function
limit
calculus
derivative
calculus computations

Suppose $ f$ is a continuous function such that the limit $ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}}{h}$ exists.
Then is it true that $ f$ is differentiable at $ a$?

[hide="Comment"][b]comment[/b]: Actually , the "nearly" converse of this statement can be proved,
which is: if $ f$ is differentiable at $ a$, then the limit $ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}}{h}$ exists and equals $ f'(a)$
It can be done as follows:
$ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} h) \minus{} f(a)}} {h} \plus{} \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}} {h}$

$ \equal{} \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h) \plus{} f(a \plus{} h) \minus{} f(a)}} {h}$
$ \equal{} \mathop {\lim }\limits_{2h \to 0} \frac {{\left( {f(a \plus{} 2h) \minus{} f(a)} \right)2}} {{2h}} \equal{} 2f'(a)$

$ \therefore$$ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}} {h} \equal{} f'(a)$
[/hide]

i have a doubt:1.if a function is differentiable at a point then is it true that the limit at that point exists
2.for a discontinuous function  does limit exist at the point of discontinuity

[quote="aadil"]for a discontinuous function  does limit exist at the point of discontinuity[/quote]  A continuous function is one for which $ \lim_{x \to a} f(x) \equal{} f(a)$ for all $ a$.  A discontinuous function can fail be be continuous either because the limit fails to exist or because the limit exists but is not equal to the function value.  The latter is a "removable discontinuity" -- for example, the function $ \chi$ defined by
\[ \chi(x) \equal{} \begin{cases} 1, & x \in {\bf Z} \\
0, & \textrm{otherwise}\end{cases}\]
is discontinuous but $ \lim_{x \to a} \chi(x) \equal{} 0$ for all $ a$.

The answer to your first question is that differentiability implies continuity (but the converse is false).  Both of these answers can be found in any standard calculus text.

@JBL: I did not  say that continuity implies differentiability.

The function in the question  is continuous and the limit $ \mathop{\lim }\limits_{h\to 0}\frac{{f(a\plus{}2h)\minus{}f(a\plus{}h)}}{h}$ exists([hide="Note"]note that this is not the limit used to define the derivative at $ a$[/hide]),
I wanted to know if these conditions imply that $ f$ is differentiable at $ a$.

[hide="Comment on comment"][quote="apratimdefermat"]
It can be done as follows:
$ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} h) \minus{} f(a)}} {h} \plus{} \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}} {h}$

$ \equal{} \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h) \plus{} f(a \plus{} h) \minus{} f(a)}} {h}$
$ \equal{} \mathop {\lim }\limits_{2h \to 0} \frac {{\left( {f(a \plus{} 2h) \minus{} f(a)} \right)2}} {{2h}} \equal{} 2f'(a)$

$ \therefore$$ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}} {h} \equal{} f'(a)$
[/hide][/quote]There's a slight flaw in this proof. In adding the two limits, you already assumed that $ \mathop {\lim }\limits_{h \to 0} \frac {{f(a \plus{} 2h) \minus{} f(a \plus{} h)}} {h}$ exists (because the addition law requires both limits exist), even though you have yet to prove that.

apratimdefermat, none of my comments were directed to you.

Elixir, it's also true that if $ \lim_{x \to a} f(x) \equal{} L$ and $ \lim_{x \to a} g(x)$ does not exist then $ \lim_{x \to a} f(x) \plus{} g(x)$ does not exist.  (This is almost trivially equivalent to the addition rule as you've stated it: if $ \lim f(x)$ and $ \lim f(x) \plus{} g(x)$ exist, then $ \lim \minus{}f(x)$ exists and so does $ \lim (f(x) \plus{} g(x)) \minus{} f(x) \equal{} \lim g(x)$ exists.)  In any case, we get to start by computing, and if the answer exists, that retroactively shows that the limit we added exists, as well as giving its value.

Here's an argument:

Start by fixing $ h$ and consider the sum

$ \frac{f(a\plus{}h)\minus{}f(a\plus{}h/2)}{h}\plus{}\frac{f(a\plus{}h/2)\minus{}f(a\plus{}h/4)}{h}\plus{}\frac{f(a\plus{}h/4)\plus{}f(a\plus{}h/8)}{h}\plus{}\cdots$

The sum telescopes, and by the continuity of $ f$ (the continuity of $ f$ is vital to this theorem), this sum become $ \frac{f(a\plus{}h)\minus{}f(a)}{h}.$ So then,

$ \frac{f(a\plus{}h)\minus{}f(a)}{h}\equal{}\frac12\cdot\frac{f(a\plus{}h)\minus{}f(a\plus{}h/2)}{h/2}\plus{}\frac14\cdot\frac{f(a\plus{}h/2)\minus{}f(a\plus{}h/4)}{h/4}\plus{}\cdots$

We are given that $ \lim_{h\to0}\frac{f(a\plus{}2h)\minus{}f(a\plus{}h)}{h}\equal{}L$ exists. Therefore, given any $ \epsilon >0$ there exists $ \delta$ such that if $ |2h|<\delta,$ then $ \left|\frac{f(a\plus{}2h)\minus{}f(a\plus{}h)}{h}\minus{}L\right|<\epsilon.$

For such $ h,$ it follows that $ \left|\frac1{2^k}\left(\frac{f(a\plus{}2^{1\minus{}k}h)\minus{}f(a\plus{}2^{\minus{}k}h)}{2^{\minus{}k}h}\minus{}L\right)\right|<2^{\minus{}k}\epsilon.$

So $ \frac{f(a\plus{}h)\minus{}f(a)}{h}\minus{}L\equal{}\sum_{k\equal{}1}^{\infty}\frac1{2^k}\left(\frac{f(a\plus{}2^{1\minus{}k}h)\minus{}f(a\plus{}2^{\minus{}k}h)}{2^{\minus{}k}h}\minus{}L\right)$

$ \left|\frac{f(a\plus{}h)\minus{}f(a)}{h}\minus{}L\right|\le\sum_{k\equal{}1}^{\infty}\left|\frac1{2^k}\left(\frac{f(a\plus{}2^{1\minus{}k}h)\minus{}f(a\plus{}2^{\minus{}k}h)}{2^{\minus{}k}h}\minus{}L\right)\right|$

$ <\sum_{k\equal{}1}^{\infty}2^{\minus{}k}\epsilon\equal{}\epsilon$

Therefore, the given hypotheses imply that $ f$ is differentiable.

Note that this proof would collapse without the assumption that $ f$ is continuous, and indeed, this can be shown to be false for any step function (piecewise constant function) at a point of discontinuity.

Somehow I had started down this path and then decided it didn't work out ... alas.

Thank you Dr. Kent for your excellent argument.

I had started down this path too , but I was stuck somewhere.

[hide="Well, here it is "]$ \mathop {\lim }\limits_{h \to 0} \frac{{f(a \plus{} 2h) \minus{} f(a \plus{} h)}}{h} \equal{} L$

$ \mathop {\lim }\limits_{h \to 0} \frac{{f(a \plus{} h) \minus{} f(a \plus{} \frac{h}{2})}}{h} \equal{} \frac{L}{2}$

$ \vdots$

$ \mathop {\lim }\limits_{h \to 0} \frac{{f(a \plus{} \frac{h}{{2^{n \minus{} 1} }}) \minus{} f(a \plus{} \frac{h}{{2^n }})}}{h} \equal{} \frac{L}{{2^n }}$

Adding these equations we get,
$ \mathop {\lim }\limits_{h \to 0} \frac{{f(a \plus{} 2h) \minus{} f(a \plus{} \frac{h}{{2^n }})}}{h} \equal{} L\left( {1 \plus{} \frac{1}{2} \plus{} \frac{1}{{2^2 }} \plus{} ... \plus{} \frac{1}{{2^n }}} \right)$

And finally, I ended up here ,(this is where I was stuck)
$ \mathop {\lim }\limits_{n \to \infty } \left( {\mathop {\lim }\limits_{h \to 0} \frac{{f(a \plus{} 2h) \minus{} f(a \plus{} \frac{h}{{2^n }})}}{h}} \right) \equal{} 2L$

Suppose, $ g(h,n) \equal{} \frac{{f(a \plus{} 2h) \minus{} f(a \plus{} \frac{h}{{2^n }})}}{h}$
then we have
$ \mathop {\lim }\limits_{n \to \infty } \left( {\mathop {\lim }\limits_{h \to 0} g(h,n)} \right)\equal{}2L$

I [b]doubt[/b] if I am right in doing this:

$ \mathop {\lim }\limits_{n \to \infty } \left( {\mathop {\lim }\limits_{h \to 0} g(h,n)} \right)$

$ \equal{} \mathop {\lim }\limits_{h \to 0} \left( {\mathop {\lim }\limits_{n \to \infty } g(h,n)} \right)$

$ \equal{} \mathop {\lim }\limits_{h \to 0} \left( {\mathop {\lim }\limits_{n \to \infty } g(h,n)} \right)$

$ \equal{} \mathop {\lim }\limits_{h \to 0} \left( {\phi (h)} \right)$

$ \equal{}\mathop {\lim }\limits_{h \to 0} \frac{{f(a \plus{} 2h) \minus{} f(a)}}{h}$

$ \equal{} 2L$

$ \therefore \mathop {\lim }\limits_{l \to 0} \frac{{f(a \plus{} l) \minus{} f(a)}}
{l} \equal{} L$[/hide]

[quote="JBL"]apratimdefermat, none of my comments were directed to you.

Elixir, it's also true that if $ \lim_{x \to a} f(x) \equal{} L$ and $ \lim_{x \to a} g(x)$ does not exist then $ \lim_{x \to a} f(x) \plus{} g(x)$ does not exist.  (This is almost trivially equivalent to the addition rule as you've stated it: if $ \lim f(x)$ and $ \lim f(x) \plus{} g(x)$ exist, then $ \lim \minus{} f(x)$ exists and so does $ \lim (f(x) \plus{} g(x)) \minus{} f(x) \equal{} \lim g(x)$ exists.)  In any case, we get to start by computing, and if the answer exists, that retroactively shows that the limit we added exists, as well as giving its value.[/quote]But you can't prove something it exists by assuming that it exists. That's circular reasoning.

In any case, the problem has been solved. :)

That"s not at all circular reasoning though it seems like one.
Actually what I showed was that if $ A$ exists and $ B$ exists ,
so does $ A\minus{}B$, which is justified... :wink:

[quote="apratimdefermat"]That"s not at all circular reasoning though it seems like one.
Actually what I showed was that if $ A$ exists and $ B$ exists ,
so does $ A \minus{} B$, which is justified... :wink:[/quote]The first step of your proof, though, was to add that limit to the derivative, even though you weren't sure that limit existed yet (which is what you were trying to prove). It would have been better to start by subtracting the two limits you know exist, namely $ f'(x)$ and $ \equal{}\mathop{\lim }\limits_{2h\to 0}\frac{{\left({f(a\plus{}2h)\minus{}f(a)}\right)2}}{{2h}}\equal{} 2f'(a)$.

But no matter. It was just my teacher rubbing off on me. I became a little finicky with proofs because of this time where he rejected my classmate's proof for doing the same thing: performing limit operations on limits that haven't been proven to exist.

That's a good precautionary principle, but it will occasionaly reject perfectly legitimate arguments.  The statement
\[ \lim f(x) \plus{} \lim g(x) \equal{} \lim(f(x) \plus{} g(x))\]
is not always true: it's possible for one side to not exist while the other exists.  (E.g., let $ f(x)$ be any badly behaved function and $ g(x) \equal{} 1 \minus{} f(x)$.)  [b][i]However[/i][/b], if $ \lim f(x)$ exists then this statement becomes true regardless of whether or not $ \lim g(x)$ exists (assuming we agree to the perfectly reasonable "DNE = DNE").  Discovering at some later step that $ \lim f(x) \plus{} g(x)$ exists simultaneously proves that $ g(x)$ exists and allows you to compute its value,* but you don't need to know that it exists in advance in order to make this computation valid (provided, again, that you already knew $ \lim f(x)$ exists).

* This is not that unusual, since the way a standard epsilon-delta proof of a limit works, you actually need to know what the limiting value is to write down the proof that the limit exists (at least until you acquire some better tools).


induction
number theory solved
number theory

Let {a(n)} be a sequence such that a(1)=1,a(2)=2,a(3)=3 , and a(n) = (a(n-1)*a(n-2) + 7)/a(n-3)  for n > 3 .
Prove that all a(n) are integer and non divisible by 7.
How do you generalize this result ?!
..

Ok, I got it : Clearly, each of the term of the sequence are positive real numbers.
The key obesrvation is to compute the numbers of the form (a_(n+1) + a_(n-1))/a_n.
W
We may rewrite the given relation as
a_n*a_(n-3) = a_(n-1)*a_(n-2) + 7, so that :
a_n*(a_(n-1) + a_(n-3)) = a_(n-2)(a_(n+1) + a_(n-1)).  (1)

Direct checking leads to :
a_(2n+1) + a_(2n-1) = 2a_(2n)
a_(2n) + a_(2n+2) = 5a_(2n+1)
for of n = 1,2,3.
An easy induction using (1) shows that these relations remains true for each integer n > 0.
It follows easily that a_(2n+1) = 2a_(2n) - a_(2n-1) is an integer, and that a_(2n+2) = 5a_(2n+1) - a_(2n) is an integer.
Thus, each term of the sequence (a_n) is a positive integer.

Suppose, for a contradiction, that at least one of the member of this sequence is a multiple of 7. Let n be the smallest integer such that a_n = 0 mod[7].
It is easy to see that n > 4, and since a_n*a_(n-3) = a_(n-1)*a_(n-2) + 7, it follows that a_(n-1)*a_(n-2) = 0 mod[7].
Since 7 is a prime, we deduce that a_(n-1) or a_(n-2) is divisible by 7, which contradicts the minimality of n.

