function
calculus
derivative
inequalities proposed
inequalities

Let $ a,\ b,\, c$ be positive real numbers such that $ a \plus{} b \equal{} 1,\ a^3 \plus{} b^3 \plus{} c^3 \equal{} 1$.
Find the maximum value of $ a^2 \plus{} b^2 \plus{} c^2$.

hello kunny, from $ b\equal{}1\minus{}a$ and $ a^3\plus{}(1\minus{}a)^3\plus{}c^3\equal{}1$ we get the function
$ f(a)\equal{}2a^2\minus{}2a\plus{}1\plus{}\left(\sqrt[3]{3a\minus{}3a^2}\right)^2$. Further we have
$ f(0)\equal{}f(1)\equal{}1$ and $ a_{max}\equal{}1$. So we get our searched maximum as
$ \frac{1}{2}\plus{}\frac{1}{4}\sqrt[3]{9}\sqrt[3]{4}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]
$ f(0) \equal{} f(1) \equal{} 1$ and $ a_{max} \equal{} 1$. So we get our searched maximum as
$ \frac {1}{2} \plus{} \frac {1}{4}\sqrt [3]{9}\sqrt [3]{4}$.[/quote]
hello Dr Graubner, could you please explain this part more clearly?

Hello, sorry, it was a typo, it must be $ a_{max}\equal{}\frac{1}{2}$. Was this your question?
Sonnhard.

[quote="Dr Sonnhard Graubner"]Hello, sorry, it was a typo, it must be $ a_{max} \equal{} \frac {1}{2}$. Was this your question?
Sonnhard.[/quote]
Hello, how do I know that $ f$ is maximum when $ a\equal{}\frac{1}{2}$

hello, thank you for your question. The first derivative of $ f(a)$ is:
${ f^'(a) = 4a - 2 + 3^{\frac {2}{3}}\frac {2}{3}\frac {1 - 2a}{\sqrt [3]{a - a^2}}}$.
Now we have $ f^'(a) = 0 \Leftrightarrow$:
${ \frac {1 - 2a}{3^{\frac {1}{3}}\sqrt [3]{a - a^2}}} = 1 - 2a$.
By solving this equation we get the desired solution.
Sonnhard.

$ a \plus{} b \equal{} 1 \Rightarrow a^3 \plus{} b^3 \minus{} 1 \equal{} \minus{} 3ab$ or $ a^3 \plus{} b^3 \equal{} 1 \minus{} 3ab$

$ \therefore \ a^3 \plus{} b^3 \plus{} c^3 \equal{} 1 \Rightarrow c^3 \equal{} 3ab$ or $ c \equal{} \sqrt [3] {3ab}$

Now, $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1 \minus{} 2ab \plus{} (3ab)^{\frac {2}{3}}$

Let $ \sqrt [3] {ab} \equal{} x$ We are asked to maximise $ f(x) \equal{} 1 \minus{} 2x^3 \plus{} 3^{\frac {2}{3}} x^2$

This is an increasing function when $ 0 < x < \frac {1}{\sqrt [3] 3}$
Remember that the maximum value of ab is $ \frac{1}{4}$. Hence, $ \sqrt [3] {ab} < \frac {1}{\sqrt [3] 4} < \frac {1}{\sqrt [3] 3}$

So, it is an increasing function which attains max when $ a \equal{} b \equal{} \frac {1}{2}$ and hence $ c \equal{} \sqrt [3] {\frac {3}{4}}$

The maximum value attained is hence $ \frac {1}{2} \plus{} \left( \frac {3}{4} \right)^{\frac {2}{3}}$


find the exact value of:
$ \tan ^2 1\plus{}\tan ^2 2 \plus{} \tan ^ 3\plus{}...\plus{}\tan ^2 89$
$ \sin ^2 1\plus{}\sin ^2 2 \plus{} \sin ^ 3\plus{}...\plus{}\sin ^2 89$
$ \tan (1000!)$

[quote="mathson"]find the exact value of:
$ \tan ^2 1 \plus{} \tan ^2 2 \plus{} \tan ^ 3 \plus{} ... \plus{} \tan ^2 89$
$ \sin ^2 1 \plus{} \sin ^2 2 \plus{} \sin ^ 3 \plus{} ... \plus{} \sin ^2 89$
$ \tan (1000!)$[/quote]
You mean
$ \tan ^2 1 \plus{} \tan ^2 2 \plus{} \tan ^ 2 3 \plus{} ... \plus{} \tan ^2 89$
$ \sin ^2 1 \plus{} \sin ^2 2 \plus{} \sin ^ 2 3 \plus{} ... \plus{} \sin ^2 89$?

$ \tan 1000!\equal{}\tan 180n$, where $ n$ is a large positive integer, so $ \tan 1000!\equal{}0$.

sin(x)=cos(90-x)
sin^2(x)=cos^2(90-x)

so

sin^2(1)+sin^2(89)=1
sin^2(2)+sin^2(88)=1
......
sin^2(45) does not hav a pair, but we know its value

the sum of all the sines is 1*44+sin^2(45)=89/2


Έστω $ x\equal{}y\plus{}d$ με $ d\geq 1$ τότε $ (3d\minus{}1)y^2\plus{}(3d\minus{}1)yd\plus{}d^3\equal{}61$

Πρέπει $ d^3<61$ άρα $ d\equal{}1,2,3$ από τις οποίες μόνο μία δίνει y φυσικό , η d=1 και τότε χ=6 , y=5 and done

[quote="maniopas"]Να λύσετε την εξίσωση $ x^3 \minus{} y^3 \equal{} xy \plus{} 61$ για φυσικούς αριθμούς $ x, y$.
(15η Πανενωσιακή Ολυμπιάδα, 1981)

Είναι ενδιαφέρουσα, αν και κάπως εύκολη.[/quote]

EULER

ti paei na pei EULER re paidia????? :oops:

Προφανως χρησιμοποιοι την ταυτοτητα EULER:$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)$.

[quote="silouan"]Πράγματι βγαίνει και με τον Euler . Απλά πολλαπλασιάστε με το 27 και τα δύο μέλη . (Αν και νομίζω ότι η λύση που έβαλα είναι αρκετά ικανοποιητική  :P )[/quote]

Η λύση που έβαλες είναι η καλύτερη δυνατή. (Την ίδια έκανα πάνω - κάτω κι εγώ.  :lol: )

Ο άλλος όμως τρόπος (πολλαπλασιασμός με 27 και μετά Euler) γίνεται:
$ x^3 \minus{} y^3 \equal{} xy \plus{} 61 < \equal{} > 27(x^3 \minus{} y^3) \equal{} 27xy \plus{} 61\cdot 27 < \equal{} >$
$ (3x)^3 \plus{} ( \minus{} 3y)^3 \equal{} 3(3x)( \minus{} 3y)(1) \plus{} 1647 < \equal{} >$
$ (3x)^3 \plus{} ( \minus{} 3y)^3 \plus{} 1^3 \minus{} 3(3x)( \minus{} 3y)(1) \equal{} 1647 \plus{} 1 < \equal{} >$
$ (3x \minus{} 3y \plus{} 1)(9x^2 \plus{} 9y^2 \plus{} 1^2 \minus{} (3x)( \minus{} 3y) \minus{} 3x \minus{} ( \minus{} 3y)) \equal{} 1648 < \equal{} >$
$ (3x \minus{} 3y \plus{} 1)(9x^2 \plus{} 9y^2 \plus{} 9xy \minus{} 3x \plus{} 3y \plus{} 1) \equal{} 2 \cdot 2 \cdot 2 \cdot 2 \cdot 103$

Μετά, αν εξετάσω τα δυνατά συστήματα γίνεται χάος (βγαίνει, αλλά θέλει πράξεις με κάπως μεγάλους αριθμούς και πολλές περιπτώσεις). :o

Μήπως είναι δυνατή κάποια διαφορετική παραγοντοποίηση (εφαρμόζοντας πάντα Euler), με την οποία η λύση βγαίνει πιο γρήγορα και απλώς δεν τη βλέπω;  :huh:

[quote="maniopas"][quote="silouan"]Πράγματι βγαίνει και με τον Euler . Απλά πολλαπλασιάστε με το 27 και τα δύο μέλη . (Αν και νομίζω ότι η λύση που έβαλα είναι αρκετά ικανοποιητική  :P )[/quote]

Η λύση που έβαλες είναι η καλύτερη δυνατή. (Την ίδια έκανα πάνω - κάτω κι εγώ.  :lol: )

Ο άλλος όμως τρόπος (πολλαπλασιασμός με 27 και μετά Euler) γίνεται:
$ x^3 \minus{} y^3 \equal{} xy \plus{} 61 < \equal{} > 27(x^3 \minus{} y^3) \equal{} 27xy \plus{} 61\cdot 27 < \equal{} >$
$ (3x)^3 \plus{} ( \minus{} 3y)^3 \equal{} \boxed{3(3x)( \minus{} 3y)(1)} \plus{} 1647 < \equal{} >$
$ (3x)^3 \plus{} ( \minus{} 3y)^3 \plus{} \boxed{1^3 \minus{} 3(3x)( \minus{} 3y)(1)} \equal{} 1647 \plus{} 1 < \equal{} >$

Μήπως είναι δυνατή κάποια διαφορετική παραγοντοποίηση (εφαρμόζοντας πάντα Euler), με την οποία η λύση βγαίνει πιο γρήγορα και απλώς δεν τη βλέπω;  :huh:[/quote]

maniopa, όταν πας απ'την πρώη στη δεύτερη γραμμή κάνεις ένα λάθος! Πρέπει να είναι -1 ο παράγοντας κι όχι 1.
Ετσι βγαίνει:
$ (3x)^3 \plus{} ( \minus{} 3y)^3 \plus{} (\minus{}1)^3 \minus{} 3(3x)( \minus{} 3y)(\minus{}1) \equal{} 1647 \minus{} 1\equal{}1646\equal{}2*823$ (δεν είμαι σίγουρος αλλά μου φαίνεται ότι το 823 είναι πρώτος)

Σωστά! (Ευχαριστώ για τη διόρθωση. Θα έσπαγα ακόμα το κεφάλι μου να βρω άλλη παραγοντοποίηση.  :blush:)  Άρα βγαίνει (για να τελειώσω αυτόν τον τρόπο λύσης):

$ (3x \minus{} 3y \minus{} 1)(9x^2 \plus{} 9y^2 \plus{} 9xy \plus{} 3x \minus{} 3y \plus{} 1) \equal{} 2 \cdot 823 < \equal{} >$
$ 3x \minus{} 3y \minus{} 1 \equal{} 2 , 9x^2 \plus{} 9y^2 \plus{} 9xy \plus{} 3x \minus{} 3y \plus{} 1 \equal{} 823 < \equal{} >$
$ y \equal{} x \minus{} 1, 9x^2 \plus{} 9y^2 \plus{} 9xy \plus{} 3x \minus{} 3y \plus{} 1 \equal{} 823 \equal{} >$
$ 9x^2 \plus{} 18x \plus{} 9 \plus{} 9x^2 \plus{} 9x^2 \plus{} 9x \plus{} 3x \plus{} 3 \minus{} 3x \plus{}1 \equal{} 823 < \equal{} >$
$ 18x^2 \plus{} 27x \minus{} 810 \equal{} 0 < \equal{} > 2x^2 \plus{} 3x \minus{} 90 \equal{} 0$
$ D \equal{} 9 \plus{} 8 \cdot 90 \equal{} 729 \equal{} 27^2 > 0$
άρα
$ x \equal{} \frac{\minus{}3 \minus{} 27}{4} \equal{} \minus{}\frac{15}{2} < \equal{} >$ άτοπο
ή
$ x \equal{} \frac{\minus{}3 \plus{} 27}{4} \equal{} 6 < \equal{} > y \equal{} 5$  :D


Identify compounds A to I in the following sequence of rns  :o

[hide="Compound B"]3-Phenylpentan-2-ol.[/hide]

[hide="Compound C"]3-Bromo-3-phenylpentan-2-ol.[/hide]

Can you tell me the exact function of 9-BBN??(is it something similiar to hydroboration?)  :maybe:  :maybe:

9-BBN stands for 9-borabicyclononane, and yes, it is a selective hydroboration agent (because it is a large molecule). Look [url=http://en.wikipedia.org/wiki/9-Borabicyclononane]here[/url].

[hide="Compound D"]3-Phenylpentan-2,3-diol.[/hide]

[hide="Compound E"]2-Phenyl-2-methylbutanal.[/hide]

[hide="Compound J"]The epoxide.[/hide]

[hide="Compound K"]3-Phenylpentan-3-ol.[/hide]

[quote="Carcul"]
 K=3-Phenylpentan-3-ol[/quote]

Why not 3-Phenylpentan-2-ol ?  :maybe:  :maybe:

[quote="Carcul"]
E=2-Phenyl-2-methylbutanal [/quote]

Can you explain how you got E from D?  :maybe:

According to me it is pinacol-pinacolone rearrangement  :idea:

So E might be 2-phenyl-pentan-3-one?  :maybe:

Please correct me if I am wrong!  :(

also, can you tell me what is the function of NH2OH(in the conversion from K to L) here? i will try to complete the prob. then!  :blush:  :maybe:  :ninja:

[quote="Da Vinci"]Why not 3-Phenylpentan-2-ol?[/quote]

Because hydride reduction of an epoxide is an SN2 reaction, with the nucleophile (hydride) attacking the less crowded carbon.

Yes, from D to E I supposed a pinacol rearrangement. Look at the mechanism of this rearrangement: its much more likely that it will be the hydroxy of carbon 3 to leave as a molecule of water.

[quote="Da Vinci"]also, can you tell me what is the function of NH2OH (in the conversion from K to L) here?[/quote]

I have no idea for the moment. But this reaction scheme looks already strange, so everything is possible.

[quote="Carcul"]
Yes, from D to E I supposed a pinacol rearrangement. Look at the mechanism of this rearrangement: its much more likely that it will be the hydroxy of carbon 3 to leave as a molecule of water.
[/quote]

I protonated the hydroxy at C-2 position because the migratory aptitude of phenyl>methyl  :D  :P

But the insipient carbocation is more stable for C3.

In my opinion there are two oxidation steps missing on this reaction scheme. This because:

1) as far as I know, alcohols do not react with hydroxylamine.

2) reaction with hydroxylamine followed by acidification suggests the Beckmann rearrangement, whose starting compound is a ketone.

