Can you take it more further, considering that you are right about the method of approaching?

I don't think inequality is the best method for #20. One thing is this: the question specified that $ x,y,z$ are integers so we have to consider that it's possible with integers.

Thus, here is my approach

[hide="Solution"]
The key part in this problem is to note that there are only limited number of perfect squares less than 174. They go from 1^2 to 13^2. Now, starting from 13^2, we see if it's possible produce sum of 174.

It seems to me pretty clear that two possible triples are (4,49,121) and (25,49,100). This is done by brute-forcing and I don't know what's easier than that. But you can make it fast by adding the units digit first and see if that works out (so you can eliminate ones that don't and calculate the remaining ones).

So, my answer is 22 (7+5+10 = 22). [/hide]

Tell me if that's incorrect.  :)


articles
geometry
LaTeX
AoPS Books
\/closed

Hi I'm new to this site.   I was wondering: How would one make the most of this site? I know that there are a lot of problems and information, but is there a systematic way of doing them??

We know this is an issue with the new users, because of the huge amount of information available. Depending on your level there are a few choices you can do to ease your way around. So where would you place yourself?

PS Another quick way to get help is usually to address your self in the National forum, which in your case seems to be [url=http://www.mathlinks.ro/Forum/index.php?f=320]Canada[/url].

Hellstar,

You might just try playing around in the middle or high school forums (whichever fits your age/difficulty range best) and seeing which forum challenges you most at your own level.  When you're getting a lot of them right and a lot of them wrong, you've probably found you're home.

Check out the resources too.  There are Math Jams (free classes), articles in the resources area, as well as the AoPS books and classes.  There is even a LaTeX guide for those learning the math formatting language.

Thanks! I thought I ran out of math resources but I guess I thought wrong...


vector
Functional Analysis
advanced fields
advanced fields unsolved

Let $A$ be bounded surjective linear operator from Banach space $E$ to $E_1$.Prove that there exists a constant $\alpha>0$ such that if $B \in L(E,E_1)$ and $||A-B|| < \alpha$, then $B$ is surjective.

There should be some extra conditions on $A$. What are they? :)

I'll replace that $E_1$ with an $F$, because it just looks nicer :).

First of all, assume $F$ is just $E/G$, where $G\le E$ is a closed subspace of $E$, and $A$ is the canonical projection $\pi$ onto $E/G$. We can reach this situation as follows: $G$ will be $\ker A$. Since $A$ is onto, it induces an invertible transformation $\widetilde A: E/G\to F$. The composition $\widetilde A^{-1}\circ A: E\to E/G$ is precisely $\pi$. Hence instead of $A: E\to F$ we work with $\pi: E\to E/G$, and instead of $B: E\to F$ we work with $\widetilde A^{-1}\circ B: E\to E/G$, which we denote by $B$ again, in order to keep things simple. When $\|A-B\|$ is small $\|\pi-B\|$ will be small as well, and the new $B$ is onto iff the old one is.

We now want to prove that there is an $\varepsilon>0$ such that $\|\pi-B\|<\varepsilon$ implies the surjectivity of $B: E\to E/G$. In general, for $x\in E,\ \pi(x)$ will be denoted by $\bar x$.

Take $\varepsilon=\frac 12$ (this should be more than enough :)). Let $C=\pi-B$ (so $\|C\|<\frac 12$), and fix some $\bar x\in E/G$. We want $y\in E$ such that $(\pi-C)y=\bar x$. This $y$ will be of the form $x_0+x_1+\ldots$ such that $\|x_i\|\le\frac 12\|x_{i-1}\|,\ \forall i\ge 1$. In order to have $(\pi-C)(x_0+x_1+\ldots)=\bar x$, it suffices to have $\pi x_0=\bar x$, then $\pi x_1=Cx_0$, then $\pi x_2= Cx_1$, and so on. $x_0$ can be chosen arbitrarily. Since $\|C\|<\frac 12$ and in $\pi^{-1}(Cx_0)$ there are vectors with norm arbitrarily close to $\|Cx_0\|$, we can choose $x_1$ with norm $<\frac 12\|x_0\|$, and I think it's clear how to continue.


$2 * 1$ dominos are used to tile an $m * n$ square, except for a single $1 * 1$ hole at a corner. A domino which borders the hole along its short side may be slid one unit along its long side to cover the hole and open a new hole. Show that the hole may be moved to any other corner by moves of this type.

I remember posting this myself, and there was at least one other thread discussing it. Try searching the 'Combinatorics' section for Russian problems. There are plenty of good keywords (like "domino", for instance).


greatest common divisor
number theory
prime numbers
prime factorization
relatively prime
least common multiple

Let $a,b,c$ are integer numbers. Let $a.c$ is divisible by $b$. Prove that $(a,b).(b,c)$ is divisble by $b$, too, where $(x,y)$ denotes the greatest common divisor of $x$ and $y$.

