Let $ ABC$ be an acute-angled triangle. Let $ E$ be a point such $ E$ and $ B$ are on distinct sides of the line $ AC,$ and $ D$ is an interior point of segment $ AE.$ We have $ \angle ADB \equal{} \angle CDE,$ $ \angle BAD \equal{} \angle ECD,$ and $ \angle ACB \equal{} \angle EBA.$ Prove that $ B, C$ and $ E$ lie on the same line.

$ \frac {AD} {DC}\equal{}\frac {BD} {DE} \Rightarrow \triangle ADC \sim \triangle BDE \Rightarrow \angle DCA\equal{}\angle DEB$ and $ \angle DAC\equal{}\angle DBE$

Denote by $ P\equal{}BE \cap AC$

$ \angle ACB\equal{}\angle ABP \Rightarrow \triangle APB \sim \triangle ABC \Rightarrow \angle ABC\equal{}\angle APB\equal{}\angle ADB$

In the quadrileteral $ ABCE$ the sum of the angles is equal with $ 180^{\circ}$

$ (\angle CEA\plus{}\angle EAB\plus{}\angle ABC)\plus{}\angle BCE\equal{}360^{\circ} \Rightarrow 180^{\circ}\plus{}\angle BCE\equal{}360^{\circ} \Rightarrow$

$ \Rightarrow \angle BCE\equal{}180^{\circ}$

[quote="moldovan"]

In the quadrileteral $ ABCE$ the sum of the angles is equal with $ 180^{\circ}$

$ (\angle CEA\plus{}\angle EAB\plus{}\angle ABC)\plus{}\angle BCE\equal{}360^{\circ} \Rightarrow 180^{\circ}\plus{}\angle BCE\equal{}360^{\circ} \Rightarrow$

$ \Rightarrow \angle BCE\equal{}180^{\circ}$[/quote]
Can you explain the remark above for me ? :)

[quote="sunken rock"]The second angle equality was sufficient for the requirement of the problem, that makes $ABCD$ cyclic, done.

Best regards,
sunken rock[/quote]
What is the angle equality which you talk about? :(
Can you write it? :(

Pls, disregard my previous post, I was not ON!

Best regards,
sunken rock

Let $AC\cap BE=X, AB\cap CE=Y$.
By first two angle conditions $\triangle DAB\sim \triangle DCE$, means $D$ is Miquel point of $ABEC$. So $ECXD, ACYD$ are cyclic. Since $D\in AE$, by Miquel on $\triangle ABE$ we have $BXCY$ - also cyclic. So $\angle ACY = \angle XBY=\angle EBA=\angle ACB$ , which is what we need.


Hey, have you looked at the shortlisted geometry problems? Does anybody have a nice sol for G2? I have one, but I would like to see others. Here it is (the problem): The sides of the triangle ABC subtend equal angles at the point F inside the triangle (I hope this means that F is the Torricelli-Fermat point). The cevians BF and CF meet AC and AB at D and E respectively. Prove that AB+AC>=4DE.

I'll post my solution later.

I can solve it by some computations.It's not too difficult , is it?

Here's what I did :

We know that F is Torricelli's point, so it's the intersection between the circumcircles of the equilateral triangles constructed on the sides of ABC, in the exterior. Let ACM and ABN be 2 such equilateral triangles. F is also the intersection between BM and CN (these are known theorems, their proofs may be found in many places). We have to use the following lemma: ACM is an equilateral triangle, F is on its circumcircle, On the opposite side of AC than M, FM intersects AC at D. We can easily prove that FD/FM<=1/4 (think that if F is the midpoint of the arc AC then we haveequality, and look at what happens if we move from that location).

In the triangle FNM we have FD/FM, FE/FN<=1/4, so 4DE<=MN<=MA+AN=AC+AB because triangles ACM and ABN are equilateral. These completes the proof. I sure hope it's correct.


integration
limit
function
search
real analysis
real analysis unsolved

Let $ {f_n}$  be a sequence of nonnegative $ L^p$ functions(i.e.$ f_n\in L^p$) ($ p>0$)such that $ f_n\to f$ in $ L^1$. When can we have $ \int{f_n}^p\to \int{f}^p$?

(1) what measure space?

(2) you allow p<1 as long as it is >0 ?

If we are speaking of $ L_p(X,\mu)$ with and the measure $ \mu$ is finite, then a sufficient condition is $ \lim_{c\to\infty} \sup_n\int |f_n|^p 1_{|f_n|\ge c} \equal{} 0$ (but not necessary; however, if $ f \equal{} 0$, it is also necessary). This is called uniform integrability. In turn, a sufficient condition for the uniform integrability is $ \sup_n\|f_n\|_q < \infty$ for some $ q > p$. So if you have $ p < 1$ in your problem, then it is fulfilled.

1.The underlying measure space need not be finite.
2. I want the solution mainly for  $ p < 1$.

zhoraster, is the result true for infinite measure space also?

[quote="zhoraster"] So if you have $ p < 1$ in your problem, then it is fulfilled.[/quote]
How is this true in my problem?

[quote="limsup"]How is this true in my problem?[/quote]

Yes I understand now. When $ p<1$, I  will have $ \sup_{n}\|f_{n}\|<\infty$ because $ f_{n}\to f$ in $ L^{1}$. But I don't know whether the result would be true for infinite measure spaces also :(

I believe Kent had gave very nice proof for the case $ 0<p<1$.  The proof is based on the properties of convex function $ f(t)\equal{}t^p$. You can search on the forum ! :lol:

[quote="1234567a"]I believe Kent had gave very nice proof for the case $ 0 < p < 1$.[/quote]

I could not find that particular post. Could you please help me in getting that post?

Thanks.

The uniform integrability concept if relevant only for a finite measure: it relies mainly on the fact that a constant function is integrable.


It's $\infty$.
See:
http://www.mathlinks.ro/Forum/viewtopic.php?t=44490
http://www.mathlinks.ro/Forum/viewtopic.php?t=10626

Also some more precise things:

Merten's Theorem (well only one of some more):
$\sum_{p \leq x} \frac{1}{p} = \log ( \log (x )) + c + O\left( \frac{1}{\log(x)} \right)$ where $c$ is some constant.

A Theorem by Dirichlet (this impleas the existence of $\infty$ many prime $\equiv a \mod m$):
Let $m,a$ be coprime. Then:
$\sum_{p \equiv a \mod m} \frac{1}{p^x} = \frac{-1}{\varphi(m)} \log(x-1) + O(1)$.


Σε ένα μη εκφυλισμένο τρίγωνο μια διχοτόμος είναι κάθετη σε μια διάμεσο. Να βρεθούν τα μήκη των πλευρών του αν αυτά είναι διαδοχικοί ακέραιοι.

Αν υποθέσουμε ότι  ΑΒ < ΒΓ< ΑΓ,  έχω βγάλει ότι

ΑΒ = 2,  ΒΓ = 3,  και  ΑΓ = 4  όπου η διάμεσος και η διχοτόμος άγονται από τις

κορυφές  Α  και  Β  αντίστοιχα . :)

Μανιέ πές μου άν έχει γίνει κάποιο λάθος στα αποτελέσματα. Η λύση σου ποιά είναι ?

Έστω ABC το τρίγωνο και (χωρίς βλάβη της γενικότητα2) AD η διχοτόμος
επειδή δε γίνεται να είναι AD κάθετη στη διάμεσο AK, μπορούμε να θέσουμε (χωρίς βλαβη της γενικότητας) ότι AD είναι κάθετη στη διάμεσο BM.

Το τρίγωνο ABM είναι ισοσκελές άρα $ AB \equal{} AM \equal{} \frac{AC}{2}$

άρα $ c \equal{} \frac{b}{2} \Leftrightarrow b \equal{} 2c (I)$

1. αν $ c \equal{} b \minus{} 1$ τότε $ (I) \Leftrightarrow b \equal{} 2(b \minus{} 1) \Leftrightarrow b \equal{} 2$
    άρα $ c \equal{} 1$ και (επειδή $ a \not\equal{} 0$) $ a \equal{} 3$, για τα οποία δεν ισχύει η τριγωνική ανισότητα
    αφού είναι $ 1 \plus{} 2 \equal{} b \plus{} c \equal{} a \equal{} 3$ άρα άτοπο

2. αν $ c \equal{} b \plus{} 1$ τότε $ (I) \Leftrightarrow b \equal{} 2(b \plus{} 1) \Leftrightarrow b \equal{} \minus{}2 < 0 \Leftrightarrow$ άτοπο

3. αν $ c \equal{} b \minus{} 2$ τότε $ (I) \Leftrightarrow b \equal{} 2(b \minus{} 2) \Leftrightarrow b \equal{} 4$
   άρα $ c \equal{} 2$ και $ a \equal{} 3$

4. αν $ c \equal{} b \plus{} 2$ τότε $ (I) \Leftrightarrow b \equal{} 2(b \plus{} 2) \Leftrightarrow b \equal{} \minus{}4 < 0 \Leftrightarrow$ άτοπο

άρα οι πλευρές του τριγώνου έχουν μήκη $ 2 , 3 , 4$


The first nineteen terms (a1 through a19) of an arithmetic series sum to 100. The second nineteen terms (a20 through a38) sum to 210. Find a1 and a20.

Thanks..

Very similar to the problem that I had, and the solution that I posted...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136868

[hide="long version"]
[b]1[/b]
With d as the common difference,
$\sum_{n=1}^{19}a_{n}=\sum_{n-1}^{19}(a_{1}+(n-1)d)$
which equals

[b]2[/b]
$19a_{1}+d\sum_{n=1}^{19}(n-1)=19a_{1}+d\frac{(19-1)19}{2}=100$.
Similarly

[b]3[/b]
$\sum_{n=1}^{38}a_{n}$

[b]4[/b]
$=\sum_{n=1}^{38}(a_{1}+(n-1)d)$

[b]5[/b]
$=38a_{1}+d\sum_{n=1}^{38}(n-1)$

[b]6[/b]
$=38a_{1}+d\frac{(38-1)38}{2}$

[b]7[/b]
$=100+210=310$.
Thus we have

[b]8[/b]
$19a_{1}+171d=100$ and $38a_{1}+703d=310$.

[b]9[/b]
Double the first of these and subtract from the second to get
$361d=110$, so $d=\frac{110}{361}$. ??? what did I do wrong...[/hide]

[hide="Shortened Version"]
Sum of first 19 terms:
$19a_{1}+d\frac{(19-1)19}{2}=100$

Sum of all 38 terms:
$38a_{1}+d\frac{(38-1)38}{2}=100+210=310$.

System of equation to find a and d:
$19a_{1}+171d=100$ and $38a_{1}+703d=310$.

Double the first of these and subtract from the second to get
$361d=110$, so $d=\frac{110}{361}$.[/hide]

[quote="jeez123"][hide]so $a=\frac{910}{361}$ and $d=\frac{110}{361}$

If that is correct then the first term should be about [b]2.52[/b] and the 20th term will be about [b]8.31[/b].[/hide][/quote]

jeeze123... how did u do that??

Since cincodemayo said what I got for d is right...

I plug the value of d into the equation:
$38a_{1}+703d=310$
$38a_{1}+703(\frac{110}{361})=310$
$a=\frac{910}{361}$
well, I got a decimal first and then I punched Math > Frac in my calc to get the fraction.

So we got a and d, now the equation is
$a_{n}=a_{1}+(n-1)d$

so just plug in $1$ for $n$ and the value for $a_{1}$ and $d$, same thing with $20$. So there.

P.S. I just edit and shortened my solution on my first reply.

take a look at tOraji... method http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136868


topology
analytic geometry
linear algebra
matrix
function
real analysis
real analysis unsolved

Prove the following theorem:
Let O(n) denote the set of all orthogonal n by n matrices, considered as a subspace of R(N), where N=n^2. Show O(n) is a compact manifold without boundary. What is its dimension?

I don't even know what coordinate patch to use here. Also, what does it mean to say a manifold in R(n) is compact?

The topology is inherited from the standard topology in $\mathbb{R}^{N}$- and a set is compact in that topology if and only if it is closed (in $\mathbb{R}^{N}$) and bounded.

The orthogonal matrices are the zero set of the matrix-valued smooth function $F(X)=X^{T}X-I$. You'll want to apply a theorem in the IFT family- see your other thread.

I see that but I still don't see what it means to speak of open sets and continuously differentiable functions in this space. Are we considering elements to be n^2-tuples of real numbers or is this an actual linear space of matrices (in which case I don't understand the concept of differentiability)? In that case F would be fairly complicated but I don't think computing it explicitly for each component is relevant here.

I'll think about it some more in the meanwhile. Thanks

I think that it's clear from the question that the matrices are indeed regarded as $n^{2}$-tuples of real numbers.


I don't know if this post is in the right thread or not but I'll ask anyway.

This is the problem i need to solve:

Team A is playing Team B.  Team A offense averages 29.5 points against teams that on average give up 25.76 points on defense.  Team B gives up on defense an average of 32.1 points against teams that average 25.48 points on offense.  What is the best estimate of how many points Team A will score???

I have 2 ways to solve this problem but I'm getting different answers.  I always set these up as fractions.  Like 29.5/25.76 and 32.1/25.48   as actual on top and opponents average on bottom.