Why not E= phenyl sec-butyl ketone ? It seems by far the most stable carbocation with OH on one side and phenyl on the other? (and ethyl gp migrating)

That doesn't make sense, as a phenyl group has superior migratory aptitude compared with ethyl.

[quote="Carcul"]
2) reaction with hydroxylamine followed by acidification suggests the Beckmann rearrangement,[b] whose starting compound is a ketone[/b].[/quote]

Starting compound can be aldehyde also  :D  :D   :maybe:

[quote="Carcul"]Yes it can, but hydrogen almost never migrates. What I wanted to say was that the starting compound cannot be for sure an alcohol.[/quote]

Why should the hydrogen migrate?  :maybe:

The OH group on nitrogen is anti to the bulky group(because of steric hinderance). :maybe:  :maybe:


hello, i have got $ x\equal{}y\equal{}z\equal{}1$;$ x\equal{}\minus{}1/2,y\equal{}\minus{}2,z\equal{}1$;$ x\equal{}\minus{}\frac{1}{1\plus{}z},y\equal{}\minus{}\frac{1\plus{}z}{z}$
Sonnhard.

[quote="luiseduardo"]Resolve the question in Real Numbers:

[img]http://img687.imageshack.us/img687/9803/cals.jpg[/img][/quote]
[url=http://www.servimg.com/image_preview.php?i=430&u=12899977][img]http://i99.servimg.com/u/f99/12/89/99/77/almonk12.gif[/img][/url]

Let $ x \equal{} \frac {a}{b}$, $ y \equal{} \frac {b}{c}$, and $ z \equal{} \frac {c}{a}$, so $ m \equal{} \frac {a \plus{} c}{b} \equal{} \frac {a \plus{} b}{c} \equal{} \frac {b \plus{} c}{a} \equal{} \frac {(a \plus{} c) \plus{} (a \plus{} b) \plus{} (b \plus{} c)}{b \plus{} c \plus{} a} \equal{} 2$ if $ a \plus{} b \plus{} c\ne0$.* Then $ a \plus{} c \equal{} 2b$ and $ a \plus{} b \equal{} 2c$, so $ c \minus{} b \equal{} 2b \minus{} 2c\implies b \equal{} c\implies a \equal{} 2b \minus{} c \equal{} b \equal{} c$, and we have $ \boxed{(x,y,z) \equal{} (1,1,1)}$.

*If $ a \plus{} b \plus{} c \equal{} 0$ then this is equivalent to $ m \equal{} \frac {a \plus{} c}{b} \equal{} \frac { \minus{} b}{b} \equal{} \minus{} 1$. Hence we can let $ \boxed{(x,y,z) \equal{} \left(\frac {a}{b}, \minus{} \frac {b}{a \plus{} b}, \minus{} \frac {a \plus{} b}{a}\right)}$ as long as $ a,b,a \plus{} b\ne0$, which satisfies the original conditions.


inequalities
blogs
inequalities proposed

Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. If $ 0\le k \le 2,$ then
$ \left( \frac {a}{b \plus{} kc}\right)^2 \plus{} \left(\frac {b}{c \plus{} kd}\right)^2 \plus{} \left(\frac {c}{d \plus{} ka}\right)^2 \plus{} \left( \frac {d}{a \plus{} kb}\right)^2 \ge \frac {4}{(1 \plus{} k)^2}.$

[b]Remark. [/b]This inequality does not hold for $ k>2.$

After some days, I will publish the solution on my blog: http://can-hang2007.blogspot.com/2010/01/inequality-45-v-q-b-can.html

I am waiting for someone to post their nice solutions.

[b]Is it true the next statement ?[/b] :)

For $ x,y,z,t$ positive real numbers, and $ 0\leq k\leq 2$, the following inequality holds

$ \frac{x}{y\plus{}kz}\plus{}\frac{y}{z\plus{}kt}\plus{}\frac{z}{t\plus{}kx}\plus{}\frac{t}{x\plus{}ky}\geq\frac{4}{1\plus{}k}$

I have proved the above inequality when $ 0\leq k\leq 1$, but I don't think it's true when $ 1\leq k\leq2$.

[quote="marin.bancos"][b]Is it true the next statement ?[/b] :)

For $ x,y,z,t$ positive real numbers, and $ 0\leq k\leq 2$, the following inequality holds

$ \frac {x}{y \plus{} kz} \plus{} \frac {y}{z \plus{} kt} \plus{} \frac {z}{t \plus{} kx} \plus{} \frac {t}{x \plus{} ky}\geq\frac {4}{1 \plus{} k}$[/quote]
Dear marin.bancos, I aslo think the same as BakerBakura. This inequality does not holds for $ 1<k \le 2.$ Indeed, taking $ x\equal{}z\equal{}1$ and $ y\equal{}t \ne 1,$ the above inequality becomes
$ \frac{x}{y\plus{}kx}\plus{}\frac{y}{x\plus{}ky} \ge \frac{2}{k\plus{}1},$
or
$ \frac{(x\minus{}y)^2(1\minus{}k)}{(k\plus{}1)(x\plus{}ky)(y\plus{}kx)} \ge 0,$
which is clearly false for $ k>1.$

See my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-45-v-q-b-can.html


calculus
integration
calculus computations


The right triangles $ AFC,AFG$ have also $ \angle CAF\equal{}\angle GAF$ and $ AF$ common. So they 're congruent, which yields $ AC\equal{}AG$ and $ FC\equal{}FG$

So the triangle $ ACG$ is isoceles $ \angle COG \equal{} 36^{\circ}\Rightarrow \angle CGA \equal{} 72^{\circ}$

But $ \angle GOC \equal{} 72^{\circ}\Rightarrow  COG$ is isosceles $ (CO\equal{}CG\equal{}R)$

[img]10431[/img]

On the other hand, $ \angle OCE \equal{}\angle OCB\plus{}\angle BCD \equal{} 54\plus{}18\equal{}72^{\circ}$

So the triangle $ ECO$ is isosceles too $ (EC\equal{}EO)$

Now, the triangles $ ECO,ACG$ are congruent (they're similar with equal bases).
So $ OE\equal{}AG$ and finally $ EG\equal{}OA\equal{}R$


At my school we can take honors classes if we are recommended by a teacher.  This year, out of Math, Physics, English, German, and History, i was recommended to all but english.  I was just wondering if that was good or not.  How many honors courses are you guys taking?

Next year I think that all of my high-school courses (orchestra, english, history, french, physics) will be either honors or AP.  This is more-or-less what I did this year, and it went fairly well.  I think that whether you choose to take honors classes or not depends mostly on the amount of work you are willing to do--do you really [i]want[/i] to work harder and cover topics faster?

My school ran out of math classes, so I'm taking math at the local college (probably Calc III and Intro to Abstract Algebra).

well, in my school, we have GATE(gifted and talented education) classes and its basically same as honors course, although they are only for math and english classes.
To take honors course or gate class (in my case), we have to take assessment test....
and I didn't do very well on that test and didn't get into english honors course.
Luckily, we have an alternate way which is to get parent's permission and get into honors class..

Does your school have alternate way?

My school has no alternate way.  We just have to get recommended by the teacer.


A billiard ball is hit at a 45-degree angle from a corner of a 4 foot by 7 foot billiards table. How many times will the ball rebound off an edge of the table before landing in a corner?

[hide="answer"]
they're relatively prime, so...
7+4-2
9
[/hide]

[quote="JesusFreak197"]Actually, the answer should be $\boxed{7}$. Obviously, it will go four feet up or down for every four feet left or right. This means that you simply need $LCM(7,4)=28$, $\frac{28}{4}=7$ rebounds and it will end up in a corner again.[/quote]

The answer sheet i have says 9  :(
does anyone have a solution for $\boxed{9}$?

Oh, sorry, I didn't quite catch the question. I was answering something else. :roll:

Treething's answer could use a little more explanation, though. I wasn't taking into account the rebounds off of the far edges of the table (the length $4$ edges), so that would add $3$ rebounds, and then you don't count the endpoints, so then you only have $6$. Therefore, $6+3=9$.

If they are relatively prime, then it is always the sum -2
If it isn't divide out the common factors and then the sum-2 is the answer


I think whenever the GameBot posts FTW and Alcumus problems, it should include the answer with it. That way, the people who are posting the solution would know whether their solution was right and will be, in turn, verifying the answer of the GameBot.

Dumb people like me get scared whenever I post a solution on Alcumus or FTW that it is incorrect and I am teaching the asker incorrect processes.

I think all would benefit if an answer was included when the GameBot posts a problem.

[quote="akhil0422"]

Dumb people like me get scared whenever I post a solution on Alcumus or FTW that it is incorrect and I am teaching the asker incorrect processes.
[/quote]

But other people read the forum too, and they can correct you if you make a mistake. Anyway, if you're really uncertain about whether your solution is correct or not, put something such as "I'm not sure if this is right" at the top.

I have question about GameBot.

What is his purpose there? Has it ever repeated a problem?

No. When you ask a question that has already been asked, it directs you to the old thread. Its purpose is to ask the problems you don't understand on FTW..

...and, I will ask again, since it wasn't fully answered the previous time I asked, why in the world does this robot have an e-mail address?

Each account needs to have a email address, even when the admins create it for bots.

For example, the accounts that are used in the AoPS Classroom Preview (ClassroomUser and Classroom Teacher) have email accounts.  (ClassroomUser has an "email" of dummy@spamherelots1314141241.com and Classroom Teacher has one of teacher@thisisnotmyrealemailwee.com).

Does GameBot have a password?  Can you actually log onto [i]its[/i] account?

What's the point of hacking into GameBot anyway? You don't get extra priviliges and even if you do (priviliges that I don't know of) then the admins will see something wrong right way (GameBot posting in the MOEMS forum, maybe?) and delete GameBot and create a new problem-posting bot like BotGame (I dunno; it was random).

Admins can change the password of any account :wink:


inequalities
geometry
inequalities proposed

Find the maximal area of a triangle with the following property:

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

nice ineq but easy,hehe
with the property,we have $a+b+c \leq 3$
the rest is easy.

Zhaobin,how did you prove that $ a+b+c<=3$? I proved it with Chebysev's inequality.

just prove:$3(a^3+b^3+c^3) \geq (a+b+c)(a^2+b^2+c^2)$
yes,Chebysev's inequality is one of method.
but there is some other ways.
for example:power inequality.


geometry
trapezoid
analytic geometry
inequalities

I have a very ugly (and straightforward) solution to this problem, using coordinates and a lot of messy algebra.

Any nice geometric way to do this?

Let $ABCD$ be a trapezoid whose area is $32$ square meters. Suppose the lengths of the parallel segments $AB$ and $DC$ are $2$ meters and $6$ meters, respectively, and $P$ is the intersection of the diagonals $AC$ and $BD$. If a line through $P$ intersects $AD$ and $BC$ at $E$ and $F$, respectively, determine, with a proof, the minimum possible area for quadrilateral $ABF E$.

[quote="white_horse_king88"]A geometric thought.

[hide] I'll attempt to show the minimum area of $ABFE$ is achieved at $AB\parallel EF\parallel DC$.

Let's draw $GH\parallel AB\parallel DC$ for the sake of argument.

Draw a line parallel to $BC$ from $G$ to $G'$ on side $DC$, and another from $H$ to $H'$ on $DC$ parallel to $AD$.

I think it suffices to show that $[ABFE]$'s area is minimized when $EF=GH$.

For $EF\not= GH$, note that all possible segments $EF$ intersect $GG'$ or $HH'$.

If $EF$ intersects $GG'$ at some point $E'$, we see that $GPE' \cong HPF$. So $[GPE']+[GE'E] > [HPF]$, and thus $[ABFE] >[ABGH]$.

The same is true if $EF$ intersects $HH'$ at $F'$. $HPF' \cong GPE$, so $[HPF']+[HF'F] > [GPE]$, and thus $[ABFE] > [ABGH]$.

So for the case $EF = GH$, we find the area to be $5$. Maybe.

[/hide][/quote]

Wow, that is smart.

I wish I thought of that in the real test.

I said something like that though, but not as well as you did.

blah...extend $EF$ to meet $DC$ at $G$ (wlog beyond C) and denote $\frac{BF}{FC}=x$ and $\frac{EA}{DE}=y$

menelaus on transversal FGP on triangle BCD: $\frac{BF}{FC}*\frac{CG}{GD}*\frac{DP}{PB}=1$

menelaus on transversal PGE on triangle ACD: $\frac{AP}{PC}*\frac{CG}{GD}*\frac{DE}{EA}=1$

This implies $9xy=1$ The area of the quadrilateral is $[APD](\frac{x}{x+1}+\frac{y}{y+1})+[APB]$ where $xy=\frac{1}{9}$, good ol' inequalities finish this...min when $x=y$, giving $5$ as the answer


A leaf is torn from a paperback novel. The sum of the numbers on the remaining pages is $15000$. What are the page numbers on the torn leaf?

Whats a leaf?  Do you mean a page, but then you used the word 'page' elsewhere ...

[quote="mathmanman"]Yes, I think he meant a "page"...

[hide="hint"] Use : $1+2+...+n = \frac{n(n+1)}{2}$[/hide]

:)[/quote]

mathmanman please tell how to solve it.
Because the formula you told is known to us. How to apply it here?

[hide] Let $k$ be the smaller number on the page we remove, and $n$ be the total number of pages. Then we're given that $n^2/2 + n/2 - k - (k+1) = 15000$. Lets just pick first $n^2 + n > 30000$, which would probably be like $n = 173$ ($\sqrt{3} \approx 1.732$). That gives $2k + 1 = 51$, so $k = 25.$ So the two page numbers are 25 and 26.[/hide]

Oh
now I get it. It is just maths :roll:

they at least have to give the number of pages in the book.

