MATHCOUNTS
geometry
analytic geometry
parallelogram

So I encountered two problems recently (they are not problems from MathCounts but are very similar) where I used the shoelace method and it didn't seem to work.  For one of them, I tried using a different method after using the shoelace method because my answer was not one of the multiple choice answers.  For the second one, I didn't catch my mistake.

So, the first problem is:
"The area of the triangle bounded by the lines y=x, y=-x, and y=6 is:"
Then there was multiple choice.  Obviously, the triangle's vertices' coordinates are (-6,6), (0,0), and (6,6).  Also, I realize that you can solve this by drawing a 45-45 right triangle of leg length 6 and by deducing that the leg length of the original triangle is therefore 6 sqrt 2 (then square that and divide by 2 to get 36 as the area).
However, I initially tried the shoelace method, like this:
-6 6
0 0
6 6
-6 6
and I did the following shoelace operations: (left to right) (-6x0)+(0x6)+(6x6), minus (right to left) (6x0)+(0x6)+(6x-6).  All that gets me is 72.  You may be saying that I could divide by 2 or should divide by 2 because it's a triangle.  However, the shoelace method isn't supposed to be like that.  It's evidenced in the next problem...

The second problem is:
"The area of quadrilateral ABCD=10.  The vertices of ABCD are A(4,1), B(7,4), C(9,3), and D(X,0).  If point D is in Quadrant I and X>0, find the value of X."
Again, I tried the shoelace method and failed, but I didn't notice I had failed.
I did this:
4 1
7 4
9 3
X 0
4 1
And solved 10=|((4x4)+(7x3)+(9x0)+(Xx1))-((1x7)+(4x9)+(3xX)+(0x4))|=|37+X-43-3X|=|-6-2x|, where || means absolute value.  After solving it all, I get -8 or 2, for which I know 2 must be the answer since the vertex D is in Quadrant I.  After I found out that this answer was incorrect, I just drew the picture really fast and I noticed that even with rough drawings, one can see that D cannot have the coordinates (2,0) and should have the coordinates (6,0).  However, the greatest mystery to me is how X=6 doesn't work in the Shoelace Method.

Someone please help, I'm confused...

Umm first of all, the second answer is 7...not 6

Second, and more importantly, your version of the "shoelace method" is wrong.

Oh, you forgot that the first one is (-6X0)+(0X6)+(6X6)+[b](-6X6)[/b]. you forgot the last part of the shoelace.  The same goes for the right-to-left direction.

Shoelace... is derived using matrices... In doing so, it ends up having to divide by 2 EVERY TIME.

Here's a site that shows some examples, but does not derive it.

[url]http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/url]

Where can I get more info on this shoelace method? The pdf doesn't work for me.

the future, you're flat-out wrong, the answer to the second one is 6.  If you use the distance formula for D(6,0), AB=CD, as it should be in all parallelograms ABCD.  Attempting to have D(7,0) as the answer does not result in AB=CD, so therefore it is wrong.  Check your math before posting it.
However, the second part of your post (which is the fact that my method of the shoelace method is wrong) may be true.

infinity4ever, what you said seems to be true at first, but at a closer glance adding that 'last step' you suggested does absolutely nothing but add another calculation.  This is because in the shoelace method, one lists all the points of a polygon clockwise or counterclockwise, and finishes by repeating the last point.  This means that the last 2 pairs of numbers will always be the same.  I'll show you why you're wrong with a hypothetical triangle of vertices A(2,1) B(3,8) and C(4,5).  Here's the setup:
2 1
3 8
4 5
2 1
When you apply your so-called 'last step,' this is how it looks:  ((2x8)+(3x5)+(4x1)+(2x1))-((1x3)+(3x4)+(5x2)+(2x1)).  Notice that the last parts of both parts of the subtraction end with '(2x1))' which means that adding a last step does nothing because it's like adding b to a-c.  a-c=(a+b)-(c+b), where b is your 'last step.'  Therefore you haven't pointed out the flaw in my method.

mathcrazed, you have given me the most useful post.  The ironic thing is that I looked into the link you sent me, but that is the exact same link I used to learn the shoelace method.  However, the only thing I forgot when I memorized it was that you have to divide by 2.  So, yes, my method was wrong, but only because I forgot that final division step to do.

lol im pretty positive the second one is 7 (like 100%)

It never said that quadirlateral ABCD is a parallelogram...

it just said the area is 10.

This was your question
[quote]"The area of quadrilateral ABCD=10. The vertices of ABCD are A(4,1), B(7,4), C(9,3), and D(X,0). If point D is in Quadrant I and X>0, find the value of X." [/quote]

Before learning the shoelace method, I think you might want to consider the fact of learning that all parallelograms are quadrilaters BUT NOT all quadrilaterals are parallelograms. So yea if D were (6,0) it is a parallelogram (but who cares?), it doenst give an area of 10.

Again, shoelace is only useful when you have no other options, I dont recommend it in any other circumstance.

[quote="MoreSubtleThan1+1"]  If you use the distance formula for D(6,0), AB=CD, as it should be in all parallelograms ABCD.  Attempting to have D(7,0) as the answer does not result in AB=CD, so therefore it is wrong.  Check your math before posting it.
However, the second part of your post (which is the fact that my method of the shoelace method is wrong) may be true.
[/quote]

I dont understand that solution.

the future, you are correct in that it doesn't say it's a parallelogram.  It actually turns out that neither you or I seem to be correct.  Upon closer investigation, X appears to be 14/3.  I tried using the CORRECT form of the shoelace method on it, and it seems to work.

And as for you, firecricket91, I'll explain that solution to you (which is false, actually).  The solution is the way to find the value of X such that point D is situated in a place that makes ABCD a parallelogram.  Now, as the future pointed out, ABCD isn't necessarily a parallelogram.
But for the purposes of explaining my method to you, let's assume that the problem says this:
'The vertices of parallelogram ABCD are A(4,1), B(7,4), C(9,3), and D(X,0). If point D is in Quadrant I and X>0, find the value of X.'
Now, a basic fact about parallelograms is that in all parallelograms, opposite sides are congruent.  The reverse of this fact is also true; if a quadrilateral's opposite sides are congruent, then it is a parallelogram.  Buy a geometry book if you don't agree with these facts; they'll prove it to you.  Now, any parallelogram will look roughly like this:
A        B

D         C
I know that parallelogram ABCD doesn't look like that, but the important thing is that the order of the points of the above diagram will look like parallelogram ABCD (that is, the vertices of the parallelogram will go in order from A to D either clockwise or counterclokwise).
Now, we have to find the value of X so that either AD=BC, or AB=DC (remember that opposite sides must be congruent in a parallelogram).  By the distance formula, AB=sqrt((4-7)^2+(1-4)^2)=sqrt(18).
Since DC should equal AB,  DC=AB=sqrt(18), and by the distance formula, DC=sqrt((x-9)^2+(0-3)^x)=sqrt(x^2-18x+90).  With the two equations in the previous sentence being equal to each other, sqrt(18)=sqrt(x^2-18x+90).  Taking the square of both sides, 18=x^2-18x+90, and subtracting 18, x^2-18x+72=0.  By intuition, I can simplify x^2-18x+72 into (x-12)(x-6)=0; trust me, if you expand, it'll still be equal to x^2-18x+72.  This gives solutions of x=12 or x=6 in order to have (x-12) or (x-6) be equal to 0.
Now, by plotting A, B, and C, we can tell that x=12 will not make D a vertex of parallelogram ABCD.  Therefore, x=6 will (even draw out the diagram).

There.

"The area of quadrilateral ABCD=10. The vertices of ABCD are A(4,1), B(7,4), C(9,3), and D(X,0). If point D is in Quadrant I and X>0, find the value of X."

actually... $ x\equal{}7$.
The future is right, and this was from the purdue math thing?
Let's do this shoelace method style! XD

$ 4   1$
$ 7   4$
$ 9   3$
$ X   0$
$ 4   1$

By shoelace, $ |(4*4\plus{}7*3\plus{}X)\minus{}(1*7\plus{}4*9\plus{}3X)|\equal{}20$
Thus, $ |37\minus{}43\minus{}2x|\equal{}20$
Obviously, if $ X>0$, the left hand side is negative.

Thus, $ 2x\plus{}6\equal{}20$
$ x\equal{}7$

GG KKTHXBAI.

BTW: mathcrazed i hear, you did pretty well on that purdue thing, and so help me MoreSubtleThan1+1, if you tell anyone my score i will slap you at school.

isn't shoelace method also called surveyor's method or something?


An even number is a perfect square only if it is divisible by 4. The sum of the squares of two odd numbers is always congruent to 2 modulo 4 and so can’t be a perfect square. For if $ 2a\plus{}1$ and $ 2b\plus{}1$ are odd numbers, $ (2a\plus{}1)^2\plus{}(2b\plus{}1)^2\equal{}4(a^2\plus{}a\plus{}b^2\plus{}b)\plus{}2$.

[quote="guile"][quote="Jane_Bennet"]An even number is a perfect square only if it is divisible by 4. The sum of the squares of two odd numbers is always congruent to 2 modulo 4 and so can’t be a perfect square.[/quote]

Any mathematical proof?[/quote]

Even number are of the form $ 2m$, odd numbers $ 2m\plus{}1$

Hence $ (2m)^2 \equal{} 4m^2$ which is divisible by $ 4$.

$ (2m\plus{}1)^2\plus{}(2n\plus{}1)^2 \equal{} 4m^2\plus{}4m \plus{} 1 \plus{} 4n^2\plus{}4n\plus{}1 \equal{} 4(m^2\plus{}m\plus{}n^2\plus{}n)\plus{}2$

[quote="guile"][quote="Jane_Bennet"]An even number is a perfect square only if it is divisible by 4. The sum of the squares of two odd numbers is always congruent to 2 modulo 4 and so can’t be a perfect square.[/quote]

Any mathematical proof?[/quote]
Sorry, I edited my post to add the proof while you were replying.

