Let $ AB$ be the diameter of a circle with a center $ O$ and radius $ 1$. Let $ C$ and $ D$ be two points on the circle such that $ AC$ and $ BD$ intersect at a point $ Q$ situated inside of the circle, and $ \angle AQB\equal{} 2 \angle COD$. Let $ P$ be a point that intersects the tangents to the circle that pass through the points $ C$ and $ D$.

Determine the length of segment $ OP$.

Let $ \angle COD \equal{} x$, $ \angle AQB \equal{} 2x$, $ \angle CAO \equal{} \angle ACO \equal{} m$, $ \angle DBO \equal{} \angle BDO \equal{} n$. From $ \triangle AQB$, $ m \plus{} n \equal{} 180 \minus{} 2x$. But the angles of ODQC must add up to 360, note that $ \angle DQC$ counts as more than 180 degrees since the quad is concave, so we get $ m \plus{} n \equal{} x$. So $ x \equal{} 60$, so triangles POD and COP are 30-60-90 triangles with DO=CO=1, so $ PO \equal{} \frac {2\sqrt {3}}{3}$.


Let a triangle $ABC$
Prove that:
$Sin^3A+Sin^3B+Sin^3C<2$

[quote="Kapi"]Let a triangle $ABC$
Prove that:
$sin^3A+sin^3B+sin^3C<2$[/quote]

Make use of well-known inequalities :
$sinA+sinB+sinC \leq \frac{3  \sqrt{3}}{2}$
$sin^2A+sin^2B+sin^2C \leq \frac{9}{4}$

and then chebysev's:

$sin^3A+sin^3B+sin^3C \leq \frac{(sinA+sinB+sinC )(sin^2A+sin^2B+sin^2C)}{3} \leq \frac{9 \sqrt {3}}{2}<2$

[quote="siuhochung"]I think you reversed the sign of chebyshev.[/quote]

Oh yes........silly me    :blush:

I have a proof :through p,R,r
[tex]sinA^3+sinB^3+sinC^=\frac{p(p^2-3r^2-6Rr)}{4R^3}<2[/tex]
And use Gerretsen's ineq:[tex]p^2\leq\ 4R^2+4Rr+3r^3(*)[/tex]
I only prove:
[tex]p(4R^2-2Rr)\leq\ 8R^3[/tex]
[tex]p(2R-r)\leq\ 4R^2[/tex]
Multiping , and use (*) one time we have:
[tex]8Rr^3\geq\ 3r^4[/tex]
[tex]8R\geq\ 3r[/tex]
True by:[tex]R\geq\ 2r[/tex]
[tex]8R\geq\ 16>3r[/tex]

[quote="keira_khtn"]I have a proof :through p,R,r
[tex]sinA^2+sinB^2+sinC^2=2p^2-8Rr-2r^2[/tex][/quote]
i don't understand your idea :blush:

Nice and suprising inequality.
I prove as follow:

$(i)$If $A \le B \le C$ then

$sin^3A+sin^3B \le 2sin^3\frac{A+B}{2}$

$(ii)$ For all $A \le \frac{\pi}{2}$ :

$2sin^3A+sin^32A \le 2$


function
symmetry
analytic geometry
Vieta
algebra unsolved
algebra

Let $ f(x)\equal{}x^3\minus{}6x^2\plus{}17x$.If $ f(a)\equal{}16,f(b)\equal{}20$,try to find the value of $ a\plus{}b$

hello, calculating the roots of the equations
$ a^3\minus{}6a^2\plus{}17a\equal{}16$
$ b^3\minus{}6b^2\plus{}17b\equal{}20$
with the formulas of Cardano and adding the solutions i have got $ a\plus{}b\equal{}4$.

[quote="Dr Sonnhard Graubner"]hello, calculating the roots of the equations
$ a^3 \minus{} 6a^2 \plus{} 17a \equal{} 16$
$ b^3 \minus{} 6b^2 \plus{} 17b \equal{} 20$
with the formulas of Cardano and adding the solutions i have got $ a \plus{} b \equal{} 4$.[/quote]

Can you give me the full solution?

Who can find $ a\plus{}b\equal{}?$ by other way?

[quote="skywalkerJ.L."]Let $ f(x) \equal{} x^3 \minus{} 6x^2 \plus{} 17x$.If $ f(a) \equal{} 16,f(b) \equal{} 20$,try to find the value of $ a \plus{} b$[/quote]

Let $ a\plus{}b\equal{}k$, we have $ a^3\plus{}b^3\equal{}k^3\minus{}3abk$.
$ \therefore f(a)\plus{}f(b)\equal{}36\Longleftrightarrow a^3\plus{}b^3\minus{}6(a^2\plus{}b^2)\plus{}17(a\plus{}b)\equal{}36$
$ \therefore k^3\minus{}3abk\minus{}6(k^2\minus{}2ab)\plus{}17k\equal{}36$
$ \Longleftrightarrow k^3\minus{}6k^2\plus{}17k\minus{}36\minus{}3(k\minus{}4)ab\equal{}0$
$ \Longleftrightarrow (k\minus{}4)(k^2\minus{}2k\plus{}9\minus{}3ab)\equal{}0$
$ \Longleftrightarrow k\equal{}4\ or\ k^2\minus{}2k\plus{}9\minus{}3ab\equal{}0$.

We need to examin for $ k^2\minus{}2k\plus{}9\minus{}3ab\equal{}0$.

[quote="skywalkerJ.L."] [color=darkred]Let $ f(x) \equal{} x^3 \minus{} 6x^2 \plus{} 17x$. If $ f(a) \equal{} 16,f(b) \equal{} 20$ , then try to find the value of $ a \plus{} b$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Observe that the function $ f$ is strict increasing on $ \mathbb R$ and $ (\forall )\ x\in\mathbb R\ ,\ f(4 \minus{} x) \equal{} 36 \minus{} f(x)$ .

Hence its graph has the point of symmetry $ (2,18)$ . In the particular case $ x: \equal{} a$ obtain

$ f(4 \minus{} a) \equal{} 36 \minus{} f(a) \equal{} 36 \minus{} 16 \equal{} 20 \equal{} f(b)\ \stackrel{(f\nearrow )}{\ \implies\ }\ 4 \minus{} a \equal{} b\ \implies\ \boxed {\ a \plus{} b \equal{} 4\ }$ .[/color]

Let $ g(x)\equal{}f(x\plus{}2)$ then $ g(x)\equal{}x^3\plus{}5x\plus{}18$. $ 16\equal{}f(a)\equal{}g(a\minus{}2)\equal{}g(c)$.

$ g(c)\plus{}g(d)\equal{}c^3\plus{}d^3\plus{}5(c\plus{}d)\plus{}18\plus{}18\equal{}16\plus{}20$
$ (c\plus{}d)(c^2\minus{}cd\plus{}d^2\plus{}5)\equal{}9$.

The second factor has no real solutions...

Let $ g(x)\equal{}(2\plus{}x)^3\minus{}6(2\plus{}x)^2\plus{}17(2\plus{}x)\minus{}18\equal{}x^3\plus{}5x$, then $ g(a\minus{}2)\equal{}\minus{}2,g(b\minus{}2)\equal{}2\equal{}\minus{}g(b\minus{}2)$. Because $ g'(x)>0$ and $ g(\minus{}x)\equal{}\minus{}g(x)$
we get $ a\minus{}2\equal{}\minus{}(b\minus{}2)$ or $ a\plus{}b\equal{}4$.

[quote="Virgil Nicula"][quote="skywalkerJ.L."] [color=darkred]Let $ f(x) \equal{} x^3 \minus{} 6x^2 \plus{} 17x$.If $ f(a) \equal{} 16,f(b) \equal{} 20$ . Try to find the value of $ a \plus{} b$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Observe that the function $ f$ is strict increasing on $ \mathbb R$ and $ (\forall )\ x\in\mathbb R\ ,\ f(4 \minus{} x) \equal{} 36 \minus{} f(x)$ .

Hence its graph has the point of symmetry $ S(2,18)$ . In the particular case $ x: \equal{} a$ obtain

$ f(4 \minus{} a) \equal{} 36 \minus{} f(a) \equal{} 36 \minus{} 16 \equal{} 20 \equal{} f(b)\ \stackrel{(f\nearrow )}{\ \implies\ }\ 4 \minus{} a \equal{} b\ \implies\ \boxed {\ a \plus{} b \equal{} 4\ }$ .[/color][/quote]

Nice Solution, Virgil Nicula.

To be honest, first I imagined the inflection point.

$ f(x) \equal{} (x \minus{} 2)^3 \plus{} 5(x \minus{} 2) \plus{} 18$ gives the inflection point $ (2,\ 18)$, remark that $ \frac {f(a) \plus{} f(b)}{2} \equal{} 18$, we can imagine that $ a,\ b\ (a<b)$ are the coordinates of the points of the local maximum, local minimum respectively. Then $ f'(x) \equal{} 3x^2 \minus{} 12x \plus{} 17$, by the property of the cubic function, using Vieta's formula we obtain
$ a \plus{} b \equal{} \boxed{4}$.


Ok, so last year i've handwritten my solutions and sent them through the mail, but this time I want to upload them.
Can you handwrite your solutions, scan those to a single pdf, and then upload them?
I've read the rules (below) and not quite sure if you can do that when submitting solutions by web upload..
[quote]Submitting solutions by mail
You may handwrite or type your solutions. We strongly recommend using these blank solution sheets. (However, if you wish to use blank paper, you may do so.)

•Write on only one side of the page.
•All pages must be stapled together, with the problems in order (Problem 1 should be first, then Problem 2, etc.).
•Keep a copy of your solutions for yourself, but send us the original.

When mailing, please note that the mail submission deadline is a postmark deadlines -- it is OK if we receive the solution after the deadline as long as the postmark is on or before the deadline. Mail solutions to:

USAMTS
PO Box 2090
Alpine, CA 91903-2090

Submitting solutions by web upload
There is a form on your My USAMTS page from which you can submit your solutions directly via the website. To do this, your solutions must be in a single PDF file. Absolutely no other file formats will be accepted.

IMPORTANT!  The deadline for submitting solutions via the website is

3 PM Eastern / Noon Pacific
on the due date.

Writing your solutions using LaTeX (see below) is the best way to make a nice PDF file. Alternatively, you can convert many other document types (such as Microsoft Word documents) to PDF. You can download one of a number of free PDF conversion programs -- we recommend CutePDF Writer. Or you can use a free web-basd PDF conversion service such as PDF Online. Additionally, most scanner software has an option to scan documents into a PDF file.[/quote][/quote]

Sure, you can do that.  The rules don't specify that the PDF file has to be typeset in LaTeX or anything else.  Many students handwrite, scan, then upload.  Check the scan first though: make sure it is dark and legible for our graderss.

Can you web submit your solutions and mail your entry form and permission and stuff?

Is there a way way to electronically submit entry forms?

Entry forms must be mailed or faxed.  Follow the instructions on the form.  It cannot be uploaded to us via the website.  If you are planning to upload your Round 1 solutions, then you should separately mail or fax your entry form.

Since I'm turning 13 in a little over a month, do I still need to fill out the Under 13 form? The first round problems are due after my birthday, so it seems like there's no point, but I just want to make sure.

EDIT: I swear I made this a new topic, but it got screwed up and went as a reply. Sorry about that and thanks. :oops:

If you are under 13 at the time that you first register for the site, you need to either (a) send us the permission form, or (b) wait until you turn 13 and send us (at usamts@usamts.org) an email informing us that you're now 13, and we can change the status of your account from "under 13" to "13 and over".  It's slightly easier for us if you choose option (a), but it really doesn't matter.


$ a_i$ must be larger than $ 1$, else $ a_i^2 \minus{} 1 < 0 < a_i$

(or I'm misunderstanding something)

For example, if $ a_i\equal{}\left (\frac{1}{2} \right )^i$, then the second inequality doesn't work.

[quote="EastyMoryan"]$ a_i$ must be larger than $ 1$, else $ a_i^2 - 1 < 0 < a_i$

(or I'm misunderstanding something)

For example, if $ a_i = \left (\frac {1}{2} \right )^i$, then the second inequality doesn't work.[/quote]

but

$ \frac{1}{2}(\sum{a_{i}})(\sum{{a_{i}}^{2}-1})=\frac{1}{2}(\sum{{a_{i}}^{2}})(\sum{\frac{1}{{a}_{i}})-\frac{1}{2}(\sum{a_{i}})}$ ?

why larger than 1?

The second inequality states that $ \frac{1}{2}\left (\sum{a_{i}}\right )\left (\sum{{a_{i}}^{2}\minus{}1}\right )\geq\sum{a_{i}}$
This is equivalent to:

$ \left (\sum a_i^2\minus{}1\right ) \ge 2$
Except that if $ 0<a_i<1\to 0<a_i^2<1\longrightarrow a_i^2\minus{}1<0<2$


modular arithmetic
algebra unsolved
algebra

[quote="AndrewTom"]Find, showing your method, a six-digit integre $ n$ with the following properties: (i) $ n$ is a perfect square, (ii) the number formed by the last three digits of $ n$ is exactly one greater than the number formed by the first three digits of $ n$. (Thus $ n$ might look like $ 123124$, although this is not a square.)[/quote]
Let $ x^2\equal{}\overline{abcdef}$ and $ y\equal{}\overline{abc}$. we get $ x^2\equal{}1000y\plus{}y\plus{}1\equal{}1001y\plus{}1$

And so we are looking for numbers whose square is $ 1$ $ \pmod {1001}$, so $ x^2\equal{}1\pmod 7$ and $ x^2\equal{}1\pmod {11}$ and $ x^2\equal{}1\pmod {13}$ and $ x\in[317,999]$

So $ x\equal{}\pm 1\pmod 7$ and $ x\equal{}\pm 1\pmod {11}$ and $ x\equal{}\pm 1\pmod {13}$ and $ x\in[317,999]$ and so just $ 8$ tests :

Trying $ x\equal{} 1\pmod 7$ and $ x\equal{} 1\pmod {11}$ and $ x\equal{} 1\pmod {13}$ gives $ 1$ as first value, too low

Trying $ x\equal{} 1\pmod 7$ and $ x\equal{} 1\pmod {11}$ and $ x\equal{} \minus{}1\pmod {13}$ gives $ 77m\plus{}1\equal{}13n\minus{}1$ and so $ 13n\minus{}77m\equal{}2$ which is easy to solve, giving $ 155$, too low

Trying $ x\equal{} 1\pmod 7$ and $ x\equal{} \minus{}1\pmod {11}$ and $ x\equal{} 1\pmod {13}$ gives $ 91m\plus{}1\equal{}11n\minus{}1$ and so $ 11n\minus{}91m\equal{}2$ which is easy to solve, giving $ 274$, too low

Trying $ x\equal{} 1\pmod 7$ and $ x\equal{} \minus{}1\pmod {11}$ and $ x\equal{} \minus{}1\pmod {13}$ gives $ 7m\plus{}1\equal{}143n\minus{}1$ and so $ 143n\minus{}7m\equal{}2$ which is easy to solve, giving $ 428$, which is OK.