Method 1.     Find out the differences.  Team A outscores their oppontents averages by (29.5 - 25.76) = 3.74 and then Team B defense the same thing (32.1 - 25.48) = 6.62  Add these two together 3.74 + 6.62 = 10.36  and then add 10.36 to the average of the denominators which is (average of 25.76 and 25.48=25.62) which gives you [b]35.98[/b]

Method 2.  Cross Multiply and take averages.  Team A averages 29.5 against teams that give up 25.76  But Team B doesn't give up 25.76 they give up 32.1 so Team A has 36.76  But then Team B gives up 32.1 against teams that put up 25.48 but Team A doesn't put up 25.48 they put up 29.5 so Team B defense has 37.17  Take the average of the two (36.76 and 37.17) and you get [b]36.96[/b]

35.98 and 36.96 are too far apart for me.  Maybe there is another more accurate way to solve this than what i'm doing.  there shouldn't be two different answers for 1 problem.  I understand it's averages, so maybe a range of numbers are possible.  Please help me out i need the most accurate way to find out the solution whether it's an actual number or a range.

I'm not completely sure about this, but I think the number of points team b scores doesn't matter here.

$ \frac{29.5}{25.76}\equal{}\frac{x}{25.48}$

$ \frac{(29.5)(25.48)}{25.76}\equal{}x$

According to my calculations,

$ x\equal{}\frac{5369}{184}$

which is approximately [b]29 points[/b].

It just seems to take a simple proportion.

Probably not right but, I would just take the average of the number of points Team A would score, and the number of points Team B would give up so:

29.5+32.1=61.6

61.6/2=30.8


linear algebra
matrix
search
linear algebra unsolved

Is any matrix in $ \mathcal{M}_n(\mathbb{R})$ the product of at most $ m$ symmetric real matrices? If yes, give a sharp bound for $ m$.

Answered [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=186254]here[/url]. Proof referenced but not given.

It's a result of Frobenius (1910). Another reference for ${ K=\mathbb{R}}$ or $ K=\mathbb{C}$ is:
 A. J. Bosch (1986). "The factorization of a square matrix into two symmetric matrices". American Mathematical Monthly 93: 462-464.
I have no reference for the other fields.

it has already been discussed, with no complete proofs:
http://www.mathlinks.ro/viewtopic.php?search_id=1386815859&t=42300
http://www.mathlinks.ro/viewtopic.php?search_id=1386815859&t=67449

Actually, it's been already solved on the forum, guys. I perfectly remember (but can no longer find...)  having posted a problem in which it was asked to prove that over any field any matrix is similar to its transpose via a symmetric matrix. This is done using cyclic spaces theorem (or rational canonical form, to make it english...) and then reducing it to cyclic spaces, for which you can find two proofs on the forum, one by Grobber, one by me. Enjoy searching!

Harazi,
Are you implying that for any field $ \mathbb{F}$, any matrix in $ \mathcal{M}_n(\mathbb{F})$ is the product of at most $ 2$ symmetric matrices?

Yes. See the link http://www.mathlinks.ro/viewtopic.php?search_id=691526376&t=73309


hi, i'm new around here,and i thought i could start with  a nice equation :blush:  as the following:
   prove that for any  integer $x$ ,were $|x|>1$ ,there isn't a positive integer $n$  from the:
 \[2003+2^2x+3^3x^2+...+(n+1)^{n+1}x^n+2011x^{n+1}=0\] :)

What does it mean?
This equation can have roots only on $[-1,1]$?  I don't believe it :?

As moderater of this forum I'm gonna have to say, this is too messy for this place. I'll put it in Getting Started.

sorry  |x| can also be 1,but i was interested  in a solution wich  x could take.you see, solving this equation as a diophantic one finely you would get to the following solutions for x:+1,-1,+2003,-2003,+2003/2011,-2003/2011,+1/2011 and -1/2011.
 now notice that  theese aren't  all roots of the equation ,so by    cheking up all of the possible solutions you'll find the root
 in this case n does not exists

Sorry, I missed words "$x$ is an integer"  :blush:


Consider the following fairly general situation: I have some property $ P$ of $ k$-tuples and I have a set $ S$ of sequences.  I wish to generate all the elements $ S$ having no subsequence of length $ k$ with property $ P$.  I generate $ S$ by a depth-first search (i.e., I build each element of $ S$ one letter at a time), and I test whether a sequence has property $ P$ by testing every possible $ k$-subsequence.  At what stages during the process of generating all these permutations is it best to test in order to generate my set as time-efficiently as possible?
(Things it's fair to assume: testing a given $ k$-tuple is constant time, and there is a fixed probability $ 0 < p < 1$ that any given $ k$-tuple will have property $ P$.  Also, the number of elements of $ S$ with no $ k$-subsequence having property $ P$ will be exponentially small compared to the size of $ S$, e.g. exponential versus either an exponential with larger base or, more likely, exponential versus something like factorial.)

Here's a specific example*:  I want to generate the set of permutations of length $ n$ (or the set of words of length $ n$ over some finite alphabet, or the set of alternating permutations of length $ n$, etc.) with no three-term subsequence in which the smallest entry is first and the largest entry is second.  I build my permutations up letter by letter, and each time I add an additional letter I have the option of testing to see whether the prefix I've generated so far has any subsequence of length 3 with this form.  One possibility is that I could wait until I've generated the entire permutation before checking -- in this case, I have to run $ \binom{n}{3}$ tests for each of $ n!$ permutations.  Alternatively, I could test every single time I add a letter.  On one hand, for a permutation without any subsequence of length 3 having the desired form, I end up running $ \sum_{j \equal{} 1}^n \binom{j}{3} \equal{} O(n^4)$ tests, and each additional letter seems not to add too much information, but on the other hand I get to avoid testing a very large number of permutations (e.g., it only takes a single test to exclude all $ (n \minus{} 3)!$ permutations that begin $ 132\ldots$).

* But not one I actually need help with, i.e., this example has structural properties that allow for better approaches than what I've described, but the examples for which I might actually want to implement this don't have those structural properties.

Surely in all cases you only need to test subsequences involving elements you haven't already tested? Ie, if you are testing after adding each element, you can assume that element is the 3rd one in the subsequence (thus only getting $ \sum_{j \equal{} 1}^n \binom{j\minus{}1}{2} \equal{} O(n^3)$), which means that you never do more tests than if you test it at the end).

Whatever you do, you'll need to test each possible subsequence once, so the earlier you do it the better.

Maybe I'm misunderstanding something.

[quote="TripleM"]Maybe I'm misunderstanding something.[/quote]  No, I just missed the observation of your first and second sentences.  :oops:  So, you're absolutely right -- I just have to change the way I've decided to test a sequence.

As an academic exercise, we can still make a nontrivial (I hope) question out of this: suppose that we can't do what you've suggested, i.e., suppose the tester for a given sequence of length $ n$ and tuples of length $ k$ is a black box that runs in time $ O(n^k)$.  Then is "always test" still the best regieme?

I'm sure there is definitely some non-trivial stuff here, as I definitely remember getting a speed improvement when trying to solve a problem using a similar idea of not testing every single time, which is why I triple checked what I wrote before sending that reply. I just can't remember what the circumstances were!


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Prove, that integral
\[ \int_{0}^{\infty}{\frac{\ln x}{x^2\plus{}a^2}dx}\]
converges and calculate it.

[hide]Make x = a tan(t) to get $ I \equal{} \int_0^{\pi /2}\frac{\ln(a \tan t)}{(a \tan t)^2 \plus{} a^2} \cdot a \sec^2t\,dt \equal{} \int_0^{\pi /2} \frac{1}{a} \ln(a \tan t)\,dt$

$ \equal{} \frac{\pi}{2a} \ln a \plus{} \frac{1}{a} \int_0^{\pi /2} \ln(\tan t)\,dt \equal{} \frac{\pi}{2a} \ln a$[/hide]

[quote="Yuriy Solovyov"][color=darkred]Prove that integral $ I\equal{}\int_{0}^{\infty}\frac {\ln x}{x^2 \plus{} a^2}\ \mathrm {dx}$  converges and calculate it.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Using the substitution $ x: \equal{}ax$ obtain $ I\equal{}\frac 1a\cdot\left(J\plus{}\ln a \cdot\int_0^{\infty}\frac {1}{x^2\plus{}1}\ \mathrm {dx}\right)$ , where $ J\equal{}\int_0^{\infty}\frac {\ln x}{x^2\plus{}1}\ \mathrm {dx}$ .

Since $ \int_0^{\infty}\frac {1}{x^2\plus{}1}\ \mathrm {dx}\equal{}\frac {\pi}{2}$ and using the substitution $ x: \equal{}\frac 1x$ obtain $ J\equal{}\minus{}J$ , i.e. $ J\equal{}0$ , means that $ I\equal{}\frac {\pi}{2}\cdot\frac {\ln a}{a}$ .[/color]


You're given 9 empty boxes and the digits from 1 to 9. Each digit may only be used once and only one digit can be used in one box. Find some permutation of the digits such that \[\frac{\square}{\square}\times \square+\frac{\square \times \square \times \square}{\square}+\square \times \square =100.\] Have fun. :D

I actually solved this with some casework. It took like 10 minutes.  :blush:


$ 1.)$ Let $ a,b,c \in (0;1)$ ; . with $ 0 < a \leq b \leq c < 1$ .

Compare the numbers :
 $ A \equal{} \log_ba \plus{} \log_cb \plus{} \log_ac$ ; and $ B \equal{} \log_ab \plus{} \log_bc \plus{} \log_ca$  .

[hide="hint"]Use base change rule and rearrangement inequality[/hide]

I can't see how can I use rearrangement inequality there . Could you post your full solution ?

I [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184474]already answered this question[/url].

You're right. we cannot  use rearrangement inequality.


The sequence of digits

123456789101112131415161718192021...

is obtained by writing the positive intergers in order. If the 10nth digit inthis sequence occurs in the part of the sequence in which the m-digit numbers are placed, define f(n) to be m. For example, f(2) = 2 because the 100th digit enters the sequence in the placement of the two-digit interget 55. Find, with proof, f(1987)

Source of the problem:
Putnam 1987 A2.

[hide="Answer:"]
1984[/hide]

All I found is that any n-digit number $x$ starts at the place $\boxed{n\cdot x-\frac{10^{n}-10}{9}}$


The Ferret Wars is a rpg. To play, you will have 20 skill points, and 4 categories to divide into. Power, Intelligence, Agility, and Five Senses. There as many spots open as possible. If you sign up, pick whether you wish to be a ferret, or a rabbit. Also pick you're skill point category divination. The game officially starts December 9. I will only allow the 5 slots to be filled after that point.

Oops, I forgot to mention the other Guidelines, ferrets and rabbits are fighting in an all-out war. The rabbits have gained an upperhand. Only the ancient Crystal of Klyne can save the ferrets. Those who join the ferrets will either defend Ferretopia, or journey with Todd(I am Todd) to retrieve the crystal. Those who join the rabbits will either seek to stop Todd, or try to conquer Ferretopia. For questions, please contact me via this forum.

I join the ferrets.
Skills:
Power:2
Intelligence:6
Agility:8
Five senses:4
Journey with Todd.

join ferrets, journey with Todd

Power-5
Intelligence-6
Agility-5
Five senses-4

join ferrets

Power-3
Intelligence-11
Agility-3
Five senses-3
 journey with Todd

Join the ferrets

Power-7
Intelligence-2
Agility-8
Five Senses-3

join ferrets

go with Todd

Power-3
Intelligence-2
Agility-8
Five Senses-7

go ferrets!!! :)

Power: 7
Intelligence: 6
Agility: 4
5 Senses: 3

EDIT: and journey with Todd

join rabbits

Power: 3
Intelligence: 10
Agility: 2
5 Senses: 5

Wait!!! If you join either side, you must pick what you wish to do. Read my rules for instructions.
Todd's or my stats are all 5.

Join ferrets:
Power:5
Intelligence:5
Ability:5
Five Senses:5
uh
I don't really get this game..
Ok..jourey with Todd?
EDITED.Happy now?

[quote="k00lperson"]Join ferrets:
Power:8
Intelligence:10
Ability:8
Five Senses:5
uh
I don't really get this game..
Ok..jourey with Todd?[/quote]

Uh, you used 31 points, not 20, which you are supposed to use.

This game is popular. :)

[quote="Yoshi"]Join ferrets.

I'm going to defend!

Power: 1
Intelligence: 6
Agility: 8
Five Senses: 4[/quote]

and Yoshi didn't use one of his points

[quote="15!!!"][quote="Yoshi"]Join ferrets.

I'm going to defend!

Power: 1
Intelligence: 6
Agility: 8
Five Senses: [b]5[/b][/quote]

and Yoshi didn't use one of his points[/quote]

Ok. Fine. Happy?

I'll rabbit attack polecat town since it's easier to type :P

AAATTTAAACCCCKKK, MMMMYYY AAARRRRMM (cough) MMYYYY!!!!!

wait... where'd they go  :o

AAAHHHH!!! RETREAT!!!!

The first attack on polecat town did not go so well. The landscape has several trees, shrubs, and a peculiar looking giant stone. :huh:

That night, a rabbit spy comes to see what is on the giant stone. Could it be a shape? An inscription? Who knows. Maybe something is under it. A scouting party will try to lift it the next morning. :P

Thanks for the offer, but I will not mod this.

[quote=butler]uh. bluecarneal will mod instead.its like with any other rpg, or game. Just dont end it before i get back plz.[/quote]

ok...i guess i'll be modding this then?
Ok..just play on u guys..idk what to do..so JUST KEEP GOING!

Rabbits
Stop Todd and his gang of thugs.
Power:5
Intelligence:5
Ability:5
Five Senses:5

alright.....CONTINTUE PLAYING..DON'T LET THIS DIE..

I've seen that butler is back and active, so *Bump*

*plays*

but seriously, in order to play, you must provide a storyline.

okay okay okay!!! um........

