calculus
inequalities unsolved
inequalities

I think it's a very very weak ineq  :huh:
Let $ a \equal{} \frac {1}{x}$,...=> $ x \plus{} y \plus{} z \equal{} 1$
We'll have to prove: $ \sum\ \sqrt {\frac {x^2y^3}{x^2 \plus{} y^3}} \plus{} \sum\ \sqrt {\frac {x^3y^2}{x^3 \plus{} y^2}} \leq\ 1$
Using Am-Gm and let $ x \equal{} a^2,...$ => $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
We only need to prove: $ \sum\ \sqrt {a^2b^3} \plus{} \sum\ \sqrt {a^3b^2} \leq\ \sqrt {2}$ obviously trues because
$ \sum\ a^2b^2(a \plus{} b) \leq\ ...$

[quote="nguoivn"]
Using Am-Gm and let $ x \equal{} a^2,...$ => $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
We only need to prove: $ \sum\ \sqrt {a^2b^3} \plus{} \sum\ \sqrt {a^3b^2} \leq\ \sqrt {2}$ obviously trues because
Sorry，I don't know how to use AM-GM here...

[quote="Jianxing113725"][quote="nguoivn"]
Using Am-Gm and let $ x = a^2,...$ => $ a^2 + b^2 + c^2 = 1$
We only need to prove: $ \sum\ \sqrt {a^2b^3} + \sum\ \sqrt {a^3b^2} \leq\ \sqrt {2}$ obviously trues because.[/quote]
Sorry，I don't know how to use AM-GM here...[/quote]
I think He means to say that it is required to prove:
$ \sum\ \sqrt {\frac {x^2y^3}{x^2 + y^3}} + \sum\ \sqrt {\frac {x^3y^2}{x^3 + y^2}} \leq\ 1$.
Let $ x=p^2, y=q^2, z=r^2\implies p^2+q^2+r^2=1$.
From AGM,
 $ \sum\ \sqrt {\frac {x^2y^3}{x^2 + y^3}} + \sum\ \sqrt {\frac {x^3y^2}{x^3 + y^2}} \leq \sum \sqrt{\frac{x^2y^3}{2xy^{3/2}}}+\sum\sqrt{\frac{x^3y^2}{2yx^{3/2}}}$
$ =\sum\frac{p^2q^3}{\sqrt{2p^2q^3}}+\sum\frac{p^3q^2}{\sqrt{2p^3q^2}}$
Hence the RTP ineq ${{ \implies \sum\sqrt{p^2q^3}+\sum\sqrt{p^3q^2}}\le \sqrt{2}}$........

I hope I did not make any typo! :blush:

[quote="nguoivn"]I think it's a very very weak ineq  :huh:
Using Am-Gm and let $ x \equal{} a^2,...$ => $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
We only need to prove: $ \sum\ \sqrt {a^2b^3} \plus{} \sum\ \sqrt {a^3b^2} \leq\ \sqrt {2}$ obviously trues because
$ \sum\ a^2b^2(a \plus{} b) \leq\ ...$[/quote]

Try $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt{3}}$,then we have
\[ \sum{\sqrt{a^2b^3}}\plus{}\sum{\sqrt{a^3b^2}}\equal{}6\cdot \sqrt{\frac{1}{9\sqrt{3}}}\equal{}1.5196>\sqrt{2}\]
 :!:  :!:  :!:

[quote="Tourish"][quote="nguoivn"]I think it's a very very weak ineq  :huh:
Using Am-Gm and let $ x \equal{} a^2,...$ => $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
We only need to prove: $ \sum\ \sqrt {a^2b^3} \plus{} \sum\ \sqrt {a^3b^2} \leq\ \sqrt {2}$ obviously trues because
$ \sum\ a^2b^2(a \plus{} b) \leq\ ...$[/quote]

Try $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt {3}}$,then we have
\[ \sum{\sqrt {a^2b^3}} \plus{} \sum{\sqrt {a^3b^2}} \equal{} 6\cdot \sqrt {\frac {1}{9\sqrt {3}}} \equal{} 1.5196 > \sqrt {2}\]
:!:  :!:  :!:[/quote]
Thank you, Tourish. I realized my mistake. I used AM-GM wrong because $ x^2$#$ y^3$ when $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$
Here is my new proof:
Using AM-GM: $ 3.\frac {x^2}{3} \plus{} y^3 \ge 4.\sqrt [4]{\frac {x^6y^3}{27}}$
So, the proof'll completer if we can show that:
$ \sqrt [8]{x^2y^9} \plus{} \sqrt [8]{y^2z^9} \plus{} \sqrt [8]{z^2x^9} \plus{} \sqrt [8]{x^9y^2} \plus{} \sqrt [8]{y^9z^2} \plus{} \sqrt [8]{z^9x^2} \le \frac {2}{\sqrt [8]{27}}$
We can prove the harder:
$ \sqrt [8]{x^2y^9} \plus{} \sqrt [8]{y^2z^9} \plus{} \sqrt [8]{z^2x^9} \le \frac {1}{\sqrt [8]{27}}$
With $ x \plus{} y \plus{} z \equal{} 1$. This is a homogeous ineq, so we can assume that $ x \plus{} y \plus{} z \equal{} 3$ (for the calculus become simpler).
Now, we need to prove: $ \sqrt [8]{x^2y^9} \plus{} \sqrt [8]{y^2z^9} \plus{} \sqrt [8]{z^2x^9} \le 3$
Using AM-GM for 8 numbers:
$ \sqrt [8]{x^2y^9} \equal{} \sqrt [8]{xy.xy.y^2.y.y.y.y.y} \le \frac {2xy \plus{} y^2 \plus{} 5y}{8}$
.....
Adds 3 ineqs above, we have done  :)

I've been searching for a better proof to this problem for so long. :D


Write a recursive formula for this scenario:

On October 1, 2002, Sal invested 24,000 in a bank account earning 3.4 percent interest anually, compounded monthly. A month later, he withdrew 100 dollars and continued to withdraw 100 dollars on the first of every month thereafter.

Use the regular interest equation and apply the new info.

I=p( 1+r)^t

I -100(12t)=0

24,000(1.034)^t-1200t=0?

We are trying to find how long until he runs out of money, right?

Question building off of this:

What monthly withdrawal amount would maintain a constant balance of 24,000 dollars in the long run? How would you do this?

So you want 24.000 at the end of each year

branching off my original equation, x is now the principle withdrawal and t is now 1 year, I suppose for each year:

24,000(1.034)-12x= 24,000.
24,000(1.034)-24,000=12x
24,000(1.034-1)/12=x
2,000(0.034)=x
x=68


Sequence
recurrence relation
algebra
IMO Shortlist

Let $ a_{0} \equal{} 1994$ and $ a_{n \plus{} 1} \equal{} \frac {a_{n}^{2}}{a_{n} \plus{} 1}$ for each nonnegative integer $ n$. Prove that $ 1994 \minus{} n$ is the greatest integer less than or equal to $ a_{n}$, $ 0 \leq n \leq 998$

$ a_{n \plus{} 1} \equal{} a_{n} \minus{} 1 \plus{} \frac {1}{a_{n} \plus{} 1}$.

We see, then, that

$ a_{k} \equal{} 1994 \minus{} k \plus{} \frac {1}{a_{k \minus{} 1} \plus{} 1} \plus{} \frac {1}{a_{k \minus{} 2} \plus{} 1} \plus{} ... \plus{} \frac {1}{1994 \plus{} 1}$

It remains to establish an upper bound for the "fractional part."  It is plain from our expression that $ a_{k} > 1994 \minus{} k$; hence,

$ \frac {1}{a_{k \minus{} 1} \plus{} 1} \plus{} \frac {1}{a_{k \minus{} 2} \plus{} 1} \plus{} ... \plus{} \frac {1}{1994 \plus{} 1} < \frac {1}{1995 \minus{} k} \plus{} \frac {1}{1995 \minus{} (k \minus{} 1)} \plus{} ... \plus{} \frac {1}{1995}$

Since these "fractional parts" are strictly increasing, it suffices to evaluate it for $ k \equal{} 998$.  In this case, our upper bound is

$ \frac {1}{997} \plus{} \frac {1}{998} \plus{} ... \plus{} \frac {1}{1995}$

We want to show that this is strictly less than $ 1$.  There are probably several ways to do this, but I'm going to use the fact that

$ \lim_{n \to \infty}\left( \frac {1}{n \plus{} 1} \plus{} ... \plus{} \frac {1}{2n}\right) \equal{} \ln 2 < 1$

[quote]I'm going to use the fact that

$\lim_{n \to \infty}\left( \frac{1}{n+1}+...+\frac{1}{2n}\right) = \ln 2 < 1$[/quote]

Sorry to nit-pick, but this fact isn't the important one: what's relevant is that the sum increases monotonically and doesn't get bigger than one.  Its limiting behavior is irrelevant.  (I know you know this, I just thought it should be said.)

Anyway, the claim that the expression is $< 1$ is a trivial one (just note that $\frac{1}{n+k}< \frac{1}{n}$ and you are done.)


What do you think is the best way of teaching? Would it be of a Socratic way where you keep on pounding the one who is learning with questions? Or is it other

It's the AoPS way, of course! Try problems yourself first, because even if have some vague idea, it will still help you understand the concept much better if you solve it yourself.

I think that the best way to learn math is to read as I find that I learn much faster and have a much stronger understanding of of math topics when I teach myself from books and personal research versus taking notes in class (in fact, the only notes I take now are to record which subjects I want to research). Of course, once you learn the theory, you must also learn how to apply and this website is a great resource for that  :D

[quote="ashpotter"]What do you think is the best way of teaching? [/quote]
I see no specification of the word "math" here :)

I think the first two things to take into consideration is that 1) different teaching styles are better suited to different subjects, and 2) different teaching styles are better suited to different [i]students.[/i]  So the question as stated is very general and it's hard to say much about it.  If you could narrow down your question, we could discuss something more specific :)

ok then, what do you think is the best way to teach a foreign launguage?

Language is an interesting special case.  You've taken 1) into consideration, but what about 2)?

Most psychologists agree there is a [i]critical period[/i] in a child's development where she is most receptive to language; beyond that point, acquisition becomes much harder (though by no means impossible), so the answer to this question (although I don't know it) is certainly very different for a child than for an adult.  It seems to me like immersion is a good strategy in both cases, though (for example, I grew up as a toddler in Singapore, but my Singaporean accent was completely removed by watching hours of English-language television here in America).

Taking into account t0r's (2),

I have to give a presentation on my research-type work at school for the rest of my independent study group.  Unfortunately, I'll have to take it down several levels, because most people there are not very keen on math (though some are).  The teacher recommended that I use visuals: not just chalkboard visuals, but projector or computer simulation visuals.

I guess a lot of people there need to learn visually, whereas people on AoPS are usually also very comfortable with abstract concepts and equations.  On AoPS, people have also learned how to switch quickly between the two (in math, of course).  I guess this kind of ability to switch back and forth is what I've sort of developed over time, so I need to teach in a way which addresses both kinds of learning styles.


Does there exist a Fibonacci number whose decimal representation ends in 999999999? (in nine nines)

well locking mod 10^n the F sequence we see it is periodic, then ...1,1,2..... appeaes in somewhere in the sequence, but before ....1,1,2... there is the subsequence.....-1,1,0,1,1,2...... so result ius obvius now

how do you "see" it is periodic? how do you "prove" that it's a closed cycle? (that it doesn't go like 1,1,2,5,7,5,7,5,7,... or anything? ;)

A little rigorousness won't hurt you :P

you are right, it is periodic because a pair of consecutive fibbonaccis determines the sequence from there and backwards, like there are 10^n*10^n such pairs at most,(mod 10^n) a pair eventually repeats, let F_k and F_k+1 be the first repeated pair, since it determines the sequence backwards, 1,1 must appear somewhere between the first and secound appearence of the pair, or it wouldnt appear initially. so the first pair to repeat is 1,1 and before it is a -1 as we desired, and as i shwed in my first post.

Sorry if i didnt explain myself enough.


$ x_{1}\equal{}10\sqrt{5}$ and $ x_{n}(x_{n\plus{}1}\minus{}x_{n})\equal{}10$  then what is the greatest integer that not exceeding  $ x_{101}$ ?

Let $ x_n\equal{}\sqrt{10 y_n}$, then $ y_{n\plus{}1}\equal{}y_n\plus{}2\plus{}\frac{1}{y_n}, y_1\equal{}5$, therefore $ y_n\equal{}2n\plus{}\frac 12 ln n\plus{}c_n, c_n\to c$.

Rust, can you go on solution?  I realy did not understand anything.

$ x_{n\plus{}1}\equal{}x_n\plus{}\frac{10}{x_n}>x_n\to x_{n\plus{}1}^2\equal{}x_n^2\plus{}20\plus{}\frac{100}{x_n^2}\to y_{n\plus{}1}\equal{}y_n\plus{}2\plus{}\frac{1}{y_n}.$
We get $ y_1\equal{}50.$
If $ y_n\equal{}2n\plus{}\frac 12 lnn \plus{}c_n$, then $ c_1\equal{}48$ and $ c_{n\plus{}1}\equal{}c_n\minus{}\frac 12 ln(1\plus{}\frac 1n)\plus{}\frac{1}{2n\plus{}\frac 12ln n\plus{}c_n}$.
Because $ c_{n\plus{}1}\minus{}c_n\equal{}O(\frac{ln(n)}{8n^2})$ we get $ c_n\to c$.


inequalities
trigonometry
triangle inequality

for a triangle ABC with sides a,b,c
show that the following inequality holds,
$ \sum_{cyclic} a(b^2 \plus{}c^2 \minus{} a^2) \le 3abc$

It's a special case of Schur's inequality!

Here's an easy proof for the case you've mentioned.

Rewrite the inequality as: \[ \ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \le abc \]

Now, as $ a,b,c$ are sides of a triangle, there exist positive reals $ x,y,z$ such that $ a\equal{}x\plus{}y$, $ b\equal{}y\plus{}z$ and $ c\equal{}z\plus{}x$.