[quote="guile"]Prove that the sum of the squares of two odd integers cannot be a perfect square.[/quote]

[hide="Two lines"]For all odd $ n$ we have $ n^2\equiv 1\pmod 8$. Thus for odd $ x,y$ we have $ x^2\plus{}y^2\equiv 1\plus{}1\equiv 2\pmod 8$. But $ 2$ is not a quadratic residue $ \mod 8$. Hence the conclusion.[/hide]

[quote="nayel"][quote="guile"]Prove that the sum of the squares of two odd integers cannot be a perfect square.[/quote]

[hide="Two lines"]For all odd $ n$ we have $ n^2\equiv 1\pmod 8$. Thus for odd $ x,y$ we have $ x^2 \plus{} y^2\equiv 1 \plus{} 1\equiv 2\pmod 8$. But $ 2$ is not a quadratic residue $ \mod 8$. Hence the conclusion.[/hide][/quote]

Same except replace 8 with 4. I suspect that those are the only two proofs.


inequalities
algebra
polynomial
Support
IMO
IMO 2005

[color=red][[b]Moderator edit.[/b] A number of solutions of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44480]IMO 2005 problem 3[/url] used the Muirhead inequality, which is also called bunching inequality. This topic discusses the use of this inequality in olympiad inequality problems.][/color]

Mathlinkers are famous because of their extraordinary abilities to solve inequalities :D

A quote from the report on IMO 2005 by Geoff Smith, leader of the UK team:

[i]The harder questions are selected next, and since there is a good combinatorics question, the obvious thing is to select what has been classified as the hardest algebra question. Many leaders don’t know the technique known as Muirhead’s inequality, and some of those who do seem to think that it is an advanced exotic technique. In fact many of the IMO students are networking to great effect, partly with the help of the twin internet sites Mathlinks and AoPS – the Art of Problem Solving, and they are sharing ideas and techniques independent of their national training regimes. The IMO problem selection committee and jurors will have to monitor these sites to see what ideas the students are sharing. It turns out that Muirhead’s inequality might be better known among the students than parts of the jury. [/i]

The whole report is available at [url]http://www.imo-register.org.uk/2005-report.html[/url]

[quote="Arne"]Mathlinkers are famous because of their extraordinary abilities to solve inequalities :D

A quote from the report on IMO 2005 by Geoff Smith, leader of the UK team:

[i]The harder questions are selected next, and since there is a good combinatorics question, the obvious thing is to select what has been classified as the hardest algebra question. Many leaders don’t know the technique known as Muirhead’s inequality, and some of those who do seem to think that it is an advanced exotic technique. In fact many of the IMO students are networking to great effect, partly with the help of the twin internet sites Mathlinks and AoPS – the Art of Problem Solving, and they are sharing ideas and techniques independent of their national training regimes. The IMO problem selection committee and jurors will have to monitor these sites to see what ideas the students are sharing. It turns out that Muirhead’s inequality might be better known among the students than parts of the jury. [/i]

The whole report is available at [url]http://www.imo-register.org.uk/2005-report.html[/url][/quote]
 :rotfl:  :rotfl:  :rotfl:  :rotfl: I do think muirhead is famous although we don't have that in our training.

I have one stupid question: why should inequalities that can be solved immediately with this bloody ugly and efficient method be given in such prestigious contests? I mean, is it so difficult to find something that avoids this method? :?  :?  :blush:

Yes, muirhead is bunching. Tell me harazi, how would you avoid this?

[quote="harazi"]inequalities that can be solved immediately with this [b]bloody ugly and efficient[/b] method be given in such prestigious contests? [/quote]
:coolspeak:
I can see that harazi loves the Muirhead.

I by myself agree  with harazi.
There exist inequalities that can resist against the Murihead.

[quote="Peter DVD  :lol:  "]How would you avoid it,harazi? [/quote]
yeah ,harazi .show us your idea about the real worthy inequalities which can not be solved with Muirhead?

I think there are many examples in the forum. For example, the well known inequality $(a^2+b^2+c^2)^2 \ge 3(a^3b+b^3c+c^3a)$. I don't think it can be solved by muirhead.

Because that one is not symmetric. If you only consider non-symmetric ones, then you'll have few choice I think. (and many many other methods will fail as well... only schur and some alikes remain I think? besides the usual basics like AMGM...)

Muirhead obviously won't deal with non-homogeneous inequalities either... such as $a^2+1\geq 2a$ to take a simple example. Of course in that case you can homogenise with an auxiliary variable, but I'm not sure that always works.

Muirhead does work there: \[a^2\cdot 1^0 + a^0 \cdot 1^2 \geq a^1 \cdot 1^1 + a^1\cdot 1^1\]

Yes, that's what I meant by an auxiliary variable... but I don't think it works in all cases.

why not? $a^2+1\ge 2a$ is no more than filling in $b=1$ in $a^2+b^2\ge 2ab$?

Could you give an example of what you mean?

Here is a nice one: http://www.mathlinks.ro/Forum/viewtopic.php?t=6138 .

[quote][color=blue]If a, b, c are any three positive reals, then prove the inequality

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$\geq 2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)$.[/color][/quote]

Upon expanding, it becomes

$\left[4,3,0\right]+\left[3,2,2\right]\geq\left[4,2,1\right]+\left[3,3,1\right]$.

Try to bunch it out. I am not sure that it's impossible (with Muirhead, of course it is, but I mean Schur), since I am not really used with bunching in practice (well, you see that I didn't do the IMO one), but it would be a bit strange if it would work.

By the way, I look forward to hear Geoff's opinion on Vornicu-Schur next year. :D

  Darij

Muirhead doesn't work there, for example:
$ \sum a^3(b \plus{} c) \le \frac {1}{8}(a \plus{} b \plus{} c)^4$;
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \sum a^2(b \plus{} c)$ (Schur inequality for 3 variables),
where $ a,b,c$ are non-negative numbers.

And then there are the inequalities with bizarre equality cases... I think it was Simon Marshall (the New Zealand deputy) who had a nice-looking symmetric three-variable inequality whose equality cases were irrational roots of a sixth-degree polynomial or something. It was a very, very hard problem, I heard. I didn't try it. But there are some three-variable inequalities which have equality iff exactly two of the variables are zero, for example. Obviously Muirhead can't work on any of these.

[quote="Peter VDD"][quote="darij grinberg"]$\left[4,3,0\right]+\left[3,2,2\right]\geq\left[4,2,1\right]+\left[3,3,1\right]$.

Try to bunch it out. I am not sure that it's impossible (with Muirhead, of course it is, but I mean Schur)...[/quote]

--> You think it is impossible with muirhead?
Doesn't this trick work? Just cancelling [3,1,0] from all sides gives us the trivial [1,2,0]+[0,1,2]>=[1,1,1]+[0,2,1], which is nothing else than [2,1,0]>=[1,1,1]?[/quote]
sorry, but can you cancel things on both sides?!?!?!?!

[quote="Peter VDD"]Not just random things, that's for sure.

But I believe cancelling the highest possible factor everywhere, will result in a 'degenerate' inequality, which means, usually you get equality, but if you still have the inequality (like here) then it proves the original ineq. Not sure about this though. Anyone knowing more about this?

[but even if I'm right, it is not a proof that goes everywhere, since in many cases you end up with equality...][/quote]

Well, I just wonder whether this action is valid. You are considering symmetric sums, so the terms don't have a common factor like [3,1,0]. In addition I really haven't come across this cancellation before. Can you explain the mechanism behind in details? Thanks.

sorry, i totally don't understand what you are talking about.....
cancellations of (1,1,1) are obvious, but besides that i understand nothing. :(  :(

Until he presents something with an ounce of mathematical rigor in it, I am inclined to disregard it.

Peter, please, it looks right, but I am sure it is "illegal" (unless you manage to prove your claim).

[quote="Arne"]Well, if the jury doesn't know that this method exists, then they don't know that they actually have to look for something else :)[/quote]

I'm sure some team leaders were aware that Q3 could be done by majorisation-whacking (what we on this lil' isle call a whack-able inequality), and voted against it. They were clearly in the minority. And it seems that quite a number of contestants realised that fact very quickly and got that question out of the way with little fuss.

Before this IMO I was saying that we'll never see a symmetric inequality for Q3 or Q6. Hmph.

[quote="Yossarian"]

I'm sure some team leaders were aware that Q3 could be done by majorisation-whacking (what we on this lil' isle call a whack-able inequality), and voted against it. They were clearly in the minority. And it seems that quite a number of contestants realised that fact very quickly and got that question out of the way with little fuss.
[/quote]

Or maybe they just realized that they students would do better than many others because they know how to "bunch", so they just kep quiet and accepted the question? I don't know :)

Sorry, I forgot to answer to some questions. How to avoid Muierhead? Please, tell me how many problems I posted can be solved with Muierhead? I bet not even 10%. I mean, one can easily create cyclic, non-symmetric, n-variables inequalities with nice ideas that cannot be solved with Muierhead not in 1000000 years (look at mathlinks contest for example or look at Vasc's problems or some of my problems or Namdung's or many other mathlinkers who posted magnificient problems on the forum, thousands of times smarter and difficult than this problem). It is not fortunate that from all inequalities proposed for IMO the only ones that figure out in the contest are those that are straightforward for bloody killers. This is really bad, since each year on the shortlist there are great problems on inequalities, but they are simply neglected because they are considered "different". What a pity...

[quote="Arne"]Or maybe they just realized that they students would do better than many others because they know how to "bunch", so they just kep quiet and accepted the question? I don't know :)[/quote]

And I indeed think this is the answer to harazi's question. Many of the big teams are "well-trained", which means they cannot miss a simple muirhead, but some could miss an advanced inequality requiring certain 'new' brain-clicks. So it seems not unlogical some team leaders will support the taking of this ineq.

[quote="Vasc"]Muirhead doesn't work there, for example:
[...]
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \sum a^2(b \plus{} c)$ (Schur inequality for 3 variables),
where $ a,b,c$ are non-negative numbers.[/quote]

It does. Just multiply Schur out by $ \sum_{sym}a^3b \plus{} \sum_{sym}a^2b^2 \plus{} \sum_{sym}a^2bc$, which is of course positive.


topology
function
geometry
advanced fields
advanced fields theorems

We well know the way to define an indentification topology from topology space $X$:
Let $Y$ is partition(the family of disjoint nonempty subsets) of $X$ and consider the mapping $\pi: X\to Y$ send $x\in X$ to subsets (member of) $Y$ contain $x$ and then indentification topology is space $(Y,\tau)$ with $\tau$ is the lagrest topology for $\pi$ continuous.

But I can't imagine the contruction of Mobius strip by this way by indentification topology we contruct Mobius strip as follow
Let $X=[0,1]\times [0,1]\subset \Bbb{R}^{2}$ and let $Y=\{\{(0,y)| 0\le y\le 1\},\{(1,1-y)|0\le y\le1\},\{(x,y)|0<x<1,0\le y\le 1\}\}$ now we call Mobius is $Y$ with indentification topology from $X$.

But we well know Mobius strip is only subspace of $\Bbb{R}^{3}$ with topology of Mobius induces from topology of $\Bbb{R}^{3}$ can any body show me two ways is the same ?

Another question: In some books they consider indentification topology and quotient topology as the same, but we known quotient topology needs the equivalent relation can anybody show me this relation of indentification topology ?

A piece of paper, scissors, and glue are all you need to answer the first question.

For the second: two elements are equivalent iff they belong to the same subset (aka an equivalence class).

Talking as you then everybody can say when they understood somethings on topology, it isn't the thing I need when I post this question!
Can you show by formula two Mobius is contructed by two way are homeomorphic, and I don't think with "paper, scissors, and glue" you can describe the mapping and the largest topology for it continuous! :rotfl:

Take a typical open set (ie a basis element) for one topology that contains a point $x$ and show that a basis element for the other topology is inside of it and still contains $x$.

mlok was suggesting that you make a Mobius strip because if you do that it's easy to visualize why the topologies are the same.