The rabbits are attacking Polecat Town. The rabbits attempt to barrage the town, but the ferrets return fire with stones. The previous settings are the same.

Meanwhile, Todd and his friends have been captured by an evil wolverine, and have been thrown into a dungeon. There are several odd objects. A grinning skull is atop the chair sitting in the middle of the room. Also, a glint of light catches Todd's eye in a corner.

The lone rabbit searches for a way to stop Todd. He comes along a high tower with a menacing catapult at the top.

I will let you all ponder what to do, I will provide some more detail Monday.

*quits* Nothing has happenned for long periods of time... And I don't want to bother checking this unless it is very active...

AT POLECAT:

The scouting party of rabbits sneaks in and looks under the stone, however, the stone flips, squashes some scouts, (ouch) and starts falling down the cliff towards polecat town.

Luckily, we found out what was under the rock.

A box.

A box?! A simple box?!?! Yet, the surviving scouts could not open it. They take the box away back to rabbit base.


function
real analysis
real analysis unsolved

Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function such that  $f(a)=f(b)\Rightarrow f'(a)=f'(b)$. Prove $f$ is monotone (non-decreasing or non-increasing).

How do you solve it in the case where $f$ is not monotone on any open interval ? If we have $f(a)=f(b) \Leftrightarrow f'(a)=f'(b)$ or if $f$ is $C^{1}$ then this is easier but with the initial wording I am stuck on that case.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67252[/url] for a full solution of an equivalent problem.


Dear all:

This is the first time i come here. I am not quite sure how this works .So now i am trying.

I had a bit hard time to do the generating function problems. Here is one of them.

Find the ordinary generating function of the sequence (a0,a1,a2......) where ar is the number of ways of selecting r objects from a set of six distict objects,where each objects can be selected no more than thrice.

Any help would be appreciated!!

Hello Kathy,

This topic should be placed in a math forum of the appropriate difficulty level. I am moving it to Intermediate for now.

how to determine the number of outcomes when three indistinguished dice are rolled from this answer?

Thanks!!

Plug in $x=1$, that will give you the sum of the coefficients.

So the answer is $4^6$


algebra
polynomial
Vieta
system of equations

Let $p,q,r$ be distinct real numbers which satisfy the equations
$q=p(4-p)$
$r=q(4-q)$
$p=r(4-r)$
Find all possible values of $p+q+r.$

does it have another solution beside $(0,0,0)$ and $(3,3,3)$ :?:

[hide]Using Vieta, and after quite long and boring calculations I got $p+q+r = 6, 7, 8$.[/hide]

Let $f(x)=x(4-x)$. Then $p,q,r$ must be roots of the equation $f(f(f(x)))=x$. But
\[x-f(f(f(x)))=x(x-3)(x^3-7x^2+14x-7)(x^3-6x^2+9x-3)\]
Now, $x=0$ and $x=3$ are the fixpoints of $f$, so $p,q,r$ must be roots of the cubics.
But it's easy to check that if $x$ is a root of one of the cubics, then $f(x)$ is too, so either $p,q,r$ are the three roots of the first cubic, in which case $p+q+r=7$ or of the second cubic in which case $p+q+r=6$.

The Vieta way suggested in #4 is in principal easier of course, you only get that wrong value $p+q+r=8$ which you then need to exclude.
[hide=By Vieta]
Let $(z-p)(z-q)(z-r)=z^3-az^2+bz-c$. Then adding the three equations we have $3a=a^2-2b$. Multiplying them (and noting that $p,q,r \ne 0$) we have $16a-4b+c=63$. Finally, letting $pq+pr+qr$ on the LHS, we find $15b=4ab-12c+2ac-b^2$.
Solving the first two equations for $b,c$ and pluggin them into the third one, you get a fourth degree polynomial with roots $a=6,7,8,9$.
Now, $a=9$ is the case $(3,3,3)$ which must be excluded. $a=6$ and $a=7$ are solutions, as in my solution above, but $a=8$ is not a solution which is the nasty thing about this way of solving the problem...
[/hide]


Solve the system of equations.
$ a^2 \minus{} b \equal{} 1 \\b^2 \minus{} 2b \minus{} c \equal{} 2\\c^2 \plus{} 2c \minus{} a \equal{} 1$
I have no idea how to deal with it :(

:o
How should we go about factorizing $ a^{8}\minus{}8a^{6}\plus{}20a^{4}\minus{}16a^{2}\minus{}a\plus{}2$?

I know we can easily see the expression equals $ 0$ when $ a$ equals $ 2$ or $ \minus{}1$, but how about the rest?

Of course try with factorization : $ (a^3\plus{}xa^2\plus{}ya\plus{}z)(a^3\plus{}ta^2\plus{}ua\plus{}v)$, try solving this system... It can be difficult, so it's not advised, but it's worth a try.

[img]http://alt1.mathlinks.ro/latexrender/pictures/7/c/5/7c57b3d7a035c39de0ad1bb22401ca64b6eebd32.gif[/img]

how did u solve it?  :maybe:

$ x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1 \equal{} \left\{\begin{array}{l}
(x^2\plus{}ax\plus{}1)(x^4\plus{}bx^3\plus{}cx^2\plus{}dx\minus{}1) \\
\text{or} \\
(x^2\plus{}ax\minus{}1)(x^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}1) \\
\text{or} \\
(x^3\plus{}ax^2\plus{}bx\plus{}1)(x^3\plus{}cx^2\plus{}dx\minus{}1)
\end{array}\right.$

If $ x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1\equal{}(x^2\plus{}ax\plus{}1)(x^4\plus{}bx^3\plus{}cx^2\plus{}dx\minus{}1)$,
$ a\plus{}b\equal{}1$. Because coefficient of $ x^5$.
So, $ a\equal{}\minus{}1$. Because coefficient of $ x$.
But, $ x^2 \minus{} x \plus{} 1 \not |x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1$.
Therefore incorrect.

If $ x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1\equal{}(x^2\plus{}ax\minus{}1)(x^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}1)$,
$ a\plus{}b\equal{}1$. Because coefficient of $ x^5$.
So, $ a\equal{}1$. Because coefficient of $ x$.
But, $ x^2 \plus{} x \minus{} 1 \not |x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1$.
Therefore incorrect.

If $ x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1\equal{}(x^3\plus{}ax^2\plus{}bx\plus{}1)(x^3\plus{}cx^2\plus{}dx\minus{}1)$,
$ a\plus{}c\equal{}1$. Because coefficient of $ x^5$.
$ \minus{}b\plus{}d\equal{}1$. Because coefficient of $ x$.
So, $ a\plus{}b\equal{}\minus{}3$. Because, coefficient of $ x^3$.
Therefore, coefficient of $ x^4$ is $ a(a\plus{}1)\equal{}0$ and coefficient of $ x^2$ is $ a(a\plus{}3)\equal{}0$.
So a=0.
Hence, $ (a, b, c, d) \equal{} (0, \minus{}3, 1, \minus{}2)$.

Thus, $ \boxed{x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1\equal{}(x^3\minus{}3x\plus{}1)(x^3\plus{}x^2\minus{}2x\minus{}1)}$.

[quote="mathwizarddude"]I think you forgot to include $ dx$ for the coefficient of $ x$ in the first two cases.[/quote]

Sorry.

If $ x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1 \equal{} (x^2 \plus{} ax \plus{} 1)(x^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \minus{} 1)$,
$ a \plus{} b \equal{} 1$. Because coefficient of $ x^5$.
$ \minus{}a \plus{} d \equal{} 1$. Because coefficient of $ x$.
So, coefficient of $ x^4$ is $ \minus{}a^2 \plus{} a \plus{} c \plus{} 1 \equal{} \minus{}5$ and coefficient of $ x^2$ is $ a^2 \plus{} a \plus{} c \minus{} 1 \equal{} 7$.
But, coefficient of $ x^3$ is [b]not[/b] $ \minus{}3$.
Therefore incorrect.

If $ x^6 \plus{} x^5 \minus{} 5x^4 \minus{} 3x^3 \plus{} 7x^2 \plus{} x \minus{} 1 \equal{} (x^2 \plus{} ax \minus{} 1)(x^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} 1)$,
$ a \plus{} b \equal{} 1$. Because coefficient of $ x^5$.
$ a \minus{} d \equal{} 1$. Because coefficient of $ x$.
So, coefficient of $ x^4$ is $ \minus{}a^2 \plus{} a \plus{} c \minus{} 1 \equal{} \minus{}5$ and coefficient of $ x^2$ is $ a^2 \minus{} a \minus{} c \plus{} 1 \equal{} 7$.
But, coefficient of $ x^3$ is [b]not[/b] $ \minus{}3$.
Therefore incorrect.


conics
ellipse
geometry
circumcircle
trigonometry
geometry solved

Consider the elipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Consider three points on the elipse $A,B,C$ such that the tangents through this points to the elipse are paralel to the opposite site of the triangle.
a) What is the maximum of $S_{ABC}$
b) Find the distance from the origin to the Toricelli point of $ABC$

The first question seems very strange, because it looks to me like with the given hypothesis, the area of $ABC$ is always the same. :? Maybe I'm just wrong.

I guess you're not wrong, that was a trick(from the proposer) as far as I understood!

Here's how I got that: the hypothesis says that the center of the ellipse (the origin) is the centroid of the triangle. Now consider the circle having the large axis of the ellipse as its diameter and standing in a plane from which it's projected onto the ellipse. Let $A',B',C'$ be the points in that plane whose projections are $A,B,C$. $O$ is both a circumcenter and a centroid for $A'B'C'$, so $A'B'C'$ is an equilateral triangle inscribed in a circle of constant radius (which is $a$, by the way). This means that its area is always the same, and when we project it on the plane of the ellipse, the area of $ABC$ is equal to the area of $A'B'C'$ times the cosine of the angle between the planes. This means that the area of $ABC$ is always the same.

The second question is also a bit confusing. I wonder: does it mean that that distance is always the same?

Well I guess you have to find it first and then say is it constant or what ;)

grobber, could you please post the proof of the statement that $O$ is the centroid?
there's always the analitical way but I'm sure there's a standard statement that I can't see now  :?
thx in advance

Hmm... We can simply apply stretching wrt to $y$ axis which map ellips to a circle. Then $ABC$ transforms to $A'B'C'$ with the same properties about tangents. But it follows that $A'B'C'$ is equilateral and its centroid is $O$. Applying inverse transformation the centroid $O$ transforms to the centroid of $ABC$, but under our stretching $O$ maps to $O$.

Oh. I forgot to ask what is a Torichelli point in this problem?

Looks like extensive knowledge of triangle geometry is becoming a necessary condition of receiving full marks on the Moldavian NMO. Well, when it will become a sufficient one, I will think about immigrating to Moldavia...

As for this particular problem, part [b]a)[/b] is about the [url=http://mathworld.wolfram.com/SteinerCircumellipse.html]Steiner circumellipse[/url] of a triangle. This is the ellipse passing through the vertices of the triangle and having the centroid of the triangle as its center. Or, equivalently, it is the ellipse passing through the vertices of the triangle and satisfying the property that the tangents at these vertices are parallel to the opposite sides of the triangle.

As for part [b]b)[/b], I am not really sure about what it was supposed to mean, but if it means that the distance from the Toricelli point (@Myth: this is what you call Fermat point, or Fermat-Toricelli point, or first Fermat point to be precise) to the origin (i. e. to the centroid of the triangle) depends only on the ellipse (and not on the concrete position of the points A, B, C on this ellipse, of course provided that the condition about the parallelism of the tangents to the opposite sides is satisfied), then it's wrong. But maybe the question was to find the maximal or the minimal such distance, or something like that...

  darij


A garden centre advertises young plants to be used as hedging. After planting, the growth G metres (i.e. increase in height) after t years is modelled by the differential eqaution:

dG/dt = (25k - G)/25

where k is a constant and G = 0 when t = 0.

(i) On the plant labels it states that the expected growth after 10 years is approximately 1 metre. Is this claim justified given that a plant grows 0.6 metres by the end of 5 years?

(ii) Given that the inital height of the plants was 0.3m, what is the likely long-term height of the plants?

sludgethrower, would you please slow down your rate of posting? This many problems, (which mostly appear to be typical textbook exercises) all at once is crowding the forum and making it difficult for readers to find problems they want to respond to. Wait a couple of days before you post any more.

ok...sorry...I wont post anymore....thanks for telling me...


Still hi guys,

here's the topic of the "[b]Oliforum contest[/b]", the competition for training that started some months ago  :lol: . You can find problem, solutions and ranks in the following links: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=227127]first[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]second round[/url].
i'll sumarize all details about it, be carefully.

[b]1-[/b] The contest is made of 3 rounds: this is the last one.

[b]2-[/b] Each round is made of only 3 problems, and i will give points 0 to 7: 6 for all correct solutions, 7 for [i]only[/i] elegants and correct solutions (similar to ML contest).

[b]3-[/b] It is tought for high-school student, and also for undergraduate or higher student, but you cannot use non-elementary-theory, otherwise the solution is invalid.

[b]4-[/b] All begins from italian-oliforum and the final rank is unique. You can find the partial rank at the end of second round at this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]link[/url].

[b]5-[/b] The round starts on 12:00 of 6 december and it ends on 24:00 of 21 seprember, so saturday and sunday (you have 36 hours in total!).
        [b]NB.[/b] All times (date and hours) are according to [i]ROME MERIDIAN (+2GTM), [/i]please attention!

[b]6-[/b] Enrollment is not necessary, it sufficies to send me solution.