Hence a solution : $ \boxed{428^2\equal{}183184}$

It would be easy to test the 4 remaining cases to find the 3 other solutions :
$ 573^2\equal{}328329$
$ 727^2\equal{}528529$
$ 846^2\equal{}715716$

[quote="AndrewTom"]Hi pco.

Why just $ 8$ tests and not $ 9$?[/quote]
$ x\equal{}1\pmod 7$ and $ x\equal{}1\pmod{11}$ and $ x\equal{}1\pmod{13}$
$ x\equal{}1\pmod 7$ and $ x\equal{}1\pmod{11}$ and $ x\equal{}\minus{}1\pmod{13}$
$ x\equal{}1\pmod 7$ and $ x\equal{}\minus{}1\pmod{11}$ and $ x\equal{}1\pmod{13}$
$ x\equal{}1\pmod 7$ and $ x\equal{}\minus{}1\pmod{11}$ and $ x\equal{}\minus{}1\pmod{13}$
$ x\equal{}\minus{}1\pmod 7$ and $ x\equal{}1\pmod{11}$ and $ x\equal{}1\pmod{13}$
$ x\equal{}\minus{}1\pmod 7$ and $ x\equal{}1\pmod{11}$ and $ x\equal{}\minus{}1\pmod{13}$
$ x\equal{}\minus{}1\pmod 7$ and $ x\equal{}\minus{}1\pmod{11}$ and $ x\equal{}1\pmod{13}$
$ x\equal{}\minus{}1\pmod 7$ and $ x\equal{}\minus{}1\pmod{11}$ and $ x\equal{}\minus{}1\pmod{13}$

What could be the $ 9^{th}$ ?

Sorry pco; I wasn't thinking straight. Thanks for your reply.


i have a question. i know how you would find the answer (exact value)  to ${%Error. "display" is a bad command.
\prod_{n=1}^{89}}\sin(2n-1)$ (angles are in degrees)
but i am wondering how you would find the answer (exact value) to $\displaystyle\prod_{n=1}^{89}\sin(2n)$

The problem looks very :) familiar !

The answer, of course is  $\frac{45} {2^{87}}$  :)

[hide]Okay you can look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53316&start=20]Here[/url][/hide]


A riverboat theater offers bus tours to groups on the following basis.  Hiring the bus costs the group 360 (in dollars), to be shared equally by the group members. Theater tickets, normally 30 each, are discounted by .25 times the number os people in the goup. How many members must be in the group so that the cost of the theater tour (bus fare plus theater ticket) is less than 39 per person?

I set up the following inequality:

(30x - .25x) + (360 / x) < 39x

Did I set it up right?

I would first set up an equality relationship and then figure out what you need for it to be less (they are easier to work with than inequalities).
So, solve this relationship:
(30x - .25x) + (360 / x) = 39x
And then the answer ought to be whatever x comes out to be minus 1.


What is the usage of natural log and logarithms?

It's the inverse of exponents. Like if u use a log of a number to a base, the logarithm is the number u have to raise the base to the power to get the number. (e.g. log2 4=2) And ln(x) or natural logarithms, is with base e. (e.g. ln e=1)

Logarithms were originally developed to simplify multiplication.  Rather than multiplying two numbers $ a, b$, one looked up their logarithms in a table, used the identity

$ \log ab \equal{} \log a \plus{} \log b$

and looked up the number $ ab$ corresponding to the added logarithm $ \log a \plus{} \log b$.

Nowadays, logarithms are a fundamental tool in calculus (that is, analysis).  If that's the kind of answer you're looking for.

Subject to a few minor restrictions, logarithms are the only continuous functions from the positive reals to the reals that satisfy $ f(ab) \equal{} f(a) \plus{} f(b)$. This is the reason they were used as described by t0rajir0u.


I think the correct is

For $ a,b,c$ positive reals such that $ a+b+c=3$
\[ \frac{a}{b^{2}+2}+\frac{b}{c^{2}+2}+\frac{c}{a^{2}+2}\ge 1 \]

and also the folllowing general statement holds
For $ a,b,c$ positive reals such that $ a+b+c=3$ and $ 0<p\leq 3$
\[ \frac{a}{b^{2}+p}+\frac{b}{c^{2}+p}+\frac{c}{a^{2}+p}\geq\frac{3}{1+p}\]

I found the topic:

"From one of australian team Created by Vasc"

but I wait a simple solution for Silouan generalization.

Thank you very much.

[quote="manlio"]but I wait a simple solution for Silouan generalization.
Thank you very much.[/quote]

In Vasc's book there is a great solution using only AM-GM and an inequality known again from him .
But If I remember well Vasc also posted a prof here in ML using Cauchy-Swartz
and $ (a^{2}+b^{2}+c^{2})^{2}\geq 3(a^{3}b+b^{3}c+c^{3}a)$

(It was a topic of Hungthn )

[quote=manlio]The Vasc inequality:

For $ a,b,c$ positive reals such that $ a+b+c=3$
\[ \frac{a}{b^{2}+2}+\frac{b}{c^{2}+2}+\frac{c}{a^{2}+2}\ge 1 \][/quote]

For numbers positive reals $a,b,c$ such that $a+b+c = 3$, [url=https://artofproblemsolving.com/community/c6h1840932p12368123]prove that
[/url]
$$\frac{a}{b^3+2} + \frac{b}{c^3+2} + \frac{c}{a^3+2} \geq 1 .$$

For $ a,b,c$ positive reals such that $ a+b+c=3$, prove or disprove
\[ \frac{a}{b^{n}+2}+\frac{b}{c^{n}+2}+\frac{c}{a^{n}+2}\ge 1 \]
Where $2\leq n\in N .$

[quote=sqing]
For $ a,b,c$ positive reals such that $ a+b+c=3$, prove or disprove
\[ \frac{a}{b^{n}+2}+\frac{b}{c^{n}+2}+\frac{c}{a^{n}+2}\ge 1 \]
Where $2\leq n\in N .$[/quote]

For the positive integer $n \ge 4$, the inequality is not true.
Try $a=c \rightarrow 1.5^-$ and $b\rightarrow 0^+$.

[quote=sqing]

For numbers positive reals $a,b,c$ such that $a+b+c = 3$, [url=https://artofproblemsolving.com/community/c6h1840932p12368123]prove that
[/url]
$$\frac{a}{b^3+2} + \frac{b}{c^3+2} + \frac{c}{a^3+2} \geq 1 .$$[/quote]
Thanks. Is true ? If inequality is true, they are looking for a solution.

[quote=NaPrai]
[/quote]
For $ a,b,c$ positive reals such that $ a+b+c=3$, inequality
\[ \frac{a}{b^{n}+2}+\frac{b}{c^{n}+2}+\frac{c}{a^{n}+2}\ge 1 \] for n = 3 is true ? I'm not solution.

[quote=florin_rotaru]
[/quote]
For $ a,b,c$ positive reals such that $ a+b+c=3$, inequality
\[ \frac{a}{b^{n}+2}+\frac{b}{c^{n}+2}+\frac{c}{a^{n}+2}\ge 1 \] for n = 3 is true ? I'm not solution.
[quote=arqady]
See here:
https://artofproblemsolving.com/community/c6h1256456[/quote]
Thank you very much.


modular arithmetic
number theory
prime numbers
number theory unsolved

Is this the problem you intend:
For any integer $ n$, there exists a prime $ p$, so that each of the numbers $ p\minus{}1, p\plus{}1, p\plus{}2$ have more than $ n!$ divisors?
It seems like a very strange problem, and if that is what you intend, maybe this should go into Open problems section.

I think the problem might be "some $ x \in \{p\minus{}1, p\plus{}1, p\plus{}2\}$ has more than $ n$ divisors."

It 's trivial with the use Chinese Remainder Theorem + Dirichlet theorem

could you please post the solution?? thankyou very much

Nobody can post a solution before you state the problem more clearly!

Let $ 2,p_1,...,p_N$ prime divisors $ p\minus{}1$, $ 2,q_1,...,q_N$ prime divisors $ p\plus{}1$ and $ r_1,..,r_N$ prime divisors $ p\plus{}2$.
It give $ p\equal{}S\mod P\equal{}2p_1...p_Nq_1....q_Nr_1...r_N$ if $ p_1,..,p_N,q_1,...,q_N,r_1,...,r_N$ are distinct odd primes.
By Dirichlet's theorem we find prime, suth that $ p\equal{}S\mod P$ (or equavalent $ 2p_1...p_N|p\minus{}1, \ 2q_1...q_N|p\plus{}1, \ r_1...r_N|p\plus{}2$.

[quote="kihe_freety5"]prove that with each value of given n we can determine  p which is prime number such that if x is in{p-1;p+1;p+2 } then the number divisor x is greater than n![/quote]
sorry,
the problem is :
prove that with each value of given n we can determine  p which is prime number such that if x is in{p-1;p+1;p+2 } then the number divisor x is greater than n
 I wrote "n!"; it is "n"

sorry!!!  :(

[quote="kihe_freety5"]prove that with each value of given n we can determine  p which is prime number such that if x is in{p-1;p+1;p+2 } then the number divisor x is greater than n[/quote]
Take three pairwise co-prime numbers $ m_1$, $ m_2$, $ m_3$ each of which has greater than $ n$ divisors (such numbers can be formed by products of distinct primes, for example), solve the system of congruences:
\[ \begin{cases} x\minus{}1\equiv 0\pmod{m_1}\\ x\plus{}1\equiv 0\pmod{m_2}\\ x\plus{}2\equiv 0\pmod{m_3}\end{cases}\]
and get a solution in the form $ x\equiv t\pmod{m_1 m_2 m_3}$, and finally use Dirichlet theorem to conclude that there is a prime number that is a solution to this system. This prime is exactly what you are looking for.


analytic geometry
LaTeX
linear algebra
matrix

[quote="pda"]An ant wanders over the coordinate plane. Whenever it finds itself stationary at a point (x,y), it walks without stopping to the point (-3x-y,7x+ky), where k is an integer. What is the value of k if, wherever you place the ant on the plane, it returns to that point after a finite number of stops?[/quote]
[hide="Click here for a messy solution"]Let's look at the first few points over where the ant wanders to:
$ (x,y)$

$ (-3x-y,7 x+k y)$

$ (2x+(3-k)y, (7k-21)x+(k^{2}-7)y)$

$ ((15-7k)x-(k-1)(k-2)y) , 7(k-1)(k-2)x+(k^{3}-14k+21) y)$

$ \vdots$

Clearly, it is impossible after the second location to concur with the first location, since none of the two x-coordinates involve $ k$.

Trying to match up the first with the third, we get $ k=2$, but that fails when trying to match with the y-coordinates.

Trying the math up the first with the fourth, we get $ \boxed{k=2}$, which works.
[/hide]

Ugh, ugly computation. Is there an easier method?
This problem does look like it has been carefully rigged...

[quote="JBL"]LaTeX hint: using arrays and double dollar signs can make things much prettier.

[quote="vishalarul"]Clearly, it is impossible after the second location to concur with the first location, since none of the two x-coordinates involve $ x$.
...
Ugh, ugly computation. Is there an easier method?[/quote]

This is not clear at all to me -- in fact, I have no idea what you're talking about.

This problem lives most naturally as a simple exersize in linear algebra.[/quote]

Sorry, I meant $ k$.

I just started looking at some points to where the ant would go and started to match up the coefficients.

[hide="If I were to use linear algebra,"]Then the problem asks for what $ k$, there is a finite $ n$ such that:

$ \begin{pmatrix}-3 &-1 \\ 7 & k \end{pmatrix}^{n}= I_{2}$

But I don't know how to continue from here.  :( [/hide]

If two matrices are equal, their determinants are equal.  Is that enough to finish it for you?  (Don't forget to watch out for the complex numbers.)


calculus
integration
function
calculus computations

Hello!

Does there exist a function $ f: [0,1]\to {\Bbb R}$ such that
\[ f(x)\equal{}1\plus{}x\int_0^1 f(t)\,dt\plus{}x^2\int_0^1 f(t)^2\,dt.\]

[quote="yuyang"]Hello!

Does there exist a function $ f: [0,1]\to {\Bbb R}$ such that
\[ f(x) \equal{} 1 \plus{} x\int_0^1 f(t)\,dt \plus{} x^2\int_0^1 f(t)^2\,dt.
\]
[/quote]

We find $ f(x)$ as $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$.

Substitute this form to equation. We receive
$ ax^2 \plus{} bx \plus{} c \equal{} (a^5/5 \plus{} ab/2 \plus{} bc \plus{} c^2 \plus{} (b^2 \plus{} 2ac)/3)x^2 \plus{} (a/3 \plus{} b/2 \plus{} c)x \plus{} 1$.

We have system
$ a^5/5 \plus{} ab/2 \plus{} bc \plus{} c^2 \plus{} (b^2 \plus{} 2ac)/3 \equal{} a$,
$ a/3 \plus{} b/2 \plus{} c\equal{}b$,
$ c \equal{} 1$.
This system have only complex solution.

If we solve equation $ f(x) \equal{} 1 \plus{} x\int_0^1 f(t)\,dt \plus{} x^2\int_0^1 f(t)\,dt$, then we receive $ f(x)\equal{}6x^2\plus{}6x\plus{}1$.

Dear V.V. : is $ f(x)\equal{}6x^2\plus{}6x\plus{}1$ the only solution to this integral equation?