Now, let p be any prime, a,b,c be positive integers such that a+c = kb for some integer k.
Define the sequence (a_n) by a_1 = a, a_2 = b, a_3 = c, and a_n = (a_(n-1)*a_(n-2) + p)/a_(n-3).
The relations (1) are still valid, so the whole reasoning still holds if a_4 is an integer such that a_4 + a_2 is a multiple of a_3, that is (bc+p)/a = (kb 2  + p)/a - b is an integer.
It suffices to choose a,b,k,p such that (kb 2  + p)/a is an integer which is a multiple of c = kb-a.
The above case corresponds to a = 1 (for which the conclusion is clear), p = 7, b = 3, and k = 2.

Pierre.

Hello world, it's not too bad, but have you got  a better generalization ?
Observe the case a(1)=3,a(2)=4,a(3)=1 and a(n)=(a(n-1)*a(n-2) + 35)/a(n-3)
How do ou prove that all a(n) are integer and coprime with 35 ?
..

The fact that a_n is an integer do not use the fact that p is prime. The fact that a_n is coprime with p just use that fact that a_1*a_2 is coprime with p.

So, for your last proposition, the reasoning above can be used.

Pierre.

You are right. I did not say that I solved the general problem with the minimal conditions. You may find many raffinements to the conditions used in the proof above. I just gave some of them.

Pierre.

[quote="pbornsztein"]You are right. I did not say that I solved the general problem with the minimal conditions. You may find many [b]raffinements[/b] to the conditions used in the proof above. I just gave some of them.

Pierre.[/quote]
[b]refinements[/b]

Well... I used the french writing....sorry....

Hmmm...
Myth says that his problems deserve more elegant solutions than mine.
Diogene says that my generalisations are not enough general.
3x-lich says that my english is not correct.

So my question is 'why always me????' :D

Pierre.

Why always you? Because Pierre is famous in MathLinks!!! :D  :D  :D


trigonometry
modular arithmetic
limit
calculus
calculus computations

$ \begin{align}\sum_{k=0}^n f_k g_k &= f_n \sum_{k=0}^n g_k - \sum_{j=0}^{n-1} \left( f_{j+1}- f_j\right) \sum_{k=0}^j g_k$

Let's say I use this identity on a sum, and the left term simplifies how I want it to, but the right term becomes more ugly than the original sum was. Is there a way that I could continue to use this identity on the right term an infinite amount of times? It would result in some double summation with the outside sum being an infinite sum, is there a generalized form of this?

Remember that in most of the uses we have for this, we are not trying to compute an exact value for a sum but rather we are engaged in some sort or estimation. In that context, it is often sufficient to know that the partial sums $ \sum_{k\equal{}0}^jg_k$ are bounded.

Example 1: prove that $ \sum_{n\equal{}1}^{\infty}\frac{\sin nx}n$ converges for all $ x\not\equiv 0\pmod{\pi},$ even though it doesn't converge absolutely. Plan: use a summation by parts, summing the $ \sin nx$ and differencing the $ \frac1n.$ For each fixed $ x,$ the sums of the sines are bounded in $ n,$ whereas the differences of the $ \frac1n$ give an absolutely convergent series.

Example 2: Prove Abel's theorem: If $ \sum_{k\equal{}0}^{\infty}a_k\equal{}L,$ then $ \lim_{r\to1^\minus{}}\sum_{k\equal{}0}^{\infty}a_kr^k\equal{}L.$

I'll let you figure out what the plan is, but note that we do know what the behavior is of $ \sum_{k\equal{}0}^na_k$ for large $ n.$


A candy bar weighs $ 4.48$ oz. If a person eats half the remaining candy bar with each bite, how many bites have been taken when there is exactly $ 0.07$ oz remaining?

From the information given, you can write the equation:

4.48 * (1/2) ^ x = .07      where x = the number of bites.

Since 1/2 of the bar is taken away each bite, this makes sense.

Then, just simplify:

(1/2) ^ x = 1/64      <--------or .015625 in decimal form, but its easier to see the relationship in fraction form.

We can then gather that x = 6.

Therefore, there were 6 bites that the person took.


Online Math Open
number theory
greatest common divisor

Βάζω τis επόμενες πολύ καλές άσκησεις  (κατά τη γνώμη μου ) που είναι σχετικά εύκολες .

1) Να λυθεί στους ακεραίους η εξίσωση

$x^{5}-1=y^{2}$

2)Αποδείξτε ότι δεν υπάρχουν θετικοί ακέραιοι $a,b,c$  ώστε ο αριθμός

$K=\frac{a^{2}+b^{2}+c^{2}}{3(ab+bc+ca)}$  να είναι ακέραιος .

αφου θυμηθηκαμε τα παλια :lol:  ας δωσω καποια hint για τις 2 αυτες ασκησουλες!!! :)
η πρωτη γραφεται $ x^5=(y+i)(y-i)$..τωρα αρκετα ευκολα δουλευοντας στο $ Z[i]$ παιρνουμε τις λυσεις(ειναι θεμα πραξεων αν ξερει κανεις τα στοιχειωδη χαρακτηριστικα της αλγεβρικης θεωριας αριθμων)!!!!!!!!!!

για την δευτερη,και μαλλον πιο δυσκολη,γραφουμε:εστω $ (a+b+c)^2=(3m+2)(ab+bc+ca)$ και τωρα θεωρουμε εναν πρωτο διαιρετη $ p=2 mod3$ του $ 3m+2$.υποχρεωτικα καποιος πρωτος τετοιας μορφης θα ειναι υψωμενος σε περιττο εκθετη γιατιαν ολοι ηταν σε αρτιο θα ειχαμε $ 3m+2=1 mod3$ ατοπο.τωρα παρατηρουμε πως αφου ο LHS ειναι τελειο τετραγωνο ειναι  και $ a+b+c=ab+bc+ca=0 mod 3$ και απο αυτες προκυπτει οτι $ a^2+ab+b^2=0modp$  kai pollaplasiazontas me 4 exoyme $ [(2a+b)b{}^-{}^1]^2=-3 mod p$ opou $ b{}^-{}^1$ ο αντιστροφος mod p....εδω με απλη χρηση του legendre καταληγουμε σε ατοπο αφου $ p=2mod3$....

Nice.
Γενικευεται ωραια παντως η πρωτη ασκηση με χρηση του $ Z[i]$ λεγοντας οτι  η εξισωση $ x^2\plus{}1\equal{}y^p$ δεν εχει non trivial λυσεις γενικα με p περιττο πρωτο!

[quote="lefmast"]αφου θυμηθηκαμε τα παλια :lol:  ας δωσω καποια hint για τις 2 αυτες ασκησουλες!!! :)
η πρωτη γραφεται $ x^5 = (y + i)(y - i)$..τωρα αρκετα ευκολα δουλευοντας στο $ Z[i]$ παιρνουμε τις λυσεις(ειναι θεμα πραξεων αν ξερει κανεις τα στοιχειωδη χαρακτηριστικα της αλγεβρικης θεωριας αριθμων)!!!!!!!!!!

για την δευτερη,και μαλλον πιο δυσκολη,γραφουμε:εστω $ (a + b + c)^2 = (3m + 2)(ab + bc + ca)$ και τωρα θεωρουμε εναν πρωτο διαιρετη $ p = 2 mod3$ του $ 3m + 2$.υποχρεωτικα καποιος πρωτος τετοιας μορφης θα ειναι υψωμενος σε περιττο εκθετη γιατιαν ολοι ηταν σε αρτιο θα ειχαμε $ 3m + 2 = 1 mod3$ ατοπο.τωρα παρατηρουμε πως αφου ο LHS ειναι τελειο τετραγωνο ειναι  και $ a + b + c = ab + bc + ca = 0 mod 3$ και απο αυτες προκυπτει οτι $ a^2 + ab + b^2 = 0modp$  kai pollaplasiazontas me 4 exoyme $ [(2a + b)b^ - ^1]^2 = - 3 mod p$ opou $ b^ - ^1$ ο αντιστροφος mod p....εδω με απλη χρηση του legendre καταληγουμε σε ατοπο αφου $ p = 2mod3$....[/quote]

Poli oraia, gia na einai apolita sosto omos prepei na aplopoieiseis ton gcd twn 3 akeraiwn kai na anaxtheis stin periptosi opou gcd(a,b,c)=1, gia na mporeseis meta na antistrepseis (modp) ton enan apo tous 3 akeraious (wlog ton b) pou den diaireite me p

[quote="silouan"]2)Αποδείξτε ότι δεν υπάρχουν θετικοί ακέραιοι $a,b,c$  ώστε ο αριθμός

$K=\frac{a^{2}+b^{2}+c^{2}}{3(ab+bc+ca)}$  να είναι ακέραιος .[/quote]
πάλι [url=http://www.mathematica.gr/forum/viewtopic.php?f=111&t=36136]εδώ[/url]


prove that if you have a sheet of infinite graph paper and color the vertices 2 colors, prove that you can find a rectangle that's vertices are all one color

[hide]Define C(x,y) as the color of the point (x,y). Draw the lines y=0, y=1, and y=2. If there are three numbers x for which C(x,0)=C(x,1), then by the pigeonhole principle, a rectangle must be formed (two pairs must have same color). The same happens using y=1 and y=2 instead. Therefore, given C(x,1), there are only two (maximum) values of x for which C(x,0) is unequal to C(x,2). Take two of the places where they are equal and form a rectangle from them.

Therefore, not only is it true for an infinite plane, but also for a constant 3-by-3 square.[/hide]

really?
what if we color the following red:
(3,0)(2,1)(1,2)(0,3)(3,1)(2,2)(1,3)(0,0) and the rest blue?

Oops. Well, it works for a five-by-five square (note that this goes from 0-5 not 0-6--i mentioned this because you wrote up to 3 for the 3-by-3 square). Anyway, if you want to suggest a 5-by-5 square, I will gladly find the rectangle.

[hide]Consider a 3x7 lattice. If there are two columns that are colored with the same permutation of colors, like RBR in the first and RBR in the fifth, then there is a rectangle with same colored vertices, in this case, a 2x4 red rectangle. So if there exists no such rectangle, each column must have a different permutation. Lucky for us we can choose the colors in 8 different ways. Unfortunately, 2 of those ways have all points the same color, so we can't use those. So we have 6 possible permutations of colors that must be applied to 7 columns. Pigeonhole says that at least two have the same arrangement, so there must be a rectangle with same colored vertices in this lattice. So there exists a rectangle in the infinite sheet of graph paper with same colored vertices.[/hide]


Given any  integer $ n$ . Prove that for any open interval $ (a,b)$ there exists integers $ p$ and $ q$ with $ q \geq n$ and $ (p,q)\equal{}1$ such that $ \frac{p}{q} \in (a,b)$.

sounds obvious, but can someone give me a rigorous proof?
I must have missed something.  :maybe:

we try to prove this by contradiction.
if possible let it be true that there does not exist any rational p/q in (a,b) with q>=n.
so every rational in (a,b) must be of the form p/q where (p,q)=1
and 0< q<n;
now there can only be finitely many such rationals in (a,b).
but we know that in any open interval (a,b) with a not equal to b,
there are countably many rational numbers.
so we have a contradiction.

(is this okay??  :maybe: )

[quote]so every rational in (a,b) must be of the form p/q where (p,q)=1
and 0< q<n; [/quote]

i sense it is not the complete implication of the negated statement. but i think it is not a big deal here. Thanks.
i must have forgotten the contradiction method. :blush:


LaTeX
number theory unsolved
number theory

Prove that exist infinite a and b such that $ a^{2}006\plus{}b^{2}007\plus{}1: ab$(divide)

But, TTsphn, His problem is $ a^{2006}\plus{}b^{2007}$ not $ a^{2006}\plus{}b^{2006}$. What do you mean nguoigo  :(

This problem has same method
Lemma 1:There exist infinite $ (x,y)$ such that
$ x|y^{2}+1$
$ y|x^{2}-x+1$
$ \left rigtarraw xy|x^{2}+y^{2}-y+1$
Consider equation
$ x^{2}+y^{2}-y+1=2xy$
$ \leftrightarrow(x-y)^{2}=y-1$
Chose $ y=m^{2}+1$
Then the eqution has infinite solution
Lemma 1 was prove
From lemma 1
$ x|y^{2}+1$
$ y|x^{3}+1$
Use this has done.
$ x^{3}+1|x^{2007}+1$
$ x^{2}+1|x^{2006}+1$

I have a conjure(If it true I think very easy to prove with Viete).
$ xy|x^{2}\plus{}y^{2}\minus{}y\plus{}1$ then
$ 2xy\equal{}x^{2}\plus{}y^{2}\minus{}y\plus{}1$

I think it's true
Consider the equation $ x^{2}\minus{}kxy\plus{}y^{2}\minus{}y\plus{}1 \equal{} 0\ \ (*)$
When $ y>1$, we have:
$ (x,y)$ satisfies $ (*)$ and $ x\geq y$, then $ \left(\frac{y^{2}\minus{}y\plus{}1}{x}, y\right)$ satisfies $ (*)$, and $ \frac{y^{2}\minus{}y\plus{}1}{x}< y$ (really $ <$ when $ y>1$)
$ (x,y)$ satisfies $ (*)$ and $ x < y$, then $ \left( x,\frac{x^{2}}{y}\right)$ satisfies $ (*)$ and $ x >\frac{x^{2}}{y}$
So, from a pair $ (x_{o},y_{o})$ satisfies $ (*)$, we can jump down and down to construct another pair $ (1,1)$ satisties $ (*)$. Now easy to have that $ k\equal{}2$.


induction
combinatorics unsolved
combinatorics

In a group of $ n$ people, each person points his finger to one other person. Prove that the people can be separated into three groups, so that in each group, there are no two people, one of whom points his finger to the other.