----
edit: the point is that you don't always have to do everything "by formula." If you can argue in words that something must be true, that could be much more concise and clear than trying to explicitly produce numerical examples for all of the types of basis elements in the proof that the topologies are the same.
----

I think equivalence relations are in 1-1 correspondence with identification topologies (so they can all be called quotient topologies). This is because given any injective function $f: X\to Y$ where $X$ is a topological space and $Y$ is a space, you just let the open sets of $Y$ be the images of open sets of $X$ and that's the quotient topology.

Hopefully I'm not talking nonsense. I've had a long day... ;)

Thanks for your careful answer Xevarion, I feel i understand some more ! An other question can we parametrize a Mobius strip from a set $U\subset \Bbb{R}^{2}$ or in other words do you know equation (parametrisation or Cartesian) of Mobius strip in $\Bbb{R}^{3}$?

Both [url=http://en.wikipedia.org/wiki/M%C3%B6bius_strip#Geometry_and_topology]Wikipedia[/url] and [url=http://mathworld.wolfram.com/MoebiusStrip.html]Mathworld[/url] have parametrizations.

But I didn't know these myself, no. :P


In triangle $ABC$ the bisector of the angle at $B$ meets $AC$ at $D$ and the bisector of the angle at $C$ meets $AB$ at $E$.  These bisectors intersect at $O$ and the lengths of $OD$ and $OE$ are equal.  Prove that either $\angle BAC = 60^\circ$ or triangle $ABC$ is isosceles.

A village is constructed in the form of a square, consisting of 9 blocks, each of side length $l$, in a 3 x 3 formation.  Each block is bounded by a bitumen road.  If we commence at a corner of the village, what is the smallest distance we must travel along bitumen roads, if we are to pass along each section of bitumen road at least once and finish at the same corner?

Find all solutions of
\[ 2^n + 7 = x^2 \]
in which $n$ and $x$ are both integers.  Prove that there are no other solutions.

Eight football teams participate in a tournament of one round (each teams plays each other team once).  There are no draws.  Prove that it is possible at the conclusion of the tournament to be able to find 4 teams, say $A$, $B$, $C$ and $D$ so that $A$ defeated $B$, $C$ and $D$, $B$ defeated $C$ and $D$, and $C$ defeated $D$.

Sorry to bring back an old post, but I don't see how #3 is hard...[hide="solution"]mod 4, the equation for $n>1$ is $0+(-1)=x^{2}mod 4$ and it is clear that $-1$ is not a square mod 4[/hide]

dude you dont live in AL and its over two years old.


algebra
polynomial
number theory solved
number theory

Prove that if a^n-1 divides b^n-1 for all  n positive integer, then a must be b^k for some  positive interger (a,b posituive integers)

This was definitely posted before, but I have no idea what the topic was called.

well it's no much pain to solve it again:

n=1 means b|a

now if b had any factors that a didn't (or higher prime powers) then for big enough n the statement $b^{n-1}|a^{n-1}$ becomes false (obviously)

:D Well Peter I think you misunderstood the statement, which is $a^n-1$ divides $b^n-1$ and not $a^{n-1}$ divides $b^{n-1}$.
So, the problem becomes much more difficult...

It is an hard problem which appeared in the A.M.M.

Pierre.

ah I see :)

That's a lot harder indeed. I'll try it though.

Try it just if you don't have other things to do. I posted in " A masterpiece". And I remember you about my post " Related to the masterpiece". Maybe someone will solve that problem, too. I solved this beautiful problem using the same idea as in the polynomial $ p(2^n) $ that Peter Scholze solved. In this problem, the double recurrences are much easier. But the problem is extraordinary difficult and beautiful.

'A masterpiece' is here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=4556&highlight=masterpiece

That's funny, I didn't remember me writing a solution :maybe:

Pierre.


Urgently waited.
My Solutions is $ n!$.But I'm not sure.

obsreve that in each columm there will be only 1 rook.consider the first columm.we have n places to put the first rook, n-1 ro the second...
Therefore the answer is n!.

[quote="JBL"]No one reads the final clause of the problem?[/quote]

ow, i didn´t see that.


Post the names and scores of your state's Top 4 here.  [b]This thread is for scores only, anything you want to discuss should be discussed in the "Which State Will Win at Nationals Thread."  [/b]
Please format any posts in this thread like so:

[u]Full State/Province Name[/u]

1st Place:  John Doe - 43
2nd Place: Sarah Doe - 42
3rd Place: John Deere - 40
4th Place: Sally Deere - 37

PS: I wouldn't mind if a Moderator would sticky this thread...

[u]California:[/u]

1st place: (43)
2nd place: Me (40)
3rd place: (39)
4th place: (38)

1st place team: (54.75)

[u]North Carolina[/u]

1st place: Eric (Pom) McCabe 39
2nd place: John Berman         38
3rd place: Bryce Taylor           37
4th place: Kevin Lang             36

Arkansas
1st: 37  David Aldrich
2nd: 37 Karan Batra(me)
3rd: 37 Timothy Mellor
4th: 37 Jack Chen

GA

1st 34 Xantos C. Guin
2nd 34
3rd 33
4th 31 solafidefarms

some people might no want their real name up here.

The ones on my team wouldn't know, because they don't have this, but I guess Ishould have asked them first.

Why does everything always turn into a pointless discussion? Nevertheless, I feel obliged to respond: If their names aren't posted here, we may as well just go look them up on the Mathcounts website. They have the entire roster. Also, just posting your score implies the approximate value of the other scores. You may as well just never say what your scores are or what other people on the team got (no matter who they are). Like, mysmartmouth, I would really like this to be a non-discussion forum. I just had to state my opinion. Please try not to discuss stuff in here, and if a moderator objects to the subject, they can certainly change it.

Thank you, I agree-- It's kind of worthless guaging people's scores, because looking at other people's scores at states won't make you any better at nationals.  Why don't you just wait till nationals and see?  Nobody can predict the future.  And about this whole "individual who will win nats" who cares??  We'll find out soon enough.  Besides, I bet these people you're talking about would be happy to know they're being betted on like a hand of poker.

Mr.Ocax

*Edit* I suppose you guys do have freedom of speech, etc., so by all means, don't let this stop you (it won't).  Just stating my opinion like jb05

texas:

mark zhang 46
kevin chen 40
jeffrey chan (kingof21penguins) 38
dennis mou (me--  :( ) 37

[quote="mysmartmouth"]Take a look at this attachment, and let me know of anything I missed, or anything I need to add.  No word on Illinois, and I would like to see their scores...

Edit:  NC's box got butchered, but no harm done, I will fix it in Version 2! ;)[/quote]

umm... msm are you studying for nationals or wasting your time making useless tables that MATHCOUNTS already made?

[quote="chess64"]umm... msm are you studying for nationals or wasting your time making useless tables that MATHCOUNTS already made? http://artofproblemsolving.com/Forum/privmsg.php?mode=reply&p=58971 just wanted to let you know about that  :P[/quote]

Let me know about what?

I'm not in Nats this year due to circumstances beyond my control, so I have plenty of time to do stupid things like this.  Besides, I've got a buch of money riding on the Nats... ;)

[url=http://www.mathcounts.org/webarticles/articlefiles/536-State%20Winners%20for%20Web%20PDF.pdf]Here is the official list of winners, but no scores.[/url]  Also, People shouldn't object to having their names displayed on AoPS, because their names are already on MathCounts's website.

Version 2:  Ready!

Edit:  To find out your state scores, you can try to email your state coordinator.  Look [url=http://www.mathcounts.org/webarticles/articlefiles/StateCompDates05.pdf]here[/url] for details.

Just to let you know, the names of all the State Champions are here:

http://www.mathcounts.org/webarticles/articlefiles/536-State%20Winners%20for%20Web%20PDF.pdf

 :)

Edit: Oh, whoops, didn't see the post above! :blush:

Yeah, our chapter is the best in Mass, many teams have left it so they have a chance.

This year three people from it made nationals, last year all four were from it, and the year before three were from it.

[quote="Xantos C. Guin"]Delaware doesn't have chapter?[/quote]Yeah. I lucked out though cuz I took the chapter test and scored a 20. I think I was drunk

[quote="jb05"]I find cases like yours very interesting. Do you (or anybody else who went from so-so at regional to excellent at state) think you made tons of progress between competitions, or were you unlucky at regional/lucky at state? I have always been surprised when people mention this because I was 1st in the region twice in a row and second in the state this year.[/quote]

I think I was rather unlucky at chapter and maybe a little lucky at state. I guessed randomly (not joking) on #28, which not only tied my score for fourth place, but broke the tie in favor of me.

[quote="Phelpedo"]I think I was rather unlucky at chapter and maybe a little lucky at state. I guessed randomly (not joking) on #28, which not only tied my score for fourth place, but broke the tie in favor of me.[/quote]

That is utterly amazing. If you believe in fate, you are going to nationals for a reason! (Please tell me that #28 was like "5" and not like "(14+sqrt31)/17.")

New Jersey
43
37
34
34-me

I was   originally 3rd until I got knocked to fourth by countdown. Oh man, thankfully though, I still made Nationals.

So it's official in NJ? FOr some reason the official standing seems to be that you were 3rd.

[quote="mysmartmouth"][quote="Phelpedo"]It was 182370.029814

Just kidding, the only answers possible were 0,1,2, and 3, and as soon as I saw the problem, some divine instinct made me put down 3 without even trying to actually solve it.[/quote]

Here is the problem:

The Fibonacci sequence is the sequence 1, 1, 2, 3, 5, ... where each term is the sum of the previous two terms. What is the remainder when the 100th term of the sequence is divided by 4?

(Anyone got a solution?)[/quote]

Write it in mod 4...

1 1 2 3 1 0 1 1 2 3 1 0... Whoa, a pattern!! Unbelievable!!!

100 is 4 mod 6, so it's the 4th number of that pattern, which is 3.

Yeah but I didn't even try to solve it, I just looked at it and guessed. Then time was called a millionth of a nanosecond after I put it down and it won me the spot to nationals. Just kidding, that was just for drama.

See, I had done tons of Fibonacci stuff at school for project day. Especially with Mod 4, so the second I saw this I flipped out. Easiest.Sprint.Round.Question.Ever Except for maybe that one about the distance between two cities.

Usually a problem like this involves a pattern.

[quote="NoSoupForYou"]Just wondering, will the Fibonacci sequence create a pattern in [i]any [/i]mod? So far a pattern was formed in every mod I tried...[/quote]
Yes, it will.  Think about it this way...  If the modulus is 4, we know every term of the sequence (modulo 4) will be 0, 1, 2, or 3.  If you ever get two consecutive terms that are a duplicate of a previous pair of previous terms, the sequence will repeat forever.  The maximum number of distinct pairs is $4\times 4=16$, so we know the sequence must repeat with a cycle of 16 or fewer.


modular arithmetic
algebra proposed
algebra

Show that the following equation has no real solution:

$ [x]\plus{}[2x]\plus{}[4x]\plus{}[8x]\plus{}[16x]\plus{}[32x]\equal{}12345.$

[quote="moldovan"]Show that the following equation has no real solution:

$ [x] \plus{} [2x] \plus{} [4x] \plus{} [8x] \plus{} [16x] \plus{} [32x] \equal{} 12345.$[/quote]

$ 12345 \equiv 60 \pmod{63}$

Let $ y\equal{}\{ x \}$, then it becomes to prove $ [y]\plus{}[2y]\plus{}[4y]\plus{}[8y]\plus{}[16y]\plus{}[32y]\equal{}60$ has no real solution.

