For $ a,b,c\ge0$ Prove that $ \frac{a^{3}}{bc}\plus{}\frac{b^{3}}{ac}\plus{}\frac{c^{3}}{ab}\ge a\plus{}b\plus{}c$

[hide="1st Solution"]
$ \frac {a^{3}}{bc} \plus{} b \plus{} c\ge 3a$

$ \frac {b^{3}}{ac} \plus{} c \plus{} a\ge 3b$

$ \frac {c^{3}}{ab} \plus{} a \plus{} b\ge 3c$
Adding all the 3 we get desired result

[/hide]

[hide="2nd Solution"]Use Cauchy - Buniakowski - Schwartz :

$ LHS \equal{} \sum_{cyc}\frac {a^4}{abc}\ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{3abc}$

So it'll be enough to prove that :

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2\sum_{cyc}a^2b^2\ge 3abc(a \plus{} b \plus{} c)$

But $ a^4 \plus{} b^4 \plus{} c^4\ge abc(a \plus{} b \plus{} c)$

$ \sum_{cyc}a^2b^2\ge abc(a \plus{} b \plus{} c)$ , which are trivial .[/hide]

[hide="3rd Solution"]The given inequality can be transformed into $ \frac{a^4}{abc}\plus{}\frac{b^4}{abc}\plus{}\frac{c^4}{abc}\ge a\plus{}b\plus{}c$.

The inequality $ a^4\plus{}b^4\plus{}c^4\ge a^2bc\plus{}ab^2c\plus{}abc^2$ is true by Muirhead's Inequality since $ (4,0,0)$ majorizes $ (2,1,1)$. $ \blacksquare$[/hide]

Hopefully I used Muirhead correctly here :D

[hide="4th Solution"]We use Holder :

$ \left(\frac {a^3}{bc} \plus{} \frac {b^3}{ca} \plus{} \frac {c^3}{ab}\right)(ab \plus{} bc \plus{} ca)(1 \plus{} 1 \plus{} 1)\ge (a \plus{} b \plus{} c)^3$

$ \Rightarrow LHS\ge \frac {(a \plus{} b \plus{} c)^3}{3(ab \plus{} bc \plus{} ca)}\ge \frac {(a \plus{} b \plus{} c)^3}{(a \plus{} b \plus{} c)^2}\ge a \plus{} b \plus{} c$[/hide]

[hide="5th Solution"]

$ \frac{a^{3}}{bc}\plus{}\frac{b^{3}}{ac}\plus{}\frac{c^{3}}{ab}\ge\frac{(a^{3}\plus{}b^{3}\plus{}c^{3})}{3}(\frac{1}{bc}\plus{}\frac{1}{ac}\plus{}\frac{1}{ab})$

$ a^{3}\plus{}b^{3}\plus{}c^{3}\ge 3abc$ put this value and get the desired result
[/hide]

[hide="6th Solution"]By rearrangement: $ a^3 \geq b^3 \geq c^3$ and $ \frac {1}{bc} \geq \frac {1}{ac} \geq \frac {1}{ab}$[/hide]

[hide="7th Solution"]Use AM-GM :

$ \frac{a^3}{bc}\plus{}\frac{b^3}{ca}\ge 2\frac{ab}{c}$ and similars

$ \Rightarrow \frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^3}{ab}\ge \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}$

Again AM-GM :

$ \frac{ab}{c}\plus{}\frac{bc}{a}\ge 2b$ and similars

$ \Rightarrow \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}\ge a\plus{}b\plus{}c$[/hide]

[hide="8th solution"]

well this is quite the same as 6th ,however ...

WLOG $ a\geq b \geq c$

Now by Chebyshev we have

$ LHS \geq \frac{1}{3}(a+b+c) \sum_{cyc}{\frac{a^2}{bc} \geq \frac{1}{3}(a+b+c)\frac{(a+b+c)^2}{ab+bc+ca} \geq a+b+c}$[/hide]

[hide="9th Solution"]Recall from 7th Solution that $ \frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^3}{ab}\ge \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}$

Then by Cauchy :

$ \left(\frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^3}{ab}\right)^2\ge \left(\frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^3}{ab}\right)\left(\frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}\right)\ge (a\plus{}b\plus{}c)^2$[/hide]

I call SOS:

$ a^4\plus{}b^4\plus{}c^4\ge a^2bc\plus{}b^2ca\plus{}c^2ab$

$ \frac{1}{2}\sum_{cyc}(a\minus{}b)^2(a^2\plus{}b^2\plus{}c^2\plus{}2ab)\ge 0$

[hide="11th solution"]
By Cauchy,

$ (1\plus{}1\plus{}1)(x^4\plus{}y^4\plus{}z^4) \geq (x^2\plus{}y^2\plus{}z^2)^2$

Similarly,

$ (1\plus{}1\plus{}1)(x^2\plus{}y^2\plus{}z^2) \geq (x\plus{}y\plus{}z)^2$

Therefore $ (x\plus{}y\plus{}z)^2 \leq 3(x^2\plus{}y^2\plus{}z^2)$

$ (x\plus{}y\plus{}z)^4 \leq 9(x^2\plus{}y^2\plus{}z^2)^2 \leq 27(x^4\plus{}y^4\plus{}z^4)$

By AM-GM,

$ \frac{x\plus{}y\plus{}z}{3} \geq \sqrt[3]{xyz} \implies (x\plus{}y\plus{}z)^3 \geq 27xyz$

It then follows that

$ 27(x^4\plus{}y^4\plus{}z^4) \geq 9(x^2\plus{}y^2\plus{}z^2)^2 \geq (x\plus{}y\plus{}z)^4 \geq 27xyz(x\plus{}y\plus{}z)$

$ 27(x^4\plus{}y^4\plus{}z^4) \geq 27xyz(x\plus{}y\plus{}z)$

$ x^4\plus{}y^4\plus{}z^4 \geq xyz(x\plus{}y\plus{}z)$

Dividing by xyz, we win.

[/hide]

[hide=" 12th Solution"]We use 3 applications of AM-GM.

$ \frac{a^4\plus{}a^4\plus{}b^4\plus{}c^4}{4} \ge \sqrt[4]{a^8b^4c^4}$, or rather:

$ \frac{2a^4\plus{}b^4\plus{}c^4}{4} \ge a^2bc$

Similarly, $ \frac{2b^4\plus{}a^4\plus{}c^4}{4} \ge b^2ac$ and $ \frac{2c^4\plus{}b^4\plus{}a^4}{4} \ge c^2ab$

Adding these three inequalities we obtain:

$ \frac{4a^4\plus{}4b^4\plus{}4c^4}{4} \ge a^2bc\plus{}b^2ca\plus{}c^2ab$ or rather:

$ a^4\plus{}b^4\plus{}c^4 \ge a^2bc \plus{} b^2ca \plus{} c^2ab$

Dividing both sides by $ abc$ we obtain:

$ \frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^2}{ab} \ge a\plus{}b\plus{}c$ as desired.[/hide]

[hide="13th solution"]$ LHS \equal{} \frac {\sum a \cdot a^3}{abc} \ge \frac {(\sum a)(\sum a^3)}{3abc}$   By Chebychev

But $ \sum a^3 \ge 3abc$ by AM-GM   $ \implies LHS \ge \sum a$

Hence proved[/hide]

[quote="Mewto55555"]
$ (x \plus{} y \plus{} z)^4 \leq 9(x^2 \plus{} y^2 \plus{} z^2)^2 \leq 27(x^4 \plus{} y^4 \plus{} z^4)$
[/quote]

That is one row with Power mean...

Thanks to all those who posted their solutions.
13 solutions :omighty: Great work.!!

More solutions are welcomed!!

[quote="enndb0x"]$ (3, \minus{} 1, \minus{} 1) \succ (1,0,0)$[/quote]
[b]Muirhead  (three variable)[/b]

Let $ a_1$, $ a_2$, $ a_3$, $ b_1$, $ b_2$, $ b_3$ be real numbers such that

$ a_1 \ge a_2 \ge a_3 \ge 0$,
$ b_1 \ge b_2 \ge b_3 \ge 0$,

$ a_1 \ge b_1$,
$ a_1\plus{}a_2 \ge b_1\plus{}b_2$,
$ a_1\plus{}a_2\plus{}a_3 \equal{} b_1\plus{}b_2\plus{}b_3$.

Let $ x$, $ y$, $ z$ be positive real numbers. Then, we have
\[ \sum_{sys}x^{a_1}y^{a_2}z^{a_3} \ge \sum_{sys}x^{b_1}y^{b_2}z^{b_3}.\]

OK, but it's clearly true, multiply by abc  :wink:
You'll get $ (4,0,0)\succ (2,1,1)$

[quote="Kouichi Nakagawa"][quote="enndb0x"]$ (3, \minus{} 1, \minus{} 1) \succ (1,0,0)$[/quote]
[b]Muirhead  (three variable)[/b]

Let $ a_1$, $ a_2$, $ a_3$, $ b_1$, $ b_2$, $ b_3$ be real numbers such that

$ a_1 \ge a_2 \ge a_3 \ge 0$,
$ b_1 \ge b_2 \ge b_3 \ge 0$,

$ a_1 \ge b_1$,
$ a_1 \plus{} a_2 \ge b_1 \plus{} b_2$,
$ a_1 \plus{} a_2 \plus{} a_3 \equal{} b_1 \plus{} b_2 \plus{} b_3$.

Let $ x$, $ y$, $ z$ be positive real numbers. Then, we have
\[ \sum_{sys}x^{a_1}y^{a_2}z^{a_3} \ge \sum_{sys}x^{b_1}y^{b_2}z^{b_3}.\]
[/quote]
Sorry "Kouichi Nakagawa" I didn't get your point for quoting "enndb0x" is it because of "$ a_1 \ge a_2 \ge a_3 \ge 0$,
$ b_1 \ge b_2 \ge b_3 \ge 0$," but I think that Murihead works with negative stuff as well isn't it  :wink:
Please explain to me I'm so confused  :) .

Hello,

I don't want to double post but can someone really clarify this (the post before this) for me please.

Thanking you in advance  :)

Sorry, I didn't read all the above methods.

[b][u]Method VN1[/u].[/b] I"ll use only a simple well-known inequality $ x^2 \plus{} y^2 \plus{} z^2\ge xy \plus{} yz \plus{} zx$

for any real numbers $ x$ , $ y$ , $ z$ . [u]Indeed[/u], $ \sum\frac {a^3}{bc}\ge \sum a\Longleftrightarrow a^4 \plus{} b^4 \plus{} c^4\ge abc(a \plus{} b \plus{} c)\ (*)$ . Therefore,

$ a^4 \plus{} b^4 \plus{} c^4\ge a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} (ab)^2$ $ \plus{} (bc)^2 \plus{} (ca)^2\ge ab\cdot bc \plus{} bc\cdot ca \plus{} ca\cdot ab \equal{} abc(a \plus{} b \plus{} c)$ , i.e  $ (*)$ .

[b][u]Method VN2[/u].[/b] I"ll use only a very simple well-known inequality $ x \plus{} y\ge\sqrt {xy}$ for any positive numbers $ x$ , $ y$ . [u]Indeed[/u],

$ \sum\frac {a^3}{bc} \equal{} \frac 12\cdot\sum\left (\frac {b^3}{ac} \plus{} \frac {c^3}{ab}\right)$ $ \ge \frac 12\cdot 2\cdot\sqrt {\frac {b^3}{ac}\cdot\frac {c^3}{ab}} \equal{} \sum \frac {bc}{a}$ $ \equal{} \frac 12\cdot\sum a\cdot \left(\frac {c}{b} \plus{} \frac {b}{c}\right)\ge \frac 12\cdot 2\cdot \sum a \equal{} a \plus{} b \plus{} c$ .

[hide="14th solution"]First substitution : $ a_1 \equal{} bc\ ,\ b_1 \equal{} ca\ ,\ c_1 \equal{} ab$.

$ \frac {a^3}{bc} \plus{} \frac {b^3}{ca} \plus{} \frac {c^3}{ab}\ge a \plus{} b \plus{} c\Leftrightarrow \frac {a_1\sqrt {a_1b_1}}{c_1\sqrt {c_1}} \plus{} \frac {b_1\sqrt {b_1c_1}}{a_1\sqrt {a_1}} \plus{} \frac {c_1\sqrt {c_1a_1}}{b_1\sqrt {b_1}}\ge$ $ \sqrt {\frac {a_1b_1}{c_2}} \plus{} \sqrt {\frac {b_1c_1}{a_1}} \plus{} \sqrt {\frac {c_1a_1}{b_1}}$

Second substitution : $ x \equal{} \sqrt {a_1}\ ,\ y \equal{} \sqrt {b_1}\ ,\ z \equal{} \sqrt {c_1}$

$ \frac {a_1\sqrt {a_1b_1}}{c_1\sqrt {c_1}} \plus{} \frac {b_1\sqrt {b_1c_1}}{a_1\sqrt {a_1}} \plus{} \frac {c_1\sqrt {c_1a_1}}{b_1\sqrt {b_1}}\ge \sqrt {\frac {a_1b_1}{c_2}} \plus{} \sqrt {\frac {b_1c_1}{a_1}} \plus{} \sqrt {\frac {c_1a_1}{b_1}}$ $ \Leftrightarrow \frac {x^3y}{z^3} \plus{} \frac {y^3z}{x^3} \plus{} \frac {z^3x}{y^3}\ge \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}$

for which can be found again very many solutions...