[hide="Solution"]If there are $ n$ pages in the book, then clearly some pages $ k\minus{}1$ and $ k$ are removed, where $ k\minus{}1\ge1$ and $ k\le n$. Clearly, then, we have:

$ (n)(n\plus{}1)/2\minus{}(n)\minus{}(n\minus{}1) \le 15000 \le n(n\plus{}1)/2\minus{}(2)\minus{}(1)$

and furthermore, there is some integral k in the interval $ (1,n]$ such that $ (n)(n\plus{}1)/2\minus{}(k)\minus{}(k\minus{}1)\equal{}15000.$ From the inequality, we have:

$ n^2\plus{}n\minus{}4n\plus{}2 \le 30000 \le n^2\plus{}n\minus{}3$

$ (n\minus{}2)(n\minus{}1) \le 30000 \le (n\minus{}2)(n\plus{}3)$

The first inequality is $ n^2\minus{}3n\minus{}29998 \le 0$ which has solutions in the positive reals $ 0 < n \le \frac{3\plus{}\sqrt{120001}}{2}$ or in the positive integers $ 1 \le n \le 174$. Similarly, the second inequality ($ n^2\plus{}n\minus{}30003 \ge 0$) has solutions $ n \ge \frac{\minus{}1\plus{}\sqrt{120013}}{2}$ or $ n ge 173$. Together, we have $ 173 \le n \le 174$. So there are either $ 173$ or $ 174$ pages in the book.

If there are 173 pages, then $ (173)(174)/2\minus{}(k)\minus{}(k\minus{}1)\equal{}15000 \Rightarrow 51 \equal{} 2k\minus{}1 \Rightarrow k\equal{}26$.
If there are 174 pages, then $ (174)(175)/2\minus{}(k)\minus{}(k\minus{}1)\equal{}15000 \Rightarrow 225\equal{}2k\minus{}1 \Rightarrow k\equal{}113$.

Indeed, the two solutions are the two ordered pairs $ (25,26)$ and $ (112,113)$, corresponding to the books containing $ 173$ and $ 174$ pages, respectively.Unfortunately, there is a dependence on how the pages are numbered (If the back of the front cover is page $ 1$, then the only possible solution is $ (112,113)$, because a single physical page, save the first, includes an even page number and an odd page number, in that order. If the page immediately following the front cover is denoted page $ 1$, then the only possible solution is $ (25,26)$, by the same logic.   :wink: [/hide]

[quote="aadil"]why are you complicating your proof.it is quite simple[/quote]

The initial inequality (both sides) is necessary to encompass the statement of the problem. I then solved both inequalities precisely (instead of using $ \sqrt{30000}\approx173$ and plugging in, although this can be made rigorous too) for full rigor. When I got the final "Diophantine inequality," I did casework to solve for $ (n,k)$. The final explanation was necessary to explain the parity of the two ripped-out pages. None of the steps were extraneous (except maybe factoring :D), but no steps were skipped. It wasn't very complicated...

[quote=murgi][quote="palluchaw"]If n=174 ( no. of pages in the book) then the number on the torn leaf will be 112 and 113.[/quote]

post ur solution.....I got a answer 25 and26[/quote]

My proof also gets $112$ and $113$.

There will be an even number of total pages since for every page there will be a back. Using $n(n+1)>2*15000=30000$, we get n is at least $74$. This totals $n(n+1)/2$  to be $15225$, or the sum of the front and back of the leaf is $225$. Let $x$ be the number on the front. Then $x+x+1=2x+1=225$; or x=$112$. So the numbers will be $112$ and $113$.

Of course, if you assume $n$ can be odd or $n=73$, then I also get 25 and 26. It is just the assumption.

Let the page numbers go from 1 till n; And let pages {k,k+1} be removed.
Start by searching near misses of
n(n+1)/2 ~ 15000;
i.e. n^2 ~ 30000;
By which we get n ~ 100*sqrt(3) ~ 173
Trying n=173 in S=n(n+1)/2, we get S=15051.
This is 51 more than 15000.
Solving for k in
k+(k+1)=51
We get k=25
Hence, Pages 25 and 26 are torn from the 173-page book.

Note that n=174 does also work, then k=112;

But practically in a Book, there are even number of pages; Since each sheet has 2 pages, the number of pages would be 2*(Number of sheets), which is even.
So, the torn pages are 112 and 113 torn from a 174-page book.


induction
ratio
inequalities
limit
real analysis
real analysis unsolved

$ x_{n \plus{} 1} \equal{} 2 \minus{} \frac {1}{x_n}$

I'm trying to do this by induction, I can actually show this for the case k=n. I cannot however do it for the case =1.

Also $ x_1 > 1$

Can anyone help me show $ x_1$[u]>[/u]$ x_2$?

For $ n\equal{}1$, you say that $ x_{1}>1$.  By definition,
$ x_{2}\equal{}2\minus{}\frac{1}{x_{1}}\equal{}\frac{2x_{1}\minus{}1}{x_{1}}$

To show that the sequence is decreasing (or do you mean non-increasing because the $ \geq$ means non-increasing), you can show that the ratio of two successive terms is less than or equal to 1.

$ \frac{x_{2}}{x_{1}}\equal{}\frac{2x_{1}\minus{}1}{x_{1}}\cdot\frac{1}{x_{1}}\equal{}\frac{2x_{1}\minus{}1}{x^{2}_{1}}$

For the ratio to be less than 1, $ x^{2}_{1}>2x_{1}\minus{}1\Rightarrow x^{2}_{1}\minus{}2x_{1}\plus{}1>0$.  Solving,
$ x_{1}>\frac{2\pm\sqrt{4\minus{}4}}{2}\equal{}1$ so $ x_{1}>1$,
which is what we want.  So $ (x_{n})$ is decreasing.

But then again, I could be completely wrong.

Could someone help me show that 1 is the infinum of this set?

$ x_{n\plus{}1} \minus{} x_n \equal{} 2\minus{}x_n\minus{}\frac{1}{x_n}$

does this factor in a useful way?

$ x_n(x_{n \plus{} 1}\minus{}x_n) \equal{} (x_n \minus{} 1)^2$

I don't see how it's useful though

$ y_n\equal{}x_n\minus{}1>0$

$ y_{n\plus{}1}\plus{}1\equal{} 2\minus{}\frac{1}{y_n\plus{}1}\equal{}\frac{2y_n\plus{}1}{y_n\plus{}1}\equal{}1\plus{}\frac{y_n}{1\plus{}y_n}$

$ y_{n\plus{}1}\equal{}\frac{y_n}{1\plus{}y_n}<\max (y_n,1)$

$ f(x) \equal{} 1\minus{}\frac{1}{1\plus{}x}, f'(x) \equal{} \frac{1}{(1\plus{}x)^2} > 0, \forall x$

so $ x<y \to f(x) < f(y)$

let $ y_2' \equal{} 1 > y_2, y_{n\plus{}1}'\equal{}\frac{y_n'}{1\plus{}y_n'} \to y_n' > y_n$

claim: $ y_n' \equal{} \frac{1}{n\minus{}1}$

$ y_{n\plus{}1} '\equal{} 1\minus{} \frac{1}{1\plus{}\frac{1}{n\minus{}1}}\equal{}1\minus{}\frac{n\minus{}1}{n}\equal{}\frac{1}{n}$

since y_2'=1, induction gives us the claim, we can use this along with the inequality so that

for any $ \epsilon > 0, \exists N > \frac{1}{\epsilon} \implies$

$ n>N \to y_n'<\epsilon \to y_n < \epsilon \to x_n < 1\plus{}\epsilon$

which makes 1 your inf

The sequence $ x_n$ is converges to $ 1$.


Let a,b,c be positive real numbers prove that
$ \frac{(a\minus{}b)(a\minus{}c)}{(b\plus{}c)^{2}}\plus{}\frac{(b\minus{}c)(b\minus{}a)}{(c\plus{}a)^{2}}\plus{}\frac{(c\minus{}a)(c\minus{}b)}{(a\plus{}b)^{2}}\geq0$

Because the inequality is homogenous we can assume that $ a\plus{}b\plus{}c\equal{}1$ ,and consider the function $ f: (0,1)\to\mathbb R$ ,$ f(x)\equal{}\frac{1}{(1\minus{}x)^{2}}$ ,$ f$ is convex.
Our inequality is equivalent to :
$ \sum f(a)(a\minus{}b)(a\minus{}c)\geq 0$ ,but is true from Vornicu-Schur inequality.

[quote="Svejk"]Because the inequality is homogenous we can assume that $ a\plus{}b\plus{}c \equal{} 1$ ,and consider the function $ f: (0,1)\to\mathbb R$ ,$ f(x) \equal{}\frac{1}{(1\minus{}x)^{2}}$ ,$ f$ is convex.
Our inequality is equivalent to :
$ \sum f(a)(a\minus{}b)(a\minus{}c)\geq 0$ ,but is true from Vornicu-Schur inequality.[/quote]
My solution is the same as yours  :wink:  :P


i'm really stuck here.  pls factor this out, and give me a clear explanation on how you got the answer..

a4  + 6a2b2 +25b4

thanks for the vivid explanation!  keep it up!


group theory
abstract algebra
combinatorics proposed
combinatorics

Nine cities are connected with one-way roads. Each city has exactly 3 outgoing roads which enter into other 3 cities and maybe some ingoing roads. Prove that it is possible to make a circular route which passes through at most 3 cities in accordance with traffic regulations.

In graph-theoretical language, we are given a digraph (directed graph) on $ 9$ vertices, with all outdegrees equal to $ 3$. Since the sum of the indegrees is equal to that of the outdegrees (each outgoing road from some city is an ingoing road for some other city), there must exist (at least) a city, say $ X$, with indegree at least $ 3$. Let those cities with outgoing roads towards $ X$ be $ A_i$, $ i \equal{} 1,2,3,\ldots,k \geq 3$. If any of the $ 3$ outgoing roads from $ X$ leads to some $ A_i$, we have a circuit $ XA_iX$ of length $ 2$, so assume all of them lead to cities $ B_j$, $ j\equal{}1,2,3$. Now, if any of the $ 3$ outgoing roads from $ B_j$ leads to $ X$, we have a circuit $ XB_jX$ of length $ 2$, while if it leads to some $ A_i$ we have a circuit $ XB_jA_iX$ of length $ 3$. But there are at most two cities left, say $ Y$ and $ Z$, not enough so that all the $ 3$ outgoing roads from $ B_j$ lead to them, therefore for each $ B_j$ there is (at least) one outgoing road leading to some other $ B_k$. This generates (at least) a circuit of length $ 2$ or $ 3$ among the $ B$-cities.

I am presenting yet another problem on such digraphs. [color=blue]Prove that from any group $ G$ of $ n \geq 4$ cities, each of them having at least $ 3$ outgoing roads to different cities in $ G$ (digraph with outdegrees at least $ 3$), one may choose a subgroup $ S$ of $ 4 \leq m \leq n$ cities, each of them having at least $ 3$ outgoing roads to different cities in $ S$ (a subdigraph with outdegrees at least $ 3$), and such that from any city in $ S$ one can reach any other city in $ S$, passing through cities of $ S$ only, in accordance with traffic regulations (strongly connected).[/color]


ARML
calculus
integration
geometry
rectangle
analytic geometry
graphing lines

[quote="ComplexZeta"]You didn't like question #6 on the actual contest?[/quote]

What do you mean?

These questions are from ARML 2001, but there are only 7 of them because question 6 from the actual contest is missing.

Ah, okay, got it.  This number 8 seems so easy now ...

Yeah, it does. At the time I thought it was the most ridiculous question ever (since I couldn't imagine how one could just figure it out), but now (possibly as a result of showing it to people regularly since June 2001) it seems utterly trivial to me.

7. (actually 8.)

We have

[tex]\displaystyle \frac{1}{f(x)} = \left(\frac{1}{1+x}\right)\left(\frac{1}{1+x^4}\right)\left(\frac{1}{1+x^{16}}\right)\cdots[/tex]

[tex]\displaystyle = \left(\frac{1-x}{1-x^2}\right)\left(\frac{1-x^4}{1-x^8}\right)\left(\frac{1-x^{16}}{1-x^{32}}\right)\cdots[/tex]

[tex]\displaystyle = \left(1-x\right)\left(\frac{1-x^4}{1-x^2}\right)\left(\frac{1-x^{16}}{1-x^8}\right)\cdots[/tex]

[tex]\displaystyle = (1-x)(1+x^2)(1+x^8)(1+x^{32})\cdots[/tex]

[tex]\displaystyle = (1-x)f(x^2)[/tex]

So

[tex]\displaystyle \frac{8}{5f\left(\frac{3}{8}\right)}[/tex]

[tex]\displaystyle = \frac{8}{5}\cdot \frac{5f\left(\frac{9}{64}\right)}{8}[/tex]

[tex]\displaystyle = f\left(\frac{9}{64}\right)[/tex]

so the answer is [tex]\displaystyle \frac{9}{64}.[/tex]

Sweet problem! :)

(edited)

[quote="takuanitromars36"]For number 6, I think I know the answer. (x, y)=(4.3, 1.7). However, the answer that somebody gave me was 6 (I think he BSed the answers that he gave me), so I got confused and put it on anyway. Is (4.3, 1.7) correct? Can you give me the answer key for this set of questions? I did it this morning, but I need answers that are legitimate and not BSed.[/quote]

There are more solutions than just (4.3, 1.7).

[hide] For 6,

let {x} denote fractional part of x.

From the first two equations, you learn that {x}=.3 and {y}=.7 . Then make the substitution {x}=x-[x] and {y}=y-[y]. If you let the integer part of x be a, and the integer part of y be b, then the problem boils down to how many solutions in non-negative integers are there to a+b=5. This is 6. [/hide]

zabelman, you made a minor error. You should recheck your work, and I'm sure you'll see what went wrong quickly. But your solution comes out much more naturally than the official one.

i believe this is from the 2001 ARML... if i remember correctly, the answer to #8 is 9/64

zabelman, replace f(sqrt(X)) with f(x^2)

Thanks bugz and CZ, I fixed it. What did the official solution look like?

#3.

Compute the largest real value b such that the solutions to [tex](\log_{2^{10}} x^{2b})^2 = \log_{2^{10}} x^4[/tex] are integers.

[tex](\log_{2^{10}} x^{2b})^2 = \log_{2^{10}} x^4[/tex]
[tex]({2b \log_{2^{10}} x)^2 = 4 \log_{2^{10}} x[/tex]

Let [tex]y = \log_{2^{10}} x[/tex]

[tex](2by)^2 = 4y[/tex]
[tex]4b^2 y^2 = 4y[/tex]
[tex]4y(b^2 y - 1) = 0[/tex]
[tex]4y = 0 \quad \textrm{or} \quad b^2 y = 0[/tex]
So [tex]y=0[/tex] or [tex]\log_{2^{10}} x = \frac{1}{b^2}[/tex]
[tex]x = (2^{10})^{(1/b^2)} = 2^{(10/b^2)}[/tex], which is integral if 10/b^2 is a non-negative integer.  Since [tex]\frac{10}{b^2} \neq 0[/tex] for a finite b, [tex]\frac{10}{b^2}[/tex] is at least 1.  If [tex]\frac{10}{b^2} = 1, b = \pm \sqrt{10}[/tex].  So the maximum value of [tex]b[/tex] is [tex]\sqrt{10}[/tex].