Substituting this in the original inequality, we have: \[ \ 8xyz \le (x\plus{}y)(y\plus{}z)(z\plus{}x) \] which is a direct consequence of the AM-GM inequality.

thanx antiparticle,for the solution
i hava different one here is my solution to the problem,
expanding the expression in the $ RHS$ grouping,we get
T.P.T $ (\sum_{cyc} {a^2(b\plus{}c)}) \minus{} \sum_{cyc} a^3 \le 3abc$
or T.P.T $ \sum_{cyclic}a^2(b \plus{} c \minus{} a) \le 3abc$
now,substitute
 $ b \plus{} c \minus{} a \equal{} x$,
 $ a \plus{} c \minus{} b \equal{} y$,
 $ a \plus{} b \minus{} c \equal{} z$ (note that $ x,y,z$ are positive reals because a,b,c are sides of a triagle
                          and follow triangle inequality)
so we have $ a \equal{} \frac {y \plus{} z}{2}$
and similarly for $ b$ and $ c$
replacing in the inequality we get,
$ \sum_{cyclic}x(y \plus{} z)^2 \le 3(x \plus{} y)(y \plus{} z)(z \plus{} x)$
or,
T.P.T, $ 2 \sum_{cyclic}x(y \plus{} z)^2 \le 3( \sum_{cyclic}(xy(x \plus{} y) \plus{} 2xyz)$
or,
T.P.T$ 2\sum_{cyclic}xy(x \plus{} y) \plus{} 12xyz \le 3[\sum_{cyclic}xy(x \plus{} y) \plus{} 2xyz]$
or ,
T.P.T, $ \sum_{cyclic}xy(x \plus{} y) \ge 6xyz$
which is true by applying AM-GM for 6 reals(since sides of a triangle)
hence proved

In fact, this inequality is true for all positive $ a,b,c$ :P

[quote="geniusbliss"]for a triangle ABC with sides a,b,c
show that the following inequality holds,
$ \sum_{cyclic} a(b^2 \plus{} c^2 \minus{} a^2) \le 3abc$[/quote]

Notice that $ b^2\plus{}c^2\minus{}a^2\equal{}2bc\cos{A}$, by Law of Cosines.

So in essence, we have to prove that $ \cos{A}\plus{}\cos{B}\plus{}\cos{C} \le \frac{3}{2}$.

This follows from Jensen's Inequality.

[b]Check out Schur's inequality [url=http://www.artofproblemsolving.com/Wiki/index.php/Schur's_Inequality]here[/url].[/b]

[quote="mathgeniuse^ln(x)"]
Notice that $ b^2 \plus{} c^2 \minus{} a^2 \equal{} 2bc\cos{A}$, by Law of Cosines.

So in essence, we have to prove that $ \cos{A} \plus{} \cos{B} \plus{} \cos{C} \le \frac {3}{2}$.

This follows from Jensen's Inequality.[/quote]

It follows from Jensen's as long as the triangle is [b]acute angled[/b]. I have posted three proofs of the trigonometric inequality you've mentioned [b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=260543]here[/url][/b].

[quote="geniusbliss"]rolfer do you have a proof for arbitary positve reals a,b,c?
please post it
mathgeniuse^ln(x),
you know you can prove
$ cosA \plus{} cosB \plus{} cosC \le \frac {3}{2}$
through trigonometry and by getting it in the form
like how antiparticle got

antiparticle how is it a special case of schur's inequality???
please elaborate[/quote]

True, I didn't see that it could be any triangle.

$ \cos{A}\plus{}\cos{B}\plus{}\cos{C}\equal{}1\plus{}\frac{r}{R} \le \frac{3}{2}$, as we know that $ OI^{2}\equal{}R^2\minus{}2Rr \ge 0$, so $ R \ge 2r$.

Thanks for the catch.


How many different combinations of coins could a person have if she has exactly 21 cents?

i got 9. hmm, i memorized tht 20-24 were 9 combinations soo yea. thts how i do the coins.

We could make a chart... Hah... D for Dimes N for Nickels and P for Pennies. We can't have Quarters, it would be too high...
D N P
2 0 1
1 2 1
1 1 6
1 0 11
0 4 1
0 3 6
0 2 11
0 1 16
0 0 21
So there are 9 combinations as you said...


probability
limit
real analysis
real analysis unsolved

we take arbitrarily 2 strict positive integers , what the probability that they are prime between each other?
(i really don't where i should put this problem)

This problem is not well-defined, because there is no natural or canonical way to make the positive integers into a probability space. There are several possibilities for what you meant. The following is the most famous:

Fix $n$. Choose $j$ and $k$ independently from the set $\{1,2,\dots,n\}$ according to an equally likely probability distribution. Let $P_n$ be the probability that $j$ and $k$ are relatively prime.

Compute $\lim_{n\to\infty}P_n.$

This is a well-known problem, and the exact answer is one you will notice (and immediately question it's relationship to some other famous things). It's been dealt with before on this site.

For example here (I already gave this link in one of the other topics): http://www.mathlinks.ro/Forum/viewtopic.php?t=31028

But there are at least three topics with exactly the same theme...


[quote="τρικλινο"]γεια χαρα.

Μπορει καπιος απο εσας ρε παιδια,να μου γραψει σε ποσοδεικτες τον ορισμο της τετραγωνικης ριζας που εβρικα στο βιβλιο της

τριτης γυμνασιου ????

Ο ορισμος ειναι ο εξης¨

Η τατραγωνικη ριζα ενος μη αρνητικου αριθμου α ( α>=0) ειναι ο μη αρνητικος αριθμος β (β>=0) ο οποιος οταν υψωθει στο τετραγωνο μσς δινει τον αριθμο α.

Δηλαδη να γραφτει ο ορισμος σε καθαρη φορμαλιστικη μορφη[/quote]
$ a,b \in R, \forall a\ge 0,\exists b \ge 0 : b^2\equal{}a$

[quote="τρικλινο"][quote="r_boris"][quote="τρικλινο"]γεια χαρα.

Μπορει καπιος απο εσας ρε παιδια,να μου γραψει σε ποσοδεικτες τον ορισμο της τετραγωνικης ριζας που εβρικα στο βιβλιο της

τριτης γυμνασιου ????

Ο ορισμος ειναι ο εξης¨

Η τατραγωνικη ριζα ενος μη αρνητικου αριθμου α ( α>=0) ειναι ο μη αρνητικος αριθμος β (β>=0) ο οποιος οταν υψωθει στο τετραγωνο μσς δινει τον αριθμο α.

Δηλαδη να γραφτει ο ορισμος σε καθαρη φορμαλιστικη μορφη[/quote]
$ a,b \in R, \forall a\ge 0,\exists b \ge 0 : b^2 \equal{} a$[/quote]

ευχαριστω φιλε,αλλα η φορμουλα που εγραψες λεει¨

για καθε μη αρνητικο αριθμο α ( α>=0) υπαρχει ενας μη αρνητικος αριθμος β (β>=0) ετσι ωστε β^2 =α.

Δηλαδη εγραψες το θεωρημα υπαρξεως πανω στο οποιον στηριζεται ο ορισμος της τετραγωνικης ριζας[/quote]

$ a,b \in R, \forall a\ge 0,\exists b \ge 0 : b^2 \equal{} a, b: \equal{}\sqrt{a}$

entaksei tora?

Γενικά τη νιοστή ρίζα μπορείς να την ορίσεις και πιο φορμαλιστικά ως εξής :
Δοθέντος θετικού αριθμού $ x$ έστω $ A$ το σύνολο των πραγματικών αριθμών $ t$ με $ t^n < x$ . Τώρα πάρε $ y \equal{} \sup A$ και $ y^n\equal{}x$ άρα $ y\equal{}\sqrt[n]{x}$

[quote="silouan"]Γενικά τη νιοστή ρίζα μπορείς να την ορίσεις και πιο φορμαλιστικά ως εξής :
Δοθέντος θετικού αριθμού $ x$ έστω $ A$ το σύνολο των πραγματικών αριθμών $ t$ με $ t^n < x$ . Τώρα πάρε $ y \equal{} \sup A$ και $ y^n \equal{} x$ άρα $ y \equal{} \sqrt [n]{x}$[/quote]

Με άλλα λόγια $ \sqrt [n]{x}: \equal{}\sup_{y\in\mathbb{R}}\{y: y^n < x\} \equal{} \sup_{y\in\mathbb{R}}\{y: y^n \equal{} x\}$.

Hehe, έχει πλάκα το ότι αυτός ο ορισμός οδηγεί σε $ \sqrt [n]{x} \equal{} \minus{} \infty$ για $ x < 0$ (εξαιτίας της τυπικής σύμβασης $ \sup\varnothing \equal{} \minus{} \infty$). Ποιος είπε ότι χρειάζεσαι μιγαδικούς για να ορίσεις τις ρίζες αρνητικών αριθμών, eh? :P

(Tongue in cheek!!! :D)

Cheerio,

Durandal 1707

Συγνωμη φιλοι αλλα ισως πρεπει να σας δωσω ενα παραδειγμα τι ενοω φορμαλιστικος ορισμος.

Ας παρουμε τον ορισμο του οριου :  lim f(x) =m  as x---> to a :

Oριζουμε λοιπον το οριον μιας συναρτησις f(x) ,καθως το x πλησιαζει το a, τον αριθμο m εαν και μονο εαν  ( iff) :

Για καθε e>0 υπαρχει ενα d>0 ετσι ωστε ,για καθε χ που ανοικει στο πεδιον ορισμου της f(x) και 0<lx-al<d τοτε  l F(x)-ml<ε.

Τωρα Η ΑΚΡΙΒΗΣ ΜΕΤΑΦΟΡΑ ΣΕ ΣΥΜΒΟΛΑ ΕΙΝΑΙ:

 limf(x)=m (x--->a) <====> $ \forall e$[ e>0 ===> $ \exists d$ ( d>0 & $ \forall x$ ( xεD(f) & 0<l x-al<d ===>lf(x)-ml<e))]

Συγνωμη για τα e και d αντι ε και δ,αλλα δεν ξερω καλλα το la tex σε αυτο το φορουμ.

Βεβαια ο ποιο πανω ορισμος  του οριου δεν 100% ακριβης ,αλλα για παραδειγμα ειναι αρκετος.

Στον ορισμο  της 3ης γυμνασιου δεν υπαρχει η λεξις ( υπαρχει ενα β>=0),αλλα ο ορισμος λεει οτι η τετραγωνικη ριζα του α>=0 ειναι ο αριθμος β>=0 τετοιος ωστε................................................

$ [b \equal{} \sqrt{a}, a \ge 0] \Leftrightarrow[ b^2 \equal{} a, b \ge 0]$

[quote="maniopas"]$ [b \equal{} \sqrt {a}, a \ge 0] \Leftrightarrow[ b^2 \equal{} a, b \ge 0]$[/quote].

maniopa ,ευχαριστω ,αλλα επροσπαθησα να αποδειξω  $ \forall a$( a>=0 ====> $ \sqrt a$>=0) χρησημοποιοντας την φορμουλα σου και δεν μπορεσα .Ισως εσυ μπορεις

Αν δείτε, η formula μου, έχει εξ' ορισμού ότι η τετραγωνική ρίζα μεγαλύτερη ή ίση του 0
(ορίζω ότι $ b\equal{}\sqrt a$ και στο δεύτερο με΄λος λέω ότι $ b \ge 0$)


for all chords of a circle passing through a single point, show the point of intersections of the tangents to the circle at the end points of each chord are colinear.  maybe you guys could include a non-projective proof as well?  thanks

If P is the common point to $A_1B_1$, $A_2B_2$, ... $A_jB_j$ and $P_1$, $P_2$, ... $P_j$ are the points where tangents at $A_jB_j$ to the circle meet, then we have that

$OM_j\cdot OP_j = r^2$ where O is the center of circle and $M_j$ the midpoint of $A_jB_j$.

It holds that for example $M_1M_2P_2P_1$ and $M_2M_3P_3P_2$ are ciclic. Also $OPM_1M_2M_3$ is ciclic then is easy prove the claim.

I think the projective soln you're referring to is this:

For a certain chord the point of intesection of the tangents to the circle at the ends of the chord is the pole of that chord with respect to the circle. By duality, the locus of the poles of all the chords passing through that specific point is the polar of the point, so all those pts must be on a single line = the polar of our point.


trigonometry
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calculus computations

how does lim as x approaches 0 of ((sin2x)/(sin3x))=2/3?

what is the sandwich theorem and could please give an example of it?

thank you

Hint: Note that $ \sin x$ is closer to $ x$ whenever $ x$ is closer to 0.

Also look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1149011684&t=179706

Often also called the 'squeeze theorem', it states that if three functions $ f_l,f,f_r$ satisfy $ f_l(x)\leq f(x) \leq f_r(x)$ and $ \lim_{x\to a}f_l(x) \equal{} \lim_{x\to a}f_r(x) \equal{} L$, then it is also true that $ \lim_{x\to a}f(x) \equal{} L$.

By angle additional formulas we can show $ \frac {\sin 3x}{\sin 2x} \equal{} \frac {3\sin x \minus{} 4\sin ^3 x}{2\sin x \cos x} \equal{} \frac {3}{2}\frac {1}{\cos x} \minus{} 2\frac {\sin ^2 x}{\cos x}$.  Now define the following three functions
\[ f(x) \equal{} \frac {3}{2}\frac {1}{\cos x} \minus{} 2\frac {\sin ^2 x}{\cos x} \\
 \\
f_l(x) \equal{} \frac {3}{2} \minus{} 2\frac {\sin ^2 x}{\cos x} \\
 \\
f_r(x) \equal{} \frac {3}{2}\frac {1}{\cos x}
\]
Show that $ f_l(x)\leq f(x)\leq f_r(x)$ and that $ \lim_{x\to 0}f_l(x) \equal{} \lim_{x\to 0}f_r (x) \equal{} \frac {3}{2}$, and invoke the squeeze theorem.  If $ \lim_{x\to 0}f(x)\equal{}\frac{2}{3}$, then what is the original limit you're looking for?