[quote="yuyang"]Dear V.V. : is $ f(x) \equal{} 6x^2 \plus{} 6x \plus{} 1$ the only solution to this integral equation?[/quote]

Yes, it is.

Let $ a \equal{} \int\limits_0^1 f(t)\,dt$.
Then $ f(x) \equal{} ax^2 \plus{} ax \plus{} 1$. We substitute this function to equation and we find $ a$.


Two distinct lines pass through the center of three concentric circles of radii $3$, $2$, and $1$. The area of the shaded region in the diagram is $8/13$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)

[asy]
defaultpen(linewidth(0.8));
pair O=origin;
fill(O--Arc(O, 2, 20, 160)--cycle, mediumgray);
fill(O--Arc(O, 1, 20, 160)--cycle, white);
fill(O--Arc(O, 2, 200, 340)--cycle, mediumgray);
fill(O--Arc(O, 1, 200, 340)--cycle, white);
fill(O--Arc(O, 3, 160, 200)--cycle, mediumgray);
fill(O--Arc(O, 2, 160, 200)--cycle, white);
fill(O--Arc(O, 1, 160, 200)--cycle, mediumgray);
fill(O--Arc(O, 3, -20, 20)--cycle, mediumgray);
fill(O--Arc(O, 2, -20, 20)--cycle, white);
fill(O--Arc(O, 1, -20, 20)--cycle, mediumgray);
draw(Circle(origin, 1));draw(Circle(origin, 2));draw(Circle(origin, 3));
draw(5*dir(200)--5*dir(20)^^5*dir(160)--5*dir(-20));[/asy]

$ \textbf{(A)} \frac{\pi}8\qquad
\textbf{(B)}\frac{\pi}7\qquad
\textbf{(C)}\frac{\pi}6\qquad
\textbf{(D)}\frac{\pi}5\qquad
\textbf{(E)}\frac{\pi}4 $

[hide]Let acute angle be $\theta$ so that obtuse angle is $\pi-\theta$

Area of shaded region within the acute angles:

\[A_{1}=2\cdot\frac{\theta}{2\pi}\left(3^{2}-2^{2}+1^{2}\right)\pi=6\theta\]

Area of shaded region within the obtuse angles:

\[A_{2}=2\cdot\frac{\pi-\theta}{2\pi}\left(2^{2}-1^{2}\right)\pi=3(\pi-\theta)\]

Given that $A_{1}+A_{2}=\frac8{13}A_{\text{unshaded}}$,

\[A_{\text{unshaded}}=\frac{13}8(A_{1}+A_{2})=\frac{39}8(\pi+\theta)\]

\[\begin{aligned}3^{2}\pi=A_{1}+A_{2}+A_{\text{unshaded}}&=\left(3+\frac{39}8\right)(\pi+\theta)\\ 9\pi&=\frac{63}8(\pi+\theta)\\ \theta&=\frac17\pi \end{aligned}\][/hide]

[hide]Radians are too fancy for me, but basically, if the acute angle is $\frac{\pi}{x}$, the shaded region is $(1*\frac{1}{x})+(3*\frac{x-1}{x})+(5*\frac{1}{x})$ and the unshaded region is $(1*\frac{x-1}{x})+(3*\frac{1}{x})+(5*\frac{x-1}{x})$. This simplifies down in a ratio to $\frac{x+1}{2x-1}=\frac{8}{13}$. Cross multiply $13x+13=16x-8$ and solve for x to get $x=7$ meaning the answer is $\boxed{B}$.[/hide]

[quote="patelparth"]In A1, and A2, why do you put the 2(pi) in the denominator?[/quote]A whole circle has $ 2\pi$ radians. So if I have $ \frac{\theta}{2\pi}$ in A_1, it means the portion of the circle.

Example: Half a circle is $ \theta\equal{}\pi$ so $ \frac{\theta}{2\pi}\equal{}\frac12$, which is half of a circle as expected. Try, $ \theta\equal{}\frac{\pi}2$, which is a right angle. You should get a quarter of a circle.

[quote]
Radians are too fancy for me, but basically, if the acute angle is $ \frac{\pi}{x} $ , the shaded region is $ (1*\frac{1}{x})+(3*\frac{x-1}{x})+(5*\frac{1}{x}) $ and the unshaded region is $ (1*\frac{x-1}{x})+(3*\frac{1}{x})+(5*\frac{x-1}{x}) $  . This simplifies down in a ratio to $ \frac{x+1}{2x-1}=\frac{8}{13} $ . Cross multiply $ 13x+13=16x-8 $ and solve for x to get $ x=7 $   meaning the answer is $ \boxed{B} $ .[/quote]

Can someone tell me how the areas of the shaded and unshaded regions were calculated? I'm not really sure how Ignite168 got to it.

Thanks!

[img]http://grab.by/rGMq[/img]

Can somebody explain where the $1/2$ comes from in $2(\frac12 \theta (1)^2 )?$

Note that the formula for a circular sector with angle $\theta$ and radius $r$ is $\frac{\theta}{2\pi} \cdot \pi r^2=\frac{1}{2} \theta \cdot r^2$.

Why can't we just use degrees instead of radians?

I remember this problem.
It was in Alcumus (Funky Circle Areas).

[quote=mathman46]What is Alcumus?[/quote]

It's a place on AoPS where you can do practice problems. The difficulty of the problems ranges from MathCounts to hard AMC 10 problems. To access it, go to the resources tab and click on Alcumus.

Wow the #$21$ for $2004$ is surprisingly easy since you really don't have to make any "problem solving leaps" because this is so bashable.. Oh right AMC was always easy before $2001$.


function
algebra
domain
complex analysis
complex analysis unsolved

Let $ f(z)$ and $ g(z)$ be analytic functions on the bounded domain $ D$ that extend continuously to $ \partial D$ and satisfy $ |f(z)\plus{}g(z)|<|f(z)|\plus{}|g(z)|$ on $ \partial D$. Show that $ f(z)$ and $ g(z)$ have the same number of zeros in D, counting multiplicity.

The hint here says that neither $ f(z)$ nor $ g(z)$ has zeros on $ \partial D$, so each has at most finitely many zeros in $ D$. Estimate shows $ \frac{f(z)}{g(z)} \not \in (\minus{}\infty, 0]$ for z near $ \partial D$, so $ \text{Log}(f(z)/g(z))$ is continuously defined near $ \partial D$. Increase in $ \text{arg} f(z)/g(z)$ around any closed path near $ \partial D$ is zero.

We have covered Rouche's Theorem too. I do not see how to prove this right now. I need some pointers on where to go. Thanks.

I think the assumption shows $ f/g$ can't be in $ [0,\plus{}\infty)$ but this does not make much difference: you can replace $ g$ by $ \minus{}g$.
What you should prove is that both $ \arg f$ and $ \arg g$ increase by the same amount as you travel around $ \partial D$, and that follows from the fact that $ \arg (f/g)$ is continuously defined. [One approach: translated $ D$ so that $ 0\in D$. If $ \arg g$ increase by $ 2\pi k$ along $ \partial D$, then $ \arg (g/z^k)$ is continuously defined there. But then $ \arg (f/z^k)$ can also be continuously defined, namely as $ \arg (f/z^k)\equal{}\arg(f/g)\plus{}\arg (g/z^k)$.


analytic geometry
vector
linear algebra
matrix

In art of problem solving book II, pg. 148 section 11.2, it says that the coordinates of the cross product of vector V, (X1 Y1 Z1) and vector W, (X2 Y2 Z2) is equal to ((Y1Z2 - Y2Z1) (Z1X2 - Z2X1) (X1Y2 - X2Y1)).  I've been trying for some time now to prove it, but with no luck.  Any ideas?

Am I missing an important point of this question?
The statement follows from the [b]definition[/b] of a cross product.

$v \times w \;= \;\begin{vmatrix}i & j & k\\ x_1& y_1 & z_1\\ x^2 & y_2 & z_2\end{vmatrix} \;= \; (y_1z_2-y_2z_1)i + (x_2z_1-x_1z_2)j + (x_1y_2-x_2y_1)k $

The coordinates of $v\times w$ are: $(y_1z_2-y_2z_1,\;x_2z_1-x_1z_2,\;x_2y_2-x_2y_1).$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Or perhaps we are to prove it by the perpendicularity of the cross product?
[hide]

$\text{Let }u = (a,b,c)\text{ be the cross product }v \times w.$

$\text{Then }u \perp v\text{ and }u \perp w\text{ . . . hence: }u\cdot v = 0,\;u\cdot  w = 0.$

$\text{We have: }u\cdot v = (a,b,c)\cdot(x_1,y_1,z_1) = ax_1 + by_1 + cz_1 = 0$
$.\quad\;\;\text{and: }u\cdot w = (a,b,c)\cdot(x_2,y_2,z_2) = ax_2 + by_2 + cz_2 = 0$

$\text{Solving for }b\text{ and }c\text{ in terms of }a\text{, we get:}$

$.\qquad b \;= \;a\cdot\frac{x_2z_1-x_1z_2}{y_1z_2-y_2z_1}\qquad c \;= \;a\cdot\frac{x_1y_2-x_2y_1}{y_1z_2-y_2z_1}$

$\text{Let }a = y_1z_2-y_2z_1\text{ . . . then: }b = x_2z_1-x_1z_2,\;\;c = x_1y_2-x_2y_1$

$\text{Therefore: }v\times w \;= \;(y_1z_2-y_2z_1,\;x_2z_1-z_1z_2,\;x_1y_2-x_2y_1)$[/hide]

Just check for yourself that $v\cdot (v\times w)=w\cdot (v\times w)=0$


Here's problem I want everyone's opionion on.

Here's the problem. A lot of you probably have seen this problem before.

1=5
2=25
3=125
4=625
5=?

Some people might think 5=625*5, becaue of the powers of five. However the trick is since 1=5, then 5=1. Do everyone think this should be true. Should these numbers still follow the rules of the reflexive property? I guess if you want to say something about powers of five you should really say congruent instead of equal to. What do everyone think?

If 1=5, then we can do anything we want. There is a famous story involving Bertrand Russell. Russell claimed that if he was allowed to assume that 1=2, he could prove that anything is true. Then a skeptic challenged him: ``If 1=2, prove that you are the Pope.'' Russell thought for not more than a few seconds before replying ``The Pope and I are two, therefore the Pope and I are one.''

Exactly.. if you say 1=5, then you could add 4 to both sides and get 5=9. So 1=9 too. Or you could subtract 1 to get 0=4 and then all sorts of weird stuff. Unimaginable.

On the other hand, if you have f(1) = 5, f(2)=25, .. then its OK.

to add 4 to both sides, you have to assume 4=4. is reflexivity necessarily true? shouldnt 4=625, so that if you add 4 to a side, you add 625 to the other?


I'm looking for a copy of Olymon for January, 2005. Can someone post an image (or possibly a pdf) of the problems here.

It's not out yet. In fact the December 2004 edition is not out.

I heard that there will be no December Olymon but that there is a January one that has been given to people in the math circles in Toronto.


Does anyone know where I can get past contests about physics that are available in Canada?

P.S. - not the building kind, just the written ones
Thanks

You can try the IPhO although these questions tend to be quite difficult.


[quote="probability1.01"]It seems like you can't have caps in the subject line beyond the first letter. For example, I had entered "CAPS IN SUBJECT" as the subject of this topic. Was it always this way?[/quote]It's a precaution against people trying to CAPS LOCK-ify their topics in order to get attention.


Only one number between 33,000 and 34,000 with units digit 6 is a perfect square. What is the sum of that number's tens and hundreds digits?

Taking the square root of 34000 yields a number a little higher than 184, and $ 184^{2}\equal{}33856$.  Hence the sum of the hundreds and tens digits is $ 8\plus{}5\equal{}13$.

notice the subject: 2000 School [b] TARGET [/b].

Even without it, though, you can get it to $ \sqrt{33000}\equal{}10\sqrt{324\plus{}6}\equal{}10(18.\sim )$, which gives you few numbers to try.

[quote=james4l]Also note that the ones digit must be a 4 or a 6, further reducing possibilities.[/quote]

It says in the problem that the ones digit is 6 to begin with.


Show, that

$ (A \times B) \setminus (A\times C) \supseteq A\times (B \setminus C)$

Let $ x\in A\times(B\setminus C)$. Then, $ x\equal{}(a,b)$, where $ a\in A$ and $ b\in B\setminus C$. So, $ x\in A\times B$. Let's assume $ x\in A\times C$. That'd mean $ x\equal{}(a,c)$ for $ c\in C$ and we have an obvious contradiction.

However, I'm sort of confused here... wouldn't a stronger claim ($ (A\times B)\setminus (A\times C)\equal{}A\times (B\setminus C)$) also be easily provable or am I missing something here?

Right on. The Cartesian product $ \times$ is distributive with respect to $ \setminus$, $ \cup$ and $ \cap$, set difference, union, and intersection.


function
quadratics
inequalities
calculus
search
inequalities theorems

[b]The sum of the minima of n functions is less or equal to the minimum of the sum of these functions.[/b]

This fact, which was taught me by Arthur Engel (but which he didn't mention in his Problem-Solving Strategies book), is a trivial consequence of the definition of a minimum, but it turns out to be a helpful principle for proving inequalities.

For instance, the quadratic function $f\left(x\right) = a x^2 + 2 b x$ has the minimum $\displaystyle -\frac{b^2}{a}$. Now let $a_1$, $a_2$, ..., $a_n$ be n arbitrary positive numbers, and let $b_1$, $b_2$, ..., $b_n$ be n arbitrary real numbers. Consider the quadratic functions $f_i\left(x\right) = a_i x^2 + 2 b_i x$, for all natural i with $1 \leq i \leq n$. These quadratic functions have the respective minima $\displaystyle -\frac{b_i^2}{a_i}$, while their sum is the quadratic function $f_{\Sigma}\left(x\right) = \left(a_1 + a_2 + ... + a_n\right) x^2 + 2 \left(b_1 + b_2 + ... + b_n\right) x$ and has the minimum $\displaystyle -\frac{\left( b_1 + b_2 + ... + b_n\right)^2}{a_1 + a_2 + ... + a_n}$. Hence, since the sum of the minima of our n functions is less or equal to the minimum of the sum, we get

  $-\frac{b_1^2}{a_1} -\frac{b_2^2}{a_2} - ... -\frac{b_n^2}{a_n} \leq -\frac{\left( b_1 + b_2 + ... + b_n\right)^2}{a_1 + a_2 + ... + a_n}$.

