function
analytic geometry
algebra
polynomial

A function $f$ which takes real values, is symmetric about the y-axis satisfies and satisfies for $x,y,x+y \neq 0:$ \[f \left( \dfrac{1}{x+y} \right) = f \left( \dfrac{1}{x} \right) + f \left( \dfrac{1}{y} \right) + 2xy - 1.\] Find all functions $f.$

If a function is symmetric about the y-axis, it satisfies [tex]f(x) = f(-x)[/tex] for all x in its domain. Then we have:
[tex]f(\frac{1}{x+y}) = f(\frac{1}{x}) +f(\frac{1}{y}) +2xy-1[/tex]
[tex]f(-\frac{1}{x+y}) = f(-\frac{1}{x}) +f(-\frac{1}{y}) +2xy-1[/tex]
But we have [tex]f(\frac{1}{x+y}) = f(-\frac{1}{x+y}) [/tex], so we get
[tex] f(\frac{1}{x}) +f(\frac{1}{-x})  = f(-\frac{1}{y}) +f(-\frac{1}{y})[/tex].
Then we get that [tex]f(\frac{1}{x}) +f(\frac{1}{-x})  = c[/tex] where c is a constant. Again, we use that [tex]f(x) = f(-x)[/tex] , and then put in [tex]x^{-1} \longrightarrow x[/tex]. Thus, [tex] f(x) = c/2 [/tex]. But plugging this into the original equations shows there are no such functions.

My function equation skills are quite bad, so please correct any errors  :D

There is a function which satisfies the conditions. Please have another try.  :)

By inspection, $f(x) = 1/x^2 + 1$ is a solution.

[quote="tetrahedr0n"]If a function is symmetric about the y-axis, it satisfies [tex]f(x) = f(-x)[/tex] for all x in its domain. Then we have:
[tex]f(\frac{1}{x+y}) = f(\frac{1}{x}) +f(\frac{1}{y}) +2xy-1[/tex]
[tex]f(-\frac{1}{x+y}) = f(-\frac{1}{x}) +f(-\frac{1}{y}) +2xy-1[/tex]
But we have [tex]f(\frac{1}{x+y}) = f(-\frac{1}{x+y}) [/tex], so we get
[tex] f(\frac{1}{x}) +f(\frac{1}{-x})  = f(-\frac{1}{y}) +f(-\frac{1}{y})[/tex].
[/quote]

Instead of this, you should have gotten
\[f \left( \frac{1}{x} \right) - f \left( -\frac{1}{x} \right) =
f \left( -\frac{1}{y} \right) - f \left( \frac{1}{y} \right).\]
But you know $f$ is symmetric, so both sides are 0.  So this approach doesn't tell you anything.

Ah, my solution completely fails. Thank for pointing that out.

it's a little confusing using x and y where you state the y-axis. Is this the same y? Can you rewrite it in function of a,b if possible, :)

Peter, let me try to clear it up.

The cartesian coordinate plane has two lines that define it; the x-axis and the y-axis.  They intersect at the origin.  The equation of the x-axis is y=0 and the y-axis is x=0.  A point in this plane is denoted by (x,y).  The x part of (x,y) tells you how far away the point is from the y-axis and the y tells you how far away from the x-axis the point is.

A function takes x-values and gives back y values.  This means that f(x)=y.  So if you want to graph the function you would plug in an x value and you would then get a y value. So then you plot the point (x,y).  You could keep doing this until you have a suitable graph.

So there is a connection between the x's and y's  and the y-axis.

I hope that this helps.

Ask again if you still don't understand.

No need to teach me about the carthesian plane Jolm! :D

What I mean is: he's speaking about f(x) AND about f(y), and that is confusing since y=f(y) for this point, so he uses a dubious notation.  Besides, there is [b]no reason at all[/b] even to mention the plane here :)

Just watch as it goes:
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy:
\[\forall a \in \mathbb{R}: f(a)=f(-a)\]
\[\forall a,b \in \mathbb{R}_0, a\not=-b : f\left(\frac{1}{a+b}\right) = f\left(\frac{1}{a} \right) + f\left( \frac{1}{b}\right) + 2ab - 1\]

I was pretty sure that you knew about the cartesian plane  :D but I thought that maybe you had a different convention in your country.  I am not too familiar with other countries' curricula.  I just explained it in the case that that was what you had trouble with.  I see what you mean now.

[quote="Peter VDD"]\[\forall a,b \in \mathbb{R}_0, ...\][/quote][quote="orl"]I hope you don't mind adding: $a,b \neq 0.$[/quote]

:roll:

Jolm: yeah I understand, still I though you'd have figured out that when saying f(x)=y then f(y) would get a bit problematic :)

Could you explain that in more detail BH?

[quote="anipoh"]From:
$f\left(\frac{1}{a+b}\right) = f\left(\frac{1}{a} \right) + f\left( \frac{1}{b}\right) + 2ab - 1$
Since the R.H.S. has a $2ab$ term and a constant only, can we assume that the function f is always of the form $f(x) = m/x^2 + n$, where $m$ and $n$ are constants?[/quote]

No, you can never assume that. Just like you cannot assume that a function from N to N will be a polynomial, or anything else. Just assume what is given :)


[u][b]Formal Instructions:[/u][/b]
The aim of the game is to be the last player remaining.
Every round, there is a time limit.
Before the time limit ends, all players must submit via PM a number between 0 and 100 to me.  There is a specific submission format written below and you must abide by this format or else your submission WILL NOT BE COUNTED.
And the end of the time limit (called a round), I tally up the submissions and scores in this fasion:
- I post all the submissions (a fail to submit counts as 100).
- I post what half the average submission was.
- I post all players who submitted under half the average.  These players are OUT.
- Everybody gains points equal to their distance from half the average.  Note: you do not want points!
- If anybody has over 100 points total, they are declared OUT.
- Submissions for the next round must be in [b]within 24 hours of the previous round[/b].  Yes, that is a new rule.
When there is one player remaining, they win.

[u][b]Submission format (has changed from Game 5)[/u][/b]
Submission Title: "[u]Number[/u], [u]Username[/u] (Round [u]#[/u])
Submission Body: Any comments.  If you would like me to read the body, out a * after the Submission Title.

[u][b]Previous games[/u][/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1734033730&t=149694]Game 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=151415]Game 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=154425]Game 3[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=168049]Game 4[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170873]Game 5[/url]

[u][b]Sign-up list (12)[/u][/b]
AstroPhys
Hamster1800
xpmath
perfect628
shaggy75
Nerd_of_the_Ages
DonkeyKong
Randomdragoon
abacadaea
Klebian
hunter
mz94

Above is the signup list from last game.  If you would like to be moved off, post.  Otherwise, the game starts in 3 days.

[b][u]Note to returning players[/u][/b]
The rules have changed.  Please read them.

/out.

The message is too small. Please make the message longer before submitting.

[u][b]Sign-up list (13)[/u][/b]
AstroPhys
Hamster1800
xpmath
perfect628
shaggy75
Nerd_of_the_Ages
DonkeyKong
Randomdragoon
abacadaea
Klebian
hunter
mz94
erabel

Wow, big game....  Sign-ups still allowed.  The game will start on Saturday.

[b]Sign-up list (14)[/u][/b]
AstroPhys
Hamster1800
xpmath
perfect628
shaggy75
Nerd_of_the_Ages
DonkeyKong
Randomdragoon
abacadaea
Klebian
hunter
mz94
erabel
mnmath

That's enough.  [b]Round 1 Due sometime tomorrow (Saturday)[/b].

There are quite possibly some people on that list (from last game) who do not know they are on this list

Can you just remove them from the game instead of giving them 100s if they don't submit this round?
Otherwise it's pretty cheap...

Okay, if anybody does not submit the first round, they are [i]removed from the game[/i].

Darn, I was under the impression that the 100s would still count towards the average..(in which case I would have won first round :))

[quote="miyomiyo"]Only 6 submissions....
[b][size=200]ROUND 1 RESULTS[/size][/b]
[hide="Submissions"]Hamster: 37
Klebian: 30
xpmath: 30
perfect: 25
abacadaea: 25
shaggy: 20[/hide][hide="Half The Average"]Half the average was a stunning [b]14[/b][/hide][hide="Below Half the Average"]Nobody[/hide][hide="Scores"]Hamster: 24
Klebian: 17
xpmath: 17
perfect: 12
abacadaea: 12
shaggy: 7[/hide][hide="Over 100"]Nobody[/hide][hide="Players Remaining"][u](6/14)[/u]
Hamster
Klebian
xpmath
perfect
abacadaea
shaggy[/hide][/quote]

Wait a sec!!! No submissions are counted as 100 no more??

I had NOADSO(Northern Oklahoma A...something District String Orchestra) on Sat. , so I couldn't get online and thought you would count me as 100...

Am I still in the game?
P:S I am submitting this time..

[quote="miyomiyo"][quote="DonkeyKong"]Wait a sec!!! No submissions are counted as 100 no more??[/quote][quote="miyomiyo"]Okay, if anybody does not submit the first round, they are removed from the game.[/quote]
[/quote]

But I really couldn't get on, though...

It lasted through 8 AM in the morning and I came back home at 9 PM and had to sleep then....

So, like are we playing the game or not?

Sorry, I'm too busy to mod this game.   :(

Feel free to make your own HTA games...


Find:
$\left[\frac{a-c}{(b-c)(a-b)}+\frac{c-b}{(a-b)(c-a)}-\frac{b-a}{(c-a)(b-c)}\right]\cdot\left(\frac{a-b}{2}\right)$

In what contest would you see that kind of problem? The solution's quite simple and clever though. Maybe good enough to memorize.

[hide]The expression on the left can be somewhat simplified as
$\frac{-(c-a)^2-(b-c)^2+(a-b)^2}{(b-c)(a-b)(c-a)}$
When multiplied by (a-b)/2, the a-b in the denominator is canceled out by the one in the numerator, so
$\frac{-(c-a)^2-(b-c)^2+(a-b)^2}{2(b-c)(c-a)}$
Expand the numerator.
$\frac{-c^2+2ac-a^2-b^2+2bc-c^2+a^2-2ab+b^2}{2(b-c)(c-a)}$
All the squared terms cancel out, leaving
$\frac{2ac+2bc-2ab}{2(b-c)(c-a)}$
We can divide out a 2 from the numerator and denominator (and expand the denominator to see if anything pops out):
$\frac{ac+bc-ab}{(b-c)(c-a)}$
The numerator doesn't factor as far as I can tell. So, the above is the final answer.
[/hide]

[quote="LynnelleYe"][hide]
$\frac{-c^2+2ac-a^2-b^2+2bc-c^2+a^2-2ab+b^2}{2(b-c)(c-a)}$
All the squared terms cancel out, leaving
$\frac{2ac+2bc-2ab}{2(b-c)(c-a)}$
[/hide][/quote]

[hide]You might have missed the fact that the -c^2 terms don't cancel out, they add to -2c^2. Good work, but paraphrasing rusczyk, can you find a slicker way?[/hide]

Theres a small mistake there, when its fixed you'll see that it does factorise in the end :)

Originally I just expanded and condensed the stuff until I eventually got a solution of:[hide]-1[/hide]

[quote="EFuzzy"]
Good work, but paraphrasing rusczyk, can you find a slicker way?[/quote]

Then I read that.  You can let a-c=x, b-c=y, and a-b=x-y and it becomes easier (it is still just manipulating algebra).

Is there an even slicker way?

I would like to know the title of the Chinese book that is the source of the problem. That kind of problem is rather typical of junior-high math problems in China (or Taiwan).

I believe no problem like that appear on the real math competitions rarely. But, it's true that many geometry and algebra problems involve clever manipulation of x's and y's or a's,b's, and c's, it's good to play with them.

 :)

[hide]No, I don't think the answer is -1.  :) [/hide]

Hint:
[hide]In the first term, where will adding a "-b+b" term help? Is it legal?[/hide]

The title of the book was 中学数学演算 一题多解

[quote="EFuzzy"]The title of the book was 中学数学演算 一题多解[/quote]

Thank you. I take it that the book regards a problem of this level as a JUNIOR-high math problem, right? It is at a very good level for this Getting Started forum, as it only takes relentless application of simple algebra. (Chinese-language textbooks characteristically apply simple ideas to knotty, multistep problems, so that the learners really get to know the basic ideas thoroughly.)

I'd be very happy to see you and other participants here post more problems like that as challenge problems to this forum. Thanks for sharing.

[quote="EFuzzy"]Find:
$\left[\frac{a-c}{(b-c)(a-b)}+\frac{c-b}{(a-b)(c-a)}-\frac{b-a}{(c-a)(b-c)}\right]\cdot\left(\frac{a-b}{2}\right)$[/quote]

Here is my solution.

[i]Note on moderator edit: I used my moderator powers to put the solution in hidden text, as is preferred in challenge problem threads.--tokenadult[/i]

[hide]Paying attention to cyclic order of $a,b,c$ ,that is to say $a\rightarrow b\rightarrow c\rightarrow a$ and $(a-b)\rightarrow (b-c)\rightarrow (c-a)\rightarrow (a-b)$

$\left[\frac{a-c}{(b-c)(a-b)}+\frac{c-b}{(a-b)(c-a)}-\frac{b-a}{(c-a)(b-c)}\right]$

$=-\frac{c-a}{(a-b)(b-c)}-\frac{b-c}{(c-a)(a-b)}+\frac{a-b}{(b-c)(c-a)}$

$=\frac{-(c-a)^2-(b-c)^2+(a-b)^2}{(a-b)(b-c)(c-a)}$

$=-\frac{(b-c)^2+(c-a)^2-[{(b-c)+(c-a)}]^2}{(a-b)(b-c)(c-a)}$

$=\frac{2(b-c)(c-a)}{(a-b)(b-c)(c-a)}$

Therefore the desired answer is 1.[/hide]

You can practice more at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=19463[/url]

kunny

A very slick way we can simplify the terms is as follows:

[hide]Take a look at the first term:
$\frac{a-c}{(b-c)(a-b)}$

We can add a $\"-b+b\"$ to the numerator without any harm:
$\frac{a-b+b-c}{(b-c)(a-b)}$

Now we can seperate this into seperate fractions:
$\frac{a-b}{(b-c)(a-b)}+\frac{b-c}{(b-c)(a-b)}$

$\frac{1}{b-c}+\frac{1}{a-b}$

And we can do this for all of the first 3 terms![/hide]


Hi. I’m new here. I need help with the following inequality please.