Although we really don't need this long-winded argument nor squeeze theorem for this case.  Can you think of a way to do it using only the properties of limits? [hide="hint"]The limit of a product is the product of limits provided that each exists.[/hide]

well sinx is between -1 and 1 but that wouldn't help me with sin3x and sin2x

oh,to find $ \lim_{x\to 0}f(x) = \frac {sin3x}{sin2x}$
we only need Lopitan order.used it we have $ LHS = \lim_{x\to 0}\frac {3cos3x}{2cos2x} = \frac3}{2}$


MATHCOUNTS
ratio
geometry
rectangle
function
3D geometry
percent

1. given the equation 1/m + 1/n = 2/15, where m and n are positive integers, find the value of m+n wuch that m>n and m is not a multiple of n.

2. the ratio of the length of a rectangle to its width is the same as that of the diagonal of the length. if the width is 2, how many units are in the length of the diagonal?

3. how many integers n can be written as a+b+c where a, b, and c are distinct odd integers between 2 and 20?

4. solve for x: {3+[2+(1+x :^2: ) :^2: ] :^2: } :^2:  = 144

5. l x+2 l + l y-3 l = 1 is an equation for a square. how many units are in the lengths of its diagonals?

6. what is the minimum value of the function y = x :^2:  -  6x+5?

7. the dimensions of a rectangular block are 30cm, 16cm, and 12cm. it is made from cubes which are each 1cm :^3:  in volume. if the outside of the block is painted red, what percent of the block is made up of cubes with exactly two sides painted red? express your answer to the nearest tenth.

8. if 3^(x+1) = 5 find 9^(2x+2).

9. find the smallest real value of x that satisfies the equation:
(x+5)(x :^2: -x-11)=x+5 (i can do this but i dont get the right answer if i divide each side by x+5 which should give me the right answer shouldnt it?)

10. in how many distinguishalb eways can four identical red chips and two identical white chips be arranged in a circle?

11. when the product of three consecutive positive integers is divided by their sum, the quotient is 33. what is the largest of the three integers? (you wouldn't do this algebraically right bc i got in a big mess when i tried)

12. if a die whose faces are displayed in the figure are rolled and the numbers on the four lateral faces are added, what is the probability that the sum is a prime number? express your answer as a common fraction. the numbers on the die should look like this rolled out where each number is a face and 0 represents  a space:
0 3
1 4 6 2
0 5

13. given that 7a :^2: +3a :^2: b+14ab-2ab :^2: =61 and 7b :^2: -3a :^2: b+2ab :^2: =2, find the value of l a + b l

14. what real number is not contained in the range of the function defined by f(x)=1/x - 2

15. the vertices of a triangle have coordinates (-3,1),(-2,-3), and (x,0). find x, given x is positive and the area of the triangle is 7.5 square units.

16. to the nearest second, what is the first time between 4:00 and 5:00 that the minute and hour hands of the clock are perpendicular to each other. express your answer in the form:
___(hr):___(min):___(sec)

17. a circle with a radius of one yard has within it an infinite sequence of concentric circles. each circle has a radius that is one half the raius of the next larger circle. if the shading pattern continues infinitely, what will be the total area in square yards of the shaded region? express your answer as a common fraction in terms of pi. (the shaded region is where the largest circle is shaded then the circle next to it isnt then the circle inside of that one is shaded....so its every other concentric circle is shaded)

thanks so so much! since my mc competition is friday and im leaving thursday, it would be greatly appreciated if you guys could post the solutions as quickly as possible! thanks again!

I think I know most of these problems..

But the easiest one I think is this one,

6. what is the minimum value of the function y = x  - 6x+5?

[hide]This question is one of things I learned recently in Algebra II. This equation is in form of y=ax^2+bx+c. Since it has degree of two, it means it crosses x-axis twice, or forms a parabola. Each of parabola has either minimum value or maximum value and for positive ax^2, the graph has minimum value. The minimum value or maximum value is called vertex. There are few ways to do this but mainly way is finding axis of symmetry and then submit for the vertex.

Formula for axis of symmetry is: x=-b/2a

For vertex, submit the number for axis...

Then the answer is following ordered pair: (-3,32)...

[/hide]

i dont think they're lookin for an ordered pair here. i think the answer is just the x coordinate.

Actually, they're looking for just the y value (the value of the function).

Here's the other one I think I can do:

9. find the smallest real value of x that satisfies the equation:

(x+5)(x^2-x-11)=x+5 (i can do this but i dont get the right answer if i divide each side by x+5 which should give me the right answer shouldnt it?)

[hide] You probably got the -3 for answer, right? Well, if you didn't, I can't guess what your answer might be... In this question, you must understand the idea of finding all x values for this question. For this, there are three possible solutions that are possible. They are: -5, -3, 4. Here's way to get -3, and 4.

*Divide both sides by x+5.

*Factor out (x^2-x-12)=0 (see that I removed 1 from right side to left side)

*Get (x+3)(x-4)=0

*I hope you know what's next

For -5, it's very simple. Since there are part for mulitplying (x+5), submit -5 for x-value. You can see this is the answer because          -5+5=0 is zero and anything times 0 is zero...

So, hope this helps.. [/hide]

could you guys just solve problems 1, 5, 12, 13, 14 bc i've gotten the answer from some very helpful people :D ?  thanks so much, i wanna learn these before thursday!

Elan, good luck at Florida states, you'll see my old school there too.

i might have done this one too late... but here we go
4. solve for x: {3+[2+(1+x :^2: )  ]  }  = 144
first, take the square root of both sides:
3+[2+(1+x :^2: ) :^2: ] :^2: =12
then subtract 3..
[2+(1+x :^2: ) :^2: ] :^2: =9
take square root again
2+(1+x :^2: ) :^2: =3
subtract 2
(1+x :^2: ) :^2: =1
square root again
1+x :^2: =1
subtract one
x :^2: =0
x=0

can someone do 5 for me plz? I need it to prepare for the competition this Saturday

[hide]Plug in some points to get a sense of what's happening. Realize that you can't plug in values for $x$ when $x$ is greater than $-1$ and smaller than $-3$.
Plug in $-1$ for $x$ to get 1. Then the absolute value of $y-3$ equals 0. $y=3$
Plug in $-2$ for $x$ to get $0$ for the $x$-absolute value part. So you get that the absolute value of $y-3$ is equal to $1$. This means that $y$ could equal either $4$ or $2$. You get this by simply setting up two equations.
$y-3=1$ and $y-3=-1$
Finally plug in $-3$ for $x$. You get that $y=3$. You get a square of length $\sqrt2$. Drawing it on the x-y plane, you get that the diagonals are lengths of $2$ units.[/hide]
Sorry if it's a little wordy and long.

Seriously, do you know how hard it is to red these? Can you at least put it in LaTeX?

[hide="Prob 14 (I think)"]
We notice that $\frac{1}{x}$ cannot equal 0, so therefore
$0-2$

$\boxed{-2}$
[/hide]

[quote=eqn1190]1. Given the equation $\frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ which  $m>n$ and $m$ is not a multiple of $n$.

2. The ratio of the length of a rectangle to its width is the same as that of the diagonal of the length. If the width is $2$, how many units are in the length of the diagonal?

3. How many integers $n$ can be written as $a+b+c$ where $a$, $b$, and $c$ are distinct odd integers between $2$ and $20$?

4. Solve for $x$: $(3+(2+(1+x^2)^2 )^2)^2= 144$.

5. $|x+2| + | y-3| = 1$ is an equation for a square. How many units are in the lengths of its diagonals?

6. What is the minimum value of the function $y = x^2  -  6x+5$?

7. The dimensions of a rectangular block are $30$ cm, $16$ cm, and $12$ cm. It is made from cubes which are each $1$ cm$^3$  in volume. If the outside of the block is painted red, what percent of the block is made up of cubes with exactly two sides painted red? Express your answer to the nearest tenth.

8. If $3^{(x+1)} = 5$ find $9^{(2x+2)}$.

9. Find the smallest real value of $x$ that satisfies the equation:
$(x+5)(x^2 -x-11)=x+5$ (I can do this but i don't get the right answer if i divide each side by $x+5$ which should give me the right answer shouldn't it?)

10. In how many distinguishable ways can four identical red chips and two identical white chips be arranged in a circle?

11. When the product of three consecutive positive integers is divided by their sum, the quotient is $33$. What is the largest of the three integers? (You wouldn't do this algebraically right because I got in a big mess when I tried)

12. If a die whose faces are displayed in the figure are rolled and the numbers on the four lateral faces are added, what is the probability that the sum is a prime number? Express your answer as a common fraction. Below is an example of a rolled out die.
$\ \ \ \ \boxed{3}$
$\boxed{1}\boxed{4}\boxed{6}\boxed{2}$
$\ \ \ \ \boxed{5}$

13. Given that $7a^2+3a^2b+14ab-2ab^2=61$ and $7b^2 -3a^2b+2ab^2=2$, find the value of $|a+b|$.

14. What real number is not contained in the range of the function defined by $f(x)=\frac{1}{x} - 2$?

15. The vertices of a triangle have coordinates $(-3,1)$,$(-2,-3)$, and $(x,0)$. Find $x$, given $x$ is positive and the area of the triangle is $7.5$ square units.

16. To the nearest second, what is the first time between 4:00 and 5:00 that the minute and hour hands of the clock are perpendicular to each other. express your answer in the form:
___(hr):___(min):___(sec)

17. A circle with a radius of one yard has within it an infinite sequence of concentric circles. each circle has a radius that is one half the radius of the next larger circle. If the shading pattern continues infinitely, what will be the total area in square yards of the shaded region? Express your answer as a common fraction in terms of pi. (the shaded region is where the largest circle is shaded then the circle next to it isnt then the circle inside of that one is shaded....so its every other concentric circle is shaded)

thanks so so much! since my mc competition is friday and im leaving thursday, it would be greatly appreciated if you guys could post the solutions as quickly as possible! thanks again![/quote]

By the way for number 9, if you divide by $x+5$, you get $x=-3,4$. If $x=-5$, then it still works as an equation. That is the smallest value.

10. First white chip has 1 choice.
      Second white chip has 5 choices.
      Red chips will take the leftover seats.
      So 1*5=5

16.
The hour hand is 120 ahead. The hour hand moves 0.5 a minute and the minute hand moves 6 a minute. In a minute, the minute hand will be 5.5 closer to the hour hand. 120/5.5=1200/55=240/11=21.81818181...
So 4(hr):21(min):49(sec)


analytic geometry
calculus
derivative
function
trigonometry
probability
integration

The ends $A$ and $B$ of a rod of length $\sqrt{5}$ are sliding along the curve $y=2x^{2}$. Let $x_{A}$ and $x_{B}$ be the $x$-coordinates of the ends. Find the derivative $\frac{dx_{B}}{dx_{A}}$ at the moment when $A$ is at $(0,0)$ and $B$ is at $(1,2)$.

  I put this problem on a Calculus III midterm exam. It set my personal record for the lowest average score on a problem: $1.2$ out of $7$ points.  :read:  :censored:

I think I'll ... not use it. I do like to see some right answers.

Well, how bad is it? I'm going to take the advice of a colleague of mine and be very 18th century about this - differentials, not derivatives. We have:

$(x_{B}-x_{A})^{2}+(2x_{B}^{2}-2x_{A}^{2})^{2}=5$

$2(x_{B}-x_{A})(dx_{B}-dx_{A})+2(2x_{B}^{2}-2x_{A}^{2}) (4x_{B}\,dx_{B}-4x_{A}\,dx_{A})=0$

Now let's put in $x_{A}=0,x_{B}=1.$

$2\cdot1(dx_{B}-dx_{A})+2\cdot 2(4dx_{B}-0)=0$

$2dx_{B}-2dx_{A}+16dx_{B}=0$

$18dx_{B}=2dx_{A}$

$\frac{dx_{B}}{dx_{A}}=\frac19$

So: massively intimidating setup, but the computation isn't bad at all.

To be fair, I personally don't understand what this problems means. This is probably due to two reasons. The first is that I've never had to work problems this applied, the second is that I am so opposed to 18th century 'calculus' and differential notation that I refuse to understand it.

Does calculating $\frac{dx_{B}}{dx_{A}}$ mean that we find $x_{B}$ as a function of $x_{A}$ and then calculate the derivative? Or is it implied that $x_{A}$ and $x_{B}$ are both functions of some variable $t$ so that we have some type of parametrisation of the rod $(x_{A}(t),x_{B}(t))$, so that $\frac{dx_{B}}{dx_{A}}$ is some crazy Leibniz notation.

I'm sure that either way you do it you can realize that you get the same answer, but to me, just writing $x_{A}$ and $x_{B}$ are the $x$-coordinates of the ends is not clear. It can be interpreted in many ways, since there clearly aren't fixed $x$-coodinates.

It's the first, except that it's not worth solving all the way to a formula $x_{B}=f(x_{A})$. $x_{B}$ and $x_{A}$ are related by the implicit formula in Kent's first line. The rest of the work is implicit differentiation.

The single-variable calculus explanation of implicit differentiation:
We are given that $g(x,y)=0$ for some nice function $g$. Suppose that there is some differentiable function $y(x)$ so that $g(x,y(x))=0$. Then, differentiating the expression $g(x,y(x))$ by the chain rule and whatever it takes to differentiate $g$, we get an equation in $x,y(x),y'(x)$. Since $y'(x)$ can only appear multiplying a term in which we differentiated $y$, this equation can be written $g_{1}(x,y(x))\cdot y'(x)+g_{2}(x,y(x))=0$. Solving, we have that $y'(x)=-\frac{g_{2}(x,y(x))}{g_{1}(x,y(x))}$.
In practice, we write $y$ rather than $y(x)$, but we think of $y$ as a function of $x$ in order to make this work.

This problem is set up perfectly for the method; we are given a point $(x_{A},x_{B})=(0,1)$ which satisfies the equation, and we want to find the derivative of the differentiable function which passes through that point and satisfies the equation.