Thus,

  $\frac{b_1^2}{a_1} + \frac{b_2^2}{a_2} + ... + \frac{b_n^2}{a_n} \geq \frac{\left( b_1 + b_2 + ... + b_n\right)^2}{a_1 + a_2 + ... + a_n}$.

This is the so-called [b]Cauchy-Schwarz inequality in the Engel form[/b], also called [b]Andreescu Lemma[/b]. It is equivalent to what is usually called the Cauchy-Schwarz inequality.

Finally I should mention that the minimum thesis of Arthur Engel can be applied some more times to rather complicated functions, but mostly the minima must be found by calculus then.

[b]NOTE:[/b] Some time ago it turned out that all of the above facts together with their proofs were known in Russia long before Engel and before Andreescu as well. This neither surprised nor disappointed me. What actually keeps annoying me is when I write "by the Cauchy-Schwarz inequality in Engel form, we have..." and somebody replies to me that the inequality should be called differently and that it was known before Engel. Indeed, I [b]do[/b] know that the inequality was known before Engel, and I don't claim it is Engel's invention; I just use the name "Cauchy-Schwarz inequality in Engel form" since this name is widely used (at least among the German olympiad participants) and people understand me when I call it this way. Same holds for the Karamata inequality.

(Just a historic excursus: You probably know that the terminus "Simson line" is apparently ahistoric, as [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Simson.html]Simson[/url] didn't have anything to do with this line, at least no mention of it was located in his works. The first discoverer of the line seemed to be [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Wallace.html]William Wallace[/url]. When some historian noted this, he proposed to rename the Simson line into Wallace line. Yet this new name didn't become mainstream, it caused a lot of trouble; still, I sometimes talk about Simson lines and people don't understand me since they are used to calling them Wallace lines, and conversely.)

  Darij

Some time ago, I have found this thread: http://www.mathlinks.ro/Forum/viewtopic.php?p=10525 . In this thread, Fedor Petrov gave a very beautiful proof of the inequality $\sqrt[3]{ab}+\sqrt[3]{cd} \leq \sqrt[3]{\left(a+c+b\right)\left(a+c+d\right)}$. Actually, the method he used is exactly the principle that the sum of the minima is less or equal to the minimum of the sum, under a slightly different guise.

  Darij

You may also search for beautiful article in Kvant "Summa minimumov i minimum summy". I think it was at the begining of 90th.

:blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

Now I see it's not Engel's principle at all...

Р. Алексеев, Л. Курляндчик, [i]Сумма минимумов и минимум суммы[/i], Квант 3/1991, pages 49-51, 55.

http://kvant.mccme.ru/1991/03/summa_minimumov_i_minimum_summ.htm

Thanks!!

  Darij

I understand; I was also very impressed when I heard Engel's lectures about it.

  Darij

May I know what should I quote if i would like to use this result in my ineq proof?
Is there any name of this result?

[quote="Myth"]You should write "The sum of the minima of n functions is less or equal to the minimum of the sum of these functions."  :)[/quote]

I don't know if it's ok to write this directly.... so it's better to quote the source i think... however, i don't know russian words at all..

It is just well known idea. You no need to use any references.

It is actually trivial to prove.

  Darij

[quote="Myth"]This article made GREAT impression on me 13 years ago (I was 11 years old).[/quote]

Which were the 5 (10) MOST impressive articels you read in Kvant. I mean which ones did you like best ?  :)

You guys can all read russian?

And is Kvant available in English?

Нет ;) . See [url=http://www.nsta.org/quantum/foreign.asp]http://www.nsta.org/quantum/foreign.asp[/url].

And no, I can't read Russian :D .

[quote="Valiowk"]Нет ;)
And no, I can't read Russian :D .[/quote]
 :D  :D  :D


inequalities
geometry
trigonometry
inequalities unsolved

[quote="shaaam"]Let $ A,B$ and $ C$ be angles of triangle, prove that

(a) $ \frac {sinA sinB sinC}{cotA \plus{} cotB \plus{} cotC} \leq \frac {3}{8}$

(b) $ \frac {sinA \plus{} sinB \plus{} sinC}{cotA \plus{} cotB \plus{} cotC} \leq \frac {3}{2}$[/quote]

Both of them trivial.

Honey_S, can you please post solution to them?

Thank you!

$ \frac {\prod sinA}{\sum cotA} \equal{} \frac {\frac {S}{2R^2}}{\frac {\sum a^2}{4S}} \equal{} 2(\frac {S}{R})^2\frac {1}{\sum a^2} \equal{} 2(\frac {1}{4}\frac {\sum a\prod(a \plus{} b \minus{} c)}{abc})^2\frac {1}{\sum a^2} \equal{} \frac {1}{8}.\frac {(\sum a)^2}{\sum a^2}.(\frac {\prod(a \plus{} b \minus{} c)}{abc})^2 \leq \frac {3}{8}$

$ \frac{\sum sinA}{\sum cotA}\equal{}\frac{\frac{1}{2R}\sum a}{\frac{\sum a^2}{4S}}\equal{}\frac{2S}{R}\frac{\sum a}{\sum a^2}\equal{}\frac{1}{2}.\frac{(\sum a)^2}{\sum a^2}.\frac{\prod(a\plus{}b\minus{}c)}{abc}\leq\frac{3}{2}$

Replace $ cotA +cotB +cotC$ in both the inequalities by $ \frac{a^2+b^2+c^2}_{4S}$ where S is the area of the triangle.

Now use Weitzenbock's Inequality: $ a^2+b^2 +c^2$ $ \geq$ $ 4\sqrt{3}S$ and the well known bounds $ \prod{sinA}$ $ \leq$ $ \frac{3\sqrt3}{8}$  and $ \sum{sinA}$ $ \leq$ $ \frac{3\sqrt{3}}_{2}$

[quote="Michael Niland"]Replace $ cotA + cotB + cotC$ in both the inequalities by $ \frac {a^2 + b^2 + c^2}_{4S}$ where S is the area of the triangle.

Now use Weitzenbock's Inequality: $ a^2 + b^2 + c^2$ $ \geq$ $ 4\sqrt {3}S$ and the well known bounds $ \prod{sinA}$ $ \leq$ $ \frac{3\sqrt3}{8}$  and $ \sum{sinA}$ $ \leq$ $ \frac {3\sqrt {3}}_{2}$[/quote]

I totally understand how you solve this problem... But how do show

$ cotA + cotB + cotC = \frac {a^2 + b^2 + c^2}{4S}$

and how do you prove $ \prod{sinA} \leq \frac {3\sqrt3}{8}$
I know it is famous, but i want to know elementary proof of them.
Thank you for your attention!  :)

[quote="shaaam"]

I totally understand how you solve this problem... But how do show

$ cotA \plus{} cotB \plus{} cotC \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{4S}$ [/quote]

For this, just note that $ \cot A \equal{} \frac{\cos A}{\sin A} \equal{}\frac{b^2 \plus{} c^2 \minus{} a^2}{2bc \sin A} \equal{} \frac{b^2 \plus{} c^2 \minus{} a^2}{4S}$ (I assume that you know $ S_ABC \equal{} \frac 12 bc \sin A$). Summing up yields the result.

[quote="shaaam"]
and how do you prove $ \prod{sinA} \leq \frac {3\sqrt3}{8}$
I know it is famous, but i want to know elementary proof of them.
Thank you for your attention!  :)[/quote]

This inequality could be derived from the AM-GM inequality and the famous inequality: $ \sin A \plus{} \sin B \plus{} \sin C \leq \frac{3\sqrt{3}}2$

[quote="mathangel"][quote="shaaam"]

I totally understand how you solve this problem... But how do show

$ cotA \plus{} cotB \plus{} cotC \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{4S}$ [/quote]

For this, just note that $ \cot A \equal{} \frac {\cos A}{\sin A} \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc \sin A} \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{4S}$ (I assume that you know $ S_ABC \equal{} \frac 12 bc \sin A$). Summing up yields the result.

[quote="shaaam"]
and how do you prove $ \prod{sinA} \leq \frac {3\sqrt3}{8}$
I know it is famous, but i want to know elementary proof of them.
Thank you for your attention!  :)[/quote]

This inequality could be derived from the AM-GM inequality and the famous inequality: $ \sin A \plus{} \sin B \plus{} \sin C \leq \frac {3\sqrt {3}}2$[/quote]

Thank you! Yes i know basic area formulas using sine...  :blush: . I didn't realise how basic it is....

[quote="Michael Niland"]To prove $ \sum{sinA}$ $ \leq$ $ \frac{3\sqrt3}{2}$ ,see http://www.mathlinks.ro/viewtopic.php?t=209782

And to  get the upper bound on  $ \prod{sinA}$ use AM-GM inequality.[/quote]

Thank you! .. you are right.. Indeed

$ \sum{sinA} \leq \frac {3\sqrt3}{2}$

by AM-GM

$ \sum{sinA} \geq 3\sqrt [3]{\prod{sinA}}$

$ 3\sqrt [3]{\prod{sinA}} \leq \frac {3\sqrt {3}}{2}$

$ \prod{sinA} \leq \frac {3\sqrt {3}}{8}$

 :blush:


geometry
rectangle
parallelogram
geometry unsolved

in rectangle $ ABCD$ , let P be a point on side $ AD$ such that $ <BPC\equal{}90$.The perpendicular to $ BP$
through $ A$ intersects $ BP$ at $ M$ and the perpendicular to $ CP$ through $ D$ intersects $ CP$ at $ N$. Show that the center of the rectangle belongs to $ MN$.

[color=darkblue]Let $ AM$ cuts $ BC$ at $ E$.

We have $ \angle APB\equal{}90^\circ\minus{}\angle APB\equal{}\angle{DPC}\equal{}\angle NDC$
Also $ AB\equal{}CD$, therefore $ \triangle AMB \equiv CND$.

Again easy angle chasing and $ BM\equal{}DN$, prove $ \triangle BME\equiv \triangle DNP$. So $ PN\equal{}ME$ and $ PMEN$ is rectangle. Then

midpoint of $ MN$=midpoint of $ PE$ = midpoint of $ AC$

The last relation follows from the fact that  $ APCE$ is parallelogram.[/color]

Dear Mathlinkers,
1. let Q the meetpoint of AM and DN
2. By a converse of Pappus's theorem, Q is on BC
3. AQCP and PMQN are two parallelograms and we are done...
Sincerely
Jean-Lousi

The circle diameter BC intersects AD at P and Q | AP = DQ and call R the projection of Q onto BC. Obviously, ARCP is parallelogram, while ARD is a right-angled triangle, equal to CPB, hence MPNR is a rectangle. From ARCP the middle of PR is the middle of AC too, whilst from MRNP the middle of MN is the middle of PR, done.

Best regards,
sunken rock

Let O be the midpoint of MN. It's easy to see that the triangles DNO and OMB are congruent, that is, <NOD=<MOB, that is, BD passes through O. Similarly AC passes through O, and we are done.


inequalities
algebra
polynomial
inequalities open

We had a inequality 3abc \geq    \sum {sym}
a 2 b  .It is Schur unequalities.But can you have m,n  for the inequality   a^3 +b^3 +c^3 + mabc     \leq    n* \sum {sym}
a 2 b.If you have ,can you find  the greatest of n/m????????I don't know  that. Please solve it !

I don't think we can have something like this. Since if you take b=c=1 we find that a monic polynomial of degree 3 (in a) is bounded above. Anyway, for a triangle there is such an inequality:
  [tex] \sum a^3+9abc \leq 2 \sum a^2(b+c).[/tex]

Starting from

a(a+b-c)(a-b+c) + b(b+c-a)(b-c+a) + c(c+b-a)(c-b+a) >=0

we obtain suma^3 +6abc >= suma^2(b+c).

And for your inequality, Harazi, which is the start point?

Thank you very much.


A guided missile is intially moving at 500m/s [W] when it begins to accelerate. Twenty seconds later, the missile is moving at 600 m/s[W 30 S ] What is the displacement of this missile during the 20 second acceleration?

By Newtonian mechanics, we have $ v^{2}\equal{} u^{2}\plus{}2as$ and $ a \equal{}\frac{v\minus{}u}{t}$, combine together and we will have $ v^{2}\equal{} u^{2}\plus{}2(\frac{v\minus{}u}{t})s$.
Substitute the numbers in and we will get: $ 600^{2}\equal{} 500^{2}\plus{}2\frac{(600\minus{}500)}{20}s$

Hence $ s \equal{} 11000$m


[b]Using modular arithmetic[/b],prove that $13^3$ is of the form $9n+-1$

[hide="Not really a solution...but I am learning"]
$9n-1=13^3 \\ n=\frac{13^3+1}{9} \\$[/hide]

[quote="Andreas"][hide]$13 \equiv 4 \mod 9$, $13^3 \equiv 1 \mod 9$. So $13^3 = 9n + 1$.[/hide][/quote]

I had the same as you, but what's with the +-1?


Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=301436[/url]


geometry
function
algebra
polynomial
topology
advanced fields
advanced fields unsolved

Let $ H_{i}$ and $ H_{j}$ be the hyperplanes in $ \mathbf{P}^{n}$ defined by $ x_{i} \equal{} 0$ and $ x_{j} \equal{} 0$, with $ i \neq j$. Show that any regular function on  $ \mathbf{P}^{n} \minus{} (H_{i} \cap H_{j})$ is constant.

A regular function on $ \mathbf{P}^n \minus{} H_i$ takes the form $ \frac{f_i}{x_i^{k_i}}$, where $ f_i$ is a homogeneous polynomial of degree $ k_i$. Same for $ j$. So a regular function on $ \mathbf{P}^n \minus{} H_i \cap H_j$ is obtained by having two functions $ \frac{f_i}{x_i^{k_i}}$, $ \frac{f_j}{x_j^{k_j}}$ that are compatible on $ \mathbf{P}^n \minus{} H_i \cup H_j$. This forces $ k_i \equal{} k_j \equal{} 0$, i.e. it is a constant.

Does this argument works for this case?

 If $ Y \subseteq \mathbf{P}^{n}$ is a projective variety then every regular function on $ Y$ is constant.