Let $ a,b,c$ be positive real numbers with $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that $ (a \plus{} b \plus{} c)^2[1\plus{}abc(a \plus{} b \plus{} c) ]\geq4$.

This is part of another problem that I’m trying to solve. I’ve got the problem down to the above inequality, and if I can solve it I’ll have solved my original problem. :wink:

[quote="Jane_Bennet"]Hi. I’m new here. I need help with the following inequality please.

Let $ a,b,c$ be positive real numbers with $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that $ (a \plus{} b \plus{} c)^2[1 \plus{} abc(a \plus{} b \plus{} c) ]\geq4$.
[/quote]
$ (a \plus{} b \plus{} c)^2[1 \plus{} abc(a \plus{} b \plus{} c) ]\geq4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2\plus{}2ab)\cdot\sum_{cyc}(a^2b^2\plus{}3a^2bc)\geq4(ab\plus{}ac\plus{}bc)^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4b^2\plus{}a^4c^2\minus{}2a^3b^3\plus{}3a^4bc\minus{}a^3b^2c\minus{}a^3c^2b\minus{}a^2b^2c^2)\geq0,$ which true by Muirhead.

[quote="Jane_Bennet"]Sorry, what is Muirhead? :([/quote]
Muirhead Inequality
if $ a_1, a_2, ..., a_n, b_1, b_2, ....b_n$ are integer number such that:
1) $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k \ge b_1 \plus{} b_2 \plus{} ... \plus{} b_k$ for $ n \minus{} 1 \ge k \ge 1$ and
2) $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} b_1 \plus{} b_2 \plus{} ... \plus{} b_n$ so:

$ \sum_{sym}x_1^{a_1}x_2^{a_2}...x_n^{a_n} \ge \sum_{sym}x_1^{b_1}x_2^{b_2}...x_n^{b_n}$ for all $ x_1, x_2, ..., x_n \in R^{ \plus{} }$

No! It's very easy. I tell the truth!
I put $ p\equal{}a\plus{}b\plus{}c;q\equal{}ab\plus{}bc\plus{}ca(\equal{}1);r\equal{}abc$
Rewrite as follows:
$ p^2(1\plus{}pr)\geq{4}$ very simple.


Show that no integer of the form 4a(8k + 7) is a sum of three squares.

Is any integer not of the form 4a(8k + 7) a sum of three squares?

I think the last part should be "... no sum of 3 ..." ?

anyway, a)
no number of form 8k+7 is writable as sum of 3 squares, so the 4^n factor will not chance anthing to this :?

b) I suppose a is an integer? elseway it's pretty easy
anyway, I think 22 is a good counterexample


geometry
rectangle
induction
combinatorics proposed
combinatorics

In a $2n\times 2n$ table we have put some $n\times1$ rectangles which they don't overlap and we can't put an $n\times 1$ rectangle anymore. Find the maximum number of uncovered rectangles.

The answer is n+2.Prove that by induction on general statement

[quote="Megus"]I suppose $2n^2-n$ is the answer - right?[/quote]
Yes, I think it's the right answer,too :)  and we need $2n+1$ rectangles$n\times 1$to cover it :roll:


Prove that: \[ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\binom{n}{3}\plus{}\ldots\plus{}\binom{n}{k}\plus{}\ldots\plus{}\binom{n}{n}\equal{}2^n\]  \[ k \le n\]

$ (1+x)^n=\sum_{r=0}^n {}_nC{}_r x^r$.

The number of the sub-set which is made from distinct $ n$ elements.

[hide="Proof 1"]The LHS is the sum over all $ i$ the number of subsets of $ \{1, 2, \ldots, n\}$ with $ i$ elements, i.e., simply the number of subsets of $ \{1, 2, \ldots, n\}$. However, $ 2^n$ gives this number as well (associate with each subset an n-bit long binary number, with the $ i$th bit corresponding to whether $ i$ is in that subset or not. [/hide]

[hide="Proof 2 (slightly overkill. just slightly.)"]$ e^{2x} \equal{} 1 \plus{} \sum_{k\equal{}1}^{\infty} \frac{2^k x^k}{k!} \equal{} (e^x)(e^x) \equal{} \sum_{k\equal{}1}^{\infty} \sum_{j\equal{}0}^k \frac{x^k}{j! (k \minus{} j)!}$ (by simply multiplying their Taylor series together.) Equating the $ n$th coefficient gives us $ \frac{2^n}{n!} \equal{} \sum_{k\equal{}0}^n \frac{1}{k! (n\minus{}k)!}$, i.e., $ 2^n \equal{} \sum_{k\equal{}0}^n {n \choose k}$. [/hide]

In the #2 post put x=1 and the answer will be OK

Yes. :)

In short, Binominal Theorem : $ \boxed{(a+b)^n=\sum_{k=1}^n {}_nC{}_k a^{n-k}b^k}$

[quote="hasan4444"]Prove that:
\[ \binom{n}{0} \plus{} \binom{n}{1} \plus{} \binom{n}{2} \plus{} \binom{n}{3} \plus{} \ldots \plus{} \binom{n}{k} \plus{} \ldots \plus{} \binom{n}{n} \equal{} 2^n\]

\[ k \le n\]
[/quote]

I'm confused, isn't it implied by the ellipsis that $ k \le n$ ? Also, how could $ \binom {n}{k}$ ever have $ k > n$? How can you do n choose k if k is greater than n?

Yes "zserf" I'm also confused by what you are saying I mean logically $ k<n$ so what is left to prove but maybe that's one of the weird problems I've seen a problem before that says prove that there is no natural number between 0 and 1  :rotfl:

[hide="late proof"] Let x and y equal 1 in the binomial theorem. Thus, we get: $ 2^n \equal{} \binom{n}{0} \plus{} \binom{n}{1} \plus{} \binom{n}{2}... \plus{} \binom{n}{n}$ QED.[/hide]


geometry
circumcircle
analytic geometry
graphing lines
slope
Pythagorean Theorem
perpendicular bisector

This is a theorem which I found when I was (I think) 15. I don't know if it already existed or exists now. If it does, and anyone recognizes it, I really wish to know. I already have the proof myself.

Given an isosceles triangle T, circumcircle C1 and the smallest possible circle C2 at the center of C1 which touches T.
With:
$ 3b \equal{} 2h$
prove that:
$ Area\ C1 \equal{} 10 \cdot Area\ C2$

[img]http://i24.tinypic.com/ibf78j.gif[/img]

[edit]error in formula[/edit]

[hide="here goes nothing"]
Call the triangle ISO with vertex I.  Draw IA perpendicular to SO at A.  Then $ SA\equal{}\frac{b}{2}$.  Call the circumcenter of ISO X.  X is obviously on IA.  Then call $ AX\equal{}y$ and $ AS\equal{}x$
We can use the Pythagorean Theorem on SAX to get $ x^2\plus{}y^2\equal{}9x^2\plus{}y^2\minus{}6xy$
Simplify
$ 6xy\equal{}8x^2$
$ 3y\equal{}4x$
So the radius of the smaller circle is $ \frac{4x}{3}$ and the radius of the larger circle is $ \frac{5x}{3}$ which means either of the following:
1. I made a calculation error
2. The theorem is false

However, I did notice that $ IS\equal{}x\sqrt{10}$, but I don't know how that would work in.
[/hide]

[hide="here defies nothing"]
I can follow you until the third line. Where did you get $ 9x^2 \plus{} y^2 \minus{} 6xy$? This implies that $ SX \equal{} \sqrt {9x^2 \plus{} y^2 \minus{} 6xy}$ which simplifies to $ SX \equal{} 3x \minus{} y$ Is this how you reasoned? if yes, then what was your motivation? [/hide]

Hint: My image is accurate to the pixel. Measure it for weak proof.

[edit]$ typo^2$[/edit]

[hide="here defies defiance"]
You made a mistake.  It should be $ SX\equal{}3x\minus{}y$, which is obvious when you note that $ AI\equal{}3x$ and $ AI\minus{}AX\equal{}IX\equal{}SX$
[/hide]

[hide]
Call the vertex of $ T$ as $ A$, the other two $ B, C$. Draw an altitude from $ A$ to $ BC$, with the foot $ H$. Then, $ AH \equal{} h$, and $ BH \equal{} \frac {b}{2}$. Because $ 3b \equal{} 2h,$ so $ AH \equal{} \frac {3}{2}b \equal{} 3BH$, and so $ AB \equal{} \sqrt {10}BH$
Call the radius of $ C1$ and $ C2$ as $ r1$ and $ r2$, respectively, and their center $ O$. Draw a perpendicular from $ O$ to $ AB$, with foot $ D$.
Then, $ r2 \equal{} OD, r1 \equal{} OA$. Because $ \frac {OA}{OD} \equal{} \frac {AB}{HB} \equal{} \sqrt {10}BH, S_{C1} \equal{} 10S_{C2}.$

But this is trivial.

[/hide]

[hide="here goes defiance"]Okay, I could verify all your calculations, except the radius of the smaller circle.[/hide]

timwu got it right!

[hide="My proof"]Okay, here's my proof. (let's keep timwu's notation)

Call the triangle ABC clockwise with AB=AC. Draw all three perpendicular bisectors. The feet of the bisectors are H, D and E. H bisects BC, D bisects AB, E bisects AC. With BC horizontal and A above BC, The slope $ S_1$ of AC is $ \frac {h}{\frac {1}{2}b} \equal{} 3$, making the slope $ S_2$ of OE $ \minus{} S_1^{ \minus{} 1} \equal{} \minus{} \frac {1}{3}$

For the sake of simplicity, let $ h \equal{} 18u$ and $ b \equal{} 12u$ for any $ u \in \mathbb{R}^ \plus{}$.
[hide="click for what happens when I dont use this"]
When perpendicular bisector OE goes right $ x \equal{} \frac {1}{4}b$,
it goes down $ y \equal{} |S_2 \cdot x| \equal{} \frac {1}{12}b$ giving total length of $ R_2$ (OE) = $ \sqrt {(\frac {1}{4}b)^2 \plus{} (\frac {1}{12}b)^2} \equal{} \frac {1}{12}b\sqrt {10}$

$ r_1$ (OA) = $ \frac {1}{2}h \plus{} \frac {1}{12}b \equal{} \frac {3}{4}b \plus{} \frac {1}{12}b \equal{} 10\frac {1}{12}b$

$ \underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$[/hide]
When perpendicular bisector OE goes right $ x \equal{} \frac {1}{4}b \equal{} 3u$,
it goes down $ y \equal{} |S_2 \cdot x| \equal{} 1u$ giving total length of $ R_2$ (OE) = $ \sqrt {(3u)^2 \plus{} u^2} \equal{} u\sqrt {10}$

$ r_1$ (OA) = $ \frac {1}{2}h \plus{} y \equal{} 10u$

$ \frac {r_1}{r_2} \equal{} \sqrt {10}$

$ \frac {\pi r_1^2}{\pi r_2^2} \equal{} 10$

I still wish to know if this theorem is known elsewhere or not, because I think it is a very elegant construction with only integer constraints. {1; 2; 3; 10}
$ \underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$
[img]http://i23.tinypic.com/2a0tw8.gif[/img]
[/hide]


equation
number theory
algebra
Diophantine equation
IMO Shortlist

Prove that the equation \[ x^n + 1 = y^{n+1}, \] where $n$ is a positive integer not smaller then 2, has no positive integer solutions in $x$ and $y$ for which $x$ and $n+1$ are relatively prime.

(y-1)(y^n+...+y+1)=x^n

(1) gcd(y-1,y^n+...+y+1)=1
proof:
assume that gcd(y-1,y^n+...+y+1)>1, let p be a common prime divisor

on one hand y==1(mod p)
==> y^n+...+y+1==n+1==0(mod p)

on the other hand p|x

contradiction because gcd(x,n+1)=1

(2) from (1) it follows that y^n+...+y+1=z^n for some positive integer z, which is impossible because y^n < y^n+...+y+1 < (y+1)^n

Therefore there are no positive integer solutions.

See also here
http://www.mathlinks.ro/Forum/viewtopic.php?t=45697

It can be also solved by using Euler's totient theorem  very easily. First assume x to be prime or prime to the power of something. Then applying inequality.

Suppose $n=p-1$ for any odd prime $p$. Then we have,
$$x^{p-1} \equiv 1 \pmod {p}\implies y^p-2 \equiv 0\pmod{p}$$
Then we have,
$$y \equiv 3 \pmod {p}$$
If $p$ is odd we must have $x$ even implying $y$ even. Contradiction.
If $p=2$, we have
$n<2$. Contradiction.

If $n=p$, for some odd integer $p$.
$$x^{p}+1 \equiv y^{p+1} \pmod{p}$$
We must have $x$ odd.Then, we see that $y$ is even.
$$x^{\phi{p}} \equiv 1 \pmod{p}$$
I can't see motivation to proceed. Can anyone help me?