$ \frac {x^3y}{z^3} \plus{} \frac {y^3z}{x^3} \plus{} \frac {z^3x}{y^3}\ge \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}\Leftrightarrow x^6y^4 \plus{} y^6z^4 \plus{} z^6x^4\ge x^4y^4z^2 \plus{} x^4y^2z^4 \plus{} x^2y^4z^4$ , true by Muirhead .[/hide]

[hide="15th solution"]A directly substitution , let $ x \equal{} \frac {a}{bc}\ ,\ y \equal{} \frac {b}{ca}\ ,\ z \equal{} \frac {c}{ab}$

Then the inequality becomes :

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge \frac {1}{\sqrt {xy}} \plus{} \frac {1}{\sqrt {yz}} \plus{} \frac {1}{\sqrt {zx}}\Leftrightarrow \left(\frac {1}{\sqrt {x}} \minus{} \frac {1}{\sqrt {y}}\right)^2 \plus{}$ $ \left(\frac {1}{\sqrt {y}} \minus{} \frac {1}{\sqrt {z}}\right)^2 \plus{} \left(\frac {1}{\sqrt {z}} \minus{} \frac {1}{\sqrt {x}}\right)^2\ge 0$[/hide]

[hide="16th solution"]
Using chebycheff we get $ \frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}\ge\frac{(a^{3}+b^{3}+c^{3})}{3}({\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}})$ using

power mean in $ a^{3}+b^{3}+c^{3}\ge \frac{(a+b+c)^{3}}{9}$ putting this in above we are left to prove $ (a+b+c)^{2}\ge3(ab+bc+ac)$ which is trivial!!

[/hide]

[b]A similar inequality[/b] : in a triangle $ ABC$ , $ \left(\frac {p}{R\plus{}r}\right)^2\ \le\ \sum\frac {a^2}{bc}\ \le\ \frac {2R\minus{}r}{r}$ .

[quote="Virgil Nicula"][b]A similar inequality[/b] : in a triangle $ ABC$ , $ \left(\frac {p}{R \plus{} r}\right)^2\ \le\ \sum\frac {a^2}{bc}\ \le\ \frac {2R \minus{} r}{r}$ .[/quote]

For the second ineq.:

$ \sum\frac{a^2}{bc}\equal{}\frac{a^3\plus{}b^3\plus{}c^3}{abc}\equal{}\frac{2p(p^2\minus{}6Rr\minus{}3r^2)}{4RS}\equal{}\frac{p^2\minus{}6Rr\minus{}3r^2}{2Rr}$ .

$ \implies \sum\frac{a^2}{bc}\le \frac{2R\minus{}r}{r}\Longleftrightarrow \frac{p^2\minus{}6Rr\minus{}3r^2}{2R}\le 2R\minus{}r\Longleftrightarrow$ $ p^2\le 4R^2\plus{}4Rr\plus{}3r^2$, true (Gerretsen's ineq.)

on wiki it says that there is no lower bound on any of the a_n and b_n

a^3/ab+b^3/ac+c^3/ab.
we use AM-GM on terms
b^3/ac+c^3/ab>=2*bc/a
So
we know that a^3/ab+b^3/ac+c^3/ab>=ab/c+bc/a+ac/b
AM-GM on this we get:
ab/c+bc/a>=2b
so we know that this is >= a+b+c.

[quote] [color=darkred][b][u]An easy extension[/u].[/b] For any positive numbers $ a$ , $ b$ , $ c$ and $ n\in \mathbb N$ there is the relation

$ \boxed {\ \frac {a^{n \plus{} 2}}{bc} \plus{} \frac {b^{n \plus{} 2}}{ca} \plus{} \frac {c^{n \plus{} 2}}{ab}\ \ge\ a^n \plus{} b^n \plus{} c^n\ \ge\ \frac {(a \plus{} b \plus{} c)^n}{3^{n \minus{} 1}}\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] [/color]$ \left\|\ \begin{array}{ccc} 3\cdot\left(a^{n \plus{} 3} \plus{} b^{n \plus{} 3} \plus{} c^{n \plus{} 3}\right) & \ge & \left(a^n \plus{} b^n \plus{} c^n\right)\cdot \left(a^3 \plus{} b^3 \plus{} c^3\right) \\
 \\
3^{n \minus{} 1}\cdot \left(a^n \plus{} b^n \plus{} c^n\right) & \ge & (a \plus{} b \plus{} c)^n \\
 \\
a^3 \plus{} b^3 \plus{} c^3 & \ge & 3\cdot abc\end{array}\ \right\|\ \bigodot\ \implies$

$ a^{n \plus{} 3} \plus{} b^{n \plus{} 3} \plus{} c^{n \plus{} 3}\ \ge\ abc\cdot \left(a^n \plus{} b^n \plus{} c^n\right)$ $ \implies\ \frac {a^{n \plus{} 2}}{bc} \plus{} \frac {b^{n \plus{} 2}}{ca} \plus{} \frac {c^{n \plus{} 2}}{ab}\ \ge\ a^n \plus{} b^n \plus{} c^n$ .

[b][u]Remark[/u].[/b] I used thr remarkable inequality $ \boxed {\ \left(\frac {a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n}{n}\right)^p\ \le\ \frac {a_1^p \plus{} a_2^p \plus{} \ldots\ \plus{} a_n^p}{n}\ }\ \ (*)$ .
[hide="Proof."]
$ \left\|\ \begin{array}{c} \sum a_k\cdot\sum a_k\ \le \ n\cdot\sum a_k^2 \\
 \\
\sum a_k\cdot\sum a_k^2\ \le \ n\cdot\sum a_k^3 \\
............................................. \\
\sum a_k\cdot\sum a_k^{p \minus{} 2}\ \le \ n\cdot\sum a_k^{p \minus{} 1} \\
 \\
\sum a_k\cdot\sum a_k^{p \minus{} 1}\ \le \ n\cdot\sum a_k^p\end{array}\ \right\|\ \bigodot\ \implies$ $ \left(\sum a_k\right)^p\ \le\ n^{p \minus{} 1}\cdot\left(\sum a_k^p\right)\ \implies\ (*)$ .[/hide]
[quote] [color=darkred][b][u]An another extension[/u].[/b] For any positive numbers $ a$ , $ b$ , $ c$ and $ \{m,n\}\subset \mathbb N$ there is the relation

$ \boxed {\ \frac {a^{n\plus{} 2m}}{b^mc^m} \plus{} \frac {b^{n\plus{}2m}}{c^ma^m} \plus{} \frac {c^{n\plus{}2m}}{a^mb^m}\ \ge\ a^n \plus{} b^n \plus{} c^n\ \ge\ \frac {(a \plus{} b \plus{} c)^n}{3^{n \minus{} 1}}\ }$ .[/color] [/quote]


Hello,

Can someone please tell me where and when the Toronto Math Circles will be this year?

Thank you.

Does Anyone Know if there are Mississauga Math Circles???

EDIT: 99th Post!!!

[quote="mathboyzld"]I think it will be Saturdays from 1:00-3:00 pm at the Fields Institute of Mathematics at 222 College Street, east of Spadina.  First day is September 29.[/quote]
Thanks! :)

really? that's great. it's closer to my home

Is it OK if I missed the first 2 meetings, and can come this week??? Also, what things should I bring?

Yes, it is fine. Just bring a pen/pencil, calculator (not even necessary), and paper.

Yeah. And they give free refreshments, which is really nice. :)

Sorry to revive this thread, but I want in on this Math Circles. Date and place for Toronto for 2008?


Prove there doesn't exist consecutive perfect numbers

[quote="N.T.TUAN"][hide="A hint"]every even perfect number is of the form $ 2^{n\minus{}1}(2^{n}\minus{}1)$ where $ 2^{n}\minus{}1$ is prime.[/hide][/quote]

yes, I got that far... but still couldn't do it. Could someone tell me the full solution please? Thanks!

a perfect number is one devisors of which add to this number? i was always under the impression  that all of them are even. can someone give me any usefull links on perfect nubers please:D

as for the link http://en.wikipedia.org/wiki/Perfect_number

as for the solution: the potential pair of consecutive numbers are $ 2^{n\minus{}1}(2^{n}\minus{}1)\,,\,2^{n\minus{}1}(2^{n}\minus{}1)\pm 1$ with $ 2^{n}\minus{}1$ a prime number and $ n>2$. By Touchard/Holdener every odd perfect number has residue 1 modulo 12 or 9 modulo 36. Therefore, $ 2^{n\minus{}1}(2^{n}\minus{}1)\pm 1\equiv 1\bmod{4}$ must hold and only the "+"-case needs to be investigated. Moreover, we have  $ 2^{n\minus{}1}(2^{n}\minus{}1)\plus{}1\equiv (\minus{}1)^{n}\bmod{3}$, hence $ n$ must be even. But then the prime $ 2^{n}\minus{}1$ is divisible by the proper divisor $ 2^{\frac{n}{2}}\minus{}1$, a contradiction.


probability
calculus
probability and stats

Hello and thank you for reading my post.

Here's my question :

The tests used to diagnose the presence of the virus HIV which causes the AIDS are often imperfect, and can give results called positive forgeries (the person does not have the virus, but the test detects it) or negative forgeries (the person has the virus, but the test does not detect it). In the case of the HIV, the tests used have a rate of positive forgeries of 0.5% and a rate of negative forgeries of 2%. It is supposed that 0.04% of the Canadian men are infected by the HIV. Find the probability that a man tested positive for the HIV is really infected by the virus.

Let $ T$ be the event of a positive test and $ H$ be the event associated to the HIV infection, then you required to calculate

$ \mathbb{P}(H|T)\equal{}\frac{\mathbb{P}(H)\minus{}\mathbb{P}(\bar{T}|H)\mathbb{P}(H)}{\mathbb{P}(H)\minus{}\mathbb{P}(\bar{T}|H)\mathbb{P}(H)\plus{}\mathbb{P}(T|\bar{H})(1\minus{}\mathbb{P}(H))}\equal{}\frac{0.004\minus{}0.02\cdot 0.004}{0.004\minus{}0.02\cdot 0.004\plus{}0.005\cdot 0.996}$

Hello hjbrasch and thank you for your reply.

I'm trying to use the Bayes'Theorem to solve the problem and my answer is 1.95%.
According to your calculs, I should have 7% which seems to be the right answer.

Can you explain the steps of your solutions ?

My steps are :

P(H,T) = P(T|H) * P(H)  / P(T)

= P(T|H) * P(H) /(P(T|H) * P(H) + P(T|H^c) * P(H^c))

= 0.995 * 0.0004 / (0.995*0.0004 + 0.02*0.9996)
= 1.95 %

By the way, the probability is 0.04 % = 0.0004 and you wrote 0.004 (no big deal, just to let you know).

$ \begin{array}{ll} \mathbb{P}(H|T) & = \frac{\mathbb{P}(H\cap T)}{\mathbb{P}(T)}=\\[0.2cm]
 & =\frac{\mathbb{P}(H)-\mathbb{P}(H\cap \bar{T})}{\mathbb{P}(T\cap H)+\mathbb{P}(T\cap \bar{H})}\\[0.2cm]
 & =\frac{\mathbb{P}(H)-\mathbb{P}(\bar{T}|H)\mathbb{P}(H)}{\mathbb{P}(H)-\mathbb{P}(\bar{T}|H)\mathbb{P}(H)+\mathbb{P}(T\cap \bar{H})}\\[0.2cm] & =\frac{\mathbb{P}(H)-\mathbb{P}(\bar{T}|H)\mathbb{P}(H)}{\mathbb{P}(H)-\mathbb{P}(\bar{T}|H)\mathbb{P}(H)+\mathbb{P}(T|\bar{H})\mathbb{P}(\bar{H})}\\[0.2cm] & =\frac{\mathbb{P}(H)-\mathbb{P}(\bar{T}|H)\mathbb{P}(H)}{\mathbb{P}(H)-\mathbb{P}(\bar{T}|H)\mathbb{P}(H)+\mathbb{P}(T|\bar{H})(1-\mathbb{P}(H))}
\end{array}$


You are given an unfair $n$-sided die, labeled from $1$ to $n$, with each side weighted according to its value. For example, the $3$ side is three times as likely to be rolled than is the $1$ side. What is the expected value of the sum of three rolls?

[hide] The probabilty for 1 to be rolled is $\frac{1}{1+2+3...+n}=\frac{1}{{n(n+1)}{2}}$.  Similary, for 2, it is $\frac{2}{1+2+3...+n}=\frac{2}{{n(n+1)}{2}}$.  Thus, for x, it is $\frac{x}{\frac{n(n+1)}{2}}$.

Thus, the expected value of hitting x is $\frac{(x)(x)}{\frac{n(n+1)}{2}}=\frac{x^{2}}{\frac{n(n+1)}{2}}$.