There's one hole, of course. The above argument does not show that a differentiable function satisfying the equation exists. When we return to this topic in multivariable calculus, we see that the $g_{1}$ and $g_{2}$ are partial derivatives. We also learn the Implicit Function Theorem, which says that there is a locally unique differentiable solution wherever $g_{1}$ is nonzero (invertible).

[quote="Kalle"]I am so opposed to 18th century 'calculus' and differential notation that I refuse to understand it. [/quote]
And I'm not so quick to condemn a body of ideas that worked quite well for well over a century. Yes, there was a need for Cauchy and increased rigor (leading inevitably to the finding that Cauchy wasn't rigorous enough). Yes, I can quote Weierstrass as well as you can. But calculus exists to calculate things, and for that, you need to adopt a flexible attitude towards variables and notations.

I understand "differentials" as notational stand-ins behind which lurk such notions as the chain rule, Riemann sums, and Taylor polynomials. (No, I'm not a fan of non-standard analysis, either.) And yes, as jmerry indicated, this problem invokes the Implicit Function Theorem, which I'll paraphrase as "you can do smoothly what you can do linearly."

Here's the test for the usefulness of using Leibniz notation, provided you use it well: that other topic, with Euclid 3.14's botched computation of the volume of a sphere. Kalle - your own contributions that topic tend to cast a pall of uncertainty and doubt over the uses of such notation - but did you help explain anything there? I think you can tell the difference between an arclength segment of $a\,d\theta$ and the thickness of a "disk" $a\sin\theta\,d\theta$ without being unduly bothered by the fact that it isn't exactly the frustrum of a cone.

I think that such "applied" problems are not out of place in a "calculus for engineers" class like mine. They all know how to differentiate, integrate, and even solve differential equations with TI-89. Not so many know [b]what[/b] to differentiate... The Implicit Function Theorem is stated and used in that class, although not proved.

My solution: the coordinates of $A$ and $B$ are $(x_{A},2x_{A}^{2})$ and $(x_{B},2x_{B}^{2})$, respectively. Since the length of the rod is $\sqrt{5}$, we have $(x_{A}-x_{B})^{2}+(2x_{A}^{2}-2x_{B}^{2})^{2}=5$. Let $F(x_{A},x_{B})=(x_{A}-x_{B})^{2}+4(x_{A}^{2}-x_{B}^{2})^{2}$, so that $F(x_{A},x_{B})=5$ holds identically. By the Implicit Function theorem
$\frac{dx_{B}}{dx_{A}}=-\frac{\partial F/\partial x_{A}}{\partial F/\partial x_{B}}=-\frac{2(x_{A}-x_{B})+16x_{A}(x_{A}^{2}-x_{B}^{2})}{-2(x_{A}-x_{B})-16x_{B}(x_{A}^{2}-x_{B}^{2})}=\frac{1+8x_{A}(x_{A}+x_{B})}{1+8x_{B}(x_{A}+x_{B})}$
which is $\frac19$ with the given data.

An interesting observation: $\frac{dx_{B}}{dx_{A}}$ can be negative. You can push one end of the rod to the right, and the other one will move to the left.

Another interesting point: an explicit formula for $x_{B}=f(x_{A})$ takes over two pages (generated in Mathematica).

[quote="mlok"]Not so many know [b]what[/b] to differentiate... [/quote]
The whole challenge of teaching calculus (or differential equations, or linear algebra, or probability, or anything else we teach all of the users of mathematics) summed up in one sentence.

It's also the source of a huge advantage that the contest problem-solvers often have over everyone else. The former are used to the idea that the statement of the problem does not contain "how to" instructions; it's up to the solver to decide what machinery to bring to bear.

Despite the differences in notation and appearance, the only practical difference between mlok's solution and mine is that he obtained a general implicit form for the derivative before plugging in the specific data, while I put in that specific data at an earlier stage.

[quote="Kent Merryfield"]
And I'm not so quick to condemn a body of ideas that worked quite well for well over a century. Yes, there was a need for Cauchy and increased rigor (leading inevitably to the finding that Cauchy wasn't rigorous enough). Yes, I can quote Weierstrass as well as you can. But calculus exists to calculate things, and for that, you need to adopt a flexible attitude towards variables and notations.
.[/quote]
I am not opposed to the ideas themselves, but I feel that "calculus by notational abuse" is never presented in a clear way. If a book begins by saying "rather than writing u(s(x,y),t(x,y)) we shall in the future interpret u as a function of both variables x,y and variables s,t" I will have to live with that. However, I don't want to have to live with the exposition going on to completely ignore the whole problem complex of the inverse function theorem (simply assuming that all smooth coordinate changes have smooth inverses), without even mentioning this.

I don't know what I am trying to say. I think I am just frustrated that notational abuse is not entirely clearly presented. I would like it to be defined exactly what we mean by the notation, and also explanations of what the notation means in terms of 'modern stringence'. Often the student is asked to do this on his own. I am usually not smart enough to figure out on my own what the notation should mean, and perhaps this is what makes it so frustrating for me. I can guarantee you that a lot of students simply choose to learn how the notation works when it's not clearly laid out what it actually means.

[quote="Kent Merryfield"]
Here's the test for the usefulness of using Leibniz notation, provided you use it well: that other topic, with Euclid 3.14's botched computation of the volume of a sphere. Kalle - your own contributions that topic tend to cast a pall of uncertainty and doubt over the uses of such notation - but did you help explain anything there? I think you can tell the difference between an arclength segment of $a\,d\theta$ and the thickness of a "disk" $a\sin\theta\,d\theta$ without being unduly bothered by the fact that it isn't exactly the frustrum of a cone
[/quote]
Yeah, ok, I suppose I wasn't precisely helpful in that thread. But I do remain the point of view that textbooks are doing the student a disservice by not being clear that there is more to the whole truth when it comes to integration theory. Another thing that comes to mind is how flippant many textbooks are in complex analysis when it comes to cauchy's integral theorem/green's theorem and the jordan curve theorem. Often the huge difficulties behind these theorems are wiped under the mat and the theorems are presented in a quite unprecise way. Shouldn't it be the book's responsibility to inform the reader about the problems behind cauchy's theorem, and that it in fact requires the modern workings of topology and algebraic topology to sort out?

When something is not done correctly I would like warning labels, but authors never seem to want to admit when they are cheating.


Princeton college LaTeX geometry rectangle

A brilliant mathematician from Princeton decided to draw a line on a paper.  He did so, which appeared like this: __________
He looked to find the width of the line, but his ruler's intervals are way too big.  So, he draws lines with equal width on that same paper, such that the distance between two consecutive lines is equal.  His paper now looks like this:
________
________
________
________
________
________
________
If each space between any pair of consecutive lines is 1/4 inches, and the width of the whole picture is 7/4 inches, find the width of the first line drawn by the mathematician.(Source: Me, again)

additionally, while this has no relevance to the problem, can anyone tell me how to make subscripts without using LATEX?

Width meaning length?  And what do you mean "length of the whole picture"?  Which direction?

By width, I don't mean how long ________ that line is, but how wide it is (very miniscule, obviously).  In the problem, I say "width of the whole picture," but by that I mean the downwards length.

Technically, a line has no width, but I think I get what you mean.

Let's see, the entire length is $ \frac{7}{4}$.  So now subtract out the $ 6\times\left(\frac{1}{4}\right)$ yielding $ \frac{1}{4}$, and then divide by $ 7$ and we are indeed left with $ \frac{1}{28}$.

And a point by definition has no size.  But the universe started as an infinitely small and infinitely dense point.  Think about it!!! :D

[quote="Math Geek"][quote="Z-coli"]Correct, and @ alanchou, I'm sure that if you draw a line it will have some width.[/quote]

Um, not really.  A line by definition has no width.[/quote]

Drawing a line is impossible.

But i believe Z-coli meant drawing a representation of a line will have a width.
which is correct!

Technically, if you draw a line with width, it would be considered a polygon or a rectangle that is very thin thus a line doesn't have width.
:ninja:  :ninja:

You don't actually "draw" a line; you draw a representation of a line.  Similarly, if I were to draw a "circle" on a piece of paper, there isn't really a circle, just a representation of one.

The reason you can't draw a line is because it has no width, and a line goes on forever. So what you could say is that you drew a representive of a line segment.

and i agree with the answer 1/28 saying that the representitive of the line segment has some kind of width right now.

Look, if you draw a "representation" of a line, it will have to have some width since the human hand cannot draw something with no width.  Unless, of course, they don't draw anything.

[quote="JRav"]Let's see, the entire length is $ \frac {7}{4}$.  So now subtract out the $ 6\times\left(\frac {1}{4}\right)$ yielding $ \frac {1}{4}$, and then divide by $ 7$ and we are indeed left with $ \frac {1}{28}$.[/quote]
yes what he said..

but my point was that a representation of a line still would go on forever, because the definition of a line says that it goes on forever.

Firstly, if you read the problem, the length has no bearing on it. And second, I think we all know what the original problem meant; if the word line hadn't been used (rectangle could have been used), we wouldn't have all these posts. I also believe that at this level, we are all familiar with the definition of a line, so perhaps this argument should be in a different forum, where it doesn't bother people actually solving the problem.

[quote="Z-coli"]
additionally, while this has no relevance to the problem, can anyone tell me how to make subscripts without using LATEX?[/quote]

Without LaTeX? I don't know about BBCode, but there is the <sub> tag in HTML. I don't know if anyone uses HTML vs. BBCode here though...

It's simple with LaTeX, though
[code]
$x_{1}$
[/code]
gives $ x_{1}$

1/28 of an inch

7/4-6/4 leaves us with 1/4

assuming that all the lines are of equal width we are left with

1/4 divided by 7 = 1/28 of an inch

^^^Read the previous posts, it's been solved many times.

Why are we still discussing this?  The problem was solved weeks ago.


Let $a$ and $b$ be integers. Show that $a$ and $b$ have the same parity if and only if there exist integers $c$ and $d$ such that $a^2 +b^2 +c^2 +1 = d^2$.

[quote="zakor"]Let $ a$ and $ b$ such that there exist $ c,d$ with $ a^{2}+b^{2}+c^{2}+1=d^{2}$. Assume that $ a$ and $ b$ have different parity. Then $ d^{2}-c^{2}$ is 2 mod 4, impossible. Now, for the converse, observe that  $ a^{2}+b^{2}+1$ is either $ 4k+1$ or $ 4k+3$. Therefore we may take $ (c,d)=(2k, 2k+1)$ or $ (c,d)=(2k+1, 2k+2)$. Hope I'm right because this is my first post :)[/quote]

first we prove the "if" part -- if possible , assume that $a,b$ are of different parity . then $a^2+b^2=1(mod4)$
so , $d^2-c^2=2(mod4)$ ,which is impossible.hence done
then we prove the ""only if"" part -- a,b are of same parity .then take $d=\frac{a^2+b^2+2}{2}$
and $c=\frac{a^2+b^2}{2}$


function
geometry
calculus
calculus computations

You are designing a new cylindrical can to hold 1 liter (1000 cubic centimeters) of Jake's Special Paint. Find the dimensions that will minimize the cost of metal to manufacture the can.

[hide="Click here"]Let y be the height of the cylinder and x the radius of its base (these quantities are measured in dm). The function f(x,y) that gives the total surface area of the can is

$ f(x,y) \equal{} 2 \pi xy \plus{} 2 \pi x^2$.

On the other hand, its volume V is given by $ V \equal{} \pi x^2y$. Our constraint is V = 1, or $ \pi x^2y \equal{} 1 \equal{} > y \equal{} \frac {1}{\pi x^2}$. Substituting this on the expression for f(x,y) we get a function of only one variable x:

$ f(x) \equal{} \frac {2}{x} \plus{} 2 \pi x^2$.

Now our purpose is to minimize this function.

[hide="Answer"]We have $ f'(x) \equal{} \minus{} \frac{2}{x^2} \plus{} 4 \pi x \equal{} \frac{4 \pi x^3 \minus{} 2}{x^2}$. Now f'(x) = 0 implies

$ 4 \pi x^3 \minus{} 2 \equal{} 0 \equal{}> x \equal{} \frac{1}{\sqrt[3]{2 \pi}}$.

This value of x is indeed a minimum of f(x) because $ f''\left(\frac{1}{\sqrt[3]{2 \pi}}\right) \equal{} 12 \pi > 0$. So the dimensions that minimize the surface area of the can (and hence the cost of the metal needed to built it) are

$ x \equal{} \frac{1}{\sqrt[3]{2 \pi}} \approx 0.542\, dm \equal{} 5.42\, cm$ and $ y \equal{} \left(\frac{2}{\sqrt{\pi}}\right)^{2/3} \approx 1.08\, dm \equal{} 10.8\, cm$.
[/hide]
[/hide]

For a Calculus I (AP Calculus AB) answer, click on Carcul's hidden text.

Here's a Calculus III (multivariable calculus) answer using the technique of Lagrange multipliers. Lagrange multipliers is a method extremely well-matched to this sort of problem, as the answer tends to come out directly as the "shape" of the optimum. Also there is no practical distinction between the original problem and its dual problem.

The cylinder has radius $ r$ and height $ h.$ We wish to maximize the area $ A \equal{} 2\pi r^2 \plus{} 2\pi rh$ subject to the constraint $ \pi r^2h \equal{} V \equal{} 1000\text{cm}^3.$

The Lagrange multiplier equations:

Take $ \frac {\partial}{\partial r}: \ 4\pi r \plus{} 2\pi h \equal{} \lambda 2\pi rh$

Take $ \frac {\partial}{\partial h}: \ 2\pi r \equal{} \lambda \pi r^2.$

From the second equation, $ \lambda \equal{} \frac2r.$ Substitute that in the first equation to get

$ 4\pi r \plus{} 2\pi h \equal{} 4\pi h,$ from which we get $ h \equal{} 2r.$

So that's the shape of the solution. The height of the cylinder is exactly the same as the diameter; viewed directly from the side, the cylinder would appear to be a square.