Edited: Fixed exponent problems

[tex]f(x) = mx[/tex]
[tex]f^{-1}(x) = 1/m*x[/tex]

Since the line f(x) has slope m, we can form a right triangle from it with base 1 and height m.  (see diagram).  Call the angle between the base and the hypotenuse A, as shown.  Then tan A = m/1 = m

Since the line [tex]f^{-1}(x)[/tex] has slope 1/m, we can form a right triangle from it with base 1 and height 1/m (again, see diagram).  Call the angle between the base and the hypotenuse B, as shown.  Then tan B = (1/m)/1 = 1/m.

The angle [tex]\theta[/tex] is equal to A - B, so  [tex] \tan( \theta )[/tex] = [tex] \tan(A-B) [/tex].  By the difference of a tangent formula,

[tex]\displaystyle \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}[/tex]

We found expressions for tan A and tan B in terms of m.  Substituting these in, we get

[tex]\displaystyle \tan(A-B) = \frac{m - 1/m}{1 + m \cdot 1/m} = \frac{m-1/m}{2}[/tex]

Since [tex] \tan( \theta ) = \frac{5}{12}[/tex], we have

[tex]\frac{m-1/m}{2} = \frac{5}{12}[/tex]
[tex]m-1/m = \frac{5}{6}[/tex]
[tex]m^2 - 1 = \frac{5}{6} m[/tex]
[tex]6m^2 - 5m - 6 = 0[/tex]
[tex](2m-3)(3m+2) = 0[/tex]
m = 3/2, -2/3.  Since m > 0, m = 3/2, so 1/m = 2/3.  3/2 + 2/3 = 13/6.

OR

[tex]m-1/m = \frac{5}{6}[/tex]
[tex](m-1/m)^2 = \frac{25}{36}[/tex]
[tex]m^2 - 2 + 1/m^2 = \frac{25}{36}[/tex]
[tex]m^2 + 2 + 1/m^2 = \frac{25}{36} + 4[/tex]
[tex](m+1/m)^2 = \frac{169}{36}[/tex]
[tex]m+1/m = \frac{13}{6}[/tex]

That was a fun problem!

[Edited: Fixed factorization, and added another way.  Thanks, gauss!]

[quote="takuanitromars36"]1. If 4/2001<a/(a+b)<5/2001, compute the number of integral values that b/a can take on.[/quote]

[color=red]4/2001 < a/(a+b) < 5/2001

1/(4/2001) > 1/(a/(a+b)) > 1/(5/2001)

2001/4 > (a+b)/a > 2001/5

500.25 > 1 + b/a > 400.2

499.25 > b/a > 399.2

b/a can be any integer from 400 to 499 inclusively.  There are 100 integral values possible for b/a.[/color]

[quote="mathfanatic"]
[tex]6m^2 - 5m - 6 = 0[/tex]
[tex](2m-3)(3m-2) = 0[/tex]
m = 3/2, 2/3.  In either case, m + 1/m = 13/6.

[/quote]

Your final answer is right, but you factored this wrong.

By the way, can you find another way to get from m - 1/m = 5/6 to the value of m + 1/m?

Here is the official solution to #8 (more or less):

[tex]\displaystyle f(x)f(x^2)=(1+x)(1+x^2)(1+x^4)\cdots=\frac{1}{1-x}[/tex]

[tex]\displaystyle f\Bigg(\frac{3}{8}\Bigg)f\Bigg(\frac{9}{64}\Bigg)=\frac{8}{5}[/tex]

[tex]\displaystyle f^{-1}\Bigg(\frac{8}{5f(3/8)}\Bigg)=\frac{9}{64}[/tex]

There were probably a few more words.

1.

[tex]\displaystyle{4<\frac{2001*a}{a+b}<5}[/tex]
[tex]\displaystyle{4*a+4*b<2001a<5*a+5*b}[/tex]
[tex]\displaystyle{4b<1997a<a+5b}[/tex]
[tex]\displaystyle{4*\frac{b}{a}<1997<1+5*\frac{b}{a}}[/tex]
[tex]\displaystyle{399.2<\frac{b}{a}<499.25}[/tex]

So there are 100 integral values for b/a.  I hope.

Does anyone know if I'm right for that last one?

guess I get the easy ones:

#2: [hide]304 factors as 2^4*19, and drawing this on a lattice, we see that we want two numbers x*y that multiply to 304, x<y, and we want to minimize (x+1)/x, so the smaller number needs to be as large as possible. So, 16x19, and 17x19=323.[/hide]

#4: [hide]Were the dodecagon centered at (0,0), all the xs and ys would cancel out in the sum. This is easy to see geometrically. So, since the center of the dodecagon is at (15,7), the answer is (15-7)*12=96.[/hide]

oh by the way, what was the real number 6?

[quote="WarpedKlown1335"]Yeah you are.  Your solution matches mine.[/quote]

Although I do believe that both you and he are right, the fact that your answers are the same does not guarantee that you are correct.  It would be arrogant to assert so.

[quote]Can somebody explain 2 and 3 to me again? I do not understand the solutions that rand()%0 and mathfanatic put for 2 and 3, respectively.[/quote]

I'll try and elucidate my writings, since I wasn't very clear and made a small TeXing mistake.

First of all, do you know what logarithms are?  In case not, I'll give a brief overview:

Logarithms are the opposite of exponents.  The expression [tex]\log_b x[/tex] means "find the number such that b raised to that number is equal to x."  For instance, [tex]\log_2 8 = 3[/tex] because 2^3 = 8.  According to this definition, then,

[tex]\log_b x = a \quad \textrm{if and only if} \quad b^a = x[/tex] (1)

One of the properties of logarithms is that
[tex]\log_b x^a = a \log_b x[/tex]. (2)

That is, the logarithm of a number to a certain power is the same as that power times the logarithm of the number.

But back to the problem, which was "Compute the largest real value b such that the solutions to [tex](\log_{2^{10}} x^{2b})^2 = \log_{2^{10}} x^4[/tex] are integers."

Suppose we were given the equation [tex](\log_{2^{10}} x^{2b})^2 = \log_{2^{10}} x^4[/tex] with some value of b already filled in, and we were asked to solve the equation.  In such a case, the first thing we'd do would be to simplify the equation as much as possible.  So we do this.

Applying (2) as discussed above,

[tex]\log_{2^{10}} x^{2b} =  2b \log_{2^{10}} x[/tex]
and
[tex]\log_{2^{10}} x^4 = 4 \log_{2^{10}} x[/tex]

So the original equation becomes

[tex]({2b \log_{2^{10}} x)^2 = 4 \log_{2^{10}} x[/tex]

Now, [tex]\log_{2^{10}}[/tex] is pretty complicated, and it exists several times in the equation.  To simplify things, why don't we just replace it with a single variable, like y?  If we let [tex]y = \log_{2^{10}} x[/tex], then the equation becomes:

[tex](2by)^2 = 4y[/tex]

Expanding the left side:

[tex]4b^2 y^2 = 4y[/tex]
[tex]4y(b^2 y) = 4y[/tex]
[tex]4y(b^2 y) - 4y = 0[/tex]
[tex]4y(b^2 y - 1) = 0[/tex]

Remember, we're still pretending like b is some number that we know, and we're trying to solve the equation for x.  What would we do here?

Well, we have two factors that multiply to 0.  So at least one of the factors must be 0.  Hence either  [tex]4y = 0[/tex] or [tex]b^2 y - 1 = 0[/tex]

If [tex]4y=0[/tex],
then [tex]y=0[/tex].
Substituting [tex]\log_{2^{10}} x[/tex] back in for y, we get

[tex]\log_{2^{10}} x = 0[/tex]

Rewriting this according to (1), we get
[tex](2^{10})^0 = x[/tex]
[tex]1 = x[/tex]

So x = 1 is a solution to the equation [tex]\log_{2^{10}} x^{2b} =  2b \log_{2^{10}} x[/tex], regardless of what b is.  That means that we don't ever need to worry about making one of the roots integral - it always is!  Instead, we can focus on the second root, which occurs when [tex]b^2 y - 1 = 0[/tex]

If [tex]b^2 y - 1 = 0[/tex], then
[tex]y = \frac{1}{b^2}[/tex]

Substituting [tex]\log_{2^{10}} x[/tex] back in for y, we get

[tex]\log_{2^{10}} x = \frac{1}{b^2}[/tex]

Rewriting according to (1), we have
[tex]x = (2^{10})^(1/b^2)[/tex].
[tex]= 2^{(10/b^2)}[/tex].

Suppose that [tex]2^{(10/b^2)} = k[/tex], where k is some integer.  Then [tex]\log_2 k = \frac{10}{b^2}[/tex].  So 10/b^2 must be a logarithm of the form log_2 k, where k is an integer.  Since 2^(any power) must be positive, k must be positive.  So we conclude that k is a positive integer.

As b increases, 10/b^2 will get smaller.  Since we want the biggest value of b, we want the smallest possible value of 10/b^2.  This occurs when k is smallest.  So we start with the smallest value of k, which is 1.

If [tex]k = 1, \frac{10}{b^2} = \log_2 k = \log_2 1 = 0[/tex].  But 10/b^2 cannot be 0, since this would require b to be infinitely large.

If [tex]k=2, \frac{10}{b^2} = \log_2 k = \log_2 2 = 1[/tex].
So [tex]\frac{10}{b^2} = 1[/tex]
[tex]10 = b^2[/tex]
[tex]b = \pm \sqrt{10}[/tex]

So the maximum value of b is [tex]\sqrt{10}[\tex].

rand()%0's solution is wrong. The answer is 288.

If you call the lengths of the sides of rectangle R, a and b, then the number of lattice points in its interior and on its sides is (a+1)(b+1). If you double the length of the shortest side (lets say a) to make a larger rectangle, then the number of lattice points in its interior and on its sides is (2a+1)(b+1).

So (a+1)(b+1) + 304 = (2a+1)(b+1)
ab + a = 304
a(b+1) = 304

That means a and b+1 are factors of 304 = 16*19, and a < b. To make ab = 304 - a as small as possible, a has to be the largest factor possible still less than b. That would make a = 16, and ab = 304 - 16 = 288.

You've got to post the questions if you want help.

oops. No, my solution was right, I just didn't answer the right question. 17x19=323 was the number of lattice points in the original polygon, not the area. The area is (17-1)(19-1)=16x18=288.

wow i just realized 1998 is one of the few years i dont have


I found this problem on a website:

Show a 3D closed convex object, such that the set of its extremal points is not closed.

I'd love to try to solve this, but i don't know what these people mean by 'extremal points'. Can ne1 tell me what extremal points of a figure mean (I mean a rigorous definition).

Thanks in advance! :D

Points of locally maximal curvature in both principal directions of the surface.

[url]http://www-mitpress.mit.edu/e-journals/Videre/001/articles/Pennec/PennecVidereDemo/FeatureExtraction.ExtremalPoints.html[/url]

Thanks for the help. I looked it up myself (using google), but I think 'extremal point' is too ,general and I found all sorts of things, except for what I was looking for. :D


calculus
algebra
polynomial
function
real analysis
real analysis unsolved

Let s= 23, p= 560.
Let a(0)= s, a(1)= p, and a(n+1) = a(n) + a(n-1),  n=1, 2, 3..
set X(n)= a(n) / a(n+1)
Use calculus technique to answer:
Does this sequence converge or diverge. If it converges, find its limit.

hello, for your sequence exists a general formula
$ a_n\equal{}\left(\frac{23}{2}\plus{}\frac{1097}{10}\sqrt{5}\right)\left(\frac{1}{2}\plus{}\frac{1}{2}\sqrt{5}\right)^n\plus{}\left(\frac{23}{2}\minus{}\frac{1097}{10}\sqrt{5}\right)\left(\frac{1}{2}\minus{}\frac{1}{2}\sqrt{5}\right)^n$
Sonnhard.

how you get the general formula? can you show me the steps clearly? let says i get another a(n), how i solve it?

You can solve a linear recurrence of the form $ a_{n \plus{} 1} \equal{} \sum_{i \equal{} 0}^{k \minus{} 1} c_{n \minus{} i}a_{n \minus{} i}$ by finding all solutions to the polynomial equation  $ x^k \equal{} \sum_{i \equal{} 0}^{k \minus{} 1} c_{n \minus{} i}x^{k \minus{} i \minus{} 1}$. For each such solution $ \lambda_j$, $ a_n \equal{} \lambda_j^n$ will solve your recurrence. As long as the $ \lambda_j$ are all unique (if their not there is a bit more work to do), you can use the initial conditions to find a linear combination of the $ \lambda_j$ that solves the general recurrence and the initial conditions.

See more [url=http://www.artofproblemsolving.com/Wiki/index.php/Binet%27s_Formula]here[/url] and [url=http://en.wikipedia.org/wiki/Recurrence_relation]here[/url].

EDIT: corrected issues with sums

FibonacciFan, you have some indexing issues in your sums ;)

Here's a related question: suppose we have a sequence $ (b_n)$ defined by $ b_0 \equal{} b > 0$ and $ b_{n \plus{} 1} \equal{} \frac{1}{1 \plus{} b_n}$.  Does $ (b_n)$ converge?

[quote="JBL"]Suppose we have a sequence $ (b_n)$ defined by $ b_0 \equal{} b > 0$ and $ b_{n \plus{} 1} \equal{} \frac {1}{1 \plus{} b_n}$.  Does $ (b_n)$ converge?[/quote]
Clearly $ b_i \in [0,1]$ for $ i \ge 1$. Since the iteration function $ f(x) \equal{} \frac{1}{1\plus{}x}$ maps this interval to itself and $ |f'(x^*)| < 1$ at the unique fixed point $ x^* \equal{} \frac{\sqrt{5}\minus{}1}{2} \in [0,1]$, the answer is yes.


I posted something on a thread.  I realized it was stupid so i tried to delete it.  It said "are you sure you want to delete this?"  i clicked yes. Then i went back to the thread.  Instead of it being deleted, it was written twice, and i couldnt delete it.