Let $ f(y)\equal{}[y]\plus{}[2y]\plus{}[4y]\plus{}[8y]\plus{}[16y]\plus{}[32y]$ we know that $ f(y)$ is strictly increasing.

For $ 1>y \ge \frac{31}{32}$, $ f(y)\equal{}57$.
For $ \frac{31}{32}$, $ f(y)<57$.

 So $ [y]\plus{}[2y]\plus{}[4y]\plus{}[8y]\plus{}[16y]\plus{}[32y]\equal{}60$ has no real solution ,done.


trigonometry
algebra
polynomial
algebra unsolved

Are the numbers $sin1^{\circ}$ and $cos1^{\circ}$ irrational.
How about $sincos1^{\circ}$ and $cossin1^{\circ}$.

$\cos nx$ is a polynomial in $\cos x$ (easily proven by De Moivre's formula). Hence if $\cos 1^\circ$ is rational, then $\cos 30\cdot 1^\circ$ is also rational. But $\cos 30^\circ=\sqrt{3}/2.$

Similar for $sin 1^\circ$.


calculus
integration
geometry
3D geometry
number theory

Determine all integers a, b, c, which satisfy the following equation:
[tex]a^3-b^2=c^2[/tex]

[hide]
Well  I'm pretty sure a number x can be written as a sum of integral squares iff all prime divisors of the form 4k+3 divides x an even number of times.

So.... if a can be written as a sum of squares, so can a^3, and vice-versa. In other words, we can get LOTS of solutions....

[/hide]

Maybe this is too stupid but since no restriction is given such that a,b, and c are distinct and none of them is nonnegative, as long as one of them is zero and other two are same, there are infinitely many solutions.

Probably wrong though...

lol Silverfalcon... That's a pretty good observation!  :lol:

There is another way to say this has infinitely many solutions.

I'll keep this as a secret tilll Friday though. This question is one of the question one mock AMC G.

 :D

Easier, of course! :)

[quote="Nukular"][quote="Myth"]
Prove that the following equations have infinitely many solutions in natural numbers:
...
4) $a^3-b^3=c^3$.
[/quote]
Why do I find that one a little hard to swallow? :)[/quote]
 :what?: Do you think so? I am really surprized!  :rotfl:

lol... Was that supposed to be a trick or something?

For the first one, it does look like it has infinitely many solutions except I'm not so quite sure on how. Remember, I'm still spending my most of time on Getting Started or Intermediate. :D

We can get: $a^3 = c^2 - b^2$ which would let us to

$a^3 = (c+b)(c-b)$

So, if I can somehow prove that there exists perfect cube that can be expressed as a product of two integers' sum and difference, it's proved.

I got a=3,c=6,b=3 but I can't think of anymore now.

To Myth, you can tell it but I'm just saying that I have one way to prove the original question using the fact "no restriction."

 ;)

[quote="Myth"]You mean, we are not allowed to send solutions till Friday?

I propose another similar problem:
Prove that the following equations have infinitely many solutions in natural numbers:
1) $a^3+b^2=c^2$;
2) $a^3+b^3=c^2$;
3) $a^3-b^3=c^2$;
4) $a^3-b^3=c^3$.

 ;)[/quote]
Why wait till Friday?!?!?
[hide=""CLICK ME CLICK ME""]
1) $a^3+b^2=c^2\Rightarrow a^3=(c-b)(c+b).$ One set of solutions is $c=\frac{a(a+1)}{2}, b=\frac{a(a-1)}{2}$.
2) Let $a=2k^2,b=k^2$. So $c=3k^3$.
3) Let $a=10k^2,b=6k^2$. So $c=28k^3$.
4)  :huh:  :what?:  :rotfl:  :rotfl:  :rotfl:
For the original problem
[hide=""Click me to open sesame a useless set of integers""]
Let $a=5k^2,b=2k^3,c=11k^3$[/hide]
[/hide]

[quote="3X.lich"][quote="Myth"]4) $a^3-b^3=c^3$.[/quote]
4)  :huh:  :what?:  :rotfl:  :rotfl:  :rotfl: [/quote]
I am glad you like it  :D  :D  :D

Although 4 is a special case of FLT, it is nonetheless well known and open to the methods of traditional number theory, if not particularly easy.

For original problem:

[quote]$a^3-b^2 = c^2$[/quote]

If we can say that a=c, we have the old AHSME problem. (This was also in mock AMC G.)

So, this is what we have then:

$a^3-b^2 = a^2\\
a^3-a^2 = b^2\\
a^2 (a-2) = b^2$

If there exists $k^2 = (a-2)$, then we can state that it has infinitely many solutions.

For example, if $k^2 = 25$, then:

$a^2 * 25 = b^2$ and $25 = a-2$ which leads us that a = 27. So, $a^2 = 27^2$ So, then $b^2 = 27^2 * 25$ and $b= 5*27$.

Using same way, we can state that this has infinitely many solutions.

This was only possible because there was no restriction. Is this okay? :P

Re: Silverfalcon
So your solution is $a=27,b=5\times27,c=a?$
If that is the case, $a^3-b^2=1458\neq c^2$ :?
So what went wrong?
[hide=""Hint""]One digit[/hide]

I should probably point out also, Silverfalcon, that the problem asked you to find [b]all[/b] solutions - not to prove there are infinitely many.

Oh, for my answer. I should have changed in form of:

$a^2+b^2 = a^3$

Then:

$b^2 = a^3-a^2\\
b^2 = a^2 (a-1)$

Then call the $k^2 = a-1$ and then:

$k^2 = a-1 \Rightarrow k^2+1 = a$

If we say that $k^2 = 25$ then: $25+1 = a = 26$

Thus, a=26 and b = 26*5.

To TripleM, I know the question necessarily didn't ask to prove it but if I can prove it and pick out the answer, it's going to be right, isn't it? I know it seems that I did the extra work but anyway, the answer was right. :) At least that's what other people and I posted. :P

Finding all solutions is stronger than proving there are an infinite number.  If you could write down what all the solutions are in terms of the others then it will follow that they are infinite, as well as telling you what they are.

[quote="Silverfalcon"]To TripleM, I know the question necessarily didn't ask to prove it but if I can prove it and pick out the answer, it's going to be right, isn't it? I know it seems that I did the extra work but anyway, the answer was right. :) At least that's what other people and I posted. :P[/quote]

As Wumbate said, finding all solutions is a lot harder than proving there are infinitely many. As a relatively simple example, if you want to find all pythagorean triples, its easy to prove there are infinitely many solutions (3k, 4k, 5k) but its a lot harder to find ALL solutions.

May be I am missing something simple but isn't  a^2+b^2 = a^3  is little different than  a^2+b^2 =[b] c^3 [/b]

In that case, for a given  c^3, count the  factors of the type 4k+1, subtract the factors of the type (4k-1 ) and multiply the result by 4, that will give you the number of solution for any given c^3.

The trick will also work for Myth Problem 4.  You just have to use   the  expression '4k+1' insted of 4k-1 in the above expression  :)

Hmm.. a^2+b^2 = c^3 is definitely different to a^2+b^2=a^3 but what I said earlier was that since no restriction was given, I can say a=c. At least, that seemed allowed. :D

And oh yeah, I see now. Listing the solutions seem a lot harder than proving it.

I thought proof was always hardest thing in math. ;)  :P


A box of donuts can be equally divided (without cutting pieces) among eight, twelve or fifteen people. What is the least number of donuts that the box could contain?

You are looking for a common multiple of 8, 12, and 15. 12 and 15 have a least common multiple of 60. 8 and 60 have a least common multiple of 120, which is the answer.


37 football players want to play a game. They divide into even teams of 18 players each, with 1 person being referee. No matter who is chosen as referee, the total weights of each team are the same. Prove that each player must weigh the same amount.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274682

You're going through Zeitz or some similar text?  Many, many problems from such books have been posted here before -- do a search!  (You could have found the thread I just linked to by searching for the words "weight" and "football," for example.)

These are problems from a recent competition, I just wanted some other opinions on how to solve the problem. Thankyou for the link :)

[quote="JBL"]What competition?  (It seems they couldn't be bothered to produce original questions.)[/quote]

They're from the Melbourne University Maths competition (for high-school students) in Australia. The questions may not be very hard or original because it is assumed that most kids who take it have not seen too many questions like this before.

Also it's not as serious as the AMC, which has lots of funding so their questions are better.


Whats a generating function?...

I looked on the wiki and i didn't really get it. Can someone give me an example (I can't understand things well with symbols.)

EDIT - If a mod comes across this thread, could they please delete it, I meant to post this in my other thread.

[quote="123456789"]Let's say $ 6$ people each have three $ \$1$ bills. You want to find how many ways there are for them to pay for a $ \$10$ object.
Since each person can contribute either $ \$0$, $ \$1$, $ \$2$, or $ \$3$, you can write the generating function as:
$ (x^{0} \plus{} x^{1} \plus{} x^{2} \plus{} x^{3})^{6} \equal{} (1 \plus{} x \plus{} x^{2} \plus{} x^{3})^{6}$

Basically, the coefficient of the $ x^{10}$ term in the complete expansion of the function determines the amount of possible ways for those $ 6$ people to pay for the object, because all combinations of exponents that sum to $ 10$ are used.[/quote]
Actually,your problem is the small case of Euler's problem, which is usually spoken like this:"How many ways you can divide $ \$$1 into pennies, quaters, dimes and cents?
PS @ejr322:Yes, this is the most difficult method in combinatories(in my opinion). Even now when I read many good lectures, I still not good at solving combinatories by this method! My advice to you: read as many as you can from Helbert's book and all lectures about this until you get the actual idea of every problems you face in there! Ask friends as I did is also the good way to learn it fast! Gathering more experience from problems you face everyday, you will succeed in using it someday. Don't give up and be patient as I was! :)


Show that any number in the form

$ 2^k(2^{k\plus{}1}\minus{}1)$

is perfect if $ (2^{k\plus{}1}\minus{}1)$ is prime.