Note that this is equivalent to $x^n=y^{n+1}-1=(y-1)(y^n+y^{n-1}+...+y+1)$.

[b]Lemma: [/b] $\gcd{(y^n+y^{n-1}+...+y+1,y-1)}=\gcd{(y-1,n+1)}$.
[b]Proof of Lemma: [/b] Using Euclidian Algorithm,
\begin{align*}\gcd{(y^n+y^{n-1}+...+y+1,y-1)}&=\gcd{((-y^n+y^{n-1})+y^n+y^{n-1}+...+y+1,y-1)}\\&=\gcd{(y^{n-1}+(y^{n-1}+...+y+1,y-1))}.\end{align*}
It's easy to see that we can continuously replace $y^k$ with $y^{k-1}$ while keeping the $\gcd$, so we result with $n+1$ (as we started with $n+1$ terms. $\blacksquare$

From the Lemma and the problem condition that $\gcd{(x,n+1)}=1$, $\gcd{(y^n+y^{n-1}+...+y+1,y-1)}=1$. We see that $y-1\mid x^n$, so unless $y-1=1$ or $y^n+y^{n-1}+...+y+1=1$, the latter of which is absurd, there does not exist a solution. So we conclude that $y=2,$ if any solution exists. Now we wish to find solutions to
$$x^n+1=2^{n+1}.$$ It's obvious that $x$ must be odd for this to be true. But then note that $x\geq 3\implies x^n\geq 2^{n+1}$ for $n\geq 2$ and $x\geq 5\implies x^n\geq 2^{n+1}$ for $n\geq 1$, so if any solution exists then (by the problem condition for $n$) $n=2$ and $x=3$ or $x=1$, both of which clearly do not work, and we may conclude.


geometry
3D geometry
tetrahedron
area of a triangle
Heron\u0027s formula
geometry proposed

Heron's formula expresses the area of a triangle in terms of the lengths  of its sides.
I) Is there an analogus formula for the volume of a tetrahedron in terms of area of its faces?
II) Is there an analogus formula for the volume of a tetrahedron in terms of the lengths of its edges that is symmetric in the edges?
THe Cayley formula
\[
V=\frac{1}{2^{3/2}3!}
\left (\begin{array}{ccccc}
0 & 1 & 1 & 1 & 1\\
1 & 0 & AB^2 & AC^2 & AD^2\\
1 & AB^2 & 0 & BC^2 & BD^2 \\
1 & AC^2 & BC^2 & 0 & CD^2\\
1 & AD^2 & BD^2 & CD^2 & 0
\end{array}\right)^{1/2}
\]
gives the volume of a tetrahedron with vertices A, B, C, D, but is not symmetric in those lengths.

For (1), for any triangle there is a tetrahedron with all faces congruent to the triangle, but I don't think these all have same volume.

For (2), what do you mean by symmetrical in the edges?  The formula should be symmetrical in the vertices, but some permutations of the edges change the volume.

There is no such formula in either case. A tetrahedron with one face equilateral with side of lenght 2 and the other three side of length 3 has volume$\sqrt{23}/3$.Changing the length ( one side 3 and the other 2) changes the volume.The new volume is $3\sqrt{3}/4$. Then there is not symmetric formula in terms of edge lengths.


[color=darkred]Let $ABC$ be a triangle. Denote the points $\{\begin{array}{c}B'\in (AC)\\\ C'\in (AB)\end{array}$ so that $BC'=CB'$.

Prove that the second intersection $X$ between the circumcircles of the triangles

$ABB'$ and $ACC'$ belongs to the bisector of the angle $\widehat{BAC}$.[/color]

[b][u]Please, I wish and another proofs ! Thank you.[/u][/b]

[hide="The my proof."]
[color=darkblue]The quadrilaterals $\{\begin{array}{c}ABXB'\\\ ACXC'\end{array}$ are cyclically $\Longrightarrow$ $\{\begin{array}{c}\widehat{XB'C}\equiv\widehat{XBC'}\\\ \widehat{XCB'}\equiv\widehat{XC'B}\end{array}\|$ $\Longrightarrow$

$\{\begin{array}{c}XB'C\sim XBC'\\\ B'C=BC'\end{array}\|$ $\Longrightarrow$ $XB'C\equiv XBC'$ $\Longrightarrow$ $XB'=XB$ $\Longrightarrow$

$\widehat{XB'B}\equiv\widehat{XBB'}$ $\Longrightarrow$ $\boxed{\ \widehat{XAB}\equiv\widehat{XAC}\ }$.

[b][u]Remark.[/u][/b] The line which passes by the midpoints of the segments $[BC]$, $[B'C']$

is parallelly with the bisector of the angle $\widehat{BAC}$ ![/color][/hide]

Your proof is very beautiful, after seeing it I can't think of any other way to do this problem... it would be so ugly :(
Here is a diagram so everyone can enjoy your proof:
[img]6919[/img]

[color=blue]my proof[/color][hide]
Let $s1$ the perp. bisector of $CC'$ and $s2$ the perp. bisector of $BB'$ intersect at $X$.
As $XC=XC' (X \in s1)$ and $XB=XB' (X \in s2)$ and $CB'=BC'$ $\Delta XCB' \equiv \Delta XC'B$. So perpendiculars from $X$ to $CB'$ and to $BC'$ are equal. This means $X$ [b]lies on the angle bisector[/b] of $\angle BAC$.
The perp. bisector of $AX$ intersects $s1$ in $O1$. Clearly $O1$ is equidistant from $C,X,C',A$ and is the center of the circumscribed circle. The same reasoning shows that $O2$ the intersection of perp. bisector of $AX$ with $s2$ is the center of the circumscribed circle around $B,X,B',A$. [/hide]


Find all $ x\in\mathbb R$ such that        $ \frac{3}{2}(5^{x}\plus{}1)\equal{}2^{x}\plus{}3^{x}\plus{}4^{x}$ .

[quote="Rust"]Only $ x \equal{} 0$ and $ x \equal{} 1$.[/quote]
Can you prove it? :)

Let
$ f(x) \equal{}\frac{3}{2}(5^{x}\plus{}1)\minus{}2^{x}\minus{}3^{x}\minus{}4^{x},$
$ g(x) \equal{} 5^{\minus{}x}f'(x) \equal{}\frac{3}{2}ln5\minus{}ln2 (\frac{2}{5})^{x}\minus{}ln3(\frac{3}{5})^{x}\minus{}ln4(\frac{4}{5})^{x}.$
Obviosly $ g'(x) > 0\ \forall x\in R$ and $ g(0) < 0,g(\infty ) \equal{}\frac{3}{2}ln 5 > 0$. Therefore exist $ x_{0}$, suth that
$ g(x) > 0\to f'(x) > 0$ if $ x > x_{0}$ and $ g(x) < 0\to f'(x) < 0$ if $ x < x_{0}$.
Therefore $ f(x) \equal{} 0$ have no more 2 solution $ x_{1}< x_{0}$ and $ x_{2}> x_{0}$.
We have solutions $ x_{1}\equal{} 0, x_{2}\equal{} 1$, therefore they are all solutions.

[quote="Rust"]Only $ x \equal{} 0,x \equal{} 1,x \equal{} 2$.[/quote]

??? Why ?
You ????

Solve by anologue. But these case is longer.


sooorry,people if i am dumb, but i aint gettin it, will someone please help me

2 identical buggies each of mass $150 kg$ move one after the other without friction with same velocity $4 m/sec$. A man of mass $m$ rides the rear buggy . At a certain moment the man jumps into the front buggy with a velocity $v$ relative to his buggy. As a result of this process rear buggy stops .If the sum of kinetic energies of man and front buggy just after collison , differs from that just before collison by $2700 joules$,calculate values of $m$ and $v$

What you've got here is a system of three nonlinear equations in three unknowns. Two of the unknowns are $m$ and $v.$ I'm not sure what the most efficient choice of the third unknown, but let's say that the final speed of the front buggy is $4+u.$ There are three states of the system:

State 1: Man riding in back buggy.

Momentum = $(150+m)\cdot4+150\cdot4$
(Energy irrelevant)

State 2: Man in mid-air between buggies.

Momentum = $0+m(4+v)+150\cdot4$
Energy = $(0 + m(4+v)^2 + 150\cdot4^2)/2$

State 3: Man riding in front buggy.

Momentum = $0+(150+m)(4+u)$
Energy = $(0+(150+m)(4+u)^2)/2$

Momentum is conserved; energy is not. Two of our three equations come from conservation of momentum:

$(150+m)\cdot4+150\cdot4=0+m(4+v)+150\cdot4=0+(150+m)(4+u)$

which can be simplified to $mv=600=(150+m)u.$

The energy condition says that

$(0 + m(4+v)^2 + 150\cdot4^2)/2=(0+(150+m)(4+u)^2)/2+2700$ or

$m(4+v)^2+150\cdot16=(150+m)(4+u)^2+5400.$

I'll let you wrestle with these three equations, but that's the setup.


I've translated the Chinese TST for 2003. Thanks to zhaoli for providing the source.

Thanks for your excellent work.
Have you received the TST problems from 1986-2002?
If you had time, would you please translate them into English and post them in Mathlinks?

Hmm. They are already translated. I will post them when I have time.

The credit for translating goes to Charmaine Sia and Yue Yang, both extremely successful and experienced IMO contestants. :)

If I have spare time, I am going to translate the other problems of China TST 2003.

[quote="zhaoli"]In China training camp, there are 8~9 sets of problems to select IMO contestants.
Usually, only the last two sets having six problems is officially called as TST, the others are called quizes.
 :)[/quote]

You can take your time and translate the TSTs for 2004 and after that. The problem sets I have mostly 6 problems a year. If you sent a more comprehensive problem set to Billzhao he can translate his time. Otherwise it is not worth going for it as I already have it in English.  ;)

[quote="filletwho"]Did you post them? I can't find them...[/quote]

No, not yet. Please be patient.  :)

I can be patient, very patient. so are they up now? howabout now? jk :D


induction
number theory unsolved
number theory

Let {$ a_n$}={$ 2,4,8,1,3,6,...$} be the infinite integer such that $ a_n$ ids the leftmost digit in the  decimal of $ 2^n$ and {$ b_n$} ={$ 5,2,1,6,3,1,...$} be the infinite inte integer  such that $ b_n$ ids the leftmost digit in the  decimal of $ 5^n$ .Prove that foer any block of consecutive terms in {$ a_n$} there is a block of consecutive terms in {$ b_n$} in the reverse order.

[hide]It is a well known theorem of Jordan, that given any $ t$ such that $ t$ is not a power of 10, for any finite sequence of digits $ \{d_i\}$, there exists an $ n$, such that $ t^n$ begins with that sequence of digits.

Using this theorem, we assert that if the given sequence is $ \{a_i,a_{i\plus{}1},...,a_k\}$, then there exists an $ n$ such that $ 5^n$ begins with the sequence of digits forming $ 2^k$.  Thus, let $ 5^n\equal{}2^k\cdot10^m\plus{}c$, where $ c<10^m$.  We claim that the desired sequence is $ \{b_n,b_{n\plus{}1},...,b_{n\plus{}k\minus{}i}\}$.  We prove that $ b_{n\plus{}j}$ begins with the sequence of digits forming $ 2^{k\minus{}j}$ by induction on $ j$.  The claim is obvious for $ j\equal{}0$ since $ b_n\equal{}2^k\cdot10^m\plus{}c$.  Now assume that the claim is true for $ j\equal{}j_0$.  Then we have that $ b_{n\plus{}j_0}\equal{}2^{k\minus{}j_0}\cdot10^{m_0}\plus{}c_0$ for $ c_0<10^{m_0}$.  Notice that

$ b_{n\plus{}j_0\plus{}1}\equal{}5b_{n\plus{}j_0}\equal{}\frac{10}{2}b_{n\plus{}j_0}\equal{}2^{k\minus{}j_0\minus{}1}\cdot10^{m_0\plus{}1}\plus{}\frac{10c_0}{2}$

Since $ c_0<10^{m_0}$, we have that $ \frac{10c_0}{2}<10^{m_0\plus{}1}$, thus $ b_{n\plus{}j_0\plus{}1}$ begins with the sequence of digits forming $ 2^{k\minus{}j_0\minus{}1}$ and the claim is proven.[/hide]


inequalities
induction
limit
factorial
inequalities proposed

Prove that:

\[ \boxed{\left(1\plus{}\frac{1}{1^2}\right)\left(2\plus{}\frac{1}{2^2}\right)\left(3\plus{}\frac{1}{3^2}\right)...\left(n\plus{}\frac{1}{n^2}\right)<\sqrt{6} \cdot n!}\]

Any ideas? It seems to be a hard inequality. I don't think that induction is the best choice.

If we divide by $ n!$, we get $ (1 \plus{} \frac {1}{1^3})(1 \plus{} \frac {1}{2^3})...(1 \plus{} \frac {1}{n^3}) < \sqrt {6}$. Now I think that $ lim_{n \rightarrow \plus{} \infty}LHS \equal{} 2 < \sqrt {6}$...
I may be wrong, please correct me in such a case...

hello, the searched limit of your product is
$ \lim_{n \to \infty}\prod_{i\equal{}1}^{n}1\plus{}\frac{1}{i^3}\equal{}\frac{1}{(\frac{1}{2}\minus{}\frac{\sqrt{3}}{2}i)!(\frac{1}{2}\plus{}\frac{\sqrt{3}}{2}i)!}$
Sonnhard.

Is the factorial defined for complex numbers? Please provide details.