Try editing the post and checking the box that says "Delete this post"

Self-deletion of a post doesn't seem to work, according to my experiment, if there has been another reply to the thread. But moderators can always delete posts, and I am happy to delete anything from Getting Started or from College that users request me to delete. So the moral of the story is, if you have second thoughts about one of your own posts, delete it SOON, or else ask a moderator to delete it for you.

I guess as a last resort, you could always edit your post and delete whatever you said that you didn't like.. the whole thing if need be...


algebra
function
domain
complex analysis
complex analysis unsolved

Suppose $f$ is analytic and never zero on a domain $D$. Show that $|f|$ has no local minima in $D$. It says to use the conclusion of a different excerise: $|f(z)|\leq \text{max}_{0\leq t\leq 2\pi}|f(z+re^{it})|$ for sufficiently small $r$. Now I've proved that excercise. Plus the excercise that there is no strict local maximum.

I think the problem is that I don't believe it. To do maximum, you can assume there is one and get a contradiction from the estimate. The estimate says that the function is smaller than something on a circle around that point (essentially). This doesn't seem to indicate no minima. In fact, I feel like it suggests it.

I'd like to work out the proof, so you don't have to post one, but some insight into why it is believable would be great to get me started, (or maybe a slight hint).

Aha! Got it. Corollary: So since there are no local max or min, $f$ is constant on $D$?

Analytic functions are so well-behaved.

Not at all: $|f|$ is not constant unless $f$ is. Recall that $D$ is an open set, hence not compact.


A bookstore employee is given a $ 30\%$ discount off the retail price of any book. How many dollars does she pay for a book with a retail price of $ \$40$?

a 30% discount means that 70% of the retail price is paid.
.7*40=28

ANSWER:28

If the employee gets a 30% discount, that means she gets 70% of the retail price. Thus:

$ 0.7 \times 40 \equal{} \boxed{28}$.

EDIT: Beaten to it.

The employee paid 70% of the retail price so you have to do: 7*40=28


probability
function
integration
expected value
calculus
calculus computations

The time required to complete a certain task is a random variable X that is exponentially distributed with a mean 60 hours.
a)  What is the probability that the task will take moe than 30 hours to complete?
b)  Suppose the cost of performing this task in dollars is given by C=10X+200.  Find the expected value and variance of the cost?

If $X$ is exponentially distributed with mean $\beta,$ then its density is $f(x)=\frac1{\beta}e^{-x/\beta}$ for $x>0$ and zero otherwise.

In that case, the complement of the cumulative distibution function is

$P(X>x)=\int_{x}^{\infty}\frac1{\beta}e^{-t/\beta}\,dt= e^{-x/\beta}.$

So there's the answer to your question (a):

$P(X>30\text{ hr})=e^{-30/60}=e^{-1/2}.$

As for part (b), use this:

If $U=aX+b,$ then $E(U)=aE(X)+b$ and $\text{Var}(U)=a^{2}\text{Var}(X).$

Also use that an exponential distribution with mean $\beta$ has variance $\beta^{2}.$


geometry
circumcircle
LaTeX
parallelogram
geometric transformation
homothety
power of a point

Let (O) and (O') be two circles intersecting each other at two distinct points K and L. Let AB be one of the two common tangents of (O) and (O') (with A lying on (O), and B lying on (O')). Let A' and B' be points such that L is the midpoint of the segments AA' and BB'. The circumcircle of triangle ABA' cuts (O) at the second intersection V. The circumcircle of triangle ABB' cuts (O') at the second intersection U. The line A'V cuts (O) at I, the line B'U cuts (O') at J. Prove that IJ is the other common tangent of the circles (O) and (O').

$\LaTeX$ing the problem:

Let $(O)$ and $(O')$ be two circles intersecting each other at 2 distinct points $K, L$. $AB$ is one of the two common external tangents of $(O)$ and $(O')$, ($A \in (O),\ B \in (O')$). $A', B'$ are points such that $K$ is midpoint of $AA'$ and $BB'$. The circumcircle of $\triangle ABA'$ cuts $(O)$ at the second intersection $V$. The circumcircle of $\triangle ABB'$ cuts $(O')$ at the second intersection $U$. $A'V$ cuts $(O)$ at $I$, $B'U$ cuts $(O')$ at $J$. Prove that $IJ$ is the other common tangent of $(O)$ and $(O')$.

Let $r, r'$ be the radii of the circles $(O), (O')$ and WLOG, assume that $r \ge r'$. Let $CD$ be the other common external tangent of the circles $(O), (O')$, with the tangency points $C \in (O),\ D \in (O')$. The quadrilateral $ABA'B'$ is a parallelogram, because its diagonals $AA', BB'$ intersecting at the point $K$ cut each other in half. The radical axis $KL$ of the circles $(O), (O')$ cuts the segment $AB$ of their common external tangent at its midpoint $M$. Hence, $KM$ is the common midline of the triangles $\triangle ABB', \triangle ABA'$ parallel to their sides $AB', BA'$, i.e., $KM \parallel AB' \parallel BA'$. Since the radical axis $KL \equiv KM$ is perpendicular to the center line, $AB' \perp OO', BA' \perp OO'$. It follows that the lines $AB', BA'$ pass through the tangency points $C, D$, respectively. Let $H$ be the external homothety center of the circles $(O), (O')$, the intersection of their common external tangents $AB, CD$.

The angles $\angle BVD = \angle HBD$ are equal, because the angle $\angle BVD$ spans the major arc $BD$ of the circle $(O')$ and the line $BH$ is a tangent to this circle at the point $B$. The angles $\angle AVB = \angle AB'B = \angle A'BB' \equiv \angle DBB'$ are equal, the 1st pair spanning the same arc $AB$ of the circumcircle  $(\triangle ABB')$ and the 2nd pair are alternate angles of the transversal $BB'$ cutting the parallel lines $AB' \parallel BA'$. The angles $\angle B'VA = \angle B'BA$ are equal, because they span the same arc $B'A$ of the circumcircle $(\triangle ABB')$. As a result,

$\angle B'VD = \angle B'VA + \angle AVB + \angle BVD =$

$= \angle B'BA + \angle DBB' + \angle HBD = \angle HBA = 180^o$

Consequently, the points $B', V, D$ are collinear and the points $D \equiv J$ are identical.

Similarly, the angles $\angle AUC = 180^o - \angle HAC$ add up to a straight angle, because the angle $\angle AUC$ spans the minor arc $AC$ of the circle $(O)$ and the line $AH$ is a tangent to this circle at the point $A$. The angles $\angle BUA = \angle BA'A = \angle B'AA' \equiv \angle CAA'$ are equal, the 1st pair spanning the same arc $BA$ of the circumcircle  $(\triangle ABA')$ and the 2nd pair are alternate angles of the transversal $AA'$ cutting the parallel lines $AB' \parallel BA'$. The angles $\angle A'UB = \angle A'AB$ are equal, because they span the same arc $A'B$ of the circumcircle $(\triangle ABA')$. As a result,

$\angle A'UC = \angle A'UB + \angle BUA + \angle AUC =$

$= \angle A'AB + \angle CAA' + 180^o - \angle HAC = 180^o - \angle HAB = 180^o$

Consequently, the points $A', U, C$ are collinear and the points $C \equiv I$ are identical.

Another solution of the problem in Treegoner's formulation (not in Yetti's one), heavily based on the "working backwards" strategy:

Let O and O' be the centers of the circles (O) and (O'). As we know, one of the two external common tangents of these two circles (O) and (O') touches (O) at A and (O') at B. Let their other external common tangent touch (O) at C and (O') at D. We will prove that I = C; then, by analogy, we will have J = D, and thus, the line IJ will be the second external common tangent of the circles (O) and (O').

Hence it remains to prove that I = C. Since both points I and C lie on the circle (O), the only thing we must show is that C lies on the line A'V.

We have $AC\perp OO^{\prime }$ and $BD\perp OO^{\prime }$ by symmetry. Therefore, AC || BD.

Let M be the midpoint of the segment AB. Then, MA = MB, so that $MA^2 = MB^2$. Therefore, the point M has equal powers with respect to the two circles (O) and (O'), and thus lies on their radical axis. The point L also lies on their radical axis (since L is a common point of (O) and (O')). Hence, the line ML is the radical axis of the two circles (O) and (O'), and thus $ML\perp OO^{\prime }$. Now, since the point L is the midpoint of the segment AA', and the point M is the midpoint of the segment AB, we have ML || BA', and hence $BA^{\prime }\perp OO^{\prime }$. But we also have $BD\perp OO^{\prime }$; thus, BA' || BD; this means that the lines BA' and BD coincide, i. e. the point A' lies on the line BD.

With directed angles modulo 180°, we now have

  < CVA = < CLA        (since the points C, A, V, L are concyclic)
        = < (AC; AB)       (angle between secant and tangent)
        = < (BD; AB)       (since AC || BD)
        = < A'BA
        = < A'VA       (since the points A, B, A', V are concyclic).

Therefore, the point C lies on the line A'V, and the solution is done.

  Darij


calculus
integration
rotation
geometry
3D geometry
sphere
calculus computations

Hi, can anyone help me set up a formula to find the volume of:

$x^2+(y-1)^2=1$ when it is rotated around the y-axis?

I first put everything in terms of x, so $x^2=1-(y-1)^2$, so $x=\pm\sqrt{1-(y-1)^2}$. Since this graph is symmetric about the y-axis (a circle with center $(0,1)$. Then I only take the positive x values since I can integrate with respect to $y$:

$\displaystyle\int^2_0 \sqrt{1-(y-1)^2}\,dy$, which using a calculator I find the value is $4.93$... However, Since the volume is simply that of a sphere, I can also find the volume using $\frac{4}{3}(\pi)r^3$, with $r=1$, which equals $\frac{4}{3} \pi=4.18...$. How come these two values are different? Is there something wrong that I did?

Help is greatly appreciated.

If you're using 'washers' to integrate (I assume you're not using multiple integrals) then we get:

$V=\pi\int^2_0 (1-(y-1)^2)dy=\frac {4}{3}\pi$

which is expected.

If you're using shells, then we get (translating it since it doesn't make a difference)

$\frac {1}{2}V=2\pi\int^1_{0} (x\sqrt {1-x^2})dx=\frac {2}{3}\pi$

Does this help at all?

Yeah, sorry, I did use washers (or disc method, as the book calls it), put I forgot to square the expression since we are looking for $\pi(r)^2$. Thanks.


number theory
relatively prime
Additive Number Theory

Let $n$ be an integer of the form $a^2 + b^2$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \leq \sqrt{n}$, then $p$ divides $ab$. Determine all such $n$.

The key is to use the fact that a and b are relatively prime. We show in fact that they must differ by $ 1 (or 0)$. Suppose first that a = b. Then since they are relatively prime they must both be 1. That gives the first answer above. So we may take a > b. Then $ (a \minus{} b)^2 < a^2 \plus{} b^2 \equal{} n$, so if $ a \minus{} b$ is not $ 1$, it must have a prime factor which divides ab. But then it must divide a or b and hence both. Contradiction. So $ a \equal{} b \plus{} 1$.

Now $ (b \minus{} 1)^2 < b^2 < n$, so any prime factor p of b - 1 must divide $ ab \equal{} b(b \plus{} 1)$. It cannot divide b (or it would divide b and b - 1 and hence 1), so it must divide b + 1 and hence must be 2. But if 4 divides b - 1, then 4 does not divide b(b - 1), so b - 1 must be 0, 1 or 2. But it is now easy to check the cases a, b = (4, 3), (3, 2), (2, 1).

Chien wrote that (a,b)=(4,3). But then n=25 according to the problem 5|12 which is a contradiction..

I think that $n$ should be a prime number. Let $n$ be composite. If $d$ is a divisor of $n$ then $n/d$ is also a divisor of $n$. Now let $d<\sqrt{n}$. Let $q$ be a prime which divides $d$ . then $q|ab$ and also $q|n$ which follows that $q|a$ and $b$ which is a contradiction. Hence $n$ must be a prime number.
Now $p=a^{2}+b^{2}$ and $ab=kq_{1}q_{2}...q_{l}$ where $q_{i}$ is a prime less than or equal to $\sqrt{n}$.
 now $|a-b|<\sqrt{p}$. But we see that $q_{i}$ does not divide $|a-b|$. This forces that either $|a-b|=1$ or there is only one prime less than $\sqrt{n}$. This prime must be $2$. We know that a prime can be expressed as the sum of two squares iff it is of the form $4k+1$. we see that there is only prime, $5$, satisfying these properties. So $n=5$ is an answer. But if $p=a^{2}+(a+1)^{2}$ then it follows that any prime less than $\sqrt{n}$ must be less than $a+1$ otherwise if q be a prime possessing this then q will not divide $a(a+1)$. Now it follows that $a(a+1)$ is divisible by any prime less than  $a+1$ is possible only when $a=2$( one may use Bertrand Conjecture to prove this). If $a=2$ then $n=13$.
Hence the answer is $n=5,13$


Consider all fractions that have positive denominators between $ 1$ and $ 100$. Write a program that finds the two such fractions that are cloest to $ \frac{17}{76}$: one from above and one from below.

Write a program?  You can use the Euclidean algorithm to find them by hand.

[hide="TI Basic code"][code]:1->x
:{0,0->L1
:For(i,1,100
:For(j,1,i
:If x>abs(j/i-17/76
:Then
:{i,j->L1
:i/j->x
:End
:End
:End
:ClrHome
:Output(1,1,L1(2
:Output(2,1,"--
:Output(3,1,L1(1[/code]
[/hide]

It can be easily modified to find the closest above and below instead of just the closest.