[hide="hint"] Define $ s(n)$ to be the sum of divisors of $ n$. Then $ s(n)$ is multiplicative. In other words, if $ n \equal{} ab$ for relatively prime $ a, b$, then $ s(n) \equal{} s(a)s(b)$. This is pretty simple to prove.[/hide]

[hide="solution"] Using the hint, the sum of the proper divisors of $ 2^k(2^{k \plus{} 1} \minus{} 1)$ is
\[ (1 \plus{} 2 \plus{} 2^2 \plus{} ... \plus{} 2^k)(1 \plus{} 2^{k \plus{} 1} \minus{} 1) \minus{} 2^{2k \plus{} 1} \plus{} 2^k\]

\[ \equal{} (2^{k \plus{} 1} \minus{} 1)(2^{k \plus{} 1}) \minus{} 2^{2k \plus{} 1} \plus{} 2^k\]

\[ \equal{} 2^{2k \plus{} 2} \minus{} 2^{k \plus{} 1} \minus{} 2^{2k \plus{} 1} \plus{} 2^k\]

\[ \equal{} 2^{2k \plus{} 1} \minus{} 2^{k}\]

\[ \equal{} 2^k(2^{k \plus{} 1} \minus{} 1)\]
as desired.[/hide]

Just a note to inexperienced problem solvers: the condition that "$ 2^{k\plus{}1}\minus{}1$ is prime" is necessary and it's implicitly used in xpmath's solution. Without this condition, the proof fails.


geometry
perimeter
geometric transformation
rotation
exterior angle

How can you prove the formula for the sum of the angles in an n-sided polygon?

While I don't know how to prove it, can you prove that a polygon with less than 3 sides can be made?  How many degrees are in a three-sides polygon.  That is why we subtract 2 in the equation, to correct for the number of sides

Equation: $ 180(s\minus{}2)$ where $ s$ is the number of sides in the polygon

We can partition an $ n$-sided polygon into $ (n\minus{}2)$ triangles. Why? Think about it! From one vertex, you can draw a total of $ n$ "diagonals" to other points on the polygon, but two of them are the endpoint of side lengths, and the other one is the original point! Therefore, you can draw $ (n\minus{}3)$  diagonals which partition an $ n$-sided polygon into $ (n\minus{}2)$ triangles. Since each triangle has a total of $ 180^{\circ}$, the total degree sum of the angles is $ (n\minus{}2)(180^{\circ})$

Here is a picture for you to make it a little easier to understand.

 [geogebra]cb8b75f8fb0201fdb364bc5c6f84e909b33c3aca[/geogebra]

The hexagon has $ 6$ sides, $ (6\minus{}3)\equal{}3$ diagonals, and $ (6\minus{}2)\equal{}4$ triangles. Therefore, the total degree sum of the angles is $ (4)(180^{\circ})$ (in this example).

Thanks. I'm just wondering if there is an algebraic way of doing it instead of looking at pictures (there probably isn't). That was very helpful anyway.

[quote="t0rajir0u"]That depends on what you mean by "algebraic," but the way I like to do this proof is to look at the [b]exterior[/b] angle (instead of the interior angle).  The exterior angle is defined as $ 180^{\circ}$ minus the interior angle; in other words, if you're walking along the $ n$-gon, it's the angle you have to turn in order to keep traveling along the perimeter at a vertex.

The exterior angles have to sum to $ 360^{\circ}$.  Why?  How does this prove the result?[/quote]
I have already thought about that way but I still can't prove the part where the exterior angles add up to 360. I know it works, but I can't prove it.

[quote="isabella2296"]-- Before posting, think hard whether your post is necessary or not. Remember, you do not have to press the "Submit" button. [/quote]

Think about your post. Was it really necessary?

[quote="batteredbutnotdefeated"]awww... Rofl got there first... 2 fast and 2 smart...

EDIT: errr... Sorry for the spam...[/quote]

He shouldn't have deleted his own proof.

You might've come up with some other way to prove it.

[quote="t0rajir0u"][quote="PowerOfPi"]I still can't prove the part where the exterior angles add up to 360.[/quote]
Suppose you walk along the circumference of an $ n$-gon once.  Keep track of the total angle you had to turn to get back to where you started.  What is it?[/quote]
I know it adds up to 360, but others might not accept because there is no mathematical proof behind it.

The intuition is more important than the proof.  If you really want the proof, I could just say "the winding number of a Jordan curve is $ 1$," but that doesn't help you understand the result any better.  Do you understand that walking along the circumference is equivalent to a $ 360^{\circ}$ rotation?

My other point here is that I think this result is much easier to remember than the original result.


Yes, nice proof. In the same way, we can prove
\[ \sum x(b + c + d) \geq2\sqrt {(ab + bc + cd + da + ac + bd)(xy + yz + zt + tx + ty + yz)}.
\]
I'm thinking about the stronger inequality:
\[ x^2(b + c) + y^2(c + a) + z^2(a + b)\geq2\sqrt {(x^4 + y^4 + z^4)(ab + bc + ca).
}\]

[quote="pvthuan"]
I'm thinking about the stronger inequality:
\[ x^2(b + c) + y^2(c + a) + z^2(a + b)\geq2\sqrt {(x^4 + y^4 + z^4)(ab + bc + ca).
}\]
[/quote]
Try $ a=b=c.$ :wink:

I posted it here before:
[url]http://www.mathlinks.ro/viewtopic.php?t=176151[/url]

Oh, yes thanks arqady. Can you sharpen it?

[quote="colorfuldreams"][quote="manlio"]For $ a,b,c$ and $ x,y,z$ nonnegative real numbers, prove that

$ x(b\plus{}c) \plus{} y(c\plus{}a) \plus{} z(a\plus{}b) \geq 2\sqrt{(xy\plus{}yz\plus{}zx)(ab\plus{}bc\plus{}ca)}$[/quote]

the strong one

$\sum{\frac{a}{b+c}(y+z)} \ge \sqrt{3(xy+yz+zx)}$[/quote]
[hide="prove"](this prove is wrong,just ingore it, :oops: )
\[ \frac{a}{b+c}(y+z)+\frac{b}{a+c}(x+z)+\frac{c}{a+b}(x+y)\geq
\frac{2}{3}(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})(x+y+z) \] by
\[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2} \]
it enough to check that \[ (x+y+z)\geq \sqrt{3(xy+yz+xz)} \] Which
is obvious.
[/hide]

Good question... anyway, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=360964]here[/url].

[quote="arqady"][quote="pxchg1200"][quote="colorfuldreams"]
the strong one

$\sum{\frac{a}{b+c}(y+z)} \ge \sqrt{3(xy+yz+zx)}$[/quote]

\[ \frac{a}{b+c}(y+z)+\frac{b}{a+c}(x+z)+\frac{c}{a+b}(x+y)\geq
\frac{2}{3}(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})(x+y+z) \][/quote]
Why it's true? :maybe:[/quote]
Sorry for my mistake, :oops: I find that is not true,just let $ x=a=1,y=b=2,z=c=3$,Thanks aqrady to point out my mistake,I will try again..

[quote=manlio]For $ a,b,c$ and $ x,y,z$ nonnegative real numbers, prove that

$ x(b\plus{}c) \plus{} y(c\plus{}a) \plus{} z(a\plus{}b) \geq 2\sqrt{(xy\plus{}yz\plus{}zx)(ab\plus{}bc\plus{}ca)}$[/quote]
Let $ a,b,c;  x,y,z$ be positive real numbers such that $ab + bc + ca = xy + yz + zx = 1$.
Prove that$$a(y+z)+b(z+x)+c(x+y)\ge 2$$
(Mathematical Reflections 2 (2013) S264)

[quote=manlio]For $ a,b,c$ and $ x,y,z$ nonnegative real numbers, prove that

$ x(b\plus{}c) \plus{} y(c\plus{}a) \plus{} z(a\plus{}b) \geq 2\sqrt{(xy\plus{}yz\plus{}zx)(ab\plus{}bc\plus{}ca)}$[/quote]
Let $x$, $y$ and $z$ positive real numbers such that: $x+y+z=1$. Then for all real numbers $a$, $b$ and $c$ we have$$(x(a+b)+y(b+c)+z(a+c))^2 \geq 4(xab+ybc+zac)$$[url=https://artofproblemsolving.com/community/c6h1495823p8801807]here[/url]

If $x+y+z=0$ the inequality is obvious. Thus we can assume that $x+y+z=1$ and $a+b+c=1.$ Then inequality becomes
$1-(ax+by+cz)\geq{\sqrt{(2xy+2yz+2zx)(2ab+2bc+2ca)}}=\sqrt{(1-(x^2+y^2+z^2))(1-(a^2+b^2+c^2))}$
On the other hand, we have $(a^2+b^2+c^2)(x^2+y^2+z^2)\geq(ax+by+cz)^2.$
So let $a^2+b^2+c^2=M^2$ and $x^2+y^2+z^2=N^2$ then inequality becomes
$(1-M^2)(1-N^2)\leq (1-MN)^2$ which is obviously true.
$\square$


Linear Equation

ax+b=c

EX

3x-4=8
 [u]   +4 +4[/u]

Add four on both sides
cancel the negative four and positve four add the eight and positive.

Now its

3x=12

Divide

3x=12
3      3

Another EX

3(y-4) + 2y=8

3y-12+2y=18

Combine 3y and 2y because

5y-12=18
[u]  +12  +12[/u]

the 12's cancel and we add 18 to 12.

now I have 5y= 30

the 5's cancel and I divide 30 and

y=6

[u]Check[/u]

in the original problem I replace y with 6.

3(y-4)+ 2y=8

3(6-4)+ 2(6)=8

I work the Parenthesis first

3(2)+2(6)= 18

Multiply

6+12= 18

Add

18=18

[quote="chess64"]Please only post questions or problems that you would like us to solve, not just your own solutions to problems that you have already solved. ;)[/quote]

My mistake, I won't do it again.

[quote="biffanddoc"]Also, this is a little easy for getting started. ;)[/quote]

I was going to go alot further in the chapter because it's the beging of Chapter 2.1, but eh, I guess its good I didn't. I'll only post when I need help.

I think I understand the confusion.  This forum has nothing to do with AoPS books, however, it's probably a good place to go if you need help. :)


\documentclass{book}  or  \documentclass{report}

How  to get the output that there is   no  page  number  on  the  first  page  of  each chapter ?

Thank you!

\thispagestyle{empty} after each chapter heading or using a recent version of titlesec (v2.8, 2007-08-12) \assignpagestyle{\chapter}{empty}

Thank stevem !

I read   the  fancyhdr documentation,  and    I think that   there   is   another   way    to   accomplish   this:

redefining plain style:

\usepackage{fancyhdr}
\fancypagestyle{plain}{%
\fancyhf{}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{0pt}
}

See section  7   of  the  fancyhdr documentation

[color=red]Am  I   right?[/color]


output a counter example

                        [a= 1, b= 1, c= -1].

x,y,z be nonegative real numbers, the inequality holds.

[quote="Alexander Khrabrov"]If it's true, it can be prove using Muirhead and Schur inequalities probably.[/quote]but how? I don't understand Muirhead very well


1/(1+1)=1/2=1-1+1-1...=0.

i just saw this on some math brochure, anyone else seen this? I never thought of this before, but its interesting.

You can't use that formula for the sum of an infinite geometric series because it only converges when |r|<1, and here r=-1.


Forum for a new game... max of 5 people sign-up.  When all 5 are signed up, I will distribute rules. :starwars:

\in   congrats on creating the 3300th topic.

You created the 3300th topic in games/ff

and I join.

i join. what's this game going to be about?

/In
[hide="For Starwars smiley spam"]
:starwars::starwars::starwars::starwars::starwars::starwars::starwars::starwars::starwars::starwars:[/hide]
[hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."]
[/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide][hide="."][/hide]
Too many smilies in your post: 54.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.

