I ****************failed**************** the try out.

SRSLY, I didn't even answer 2/3 of the questions and I think of the ones I answered, I got less than 1/2 right.

I'm aiming for D team which I hear there will be this year.

well i'm not in anything of those things.. what excatly do you do in math counts.. :)

H provides all these academic activites???

Math team isn't exactly an uncommon activity.

Varsity A Team
2 12th graders
2 11th graders
1 9th grader

Varsity B Team
2 12th graders
1 11th grader
1 10th grader
1 9th grader

Maximum for A and B team is 2 seniors each so that's why there aren't more.

Exactly. That seventh grader blows at MathCounts too.

what's ur school average for chapter pctces, i get 44ishes
-jorian


I'm kinda stuck here.  :blush:
I have [b]huge[/b] whitespace at the bottom of my document. I tried to insert a table, but it goes on to the next page. I've decreased the top margin, but the only thing that happens is the space at the bottom gets larger and the next text is thrown onto another page.

Thanks. :)

Are you using tabular or array? Have you tried using b or t as detailed in http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#tables?

You can also use \begin{table}[option] where as well as b and t you can use h (=here) and !h (I insist that it goes here).

If still no joy try putting the table inside a minipage whose size can be controlled.

I'm using tabular.
If I were to use \begin{table}[option], where would I place the !h ? Can I place the !h right after the {tabular}?

In general, how would one change the bottom margin? Throughout my whole paper, it is like 5 inches. -_-

'option' should be replaced by b, t, h or !h to get for example [code]\begin{table}[!h]
 \begin{tabular}{cc}
 body of table
 \end{tabular}
more words can go here
\end{table}[/code]
tabular only takes the b or t options.

LaTeX lets you set all sorts of things that affect the margins but far and away the simplest is to use the geometry package (nothing to do with studying geometry :)).
[code]\usepackage[margin=1.5in]{geometry}[/code] produces a margin of 1.5in all the way round (left, right, top & bottom). [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]Geometry manual[/url]

Sweet! Geometry package works. :)
Thanks a lot man. I should be all right now.

Too bad my .dvi file is crashing.  :rotfl:


geometry
calculus
integration
function
analytic geometry
linear algebra
matrix

The hyperbolic space $ H_n$ is the unit disc $ \{(x_1, . . . , x_n) |\sum_i x^2_i < 1\}$ with the hyperbolic metric $ g_H \equal{} \left(\frac {2}{1 \minus{} \sum_i x^2_i}\right)^2g_E$, where $ g_E$ is the Euclidean metric. Consider the parametrized surface $ S \equal{} F(D^2)\subset  H^3$, where $ F(u, v) : \equal{} (u, v \minus{} u, v)$ and $ D^2 \equal{} \{(u, v) | u^2 \plus{} v^2 < \frac {1}{4}\}$. Let $ \Omega$ be the volume form on S associated to the metric, use Stokes theorem to express area of $ S$ as line integral over $ \delta D$.

This is one of the math PhD prelims exam question that is asked previous year in our university in Diff Geometry. Any help would be appreciated.

Especially, my concern is what is metric? Isnt it suppose to be a symmetric definite bilnear form on the tangent space? But, in above example $ g_H$ seems to be function of only one variable?

This is sometimes called a conformal change of metric. The symmetric bilinear form is $ g_H(x)\sum dx_i^2$. Another way to put it: in terms of coordinates, a Riemannian metric is defined by a positive definite matrix. Here we have the scalar matrix $ g_H(x)I_n$.


My marks are (from P1 to P6 out of 10 with half-marks):
0.5; 5.5; 4.0; 1.0; 3.0; 5.0

And will you share your opinions?

may I know if you are the russia contestant?

Yes, RUS6 in Mexico and RUS4 in Slovenia.

Whoa. Then u are the guy who got 41 in Mexico and 42 in Slovenia. :o That's insane. Great job! :thumbup:

I assume you are grading beauty rather than difficulty, since the latter would be more appropriately evaluated on an upwardly open scale...

Anyway, my opinions:

1) 1.0 (a quickie without mathematical significance, appropriate as a warm-up exercise in a book)
2) 3.0 (not pretty interesting question, and not quite conceptual solution)
3) 5.0 (nice inequality, not that nice as an olympiad problem because of the lot of computation involved)
4) 0.5 (spam, takes about 6 times as long to write up as to solve)
5) 6.0 (very good competition problem)
6) 7.5 (I can't resist giving it a great score because of its beauty, but it is not really fair as an olympiad problem because its difficulty decreases significantly with theoretical knowledge)

By the way, congratulations, Alexander!

  darij

I have to say that I agree with Darij for the marks above !
As for the sixth problem, well there have been many such problems that could be solved more easily with the help of advanced knowledge [2003/6 for instance].The solution presented [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101488]here[/url] reveals its beauty[2006/6] more than the official solutions did... :)


Perhaps to avoid the repetition of questions, I suggest making the questions dynamic.

The parameters of the question should be programmed to have random values, or a fixed value chosen from a pre-specified array.

This will not only increase the replayability of the game, but ensure that the chance that a person answered a question first due to an impressive memory is low.

Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179861]Question Variety[/url] topic.


calculus
integration
trigonometry
function
symmetry
complex analysis
calculus computations

Evaluate
$ \int_{0}^{\pi}\sin^{4}(x\plus{}sin 3x) dx$
 :)
Thanks in advance for your nice solution

[quote="Rose_joker"]Evaluate
$ \int_{0}^{\pi}\sin^{4}(x \plus{} sin 3x) dx$
 :)
Thanks in advance for your nice solution[/quote]

Bessel J Function have:
  $ J_n(x) \equal{} \frac{1}{\pi} \int_{0}^{\pi} \cos (n \tau \minus{} x \sin \tau) d\tau$.

So... do you have any way to make that relevant, or are you just stating a random fact. That's not at all the same inside, where it matters.

The obvious ideas of symmetry don't go anywhere; this is neither even nor odd around its center, and not particularly close to either. Antiderivatives are obviously hopeless. What's left? This is still a nice interval, and I'll look for complex analysis methods.

First, $ \sin^4$ is even and $ x+\sin 3x$ is odd, so $ \int_{-\pi}^{\pi}\sin^4(\theta+\sin 3\theta)\,d\theta=2\int_0^{\pi}\sin^4(x+\sin 3x)\,dx=2I$. Let's turn that into an integral around the circle $ z\e^{i\theta}$:
$ 2I=\int_{-\pi}^{\pi} \frac1{(2i)^4}\left(e^{i\theta+i\sin(3\theta)}-e^{-i\theta-i\sin(3\theta)}\right)^4\,d\theta$
$ 32I=\int_{-\pi}^{\pi}\left(e^{i\theta+i\sin(3\theta)}-e^{-i\theta-i\sin(3\theta)}\right)^4\,d\theta$
Substitute $ z=e^{i\theta}, d\theta=\frac1{iz}\,dz$:
$ 32I=\int_{|z|=1}\left(z\cdot e^{(z^3-z^{-3})/2}-z^{-1}e^{(-z^3+z^{-3})/2}\right)^4\frac{1}{iz}\,dz$
Writing $ w=z^3-z^{-3}$, that becomes
$ 32i\cdot I=\int_{|z|=1}\left(z e^w-z^{-1}e^{-w}\right)^4\cdot z^{-1}\,dz$
$ 32i\cdot I=\int_{|z|=1}z^3 e^{4w}-4z e^{2w}+6z^{-1}-4z^{-3}e^{-2w}+z^{-5}e^{-4w}\,dz$
The function we're integrating has one singularity inside the circle, at the origin. We need to calculate the residue there. Unlike must residue problems, this one is an essential singularity rather than a pole. The residue of the $ 6z^{-1}$ term is obviously $ 6$, but the other terms require a closer look at the Laurent series for $ e^{kw}$.

To calculate that Laurent series, we use the power series for the exponential. $ e^{kw}=1+kw+\frac{(kw)^2}2+\frac{(kw)^3}{6}+\cdots$
$ =1+\frac{k}{2}(z^3-z^{-3})+\frac{k^2}{8}(z^3-z^{-3})^2+\frac{k^3}{48}(z^3-z^{-3})^3+\cdots$
Any particular term in that series is an infinite sum. Very nasty. But wait- we don't actually have to calculate those terms. Only the $ z^{3j}$ terms are nonzero; when we multiply by $ z^3$, $ z^1$, $ z^{-3}$, or $ z^{-5}$, the $ z^{-1}$ term will still be nonzero. All four of the terms with exponential pieces have zero residue. That leaves
$ 32i\cdot I = 2\pi i\cdot 6$, and $ I=\frac{12\pi}{32}=\frac{3\pi}{8}$.

Numerical approximation gives an integral of approximately $ 1.17809725$, matching the above answer.

Now, does someone have a more elementary approach? It's already rare for a high school contest to bring in calculus, and residue calculations at an essential singularity are really way out there.

[quote="kunny"]I got also the answer by basic manipulation. :lol:

$ \sin ^ 4 \theta \equal{} \frac 38 \minus{} \frac 12 \cos 2\theta \plus{} \frac 18\cos 4\theta$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=313920]to be continued to this[/url][/quote]
Can you explain more in detail?
Thanks in advance.

is it really high school math problem in Thaîland ?