Hint: Rephrase the problem in terms of a directed graph, wherein each vertex has out-degree 1. Then, it suffices to check the result for arbitrary directed cycles.

To be honest, I don't see why directed cycles should be enough here. All out-degrees are 1, but in-degrees may differ. But rephrasing is a good idea; I would rephrase it as follows:

[color=blue][b]Problem.[/b] Given a map $ f: \left\{1,2,...,n\right\}\to\left\{1,2,...,n\right\}$ without fixed points, find a partition of the set $ \left\{1,2,...,n\right\}$ into three subsets $ A$, $ B$, $ C$ such that $ f\left(A\right)\cap A \equal{} \emptyset$, $ f\left(B\right)\cap B \equal{} \emptyset$, $ f\left(C\right)\cap C \equal{} \emptyset$.[/color]

  darij

Directed cycles are sufficient, because every (weakly) connected component is a single directed cycle with branches
coming off of it, and these "branches" can be removed by edge contractions, and so can be trivially colored.

EDIT: I will elaborate.

The problem is asking to show these graphs are tricolorable, since I seem to be falling back on that terminology. If there
is any vertex $ v$ with in-degree zero, then the claim can be reduced to coloring the rest of the graph (without $ v$), since $ v$ is trivially colored once the rest of the graph is. Therefore, the problem is reduced to the case where there
are no vertices of in-degree zero; this means we have graphs consisting entirely of directed cycles.

(Motivation is largely that, topologically, the only obstructions to an arbitrary labeling of a graph's edges being a coboundary are the cycles of that graph.)

[quote="darij grinberg"]Ah, I see your argument. I was thinking of an induction approach, but you have handled it better.

  darij[/quote]

Induction: suppose n-1 or less work. Look at n people, there must be a person P that at most 1 other person points at him. Take P out and group the other n-1 persons into 3 groups. Now since P points at one person A and is pointed by at most one person B, put P into the group without A or B. Done.


ποια ειναι η βασικη διαφορα μεταξυ αξιωματος και ορισμου??

Εαν ο ρολος του ορισμου στην αποδειξη ενος προβληματος ειναι περιπου ο ιδιος με αυτον του ΑΞΙΩΜΑΤΟΣ γιατι να μην ονομαζουμε τον ορισμο αξιωμα???

Δεν είμαι σίγουρος αν αυτά που θα γράψω θα καλύψουν την ερώτησή σου. (Είμαι σίγουρος πως μπορούν να λεχθούν πολύ περισσότερα για το θέμα.)

Μπορούμε να ξεχωρίσουμε δύο διαφορετικά είδη ορισμού.

Για παράδειγμα, στον κατηγορηματικό λογισμό που συζητούσαμε σε άλλο θέμα, δεν χρειάζεται να χρησιμοποιήσουμε το σύμβολο $ \exists$.  Αντί να γράφουμε $ \exists x p(x)$ μπορούμε να γράψουμε $ \neg (\forall x (\neg p(x)))$. Ουσιαστικά αυτός είναι και ο ορισμός (της χρήσης) του συμβόλου $ \exists$. Ο μόνος λόγος που ορίζουμε και χρησιμοποιούμε το σύμβολο είναι για να κάνουμε τις προτάσεις μας πιο σύντομες αλλά κυριως πιο κατανοητές.

Σαν ένα άλλο παράδειγμα, όταν δουλεύουμε στην θεωρία των φυσικών αριθμών ορίζουμε ένα αριθμό p να είναι πρώτος αν και μόνο αν $ p > 1$ και $ p \equal{} mn \Rightarrow (m\equal{}1) \vee (n\equal{}1)$.  Πάλι αυτός ο ορισμός χρησιμοποιείται για καλύτερη κατανόηση.

Υπάρχει όμως και ένα άλλο είδος ορισμού στο οποίο νομιζω πως αναφέρεσαι. Για παράδειγμα στην θεωρία συνόλων ορίζουμε το διατεταγμένο ζεύγος $ (x,y)$ ως το σύνολο $ \{x,\{x,y\}\}$.

Για αυτόν τον ορισμό, παραθέτω από το Naive Set Theory του Halmos (σελίδα 25):

... we need to know that ordered pairs are determined by and uniquely determine their first and second coordinates, that Cartesian products can be formed, and that every set of ordered pairs is a subset of some Cartesian product; which particular approach is used is immaterial...there is nothing wrong or anything misleading [with this definition]; the relevant properties of the concept we have defined are all correct ... and all the correct properties are present. The trouble is that the concept has some irrelevant properties that are accidental and distracting ... The fact that $ \{a,b\} \in (a,b)$ ... seems accidental; it is a freak propert of the definition rather than an intrinsc property of the concept. The charge of artificiality is true, but it is not too high a price to pay for conceptual economy. The concept of an ordered pair could have been introduced as an additional primitive, axiomatically endowed with the right properties, no more no less. In some theories this is done. The mathematician's choice is between having to remember a few more axioms and having to forget a few accidental facts; the choice is pretty clearly a matter of taste...

[quote="Demetres"]. The concept of an ordered pair could have been introduced as an additional primitive, axiomatically endowed with the right properties, no more no less. In some theories this is done. The mathematician's choice is between having to remember a few more axioms and having to forget a few accidental facts; the choiceα  is pretty clearly a matter of taste...[/quote]

Δηλαδη,Δημητρη, εαν στην αξιωματικη θεμελιωση των πραγματικων αριθμων προσθεσουμε το συμβολο της τετραγωνικης ριζας σαν ενα primitive συμβολο και γραψουμε μερικα αξιωματα για αυτο αυτοματως μετατρεπουμε τον ορισμο σε αξιωμα,και αρα δεν
υπαρχει διαφορα μεταξυ ορισμου και αξιωματος???

Διαφοροτροπως ποια ειναι η βασικη διαφορα μεταξυ ορισμου και αξιωματος???

Με τι κριτηρια σε μια αχιωματικη θεμελιωση θα μπορουσαμε να χρησιμοποιησουμε τον ελαχιστο αριθμο των primitive symbols.

Πρεπει να δωσουμε ενα ορισμο του ορισμου.

Σε μια 1st order theory ποτε και πως εμφανιζεται ενας ορισμος??

Νομίζω πως ναι θα μπορούσαμε να ορίσουμε από την αρχή την τετραγωνική ρίζα σαν primitive symbol και να θέσουμε τα αξιώματα που θέλουμε να ικανοποιεί. Ο λόγος κατά την γνώμη μου που δεν το κάνουμε είναι ο εξής:

Γιατί θέλουμε να ορίσουμε την τετραγωνική ρίζα; Διότι αποδείξαμε ένα θεώρημα που λέει ότι για κάθε μη αρνητικό αριθμό $ x$ υπάρχει μοναδικός μη αρνητικός αριθμός $ y$ έτσι ώστε $ x \equal{} y^2$. Αν δεν είχαμε αυτό το θεώρημα, τότε δεν θα είχαμε κανένα λόγο να ορίσουμε την τετραγωνική ρίζα. Και αν από την αρχή προσθέταμε την τετραγωνική ρίζα σαν primitive symbol με τα "γνωστά" αξιώματα, αλλά το παραπάνω θεώρημα δεν ίσχυε, τότε η θεωρία θα κατέρρεε.

Άρα νομίζω πως είναι "λογικό" να προηγείται η απόδειξη του θεωρήματος του "ορισμού" της τετραγωνικής ρίζας. Αλλά για να αποδείξουμε το θεώρημα πρέπει ήδη να έχουμε στην διάθεσή μας την θεωρία.

Αν προσπαθούσα να δώσω ένα ορισμό του ορισμού σε first order theory θα έλεγα το εξής:

Για κάθε πρόταση $ A(x_1, \ldots , x_n)$ με δεσμευμένες μεταβλητές $ x_1, \ldots , x_n$ μπορούμε να ορίσουμε ένα νέο predicate symbol P με arity n και την πρόταση $ P(x_1, \ldots , x_n) \Leftrightarrow A(x_1, \ldots , x_n)$.

Για κάθε πρόταση $ A(x, x_1, \ldots , x_n)$ με ελεύθερες μεταβλητές $ x,x_1, \ldots , x_n$, αν ισχύει η πρόταση $ \exists x (A(x, x_1, \ldots , x_n) \text{ AND } (A(y, x_1, \ldots , x_n) \Rightarrow (x\equal{}y)))$ τότε μπορούμε να ορίσουμε ένα νέο function symbol f με arity n με $ f(x_1, \ldots , x_n) \equal{} x \Leftrightarrow A(x, x_1, \ldots , x_n)$.

Δεν ξέρω αν είναι ορθός αυτός ο ορισμός του ορισμού (η αν μπορεί να γίνει ορθός με μερικές τροποποιήσεις). Η ιδἐα όμως νομίζω πως είναι ότι προσθέτουμε νέα σύμβολα, είτε predicate ειτε function, έτσι ώστε να κάνουμε κατανοητές τις προτάσεις μας. Είτε το κάνουμε αυτό είτε όχι θα μπορούμε να αποδείξουμε ακριβώς τα ίδια πράγματα στην θεωρία.

Δεν νομίζω ότι προσπαθούμε πάντα να χρησιμοποιούμε τον ελάχιστο αριθμό primitive symbols. Δεν έχω κάποιο παράδειγμα στο μυαλό μου. Αυτό που έχω υπόψη μου είναι ότι δεν χρησιμοποιούμε τον ελάχιστο αριθμό αξιωμάτων στην θεωρία συνόλων. Π.χ. το "there exists a set" και το "there exists an infinite set" είναι και τα δύο αξιώματα της θεωρίας συνόλων αλλά ουσιαστικά το πρώτο δεν χρειάζεται. Ο μόνος λόγος που υπάρχει (κατά την γνώμη μου) είναι διότι είναι πιο "απλό" από το δεύτερο.

[quote="τρικλινο"][b]ΣΩΣΤΑ,εννοω συμφωνω εγω , ΚΑΙ μερικα ερευνητεα[/b]

Αρα υπαρχει  διαφορα αξιωματος και ορισμου?? ποια ειναι?? Το θεωρημα υπαρξεως του ορισμου??

ΚΑΙ ο ρολος του ορισμου σε καθε αποδειξη σε τι διαφερει απο αυτον του αξιωματος??

..............................καλη χρονια......................................................................................[/quote]

Δεν μπορώ να το ψάξω στην βιβλιοθήκη τώρα αλλά όπως είπα και στο προηγούμενο post, νομίζω ότι η διαφορά είναι ότι ο ορισμός μπαίνει καθώς η θεωρία εξελίσσεται. Ο ρόλος του ορισμού στην απόδειξη είναι κατά την γνώμη μου μόνο και μόνο για συντόμευση της απόδειξης. Αντίθετα το αξίωμα, εκτός αν αποδεικνύεται από άλλα αξιώματα είναι απαραίτητο σε κάποιες αποδείξεις.

Δεν θυμάμαι να έχω δει σε κάποιο βιβλίο λογικής τον ορισμό του ορισμού ή κάποια συζήτηση για την διαφορά μεταξύ ορισμού και αποδείξεως για να μπορώ να πω κάτι περισσότερο από ότι έχω ήδη γράψει.

[quote="τρικλινο"][quote="Demetres"]

. Ο ρόλος του ορισμού στην απόδειξη είναι κατά την γνώμη μου μόνο και μόνο για συντόμευση της απόδειξης. Αντίθετα το αξίωμα, εκτός αν αποδεικνύεται από άλλα αξιώματα είναι απαραίτητο σε κάποιες αποδείξεις.

[/quote]

[b]Ενα παραδειγμα θα βοηθουσε παρα πολυ[/b][/quote]

Για παράδειγμα, στην θεωρία συνόλων το $ 1 \plus{} 1 \equal{} 2$, είναι θεώρημα. Αλλά τα 1,2 και + δεν είναι σύμβολα της θεωρίας συνόλων. Αυτή η πρόταση θα μπορούσε να γραφτεί μόνο με σύμβολα της θεωρίας συνόλων χωρίς να χρειαστεί να ορίσουμε τα 1,2 και +. Θα ήταν όμως μια ορθή αλλά "ακαταλαβίστικη" πρόταση.

Από την άλλη, το αξίωμα της θεωρίας συνόλων που λέει ότι ύπαρχει ένα άπειρο σύνολο, είναι απαραίτητο στην θεωρία συνόλων για να αποδειχθεί η ύπαρξη του συνόλου $ \mathbb{N} \equal{} \{1,2,3,\ldots\}$. Εαν δεν υπάρχει αυτό το αξίωμα (ή κάποιο παρόμοιο) στην θεωρία τότε δεν μπορούμε να πούμε: "Ορίζω $ \mathbb{N} \equal{} \{1,2,3,\ldots\}$", διότι η ίδια η θεωρία δεν μου εγγυάται ότι αυτό θα είναι πράγματι σύνολο.

Το ίδιο συμβαίνει και με την τετραγωνική ρίζα των θετικών αριθμών. Πρώτα αποδείξαμε την ύπαρξη (και την μοναδικότητα) και μετά ήρθαμε και ορίσαμε την τετραγωνική ρίζα.


I have only just registered on this site and am not sure if I am posting my question in the right area but here goes!

I am currently doing a mathematics course and have come up against finding a closed form for a linear recurrence sequence. I am comfortable with closed forms for arithmetic and geometric sequences but cannot get my head around those for linear recurrence sequences.

Can anyone give me any advice on how to solve this?! Unfortunately, I cannot give an example as the only example I have is the question itself and I would rather learn how to solve these problems rather than rely on someone giving me the answer.

Thanks  :)

[quote="kunny"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109292=]How about this?[/url][/quote]

Kunny,

I shall have a read through the link, thank you for the quick response  :) !