I take $ Y \cap U_{i}$ where $ U_{i}$ is the open set $ x_{i} \neq 0$, then every regular function there is of the form: $ \frac{f_{i}}{x_{i}^{k_{i}}}$, and they have to be compatible in the set
$ \bigcap_{i\equal{}0}^{n}(Y \cap U_{i})$ ?

No.

In the original case $ f_i x^{k_j} \equal{} f_j x^{k_i}$ on a Zariski-open subset of $ \mathbb{P}^n$ would force them to be equal polynomials.

However if you restrict it to a projective subvariety, you can only say that $ f_i x^{k_j} \minus{} f_j x^{k_i}$ vanish on the subvariety. They may not be the same polynomial. This way you cannot get a "global description" for the regular function and thus it can't force $ k_i \equal{} k_j \equal{} 0$.


I'm guessing you want there to be an integer solution $x$.

Yes. It has to be, because that's what the coices are :D

Its from some old AMC

[quote="jli"][quote="#H34N1"]For which of the following values of $k$ does the equation $\frac{x-1}{x-2}=\frac{x-k}{x-6}$?

A) 1
B) 2
C) 3
D) 4
E) 5[/quote]

[hide]i dont think you finished the question.

heres the work so far though

$\Rightarrow x^{2}-7x+6=x^{2}-(2+k)x+2k$

$\Rightarrow 2k=(k-5)x$[/hide][/quote]
The question was for what value of k does the equation have no solution.  Here is my solution to it:
[hide] Cross-multiply to get $x^{2}-7x+6=x^{2}-(2+k)x+2k\Rightarrow x(k-5)=2k-6.$  This implies that $x=\frac{2k-6}{k-5}$.  If the denominator is 0, then x has no solution.  So, k-5=0, hence k=5.  The answer is $E$. [/hide]


geometry
analytic geometry
graphing lines
slope

The equation of a line passing through [b](-a,0)[/b] which makes with the axes a triangles of area [b]T[/b] is
[b](a)2Tx+a^2y+2aT=0
(b)2Tx-a^2y+2aT=0
(c)2Tx-a^2y-2aT=0
(d)none of these.
[/b]

Answer is (a).Please give me its solution.Thanks a lot.

hello, we have $ y\equal{}mx\plus{}n$ and $ P(\minus{}a,0)$ from here we get $ n\equal{}am$ and the equation of your line is
$ y\equal{}m(x\plus{}a)$ and the area $ T$ of the triangle is given by $ T\equal{}\frac{1}{2}ma^2$, from here we get the slope of the straight line $ m\equal{}\frac{2T}{a^2}$ setting this in the equation above we get
$ y\equal{}\frac{2T}{a^2}(x\plus{}a)$ or $ 0\equal{}2Tx\plus{}2Ta\minus{}a^2y$
Sonnhard.

Let the line be $ y \equal{} mx \plus{} c$ ---(1)

It passes through $ ( \minus{} a,0)$

Substitute in (1)

$ \Longrightarrow m \equal{} \frac {c}{a}$

Equation of the line is $ y \equal{} \frac {c}{a}x \plus{} c$ --(2)

It forms a triangle of area $ T$ with the axes,

$ \Longrightarrow$ the vertices of the the triangle are $ (0,0) , ( \minus{} a,0) ,(0,c)$

(third vertex can be found by substituting $ x \equal{} 0$ in (2) )

$ T \equal{} \frac {1}{2}| \minus{} a(0 \minus{} c) \plus{} 0(c \minus{} 0) \plus{} 0(0 \minus{} 0)|$

$ T \equal{} \frac {1}{2}ac$

$ c \equal{} \frac {2T}{a}$ ---(3)

Substitute (3) in (2)

$ y \equal{} \frac {c}{a}x \plus{} c$

$ y \equal{} \frac {2T}{a^2}x \plus{} \frac {2T}{a}$

$ a^2y \equal{} 2Tx \plus{} 2aT$

$ \boxed {2Tx \minus{} a^2y \plus{} 2aT \equal{} 0}$

Hello , thanks a lot to both of you for your answers.I got the solution.You solved my problem. Thank you so much to both of you.


Let $ n > 1$ be a natural number, and $ p$ a prime number. Prove, that if $ p\mid n^3 \minus{} 1$ and $ n\mid p \minus{} 1$, then $ 4p \minus{} 3$ is a perfect square.

[quote="cnyd"]$ p > 3$ $ \implies$  $ p|(n \minus{} 1)$  V
$ p|n^2 \minus{} n \plus{} 1$
$ i)p|n \minus{} 1 ,n|p \minus{} 1, n \equal{} pk \plus{} 1,pk \plus{} 1|p \minus{} 1, pk \plus{} 1\leq p \minus{} 1$ $ \implies$  $ k \equal{} 0, n \equal{} 1$ Contradiction!
$ i)p|n^2 \minus{} n \plus{} 1 , n|p \minus{} 1 p \equal{} nk \plus{} 1 , p|n^2 \minus{} n \minus{} nk p|n(n \minus{} k \minus{} 1) , (p,n) \equal{} 1 ,p|n \minus{} k \minus{} 1$
$ nk \plus{} 1\leq n \minus{} k \minus{} 1 , nk \plus{} 2\leq n \minus{} k$   contradiction!
$ n \minus{} k \minus{} 1 \equal{} 0 n \equal{} k \plus{} 1, k(k \plus{} 1) \plus{} 1 \equal{} k^2 \plus{} k \plus{} 1, 4k^2 \plus{} 4k \plus{} 1 \equal{} (2k \plus{} 1)^2$
$ p \equal{} 3$ $ \implies$  $ 4.3 \minus{} 3 \equal{} 9 \equal{} 3^2$   :wink:[/quote]

How did you get p|n^2 - n +1 :?:

[quote="amparvardi"]It's Iran NMO 2005.[/quote]
But I found it in a book `Problems For Math Olympians` (by Andre Negut)

[quote="mavropnevma"]Negut book has been published in 2005. He indeed offers as source MOSP 2004 (but he is known to have been wrong on occasion - I am a friend of his, so I can afford to quip!).[/quote]

But it is also in MOSP 2004 (black group).

[quote="maxal"]The stronger version of this problem was published in Russian journal Kvant issue 5, 2001:
http://kvant.mccme.ru/pdf/2001/05/kv0501probl.pdf
See problem M1787.[/quote]

Translated:
Let $p$ and $q$ be integers greater than $1$. Assume that $p\mid q^3-1$ and $q\mid p-1$. Prove that $p=q^{3/2}+1$ or $p=q^2+q+1$.
(N. Osipov)


linear algebra
matrix
invariant
vector
linear algebra solved

Finally tex on the board.Well done Valentin.

Given $ \phi  \in End(\mathbb{R}^5 ) $ and its mapping matrix  $
A: = \left( {\begin{array}{*{20}c}
   1 & 0 & 0 & 0 & 0  \\
   1 & 1 & 0 & 0 & 0  \\
   0 & 1 & 1 & 0 & 0  \\
   0 & 0 & 0 & 2 & 0  \\
   0 & 0 & 0 & 1 & 2  \\
\end{array}} \right)
$ in relation to the Standard Basis.

Determine all $\phi$-invariant Subspaces related to $\mathbb{R}^5 $

Note:This exercise was posted also on other forums, but i didnt receive any answer.

You can write $\phi$ as $u\oplus v: \mathbb R^3\oplus\mathbb R^2\to\mathbb R^3\oplus\mathbb R^2$, where $u: \mathbb R^3\to\mathbb R^3$ has the matrix \[ \left(\begin{array}{ccc}1&0&0\\1&1&0\\0&1&1\end{array}\right), \] and $v: \mathbb R^2\to\mathbb R^2$ has the matrix \[ \left(\begin{array}{cc}2&0\\1&2\end{array}\right). \]

Let $U\oplus V$ be a subspace of $\mathbb R^3\oplus\mathbb R^2$ invariant under $u\oplus v$. We then have $(u\oplus v)(U\oplus V)=u(U)\oplus v(V)\subset U\oplus V$. It is easy to prove that $U$ is invariant under $u$ and $V$ is invariant under $v$.
Conversely if $U$ and $V$ are invariant subspaces of $u$ and $v$ respectively, the $U\oplus V$ is an invariant subspace of $u\oplus v$.

It is therefore sufficient to find the invariant subspaces of $u$ and $v$.

We start with $v$. Its invariant subspaces of dimensions 0 and 2 are obviously 0 and $\mathbb R^2$ respectively.
To find its invariant subspaces of dimension 1 we must find its eigenvectors. They are given by the equation
\[ \left(\begin{array}{cc}2&0\\1&2\end{array}\right)\cdot\left(\begin{array}{c}y_1\\y_2\end{array}\right)=2\left(\begin{array}{c}y_1\\y_2\end{array}\right). \]
because $2$ is the only eigenvalue of $v$. The system is $\left\{\begin{array}{c}2y_1=2y_1\\y_1+2y_2=2y_2\end{array}\right.\Rightarrow y_1=0$.
Therefore $\mathbb R\left(\begin{array}{c}0\\1\end{array}\right)$ is the only invariant subspace of dimension 1.

Now for $u$ just like before, the invariant subspaces of dimension 0,1 and 3 are 0, $\mathbb R\left(\begin{array}{c}0\\0\\1\end{array}\right)$ and $\mathbb R^3$ respectively.
Let $U=<(a,b,c),(a',b',c')>$ be a 2 dimensional invariant subspace of $u$. Then \[ u(a,b,c)=(a,a+b,b+c)\in U\Leftrightarrow \det\left|\begin{array}{ccc}a&a'&a\\b&b'&a+b\\c&c'&b+c\end{array}\right|=0\Leftrightarrow \det\left|\begin{array}{ccc}a&a'&0\\b&b'&a\\c&c'&b\end{array}\right|=0\Leftrightarrow\left\{\begin{array}{c}-a(ac'-a'c)+b(ab'-a'b)=0 \\ -a'(ac'-a'c)+b'(ab'-a'b)=0\end{array}\right. \]. The system leads to $\left\{\begin{array}{c}ab'-a'b=0\\ ac'-a'c=0\end{array}\right.$.
One of the solutions of the system is $a=a'=0$ and $(a,b,c),(a',b',c')$ linearly independent which leads to $U=\mathbb R(0,1,0)+\mathbb R(0,0,1)$.
Assume $a\neq 0$. Then $a'=\frac{a'}aa$, $b'=\frac{a'}ab$ and $c'=\frac{a'}ac$, so $(a,b,c)$ and $(a',b',c')$ are linearly dependent which contradicts $dim_{\mathbb R}U=2$.
Therefore the invariant subspaces of $u$ are $0$, $\mathbb R(0,0,1)$, $\mathbb R(0,1,0)+\mathbb R(0,0,1)$ and $\mathbb R^3$.

The invariant subspaces of $u\oplus v$ are obtained by summing any 2 invariant subspaces of $u$ and $v$ respectively.

[quote="alekk"]you can also use relation between invariant planes and eigenvector of $A^t$[/quote]

Which is?

In an euclidian space of dimension $3$:
suppose that there is a plane $P=(u,v)$ such that $AP \subset P$.
Suppose that $w$ is an orthogonal vector to $P$ (i.e. $(w|x)= 0$ for any $x \in P$. Then, for any $x \in P$,$(A^tw|x)=(w|Ax)=0$. Hence $A^tw$ is orthogonal to $P$ :$A^tw$ is collinear to $w$. $w$ is an eigenvector of $A^t$. So, if one wants to find a stable plane of $A$, one can look at the eigenvectors of $A^t$.


In reading a solution for an ISL geometry problem, I saw that grobber used the following notation:

(TR, TX; TA, TM) = -1

Can anyone shed some light on what that means?

That means that cross ratio of points $R,X,A,M$ equals $-1$.


Do you perfer facebook, myspace or other?

Personally I like Facebook because its easier to use and its less chaotic, in my opinion.

Facebook; Myspace has too many popups and ads.

I use Facebook, but I forgot why did I create my account. There are a lot of virus containing the word "Myspace", and I have no idea about it, so Facebook.

facebook. myspace is for like 5th graders

Facebook since no one uses myspace. It also looks better and has better applications.I have a myspace though.

I've never even been on myspace, but I do have a facebook account, and facebook is really popular in my country.

[quote="TZF"]Facebook legit owns Myspace, k?[/quote]

Although Myspace does allow custom profiles (although I never figured out where but I suspect CSS?)

I just got my facebook account like 2 weeks ago.

[quote="TZF"]Facebook legit owns Myspace, k?[/quote]

exactly. thank you. :)

and more and more people are switching to facebook.

My sister does Facebook and Myspace, I'll try to ask her what her favorite one is.

I know a lot of people who have MySpace...(i don't have one) but noone I know has Facebook!

[quote="7h3.D3m0n.117"][quote="myyellowducky82"]...but noone I know has Facebook![/quote]

Because you have to be in high school to "legally" have a facebook. Which means some of you AoPSers aren't following the rules  :P[/quote]
True. But all of the highschoolers i know don't have facebook

facebook was really good, but now the new facebook is really bad and pretty annoying

Facebook, Myspace is a giant memory suck thanks to profile customizations. The new Facebook however, seems kind of awkward for me to use, since I got my account way before they changed. But I do like the 中文（简体） option!

you know you guys can switch back??
everyone else that I know did.

just search a group with a name like
SWITCH TO THE OLD FACEBOOK
or something
and then follow the directions.

I [i]think[/i] that [url=http://apps.new.facebook.com/?fbnew_opt_out=1]this[/url] is the right link.

Don't shoot me if it doesn't work

[quote="TZF"]Don't shoot me if it doesn't work[/quote]

Gahh - it didn't work  :| . Where's my R700 ACOG?  :P Just kidding.

I have a 22" widescreen so the new Facebook isn't really a problem for me. Actually, I kind of like the new features. The old facebook was kind of laggy and slow.

I heard that people really hate the new facebook, but I got my account right after it changed.

i have facebook. i dont have myspace
i dont like the new facebook, but i liked the old one.
i didnt check my fb for two months and i had like 20 friends i didnt add or accept.

im a completely fb addicted userrr, myspace. used to be really addicted but i use fb more frequently.
i also have friendster, cyworld[both korean&world version] used to have xanga when i was like 12-13ish.
errr what else is out thereee.
i have a bunch more that i never really use.
but FACEBOOK OWNS EVERYTHING
the end.