I join...

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

@butler... if you don't liike pms beacause it sends a message alert to your email, you can change that in your profile.

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

Why must it be 5?

If you change it with a spot for me, /in.

Whatever, everyone who has asked can join.  This is about Ancient Greece.  Character classes coming up in a second.  8 people are in.

These are the classes and how many people can be in 1 class

Warrior-3 people
Priest-3 people
Scout-2 people
Assain-2 people
I will release stats after everyone has chosen.

Is the priest like a sorcerer or something like that? If so, I pick priest. If not, I pick warrior

warrior
 :starwars:  :starwars:  :starwars:

I want to know what a priest or scout is before I pick (such as specialty)

Sorry Butler- 2 assains only.  Priests can regenerate people and are like sorcerers.  Scouts can become invisible.  Continue picking. [hide="Butler only"]
Sorry Butler, pick another class[/hide]

I guess I'll be a priest then. Can I be a priestess instead though?


In triangle ABC, D is the midpoint of BC and E is the midpoint of CA. AD and BE intersect at F, and the midpoints of AF and BF are X and Y, respectively. Show that EDYX is a parallelogram.

[hide="hint"]The medians of a triangle intersect at the centroid. The centroid is the point which is 1/3rd of its way up the height. Therefore if x is the midpoint of the AF, AX=AF=FD. [/hide]

$ XY\parallel AB$ and $ ED\parallel AB$, so $ XY\parallel ED$. Draw $ FC$. $ FC\parallel DY$ and $ FC\parallel EX$, so $ DY\parallel EX$. Therefore $ EDXY$ is a parallelogram.


geometry
perimeter
inequalities
triangle inequality
inequalities unsolved

(I'm posting this on Arne's behalf, since his PC is slightly wacko :( .)

In a unit square one draws 100 segments from the center to the perimeter such that the square is divided into 100 parts with equal perimeter $p$. Show that $1.4 < p < 1.5$.

Hi Valiowk!

It is quite easy problem ;)
Let O be center of square.
Consider part, which contains vertex A of square. Using triangle inequality we obtain that $p\geq 2\cdot OA= 0.5\cdot\sqrt 2+0.5\cdot\sqrt 2=\sqrt 2=1.41...$.
Since perimeter is divided into 100 'segments' there are 'segment' whose length at most 4/100. Consider corresponding part of square. Its perimetr at most $0.5\cdot\sqrt 2+0.5\cdot\sqrt 2+0.04=1.45...$.

I have stupid question:
Is it possible to find such 100 segments?


Angle x is 125 degrees. The attached image contains the solution. Is it correct?

Well, ED=EA does not imply $\angle{EAD}=\angle{DEA}$

In fact, the solution is wrong, and the angle < DEC is approximately 118,9°, so I doubt the problem was correct. However, the solution works for the following problem:

[color=blue][b]Problem.[/b] Let ABC be a triangle with < CAB = 110° and < ABC = 55°. Let D and E be two points on the sides BC and CA of this triangle such that AB = AE and AD = BD. Prove that < DEC = 125°.[/color]

[i]Solution.[/i] Since AD = BD, the triangle ABD is isosceles, and thus < BAD = < ABD, so that < BAD = < ABC = 55°. Hence, < EAD = < CAB - < BAD = 110° - 55° = 55° = < BAD. Also, AB = AE and AD = AD. Thus, the triangles ABD and AED are congruent, so that < AED = < ABD. Hence, < DEC = 180° - < AED = 180° - < ABD = 180° - < ABC = 180° - 55° = 125°. Problem solved.

  darij

[quote="darij grinberg"]In fact, the solution is wrong, and the angle < DEC is approximately 118,9°, so I doubt the problem was correct. However, the solution works for the following problem:
[/quote]
Same here.
$\angle{DEC}=118.9^{o}$ means that we can only solve it by trig?

If angle x is 125 then $AC\bot BE$, and that is not true, you can calculate angles $EBA$ and $BAD$ and that angles don't give 90 in their sum


We made a few code changes to the weblog system.  Nothing major has changed, but there might be some small glitches in it still.  Please post here if you encounter any bugs.

Yes! The blog quote is fixed. It looks very nice and colorful. Great fixes! Thank you!  :)

Everyone's blog I've looked at recently lost a few of their shouts(my own, LadyKn1ght, and NoSoupForYou)

Yes, what happened to my shouts!??  :mad:

I lost about 75% of them!!

Hurm - I'll take a look.  They're still in the database, I think, so we should be able to retrieve them.

I can't seem to change my blog's colors via the blog control panel. Why?

Yeah, my blog is now ugly.  :cursing:

No offense. :D:D:D


inequalities
trigonometry
geometry
circumcircle
incenter
inequalities unsolved

Prove the following when $ a,b,c$ are the sides of a triangle....
\[ \sqrt {\frac{{bc(b \plus{} c)^2 }}
{{b \plus{} c \minus{} a}}}  \plus{} \sqrt {\frac{{ca(c \plus{} a)^2 }}
{{c \plus{} a \minus{} b}}}  \plus{} \sqrt {\frac{{ab(a \plus{} b)^2 }}
{{a \plus{} b \minus{} c}}}  \geqslant (a \plus{} b \plus{} c)^{3/2}\]

这个问题好像应该是：
\[ \sqrt{\frac{{bc(b\plus{}c)^{2}}}{{b\plus{}c\minus{}a}}}\plus{}\sqrt{\frac{{ca(c\plus{}a)^{2}}}{{c\plus{}a\minus{}b}}}\plus{}\sqrt{\frac{{ab(a\plus{}b)^{2}}}{{a\plus{}b\minus{}c}}}\geqslant\frac{2}{\sqrt{3}} (a\plus{}b\plus{}c)^{3/2} .\]
它等价于
\[ \sum {\cos \frac{{B \minus{} C}}{2}}  \geqslant \frac{2}{{\sqrt 3 }}\sum {\sin A} .\]

\[ \sqrt {\frac{{bc{{(b \plus{} c)}^2}}}{{b \plus{} c \minus{} a}}}  \equal{} \frac{{bc}}{{{t_a}}}\sqrt {a \plus{} b \plus{} c}  \equal{} 2R\sqrt {a \plus{} b \plus{} c} \cos \frac{{B \minus{} C}}{2}\]

\[ \begin{gathered}
  \sin A \plus{} \sin B \plus{} \sin C \leqslant \frac{{3\sqrt 3 }}
{2}, \hfill \\
  \cos \frac{{B \minus{} C}}
{2} \plus{} \cos \frac{{C \minus{} A}}
{2} \plus{} \cos \frac{{A \minus{} B}}
{2} \leqslant 3 \hfill \\
\end{gathered}\]
But that doesn't imply that
\[ \sum{\cos\frac{{B\minus{}C}}{2}}\geqslant\frac{2}{{\sqrt 3 }}\sum{\sin A}.\] :?:  :wink:

This is how I did it:

 [geogebra]bf52141e641be4a255c8074f4da5ab44e9208a37[/geogebra]
Now,

the inequality is reduced to proving
$ AD\plus{}BE\plus{}CF\ge AB\plus{}BC\plus{}CA$ (which is obvious by Ptolemy)

where $ AD,BE, CF$ are the segments joining the vertices $ A,B,C$ to the circumcircle through the incenter

\[ \sum{\cos\frac{{B\minus{}C}}{2}}\geqslant\frac{2}{{\sqrt 3 }}\sum{\sin A}.\]
it can be proved.
please see CRUX-613.

[quote="Xaenir"]as a matter of fact i don't understand kanji.... :D[/quote]
It’s Chinese, not Japanese. :P

[quote="wxd686"][color=black]The inequality probably ought to be this：[/color]
\[ \sqrt {\frac {{bc(b \plus{} c)^{2}}}{{b \plus{} c \minus{} a}}} \plus{} \sqrt {\frac {{ca(c \plus{} a)^{2}}}{{c \plus{} a \minus{} b}}} \plus{} \sqrt {\frac {{ab(a \plus{} b)^{2}}}{{a \plus{} b \minus{} c}}}\geqslant\frac {2}{\sqrt {3}} (a \plus{} b \plus{} c)^{3/2} .\]
[color=black]It’s equivalent to[/color]
\[ \sum {\cos \frac {{B \minus{} C}}{2}} \geqslant \frac {2}{{\sqrt 3 }}\sum {\sin A} .\]
[/quote]

Thanks for  ValentineA ! :P
You help me translate it from Chinese into English.

:D
Actually kanji and chinese look a lot similar , since kanji includes those symbols which came to be included in Japanese from chinese...

We use $ m_a,  m_b,  m_c$ for the medians, Prove or disprove that
\[ \sum {\frac {{bc}}{{{m_a}}}} \geqslant \frac {2}{{\sqrt 3 }}(a \plus{} b \plus{} c).\]


Putnam
number theory proposed
number theory

Show that the following diophantine equation has an unique solution:
$a^{n+1}-(a+1)^n=2001$

This is Putnam, 2001, A5:
[url]http://www.kalva.demon.co.uk/putnam/psoln/psol015.html[/url]


[quote="AIME15"]A ream of paper containing $ 500$ sheets is $ 5$ cm thick.  Approximately how many sheets of this type of paper would there be in a stack $ 7.5$ cm high?
\[ \textbf{(A)}\ 250 \qquad \textbf{(B)}\ 550 \qquad \textbf{(C)}\ 667 \qquad \textbf{(D)}\ 750 \qquad \textbf{(E)}\ 1250
\]
[/quote]

[hide]$ \frac{500}{5}\equal{}\frac{x}{7.5}\implies x\equal{}\boxed{750}$[/hide]

you just need to multiply 500 by 7.5/5 which is 1.5 so it's 750


Is there any one how can help me to find some books in pdf version for "Dynamics of discrete systems" :?:

sorry ramtin but what is it about??
but if you want I can give it to you about a month later


ARML
MATHCOUNTS
email
probability
expected value

Well, what did you think?

I only got four right, because I thought 51 was divisible by 6. :blush:  Oh well.  How'd everyone else do?

i made a LOT of stupid mistakes and  i mean a lot. there were also like 3 or 4 questions i had seen before, and only got 1 of them right .  :(

But I got 8 which i believe was the 2nd highest score (tied like 3 people) and the highest score was 10 by Junda....

That power round though was UNBELIEVABLY HARD

i think i know what the team A probably is gonna be based on what i saw today

[quote="joml88"]What were the power round questions about? (just give me the general topic, etc...)[/quote]

clusters or something, i don't know......i just know i made no contrbution due to my lackluster skills

[quote="joml88"]What were the power round questions about? (just give me the general topic, etc...)[/quote]

They were about expected value, and clusters of cars that had different speeds.

[quote="Iversonfan2005"]i didn't understand a single thing about it....