Thus, the sum of the expected values for x between 1-n is $\frac{1^{2}+2^{2}+3^{2}...+n^{2}}{\frac{n(n+1)}{2}}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}$

Now, this is for each roll.  Thus, for 3 rolls, you would have to multiply this value by 3 to get our answer as $2n+1$ [/hide]

That's correct, QuattoMaster.  :)

(I find it interesting how something like this works itself out into a nice, simple form.  :) )


I know that this is classic cyclic inequality,but anyway zaya_yc could you post your solution.I try to solve with am-gm inequality:
[hide]But my solution is true only for $ 24y\geq z$.
Suppose $ x \equal{} max\{x,y,z\}$.Denote $ S \equal{} x^{3}y^{2}\plus{}y^{3}z^{2}\plus{}z^{3}x^{2}$.Since $ x\geq y,x\geq z\Rightarrow y,z\leq\frac{1}{2}$.
$ S \equal{} x^{3}y^{2}\plus{}y^{3}z^{2}\plus{}z^{3}x^{2}\leq x^{3}y^{2}\plus{}x^{2}y^{2}z\plus{}\frac{1}{2}(z^{3}x^{2}\plus{}z^{2}x^{3}) \equal{} x^{2}(x\plus{}z)(y^{2}\plus{}\frac{z^{2}}{2})\leq$(here is something like a mistake)
$ \frac{4}{9}(x^{2}(x\plus{}z)(\frac{3}{2}y\plus{}z)^{2})\leq\frac{4}{9}(\frac{2x\plus{}2(\frac{3}{2}y\plus{}z)\plus{}(x\plus{}z)}{5})^{5}\equal{}\frac{108}{3125}$
I don't  know maybe is this one true:
$ S \equal{} x^{3}y^{2}\plus{}y^{3}z^{2}\plus{}z^{3}x^{2}\leq\frac{4}{9}(\frac{2x\plus{}2(\frac{3}{2}y\plus{}z)\plus{}(x\plus{}z)}{5})^{5}$.
I am too lazy to check it. :)[/hide]

$ x,y,x\geq 0$ and x+y+z=1 . n , m different; prove:
$ x^{n}y^{m}\plus{}y^{n}z^{m}\plus{}z^{n}x^{m}\leq\frac{n^{n}m^{m}}{(n\plus{}m)^{n\plus{}m}}$

Is there some technique or methods to prove inequalities like this?
You don't post your solution? :maybe:

The solution for the original problem, so $ S(x,y,z)\le\frac{108}{3125}$
We can suppose that $ max\{x,y,z\}\equal{} x$. We will prove that $ S(x,y,z)\le S(x,y\plus{}z,0)$ so
$ x^{3}y^{2}\plus{}y^{3}z^{2}\plus{}z^{3}x^{2}\le x^{3}(y\plus{}z)^{2}$ which is pretty clear (in the general case as $ m$ will be larger this will be weaker). So we can suppose that $ z \equal{} 0$, which means that we have to prove that $ \left(\frac{x}{3}\right)^{3}\left(\frac{y}{2}\right)^{2}\le\frac{1}{5^{5}}$ with $ x\plus{}y \equal{} 1$, which is AM-GM(in the general case it will be too):
$ \left(\frac{x}{3}\right)^{3}\left(\frac{y}{2}\right)^{2}\le\left(\frac{3\frac{x}{3}\plus{}2\frac{y}{2}}{2\plus{}3}\right)^{2\plus{}3}\le\frac{1}{5^{5}}$. We have equality with $ (\frac{3}{5},\frac{2}{5},0)$ and cyclic permutations.
The general case would be with $ n > m\ge 2$ naturals, and the solution is just the same.


I derived this formula today. Prove that if $a,b>0$ then $\log(a+b)=\log(\frac{ab+\sqrt{(ab)^2-4(a+b)}}{2})+\log(\frac{ab-\sqrt{(ab)^2-4(a+b)}}{2})$

[hide="Big hint"] Consider the roots of the quadratic $Q(x) = x^2 - abx + (a + b)$.  :) [/hide]

[hide]ummm is it just me or when you use the identity log(a) + log(b) = log(ab)  on the right side everything cancels out and you get log(a+b)=log(a+b)?[/hide]

Yeah, but you would also have to show that the arguments of the logarithms are positive for the proof to be rigorous. I just thought it was a nice identity.

[quote="drunner2007"]Yeah, but you would also have to show that the arguments of the logarithms are positive for the proof to be rigorous. [/quote]

My method works out fine for this :)


Find four square numbers such that the sum of every two of them is a square.

According to R.K.Guy, 'Unsolved problems in number theory', pb. D-18, this is an open problem.

Pierre.

excuse me pierre, but how old is that book? perhaps i'm wrong, but i think that it's not so dificult to solve that problem (and also give all possible solutions expressed in terms of 4 unknowns)... maybe a little knowledge about biquadratic equations, which i don't have :S

I'm reffering to the second edition of the book (edited in the 90's). There is a third recent edition...
Perhaps it is now solved even if it would surprise me.
If you think it is easy come on, solve it. Sometimes open problems are open because they are known as open so that no one try to solve them...

Pierre.


Find max
$sinA\sqrt{sinBsinC}$ (triangle ABC)

$\sin(\gamma) \sin(2\gamma)$ with $\gamma = \cos^{-1}(\frac{\sqrt{3}}{3})$, which occurs for $B=C=\gamma$?
Simplifying we get $\frac{4\sqrt{3}}{9}$.

okay heres my attempt.. if anyone can help me out thatd be great
[hide="answer"]
$\sin{A}\sqrt{\sin{B}\sin{C}}$
by AM GM

$\left(\frac{\sin{A}+\sqrt{\sin{B}\sin{C}}}{2}\right)^{2}\ge \sin{A}\sqrt{\sin{B}\sin{C}}$

thus there is equality when:
$\sin{A}= \sqrt{\sin{B}\sin{C}}$
or when $\sin^{2}A = \sin{B}\sin{C}$, thus $A = B = C = 60^{\circ}$

Soo

${\sin{A}\sqrt{\sin{B}\sin{C}}\Rightarrow \frac{\sqrt{3}}{2}}\cdot\sqrt{\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}}\Rightarrow \boxed{\frac{3}{4}}$

[/hide]

@ GlenthemanN

HINT.

Ok, you want to find $\max F(x)$. You prove that $G(x) \geq F(x)$ and for a certain $x_{0}$ the identity $G(x_{0})=F(x_{0})$ holds, where $G$ is a non-constant function (it is important).
Does it follow that $F(x_{0})=\max F$?

hmm I guess it woudlnt, but I was thinking and do I need to instead find the harmonic mean?
since its $am \ge gm \ge hm$ correct?

edit:: upon further thinking I believe I am on the completely wrong track :o

could someone explain uldis' reply?

Thanks :)


What is the base-ten number represented by the base-eight
number 31?

This $ 3\times 8 \plus{} 1 \equal{} 25$.


linear algebra
matrix
linear algebra unsolved

Is it true that every element of $ \mathrm{SL}_2(\mathbb R)$ is the product of two positive semidefinite (but not necessarily symmetric) matrices with real entries? By positive semidefinite I mean $ x^TAx\ge 0$ for all $ x$.

Cheap observation: there was no point in saying "semidefinite." If a matrix in $ \text{SL}_2(\mathbb{R})$ is written as a product of two other matrices, then both of those other matrices are nonsingular. And a nonsingular positive semidefinite matrix is positive definite.

Actually, I do need "semidefinite". For example, $ \begin{pmatrix} \minus{} 1 & 0 \\
0 & \minus{} 1\end{pmatrix} \equal{} \begin{pmatrix}0 & \minus{} 1 \\
1 & 0\end{pmatrix}\begin{pmatrix}0 & \minus{} 1 \\
1 & 0\end{pmatrix}$, but it is not a product of any two positive definite matrices. (The matrices are [b]not[/b] assumed symmetric.)

[color=green][I just edited the word "not" into your first post. - K.M.][/color]


Putnam email trigonometry calculus derivative induction function

right now, I'm signed up for

math 135
math 137
physics 121/131L
chem 120/120L
CS 135
physics 10

The physics 10 class is a no-credit seminar-type course, so I don't think that'll be an issue, but I'm starting to worry about the labs now. When I was picking my classes, I was still undecided if I should try and switch into the advanced math classes, but I'm pretty sure that I want to try them now. Assuming that they let me into the advanced section, will my course load be way too much?

get rid of the labs and seminar and you'll be fine.

...well the reason why I signed up for these courses in the first place was that I was interested in taking all of them. I guess dropping all the labs and the seminar would take a pretty big load off my shoulders, but that's not really what I was going for here...

My initial concern was that because I have the extra labs and was considering  switching into the advanced section, I wouldn't have have enough time to devote time to study for each of these classes in depth.

advanced math VS labs & seminar

choose one

As far as I know, the advanced classes are worth the same number of credits as the regular section, so there really is no reason for me to drop both my labs and physics 10. I don't see why you see the science labs and the math classes as being mutually exclusive. At the most, I was considering dropping the chemistry lab and seeing how it went from there. Furthermore, I [i]know[/i] that you're not an upperyear student at the school, and therefore can't have that much clearer of an idea of what the course load is like for any of these classes, so what gives?

There are courses just for fun? Where is the list of them? I want in on some!

Remineds me: Someone tell me where/when the putnam'practice'clubin' happens ( or whatever its called ).

Don't worry Sisyphus. Take the advance classes, and show up to the seminar when you can. If its no credit, then its a no-issue. Also, you really should go to the advance classes, so that I have someone at my skill level to sit with  :rotfl:  :D

"courses for fun".....I've been known to be slightly masochistic... :D

Are you really planning on writing the Putnam? I thought that was more of an upper-year thing in university.

I'm almost positive that I'll switch into the advanced classes, only problem is the aforementioned classload. =\

OK... first of all, your courseload shouldn't be a problem.  Don't drop anything until you're sure that you want to.  You have a few weeks until the withdraw penalty kicks in, so you can get out of a class/lab if you discover that the workload is too great.

I DEFINITELY recommend doing the advanced math.  However, if you find this to be too difficult, you can always switch back into honours.  The other way around is more difficult.  I do, however, know people who started off in 147, switched out of it to 137, and then switched back to 147.

For putnam... if you're in advanced math, you'll get an email about it.  It's a contest, written every December, that is written by a few thousand students at top universities (almost exclusively North America).  There are cash prizes for the top few students, and the top 5-ish people are crowned "putnam fellows" and forever worshipped by the academic world.  Becoming a putnam fellow is harder than getting a gold on IMO.

Generally, people interested in putnam meet one night a week, do problems for a few hours, share solutions, eat donuts and drink juice, laugh a lot, and improve at problem-solving.  Putnam is very approachable to students of all levels... in fact, most of the questions can be done by a student who has completed first-year math at Waterloo.  There's essentially nothing on Putnam that's harder than second-year math.  DO NOT wait until you're an upper-year student to start writing putnam... there will be plenty of problems that you can approach, even with very little calc/algebra knowledge.

Putnam consists of 2, 6-question, 3-hour papers, written in the morning and afternoon of the contest day.  For people used to Olympiad-type stuff (where you get 4.5 hours for 3 problems), the time limit is a lot tighter.  Fortunately for the first-years, about 1/2 of the problems will be completely over your head, and you'll be able to ignore those and spend your time working on the subset of the problems that you could potentially solve.

Putnam is difficult though... I got 5/12 problems last year as a first-year and that was good enough for top 75 worldwide.  The median score is usually zero (however... NOT AT WATERLOO!!!)

So yeah... for those interested, you'll get an email about it sometime in the fall, and if not, we'll probably discuss the putnam sessions in more detail when they start happening.

so how many students come to the putnam thing every week? and who gets selected out of those students?

[quote="Sisyphus"]OK, I wasn't sure if it was possible to drop courses like that once the term started, so I guess I'll see how it goes for the first little while with my current classes. Also, I remember reading here that the advanced classes are unusually big this year -- do you think that this'll affect my chances of switching in? (I know that I'll find out in 20 odd days or so, but nonetheless, I've been somewhat anxious about this lately  :P )

Thanks for the response Elyot.

About the putnam, can anyone write it, or do you have to go through qualifying stuff before it? If it's open to anyone, I think I'll look into it, though I'm garbage at problem solving.[/quote]

What you need to do is get onto the quest site and look at the important dates section... here's a link for you:

[url]http://quest.uwaterloo.ca/undergraduate/dates.html#fall[/url]

That tells you when all the withdraw dates are with the various levels of penalties.

Yes, anyone can write putnam, but only 3 people are selected for the school's "putnam team", which competes against the other teams from other schools.  The team selection actually has to be done before the contest is written, and it's sometimes tough to pick the best 3 scores beforehand.  A school's team score is the sum of the RANKS of the three team members, and a low score is better.  So if 3 people on waterloo's team came 10th, 42nd, and 1285th, then the team score is 1337.  If one person messes up, the score is really dragged down, so it's VERY important for the coach to pick those 3 people carefully.

[quote="Elyot"]
I DEFINITELY recommend doing the advanced math.  However, if you find this to be too difficult, you can always switch back into honours.  The other way around is more difficult.  I do, however, know people who started off in 147, switched out of it to 137, and then switched back to 147.
[/quote]

At this point, I'm not [i]that[/i] worried about taking the advance courses. My reasoning: I can't picture very many students who go in excited to learn the hard stuff ( my position ), and then find themselves supprised. Maybe some find themselves unprepared, but I'm not even starting from scratch. Thanks to the boards here and other reasorces, I now have tried many ( and a handfull of times succeedd) on some hard problems. Aops and such have boosted my confidence moderetly  :)

[quote="Sisyphus"]a professor (Prabhakar Ragde) posted this in response to a question on the frosh forum:

You're not going to be able to switch 135 sections for that reason, [b]and you're not going to be able to get into 145/147, which is already overfull.[/b] John Lawrence is a bit reserved, but get to know him and you find a sly wit, plus he is solid on the material and everything that follows from it for several years after that.

uh oh? does this spell the end?[/quote]

ahhh. what?

Not cool. This was the wrong things to read just before bed.

Hey!  I've finished my first year of studies at Waterloo in advanced math, and I'm getting ready for more this fall!  I totally agree with Elyot.  Anyone who enjoys math should get into the advanced section and just try it out, and get a feel for it.  If you find the material too hard, or too dull, or you simply don't have enough time to keep up with the assignments (a common scenario), dropping to 130-level math is no problem at all.  In fact, a majority of students take that route!