To finish this off, put $ h \equal{} 2r$ into the constraint equation $ \pi r^2h \equal{} V$ to get that $ r \equal{} \left(\frac {V}{2\pi}\right)^{\frac13}$ and $ h \equal{} 2\left(\frac {V}{2\pi}\right)^{\frac13}.$ (Note: this agrees exactly with what Carcul said.) We also need a little bit of an argument to show why this unique critical point is in fact a minimum of the area. To that end, we need to show both that large $ r,$ small $ h,$ and large $ h,$ small $ r$ lead to large area.

The dual question would be this: among all cylinders which use the same specified amount of material, which one has the maximum volume? And the answer would be precisely the same: the cylinder whose diameter equals its height.

One should note that observation of actual cylindrically-packaged goods in the marketplace shows that cylinders this stocky are relatively rare, and that cylinders taller than wide are more common. Soft drink cans are taller than wide, as are canned soup and a variety of other grocery products. But canned tuna fish is wider than it is tall. Small cans of corn and peas may be close to the optimum of this problem; so also may be paint cans (the subject of this problem).

[color=blue][I'll leave to someone else the post basing the proof of this problem on the AM-GM inequality.][/color]


logarithms
trigonometry
integration
calculus
calculus computations

Why do I have the strange result?

\[ \cot \left( \minus{}x\plus{}\frac{y}{2}\plus{}a \right)\equal{}\frac{1}{4}.\log \left| \frac{1\plus{}\sin 2\left( \minus{}x\plus{}\text{y/2}\plus{}a \right)}{1\minus{}\sin 2\left( \minus{}x\plus{}\text{y/2}\plus{}a \right)} \right|\text{  ?}\]
The problem is rising from difference y and y/2.
Solve the differential equation:    $ {{y}^{'}}\equal{}2{{\cos }^{2}}\left( \minus{}x\plus{}y\plus{}a \right)$
Solution

$ \text{*  }{{y}^{'}}\equal{}2{{\cos }^{2}}\left( \minus{}x\plus{}y\plus{}a \right)\text{, }{{y}_{0}}\equal{}y\left( {{x}_{0}} \right)\text{, Put }u\equal{}\minus{}x\plus{}y\plus{}a\Rightarrow {{u}^{'}}\equal{}\minus{}1\plus{}{{y}^{'}}$

$ \Rightarrow {{y}^{'}}\equal{}{{u}^{'}}\plus{}1\Rightarrow {{u}^{'}}\plus{}1\equal{}2{{\left( \cos u \right)}^{2}}\Leftrightarrow {{u}^{'}}\equal{}2{{\left( \cos u \right)}^{2}}\minus{}1\equal{}\cos 2u$

$ \Rightarrow \int\limits_{{{u}_{0}}}^{u}{\frac{du}{\cos 2u}}\equal{}\int\limits_{{{x}_{0}}}^{x}{dx}\Rightarrow x\minus{}{{x}_{0}}\equal{}\frac{1}{2}\int\limits_{{{u}_{0}}}^{u}{\frac{\cos 2u.d\left( 2u \right)}{{{\left( \cos 2u \right)}^{2}}}}\equal{}\frac{1}{2}\int\limits_{{{u}_{0}}}^{u}{\frac{d\left( \sin 2u \right)}{1\minus{}{{\left( \sin 2u \right)}^{2}}}}$

$ \equal{}\frac{1}{2}\int\limits_{{{t}_{0}}}^{t}{\frac{dt}{1\minus{}{{t}^{2}}}}\equal{}\frac{1}{4}\left( \int\limits_{{{t}_{0}}}^{t}{\frac{dt}{1\minus{}t}}\plus{}\int\limits_{{{t}_{0}}}^{t}{\frac{dt}{1\plus{}t}} \right)\equal{}\frac{1}{4}\left( \minus{}\ln \left| 1\minus{}t \right|\plus{}\ln \left| 1\minus{}{{t}_{0}} \right|\plus{}\ln \left| 1\plus{}t \right|\minus{}\ln \left| 1\plus{}{{t}_{0}} \right| \right)$

$ \equal{}\frac{1}{4}.\left( \ln \left| \frac{1\plus{}t}{1\minus{}t} \right|\minus{}\ln \left| \frac{1\plus{}{{t}_{0}}}{1\minus{}{{t}_{0}}} \right| \right)\equal{}\frac{1}{4}.\left( \ln \left| \frac{1\plus{}\sin 2u}{1\minus{}\sin 2u} \right|\minus{}\ln \left| \frac{1\plus{}\sin 2{{u}_{0}}}{1\minus{}\sin 2{{u}_{0}}} \right| \right)$

$ \Rightarrow x\minus{}{{x}_{0}}\equal{}\frac{1}{4}.\left( \ln \left| \frac{1\plus{}\sin 2\left( \minus{}x\plus{}y\plus{}a \right)}{1\minus{}\sin 2\left( \minus{}x\plus{}y\plus{}a \right)} \right|\minus{}\ln \left| \frac{1\plus{}\sin 2\left( \minus{}{{x}_{0}}\plus{}{{y}_{0}}\plus{}a \right)}{1\minus{}\sin 2\left( \minus{}{{x}_{0}}\plus{}{{y}_{0}}\plus{}a \right)} \right| \right)$
4/ Solve the differential equation:    $ {{y}^{'}}\equal{}{{\cos }^{2}}\left( \minus{}x\plus{}y\plus{}a \right)$
Solution

\[ \text{*  }{{y}^{'}}\equal{}{{\cos }^{2}}\left( \minus{}x\plus{}y\plus{}a \right)\text{, }{{y}_{0}}\equal{}y\left( {{x}_{0}} \right)\text{, Put }u\equal{}\minus{}x\plus{}y\plus{}a\Rightarrow {{u}^{'}}\equal{}\minus{}1\plus{}{{y}^{'}}\]

\[ \Rightarrow {{y}^{'}}\equal{}{{u}^{'}}\plus{}1\Rightarrow {{u}^{'}}\plus{}1\equal{}{{\left( \cos u \right)}^{2}}\Leftrightarrow {{u}^{'}}\equal{}{{\left( \cos u \right)}^{2}}\minus{}1\equal{}\minus{}{{\left( \sin u \right)}^{2}}\]

\[ \Rightarrow \int\limits_{{{u}_{0}}}^{u}{\frac{du}{{{\left( \sin u \right)}^{2}}}}\equal{}\minus{}\int\limits_{{{x}_{0}}}^{x}{dx}\Rightarrow {{x}_{0}}\minus{}x\equal{}\int\limits_{{{u}_{0}}}^{u}{\left( {{\left( \cot u \right)}^{2}}\plus{}1 \right).du}\equal{}\minus{}\left[ \cot u \right]_{{{u}_{0}}}^{u}\equal{}\minus{}\cot u\plus{}\cot {{u}_{0}}\]

\[ {{\left( \cot u \right)}^{'}}\equal{}{{\left( \frac{\cos u}{\sin u} \right)}^{'}}\equal{}\frac{{{\left( \cos u \right)}^{'}}.\sin u\minus{}\cos u.{{\left( \sin u \right)}^{'}}}{{{\left( \sin u \right)}^{2}}}\equal{}\frac{\minus{}\left( {{\left( \sin u \right)}^{2}}\plus{}{{\left( \cos u \right)}^{2}} \right)}{{{\left( \sin u \right)}^{2}}}\equal{}\minus{}\left( 1\plus{}{{\left( \cot u \right)}^{2}} \right)\]

\[ x\equal{}\cot u\minus{}\cot {{u}_{0}}\plus{}{{x}_{0}}\equal{}\cot \left( \minus{}x\plus{}y\plus{}a \right)\minus{}\cot \left( \minus{}{{x}_{0}}\plus{}{{y}_{0}}\plus{}a \right)\plus{}{{x}_{0}}\]

\[ \text{Put }{{\text{t}}^{'}}\equal{}2{{y}^{'}}\Rightarrow t\equal{}2y\Rightarrow {{t}^{'}}\equal{}2{{\cos }^{2}}\left( \minus{}x\plus{}\frac{t}{2}\plus{}a \right)\]

\[ \Rightarrow x\equal{}\cot \left( \minus{}x\plus{}\frac{t}{2}\plus{}a \right)\minus{}\cot \left( \minus{}{{x}_{0}}\plus{}\frac{{{t}_{0}}}{2}\plus{}a \right)\plus{}{{x}_{0}}\text{,  Put }t\equal{}y\Rightarrow {{y}^{'}}\equal{}2{{\cos }^{2}}\left( \minus{}x\plus{}\frac{y}{2}\plus{}a \right)\]

\[ \Rightarrow x\equal{}\cot \left( \minus{}x\plus{}\frac{y}{2}\plus{}a \right)\minus{}\cot \left( \minus{}{{x}_{0}}\plus{}\frac{{{y}_{0}}}{2}\plus{}a \right)\plus{}{{x}_{0}}\]

\[ {{y}^{'}}\equal{}2{{\cos }^{2}}\left( \minus{}x\plus{}\frac{y}{2}\plus{}a \right)\]

\[ \Rightarrow x\equal{}\frac{1}{4}.\left( \log \left| \frac{1\plus{}\sin 2\left( \minus{}x\plus{}\text{y/2}\plus{}a \right)}{1\minus{}\sin 2\left( \minus{}x\plus{}\text{y/2}\plus{}a \right)} \right|\minus{}\log \left| \frac{1\plus{}\sin 2\left( \minus{}{{x}_{0}}\plus{}{{\text{y}}_{\text{0}}}\text{/2}\plus{}a \right)}{1\minus{}\sin 2\left( \minus{}{{x}_{0}}\plus{}{{\text{y}}_{\text{0}}}\text{/2}\plus{}a \right)} \right| \right)\plus{}{{x}_{0}}\]

\[ \Rightarrow \cot \left( \minus{}x\plus{}\frac{y}{2}\plus{}a \right)\equal{}\frac{1}{4}.\log \left| \frac{1\plus{}\sin 2\left( \minus{}x\plus{}\text{y/2}\plus{}a \right)}{1\minus{}\sin 2\left( \minus{}x\plus{}\text{y/2}\plus{}a \right)} \right|\text{  ?}\]

Even if your computations are correct (please excuse me, but I doubt this), you cannot claim this equality for [i]all[/i] $ x,y$.
You can only claim it (in fact, a slightly different equality with $ x_0,y_0$) for those $ x,y$ [i]which solve the resulting equation[/i].

UPD Of course, you are totally wrong with this conclusion. First part is ok. In second, you solve another differential equation
\[ {{y}^{'}} \equal{} 2{{\cos }^{2}}\left( \minus{} x \plus{} \frac {y}{2} \plus{} a\right)\]
and then you are trying equate the solution of this equation and something which is not related to the solution of original problem at all.

Pardon me, this is a complete nonsence.


For $2x\ge y$, $2y\ge z$ and $2z\ge x$, we can prove by Cauchy-Schwarz Inequality that
$\sum \frac{2x^{2}-xy}{x^{2}+y^{2}}\ge \frac 3{2}$.

I have found making
$\sum (2x^{2}-xy) (x^{2}+y^{2}) \sum \frac{2x^{2}-xy}{x^{2}+y^{2}}= (\sum (2x^{2}-xy))^{2}$ this expression

$\sum(x-y)^{2}(x^{2}+y^{2}+2z^{2}-3xy) \ge 0$

but I cannot go on :blush:

[quote="manlio"]For $2x\ge y$, $2y\ge z$ and $2z\ge x$,...
$\sum \frac{2x^{2}-xy}{x^{2}+y^{2}}\ge \frac 3{2}$.
[/quote]
I think, it's true for all $x,$ $y$ and $z$ distinct from zero.

Arqady, have you tried to solve it by Cauchy as Vasc said it is possible?

Many thanks for a help. :(

[quote="manlio"]Thank you very much, but I am not sure if you can choose $x\geq y\geq z$ as the inequality is only cyclic and not simmetric.

Please, can you make plausible $x\geq y\geq z$ ?[/quote]
$\sum_{cyc}(x-y)^{2}(x^{2}+y^{2}+2z^{2}-3xy)\geq0$ is symmetric. :wink:


Let $ \varphi:=\displaystyle \frac{1+\sqrt{5}}{2} $  and  $[\; \cdot \; ] $ denotes the integer part. If $ j:=\displaystyle
\left[\frac{n}{\varphi}\right] \; , \;  $  prove that
\[
\sum\limits_{k=0}^{n}{n\choose k}\varphi^{k}|k\varphi -n |\le 2(n-j){n
\choose j}\varphi^{j}\; ,\; \; \; n\in {\mathbb N}^* .
\]

This inequality is so beautiful. Have you come up with any solution?


Silviu poţi să foloseşti direct \lim fără \mathop{\lim} ;)

Cauta te rog in algebra solved problems. E problema pusa de Namdung, imi amintesc ca sigur am rezolvat-o simplu, dar nu mai stiu cum. Parca cu Cesaro-Stolz.

i) $\lim a_n=0. a_{n+1}-a_n<0, a_n\in(0,1)\Rightarrow$ exista $\lim a_n=r\Rightarrow r^2=0\Rightarrow r=0$.

ii) $\lim na_n=1\Leftrightarrow \lim \frac n{\frac 1{a_n}}=\lim \frac 1{\frac{a_{n}-a_{n+1}}{a_na_{n+1}}}=\lim \frac {a_{n+1}}{a_n}=\lim (1-a_n)=1$.

iii) $x_n=\sqrt{n}(na_n-1)$ are limita si aceasta este 0. Deorece $\lim na_n=1$ e de ajuns sa aratam ca $\lim \frac {\frac{1-na_n}{a_n}}{\frac 1{\sqrt{a_n}}}=0$ sau cu Cesaro-Stolz ca $\lim \frac {\frac {a_n}{a_{n+1}}-1}{\frac {\sqrt{a_n}-\sqrt{a_{n+1}}}{\sqrt{a_na_{n+1}}}}=\lim \frac {a_n-a_{n+1}}{\sqrt{a_n}-\sqrt{a_{n+1}}}=\lim (\sqrt{a_n}+\sqrt{a_{n+1}})=0$, deorece $\lim \frac {a_n}{a_{n+1}}=1$.

iv) $\lim\frac {n(1-na_n)}{\ln n}=1$. Cu Cesaro-Stolz aceasta limita este \[\lim \frac {1+n^2a_n-(n+1)^2a_{n+1}}{\frac 1n\ln (1+\frac 1n)^n}=\lim n\{[a_n(n+1)-1]^2+a_n\}=\lim na_n+(\lim \sqrt n(na_n-1)+\sqrt na_n)^2=1,\] folosind ii) de doua ori si iii) o data.