[quote="J.Y.Choi"]$ \sin^4{\theta} \equal{} \left(\frac {1 \minus{} \cos{2\theta}}{2}\right)^2 \equal{} \frac {1}{4} \minus{} \frac {1}{2}\cos{2\theta} \plus{} \frac {1}{4}\cos^2{2\theta} \equal{} \frac {1}{4} \minus{} \frac {1}{2}\cos{2\theta} \plus{} \frac {1}{4}\frac {1 \plus{} \cos{4\theta}}{2} \equal{} \frac {3}{8} \minus{} \frac {1}{2}\cos{2\theta} \plus{} \frac {1}{8}\cos{4\theta}$.[/quote]
No, no I didn't mean this. I want to know how to do the integral after knowing the identity.


Please decode. You may cooperate, and PM your solution attempts to me.

Is there a mistake with the 4th vertical line?

Yeah there is - it should have 2 more spaces.

Also, there is a mistake in the fourth box. The font, Tahoma, was not intended to be Tahoma. I have changed that as well.

That location would be Eddy St, or on the abc car company or something?

Meh, this is dead; the answer is as follows:

[hide]The dividers spell INITIAL (used to be INIRIAL whoops)

[hide="Individual Answers"]So, we take each answer's initials. The answers are as follows:
up up down down left right left right b a is a variation of the KONAMI CODE
At the bottom of the second box is a RADICAL or ROOT sign.
There are 25+9+4+1=39 dots representing protons. The element is YTTRIUM.
The new font is PALATINO (linotype), bolded.
Upon filling in the fifth box you get the word TRANSPARENT.
The sixth box is a picture of the ODYSSEY (^2).
The last box shows coordinates for SAN FRANSISCO.

Combining first letters of each answer, we get KRYPTOS.[/hide][/hide]


induction
number theory proposed
number theory

$F_{n}=F_{n-1}+F_{n-2}$ is the $nth$ term in the  $Fibonacci$ serie.

Lucas numbers are define as follow :
$L_{n}=F_{n+1}+F_{n-1}$

Prove : $F_{2n}=F_{n}*L_{n}$

I don't get your clue...What you mean for it?...I took that excercice from the induction section of a book on Number Theory...

I think the best way to solve this problem is calculating$L_{n},F_{n}$
I think the proof by this way may be much shorter .


calculus
integration
parameterization
real analysis
real analysis unsolved

[quote="harazi"]Find for any real numbers $a,b$ the value of $\int_{-\infty}^{\infty}{\frac{e^{iax}dx}{x^2+b^2}}$[/quote]Fourier transform of $\frac{1}{x^2+1}$, up to scaling. It's in the tables.

Or, if you forgot it, use contour integration and the residue theorem  to get the answer in next to no time...

Let's make it more entertaining. Instead of having $a$ and $x$ be real numbers, let's have each of them be members of $\mathbb{R}^n$ for some $n>1.$ Now compute this:

$\int_{\mathbb{R}^n}\frac{e^{i(a\cdot x)}\,dV(x)}{(|x|^2+1)^{\frac{n+1}2}}$

If $n=1,$ then this is essentially the problem harazi posted - which is why I suggested making $n>1.$

(If you're more comfortable with a $2\pi$ in the exponent of the exponential, then put it there.)

(By $dV(x),$ all I mean is standard $n$-dimensional measure, or $dx_1dx_2\cdots dx_n.$)


Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that
\[ \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2d}\plus{}\frac{c}{d\plus{}2a}\plus{}\frac{d}{a\plus{}2b}\geq1\]

All sums are cyclic.

$ \sum\frac{a}{b\plus{}2c}\ge \frac{(a\plus{}b\plus{}c\plus{}d)^2}{ab\plus{}bc\plus{}cd\plus{}da\plus{}4(ac\plus{}bd)}$

It remains to show that $ (a\plus{}b\plus{}c\plus{}d)^2\ge ab\plus{}bc\plus{}cd\plus{}da\plus{}4(ac\plus{}bd)$

But $ (a\plus{}b\plus{}c\plus{}d)^2\ge 2(ab\plus{}bc\plus{}cd\plus{}da)\plus{}4(ac\plus{}bd)$

$ \iff a^2\plus{}b^2\plus{}c^2\plus{}d^2\ge 2(ac\plus{}bd)$

Which follows from $ a^2\plus{}c^2\ge 2ac$ and $ b^2\plus{}d^2\ge 2bd$.

Equality holds if $ ab\plus{}bc\plus{}cd\plus{}da\equal{}2(ab\plus{}bc\plus{}cd\plus{}da)$ and $ a\equal{}c$ and $ b\equal{}d$ i.e. $ a\equal{}c\equal{}0, b\equal{}d$ or $ a\equal{}c, b\equal{}d\equal{}0$. But as $ a,b,c,d$ are positive, equality cannot hold.

[quote="ElChapin"][quote="nayel"]

$ \sum\frac {a}{b \plus{} 2c}\ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{ab \plus{} bc \plus{} cd \plus{} da \plus{} 4(ac \plus{} bd)}$

[/quote]

WoW!! Untill now I hadn't seen Cauchy applied in such way
Very useful, very educative!  :wink:

Great solution! Thanks for the "trick".

 :)[/quote]

I'm sorry, I don't see how this follows :blush: Could you please explain? Thanks!

$ \sum_{cyc}a(b\plus{}2c)\sum_{cyc}\frac{a}{b\plus{}2c}\geq (a\plus{}b\plus{}c\plus{}d)^2\iff \sum_{cyc}\frac{a}{b\plus{}2c}\geq \frac{(a\plus{}b\plus{}c\plus{}d)^2}{\sum_{cyc}a(b\plus{}2c)}$ (Cauchy in Engel Form)


Draw a regular polygonon of $5$ sides, and draw all the diagonals: Then, how many triangles are there??????

hide your answers please.

[hide="i get"]35.
i just counted so it could be incorrect. i'd like to know if there's a quck method to get the answer.[/hide]

You all lose, I got
[hide="answer"]
35
[/hide]

fireemblem got the same answer as me but he doesn't count =/

Yes, maybe with a picture it'd be easier

[quote="ch1n353ch3s54a1l"][hide]Um I counted and got 33 >.>
I think there might be a general formula for an n-gon but...i dont know. [/hide][/quote]

There IS a formula that was devised by people at UCLA I think.

But the formula is like 5 lines long. o.O

The same math teacher also told me that the formula was put on a cup that said:

"God said [insert 5-line long formula here], and there was light"  :rotfl:

A formula that is 5 lines long, that is just scary.  I guess it is the best way to just draw the picture and figure out the answer.


Solve the equation z^8 + 4z^6 − 10z^4 + 4z^2 + 1 = 0.

[hide="Try..."] The substitution $ u = z^{2}+\frac{1}{z^{2}}$. [/hide]


[hide="Hint to the cocky solution"] Show that the sum of the reciprocals of the primes diverges (using the Zeta function or similar), and then this follows trivially.  :D [/hide] [hide="Hint to the common elementary solution"] $\frac{1}{2^{n}+k}< \frac{1}{2^{n}}$ [/hide] [hide="Hint to the solution by calculus"] This is an LRAM. ;) [/hide]
[b]Follow-up:[/b]  Show that the harmonic series is never an integer for $n > 1$.

Do you know how to do the follow-up? And while were at it, find a limit value for the sum
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...$

[quote="vishalarul"]Prove that the harmonic series $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ diverges.[/quote]
many many ways to show that $\lim_{n\to\infty}{\sum_{x=1}^{n}{\frac1x}}=\infty$ as t0rajir0u said.

[quote="t0rajir0u"][b]Follow-up[/b]: Show that the harmonic series is never an integer for $n>1$.[/quote]

if $h_{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
Let $k$ be the largest power or $2$ less than or equal to $n$

Now if sum is $h_{n}=p$, multiply both side by $\frac{k}{2}*$(product of all odd numbers up to $n$ ) then except for the term corresponding to $\frac{1}{k}$ every other term would be an integer so: contradiction.
hope this helps... ;)

in otherwords
IF, the sequence of Harmonic numbers DOESN'T start with 1 , then the sum would be always and odd number over an even number, therefore the sum can never be an integer

[quote="vishalarul"] find a limit value for the sum
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...$[/quote]

[hide="Small hint"] The sum telescopes ;) [/hide] [hide="Big hint"] $\frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}$ [/hide]

This topic has appeared WAY too many times; both the original question and the follow-up.

For instance:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=series+harmonic&t=91309

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=series+harmonic&t=63991

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=series+harmonic&t=48309

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=series+harmonic&t=31036

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=series+harmonic&t=19572

Satisfied? And that's only in the Intermediate Forum.


quadratics
modular arithmetic
number theory unsolved
number theory

I think this problem is pretty hard. I took a long time trying to solve it but couldn't figure it out

Let $ p$ be a prime of the form $ 4n \plus{}1$. For every integer $ a$ with $ (a,p) \equal{} 1$, define

$ S(a) \equal{} \sum^p_{x\equal{}1}(\frac{x(x^2\plus{}a)}{p})$

(a) Prove that $ S(a)$ is even, and that
$ S(at^2) \equal{} (\frac{t}{p})S(a)$ holds for every integer t.