[hide="Java Code"][code]double lowest = 1.0;
int tempnum, tempdenom;

for(int denom = 1; denom<=100; denom++)
{
     for(int num = 1; num<denom; num++)
     {
          if(Math.Abs((double) num/denom)<x)//I can't remember what the method is, exactly
          {
               tempnum=num;
               tempdenom=denom;
               x=(double) num/denom;
          }
     }
}
//I think the casting will work correctly, but again, I'm not sure
System.out.println(num + "/" + denom);[/code]
[/hide]

[quote="8parks11"]Consider all fractions that have positive denominators between $ 1$ and $ 100$. Write a program that finds the two such fractions that are cloest to $ \frac {17}{76}$: one from above and one from below.[/quote]
[hide="vb6"]
I tried to make this friendly to read while keeping efficiency, so it might not be the most efficient possible.
[code]Option Explicit
Private Type Fraction
    Numerator As Long
    Denominator As Long
End Type
Function ClosestFractions() As String
    Const N& = 17
    Const D& = 76
    Dim DTest   As Long
    Dim LBGF    As Fraction, UBLF   As Fraction
    Dim LDiff   As Single, UDiff    As Single
    LDiff = 1
    UDiff = 1
    For DTest = 1 To 100
        If DTest = D Then DTest = DTest + 1
        If (N / D - Fix(DTest * N / D) / DTest) < LDiff Then
            UBLF.Numerator = Fix(DTest * N / D)
            UBLF.Denominator = DTest
            LDiff = N / D - Fix(DTest * N / D) / DTest
        End If
        If ((Fix(DTest * N / D) + 1) / DTest - N / D) < UDiff Then
            LBGF.Numerator = Fix(DTest / 76 * 17) + 1
            LBGF.Denominator = DTest
            UDiff = (Fix(DTest * N / D) + 1) / DTest - N / D
        End If
    Next DTest
    ClosestFractions = ("Lower Fraction: " & (CStr(UBLF.Numerator) & _
        "/" & CStr(UBLF.Denominator)) & vbCrLf) & ("Upper Fraction: " & _
        (CStr(LBGF.Numerator) & "/" & CStr(LBGF.Denominator)))
End Function
[/code]
N and D might be arguments instead of constants if other fractions are to be used for other problems asking for the same thing.
However, if N and D become arguments, this line should be added after the beginning of the For loop:
[code]        If D Mod DTest = 0 Then DTest = DTest + 1
[/code]

This will ensure that it will work properly as long as N and D are relatively prime.
[/hide]

[hide="the test of it"]
$ \text{Debug.Print ClosestFractions}$ outputted the following:
[code]Lower Fraction: 19/85
Upper Fraction: 15/67
[/code]
[/hide]

Seeing as how this was posted in HSB and not in the Computer Science forum, [hide="I might as well give the mathematical solution"] $ 76 \equal{} 4 \cdot 17 \plus{} 8$
$ 17 \equal{} 2 \cdot 8 \plus{} 1$
$ 8 \equal{} 8 \cdot 1 \plus{} 0$

We conclude that

$ 9 \cdot 17 \minus{} 2 \cdot 76 \equal{} 1 \Leftrightarrow$
$ 85 \cdot 17 \minus{} 19 \cdot 76 \equal{} 1 \Leftrightarrow$
$ \frac{17}{76} \minus{} \boxed{\frac{19}{85}} \equal{} \frac{1}{76 \cdot 85}$

Similarly,

$ \minus{}9 \cdot 17 \plus{} 2 \cdot 76 \equal{} \minus{}1 \Leftrightarrow$
$ 67 \cdot 17 \minus{} 15 \cdot 76 \equal{} \minus{}1 \Leftrightarrow$
$ \frac{17}{76} \minus{} \boxed{\frac{15}{67}} \equal{} \minus{} \frac{1}{76 \cdot 67}$

:) [/hide]

Mathematical Approach

Two basic theorems that describe characteristics of successive terms in a Farey series $ F_n$ of order n are:

(1) If $ \frac {h}{k}$ and $ \frac {h'}{k'}$ are two successive terms of the Farey series, then $ kh - hk' = 1$
(2) If $ \frac {h}{k}$ and $ \frac {h'}{k'}$ are two successive terms of the Farey series, then $ k + k' > n$.

By using theorem $ (1)$ with $ \frac{h}{k}$ set to $ \frac{17}{76}$ where $ h$ is $ 17$ and $ k$ is $ 76$,
$ 76h - 17k' = 1$

This is similar to solving $ 76h' = 1 mod 17$ .

This is solved when $ h' = 15 mod 17$ Substituting 15 into the first equation yields $ k' = 67$ , and $ \frac{h'}{k'} = \frac{15}{67}$
Thus
$ \frac {17}{76} < \frac {15}{67}$
Next, by using theorem $ (1)$, but with $ \frac{h'}{k'}$ as $ \frac{17}{76}$,
$ 17k - 76h = 1$

This is similar to solving $ 17k = 1 mod 76$ . This is satisfied when $ k$ is equal to $ 9 mod 76$ .

However, when $ k = 9$, it contradicts theorem $ (2)$. Thus $ k$ is not equal to $ 9$ . When $ k = 85$, it satisfies both theorems.

When substituted into the equation, $ \frac {h}{k} = \frac {19}{85}$
Thus, $ \frac {17}{76} < \frac {15}{67} < \frac {19}{85}$


1) If p≥5 is a prime number show that (p^2)+2 is composite.
[ hint: p takes one of the forms 6k+1 or 6k+5]

2) If p ≠ 5 is an odd prime prove that either [(p^2)-1] or [( p^2) +1] is divisible by 10 .

The second one can be proved by considering p = 5k + 1, 2, 3, 4.
 :D These two are quite nice man!

[hide="Number 1"]If $ p\equal{}6k\plus{}1$, then $ p^2\plus{}2\equal{}36k^2\plus{}12k\plus{}3\equiv 0\pmod 3$, so it is obviously composite.
If $ p\equal{}6k\plus{}5$, then $ p^2\plus{}2\equal{}36k^2\plus{}60k\plus{}27\equiv 0\pmod 3$ as well.[/hide]

Please don't post topics twice.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262704


geometry
circumcircle
analytic geometry
conics
hyperbola
perpendicular bisector
geometry proposed

A triangle $ ABC$ is given in the plane. Let $ M$ be a point inside the triangle and $ A'$, $ B'$, $ C'$ be its projections on the sides $ BC$, $ CA$, $ AB$, respectively. Find the locus of $ M$ for which $ MA \cdot MA' \equal{} MB \cdot MB' \equal{} MC \cdot MC'$.

Wonder why the problem restricts point M in the interior of ABC ?. The locus of $M$ would be the union of the circumcircle of $\triangle ABC$ and its orthocenter. Let $(x:y:z)$ be the barycentric coordinates of $M$ with respect to $\triangle ABC.$

For instance, condition $MB \cdot MB'=MC \cdot MC'$ implies alone

$\frac{(a^2z^2+2S_Bzx+c^2x^2)y^2}{b^2}=\frac{(b^2x^2+2S_Cxy+a^2y^2)z^2}{c^2}  \ \Longleftrightarrow$

$ (a^2yz+b^2zx+c^2xy)(c^2xy+c^2yz-b^2yz-b^2xz)=0$

Thus, $M$ either lies on the circumcircle $\mathcal{O} \equiv a^2yz+b^2zx+c^2xy=0$ or it lies on the rectangular circum-hyperbola $\mathcal{H}_a,$ isogonal conjugate of the perpendicular bisector of $\overline{BC}.$ Hence, cyclic conditions yield that locus of  $M$ is the union of $\mathcal{O}$ and  the orthocenter of ABC.


hi, please help me on the following problems.
1.  For some two digit number(s), when you find the sum of 4 times the tens digit and 7 times the units digit you get the orginal two digit number back again.  What is(are) that(those) number?

Let the number be $ \overline{ab}\equal{}10a\plus{}b$. Then $ 10a\plus{}b\equal{}4a\plus{}7b\implies6a\equal{}6b\implies a\equal{}b\implies\boxed{11k}$ for $ 1\le k\le9$.

Use the same approach I used in the other thread.

[hide="Solution"]Write the original number as $ 10a \plus{} b$
$ 10a \plus{} b \equal{} 4a \plus{} 7b$
$ 6a \equal{} 6b$
$ a \equal{} b$
$ a, b \in \{1, 2, 3, 4, 5, 6, 7, 8, 9 \}$
So the solutions are $ 11, 22, 33, 44, 55, 66, 77, 88, 99$[/hide]

Edit: Ninja'd


I'm so frustrated with math class. I can't solve this system. I need help. Here it is:
x-y-z=1
2x+3y-z=3
-x+2y-4z=4

It's okay if none of you can solve it. :thumbup:

You guys give up? I don't have the answer!  :(
BLLLLLEEEEEEAAAAAAAAAAAAT!

Making the coefficents of $x$ (almost) the same:
$2x-2y-2z=2$
$2x+3y-z=3$
$-2x+4y-8z=8$
Adding the last two, subtracting the first from the second:
$7y-9z=11$
$5y+z=1$
Multiplying to make coefficents of $z$ (almost) equal:
$7y-9z=11$
$45y+9z=9$
Adding:
$52y=20$
$y=\frac{5}{13}$
Substituting back in:
$\frac{35}{13}-9z=\frac{143}{13}$
$9z=\frac{-108}{13}$
$z=\frac{-12}{13}$
Substituting back in:
$2x-\frac{10}{13}+\frac{24}{13}=\frac{26}{13}$
$x=\frac{6}{13}$

Checking back in the original equations, the triple $x=\frac{6}{13}, y=\frac{5}{13}, z=\frac{-12}{13}$ works.

[This should have been posted in High School Basics, or possibly Intermediate]

I was gonna put it in that game forum, but i resisted the urge of thinking it was to ez.
 :coolspeak: thanks. :coolspeak:

In the basics forum, most people do not think about matrices.

[quote="Hoborobo"]Nah, this was basics.

[hide="hint"]Just the coefficients of x,y,z into a 3x3 matrix [A]. Then plug (1,3,4) into a 3x1 matrix [B]. Then proceed to do $[A]^{-1}[B]$ and win the internets[/hide][/quote]

this is otherwise known as Cramer's Rule and if you need help on it go here:

[url]http://en.wikipedia.org/wiki/Cramer%27s_rule[/url]

The things you should know before you read this:

Read the Example section only! The others are from a different level of math

$\text{Det}$ means determinant.

[quote="asmallgoat"]WHo is Cramer? Is he some math guy I'm supposed to know? I would like to meet him if he is because I want to be a math person when I grow up :w00tb:  :w00tb:  :play_ball:[/quote]
I have a feeling that he is dead. Cramer's Rule is a method for solving n equations with n variables using an n by n matrix.

[quote="asmallgoat"]WHo is Cramer? Is he some math guy I'm supposed to know? I would like to meet him if he is because I want to be a math person when I grow up :w00tb:  :w00tb:  :play_ball:[/quote]

He is a little dead.

And yeah, Cramer is the matrix dude.

[quote="asmallgoat"]WHo is Cramer? Is he some math guy I'm supposed to know? I would like to meet him if he is because I want to be a math person when I grow up :w00tb:  :w00tb:  :play_ball:[/quote]
Wikipedia is your friend.
Anyway, PLEASE don't title topics like "IMPOSSIBLE" or "VERY HARD PROBLEM" or stuff like that, when the problem is....um not hard at all(even if you think it's hard, just avoid titling topics like that), since it might chase away new members who were thinking of joining aops who might see this and think "wow aopsers are dumb"

[quote="mathgeniuse^ln(x)"]well you are implying something when you say, "aopsers are dumb"[/quote]
implying that aopsers are smart? Generally speaking, they are.(not all tho =()
Well what I was saying was that a new member who is just about to join aops  might not join because of a thread like this


I have a very basic question about logs. I'm sure my textbook is wrong but I'm posting here just to check.
The power law of logs says that if $ x \equal{} a^{n}$, then $ \log_{}{x}\equal{}n\log_{}{a}$. I also know that the log of a negative number is undefined.
There are 2 solutions in my book that I'm not satisfied with:
Q1 'Express without logs: $ 2\log_{x}(2y\minus{}x)\equal{}0$'
My book's solution is: $ (2y\minus{}x)^{2}\equal{}1$
My solution is: $ 2y\minus{}x\equal{}1$
Surely, if $ (2y\minus{}x)^{2}\equal{} 1$ is a solution, then this suggests that $ 2y\minus{}x$ can be $ \minus{}1$, but then $ \log_{x}(2y\minus{}x)$ would be undefined.
Q2 ' Solve the equation: $ 2\log_{2}(x\minus{}1)\equal{}1$'
My book's solution is: $ \frac{1}{2}, 2$
My solution is: $ 2$
Same reasoning as question above.
I would appreciate anyone's response.

[hide] $ 2y\minus{}x$ is equal to $ \minus{}1$ since the inside of the log's inside squared. Thus $ \log{(\minus{}1)^{2}}\equal{}\log{1}$
As for the second one.
$ 2\log_{2}{(x\minus{}1)}\equal{}\log_{2}{(x\minus{}1)^{2}\equal{}1}$
exponentiate both sides with base 2.
$ 2^{\log_{2}{(x\minus{}1)^{2}}}\equal{}2^{1}$
$ (x\minus{}1)^{2}\equal{}2$
$ x^{2}\minus{}2x\minus{}1\equal{}0$
I dont see why $ 2,1/2$ are solutions!
[/hide]

Fraenkel

Thanx for responding.
I see what you're saying, but if we accept that $ \log{}(\minus{}1)^{2}\equal{}\log{}1$ then where does that leave $ \log{}(\minus{}1)^{2}\equal{} 2\log{}(\minus{}1)$?
It would mean that in Q1: 'Express without logs: $ 2\log_{x}(2y\minus{}x)\equal{}0$'
we couldn't divide both sides by 2 to get: $ \log_{x}(2y\minus{}x)\equal{}0$ since this would lead to $ 2y\minus{}x\equal{}1$ and not $ (2y\minus{}x)^{2}\equal{}1$

I know. That confuses me too.
My best guess is that since $ n\log{a}\ne\log{a^{n}}$ if $ a<0$ and $ n\equal{}2k$ dividing by n becomes invalid.

Can someone who is an expert at this explain?

[quote="Fraenkel"]I know. That confuses me too.
My best guess is that since $ n\log{a}\ne\log{a^{n}}$ if $ a < 0$ and $ n \equal{} 2k$ dividing by n becomes invalid.

Can someone who is an expert at this explain?[/quote]

In general, log rules tend to screw up where the base or argument is negative.

Thanx for responses.
Sorry, I made an error in Q2. It should have read:
'Solve the equation: $ \log_{2}{x}\minus{}2\log_{2}{(x\minus{}1)}\equal{}1$'
but my query is still the same.