If $ a_n={\left(2+\frac{1}{2^{2}}\right)\left(3+\frac{1}{3^{2}}\right)...\left(n+\frac{1}{n^{2}}\right)}$ and $ b_n={\left(2-\frac{1}{2^{2}}\right)\left(3-\frac{1}{3^{2}}\right)...\left(n-\frac{1}{n^{2}}\right)}$ then the following inequalities are true:

$ \boxed{\star}$ $ a_n \cdot b_n <(n!)^2$

$ \boxed{\star}$ $ \frac{a_n}{b_n}<\frac{3}{2}$


Hi everybody,

I'm working on a game.
We have a board 3x3.
We want to color (using one color) 5 squares
We have 126 possibilities but....
I want to eliminate all the symetrical colored squares (3x3). The "cross" have no symetry, you can put it in any shape you want you will have only one. But the "L" you can shape it four ways, so there is 3 ways to eliminate.
I can do it by checking all the squares (126).
What if I have a square with n x n (where n is bigger as 1000)?
Is there any formula to know it?
Thnk you for any help

Here is a picture of 2 different shapes
Each one have 4 ways to color it


topology
real analysis
real analysis theorems

I am afraid that I am somehow misunderstanding the proof of Urysohn's Lemma. The particular book I am using is Gamelin and Greene's "Introduction to Topology".

Let $E,F$ be disjoint closed sets of our normal topological space $X$. The strategy in the text is to construct for each dyadic rational $r \in (0,1)$ an open set $U_{r}$ such that
$\overline{U_{r}}\subset U_{s}\; \; \; 0 < r < s < 1$
$E \subset U_{r}$
$U_{r}\subset X\setminus F$
And then we let $f(x) = \sup\{r \, | \, x \,\text{not belonging to}\, U_{r}\}$

The way this is done is by first choosing $U_{1/2}$ such that
$E \subset U_{1/2}\subset \overline{U_{1/2}}\subset X\setminus F$
Then we continue with dyadics of the form $\frac{p}{2^{2}}$ and we get
$E \subset U_{1/4}\subset \overline{U_{1/4}}\subset U_{1/2}\subset \overline{U_{1/2}}\subset U_{3/4}\subset \overline{U_{3/4}}\subset X\setminus F$
And so on.

I am trying to understand why we choose dyadics. I read some comment on Wikipedia that the dyadics are somehow a "small" set and that's why they are used. But I don't see why the proof doesn't work exactly the same if we just index all rationals (or any countable dense set) in $(0,1)$ and do the exact same thing with one number at a time. It seems to me that everything works out fine if you do this. But then it seems very weird to use dyadics instead of just the rational numbers. Am I misunderstanding something?

Dyadic rationals appear naturally when at each step you are inserting a new number between two numbers that are already there.


Solve the following equation: $(ax-i\sqrt{1-x^2})^n+(ax+i\sqrt{1-x^2})^n=0, \ a\in\mathbb{R}$

Actually, I was wrong the first time.  But by DeMoivre's we get that x = 0.

Can you be more explicit  :? How I use DeMoivre?

So just put $n=1$, and $a \neq 0$ and pop

Well I was actually wrong, but sub $x = \cos \theta$ and you have

$(a \cos \theta - i \sin \theta )^n + (a \cos \theta + i \sin \theta )^n = 0$

Shouldn't be hard from there.

[quote="seamusoboyle"]Errrr, problem. That $a$ messes everything up....[/quote]

Good point.  Doing too much MOP homework has fried my brain.  I think binomial expansion might help...


AMC USA(J)MO USAMO function inequalities calculus derivative

I saw this problem in Engel, USAMO 1980;

$If a,b,c$ are 3 reals that can take any value from 0 to 1 (both inclusive); prove that $(a/(b+c+1))+(b/(c+a+1))+(c/(a+b+1))+(1-a)(1-b)(1-c)$ can take a maximum value of 1.
He has proved it by the following method:

" LHS = f(a,b,c) ; f is strictly convex in each variable, and so is defined on a closed convex cube. "
I understand this. But then he says, "thus, the function assumes extremal values at extremal points". How is this ?

He cites some theorem of Weierstrass. Please help. I need to get this cleared quick. Thanks, in advance.

If a function is strictly convex then its extrema are not saddle points. In other words, the gradient of the graph doesn't go from positive to zero and then back to positive, it must always go from negative to positive because it is convex.

Therefore any stationary point will be extremal.

can anyone provide a solution which doesnt use function,jensen inequality :maybe:

Something nicer posted here:
http://www.mathlinks.ro/viewtopic.php?p=758089#758089

But  am still confused about the double derivative $ f(a,b,c)$

Can anyone post the complete derivation process? :maybe:


geometry
geometric transformation
reflection
incenter
geometry unsolved

Let $ O$ be the incenter of triangle ABC.  Let $ O_1$, $ O_2$, and $ O_3$ be the reflections of O across $ AB$, $ BC$, and $ CA$ respectively. Also, let $ A'$ be the centre of the excircle of ABC tangent to $ BC$, $ B'$ be the centre of the excircle of ABC tangent to $ CA$, and $ C'$ be the centre of the excircle of ABC tangent to $ AB$.

1. Prove that $ C'O_1$, $ A'O_2$, and $ B'O_3$ meet at a point.
2. Let the point determined in question 1 be $ H$. Prove that triangles $ A'B'C'$ and $ O_1O_2O_3$ are similar, and that $ \frac{C'O_1}{O_1H} \equal{} \frac{A'O_2}{O_2H} \equal{} \frac{B'O_3}{O_3H}$.

This looks like it could have been a problem on one of the older Olympiads, but I don't know the solution. It seems trivial at first, but I still couldn't find it.

DE//B'C'//I2I3
they are concurrent at their homothetic center

[quote="plane geometry"]they are concurrent at their homothetic center[/quote]

So wait, explain the diagram...? I don't see any concurrent lines, nor do I see the homothetic center...

[quote="livetolove212"]Note that $ I_2I_3//B'C', I_1I_3//A'C', I_1I_2//A'B'$ and we are done![/quote]

Ah. I thought that might have been it once you said that...

...but how is that obvious, besides seeing it through the diagram?

I know that EF is parallel to I2I3 by definition, but how do you show that EF is parallel to BC? Unless it's something really obvious that I'm missing here.

Never mind, I got it.

*EDIT* ...wait. no, I [i]didn't[/i]! Argh.

*EDIT2* Okay, now I have it for real.


A cheetah runs at a speed of 70 mph, while a snail crawls at a rate of 0.005 mph. How many miles will a cheetah run in the same amount of time that it takes a snail to crawl 5.28 feet?

A cheetah runs at 70/.005 = 14000 times the rate of the snail.

When the snail travels 5.28 feet, the cheetah will travel 14*5.28*1000=14*5280 feet.  Since there are 5280 feet in one mile, the cheetah will run 14 miles.


When will the admission test B be posted? I thought it was supposed to be today?

Also, is there a way we could postpone getting items (for example recommendations; those will take a while) but still send in the test by the due date? I'm on spring break right now and it will be hard to coordinate everything.

I think it is supposed to be posted the 30th...or at least that is what the website said 11:20 AM ET.

It is currently online, so don't worry.  Also, it looks to be about the same difficulty as the first (which was not hard).

yeah it is now up:
http://www.awesomemath.org/tests/TestB.pdf

nice test...i might apply for this...i ono i have a few other things going on...but awesome math is a strong possibility...

EDIT: it is 4:35 am right now...but i am pretty sure i [talked about the admission test]!

zacacox says: Side effects of talking about the Admission Test before it's over include swelling, muscle soreness, and dryness of mouth. Please consult your doctor before talking about any new Admission Test before it's over.

I think I figured out where they [zacacox 1, premature Admission Test banter 0]

But I do want to get in...

[quote="PenguinIntegral"]
Also, is there a way we could postpone getting items (for example recommendations; those will take a while) but still send in the test by the due date? I'm on spring break right now and it will be hard to coordinate everything.[/quote]

how many problems do we solve for awesome math?

[quote="not_trig"][quote="Max300"]LMAO  :rotfl:
you're hilarious zac

i think our standards for your start of camp speech have greatly risen. "DO NOT START A FIRE FOR ANY REASON WHATSOEVER"  never gets old   :D[/quote]

rofl i agree!

GO ZAC!! (even though you were never there at the dorm)[/quote]

lol...was anyone else here in zac's cabin?

I was - for 1 day.  Then I moved to Dvince's place.  hey all.  nice to talk to you again.

[hide=":P"]
Best line of AMP 2006: Daniel Vitek to Damien Jiang

Damien, make like a tree and GET THE F*** OUT OF HERE!!!

Second-best line of AMP 2006: Daniel Vitek to David Vincent
Dude, there's like s*** on the floor.  (which there was - I had to clean it up though it wasn't mine  :stretcher:  :mad:  :bomb:  :cursing:  :censored:  :furious:  :cleaning: )

okay, so maybe these weren't the best, but they have to be up there...
[/hide]


[quote="mdk"]Let n be a positive integer. A bit string of length n is a sequence of n numbers consisting of 0’s and 1’s. Let f (n) denote the number of bit strings of length n in which every 0 is surrounded by 1’s. (Thus for n = 5, 11101 is allowed, but 10011 and 10110 are not allowed, and we have f (3) = 2,
f (4) = 3.) Prove that f (n) < (1.7)^n for all n.[/quote]

[hide="Hint"]Establish an induction formula between $ f(n)$, $ f(n\minus{}1)$ and $ f(n\minus{}2)$[/hide]

Hope this is correct...
[hide]
Let $ A_{n}$ be the number of bit strings length n that end in a 1.
Let $ B_{n}$ be the number of bit strings length n that end in a 0.

It is easy to see that $ A_{n}\equal{} A_{n\minus{}1}\plus{}B_{n\minus{}1}$ and $ B_{n}\equal{} A_{n\minus{}1}$, so:

$ A_{n}\plus{}B_{n}\equal{} (A_{n\minus{}1}\plus{}B_{n\minus{}1})\plus{}A_{n\minus{}1}$
$ F(n) \equal{} F(n\minus{}1)\plus{}F(n\minus{}2)$.

This implies that F(n) is the $ n^{th}$ fibonacci number, which can be explicitly written as $ \frac{\phi^{n}\minus{}\overline{\phi}^{n}}{\sqrt{5}}$ with $ \phi$ being the golden ratio $ \frac{1\plus{}\sqrt{5}}{2}$ and $ \overline{\phi}$ its conjugate $ \frac{1\minus{}\sqrt{5}}{2}$.

$ \frac{\phi^{n}\minus{}\overline{\phi}^{n}}{\sqrt{5}}<\phi^{n}$, and we know that $ \phi < 1.7\Leftrightarrow\phi^{n}< 1.7^{n}$, thus $ \frac{\phi^{n}\minus{}\overline{\phi}^{n}}{\sqrt{5}}< 1.7^{n}$.

[/hide]


Determine the number of ordered pairs of positive integers $ (a,b)$ such that the least common multiple of $ a$ and $ b$ is $ 2^{3}5^{7}11^{13}$.

[hide]One of $ a$ and $ b$ must be divisible by each of $ 2^{3}$, $ 5^{7}$, and $ 11^{1}3$.  The other number can have an exponent less than or equal to the first.

The factor combinations with powers of 2 are (disregarding order):

$ (2^{0},2^{3})$
$ (2^{1},2^{3})$
$ (2^{2},2^{3})$
$ (2^{3},2^{3})$

They can be distributed among $ a$ and $ b$ in 7 ways.  Similarly, the powers of 5 can be split in 13 ways, and the powers of 11 in 25 ways, for a total of $ 125\times13\equal{}1625$ ordered pairs.[/hide]


[quote="Inequalities Master"]Let $ a,b,c>0$ s.t. $ a\plus{}b\plus{}c\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$. If $ a\leq b\leq c$, then:
$ ab^2c^3\geq1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=358621


trigonometry
geometry
3D geometry
prism
geometric transformation
reflection

[quote="ROLEX4life"]Does anyone here have AP Physics B with the "College Physics" text by Serway and Faughn 5th edition? There are 2 problems that i am really struggling with but in order for someone to be able to help me, they'd need to see the diagrams...[/quote]I taught freshman physics from the 5th edition of Serway and Faughn.  What's your question?

[quote="ROLEX4life"]I have problems with Questions 38 and 41 from Chapter 22. I don't know how to figure out the indices of refraction and/or the incident angle. Both those problems refer to the diagrams on page 742 just for quick reference. Thanks for your help in advance.[/quote]Both problems deal with the same phenomenon: Total internal reflection.

Keep in mind Snell's law: $n_i\sin\theta_i = n_r\sin\theta_r$

If you solve this for $\sin\theta_r$, then $\sin\theta_r =\frac{n_i\sin\theta_i}{n_r}$

$\sin\theta_r \not > 1$; therefore, if this condition is violated the light rays cannot pass into the second medium, whatever it might be.

For question # 38 Serway and Faughn ask you to calculate the minimum index of refraction for the prism.

For which we find $\sin\theta_{air} = \frac{n_g \sin\theta_g}{n_{air}} > 1$

It is immediately apparent then that $n_g > \frac{n_{air}}{\sin\theta_g}$.

In question #41 Serway & Faughn ask you to find the maximum index of the fluid surrounding a prism having $n_g = 1.80$.

Same problem  $\frac{1.80 \sin\theta_g}{n_f} > 1$

From which we see  $1.80 \sin\theta_g > n_f$

What's left, and I have to leave something for you to do, is a geometry problem.

[quote="ROLEX4life"]don't u mean that  n < (air)/ sin 0(g) ?

I'm still confused (sorry, i'm stupid).