Let $p_{1}, p_{2},\dots,p_{n}$ be all the different prime numbers that appear in the prime factorization of either $a$ or $b$ or $c$. Then we can write

$a=p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{n}^{a_{n}}$
$b=p_{1}^{b_{1}}p_{2}^{b_{2}}\dots p_{n}^{b_{n}}$
$c=p_{1}^{c_{1}}p_{2}^{c_{2}}\dots p_{n}^{c_{n}}$

where some of the exponents $a_{i}, b_{j}, c_{k}$ can be zero.

Then the initial condition becomes

$(\forall i) a_{i}+c_{i}\geqslant b_{i}$

And we must prove

$\min\{a_{i},b_{i}\}+\min\{b_{i},c_{i}\}\geqslant b_{i}$

But that's very simple: The possible values for LHS are

$a_{i}+b_{i}, a_{i}+c_{i}, 2b_{i}, b_{i}+c_{i}$

The first, the third and the fourth are obviously not less than $b_{i}$, since all the numbers involved are non-negative, and the second is not less than $b_{i}$ by the given condition. QED

Yes, but i think this problem can prove by easy way. Because we have a simple case of one is: "If $ab$ is divisible c and $(a,c) = 1$ then $b$ is divisible $c$. I think we can follow the proof of above problem that prove my my problem but now i have still stucked.

[hide]
There exist integers $m,n,x,y$ such that $am+bn=(a,b)$ and $bx+cy=(b,c)$ by a well-known theorem. So their product is \[(am+bn)(bx+cy)=b(amx+bnx+cny)+(ac)my\] Given $ac$ divisible by $b$, the above expression is also divisible by $b$.
[/hide]

Yes, if we uses that property then result is obvious. But I wait for the solving is similar with proof Fundamental theorem of arithmetic:
[b]Theorem[/b]: A natural number which divides the product of two natural number and is relatively prime to one of them is a divisor of the other.
Proof
Let $a$ and $b$ be two ralatively prime numbers and $c$ a natural number such that $b|ac$. The number $ac$ is divisible by each of the numbers $a$ and $b$, therefore, it is also divisible by their least common multiple $[a,b]=\frac{ab}{(a,b)}=ab$. Thus $ac=tab$, where $t$ is an integer, whence $c=tb$, and therefore $b|c$


function
vector
geometry
combinatorics proposed
combinatorics

The numbers $ 1,\sqrt{2},\sqrt{3},\sqrt{2}\plus{}\sqrt{3}$ are written at the blackboard. It is allowed to increase or to decrease any number by the difference of any other two numbers multiplied by any rational number. Is it possible to obtain numbers $ 1,\sqrt{2},\sqrt{3},2\sqrt{2}\plus{}3\sqrt{3}$ at the blackboard after several such operations?

[u]Solution.[/u] Consider the set $ S \equal{} \{a \plus{} b\sqrt {2} \plus{} c\sqrt {3}|a,b,c\in\mathbb{Q}\}$.  The function $ f: S\rightarrow\mathbb{Q}^3$ given by $ a \plus{} b\sqrt {2} \plus{} \sqrt {3}\mapsto (a,b,c)$ is a bijection.  $ f$ maps $ \{1,\sqrt {2},\sqrt {3},\sqrt {2} \plus{} \sqrt {3}\}$ to $ U \equal{} \{(1,0,0),(0,1,0),(0,0,1),(0,1,1)\}$, and maps $ \{1,\sqrt {2},\sqrt {3},2\sqrt {2} \plus{} 3\sqrt {3}\}$ to $ V \equal{} \{(1,0,0),(0,1,0),(0,0,1),(0,2,3)\}$.  The given operation translated to correspond with the new setting of the problem (on the vector space $ \mathbb{Q}^3$) does not change the area of triangles, and so does not change the volume of tetrahedrons. But $ \text{vol} (\text{conv}(U))\neq \text{vol}(\text{conv}(V))$, and so the desired endstate cannot be reached. $ \Box$


Support
geometry
geometric transformation
reflection
blogs

I am so lost by this question.  Does anyone understand?

What is the primary mechanism by which CO2 contributes to global warming?

a.  by allowing more infrared radiation to reach Earth's surface
b.  by decreasing the amount of O2 in the atmosphere
c.  by increasing the amount of O2 in the atmosphere
d.  by incraseing planetary photosynthesis it allows for more life on Earth, which means an increase in the production of metabolic heat
e.  by preentin the radiation of heat from Earth to space

I am sorry, the word was preventing, not preentin.  Thank you!