Today I complete [i]three[/i] years on this site, though I've only been active perhaps for the first summer or so.
Anyway, I decided to mark the occassion with this new game.

The rules of each atomic 2-player game are as follows:
Call the players A and B.
Each has a binary decision to make: he (ignore any underlying sexist intents  in this choice of pronouns :roll: ) can either cooperate with the other player, or defect.
Accordingly, each gets a payoff based on the joint decision:
If both cooperate, each gets 3 points.
If both defect, each gets 2 points.
If A cooperates and B defects, A gets 1 and B gets 4.
If you can't figure out the fourth case yourself (given that the game is unbiased), [i]now[/i] is the time it'd be wise on your part to leave the topic. :P

[For the interested kids, here's the underlying game-theoretic motivation: imagine two prisoners convicted as partners in the same crime. If neither pleads guilty, each will get a sentence, but it will be shorter than the one he'd get if both pleaded guilty. If one pleads and the other stays mumb, the former will get off with a light sentence while the latter will porbably face something very unpleasant. Note that irrespective of what the other decides, one always gets a lighter sentence on confession. That is what is called the Prisoners' Dilemma in game theory, and is often used to extract confessions in such cases. If you think the fate of the game is decided, think again in the context of repitition.]

For the first season I'll take an even number of signups between ten and twenty, depending on the general enthusiasm (if I don't get ten, sadly the game is off). The contestants will be paired off and each given a screenname (the name of some european soccer club) so that nobody knows who his partner is. Pairs once made will remain through the season, though if one of a pair abandons the game, the other will be taken off too.

There'll be ten rounds in the season. In each round, there will be an atomic game for each of the pairs, with the players sending in their decisions by PM. The scores will be revealed after each round. Meanwhile, this thread can be used for discussing strategies, though any attempt to connect to your partner will result in disqualification. The winner will be the player with the highest total points after the final round is over.

[b]Signups for Season 1 begin now. Please sign up in this thread.[/b]

I'll begin as soon as I think I have enough players.

About [i]time[/i] the old-school AoPSers came back!
I remember the Spam & Fun Factory, the early and not-so-big mafia games with no theme, the count down/up games, etc. Good times.

Eye yam min.

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

Me join.  Me like wifom.

Eh, for both defect, shouldn't both LOSE 2 points or something?  that's how i played, but w/e it's your game :)

First the list:
Melissa
aadil
1=2
BOGTRO
Joe10112
hedgehog12
Garyzx
nt811581147
AIME15

[i]one[/i] player to go before quorum is attained.

As for AIME15's queries, the very essence of the game is that in a single game, defecting is always the better option (making this different from, say, rock paper scissors). You can read up a bit on the subject over the net - I personally find it quite absorbing. I was introduced to it on the sixth of may, 2005, through richard dawkins' movie - nice guys finish first - one I'd highly recommend to anyone interested in getting a primary idea of game theory in particular, and anybody fond of intellectual excursions in general. I'm quite eager to see the kind of discussions of strategy that take place.

And finally, a warm hello to old timers melissa and matthew babbitt - it's nice to see you guys again!

lol I was going to make something like this, but I guess I was beaten to it  :D (no join)

very soon - quorum has been attained.

Melissa
aadil
1=2
BOGTRO
Joe10112
hedgehog12
Garyzx
nt811581147
AIME15
dojo

I'll just wait for some other participants if any and start sometime tomorrow maybe.

Cool, the people from the old Mafia games I wasted who-knows-how-many hours reading.  :D

Sounds fun, I'm in.

Since I have an even number of signups now, I think I'll start.

[b]Registration for Season 1 is now closed.[/b]

Final list:

Melissa
aadil
1=2
BOGTRO
Joe10112
hedgehog12
Garyzx
nt811581147
AIME15
dojo
isabella2296
eggylv999

I'll send out the PMs as soon as I can.

Sent in the PMs.

Here are the fixtures:

Bayern vs Liverpool
Paris SG vs Spartak
Porto vs Roma
Panathanaikos vs Dortmund
Valencia vs Galatasaray
Deportivo vs Inter

(Regarding "why European clubs?", because I like them, and also because they are a nice neutral stage in context of the American-Asian rivalries not uncommon around here.  :P )

By the way, I've modified the payoffs. They now stand as:
Both cooperate: 3-3 split (as before)
Both defect: 1-1 split (less than before)
Only one cooperates: 0-5 (more disparity)

OK so the only way to make real money is if everybody cooperates(a total of 6 points is given then). I say that we should all get along.

You may discuss [i]general[/i] strategies in thread, without directly revealing what you're doing.
If you think about it, there can be plenty to discuss.

General strategy: No one defects.

However, then no teams win, and it's not fun.

well, just flip a coin! heads-cooperate, tails-defect. thats wat i did......

still to send in round 1 choices:

Melissa
aadil
hedgehog12
nt811581147

COME ON!!! Lets get on w/ the game!!

(wait, is there a deadline? if there is, i take back wat i just said :) lets just wait. ((if there is a deadline))

And finally, the results of Round 1:

(D) Bayern Munich vs Liverpool FC (C)
(C) Paris St. Germain vs Spartak Moscow (D)
(D) FC Porto vs AS Roma (C)
(C) Panathinaikos FC vs Borussia Dortmund (C)
(D) Valencia FC vs SK Galatasaray (C)
(C) Deportivo la Coruna vs Inter Milan (C)

We've had [i]four[/i] successful defections in the first round! The scores stand (leaders in bold):

[b]Bayern - 5[/b]
Liverpool - 0
Paris SG - 0
[b]Spartak - 5
Porto - 5[/b]
Roma - 0
Panathanaikos - 3
Dortmund - 3
[b]Valencia - 5[/b]
Galatasaray - 0
Deportive - 3
Inter - 3

[b]Submissions for Round 2 begin now![/b]

ummmm

well,

i believe i sent in my choice

anyone else send theirs in?

I believe this thread is now dead......

(?)   :maybe:

I was gone for such a long time and still all the choices are not in.
I'll have to send in five prod PMs today. :(

what is a "prod PM"?

oh nvm got it!!!!!!!!!!

reminding someone about the game they are playing via PM


Let $S$ be a set of $2n+1$ points in the plane such that no three are collinear and no four concyclic. A circle will be called $\text{Good}$ if it has 3 points of $S$ on its circumference, $n-1$ points in its interior and $n-1$ points in its exterior.
Prove that the number of good circles has the same parity as $n$.

[quote="who"]infact no. of such circles so called halving circles is n^2 .this is due to federico ardilla.[/quote]
Do you have a proof of this fact? Please give a link of sth?

I believe there is also a solution using inversion as Professor Andreescu once gave.

In the article mentioned above, Ardila says about this problem (which he proposed): "[It] follows easily from the nontrivial observation that, for any $A$ and $B$ in $S$, the number of halving circles that go through $A$ and $B$ is odd."

Indeed, if that is the case, then we can work mod 2, so that the total number of such circles is $\frac{1}{3}\binom{2n+1}{2}\pmod{2}$ which is equal to $n\pmod{2}$.

Now we prove the claim made in the first paragraph. I will rename the pair of points in $S$ to $P$ and $Q$, since that is what the source used I am copying from. Divide the $2n-1$ other points into two groups depending on which side of $PQ$ they lie. By the condition that no three points are collinear, each of these points fall into one of these groups. Label the points in these groups $A_1,A_2,\ldots,A_m$ and $B_1,\ldots,B_p$ respectively, where $m+p=2n-1$. Let the angles $\angle PA_1B, \ldots, \angle PA_mQ$ be $\alpha_1,\ldots,\alpha_m$, respectively and the angles $\angle PB_1Q, \ldots, PB_pQ$ be $\pi-\beta_1,\ldots, \pi-\beta_p$, respectively. Without loss of generality we assume that $\alpha_1>\alpha_2>\cdots>\alpha_m$ and $\beta_1>\beta_2>\cdots>\beta_p$. Note that none of these angles can be equal, otherwise we would have a circle passing through four of the points. No order all the angles $\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_p$ from largest to smallest.

The circle through $P$, $Q$ and $A_1$ contains the point $A_k$ (in its interior) if and only if $\alpha_k>\alpha_i$ if and only if $k<i$. Similarly $B_l$ lies in the interior of circle $PQB_j$ if and only if $l>j$. Also, $B_j$ lies in the interior of circle $PQA_i$ if and only if $\alpha_i+\pi-\beta_j>\pi$ if and only if $\alpha_i>\beta_j$. Similarly, $A_i$ lies in the interior of circle $PQB_j$ if and only if $\alpha_i>\beta_j$.

Now consider and ordering of all the angles and suppose that there is an instance where $\cdots>\alpha_i>\beta_j>\cdots$ are consecutive in this ordering. We claim that circles $PQA_i$ and $PQB_j$ have the same number of interior points. Indeed, circle $PQA_i$ contains the points $A_1,\ldots,A_{i-1}$ and $B_j,\ldots, B_p$, while circle $PQB_j$ contains the points $A_1,\ldots, A_i$ and $B_{j+1},\ldots, B_p$.

Consider the following transformation of $S$: Replace $A_i$ by another point $A'_i$ on the same side of $PQ$ such that $\alpha'_i:=\angle PA'_iQ=\beta_j$ and replace $B_j$ by another points $B'_j$ on the same side of $PQ$ such that $\pi-\beta'_j=\angle PB'_jQ=\pi-\alpha_i$. The only change in the total ordering is that $\alpha'_i$ and $\beta'_j$ are in the opposite order to $\alpha_i$ and $\beta_j$. So now we have $\cdots>\beta'_j>\alpha'_i>\cdots$.

When $k\ne i$ and $l\ne j$, performing this transformation does not change the number of points in the interior of circle $PQA_k$ or $PQB_l$. Thus the only circles that may change from good to bad or bad to good are $PQA_i$ and $PQB_j$.

Before the transformation circles $PQA_i$ and $PQB_j$ contained the same number of interior points, and thus either both were good or both were bad. A similar counting shows that circles $PQA'_i$ and $PQB'_j$ have the same number of interior points after the transformation, and thus also are either both good or both bad. So performing this transformation changes the number of good circles by either $+2$ or $-2$, and thus preserves the parity.

Now we repeat this transformation until we end up with a configuration where $\beta'_1>\beta'_2>\cdots>\beta'_p>\alpha'_1>\cdots>\alpha'_m$. For each $i$ the circle $PQA_i$ contains the $i-1$ points $A_1,\ldots, A_{i-1}$, while for each $j$, the circle $PQB_j$ contains the $p-j$ points $B'_{j+1},\ldots, B'_p$. So, if $m\ge n$, then there is exactly one good circle $PQA_i$, namely when $i=n$; and if $p\ge n$, there is exactly one good circle $PQB_j$, namely when $j=p+1-n$. Since $m+p=2n-1$, exactly one of the relations $m\ge n$ or $p\ge n$ can hold, and therefore this arrangement has exactly one good circle.


If N is a three-digit integer with  units digit 4, what is the probability that N is divisible by 6? Express your answer as a common fraction.

Because the ones digit is $ 4$, the sum of the other two must be $ 2 \text{mod} 3$. Counting them up, we have a total of $ 30$ 3-digit multiples of $ 6$ ending in $ 4$. Because tehre are a total of $ 9 \cdot 10 \equal{} 90$ three-digit numbers with ones digit $ 4$, our probability is $ \frac{30}{90} \equal{} \boxed{\frac{1}{3}}$.

Since we know the last digit is 4, we know that the number must be even.

Also, we know that 1/3 of the 3 digit numbers are divisible by 3...

So that would lead us to guess 1/3 (in a countdown round), but now for some solid hardcore work.

let the number be _ _ 4. So we know already that there are 9 possibilities for the first digit (1-9), and 10 for the second(0-9): there are 90 total possibilities of numbers with the last digit of 4.

Now we can pretty much just list the possibilities for divisibility by 6:

114, 144, 174, ..., 984

AHA! they're increasing by 30! we know how to count the number of terms in an increasing arithemetic sequence (right?). So there are (984-114)/30 +1= 29+1=30 numbers that are divisible by 6, leaving us with a probability of 30/90=1/3.


Show that there is no r.v. $ Y$ such that $ \frac {S_n}{\sqrt {n}}\to Y$ in probability, where $ S_n \equal{} \sum X_i$ with $ \left(X_i\right)_{n\geq1} \minus{} \mbox{iid}, EX \equal{} 0, Var(X) \equal{} 1$.
One can find the argument using the fact that convergence is the same as being Cauchy sequence, but i think there must be something simple..

Maybe this is instructive - let $ X_i$ be iid $ N(0,1)$ random variables, so that $ Y_i \equal{} S_i/\sqrt{i}$ has the same distribution as $ Y$ (by the central limit theorem).

On the other hand, $ Y_n\minus{}Y_m$ ($ m<n$) is normally distributed with mean $ 0$ and variance $ 2(1\minus{}\sqrt{m/n})$, so it can't be Cauchy in probability.  I'm not sure what to do in the general case.


function
calculus
derivative
inequalities
real analysis
real analysis unsolved

Find all $c\in \mathbb{R}$ such that:
$\frac{e^{x}+e^{-x}}{2}<e^{cx^{2}}$  $\forall x\in \mathbb{R}$

Since the two functions always agree at $x = 0$, you probably want that to be a $\leq$ sign.  The answer is $c \geq \frac{1}{2}$ -- both functions agree and have the same derivative at $x = 0$.  The second derivative of the left hand side at $x =0$ is 1 while the second derivative of the right hand side at $x = 0$ is $2c$, so if $c < \frac{1}{2}$ then near zero the inequality is reversed.  This shows that $c \geq \frac{1}{2}$ is necessary.  To show it is sufficient, compare the Taylor coefficients of the two functions.  (Maybe there are simpler ways for this, as well.)