Need some help with this one please.

Quoted from an EC problem my teacher gave me:

"The ideal gas law is PV=nRT where P is pressure, V is volume, n is the number of moles of gas (mole is a big number of molecules), R is constant, and T is temperature in degrees Kelvin.

1. What happens when you heat a closed system? That is, what happens when temperature rises, gas cannot enter or escape, and the volume is fixed?
2. What happens when temperature drops and pressure remains the same?
3. Why does hot air rise?
4. Ever notice that it's usually colder at the top of a mountain that at lower elevations? In fact, it is 7 degrees C colder per vertical kilometer. This is known as the adiabatic lapse rate. Adiabatic expansion is the expansion of a gas at constant energy, resulting in a drop in temperature. Sketch out this situation using P1, V1, n1, T1, and P2, V2, n2, T2. (Meaning P to the 1st power etc.) Try taking a set volume and moving it upwards. How does pressure change at high altitudes?
5. Look up a value for R. Be sure to notice the units. Using appropriate units or conversions, figure out how many moles of gas (i.e. solve for n) are present at 1 atom pressure in 10 L of gas at 20 degrees C (be sure to convert to Kelvin.)"

1. This is Gay-Lussac's Law: if n and V are to remain constant, then heating a sample of gas will increase its pressure,

$\frac{P_{1}}{T_{1}}= \frac{P_{2}}{T_{2}}$.

2. This is Charles's Law: if the pressure is to remain constant, then cooling a gas will have the efect of reducing its volume,

$\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$.

3. Considering that air behaves like a perfect gas, and rearranging the equation of state of perfect gases we get $MP = dRT$, where M is air's molear mass and d its density. So, if M and P are constant we notice that heating air has the effect of decreasing its density. So, we have air of low density "immersed" in normal air of higher density, and according to Archimedes's Law to the hot air is then applied a vertical force directed upwards that makes it rise.

5. According to the equation of state of perfect gases we have $n = \frac{PV}{RT}$. 20ºC = 293.15 K, and R = 82.058 $cm^{3}atm mol^{-1}K^{-1}$. So,

$n = \frac{1atm \cdot 10000cm^{3}}{82.058cm^{3}atm mol^{-1}K^{-1}\cdot 293.15K}= 0.416 mol$

Carcul


Find the length of curve y= 7/(x^3) from 2 to 4.

Do you know calculus?  That is how I would solve the problem.  Use the arclength formula:
$ L\equal{}\int_{a}^{b}\sqrt{1\plus{}\left(\frac{dy}{dx}\right)^{2}}\mbox{ d}x$


If $x,y,z>0$ and $\frac{1}{3}\leq xy+yz+xz \leq 3$ find the range of $xyz$ and $x+y+z$.

To find range of $xyz$ use AM-GM and chained inequalities; virtually every equality case is the same so this works.

For $\sum x$ use $\sum x^{2}+2\sum xy =(\sum x )^{2}$ and chained inequalities again.

I don't see any difficulties yet... (but I won't post my solution yet either)


vector
analytic geometry
combinatorics solved
combinatorics

Given a set $\mathcal M$ of points in the plane, we draw some vectors, each having two of the points in $\mathcal M$ as endpoints. We know that for each point in $\mathcal M$ the number of outgoing vectors equals the number of incoming vectors. Show that the sum of all the vectors is $0$.

Consider the directed graph whose vertices are the points and the vectors are the edges.
Since for each vertex $v$, we have $d^+(v) = d^-(v)$ it is well-known that this graph is eulerian (i.e. has an eulerian closed path). This closed path leads to the desired result.

Pierre.

Alternately, we can assign each point $P$ a vector quantity that represents its coordinate, so then the vector from $P$ to $Q$ is $Q-P$.

So if $d_1$ is the in degree of a point $P$ and $d_2$ is its out degree, note $P$ is subtracted off $d_2$ times and added in $d_1$ times in the whole sum; since $d_1 = d_2$ for all points $P$ the sum is equal to zero.


geometry conics ellipse calculus integration parameterization vector

Given the major and minor axis of an ellipse as a and b respectively, find the area of the elliptical pie wedge bound by two rays originating at the ellipse center and extending into the first quadrant at angles  :theta:1 and  :theta:2. The ellipse center is at (0,0).

My first intinct was to take the integral of a the line segment from the center of the ellipse to the ellipse from  :theta:1 to  :theta:2 with d:theta:  being a very small arc length. For this I used the following:

x = a cos :theta:
y = b sin :theta:
r =  :sqrt:(x2+y2)
eccentricity, e = :sqrt:(1-b2/a2)
arc length, ds = a*integral(1-(e sin :theta:)2)d:theta:  from :theta:1 to :theta:2

So the area of the small pie wedge is rds and the total area is integral(rds) from :theta:1 to :theta:2, or

a*integral(((a cos :theta:)2+(b sin :theta:)2)(1-e2 sin2 :theta:))d:theta: from :theta:1 to :theta:2

I'm having alot of trouble with this integral. I've tried using both the Simpson's Rule and Runge Kutta method to find a numerical solution, but keep on coming up short.

You seem to be confused on several accounts. First of all, $\theta$ in the parametrization of the ellipse is [b]not[/b] the angle between the ray originating at the center of the ellipse and going to the point on the ellipse and the major axis. Secondly, the area is not $\int rds$. The angle between the radial and the tangent vector is important too (not  mentioning forgotten $\frac12$). Thirdly, you forgot all the square roots in your integral (doesn't matter much because it is a wrong integral anyway) and, finally, Runge-Kutta (as far as I know) is a method for solving differential equations, not for finding definite integrals.

While it is possible to correct all this, there is a simpler approach. If you shrink your ellipse $a$ times along the major axis and $b$ times along the minor axis, it'll become a unit circle. The elliptical wedge will become a circular wedge whose area is easy to find. The only problem will be to recalculate the angles. Try to do it yourself. The final answer is hidden [hide="here"] $\frac12 ab\left[\arctan \left(\frac ab\tan \theta_2\right)-\arctan \left(\frac ab\tan \theta_1\right)\right]$ [/hide]

fedja,

First of all, thank you so much for your help.

As is generally the case, my error originated with a small mistake.  I found the parametric equations on a math website and didn't study the picture close enough to realize that the  :theta: wasn't the angle from the center to the point.  I'm not sure how I forgot the 1/2 in the equation.  As for the square roots, I had copied my post from another post I made.  Apparently the other forum doesn't use the same  : sqrt: callout.

I do have one followup question.  As :theta: approaches  :pi:/2 the Tan function goes to  :inf:.  Since I will be taking the arctan of this, is it appropriate to substitute :pi:/2 for arctan(2tan :pi:/2) in this case?

Once again, thank you for your reply and help.

Mike

[quote="crazybass2"]
I do have one followup question.  As :theta: approaches  :pi:/2 the Tan function goes to  :inf:.  Since I will be taking the arctan of this, is it appropriate to substitute :pi:/2 for arctan(2tan :pi:/2) in this case?
[/quote]
Yes, in this case you'll get $\pi/2$ again.


See title. Wondering what you're taking.

I have to take 12s only because I'm in the 11th grade  :( I could have totally gotten 120+ on the 10.

10A and 12B. Barely made AIME on 10A, looking for hopefully the same result on the 12B

10A, 12B

I hope the 12B won't be too difficult...

Dunno...Got 144 on 10A (stupid mistake)...so maybe taking 10B....I'm in 10th grade so I REALLY want that perfect score...It's probably going to be quite harder though....

I might take the 12B just for experience though

10A...not sure if I should take 10B or 12B.  I got 138 on the 10A, but everyone says it was really easy so I might not be able to do better on the 10B.  Also, the 12A was apparently really hard, so relatively speaking I'd do better on the 12B?

If you got 144 on 10A, would you spend time to do 10B or would you focus on AIME?

[quote="Yongyi781"]My school doesn't offer the B's at all. Besides, my score is already set. Nothing I do on the B can change my score.[/quote]

Couldn't you take 12B at another school?

120 on amc 12a as 10th grader: focus on aime or do 10b?

Depends on your goals...If you want to make USAMO...you should try to get at least 7-8 on AIME...

With your 12 score, you could probably get upper 130s at least on the 10. So...Take the 10B, but don't study THAT much for it. The index for 10 is usually lower than that for 12, so quite a bit easier to get in. Still focus on AIME though.

12A and 12B originally.

got a 133.5, and had some other thoughts.

12A and 10B.

This will be my first/last 10 ever.


Given $ n \geq 3$, $ n$ is a integer. Prove that:
\[ (2^n \minus{} 2) \cdot \sqrt{2i\minus{}1} \geq \left( \sum_{j\equal{}0}^{i\minus{}1}C_n^j \plus{} C_{n\minus{}1}^{i\minus{}1} \right) \cdot \sqrt{n}\]
where if $ n$ is even, then $ \displaystyle 1 \leq i \leq \frac{n}{2}$; if $ n$ is odd, then $ \displaystyle 1 \leq i \leq \frac{n\minus{}1}{2}$.

[quote=MathPanda1]Sorry, but what does $C_n^j$ and $C_{n-1}^{i-1}$ mean? Is it like $\binom{j}{n}$ or $\binom{n}{j}$?[/quote]

$C_n^j=\binom{n}{j}$  and  $C_{n-1}^{i-1}=\binom{n-1}{n-i}$

Note that for each positive integer $i$ that $2\leq i\leq \lfloor \frac{n}{2}\rfloor -1$, the inequality $\frac{\binom{n-1}{i-1} +\sum_{j=0}^{i-1}{\binom{n}{j}}}{\sqrt{2i-1}} \leq \frac{\binom{n-1}{i} +\sum_{j=0}^{i}{\binom{n}{j}}}{\sqrt{2i}}$ is equivalent to $$(\sqrt{2i+1}-\sqrt{2i-1} )\sum_{j=0}^{i-1}{\binom{n}{j}} \leq \sqrt{2i-1}\binom{n-1}{i} +\sqrt{2i-1}\binom{n}{i}-\sqrt{2i+1}\binom{n-1}{i-1}.$$
We've $LHS\leq \frac{2}{\sqrt{2i+1}+\sqrt{2i-1}}\sum_{j=0}^{i-1}{\binom{n}{i}} \leq \frac{\sqrt{2i+1}}{2}\binom{n}{i}$.
And $\sqrt{2i-1}\binom{n-1}{i} +\sqrt{2i-1}\binom{n}{i}-\sqrt{2i+1}\binom{n-1}{i-1} -\frac{\sqrt{2i+1}}{2}\binom{n}{i} =\frac{(n-1)!}{i!(n-i)!} \times \Big( (n-i)\sqrt{2i-1} +n(\sqrt{2i-1}-\frac{\sqrt{2i+1}}{2}) -i\sqrt{2i+1}\Big)$.
The last term is non-negative since $2i(\sqrt{2i-1}+\sqrt{2i+1})\leq n(\sqrt{2i-1}+\sqrt{2i+1})\leq n(4\sqrt{2i-1}-\sqrt{2i+1}).$ (the last bound is equivalent to $3\sqrt{2i-1}\geq 2\sqrt{2i+1}$ which is true for all $i\geq 2$.)
Hence the first inequality we stated is true.
This reduce the problem to the case when $i\in \{ 1,\lfloor \frac{n}{2}\rfloor \}$, which is easy.


function
algebra
polynomial
calculus
calculus computations

$ f(x)$ is a function s.t. $ f'''(x)\in C[\minus{}1,1]$, and $ f(\minus{}1)\equal{}0, f(1)\equal{}1, f'(0)\equal{}0$

Prove: $ \exists \xi \in (\minus{}1,1)$ s.t. $ f'''(\xi)\equal{}3$

I would begin by subtracting $ (x^3\plus{}1)/2$ from $ f$, so that the problem reduces to: if $ g(\minus{}1)\equal{}g(1)\equal{}g'(0)\equal{}0$, then $ g'''$ doesn't have constant sign.

Now write second-degree Taylor's polynomial for $ g$ centered at $ 0$...
[hide]since the polynomial is even, the remainder takes the same value at $ 1$ and at $ \minus{}1$. But the remainder is related to values of $ g'''$[/hide]


projective geometry
Putnam
calculus
calculus computations

Does anyone here know how to use Pappus theorem for finding the volume of revolutions in the Cartesian plane?

I'll look into it more, but would you please tell me where I could find a proof of this?

Try looking for it in the index of a "mainstream" calculus textbook. The book I'm holding right now (Stewart's [i]Essential Calculus[/i]) has it stated, boxed off and (more or less) proved in the text of the section in which centers of mass, work, etc. are discussed (which is in the same chapter as the volumes of the solids of revolution.) Done in that context, it's basically just a computation.

Hmm, cool, I'll check that out. Thanks a bunch.


LOGIC Question!!!
There are 6 gold coins and one is different from others.There is balance(with 1 hand and shows the weight).You can put coins on balance only three times.Find the different one and find its weight(heavy from others or not)???
It is an amazing question!

Please avoid double posting
http://www.mathlinks.ro/Forum/viewtopic.php?p=1298941#1298941

In another hand, you are right : it is not a number theory question...

Pierre.


$a_{1},a_{2},...,a_{n}$ are real numbers different from $0$. Find the real numbers $r_{1},r_{2},...,r_{n}$ such that :
$\underset{i=1}{\overset{n}{\sum }}r_{i}\left( x_{i}-a_{i}\right) \leq \sqrt{\underset{i=1}{\overset{n}{\sum }}x_{i}^{2}}-\sqrt{\underset{i=1}{\overset{n}{\sum }}a_{i}^{2}}$
For all the reals $x_{1},x_{2},...,x_{n}$


Yeah, what happened last night? The site was completely down (for me atleast) and we couldn't have class.  :ninja:

We had a small problem with our server, we tried to fix it, but it took a lot longer than we thought. We apologize for that, and it's all good now.

It was still have a problem getting to the site today too.  A few moment ago, IE said the page could not be found and after a few times of refreshing and refreshing the cache it loaded.  Though it would not let me navigate the site so I refreshed in the same manner again.  Now it seems to work.

Yep, $ \approx$ 2:00 - 2:20 (UTC-5) down, for me.