[b]7-[/b] I’ll take in consideration time of arriving solutions [i]only[/i] for the first place [i]if [/i] there are more 21/21 points.

[b]8-[/b] [i]Way of sending solutions[/i].
-          You need to send solutions to both email address: pao_leo88@hotmail.it and leonettipaolo@gmail.com in a [i]unique pdf file[/i] (so, with attachment)  of a right size;
-          The pdf file must be written with [i]Latex [/i](or a way that I can understand);
-          You [i]rename the pdf file [/i]as your nickname (the same of yours in ML);
-          You need to write nothing in the email;
-          Try to be [i]clear[/i] in solutions! I’ll estimate it!

 {If [i](and only if)[/i] you are not able to create a pdf in latex model, you can send me a email in .doc format with solutions, as latex or in a way that i (and all correctors) can understand}.

Later, I'll make a list below this topic where i'll write all names of who sent me solutions in right way.

For all doubts and questions, ask me in mp please, for clearity of all :wink:

I'll put below this topic the list of problems on the morning of saturday 6 december, at 12:00 (+2GTM).
Later, I'll make a list still below this topic where i'll write all names of who sent me solutions in right way.

Thanks for the attentions
And good luck

[i]Paolo Leonetti[/i]
[url=http://olimpiadi.dm.unibo.it/oliForum/viewtopic.php?t=12108]Italian version[/url]

[b]Problem 1[/b]-([i]Anass BenTaleb, Ali Ben Bari High School - Taza,Morocco[/i])
Let $ a,b,c$ positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 3$, show that:
$ \displaystyle a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \ge \frac {a(3 \plus{} bc)^2}{(c \plus{} b)(b^2 \plus{} 3)} \plus{} \frac {b(3 \plus{} ca)^2}{(a \plus{} c)(c^2 \plus{} 3)} \plus{} \frac {c(3 \plus{} ab)^2}{(b \plus{} a)(a^2 \plus{} 3)}$

[b]Problem 2[/b]- ([i]Daniel Kohen, University of Buenos Aires - Buenos Aires,Argentina[/i])
Find all non-negative integers $ x,y,z$ such that $ 5^x \plus{} 7^y \equal{} 2^z$.

[b]Problem 3[/b]-([i]Paolo Leonetti, Università Bocconi - Pescara,Italy[/i])
Let $ 0 < a_1 < a_2 < a_3 < ... < a_{10000} < 20000$ be integers such that $ gcd(a_i,a_j) < a_i, \forall i < j$ ; is $ 500 < a_1$ [i](always)[/i] true ?

Good work.. :wink:

edit: in the last problem i had added "always", altough i think you understand..

Contestants who give me solutions:

Honey_S (partial)
Mehdi Cherif (partial)
Inequality_Master (partial)
fede90 (partial)
dima_ukraine (partial)
Rigel (partial)
[...]

it answered also Akashnil and giove and DDs wife..( i cannot edit the post before..)

as soon as possibile we 'll do ranks and contact all you  :wink:

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

Link to problems:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=244069]Problem1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=244070]Problem1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=244071]Problem1[/url]

I have also added other two that had not be used:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=243776]Fake edition#1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=243778]Fake edition#2[/url]

Sorry for not participating  :(  :(
I just realized that I lost the final round because I had some problems and I had not opened my mailbox for 10 days :(

I posted my solution to the first problem at the inequalities section (the link above)

We are very sorry to realize that our solution of number 2 was wrong..
we've got only some steps, but not the whole solution; we're thinking about a new problem to substitute to it, it will follow more information..

[b]Solution Problem 1[/b]
$ \sum_{cyc}{a^2}\equal{}\sum_{cyc}{a(\frac{a\plus{}c\minus{}c}{a\plus{}b})(a\plus{}b)}\equal{}$ $ \sum_{cyc}{\frac{a(a\plus{}c)(a\plus{}b)}{a\plus{}b}\minus{}ac} \ge \sum_{cyc}{\frac{a(a\plus{}\sqrt{bc})^2}{a\plus{}b}\minus{}ac}\ge$ $ \sum_{cyc}{\frac{a(a\plus{}\frac{2}{\frac{1}{b}\plus{}\frac{1}{c}})^2}{a\plus{}b}\minus{}ac}$ $ \equal{}\sum_{cyc}{\frac{a(ab\plus{}ac\plus{}2bc)^2}{(c\plus{}b)^2(a\plus{}b)}\minus{}ac}\equal{}$ $ \sum_{cyc}{\frac{a(3\plus{}bc)^2}{(c\plus{}b)(b^2\plus{}3)}\minus{}ac}$
The first follows by CS, the second one by GM-HM

[b]Solution Problem 3[/b]
It is enough to show that if we have $ 2n>3^k$ distinct positive integers s.t. noone divides another one, so the smallest element is at least $ 2^k$.
Let $ \upsilon_p(x)$ be the $ p\minus{}$adic valuation of $ x$, so naturally there exist a bijective function from $ \{\frac{a_i}{\upsilon_2(a_i)}\}$ to $ \{2i\minus{}1\}$, where $ 1 \le i \le 10^4$. Consider the numbers s.t. $ \upsilon_p(a_i)\equal{}0, \forall p>3$: if $ \upsilon_2(a_i) \le \upsilon_2(a_j)$ and $ \upsilon_3(a_i) \le \upsilon_3(a_j)$ we have contradiction: it follows that $ \upsilon_2(a_i) \plus{} \upsilon_3(a_i) \ge k$, so the [i]only[/i] number $ a_h$ with $ \frac{a_h}{2^{\upsilon_2(a_h)}}\equal{}1$ has the property that $ \upsilon_2(a_h) \ge k$. Now it is enough to prove that $ h \equal{}1$, but it is quite easy considering the number in the form $ 2^a3^bp$ and using similar tecnique as above.

[b](Some steps on problem 2)[/b]
If $ x$ and $ y$ are $ 0$ then $ z\equal{}1$. If $ x\equal{}0$ and $ y>0$ $ 1\plus{}7^y\equal{}2^z$ If $ y$ is even then $ 2^z\equiv2\bmod{4}$ so $ z\equal{}1$ and we obtain the previous solution. If $ y$ is odd then $ 7^{y}\plus{}1 \equiv8\bmod{16}$ so $ z\equal{}3$ and $ y\equal{}1$. If $ y\equal{}0$ and $ x>0$ Looking $ mod 4$ we find no new solutions.
Now let $ x$,$ y$,$ z$ be positive. We can check some small values of $ z$ and it will not be too difficult to see that $ z>4$
By looking $ mod3$ you easily obtain that $ x$ is even and $ z$ is odd. By looking $ mod4$ we get that $ y$ is odd. Then looking $ mod 16$ we find that $ x\equal{}4a\plus{}2$ for some $ a$ nonnegative integer. We could now the following: $ 5^{4a\plus{}2} \plus{} 7^{2b\plus{}1}\equal{}2^{2c\plus{}1}$, $ 5^{4a\plus{}2}\equal{}2^{2c\plus{}1}\minus{}7^{2c\plus{}1}\plus{} 7^{2c\plus{}1}\minus{}7^{2b\plus{}1}$. The left side is multiple of $ 25$ at least. $ 2^{2c\plus{}1}\minus{}7^{2c\plus{}1}$ is a multiple of $ 5$ so $ 7^{2c\plus{}1}\minus{}7^{2b\plus{}1}\equal{}7^{2b\plus{}1}(49^{c\minus{}b}\minus{}1)$ is also a multiple of $ 5$. We find that $ c\minus{}b$  But then it is easy to see that $ 7^{2c\plus{}1}\minus{}7^{2b\plus{}1}$ is a multiple of $ 25$ so back into $ 5^{4a\plus{}2}\equal{}2^{2c\plus{}1}\minus{}7^{2c\plus{}1}\plus{} 7^{2c\plus{}1}\minus{}7^{2b\plus{}1}$. We obtain that $ 2^{2c\plus{}1}\minus{}7^{2c\plus{}1}$ is a multiple of $ 25$ so by the lifting the exponent lemma we have that $ 2c\plus{}1\equal{}5(2d\plus{}1)$  ($ 2c\plus{}1$ is a multiple of $ 5$). The equation turns into: $ 25^{2a\plus{}1}\plus{}7^{2b\plus{}1}\equal{}32^{2d\plus{}1}$. If $ d\equal{}0$ we find the solution $ a\equal{}0$ $ b\equal{}0$ which is plugging back into the original $ (2,1,5)$.

We have to prove that $ 5^{2(2a\plus{}1)}\plus{}7^{2b\plus{}1}\equal{}2^{5(2d\plus{}1)}$ has no positive solutions. Let $ j$ the number s.t. $ j^2\minus{}2\equal{}0$ and begin to work in $ \mathbb{Z}[j]$. the equation becomes $ (5^{2a\plus{}1}\minus{}j^{5(2d\plus{}1)})(5^{2a\plus{}1}\plus{}j^{5(2d\plus{}1)})\equal{}\minus{}7^{2b\plus{}1}$. We know that $ \mathbb{Z}[j]$ is a euclidean ring, so the unique factorization theorem works (unless the order of primes and adding units).
Lemma. Let $ x \in k$ ring. If $ N(x)$ is a prime integer, so $ x$ is prime in $ K$, (easily by contradiction). Since $ 7\equal{}(3\plus{}j)(3\plus{}j)$ and $ N(3\plus{}j)\equal{}N(3\minus{}j)\equal{}7$, so we can factorize RHS as $ \minus{}(3\plus{}j)^{2b\plus{}1}(3\minus{}j)^{2b\plus{}1}$. Now let $ p\equal{}3\plus{}j$ prime in $ \mathbb{Z}[j]$ and $ a\equal{}5^{2a\plus{}1}\minus{}j^{5(2d\plus{}1)}$. The original equation is $ a \overline{a}\equal{} \minus{} (p \overline{p})^{2b\plus{}1}$. $ a$ cannot be a unit, indeed otherwise $ N(a\overline{a})\equal{}N(a)N(\overline{a})\equal{}1$,absurd.
We can note that $ gcd(p,\overline{p})\equal{}gcd(a,\overline{a})\equal{}1$. So, unless adding units, $ a\equal{}p^{2b\plus{}1}$ or $ a\equal{}(\overline{p})^{2b\plus{}1}$.
So, non radical parts needs to be equal: we obtain: $ 5^{2a\plus{}1}\equal{} \frac{(3\plus{}j)^{2b\plus{}1}\plus{}(3\minus{}j)^{2b\plus{}1}}{2}$. But it is know that $ x\plus{}y|x^{2n\plus{}1}\plus{}y^{2n\plus{}1}$ so it follows that $ 3 | 5^{2a\plus{}1}$, contradiction. (but it is wrong because in $ \mathbb{Z}[j]$ every element has infinite associate others to it!  :(  )

When do we have this next Oliforum Contest? Please tell us.

Someone please reply. Will this contest be held next year too? {That is this September}.

@Agr_94, yes it can be  :wink:

however, for all contestant, on 19/03 i'll post another problem for substituting that one of number theory..
i'll post at the usual hour (12:00 at rome meridian!) and you'll have 12 hours to complete and send me the solution

see you for the last problem
bye  :wink:

[b]Problem.[/b]
Let $ S$ be a set of $ 23 \cdot 26$ distinct positive integers that are not greater than $ 2323$.
Show that there exist $ a_1 < a_2 < a_3 < ... < a_9$ element of $ S$ such that the system:
$ a_1x \plus{} a_2y \plus{} a_3z \equal{} 0$
$ a_4x \plus{} a_5y \plus{} a_6z \equal{} 0$
$ a_7x \plus{} a_8y \plus{} a_9z \equal{} 0$
has a solution in nonzero integers.

please send me a solution until 24 hours from now (that is 12.00 at Rome meridian)  :wink:

[color=red][size=117]Edit: please not that i have changed the bound from $ 23 \cdot 23$ to $ 23 \cdot 26$[/size][/color]


logarithms
function
integration
calculus
calculus computations

$x_{0}=10^{-3}$
$x_{n+1}=x_{n}+x_{n}^{2}(\ln{x_{n}})^{\frac{1}{3}}$
Find equivalent of $x_{n}$ when $n\to oo$
Idea:
easy seen: when $n\to oo$ then $x_{n}\to 0$ and and function $x(n)$ is like "continions", have derivative. therefore $x_{n+1}-x_{n}=x'_{n}$ and
$\frac{x'_{n}}{x_{n}^{2}(\ln{x_{n}})^{\frac{1}{3}}}=1$therefore
$\int_{0}^{n}\frac{x'_{n}}{x_{n}^{2}(\ln{x_{n}})^{\frac{1}{3}}}dn=n$
Therefore
$\int_{0}^{n}\frac{dx_{n}}{x_{n}^{2}(\ln{x_{n}})^{\frac{1}{3}}}=n$
therefore if $x_{n}=y$ then we have
$\int_{x_{0}}^{x_{n}}\frac{dy}{y^{2}(\ln{y})^{\frac{1}{3}}}=n$
where $x_{n}\to 0$
question:
How to find equivalent of this integral?
if $\ln{y}=t$ then we have

$\int_{\ln{x_{0}}}^{\ln{x_{n}}}e^{-t}t^{-\frac{1}{3}}dt=n$
next?
I know it is  simpl for them who know well some theorem but I even don't begin study it  :huh:

[quote="fedja"][quote="Extremal"]
$\int_{x_{0}}^{x_{n}}\frac{dy}{y^{2}(\ln{y})^{\frac{1}{3}}}=n$
where $x_{n}\to 0$
question:
How to find equivalent of this integral?
[/quote]
Note that $\ln y$ is a [b]very[/b] slowly changing function, so you can put it outside the integral sign and say that the integral is equivalent to $\frac 1{(\ln x_{n})^{1/3}}\int_{x_{0}}^{x_{n}}\frac{dy}{y^{2}}$. To justify this trick formally, split $\int_{x_{0}}^{x_{n}}= \int_{x_{0}}^{x_{n}^{\alpha}}+\int_{x_{n}^{\alpha}}^{x_{n}}$ where $\alpha<1$ is very close to $1$ or integrate by parts.[/quote]
Thanks, I think so!


modular arithmetic
number theory unsolved
number theory

A positive integer is called PERU if for some odd k > 1,
it is a sum of the squares of k consecutive positive integers.