I end up with 1.8 = sin (incident angle)/ sin (exit angle) but wouldn't the incident angle be 90 degrees since the diagram shows the ray entering parallel to the normal? I can't figure out the geometry set up. Also, how did you find that n(g) = 1.8? Do you always assume a prism has n = 1.8?

Also, could you help me with number 45 part B and number 51? I'm really sorry. I truly want to understand this stuff but it's just too hard. I get the concepts but i can't apply them to problems. Also, my stupid teacher doesn't even go over any problems in class. He ONLY explains the most BASIC concepts. All he has actually taught us is how to draw ray diagrams for mirrors and lenses, the law of reflection, and snell's law. I'm not kidding. You have to figure out the problems for yourselves....

man i'm so screwed in this class....:([/quote]First, you're not stupid; you're a student. These things are difficult. If you have trouble with them it comes from lack of experience, not dullness of mind.

Second, I meant what I wrote. In both cases, the glass is the incident medium. In the first problem the light tries to refract from the glass into air, and in the second problem, from glass into some unknown fluid.

The result is that in the first problem $n_g$ greater than some number (its minimum value is that number) and in the second problem you get $n_f$ less than some other number (its maximum value is this other number).

No, you don't assume that a prism always has $n = 1.8$. Remember, in the first problem the index of the prism is what you're trying to solve for. Index of refraction is a material property. If the index of a prism was always 1.8 that would mean that all prisms were made of the same material, which obviously isn't true.

For your second problem, #41 I believe, the fact that $n_g = 1.8$ was given in the problem. You were supposed to find the maximum index of the surrounding fluid.

Other problems will have to wait as my copy of S & F is at home, and I'm quite busy at work right now.

[quote="ROLEX4life"]Ok, thanks for your help and patience with me  :P

btw i figured out number 51 so you don't need to explain that...[/quote]S & F, Problem 22.45b, at what value of $n_2$ does total internal reflection cease?

$n_1\sin\theta_1 = n_2 \sin\theta_r$

When refraction actually occurs $\sin\theta_r \le 1$

$\Rightarrow n_1\sin\theta_1 \le n_2$.

Total internal reflection ceases when $n_2 = n_1 \sin\theta_1$.


Gauss
function
integration
trigonometry
complex numbers
complex analysis
complex analysis unsolved

Find the complex numbers $ z$ for which the series
\[ 1 \plus{} \frac {z}{2!} \plus{} \frac {z(z \plus{} 1)}{3!} \plus{} \frac {z(z \plus{} 1)(z \plus{} 2)}{4!} \plus{} \cdots \plus{} \frac {z(z \plus{} 1)\cdots(z \plus{} n)}{(n \plus{} 2)!} \plus{} \cdots\]
converges and find its sum.

there is many way how to prove that series converges for $ |z| < 1$
 for example one can use Gauss formula for Gamma function and obtain that $ z(z + 1)(z + 2)...(z + n) \approx C(z) n^{z}n!$
 and when $ n \to \infty$  we have that $ C(z) \to \Gamma(z)$
 (you can easily make all this a proper proof, I think..)
The main result is to find sum .
Lets denote your sum by $ S(z)$ and consider $ S(z) - 1$
we can write it as
$ S(z) - 1 = \sum_{n\geq 0} \frac {\prod_{i = 0}^{n}(z + i)}{(n + 2)!}$
use Gamma function property that $ \Gamma(z + 1) = z\Gamma(z)$ and we see that :
${ \Gamma(z)\prod_{i = 0}^{n}(z + i)} = \Gamma(z + n + 1)$
thus we obtain :
\[ \Gamma(z)\Gamma(2 - z)(S(z) - 1) = \sum_{n\geq 0} \frac {\Gamma(z + n + 1)\Gamma(2 - z)}{\Gamma(n + 3)} = \sum_{n\geq 0}\beta (z + n + 1,2 - z)
\]

\[ = 2\sum_{n\geq 0} \int_{0}^{\pi/2}\sin^{2z + 2n + 1}\alpha \cos^{3 - 2z}\alpha d\alpha
\]

\[ = 2\int_{0}^{\pi/2}\cos^{3 - 2z}\alpha \sin^{2z + 1}\alpha \sum_{n\geq 0}\sin^{2n}\alpha d\alpha = 2\int_{0}^{\pi/2}\cos^{1 - 2z}\alpha \sin^{2z + 1}\alpha d\alpha = \Gamma(1 + z)\Gamma(1 - z)
\]
thus :
\[ S(z) = \frac {\Gamma(1 + z)\Gamma(1 - z)}{\Gamma(z)\Gamma(2 - z)} + 1 = (1 - z)^{ - 1}
\]

It converges at negative integers too... ;)

Sure  :blush: , not only it converges also for all $ Re z \leq \minus{}1$


I need help in understanding these proofs. Can someone please explain?

Define $T: V\rightarrow V$, a linear transformation mapping $V$ to $V$, where $V$ is a vector space of finite dimension.

a) If $T$ is one-to-one, prove that $T$ is also onto.

b) If $T$ is onto, prove that $T$ is one-to-one.

This is a theorem about matrices. Choose a basis for the $n$-dimensional vector space $V.$ Expressed in terms of that basis, a linear transformation is given by an $n\times n$ matrix.

Then the result follows from rank + nullity = dimension. This fundamental theorem of matrices can be proved using the mechanics of gaussian elimination and by counting pivot elements and non-pivot columns.

There are some other ways of going through this. You could use the following assertions:

$T$ is one-to-one iff it maps every independent set to an independent set.

$T$ is onto iff it maps every spanning set to a spanning set.

Any two bases for $V$ have the same number of elements.

Two side notes:

In your statement of the problem, you said that $V$ was a vector space but didn't say over what field. You didn't have to; the theorem is the same, and with the same proof, over all fields.

You did say that $V$ was finite dimensional. This is vital, as the statement would be false in an infinite-dimensional space.


Solve that equation of system:

$ \left\{ \begin{array}{l}
 x \plus{} \frac{{2xy}}{{\sqrt[3]{{x^2  \minus{} 2x \plus{} 9}}}} \equal{} x^2  \plus{} y \\
 y \plus{} \frac{{2xy}}{{\sqrt[3]{{y^2  \minus{} 2y \plus{} 9}}}} \equal{} y^2  \plus{} x \\
 \end{array} \right.$

hello, i have only found the solutions $ x\equal{}0,y\equal{}0$ or $ x\equal{}1,y\equal{}1$.
Sonnhard.


if the $ gcd (a,b) \equal{} 1$, prove that the $ gcd (a\plus{}b, a\minus{}b) \equal{} 1$ or $ 2$.

I am no math wiz or anything close to it. I have my own post on here I am waiting for an answer to, Here is as far as I get. Someone wanna pick up where I left off?

Let a = 80, b = 83, c = a+b, d = a-b
gcd(a,b) = 1
let e = gcd(c,a), e  = 1
let f = gcd(d,b), f = 1

gcd(e,f) = 1, therefore, gcd(c,d) = 1 or restated, gcd(a+b,a-b) = 1

again, correct me if I am wrong.

Lemma: $ (a,kb)|k*(a,b)$ where $ a$, $ b$, and $ k$ are integers.

So $ (a\plus{}b,a\minus{}b)\equal{}(a\plus{}b,2a)|2*(a\plus{}b,a)\equal{}2*(b,a)\equal{}2$. Then since the gcd divides two, it is equal to one or two.

Let the GCD be $ k$
then $ k|2a$ and $ k|2b$
as $ gcd(a,b)\equal{}1$
$ k\equal{}1 or 2$

[hide="Another approach"] [b]Lemma:[/b]  If $ ax \plus{} by \equal{} d$ then $ \gcd(x, y) | d$.

[b]Bezout's Lemma:[/b]  $ \forall x, y \in \mathbb{Z} : \exists a, b \in \mathbb{Z} : ax \plus{} by \equal{} \gcd(x, y)$.

But of course,

$ (a \plus{} b) \plus{} (a \minus{} b) \equal{} 2a$
$ (a \plus{} b) \minus{} (a \minus{} b) \equal{} 2b$

So if $ ac \plus{} bd \equal{} \gcd(a, b) \equal{} 1$ then

$ (a \plus{} b)(c \plus{} d) \plus{} (a \minus{} b)(c \minus{} d) \equal{} 2$

Hence $ \gcd(a \plus{} b, a \minus{} b) | 2$. [/hide]


Fix two integers $ n > m > 1$, and $ m$ integers $ 1 \le x_1 \le x_2 \le \ldots \le x_{m} \le n \minus{} 1$ such that $ n \mid x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m$.

Prove, or disprove, that there exist always a permutation $ \{\sigma(x_1),\ldots,\sigma(x_m)\}$ of $ \{x_1,\ldots,x_m\}$ such that the numbers $ y_j: \equal{} j \plus{} \sigma(x_j)$ [color=orange](edited)[/color] for $ j \equal{} 1,2,\ldots,m$ are all different modulo $ n$ [color=orange]and $ n \nmid y_j$ for all $ j\equal{}1,2,\ldots,m$ (edited).[/color]

[size=75]Ps. This problem arises from a discussion from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=308048]here[/url]  :)[/size]

Pick $ n \equal{} 5$, $ m \equal{} 4$, and $ \{x_i\} \equal{} 1,1,4,4$. WLOG $ \sigma(x_1)$ is a $ 1$, then $ \sigma(x_2)$ must be $ 1$ (otherwise the $ \{y_i\}$ hit $ 0$ too early), then $ \sigma(x_3)$ has to be a $ 4$ and $ y_3 \equal{} 1$ is repeated. Perhaps the sequence should be strictly increasing?

Omg, very sorry. I have wrong copying the definition of $ y_j$  :o

The correct statement is:

[i]"Fix two integers $ n > m > 1$, and $ m$ integers $ 1 \le x_1 \le x_2 \le \ldots \le x_{m} \le n \minus{} 1$ such that $ n \mid x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m$.

Prove, or disprove, that there exist always a permutation $ \{\sigma(x_1),\ldots,\sigma(x_m)\}$ of $ \{x_1,\ldots,x_m\}$ such that the numbers $ y_j: \equal{} j \plus{} \sigma(x_j)$ for $ j \equal{} 1,2,\ldots,m$ are all different modulo $ n$, and $ n \nmid y_j$ for all $ j \equal{} 1,2,\ldots,m$."[/i]   :P


prove that : $ \forall n>2: \exists x_n,y_n: 2^n\equal{}7x_n^2\plus{}y_n^2$

sorry if it posted before...


(1) Find the equation of the circle with centered $ (5,\ 2)$, which touches the line $ 3x\minus{}4y\minus{}22\equal{}0.$

(2) Find the range of $ m$ for which a circle $ x^2\plus{}y^2\minus{}5\equal{}0$ has distinct points of intersection with the line $ 2x\plus{}y\plus{}m\equal{}0.$

[quote="kunny"](1) Find the equation of the circle with centered $ (5,\ 2)$, which touches the line $ 3x \minus{} 4y \minus{} 22 \equal{} 0.$

(2) Find the range of $ m$ for which a circle $ x^2 \plus{} y^2 \minus{} 5 \equal{} 0$ has distinct points of intersection with the line $ 2x \plus{} y \plus{} m \equal{} 0.$[/quote]

[b](1)[/b] Distance from the center to the line $ \equal{}|\frac{3(5)\minus{}4(2)\minus{}22}{\sqrt{3^2\plus{}4^2}}|\equal{}3$

So the equation : $ (x\minus{}5)^2\plus{}(y\minus{}2)^2\equal{}3^2$

$ x^2\plus{}y^2\minus{}10x\minus{}4y\plus{}20\equal{}0$

[b](2)[/b] Put $ y\equal{}\minus{}2x\minus{}m$ into the equation of the circle.

$ 5x^2\plus{}4mx\plus{}(m^2\minus{}5)\equal{}0$

$ Discrimant \equal{}(4m)^2\minus{}4(5)(m^2\minus{}5)>0$
$ \Leftrightarrow 25\minus{}m^2>0$
$ \Leftrightarrow \minus{}5<m<5$ is the range.


Αν α,b,c,x,y > 0 και ισύει [b][u]για κάθε x,y >0[/u][/b], ... $ a(x^2 \plus{} y^2 ) \plus{} \frac {b}{x \plus{} y}\ge c \minus{} 2xya$,

να δ.ο.  ... $ 27ab^2\ge4c^3$

Κανένα άλλο περιορισμό για τα x,y δεν έχει;  :ninja:

Η ασκηση μου φαινεται λαθος...vangelismak εισαι σιγουρος?

Παιδιά μονο αυτός είναι ο περιορισμός ! Δεν νομίζω ότι έχει κάποιο λάθος  :(
Νομίζω ότι με συμβατικές μεθόδους δεν βγαίνει κάτι. Μάλλον πρέπει να καταφύγουμε στη λύση της ανάλυσης.

Να η δική μου λύση με τη χρἠση ανάλυσης . Η δοθείσα σχέση είναι ισοδύναμη με

$ a(x\plus{}y)^3\minus{}c(x\plus{}y)\plus{}b\geq 0$

Θεωρούμε την συνάρτηση $ f(u)\equal{}au^3\minus{}cu\plus{}b$  . Τότε $ f(u)\geq 0$ για κάθε u>0 . Αυτή η συνάρτηση για u>0 παρουσιάζει ελάχιστο στο $ u_1\equal{}\sqrt{\frac{c}{3a}}$ . Όμως θα πρέπει $ f(u_1)\geq 0$ άρα

$ \sqrt{\frac{c}{3a}}(c\minus{}\frac{c}{3})\leq b$  ή ισοδύναμα $ 27ab^2\geq 4c^3$   :)

***
One line solution by Michael Savvas uses only AM-GM  :wink:

One line solution ?  :wow:

Θα δώ αν μπορώ να κάνω κάτι ... :mad:

 :P


AMC
AIME
geometry
topology
inequalities
Ross Mathematics Program
Harvard

From what I heard, all people who scored over 50 in COMC gets a pamphlet to 'try out' for the Math Camp in Maine.