This problem is part of a list at
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151928[/url]


How many non-overlapping 9-inch by 9-inch square tiles does it take to a 12-foot by 15-foot room?

Wouldn't it be 400? If you convert the feet to inches you get 144*225 divide that by 9*9 and you get 400

$ 12\cdot 12\equal{}144$ inches in 12 feet and $ 15\cdot 12\equal{}180$ inches in 15 feet, so you need \[ \frac{144\cdot 180}{9\cdot 9}\equal{}\boxed{320}\]


For high school juniors and seniors in the NYC metro area:  Would you and your friends like to win $\$$20,000 in scholarships?  Check out the [url=http://m3challenge.siam.org]Moody's Mega Math Challenge[/url].  It's an applied math competition for teams of 3 to 5 students.  It takes place on March 4 and 5 of this year.

Good luck!

Reminder:  If you would like to register for this contest, the registration deadline is February 19.


A set of $ n(n\plus{}1)/2$ distinct values is randomly arranged in a triangle of $ n$ rows where the first row contains one value, the second row two values, and so on. Let $ M_k$ denote the largest value in the $ k$th row. Find the probability that $ M_1 < M_2 < \dots < M_n$.

[hide="Hint"]
Note that this happens precisely when
1. the largest element is placed in the last row, and
2. the elements above the last row satisfy the same condition, with $ n$ smaller by 1.
You can now derive a simple formula for $ p_n$ in terms of $ p_{n\minus{}1}$. The answer is $ p_n \equal{} \frac{2^n}{(n\plus{}1)!}$.
It will be very interesting to find a direct, "bijective" proof of this.
[/hide]

[hide="sketch"]Well, let $ P(k)$ denote the probability of $ M_1 < M_2 < \dots < M_k$.  Then we want to know $ P(n)$.
However, the one requirement is that the largest of what is left will go in the last row.
Thus $ P(k) \equal{} P(\mbox{largest} \in M_k) P(k\minus{}1)$ for $ k>1$ (obviously $ P(1) \equal{} 1$).

I believe $ P(\mbox{largest} \in M_k) \equal{} \frac{2}{k\plus{}1}$, so from here, we can see $ P(n) \equal{} \frac{2^n}{(n\plus{}1)!}$[/hide]

darn, someone beat me to it.


induction
number theory unsolved
number theory

if $p$ is a prime ,$n>p\in\mathbb{N}$ , prove that we can always find a prime divisor $>p$ in the sequence

$n,n+1,n+2,\cdots ,n+p-1$

at least the proof should be quite similar to bertrand???

This is a special case of Sylvester's theorem. It was proved using elementary methods by Erdos.
(The proof is rather long, and not so obvious :))

[quote="Gabriel(fr)"]We can suppose that $pk+1 \le n \le (k+1)p$ for a fixed $k \in \mathbb{Z_{+}}$ , then,  $(k+1)p \le n+p-1 \le (k+2)p-1$, consequently, all the sequences that have  $n \in [pk+1,p(k+1)]$ as first term, have $(k+1)p$ as a term, and is easy to prove by induction that for all $k \ge 1$ this works.[/quote]

I cant see how this implies there exist a prime divisor $>p$ though ..... can you explain ? :huh:

my conjecture about the problem :
We can always find a prime divisor $>p$ in the sequence

$n,n+a,n+2a,\cdots ,n+(p-1)a$ where $(a,n)=1$ and $a,n\in\mathbb{N}$

How true is this?


function
trigonometry
integration
limit
real analysis
real analysis unsolved

I have to prove that for $ a \leq 1$ and $ k \in \mathbb{N}$ is odd, the function $ g(t) \equal{} t^a (\sin{1/t})^{1/k}$ is not of bounded variation on [0,1].

Is there an easier way to prove this without having to resort to the definition of bounded variation and negating it?[/code]

The problem is easy if you so resort. Find obvious sequences $ b_1 > a_1 > b_2 > a_2 > ...$ that make $ \sin{1/t} \equal{} 1, 0, 1, 0, ...$  and ponder $ \sum_{n\equal{}1}^\infty |g(b_n) \minus{} g(a_n)|$.

[quote="sphere929"]That's what I was trying to avoid doing.  I did see in a textbook that $ AC[0,1]\subset BV[0,1]$ so it might be easier to show that the function is not in $ AC[0,1]$ by showing that $ g(x)\neq \int_0^x g'(t)dt$ since $ \lim_{t\to 0} g(t)$ DNE[/quote]

Put down that beer; there are lots of functions in $ BV \minus{} AC$.

I'm not sure why you want to avoid a few minutes of work.

WWW's suggestion is (in my opinion) the best way to go about the problem.  When you've got something that oscillates, like sine or cosine, locate the extreme values and make sure your partition contains these points.  Then get some estimate on the number of extrema and approximate that sum.

[quote="mlok"][quote="sphere929"] $ \lim_{t\to 0} g(t)$ DNE[/quote]
It does exist when $ a > 0$, which does not change the fact that the function has infinite variation for $ a\le 1$.

Otherwise, WWW has a good point: the purpose of exercises like this is to get your hands dirty and engage with the definition directly. One can't do interesting math just by gluing theorems together.[/quote]

you're right, i meant to put that the limit is infinite...

As for why I want to save a few minutes a work, it's because questions like these have appeared on previous qualifying exams and every second counts on those...

That's why you'd want to get good at doing them now. :wink:

yeah, i just realized what a silly mistake i made earlier with proving that g wasn't AC and thus not BV... now that i have the definition down, i won't make mistakes like that again...

Thank you all for the suggestions!


I would really appriciate if some kind hearted mathematician would help me out on this: <even a hint, if not the full solution, will be appriciated>

Show that given any integer 0 < n <= 2^1000000 we can find at set S of at most 1100000 positive integers such that S includes 1 and n and every element of S except 1 is a sum of two (possibly equal) smaller elements of S.

Thanks

well, I doubt if this would help, but clearly you are not the only one searching for the solution to this one: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1100000+positive+integers&t=41233
 :D

Thanks!

I like the solution...couldnt think of that at all :P


induction
search
number theory unsolved
number theory

Let $ L_0 = 2, L_1 = 1$ and ${ L_{n+2} = L_{n+1}+ L_n}$
be the famous Lucas’s sequence. Prove that the only $ n > 1$ such that $ L_n$ is a
perfect square is $ n = 3$.

Hint:
[hide] Prove by induction the following:
For $ n\geq10, \L_n\gneq n^2$.[/hide]

hana1122: This was discussed before (and linked to some paper). A search here or with google should do.

raf92: That doesn't help as the problem never talks about $ L_n \equal{}n^2$, but about $ L_n\equal{}k^2$ with $ k$ possibly $ \neq n$.

I'm sorry, I made a mistake when I read the topic.

[quote="ZetaX"]hana1122: This was discussed before (and linked to some paper). A search here or with google should do.[/quote]
Thank you. I found it.


[quote="REMmargorp"]I'm new to this forum, so I don't know how to write the equations fancily yet.[/quote]Welcome! :)
You can quickly learn $\LaTeX$ (or how to write "fancy" equations) by reading carefully this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 downloading the file in it (for future reference of commands) and practicing a little bit in our [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url] :)

[quote="Valentin Vornicu"][quote="REMmargorp"]I'm new to this forum, so I don't know how to write the equations fancily yet.[/quote]Welcome! :)
You can quickly learn $\text{\LaTeX}$ (or how to write "fancy" equations) by reading carefully this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 downloading the file in it (for future reference of commands) and practicing a little bit in our [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url] :)[/quote]

Well, I have never before read the document that Valentin is mentioning, but it seems like a consise abstract of all you need to know to begin with $\text{\LaTeX}$ (BTW, very well done Valentin!). Anyway, I always recomend [url=http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The not so short introduction to $\text{\LaTeX}2_\varepsilon$[/url], it is the best book for $\text{\LaTeX}$ that you can find for free. It has changed a little since the first time that I read it (1996) then I can't tell you what sections to skip. Either way, someone who is into math edition should read this complete at least one (I have done it) and it is a good reference. If you want reference to simbols, I recomend [url=http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]The Comprehensive $\text{\LaTeX}$ Symbol List[/url]

Best,


I was looking around for videos about general mathematics and calculus and I came to this place http://www.isharecreativity.com  this place help me with my homework.

Does anyone ever read anything before signing up on these forums?

[size=200] NO ADVERTISING!!![/size], whatever your intention is.


inequalities logarithms integration induction limit

Let $ n \ge 1, n\in N$. Prove that:
$ n! \ge (\frac{n}{e})^n$

[quote="NguyenDungTN"]Let $ n \ge 1, n\in N$. Prove that:
$ n! \ge (\frac {n}{e})^n$[/quote]
$ \ln n! \equal{} \ln2 \plus{} \ln3 \plus{} ... \plus{} \ln n\geq\int_1^n\ln xdx \equal{} n\ln n \minus{} n \plus{} 1 > n\ln n \minus{} n.$

[quote="NguyenDungTN"]Let $ n\in \mathcal N^*$. Prove that $ n! \ge \left(\frac {n}{e}\right)^n$ .[/quote]
Denote $ a_n\equal{}\frac {n!e^n}{n^n}$ . Observe that for any $ n\in\mathcal N^*$ , $ \frac {a_{n}}{a_{n\plus{}1}}\equal{}\frac 1e\cdot\left(1\plus{}\frac 1n\right)^n<1$ .

Therefore the sequence $ a_n$ is increasing, i.e. $ 1<e\equal{}a_1<a_2<\ldots <a_n<a_{n\plus{}1}<\ldots$

In conclusion, for any $ n\in\mathcal N^*$ , $ a_n>1$ , i.e. $ n! \ge \left(\frac {n}{e}\right)^n$ .

[b]Remark.[/b] For any natural number $ n\ge 6$ there is the chain $ \left(\frac n3\right)^n<n!<\left(\frac n2\right)^n$ .

Try induction:

With $ e \equal{} \lim_{n\rightarrow\infty}{(1 \plus{} \frac {1}{n})^n}$

$ n \equal{} 1$ it is clear

Suppose it is true for $ n$ so: $ n! > \frac {n^n}{e^n} \Rightarrow$

$ (n \plus{} 1)! > (n \plus{} 1)\frac {n^n}{e^n} > \frac {(n \plus{} 1)^{n \plus{} 1}}{e^{n \plus{} 1}} \Leftrightarrow en^n > (n \plus{} 1)^{n} \Leftrightarrow e > (1 \plus{} \frac {1}{n})^n$

 :)


geometry
geometric transformation
reflection
analytic geometry

If I'm given a point (3,5) the reflection on the line y = x is (5,3). What is a simple strategy for getting the reflection on any arbitrary line for (x1, y1). I calculated the coordinates of the other point by drawing a line perpendicular to the original line that goes through the point (x1, y1). Then, duplicate the change in x to the new x, and the change in y to the new y. For example:

(3, 5) is the original point. (x1, y1)

y = -x is the original line.

y = x + 2 is the perpendicular line that goes through (3, 5).

(-1, 1) is the interception point of the two lines. (x2, y2)

x2 = x1 - 4. y2 = y1 - 4.

Therefore the reflection point must be (-1-4, 1-4) = (-5, -3)

What is a simpler way of solving a problem like this, for an arbritary point and line?

There is a formula for the distance from a point to a line.  That may be useful for your question.

Let (x1, y1) be a point.  Let ax + by + c = 0 be a line.  Then the distance from the point to the line is
|ax1 + by1 + c| / :sqrt:(a2 + b2).


induction
inequalities proposed
inequalities

[quote="Albanian Eagle"]Let $x_{1},x_{2},..,x_{n}$ and $y_{1},y_{2},...,y_{n}$ be positive real numbers such that $kx_{k}y_{k}\geq 1$ for $k=1,2,...,n$ Prove that
$\sum_{k=1}^{n}\frac{x_{k}-y_{k}}{x_{k}^{2}+y_{k}^{2}}\le \frac{1}{4}n\sqrt{n+1}$
When does equality hold?[/quote]
I don't think it is too hard.
My solution:we can prove that $\frac{x_{k}-y_{k}}{x_{k}^{2}+y^{2}_{k}}\leq \frac{1}{2\sqrt{2x_{k}y_{k}}}\leq \frac{\sqrt{k}}{2\sqrt{2}}$
then we only need to prove that ${\sum_{k=1}^{n}}\frac{\sqrt{k}}{2\sqrt{2}}\leq \frac14 n\sqrt{n+1}$
by indution :
$\sum_{k=1}^{n+1}{\frac{\sqrt{k}}{2\sqrt{2}}}\leq \frac14 n\sqrt{n+1}+\frac{\sqrt{k}}{2\sqrt{2}}\leq \frac14 (n+1)\sqrt{n+2}$( because it is equivlent to $(3-2\sqrt{2})n>0$
I think the equality can not be attained.

Yes it is not too hard, but equality is attained at $n=1,x_{1}=\frac{\sqrt{6}+\sqrt{2}}{2},y_{1}=\frac{\sqrt{6}-\sqrt{2}}{2}$ :D

Yes ,you are right .I didn't consider the n=1 case.


I've split it up into 3 pieces, 2 triangles and a rectangle, but I can't figure out the side length of the rectangle.  ANyone have a different way, or a way to complete on my idea?

You should be able to find the lengths by doing this. You would drawn 2 altitudes: D to AB and B to DC. From this you would get a 45-45-90 triangle, and and 30-60-90 triangle. Plus a rectangle which im assuming you did. Start with the 30-60-90 tri. We no that the sides are in the ratio of $ x: x\sqrt {3}: 2x$ in respect to the sides opposite of the 30:60:90 degrees. So this means that height of the triangle of $ \frac {20}{2} \equal{} 10$. The base is $ 10\sqrt {3}$. The area of that triangle is then $ 50\sqrt {3} in^2$.