For example:
  14 = 1^2 + 2^2 + 3^2, and
415 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 are PERU.

Find the smallest PERU integer which is also an odd square.

This is a bit very easy isn't it?

$25=16+9$ so that one is ok

The lower ones are not:
$9\not=4+1$
$9\not=4$

The number has to be the sum of $k$ squares for an odd $k>1$..

ic
because squares are 1 mod 8, and only the odd squares contribute 1, and k is odd itself, and if k were 3 mod 8 then some even square would add 4 mod 8, [b]so k must be 1 mod 16[/b].  (sorry for the bad explaining)

Please note that we can easily maniupulate sum of consecutive squares with the (n+n)/2 formula.

We "see" a solution 7+...+39=143

17 and 33 are to be researched, 1+...+49 is way bigger than 143 so doesn't need any research. 1+...+7 is smaller than even 39 so k=33 doesn't bring any better solutions.

So all what is left is showing that there are no smaller ones with k=17. (in fact, there are just none with k=17 :))

This part is left up to the reader :P ;)

Hello.

Where is this problem from? or what subject is it related to?
What can I refer to for this problem?

I want a full solution for this problem.
 :help:

Let $S(n.k)$ be the sum of the $2k+1$ consecutive squares from $n^2$ to $(n+2k)^2$, i.e.

\begin{eqnarray*}
S(n,k) &=& \sum_{m=0}^{2k} (n + m)^2 \\
         &=& \sum_{m=0}^{2k} n^2 + 2nm + m^2 \\
         &=& \sum_{m=0}^{2k} n^2 + 2n\sum_{m=1}^{2k}m + \sum_{m=1}^{2k}m^2 \\
         &=& (2k+1)n^2 + 2n \cdot {\textstyle \frac{2k(2k+1)}{2} + \frac{2k(2k+1)(4k+1)}{6}} \\
         &=& (2k+1)[n^2 + 2kn + {\textstyle \frac{k(4k+1)}{3}}] \\
         &=& (2k+1)[(n+k)^2 + {\textstyle \frac{k(k+1)}{3}}]
\end{eqnarray*}

Next assume $S(n,k)=u^2$ is an odd square, i.e. $u$ is odd.  According to the calculations above

$(1) \;\;  (2k+1)[(n+k)^2 + {\textstyle \frac{k(k+1)}{3}}] = u^2$.

Multiplying both side of (1) with 3, we obtain

\begin{eqnarray*}
&& (2k+1)[3(n+k)^2 + k(k+1)] = 3u^2 \\
&\Rightarrow \;\;& (n+k)^2 \equiv u^2 \!\!\!\! \pmod{2} \mbox{ since } k(k+1) \mbox{ is even} \\
&\Rightarrow \;\;& n+k \equiv 1 \!\!\!\! \pmod{2} \mbox{ since } u \mbox{ is odd}
\end{eqnarray*}

Hence $(n+k)^2 \equiv 1 \pmod{8}$, which inserted in (1) yield

$(2k+1)(k^2 + k + 3) \equiv 3 \!\! \pmod{8},$

which implies

$(2) \;\; k \equiv 0,5  \!\! \pmod{8}.$

Therefore $k \geq 5$.  Inserting $k=5$ in (1), the result is

\begin{eqnarray*}
&& 11[(n+5)^2 + 10] = u^2 \\
&\Rightarrow \;\;& (n+5)^2 + 10 = 11v^2 \mbox{ where } u=11v \\
&\Rightarrow \;\;& (n+5)^2 - 1 \equiv 0 \!\!\!\! \pmod{11} \\
&\Rightarrow \;\;& (n+4)(n+6) \equiv 0 \!\!\!\! \pmod{11} \\
&\Rightarrow \;\;& n \equiv 5,7 \!\!\!\! \pmod{11} \\
&\Rightarrow \;\;& n \equiv 16,18 \!\!\!\! \pmod{22} \mbox{ since } n+k = n + 5 \mbox{ is odd, i.e. } n \mbox{ is even.}
\end{eqnarray*}

Using the formula $(n +5)^2 + 10 = 11v^2$, and choosing the two smallest possible values for $n$, we find

$n=16 \;\; \Rightarrow \;\; v^2 = 41$
$n=18 \;\; \Rightarrow \;\; v^2 = 49 = 7^2$

Hence $18^2 + 19^2 + \cdots + 28^2 = S(18,5) = u^2 = (11v)^2 = (11 \cdot 7)^2 = 77^2$ is the smallest odd square when $k=5$.

Finally assume $u < 77$.  Obviously $k>5$.  Futhermore, according to (1)

\begin{eqnarray*}
&& (2k+1)[3(n+k)^2 + k(k+1)] = 3u^2 \\
&\Rightarrow \;\;& (2k+1)[3(n+k)^2 + k(k+1)] \leq 3 \cdot 75^2 \\
&\Rightarrow \;\;& (2k+1)[3(k+1)^2 + k(k+1)] \leq 16875 \mbox{ since } n \geq 1 \\
&\Rightarrow \;\;& (k+1)(2k+1)(4k+3) \leq 16875 \\
&\Rightarrow \;\;& 8k^3 < 16875 \\
&\Rightarrow \;\;& k < 13
\end{eqnarray*}

Hence $5 < k < 13$. Thus $k=8$ by (2), which inserted in (1) result in the equation

\begin{eqnarray*}
&& 17[(n+8)^2 + 24] = u^2 \\
&\Rightarrow \;\;& (n+8)^2 + 24 = 17v^2 \mbox{ where } u=17v \mbox { for an odd integer } v>1 \\
&\Rightarrow \;\;& (n+8)^2 + 24 = 17 \cdot 3^2 \mbox{ since } u = 17v < 77, \mbox { i.e. } v < 5 \\
&\Rightarrow \;\;& n+8 = \sqrt{129} \\
&\Rightarrow \;\;& n \not \in \mathbb{N}
\end{eqnarray*}

This contradition leads us to the following conclusion:

[b]The smallest PERU integer with is also an odd square is $5929 = 77^2$.[/b]


Let $ x_1$, $ x_2$, ..., $ x_{n}$ be real numbers such that $ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}1$ and such that $ x_{i}<1$ for every $ i\in\left\{1,2,...,n\right\}$. Prove that

$ \sum_{1\leq i<j\leq n} \frac{x_ix_j}{\left(1\minus{}x_i\right)\left(1\minus{}x_j\right)} \geq \frac{n}{2\left(n\minus{}1\right)}$.

[i]Remark.[/i] In Vasile Cîrtoaje, [i]Algebraic Inequalities - Old and New Methods[/i], Gil 2006, this problem appears with a much stronger condition (namely, requiring that all $ x_i$ are $ \geq 0$), which makes it a lot easier. You can find my solution for the general case at http://www.cip.ifi.lmu.de/~grinberg/MTProblem1.pdf . But can you find a different one?

  darij

Note that :
$ \frac{(\sum_{1\leq i < j\leq n}  {x_ix_j})^2}{\sum {x_ix_j}\left(1 \minus{} x_i\right)\left(1 \minus{} x_j\right)} \geq \frac {n}{2\left(n \minus{} 1\right)}$.

 :lol:

Have you read the remark? You have used the $ x_i\geq 0$ assumption that, as I said, makes the problem much easier. Now try to solve the general case on 1 line ;)

  darij

@Darij

Woah.....

Ur solution is very long , but your ineq is realy strong :)


1.  Find all real roots of the equation $\sqrt{x^2-p}+2\sqrt{x^2-1}=x$
where $p$ is a real parameter.

What u do here is just squaring, and solve for p in terms of x, right?

Actually, when you get through with the squaring, you'll be able to solve for $x$ in terms of $p$.

[hide]
After squaring and simplifying the equations, u get:
$4x^4+16+p^2-16x^2-8p+4px^2=4x^3-4xp$.
It is much easier to solve for $p$, since its highest degree is 2.
Is it possible to solve for $x$ directly from the above equation?
[/hide]

I didn't get the same thing when I squared out:

[hide]Squaring once, I get
$(x^2 - p) + 4(x^2 - 1) + 4\sqrt{(x^2 - p)(x^2 - 1)} = x^2$ so that $4\sqrt{(x^2 - p)(x^2 - 1)} = (p + 4) - 4x^2$.  Then squaring again,  $16(x^4 - (p + 1)x^2 + p) = 16x^4 - 8(p + 4)x^2 + (p + 4)^2$ and we're left with a quadratic in both $p$ and $x$.[/hide]

pretty easy
[hide="solution"]
if $p<0$, we have
$\sqrt{x^2-p}+2\sqrt{x^2-1}\geq \sqrt{x^2-p}>x$

so in order for the quation to have a solution, we must have $p\geq 0$.  Now rewrite it as:
$2\sqrt{x^2-1}=x-\sqrt{x^2-p}$

squaring yields:
$2x^2+p-4=-2x\sqrt{x^2-p}$
squaring again and solving for $x^2$;

$x^2=\frac{(p-4)^2}{8(2-p)}$
thus in order to have a solution $0\le p <2$, and then the only possible solution is $x=\frac{4-p}{\sqrt{8(2-p)}}$.  We now substitue this value of $x$ in the original equations.  multiplying all terms by $\sqrt{8(2-p)}$ yields:
$|3p-4|+2p=4-p\Rightarrow |3p-4|=-(3p-4)$
this holds iff $3p-4\le 0\Rightarrow p\le \frac{4}{3}$ so the equation has solution when $0\le p\le \frac{4}{3}$ and then $x= \frac{4-p}{\sqrt{8(2-p)}}$
Q.E.D.[/hide]


Hey guys do you know any good website for number theory?

If you have a pdf file of the said topic and you are kind enough, you can send me the resources at gbautista03@yahoo.com. Just put Number Theory from AOPS, since i dont open spams.  Thanks a lot  :blush:

So guys, if you know them, can you please post the links and files here?
Thanks.

Well I now this really cool website which has a load of maths problems which are really hard, but I think you already know it,
It is mathproblems.info plese try it out.
It is really cool :lol:
Does anyone know any more??? :?  :(

http://www.mathproblems.com  :rotfl:  :rotfl:  :rotfl:

[quote="G-UNIT"]www.mathproblems.com  :rotfl:  :rotfl:  :rotfl:[/quote]
G-unit It is mathproblems.info you got it wrong.
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :P


E and F are respective  midpoints of the diagonals AC and  BD of a cyclic quadrilateral  ABCD.Prove  that  if <BEC=<DEC  ,then  <AFD=<CFD.

It was posted [url=http://www.mathlinks.ro/viewtopic.php?p=22938#22938]here[/url].


algebra
polynomial
quadratics
number theory proposed
number theory

Let $P(x,\, y)=2x^{2}-6xy+5y^{2}$. Let us say an integer number $a$ is a value of $P$ if there exist integer numbers $b$, $c$ such that $P(b,\, c)=a$.

a) Find all values of $P$ lying between 1 and 100.

b) Show that if $r$ and $s$ are values of $P$, then so is $rs$.

Question about part b:  Several other quadratic forms have this property; for example, $x^{2}+dy^{2}$ (for squarefree $d$) and $x^{2}+xy+y^{2}$.  With what conditions on $a, b, c$ does the quadratic form $ax^{2}+2bxy+cy^{2}$ have the same property?

b) We have $p(x,y)=(x-2y)^{2}+(x-y)^{2}$. Now note that
1-For all $a,b\in\mathbb{Z}$ there exist $x,y\in\mathbb{Z}$ such that $a=x-2y$ and $b=x-y$.
2-For all $a,b,c,d\in\mathbb{Z}$ we have $(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}$.

We're done.


Is it just me, or have the topics in G & FF gotten a lot more boring?
I used to post in here a lot more, because there were actually interesting topics...
what do you think?

(Btw, I wasn't sure if this should go in here or round table)

Eh, meant to add an "about the same" option to the poll, but w/e, too late now..

i wud vote, but i havent been on for long :D  :D  :


Prove that algebraic number numbers is countable.

Rudin's hint :
There is only finite numbers of equations such that $n+|a_0|+|a_1|+\cdots+|a_n|=N$.

i'm just want to ask: How to prove it using this hint? :blush:

And every such equation has only finite numbers of roots. Then set of  algebraic numbers is countable union of finite sets. Thus it's countable. :)


calculus integration trigonometry function inequalities parameterization algebra

$ \int_{0}^{\pi}\frac {\cos nx}{5 \plus{} 4\cos{x}}dx$

I know that the solution is $ \frac {( \minus{} 1)^n}{3\cdot2^n}$ but don't know how to prove it.

(It's a part of a bigger Fourier analysis problem)

EDIT: thank you fedja, it was a typo.

So far it just diverges as written. Did you mean $ \int_0^{{\color{red}2\pi}}$?

I've edited the statement, thank you for noting the typo.