Incercati sa demonstrati ca $na_n$ este crescator  ;)


Julie baked cupcakes for her family at home and for a party at school. She iced 4 cupcakes with red frosting, 2 cupcakes with orange frosting, 2 with yellow, 2 with green, 3 with blue and the last 3 with violet frosting. Each cupcake is iced with exactly one color of frosting. Julie plans to take exactly 10 of the cupcakes to her party, and will take either all of the cupcakes of a particular color or none of the cupcakes of that color. How many different combinations of cupcakes could she take to her party?

http://www.mnspe.org/mathcounts/2007tests/510-07_S_Solutions.pdf

Go to number 24 on the sprint round

For those of you too lazy to look at the MathCounts link given above, the answer is [b]5[/b].


function
trigonometry
calculus
derivative
topology
real analysis
real analysis unsolved

is there any $ f: R\to R$, differentiable, such that $ f'$ is discontinuous on $ Q$ ?

Do you mean differentiable on the entire interval or at some point?  Also, does $ Q$ represent the rationals?  If so, I can't think of a such a function.  From wolfram: If  is differentiable at point $ x$, then it is also continuous at $ x$.  So therefore, I don't believe said function exists.

read more carefully ;) not $ f$ but $ f'$ should not be continuous !

1. $ f'(x)$ should exist for every real $ x$

2. the set of rationals $ \equal{}$ the set of points of discontinuity of $ f'$

Define $ s(x) \equal{} \begin{cases} x^2 \sin (1/x) & x \neq 0 \\
0 & x \equal{} 0 \end{cases}$ on $ ( \minus{} 1, 1)$. Then $ s$ is a differentiable function whose derivative is bounded and discontinuous at $ x \equal{} 0$. Extend $ s$ to the differentiable function $ \bar{s}$ defined on $ \mathbb{R}$ so that both $ \bar{s}$ and $ \bar{s}'$ are bounded on $ \mathbb{R}$ and $ \bar{s}'$ is discontinuous only at $ x \equal{} 0$.

Enumerate the rational numbers to write $ \mathbb{Q} \equal{} \{ r_1, r_2, r_3, \cdots \}$ and define $ f_n(x) \equal{} \frac {1}{2^n} \bar{s} (x \minus{} r_n)$. Also, let $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty} f_n (x)$.

By Weierstrass M-test, both $ \sum_{n \equal{} 1}^{\infty} f_n (x)$ and $ \sum_{n \equal{} 1}^{\infty} f'_n (x)$ converges uniformly on any bounded interval, so $ f$ is differentiable and $ f'(x) \equal{} \sum_{n \equal{} 1}^{\infty} f'_n (x)$. It is not hard to see that $ f'(x)$ is discontinuous at each rational point and continuous at each irrational point.

That looks like a solution pulled from Spivak's book. :wink:

To have a derivative continuous exactly on $ S \subset \mathbb{R}$ is possible iff $ S$ is dense and $ G_{\delta}$. The if part is done like sos440 does, using the Volterra functions, which are functions whose derivative is discontinuous exactly at a closed set without interior. The only if part are standard results which can be found in various threads on this forum


Complementary Counting Part 2

We tackle another challenging problem by counting what we don't want.

[quote="solved"]For some reason, the video is not loading correctly. I have watched some others and they are working fine. This one is not  :( .[/quote]

It works for me...must be you computer or internet connection...

Occasionally, a video won't load. You could try waiting or try accessing the video at another time.

Man... That's a really good way of thinking about it! I couldn't have figured that out by myself x_x

you could Go to the library and watch it...
ok i know thats a bad idea, try to go to a different comp?? :blush:

Did You see how he calculated 26^5-26*25^4!!! he done it mentally!! oh my gosh, i wish i could do that!!!!

i think he was looking at a calculator ?


This forum is dead. I'm not going to lie. I feel bad for the mods.

I'm gonna start a small competition, I only need like 5 people. I'll post 5 questions every 3-4 days, with different point values. Once I get enough people, I'll start posting questions. Don't worry, these will be MOEMS level.

Sample question:

What is the average of the exterior angles in a convex pentagon?

man thats easy
72

------------
James Zhang

most of them are going to be harder than that but whatever.

I hope this isn't old...

Sign me up! I'm ready *does Rikku's-ffx2;gunner;winning dance*

Well we have 5. More people can join along the way. Now for the 1st round, 30 points maximum.

(4) Find the sum of all numbers less than 100 that are divisible by 7.
(4) I. All crocodiles are scary.
    II. All scary creatures are always big.
    III. People are small.
Which of the following is true?
A. People are scary.
B. Crocodiles are not small.
C. Crocodiles are not scary.
D. People are not scary
(6) A circle is inscribed in a triangle. If D is a tangent point to the triangle, E is on both the triangle and circle, A is a vertex of the triangle, and AD = 6, find 2AE.
(7) At a party, there are 10 men and each brings their wife. There are only 20 people at the party. Each of the men shakes hand with each of the men with each of the women willing, other than their wife. Half of the women are not willing to shake hands with anyone. The other half shakes hands with every woman willing and every man other than their husband. How many handshakes take place?
(7) Brent has a conjecture. He thinks that if there is a vowel on one side of the card, there must be a composite number on the other. There are 6 cards. Each has one side showing. The letters/numbers showing are:

A, B, C, D, E, 1, 2, 3, 4, 5, 6. Which of these does he have to turn over to test his conjecture?

BONUS QUESTION, 5 POINTS.

(B5) How many factors does the number 360000 have?

i have a question on the second problem.  couldn't [hide]both b and d[/hide] be true?

also, on the one before the bonus, is there an F along with the other letters? and what do you mean by "test his conjecture?" couldn't he flip over any of them to test his conjecture? maybe it was something like "prove his conjecture."

this thing has me all confused. :(

[quote="jclarkemathL314159"]Each of the men shakes hand with each of the men with each of the women willing, other than their wife. Half of the women are not willing to shake hands with anyone. [/quote]
Do you mean with each of the men [i]and[/i] each of the women willing? or does the woman have to be willing in order for her husband to shake any hands?

Can I join? Is there an age limit? I'm going into high school next year. :?

[quote="lotrgreengrapes7926"][quote="Captain Sugar"]i have a question on the second problem.  couldn't [hide]both b and d[/hide] be true? [/quote]
Only one of the four choices is true.   ;) :D[/quote]

i know that, but it looks as if both of those choices could be true.

Oooh, I hope it's not too late to sign up...can I?  :oops_sign:

i mean, since all crocodiles are scary, and all scary things are big, crocodiles are not small (B). but since people are small, and all scary things are large, people are not scary (D). couldn't it be both? is it a trick question?  :huh:

erm..yesh

that first one is really confusing me and sending me off track with the others; hehe

Sorry, I can't sign up, I am certain not to make it into the next round.


function
algebra
polynomial
logarithms
number theory unsolved
number theory

For those who like challenges, here is a (possibly extremely difficult) conjecture that occured while I was conducting a research on the number of non-zero coefficients of certain polynomials. If true, this would imply certain very nice results  :) , so this problem is of a particular interest.

Here it is:

Let $x_1,x_2,\ldots,x_n \in \mathbb{R}$ be arbitrary real numbers. Let
$\delta (i,j) = 1$, if $i=j$, and $0$, otherwise, be the Kronecker delta function
 For $1 \leq i \leq j \leq n$, define $\lambda (i,j) = x_i + x_j + \delta (i,j)$. Prove or disprove that there always exist at least $n$ values $k \in \{2,3,\ldots,2n\}$ for which $\lambda (i,j)$ attains a unique maximum among all points $(i,j)$ with $i \leq j$ and $i+j=k$.

Yes, at the time I wrote this, I didn't know the fact that for a polynomial $f \in \mathbb{C}[x]$, the number of nonzero coefficients of $f^2$ may well be (much) less than the number of nonzero coefficients of $f$ --- and I wrote this "conjecture" in an attempt to approach the wrong assertion that $n(f^2) \geq n(f)$ (for, if the "conjecture" were true, then -- by considering the $2$-adic valuation, extended to $\mathbb{C}$, and applying it to the coefficients of $f$, it would easily follow that $n(f^2) \geq n(f)$ -- but, as we know, this is far from the truth: a result of Schinzel asserts that if $Q(n)$ is the minimal number of terms of $f^2$ for a polynomial $f$ with $n$ terms, then $\log\log{n} < Q(n) < n^{1-c}$ for some $c > 0$. But this is far from optimal...)

At least we know: if we denote by $q(n)$ the minimal possible number of indices $k$ with [...] (the described property), then we have the bound $Q(n) \geq q(n)$. This gives one approach for deriving a lower bound on $Q(n)$ -- by looking at $q(n)$.

--Vesselin


function
superior algebra
superior algebra unsolved

I think there exist $ a,b>0$ s.t. if $ G$ is a finite group with $ m$ equivalence classes, then $ |G|\leqslant a^{b^m}$.
Can we find a much better bound for $ |G|$?

Consider a finite group $ G$ acting on a set $ X$. The orbit of a point $ x$ in $ X$ is the set of elements of $ X$ to which $ x$ can be moved by the elements of $ G$.
The orbits are then the equivalence classes under this relation.

Can you phrase your question more precisely? If you mean that you have a group G, acting on some set X, with m orbits, then you cannot even bound |G| by m. For example you may take X to be a point set, and G acts trivially on X, so that m must be 1. Then G could be very large and this is all unrelated to m.

[quote="Soarer"]If you mean that you have a group G, acting on some set X, with m orbits, then you cannot even bound |G| by m. [/quote]
That's what I meant. I thought we could bound $ |G|$ as a function of $ m$.


probability
integration
function
real analysis
topology
real analysis unsolved

Let $(X, d)$ be a metric space and $P, Q: \mathcal B\to [0,1]$ two probabilities such that
\[\int fdP=\int fdQ, \]
for any function $f: X\to\mathbb{R}$ continuous and bounded.
  Prove that $P=Q$.

It's enough to prove that $P=Q$ on closed sets. For a closed set $A\subset X$ consider $f_{n}(x)=(1-n\,\mathrm{dist}(x,A))^{+}$ and apply Lebesgue domination.

You only need to show that the measures agree on a collection of sets which will generate the Borel sets. Closed, open whatever.

Then, pick some continuous functions that will approximate the characteristic functions of the sets you want, and apply some kinda convergence theorem. The given condition about the functions will tell you that the measure of those sets is the same.

I posted not too long ago a thread about two ($\sigma$-finite) Borel measures being the same if they agree on the open sets. The argument for closed sets would be similar. (In fact, since you are dealing with probability measures, $1-\ldots$)


pigeonhole principle
AMC
AIME
USA(J)MO
USAMO
articles
Putnam

What exactly is the pidgeonhole principle, can someone give me some sample hard AIME/ easy USAMO problems with this priciple (maybe 2-3)?

The pigeonhole principle (note the spelling), made famous by Dirichlet, says simply that if you have n things ("pigeons"), and k places to put them ("pigeonholes"), and n is greater than k, that at least one place will have more than one thing put in it. Once mathematicians get to work with this principle the "things" and "places" can get to be quite abstract, and the art of problem-solving includes recognizing when this common-sense principle applies. There are articles about Dirichet's pigeonhole principle on [url=http://www.cut-the-knot.org/do_you_know/pigeon.shtml]Cut the Knot[/url] and on [url=http://mathworld.wolfram.com/DirichletsBoxPrinciple.html]Mathworld[/url].

Hope this helps!

A problem from the Putnam a couple of years back:

Prove that, given 5 points on the surface of a sphere, there exists a closed hemisphere which contains at least 4 of them.

Here's another pigeonhole problem.

The prime factorizations of [tex]r+1[/tex] positive integers ([tex]r \ge 1[/tex]) together involve only [tex]r[/tex] primes. Prove that there is a subset of these integers whose product is a perfect square.

[quote="Ragingg"]And a more basic one....

Prove that if you have 5 points in a square with side length 2, the greatest distance between any two points is sqrt(2)[/quote]

try again.  there is a legitimate problem hidden in here somewhere, but what you've written is untrue when taken literally

Should it be...

If 5 points are chosen on a square of side length 2, there exist 2 points whose distance is less than or equal to $\sqrt{2}$.

Osiris wrote:If 5 points are chosen on a square of side length 2, there exist 2 points whose distance is less than or equal to .

[hide]Draw two lines connecting the midpoints of the opposite sides of the square. This divides the square into 4 smaller congruent squares with side lengths of 1. By the Pigeonhole Principle, at least two of the points must be in one of the four squares. The farthest distance apart these two points can be is on opposite ends of a diagonal of the square. Because the diagonal has a length of , they can be no father than  apart.[/hide]

[hide=pigeonhold principle]
if we have n items and k boxes where n > k, at least 1 of the boxes must contain more than 1 item
[/hide]

[quote=tigeryong]this title is spelt wrong.

the farthest that 2 points could be is sqrt2 bc they would be on a diagonal of the square[/quote]

The square has side length $2$ and diagonal $2\sqrt{2}$. If you mean the diagonal of the quarter-square, that's somewhat non-rigorous until formalized with pigeonhole principle.
[hide=solution]
Cut the $2\times2$ square into four $1\times 1$ squares. By pigeonhole principle, there must be at least two points in the same $1\times 1$ square. Since the longest distance between two points in a square is the diagonal, we can say that there exists at least two points whose distance is at most $\sqrt{2}$. This is still not as rigorous as possible for points that lie exactly on the square borders, but it's mostly fine.
[/hide]


The Baseball All-Star Game's tonight! Who does everyone think will win?
I think the National League will win, even though I want the American League to win.