(b) If r is some quadratic residue modulo $ p$, and $ n$ is some quadratic nonresidue modulo $ p$, show that
$ p \equal{} (\frac{1}{2}S(r))^2 \plus{} (\frac{1}{2}S(n))^2$

$ \sum_{x\equal{}1}^p x^3 \equal{} \left[\dfrac{p(p\plus{}1)}{2}\right]^2$ and $ \sum_{x\equal{}1}^p x \equal{} \left[\dfrac{p(p\plus{}1)}{2}\right].$  Therefore $ S(a) \equal{} \frac{1}{4}p(p\plus{}1)^2 \plus{} \frac{1}{2}a(p\plus{}1).$

Consequently, both assertions in part a are wrong: counter example is $ p\equal{}5,a\equal{}2,t\equal{}1.$

[quote="zugzwang"]$ \sum_{x \equal{} 1}^p x^3 \equal{} \left[\dfrac{p(p \plus{} 1)}{2}\right]^2$ and $ \sum_{x \equal{} 1}^p x \equal{} \left[\dfrac{p(p \plus{} 1)}{2}\right].$  Therefore $ S(a) \equal{} \frac {1}{4}p(p \plus{} 1)^2 \plus{} \frac {1}{2}a(p \plus{} 1).$

Consequently, both assertions in part a are wrong: counter example is $ p \equal{} 5,a \equal{} 2,t \equal{} 1.$[/quote]

u are maybe right since we have to use $ \left(\frac{x(x^2\plus{}a)}{p} \right)$ not $ (\frac{x(x^2\plus{}a)}{p})$ for showing jacobi symbol,
but violetcraze meant jacobi symbol...
anyway ,
here is a proof of the first part (which is really easy,but for (b)... :( )
a)$ S(a)$ is even because it is either the sum of one zero and some $ p \minus{} 1$ odd numbers and since $ p \minus{} 1$ is even.... or the sum of three zero and $ p \minus{} 3$ odd numbers.
consider $ t'$ such that $ p | tt' \minus{} 1$ then we get
$ \left(\frac {t'}{p} \right) \equal{} \left(\frac {t}{p} \right)$
now multiply both sides of the equality
$ S(at^2) \equal{} \sum^{p}_{x \equal{} 1} \left(\frac {x(x^{2} \plus{} at^2)}{p} \right)$
by $ \left(\frac {t'^3}{p} \right) \left(\frac{t}{p} \right)$ and we get what we want.
b)we try to use the last part for solving this part,note that $ t^2 \equiv r \pmod {p}$ for some $ t$ and we get $ S( 1 \times t^{2}) \equal{} \left (\frac {t}{p} \right)S( 1)$.
also note that if $ n_1,...,n_{\frac{p\minus{}1}{2}}$ are the quadratic nonresidues modulo $ p$ then similarly we can easily show that
$ S(n_1)^2\equal{}S(n_2)^2\equal{}...\equal{}S(n_{\frac{p\minus{}1}{2}})^2$
we know that the representation of $ p$ of the form $ a^2 \plus{} b^2$ is unique so  :idea: ...
we must prove the equality $ \{(2a)^2,(2b)^2 \} \equal{} \{(S(1))^2,(S(n_i))^2 \}$
my idea is to find a congruency modulo $ p$ for which the number of its solutions is $ a$ or $ 2a$ or...  and use double-counting
and get the desired equality(similarly for $ b$).
for example if we want to prove $ \sum_{k \equal{} 0}^{p \minus{} 1} \left(\frac {k^2 \plus{} a}{p} \right) \equal{} \minus{} 1$ where $ p \nmid a$ we use the congruency
$ x^2 \minus{} y^2 \equiv a \pmod {p}$
and counting the number of its solutions in two ways.
but i didn't succeed ....  :(


[quote="parinaz.sa"]if$x_1,x_2,x_3,x_4$are positive numbers and we have $x_1.x_2.x_3.x_4=1$prove:
$\displaystyle\sum_{i=1}^{4}x_i^3\geq max\{\displaystyle\sum_{i=1}^{4}x_i, \displaystyle\sum_{i=1}^{4}\frac{1}{x_i}\}$.[/quote]

Have have :

$\displaystyle\sum_{i=1}^{4}x_i^3+8=\displaystyle\sum_{i=1}^{4}(x_i^3+1+1) \ge 3\displaystyle\sum_{i=1}^{4}x_i \ge \displaystyle\sum_{i=1}^{4}x_i+8$

And :

$\displaystyle\sum_{i=1}^{4}x_i^3\geq x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_1x_2x_4 =\displaystyle\sum_{i=1}^{4}\frac{1}{x_i}$


Let $ a, b, c$ be real positive numbers such that $ a \plus{} b \plus{} c \equal{} abc$. Prove or disprove that
$ \frac{a\plus{}1}{\sqrt{a^2 \plus{} b^2}}\plus{} \frac{b\plus{}1}{\sqrt{b^2 \plus{} c^2}} \plus{} \frac{c\plus{}1}{\sqrt{c^2 \plus{} a^2}}\le \frac{3\sqrt{2} \plus{} \sqrt{6}}{2}$ .

___________________
 :blush:  :idea:

[quote="Ligouras"]Let $ a, b, c$ be real positive numbers such that $ a \plus{} b \plus{} c \equal{} abc$. Prove or disprove that
$ \frac {a \plus{} 1}{\sqrt {a^2 \plus{} b^2}} \plus{} \frac {b \plus{} 1}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {c \plus{} 1}{\sqrt {c^2 \plus{} a^2}}\le \frac {3\sqrt {2} \plus{} \sqrt {6}}{2}$ .

___________________
 :blush:  :idea:[/quote]

Who can solve this?  :(    :(


function
number theory unsolved
number theory

Find all functions $ f: \mathbb{N}\rightarrow\mathbb{N}$ such that

$ f(f(f(n)))\plus{}6f(n)\equal{}3f(f(n))\plus{}4n\plus{}2001$

It is easy to check that f(n)=n+667 satisfies. Indeed we can show it is the only possible solution. I give a sketch of proof below.

Let k be a fixed natural number. Define a series as
a_0=k,a_1=f(a_0),a_2=f(a_1),...,a_n=f(a_{n-1})
Then we have
a_{n+3}+6a_{n+1}=4a_{n+2}+3a_{n}+2001, from which we can get equation for a_{n} as
a_n = 667n+c1+c2*x^n+c3*y^n, where c1,c2,c3 depends on a_0,a_1,a_2, and x,y are roots of x^2-2x+4.
Then the only way to make sure the series has no negative number is to have
c2+c3=0,c2*x+c3*y=0,  or c2=c3=0.

Then we can show a_1=f(k)=k+667. This works for any k.


Let $a,b,c >0 $ and $a=min(a,b,c)$ . Prove that $a^3+b^3+c^3-3abc \geq 2( \frac{b+c}{2}-a)^3$ .

cheers!

After some computation you get

4a^3+b^3+c^3 +2a(b^2+c^2) \geq bc(b+c) +4a^2(b+c)

but b^3+c^3 \geq bc(b+c)

so it remains to prove

2a^2+b^2+c^2  \geq 2a(b+c)   with b+c \geq 2a

and now how do you continue ??????

Brute force again !

Developing ineq gives :

4a^3 + b^3 + c^3 + 2b^2a+2c^2a >= bc + bc + 4ba + 4ca

b^3 + c^3 >= bc + bc is well known

4a^3 + 2b^2a+2c^2a >= 4ba + 4ca is 2a((a-b)^2 + (b-c)^2) >= 0  :D

PS To answer Manlio's private message : yes  :)


He ate $ 0.64x \times 4\equal{}2.56x$ calories from carbohydrates, and $ 0.36x \times 9\equal{}3.24x$ calories from fats.  The percentage desired is $ 100 \left (\frac{2.56x}{2.56x\plus{}3.24x}\right ) \approx \boxed{44 \%}$.

But on the problem the answer is 80%. How is that?

[quote="GameBot"]Carbohydrates provide four calories per gram, and fats provide nine calories per gram.  If $ 36\%$ of the calories Curtis ate at lunch came from fats and $ 64\%$ came from carbohydrates, what percent of the grams eaten came from carbohydrates?[/quote]

Say that Curtis consumed 100 calories. 36 of these calories came from fats, and 64 from carbohydrates. Thus, Curtis consumed 16 grams of carbohydrates and 4 grams of fats. The desired percentage is thus $\frac{16}{4+16}=80 \%$.

Or say that Curtis consumed c calories. He thus consumed $\frac{c}{25}$ fats and $\frac{4c}{25}$ carbohydrates. The desired percentage is $80$.


calculus
integration
number theory
relatively prime

Prove that consecutive numbers are always relatively prime.

[quote="akhil0422"]The difference between consecutive numbers has to be one. All numbers from two upwards have a difference of 2 and above between each number. Thus, it is not possible for consecutive numbers to not be relatively prime (excluding 1 and 2).[/quote]
I don't understand your second sentece, and I don't understands how it proves it either.

It proves that two consecutive numbers [b]can't not[/b] be relatively prime.