BTW, my unsuccessful approaches included:

*taking the exp form of cosines and using it to transform the subintegral function in a rational one.
*Chebyshev polynomials.
*Complex integration on various contours.

$ \frac{1}{5\plus{}4\cos x}\equal{}\frac15\cdot\frac{1}{1\plus{}\frac45\cos x}$

$ \equal{}\frac15\sum_{k\equal{}0}^{\infty}(\minus{}1)^k\left(\frac45\right)^k\cos^kx \equal{}\sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^k4^k}{5^{k\plus{}1}}\,\cos^kx.$

Now: can you compute $ \int_0^{\pi}\cos nx\cos^kx\,dx\ ?$

I'm afraid I still need some further assistance on this. Professor Merryfield's geometric-series transformation did simplify the integral- and I see that integrating the series term by term leads to the solution, but

    $ \int_0^{\pi}\cos nx\cos^kx\,dx$

is still something I can't solve-it brought me from integrating a function with one integer parameter to a function with two integer parameters.

My idea was to make it recursive in respect to n (using sum and difference formulas) or bash it with Chebyshev polynomial expansion of $ \cos nx$. But none of it worked.

Go ahead and integrate $ 0$ to $ 2\pi$ and divide by 2 - that's just even/odd symmetry. Write $ \cos nx\equal{}\frac{e^{inx}\plus{}e^{\minus{}inx}}2$ and write
\[ \cos^kx\equal{}\frac1{2^k}\sum_{j\equal{}0}^n\binom{k}{j}e^{i(k\minus{}2j)x}\]
Multiply those together and integrate. Nearly all of the terms integrate to zero; you just have to pay attention to the ones you've got left.

While Kent's approach certainly leads to an answer, it is a bit easier to notice that what we are looking for is (up to a numerical coefficient) the $ n$-th coefficient in the Laurent series for $ \frac {z}{2z^2\plus{}5z\plus{}2}$ on the unit circumference and to use partial fractions. By the way, $ \pi$ should appear as a factor in the answer, so the suggested answer can be right only if there is a factor like $ \frac 1\pi$ in front of the integral.

Thank you, Kent. I figured out that the only "survivors" left after multiplication are elements in expansion which have the power $ n$. So the result is something like:

$ 2\pi C(k,(k\minus{}n)/2)\cdot 2^{\minus{}k}$ if $ k$ and $ n$ have the same parity and k is larger than $ n$. But how can I put it in the original sum so I can get anywhere near a nice result-and more important, how will I evaluate the sum?

@fedja: yes, I omitted the $ \frac{1}{\pi}$ factor. And as a matter of fact, I've seen the complex analysis solution today, but wanted to see if there's a more elementary approach.


Consider a point $P$ inside square $ABCD$ s.t. $\angle PAB= \angle PBA= 15 ^{\circ}$. Show that $\triangle PCD$ is equilateral

$D \in AB$ such that $PD \perp AB$. Then $DB=DA=(\sqrt{3}+1)k$;$PD=(\sqrt{3}-1)k$.
Let $E$ be the intersection of the line $DP$ with $CD$. Then $PE=\sqrt{3}+3$;$EC=ED=\sqrt{3}+3$.
We can see that $\measuredangle PCD=\measuredangle PDC=60^\circ$. Then $PCD$ is equilateral.

Let $Q$ be point inside the square that $\triangle DQC$ is equilateral. If we prove that $QAB=QBA=15$ ($XYZ$ means the measure of angle $XYZ$ measured in degrees) it will mean than $P=Q$ and we will be done.
$PDA=PCB=30$, from $AD=DP=PC=CB$ we get $DAP=PBC=75$, so $QAB=QBA=15$ which ends the proof.


if you know french you can see  http://www.animath.fr  you can find a book only for FE :)

Or you can find following the book
Equational Functions and How to solve them, by Christopher G. Small

Here is something:
http://www.imocompendium.com/index.php?options=mbb|tekstkut&page=0&art=funeqn_mr|f&ttn=Marko%20Radovanovic1|%20Functional%20Equations|N/A&knj=&p=999c3m3llp0[url]

edit: go to imo.org.yu and then training materials

[quote="surshu"]Here is something:
http://www.imocompendium.com/index.php?options=mbb|tekstkut&page=0&art=funeqn_mr|f&ttn=Marko%20Radovanovic1|%20Functional%20Equations|N/A&knj=&p=999c3m3llp0[url][/quote]
Hmm, maryam rashin said that : i want an online book for functional equations. :D Please check link :)
Edited : //http://www.imocompendium.com/index.php?options=mbb|tekstkut&p=999c3m3llp0


MATHCOUNTS LaTeX geometry inradius calculus ratio trigonometry

[u][i][b]
Date/Start Time: Thursday, January 27th @7:30 PM US Eastern Time.

CHAT WILL BE OPEN @7:20

CHAT ROOM NAME IS: Mock MathCounts Jam

MY SCREEN NAME IS: Yzaraf

QUESTIONS ARE MATHCOUNTS RELATED (Obviously)[/b][/i][/u]

At today's MathCounts Jam, the subject of getting together on AIM to do another one of these came up.  I was wondering who is interested.  I also think we should get a couple of different people to moderate and post up questions.  If your interested, post here and tell me when would be a good time, and if you are willing to be a moderator or not.

This will be done with AIM.  To download AIM go to http://aim.com.

Feel free to ask any questions.

My AIM is Yzaraf (Farazy spelled backwards)...So if you need to contact me there, feel free to do so.

Okay the time has been decided: [b]Thursday, January 27th @7:30 PM US Eastern Time[/b].

I still need a couple other moderators...

But here is how it is going to work:

One of the moderators will post a problem.  Work out the problem and send your answer as a [b]PRIVATE INSTANT MESSAGE[/b] to the moderator that gave the question.   Each respondent will score 1 point for a correct answer. Only your first submitted answer will count.

Those three people who have figured out the answer first are going to be given the chance to explain how they got the answer.  They will [b]PRIVATE INSTANT MESSAGE[/b] the moderator who gave the question their solution. Whoever gives the best solution method and explains it best gets an extra point.  If more than one out of those three people gives an equally good answer, they will both be awarded the extra point.  After they give their explanations, a moderator can give the full length explanation if they feel it necessaryPerson with the most points at the end wins.

You won't win anything...And I will post a transcript up here after the Mock Jam for reviewing.

The chat room will [b]NOT [/b]be moderated.  AIM does not allow a feature of the sort, so please be responsible!
If you do or say anything we don't like, we can boot you from the chat

Thanks!
 :P

Most of the people here are generally well behaved.

I most likely will join.

SN = Gupti Aga

[quote="Treething"]Most of the people here are generally well behaved.[/quote]

I remember one math chat where we had to change chatrooms to avoid someone with the SN Spillsdaughter or something like that.

[quote="RC-7th"]I remember one math chat where we had to change chatrooms to avoid someone with the SN Spillsdaughter or something like that.[/quote]

Sorry about that  :blush:   But she isn't on AoPS.

I'll be a moderater.  You know my sn and my number faraz so just call me or something.  Its understandable why he needs moderaters because if someone just gets carried away.  Faraz can only type soo fast.

Lol, it'd be funny if there were more moderators than participants.

You really can't get carried away because only your first answer counts. The mod could just ignore all subsequent answers.

It'd be funny if all the participants decided to only send their answers to one mod.

[quote="joml88"]I'm pretty sure that the participants are only supposed to send their answers to one mod.  Otherwise how will the mods know who got it first?[/quote]

Time Stamping

The mods have a seperate chatroom where they compare the times of every participant. However, that does put people with a slow internet connection at a disadvantage.

Okay, I think there is a misunderstanding...

We don't need any more mods by the way.

The other moderator and myself will take turns posting questions.  YOU TURN YOUR ANSWERS IN TO THE MODERATOR THAT GAVE THE QUESTION.

Next, the chat room will be open @ 7:20 PM

Try and message me sometime before so that I can add you to the chat room.  People who are late are also invited to join us.

[quote="Farazy"]Next, the chat room will be open @ 7:20 PM

Try and message me sometime before so that I can add you to the chat room.  People who are late are also invited to join us.[/quote]

Can't you just tell us the name of the chat room?

I'm in, unless I have a lot of homework that day.  My sn is dancingbanana204

I'll be in, unless I am unavoidably detained.

Aim = burnsoutbright22

[quote="Farazy"][u][i][b]
Date/Start Time: Thursday, January 27th @7:30 PM US Eastern Time.

CHAT WILL BE OPEN @7:20

CHAT ROOM NAME IS: Mock MathCounts Jam

MY SCREEN NAME IS: Yzaraf

QUESTIONS ARE MATHCOUNTS RELATED (Obviously)[/b][/i][/u]

[/quote]

Just wanted to remind everyone of the information.

I'm the other moderator.  My sn is:  BaseballJoe2002.

MOCK MATHCOUNTS JAM IS TODAY @7:30PM EASTERN TIME!

[img]http://img.photobucket.com/albums/v221/Farazy/Question1.gif[/img]

Problem #1

Session Start (Moderator1:Mock MathCounts Jam): Sun Jan 23 20:50:17 2005
Session Close (Mock MathCounts Jam): Sun Jan 23 20:50:56 2005

Session Start (Moderator1:Mock MathCounts Jam): Sun Jan 23 20:57:01 2005
Moderator1: hello
Session Close (Mock MathCounts Jam): Sun Jan 23 20:57:26 2005