[quote=mikimoto12][quote=mathleticguyyy]The American League beat the National League 8-6[/quote]

2004 all star not 2018[/quote]

yeah lel

The american usually win. Everyone was getting false hopes for the national league.


[b]Problema B.4174:[/b] $ \mbox{ Rezolvati in multimea numerelor reale sistemul: } \{\begin{array}{c} 4a + bc = 32 \\
2a - 2c - b^2 = 0 \\
a + 12b - c - ab = 6. \end{array}$

[b]Problema B.4175:[/b]
$ \mbox{Fiind date 4 puncte coplanare A, B, C si D, aratati ca daca cercurile ABC si ABD sunt ortogonale, }$
$ \mbox{atunci cercurile ACD si BCD sunt deasemenea ortogonale.}$

[b]Problema B.4176:[/b]
$ \mbox{Rezolvati ecuatia: } (\sin{x} + \sin{2x} + \sin{3x})^2 + (\cos{x} + \cos{2x} + \cos{3x})^2 = 1.$

[b]Problema B.4177:[/b]
$ \mbox{Tangentele in B si C la cercul circumscris triunghiului ABC se intersecteaza in punctul M. Paralela}$
$ \mbox{dusa prin M la AB intersecteaza pe AC in N. Sa se arate ca: } |AN| = |BN|.$

[b]Aceste probleme, participa la concursul revistei. Solutiile lor putand fi trimise pe adresa revistei pana pe 15.05.2009.
Pentru a nu prejudicia acest concurs, va rog sa nu postati nici o solutie pana la expirarea termenului de mai sus!!![/b]

[b]Please do not discuss [url=http://www.komal.hu/verseny/feladat.cgi?a=honap&h=200904&t=mat&l=en]active problems[/url].[/b]
(The deadline is 15 May.)


floor function
number theory unsolved
number theory

1)consider the minimal element. $ \rightarrow$ $ 0\in A$
2) every even number is in $ A$
3) to get $ 2n\plus{}1\in A$ consider $ \lfloor\frac{(2n\plus{}1)a}{a}\rfloor\in A$ if $ a$ is even or $ \lfloor\frac{(2n\plus{}1)a\plus{}1}{a}\rfloor\in A$ otherwise.

[quote="outback"]Let $ a$ be a natural number greater than $ 1$.
Consider a nonempty set $ A$ of natural numbers such that if $ k\in A$ then $ k \plus{} 2a \in A$ and $ \lfloor \frac {k}{a}\rfloor \in A$.

Prove that $ A$ is the set of natural numbers itself.[/quote]

This is a problem on Mathematics and Youth Magazine.

Suppose that the given set $ A$ is not the set of natural number $ \mathbb{N}$, so there must be some natural numbers which does not belong to $ A$. Let $ m$ denote the smallest number of them then $ m<2a$ (since $ m$ is the smallest natural number not belong to $ A$, if $ m\geq 2a$ then $ (m\minus{}2a)\in A$ which derived in $ m\equal{}(m\minus{}2a)\plus{}2a\in A$, contradiction)

By the way, we have $ m\minus{}1\in A$ so $ (m\minus{}1)\plus{}2ka \in A, \forall k\in\mathbb{N}$.
$ \Rightarrow\lfloor\frac{m\minus{}1\plus{}2ka}{a}\rfloor\in A,\forall k\in\mathbb{N}$

But $ m<2a\Rightarrow m\minus{}1<2a\minus{}1$ and $ \lfloor\frac{m\minus{}1\plus{}2ka}{a}\rfloor\equal{}\lfloor\frac{m\minus{}1}{a}\rfloor\plus{}2k\in A,\forall k\in\mathbb{N}$ so we just need to consider two cases:

1. $ \lfloor\frac{m\minus{}1}{a}\rfloor\equal{}1\Rightarrow2k\plus{}1\in A,\forall k\in\mathbb{N}$.
Thus, $ 2a\plus{}1\in A$ and we obtain $ \lfloor\frac{2a\plus{}1}{a}\rfloor\equal{}2\in A$ (because $ a>1$).
Finally, $ 2\in A$ yields $ \lfloor\frac{2\plus{}2ka}{a}\rfloor\equal{}2k\plus{}\lfloor\frac{2}{a}\rfloor\equal{}2k\in A,\forall k\in\mathbb{N}$.
Therefore, since $ A$ contains every even numbers and odd numbers, it is the set of natural number $ \mathbb{N}$.

2. $ \lfloor\frac{m\minus{}1}{a}\rfloor\equal{}0$, similarly we get what desired.

End of proof.


Hi.
Iranian olympiad second round took place on April 26,27.Here are the problems .

First Day

Problem 1
 Let n $n$ be a natural number ,$n>1$ and p be a prime number such that $n \mid p-1 , p \mid n^3-1$.
Prove that $4p-3$ is a perfect square .

Problem 2
 Let $ABC$ be a triangle , $A=60$. Let $D$ be transitive on $BC$. Let $O_1,O_2$ be the circumcenters of triangles $ABD,ACD$ respectively. Let $M$ be the intersection of $BO_1,CO_2$ and $N$ be the circumcenter of $DO_1O_2$ . Prove that the line $MN$ passes through a fixed point .

Problem 3
 A galaxy has more than one million stars . We write the distance between any pairs of them .
Prove that we have at least $79$ distinct numbers . (you can spot  star a point ).

Problem 4
We have a $2*n$ table and there is at least $2^n$ stones in some of its $2n$ squares. In each step if there was a square such that there were at least $2$ stones we can delete $2$ of its stones and add $1$ stone to the square which is above it or add $1$ stone to the square which is in right it. Prove that we can do this steps finite such that at last there will be at least $1$ stone in the square which is in last column and above row.

Problem 5
Let $BC$ be the diagonal of circle $C$. And $XY$ be a cord of $C$ and be perpendicular to $BC$. Let $P,M$ be two points on $XY,CY$ (or their extensions) respectively such that :
$PB||CY,MP||CX$ . Let $K$ be the intersection of $PB,CX$. Prove that $PB\perp MK$.

Problem 6
Find all $F : R^+ -\to R^+$ such that :
$(x+y)f(f(yf(x)))=(x^2)f(f(x)+f(y))$.

[quote="kvmedya"]Hi.
Iranian olympiad second round took place on May 4,5.Here are the problems .[/quote]
 :huh: What do you mean?

yeeh What do you mean ?? they took place on 4,5 may ?? Are you sure ??  :?
P.S. : The first problem was given in the Morrocan pre TST. Have a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202788#202788]here[/url]

Maybe he means last year's second round?  :?  :?  Omid, can you help us clear this issue?

Actually the exam was held on 26,27 April this year (2005).
Note that this round is a part of selection for IMO 2006.

Could somebody post a solution to the geometry problem 2, please? The sketch of my DynaGeo-program suggests that all lines pass through the midpoint of arc $BC$. But how to prove this  :?

[quote="kvmedya"]Problem 3
 A galaxy has more than one million stars . We write the distance between any pairs of them .
Prove that we have at least $79$ distinct numbers . (you can spot  star a point ).[/quote]
Yeah, I know $10^6$ is a huge constant here. So problem is quite easy.

Suppose the contrary.
Take two points $A$ and $B$. Then there are at most $78^2$ distinct pairs $(AC,BC)$, there $C\neq A,B$. It means that there are at least $\frac{10^6-2}{78^2}\approx 164$ points $C$ with the same pair of numbers $(AC,BC)$. All this points lies on the circle, but it is clear that 164 points on a circle determines at least 81 distinct distances. Contradiction.

[quote="kvmedya"]Problem 2
 Let $ABC$ be a triangle , $A=60$. Let $D$ be transitive on $BC$. Let $O_1,O_2$ be the circumcenters of triangles $ABD,ACD$ respectively. Let $M$ be the intersection of $BO_1,CO_2$ and $N$ be the circumcenter of $DO_1O_2$ . Prove that the line $MN$ passes through a fixed point .[/quote] Nice! It is really not difficult, but I like it.

We have $A)_1=DO_1$ and $AO_2=DO_2$, so $\triangle AO_1O_2=\triangle DO_1DO_2$. Moreover, by easy angle chasing we see that $\angle O_1AO_2=60^\circ$. It follows that $\angle O_1NO_2=120^\circ$, therefore $N$ lies on the circumcircle of $AO_1O_2$, and $NO_1=NO_2$. It means, $N$ is a midpoint of the arc $(*)$.
Also by simple angle chasing we obtain $\angle BMC=60^\circ$. Thus $M$ lies on the circumcircle of $AO_1O_2$ and $ABC$. Using $(*)$ we conclude $MN$ is a bisector of $\angle BMC$, so it intersect the circumcircle of $ABC$ in the middle of the arc $BC$.

first of all we prove that it is an injective function.then if put Y=x 2-x the we have( x 2 -x-1) *f(x)=f(x 2 -x) then if put x=(1+ :sqrt: 5)/2 we have 0=f(1) then it is a contradiction becuase we have F:R+----->R+ and 0 is not in R+


Let Q be the rationals. Find all functions $f : Q \rightarrow Q$such that $f(x + f(y) ) = f(x) + y$ for all $x, y$

[hide]First substitute in $x=0$ to get $f(f(y))=y+f(0)$. So $f(f(f(y)))$ is equal to both $f(y+f(0))$ and $f(y)+f(0)$, by using the substitution on first the inner $f$'s, then the outer ones.

But $f(y+f(0))=f(y)$ so $f(y)=f(y)+0\Rightarrow f(0)=0$.

Now substituting $x=-f(y)$, $0=f(-f(y)+f(y))=f(-f(y))+y\Rightarrow f(-f(y))=-y$. Applying $f$ to both sides, and since $f(f(x))=x$, $f(-y)=-f(y)$.

Another property of $f$ is $f(nx)=nf(x)$ for $n\in \mathbb{N}$ which is shown by induction. The base case, $n=1$, is clearly true. For the inductive step, $f((n+1)x)=f(nx+f(f(x)))=f(nx)+f(x)=nf(x)+f(x)=(n+1)f(x)$.

Let $f(1)=C$. Then $f(n)=Cn$ for positive integers $n$. For positive rational $x$, write $x=p/q$ where $p$ and $q$ are positive integers. Then $Cp=f(p)=f(qx)=qf(x)\Rightarrow f(x)=Cx$.

For negative rational $x$, write $y=-x$ so $f(x)=f(-y)=-f(y)=-Cy=Cx$. So $f(x)=Cx$ for all rationals.

Substituting back into the original equation, $f(x+f(y))=f(x+Cy)=Cx+C^2y$. Since this must equal $Cx+y$, $y=C^2y\Rightarrow C=\pm 1$.

This leaves the functions $x$ and $-x$, which both satisfy the equation. So these are the only two functions.[/hide]

Art of Problem Solving has a chapter in I believe Volume 2 all about this...


function
induction
graph theory
combinatorics proposed
combinatorics

Find the minimal $k$ (as a function of $n$) s.t. any graph with $n$ vertices and $k$ edges is hamiltonian.

i think this is due to Ore, not Erdos...

i don't remember the answer but i think it is $\frac{n^2-3n+6}{2}$.it is easy to see that this much is necessary;take a $K_{n-1}$ and add one extra edge from the $n^{th}$ vertex to it-this graph has (obviously) no hamiltonian cycle and has $\frac{n^2-3n+4}{2}$ edges.

Thanks. I saw it listed as being proposed by Erdos in Math. Magazine, but it was on a website, so I don't really know how reliable the source was :).

We can prove sufficiency by induction on $ n $. $ n = 3 $ is trvial; the only possible such graph being $ C_3 $. Now, any graph $ G $ of $ n $ vertices containing at least $ \binom{n}{2} - (n-3) $ edges must contain a vertex $ v $ adjacent to at least $ n - 2 $ vertices.

Case 1: $ d(v) = n - 1 $
If $ G - v $ is complete, then it must contain a hamiltonian path; otherwise, add one edge $ e $ to $ G - v $. By the induction hypothesis, $ G - v + e $ must contain a hamiltonian cycle, so $ G - v $ contains a hamiltonian path. Then adding back $ v $ extends this path to a hamiltonian cycle of $ G $.

Case 2: $ d(v) = n - 2 $
$ G - v $ must contain a hamiltonian cycle of length $ n - 1 $. Since $ n \geq 4 $, $ v $ must be adjacent to 2 vertices adjacent in the cycle, so we can extend it to a hamiltonian cycle of $ G $.

Alternatively, we can use the following result (also by Ore I think): A graph $ G $ is hamiltonian iff $ G + uv $ is hamiltonian, where $ u, v $ are non-adjacent vertices of $ G $ such that $ d(u) + d(v) \geq n $, the order of $ G $. Here, all such pairs of vertices obey this property, so adding all the edges gives us a complete graph.


ratio
geometry
3D geometry
MATHCOUNTS
factorial

This contest is only for the STUDENTS of the MOEMS forum (absolutely NO ONE over 6th grade). ALSO: NO CALCULATORS. Thank you for participating in the MYTEST314.

1. The ratio of boys to girls in a class of 35 students is 3:4 How many more girls than boys are in the class?

2. When four standard dice are rolled, what us the most likely sum of the numbers?

3. If F is defined by the set {2, 3, 4, 5, 9}, and N is defined by the set {480, 1260, 1890, 3888}, which number from set N is divisible by all 5 numbers in set F?

4. What is the 100th term in the sequence 2, 3, 5, 7, 10, 11, which consists of all the numbers that are neither perfect squares nor perfect cubes?

5. Anny, Bob, Carol, Dan, and Edith each have one kind of pet. They're five pets are a ferret, gerbil, hamster, iguana, and jackrabbit, not necessarily in that order. In how many ways can the five kids be paired with the five animals?