The answer they are looking for is e.  I agree with the others.  While CO2 does allow radiation (visible) to reach the earth, so do the other gases.  The critical aspect for the effect that is the focus of the question is the prevention of the heat (infrared) from escaping to space.

It is interesting that this question assumes, with justification or discussion that CO2 is causing "Global Warming."  This is not known and there are many factors occurring which a good scientist would agree preclude logically determining if man's activity is warming the planet.  The largest of these is the fact that the assumption of initial steady state is wrong.  The earth is at the end of an ice age.  All the ice caps will melt.  According to the record read in the deposits on earth they did last time, before large amounts of human activity.  What we don't know is how fast would the change take place without man's burning CO2 trapped in peat, coal, oil and the rest?

No one disputes that putting CO2 into the atmosphere increases the greenhouse gas effect (absent some equilibrium entrapment in the oceans or some other effect that we haven't found yet).  The question is relative size of the effect.

For example if I go to the Pacific Ocean and cast three stones, of less than one kg mass each into the ocean and notice the tide coming in, I am correct to say that my activity in casting each stone will result in a higher tide than would occur without my activity.  However, we understand this situation is such that the effect of my three stones will not be measurable, certainly not causing any additional harm to those affected by the high tide.  (Assuming that I didn't take the stones from a critical place in a breakwater or something equally absurd.)

The problem with the global warming discussion is that we don't understand the effects enough to determine if the effects from human activity is significant.  There is lots of material for those already convinced of a position to say that their position is supported, but nothing overwhelming one way of the other.

Some say that we should err on the "safe" side.  Ok, if I agree, which is the "safe" side.  The one that says we should impact the global economy, killing people in the third world (they are very near dying anyway, but every increase in the costs of energy will have a body count.  It just isn't in the news.)  I think the "safe" position is not to change things until we are reasonably sure what is happening.  If climate change is really affected by human activity, the proposals so far are meaningless.  We need to focus the world on getting essentially off burning Carbon based fuels, not rolling them back to some level where the economic impact is so low on a majority of nations that they support "socking it to the US."  Japan hasn't grown, Europe hasn't grown, Russia has imploded, China and the other developing nations are exempt.  The only nation the agreements would significantly impact is the US.

However,  we now see that the Europeans (once again) are all talk, no action.  When they are piloting their greenhouse gas reductions, they find that it will be hard and are not meeting even their minor, short term trial goals.

Much too much from one questions, but young fellows, you need to be on watch for this type of presentation that puts the major point behind the question in an assumption that you aren't even thinking about.  (Remember in the Wizard of Oz, "Pay no attention to the man behind the curtain.")

Twenty years after we started killing many by driving up the cost of refrigeration, we now have an extremely large ozone hole.  Rather than say that perhaps the Montreal Protocol was based on bad science, the position is that it must be taking longer to heal than we suspected.  The science was never there that the ozone hole was caused by man.  It existed from the first measurement.  The reason that the science took off was DuPont (who invented Freons) figured out that the old, easy to make initial Freons were off patent protection and the just invented ones were more expensive to make and had no reason anyone would buy them in large quantity.  So they got the world to make it illegal to make them.  What a windfall for big business.  They make more money and the world kills far more than have died in Iraq.  The left can go to bed not knowing that what they really did was make the world a better place for the big US business that outsmarted them and killed a bunch of people in the third world.

A relatively large factor in global warming is from livestock flatuence and burping. Whenever they fart or burp, it releases lots of methane, which is even worse than CO2. This doesn't really have to do with the question, but I thought it was interesting.

I'd say it's undisputed scientific fact that the greenhouse effect has a substantial effect on Earth's surface temperatures- the blackbody temperature for an object at approximately our distance from the sun, with the same albedo, is close to $ \minus{}20^\circ C$, and the average temperatures here are significantly higher. Most of this effect is due to carbon dioxide, simply because there's a lot more carbon dioxide than there is of the other major greenhouse gases.

The question assumes nothing about the human role- only that a certain amount of carbon dioxide in the air is necessary so the oceans don't freeze over.

exactly e :D  :D --co2 reflects back the heat

Sorry to resurrect a slightly old thread, but I thought I'd correct a small error of jmerry's: the largest single contributor to the greenhouse effect is actually water vapor.  Because the surface of the Earth is largely covered by water, however, water vapor is usually considered a climate feedback rather than a forcing -- the only way that we can change the amount of water vapor in the atmosphere is to change the temperature, because increasing evaporation or precipitation in one location will be very quickly off-set (hours to days) by increased precipitation or evaporation elsewhere.  (In fact, it is a positive feedback on temperature change: increased temperatures increase the amount of water vapor in the atmosphere, and this increases the strength of the greenhouse effect.)