As long as you have a foundation in the fundamentals (basic algebra and trigonometry), you're prepared for advanced math.  We all get different profs so we enjoy different experiences, but from what I remember, the hardest high school result we used in MATH147 was the expansion of sin(x+y) when we proved that the derivative of sinx is cosx.  It's easy to get fooled and think advanced math is easy because you've seen sequences, limits and derivatives before, but if you haven't, it's new and exciting!  If you have, you're going to understand them even better!  Once you start 1B, all the math you'll be exposed to should be new, so it's going to take a lot more time and effort to learn all the new ideas, master the proofs, and still find the time for the challenging assignments (which you learn SOOO much from)!  By 2A, the class size drops down to 10-30 and it feels like you're in one of those 4th year pure math courses!

So anyone reading this who loves math, take on the challenge of advanced math and see what you think!  In the first two lectures in both MATH 145 and 147 last year, we learned induction, which was totally new to my friend.  But he stuck with it, worked very hard, learned TONS, and now he's taking MATH 245 and 247 with me in the fall.  And be prepared to see a lot of rigorous proof.  It's not called advanced because you study different things.  You get to see what goes on behind the scenes.  In calculus, you'll get to learn what it really means for a function to be continuous, why continuous functions on a closed interval are bounded, and why they assume max and min values there... the Extreme Value Theorem!  It's awesome.

Enjoy your epsilons and deltas!

[quote="lotus87"]You get to see what goes on behind the scenes.[/quote]

 :rotfl:

[quote="killmesoft"]rzsolt you didn't choose advanced math while doing the m-section thing?[/quote]

I only got a 67 on the euclid. You need a 70 or greater to be accepted automaticaly.

[quote="killmesoft"]lol... that's bad. i got 80 and it automatically asked for adv math when choosing the m-section.[/quote]
That's fantastic.

Remind me to buy you a cookie when I see you at Waterloo

[quote="rzsolt"]But Wait, Will I even be able to get into the advance classes? This news that sisyphus has just droped, has really got me worried.[/quote]

If history repeats itself, many students will decide that they prefer the honours section over the advanced section, so expect the class to be smaller after just a few weeks.  Transferring to the 140s sometime in September seems like it's a reasonable idea since it will be very easy to catch up on five lectures or so.  A great idea would be talking to the profs and asking them for a copy of the first assignment, just to get a feel for the difficulty you should expect in advanced.

It's disappointing that only the 70+ people are automatically accepted because I know many people with average Euclid scores who have been very successful in advanced.  Don't expect contest math questions on your assignments (maybe a few in 145  :P )!  The problem solving you've developed will transfer over, but I stress that anyone who's familiar with basic algebra and trigonometry, has a great work ethic, and loves math, will do just fine!

[quote="killmesoft"]what's your definition of doing fine? is it 70s? or 80s? or 90s?[/quote]

Different for eveyone. Lots of people are happy learning tons with 70's, others feel like their failing with 80's. Realtivity,cica 1903.m  :D

[quote="killmesoft"]well my definition of "fine" would be getting 80 over in adv math. do many of them really gt over 80?[/quote]
In your 1A term, your marks in advanced math should be the same as the ones you would have received if you switched to honours.  Last fall, the class average in MATH 147 was slightly above 90% and several students received 100%.  For those worried about marks, assignments often include bonus problems, and tests are generously curved.

[quote="WindingFunction"][quote="lotus87"]
If history repeats itself, many students will decide that they prefer the honours section over the advanced section, so expect the class to be smaller after just a few weeks.  Transferring to the 140s sometime in September seems like it's a reasonable idea since it will be very easy to catch up on five lectures or so.  A great idea would be talking to the profs and asking them for a copy of the first assignment, just to get a feel for the difficulty you should expect in advanced.[/quote]

If Waterloo is anything like other universities, you'll also easily be able to sit in on the advanced sections even while you're not enrolled in the course.  That way, you won't even be missing anything.  (Although you would be attending double the number of math classes, which could take up some time...)[/quote]
That's a good idea.  I thought of it but with the classes already being over their capacity, it would be too chaotic.  Best to wait for students to drop into honours unless the classroom is large enough!

Oooh i'm taking chem too.

Er, that seemed random.  Anyone hear anything about Bio 120 while we're talking about courses?  Meh.  Sooo excited!

[quote="The Master"]rzsolt - i don't think you should have a problem getting into the advanced math courses

i'm in the advanced section right now, and i'm already thinking about switching to the honours. i'm really excited about it and i will do everything i can to keep up with it, but since i'm a pretty slow worker, i think i may have some trouble. the thing is, tennis is very important to me, and i will need to make time for that. i don't know what university will be like, but i hope to be able to continue practising everyday, or as much as i can. i've heard that it will take a lot more time working through the advanced classes compared to the honours. so i will see, but it seems likely that i will switch. i'm thinking that there will be quite a few switching to the honours from advanced, so i think you will be able to get in. and ya, like someone else suggested, try sitting in on the classes - if possible.[/quote]

 :D  Now, you want to sighn a piece of paper, saying you give your seat to me?  :lol: *paranoia*

Seriously though: thanks for the reasuring comments everyone, I read Sisyphus post at 3am last night, and had horrible nightmares becuase of it ... welll... It was probebly excentuated by the fact I was reading comics about the end of the world for the four hour before  :P

[quote="Elyot"][quote="rzsolt"]

Seriously though: thanks for the reasuring comments everyone, I read Sisyphus post at 3am last night, and had horrible nightmares becuase of it ... welll... It was probebly excentuated by the fact I was reading comics about the end of the world for the four hour before  :P[/quote]

Clearly not getting into advanced math entails the end of the world.  :roll:[/quote]

I simply shouldn't have let WATSFIC talk me into buying the 'Watchmen" graphic novel. Becuase it was too good! I stayed up all night last night reading it, and then after having nighmares about the phycis brain world ending monster all night, I was so tired at the gym this morning.... but I digress...

[quote="Sisyphus"]Sounds like it's an issue of physical capacity. I guess I can't do anything about that. This also rules out sitting in on the classes and then asking to switch in. I might check with them in person once I get to Waterloo, but I guess for now, I'll have to contend with sticking to my regular classes. Thanks everyone.[/quote]

no way man. I'm draggin you in there with me. No way your going to wimp out that easy. Or do I have to do this all by myself?

[quote="rzsolt"]
no way man. I'm draggin you in there with me. No way your going to wimp out that easy. Or do I have to do this all by myself?[/quote]

 :D

I'm still interested in the advanced section, but damn, this is proving to be a pestilence.

[quote="Sisyphus"]pestilence[/quote]

LOL!!

pes·ti·lence
n.
A usually fatal epidemic disease, especially bubonic plague.
A pernicious, evil influence or agent.

Are you sure that's the word you intended to use???

Last year there was a meeting during frosh week where the advanced section profs explained the difference between the sections and how people can transfer in or out. It's likely there will be another such meeting and that there will be a way for everyone who wants to be in the section to attend those classes. I doubt that there are even more than 100 people who will continue going to the class after the first day or two.


What is the sum of all positive divisors of 770?

I don't think that's correct because $2^{1}\cdot 5^{1}\cdot 7^{1}\cdot 11^{1}= 770$

To get the number of positive divisors, you add 1 to each exponent and find the product.
$2^{4}=16$ divisors, you only have 14.

All those numbers are integer divisors of $770$ though.  :maybe:

okay...so here's what i got
[hide];$the prime factorization of 770=7^{1}*11^{1}*2^{1}*5^{1}= (1+7)*(1+11)*(1+2)*(1+5)=1728$...this is a formula for finding the sum of the divisors of any number....it goes something like this: if the prime factorization of a number is n^2*m^3...then to find the sum of the divisors you would do this (n^0+n^1+n^2)*(m^0*m^1*m^2*m^3)...yup..that's how you do it :) [hide][/hide][/hide]

[hide]All the divisors of $770$ are all possible values of $2^{a}\cdot 5^{b}\cdot 7^{c}\cdot 11^{d}$ where $a,b,c,d$ equals $0$ or $1$.  So, the sum of all the divisors is $(2^{0}+2^{1})(5^{0}+5^{1})(7^{0}+7^{1})(11^{0}+11^{1})$, because when this product is expanded, we get all the possible divisors added together.  The total is $(3)(6)(8)(12)=1728$.[/hide]


Given a convex quadrilateral such that any three vertices belongs to an isosceles triangle. What's the shape of this quadrilateral?

Well, it could be a rhombus. Or it could be a kite ABCD such that AB, AC, and AD all have the same length. There's probably some possibilities that I'm missing, but this seems to be a start.


geometry
geometric transformation
reflection
parallelogram
Euler
cyclic quadrilateral
geometry unsolved

Let $ M$ be a point interior to triangle $ ABC$, let $ AM$ intersect $ BC$ at $ E$, let $ CM$ meet $ A$B at $ F$. Suppose that $ N$ is the reflection of $ B$ across the midpoint of $ EF$. Prove that the line $ MN$ has a fixed point when $ M$ moves in the triangle $ ABC$.

$ BENF$ is a parallelogram. Let $ ABCD$ be parallelogram and let $ DN$ cut $ CA$ at $ N'.$ Let parallels to $ NE, NF$ through $ N'$ cut $ BC, AB$ at $ E', F'.$ Let $ AE$ cut $ CD$ at $ X$ and let $ CF$ cut $ DA$ at $ Y.$

$ \frac {\overline{N'A}}{\overline{N'C}} \equal{} \frac {\overline{F'A}}{\overline{N'E'}} \equal{} \frac {\overline{F'A}}{\overline{E'C}} \cdot \frac {\overline{E'C}}{\overline{N'E'}} \equal{} \frac {\overline{FA}}{\overline{EC}} \cdot \frac {\overline{BC}}{\overline{AB}} \equal{} \minus{} \frac {\overline{FA}}{\overline{CD}} \cdot \frac {\overline{AD}}{\overline{EC}} \equal{} \minus{} \frac {\overline{YA}}{\overline{YD}} \cdot \frac {\overline{XD}}{\overline{XC}} \Longrightarrow$

$ \frac {\overline{N'A}}{\overline{N'C}} \cdot \frac {\overline{XC}}{\overline{XD}} \cdot \frac {\overline{YD}}{\overline{YA}} \equal{} \minus{} 1$

By Ceva for $ \triangle ACD,$ $ AE \equiv AX, CF \equiv CY, DN \equiv DN'$ concur at $ M.$ While $ M$ is arbitrary, $ D \in MN$ is fixed.

The points $ M,\ N,\ D,$ where $ ABCD$ is a parallelogram, are collinear and we are done.

This result has already been used as [b][size=100]Lemma  2[/size][/b], in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997]Euler lines in cyclic quadrilateral[/url] ( for a proof, see in the post #7# with the schema [b][size=100]t=107997(e)[/size][/b], in the post #8# ).

[b][size=100]REMARK.[/size][/b] – Because of this Lemma is the particular case of a well known [b][size=100]Brianchon theorem[/size][/b] (not the famous), we can think a generalization of the proposed problem as follows :

[b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ P,\ Q$ be, two fixed points on its sidelines $ BC,\ AB,$ respectively $ ($ $ B,$ between $ P,\ C$ and between $ A,\ Q$ $ ).$ We denote the points $ E\equiv BC\cap AM,\ F\equiv AB\cap CM,$ where $ M$ is an arbitrary point in the plane and let be the point $ N\equiv PF\cap QE.$ Prove that the line segment $ MN,$ passes through a fixed point.[/color][/size][/b]

Kostas Vittas.


Fie a,b,c numere reale astfel incat |ax*x+bx+c|<=1 oricare ar fi x din intervalul [-1,1]. Sa se arate ca daca q se afla in intervalul [0,1], atunci:
          q(1+q)|b|+(1-q*q)|c|<=1+q*q.
Cand are loc egalitatea?

[hide="hint, n`are legatura cu problema"]  http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php
pt latex :D
[/hide]

$|ax^{2}+bx+c|\leq 1$, oricare ar fi $x\in[-1,1]$.
Sa se arate ca daca $q\in[0,1]$, atunci:
$q(1+q)\cdot|b|+(1-q^{2})\cdot|c|\leq 1+q^{2}$.


function
calculus
derivative
real analysis
real analysis unsolved

Is the function $\sqrt{|x|}$ uniformly continuous on R? Why or why not?

Yes, it is uniformly continuous.

On the set $\{|x|\ge 1\}$ it has a bounded derivative, hence it is Lipschitz, hence it is uniformly continuous.

On the set $[-1,1],$ it is a continuous function on a compact set, hence uniformly continuous.

One can also prove this directly from the definition; however the argument will have to have both a "near zero" and a "not near zero" case.

Kent, can you please clarify what it means to be Lipschitz? Also, how can we just use the definition to prove this?

Thanks though


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

$T \in M(n,C)$  a triangular superior matrix
consider the equation $X^{m}=T$
$m>1$ a integer

1) Does the matrix $X$ is also triangular superior ?

2) Could you find all solutions of $X^{m}=T$ ?

1/ no in the general case. For example if $T=0$, any $m$-nilpotent matrix works

Solution if T is a (n,n)-dimensional non nilpotent Jordan block: T=uI+J where J^n=0, J^(n-1)<>0 and u<>0. XJ=JX and X is polynomial in J; spectrum(X)=v where v is a m-root of u; X=vI+a1.J+.... an-1.J^(n-1).
By identification in X^m=T, it is easy to be convinced  that it exists exactly 1 solution in a1,...,an-1 (when v is selected).
The single solution: X=v[I+1/u.J]^(1/m)=v[I+1/(mu).J+...] (with the binomial theorem).
Conclusion: we obtain the m solutions explicitly (all solutions are superior triangular).


analytic geometry
graphing lines
slope
calculus
derivative
logarithms

A box has height $h$. A chain of length $L$ is placed such that $L-h$ of it is on (the top) of the box and the remaining $h$ of it is hanging down the box. When the end of the chain runs through the $L-h$ length and passes the edge of the box, what will be its velocity?

i'm thinking... use energy

edit.
$\sqrt{2(L-h)(g)(h)/L}$ ?