Did anyone actually go there? I'm curious about the level of math they teach there.

Got it last year because of AIME I think, but I didn't go.  Anyways, I know someone (Chen Huang) who knows someone (Andrew Mao) who went there. lol
He didn't recommand people to go there.
Plus, as far as I know, the camp costs like 3000 bucks.  So....ya... ;) .

Er...is this is the big one in the states?  http://www.mathcamp.org?  Or are you refering to the one that Mr. Griffiths talks about sometimes?

I actually went there last summer, I mean the big one in the states :P
It's actually a pretty cool math camp which lasts for over a month and has a variety of activities like hiking and others.
They teach quite a lot of topics too including abstract algebra, projective geometry, number theory, topology and many others with cool sounding names.  I'm sure these stuff never get taught in school, (Canadian schools anyways), and they are more university oriented.  But there's also introductory and olympiad problem solving that teach you mostly how to solve equations and medium-hard inequalities respectively.

Yeah, if you're looking at the big one in the states, Ti (remember her Jenn? )  told me it was totally worth it.

Haven't heard anything about the other one though.

Myself, I might apply to the big one, we'll see.

[quote]Got it last year because of AIME I think, but I didn't go. Anyways, I know someone (Chen Huang) who knows someone (Andrew Mao) who went there. lol
He didn't recommand people to go there.
Plus, as far as I know, the camp costs like 3000 bucks. So....ya...  .[/quote]

I think Andrew went to the math camp in Maine, while the big US math camp is in Colorado. The one in Colorado costs like $\$3000, while the one in Maine costs much less. I remember Andrew said he applied for financial assistance, and only paid something like $\$500. Even without financial assitance, I don't think it will cost over $\$1500

I got a flyer for one in Portland, I assumed for my AIME. circa $3000 for a month. It looks interesting, but I would assume it would be more for grade 11s? After all, I am going off to university to do math in 5 months, and if it is a choice between spending 3000 [even though it could work out to more like 1000] and earning 2000 and getting some work experiance, I would pick the latter.

[quote="jpark"][quote="weileaf"]I think Andrew went to the math camp in Maine, while the big US math camp is in Colorado. The one in Colorado costs like $\$3000, while the one in Maine costs much less. I remember Andrew said he applied for financial assistance, and only paid something like $\$500. Even without financial assitance, I don't think it will cost over $\$1500[/quote]

Umm.. Do you know the reason Andrew doesn't recommend it to others?[/quote]

Oh yeah, forgot to mention.  Dr. Liu told me that at this camp, they often have famous people fly in, but they leave during that same day, and the camp is not as organized as it should be.  He told me no one really leaves having learnt anything, he told me definitely not to apply.

Ah, it's decided then.. I guess we won't see each other in Maine Sarah.. Oh well..  :D

[quote="geoffreynham"]I recently received in the mail a brochure for some math camp at The Ohio State University.  From the looks of the brochure, it's 8 weeks of solid number theory.  That seems rather unusual...  Has anyone heard of it?[/quote]

I assume that you are referring to [url=http://www.math.ohio-state.edu/ross/]Ross[/url].  In any case, I have heard that it is a very good program.  However, while I did not attend Ross, a senior at CC said that she attended it and she liked it enough to recommend it.  I suppose that I should note that she will be attending Harvard next year and did RSI last year.  I hope that's a good enough reference for you  :D


let, there be $ \triangle ABC$.
let, $ \triangle A'B'C'$ be another triangle. with $ B' \in AB$ ext. if necessary and similarly $ C' \in AC$. Also we have that $ AB' \equal{} AB.k$ and $ AC' \equal{} AC.k$. prove or disprove if $ B'C' \equal{} BC.k$

[quote="communist"]Let there be $ \triangle ABC$.

Let $ \triangle A'B'C'$ be another triangle with $ B' \in AB$, extended if necessary, and similarly $ C' \in AC$.

Also, we have that $ AB' \equal{} k(AB)$, and $ AC' \equal{} k(AC)$.  Prove or disprove:  $ B'C' \equal{} k(BC).$[/quote]

Instead of trying to use periods for multiplication, for abbreviations,
and for ending sentences, I did a suggested redo above.  This and
some other changes should help someone read your content
through more readily.


For the last month or so, my PMs that I sent are never received by anybody, although they are notified that I have PMed them.
Also, I receive nobody's PMs except cognos599's and undefined117's (although maybe I can receive other people's, too).
For example, logic812 and MustItAlwaysBeMe have complained about my PM system; apparently they have PMed me, but I did not receive them.
Also, I've PMed cognos599 (multiple times) and undefined117 (once), but they say they never got my PMs.

So, I guess this topic is basically directed to Valentin, or anyone else who can help me.  I know it may not be an easy task to fix this; and it's strange how I can only receive some PMs but not others.

is your inbox full??
hmm i'll pm you and see if you receive it

and try making a poll to see if other people have the same problem you do.
good luck.

No, my inbox is completely empty (although I wish it wasn't!).

And I didn't receive your PM, and I don't think that I will.

EDIT: I received PMs from xoangieexo and levans, and I replied.
Hopefully, you two have received those replies.

Could any admin help out here (don't worry levans, I know that you're an admin too)?

[quote="MathAndKnowledge"]So, I guess this topic is basically directed to Valentin, or anyone else who can help me.  I know it may not be an easy task to fix this; and it's strange how I can only receive some PMs but not others.[/quote]PM me. We might be on something strange here (that is in case you're not wasting our time and deleting the message from outbox, after the email notification has been sent ...).

I was going to edit my previous posts, but for some reason it won't let me.

Anyways, I have breaking news.

I am not very smart.  I didn't know about the whole outbox thing; I just (for the last month or so, coincidentally [note the sarcasm]) decided to delete outbox messages, because I thought they weren't necessary; the post had been sent, hadn't it?  But I guess that I was wrong, and now I know why I haven't been able to PM for the last month or so.
I've already told this to Valentin and levans, so I don't think much further help is necessary.
All I need to do now is figure out why people like logic812 and MustItAlwaysBeMe can't PM me, but others can.  I'll work it out between them and me, so thanks for the concern and willingness to help... now this topic should die.


prove that a^a > 1/2 with a is a positive real number

You can use calculus, it's pretty straightforward.

indeed,min( x.ln(x) ) = ?
derivate: ln(x) + 1 = 0 so x = e^-1
x^x = 0,6922 which is clearly greater than 1/2

I can easily use calculas to show that the minimum is (1/e)^(1/e), for which the numerical value is that given by Peter, but is there anyway to get the result without pulling out a calculator? My impression (I may be wrong, I have very limited experience in Mathematics Olympiads) is that often you are not allowed to use a calculator.

you are NEVER allowed to use a calculator I think :? not in any round

but you don't need a calculator to prove that (1/e)^(1/e) is larger than 1/2 do you? :D

Hmm... I might, actually. I'm REALLY bad at inequalities stuff.

I honestly haven't a clue how to go about it.

well i just know the value of e by heart, but you can easily see it, as elseway the thing to prove won't be correct :D I know it's weak but it works

I have seen many literatures that start out with the following inequality regarding the value of e:

Prove that 2<e<3 without knowing in advance what the value of e is.

To help out those who needs a little more information or whoever wants to know what I mean, think about the properties of e. What is so special about e? How is e defined? Think about some rough approximations. Again, no calculus is needed.  :)

Well e is defined in two ways:

(1) e= \lim (1+1/n)n  as n->oo and

(2) e=1+(1/1!)+(1/2!)+(1/3!)+...

Let the sequence {en} be defined by en=(1+1/n)n , for all naturals n>0. You can easily show that this sequence is increasing.

The first definition can be shown easily (I will only post it if asked).

Using (1), 2=e1<en<e.

Using (2), e=1+(1/1!)+(1/2!)+(1/3!)+... < 1+(1/1)+(1/2)+(1/22)+... =2+((1/2)/(1-(1/2)))=2+1=3.

So we have shown 2<e<3. But, how can we use this fact to prove that (1/e)1/e>1/2 ? Simple.
Since e>2, 2e>22=4>3>e. So 1/e>1/2e. Both sides of the inequality are positive numbers,
then (1/e)1/e>1/2. No calculators, as I promised. :D

[quote="3X.lich"] What is so special about e? How is e defined? [/quote]
my idea is that e is so special because it is the only positive number for which the function ex derivated remains the same. This can also hold for a definition. Try to prove using this definition the other properties of e (ie being the limit of the sequence en described above by 3x.lich for example).


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\geq \frac{2a}{3a^2\plus{}bc}\plus{}\frac{2b}{3b^2\plus{}ca}\plus{}\frac{2c}{3c^2\plus{}ab}$.

No one can solve?   :(  :?:

[hide="to arqady"]arqady, please post your's or Vasc's solution.  [/hide]

[quote="Allnames"]I think it can be killed by SOS[/quote]
Are you sure? Maybe after expanding? :wink:

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq \frac {2a}{3a^2 \plus{} bc} \plus{} \frac {2b}{3b^2 \plus{} ca} \plus{} \frac {2c}{3c^2 \plus{} ab}$.[/quote]

Well, this form is not that difficult, with some efforts, I can do it through expanding.
I also prove this one, which is much fun to expand.
$ a,b,c$ are positive number,
$ \frac{c^3}{a\plus{}b}\plus{}\frac{a^3}{b\plus{}c}\plus{}\frac{b^3}{a\plus{}c} \geq \frac{2a^4}{3a^2\plus{}bc}\plus{}\frac{2b^4}{3b^2\plus{}ac}\plus{}\frac{2c^4}{3c^2\plus{}ab}$


quadratics
number theory unsolved
number theory

Find all real numbers a such that all solutions to the quadratic equation $ x^2 \minus{} ax \plus{} a \equal{} 0$ are integers.

Let $ m, n$ be the roots of the original equation (not necessarily distinct) WLOG, let $ m\geq n$
Then $ m\equal{}\frac{a\plus{}\sqrt{a^2\minus{}4a}}{2}, n\equal{}\frac{a\minus{}\sqrt{a^2\minus{}4a}}{2}$.
Since $ m, n$ are integers, $ m\minus{}n \equal{} \sqrt{a^2\minus{}4a}$ is also a positive integer, name it as $ k$.
So $ a\equal{}2m\minus{}k$ is also an integer.
Now, $ a^2\minus{}4a \equal{} k^2$
$ (a\minus{}2)^2 \minus{} k^2 \equal{} 4$
$ (a\minus{}2\plus{}k)(a\minus{}2\minus{}k) \equal{} 4$,
as $ a,k$ are integers, $ (a\minus{}2\plus{}k)$, $ (a\minus{}2\minus{}k)$ have the same parity.
So (a-2+k) = 2, (a-2-k) = 2 or (a-2+k) = -2, (a-2-k) = -2.
Solve the two simultaneous equations we can get $ (a,k) \equal{} (4,0)$ or $ (0,0)$.
So $ a \equal{} 0, 4$


Let n > 1 be a positive integer, let E={+1,-1}^n, let s=(s_1,...,s_n) in E,
let f_s:=s_1 sqrt(x+1)+...+s_n sqrt(x+n) and let N(x)=prod_{s in E} f_s.

Prove that N(x) is a polynomial in Z[x] and determine its degree in x.

Good luck !

_________________
The Owls Are Not What They Seem

> is this an open question, or a own questions to which you have the answer to?

I have a partial answer to this question. I did not post it to the
"Unsolved Problems" section, because I do not consider that it matches
the definition "Problem you couldn't solve and to which you know that
there is a solution (i.e. a problem from a contest, etc.) but you
don't know it". I had the same doubt while thinking to send it to the "Proposed & Own
Problem" section, whose definition is "Problems that you have already
solved and you are interested in second opinions or solutions"...
By the way, what exactly your definition of an "open question" is ?

Alin.

an open question is obvioulsy a question that is not closed :)

however a more detalied description could be the following:

[i]An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not. [/i]

I see. Maybe you should put down this definition at
http://www.mathlinks.ro/viewforum.php?f=3
as you did for the other themes.

_________________
The Owls Are Not What They Seem


Does Alcumus get harder and harder as you progress?
As in, does it get to about AIME / just below olympiad level difficulty?

It gets to easy-medium AIME level but no harder than that.

I thought that Alcumus problems didn't have a specific level :huh:

All problems have a level :)

The problems start out at easy MATHCOUNTS level and progress to easy-medium AIME level.

Thanks for the replies.

Btw, has anybody tried out Mildorf's mock AIME's?
Should check out his [url=http://web.mit.edu/~tmildorf/www/]site[/url]

The questions on his mock exams are really hard.. supposedly even three times harder than regular AIME questions.. is it a good idea to prepare for contests by solving problems that are harder than the ones on the actual contest??

i'd say yes
if you know how to solve problems that are harder than problems you'll be encountering, most likely, you'll know how to solve the problems you will encounter

but right before the competition, i suggest doing a little of the competition problems, to get to know about how hard it is


calculus
derivative
function
calculus computations

I'll begin with the the lemniscate, who's function is otherwise known as  $ (x^2 \plus{} y^2)^2 \equal{} x^2 \minus{} y^2$.

Now, once I begin the differentiation process, I am given this:

$ 2(x^2 \plus{} y^2)(2x \plus{} 2y\frac {dy}{dx}) \equal{} 2x \minus{} 2y \frac {dy}{dx}$

How would one go about trying to remove the other dy/dx? How should I treat it as?

hello, i have $ 2(x^2 + y^2)(2x + 2y\cdot y^') = 2x - 2y\cdot y^'$.
Sonnhard.