Now for the rect. Opposite sides of a rect. are congruent, so the other side of the rect is 10. Also one side length of the 45-45-90 tri. is 10. The ratio of the sides of a 45-45-90 tri. is $ x: x: x\sqrt {2}$. This means that the top part of the 45-45-90 tri. is also 10. Since the whole top part is 15 and the triangle leg part of it is 10, the other part (which is the length of the rect. is 5...so the area of the rect. is $ 5 \cdot 10 \equal{} 50 in^2$.) The area of the 45-45-90 tri. is $ \frac {10 \cdot 10}{2} \equal{} 50 in^2$. This means that the total area or the area of the trap. is

$ 50 \plus{} 50 \plus{} 50 \equal{} 150 in^2$

I might have made mistakes or something...but above is basically the general idea on how I would do this problem.

You made a very minor mistake. All your steps were valid except for the very last one.

It is actually $ 50 \plus{} 50 \plus{} 50\sqrt{3} \equal{} 100 \plus{} 50\sqrt{3}$.


AMC
AIME
geometry
geometric transformation
reflection
trigonometry
rotation

In the adjoining figure, two circles with radii $6$ and $8$ are drawn with their centers $12$ units apart. At $P$, one of the points of intersection, a line is drawn in sich a way that the chords $QP$ and $PR$ have equal length. ($P$ is the midpoint of $QR$) Find the square of the length of $QP$.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=793[/img]

[hide="Pretty tough, but here goes"]First, notice that if we reflect $R$ over $P$ we get $Q$. Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$, we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths:

Since $P$ is the midpoint of segment $BC$, $AP$ is a median of triangle $ABC$. Because we know that $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using stewarts or whatever else you like. We get $AC = \sqrt{56}$. So now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then

$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}$.

Doing routine algebra on the above equation, we find that $x^2=\frac{65}{2}$, so $PQ^2 = 4x^2 = 130 =$ Answer (yay!  :) )[/hide]

Interesting method, zabelman.  You can also do it by [hide]drawing radii to the midpoints of the chords.[/hide]

Nice Job, Zabelman!  My solution is not as elegant, but it works

[hide]I used trig. For triangle APB, since we know all the sides of the triangle(6,8,12), we can use law of cosine to find cos(APB)=-11/24

Cos(QPA+RPB)=cos(180-APB)=11/24

Call half of QP x, so cos(QPA)=x/8, cos(RPB)=x/6

                                sin(QPA)= :sqrt: (64-x :^2: )/8

                                sin(RPB)= :sqrt: (36-x :^2: )/6

11/24=x :^2: /48+ :sqrt: (64-x :^2: )(36-x :^2: )/48

x :^2: =y

22=y+ :sqrt: (64-y)(36-y)

(22-y) :^2: =(64-y)(36-y)

y=65/2

QP=4y=130[/hide]

yeah, I did it the way JBL suggests. Not too interesting though.

Hey JBL... Is there a way on Geometer Sketchpad to fix a point at the point Q and have a line have both its vertices on the circumferences of the circle so that i can rotate the points on the circumference but the line always passes through Q?

I haven't used sketchpad for a while, but can't you just draw the circle and point Q, then choose any point on the circle, draw the line through that point and Q, and then select the point of intersection.  Wouldn't that let you move the first point you chose around the circle?

THe only problem is that it has to go through two circles...  It works for one.

I'm not entirely sure what you're asking, but if you define the point Q, and then define a line using it and a random point on the circle, does that not do what you want?  (A picture might help me understand better if I'm still missing the point of your question.)

wait, draw radii to the midpoint of the chords? hmm this problem is buggin me... another hint... or the whole solution using this method? I prolly missed something really stupid.

See if you can figure it out from this diagram ... if you need more, I'll explain.  Basically, it's a lot of Pythagorean Theorem.

By the way, I wonder if the line between the centers and the secant of both circles really do intersect like it appears in the diagram ... hmmm ... for these numbers, it does.  Probably not in general, though, unless it's a result of zabelman's reflections.

yeah, a 3 way system of pythagorean theorem, that is the exact diagram that i started solving the problem with when i saw it, i used the same method.

Hmmm, there is one (y-z)^2+4x^2=12^2 equation and I got the 2yz stuff becoming real nasty...maybe i should just chug thru the nasty stuff.  Thanks, i take it i gotta just eliminate the x's and do substitution with the y and z and no quick way.

[hide="Solution"]In the diagram below, we let $QP=2x$, $AX=a$, and $BY=b$. From right triangle $AZB$, we see that $12^2=4x^2+(a-b)^2$. From right triangle $AXP$, $64^2=a^2+x^2$ and $36^2=b^2+x^2$. After some calculations, we find that $QP=4x^2=\boxed{130}$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=795[/img][/hide]

My method was similar to JBL's up to the drawing in the perpendiculars, but from there, I made the triangle $ABP$ and did a little trig to find its height.  From there, an equality of sums of areas led to the answer.

[quote="Allan Z"]this is a way that you don't need pythagorean

if you extend the line segment that passes through the centers. you can see there is a point making two secants.

$x*2x=(6+4)*(6+4+16)$
$2x^{2}=260$
$x^{2}=\boxed{130}$[/quote]

wait...why are the centers, and the end point of the congruent segments collinear?

yeah.. because if you name the intersection of the smaller circle and the line segment that passes through the centers to be K. and the  intersection of two circles P. Connect those two points.

$P$ is the midpoint of $QR$ (given)
$K$ is the midpoint of centers $C_{1}$$C_{2}$ by comparing lengths.

so $PK$ is the median of the right trapezoid.. so you can see its a right angle.

that means if you extend the right angle to intersect the other end of the smaller circle. that intersection is collinear with the two centers.

@JBL: how do you save geometer's sketchpad pictures as jpg?

umm... press the Print Screen key and paste it into paint  :P

[quote="junggi"]wow thanks
can't believe I didn't think of that o.o[/quote]

But I think it looks better if you save it as a .GIF


When the dimensions of a particular rectangle are expressed
in feet, the ratio of the numerical value of the perimeter to
the numerical value of the area is $ \frac{3}{4}$. What is this ratio if the dimensions are expressed in inches instead of feet? Express
your answer as a common fraction.

When you convert feet (in length) to inches, you multiply by 12. When you convert feet (square) to inches you multiply by $ 12^2\equal{}144$. You basically multiply the denominator by 12, so the new ratio in inches is $ \frac{3}{4\cdot 12}\equal{}\frac{1}{16}$.


rotation
geometry
geometric transformation
LaTeX
AMC

Equilateral triangle ABP with side AB of length 2 inches is placed inside a square AXYZ with side of length 4 inches so that B is on side AX.  The triangle is rotated clockwise about B, then P, and so on along the sides of the square until P, A, and B all return to their original positions.  The length of the path in inches traversed by vertex P is equal to:
(A) $\frac{20\pi}{3}$   (B) $\frac{32\pi}{3}$   (C) $12\pi$   (D) $\frac{40\pi}{3}$   (E) $15\pi$

Ugh...
[hide]
After just tracking the rotation around three times, I add the lengths (all of which are $\frac{\pi}{3}$ or $\frac{4\pi}{3}$) to get $\frac{40\pi}{3}$. Not sure if that's right; good chance of a stupid mistake.[/hide]

[hide]One third of the time, it's the center of rotation.  In any case, the total rotation of a point which is not the center of rotation when traversing one side is $2\pi\frac{150}{360}=\frac{5}{6}\pi$.  The triangle moves around the square 3 times.  Thus, I get the answer is $8(\frac{5}{12}4\pi)=\frac{40\pi}{3}$.[/hide]

I just did the circumference thing thrice and got 40pie / 3.

I need to learn LaTeX

05.1 LAN CTYer,
David/Johnny

Hmm, interesting...but 40pie/3 isn't a choice :P

[quote="jackb1115"][quote="DavidL"]Pie/3 = 120 degrees, doncha know...[/quote]

Erm ... I think 2pi/3 is 120 degrees.

JB[/quote]

Yup. Don't worry DavidL, joml88 was just joking about your use of using "pie" for $\pi$. Proper word will be "pi" but it doesn't matter. I personally like "pie" better.  :rotfl:  :rotfl:

[hide="Answer"]As the triangle rotates to the same position on the next side, we see that two points move $120^\circ$, then two points move $30^\circ$. Therefore, every time the triangle switches sides, one point moves $30^\circ$, one point moves $120^\circ$, and one point moves $150^\circ$. Every three sides the triangle will be in the same relative position with $P$ in the interior of the square, and therefore each point will have moved $30+120+150=300^\circ$. We need to do this four times so that the triangle is back on the bottom side, so point $P$ has moved $1200^\circ$. $\frac{1200^\circ}{360^\circ}\cdot 4\pi =\frac{40\pi}{3}\Rightarrow \boxed{D}$.[/hide]

pie=8.53973...

But anyway, pie, pi, and pai are all good :rotfl:

Sorry if you don't get all that bad humor :D

ehh pie= principal of inclusion/exclusion.

Gotcha... sorry I was a bit slow there :)

Oh god, that pie joke was definatelly witty. Oh man.... =p


Find all integer matrices $A$ of finite multiplicative order ( invertible) such that $A=I(mod2)$.

[quote="harazi"]... matrices $A$ of finite multiplicative order ( invertible)... [/quote]

What is the definition of finite multiplicative order ?


My math teacher made an interesting observation in class today:

For every primitive Pythagorean triple, the longer leg and the hypotenuse differ by one.

Prove, disprove, or salvage, to borrow a phrase from a friend.

8-15-17

Do I need to say any more?   :D

lol..what level math does your teacher teach?

or i guess i should phrase that as, what math are you currently taking, but whatever.

wrong, all primitive pythagorean triples are given by the triple

$(a,b,c)=(m^2-n^2, 2mn, m^2+n^2)$ where $m,n \in \mathbb{Z}$ and $0<n<m$

the proof is in aops... to start: show that one has to be 0 mod 4 and the other has to be 1 mod 4 for it to be primitive


Hello!

I want to take the AMC 10 B, but

1. my school doesn't offer it

2. i have read the faq 4 on the website

3. the colleges listed are way too far from where i lived

please help me and if i do convince my school to take the test, am i too late in registering

Thanks a lot!

If you're really intent on doing the AMC, and you're school won't offer it because of the cost, you can offer to pay for it. Late registration is $60 and ends Feb. 20. Since a bundle of AMC 10s is $15, you're looking at $75 for the test.

Here's a link: [url]http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Registration-B.pdf[/url]

do it at some high school.
my school doesn't offer amc either so i have to take the a and b at two different high schools
very scary (and a hassle too)

Try going [url=http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Registration-B.pdf]here[/url] but dont be surprised if you have to pay a late fee.

i already went there

what i will do is that i will try to convince my director that i will find 10 people to pay for 75 dollars

so 7.50 each

if money is the issue

and thanks also for ur help!!!

nice I have the same question but even if I pay 60 dollars, there need to be someone in the school to administer right?

yeah, you just need a math teacher, or just any teacher

but what u could do is ask 10 people that might want to do it

split up the costs

its actually 75 dollars

60-late registration
15- 10 booklets

hope this helped

I went to the college list and I found a college near my school but the name on the list is fade, does that mean the college no longer offer AMC? If it's ture it would be so bad  :furious:

And there are only colleges on the list, does anyone knows where I can find what high schools offer AMC B?

im sorry to say but ur out of luck

UNLESSS!!!

u ask ur school to take amc 10 b

get about 10 students who want to take it

75 dollars

split it up so 7.50

u have a week till deadline!!

hey which will be my score in USAMO index if I take both AMC 12A and AMC 10 B, they have different index for 12 and 10 right? I'm kind of confused.

For you they will use the higher AMC score...which will be the 10, unless something weird happened.


function
limit
real analysis
real analysis unsolved

Let $ a,b \in$[b]R[/b],$ a < b$ and for $ n \equal{} 1,2,3...$. Let $ f_n: [a,b] \minus{}>$[b]R[/b] be an increasing function. Prove that if the sequence $ f_1,f_2,f_3...$ converges to $ f$ then $ f$ is increasing, and that if $ f$ is continuous then the convergence is uniform.

Thanks :)

Well, let $ x\leq y$, $ x,y\in [a,b]$, $ f_n \to f$. We have $ f_n (x) \leq f_n(y)$ for all $ n\in \mathbb{N}$, so because $ f_n\to f$ it implies that

$ f(x) \equal{} \lim f_n(x) \leq \lim f_n(y) \equal{} f(y)$.

Thanks!

Does anyone know how to do the second part of the question. That is the one I'm really having problems with.

Are you sure you want $ f_n$ to be increasing for each $ n$ instead of for each $ x$, $ f_n$ is increasing?

Sorry I wasn't clear. I meant $ f_n(x) \leq f_n(y)$ if $ x \leq y$.


invariant
algebra
polynomial
superior algebra
superior algebra unsolved

Let $ a(X)\equal{}\frac{1}{X}$ an automorphism of $ Q(X)$. What is the invariant subfield of $ Q(X)$ under $ a(X)$?
I think it must be: $ <X\plus{}\frac{1}{X}>$ but I can not prove one of the 'inclusions': why is every $ f$ in the invariant subfield of $ Q(X)$ under $ a(X)$ a combination of $ X\plus{}\frac{1}{X}$?

Set $ Y: \equal{} X \plus{} {1\over X}$.
Then the polynomial $ f(Z) \equal{} Z^2 \plus{} 1 \minus{} YZ\in Q(Y)[Z]$ has degree $ 2$ and root $ Z \equal{} X$.
So $ X$ has degree $ \le 2$ over $ Q(Y)$, i.e. $ [Q(X): Q(Y)]\le 2$.
But the fixed field $ K$ of $ a$ satifies $ Q(Y)\subseteq K\subsetneq Q(X)$, which gives $ K \equal{} Q(Y)$.