I hate proofs and stuff...although Ben Hamner was tearing em up[/quote]

I understood the questions.  However, I have trouble proving things.  That would make a great AoPS book; I know I would buy it.

you should get volume 2...they have a good deal of proofs in that. Although unfortunatley, i skipped all the proof parts :blush:

I actually have vol 2.  A lot of stuff is beyond me, but I'm going through it now.   hadn't actually read it, but now that I finished vol. 1 and MathCounts, I guess I should.

if i were you, i'd get used to the problems of getting started for a month or so and do volume 2 during the summer. Volume 2 really takes some understanding to get through but once you finish it, problems seem to be a lot easier to solve

i thought the practice went really well. made some stupid mistakes (well, everyone did so that doesn't really mean anything) and got a 7 on the individual. the power round topic was kinda cool but i'm not very good w/ proofs so i couldn't do too much w/ it.
too bad we only plan one of these a year. hint for next year, get the qualifing exam done early and have practices every 4-6 weeks from november thru april (maybe excluding february). that would really help the team.

Hey did anyone else know that there are only 49 people on the All State team, yet 32 want to go to ARML!  :)  I guess more people are interested in math this year :yup:

[quote="DummAddict"]I did ok on the individual...got 7 but made a few stupid mistakes. i agree there should be more practices, this is my first year on state and i learned more in that practice than i would have imagined, it helps alot[/quote]

wow, if you learned "more than you would have imagined" in that one practice, then you'll learn even more than you would imagine on this forum

[quote="Iversonfan2005"]
wow, if you learned "more than you would have imagined" in that one practice, then you'll learn even more than you would imagine on this forum[/quote]

The main problem with this forum that I have noticed goes like this:

I see a problem someone has posted, and don't understand it.  I look at another user's solution, and now I understand the problem.  However, there are no more problems in the same thread of the EXACT SAME TYPE that I can do by myself to ensure that I understand the concept.  A good idea would be to post two very similar problems per thread.  Just an idea; it would sure help me.

As to the above comment about more practices, most students will have trouble making more practices.  Also, if we do the preliminary too early, we miss out on some of the best students.  On top of that, I organize the whole thing by myself.
However, I want to move towards doing more online practices.  I am about to send out an email to all the team members telling them where to find the South Carolina section on this website.  Also, I think we should be able to do the math camp thing this year, but not in the way that we originally talked about.  I am willing to house four students at my house.  I have also heard that Mrs. Reeves was willing to house a few students.  I think we can meet at my school, but I am not sure.  This will cut down on costs significantly.  I think the USC idea will cost too much.  On top of that, I think we will be able to leave a day early to have a day of practice at Penn State.

[quote="prybarczyk"] Also, I think we should be able to do the math camp thing this year, but not in the way that we originally talked about..... On top of that, I think we will be able to leave a day early to have a day of practice at Penn State.[/quote]

Awesome!

[quote="furious"][quote="prybarczyk"] On top of that, I think we will be able to leave a day early to have a day of practice at Penn State.[/quote]
have you come up with a solution to ben hamner's graduation date then?[/quote]

Dude comeon...last year we heard some kid from Lehigh say they gave up prom for ARML  :rotfl:

Is he valedictorian though?  That could make it a little more important :D

I gave up sleep for ARML.  NOW THATS SACRIFICE.

Problem:- Graduation at Academic Magnet

Solution:- I believe Bob Hess will drive Ben Hamner up to Penn State on Friday.


Prove that for any k there exist natural m so that $3^k|m^3+1999.$

[quote="grobber"]Once we have such an $m$ for $k$, we can try searching for one for $k+1$. If $3^{k+1}|m^3+1999$, we are done. Otherwise, look for solutions of the type $m+t3^{k-1}$, where $t$ takes one of the values $1,2$. I'm in a hurry right now, but this looks like it should work.[/quote]
Of course.

If I'm not wrong,this problem was used in Japan 1999 :?.Thic solution is as similar as yours grobber ;)

But I remember a slightly tougher problem from Japan. We had to show that for each $k$ there is an $m$ s.t. $3^k|\mathcal E$ but $3^{k+1}\not |\ \mathcal E$, where $\mathcal E$ was some expression depending on $m$ (something very similar to what we have here, but I can't remember the exact values; I don't think the free term was $1999$).

Yup,you're right.The free term was $ 17 \neq 1999$.Sorry for my bad memory :(


let $A$ and $B$ be fixed point.find the locus of $P$ such that $\frac{PA}{PB}=k$ and $k$ is a fixed value. :D

if $k \neq 1$ it's an [i]apollonius' circle[/i], if $k = 1$ is the line bissector of $AB$.

http://www.jimloy.com/cindy/apoll.htm

this can be proven if you use a cooridinate grid and set the distances in a ratio and simplify


Determine all triples $(x,k,n)$ of positive integers such that

$3^k - 1= x^n$

n=1 give $(3^k-1,k,1)$.
If k is odd, then $ord_2(3^k-1)=1$, therefore n=1 (don't give new solutions).
If k is even, then $ord_2(3^k-1)=m=2+ord_2(k),$ therefore n|m.
$(3^{k/2}-1)(3^{k/2}+1)=(x^n) \Longrightarrow \ x|2,n=2+ord_2(k)$. Give one solution  $(2,2,3) \ 3^2-1=2^3$.

Rust, ok. Because $3^{m}\equiv 3 (mod 8)$, $m$-odd.  It results, that $2|k$.  So, $q=1$. $p=n-1$. But why $x$ must be a divisor of 2?

Sufficiently to prove for all prime p:
$3^p-1=2y_1^n,3^p+1=4y_2^n \Longrightarrow \ n=1 \ or \ y_1=y_2=1$.

[quote="Rust"]$ord_2(3^k-1)=1$[/quote]

What do you mean by that?

He means $v_2(3^k-1)$.
For that, see the formulary.

Additionally: see http://www.mathlinks.ro/Forum/viewtopic.php?t=28544

[quote="Loser"][quote="seamusoboyle"]It has recently been proven that $3^2$ an $2^3$ are the only two powers that differ by 1, with the power bigger than 1. Kinda ruins the problem, which is very nice![/quote]

Hehehe, I know it's well known but how can you prove this ;)[/quote]

Well since you ask...


function
algebra
domain
calculus
derivative
real analysis
real analysis unsolved

if $f$ is differentiable on $I$,and such that for $\any x \in I$,we have :
$f'(x)=g(f(x))$,
($g$ is a function on $I$)

prove that $f$ is monotonic

it also seems easy,I have an approach,But I can't prove it strictly,can you help me :)

This is an autonomous differential equation. I want to invoke the existence-uniqueness theorem for differential equations, but to do so, I would require a much stronger conditionon $g$ than just that $g$ is a function. $g$ satisfying a Lipschitz condition on the range of $f$ would suffice.

Note also that domain of $g$ shouldn't be $I$ but rather $f(I).$

Maybe it's about time people stopped posting this one all over the place :).

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=differential+equation&t=44962[/url]

Also note that there is another topic concerning the same problem on the same page of the "Unsolved and Proposed Problems" section (although there's no solution there):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72088[/url]

I humbly disagree with the first remark: this problem is different from the continuous case.  :)  At least, the proof in the link grobber gave doesn't work here. As to xirti's problem, it is, indeed, exactly the same. :P

Why humbly? :) I'm sorry for that. For some reason, I was sure this was the problem posted by Harazi. Anyway, the problem isn't difficult at all. I hope this is Ok.

Take $I=[a,b],\ a<b$ such that $f(a)=f(b)$, and assume that $M=\max_If>m=\min_If$. $g$ cannot be identically zero on $[m,M]$, so we can find $y\in[m,M]$ with $g(y)\ne 0$. Notice that this $y$ can actually be chosen in $(m,M)$, since otherwise $f'$ would have at most three values on $I$, namely $g(M),g(m)$, and $0$, which is impossible because $f'$ has the intermediate value property.

Now, for each such $y\in(m,M)$, we can find $u<v\in I$ such that $f(u)=f(v)=y$. By the choice of $y$, we also have $f'(u)=f'(v)\ne 0$. Just as in the other topic, we use the fact that in such a situation, there will always be a root of $f(x)=y$ strictly between $u$ and $v$. Take $\bar u$ to be the closest such root to $u$. Since $f'(u)\ne 0,\ \bar u\in(u,v)$ does exist. We are now in the situation where $u<\bar u, f(u)=f(\bar u)=y,\ f(x)\ne y,\ \forall x\in(u,\bar u)$, and $f'(u)=f'(\bar u)\ne 0$, which contradicts the fact that there is a root of $f(x)=y$ strictly between $u$ and $\bar u$.

[quote="grobber"] Take $\bar u$ to be the closest such root to $u$. Since $f'(u)\ne 0,\ \bar u\in(u,v)$ does exist. We are now in the situation where $u<\bar u, f(u)=f(\bar u)=y,\ f(x)\ne y,\ \forall x\in(u,\bar u)$, and $f'(u)=f'(\bar u)\ne 0$, which contradicts the fact that there is a root of $f(x)=y$ strictly between $u$ and $\bar u$.[/quote]
grobber,this is the point I didn't find.
Thanks you and all of you.
I agree this problem is equal to xirti's problem :P

[quote="grobber"] Take $\bar u$ to be the closest such root to $u$.[/quote]

Why if $\bar u =u$?

As I see, in your post, grobber, you proved that $f(x)\neq f(y)$ if $f'(x)\neq 0$. Am I wright?  Why if $f'(x)=f'(y)=0$ and $f(x)=f(y)$? You must reach a contradiction in this case too.

 Sorry if there are sily questions, but can you explain more? :blush:

[quote="xirti"][quote="grobber"] Take $\bar u$ to be the closest such root to $u$.[/quote]

Why if $\bar u =u$?
:[/quote]
This is because $f'(u) \neq 0$
[quote="xirti"]
As I see, in your post, grobber, you proved that $f(x)\neq f(y)$ if $f'(x)\neq 0$. Am I wright?  Why if $f'(x)=f'(y)=0$ and $f(x)=f(y)$? You must reach a contradiction in this case too.

 Sorry if there are sily questions, but can you explain more? :blush:[/quote]
I think grobber prove there exist $f'(u) \neq 0$,then we consider that f(u),because $f(u) \in (m,M)$,that we can get f(v) such that $f(v)=f(u)$,then $f'(u)=f'(v) \neq 0$.I hope it is ok.

[quote="xirti"]I understand.. but, if $\sup\{ x\le u: f'(x)=0\}=\inf \{x\ge u: f'(x)=0\}=u$ then  $\bar u$ do not exist :!: e.g.
\[ f'(x)=\begin{cases} \frac 1 2+sin\frac 1 x \text{ if } x\neq 0\\ \frac 1 2 \text{ for } x=0\end{cases}  \][/quote]
Yes,sorry.This is a problem.
Can grobber explain it?

Hmm...I do not understand xirti's question: $\bar u$ is the closest to $u$ point where $f$ is the same as at $u$. And we know that $f$ is continuous and $f'(u)\ne 0$, so $f$ is different from $f(u)$ in a small neighborhood of $u$. So, what's the problem? Note that grobber doesn't try to find the closest point with the same derivative (this, indeed, would be impossible) but just the closest point with the same value, so xirti's example is actually not a counterexample to anything. Am I missing something? :?