It also somewhat depends on what happens to the chain after it reaches the bottom of the box. Does it keep falling or is there some table / floor on which the chain stops?

I was thinking of energy as well but after the chain is off the box,it will fall under gravity which we can find but if the question is asking the velovity right after the chain which off the box,then also if there is no friction wouldn't it all still be under free gravity?

well, I'm assuming that the box is resting on a table...
unless its a *floating* box

well, let's say the box is floating..  :)

If the box is floating you may still use the energy, only with different outcome. If you set zero potential at the top of the box then at the beginning, the chain has

$h\rho\times g\times -h/2$

potential energy and at the end it is

$L\rho\times g\times -L/2$.

Difference must equal the kinetic energy:

$1/2\times L\rho\times v^2=(-h^2-(-L^2))\rho g/2$
$v=\sqrt{\frac{(L^2-h^2)g}{L}}$

Did I get it right?

Oh - and for some reason, my solution for the other case is $\sqrt{2(L-h)g}$. Perhaps you forgot that the L-h length of the chain has zero kinetic when it's at the end lying on the table?

Hey gyus, you're playing with energy as if it was an 'untouchable' quantity. While I think bus's solution to the first case is OK (even though I still have some doubts), the second one can not be! Imagine a situation such that h=1 cm and set L, say, 101 cm. And then set L 1001 cm. It doesn't seem right to me.
When the end of the chain hits the ground, it DOES lose some (a bit)(in fact, all) of its kinetic energy. I'd rather use the force approach. The hanging part of the chain has a constant length of $h$, therefore the force on the chain is $F = \rho .h.g$. Now, I think the outcome of this problem depends on how the chain is resting on the top of the box in the beginning. If it's all just at the egde of the box, it's different than if it's laid in a line... I'll try to do the math but it seems quite complicated.

Another thing that bothers me. Suppose you have the chain laid out in a line. Then, when it starts to slide down, the sliding part has some momentum in the horisontal direction. Where it disappears when the whole chain comes down? There is no force pushing to stop the chain going behind the edge. Therefore I think that even bus's first solution using energies is not quite kosher.

OK. I've worked through it, and I don't guarrantee it's right, but if all the chain is rolled up on the edge of the table, the answer is:
\[ v = \sqrt{gh}\sqrt{1-e^{-\frac{2(L-h)}{h}}} \]
notice that the velocity of the chain approaches the critical value $\sqrt{gh}$, when all the pulling force is used to accelerate new pieces of chain which are initially at rest on the table.

[quote="kubus"]Imagine a situation such that h=1 cm and set L, say, 101 cm. And then set L 1001 cm. It doesn't seem right to me.
When the end of the chain hits the ground, it DOES lose some (a bit)(in fact, all) of its kinetic energy.

Another thing that bothers me. Suppose you have the chain laid out in a line. Then, when it starts to slide down, the sliding part has some momentum in the horisontal direction. Where it disappears when the whole chain comes down? There is no force pushing to stop the chain going behind the edge. Therefore I think that even bus's first solution using energies is not quite kosher.[/quote]

I think your example works ok with the equation I got.
The thing that I'm doing is this.  We're looking at the energy difference from the beginning and the end.  That will give the KE of the whole chain(if part of it were not stopped).  However, because a chain doesn't exert compressive forces, the front of the chain(which has been stopped by the ground) doesn't slow down the rest of the chain any, so we can use the KE expression to find the velocity.  The rest of the chain should be moving at the same velocity as it would be if there was a smooth, frictionless ramp on the bottom that diverted all the bottom so that it would still move when it hits bottom.

About the conservation of momentum- I think that if the surface the box is resting on is frictionless, the box would slide the other way, such that the center of mass doesn't move.  But because of friction, the box doesn't move- the whole system kind of does, in a small way.

Kubus, how did you derive your result?  What do you mean by "rolled"?

I see what you guys are saying... and when I realized that the chain loses some of its energy, daveed's solution with the frictionless ramp seemed to be all right... until I read the second part of your argument kubus. Daveed, you can see what kubus means by "rolled" and "laid out straight" on my picture - is that right kubus? So, the difference is that in the "rolled" case the gravitational force always accelerates only $h$ length of the chain and in the other case, the gravitational force accelerates the whole chain at the same time (although the force remains the same). This could possibly lead to different results. So what's wrong with the energy + frictionless ramp approach? The problem is that the parts of chain that were already deflected by the ramp at the bottom of the box all move with different velocities. It seems like you, daveed, thought that all of the chain will be at the end moving with the same speed $v$ and you used this to calculate your $v=\sqrt{2(L-h)gh/L}$. (But I admit that it's one step closer to the right solution than what I said :).) On the other hand, kubus took the acceleration of the chain and integrated it. I checked your solution kubus and it's correct - although I have no idea how you found it :).

Oh, and I wouldn't worry about the momentum - suppose you have a sloped surface and a ball rolling down this surface. You would surely calculate it's velocity using energy and never worry about the fact that it has gained some horizontal momentum which came from nowhere.

Bus, thanks for doing all the dirty work for me;) & a nice picture! And thanks for checking the solution .. I guess you've worked out the differential equation of motion if you have checked my solution. Anyway, it was a long way to the solution but all very, very interesting! I suggest you go and try it! Being lazy and hiding it behind the fact that I don't want to take away from you the feeling of discovering the solution yourself, only a few guidelines:

$F = dp/dt$

F is constant, work out the derivative (formally, you'd say using the product rule;), get to

[hide]
$a = (F - v^2 \rho)/(h \rho)$

[hide="solving"]
$2\sqrt{g/h}T = \ln{\frac{\sqrt{gh}+v}{\sqrt{gh}-v}}$

which forces you to integrate once more, 'cause you know the distance not the time after which the end of the chain leaves the table, with a little of fussy math and a handy substitution when integrating,
[hide]
... you come to the desired result and plug everything back in, sorry, I cannot figure it out from my messy work from yesterday, the next figure in my notebook that I'm able to read is the result;/ If you have any problems, I'll try to remember or do it again:)
[/hide]
[/hide]
[/hide]

Hey, thanks, any time. I only got to

$\frac{\text{d}v}{\text{d}l}=-\frac{v}{h}+\frac{g}{v}$

and didn't know what to do next so I checked your solution :). So, that's it.

And don't forget - beware of flying spiders!


Find the smallest positive integer which is divided by 89 and consists of 6, 7.

example) 66767 consists of 6, 7 but is not divided by 89, so 66767 is not our consideration.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230039[/url]


Let $ t>0$. Prove that $ 2^t\plus{}2^{1/t} \leq 2^t \cdot 2^{1/t}$.

[i]Sorry if I posted the problem to the wrong forum.  :oops:   Nevertheless could you help me, please?  :blush: [/i]

Rewrite the inequality as $ \frac{1}{2^{\frac{1}{t}}\minus{}1} \le 2^t\minus{}1$. It's enough to prove the inequality for $ t \ge 1$. We have equality for $ t\equal{}1$. Moreover, let $ f(t)\equal{}2^t\minus{}1$ and $ g(t)\equal{}\frac{1}{2^{\frac{1}{t}}\minus{}1}$. We have $ f'(t) > g'(t)$ for every $ t > 1$. Hence the result.


Εστω η δυο φορες παραγωγισιμη συναρτηση φ(0,+απειρο)-->R
με φ(1)=e και φ'(1)=0
και ισχυει:(χ^2)*φ''(χ)+2*χ*φ'(χ)=χ*(e^χ)  για καθε χ>0
1)δειξε οτι φ(χ)>=e
2)αν για τους θετικους αριθμους α,β,γ ισχυει α*β*γ=1 δειξε οτι
α*e^(β+γ) + β*e^(α+γ) + γ*e^(α+β)>=3*e^2

Μετά από πράξεις βρίσκουμε ότι $f(x)=\frac{e^{x}}{x}$ η οποία εμφανίζει ακρότατο για χ=1 (δες παράγωγο).
Έτσι $f(x)\geq f(1)=e$  άρα το ζητούμενο.
Όσο για το δεύτερο τώρα .
Από την ανισότητα ΑΜ-ΓΜ έχουμε.
$ae^{b+c}+be^{c+a}+ce^{a+b}\geq 3\sqrt[3]{abce^{2(a+b+c)}}$

Όμως $a+b+c\geq 3$  άρα το ζητούμενο/.........

οσο αφορα το δευτερο σκελος μια εναλλακτικη λυση ειναι η εξης
αφου φ(χ)>=e για καθε χ>0 τοτε φ(α)>=e , φ(β)>=e , φ(γ)>=e
φ(α)*φ(β)+φ(β)*φ(γ)+φ(γ)*φ(α)>=3*e^2 και α*β*γ=1......


modular arithmetic
number theory
greatest common divisor

Two measuring sticks use the same unit, but the scale on one stick has $ 635$ marks per unit and the scale on the other stick has $ 465$ marks per unit. Suppose the sticks are placed together so that the $ 0$ marks on each scale line up. What is the number of the next mark on the $ 465$ scale that lines up exactly with the mark on the $ 635$ scale.

[hide="Hint"]In order for the two tick marks to match up, the number must be divisible by both 635 and 465.  For example, suppose, to make things simpler, there were 4 marks/unit and 6 marks/unit.  The first time they match up is at 12.[/hide]

[hide="Solution"]Try a small case. Suppose one stick is marked once per unit, and the other is marked twice per unit. If we travel $ m$ marks up the first stick, we've traveled $ m$ units. If we travel $ n$ marks up the second, we've traveled $ n/2$ units. $ m\equal{}n/2\implies 2m\equal{}n$, of which the smallest nonzero solution, $ m\equal{}1$, yields $ n\equal{}2$, and the answer would be $ 2$.

In our case, traveling $ m$ units up the 465-marked stick means we've traveled $ m/465$ units, and $ n$ units up the 635-stick means we've traveled $ n/635$ units.
$ m/465\equal{}n/635\implies 127m\equal{}93n$. Taking it $ \mod 93$ we have $ 34m\equiv 0\pmod {93}$.
$ \text{gcd}(34,\ 93)\equal{}1$, so we have $ m\equal{}\boxed{93}$.[/hide]


[quote="jhcreinhardt"]$a+b+c=-1$
$a^{2}+b^{2}+c^{2}=17$
$a^{3}+b^{3}+c^{3}=11$

Find $abc$

edit: $a,b,c$ are reals[/quote]

[hide]Consider the polynomial $x^{3}+x^{2}-8x-12=0$ which has roots $a,b,c$. Then $abc=\boxed{12}$[/hide]

I just used newton's sums to find the coefficients. You could also do it that way..

WLOG, let the polynomial be monic.  find squared and first degree terms using vieta's.  since the roots are a, b, c, create three equations, plugging in a, b, c for x.  solving the system gives you the constant term, which tells you the product abc.


An $n$-digit positive integer is CUTE if its $n$ digits are an arrangement of the set $\{1,2,...,n\}$ and its first $k$ digits form an integer that is divisible by $k$, for $k = 1,2,...,n$. For example, 321 is a cute 3-digit integer because 1 divides 3, 2 divides 32, and 3 divides 321. How many cute 6-digit numbers are there?

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$

abcdef is cute

(1) b must be even
(2) a+b+c must be 0 mod 3
(3) 2c+d must be 0 mod 4
(4) e must be 0 mod 5
(5) f must be even and a+b+c+d+e+f must be 0 mod 3

----------
(4):  e = 5
(3):  d must be even

Now (b,d,f) an arrangement of (2,4,6); (a,c) an arrangement of (1,3).

The arrangements that satisfy (2) have b=2.  Using (3), d = 6.

In total, the numbers are 321654, 123654.  Answer C.


What is the value of 2a+3b+4c+5d in the system below?

a+2b+ c+ d=11
  -3b+4c- d=4
     b- c+2d=7

Solving the system in terms of d, a=18d-78, b=32-7d, c=25-5d. Subbing in, the answer is 50.

I got 40 for the answer. Just multiply the first and last equations by 2, and keep the third equation like it is. Then add them together.

Yes, you are right. I wrote $ 78 \cdot 2 \equal{}146$ on accident.

hello, solving the given system we get $ a\equal{}12\minus{}\frac{18}{5}c,b\equal{}\minus{}3\plus{}\frac{7}{5}c,c\equal{}c$ and $ d\equal{}5\minus{}\frac{1}{5}c$, inserting the solutions in the given term we have $ 2(12\minus{}\frac{18}{5}c)\plus{}3(\minus{}3\plus{}\frac{7}{5}c)\plus{}4c\plus{}5(5\minus{}\frac{1}{5}c)\equal{}40$, thus our answer is $ \text{40}$.
Sonnhard.


calculus
integration
real analysis
real analysis unsolved

is it possible to prove it through transformation to a suitable integral ? :maybe:

$ \frac{2\sqrt{2}}{9801}\sum^\infty_{k=0}\frac{(4k)! \cdot (1103+26390k)}{(k!)^{4}\cdot 396^{4k}}\;\;=\;\;\boxed{\frac{1}{\pi}}$

Ramanujan deviced this formula based on the
behavior of e^(pi*sqrt(163))


LaTeX
inequalities unsolved
inequalities

I dont knoew how to use latex hence i am sending solution using paint

what you have done i cant understand please tell the detailed solution.