You have not differentiated correctly.  Using chain rule, you should get $ 2(x^2+y^2)(2x+2yy')=2x-2yy'$...

[quote]You have not differentiated correctly. Using chain rule, you should get $ 2(x^2+y^2)(2x+2yy')=2x-2yy'$[/quote]

Uh, how is that different from what I posted?

[quote="Tyl"]The $ \minus{}$ sign.[/quote]

Sorry for the typo. However, I still don't know how to treat the dy/dx parts.

hello, your answer is still not ok.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, you answer is still not ok.
Sonnhard.[/quote]

Would you mind explaining how?

hello, when you differentiate $ (x^2+y(x))^2$ with respect to $ x$ you must use the chain rule, namely
$ 2(x^2+y(x)^2) \cdot(2x+2y(x)\cdot y^'(x))$
Sonnhard.

I changed my post accordingly. Now, how would I treat the dy/dx parts of the equation?

hello, no it isn't, your left hand-side is
$ 2(x^2\plus{}y^2)\cdot 2x\plus{}2y\frac{dy}{dx}$
and it must be
$ 2(x^2\plus{}y^2)(2x\plus{}2y\frac{dy}{dx})$
Sonnhard.

let $ y'$ denote $ \frac {dy}{dx}$

differentiating both sides of the equation w.r.t x using the chain rule, we get

$ (x^2 \plus{} y^2)^2 \equal{} x^2 \minus{} y^2$

$ \implies 2(x^2 \plus{} y^2)(2x \plus{} 2yy') \equal{} 2x \minus{} 2yy'$

$ \implies 2(x^2 \plus{} y^2)(x \plus{} yy') \equal{} x \minus{} yy'$

rearrangement

$ yy'(2x^2 \plus{} 2y^2 \plus{} 1) \equal{} x \minus{} 2x(x^2 \plus{} y^2)$

$ y' \equal{} \frac {x(1 \minus{} 2(x^2 \plus{} y^2)}{y(1 \plus{} 2(x^2 \plus{} y^2)}$ :)


inequalities
geometry
function
number theory
greatest common divisor
least common multiple
trigonometry

[quote]I think the cut-off was pretty high this time. Delhi seniors have performed extremely well because all 6 selected crossed 71. However, I expect the same kind of performance from seniors from other states like TN.

Agr_94 and nuclear alchemist bought up the topic of IMOTC attendees writing RMO. I completely agree with you guys. It is not nice at all that they write RMO and block seats. People who attend IMOTC are given the chance to write INMO directly even if they don't clear postals. Some people write RMO with the aim of getting perfect score, which is not good because in the process other 12th standard guys who worked so hard are being deprived of a chance to write INMO.

Some observations about the RMO and INMO exams.
1. The Sophie Germain identity was used even in INMO 2008 number theory question.
2. The geometry question on the INMO or RMO has almost always involved a construction of some sort.
3. The greatest integer function was used in INMO 2007 and INMO 2009.
4. Any inequality in the past three INMO's has been solvable by normalization.

Someone should start a new thread to discuss INMO strategy or something like that. I can't think of anything else right now. But the INMO questions are following a definite pattern over the past three years. So keep that in mind while preparing. All the best for INMO to all selected  Better luck next time for others.

And one last thing about cut-offs before signing off:
The INMO 2007 paper was easy compared to a usual standard INMO paper. As a result the cut-off was high, something like 35 or 40. Because this RMO was easier it completely makes sense that the cut-off be higher.
[/quote]

here is the new topic.... my observations:

1)INMO 2009 Q1 was very ingenius(at least their proof)...i dont mind Q's involving ingenius proofs..those which involve trival and knowledge-based identities worry me....
2)i dont know what normalization is ...and i've started preparing for INMo today itself...so i guess i'm not ought to..so can some1 PLZ tell me what it its??
3) INMO 2007 inequality Q6 maybe solved using normalization(which i dunno) but i am sure its solvable using simple convertion (taking xy=a , yz=b , zx=c etc) and then using AM-HM inequality...
4) i would love to have a INMO 2005 alpha beta one like Q in INMO'10...it was soo very awesm and required ingenuity rather than knowledge....i wud LOVE to use my brain rather than any concepts pre-stored in it...

[quote="Agr_94_Math"]Well from experience of having written INMO 2009, I would like to say you many things:

1. Do not go by any kind of pattern oriented preparation. befuddlers is a guy more experienced than me (He wrote INMO 2007 and 2008) but I disagree in this regard. Geometry problems in INMO are very nice and not the same construction to look for etc. Construction is done using intuition at that moment and that hugely depends on your temperament and skill at that time. Just try to solve many geometry problems and you will be in a good mind-set. Don't try to hunt for pattern in the kind of constructions the offiical solutions use in the INMOs.  Same goes for the greatest integer function problems. I think Ashwath Rabindranath got an excellent solution for the INMO 2007 gif (Sahil it was 2007 not 2005). The idea is completely different from what was required in INMO 2009. It depends and varies from problems  to problems.  So does it go for the inequalities. Befuddlers, I think the INMO 2007 inequality was actually made more round about by the normalisation which was not required after all. Fermat point or full expanding and killing it by AM GM was enough I think (Believe me I did the 2nd one just for fun-- it took around 25 minutes that is all). INMO 2009 was a good example of a normalisation but even then the problem was not very obvious. And solutions without normalisations were more elegant too.
As for Sophie Germain Identity, well, it is indeed a prowerful factorisation but not the most reliable for tough problems. You see , if it is not a generalised one, at times you have got to make use of the numbers and not the identity alone. The numbers can play a role in facotrisation. There are several examples . If you want, I can give you some.

2. You have to try and get exposure to tougher problems. Try to make a strategy on the basis of which areas you are good at and which you are not so strong. Since in the INMO, you get problems from all, (not like this year's RMO where they did not have a single algebra or a good geometry ).., try to strengthen your strong areas. For me, it appears that you are quite good in algebra -so try to make yourself strong even in number theory (mostly requires algebraic tricks).

3. Try to gain as much experience by trying problems than seeing solutions. Helps you more.

4. I also did not like one comment of yours somewhere when you said :" refer to challenge and thrill Pg 63 some exercise problem: Know one thing that it is not possible to remember each and every problem from every book and resources you see. Instead try to recollect ideas. In that case, I shall tell you that any geometry problem can be solved using some problem of challenge and thrill but the problems in challenge and thrill are very much harder than INMO level.

5. Try to solve more and more problems instead of learning more things(which you may feel like). Learn some necessary things (that you may already be knowing: if not, it is like the basic things that people use in olympiad-defitnely not at the level of mathlinks users. You may ask if you want in PM what they are. ).

6. Develop your thinking by solving some tough ones too.

All the best and congratulations for clearing RMO.[/quote]

seriously arvind...i was just saying see the prob in CTPCM and commenting on the ~ty with RMO prob..i would be the last one to mug or remember....and i follow ditto.recollect ideas... :D ..and sure i was talking of inmo 2005 Q2 alpha beta one...n BTW i dont have any strongs and weak....just depends whether it clicks or not..yaa sure i have most experience in inequalities...and sure i love geometry (evn if i cnt do every problem)...so i wont put one over another..btw NT is as awesm as others..BTW i'm happy u say CTPCM is above inmo :) ;)..that makes me a relieved man...and BTW i think inmo 2007 Q1..was seriously above any CTPCM ....look at the way they have used f(t) in the end ..it cud NEVER come to my mind :oops:  :oops: ..and arvind PLZ tell me what the real basics u talk about (in ineq i think they mean strengthing AM-GM , rearrangement , jensen cauchy etc)

As far as inequalities are concerned, don't go by the  level of the people here in mathlinks forum of Inequalities Olympiad Section. It is too too too good level compared to INMO, infact too much higher than IMOTC's and IMOs. Several people are so adept at solving inequalities around the world, they have strong techniques and ideas that are gained by experience and intuition. As regards inequalities, know stuff like Cauchy Schwatrz, Means, Rearrangement , Applying AM-GM by weights, etc).

and for NT , geometry(this is the real problem point)

[quote="Agr_94_Math"]Geometry: Ceva, Menelau, Ptolemy, and a few basic properties of triangle knowledge,quadrilaterals, and some theorems on tangents and secants is enough. (Tis may sound too much like a portion but not like that). More than knowledge, in geometry, trying more and more problems and doing soem things yourself will give you a better edge.

NT: Basic stuff like Modulos, GCD and LCM, Fermat's little theorem, Euler's Theorem, Wilson's Theorem, Chinese Remainder Theorem.[/quote]

Plz tell me 1 thing... I have never seen application of menelaus or ceva's ...in any inmo..what i HAVE seen is application of sine / cosine rule...pythagorus etc etc..can u plz give me an example frm inmo where these theorems may help (it is a humble request ..no offences  :roll:  :roll:  :lol: ....

[quote="sahilsharma94"]

Plz tell me 1 thing... I have never seen application of menelaus or ceva's ...in any inmo..what i HAVE seen is application of sine / cosine rule...pythagorus etc etc..can u plz give me an example frm inmo where these theorems may help (it is a humble request ..no offences  :roll:  :roll:  :lol: ....[/quote]
Well then you probably havent looked too hard....just see the official solution of q1 of INMO 2003

can any1 briefly and briskly explain invariance principle?

Oops! Sorry guys, looks like I didn't convey what I wanted to say in the right spirit.

Okay, just to clear things up:
1. When I said "look for a strategy," I meant something more like prepare and practice problems from the specific areas I mentioned. For example, people say greatest integer function problems are very easy, but I don't think anyone except Akashnil in this forum solved the GIF problem in INMO 2009.(I am only talking about the people who wrote INMO 2009 officially here). And of course constructions will be based on the flavor of the problem, but constructions I think are extremely important to solve some geometry problems.

Perhaps also knowing "higher concepts" like homothety and transformations in geometry will help. However, INMO tends to focus on angle chasing and other relatively simple methods to solve geometry problems.

And as for inequalities, there are definitely more elegant methods than normalization, but normalization is almost a sure success yielder in the inequalities on INMO because INMO inequalities are mostly symmetric (thus can be normalized).

Combinatorics on INMO 2008 and 2009 were combinatorial geometry problems( I think). But knowing bijections and recursions should help.

INMO problems tend not to ask NT problems with straight out applications of Euler's theorem etc, but gcd and other related stuff do play a part in the problem(almost always)

@sahil: As the name suggests "Invariance" means that some parameter remains constant/doesn't change during a specific operation/transformation. That parameter is said to be "invariant." You could look into Problem solving strategies by Engel for some good problems.

In general INMO problems don't require tremendous preparation. Good angle chasing skills and basic knowledge in other areas should be enough.

So, good luck guys! :D

[quote="befuddlers"]
Perhaps also knowing "higher concepts" like homothety and transformations in geometry will help. However, INMO tends to focus on angle chasing and other relatively simple methods to solve geometry problems.
[/quote]
I don't think just "knowing" these concepts will help....you may need some practice....(but I think that's what you meant in the first place...on second thoughts :) )

[quote="befuddlers"]
And as for inequalities, there are definitely more elegant methods than normalization, but normalization is almost a sure success yielder in the inequalities on INMO because INMO inequalities are mostly symmetric (thus can be normalized).
[/quote]
You may want to replace "symmetric" by "homogenous"

[quote="befuddlers"]
In general INMO problems don't require tremendous preparation. Good angle chasing skills and basic knowledge in other areas should be enough.
[/quote]
Sorry....but I don't think just basic knowledge will suffice...

[quote="sahilsharma94"]aur yeh hai rijul ka ek aur successful katta  :rotfl:  :rotfl:  :roll:[/quote]
 :P HEHEHE...!!!

[quote="Rijul saini"][quote="nuclear_alchemist"]Been long since we had a aops fight :P[/quote]
Bhadka mat be.....[/quote]

unfortunately all ppl out here wud know and agree that the pen is mightier than the sword..so no fights...sorry [size=150]nuke DEAL[/size]

Sorry for the typo. Normalization can be used for homogenous equations.

Some techniques of normalization can include restraining x,y,z in an inequality with the condition
$ x \plus{} y \plus{} z$ = 1 and so on

Hey all,

Happy New Year!  :)

Hope your INMO preparation is going well.

Here are a couple of websites which are really good.

http://mathcircle.berkeley.edu/
Go to the archives.

The IMO Compendium website is also really helpful.

Good luck!

No not really.
I mean I didn't refer to it much.
Maybe for starting complex geometry and the idea of extremal and invariance but that is all.
I believe PSS is now old for contemporary olympiads.

Others' opinions may differ of course.

[quote=Hrithik007]Can someone please explain what normalisation is?[/quote]

You can find it in any standard inequality text, for example 'Secrets in Inequalities'..but put briefly and rougly it is the following:

Say you have a homogenous inequality in 3 variables, i.e., you have $f(a, b, c) \ge 0$ and $f$ is such that $f(ka, kb, kc)=k^{d}f(a, b, c)$....for some constant $d$....Now if you are given that $a, b, c$ are positive.. then the given inequality is true iff $\left( \frac{1}{a+b+c}\right)^df(a, b, c)) \ge 0$ which by the homogenous condition is same as $f(\frac{a}{a+b+c}, \frac{b}{a+b+c}+\frac{c}{a+b+c}) \ge 0$...since $\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$..All this means that you could have assumed from the outset that $a+b+c=1$....