Other greenhouse gases behave differently: methane has a very large impact per unit as a greenhouse gas but there are much smaller amounts of it in the atmosphere (measured in parts per billion) and it breaks down relatively quickly into carbon dioxide.  The reasons that CO$ _2$ is the major greenhouse gas to worry about from the perspective of human-caused climate change is that we emit huge amounts of it (relative to, say, methane or ozone) and it is very long-lasting in the atmosphere (a hundred or more years) so its impact is longer-lasting, too.  (There are also other bad effects such as ocean acidification, but they [i]really[/i] have nothing to do with the original purpose of this thread (although maybe as much to do with the general subject of this forum) :).)

And, while I'm rambling on, I might point out that if you actually want to understand climate science in any detail (rather than, say, reading absurd screeds like wizard2378's above), a really wonderful resource is the RealClimate blog ([url=http://www.realclimate.org/]realclimate.org[/url]), run by actual climate scientists and ranging from detailed discussions of the state of knowledge of climatology to refutations of climate anti-science of that kind that wizard seems to enjoy.  (A good place to start is the [url=http://www.realclimate.org/index.php/archives/2007/05/start-here/]Start here[/url] link at the top of the page.)

it doesn't matter 4 amount--CO2 is 10 times more effective than H2O

[quote="rituraj007"]...CO2 is 10 times more effective than H2O[/quote]

Do you have a source for that figure or is that just out of your head?

Even if it's true, there is vastly more water vapor in the air than carbon dioxide. If the amounts were comparable, clouds and rain would be unthinkable.

At the surface, water vapor can be up to 4% of the air, and is probably 30 times as common as carbon dioxide on average in the troposphere.


geometry
area of a triangle
Heron\u0027s formula
AMC

A circle is inscribed in a triangle whose sides are 10,10,12. A second, smaller circle is inscribed tangent to the first circle and to the equal sides of the triangle. Compute the radius of the smaller circle.

Solution:

[hide]Let AB = 10, BC = 12, and CA = 10. Let the center of the incircle be O, and the center of the smaller circle be P. Let r be the radius of the incircle, and s be the radius of the smaller circle. By Heron's Formula,

s = (10+10+12)/2 = 16

A = Sqrt(s(s-10)(s-10)(s-12)) = Sqrt(16*6*6*4) = 48

A = rs = 16r

48 = 16r

r = 3

Draw the perpendicular from A to BC and from O to AB, with intersection points M and N, respectively. As BNOM is a kite, BN = BM = 12/2 = 6. Thus AN = 4.

Let the perpendicular from P to AB be Z. Now note that triangles APZ and AON are similar. Now fiddling around with the similarity should do it, although I am tired now and can't seem to figure out how to finish it off.[/hide]

Two criticisms of fierytycoon's solution:

[hide]1) Areas of isoceles triangles are much more easily computed using a perpendicular than herons formula.  Also

2) Once you have the radius of the incircle, you don't need to go through the stuff with the kite.  Instead, note that the two circles will have a common tangent line parallel to BC.  That will transform the little circle into the incircle of a little triangle which is similar to the big triangle.  We also can find the height of the little triangle by subtracting twice r from the height, and it follows immediately from similarity.[/hide]

[color=green]I am curious if your method would lead to an answer -- I'm not entirely convinced you ended up with equations you can solve.  That idea of drawing perpendiculars is a very useful one, though.  Just not for this particular problem.  But it pops up elsewhere quite often.[/color]

hmm
yea
Heron's formula gets quite messy tho


I tried doing a little bit of searching on google and this forum and haven't found quite what I'm looking for.

Question: How do you find the nth term of the Fibonacci Sequence?

I'm pretty sure there's a formula out there. Thanks in advance.

Try looking for Characteristic Root and Characteristic equation.

Basically substitute $r^n$ for $S_n$ and then divide by $r^{n-2}$, so your left with a quadratic.

Now $c_1r_1^{n}+c_2r_2^{n} = S_n$, where $c_1$ and $c_2$ are constants determined by initial conditions and the r's are the roots of the quadratic, the characteristic equation.

If you have only 1 unique root, then $S_n = c_1r_1^n+c_2nr^n$

These identities can be proven through induction.

[hide="click for solution"]$S_n = \frac{1}{\sqrt{5}}[(\frac{1 + \sqrt{5}}{2})^{n+1}(\frac{1 - \sqrt{5}}{2})^{n+1})]$[/hide]

$\frac{1}{x}=\frac{x}{x+1}\implies x^{2}-x-1=0\implies x=\phi\,,\,\varphi$

$x^{3}=x^{2}+x=2x+1$

$x^{4}=2x^{2}+x=3x+2$

$x^{n}=F_{n}x+F_{n-1}$

So we have

$\phi^{n}-\varphi^{n}=F_{n}\left(\phi-\varphi\right)\implies F_{n}=\boxed{\frac{\phi^{n}-\varphi^{n}}{\phi-\varphi}}$

I believe this is called Binet's Formula.