[quote="fedja"]$f'(u)\ne 0$, so $f$ is different from $f(u)$ in a small neighborhood of $u$.[/quote]
I think you are right...Sorry for my stupidness


quadratics
number theory
relatively prime
number theory proposed

Consider sequence $ a_{n}$ definded by $ a_{n}\equal{}n^{2}\plus{}n\plus{}1$. Find a subsequence $ b_{n}$ of this sequence such that $ b_{1}\equal{}a_{1}$, $ b_{2}\equal{}a_{2}$, $ b_{3}>a_{3}$, every pair of terms from this subsequence are relatively prime, and there are primes which divide no terms of this subsequence.

There are primes that don't divide any number of the form $ n^2\plus{}n\plus{}1$; $ 2,5,11,$ and all primes congruent to $ 5$ or $ 11$ mod $ 12$. (Quadratic reciprocity on the question of whether $ \minus{}3$ is a residue)

Making a relatively prime subsequence is also easy. First, $ a_1\equal{}3$ and $ a_2\equal{}7$ are relatively prime. Now, choose $ N(k)\equal{}\prod_{j\equal{}1}^{k\minus{}1}b_j$ for each $ k\ge 3$ and define $ b_k\equal{}a_{N(k)}$. We have $ b_k\equiv 1\mod b_j$ for all $ j<k$.


$ P$ is the midpoint of $ \overline{BD}$. $ AP = BP = 4$, $ \overline{AP} \perp \overline{BD}$, $ \overline{BD} \perp \overline{DC}$, $ \overline{AB} \perp \overline{BC}$. How many units, expressed in simple radical form, are in the perimeter of pentagon $ ABCDP$?
[asy]size(150); defaultpen(linewidth(0.8)); import geometry_dev;
pair A = origin, B = (4,4), C = (12,-4), D = (4,-4), P = (4,0);
draw(A--P--B--cycle);
draw(B--D--C--cycle);
markrightangle(A,B,C,10);
markrightangle(A,P,B,5);
markrightangle(B,D,C,5);
label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,E); label("$4$",A--P,S); label("$4$",B--P,E);[/asy]

Since $ P$ is the midpoint of $ \overline{BD}$, $ BP = PD = 4$.  Both $ \Delta APB$ and $ \Delta BDC$ are $ 45^{\circ} - 45^{\circ} - 90^{\circ}$ right triangles (you should be able to see why).  Therefore, $ DC = BD = BP + PD = 4 + 4 = 8$, $ AB = AP\sqrt {2} = 4\sqrt {2}$, and $ BC = DC\sqrt {2} = 8\sqrt {2}$  So our final answer is $ AP + AB + BC + DC + PD = 4 + 4\sqrt {2} + 8\sqrt {2} + 8 + 4 = \boxed{16 + 12\sqrt {2}}$
[asy]size(150); defaultpen(linewidth(0.8)); import geometry_dev;
pair A = origin, B = (4,4), C = (12,-4), D = (4,-4), P = (4,0);
draw(A--P--B--cycle);
draw(B--D--C--cycle);
markrightangle(A,B,C,10);
markrightangle(A,P,B,5);
markrightangle(B,D,C,5);
label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,E); label("$4$",A--P,S); label("$4$",D--P,E); label("$8$", D--C, S); label("$8\sqrt{2}$", B--C, NE); label("$4\sqrt{2}$", A--B, NW);[/asy]


algebra
function
domain
complex analysis
complex analysis unsolved

After Riemann mapping theorem, we know that for every simply connected domain $U\subset\mathbb C$ (not $\mathbb C$) whose boundary is a simple continuous curve, there exists a conformal mapping $f$ which maps $U$ onto the unit disk $\Delta$, and this $f$ can be continuously extended to the boundary of $U$. But how can I prove that $f$ maps $\overline U$ one-one onto $\overline\Delta$?
Is there anything wrong?

Why $f$ is injective at the boundary of $U$?
Why $f$ is surjective?


percent
geometry
perimeter
LaTeX
percent increase

The diagram consists of 4 congruent squares made with twelve
2'' toothpicks. By moving exactly three toothpicks, three congruent
squares can be formed so that each toothpick forms a side of a
square. What is the percent increase in the perimeter of the figure?
(Assume that the toothpicks intersect at their endpoints.)

[asy]size(75); real space = .1;
for(int i = 0; i<2; ++i) for(int j = 0; j<2; ++j)
draw((i+space,j)--(i+1-space,j) ^^ (i,j+space)--(i,j+1-space));
for(int i = 0; i<2; ++i)
draw((i+space,2)--(i+1-space,2) ^^ (2,i+space)--(2,i+1-space));[/asy]

If the squares dont hafta be connected, this is the diagram you end up with:

EDIT: NVM I CANT USE LATEX FOR BEANS SRY YOULL HAFTA FIND THE DIAGRAM YOURSELF! :oops:  :P

each side is 2", so we have a perimeter of 12*2=24"
the original perimeter was 8*2=16
so the increase is (24-16)/16=50%

thank you. I have seen this many times now and have never understood it

I don't get it, how is the original perimeter 16? How do you have a new perimeter of 24? Could you draw the diagram?

I can't draw a diagram either, but I don't think you need one.

You first observe that you have 12 toothpicks.  You then observe that there are only 3 squares in the end - 4 toothpicks per square, and a square has 4 sides!  This means that the new squares cannot in anyway share any sides.  Otherwise, you'll end up with some toothpicks left out where the squares share sides.

Thus, regardless of how you move 3 of the toothpicks, you're going to end up with 3 separate squares in the end.  The original perimeter had 8 toothpicks, and the new perimeter has 12 toothpicks since each toothpick accounts for the perimeter.  Thus, the percent increase is $100*\frac{12-8}{8} = \boxed{50}$.

Hi, I am confused on how to solve this problem -- I read the solutions but don't get what the new diagram would look like.

Have you read the solutions in this thread?


[quote="Ignite168"]Fully factor $(x^{8}+x^{4}y^{4}+y^{8})(x^{2}+y^{2})(x+y)(x-y)$.[/quote]

Do you mean expand.  If so...

[hide="solution"]Expand the last two to get $x^{2}-y^{2}$

Then expand the next two to get $x^{4}-y^{4}$

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$

Therefore, $x^{12}-y^{12}$[/hide]


function
advanced fields
advanced fields unsolved

Hello,

I am have been given a function and asked to find the fourier series representation of it.  What are the 'steps' involved in finding this?

Also i been asked "what is the component of f(t) of frequency X.

This problem is part of an assignment at college, and am keen to solve the problem mathematically on my own.  I simply need help with the 'steps'required to solve it.

I appreciate any help...

Is this the fourier series of a complex function or a strictly real one?

Anyways, you can find the formulae for coefficients here:

http://mathworld.wolfram.com/FourierSeries.html


How do you get $\cos {2 \alpha} = \cos^2 \alpha - \sin^2 \alpha$?

I'm not familiar with trigonometry identities so any good help would be great.

Thanks!

Well, you could use the identity for angle sums of cosines:

\[\cos(\alpha+\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta.\]

Then you would make $\beta=\alpha$:

\[\cos(2\alpha)=\cos^2 \alpha -\sin^2 \alpha.\]

The proof I know of the sum of angles formula is geometric (and takes a little casework).  There are better proofs out there if you use this site called http://www.google.com . :D

blahblahblah--are you sure the e^i :theta:  thing doesn't itself come from sum of angle formulae at least partly?  Anyway, I usually use the trig stuff to understand the complex number stuff :D

[quote="blahblahblah"]In most analysis textbooks, the trig functions are defined in terms of the complex exponential.[/quote]

Yes and see  the post above:
[quote="blahblahblah"]It's good to develop a feeling for trig identities using complex numbers.

$e^{it}=\cos t+i\sin t$
$\Longrightarrow \cos {2t}+i\sin {2t}=e^{i2t}=(e^{it})^2=(\cos^2t-\sin^2t)+i(2\cos t\sin t)$
$\Longrightarrow \cos{2t}=\cos^2t-\sin^2t$
$\Longrightarrow \sin{2t}=2\cos t\sin t$

By equating real and imaginary parts.[/quote]

Is a nice way to derive these formulae , for example  taking,
$ \cos {3t}+i\sin {3t}=  (\cos t + i \sin t)^3 $
and equating real and imaginar parts you get the formula for  $\sin {3x}$ and $ \cos {3x} $


let ABC be an eqaliraterall tiangle.M is a point in the interior of it ,inwich:

angle MAB+angle MBC+angle MCA=90.

find the locus of M.

When you say 90, do you mean $90^{\circ}$?  Because then the locus is [hide]nothing.  We know that if $M$ is in the interior of the triangle, including the sides, then $\angle{MAB}\le \angle CAB=60^{\circ}<90^{\circ}$[/hide]

I think you are wronk K81o7, indeed each angle is smaller than 90°, but the sum can be 90.

Searched points: All points M on one of the three angle bissectors of the triangle. I don't know how to prove that these are the only ones, though...  :?

Oh...heheh didn't realize it was sum, i assumed it was $MAB=MBC=MCA=90$..

See  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=54016]locus (nice, nice, ...)[/url] in Geometry, Olympiad section.

arne ,i think brocard point will help. but in this case,brocard point will be the intersection of angle bisectores.and the

locus wont be complete :?

and about what jan said:

jan its not hard to prove that.suppose X is a point in triangle wich isnt on the angle bisectores. and

XAB+XBC+XCA=90

then itwill prove easily that:angle XBC=90.


hello guys i have come to the forum at last!!!! can you guess who iam?? :P

Rushil would use his own ID na??

My Guesses

i) Prince William
ii) Hillary Clinton
iii) Abishek Dang
iv) Bjarne Stroustrup
v) Iona Potapov
vi) Achin Bansal

Ofcourse it can't be any of them because of his user ID. Physici Olympiaci... The title of the most brilliant book in physics of all time . :D:D:D For those who know the funda. :rotfl:

My guesses
gokul or prasad or ranga? (Maybe zombie)

most likely R. Gokul Subramaniam  :D
Even though it might just be Osama ( I invited him a long time ago )  :P

i am afraid Rushil might have tipped osama that i ruined his power :D

dont you worry i have osama under my control  :P
A sound click of my left hand and USA is ruined.
A sound click of my right hand and Osama calms down  :D

thanks madness that calmed me a bit  :rotfl:

Call for mod Call for moD: Please merge the the two threads with Introduction/Welcome to India thread.

[quote="hash_include"]my guess is prasad

and LOL at the iona popatov shreyas!! :D  :rotfl:

btw who is this rushil??[/quote]

It is prasad. He posted in the other thread. Anyways. Rushil is rushil goel, ex-mod of this forum, still mod of the physics forum. IIT AIR 7 (or 6 or 8) and Training camp selected to 3 olympiads 2006.

I don think rushil got to IPhO TC da.. Jus confirm..

no he didnt go to IPhOTC
he himself regretted it saying that it was and probably still is his favourite subject  :(
For Further details see the Olympiad results of 2006