Sorry. :blush:  :blush: I understand now. Thanks to both of you.

[quote="jonathanchou711"][quote="PowerOfPi"]Please rewrite your second sentence in a different way and explain.[/quote]
Dude. It sounds like you're our teacher. This place is for problems! Not tests that you give us. If you have a question about a problem, then post here. You can post problems that you want to discuss here, but put it in the title. Such as:
Discussing Pythagorean Theorem.

And akhil means that any number greater than 2 that have a same prime number must be differing by two. Which is 2. But since 1 is not a prime number, you exclude 1 and 2.[/quote]

Your flame didn't make much sense whatsoever...

Anyway, here's a proof (badly worded, I'm in a hurry) that doesn't rely so much on intuition...

Assume there are two numbers $ k$ and $ k\plus{}1$, where $ k$ is a natural number, such that $ k$ and $ k\plus{}1$ are not relatively prime.  Then they share some common divisor $ x \ge 2$, where $ x$ is an integer.  If $ k$ is divisible by $ x$, we can write $ k$ as $ ax$ for some $ a$.  The next integral multiple of $ x$ is $ x(a\plus{}1)\equal{}ax\plus{}x$, which must be $ k\plus{}1$.  However, making the substitution $ k\equal{}ax$, we have $ x\equal{}1$, which contradicts our assumption.  Thus, $ k$ and $ k\plus{}1$ do not exist.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be an infinite group generated by nilpotent normal subgroups. Prove that every maximal Abelian normal subgroup of $ G$ is infinite. (We call an Abelian normal subgroup maximal if it is not contained in another Abelian normal subgroup.)

[i]P. Erdos[/i]

Solution:

Lemma 1:

$\forall \{e\}\not= H\triangleleft G$, $H\cap Z(G)\not= \{e\}$.

See it here: http://www.artofproblemsolving.com/blog/95197

Lemma 2:

Let $H\triangleleft G: |H|<\infty$, then $Z_G(H)\triangleleft G$, $[G:Z_G(H)]<\infty$

Let's use this, let $A$ is maximal Abelian normal subgroup of $G$, $\{G_i\}_{i=1}^n$ are nilpotent normal generators of $G$, $|A|<\infty$, so from Lemma 2 we get that $A\subset Z_G(A)$, $[G:Z_G(A)]<\infty$, so $A\not= Z_G(A)$ denote $\pi_A : G\to G/A$, $\pi(Z_G(A))\triangleleft \pi_A(G)$, easy to see that $\{\pi_A(G_i)\}_{i=1}^{n}$ are nilpotent normal generators of $\pi_A(G)$, so from Lemma 1 $\pi(Z_G(A))\cap Z(\pi_A(G))\not= \{e\}$, take $\alpha \in \pi(Z_G(A))\cap Z(\pi_A(G)) : \alpha \not=e$, $<\alpha>\subset Z(\pi_A(G))$, so $<\alpha>\triangleleft \pi_A(G)$, so $A^*= \pi_A^{-1}(<\alpha>)\triangleleft G$, $A\subset A^*$, $\pi_A(A^*)\subset \pi(Z_G(A))$, so $A\subset Z(A^*)$, $A^*/A=<\alpha>$ is cyclic, so $A^*/Z(A^*)$ is cyclic, so $A^*$ is Abelian $A^*\triangleleft G$, so $|A|=\infty$. done


calculus
integration
geometry
3D geometry
sphere
probability
calculus computations

Consider a 7-dimensional "space" with $t,\;u,\;v,\;w,\;x,\;y,\;$ and $z$ axes, all perpendicular to each other in 7-space.

Consider the set of all points $(t,u,v,w,x,y,x)$ that satisfy:

$t^{2}+u^{2}+v^{2}+w^{2}+x^{2}+y^{2}+z^{2}\leq r^{2}$

If we define the distance of a point from the origin to be $\sqrt{t^{2}+u^{2}+v^{2}+w^{2}+x^{2}+y^{2}+z^{2}}$, find the root mean square distance of all points in the region to the center.

[hide="I_get..."]
$\frac{\sqrt{7}}{3}r$, but I'm not too sure about it... :maybe:
[/hide]

Your answer is correct.

Consider this problem in $\mathbb{R}^{n}.$ Let $A_{n}$ be the $n-1$ dimensional measure of the unit sphere $\{\|x\|=1\}$ and let $V_{n}$ be the $n$ dimensional measure of the unit ball $\{\|x\|\le1\}.$ Now it happens that $A_{n}$ and $V_{n}$ are well known, but we don't need their exact values.

Consider a "spherical shell" of radius $\rho$ and thickness $d\rho.$ This has "volume" (that is, $n$ dimensional measure) $dV=A_{n}\rho^{n-1}\,d\rho.$

The first thing this gives us is that $V_{n}=\int_{0}^{1}A_{n}\rho^{n-1}\,d\rho=\frac{A_{n}}n.$

Now let's compute your root mean square average:

$\int_{\|x\|\le r}\|x\|^{2}\,dx=\int_{0}^{r}A_{n}\rho^{n+1}\,d\rho =\frac{A_{n}r^{n+2}}{n+2}$

Divide that by the volume of the ball of radius $r$ to get

$\frac1{V_{n}r^{n}}\cdot\frac{A_{n}r^{n+2}}{n+2}= \frac{n}{n+2}\,r^{2}.$

To get the root mean square average, take the square root of this, getting

$\sqrt{\frac{n}{n+2}}\,r.$ That agrees with your answer for $n=7.$

what exactly is the root mean square average? could someone please provide me with a way to visualize this (no formula) if at all possible.

The root mean square is simply the square root of the average of the squares. If we had numbers $a_{1},a_{2},\dots a_{n}$ (and we wanted to use a "flat" weight) the root mean square would be

$\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}n}$

In probability, the root mean square of the deviation from the mean is the standard deviation.

For alternating current electricity, the most useful number to use for the "average" voltage is the root mean square average, since power dissipation in a resistor would be proportional to the square of the voltage.

I realize now that I shouldn't have said "root mean square average" since the words "mean" and "average" are reduntant - you need only one of those words.

Awesome, that was more or less what I did. Thanks a lot  :)


AMC
AIME
rotation
geometry
geometric transformation
linear algebra
matrix

The points $(0,0),$ $(a,11)$, and $(b,37)$ are the vertices of an equilateral triangle.  Find the value of $ab$.

Since it is equilateral, the distances between each of the points must be equal. Plugging into the distance formula:
(b-a) :^2:  +(37-11) :^2: =a+11 :^2: =b+37 :^2:
Simplifying the left two:
(b-a) :^2:  +(37-11) :^2:  =a+11 :^2:
b :^2: -2ab+a :^2: +26 :^2: =a :^2: +121
b :^2: -2ab+555=b :^2: +37:^2:
-2ab=814
ab=-407
correct?

I don't think so...

I think you made a mistake in the fourth line from the bottom.

b :^2: -2ab+555=0...not b :^2: +37 :^2:

right....*stuck* but I'm still working on it

Umm...I think I'd just use matrices.  It seems like (a, 11) is rotated 60 :^o: around the origin to (b, 37).

So using matrices*:

(0.5 -:rt3:/2)(a)
(:rt3:/2  0.5)(11)  =

(b)
(37)

0.5a - 11 :rt3:/2  = b
a :rt3:/2  + 5.5 = 37

b = 5 :rt3:

a = 21 :rt3:

ab = 315

Jeffrey, aren't there no fractional answers on the AIME though?

*The rotation matrix to rotate a given point (a, b)  :theta: is:
(cos :theta: -sin :theta:)
(sin :theta: cos :theta: )*

That was really easy for an AIME problem.

I don't have an answer key, so I don't know...

If that's the answer than on the AIME it would probably be expressed as a/b = a+b if you know what I'm saying.

Matrices, crap, I've always hated them...

Aren't answers on the AIME always integers from 0-999?

just 1 minor problem.... the sin of 60 is not sqrt(3).

Similar line (as matrices): rotation  by 60 degrees in complex  plane is writen as  (1/2)(1+ :rt3:  i)  (i^2=-1)
so  (a+11i)*(1/2)(+ :rt3: i)=b+37 i

will give a and b..

[hide="This was my solution from the AIME marathon"]Consider the equilateral triangle in the complex plane.
That is, $(a,11)$ represents the point $a+11i$ and $(b,37)$ represents the point $b+37i$.
Then $b+37i$ is a $60^\circ$ counterclockwise rotation of $a+11i$ about the origin, or $(a+11i)\text{cis}(60^\circ)=b+37i$
\[ \left(a+11i\right)\left(\frac{\sqrt{3}i}{2}+\frac{1}{2}\right)=b+37i \]
\[ \frac{a}{2}+\frac{\sqrt{3}ai}{2}+\frac{11i}{2}-\frac{11\sqrt{3}}{2}=b+37i \]Setting the real and imaginary parts equal:\[ a-11\sqrt{3}=2b \]\[ \sqrt{3}a+11=74 \]From the second equation, $a=\frac{63}{\sqrt{3}}=21\sqrt{3}$. Then we obtain $21\sqrt{3}-11\sqrt{3}=2b$ or $b=5\sqrt{3}$ after plugging back into the first equation.
Finally, $ab=\left(21\sqrt{3}\right)\left(5\sqrt{3}\right)=\boxed{315}$.[/hide]

[hide="trig solution using angle addition formulas"]
Let $\theta$ be the measure of the angle between the $x$-axis and the side of the equilateral triangle connecting the origin and the point $(a, 11)$. Also, let $x$ represent the length of one side of the equilateral triangle.