If anyone has the full Shortlist, could someone also post it to me? My e-mail address is [color=red]nguyenquockhanh12toan@yahoo.com[/color]

the Shortist for what year? The 2004 Shortlist will only be available next year this time ..

Yes, the 2004 one is secret indeed.

But if anyone has it, feel free to mail me :D

[quote="Templar"] Go to http://www.kalva.demon.co.uk/short.html, it has most of the shortlists. [/quote]
 In the case that you people haven't noticed yet, orl and other members have been working hard to make the isl/ill project available. I still do not know why you redirect people to kalva, instead of the downloads forum or the search page.

[i]Posted from a Sony Ericsson K700i [/i]

because many people don't open those ;)

I still don't get why you consider ISL/ILL problems as things to download :P

[quote="Sasha"][quote="Peter VDD"]
I still don't get why you consider ISL/ILL problems as things to download :P[/quote]
Because some of the files are really giant![/quote]

Come on, these files are not so huge again.  :)  I think most of them are less than 4 MB. Maybe my posted ISL/ILL files would have been found be more easily if they would have been linked here as well: [url=http://www.mathlinks.ro/Forum/downloads.php]Ressources[/url] .  :)

BTW, sasha, is there a policy why most of your postings are just one line long...  :D


probability
floor function
algorithm
number theory
Euclidean algorithm
real analysis
real analysis theorems

There's a theorem I've seen to the effect of: if $ a$ and $ b$ are real numbers that are linearly independent over $ \mathbb{Q}$ (like if $ a$ is rational and $ b$ is irrational for example) then the set $ \{m\cdot a \plus{} n\cdot b : m,n \in \mathbb{Z}\}$ is dense in $ \mathbb{R}$. What's this theorem called and could someone reference me to a text containing a proof? I have an amusing probability problem which I can wrap up nicely with this theorem. (I can post the problem if anyone's curious)

I don't have the name, but here's a sketch of the proof:

Let $ d\equal{}\inf\{ma\plus{}nb\mid n,m\in\mathbb{Z},na\plus{}mb>0\}$

There are two possibilities:

Case 1: $ d>0.$ Then there exist integers $ j,k$ such that $ a\equal{}jd$ and $ b\equal{}kd.$ (or, for short, $ d\mid a$ and $ d\mid b.$) That happens if and only if $ \frac{a}{b}\equal{}\frac{j}{k}\in\mathbb{Q}.$

The rest of case 1 is the first few theorems of elementary number theory. We have that $ d\equal{}\gcd(a,b)$ and the whole set $ \{ma\plus{}nb\mid m,n\in\mathbb{Z}\}$ is simply $ \{jd\mid j\in\mathbb{Z}\}$ and is discrete, not dense.

Case 2: $ d\equal{}0.$ Then $ L\equal{} \{ma\plus{}nb\mid m,n\in\mathbb{Z}\}$ is dense in $ \mathbb{R}.$

Suppose $ L$ is not dense in $ \mathbb{R}.$ Then there exists an interval $ (x,x\plus{}\delta)$ whose intersection with $ L$ is empty. But there exists $ y\in L$ such that $ 0<y<\delta.$ Now look at the points $ ky$ for $ k\in\mathbb{Z}.$ Since $ y\equal{}ma\plus{}nb$ and $ ky\equal{}kma\plus{}knb,$ we have $ ky\in L.$ Let $ k\equal{}\left\lfloor \frac{x}{y}\right\rfloor.$ Then $ x<(k\plus{}1)y\equal{}ky\plus{}y\le x\plus{}y<x\plus{}\delta.$ So we have produced an element of $ L$ in $ (x,x\plus{}\delta)$ and $ L$ is dense after all.

Again, case 2 happens if and only if $ \frac{a}{b}$ is irrational. If $ \frac{a}{b}$ is rational, then the Euclidean algorithm applied to $ a$ and $ b$ will produce $ \gcd(a,b),$ or if you're uncomfortable applying the Euclidean algorithm to non-integers, then let $ a\equal{}\frac{p}{q}b,$ then apply the Euclidean algorithm to find $ \gcd(p,q).$ Either way, we wind up in case 1. But if we're in case 1, we have that $ \frac{a}{b}$ is rational.


By Chebyshef's inequality we have
\sumx/(1 + y)(1 + z) \geq \sumx/3 \sum1/(1 + y)(1 + z).
(Since x \leq y \leq z => x \leq y \leq z and so on.)
Now put x + y + z = 3a. Then xyz = 1 => a \geq 1.
Also \sumx/3 \geq a and \sum1/(1 + y)(1 + z) = 3(1 + a)/(1 + x)(1 + y)(1 + z).
Now (1 + x)(1 + y)(1 + z) \leq (1 + a) by AM - GM.
So \sum1/(1 + y)(1 + z) = 3(1 + a)/(1 + x)(1 + y)(1 + z) \geq 3/(1 + a).
Hence \sumx/(1 + y)(1 + z) \geq 3a/(1 + a).
Now we have to prove that 3a/(1 + a) \geq 3/4.
We'll prove that 3a/(1 + a) \geq 3a/4.
Indeed, this is equivalent to 4a \geq (a + 2a + 1) or (a - 1)(3a + 1) \geq 0.
This is obviously true since a \geq 1.
So 3a/(1 + a) \geq 3a/4 \geq 3/4 (since a \geq 1).

first of all note the f(x) = (x^3)/(1+x) 2 is convex and increasing
assume x \geq y \geq z
note that
x^3/(1+z)  \geq  y^3/(1+z) and  1/(1+y) \geq  1/(1+x)
apply rearrangement ineq to get
x^3/(1+z)(1+y)+y^3/(1+z)(1+x)  \geq  x^3/(1+z)(1+x)+y^3/(1+z)(1+y)
similarly y^3/(1+y)  \geq  z^3/(1+x) and 1/(1+z)  \geq  1/(1+y)
aplly rearrangement
y^3/(1+y)(1+z)+z^3/(1+x)(1+y)  \geq  y^3/(1+y) 2 +z^3/(1+x)(1+z)
and x^3/(1+x) \geq z^3/(1+z) and 1/(1+z)  \geq  1/(1+x)
apply rearrangement
x^3/(1+z)(1+x)+ z^3/(1+z)(1+x) \geq  x^3/(1+x) 2  + z^3/(1+z) 2
add above three rearreangement inequalities to get
x^3/(1+y)(1+z)+ y^3/(1+z)(1+x)+ z^3/(1+x)(1+y)  \geq  f(x)+f(y)+f(z) \geq 3f[(xyz)^(1/3)] = 3/4
Q.E.D
cheers

Here?s another soln:

I used the lemma (this is a problem posted on this forum and solved):
For positive a,b,c,x,y,z  a^3/x+b^3/y+c^3/z >= (a+b+c)^3/3(x+y+z).
We apply this and we have to prove (x+y+z)^3/(3+2(x+y+z)+xy+yz+zx) >= 9/4 iff (3+2(x+y+z)+xy+yz+zx)/ (x+y+z)^3 <=4/9. Since x+y+z>=3 we get (1) 3/(x+y+z)^3<=1/9, (2) 2/(x+y+z)^2 <= 2/9, (3) (xy+yz+zx)/(x+y+z)^3 <= 1/9 because (xy+yz+zx)/(x+y+z)^2 <= 1/3 and 1/(x+y+z)<=1/3. We add (1), (2) and (3) and we get the ineq.


number theory
greatest common divisor
algebra
polynomial
Vieta
continued fraction
number theory unsolved

find all a's such that $ 5a^2\plus{}4$ is a perfect square.(a is an integer)

It is a square for $ |a| \equal{} 0,1,3,8,21,55,144,377,987...$ :o

I have only proved that every odd divisor of $ 5a^2 \plus{} 4$ has even digit of tenth, but it is straighforward and it doesn't solve the problem..

[quote="bboypa"]It is a square for $ |a| \equal{} 0,1,3,8,21,55,144,377,987...$ [/quote]
Do you recognize these numbers?
They form every second element of a famous sequence.

cnyd do u find all solutions??? i think u tried to make a solution in which you prove there exist 4k+0,4k+1,4k+2,4k+3 form solutions of this equation.

Use http://www.alpertron.com.ar/QUAD.HTM.

 5 x2 - y2 + 4 = 0

by Dario Alejandro Alpern

First of all we must determine the gcd of all coefficients but the constant term, that is: gcd(5, 0, -1, 0, 0) = 1.

Dividing the equation by the greatest common divisor we obtain:
 5 x2 - y2 + 4 = 0

We try to check the equation modulo the prime divisors of 5.
 5 x2 - y'2 + 4 = 0

Let x = sy - fz, so [-(as2 + bs + c)/f]y2 + (2as + b)yz - afz2 = 1.

So  5 s2 -1 should be multiple of 4.

This holds for s = 1, 3.

    * Let s = 1. Replacing in the above equation:
      -1 y2 + 10 yz - 20 z2 = 1

      We have to find the continued fraction expansion of the roots of -1 t2 + 10 t - 20 = 0, that is, (sqrt(20) - 10) / (-2)

      Simplifying, (sqrt(5) - 5) / (-1)

      The continued fraction expansion is:
      2+ //1, 3, 4//
      where the periodic part is marked in bold.

      Y0 = NUM(1) = 3
      Z0 = DEN(1) = 1
      Since X0 = Y0 - 4 Z0:

      X0 = -1
      Y0 = 3
      and also:
      X0 = 1
      Y0 = -3
      Y0 = NUM(0) = -7
      Z0 = DEN(0) = -1
      Since X0 = Y0 - 4 Z0:

      and also:
      X0 = -3
      Y0 = -7
      and also:
      X0 = 3
      Y0 = 7
      Choosing the solution with minimum y:

      and also:
      X0 = -1
      Y0 = 3

      and also:
      X0 = 1
      Y0 = -3
    * Let s = 3. Replacing in the above equation:
      -11 y2 + 30 yz - 20 z2 = 1

      We have to find the continued fraction expansion of the roots of -11 t2 + 30 t - 20 = 0, that is, (sqrt(20) - 30) / (-22)

      Simplifying, (sqrt(5) - 15) / (-11)

      The continued fraction expansion is:
      1+ //6, 4//
      where the periodic part is 4.

      Y0 = NUM(1) = 7
      Z0 = DEN(1) = 6
      Since X0 =  3 Y0 - 4 Z0:

      and also:
      X0 = -3
      Y0 = 7
      and also:
      X0 = 3
      Y0 = -7
      Y0 = NUM(2) = -3
      Z0 = DEN(2) = -2
      Since X0 =  3 Y0 - 4 Z0:

      and also:
      X0 = -1
      Y0 = -3
      and also:
      X0 = 1
      Y0 = 3
      Choosing the solution with minimum y:

      and also:
      X0 = 1
      Y0 = 3

      and also:
      X0 = -1
      Y0 = -3

Since 2 * 2 is a divisor of the constant term (4), the solutions should be 2 times the solutions of  5 u2 - v2 +1 = 0.

 5 x2 - y'2 +1 = 0

Let u = sv - Fz, so [-(as2 + bs + c)/F]v2 + (2as + b)vz - aFz2 = 1.

So  5 s2 -1 should be multiple of 1.

This holds for s = 0.

    * Let s = 0. Replacing in the above equation:
      1 v2 - 5 z2 = 1

      We have to find the continued fraction expansion of the roots of 1 t2 - 5 = 0, that is, sqrt(20) /2

      Simplifying, sqrt(5)

      The continued fraction expansion is:
      2+ //4//
      where the periodic part is 4.

      Y0 = 1
      Z0 = 0
      Since X0 = 2 U0 = 2 ( - Z0)
      and Y0 = 2 V0:

      and also:
      X0 = 0
      Y0 = 2
      and also:
      X0 = 0
      Y0 = -2
      Y0 = -1
      Z0 = 0
      Since X0 = 2 U0 = 2 ( - Z0)
      and Y0 = 2 V0:

      and also:
      X0 = 0
      Y0 = -2
      and also:
      X0 = 0
      Y0 = 2
      Choosing the solution with minimum y:

      and also:
      X0 = 0
      Y0 = 2

      and also:
      X0 = 0
      Y0 = -2

Xn+1 = P Xn + Q Yn
Yn+1 = R Xn + S Yn

In order to find the values of P, Q, R, S we have to find first an integer solution of the equation m2 + bmn + acn2 =  m2 - 5 n2 = 1.

We have to find the continued fraction expansion of the roots of 1 t2 - 5 = 0, that is, sqrt(20) /2

Simplifying, sqrt(5)

The continued fraction expansion is:
2+ //4//
where the periodic part is 4.

An integer solution of the equation m2 + bmn + acn2 =  m2 - 5 n2 = 1 is:

m = 9
n = 4

Using the formulas:
P = m
Q = -Cn
R = An
S = m + Bn
we obtain:

P = 9
Q = 4
R = 20
S = 9

Consider the equation $ (n^2 \minus{} nm \minus{} m^2)^2 \minus{} 1 \equal{} 0$ in $ \mathbb{Z}^2$, then $ n \equal{} m \implies (1,1)$ otherwise $ n > m$ and it is necessary condition that $ 5m^2\pm 4$ is a square. Now $ (n,m) \equal{} (F_{i \plus{} 1},F_i)$ is a class of solution by the identity above. And they are unique solution, by vieta jumping. Consider the minimal non fibonacci solution (a,b) with a minimal. Then (b,a-b) is another solution, but since b<a then is in fibonacci form, because if b and a-b is in fibonacci sequence, also a is in, contradiction.