Session Start (Moderator1:Mock MathCounts Jam): Thu Jan 27 19:22:54 2005
*** Moderator2 has been invited to the group chat.
*** Moderator2 has joined the chat.
*** Contestant1has been invited to the group chat.
*** Contestant1has joined the chat.
Moderator1: suprised
Moderator1: more people said they were gonna come
Contestant1: ya
*** Contestant2 has been invited to the group chat.
*** Contestant2 has joined the chat.
*** Contestant2 has left the chat.
*** ANON has been invited to the group chat.
*** contestant3has joined the chat.
*** Contestant2 has been invited to the group chat.
*** Contestant2 has joined the chat.
Contestant3: hello
Contestant2: hi
Contestant1: hello
Moderator1: hey
Contestant2: hi
Contestant3: we started?
Contestant2: no
Contestant2: couple minutes
Moderator1: yeah
Contestant3: k. then I'm not late
Moderator1: can anyone wait a couple more minutes or should we just start
Moderator1: becuase there are 3 more people that said they'd come
Contestant1: doesn't matter to me
Contestant3: doesn't matter neither. Did you contact them though?
Contestant3: like left a message or blahblah
Moderator1: yeah
*** ANON has been invited to the group chat.
*** ANON has joined the chat.
Moderator1: alright
ANON: i no no none of dese pepple
Moderator1: i know
Moderator1: it doesnt matter
Moderator1: amit and kevin
Contestant3: can I invite one of members of aops?
Moderator1: sure
Contestant2: yes yes
Moderator1: u can invite anyone
Contestant2: hello
Moderator1: amit and kevin
Contestant2: okay
*** Contestant4has joined the chat.
ANON: talk
Moderator1: no answering the question that we did today in class
Moderator1: because its not really that fair
ANON: wha
Moderator1: you'll know when u see it
Contestant3: who's distributing the problems? you, Farazy?
Moderator1: yeah
Moderator1: i go first
Moderator1: ill explain in two minutes
Contestant1: just so you know
Contestant1: I'm not really supposed to be on IM
Moderator1: lol
Contestant1: I'm at school
Moderator1: ITS A GOOD REASON!
Contestant3: now?
Contestant1: I know
Moderator1: one minute
Contestant3: where you at? PT?
ANON: lol
ANON: aight, faraz i gotta
ANON: go
ANON: ttyl though
Contestant3: how are you at school?
ANON: tell me bout it
Contestant2: dude?!
ANON: later
Contestant2: cmon!
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Contestant3: it's 7 30 night
Contestant2: darn
Contestant1: It's after school
Contestant2: we gonna start?
*** ANON2 has been invited to the group chat.
Moderator1: okay
Contestant1: it's not 7:30 where I am
Contestant3: oooo
Contestant3: it's 7:36 to my clock
Moderator1: well its ET
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Moderator1: its 7:35 to my computer
Contestant3: forgot that AoPS is from all around the universe
Moderator1: and time to start....
Moderator1: DUN DUN DUN!
Moderator1: Okay here are the rules
Moderator1: I will post questions first
Moderator1: and you will work the question out
Moderator1: and PRIVATE INSTANT MESSAGE ME THE ANSWER
Moderator1: If you get the answer correct
Contestant3: cool!!!!
Moderator1: you are awarded one point
Moderator1: well actually
ANON2: do we get a trophy?
Contestant3: what do we do if we have a point?
Moderator1: the first two to answer the question get one point
*** Contestant6 has joined the chat.
Moderator1: then, the two who got it right first will IM me an explanation to how they got the answer
ANON2: if we win nothing, im gonna go gut some fish and earn 10 bucks
*** Contestant4 has joined the chat.
ANON2: cya
Moderator1: who ever gives the best explanation gets another point
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Contestant2: we have to show work?!
Moderator1: if you want the extra point
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*** Contestant6 has joined the chat.
Contestant3: lizz-we're just going over procedures
Contestant3: if you get right, isn't it?
Moderator1: yes
Contestant4: k
Contestant3: go ahead
Moderator1: if you are one of the first TWO to get it right
Moderator1: if I feel that both explanation are equal I will award both people an extra point
Moderator1: DO NOT POST ANSWERS OR SOLUTIONS IN THE CHAT ROOM
Moderator1: only private message me the answers and solutions
Moderator1: this is where you get the questions
Contestant3: what do we do with the points?
Moderator1: certain problems contain latex or diagrams
Contestant3: quick question: after you posted the question and other ppl answered, who post next problem?
Moderator1: so i will send an image link for the diagram
Contestant2: does latex show up in these rooms?
Contestant3: nope
Moderator1: I will post the first 6 problems
Contestant3: ok
Moderator1: then the other moderator
Moderator1: will post five
Moderator1: Moderator2 is the other moderator
Contestant3: I figured
Moderator1: Points are valueless
Contestant1: lol
Moderator1: if you win you win
Contestant3: so shall we start?
Contestant2: YES!
Moderator1: and you have the pride to know you have won
Contestant3: yeah!!1
Moderator1: YES
Contestant4: sure
Contestant2: go
Contestant2: go
Moderator1: IMAGE
Moderator1: (Link: http://img.photobucket.com/albums/v221/Farazy/Question1.gif)http://img.photobucket.com/albums/v221/Farazy/Question1.gif
Contestant3: my internet is so slow
Moderator1: sorry...
Contestant4: err the picture wont work for me
Moderator1: hmm...
Moderator2: can everyone else see the picture?
Contestant6: yeah
Moderator1: i posted it on AOPS
Moderator1: check the forums
Contestant3: I did
Contestant3: I sent the answer. no response though...
Moderator1: i know
*** Contestant4 has left the chat.
Moderator1: just a sec
Moderator1: Contestant6 and Contestant3!
Moderator1: 1 point each
Contestant3: did I get it wrong?
Moderator1: yes
Moderator1: by ten
Contestant2: we had one like this in mu alpha theta
Contestant3: so what's te answer?
Moderator1: please give exlanations
Moderator1: the answer is 705
Moderator1: and kevin
Contestant3: ahh!!!!
Moderator1: WHY DIDNT YOU DO IT!
Contestant2: I was playin a game...
Contestant2: I didn't know we had started
Moderator1: but munch
Contestant2: lol
Contestant1: lol
Moderator1: Contestant3: we can simplify the right side into 11
 Contestant3: 8*(4+7) = 88
 Contestant3: 11 =  inside ()
 Contestant3: and for the left
 Contestant3: from bottom, we get 705/88
 Contestant3: and above is 65/8
 Contestant3: so, using complex fractino rule
Moderator1: Contestant3: (65*88)/(705*88) = 143/141
 Contestant3: and the top, 65/8
Moderator1: Contestant6: The left side simplifies to (8+1/8)*(8+1/88)/(8+1/8)=8+1/88. The right side simplifies to 88, so in total we have 88*88+88*1/88=704+1=705
Moderator1: okay
Moderator1: i think they both are about the same
Contestant3: that's nice question
Moderator1: unless i am mistaken
Contestant2: yeah its a good question
Moderator1: so one point to each of you
Contestant3: practically, they're same
Moderator1: yeah
Contestant3: arrr!!!
Moderator1: for detailed explanation as an image
Contestant3: miss calculations
Moderator1: go here
Moderator1: (Link: http://img.photobucket.com/albums/v221/Farazy/Answer1.gif)http://img.photobucket.com/albums/v221/Farazy/Answer1.gif
Moderator1: okay
Moderator1: get ready for the next one
Contestant3: yes
Contestant2: BRING IT!
Contestant2: now/
Contestant2: ?
Moderator1: What is the sum of all common prime factors
 of all four-digit numbers of the form A00A?
Moderator1: kevin you can do this one
Contestant2: yeah
Contestant3: hm.....
Contestant2: distinct?
Contestant2: or not?
Moderator1: yes kevin
Moderator1: ...
Contestant2: distinct?
Moderator1: YES
Contestant2: so no repeat
Contestant2: oh
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Contestant6: Do Fowi and Moderator1 go to the same school?
Moderator1: only one answer write so far
Moderator1: yeah we do
Moderator1: i cant believe he got it wrong though
Moderator1: we did one just like this
Moderator1: except it was in a different for
Moderator1: m
Moderator1: copa
Moderator1: What is the sum of all common prime factors
  of all four-digit numbers of the form A00A?
Moderator1: IM me the answer
Moderator1: you have 30 seconds
Contestant3: this makes no sense to me
Contestant3: it's a good question.
Moderator1: okay
Moderator1: times up
Contestant3: no!!!!!!
Moderator1: Contestant3: since the number is in form of A00A, the largest factor is 1001
 Contestant3: yet
 Contestant3: we're asked to find the sum of its PRIME factors and 1001 = 13*7*11
 Contestant3: 13+7+11 = 31
Contestant3: okay
Contestant1: nice
Contestant3: do I get 2 pts?
Contestant2: no that shouldn't work
Moderator1: my explanation
Moderator1: yep
*** Your previous message has not been sent.  Reason: Maximum length (232) exceeded.  You can recall the last message typed with Ctrl-Up or Up, depending on settings.
Moderator1: Lets factor what we have so far as much as we can.

 Let's begin by dividing by A, which gives us, 1001.

 1001 divides into 7, 11, and 13...

 So all of our PRIME factors are A, 7, 11, 13

 Since we are looking for all common PRIME
Moderator1: factors, take all of your constant terms and add them.

 So 7 + 11 + 13 = 31.
Contestant2: because it can come in the form a00a
Contestant2: should you would have to add 2 and 3 and 5
Contestant2: and 7
Contestant3: I miss understood questions
Moderator1: you should memorize 1001 as 7*11*13
Contestant2: waut not seven
Contestant2: I know that
Contestant3: question's asking this:
Moderator1: remember
Contestant3: y do we have to memorize it?
Contestant3: gcf(1001,2002,.....9009)
Moderator1: because it comes up alot
Contestant2: come up a lot
Contestant2: yeah
Contestant2: but...
Contestant3: and that's 1001
Contestant2: oh
Contestant3: is there any other way to understand the concept?
Contestant2: yeah
Moderator1: an easy way is 7 ll is UNLUCKY(13)
Moderator1: thats about the only conncept
Contestant3: this was just how to apply A00A form
Contestant2: okay
Moderator1: you can do it with BBB
Moderator1: which would be 40
Contestant2: because 7 11's always get robbed
Contestant2: unlucky 13
Contestant2: haha
Moderator1: moving right along
Contestant3: yes
Contestant2: ye syes
Moderator1: 5
Moderator1: 4
Moderator1: 3
Moderator1: 2
Moderator1: 1
Moderator1: What is the sum of all the elements of this finite geometric sequence
 .01, .02, .04, .08, .16, .32, ... , 20.48 ?
Moderator1: END
Contestant2: nuts
Moderator1: koopakingz0: 40.95
Contestant2: I should've known this
Contestant1: darn
Contestant3: who got it this time?
Contestant3: ah!!!!
Contestant3: hmm
Moderator1: Contestant6: 40.95
Contestant3: I spend too much time on writing on fraction
Contestant2: 2048 is a power of two
Moderator1: shut it
Contestant2: lol
Contestant3: whoaps......
Moderator1: give explanations now
Contestant3: I did it wrong
Contestant3: hm..
Moderator1: alright
Moderator1: koopa
Moderator1: do you have an explanation?
Contestant2: yeah koopa!
Contestant2: koopa power!
koopakingz0: yes? >:o
Contestant3: .01 + 20.48 concept
Moderator1: Contestant6: You use the formula: (a-ar^n)/(1-r) to get (.01-2*20.48)/(-1)=4095
 Contestant6: You use the formula: (a-ar^n)/(1-r) to get (.01-2*20.48)/(-1)=40.95
Moderator1: koopakingz0: i memorized the sum of em.........
Moderator1: although I really dont like either explanation
Moderator1: (especially koopas)
Moderator2: lol
Moderator1: i give efuzzy the point
Contestant6: A good thing to memorize is that the sum of all the previous powers of two is 1 less than the next power of two.
Contestant3: i should've paid attention to the sum formula
Contestant2: yay efuzzums powa!
Moderator1: not even that
Moderator1: tokay
*** Your previous message has not been sent.  Reason: Maximum length (232) exceeded.  You can recall the last message typed with Ctrl-Up or Up, depending on settings.
Moderator1: Let's get rid of all of the decimal places (divide by .01) and see if that makes this problem any easier for you guys.

 Alright then... Let's start adding up numbers.

 Lets start out with 1 and 2.  1+2=3
 Now lets add in 4 1+2+4=7
Moderator1: 8, 1+2+4+8=15

 Notice anything?

 The sum of all of the terms is the next term minus one.

 So 1 through 2048 would be 4096-1.  Now lets multiply the .01 back in to give us 40.95.

Contestant2: yes I remmber
Moderator1: i know u do
Contestant2: that was  good trick there
Moderator1: but they dont
Contestant3: that's nice
Contestant3: did you prove it though?
Contestant2: that annoying girls that's stalking me started bothering me
Moderator1: its a quick trick gets the answer a second faster then efuzzy though
Moderator1: yes it can be proved
Moderator2: efuzzy's method proves it...
Contestant2: try it with any other though
Contestant3: oh ok
Moderator1: put it in the formula
Contestant3: that's not exactly "proving" though
Contestant3: yea
Contestant3: that's the way
Moderator1: it comes up often, and saves you alot
Moderator1: especially on countdown
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Contestant2: yes no lets move obn/
Contestant3: I've never been to MATHCOUNTS before
Moderator2: yeah it is...it generalizes also... 1+ r+ r^2+...+r^n  = (r^(n+1)-1)
Contestant6: I think we scared koopa off...
Contestant2: hahaha
Moderator1: I stole this next question  from SAAB...
Moderator1: very common one
Contestant3: hehe
Contestant3: k
Moderator1: Question 4:

 onsider a quarter of a circle of radius  14.
 Let  r  be the radius of the circle inscribed in this quarter of a circle.
 Find  r.  Leave your answer in simplest radical form.

 Diagram
Moderator1: (Link: http://mathcounts.saab.org/mcgif/p71_1.gif)http://mathcounts.saab.org/mcgif/p71_1.gif
Contestant2: where's r?
Moderator1: R is the radius
Contestant2: oh!
Moderator1: it doesnt need to be marked
Moderator1:  Let  r  be the radius of the circle inscribed in this quarter of a circle.
*** koopakingz0 has joined the chat.
Moderator1: Moderator1: Question 4:

  onsider a quarter of a circle of radius  14.
  Let  r  be the radius of the circle inscribed in this quarter of a circle.
  Find  r.  Leave your answer in simplest radical form.

  Diagram
Moderator1: Moderator1: (Link: http://mathcounts.saab.org/mcgif/p71_1.gif)http://mathcounts.saab.org/mcgif/p71_1.gif
Moderator1: one answer in
Moderator1: im giving another minute
Moderator1: done
Contestant2: p00
Moderator1: efuzzy yet again...
Moderator1: kevin u suck
Contestant2: hahahahaha
Moderator1: they went over this in AOPS class...
Moderator1: Contestant6: you draw the radius of the big circle so that it passes through the center of the small circle. Then you can see that the radius of the big circle is equal to r*sqrt(2)+r (draw small radii to the sides of the 1/4 circle
Moderator1: to make a square), so we have r*sqrt(2)+r=14, r=14/(1+sqrt2)=14sqrt2-14
Contestant2: I'm not focuse I'm talking to other peopl
Moderator1: and efuzzy gets the other point due to defult
Moderator1: by default*
Contestant3: can you post this in LaTeX form?
Contestant3: wait
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Moderator1: here is the answer with latex and diagram
Contestant3: this is pretty hard to see with that "signs"
Contestant3: thanks
Contestant3: you had the quarter circle
Moderator1: (Link: http://mathcounts.saab.org/mcgif/p71.gif)http://mathcounts.saab.org/mcgif/p71.gif

 Lets draw in two lines to help us out. MO and PO.