[b][i][u][color=red]Post your answers as a thread to this topic, and I will pm you your answers.[/color][/u][/i][/b]

err isnt MOEMS 7th and 8th too?
you probably wont find too many <=6th graders on AoPS

Well...this isn't exactly from MOEMS. This is from the MATHCOUNTS web, it's certain problems that I thought would be easy for 6th. If i had 7th and 8th doing this, it would be a problem because they would be taking the opportunity away from the little guys. If it should be for 7th and 8th, I will post some harder problems and then release the new test.

5. [b]120[/b]

you do 5 factorial, if andy takes one pet then there is only 4 left for bob, so you multiply 5 by 4, if bob takers a pet there is only 3 left for Carol, if carol takes a pet there are only 3 left for Dan, and so on.... so you multiple 5*4*3*2*1, since each kid has its own chance to get a pet

[quote="kishiman"]1. 3/4 =x/35
multiply 3 by 35
then divide by four
to get 26.25, you must round to 27
so it would be 27:35
[b]there are 8 more girls than boys[/b][/quote]
There are 35 total students. You should never get a fractional number of students.  ;)  :D

[quote="kishiman"]1. 3/4 =x/35
multiply 3 by 35
then divide by four
to get 26.25, you must round to 27
so it would be 27:35
[b]there are 8 more girls than boys[/b][/quote]

that proportion would be $\frac{3}{7}=\frac{x}{35}$

[hide="Number 1"]The answer to number 1 is 5.[/hide]

[hide="Number 2"]The answer to number 2 is 14.[/hide]

[hide="Number 3"]The answer to number 3 is 1260.[/hide]

[hide="Number 4"]The answer to number 4 is 112.[/hide]

[hide="Number 5"]The answer to number 5 is 120.[/hide]

Oh, oops, were 8th graders not allowed to do this?  Darn it.

Hide your answers, I was puzzling them over for days and now I look through this topic and all the problems are spoiled :(

And if you don't hide them, can you at least provide a solution? They're still puzzling me :(

I hope you get the solutions, and I hope they don't puzzle you.

I already hid them, sorry.

Even though I'm in seventh grade, I'm 11 years old.
You don't have to look at my solutons.

[hide="1."]We divide 35 by 7 to get the number girls more than boys. [b]7[/b][/hide]

[hide="3."]We find the numbers that are divisible by 9: 3888, 1890, 1260
We find the numbers divisible by 4: 3888, 1260. The only one divisible by 5 is [b]1260[/b][/hide]

[hide="5."]There are 5 choices for the first pet, 4 choices for the second, 3 for third, 2 for fourth, 1 for 5th. That's 5!=[b]120[/b][/hide]


function
trigonometry
calculus
integration
algebra
domain
Ring Theory

Let R be the ring $ \mathbb{R}[\sin x, \cos x]$ that is generated over $ \mathbb{R}$ by the two trigonometric functions in the ring of functions $ f: \mathbb{R} \to \mathbb{R}$.

1. Prove that $ R = \{a_{0} + \sum_{m=1}^{n} \left(a_{m}\cos mx + b_{m} \sin mx \right): n \in \mathbb{Z}_{n \geq 1}, a_{0}, a_{m}, b_{m} \in \mathbb{R} \}$.

2. Consider $ f \in R - {0}$ and write $ f = \{a_{0} + \sum_{m=1}^{n} \left(a_{m}\cos mx + b_{m} \sin mx \right)$. Define $ \deg_{tr}(f): = \max{i}$ and $ j$ such that $ a_{i}, b_{j} \neq 0 \}$.
Show that $ \deg_{tr}(f)$ is well-defined and that $ \deg_{tr}(fg) = \deg_{tr}(f) + \deg_{tr}(g)$ for any $ f, g \in R - {0}$.

3. Prove that R is an integral domain that is not a UFD.

The first two are routine.  For the third, write $ R \simeq \mathbb{R}[x, y]/(x^2 \plus{} y^2 \minus{} 1)$.  Now $ x^2 \plus{} y^2 \minus{} 1$ is irreducible, so this is an integral domain.  To show it's not a UFD look at examples of non-unique factorization like $ \cos^2 x \equal{} (1 \minus{} \sin x)(1 \plus{} \sin x)$ and show that the factors involved are irreducible.  It might help to think about the homomorphism $ R \to \mathbb{R}(t)$ given by $ x \equal{} \frac{1 \minus{} t^2}{1  \plus{} t^2}, y \equal{} \frac{2t}{1 \plus{} t^2}$.


Vieta
number theory proposed
number theory

Find all positive integers $ (a,b,c)$ satisfying
\[ 4a^{3}+b+c=4abc+2a\]

It's an interesting problem .I solved it and I used the result of Pro5 IMO 2007.
                   $ 4a^{3}+b+c = 4abc+2a$
                  $ \Leftrightarrow 4a ( 4 a^{3}+b+c ) = 4a ( 4abc+2a)$
                  $ \Leftrightarrow (4a^{2}-1)^{2}= (4ab-1) (4ac-1)$
          hence $ ( 4a^{2}-1)^{2}$ is divisible by $ (4ab-1)$ .  Easily,hence $ a=b=c$
 This problem will be harder if we don't know IMOProblem. :)

yes it's trivial through IMO 2007 and the proof of $(4ab-1)|(4a^2-1)^2$ follows vieta's jumping.

Ну с положительными разобрались. Интересно как решения вообще выглядят.
 Оказываются так: Для уравнения $4a^3+b+c=4abc+2a$

    $a=\frac{-p+1}{2}$

     $b=\frac{p+1}{2}$

     $c=\frac{p-3}{2}$

   Где $p$ любое не чётное число.


linear algebra
matrix
superior algebra
superior algebra solved

$A\in M_n(R)$ symetric such that $A^5+A^3+A=3I_n$

Prove that $A=I_n$

It seems quite easy...

if $A$ is symetric, we can always find a base $\mathcal{B}$ of $M_n(\mathbb{R})$ so that $A_{\mathcal{B}}$ is a diagonal matrix.

Let $C = A^5 + A^3 + A$, let $c_i = c_{ii}$ a generic element of $C$ main diagonal ($C $ is obviously diagonal)
Let $a_i = a_{ii}$ a generic element of $A_{\mathcal{B}}$ main diagonal

Thus we have: $c_i = a_i^5+a_i^3+a_i = 3 \;\; \forall i=1..n$
$\Rightarrow a_i \in \{x | x$ is a solution of $x^5 + x^3 + x - 3 = 0\}$

${d \over {dx}} (x^5 + x^3 + x - 3) = 5x^4 + 3x^2 + 1$ which has no real solutions. Thus we have an only real solution for $x^5 + x^3 + x - 3 = 0$: $x = 1$

$\Rightarrow a_i = 1 \;\; \forall i = 1..n \;\; \Rightarrow A = I_n$

(I hope^^)


AMC
AIME
algebra
polynomial
Vieta
quadratics
special factorizations

Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square.

[quote="tµtµ"]Hint :

[hide]For $n > 18$,  $(n-9)^2 < n^2-19 \cdot n+99 < (n-10)^2$

$1,9,10,18$[/hide][/quote]

I don't understand the hint. Isn't that statment false for numbers greater than 18?

No.  The whole point is:  For $\ n > 18$ we $\ n^2 - 19n + 99$ cannot be a perfect square because it lies in between two consecutive perfect squares.  So we limit our guessing to $\ n \leq 18$

[quote="Gyan"]FWIW, Another standard way to do these kind of problems ...
(Compelete the square etc..)
[hide]  $(n^2-19n+99)= k^2 \implies 4n^2-4\cdot 19 n +4\cdot 99= 4k^2$
or  $(2n-19)^2 + 35 = (2k)^2$
$(2k+2n-19)(2k-2n+19) = 35$
And the problem reduces to find the factors of 35.
$35 = (\pm 7) \cdot (\pm 5)$   or $35 = (\pm 35) \cdot (\pm 1)$  etc
You get $2n - 19 = \pm 1$ and $2n-19 = \pm 17$
So the answers are $(19 \pm 1) /2$ and $(19 \pm 17) /2$ etc..
[/hide][/quote]

I am sorry that I am slow, but what does FWIW mean?

Google says it means "For what it's worth".

Yes, FWIW meant "For what it's worth" as in "Hope this information is useful" (to  at least a few people) as the method is generally well known or can  be seen in a book.

Once again, I am really slow, what does IIRC mean?  The only four letter math abbreviation I know of is WLOG, which is Without Loss of Generality.  Right?

The method I used isnt here yet so might as well post.

[hide="Solution"]
Let $n^2-19n+99=p^2$ where $p$ is a positive integer. Then $n^2-19n+(99-p^2)=0$. By Vietas, the sum of roots is $19$ and the product is $99-p^2$. $n$ must be a positive integer, so we have $9$ pairs of integers that sum to $19$ to test. We make an organized table like the following,

\[
\begin{tabular}{|l|l|l|l|}
\hline
\textbf{Pair} & \textbf{99-p\textasciicircum 2} & \textbf{p\textsuperscript{2}} & \textbf{Y/N} \\ \hline
1,18          & 18                              & 81                           & Y            \\ \hline
2,17          & 34                              & 65                           & N            \\ \hline
3,16          & 48                              & 51                           & N            \\ \hline
4,15          & 60                              & 39                           & N              \\ \hline
5,14          & 70                              & 29                           & N             \\ \hline
6,13          & 78                              & 21                           & N             \\ \hline
7,12          & 84                              & 15                           & N              \\ \hline
8,11          & 88                              & 11                           & N          \\ \hline
9,10          & 90                              & 9                            & Y            \\ \hline
\end{tabular}
\]

We find the product, and see if $p^2$ is a perfect square. We find that $n=1,18,9,10$ are the only values, so our some is $\boxed{038}$ as desired.[/hide]

The value for (7, 12) in p^2 should be 15, not 25. That might confuse some people :p

My solution:

We have $ n^2-19n+99 =m^2$, where $m$ is a positive integer. Rearranging gives us
\[n^2-19n+(-m^2+99)=0.\]
Applying the quadratic formula yields
\[n=\frac{19 \pm \sqrt{361-4(1)(-m^2+99)}}{2}.\] Now, we simplyfy:
\[n=\frac{19 \pm \sqrt{4m^2-35}}{2}.\]
In order for $n$ to be an integer, the discriminant $4m^2-35$ must be a perfect square. In other words,
\[4m^2-35=t^2,\]
where $t$ is a positive integer. Rearranging gives us
\[4m^2-t^2=35.\] Aha! Difference of squares.
\[(2m+t)(2m-t)=35.\]
Remember that both of these factors are integers, so we have a very limited amount of choices. Either
\[2m+t=35 \\ 2m-t=1\] or \[2m+t=7 \\ 2m-t=5\]
These two systems yield $m=9$ and $m=3$. Plugging $m$ back in gives us $n=9$, $n=10$, $n=18$, $n=1$. Therefore, our answer is $9+10+18+1=\boxed{38}$

[quote=4everwise][hide="Eqnarray is cool!"]We know that $n^{2}-19n+99=m^{2}$ for some positive integers $m$ and $n.$ Completing the square, \begin{eqnarray*}n^{2}-19n+99 &=& n^{2}-19n+\frac{361}{4}+\frac{35}{4}\\ &=& \left(n-\frac{19}{2}\right)^{2}+\frac{35}{4}\\ &=& m^{2}\implies \left(m+n-\frac{19}{2}\right)\left(m-n+\frac{19}{2}\right)= \frac{(5)(7)}{2^{2}}. \end{eqnarray*} We now know that $\left(m+n-\frac{19}{2}, m-n+\frac{19}{2}\right)=\left(\frac{1}{2}, \frac{35}{2}\right),\left(\frac{35}{2}, \frac{1}{2}\right),\left(\frac{5}{2}, \frac72 \right),$ or $\left(\frac72, \frac52 \right).$

This yields our solutions as $n=1,9,10,18,$ so the sum is $38.$ [/hide][/quote]

How did you factor that like so?

[b]The simplest solution we can find if we use[/b] $(n-5)^2<=n^2-19n+99<=(n-4)^2$, etc.

[quote=kylie123-]This is same as what @4everwise did but just multiplying by 4
$n^2+19n+99=k^2$
Multiply by 4 on both sides
$4n^2+76n+396=4k^2$
So $4n^2+76n+361+35=4k^2$
$(2n+19)^2+35=4k^2$
$35=(2k-2n-19)(2k+2n+19)$
So finding positive solutions
We get $n=9,10,18,1$
So sum is 38[/quote]

Is this a common factoring trick?
$35=(2k-2n-19)(2k+2n+19)$

[quote=Mathscienceclass]How do you know to multiply by 4 without guess and check?[/quote]
That is the smallest number that guarantees that once you complete the square, you get nice integer numbers.

We can write $n^2-19n+99 = p^2$, or $n^2-19n+(99-p^2) = 0$ which must have integer solutions for $n$. The discriminant is $361 - 4(99-p^2) = 4p^2 - 35$ which must be a perfect square, so we can write $4p^2 - 35 = x^2$ or alternatively $(2p-x)(2p+x) = 35$. Now, we split into cases.

Case 1: $1 \cdot 35$. Then we either have $p = 9, x = 17$ or $p = 9, x = -17$. This gives $n = 1,18$.

Case 2: $5 \cdot 7$. Then we have $p = 3$, which gives $n = 9,10$.

Case 3: $-1 \cdot -35$. Then we get $p = -9$, same as Case 1.

Case 4: $-5 \cdot -7$. Then we get $p = -3$, same as Case 2.

The solutions are $1,9,10,18$. The sum is $\boxed{38}$.


When I compile a LaTeX document that includes labels (yes, I remembered to compile it twice), my references turn out fine, but the labels on the equations are missing.

Note: when I removed the section headings, the number in the reference disappeared also.

Attached is a minimal example.

Using \ [...\ ] never gives an equation number just a displayed equation. To do that you need \begin{equation}...\end{equation}
[code]\begin{equation}
f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}h\label{tan}
\end{equation}[/code]

It compiles fine now. Thanks for your help.