Our first look, to get the gist of the problem, is when the numbers are picked, and are [b]NOT DISTINCT.[/b] In that interval, there are 13 odd numbers $(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25)$ and 12 evens:$(2,4,6,8,10,12,14,16,18,20,22,24)$

The probablility of getting an even is $\frac{12}{25}$ and for odd it is $\frac{13}{25}$

The sum is even if both are odd or both are even.

Case one: both are odd
Case two: first is odd, second is even
Case three: first is even, second is odd
Case four: both are even

As you can see, only cases one and four result in the objective.

Case one's probability is $\frac{13}{25}\cdot \frac{13}{25}$
Case four's probability is $\frac{12}{25}\cdot \frac{12}{25}$

The probability that the objective is met is $\frac{169}{625}+\frac{144}{625}$, or  $\frac{313}{625}$

[b]However, we must take into account the fact that they are distinct.[/b]

That means that we must look at the set a different way.

The first number we draw has a 13/25 chance of being odd, and a 12\25 chance of being even.

If we draw an odd, the second number is now taken from a set of 12 odd and 12 even, and so has a 50-50 chance of being even or odd.

If we draw an even first, the second number is taken from a set of 11 even and 13 odd, and so has a 13\24 chance of being odd and a 11\24 chance of being even.

Again, our target is odd + odd or even + even.

Case one: odd + odd:
Probability that the first one is odd: 13/25
Prob that second one is odd: 1/2

Case two: even + even:
Prob that the first one is even: 12\25
Prob that the second one is even: 11\24

Sum of probabilities:
$\frac{13}{25}\cdot \frac{1}{2}+\frac{12}{25}\cdot \frac{11}{24}$

$\frac{13}{50}+\frac{11}{50}$

$\boxed{\frac{12}{25}}$
$\boxed{B}$

wow...that is a long solution...we have e,e or o,o, which is just 12/15*11/24+13/25*12/25

Yes, long, but also correct.
I like to explain things a lot, making the problem really clear to the asker so that they can learn from the solution rather than simply getting the answer.

Isn't that the goal of the website?

[quote="jli"][quote="anyone"]Two distinct integers are chosen at random from the interval [1,25]. Find the probability that thier sum is even.

$A.\frac{288}{625}$   $B.\frac{12}{25}$      $C.\frac{1}{2}$ $D.\frac{13}{25}$      $E.\frac{337}{625}$[/quote]

[hide="solution"]The number of ways to choose two numbers is ${25\choose 12}=300$.  To make an even number, the two numbers are both odd or both even.

The number of ways to choose 2 odds is ${{13\choose 2}=78}$ and the number of ways to choose 2 evens is ${12\choose 2}=66$.  Therefore, $\frac{78+66}{300}=\boxed{\frac{12}{25}}$[/hide][/quote]
But the chosen numbers have to be distinct.

[quote="rnwang2"][quote="jli"][quote="anyone"]Two distinct integers are chosen at random from the interval [1,25]. Find the probability that thier sum is even.

$A.\frac{288}{625}$   $B.\frac{12}{25}$      $C.\frac{1}{2}$ $D.\frac{13}{25}$      $E.\frac{337}{625}$[/quote]

[hide="solution"]The number of ways to choose two numbers is ${25\choose 12}=300$.  To make an even number, the two numbers are both odd or both even.

The number of ways to choose 2 odds is ${{13\choose 2}=78}$ and the number of ways to choose 2 evens is ${12\choose 2}=66$.  Therefore, $\frac{78+66}{300}=\boxed{\frac{12}{25}}$[/hide][/quote]

But the numbers have to be distinct.[/quote]

i know, i didnt say they were

[quote="jli"][quote="rnwang2"]But the numbers have to be distinct.
[/quote]i know, i didnt say they were[/quote]
 :huh: But that was a condition of the problem.

[quote="rnwang2"][quote="jli"][quote="rnwang2"]But the numbers have to be distinct.[/quote]i know, i didnt say they were[/quote]
 :huh: But that was a condition of the problem.[/quote]

oh crap.  i meant i didnt say they werent.  i always screw up that stuff :blush:

[quote="jli"]no, combinations doesnt count doubles.

like if you choose a 2 person committe out of person A,B,C there is ${3\choose 2}=3$ and that makes sense.  what esle would you use?[/quote]
OMG I'm going to sleep right now.  :P

[quote="iamagenius"]wait a sec. why doesn't this work:

say you choose 1, then there are 12 (3,5,7,9...,25) other choices that can give a sum that's even

if you choose 2, there are 11 (4,6,8,...24)

if you choose 3, there are 12

...

so the probability is $\frac{1}{25}*\frac{12}{25}+\frac{1}{25}*\frac{11}{25}...$

$=\frac{1}{25}(\frac{12}{25}+\frac{11}{25}...)$

$=\frac{288}{625}$[/quote]

Well, you're choosing the second number out of $24$ numbers left, not $25$.


inequalities
function
inequalities proposed

Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. If $ k \ge 0$ is a given constant, find the minimum value of the expression
$ P(a,\; b,\; c)\equal{}\sqrt{\frac{a^2\plus{}kbc}{b^2\plus{}c^2}}\plus{}\sqrt{\frac{b^2\plus{}kca}{c^2\plus{}a^2}}\plus{}\sqrt{\frac{c^2\plus{}kab}{a^2\plus{}b^2}}.$

Is there someone interested in this inequality?

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. If $ k \ge 0$ is a given constant, find the minimum value of the expression
$ P(a,\; b,\; c) \equal{} \sqrt {\frac {a^2 \plus{} kbc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} kca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} kab}{a^2 \plus{} b^2}}.$[/quote]

$ \sqrt {\frac {a^2 \plus{} kbc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} kca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} kab}{a^2 \plus{} b^2}}\geq 3\sqrt{\frac {1 \plus{} k}{2}}.$

the inequality does not hold.

Nice problem! I found the result is:
\[ P_{min} \equal{} 2 \plus{} \sqrt {\frac {k}{2}}\quad k < 32\]

\[ P_{min} \equal{} \sqrt [3]{\frac {k}{4}} \plus{} \sqrt [3]{2k}\quad k\geq 32\]
Here is my proof:
Let $ P_{min} \equal{} C(k)$,By using $ Holder$:
\[ \left(\sqrt {\frac {a^{2} \plus{} kbc}{b^{2} \plus{} c^{2}}} \plus{} \sqrt {\frac {b^{2} \plus{} kca}{c^{2} \plus{} a^{2}}} \plus{} \sqrt {\frac {c^{2} \plus{} kab}{a^{2} \plus{} b^{2}}}\right)^2\left(\sum{(a^2 \plus{} kbc)^2(b^2 \plus{} c^2)}\right)\geq (a^2 \plus{} b^2 \plus{} c^2 \plus{} kab \plus{} kbc \plus{} kca)^3\]
Now,we have to find the $ C(k)$ such that
\[ (a^2 \plus{} b^2 \plus{} c^2 \plus{} kab \plus{} kbc \plus{} kca)^3\geq C^2(k)\left(\sum{(a^2 \plus{} kbc)^2(b^2 \plus{} c^2)}\right)\]
Let $ p \equal{} a \plus{} b \plus{} c,q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$,then we need to prove
\[ f(r) \equal{} (p^2 \minus{} 2q \plus{} kq)^3 \minus{} C^2(k)[(1 \plus{} k)(p^2q^2 \minus{} 2p^3r \minus{} 2q^3 \plus{} 4pqr \minus{} 3r^2) \plus{} 2kpqr \minus{} 6kr^2]\geq 0\]

\[ f'(r) \equal{} \minus{} C^2(k)[(1 \plus{} k)( \minus{} 2p^3 \plus{} 4pq \minus{} 6r) \plus{} 2kpq \minus{} 12kr]\]

\[ f''(r) \equal{} \minus{} C^2(k)[ \minus{} 6(1 \plus{} k) \minus{} 12k]\geq 0\]
thus we have
\[ f'(r)_{min} \equal{} f'(0) \equal{} \minus{} C^2(k)[(1 \plus{} k)( \minus{} 2p^3 \plus{} 4pq) \plus{} 2kpq\]

\[ \equal{} \minus{} pC^(k)[ \minus{} 2(1 \plus{} k)p^2 \plus{} q(4 \plus{} 6k)]\geq \minus{} pC^(k)[ \minus{} 2(1 \plus{} k)\cdot 3q \plus{} q(4 \plus{} 6k)] \equal{} 2pqC^2(k)\]

\[ \geq 0\]
so we only need to consider the case $ r \equal{} 0$ which implies that there are at least one variable are equal to 0,WLOG,let $ c \equal{} 0$
\[ P(a,b,0) \equal{} \frac {a}{b} \plus{} \frac {b}{a} \plus{} \sqrt {\frac {kab}{a^2 \plus{} b^2}} \equal{} \frac {a^2 \plus{} b^2}{ab} \plus{} \sqrt {\frac {kab}{a^2 \plus{} b^2}}\]
Denote $ \sqrt {\frac {a^2 \plus{} b^2}{ab}} \equal{} x\geq \sqrt {2}$,consider the following function
\[ f(x) \equal{} x^2 \plus{} \frac {\sqrt {k}}{x}\quad f'(x) \equal{} 0\Longrightarrow x \equal{} \sqrt [3]{\frac {\sqrt {k}}{2}}\]
$ (i)$ if$ \sqrt [3]{\frac {\sqrt {k}}{2}} < \sqrt {2}$,that is $ k < 32$,we have
\[ f_{min} \equal{} f(\sqrt {2}) \equal{} 2 \plus{} \sqrt {\frac {k}{2}}\]
$ (ii)$ when $ k\geq 32$,we have
\[ f_{min} \equal{} f(\sqrt [3]{\frac {\sqrt {k}}{2}} ) \equal{} \sqrt [3]{\frac {k}{4}} \plus{} \sqrt [3]{2k}\]
which complete the proof.
I hope there are not any logical mistake.....

[quote="Tourish"][quote="can_hang2007"]Sorry, I think your proof is wrong. I wait for another solution.[/quote]
 :blush: Can you point out which i am wrong...then I will try it another way.[/quote]

$ k \equal{} 8,a\equal{}1,b\equal{}0.2,c\equal{}0.8$

$ \sqrt {\frac {a^2 \plus{} kbc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} kca}{c^2 \plus{} a^2}} \plus{}$ $ \sqrt {\frac {c^2 \plus{} kab}{a^2 \plus{} b^2}}\equal{}5.280325472...<6.$

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. If $ k \ge 0$ is a given constant, find the minimum value of the expression
$ P(a,\; b,\; c) \equal{} \sqrt {\frac {a^2 \plus{} kbc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} kca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} kab}{a^2 \plus{} b^2}}.$[/quote]
This is a super solution (with 7 pages): http://can-hang2007.blogspot.com/2010/01/inequality-57-v-q-b-can.html


really ?? could you please post your full solution Algadin


Find the positive numbers $n$ such that $n^{4}$ is the multiple of $3n+7.$

[quote="Kurt Gödel"]If $3n+7 | n^{4}$, then $3n+7 | (3n)^{4}$, so $0\equiv (3n)^{4}\equiv (-7)^{4}\pmod{3n+7}$. Thus we really should have that $3n+7 | 7^{4}$. Now, $n$ is a positive integer, so $3n+7 \geq 3\cdot 1+7 = 10$, so only $3n+7 \in\{7^{2},7^{3},7^{4}\}$ is possible, i.e. $n\in\{14,112,798\}$. For all those numbers, it is easily checked that $3n+7 | n^{4}$.[/quote]
Ah, you used the fact that $a-b|a^{n}-b^{n}$ for $n \in \mathbb{N}$ and $a,\ b \in \mathbb{Z}$. If you had stated that explicitly in the beginning, the explanation would have looked a little nicer.

Anyways, nice problem and nice solution! Thanks :)

[quote="zpz"][quote="FieryHydra"]
Ah, you used the fact that $a-b|a^{n}-b^{n}$ for $n \in \mathbb{N}$ and $a,\ b \in \mathbb{Z}$.[/quote]

Maybe he thought it this way :
$3n+7|(3n)^{4}\Rightarrow (3n)^{4}\equiv 0\pmod{3n+7}$,
$3n+7\equiv 0\pmod{3n+7}\Rightarrow 3n\equiv-7\pmod{3n+7}$
$\Rightarrow (3n)^{4}\equiv(-7)^{4}\pmod{3n+7}$

---
zpz[/quote]

Yes, I did. But of course, that's the same as saying $(3n)-(-7) | (3n)^{4}-(-7)^{4}$.  :)

Certainly.  The statement that $a-b | f(a)-f(b)$ for $f(x) \in \mathbb{Z}[x]$ is simply the statement that $f(a) \equiv f(b) \in \mathbb{Z}[a, b]_{a-b}$.


There are $3$ players and $3$ cards.
$1{\leq}A,B,C{\leq}10$
On card 1 is written number A
On card 2 is written number B
On card 3 is written number C
(Some of the numbers on the cards can be equal)
In each round the cards are being mixed, and after that every player
get a card. Then, each player writes on the paper the number he or she sees on the card.
The game ends after several [b]rounds[/b](we don't know how much).
When the game ends each of the players calculates the [b]"sum"[/b] of the numbers which he wrote on the paper.

There was a game and the [b]"sum"[/b]s  were:
$13,15,23$.
Find $ABC$.