Ladies and Gentlemen, only a little question:
I've just bought "Problem-Solving through Problems by Loren Larson" (I'm a beginner), so I'd like to know if you suggest me to follow the book order (First complete Chapt. 1, then 2 and so on) or not... In particular, in the First Chapter Loren explains many heuristics asking the reader to apply them solving some problems: should I first study Theory Chapter (with problems) or should I shake some theory with some heuristics chapter problems?

I'm sorry for my English, I'm still learning it  :blush:

Thanks in advance

P.S. I've added a poll, please explain me why you chose a particular answer.

tmart,
PSTP (an excellent book, by the way) is not a traditional textbook, where each chapter builds on previous chapters, so do not feel obliged to read through the book in a particular order.

I would read the first chapter on heuristics first, just to get a feel for problem solving, but where you go from there is up to you.  Just dive into whatever looks interesting.

Note that the problems in the heuristics chapter don't really require any knowledge of mathematics to solve. They only require cleverness. Cleverness is more important than any specific theorem since it's easy to learn theorems and hard to learn cleverness. So I'd study the heuristics first.


linear algebra
three variable inequality
USAMO
calculus
Inequality
algebra
inequalities

This resembles APMO 2004/5 so much.

Note that if $x>0$ then $x^5-x^2+3 \geq x^3+2$, which can be seen by factoring into $(x^3-1)(x^2-1) \geq 0$.

Then the inequality reduces to
\[
\prod (a^3 +2 ) \geq (a+b+c)^3
\]

or,

\[
\prod (\frac{a^3-1}{3}+1) \geq (\frac{a+b+c}{3})^3
\]

If $a,b,c \geq 1$, or $a,b,c \leq 1$, then by Bernoulli's Inequality, we have
\[
\prod (\frac{a^3-1}{3}+1) \geq 1 + \sum \frac{a^3-1}{3} =\sum a^3/3
\]
Complete by Power-Mean

If $a \geq 1 \geq b,c$ or $a \leq 1 \leq b,c$, then by Bernoulli and Holder, we have
\[
\prod (\frac{a^3-1}{3}+1) \geq (\frac{a^3+1+1}{3})(1+\frac{b^3-1}{3}+\frac{c^3-1}{3})\geq (\frac{a^3+1+1}{3})(\frac{1+b^3+c^3}{3})(\frac{1+1+1}{3}) \geq (\frac{a+b+c}{3})^3
\]
Q.E.D.

[b]Note[/b]: Bernoulli's inequality (or perhaps an extension of it, anyone know its name?) states that if $a_1, \ldots, a_n > -1$, and that all the {a} have the same sign, then $\prod(1+a) \geq 1+\sum a$.

After Yufei's brilliant observation that x^5 - x^2 + 3 >= x^3 + 2, the proof results more directly by application of the generalized Holder Inequality to the matrix
a^3      1          1
 1         b^3      1
 1         1          c^3.  Assigning weights of 1/3 to each row gives

[(a^3+1+1)(1+b^3+1)(1+1+c^3)]^1/3>=(a+b+c)^3 as required.

Can you please send all USAMO 2004 problems?
I would be very grateful to you.

The following inequalities are also true.
Let $a,b,c>0$ ,  prove that\[(a^5-a^3+3)(b^5-b^3+3)(c^5-c^3+3)\geq 3(a+b+c)^2.\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2739210]here[/url]
Let $a ,b, c$ be positive real numbers  , prove that\[  \left(a^{2013}-a^{2011 }\plus{} 3\right)\left(b^{2013}-b^{2011}\plus{} 3\right)\left(c^{2013}-c^{2011} \plus{} 3\right)\geq 9\left(a^2 \plus{} b^2 \plus{} c^2\right) .\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=532913&p=3048963#p3048963]here[/url]

In case anyone wants some motivation as to where the inequality $a^5 - a^2 + 3 \geq a^3 + 2$, here is how we might think of something like this. Note that the RHS of the inequality is of degree 3, so we want to prove some inequality of the form $a^5 - a^2 + 3 \geq ka^3 + \ell$, for some real numbers $k$ and $\ell$. Replace $a^3$ with $x$, so we want to prove an inequality of the form $x^\frac{5}{3} - x^\frac{2}{3} + 3 \geq kx + \ell$. Note that we have equality in the original inequality at $a = 1$, so we want equality here when $x = 1$. This suggest making the line $y = kx + \ell$ tangent to this curve at the point $(1,3)$. the derivative of $x^\frac{5}{3} - x^\frac{2}{3} + 3$ evaluated at $1$ turns out just to be $1$, so the equation of the tangent line at $(1,3)$ is $y - 3 = x - 1$, or $y = x + 2$, so putting back in $x = a^3$, we want to prove that $a^5 - a^2 + 3 \geq a^3 + 2$, where is very easy to do, just by factoring, and then the final inequality we get follows easily from holders inequality.

never mind, incorrect solution was here

thanks mssmath

The Jensen solution above is definitely wrong as the sign of Jensen's on f(3x) is reversed.

[quote=TheStrangeCharm]In case anyone wants some motivation as to where the inequality $a^5 - a^2 + 3 \geq a^3 + 2$, here is how we might think of something like this. Note that the RHS of the inequality is of degree 3, so we want to prove some inequality of the form $a^5 - a^2 + 3 \geq ka^3 + \ell$, for some real numbers $k$ and $\ell$.[/quote]
I don't understand how you got this. How did you get rid of the non $a^3$ terms?

[quote=Delray]Is the observation related to the quantity in question being greater than or equal to $a^3+3$ necessary? If so, how did people manage to discover this while solving the problem?[/quote]

This is motivated by holders (three var suggests holder)

[quote=Binomial-theorem]So I had to use the hint from the AoPS book, which was as follows.  After that, the problem pretty much follows.

[hide="Hint"]

Prove that $(a^2-1)(a^3-1)\ge 0$.  [/hide]

[hide="Solution"]

$(a^2-1)(a^3-1)\ge 0$ because $(a-1)^2(a+1)(a^2+a+1)\ge 0$ because $a>0$.

Now, because $(a^2-1)(a^3-1)\ge 0$, we have $a^5-a^2-a^3+1\ge 0$.  Therefore, $a^5-a^2+3\ge a^3+2$.

Now, we desire $(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$ which is directly true by Holder's.  [/hide][/quote]

does anybody know the motivation for the hint?

@above
We wanted to remove the power 5 term along with the squares so that even  if some cubics are left we can simly use holder on it so $(a^2-1)(a^3-1)$ was the best candidate.

Observe that $(a^2-1)(a^3-1) \ge 0$, so $a^5-a^2-a^3+1 \ge 0$. This can be written as $a^5-a^2+3 \ge a^3+2$. Hence, it suffices to prove
\[\prod (a^3+2) \ge (a+b+c)^3.\]
However,
\[\prod (a^3+1+1)^{\frac{1}{3}} \ge (a+b+c)\]
by Holder's, so
\[\prod(a^3+2) \ge (a+b+c)^3\]
by cubing both sides and we're done.

I think the step to find $a^5-a^2+3 \geq a^3+2$ is pretty natural? It's pretty obvious that the fifth powers don't relate to the problem at all, so I decided to tame it with third powers so possibly I could use an inequality to finish.

But alas, I didn't find the nice solution.

[b][color=#00f]Bad Solution.[/color][/b] Observe that
\begin{align*}
(a^3-1)(a^2-1) \geq 0 \ \forall a > 0 \\
a^5-a^2+3 \geq a^3+2 \ \forall a > 0
\end{align*}
Thus it suffices to show that $\prod_{cyc} (a^3+2) \geq (a+b+c)^3$. Expanding, the inequality is equivalent to $$a^3b^3c^3+2\sum_{cyc} a^3b^3+3\sum_{cyc} a^3 - 3\sum_{sym} a^2b-6abc+8 \geq 0.$$ By Schur, $$3\sum_{cyc} a^3+9abc-3\sum_{sym} a^2b \geq 0,$$ and by AM-GM, $$2\sum_{cyc} a^3b^3 \geq 6a^2b^2c^2.$$ Thus we need $$a^3b^3c^3+6a^2b^2c^2-15abc+8\geq 0 \iff (abc-1)^2(abc+8) \geq 0,$$ which is clearly true for $abc > 0$. Equality holds when $a=b=c=1$. $\square$

[hide=Storage]$a^5+1 \ge a^3+a^2$ because $(a^2-1)(a^3-1) \ge 0$ So we have \[\prod (a^5-a^2+3) \ge \prod(a^3+2).\] Now by Hölder we get \[\prod (a^3+1+1) \ge (a+b+c)^3\] and so we are done  :D $\blacksquare$ [/hide]
[hide=Remarks]It was clear to me that you have to use Hölder's Inequality here, but for that I thought you need cubes on the left hand side and so I thought to prove $a^5+3 \ge a^3+a^2$(using AM GM) but failed, then in my AM GM proof, I noticed that I needed a $-2$ in the inequality on the left hand side and because of that I discovered the right Inequality! :D [/hide]

[b]Claim:[/b] $x^3+2 \le x^5-x^2+3$
[b]Proof:[/b] We have $$0 \le (x^3-1)(x^2-1)= x^5-x^3-x^2+1 \implies x^3+2 \le x^5-x^2+3.$$ $\Box$

[b]Claim:[/b] $\sum_{cyc} (a^3+2 ) \le (a+b+c)^3.$
[b]Proof:[/b] Expand the LHS and RHS, so the desired inequality is the same as $$a^3b^3c^3+3(a^3+b^3+c^3)+2(a^3b^3+b^3c^3+c^3a^3)+8 \ge 3(a^2b+b^2a+b^c+c^2b+a^2c+c^2a)+6abc.$$ Now we use the fact that $$a^3+a^3b^3+1 \ge 3a^2b,$$ this is direct from AM-GM. After using this, our original inequality can be simplified down to $$a^3b^3c^3+a^3+b^3+c^3+1+1 \ge 6abc,$$ but this is also straightforward from AM-GM, therefore, our original inequality is true and we are done. $\blacksquare$

Pretty much same as mathpi's solution oops. Theres probably a cleaner finish using Holder's.

It is trivial to see that $(x^3-1)(x^2-1) \geq 0 \implies x^5-x^2+3 \geq x^3+2$. It suffices to prove $\prod_{\text{cyc}} a^3+2 \geq (a+b+c)^3$. Expanding, and simplifying, we receive $-a^3b^3c^3-2a^3b^3-2a^3c^3-3a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2-2b^3c^3-3b^3+3b^2 c+3bc^2-3 c^3-8 \leq 0$. Now notice that $$a^3+a^3b^3+1 \geq 3a^2b$$ by AM-GM, which simplifies the inequality to $a^3b^3c^3+a^3+b^3+c^3+1+1 \geq 6abc$ which is a direct result of AM-GM

You could also use Holder's as the last two posts stated.

The original problem is a special case of the following generalization where
$$p=2$$.

[b]Generalization 1[/b]
Let $a_{1},a_{2},\cdots,a_{p+1}$ be positive reals and $p$ be a nonnegative integer. Then prove that

$$\prod{\left(a^{2p+1}-a^{p}+p+1\right)}\geq \left(\sum_{cyc}{a_{1}}\right)^{p+1}$$

Note that $x^5-x^2+3\ge x^3+2$ since this is equivalent to $(x-1)^2(x^2+x+1)(x+1)\ge 0$. So, it suffices to prove
\[(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3\]
and by Holder's, we have
\[(a^3+1+1)(1+b^3+1)(1+1+c^3)\ge (a+b+c)^3\]
so we are done. $\square$


analytic geometry
geometry
vector
geometry theorems

Hi all.

Can someone plz tell me about using trilinear co-ordinates in triangle geometry problems? thx.

vinoth

I think you should wait for Darij. He is the expert on this field I think.  :)

Well, I have written something [url=http://mathlinks.ro/viewtopic.php?t=4357]in this thread[/url] but it's not complete with regard to the applications in triangle geometry. I will add more if I find time (what is rather unlikely). Also, [url=http://mathworld.wolfram.com/TrilinearCoordinates.html]the Mathworld site[/url] has some interesting stuff.

  Darij

hi.

are there nice trilinear formulas for more complex stuff - e.g. is there some condition relating to whether three lines are concurrent? or if two lines are perpendicular? or two lines parellel?

thx, vinoth

If a line g has the trilinear equation kx + ly + mz = 0, where P ( x : y : z ) are the homogeneous trilinear coordinates of a variable point, then we also say that ( k : l : m ) are the [b]homogeneous trilinear line coordinates[/b] of g.

If two lines have line coordinates ( k : l : m ) and ( k' : l' : m' ), then their point of intersection has the coordinates ( lm' - ml' : mk' - km' : kl' - lk' ).

If two points have coordinates ( x : y : z ) and ( x' : y' : z' ), then the line joining these points has the line coordinates ( yz' - zy' : zx' - xz' : xy' - yx' ).

Two lines with line coordinates ( k : l : m ) and ( k' : l' : m' ) are parallel if and only if their point of intersection lies on the line at infinity, i. e. if we have

  a(lm' - ml') + b(mk' - km') + c(kl' - lk') = 0.

For perpendicular or concurrent lines, the conditions should be rather complicated. However, there is a nice relationship between trilinear coordinates and vectors: If a point P [i]not[/i] lying on the line at infinity has the homogeneous trilinears ( x : y : z ), then $ax\cdot \overrightarrow{PA}+by\cdot \overrightarrow{PB}+cz\cdot \overrightarrow{PC}=\overrightarrow{0}$.

  Darij


A regular tetahedron has edges 36 units in length. What is the altitude of the tetrahedron? Give your answer in simplest radical form.

The hight of regular tetrahedron with side length$ a$ is $ \boxed{\frac{\sqrt{6}}{3}a}$


When will the USAMTS problems be posted?

It says that you have a month to submit solutions so the first round should start early next month.