[quote="Fraenkel"]
$ \log_{2}{x}\minus{}2\log_{2}(x\minus{}1) \equal{} 1$
$ \log_{2}{x}\minus{}\log_{2}(x\minus{}1)^{2}\equal{} 1$
$ \log_{2}{(x/(x\minus{}1)^{2})}\equal{} 1$
$ x/(x\minus{}1)^{2}\equal{} 2$
$ x \equal{} 2(x\minus{}1)^{2}$
$ 0 \equal{} 2x^{2}\minus{}5x\plus{}2$
$ x \equal{}\frac{5\pm\sqrt{25\minus{}4(2)(2)}}{4}$
$ x \equal{} 1/2,2$
[/quote]

Then, we plug the roots back in and find that $ x\equal{}1/2$ doesn't work:
$ \log_{2}{x}\minus{}2\log_{2}(x\minus{}1) \equal{} 1$
$ \log_{2}{1/2}\minus{}2\log_{2}(\minus{}1/2) \equal{} 1$
$ \minus{}1\minus{}2\log_{2}(\minus{}1/2)\equal{}1$
$ \log_{2}(\minus{}1/2)\equal{}\minus{}1$
Which is not true... As Anaxerzia said, log rules get screwed up when the base/argument is negative.

There was a long discussion about logs of negative numbers a little while ago.  It never actually ended up with a conclusion...:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1614854325&t=119947[/url]


function
quadratics
algebra unsolved
algebra

For a sequence $ a_i \in \mathbb{R}, i \in \{0, 1, 2, \ldots\}$ we have $ a_0 \equal{} 1$ and  \[ a_{n\plus{}1} \equal{} a_n \plus{} \sqrt{a_{n\plus{}1} \plus{} a_n} \quad \forall n \in \mathbb{N}.\] Prove that this sequence is unique and find an explicit formula for this recursively defined sequence.

$ a_{n\plus{}1}>a_n\to  a_{n\plus{}1}>a_n\plus{}\sqrt{2a_n}$
From $ a_{n\plus{}1}\equal{}a_n\plus{}\frac 12 \pm \sqrt{2a_n\plus{}\frac 14}$ we get $ a_{n\plus{}1}\equal{}a_n\plus{}\frac 12 \plus{}\sqrt{2a_n\plus{}\frac 14}$ - unique solution.
It is easy to chek, that $ a_n\equal{}1\plus{}\frac{n(n\plus{}3)}{2}\equal{}C_n^2\plus{}C_n^1\plus{}C_n^0$ is solution.


geometry
geometric transformation
reflection
homothety
circumcircle
geometry unsolved

[hide="Nice problem"]Whoa...the line appears to pass through the Feuerbach point...

Okay, anyways...let $ A''$ be the reflection of $ A'$ over $ AI$.  Then $ r_1$ is tangent to the incircle at $ A''$, and so the pole of $ r_1$ with respect to the incircle is $ A''$.  Now because $ M_a$ is on the tangent line to the incircle from $ A'$, it follows the pole of $ M_a$ passes through $ A'$.  Coupled with the fact that $ M_aI \perp r_2$, this tells us $ r_2$ is the polar of $ M_a$.

Therefore, $ X_a \equal{} r_1 \cap r_2$ is the pole of $ A''M_a$.

It suffices to show that the polars of $ X_a, X_b,$ and $ X_c$ concur on the incircle.  These lines are $ M_aA'', M_bB'',$ and $ M_cC''$, where $ B''$ and $ C''$ are defined in a manner similar to $ A''$.

Now we do a little angle chasing.  Notice $ A'A'' \parallel{} B'C', B'B'' \parallel{} C'A', C'C'' \parallel{} A'B'$.  We will show $ B''C'' \parallel{} BC$.  What is the angle between $ B''C''$ and $ B'C'$?  It's got to be $ \frac {1}{2}(\widehat{C'C''} \plus{} \widehat{B'B''})$.  This is equal to
\[ (\angle{C'A'B'} \plus{} \angle{C'B'A'}) \plus{} (\angle{B'C'A'} \plus{} \angle{B'A'C'}) \equal{} \angle{C'B'A'} \plus{} \angle{B'C'A'}
\]
(using directed angles).  Now what's the angle between $ B'C'$ and $ BC$?  Well it's gotta be
\[ \angle{B'C'A} \plus{} \angle{ABC} \equal{} \angle{B'A'C'} \plus{} \angle{ABC}
\]

\[ \equal{} \angle{B'A'C'} \plus{} \angle{A'C'B} \plus{} \angle{ABC} \plus{} \angle{BC'A'}
\]

\[ \equal{} \angle{B'A'B} \plus{} \angle{BC'A'} \equal{} \angle{B'A'C} \plus{} \angle{BC'A'}
\]

\[ \equal{} \angle{B'C'A'} \plus{} \angle{C'B'A'}
\]
(again, using directed angles modulo $ \pi$).  Aha!  So the angle between $ B''C''$ and $ B'C'$ equals the angle between $ BC$ and $ B'C'$, in the same direction.  Hence, $ B''C''\parallel{}BC$.  Similarly, $ A''B''\parallel{}AB, C''A''\parallel{}CA$.  So then there is a homothety between $ \triangle{A''B''C''}$ and $ \triangle{ABC}$.  Hence, there is a homothety between $ \triangle{A''B''C''}$ and $ \triangle{M_aM_bM_c}$.

But there is a homothety between the incircle and nine-point circle of the triangle about their tangency point.  Because these are the circumcircles of $ A''B''C''$ and $ M_aM_bM_c$, it follows that $ M_aA'', M_bB'',$ and $ M_cC''$ concur at that homothety point, the tangency point of the two circles.[/hide]

[quote="K81o7"]

But there is a homothety between the incircle and nine-point circle of the triangle about their tangency point.  Because these are the circumcircles of $ A''B''C''$ and $ M_aM_bM_c$, it follows that $ M_aA'', M_bB'',$ and $ M_cC''$ concur at that homothety point, the tangency point of the two circles.[/quote]

A little question here... maybe my own misconception, but please forgive me anyway.

Although there's a homothety between the two triangles, it is possible that the triangle is "inversely similar", or the corresponding vertices are oriented in opposite directions. If this is the case, can you be sure that the corresponding lines concur at the homothety point?

[quote="K81o7"]
It suffices to show that the polars of $ X_a, X_b,$ and $ X_c$ concur on the incircle.  These lines are $ M_aA'', M_bB'',$ and $ M_cC''$, ...[/quote]

Anoher way to show this:

The 9-point circle $ (N)$ is tangent to both the incircle $ (I)$ and the A-excircle $ (I_a).$ Inversion with center $ M_a$ and power $ M_aA'^2$ takes both $ (I)$ and $ (I_a)$ into themselves, because they are both perpendicular to the inversion circle. It takes $ (N)$ passing through the inversion center $ M_a$ into the common internal tangent $ n_a$ of $ (I), (I_a)$ other than $ BC,$ i.e. to the reflection of $ BC$ in $ AI \equiv II_a.$ $ A''$ is the tangency point of $ n_a$ with $ (I),$ the inversion image of the Feuerbach point $ F \equiv (I) \cap (N)$ $ \Longrightarrow$ $ A'', F$ are collinear with the inversion center $ M_a.$ Similarly, $ M_bB'', M_cC''$ pass through $ F.$

[quote="massnet"]Although there's a homothety between the two triangles, it is possible that the triangle is "inversely similar", or the corresponding vertices are oriented in opposite directions. If this is the case, can you be sure that the corresponding lines concur at the homothety point?[/quote]

Yup, they still have to concur at a point.

It's just a negative homothety ratio.  If you create the segments connecting corresponding vertices, the homothety point will be in the interior of each of these segments: it will be their point of concurrency.

the line $ r_1$ is inverse of nine-pointcircle wrt a circle with center $ M_a$ and radius $ M_aA'$ hence $ M_aA''$ pass through fuerbach point  ($ K$),where $ A''$ is reflection of $ A'$ in $ AI$
line $ A'X_a$ is polar of $ M_a$ hence the polar of $ X_a$  pass through $ M_a$ and also $ A''$ hence $ X_aK$ is tangent to the incircle ,therfore $ X_a,X_b,X_c,K$ is tangent to the incircle.

[quote="massnet"][quote="K81o7"]

Yup, they still have to concur at a point.

It's just a negative homothety ratio.  If you create the segments connecting corresponding vertices, the homothety point will be in the interior of each of these segments: it will be their point of concurrency.[/quote]

By "homothety point", do you mean the point where the two circle touches. If so, I think I can draw a picture where the two triangles are inversely similar, and the segments connecting corresponding vertices are concurrent, but not at the point where the two circle touches.[/quote]

Oh, I see what you're implying.  Once you've shown the corresponding sides of $ \triangle{A''B''C''}$ and $ \triangle{M_aM_bM_c}$ are parallel, I think you can show they're in the same orientation.  Assume they are in opposite orientations.  Because $ A'A'' \parallel{} B'C'$, we have that $ A''$ does not lie on minor arc $ BC$.  Now notice that $ \triangle{A''B''C''}$ and $ \triangle{ABC}$ are in the same orientation, and have parallel sides.  Then $ AA'', BB'', CC''$ concur at some point $ P$ (the external center of similitude of the incircle and circumcircle).  So $ A''$ is the first intersection of $ AP$ with the circumcircle, which must lie on arc $ BC$.  Contradiction.

[quote="massnet"]In triangle $ ABC$ let $ M_a,M_b,M_c$ be midpoints of $ BC,CA,AB$ respectively. The incircle with center $ I$ of triangle $ ABC$ touches $ BC,CA,AB$ at $ A',B',C'$. The line $ r_1$ is the reflection of line $ BC$ in $ AI$, and line $ r_2$ is the perpendicular from $ A'$ to $ IM_a$. Denote by $ X_a$ the intersection of $ r_1$ with $ r_2$, and define $ X_b,X_c$ analogously. Prove that $ X_a,X_b,X_c$ lie on a line tangent to the incircle of triangle $ ABC$.[/quote]
This problem was proposed by [b]Jan Vonk[/b] on [i]Mathematical Reflections[/i]. And here's his solution.
[quote="Jan Vonk"]Let us call $ N$ the center of the nine-point circle of triangle $ ABC$, and $ Z$ the [i]Feuerbach point[/i]. We will show that $ X_a$ lies on the tangent to $ (I)$ at $ Z$, and similar proofs for $ X_b$ and $ X_c$ will then prove the desired result.

Let us call $ A''$ the reflection of $ A'$ in AI. Clearly, $ A'' \in (I)$. Therefore, $ r_1$ will be the tangent (or polar line) of $ A''$ with respect to $ (I)$. By the definition of $ r_2$, we know that is is the polar line of $ D$ with respect to $ (I)$. It now follows that $ M_aA''$ is the polar line of $ X_a$ with respect to $ (I)$. It suffices to prove that $ Z$ lies on $ M_aA''$.

Because $ \triangle M_aM_bM_c$ is similar to $ \triangle ABC$, can easily obtain that $ \angle NM_aM_c \equal{} \frac {\pi}{2} \minus{} \angle ABC$ $ (1)$. Now we notice that $ \angle (IA'', M_aM_c) \equal{} \angle (IA'', AC) \equal{} \frac {\pi}{2} \minus{} \angle (r_1, AC) \equal{} \frac {\pi}{2} \minus{} \angle ABC$ $ (2)$. Combining $ (1)$ and $ (2)$ implies that $ NM_a \parallel IA''$.

Because $ N$ and $ I$ are the centers of the nine-point circle and the incircle, we know that $ DA''$ goes through their external center of similitude, which in this case is evidently $ Z$ because of the tangency of the two circles due to the [i]Feuerbach theorem[/i]. This completes the proof.[/quote]


Given $ n\geq 2,n\in N$ and $ p\in R^{\plus{}}$.Find the best constant $ C_{p,n}$ such that;
                                  $ \sum_{1\leq i < j\leq n}x_{i}x_{j}(x_{i}^{p}\plus{}x_{j}^{p})\leq C_{p,n}.(\sum_{i \equal{} 1}^{n}x_{i})^{p\plus{}2}$ for all nonnegative number $ x_{1},x_{2},...,x_{n}$
When $ p\geq 2$and $ 0 < p\leq 1$ i have found that $ C_{p,n}\equal{} Min(\frac{p^{p}}{(p\plus{}1)^{p\plus{}1}},\frac{1}{2^{p\plus{}1}})$.
When $ 1\leq p < 2$ i have found that $ C_{p,n}\equal{} Min(\frac{1}{2^{p\plus{}1}},Max((1\minus{}t).t^{p\plus{}1}\plus{}t.(1\minus{}t)^{p\plus{}1}:\frac{1}{2}< t\leq 1))$
I hope someone can help me with the case $ 1\leq p < 2$ :(

This problem is very hard, Nhat. I posted before the case $ p\equal{}4$ only.

For the case $ p \equal{} 4$ i have found that $ C_{4,n}\equal{}\frac{1}{2^{5}}$(base on my result)
The solution for the case $ p\geq 2$ is incredible.It costs me a lot of time to finish.
For the case $ p \equal{} 5,6$ i have found a very nice solution.Do you want me to write this solution?

I am pleased to know your solution  :)

Actually, for $ p\ge 4$, we may assume that $ a_{1}\equal{}a_{2}\equal{}...\equal{}a_{n\minus{}2}\equal{}0$. (not sure but I think this statament holds for $ p\ge 2$). This is the same as the case $ p\equal{}4$ (so we can prove for $ p\equal{}5,p\equal{}6$ by this way). But I think solution for general case is very hard. Congratualtion you, nhat!


Duke college Support ratio algebra polynomial geometry

How do the community-based math clubs/circles in NC operate? Harold Reiter started one in Charlotte that seems to be a model.  Are there others?  What happens at typical sessions and how are they different from math team practices at high schools and middle schools?  A Wake Forest math professor and I are thinking about starting one in Winston-Salem.  Any suggestions or thoughts?

There are math circles in NC? I seem to have a lack of information

I'm almost positive there AREN'T others. For a while I really wanted one in Wilmington, but I came to the realization that those people who are really interested find ways to fuel themselves, and those who aren't don't know the difference. That and they're too much work to get together, but best of luck in starting one! That would be great to have more math circles around!

If you can't start a [i]real[/i] math circle, I think the best option is to find the (few?) kids who are interested and just meet over the weekends independently of any school activities. That seems to be beneficial, and may ultimately lead to something greater.

We're starting by getting some kids who think they want to compete at the Duke Math Meet.  Hopefully, they'll still be interested after the contest.

I'm not sure how it will evolve, but I suspect that contest preparation will be a large part of the "draw."