Your best bet is searching.  This topic is rather complicated.

Essentially, if you have a recurrence of the form $a_{n+1} = A a_n + B a_{n-1}$ (what's called a "second-order linear recurrence") then you can find a closed form for $a_n$ in terms of the roots of the quadratic equation $x^2 = Ax + B$.  (For the Fibonacci sequence, $A = B = 1$ and so the solution relies on the solutions to $x^2 = x + 1$, which are $\phi = \frac{1 + \sqrt{5}}{2}$ and $1 - \phi$.

Proving rigorously that this works requires the method of generating functions, which I don't think is a good topic for Intermediate.

just use polynomials.
let $F(x)=a_0+a_1x+a_2x^2+....$
here $a_j$ are the fibonacci numbers. note that we don't care what $x$ is, we just want to use $F(x)$ to find the $a_j$
$xF(x)=a_0x+a_1x^2+a_2x^3+a_3x^4....$
$x^2F(x)=a_0x^2+a_1x^3+a_2x^4+a_3x^5....$
now if you group these you get:
$F(x)-xF(x)-x^2F(x)=a_0+(a_1-a_0)x+ (a_2-a_1-a_0)x^2+(a_3-a_2-a_1)x^3+(a_4-a_3-a_2)x^4+.....$
the importance of the fibonacci sequence is that $(a_2-a_1-a_0), (a_3-a_2-a_1), (a_3-a_2-a_1),....=0$ so the above's equation is so simple.
$F(x)-xF(x)-x^2F(x)=a_0+(a_1-a_0)x$
we also know that $a_0=a_1=1$ so $(1-x^2-x)F(x)=1$ or
$F(x)=\frac{1}{1-x-x^2}$. now if you factor out the denominator you get:

$F(x)=\frac{1}{[1- \frac{-2x}{1-\sqrt{5}}][1-\frac{-2x}{1+\sqrt{5}}]}$
now with use of some algebra you get:
$F(x)=\frac{5+\sqrt5}{10}*[\frac{1}{[1+\frac{2x}{1-\sqrt{5}}]}]+\frac{5-\sqrt5}{10}*[\frac{1}{[1+\frac{2x}{1+\sqrt{5}}]}]$
the above parts that are in a bracket can be rewritten by the summation formula:
$F(x)=\frac{5+\sqrt{5}}{10}.\sum_{n=1}^{\infty} [\frac{-2x}{1-\sqrt{5}}]^{j}+\frac{5-\sqrt{5}}{10}.\sum_{n=1}^{\infty}[\frac{-2x}{1+\sqrt{5}}]^{j}$
now if you group the similar powers of $x$  you get:
$F(x)=\sum_{n=1}^{\infty}[\frac{5+\sqrt{5}}{10}.[\frac{-2x}{1-\sqrt{5}}]^{j}+\frac{5-\sqrt{5}}{10}.[\frac{-2x}{1+\sqrt{5}}]^{j}]$
now remember we started the problem by saying
$F(x)=a_0+a_1x+a_2x^2+....$
so now we have two differen formulas for $F(x)$ these two MUST equivelant to each other, especially the coefficents of the same powers of $x$. so we conclude that:
$a_j= \frac{5+\sqrt{5}}{10}.[\frac{-2x}{1-\sqrt{5}}]^{j}+\frac{5-\sqrt{5}}{10}.[\frac{-2x}{1+\sqrt{5}}]^{j}$
now we are almost there, all we need to do is to replace some of the stuff up there by equivalent of them.
$\frac{5+\sqrt{5}}{10}=\frac{1}{\sqrt{5}}.\frac{1+\sqrt{5}}{2}$ also $\frac{5-\sqrt{5}}{10}=\frac{-1}{\sqrt{5}}.\frac{1-\sqrt{5}}{2}$ and $\frac{-2}{1-\sqrt{5}}=\frac{1+\sqrt{5}}{2}$, and $\frac{-2}{1+\sqrt{5}}=\frac{1-\sqrt{5}}{2}$
if we replace these values for the $a_j$ formula we get:
$a_j=\frac{[\frac{1+\sqrt{5}}{2}]^j- [\frac{1-\sqrt{5}}{2}]^j}{\sqrt{5}}$