From this, we have the relations:

$\sin \theta = \displaystyle \frac{11}{x}$

$\cos \theta = \displaystyle \frac{a}{x}$

We know that the measure of each angle in an equilateral triangle is $60^\circ$, so we then have the equations:

$\sin (\theta + 60) = \displaystyle \frac{37}{x}$

$\cos (\theta + 60) = \displaystyle \frac{b}{x}$

Expanding $\sin (\theta + 60)$, and using the expressions for $\sin \theta$ and $\cos \theta$, we have:

$\sin\theta \cos 60 + \cos \theta \sin 60 = \frac{37}{x}$

$\frac{11}{2x}+ \frac{\sqrt{3}a}{2x} = \frac{37}{x}$

$11+\sqrt{3}a = 74$

$a=21\sqrt{3}$

Expanding $\cos (\theta + 60)$, we have:

$\cos \theta \cos 60 + \cos \theta \sin 60 = \frac{b}{x}$

$\frac{a}{2x} - \frac{11\sqrt{3}}{2x} = \frac{b}{x}$

$a-11\sqrt{3} = 2b$

$21\sqrt{3} - 11\sqrt{3}= 2b$

$b=5\sqrt{3}$

$ab=21\sqrt{3} \times 5\sqrt{3} = \boxed{315}$
[/hide]

[quote="M4RI0"]Why noone got $a=-21\sqrt{3}$, $b=-5\sqrt{3}$ :?:[/quote]

the mistake that has been made is that a rotation of $60$ and $-60$ degrees have not been taken into account (both of which create an equilateral triangle)

sorry to bump but I was looking for a pure algebra solution- is it possible?

Thanks

What do you mean by a "pure" algebra solution?  Any geometric problem necessarily requires a degree of geometric interpretation.

[hide="Discussion"] [b]Lemma:[/b]  $ \alpha, \beta, \gamma \in \mathbb{C}$ are the vertices of an equilateral triangle if and only if $ \alpha^2 \plus{} \beta^2 \plus{} \gamma^2 \equal{} \alpha \beta \plus{} \beta \gamma \plus{} \gamma \alpha$.

We have $ \alpha \equal{} 0, \beta \equal{} a \plus{} 11i, \gamma \equal{} b \plus{} 37i$.  Our condition is then

$ (a^2 \minus{} 121) \plus{} 22ai \plus{} (b^2 \minus{} 1369) \plus{} 74bi \equal{} (ab \minus{} 407) \plus{} (37a \plus{} 11b)i$

which gives $ a^2 \plus{} b^2 \minus{} ab \equal{} 1083$ and $ 22a \plus{} 74b \equal{} 37a \plus{} 11b \implies 15a \equal{} 63b \implies 5a \equal{} 21b$.  Let $ a \equal{} 21c, b \equal{} 5c$, which gives

$ 361c^2 \equal{} 1083 \implies c \equal{} \pm \sqrt{3}$

which is the same as the others. [/hide]

towersfreak2006, I've never seen that method of using complex numbers to rotate a point about the origin.  Is there a description of the technique in the AoPS books, because if there is I can't find it...?

[b]Lemma:[/b]  The rotation of the point $ (x, y)$ counterclockwise about an angle of $ \theta$ is given by

$ x' \equal{} x \cos \theta \minus{} y \sin \theta$
$ y' \equal{} x \sin \theta \plus{} y \cos \theta$.

Proof:  Polar coordinates.

[b]Corollary:[/b]  The rotation of the point $ x \plus{} yi$ counterclockwise about an angle of $ \theta$ is given by $ (x \plus{} yi)(\cos \theta \plus{} i \sin \theta)$.

I just used the distance formula, manipulated equations to get a^2 - b^2 = 1248 and a^2 + 121 = (a-b)^2 + 676. I then solved for b in the second equation and then plugged it into the first. Then use quadratic formula to solve for a^2 and wala, you're done.

@MCL: Unless you are very good at bashing, your method will be very difficult under testing constraints; try it for yourself. That's what makes this problem harder: anyone with a basic concept of distance can bash it in an uninsightful manner, but it requires knowledge of rotations on test day. This is how the MAA distinguishes between stronger and weaker students.

@mysmartmouth: Precalculus.

Note: complex rotations, Cartesian trig identities, and matrix rotations are equivalent to one another.

[color=darkred][b][u]PP[/u].[/b] Let $ABC$ be an equilateral triangle, where $A(0,0)$ , $B(a,11)$ and $C(b,37)$ .  Find the value of $ab$ .[/color]

[b][u]Proof[/u].[/b] Let $\beta =a+11i$ and $\gamma =b+37i$ . Thus, $\triangle ABC$ is equilateral $\stackrel{(*)}{\iff}$ $\beta^2+\gamma^2=\beta\gamma$ $\iff$ $\left\{\begin{array}{c}
a^2-ab+b^2=1083\\\\
5a=21b\end{array}\right|$ .

In conclusion, $\frac {a^2}{441}=\frac {b^2}{25}=$ $\frac {ab}{105}=\frac {a^2+b^2}{466}=$ $\frac {a^2-ab+b^2}{361}=$ $\frac {1083}{361}=3\implies \boxed{ab=315}$ .  I used the well-known property

$(*)\blacktriangleright$ [color=darkred][b][u]Lemma[/u].[/b] $\triangle ABC$ is equilateral $\iff$ $\alpha^2+\beta^2+\gamma^2=\alpha\beta+\beta\gamma+\gamma\alpha$ , where $A(\alpha)$ , $B(\beta)$ and $C(\gamma )$ .[/color]

[b][u]Proof[/u].[/b] $ABC$-[b]eq.[/b] $\iff$ $\frac {\beta -\alpha}{\gamma -\alpha}\in\left\{w,\overline{w}\right\}$ , where $\left\{\begin{array}{c}
x^2-x+1=0\begin{array}{cc}
\nearrow & w\\\\
\searrow & \overline w\end{array}\\\\
w^3+1=0\ ;\ \overline w=\frac 1w=-w^2\end{array}\right|$ $\iff$

$\left(\frac {\beta -\alpha}{\gamma -\alpha}\right)^2-\left(\frac {\beta -\alpha}{\gamma -\alpha}\right)+1=0$ $\iff$  $(\beta -\alpha )^2-(\beta -\alpha)(\gamma -\alpha)+(\gamma -\alpha )^2=0\iff$

$\alpha^2+\beta^2+\gamma^2=\alpha\beta+\beta\gamma+\gamma\alpha\iff$ $\boxed{\ (\beta -\gamma )^2+(\beta -\alpha )(\gamma -\alpha )=0\ }$ .

[color=darkred][b][u]Extension[/u].[/b] $\triangle ABC$ is $A$-isosceles with $A=\phi\iff$ $(\beta -\gamma )^2+2(1-\cos\phi )(\beta -\alpha )(\gamma -\alpha )=0$ , where $A(\alpha)$ , $B(\beta)$ , $C(\gamma )$ .[/color]

Sorry for the big bump, but this is the official thread in contest collections. (If it's not clear, this is 1994 AIME/8).

[hide=Solution]We use complex numbers. $(b,37)$ is the rotation of $(a,11)$ $60^\circ$ around $(0,0)$. Thus, we can write:

\[b+37i=(a+11i)\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=\left(\frac{1}{2}a-\frac{11\sqrt{3}}{2}\right)+\left(\frac{11}{2}+\frac{\sqrt{3}}{2}a\right)i.\]

Matching coefficients gives $\frac{11}{2}+\frac{\sqrt{3}}{2}a=37\implies a=21\sqrt{3}$. Also, $b=\frac{1}{2}a-\frac{11\sqrt{3}}{2}=5\sqrt{3}$.

Thus, $ab=21\sqrt{3}\cdot5\sqrt{3}=\boxed{315}$.
[/hide]

[quote name="Amkan2022" url="/community/p29231240"]
Solution 4 in wiki doesnt make sense to me'

"Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2$."

so why is b not 37, then?
[/quote]

That's a typo, it should be
\[a^2+11^2=(b-a)^2+26^2=b^2+37^2\]
which is actually easy to get by the Pythagorean Theorem


function
combinatorics proposed
combinatorics

This is probably known but still a nice and easy exercise:

Let $S_n$ be the number of Latin squares of order $n$ and let $R_n$ be the number of reduced Latin squares of order $n$ (Latin squares for which the first column and first row both equal $1\ 2\ 3\ \cdots\ n$ in that order).

Find $S_n/R_n$ as a function of $n$.

Since all permutations of rows and columns give a different latin square, I'd say $n!(n-1)!$?
[from $n!$ to $(n-1)!$ in the rows and then from $(n-1)!$ to $(n-2)!$ in the columns?]

Could be wrong though...

You still need to prove that you get different latin squares :)

The answer is correct though.