I had also suspected that there might not be very many math circles in NC.  Raleigh-Durham-Chapel Hill could easily support one though.

It definitely seems like RDU could, but I get the impression that the East Chapel Hillers are quite content staying in their East Chapel Hill math circle, and likewise for the other schools. They actually have enough interest in their own school for that.

There is a possibility of a Duke Math Circle in the future.  I'll keep you all updated.

[quote="gt59"]Saturday Morning Math was launched Saturday Dec, 2 at Wake Forest U. in Winston-Salem.  There were about 20 students in 6th to 10th grade and several parents in attendance.  We plan to meet once per month with the next meeting planned for January 6.  Details appear on the [url=http://web2k.wsfcs.k12.nc.us/hanesms/academics/mcounts.htm]Hanes Math site.[/url]

I've heard that Buncombe County had a math circle meeting the same day.  Anyone know what went on at the Buncombe session?[/quote]

I went to the Buncombe session  :lol:
Here are some problems we did:

Ratios practice:
# 11 and # 21 of 2000 AMC 12
Polynomials practice:
* Suppose that $a$ and $b$ are integers such that $x^{2}-x-1$ is factor of $ax^{3}+bx^{2}+1$. What is $b$?
* What is the remainder when $x^{51}+51$ is divided by $x+1$ ?

Then here is one of our HW problems:
* The radius of a given circle is 1 inch, that of a second is ½ inch, that of a third is ¼ inch and so on indefinitely. What is the sum of the areas of the circles?

I'm in the high school section and we are pretty much practicing for the AMC 12. It was sooo much fun...  :D This was great practice for me since this will be my first time going to a math competition.  :blush:
The next one is on the 12/16/06

Did anyone else go to the Buncombe session?  :huh:

Today, we met for a second time.  This time, the first part was in lecture style, but the second half was group work, so it was much better.

We now have a blackboard set up to keep the problems on, provided by Wake Forest University.

This time, we covered a bunch of things on sequences, most of which was more appropriate for everyone else (it was mainly middle school students).  Hopefully, everyone will understand sequences well and be able to use it at MathCounts.

We are from various middle and highschools in the Winston-Salem area.


[hide]Learn $\LaTeX$!
[hide="1"] $g(h(x))$
[/hide]
[/hide]
[hide="2"] Your use of the word radicand leads me to think that you don't know what it means.  The radicand is what is under the radical. i.e. $\sqrt{2}$, the radicand is $2$.

[hide="2"]$h(h(x))$[/hide][/hide]

Yes, radicand = what lies within the radical symbol.

Now, what are the steps that you used to find the answers?

[hide]
1. $p(x)=3\sqrt{x}+3=3(\sqrt{x}+1)=3h(x)=g(h(x))$.
2. $Q(x)=\sqrt{\sqrt{x}+1}+1=\sqrt{h(x)}+1=h(h(x))$.[/hide]
[quote="sharkman"](2) Q(x) = (sqrt(sqrt{x}) + 1}) + 1 [/quote]
That is actually misleading, because you opened with a ( and closed with a }.

[quote="bpms"][quote="b-flat"][quote="sharkman"]Let f(x) = x^2, g(x) = 3x and h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

(1) p(x) = 3(sqrt{x}) + 3

(2) Q(x) = (sqrt(sqrt{x}) + 1}) + 1

NOTE:

Question 2 is a radical within a radical.

So, sqrt{x} lies inside a BIGGER radical, which also includes + 1 in the radicand and OUTSIDE the radicand.

Is this clear?[/quote]

You could be ultra-clear by taking a few minutes (it's rather easy) to learn $\LaTeX$.  Try [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url] for more help.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=152626
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=151216
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=148345
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=132797
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=131688
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=131721
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=131187
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=130399
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=130251
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=130094
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=130237
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=129391
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=129133
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=128981
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=159770217&t=127739

He has been urged in all of these posts to learn latex. Why should he bother now? People still answer his questions. He doesn't have to go out of his way to help make his posts more presentable, or any of the other suggestions he gets, because people still solve his problems anyways.[/quote]

AMEN!

this topic was started after the warning so im locking it because its really easy.

and also nobody has really answered the question so far anyways.


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Direct limit is quite new for me.
In some book, I read that 'If $ R$: integral domain with the quotient field $ Q$, then $ \varinjlim \frac{1}{r} R\equal{}Q$.'

What is a direct system which has the direct limit $ Q$?
What's the order in I, if {$ \frac{1}{r} R$}$ _I$ is such a direct system?

if $ r|s$, we have $ 1/r R \subseteq 1/s R$. thus the $ 1/r R$ form a direct system with respect to the partial order $ |$.

Let {$ 1/r R$}$ _I$ is such a direct system.
If for any $ r \in I$, $ 1/r R \subseteq Q$, can we say that $ \varinjlim 1/r R \subseteq Q$?
Please, give me a full explanation.

@ Byun: What do the elements of Q look like? Can this help you to prove the inverse inclusion?


geometry
geometric transformation
rotation
parallelogram
trigonometry
inequalities
triangle inequality

find the area of an equilateral triangleif the distances of a point inside it from the three vertices are 5, 6 and 7.

there are many to do this one,but almost all involve billions(hyperbole) of variables and third degree equations.
therefore try to think of a method that is elegant and actuallly gives an answer.

There is a general formula for this kind of problems in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67254]here.[/url]

[hide="BIG hint"]
Rotations.  :wink: [/hide]

[hide="Answer"]
$\frac{ 107 \sqrt{3}}{4}$[/hide]

[quote="rishabh"]uhhhhh, cincodemayo5590, im really sorry but could you just be a little more specific about the rotations part?

once again thanks a lot for your help[/quote]

Yes, what do you rotate it about?

Edit: I think you rotate it $60^\circ$ about a side.

I can get the answer now but I get a weird answer.

$\frac{55\sqrt{3}+36\sqrt{6}}{4}$

[quote="puuhikki"][quote="mathgeniuse^ln(x)"]Yes, what do you rotate it about?[/quote]
See the link I gave.[/quote]

Yeah, I saw that, but I was wondering if there was a way to do it without formulae.

It depends what do you mean by formula. I proved the formula I gave in that link with elementary geometry and trigonometry. I think the other solutions are longer or they are using more advanced geometric facts.

[hide="This is how I did it:"]
Take the triangle, and rotate it $60^\circ$ about each side. Connect each of the rotated points with the common vertices of the original triangle and the rotated one, and connect the original point to all three vertices of the original triangle. Do you see the parallelograms that form?

We can easily show that the angles of each parallelogram which go through the common side are equal to $60^\circ$. Then, all we need to do is find the area of those three parallelograms and halve the sum, and we will have the area of the original triangle.[/hide]

Yeah, that is what I did, but I am wondering, if you made a computation error, because $\frac{107\sqrt{3}}{4}$, is actually a nice answer, but I don't know.

[hide="Computations"]
The area of a parallelogram is $ab \sin \theta$, so the area of the triangle will be $\left( \frac{5 \cdot 6+6 \cdot 7+7 \cdot 5}{2}\right) \cdot \sin 60^\circ = \left( \frac{107}{2}\right) \left( \frac{ \sqrt 3}{2}\right) = \boxed{ \frac{ 107 \sqrt 3}{4}}$.

As a check, we can see that the side length of the triangle is $\sqrt{107}\approx 10.3$, which fits the triangle inequality for the three triangles formed within the equilateral one. If you really wanted to, you could calculate three angles at the point (via Cosine Law) and make sure that they sum to $360^\circ$.[/hide]

But, if you rotate, wouldn't the side length of 5 be the one emanating from the vertex that has the segment of length five? Likewise, for 6 and 7.

In your picture it doesn't seem that when you rotate it 60 degrees, the segments coming from that vertex are the same as the segments coming from the same vertex after the rotation.

And also, my segment length is $\sqrt{55+36\sqrt{2}}\approx 10.3$

Oh wait, you're right. My diagram is off; they should be kites instead of parallelograms. Sorry!  :P

That's okay, but your idea of rotation still works, cause now you can find one of the interior angles made, and then use law of cosines to get the side and hence the area.


[delete]

I just realized some dude is posting a youtube video all over with the same thing


search
combinatorics unsolved
combinatorics

how many 7 digit numbers divisible by 3 is possible using 0,1,2,3,4,5,6.repetition is allowed.there are too many cases when i solve this question,pl suggest some another way.pl provide a simple way so that a student  of grade 12 can understand

A search for "7 digit divisible" quickly brings up [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=675695514&t=287936]this thread[/url].


symmetry
calculus
derivative
geometry
inequalities proposed
inequalities

Well it is surely the sum of the distances from $(x, y)$ to $(-1, 1)$, $(1,-1)$, and $(-2,-2)$, and we win when the three are equal. After noting that $x=y$ (by symmetry) it remains to find where $x^{2}+1 = (x+2)^{2}$, which is $x =-3/4$. This gives a minimum of $\boxed{\frac{15}{4}}$.

[quote="Xevarion"] ...we win when the three are equal[/quote]
Why?
[quote="Xevarion"] After noting that $x=y$ (by symmetry) it remains to find where $x^{2}+1 = (x+2)^{2}$, which is $x =-3/4$. This gives a minimum of $\boxed{\frac{15}{4}}$.[/quote]
If so $\frac{15\sqrt2}{4}.$ :wink:
But for $(-\frac{1}{\sqrt3},-\frac{1}{\sqrt3})$ the value is $4\sqrt{\frac{2}{3}}+\sqrt{\frac{26-8\sqrt3}{3}}<\frac{15\sqrt2}{4}.$
I think the answer is $\sqrt{14+8\sqrt3}.$ :wink:

We win when the point $(x,y)$ is the Fermat point of the triangle. That is  when $(x,y) = \left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$.

I see: $x=y$ is fine but the three terms being equal is silly (that only happens if the product is constrained I guess...).

In fact all we have to do is minimize $2\sqrt{x^{2}+1}+(x+2)$, which is (sorry, not a pretty way) easily done by taking the derivative and setting equal to 0. There is a quadratic. We get arqady's result.

 :blush:  :blush:  :blush:

edit: I see shobber is faster by knowing some nice trick. Of course if I were a grader, even if I knew what Fermat point was, I might like to see how you calculated it...

[quote="arqady"]The Torricelli's point. :wink:[/quote]
Can u tell me what's this exactly ??
Sorry, but I'm only a beginner in ineq.   :oops:

As far as I know,Toricelli point=first Fermat point

First Fermat point = Torricelli point = first isogonic center.
[url=http://mathworld.wolfram.com/FirstFermatPoint.html]Mathworld[/url]
However, naming the point they asked for and then giving the answer isn't very enlightening for those of us who don't know the geometric background... Is there a nice way to calculate it that comes from geometry?


number theory
relatively prime
number theory unsolved

Prove that in any sequence of $ 16$ consecutive integers, there exists at least $ 1$ integer that is relatively prime to the $ 15$ other ones.

there are alternating even and odd numbers which dont divide each other

aadil: What do you want to imply that way¿ It's not _that_ trivial  :wink:


Let $a,b,c$ are non-negative real number satified $a+b+c=3$. Prove that
\[\frac{a}{2b^{2}+c^{2}+1}+\frac{b}{2c^{2}+a^{2}+1}+\frac{c}{2a^{2}+b^{2}+1}\ge \frac{3}{4}\]

No, I think mine is true. Could you give some counterexamples?

[quote="The soul of rock"]Let $a,b,c$ are non-negative real number satified $a+b+c=3$. Prove that
$\frac{a}{2b^{2}+c^{2}+1}+\frac{b}{2c^{2}+a^{2}+1}+\frac{c}{2a^{2}+b^{2}+1}\ge \frac{3}{4}$
[/quote]
I think, it's wrong. Try $a=b=1.5$ and $c=0.$ :wink:

I can confirm Arqady: The inequality is wrong with either sign. Please check your software, fjwxcsl.

  Darij

[quote="fjwxcsl"]darij grinberg, the original  inequality inversely holds!

 inversely holds![/quote]

It doesn't. Take a = b = 3/4, c = 3/2.

  darij


1, Find the minimum value of:
$ A \equal{} \frac{5 x^{2} \plus{}\frac{45}{7}}{(7x \plus{}1 )^{2}}$
2, Find the maximum and minimum value of:
$ B \equal{} 7(x ^{4} \plus{} y^{4}) \plus{}16x^{2}y^{2}$
$ x, y$ are real numbers such that $ 2(x^{2} \plus{} y^{2}) \equal{} 1\plus{} xy$

[quote="Nam Luu"]
2, Find the maximum and minimum value of:
$ B = 7(x ^{4} + y^{4}) + 16x^{2}y^{2}$
$ x, y$ are real numbers such that $ 2(x^{2} + y^{2}) = 1 + xy$[/quote]

Let $ xy = t$, we have $ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = \left(\frac {1 + t}{2}{\right)^2 - 2t^2 = \frac { - 7t^2 + 2t + 1}{4}}$.

$ \therefore B: = f(t) = \frac {7}{4}( - 7t^2 + 2t + 1) + 16t^2$

$ = \frac {15}{4}t^2 + \frac {7}{2}t + \frac {7}{4}$

$ = \frac {15}{4}\left(t + \frac {7}{15}\right)^2 + \frac {14}{15}$

Now the range of $ t$ is obtained as follows:

$ 2(x^2 + y^2)\geq 4|xy|\Longleftrightarrow 1 + t\geq 4|t|$, yielding $ - \frac {1}{5}\leq t\leq \frac {1}{3}$.

$ \therefore f\left( - \frac {1}{5}\right)\leq B\leq f\left(\frac {1}{3}\right)$, yielding $ \boxed{B_{max} = \frac {2}{9},\ B_{min} = \frac {14}{15}}$

[quote="Nam Luu"]1, Find the minimum value of:
$ A \equal{} \frac {5 x^{2} \plus{} \frac {45}{7}}{(7x \plus{} 1 )^{2}}$[/quote]

$ A\equal{}\frac{5}{49}\left\{64\left(\frac{1}{7x\plus{}1}\minus{}\frac{1}{64}\right)^2\plus{}\frac{63}{64}\right\}$

[quote="Nam Luu"]1, Find the minimum value of:
$ A \equal{} \frac {5 x^{2} \plus{} \frac {45}{7}}{(7x \plus{} 1 )^{2}}$
[/quote]
 Hello
I get that the result is : 45 / 448  :maybe:  :maybe:
mrmohamed