 PO will equal x

 MO=OP=ON
Moderator1: (Link: http://img.photobucket.com/albums/v221/Farazy/Answer4.gif)http://img.photobucket.com/albums/v221/Farazy/Answer4.gif
Moderator1: thats the rest of it
Contestant3: guys listen to me
Moderator1: ill give u some time to look
Contestant3: we had quarter cirlce
Contestant3: which had a radius of 14
Contestant3: then
Contestant3: wait
Contestant3: sorry
Contestant3: blah. should've got this one
Contestant3: mybad, I know what I did wrong
Moderator1: is everyone good?
Contestant3: I was trying to make inradius of it
Contestant3: yea, I think so
Contestant3: how long we do this?
Contestant3: yup....
Moderator1: last question from me
Moderator1: kevin u can answer
Contestant1: yep
Moderator1: but u will get no point
Moderator1: because we did this one today
Moderator1: (Link: http://img.photobucket.com/albums/v221/Farazy/Question6.gif)http://img.photobucket.com/albums/v221/Farazy/Question6.gif
Contestant2: do I not answer?
Moderator1: well answer to me
Moderator1: but u wont get a point
Contestant1: I did this...a year ago..
Contestant2: meh
Moderator1: IF its right
Contestant1: I don't remember
Contestant2: I already know this too
Moderator1: but if no one gets it
Moderator1: you can explain
Moderator1: and give the answer
Contestant2: okaty
Moderator2: do they have infinite fractions on mathcounts?
Contestant3: yeah
Contestant3: they do...
Contestant6: (1+-\sqrt{5})/2
Contestant6: omg, that was stupid
Moderator2: wow...i didn't think they had those on mathcounts...cool!
Contestant6: my bad
Contestant3: I don't think they do
Contestant3: do they?
Moderator1: yes
Contestant3: Gupti Aga - Treething told me
Moderator1: ive seen it on MAO
Moderator1: and it could be on mathcounts
Moderator1: this is not a difficult problem
Moderator1: just requires logic
Moderator1: and Efuzzy
Moderator1: for the explanation
Contestant1: lol
Contestant3: I should have practiced infinite fraction
Contestant3: :-/
Moderator2: actually to be completely rigourous it requires calculus ;-)
Moderator1: Contestant6: oh, yeah: x=1+1/x, x^2+x+1=0, and use quad formula
Moderator1: again by default
Moderator1: to efuzzy
Contestant6: No, wait, oh, yeah: x=1+1/x, x^2-x-1=0, and use quad formula
Moderator1: (Link: http://img.photobucket.com/albums/v221/Farazy/Answer6.gif)http://img.photobucket.com/albums/v221/Farazy/Answer6.gif
Moderator2: the golden ratio strikes again
Moderator1: this is a really good one
Contestant2: this one again...
Contestant3: ah i see
Moderator1: the idea of subsituting infinate values has come up before
Contestant3: wasn't similar problem like this on PPPC?
Contestant2: can anyone please explain the history of the golden ratio?
Moderator1: the same principle can be applied to this problem
Moderator2: the history....that could take a while...lol
Moderator1: (Link: http://img.photobucket.com/albums/v221/Farazy/Question6.gif)http://img.photobucket.com/albums/v221/Farazy/Question6.gif
Contestant3: (Link: http://mathworld.wolfram.com/GoldenRatio.html)http://mathworld.wolfram.com/GoldenRatio.html
Moderator1: i meant (Link: http://img.photobucket.com/albums/v221/Farazy/Question7.gif)http://img.photobucket.com/albums/v221/Farazy/Question7.gif
Moderator1: the same idea different numbers
Moderator1: actually
Moderator1: now that I look
Moderator1: Efuzzy got the first one wrong.
Moderator1: (1+-\sqrt{5})/2
Moderator1: is incorrect
Contestant6: I now, I just realized that too.
Contestant2: i got it right
Moderator1: (1+-\sqrt{5})/2 it should only be plus
Contestant6: I did the exact same thing with this problem.
Moderator1: :-)
Moderator2: lol
Moderator1: the second one didnt count
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Contestant2: p00
Contestant2: no fair
Moderator1: just an example of how the method can be applied
Moderator1: ONE MORE
Moderator1: and I swear this is it
Contestant3: so what was the answer to #7?
Contestant2: 3
Moderator1: i can explain it in depth in a bit
Contestant3: please
Contestant2: lemme do it!
Contestant3: I'm not familiar with this ratio so
Moderator1: fine
*** koopakingz0 has left the chat.
Moderator1: kevin go on ahead
Contestant2: this actually has nothing to do with the golden ratio
Moderator2: yeah, it really doesn't
Contestant2: in stead begin where the equation begins repeating
Contestant2: at the second radical substitute that for x
Contestant3: so x = \sqrt {6 + \sqrt x}?
Moderator1: no
Contestant2: yeah
Moderator2: not \sqrt x
Contestant2: oh
Contestant2: no
Contestant2: yeah
Moderator2: just x = \sqrt{6+x}
Contestant2: 6+x all under
Contestant3: oh
Moderator1: and solve for x
Contestant2: yeah
Contestant3: i see
Moderator1: you factor and get 3, -2
Contestant3: but -2 is extraneous solution
Moderator1: and since it has to be positive the answer is 3
Contestant3: so 3
Moderator1: exactly
Moderator1: okay
Contestant3: I see
Moderator1: last one
Moderator1: then bballjoe
Contestant2: how many joe gonna give us?
Contestant3: I think he said about 4
Moderator1: if I subtract x from all the letters from A-Z, and multiply what is the solution?
Moderator2: 5
Moderator2: *i m giving 5 problems
Contestant3: what do you mean by sbutracting x from all the letters?
Moderator1: example (A-X)(B-X)(C-X)(D-X)...
Contestant3: oh
Contestant6: you want us to expand that?
Moderator1: i want the answer
Moderator1: okay
Moderator1: Fowi and Effuzzy win
Moderator1: and jwec does too
Moderator1: but he wasnt on time...
Moderator1: the answer is 0
Contestant3: you might wanted to clarify that more though
Contestant3: it looked like x was a number
Moderator1: does it matter?
Contestant3: it should've been like X
Contestant3: so A-Z was capital
Moderator1: true..
Contestant2: dude
Contestant3: if you want to be exact, X \neq x
Contestant2: yeah
Moderator1: We just need to look at this part: ...(W-X)(X-X)(Y-X)..., and since (X-X) is 0, the answer is 0.
Contestant3: I saw this later though
Contestant3: yea
Contestant2: when jones dide it he wrote it correct faraz
Moderator1: i know..
Moderator1: that was just a typo
Moderator1: Moderator1: example (A-X)(B-X)(C-X)(D-X)..
Moderator1: there is shows all caps
Contestant2: ohhh
Contestant3: k
Contestant2: yes yes
Moderator1: extra point to fowi, and fuzz fuzz
Contestant2: fuzz fuzz?
Contestant3: now, it's joe's turn
Moderator1: efuzzy
Contestant2: thats what you call him
Contestant2: call him fuzzmeister
Moderator1: lol
Contestant2: the fuzz
Moderator2: yep...everyone ready?
Moderator1: kevin hush up
Contestant2: k
Moderator1: and listen
Contestant2: k
Moderator2: The average of x and y is 30.  The average of 2y and 10 is z.  What is the average of x and z?
Moderator1: DONT IM ME
Moderator1: sorry
Moderator1: IM baseballjoe
Moderator1: efuzzy was first by the way
Contestant6: but wrong.
Moderator2: i have one correct so far
Moderator1: yeah
Moderator1: close though
Moderator1: really close
Moderator2: about 30 more seconds
Contestant3: This one was classic :-)
Moderator1: how is it classic?
Moderator1: its easy
Moderator2: ok, times up
Moderator1: not really classic
Moderator2: jwec gets the point and extra point
Contestant3: classic means this apears a lot
Moderator1: oh
Moderator1: lol
Contestant3: not necessarily hard
Moderator2: (Link: http://img.photobucket.com/albums/v213/joml88/solutionone.jpg)http://img.photobucket.com/albums/v213/joml88/solutionone.jpg
Moderator2: faraz are u keeping score still or should i?
Moderator1: ill give u what I got
Moderator2: ok...get ready for the next problem
Contestant3: nice
Moderator2: another 'classic'
Moderator2: What is the sum of the positive divisors of 60?
Moderator2: efuzzy and Contestant2 get it
Moderator2: answer: 168
Contestant1: yay i got it right
Contestant3: I blew this easy one :-P. Forgot 6 and 10
Moderator2: they can send solutions now for the extra point
Moderator1: you shouldnt have even needed to remember 6  and 10
Contestant2: yeahj
Contestant3: oh well
Moderator2: ok...efuzzy sent a nice solution
Moderator2: We can factorize as 2^2*3*5, we can factoring the sum of the positive divisors as such: (2^2+2^1+1)(3^1+1)(5^1+1), [try expanding it and seeing what happens]. So the answer is 7*4*6=168.
Moderator2: (thats what efuzzy said)
Moderator2: (Link: http://img.photobucket.com/albums/v213/joml88/solutiontwo.jpg)http://img.photobucket.com/albums/v213/joml88/solutiontwo.jpg
Contestant3: oh that is a nice trick
Contestant3: I didn't know that
Contestant3: interesting.
Contestant6: Look over the mathcounts toolbox
Moderator2: next problem
Moderator2: A 3-mile long train is traveling at a speed of 60 miles per hour.  If Dopey Dave fires a gun from the very end of the train straight towards the front, how long will it take for the bullet to reach the front of the train if the
Moderator2: bullet is traveling at a speed of 90 miles per hour? (Please give answer in minutes)
Contestant3: hey, I'm tired and I have trig test tomorrow
Contestant3: nice math jam and keep it up
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Contestant6: Same here, I have a project to do. I need to go now. Thanks, and have fun.
*** Contestant6 has left the chat.
Moderator2: ok answers are in
Moderator2: Contestant2 and dancingbanana  get it
Moderator1: 6 min?
Moderator2: send in solutions now if you want an extra point
Moderator2: yes 6 min
Moderator1: btw what do u use to change ur answers to gif?
Moderator1: errr i mean jpg
Moderator2: uh....i took them from adobe and put them in paint
Moderator2: then saved them
Moderator1: oooh
Session Close (Mock MathCounts Jam): Thu Jan 27 20:42:43 2005

Session Start (Moderator1:Mock MathCounts Jam): Thu Jan 27 20:43:09 2005
Moderator2: i ll give extra point to fowi
Moderator2: Contestant2: find the lcm of the twoContestant2: because it will take the bullet 2 minutes to just travel 3 miles and take 3 minutes for the train to go 3 miles so then you find the lcm
Moderator2: Contestant2: of 2 and 3
Moderator2: i did it another way
Contestant1: me too
Moderator2: (Link: http://img.photobucket.com/albums/v213/joml88/solutionthree.jpg)http://img.photobucket.com/albums/v213/joml88/solutionthree.jpg
Contestant2: I see how it is the chinese person always does it the stupid way
Contestant2: oh well I always do it the stupid way anyways
Moderator2: i dunno about taht...
Moderator1: urs does u good reasoning
Moderator2: anyway, next problm
Moderator2: How many 1- or 2-digit perfect squares are there in base b?  (Give your answer as a base 10 number in terms of b)
Contestant2: nuts I suck at bases
Moderator1: hmm i think i know how to do this
Contestant1: I'm leaving...I have to go home and do hw
Moderator1: kk
Moderator2: k
*** Contestant1has left the chat.
Moderator2: ok...well i guess i ll post the answer then
Moderator2: (Link: http://img.photobucket.com/albums/v213/joml88/solutionfour.jpg)http://img.photobucket.com/albums/v213/joml88/solutionfour.jpg
Moderator2: not many ppl are left...
Moderator2: last problem:
Moderator2: In triangle ABC, angle A = 45, angle B = 60 and angle C = 75.  If the area of the ABC equals 27 + 9 * sqrt(3), find the altitude to side AB.
Moderator1: could u expalin 4 for me
Moderator1: its still a bit confusing
Moderator2: well...let's take some examples....lets look at base 10
Moderator2: there are 9  one or two digit squares..1^2 through 9^2
Moderator2: in base five there are 4
Moderator2: 1^2 through 4^2
Contestant2: g2g bye bye
Contestant2: keep it up and I'll come
*** Contestant2 has left the chat.
Moderator2: 5^2 and above do not work because 5^2 = 100 (base 5)
Moderator1: so in b there are b-1...
Moderator1: ahh i see
Moderator1: thank you
Moderator2: no problem
Moderator2: this was pretty cool
Moderator2: see ya some other time, bye
*** Moderator2 has left the chat.
Session Close (Mock MathCounts Jam): Thu Jan 27 20:55:56 2005

SESSION TRANSCRIPT....

Look over those.  I'll post em on my website someday...


MATHCOUNTS
geometry
3D geometry
sphere
ratio

A right circular cone whose base has a radius of 4 cm and whose  height is greater than 2 cm is inscribed in a sphere with a radius of  5 cm, as shown. What is the ratio of the volume of the cone to the volume of the sphere? Express your answer as a common fraction.

You already posted and got a response [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308178]here[/url].  Locked.


calculus
integration
geometry
trapezoid
analytic geometry
real analysis
real analysis unsolved

consider $ f''(t)<0$ for all $ t\in[a,b].$ let $ T_n$ and $ M_n$ be the trapezoidal rule and the midpoint rule, prove that $ T_n\le\int_a^b f\le M_n.$

Consider one subinterval; call it $ [c,d]$. Let $ \ell$ be the secant line connecting $ (c,f(c))$ and $ (d,f(d))$. Let $ \ell'$ be the tangent to the graph of $ f$ at $ \left(\frac{c\plus{}d}{2},f(\frac{c\plus{}d}{2})\right)$. The trapezoid rule estimates the integral by the area between $ x\equal{}c$ and $ x\equal{}d$ under $ \ell$. The midpoint rule estimates the integral by the area between $ x\equal{}c$ and $ x\equal{}d$ under $ \ell'$.
By concavity, the graph of $ f$ lies below its tangents and above its secants. Sum up all of the intervals this way, and we have $ \int_a^b f\ge T_n$ and $ \int_a^b f\le M_n$.

This really deserves a picture.

There's one detail in there that should be explained a bit more. Often, you'll see the midpoint rule explained with flat bars intersecting the graph at the midpoint of each subinterval. That's how we calculate it, but it doesn't really show why the rule is as accurate as it is. Since the $ x$-coordinate is the midpoint, the area under [i]any[/i] segment through that point would be the same. We can choose one that shows better accuracy, and the tangent line is particularly good.


limit
trigonometry
function
real analysis
real analysis unsolved

[quote="Slizzel"]Evaluate $\lim_{x\to 0}{\cos\left({\frac{1}{\arctan{x}}}\right)}$ Hope it's 0 :?[/quote]

No, it doesn't exist: $\frac{1}{\arctan{x}} \to \infty$ when $x \to 0$.

Okay. I took a wrong path then. Show that $u: \mathbb{R}\to\mathbb{R}, u(x)=\{\begin{array}{cc} \cos({\frac{1}{\arctan{x}}}) \ ,&x\neq 0 \\ 0 \ & ,x=0\end{array}$ has primitives..which means that u has Darboux propriety(and it's not a continous function as i was trying to show)  :|