We just did a restart on the server. Everything should work fine now.

Man, jsut a server crash? I was expecting some uber1337 new space-age math program that gave us all Arnav's brain...  :(

I'm going to go cry my self to sleep now...

Thanks Valentin I was wondering the same thing yesterday :) .


function
limit
complex analysis
complex analysis unsolved

Let $ z_0$ be a given complex number. It is given that a function $ f: \mathbb C \backslash \{z_0\}\to \mathbb C$ has an isolated singular point at $ z_0$. Consider the function $ g : \mathbb C\backslash \{z_0\} \to \mathbb C$ given by
\[ g(z)\equal{}\cos(f(z)), \quad z \in \mathbb C\backslash \{z_0\}.\]
Can $ g$ have a pole at $ z_0$?

Thank you very much for your help.

No it can not have a pole in the point $ z_0$. To prove this one should consider three cases:
1) $ z_0$ i  s a removable singulqrity for function $ f$. Then if $ \lim f(z) \equal{} A$ then, clearly,  $ \lim g(z) \equal{} cos A$ when $ z\to z_0$. So, in this case we are done.
2)If $ z_0$ is essential singularity of $ f$ then for ''almost'' all $ A\in\mathbb {C}$ we can find the sequence $ x_n$ for which $ x_n\to z_0$ and $ \lim f(x_n) \equal{} A$.In this case $ \lim g(x_n) \equal{} g(A)$. Thus, $ z_0$ is not a pole for $ g$
3) $ z_0$ is pole for function $ f$. Let $ k$ be the order of this pole. (I'm to lazy to simplify the following arguments, this is the firsrt thing that came to my mind :))
 Denote by $ h(z) \equal{} (z \minus{} z_0)^k f(z)$. Then our function $ h$ is analytic in the some neighbourhood of the point $ z_0$ and $ h(z_0)\ne 0$. Choose some small neighbourhood( more, priciesly, open disc with circle $ \omega$ and a  small radius ) of the point $ z_0$ where $ h(z)\ne 0$ and $ |h(z)| < 1$.  Let $ \min\limits_{z\in \omega}|z \minus{} z_0|^k \equal{} b$. For $ n\in\mathbb {N}$ we have $ |h(z)| < 2\pi n |z \minus{} z_0|^k$ for $ n > \frac {1}{2\pi b}$. By Rouche theerem we get that there exists such $ t_n$ in our disc s.t $ h(t_n) \equal{} 2\pi n (t_n \minus{} z_0)^k$  or $ f(t_n) \equal{} 2\pi n$. Now we have that $ t_n\to z_0$ when $ n\to \infty$ and $ \lim g(t_n) \equal{} \lim cos(2\pi n) \equal{} 1$ when $ n\to\infty$.Hence, $ z_0$ is not a pole for our function. :)


Let an arithmetic sequence $a_1, a_2, a_3, \dotsc , a_n$ such that $a_{i+1} - a_i = r$ for every $i$.
The sum $S = \sin a_1 + \sin a_2 + \sin a_3 + \dotsc + \sin a_n = \sum_{i=1}^n \sin a_i$ is given by:
$S = \frac{\sin \left(\frac{a_1+a_n}{2}\right) \cdot \sin \frac{nr}{2}}{\sin \frac{r}{2}}$

Hint: use induction. It's pretty straightforward.

I had previosuly proven this in another topic involving trigonometric sums. :)  I'll try to find it.

Masoud Zargar


The gas milage for a certain model of car is known to have a standard deviation of 5 miles per gallon. A simple random sample of 64 cars of this model is chosen and found to have a mean gas milage of 27.5 miles per gallon. Construct a 95% confidence interval for the mean gas milage for this car model. Interpret the interval in words.

Here is what I have so far.

sample mean +/- z*(sigma/sqrt(64))

Confidence interval ends up being: (26.275,28.725)

In words this means that...

"We are 95% confident the population mean is between 26.275 and 28.725. If this study was conducted many times, I expect that 95% of all intervals would contain the population mean."

Does this seem correct?

Yes, but why do you call the average gas mileage "the population mean" all the time? :?


inequalities
geometry
circumcircle
trigonometry
cyclic quadrilateral
inequalities proposed

In triangle ABC the interior bisectors of the angles at A,B and C meet the circumcircle at P,Q and R respectively

Prove  AP+ BQ +CR > a+b+c

I have a geometric solution to this one
    Let $A_1, A_2$ be respectively the projections of A on AB and AC. It 's easy to see that $AA_1+AA_2 = b+c \geq 2AP$
     In the same manner, we obtain the same inequality with respect to AQ and AR. Therefore, $AP+BQ+CR \geq a+b+c$ :P

I almost posted a wrong solution for this one :).

$a<IB+IC$ and the like mean that it's enough to show that $IP+IQ+IR\ge IA+IB+IC\ (*)$. In triangle $PQR$, $I$ is the orthocenter, so $IP=2\lambda \cos P$ and $IA=4\lambda \cos R\cos Q$, where $\lambda$ is the circumradius of $ABC$ (I couldn't call it $R$, because we have a point with that name :)). This means that the inequality $(*)$ is equivalent to $\cos P+\cos Q+\cos R\ge 2(\cos P\cos Q+\cos Q\cos R+\cos R\cos P)$, which is, in turn, equivalent to $\cos^2P+\cos^2Q+\cos^2R\geq S^2-S$, where $S=\sum\cos P$. This follows from the following facts:

$(1)\ \sum\cos^2P\ge \frac 34$
$(2)\ 1<S\le\frac 32$
$(3)\ f(x)=x^2-x$ is increasing for $x>1$, so from $(2)$ we get $S^2-S\le \frac 34$.

It should be $AA_1 + AA_2 =b+c > 2AP$. Sorry.

I think you mean $b+c<2AP$, though, because $b,c$ are legs in some right triangles which both have $AP$ as a hypothenuse.

clairly, we have PB=PC=2RsinA/2
By ptoleme's theorem in the cyclic quadrilateral ABPC:
AP.BC=PB.CA+PC.AB=(AB+CA)BP
so AP=(AB+CA).BP/BC
but BC=2RsinA=4RsinA/2.cosA/2 :ang: 4RsinA/2=2BP
so AB+CA  :ang: 2AP
Similarly: AB+BC :ang: 2BQ and BC+CA :ang: 2CR
adding this three inequalities we get the result:
         AB+BC+CA :ang: AP+BQ+CR


In a random sample to 200 persons of a town, 120 are found to be tea drinkers.
In a random sample of 500 persons of another town, 240 are found to be tea
drinkers.
Is the propotion of tea drinkers in the town equal? Use 0.01 level of signifcance.

Can You Reply for this question?


Let $ C$ be a circle and $ P$ a given point in the plane. Each line through $ P$ that intersects $ C$ determines a chord of $ C$. Show that the midpoints of these chords lie on a circle.

Denote the center of C is O
all the midpoints lie on a circle which has OP as its diameter


integration
inequalities
real analysis
real analysis unsolved

$f_n\in L^p(\mathbb{R}) (1<p<\infty)$ and $f_n(x)\geq 0
(n=1,2,\dots)$,
Show that
$\|f_n-f\|_p \to 0 \quad (n\to\infty)$
if and only if
$\|f_n^p-f^p\|_1 \to 0 \quad (n\to\infty).$

Try to prove

$1. \int ||f| - |g||dx \leq \int ||f|^p - |g|^p|^{\frac{1}{p}}dx $ for the "if" case
 using $|x^a -y^a| \leq |x-y|^a$ if $ 0 < a < 1$ and $x, y > 0$

$2  \int ||f|^p - |g|^p|dx \leq 2pR^{p-1}\|f-g\|_{p}$ for the "only if" case
 using $|x^a -y^a| \leq a|x-y|\left(x^{a-1} + y^{a-1} \right) a$ if $ 1 < a < \infty$ and $x, y > 0$

[quote="dickchimney"]Try to prove

$2  \int ||f|^p - |g|^p|dx \leq 2pR^{p-1}\|f-g\|_{p}$ for the "only if" case
 using $|x^a -y^a| \leq a|x-y|\left(x^{a-1} + y^{a-1} \right) a$ if $ 1 < a < \infty$ and $x, y > 0$[/quote]

I don't understand the second one...

Maybe it's better now :D

$2.  \int ||f|^p - |g|^p|dx \leq 2pR^{p-1}\|f-g\|_{p}$ for the "only if" case
 using $|x^p -y^p| \leq p|x-y|\left(x^{p-1} + y^{p-1} \right) $ if $ 1 < p < \infty$ and $x, y > 0$

$|x^p -y^p\ \leq p|x-y|(x^{p-1} + y^{p-1} )$ is easy to prove. Apply this inequality, we have

$\int |f_n^p - f^p| \leq p \int |f_n -f|(|f_n|^{p-1}+|f|^{p-1})$ and apply Holder's inequality to the RHS


[quote="drunner2007"]let's say you use descartes' rule of signs on a degree 4 polynomial and you deduce that the polynomial has 4 positive roots. by positive, do we mean positive and real? can a complex number ever be considered positive or negative somehow? how does descartes' rule of signs relate to complex numbers?[/quote]

you cannot deduce that there are 4 positive real roots, you can just deduce that there are 4,2,0 positive real roots, because complex numbers come in conjugate pairs (assuming the polynomial has real coefficients)...and there is no such thing is a positive complex number...basically you don't know if those roots are complex or not, so you subtract a multiple of 2 from that


παιδια δεν ειναι : limit points,accumulation points, cluster points ( συγνωμη εκτος απο το σημειον συσσωρ. δεν ξερω πως λεγονται στα Ελληνικα τα αλλα σημεια) ενα και το αυτο πραγμα??

Hope [url=http://en.wikipedia.org/wiki/Limit_point]this[/url] helps.

Cheerio,

Durandal 1707

thanx Durandal αλλα αρχιζω να μπερδευομε σε σημειο of non return

αλλα τα βιβλια αλλα τα google αντε να βρης ακρη ,στα Ελληνικα πως λεγονται αυτα τα σημεια??


calculus
integration
quadratics
special factorizations
calculus computations

Partial fractions. And completing the square in the irreducible quadratic term. The final answer should contain a multiple of $ \frac1{x\plus{}6},$ two logarithmic terms, and one arctangent term.

Thus is also a case in which all computer algebra systems will have a considerable advantage in speed and accuracy over hand computation.


search
LaTeX
number theory unsolved
number theory

Hi,

the previous days I have tried to solve a simple problem(no luck).
it goes like this:

statement:
someone can descend some stairs one at a time,2 at a time or 3 at a time.
having a stair with n steps in how many ways can he descend it ?

my thoughts,

I thought the solution could be some simple linear recurrence but I really can't
figgure it out.
I've also made some combinatorial search to give me some good values so
I can check the sequence I find against it.
I remember that
p(0)=0
p(1)=1
p(2)=2
p(3)=4
p(4)=7
p(5)=11

I thought $ p(n) \equal{} p(n \minus{} 1) \plus{} p(n \minus{} 2) \plus{} p(n \minus{} 3)$ would be a good formula but its obvious that some
of the ways counted in p(n-1)overlap with some in p(n-2) and p(n-3).
so I thought
$ p(n) \equal{} 2p(n \minus{} 1) \minus{} p(n \minus{} 2)$ would be a good one if it was allowed only 1 and 2 steps.
I also think it has something to do with parity of n.

I look forward to hearing your thoughts on this.

Thank you

EDIT

mistake - instead of + in second latex formula.
it should substract the overlaps.

As I understand it, $ p(n) \equal{} p(n \minus{} 1) \plus{} p(n \minus{} 2) \plus{} p(n \minus{} 3)$ is correct, there is no overlapping...
Please define "descending".

descending=climb down=go down the stairs.

I know what the word means...
I wnat to know what exactly you want to count!

the distinct ways to go down the stairs.
like for a 3-step stairway

possibilities:
1step,1step,1step
2steps,1step
1step,2step
3steps

so basically I guess it has to do with partitions of a number only using 1,2,3.

With this definition, your first reccurence is correct.


A system of two particles each with spin 1/2 is described by an effective Hamiltonian
$ H\equal{}A(s_{1z}\plus{}s_{2z})\plus{}B\mathbf{s}_1\cdot \mathbf{s}_2$
where $ \mathbf{s}_1$ and $ \mathbf{s}_2$ are the two spins,  $ s_{1z}$ and $ s_{2z}$ are their $ z$-components, and $ A$ and $ B$ are constants. Find all the energy levels of this Hamiltonian.

[b]Solution.[/b]
We choose $ \chi_{SM_{S}}$ as the common eigenstate of $ \mathbf{S}^2\equal{}(\mathbf{s}_{1}\plus{}\mathbf{s}_{2})^2$ and $ S_{z}\equal{}s_{1z}\plus{}s_{2z}$. For $ S\equal{}1$, $ M_{S}\equal{}0,\pm 1$, it is a triplet and symmetric when the two electrons are exchanged. For $ S\equal{}0$, $ M_{S}\equal{}0$, it is a singlet and antisymmetric. For stationary states we use the time-independent Schrodinger equation
$ H\chi_{SM_{S}}\equal{}M\chi_{SM_{S}}$.
As
$ \mathbf{S}^{2}\chi_{1M_{S}}\equal{}S(S\plus{}1)\hbar^{2} \chi_{1M_{S}}\equal{}2\hbar^{2}\chi_{1M_{S}}, \mathbf{S}^{2}\chi_{00} \equal{} 0$,
$ \mathbf{S}^{2}\equal{}(\mathbf{s}_{1}\plus{}\mathbf{s}_{2})^{2}\equal{}\mathbf{s}_{1}^{2}\plus{}\mathbf{s}_{2}^{2}\plus{}2\mathbf{s}_{1}\cdot \mathbf{s}_{2}\equal{}\frac{3\hbar^{2}}{4}\plus{}\frac{3}{4}\hbar^{2}\plus{}2\mathbf{s}_{1}\cdot \mathbf{s}_{2}$,
we have
$ \mathbf{s}_{1}\cdot \mathbf{s}_{2}\chi_{1M_{S}}\equal{}\left (\frac{S^2}{2}\minus{}\frac{3}{4}\hbar^{2}  \right )\chi_{1M_{S}}\equal{}\left (\hbar^2\minus{}\frac{3}{4}\hbar^{2}  \right )\chi_{1M_{S}}\equal{}\frac{\hbar^2}{4}\chi_{1M_{S}}$,
$ \mathbf{s}_{1}\cdot \mathbf{s}_{2}\chi_{00}\equal{}\left (0\minus{}\frac{3}{4}\hbar^{2}  \right )\chi_{00}\equal{}\minus{}\frac{3}{4}\hbar^{2}\chi_{00}$,
and
$ S_{z}\chi_{1M_{S}}\equal{}(s_{1z}\plus{}s_{2z})\chi_{1M_{S}}\equal{}M_{S}\hbar \chi_{1M_{S}}$,
$ s_{z}\chi_{00}\equal{}0$.
Hence for the triplet state, the energy levels are
$ E\equal{}M_{S}\hbar A\plus{}\frac{\hbar^2}{4}B$, with $ M_{S}\equal{}0,\pm 1$,
comprising the three lines
$ E_{1}\equal{}\hbar A\plus{}\frac{\hbar^2}{4}B$, $ E_{2}\equal{}\frac{\hbar^2}{4}B$, $ E_{3}\equal{}\minus{}\hbar A\plus{}\frac{\hbar^2}{4}B$.
For the singlet state, the energy level consists of only one line
$ E_{0}\equal{}\minus{}\frac{3}{4}\hbar^2B$.