What do you mean precisely? Any latin square has the property that all its permutations of rows/columns are also latin squares, and which are obviously different, as I stated. (not would require twice the same symbol on a row/column)

So I do my operation on the set of [i]different[/i] latin squares, and that result is simply elimination, so the results must be different too, right? :)


1.
ABC is an equilateral triangle with side 6. Let P be a point with distances x,y,z from the three vertices (where ). Find the area of the region enclosed by all points such that x+y=z.

2. Mr. and Mrs. Brown gave a party for their friends they have not seen for a long time. Three couples came. During the party, some of the people were so happy to see each other again, that they even shook hands. (None of the men shook hands with their own wives.) Later on, Mr. Brown asked everybody how many people he/she shook hands with. He received seven different answers. How many guests did Mrs. Brown shake hands with?

Good Luck =)

2.
[hide]The maximum number of hands anyone could have shaken is 6. Therefore, somebody shook each amount of hands from 0 to 6. The person who shook 6 hands shook everybody's hand except his/her own spouse. Nobody else could have shaken 0 hands, so 0, 6 is a husband/wife pair (meaning the spouse of the person who shook 6 hands shook 0 hands). The person who shook 5 hands shook everybody's hands except for his/her spouse and the person who shook no hands. Since everybody else has shaken two hands, 1, 5 is a husband/wife pair. The person who shook 4 hands shook everybody's hands except for his/her spouse and the people who shook 0 and 1 hands. Repeating, 4, 2 is a husband/wife pair.

3 is the only left over amount, so Mrs. Brown shook 3 hands.[/hide]


Find k so that the line containing the points (k,4) and (-8,3k) is perpendicular to the line with equation 4y= 28-x

[hide="Click me!"]The slope of the line given is $-\frac{1}{4}$, so the perpendicular slope would be $4$. Applying this to slope of the line between the two points,

$\frac{4-3k}{k+8}= 4$

$4k+32 = 4-3k$

$7k=-28$

$k=-4$[/hide]


algebra
polynomial
complex numbers
superior algebra
superior algebra solved

it's a classical exercice to show that if P is a real polynomial so that P(x)>=0 for every x then there exist real polynomials A and B so that: P=A^2+B^2..

But what can we say if F(x)>=0 where F=P/Q, P and Q are real polynomials ... If i remember the conclusion is almost the same but i not quite sure.

Moreover i was wondering if there exists a sufficient condition to say that P=A^2+B^2 where A and B are complex polynomials ?? any idea ??

1. If $F\in\mathbb{R}[X]$ such as $F(x)\geq 0\ \forall x\in\mathbb{R}$ then $F=P^2+Q^2$ for some real polynomials P,Q.

Proof: If F has a non-real root $z$ then F also has the root $\overline{z}$ (the complex conjugate of z). Then $(X-z)(X-\overline{z}) | F$ and it is clear that
$\frac{F}{(X-z)(X-\overline{z})}$ is a polynomial greater or equal to $0$ on reals. So it suffices to prove the problem for polynomials with only real roots (later you will see why). Assume $F(X)=\prod_{k=1}^n (X-x_k)^{a_k}$ where $x_1<x_2<\ldots <x_n$ and the $a_k$'s are positive integers. Assume not all $a_k$'s are even and chose $r=\min_{k\in\overline{1,n}} (a_k\notin 2\mathbb{N})$.
It is easy to see that F takes negative values on $(x_r,x_{r+1})$ if $r<n$ and on $(x_r,\infty )$ if $r=n$.
So if F has only real roots and $F\geq 0$ on reals, then F is the square of some polynomial.
$(X-z)(X-\overline{z})=\left(X-\frac{z+\overline{z}}{2}\right)^2+|z|^2-\left( \frac{z+\overline{z}}{2}\right)^2$.
It is easy to see that this is a sum of squares. Using the identity $(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2$ we get that F is a sum of squares.

I'm sure you knew about that, but I need that solution for better understanding of 2.

2. If $F=\frac PQ$, P,Q are real polynomials with no common factors and $F\geq 0$ on reals except for the real roots of Q, then P,Q are sums of squares.

Proof: If $F=\frac PQ$ for ${P,Q\in \mathbb{R}[X]}$ with no common factors, and $F\geq 0$ on reals except maybe for the real roots of Q, then $F\cdot Q^2$ has the same property. So $PQ\geq 0$ on reals except for a finite set of points. By continuity, $PQ\geq 0$ on reals. Because P and Q have no common factors, we get that the real roots of P and Q have even multiplicity. So P,Q are sums of
squares.

3. Every complex polynomial can be written as sum of squares of complex polynomials.

Proof: Let F be a complex polynomial. Take $P=\frac {F+1}2$ and $Q=i\cdot\frac{F-1}2$. Then $F=P^2+Q^2$.


geometry
incenter
Angle Chasing
circumcircle
perpendicular bisector
IMO 2002
hojoo lee

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

Triangles $AOE,AOF$ are equilateral because $AE=OE=OA,\ AF=OF=OA$.

The condition $\angle AOB<120^{\circ}\iff \angle AOC>60^{\circ}$ simply means that $F$ (assume $F$ is on the same side of $OA$ as $C$) lies on the arc $AC$ which doesn't contain $E$, so $CA$ is the internal bisector of $\angle ECF$ because $A$ is the midpoint of the arc $EF$ (without the condition, $AC$ would be the external bisector). This means that $J$ lies on the bisector of $\angle ECF$.

On the other hand, we have $AD\|OJ$ and $OD\|AJ$ because they're both $\perp AB$, so $J$ is the reflection of $D$ in the midpoint of $OA$. This means $\angle EJF=\angle EDF=120^{\circ}$, and since $\angle ECF=\frac{\angle EOF}2=60^{\circ}$ and $J$ lies on the internal bisector of $\angle ECF$, it means that $J$ is the incenter (the only other position on $AC$ from which $EF$ is seen under an angle of $120^{\circ}$ is $A$, and it's clear that $J\ne A$).

[quote="orl"]The circle $ S$ has centre $ O$, and $ BC$ is a diameter of $ S$. Let $ A$ be a point of $ S$ such that $ \angle AOB < 120{{}^\circ}$.  Let $ D$ be the midpoint of the arc $ AB$ which does not contain $ C$. The line through $ O$ parallel to $ DA$ meets the line $ AC$ at $ I$. The perpendicular bisector of $ OA$ meets $ S$ at $ E$ and at $ F$. Prove that $ I$ is the incentre of the triangle $ CEF.$[/quote]
Consider the complex plane with origin $ O$. Wlog assume that $ S$ is the unitcircle. For uppercase letter points, let the lowercase letter denote their corresponding complex number. (Except for $ I$, which will has the complex number $ z_I$)

$ D$ is the midpoint of the arc $ AB$, ie. $ \angle BOD = \angle DOA \iff \frac {b}{d} = \frac {d}{a} \iff a = d^2$. (Since $ |a| = |b| = |d| = 1|$) I will prove that $ z_I = d^2 - d$. Obviously $ OI \parallel AD$ since $ z_I - o = a - d = d^2 - d$.

$ \frac {z_I - a}{c - a} = \frac {d}{1 + d^2} \in \mathbb{R}$ since $ \frac {d}{1 + d^2} = \frac {\frac {1}{d}}{\frac {1}{d^2} + 1} = \overline{\frac {d}{1 + d^2}}$ so $ I \in AC$, which concludes: $ z_I = d^2 - d$, since the point is unique.

It is easy to show that $ E$ and $ F$ are constructed by rotating $ A$ $ 60 ^\circ$ around $ O$ positivily and negatively respectively. Let $ \omega = e^{i\frac {\pi}{6}}$, then $ e = \omega^2d^2 = u^2, u = \omega d$ and $ f = \overline{\omega}^2d^2 = v^2, v = \omega^5 d$. Since $ c = w^2, w = - i$, and $ u,v,w$ the incenter of $ \triangle CEF$ is $ - uv - vw - wu$. (Wellknown. See the training materials from imomath.com)

And $ - uv - vw - wu = - \omega^6d^2 + i(\omega d + \omega^5 d) = d^2 + i^2d = d^2 - d = z_I$. So $ I$ is the incenter of $ \triangle CEF$.

enough to show $ AD\equal{}AF\equal{}AI$ and observe just B D A I C O,
easily find $ OA\equal{}AI$ by elementary angle chasing.
so we are done :lol:

[quote="grobber"]Triangles $ AOE,AOF$ are equilateral because $ AE \equal{} OE \equal{} OA,\ AF \equal{} OF \equal{} OA$.

The condition $ \angle AOB < 120^{\circ}\iff \angle AOC > 60^{\circ}$ simply means that $ F$ (assume $ F$ is on the same side of $ OA$ as $ C$) lies on the arc $ AC$ which doesn't contain $ E$, so $ CA$ is the internal bisector of $ \angle ECF$ because $ A$ is the midpoint of the arc $ EF$ (without the condition, $ AC$ would be the external bisector). This means that $ J$ lies on the bisector of $ \angle ECF$.[/quote]
My solution was the same up to that point (though I think the $ J$ should have been $ I$ throughout the proof :P.) Here's how I finished it:

We have that $ m \angle OAC \equal{} m \angle ACO \equal{} \frac{m \angle BOA}{2} \equal{} m \angle DOA$, so $ DO \parallel{} AI$. Hence, $ DOIA$ is a parallelogram, so $ DO \equal{} AI$. On the other hand, $ \triangle AOF$ is equilateral, so we have that $ AO \equal{} AI \equal{} AF$, that is, $ AI \equal{} AF$, so $ \triangle AIF$ is isosceles.