$13+15+23 = 51$ if we denote with $x$ the number of rounds then $x(A+B+C)=51$

$51=3*17$ therefore $A+B+C=3$ and $x=17$ or $A+B+C=17$ and $x=3$

The first case is wrong because it means $A=B=C=1$ and everyone has to get the same sum

Looking at the second case because there are only 3 rounds and $A+B+C=17$ which isn't equal to any of the sums that the players wrote it means each sum has two of one card and one of another card

the only numbers that satisfy this are $(9,5,3)$


If yesterday was Saturday's tomorrow and tomorrow was Wednesday's yesterday, what day would it be today?

can you show how you got that?
its correct by the way.

I doubt that someone would need to see how he got it

Yesterday was Saturday's tomorrow

Saturday's tomorrow is Sunday

Yesterday was sunday, therefore today is monday.....

Tomorrow is wednesday's yesterday

Wednesday's yesterday is tuesday

Therefore today is monday....

[quote="NoSoupForYou"]math92, please don't post 20 problems in the same forum in one day. It's generally frowned upon.[/quote]

Maybe if it's this forum, but what's wrong with posting problems in the math ones? This is a math website.

I think whats being implied is that posting 20 problems *could* be considered spamming.

[quote="NoSoupForYou"]math92, please don't post 20 problems in the same forum in one day. It's generally frowned upon.[/quote]
well, the forum seemed kinda dead so i thought i could liven it up....i didn't mean to spam...


function
trigonometry
vector
complex analysis
complex analysis theorems

To start with, it would be nice to know what theorem exactly you are talking about :roll:.

that any function which satisfies dircherlts condition can be represented as infinite sine & cosine series like
f(x)=summation of (ajcosnx+bjsinnx) where aj=jth coefficient of cos. & n is integral.

[quote="aradhita chattopadhyaya"]i cant find this anywhere. can anybody please help me? actually i put this question 1ce before but didnt find it. please reply.[/quote]

have you actually opened any textbook on Fourier series ?

well yes i did open my text books.
what i dont understand is that how can any function without vrying coefficients can be represented by sums of sines & cosines.

suppose our function under the conditions of the theorem takes the values 4 at t=t1 & 5 at t=t2. which is a function of t.
we can surely chose a0, a1, a2 as the coefficients for each cases t1 & t2 but y are they same?

[quote="aradhita chattopadhyaya"]
what i dont understand is that how can any function without vrying coefficients can be represented by sums of sines & cosines.
[/quote]
Well, does it surprise you that a vector in $ \mathbb R^n$ can be represented as a linear combination of $ n$ basis vectors? For instance, the vector $ (1,2,3,4)\in \mathbb R^4$ is a linear combination of $ (1,1,1,1)$, $ (1,\minus{}1,1,\minus{}1)$, $ (1,\minus{}1,\minus{}1,1)$, and $ (1,1,\minus{}1,\minus{}1)$. You can ask exactly the same question here. ;)


$ ABC$ is a triangle  . the bissector of  $ < B$ intersect [AC] in $ B'$ and the bissector of $ < C$ intersect [AB] in $ C'$.
prove that $ (B'C')$ intersect  the circumscribed circle of $ ABC$

What is a circonscrible circle :?: If I assume circumscribed circle then the problem is too trivial.

i edited Mr Akashnil thank's  can you post a solution of this problem

Obviously, it should be incircle!

Best regards,
sunken rock


Let $a,b,c,d > 0$ be distinct integers such that their product is a perfect square. Prove that $a^{4}+b^{4}+c^{4}+d^{4}$ can be written as the sum of five non-zero perfect squares.

[i]Dan Nedeianu[/i]


abstract algebra
superior algebra
superior algebra solved

How do you prove that $\ds HOM_R\left(\frac{R}{I},\frac{R}{J}\right)\cong\frac{J: I}{J}$ where $J: I=\{r\in R : rI \subset J\}$

[quote]How do you prove that
\[ \text{hom}_R\left(\frac{R}{I},\frac{R}{J}\right)\cong\frac{J: I}{J}  \]
where
\[ J: I=\{r\in R : rI \subset J\}  \][/quote]

This is the wrong section...whoops...I'll repost this.  This can be deleted.

Define $\varphi: J: I\to Hom_R(R/I,R/J)$ by $\varphi(r)(\bar s)=\widehat{rs}$ where $\bar s$ denotes the equivalence class in $R/I$ of the element $s$ in $R$ and $\widehat{rs}$ the equivalence class of $rs$ from $R$ in $R/J$.

1. We first prove that $\varphi$ is well defined. For this we must prove that $\bar s=\bar s'\Rightarrow \widehat{rs}=\widehat{rs'}$.
$\bar s=\bar s'\Rightarrow s-s'\in I$. Since $r\in J: I$, $r(s-s')\in J\Rightarrow \widehat{rs}=\widehat{rs'}$.
Obviously $\varphi$ is an $R$-module homomorphism.

2. We now prove that $\ker\varphi=J$. Since $R/I$ is an $R$-module generated by $\bar 1$ it is easy to see that $\varphi(r)=0\Leftrightarrow \varphi(r)(\bar 1)=0\Leftrightarrow \widehat{r\cdot 1}=0\Leftrightarrow r\in J$. Therefore $\ker\varphi=J$.

3. We prove that $\varphi$ is surjective. For this we must prove that every element $f$ in $Hom_R(R/I,R/J)$ is of the form $\bar s\stackrel f{\to}\widehat{rs}$.
This is easily seen by taking $r\in R$ such that $\hat r=f(\bar 1)$.

From the fundamental isomorphism theorem the conclusion follows.


AMC AIME MATHCOUNTS geometry USAMTS ARML USA(J)MO

This is probably a pretty cliche (i wish i knew how to make an accented e lol) topic, but for those of you with some awesome AIME improvements (like 5+), what was your improvement and what do you feel helped most to make that gain?

Last year I didn't break 100 on the AMC 10  :blush: and this year I got a 5 on the AIME.  I suppose that's pretty significant progress, but I'd like to reevaluate my strategy and figure out what the best things to work on are going to be.  As far as last year, I think the AoPS Independent Study helped me the most and second to that was just doing problems.

I got a 3 last year, and a 7 this year.  I know that's not a big improvement, but if you think about it a little more, it kind of is, because that 3 came from spending like an hour, guessing and checking on number 4.  So it's more like a 2.  And this year, I made stupid mistakes on 4 and 13, so it's more like a 9.  So if you look at it that way, I kind of improved by 7.

But anyways, this year, I just went through AoPS volume 1 over the summer, and did a whole bunch of AIME and the South African Olympiad practice tests.  I know a lot of people say that you should AoPS volume 1 for MATHCOUNTS and the AMC 10, but if you haven't looked through it yet, I'd definetley do it, because I remember I learned a lot from it, especially on the Geometry sections.

I didn't make AIME last year and this year i made it and got 6..
The only thing i did during that period is solving past AIMEs...
I think the most important thing is your confidence, cuz when I was
taking AIME this year I didn't even looked at 10-15.
After AIME , i did 11 and 14 easily....

I got much better at solving the problems from last year, but I made a lot more careless mistakes.
The improvement came from doing problems here, but really the AIME is very easy to mess up...

yeah i didnt take the amc's last year and this year i made the aime and got an 8...i think you just have to practice old tests.

I got a zero in 8th grade, followed by a 7 in 9th grade.  I think the main reason for the improvement was the exposure to more difficult and more advanced problems that came from moving from junior high to high school.  Going into my first AIME, the highest-level problems I had seen were national Mathcounts, which is not very good preparation for AIME.  The next year, I had been in a full season of a high school math league, and I think I had done quite a bit more AIME practice, too.

Well.. I improved from 3 stupid mistakes to 2 :oops:  :)

Hmm, last year I didn't take AMC.

This year I took a good few practice tests and got 122 on AMC 12. Then I took 3 AIMEs and did some other AIME problems, getting 5-6 due to stupid mistakes like not reducing fractions and mis-reading problems. Then I got an 8 on AIME II 2006  :)

[quote="matt276eagles"]Last year:  107.5 AMC 10A

This year:  144 AMC 10A, 130 AMC 12B, 10 AIME I.

So my improvement on the AIME is +10.  :D[/quote]

OMG how did you do it

Last year, I was in 8th grade, and the only competition I knew about and cared about was MATHCOUNTS.  When I took the AMC 10, I didn't know what the AIME was and I didn't care if I qualified.

Since then, I joined this website, bought the AoPS books, participated in many more competitions (USAMTS, ARML, and many local competitions), and practiced many AMC/AIME type problems.  I don't have any special practice methods unknown to everyone else.  :lol:

[quote="matt276eagles"]
Since then, I joined this website, bought the AoPS books, participated in many more competitions (USAMTS, ARML, and many local competitions), and practiced many AMC/AIME type problems.  I don't have any special practice methods unknown to everyone else.  :lol:[/quote]

Hmmm, where can you find a lot of other problems?

Last year (8th grade): I first heard about it and had no practice
AMC 10A: 144, AIME I (official):7 (1 stupid mistake), AIME II (unofficial): 10 (0 stupid mistakes)

This year (9th grade): I practiced a lot for USAMO, none for AMC, and a little from AIME
AMC 12A: 143, AMC 12B: 139.5, AIME I (unofficial): 10 (3 stupid mistakes), AIME II (official): 7 (6 stupid mistakes)

I guess I didn't really improve, but I know I improved unboundedly in USAMO though I am not allowed to take it.

My AMC 12 score improved from a 114.5 last year to a 136 this year.  And my AIME went up 4 points from freshman to sophomore year  (7 to 11), which was pretty cool.  I just did a lot on my own, like I wasn't exactly super advanced, but I knew a lot about the topics I covered.  Especially counting problems, it really helps to get good at combinatorics.

so lets see my AMC 12 ability last year: 103
this year: 150 :)

AIME problemsolving ability: 2, max 3
this year: 10

how? i joined AoPS last year's April, it was around the time they had announced 2005 USAMO qualifiers, i decided that my name [i]will [/i] be on this year's list. i started AoPS with like 5 to 10 posts a day mostly in Getting Started and Intermediate forums...
the Intermediate Forum is the best place i think to get practice for AMC and AIME (besides the past test practices which is the best)

then by the end of school last year, i was ran the "Intermediate Daily Contest" (K81o7 was a winner several times). i ran that contets for like 13/14 weeks, posting one or two  AMC/AIME problem everyday forced myself to study the questions and solutions.

in the summer i attended the "SUMMER INSTITUTE OF MATHEMTAICS AT THE UNIVERSITY OF WASHINGTON" for 6 weeks , there i met some poeple who were doing AIME problems the entire time, I did LOTS  and LOTS of AIME practice stuff over there.  after the 6 weeks program i came home and printed EVERY AIME test on Kalva, and started doing those.

then i started serious practice/review for AMC from mid Dec. also since i was the president of the math team at school i had to run different practice tests for team, which again forced me to take the tests myself, and understand the solutions so i could explain it to others. i also made like 2 Mock AMC tests, which was fun

by the mid Jan. i was just doing AMC problems, and i took the AMC 12A.  it can't be denied that a little bit of luck also helps on the AMC test :P

after the AMC, i started the practice for AIME, which mainly was doing past problems.  I never actually did an AIME for a period of 3 hours for practice, i just did one or two problems in my free time.

also, if you do AoPS a lot, don't get stuck on Getting Started/ Intermediate forums only.  make sure you go and take a look at higher lever problems once in a while.

Hmm.  Last year, I didn't take the AIME (I was in middle school, and I didn't find out about AMC until after it had already passed), but I went to Mathcounts nationals and made 20 (mode score).  I probably would have made less than five on AIME (2? 3?).

Then I started practicing mock AMCs and mock AIMEs (oddly enough, I only did the easy ones, because I didn't want to waste the hard ones right away), and got my AIME score up 5 or 6, then to 7 or 8, with a flunk (3) in between.

Then I did a few 2003 MOP homework problems and when I actually took the AIME, I found that my score had spontaneously jumped to 10.  In my opinion, the events are connected.  After working those few olympiad problems, the first c. 7–8 AIME problems seemed trivial.  (Actually, I had taken the 1983 AIME and gotten 11, but that is reputedly the easiest AIME in the history of the contest, so it didn't count.)  I haven't taken USAMO yet, so I can't say how far I'll have come from being intimidated by the problems yet :D .

Here's my progression:

7th Grade: Worked through Mathcounts Handbook through the year.  20-something score at Chapter Mathcounts, 38 or 37 (31st) at Texas State Mathcounts (weird, eh?)

8th Grade: Just worked through the Mathcounts Handbook through the year.  Messed up chapter - got a 20-something again.  Found AoPS right before state competition, but screwed over Target and got a 33 (55th) at Texas State Mathcounts.

9th Grade: Did almost no math practice until around October, when I got interested again.  Finished working through AoPS volume 1 by the end of the year, and posted a lot on the Getting Started and a bit on the Intermediate forums.  Started reading AoPS Volume 2.  That year I didn't take the AMC (school didn't offer it), but I probably would have gotten 110-115 AMC12, 5-6 on an easy AIME.

10th Grade: During the summer I went to a Math Camp at Tx. State University, did lots of AIME problems.  (Up to 1993? by  the end of summer)  Throughought the year I continued working on AIME problems (I've done all of them except AIME 2004 II), improving to around an 8 and worked through maybe 12? chapters of AoPS II.  I also solved a bunch of problems in the Intermediate Forum (graduated from Getting Started).  Got a 123 AMC 12B and 9 AIME I, making USAMO.  Hopefully I'll do well on USAMO, but I am definitely not expecting to get in MOP.