Two mirrors (= half planes with a common border line) in the space are inclined at the angle $\alpha$ to each other. A candle and an observer stand between the two mirrors. How many images of the candle does the observer see ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


geometry
rectangle
algebra
polynomial
number theory
number theory proposed

[b]2nd homework problem set of the IMO training 2004, number theory et al.[/b]

Here are the non-geometric problems of the second problem set of our IMO training. (See the geometric problems at the geometry subforum. You needn't help me with these problems since we have already compared the solutions at the seminar.)

[b]2.[/b] Ten dwarfs sit around a round table. Initially, one of the dwarfs owns 10 jewels and the other dwarfs own nothing. At the beginning of every minute, every dwarf owning more than 1 jewel gives 1 jewel to his left neighbour and 1 jewel to his right neighbour (this happens simultaneously). Is it possible that after some time, each dwarf owns exactly 1 jewel?

[Well, I have found a generalization! The table needn't be round, it can just be topologically equivalent to a circle :-)]

[b]3.[/b] A given $6 \times 6$ square is divided into eighteen $2 \times 1$ rectangles (these $2 \times 1$ rectangles can lie horizontally or vertically). Show that there is a line dissecting the square in two rectangles such that none of the $2 \times 1$ rectangles is dissected.

[Problem statement corrected.]

[b]5.[/b] On the boundary of a round sea, we have some seaports. Two seaports are connected by a shipping route if and only if their two right neighbours are not connected by a shipping route. (The "right neighbour" of a seaport P is the first seaport one comes across if one starts at P and drives along the circle in clockwise direction.) Show that every seaport can be reached from every other seaport by (at most) three ships.

[b]7.[/b] Show that for every natural number n, there exists a number m and a sequence

$a_0=0$, $a_1$, $a_2$, ..., $a_m=n$

with $\left|a_i-a_{i-1}\right|=i^2$. Also show that for almost all natural numbers n (i. e., for all natural numbers n apart from finitely many exceptions), there exists such a sequence with m < n.

[b]8.[/b] Let $f\left(x\right)=x^3+4x^2+21x+11$. Show that there exists a natural number n > 0 such hat the number

  f(f(...f(a)...)) - a    (with n iterations of f)

is divisible by 47 for all integers a.

Additional question: Find all polynomials of degree 3 having this property.

  Darij

I have a question.
 Your source is  Thorsten Kleinjung. Did he create them ?

I don't think so, as some of the problems are well-known .. :?

Kleinjung had only composed the problem set, of course most problems were taken from some books.

  Darij

In what country are these used as IMO training? Cuz our IMO training is a lot harder and we're not such a good country...

Not that much harder Peter... if our training is harder, I'm even quite sure it's not :)


[hide]
Lets say the appetaziers are 1, 2, and 3.
Lets say the main courses are 4, 5, 6, and 7.
So with 1 you can have 4 choices
with 2 you can have 4 choices
with 3 you can have 4 choices
so 4*3=12[/hide]

[quote="pascal_1623"]On a restaurant, there are three appetizers, and four main courses. How many different dinners can be ordered if each dinner consists of one appetizer and one main course?[/quote]
[hide]
3x4=12
[/hide]

i think your problem will be nicer if there is more things to choose or there is some conditions in it like this :if customer choose number a for appetizer he cant choose number b for mains.or something like that.Can you understand what i mean? ;)

Jensen, this forum is meant for problems that are not getting particularly complicated.  Read the forum posting guidelines and look around at all the other forums and you will understand.


The equation is as follows:

(x+40)(x+40)(3x-10)(3x-10) = (3x-10)(3x-10)  +  1

solve for x...

how would you solve a problem such as this?  I know that any polynomial with degree n greater than 3 cannot be solved using root extraction, so what method would give me values of x here?

[hide="hint"]Cancel the terms common to both sides of the equation.[/hide]

Actually, there's a quartic formula, but it's super-duper-ultra-mega-uber-crazy-hyper long, so it's not a very legit choice.

After all of the simplifying we have,

$ 9x^4\plus{}660x^3\plus{}9691x^2\minus{}87940x\plus{}159899\equal{}0$

After using the calculator method, we can see that the roots are (approximately):

$ 3.32564\minus{}(2.8385*10^{\minus{}16})i$

$ 3.34103\minus{}(2.8363*10^{\minus{}16})i$

$ \minus{}41\minus{}(2.07558*10^{\minus{}18})i$

$ \minus{}39\plus{}(2.29576*10^{\minus{}18})i$

[quote="Forte"]X_x Its not very appealing to the eye, I'll give it that much...

And yeah cancel common terms.[/quote]
If you meant to cancel out the 3x-10, then you can't because on the right side theres a +1 after the two (3x-10)'s.
If you meant something else please clarify for me :D

No polynomial with degree [b]5[/b] or greater can be solved using root extraction, actually.

I can't think of any way to solve it, but I can get the approximate solutions:
\begin{align*}
&\left(x+40\right)^2 \left(3x-10\right)^2=\left(3x-10\right)^2+1 \\
&\implies \left(x+40\right)^2 \left(3x-10\right)^2 \approx \left(3x-10\right)^2 \\
&\implies \left(3x-10\right)^2 \left(\left(x+40\right)^2-1\right) \approx 0 \\
&\implies \left(3x-10\right)^2 \left(x+41\right)\left(x+39\right) \approx 0 \\
&\implies \boxed{x \approx \frac{10}{3},\ -41,\ -39}
\end{align*}


Are your periods supposd to be multiplication symbols?  Everything would look prettier if you used Latex.

I think the periods are being used to symbolise the rest of the numbers on the number line between the particular ones we are looking at... I think you basically you have to dtermine if the numbers are in the right (increasing) order.

My periods are not for multiplication, they never will be, and no one else's should be used for multiplication either.  Especially once they are being used as decimal points, it is inconsistent to be used for multiplication also.  Times I expect symbols for multiplication to be used are: *, ( ), raised dot, X, for example##.  In the latter specific one, I would tell someone to avoid using those symbols with Xs and Ys because they look confusing/too similar.

My periods are there because I wanted that much spacing; if I typed it out originally like that without symbols in between, the expressions would collapse together (after submitting).

No, each problem already lists the correct relative order of the three, but not the correct relative distance from each other.  The true or false expression that applies to it is below it.

## There are other symbols for multiplication that I'm not bringing up at this time.

[quote="laughinghead505"][hide="in that case"] they are all true[/hide][/quote]

Solutions? :)

[quote="laughinghead505"][hide="in that case"] they are all true[/hide][/quote]
[hide]F, T, F[/hide]
 :wink:

laughinghead505,

your answers are what I would give if I too were guessing, but two of them are not the actual answers.

rem,

I agree with yours.

lotrgreengrapes7926,

your answers to #2 & #3 are correct.  No answer was submitted to #1 here (at that time).


Hi,

Is there any real number solutions for this equation: 10^x=x?

Thanks if anyone can assist.

There aren't any real solutions. 10^x is a monotonically increasing function. Also, 10^x is always positive. When x=0, 10^x=1, so it can't have a solution less than 1. 10^1=10. Then, the derivative of 10^x > the derivative of x, so 10^x is increasing faster, so there isn't a real solution.

So, if there isn't any real number solution for this equation, does it exist any complex number solution?

thanks.

Well, we can write x=a+bi in that case. Thus we want to find a solution for 10^(a+bi)=-a-bi, or 10^a*e^(ln(10)bi)=-a-bi. Then we can separate that out into its real and imaginary components to get

10^a*cos(ln(10)b)=-a
10^a*sin(ln(10)b)=-b.

I wouldn't have a clue were to start looking for a solution, but there probably is a solution (or more).


algorithm
number theory
Euclidean algorithm

Can $ 3b^2\plus{}7b\plus{}4$ be a square for $ b\ge 8$? Please write me an elementary solution.

[hide="partial solution"]We factor it into $ (3b\plus{}4)(b\plus{}1)$. Since $ gcd(3b\plus{}4,b\plus{}1)\equal{}1$ from the Euclidean Algorithm, both $ 3b\plus{}4$ and $ b\plus{}1$ must be perfect squares.[/hide]

[hide="Solution"]We have $ 3b^2 + 7b + 4 = (3b + 4)(b + 1)$.Since $ gcd(3b + 4,b + 1) = 1$ so If $ 3b^2 + 7b + 4$ is a perfect square then $ 3b + 4$ and $ b + 1$ are also perfect squares.So $ 3b + 4 = x^2$ and $ b + 1 = y^2$ which is equivalent to $ x^2 - 3y^2 = 1.$The last equation is on the form of $ x^2 - d.y^2 = 1$ where $ d$ is a positive integer not a perfect square, is called [url=http://books.google.com/books?id=B1c0s3ffN_0C&pg=PA272&lpg=PA272&dq=Pell-Fermat&source=web&ots=amgQJYQnM6&sig=I74lAKShcMcXqoDRViaafsqbjyQ&hl=en&sa=X&oi=book_result&resnum=2&ct=result]Pell-Fermat's equation[/url]and has many infinitely solutions.Thus,there exist infinitely many solutions to the equation $ x^2 - 3y^2 = 1 ((2,1);(7,4);(26,15)...),$so we can find infinitely many numbers $ b$ such that $ 3b^2 + 7b + 4$ is a perfect square.[/hide]

Thanks rachid! @1=2: I also got there :D

[quote="Bugi"]Can $ 3b^2 \plus{} 7b \plus{} 4$ be a square for $ b\ge 8$? Please write me an elementary solution.[/quote]
[hide="Solution"]
Let $ 3b^2 \plus{} 7b \plus{} 4 \equal{} a^2$.  Then, $ 12a^2 \equal{} 36b^2 \plus{} 84b \plus{} 48 \equal{} (6b \plus{} 7)^2 \minus{} 1$.  Let $ 6b \plus{} 7 \equal{} c$.  Then, we get the Pell's Equation $ c^2 \minus{} 12a^2 \equal{} 1$.  We can easily see that the smallest $ (a,c)$ that satisfies the equation in positive integers is $ (2, 7)$.  It is well-known that all other solutions of this Pell's Equation must be in the form of $ (a, c)$, where $ (7 \plus{} 2\sqrt {12})^n \equal{} c \plus{} a\sqrt {12}$.  Notice that $ a \equal{} \frac {(7 \plus{} 2\sqrt {12})^n \minus{} (7 \minus{} 2\sqrt {12})^n}{2}$, so $ a_n$ satisfies the recursion $ a_n \equal{} 14a_{n \minus{} 1} \minus{} a_{n \minus{} 2}$ with $ a_0 \equal{} 0$ and $ a_1 \equal{} 2$.  Similarly, we can see that $ c \equal{} \frac {(7 \plus{} 2\sqrt {12})^n \plus{} (7 \minus{} 2\sqrt {12})^n}{2}$, so $ c_n$ satisfies the recursion $ c_n \equal{} 14c_{n \minus{} 1} \minus{} c_{n \minus{} 2}$ with $ c_0 \equal{} 1$ and $ c_1 \equal{} 7$.  Now, for $ b$ to be an integer, it is necessary and sufficient for $ c \equal{} 6b \plus{} 7$ for some $ b$, which implies that $ c\equiv 1\bmod 6$.  Now, we find the period of $ c_n$ in modulo $ 6$.  We start with $ c_0, c_1\equiv 1\bmod 6$.  Then, we get that $ c_2\equiv 1\bmod 6$ again.  Then, $ c_3\equiv 1\bmod 6$, etcetera.  The period is clearly one and all $ c_i$ are congruent to $ 1\bmod 6$.  Hence, all $ c_i$ yield an integer solution for $ b$.  This means that \[ 6b \plus{} 7 \equal{} c \equal{} \frac {(7 \plus{} 2\sqrt {12})^n \plus{} (7 \minus{} 2\sqrt {12})^n}{2}\] \[ b \equal{} \frac {(7 \plus{} 2\sqrt {12})^n \plus{} (7 \minus{} 2\sqrt {2})^n \minus{} 14}{12}\] Yet, notice that $ b\ge 8$, so the condition that $ n\ge 2$ must be set, which yields all possible solutions for $ b$. [/hide]


I can't find all roots of x^y=y^x and that is a problem which I want to solve from many years ago. We fond that if x!=y than we' ve got lnx=Cx and lny=Cy for any constant C between 0 and 1/e. If you know something about that equal I'll be very happy to know it.

I'm not sure if the search function is working, but I've posted this link [b]many, many times.[/b]

http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm

Thank You about this information, but we know it. I want to see y like a function of x. Is there analytic or complex holomorph function y=y(x).

I found it in google:
x=y and
x=b^(1/b-1);y=b^(b/b-1) for every b>0 and b!=1

That's in the link I gave, if you read the whole thing.

I've read it, but there isn't answer whether that is one and only solution.

well isn 2^4=4^2? i dunno but i dun ger ur question so i might be wrong

(2,4) is included in
x=b^(1/b-1);y=b^(b/b-1) when b = 2