Now what you did with $a+b+c$ can be done with any homogenous function..meaning you can impose a condition like $g(a,b,c)=1$ for any homogenous function $g$ of degree say $r$ (Do you see why?)..eg. you can assume $ab+bc+ca=1$ and so on..but one at a time of course(it is possible to apply multiple conditions in some cases..say if the sets of variables are disjoint..e.g. in one proof of Cauchy's inequality..)...

Now, try and prove your favourite homogenous inequality by normalisation..


Buna ziua,

Cum se demonstreaza ca sinn nu are limita cand n tinde la +infinit?

Mai general, daca x(n) este un sir care are limita +infint cand n tinde la +infinit, si f(x) este periodica de perioada principala T, cum se demonstreaza ca sirul f(x(n)) nu are limita cand n tinde la +infinit?

Va multumesc si cer iertare dar nu am stiut sa bat in Latex textul

Daca $\sin n\rightarrow L$, atunci $\sin (n+1)-\sin (n-1) =2\sin 1\cos n \rightarrow L-L=0$, deci $\cos n \rightarrow 0.$ Atunci $\sin 2n =2\sin n \cos n \rightarrow 0,$ de unde rezulta $L=0.$ Contradictia se obtine observand ca
\[1=\sin^{2}n+\cos^{2}n \rightarrow 0.\]
(demonstratia e din cartea prof. Siretchi)

Rezultatul "mai general" nu este adevarat.

Domnule Bogdan Enescu, va multumesc pentru solutie!

Numai bine,

[quote="enescu"]Daca $\sin n\rightarrow L$, atunci $\sin (n+1)-\sin (n-1) =2\sin 1\cos n \rightarrow L-L=0$, deci $\cos n \rightarrow 0.$ Atunci $\sin 2n =2\sin n \cos n \rightarrow 0,$ de unde rezulta $L=0.$ Contradictia se obtine observand ca
\[1=\sin^{2}n+\cos^{2}n \rightarrow 0. \]
(demonstratia e din cartea prof. Siretchi)

Rezultatul "mai general" nu este adevarat.[/quote]

[color=red]This a great and inspirated solution(!!!) , which I see for first time.Untill now I knew only the solution based on the negation of the definition of the limit ![/color]
[u] Babis[/u]


Let $ ABC$ be a right angled triangle where $ \angle C=90^{\circ}$ and $ CD$ is the altitude. Let $ K$ be a point in plane. If $ O$ is the circumcenter of triangle $ ABK$, prove that $ AO$ is perpendicular to $ DK$ if and only if $ AC=AK$. :)

we know that $ KD \bot AO \iff KA^{2}-KO^{2}=DA^{2}-DO^{2}$
i.e. $ KA^{2}-DA^{2}=KO^{2}-DO^{2}$
but according to the power of point $ D$ to circle $ (O)$ we have:
$ |DA.DB|=|DO^{2}-OK^{2}|$
$ \Rightarrow \boxed{DA.DB=OK^{2}-DO^{2}(I)}$
and also we have $ \triangle ADC \sim \triangle CDB$ so we have $ DA.DB=DC^{2}$
thus according to $ (I)$ we get that:
$ DC^{2}=KO^{2}-DO^{2}$
thus we have:
$ KD \bot AO \iff KA^{2}-DA^{2}=DC^{2}$
but in the right angled $ \triangle ACD$ we have $ DC^{2}+DA^{2}=AC^{2}$
thus we have:
$ KA^{2}-DA^{2}=DC^{2}\iff KA^{2}=DC^{2}+DA^{2}=AC^{2}\iff KA=AC$
thus we have:
$ KD \bot AO \iff KA=AC$ as wanted...

Again a nice solution Babak :)
My solution for this one was:

Let $ DK\cap AO=H$ and let $ M$ be the midpoint of $ BK$. we'll prove that $ OMHK$ is a cyclic quadrilateral. Note that because $ ABC$ is right angled so triangles $ ABC$ and $ ACD$ are similar. so we have:  $ AC^{2}=AD\cdot AB$ but $ AK=AC$ so $ \boxed{AK^{2}=AD\cdot AB}$ but this shows the similarity of triangles $ AKD$ and $ ABK$ so we have:
$ \angle AKD=\angle ABK$, so now we can write:
$ \angle DKM=\boxed{\angle HKM=\angle AKB-\angle ABK}$

but we also have:
$ \angle AOM=\angle KOM-\angle KOA=\frac{1}{2}\angle KOB-2\angle ABK$
$ =\frac{1}{2}\left(2\angle ABK+2\angle AKB\right)-2\angle ABK=\boxed{\angle AKB-\angle ABK}$

 as a consequence $ \angle AOM=\angle HOM=\angle HKM$ and the quadrilateral $ OMHK$ is cyclic and $ \angle KHM=\angle KMO=90^{\circ}$ which means $ AO$ and $ KD$ are perpendicular to each other. :)

and now its easy to prove that if $ AO$ and $ DK$ are perpendicular then $ AK=AC$.

I read carefully both solutions. They are really very nice and entirely natural. Congrats to both of you !

  [u]Babis[/u]

midpoint of $ AK$ is $ M$.  $ N$ is intersection of $ AO,DK$. $ AO$ is perpenticular $ DK$ if and only if $ <AOM=<DKA$ and this is true if and only if  $ <ABK=<DKA$ since $ <ABK=<AOM$. $ D$ is between $ A$ and$ B$. therefore $ <ABK=<DKA$equivalenct$ AK^{2}=AD*AB=AC^{2}$ equivalent $ AC=AK$


The graphs of $ x^2 \plus{} y^2 \equal{} 4 \plus{} 12x \plus{} 6y$ and $ x^2 \plus{} y^2 \equal{} k \plus{} 4x \plus{} 12y$ intersect when $ k$ satisfies $ a \leq k \leq b$, and for no other values of $ k$. Find $ b \minus{} a$.

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 68\qquad
\textbf{(C)}\ 104\qquad
\textbf{(D)}\ 140\qquad
\textbf{(E)}\ 144$

[quote="dogseatcheese"]When two equations intersect, we usually equate them. BUt, in this case if we make 4+12x+6y=k+4x+12y, we get the wrong answer. But why?  :?:[/quote]

The two sides of the equation still depend on something else, that is:

$ 4 \plus{} 12x \plus{} 6y \equal{} k \plus{} 4x \plus{} 12y \equal{} x^2 \plus{} y^2$

which is certainly not the same as just

$ 4 \plus{} 12x \plus{} 6y \equal{} k \plus{} 4x \plus{} 12y$

[quote="dogseatcheese"]When two equations intersect, we usually equate them. BUt, in this case if we make 4+12x+6y=k+4x+12y, we get the wrong answer. But why?  :?:[/quote]

The intersection points are trivial in this problem.

[quote="JesusFreak197"][hide="Answer"]Moving the equations around a little, we have $(x-6)^2+(y-3)^2=49$ and $(x-2)^2+(y-6)^2=40+k$. The centers are at $C_1=(6,3)$ and $C_2=(2,6)$, so the distance between the centers is $\sqrt{4^2+3^2}=5$. The first circle has radius $7$, so the second circle will be internally tangent when the radius is $2$, or for $k=-36$, so that is our smallest value. The first circle will be internally tangent to the second when the radius of the second is $12$, so $k=104$. Therefore, we have $b-a=104-(-36)=140\Rightarrow \boxed{D}$.[/hide][/quote]

I don't get why the radius of circle $2$ has to be at least $2$. Anyone can explicate?

Just draw it out, then the answer is pretty obvious. The centers form a $3-4-5$ right triangle. Then the radii of the unknown circle can be expressed by $7 = 5+r$.

Similarly, to get the radius of the larger circle, we need $5+7 = r$ since then two radii and the tangent point are collinear.

Note that $(x-6)^2 + (y-3)^2 = 7^2$ and $(x-2)^2+(y-6)^2=k+40$. So, we have circle $\Gamma_1$ with center $(6,3)$ radius $7$ and circle $\Gamma_2$ with center $(2,6)$ radius $\sqrt{k+40}$. Let $A(6,3)$, $B(2,6)$, and $C=\Gamma_1 \cap \Gamma_2$. We construct $\triangle ABC$. So, we have
\[|AB| = \sqrt{(6-2)^2+(3-6)^2}=5\]
The with law of cosines, we have
\begin{align*}
BC^2 &= AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos \angle BAC \\
k+40 &= 25+49 - 70\cos \angle BAC\\
k &= 34 - 70 \cos \angle BAC
\end{align*}
Because, $-1 \le \cos \angle BAC \le 1$, so
\[-36 \le k \le 104\]
Then $b-a = 104 - (-36)=\boxed{140}$.


integration
logarithms
algebra
partial fractions
calculus
calculus computations

One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not hear the rumor.

a) Write the differential equation that is satisfied by y

>>This part I got,
          dy/dt = ky (1-y)
b) Solve the differential equation

>>I tried integrating the above equation but in the end what I got was y = (Ce^kt)+1

but the book says that the answer is y = y0/[y0 + (1-y0)e^(-kt)]
how did they get this?

Thanks

Just seperate the variables
$\frac{dy}{dt}=ky(1-y)$
$\frac{dy}{y(1-y)}=kdt$
$\int \frac{dy}{y(1-y)} = \int kdt$
Use partial fractions on the LHS.
$\frac{1}{y(1-y)} = \frac{1}{y} - \frac{1}{y-1}$
$\int \frac{1}{y} dy - \int \frac{1}{y-1} dy = \int k dt$
$\ln(y) - \ln(y-1) = kt+C$
$\ln{\left(\frac{y}{y-1}\right) = kt +C}$
$\frac{y}{y-1} = Ce^{kt}$
$y=\frac{Ce^{kt}}{Ce^{kt}-1}$
Since $C$ is just a constant we can divided the numberator and denominator by $C$ to get:
$\boxed{y=\frac{e^{kt}}{e^{kt}+C}}$.
Were there initial conditions given in the problem?
And what does $y_0$ represent?

that was all that was given, no more information was given...   so i do not understand how they got

y = y0/[y0 + (1-y0)e^(-kt)]

The usual thing that $y_0$ would mean would be an initial condition: $y_0=y(0).$

Look a couple of lines up from the bottom of Jimmy's post: he says

$\frac{y}{y-1} = Ce^{kt}$

Let $t=0$ and $y=y_0$. Then we get $C=\frac{y_0}{y_0-1}.$

Going to the next line in Jimmy's answer (where it's still the same $C$):

$y=\frac{Ce^{kt}}{Ce^{kt}-1} = \frac{\frac{y_0}{y_0-1}e^{kt}} {\frac{y_0}{y_0-1}e^{kt}-1} = \frac{y_0e^{kt}}{y_0e^{kt}-y_0+1} = \frac{y_0}{y_0+(1-y_0)e^{-kt}}$


[quote="silouan"]Οποιαδήποτε ομοιότητα με πρόσωπα ή καταστάσεις είναι εντελώς τυχαία ?  :wink:

Λοιπόν τοποθετώ το πρόβλημα G3 IMO SL 1996 στην official version του (πιστέψτε με δεν θέλετε να δείτε την official solution .Πέρνει πολλά όμοια ο άτιμος και δεν μπορώ να την καταλάβω )
(Επίσης δεν ξέρω αν κανείς σας σε άλλη άσκηση έχει δει το βασικό ίχνος του ύψους με F σε άλλη άσκηση .Αλλά να τώρα που το σκέφτομαι πριν από καμιά βδομάδα έλυνα μία με F αλλά δεν θυμάμαι που  :roll:  )

Έστω ABC οξυγώνιο τρίγωνο με $ BC > CA$ . Έστω $ O$ το περίκεντρο , $ H$ το ορθόκεντρο και $ F$ το ίχνος του ύψους από την κορυφή $ C$ .Έστω η κάθετη στην $ OF$ στο $ F$ ότι τέμνει την $ CA$ στο $ P$ . Δείξτε ότι $ \angle{FHP} \equal{} \angle{BAC}$

(Και τώρα εξηγείται γιατί πριν μια εβδομάδα σε αυτό το πρόβλημα που έλυνα κάποια σημεία ήταν άχρηστα  :huh: )
Ευτυχώς πάντως το βρήκα γιατί ένα βράδυ είχα πεταχτεί και είχα πει ότι έχω ξαναδεί λυμμένο το πρόβλημα με πεταλούδα
(see IMO compenium 3rd solution)

***Και μια τελευταία προσθήκη . Η πιο ωραία λύση που έχω δει είναι με τον κύκλο του Euler (να που λυνόταν και έτσι .Άρε Άλκηστη ατυχια...δεν πειράζει όμως keep going)[/quote]

Ανεξάντηλητος!!! Αυτό εξηγεί βέβαια και το πώς το πρόβλημα έφτασε και στο βιβλίο του Cosmin Pohoata και του Virgil Nicula για τις αρμονικές τετράδες. Πάντως πρέπει να επιβεβαιώσεις silouan ότι το είχα υποπτετυθεί από την πρώτη στιγμή, και μάλιστα είχα πει και τους λόγους για τους οποίους το πίστευα...έτσι δεν είναι?

Επίσης μου κάνει εντύπωση αυτό που λες για τον Euler...μόλις πατήσω submit θα πάω να το κοιτάξω. Ρε Σιλουαν αν το ειχες θυμηθεί νωρίτερα, θα ειχαμε πάρει τους πόντους για την Άλκηστη...αν και θα ήταν λίγο δύσκολο με τους Σκοπιανούς...

p.s. Bravo re silouan!kali douleia.


Here is a picture as attachment.
The game is simple : find all the digits such as if you read vertically, horizontally and diagonally (from the top) you have always a prime number.
The prime numbers created have to be distincts.

Ahh, a game I can understand!

Sorta...

When we put numbers in the 3x3 grid, do they have to be digits?  Or can they be numbers like 10573?