[quote="kimby_102"][quote="amirhtlusa"]i took that too....
its like 80 (i think)

just go the classes section, you can get  complete list of AoPS classes, including the AIME problem series classes[/quote]
What do u guys do in AIME problem series??
Going over past AIMEs??[/quote]

[url=http://www.artofproblemsolving.com/Classes/AoPS_C_FAQ.php#ps]...[/url]
[url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#aime]...[/url]

7th grade- didn't know difference between the AMC 8 and IMO
8th grade- got 22 on the AMC 8. Learned about the AMC 10 and 12 and the other competitons at National MC. (Thx Boy Soprano II)
9th grade- 146.5 AMC10B; 7 AIME.

The main reason for this improvement is that I learned about these contests and I did several practice competitions on AoPS. Buying an AoPS book might help me improve, but I'm too lazy to order one.

Last year: 0 on AIME :oops:

This year: 8 on AIME

I mainly did a lot of AIME problems from kalva. But the key thing that I did differently was that I didn't really time myself or anything. Rather, I skipped around and did problems that I thought were pretty cool or interesting. Maybe others will think differently, but without the timer, I could focus more on the problems alone.

If you don't study trig you'll die on the AIME.

OK that's a bit of an exaggeration (sp?) but still, even just knowing trig can increase your score by at least a few points.

2 years ago: 116 on AMC-12, 4 on AIME (6 if I didn't make any dumb mistake)
Last year: 126.5 on AMC-12, 5 on AIME (7 if I didn't make any stupid mistake)
This year: 130 on AMC-12, 6 on AIME (8 if I could add numbers correctly)

Do you see any pattern on my AIME scores? :roll:  :P
It seems that I improved on the AMC-12 but not so much on the AIME.

[quote="Aryth"]Experience pays off.
I'm a seventh grader, and in the beginning of the year I took a mock-AMC10 and got a 92  :blush:
but this year, at the real AMC 10 A, I got a 144 (BIG DIFFERENCE) and a 6 on AIME.
If you want to study for AIME, MathCounts...well, it's not the best resource. Try to find sets of past AIMEs and do those. If you're aiming for a an 8 or something like that, don't bother doing olympiad-level problems. They'll mess your skills up.
I took MathCounts state this year and screwed up horribly on Sprint round, but did fine on Target. Also, my AMC 8 score this year was a 20/25  :blush: Moral: Study for your target, not for your dreams (but keep up good spirits!)
AoPS Volume I is neccessary for AIME, and II can come in handy. If you're like me and aiming for a semi-low score, don't bother studying things such as trig, which pop up usually once in the entire thing.[/quote]
I don't see why doing olympiad topics would mess your skill.... :roll:

[quote="DPopov"]Yeah, there's no other way to get great improvements except by doing lots of problems. 1 to 13 in one year (last year, this year I actually dropped one).[/quote]
wow...
huge improvement lol
this is the first time i write AMC and i only got 8 on AIME. barely qualified :(

Please don't tell me you're frowning about qualifying for USAMO...

[quote]I don't see why doing olympiad topics would mess your skill.... Rolling Eyes[/quote]

Sorry; to clarify, studying olympiad-level topics would not assist you in the AIME as much as studying more basic (although I don't mean very basic by this) topics.

Well my advice would be not to listen to the advice of the people posting in this thread (with a few exceptions) and to [color=red][b]read this amazing thread[/b][/color]:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10806


This may sound silly, but I don't understand your inequality, can you type it using Latex?
Thank you!

Sorry I was late with the rewrital,but i made it look good!!!!! :D

Lets prove first the easy right side! :D

$\frac{2a \sqrt{b^{2}+c^{2}}}{ \sqrt{a^{2}+b^{2}+c^{2}}}<a+\sqrt{b^{2}+c^{2}}$

We clear the denomitors and split $2a\sqrt{b^{2}+c^{2}}$  into $a\sqrt{b^{2}+c^{2}}$ and $a\sqrt{b^{2}+c^{2}}$

and we need to prove that:
$a\sqrt{a^{2}+b^{2}+c^{2}}>a\sqrt{b^{2}+c^{2}}$ and $\sqrt{b^{2}+c^{2}}\sqrt{a^{2}+b^{2}+c^{2}}>a\sqrt{b^{2}+c^{2}}$ which are obvious.

As for the other side, you clear the denominators and square and get something obvious... right?

are a, b and c integers ?

$\displaystyle \frac{2a \sqrt{b^2+c^2}}{\sqrt {a^2+b^2+c^2}} = \frac {2a \sqrt {b^2+c^2}}{a \sqrt{b^2+c^2}} = 2 $ and if a, b and c are integers, it is obvious that $a+ \sqrt {b^2+c^2}>2$
and about the same for the first part of the inequality

Well, I thought about a geometrical approach, I didn't see that the algebrical side was so obvious... :)

[quote="Cauchy_Schwartz"]Well, I thought about a geometrical approach[/quote]

Can you please post it, I am interested in such an approach!

Thank you very much Cauchy-Schwartz.


Hello
I hope most of u r knowing this problem.
[b]We have 36 horses.At one time we could have race for 6 horses in our race track.Now what is least number of races we should do to find the fastest three?[/b]
Answer is 8 races.
If we make six groups,we need six races to find the fastest three in each group.But what I should do in last two races as it can happen the horse coming third in group I is better than the horse coming first in group II.
If it is posted earlier,kindly give me the link.
Thank  u

If you're posting a brainteaser, you should either hide the answer or not post it so others can attempt to figure it out first.

Hello
I thought most of u were knowing the answer already .Even then I will keep that in my mind.
Thank u

Exactly what is the point if:

1) You assume others know it [you give out the answer].
2) You assume others don't know it [you actually make the post, which wouldn't matter if people already knew it].
It's a contradiction.

Hello
Does anyone got the answer?
Thank u.

[quote="aleph0"]I think that what is going on is that kabi knows the answer but not the solution.[/quote]
Thanks there is someone who is understanding my question.I know d answer but I don't know how to get it.

[quote="kabi"]
Answer is 8 races.
If we make six groups,we need six races to find the fastest three in each group.But what I should do in last two races as it can happen the horse coming third in group I is better than the horse coming first in group II.
If it is posted earlier,kindly give me the link.
Thank  u[/quote]

I think this works:

As you said, make six groups of six and race each of them.  Keep track of which were the winners of each group, as well as the group they were originally in.  In race 7, race the winners, and take the top three from there.  Call these three horses A, B, and C, and assume without loss of generality that A was first, B second and C third.

Now we know for certain that A is the fastest horse.  To figure out who is second or third, we need one more race.  In the 8th race, put the 2nd and 3rd place from A's original group, the top two from B's group, and C (and any other horse, these are the only important ones).  The reason for racing these horses is that they are the only ones left with a shot at placing.  For example, we did not need to consider the 2nd place in C's group, because if he placed, then so would C and B, but that's too many.  So the winner and runner up in this race are the 2nd and 3rd horse overall.

Hello
thanks cyberspace.Got ur answer. :10:


i thought this was obvious  :blush:

missing link: (put this in between and it will be obvious ;) )

$3\cdot 9^{4010}+6\cdot 36^{2005}+3\cdot 4^{4010}$ (put that $3$ in front and you have something of the form $a^2+2ab+b^2$)

[quote="mathical88"]what if modulus is not allowed? how can you solve this qn by simple algrebraic manipulation? or any other methods? (just not mod)  :oops:[/quote]
divisibility almost equals mod :P
i'ld love to see a solution without it (although i'm quite sure there exist some  :ninja: )

I know how to do it another way. :D

If you have: 6^63-2^63 you can prove that it is divisible by 4 because

6^63-2^63=(6-2)(6^62+6^61*2+6^60*2^2+6^59*2^3+...+2^63) and is thus divisible by 4.

 :blush: But I don't know how you would do it for such a complicated question.

Any  :idea: minds out there? who can see how you would do this question.  :roll:

Quick manipulation!!!

Or maybe I'm just slow :(

How did you get to 3(3^2005+2^2005)^2

I don't understand :huh:


[hide="i got..."]
$\frac{1+2b-3a}{5b-5ab}$  but I'm not sure.  I'll post solution if its correct[/hide]

Hm.  I thought he meant $1125$, not $11.25$... it was messy.

[quote="xenon_nightmare152"]Do you have any more good log problems/websites?   :D[/quote]

Yes I do, i will post more problems in the forum , right?

[quote="xenon_nightmare152"]Do you have any more good log problems/websites?   :D[/quote]
try the AoPS book volume 2 :D


ratio
algebra
polynomial
geometric sequence
arithmetic sequence

Find the general form for each term in the recursion $a_{n}=Aa_{n-1}+Ba_{n-2}+C.$

[hide="A hint"]Find some $a$ which you can add to every term in order to eliminate $C$.[/hide]

[quote="FieryHydra"][b]JBL[/b], is this what you meant in your hint?

To find the general term of the recursion $a_{n}= Aa_{n-1}+Ba_{n-2}+C$, we just need to find $p,q,r$ such that $a_{n}+pa_{n-1}+r = q(a_{n-1}+pa_{n-2}+r).$

We then must have $A = q-p, B = pq$ and $C= r(q-1).$ We should be able to find $p$ and $q$ uniquely from the first two equations and then determine $r$ from the last equation.

After that we note that $b_{n}= a_{n}+pa_{n-1}+r$ is a geometric sequence with common ratio $q.$

Is this a correct approach?[/quote]

that is an interesting approach (i am not sure that there are always solutions though...)

but i think that he meant the substitution: \[b_{n}=a_{n}+k\] Where you find $k$ such that you eliminate the constant. Then you can easily use the characteristic polynomial, etc, to find the explict formula for $b_{n}$.

However, if this does not work, then you can always make the substitution: \[b_{n}=a_{n}+k\cdot n\]

[b]First possibility.[/b]

Put $a_{n}=b_{n}+K$ where $K$ is a constant chosen to cancel $C$. After rearranging we get $K(1-A-B)=C$. Hence this works only if $A+B\neq 1$, in which case we can substitute $a_{n}=b_{n}-\frac{C}{A+B-1}$ and proceed by using characteristic equation etc.

[b]Second possibility.[/b]

If $A+B=1$ and $(A,B)\neq(2,-1)$, then we have

\begin{eqnarray*}a_{n}=Aa_{n-1}+(1-A)a_{n-2}+C &\iff& a_{n}-a_{n-1}+\frac{C}{A-2}=(A-1)a_{n-1}+(1-A)a_{n-2}+\frac{C(A-1)}{A-2}\\ &\iff& a_{n}-a_{n-1}+\frac{C}{A-2}=(A-1)\left(a_{n-1}-a_{n-2}+\frac{C}{A-2}\right)\end{eqnarray*}

Then we substitute $b_{n}=a_{n}-a_{n-1}+\frac{C}{A-2}$ and solve for geometric sequence, and then $a_{n}-a_{n-1}=b_{n}-\frac{C}{A-2}$ can be solved for $a_{n}$ by summing up all the equalities and telescoping the LHS.

[i]Note to this case:[/i] This is the situation in which [b]Altheman[/b]'s proposed substitution $a_{n}=b_{n}+kn$ works.

[b]Third possibility.[/b]

If $(A,B)=(2,-1)$ then

$a_{k}-2a_{k-1}+a_{k-2}=C$

Summing up all the equalities for $k=\overline{3,n}$ and denoting $S=a_{1}+\dots+a_{n}$, we get

$(S-a_{1}-a_{2})-2(S-a_{1}-a_{n})+S-a_{n}-a_{n-1}=(n-2)C$

which after rearranging gives

$a_{n}-a_{n-1}=(n-2)C-a_{1}+a_{2}$

which obviously also can be solved by summing up the equalities and telescoping the LHS.

[i]Note to this case:[/i] Here can be employed the substitution $a_{n}=b_{n}+kn^{2}$ in order to obtain $b_{n}-2b_{n-1}+b_{n-2}=0\iff b_{n}-b_{n-1}=b_{n-1}-b_{n-2}$, which is a simple arithmetic sequence solvable immediately from the initial conditions.

Wow, that is great! Do you want to copy that to the wiki?

why did you do special cases for where $(A,B)\neq(2,-1)$ and $(A,B)=(2,-1)$?

Because in the second case there appears $A-2$ in the denominator, hence $A-2=0$ and $A+B=1$, which is the general condition of Cases 2 and 3, give $A=2, B=-1$


inequalities
induction
function
geometry
inequalities proposed

Let n be a positive integer.
Prove the inequality:             $n!>(\frac{n}{e})^n$ .

We will prove it using induction:
- of course for n=1 the inequality holds
- let's assume that $n! > (\frac{n}{e})^{n}$. We have to prove that $(n+1)! > (\frac{n+1}{e})^{n+1} \Longleftrightarrow n!(n+1) > \frac{(n+1)^{n+1}}{e^{n+1}} \Longleftrightarrow n! > \frac{(n+1)^{n}}{e^{n+1}}$. If we prove that $(\frac{n}{e})^{n} > \frac{(n+1)^{n}}{e^{n+1}}$, we are done (from our assumption). But $(\frac{n}{e})^{n} > \frac{(n+1)^{n}}{e^{n+1}} \Longleftrightarrow n^{n} > \frac{(n+1)^{n}}{e} \Longleftrightarrow e > (\frac{n+1}{n})^{n} \Longleftrightarrow e > (1+\frac{1}{n})^{n}$, which is true, since e is the limit of $(1+\frac{1}{n})^{n}$ and the function $(1+\frac{1}{n})^{n}$ is increasing.

It´s interesting you consider a function f(x)=ln(x) and a interval [1,2,3,...,n] and compare the areas above and bellow the curve. You will get a more powerfull inequalitie.

Thus, (iff my calculations are correct) we obtain $\displaystyle (n-1)! \leq \frac{n^n}{e^{n-1}} \leq n!$.