Let $ m \angle AFE \equal{} \alpha$ and let $ m \angle EFI \equal{} \beta$. $ m \angle AIF \equal{} \alpha \plus{} \beta$ since $ \triangle AIF$ is isosceles, so $ m \angle FIC \equal{} 180 \minus{} \alpha \minus{} \beta$. Also, $ \alpha \equal{} m \angle EFA \equal{} m \angle AEF \equal{} m\angle ACF$, so $ m \angle IFE \equal{} 180 \minus{} (180 \minus{} \alpha \minus{} \beta \plus{} \alpha) \equal{} \beta$. Since $ m \angle EFI \equal{} m \angle IFC$, $ I$ lies on the bisector of $ \angle EFC$. But $ I$ also lies on the bisector of $ \angle ECF$, so $ I$ is the incenter of $ \triangle ECF$.

My solution is slightly different... unfortunately I missed the fact that $DO\|AI$...

WLOG let $E$ be on minor arc $BD$. Let $OI$ hit the circle at $X$ and $Y$ so that $Y$ is on minor arc $FC$. First we note that $AE=AF$, so $I$ is on the angle bisector of $\angle{ECF}$. Since $AO=AE=AF$, it suffices to show that $AO=AI$, which would imply that $A$, the midpoint of minor arc $EF$, is the circumcenter of $\triangle{EIF}$ and $I$ is thus the incenter of $\triangle{ECF}$ (since the incenter is on this circle, and we know that $I$ is on the angle bisector of $\angle{ECF}$). Using the facts that $AD=AB$ and $AD$ is parallel to $XY$, we have (interpret all arcs as those not containing any point in the interior of minor arc $CX$)
\begin{align*}
2\angle{AOI}=2\widehat{AY}=\widehat{AY}+\widehat{DX}&=\widehat{AY}+\widehat{DB}+\widehat{BX}\\
&=\widehat{AY}+\widehat{DA}+\widehat{CY}=\widehat{DY}+\widehat{CY}=\widehat{AX}+\widehat{CY}=2\angle{AIO},
\end{align*}as desired.

Hmm.. I did exactly as Zhero did up until his angle-chase. Note that $AF=AO=AI=AE$ so $FEOI$ is cyclic with centre $A$. Then $2\angle FEI=\angle FAI=\angle FAC=\angle FEC$ so $IE$ bisects $\angle FEC$. This was much easier than G1!?

[hide][img]http://oi54.tinypic.com/52b5l4.jpg[/img]

[color=#FF0000][mod edit: http://oi54.tinypic.com/52b5l4.jpg][/color][/hide]

Since $DO\perp AB$, we conclude $DO\parallel AC$ and, with $AD\parallel OI$, $ADOI$ is a parallelogram, i.e. $AI=OD$, so $AE=AI=AF$ and $I$ is, indeed, the incenter of $\triangle CEF$.

Best regards,
sunken rock

My solution :
[img]http://i41.servimg.com/u/f41/15/61/79/59/g310.png[/img]
Let's show that $I$ is the incenter of the triangle $\triangle{CEF}$:
First, it's easy to check that $OA=OE=AE=OF=AF$ and hence the quarilateral $AEOF$ is a rhombus, and we  have $A$ is the middle of the arc $EF$ and since $A$ lies in the perpendicular bisector of $EF$, it follow from a well-known fact that $CI$ is the bisector of $\angle{ECF}$.
Otherwise, from the condition $OI\parallel AD$, we have : $\angle{AIO}=\pi-\angle{CIO}=\pi-\angle{CAD}=\angle{DBC}$. Now let $A'$ be the symetric of $A$ wrt $O$, then $\angle{AOI}=\angle{DAO}=\angle{DAA'}=\frac{\pi}{2}-\angle{ACD}=\frac{\pi}{2}-\angle{DCB}=\angle{DBC}=\angle{AIO}$ since $CD$ is the bisector of $\angle{ACB}$.
Hence wde get : $AO=AI$ which means that points $E$,$I$,$O$ and $F$ are concylic ( In particular in circle with center $A$ ), then :
$\angle{FEI}=\frac{1}{2}\angle{FAI}=\frac{1}{2}\angle{FEC}$. And so $EI$ is the bisector of $\angle{FEC}$ and so we are done !

Very easy solution:

Since <BOD = <BCA then DO||AC and therefore AIOD is a parallelogram. Therefore AI = OD = OA = OE = OF = AF = AO = AE and therefore A is the centre of the circumcircle of quadrilateral FOIE. Therefore <FIE = 180º - <FAE / 2 = 120º since < FCE = 60º  and < FAE = 120º (since FAEO is a 60º-rhombus) then if I' is the incentre of CFE, then C, I, I' are collinear since < FCI = arc AF = arc AE = < ACE and FIEI' are concyclic since < FIE = 120º = 90 + <FCE / 2 = < FI'E then I = I'. Done.

[hide="Solution"]
Note that $AO$ is the perpendicular bisector of chord $EF$, so $AEOF$ is a rhombus. Thus, $AE=AF$, so $CI$ bisects $\angle{ECF}$. Furthermore, since $D$ is the midpoint of $\widehat{AB}$, $OD\perp AB$. But $BC$ is the diameter of Circle $O$, so $AC\perp AB$. It follows that $AI||OD$, and since $AD||OI$, $ADOI$ is a parallelogram.

Let $AO$ intersect $EF$ at $G$. Since $G$ is the midpoint of $AO$, $G$ is the midpoint of $DI$. But $G$ is also the midpoint of $EF$, so $DEIF$ is a parallelogram. Therefore, $\angle{FEI}=\angle{DFE}$.

Since $DF||EI$ and $OE||AF$, let $\angle{AFD}=\angle{OEI}=x$. We have $\angle{BOD}=\angle{AOD}=2\angle{AFD}=2x$. Since $OE=OC$ (equal radii), let $\angle{OEC}=\angle{ECO}=y$. Then $\angle{EOB}=2\angle{ECO}=2y$. Observe that now,
\[ \angle{FEI}=\angle{DFE}=\frac{\angle{DOE}}{2}=\frac{\angle{BOD}-\angle{EOB}}{2}=x-y\]
and $\angle{CEI}=\angle{OEI}-\angle{OEC}=x-y$, so $EI$ bisects $\angle{CEF}$. But $CI$ bisects $\angle{ECF}$, so $I$ is the incenter of $\triangle{CEF}$. //[/hide]

This is easily done with complex numbers as well.  WLOG $ S $ is the unit circle.  WLOG let $ A, B, C, D, I, E, F $ have complex coordinates $ a^2, -1, 1, d, x, e, f $ respectively.

Clearly $ d = -ia $.  Since $ OD \perp AB $ by definition and since $ AI \perp AB $ since $ BC $ is a diameter of $ S $ we have that quadrilateral $ ADOI $ is a parallelogram and so we have that $ x = a^2 + ai $.

Now, note that $ E, F \in S $ and that the projections of either point onto $ AO $ has complex coordinate $ \frac{a^2}{2} $.  This implies that $ e, f $ are the roots of the equation $ z^2 - a^2z + a^4 = 0 $.  Therefore by Vieta's formulas $ e + f = a^2 $ and $ ef = a^4 $.

Now since the circumcircle of $ CEF $ is $ S $ it suffices to show that $ x = -\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f}) $.  Choosing the appropriate sign for the square roots given that $ \angle{AOB} < 120 $ we have that $ \sqrt{ef} = -a^2 $ and since $ (\sqrt{e} + \sqrt{f})^2 = e + f + 2\sqrt{ef} = -a^2 $ we have that $ \sqrt{c}(\sqrt{e} + \sqrt{f}) = -ai $ so $ -\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f}) = a^2 + ai $ as desired.

This is literally the worst solution possible but I felt that I had to post it for its ridiculousness.  In fact, you can easily prove that $ ADOI $ is a parallelogram with solely complex numbers (it requires solving a simple linear system of two equations).

A bit too easy for an IMO #2?

By http://www.mit.edu/~evanchen/handouts/Fact5/Fact5.pdf as usual it suffices to show that $AJ = AE = AF$, but obviously $AE = AF = AO$ so we simply wish to check that $\angle AOJ$ is isosceles which is angle chasing.

OD is perpendicular to BA and ∠BAC = 90 so OD || AI and we have OI || AD so OIAD is parallelogram.
AI = OD = OA = OE = AE = AF ---> AI = AO = AE = AF ---> IFEO is cyclic and A is center.
∠FEI = ∠FAI/2 = ∠FEC/2 ---> ∠FEI = ∠IEC ---> EI is bisector of ∠FEA.
AF = AE ---> ∠ACE = ∠ACF ---> CI is bisector of ∠EAF.
we're Done.