trigonometry
induction
irrational number
algebra unsolved
algebra

It is easy to show that equality holds when $m=\frac{1}{\sin(n+1)}.$  For example, working from the last term to the first, we get
$\frac{1}{\sin(n+1)}\cdot\frac{\sin n}{\sin^2 1}-\frac{1}{\sin n\cdot\sin(n+1)}=\frac{1}{\sin(n+1)}\cdot\frac{(\sin n-\sin 1)(\sin n+\sin 1)}{\sin n\cdot\sin^2 1}=\frac{\sin(n-1)}{\sin n\cdot \sin^2 1},$
and so on. Therefore, if $m=1,$ we get $\sin(n+1)=1,$ and thus $n=89,449,\cdots ,10529.$
If $m=2,$ we get $\sin(n+1)=\frac{1}{2},$ and thus either $n=29, 389, \cdots, 10469,$ or $n=149, 509, \cdots 10589.$
It is easy to show that the three arithmetic sequences above do not have a common term. As such, each sequence having 30 terms, we have a total of 90 acceptable values of $n.$

[quote="N.T.TUAN"]Let $m$ be equal to $1$ or $2$ and $n<10799$ be a positive integer. Determine all such $n$ for which $\sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}=m\frac{\sin{n}}{\sin^{2}{1}}$.[/quote]

$\sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}$

$=\frac{1}{\sin 1}\sum_{k=1}^{n}\frac{\sin 1}{\sin{k}\sin{(k+1)}}$

$=\frac{1}{\sin 1}\sum_{k=1}^{n}\frac{\sin (k+1-k)}{\sin{k}\sin{(k+1)}}$

$=\frac{1}{\sin 1}\sum_{k=1}^{n}\frac{\sin (k+1)\cos k-\cos(k+1)\sin k}{\sin{k}\sin{(k+1)}}$

$=\frac{1}{\sin 1}\sum_{k=1}^{n}\left(\cot k-\cot (k+1)\right)$

$=\frac{\cot 1-\cot(n+1)}{\sin 1}$

$=\frac{\sin n}{\sin^21\sin(n+1)}$

$m=\frac{1}{\sin(n+1)}$

$\sin(n+1)=1$ or $\sin(n+1)=\frac{1}{2}$

Now,the values of n can be easily computed.

The truth is, one would be hard-pressed to guess what $m$ is, and the statement of the problem makes it seem much harder than it is. In reality, I would guess that most people will try the first couple cases of $n=1$ and $n=2,$ realize what $m$ should be, and then try to prove it in general. Once you get there, the simplest and most direct way (and nothing against pankajsinha's approach, which is very nice) is to use induction. That's what I did but I tried to keep the typing to a minimum so I included only the last step of the induction process.

Solving further
$n+1=\frac{(4k+1)\pi}{2}$  or  $n+1=p\pi+(-1)^p\frac{\pi}{6}$
$k$ and $p$ being integers.

In both the cases,$n$ is an irrational number and not a positive integer as claimed in the question and $n$ being a positive integer is also some kind of requirement since the question is in terms of $\sum_{k=1}^n$.

So answer should be that for [b]no value of [/b]$n$ the statement $ \sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}=m\frac{\sin{n}}{\sin^{2}{1}} $ is possible.

Except that these numbers are in degrees, not radians... At least that's how I understood the intent...

If it's in degrees then the sum is only defined for $0<n<179$.


Let $ 1>\equal{}a,b,c>\equal{}0$

Prove that
$ \frac{a}{bc\plus{}1}\plus{}\frac{b}{ca\plus{}1}\plus{}\frac{c}{ab\plus{}1}<\equal{}2$

[hide]My solution:
From hypothesis we have two following inequalities:
(1-a)(1-b)(1-c)>=0, which is equivalent to: 1+(ab+bc+ac)>=abc+(a+b+c)(1)
(1-ab)(1-bc)(1-ac)>=0,which is equivalent to: 1+abc(a+b+c)>=(ab+bc+ac)+(abc)^2(2)
Now adding (1) and (2) side by side, we get:2+abc(a+b+c)>=(a+b+c)+(abc)^2abc(
Therefore, we have: 2+2abc>=abc(3-a-b-c)+(a+b+c)+(abc)^2>=(a+b+c)>=0(since 3-a-b-c>=0 because of the restriction)(3)
From the inequality given we will infer that: LHS=<(a+b+c)/(abc+1)(since abc+1=<bc+1,ab+1,ac+1)
Now it's sufficient to prove: (a+b+c)<=2(abc+1), which is true because of (3)(Q.E.D)
@bibobeo: Pleased check it carefully for me to see if it's correct. I haven't done inequality for a long time :oops: .Long time no see, my friend! :wink: .Nice problem though!
[/hide]

[quote="t0rajir0u"][hide="Solution"] The LHS is a sum of convex functions in any variable, therefore convex.  It attains its maximum at one (or more) of the vertices $ (a, b, c)$ where $ a, b, c \in \{ 0, 1 \}$.  At any of the vertices $ (1, 1, 0)$ and permutations it attains the maximum value $ 2$.  (Its value at the other vertices are $ 0, 1, 1.5$.) [/hide][/quote]

a,b,c here are real not integer :D

[hide="@ghik"]
Nice solution, my friend. I almost forget that technique :D[/hide]

[quote="bibobeo"]a,b,c here are real not integer :D[/quote]
His solution is still valid. Convexity allows you to only consider the endpoints.

or you can juzt use some calculations to prove that
$ f(a,b,c) \leq f(1,b,c) \leq f(1,1,c) \leq 2$
with a = max {a,b,c}

[hide]
The inequality will follow from $ a\plus{}b\plus{}c\leq 2(1\plus{}bc)$, because then this would give $ \frac{a}{1\plus{}bc}\leq \frac{2a}{a\plus{}b\plus{}c}$ and similar inequalities for the other variables.

However, the inequality $ a\plus{}b\plus{}c\leq 2(1\plus{}bc)$ is clear from $ a\leq 1$ and $ b\plus{}c\leq 1\plus{}2bc\iff 0\leq bc\plus{}(1\minus{}b)(1\minus{}c)$.
[/hide]

[hide="proof"]The inequality could look like this : $ \frac{a}{1\plus{}bc}\plus{}\frac{b}{1\plus{}ca}\plus{}\frac{c}{1\plus{}ab}\leq \frac{a\plus{}b\plus{}c}{1\plus{}bc}\leq 2$
The second becomes $ 2\plus{}2bc\geq a\plus{}b\plus{}c \implies (1\minus{}b)(1\minus{}c)\plus{}(1\minus{}a)\plus{}bc\geq 0$ wich is true.The equality case is when two of the numbers are $ 0$ and the third is $ 1$[/hide]


In textbooks explaining Mitosis and Meiosis, there are commonly four chromosomes in the nucleus of the original cell. Two are the same size, and another two are smaller. This may seem trivial and silly of a question, but are the chromosomes of the same size the same chromosome? Meaning the two big ones are say, for example, chromosome 3, and the other two are chromosome 23? And if so, where do the other two come from? That's the entire diploid thing, right? Well, where do ther other two come from?

Yes, two of the same size are the same chromosome, since a cell is diploid. The other two are presumably a different chromosome. I think the reason that textbooks include two pairs is to illustrate the crossing over that can occur in Meiosis I.


Shannon drew all of the diagonals in a regular octagon.  How many intersections were there?

cool problem, and i've found cool patterns, but no solution yet...

[hide]
For every set of 4 vertices you pick, there will be one intersection point.
Thus, the answer is $ \binom{8}{4} \equal{} \boxed{70}$[/hide]

Can the intersection points be a vertex?  I'll assume not

[hide="thoughts"]
Either make a ridiculously good diagram or use a PIE computation if possible
btw, I think this problem was posted before somewhere
[/hide]

@alkjash - that doesn't work with a regular octagon, as four diagonals intersect at the center

EDIT: I got 59 with my bad drawing.

and a hexagon only has 13 intersections.
13 isn't 6C4[/code]

hmm...

triangle -> $ \binom{3}{4}$=0
square -> $ \binom{4}{4}$=1
pentagon -> $ \binom{5}{4}\minus{}2\binom{5}{6}$=5
hexagon -> $ \binom{6}{4}\minus{}2\binom{6}{6}$=13
heptagon -> $ \binom{7}{4}\minus{}2\binom{7}{6}\minus{}\binom{7}{8}$=21

before I go on, does a regular heptagon have 21 intersection points?

alan- no on the vertex thing
i have a good drawing, but its really hard to count
remind me to show you at practice
oh, and i have no idea if it was posted before i'm sure someone wrote it before me, but i wrote the problem (or at least i haven't seen it anywhere)

oh, pfft, I mixed it up with the problem that asks for the number of regions formed.

skill alan, skill
i spose if you've posted on the problem i cant use it in the MC test i'm writing??? and i spose that if i cant solve it i cant use it

ok...

I just drew a regular heptagon (or a bad representation) and got 31 intersection points, so am I just a bad drawer or does this "pattern" not work?

and now it yields only 23 intersection points for an octagon, so yeah, this is kinda messed up...

no kidding.  if you count from one vertex, there are 27, but idk where to go from there

Yea, good point...

Well, I carefully drew it with a geometry program and only counted 49...

an answer is helpful, but i'd still like to have a real solution... but maybe its too complex

I get 58 with this picture.
[img]http://i257.photobucket.com/albums/hh202/xHassassin/Hexagon.png[/img]

can you by any chance enlarge the picture to make the intersections easier to count?

EDIT: Assuming I'm thinking correctly, the answer must be odd.

Yeah ok, however, know this.
There are four points were three diagonals intersect.
[img]http://i257.photobucket.com/albums/hh202/xHassassin/Hexagon-1.png[/img]

peijin, could you print that out to bring to practice tomorrow (or wed.) btw, its allison

meh, the diagram is a tiny bit messed up, and just because 3 lines look concurrent doesn't mean they are.  you have to prove it.

agreed, but it is a start.  did you ever figure out how many intersections in a septagon?  i think it could be a function of the diagonals.  idk, just seems like it should be.

for polygons with no lines of symmetry its just verticesC4, but an octagon has a ton.  If you think about hexagon that way, its 6C4-6/2, and 6 is lines of symmetry.  that would give 62 for ocagon, but i think you need to subtract more for triple crosses.  Maybe subtract 3 (number of triple crosses) that would get close to peijins answer.

[hide]I liked your picture Peijin, but I think this one is more accurate.  If I am going to pay the big bucks for the good geometry programs,I might as well use them!!

It appears there are 8 points where three diagonals are concurrent and one where 4 are.

Each point with 3 diagonals through it has been counted $ \binom{3}{2} \equal{} 3$ which means we overcounted each of the 8 twice.  That is 16 overcounted points. The point it the center has been counted $ \binom{4}{2} \equal{} 6$ so it was overcounted 5 times.

So the initial answer of $ \binom{8}{4} \equal{}70$(Which was my !st answer too) has been overcounted by $ 21$ leaving us $ 49$ points.

Using my picture and counting I get $ 49$ also[/hide]

PS When did AoPS become the Clarke forum???

[hide="uh..."]
Maybe, you could find the number of diagonals, choose two, and then subtract the diagonals that intersect at the vertex, the parallel diagonals, and the diagonals that are non-congruent but are parallel?  :read: [/hide]

Dude I want to join a Clarke forum...also this looks like it's from a National Team Round (don't know which year, maybe the very first one).

Anyway, try to think about why those concurrency points exist. The answer lies in the various symmetries of the octagon. I don't think there is an algorithm for doing this in the general case, and I'm pretty sure allyrb is incorrect; I believe there may still be points of concurrency in (2k+1)-gons.

I don't know if there are any points of concurrency in 2k+1 gons, but here is a nonagon and there are none.  there are exactly $ \binom{9}{4} \equal{} 126$ points of intersection in the nonagon....yes I counted them :D

Check it out.

o yeah then subtract points of concurrency  :wacko:

For each point of concurrency, you can't simply subtract each; you have to subtract the number of lines that intersect it, then add back 1.

Like if four lines passed through the same point, you would subtract 3 from $ \dbinom{n}{4}$.


Simplify the following (get it in the form $a+b\sqrt{c}$).
$(161-72\sqrt{5})^{1/4}$

And please explain it.  I got a really involved way to do it but, like I said, it's complicated and there's probably some short nice way to do it.

RC-7th wrote:[hide] is probably in the form , because 5 is a prime,but there are some exceptions to this rule.

Solving

Gives . Squaring that gives , but the first expressions negative, and we're looking for the positive root, so I'm stuck.

Edit: It seems  works. Then we find the squareroot of that[/hide]

I think you mixed up your a's and b's. Because when I did it, I got a=+/- 9 and b=-/+4 which gives two expressions:

9-4sqrt(5) and -9+4sqrt(5)

The second one obviously can't work since -9 has to equal the sum of two squares. However, 9-4sqrt(5) works fine. YOu can factor it into [2-sqrt(5)]^2. Therefore the answer to this question is: 2-sqrt(5).

Cheers!

[quote]hey shobber, did u guess?[/quote]
Well first we need to know that this kind of question can alwayse be turned in to this form. That is from $\sqrt{a+\sqrt{b}}$ to $\sqrt{(p+q)^2}$.
So in this question, we can get $\displaystyle 72\sqrt5=2\cdot{36}\cdot\sqrt5$.
Then we can try to find $P$ and $q$ such that:
\[p^2+q^2=161\]
\[p\cdot{q}=36\sqrt5\]
Solve the equation system then you can get what you want.

For me, i simplify to $\displaystyle 72\sqrt5=2\cdot{36}\cdot\sqrt5$, then i just guessed that the two numbers should be $9$ and $4\sqrt5$.

O there is another trick!
Since we want to ge $(a+b\sqrt5)^2$, so if notice no matter what value $b$ takes, the last digit of the square of this term is always $5$ or $0$!
Therefore the last digit of $a^2$ must be $6$ or $1$.
So... ...