Forgot to post this yesterday but oh well

[img width=50]https://media.discordapp.net/attachments/925784397469331477/930258088936755220/Screen_Shot_2022-01-10_at_5.33.27_PM.png?width=1324&height=1170[/img]

WLOG, let $E$ be on the same side of $C$ as $AO.$

[u]Claim 1:[/u] $CI$ bisects $\angle ECF.$

Note that $OF=OE,$ and $OF=FA,OE=EA,$ which implies that $OFAE$ is a rhombus, implying that $A$ is the midpoint of arc $EF,$ so Claim 1 is proven.

[u]Claim 2:[/u] $\angle FIE=90^\circ+\frac{1}{2}\angle ECF.$

Since $OF=OA$ in rhombus $OFAE,$ we have $\angle FOE=120^\circ$ and thus $\angle ECF=60^\circ.$

Note that $OD$ bisects $AB$ and $BC$ so $OD||AC,$ and from $OI||AD$ we have $OIAD$ parallelogram.

We have $\angle FOD=60^\circ-\angle DOA=60^\circ-\angle OAI=\angle IAE,$ so we have $\triangle FOD\cong\triangle IAE.$ Thus, $\angle OIE=\angle FDA=150^\circ=180^\circ-\angle OFE.$

This implies $FOIE$ cyclic, so $\angle FIE=\angle FOE=120^\circ=90^\circ+\frac{1}{2}\angle ECF,$ as desired.

Now, since $I$ is a unique point, and that point is the incenter, we are done.

hot take but this is geometry

The $\angle AOB<120^\circ$ condition implies that $E,F$ all lie on arc $BAC$. Since $A$ lies on the perpendicular bisector of $\overline{EF}$, it is the arc midpoint, hence $\overline{CA}$ bisects $\angle ECF$. Some angle chasing (write everything in terms of $\angle AOC$) gives $\angle DOA=\angle OAJ$, hence $DA \parallel OJ$ and $DOJA$ is a parallelogram, so $AJ=OD$. Further, by symmetry $AE=OE$ and $AF=OF$. Since these are all radii we are done by the Incenter-Excenter Lemma. $\blacksquare$

Wait this is an extremely stupid problem even if placed at IMO 1/4.

$\overline{OD}$ and $\overline{AI}$ are both perpendicular to $\overline{AB}$, so $ADOI$ is a parallelogram, and $$AI=OD=AE=AF.$$ Fact 5 implies $I$ is the incenter.

[quote=orl]The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$.  Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$[/quote]

First, note that AOE and AOF are equilateral (perp. bisectors from vertice). Trivially CI is an angle bisector since they both are inscribed angles of 60 degrees, and <ECF=60 degrees. Lemma: ADOI is a parallelogram. Proof: <DOB=2<DCB=<ACB, so by corresponding angles DO//AC, two pairs of parallel lines make a parallelogram. Now, we see that EF perp. AO, EF a chord, so EF is bisected by AO, but so is DI (diagonals are bisected in parallelogram), all concurrent at the midpoint of AO=midpoint of EF=midpoint of DI, so EDFI is a parallelogram. Now, we have 120=<EDF=<EIF, but there exists the unique point incenter iff in a triangle ABC incenter I we have <BIC=90+1/2<BAC, which is true, since 120=90+30, so I is the incenter of CEF, as desired. $\blacksquare$

Additional facts we could've derived: OI is the interior angle bisector of <FOG, since we know <EOF=<EIF, EOIF is cyclic => 1/2<FOG=<FEG=<FEI=<FOI=<FOH, as desired.

Edit: Because angle AOB never exceeds 120 degrees, it can be seen that E and F lie on opposite sides of AO

lol this is so stupid

Note that $ADOI$ is a parallelogram since
\[\angle BOD = \frac{\angle BOA}{2} = \angle BCA\]
and $AFOE$ is a rhombus, so \[AE = AF = FO = OD = AI\]
and Fact 5 finishes. $\blacksquare$

Claim: $A$ is the circumcenter of $\triangle EIF$. Let $P$ be the intersection of $AD$ and $BC$. Let $\angle ACD=\angle BCD=\alpha$. Then, $$\angle BAD=\alpha,$$ so $$\angle APB=90-3\alpha.$$ Thus, $\angle IOC=90-3\alpha,$ and since $\angle AOB=4\alpha$ we have $\angle AOI=90-\alpha$. However, $\angle AIO=90-3\alpha+2\alpha=90-\alpha$ as well, so $AO=AI$. Clearly, $AO=AE=AF$ since $\triangle AEO$ and $\triangle AFO$ are equilateral, so $AE=AF=AI$, thus $A$ is the circumcenter of $\triangle EFI$.\\

Consider $\triangle CEF$. $I$ lies on $CA$ (since $A$ is the arc midpoint), as well as the circle centered at $A$ passing through $E$ and $F$. Thus, $I$ is either the incenter or the $C$-excenter.\\

Claim: $I$ is inside the circle $S$. Note that $ADOI$ is a parallelogram since $\angle DOB=\angle ICO=2\alpha$ and we already know the other sides are parallel. Thus, $I$ is the reflection of $D$ over the midpoint of $AO$. Clearly, $\angle DAO<60$, thus $I$ lies inside the circle, hence done.

Note that triangles $AFO$ and $AEO$ are equilateral triangles so we get some easy angles. In particular, $\angle FCA=\angle ECA$.

[b]Claim.[/b] $DAJO$ is a parallelogram.
[b]Proof.[/b] $DA\parallel OJ$ by definition and
\[\angle BOD=2\angle BCD=\angle BCA \Longrightarrow DO\parallel AJ.\]

Let the midpoint of $AO$ be $M$. Then a reflection about point $M$ takes $F$ to $E$, $A$ to $O$, and $J$ to $D$. Therefore,
\[\angle FJC=180^{\circ}-\angle FJA=180^{\circ}-\angle EDO
=180^{\circ}-\left(90^{\circ}-\frac12\angle EOD\right)
=90^{\circ}+\frac12\angle EOD.\]
However,
\[\angle FEC=\angle FAC=60^{\circ}+\angle OAC=60^{\circ}+\angle ACB=60^{\circ}+\frac12\angle AOB=60^{\circ}+\angle DOA=\angle DOE\;\blacksquare\]

how did this make the shortlist

Note that $AEOF$ is a rhombus so
$$\angle ACE = \angle AFE = \angle FEA = \angle FCA,$$
and $I$ lies on the internal bisector of $\angle ECF$. But because
$$\angle DAC = 180^{\circ} - \angle DBC = 180^{\circ} - \angle ADO,$$
we find that $ADOI$ is a parallelogram, and thus $AI = DO = FO = AF = AE$. So, $A$ is the center of $(EFI)$, and since $A$ is also the midpoint of arc $EF$ not containing $C$, the conclusion follows by fact 5.

hello,

$AI=DO=AO=AE=AF$ so by Fact 5 done.

$\color{magenta}\boxed{\textbf{SOLUTION G3}}$

$\color{green} \textbf{Claim 1:}$ $CA$ is the angle bisector of $\angle ECF$
$\textbf{proof :}$ As $EF$ is the perpendicular bisector of $OA$, $A$ is the midpoint of arc $EF$ Hence done $\square$

$\color{green} \textbf{Claim 2:}$ $\triangle AOF$ is equilateral
$\textbf{proof :}$ Let $OA \cap EF \equiv X$
$\triangle AXF$ and $\triangle OXF$ are congruent. Hence $AF=OF=OA \square$

$\color{green} \textbf{Claim 3:}$ $ADOI$ is a parallelogram
$\textbf{proof :}$ $\angle DAC = 180- \angle DBC = 180 - \angle ADO$ and we have $AD \parallel OI \square$

So, $$AI = DO = FO = AF= AE$$
And by Incenter-Excenter Lemma we are done $\blacksquare$

It is easy to prove $CI$ is angle bisector. Just observe $AI=DO=AE$ and we are done.  :D

We can see that $EO = EA$ since this is the perpendicular bisector. Also, we can see that $EO = OA$. Thus, $\triangle OEA$ is equilateral. Similarly, $\triangle OAF$ is equilateral.

Now notice that $J$ lies on the angle bisector of $\angle ECF$. This is because $A$ is the midpoint of arc $EF$.

Now since $OD$ hits $AB$ at a $90$ degree angle and same with $CA$, we can see that $OD \parallel CA$. Thus $J$ is the reflection of $D$ over the midpoint of $OA$. This means that $\angle EJF = \angle EDF = 120$. Thus we can conclude. $\blacksquare$


You're missing the different orders you can get the composite and prime numbers. Hint: How many ways can you arrange the composites and primes?

You have to multiply your entire thing by $ \binom{6}{3} \equal{} 20$ because it doesn't matter which of the 6 dice are prime (as long as 3 are). Your method assumes that the first 3 dice are prime. Multipltying by 20 gets $ \frac {20}{216} \equal{} \boxed{\frac {5}{54}}$

Is that the answer?


Could anyone suggest some good books related to physics that go into depth of the topic. This is not for any entrance exam but out of pure interest in the subject..

Introduction to electrodynamics - DJ Griffith
Introduction to mechanics - Kleppner and Kolenkow

Try out Resnick Halliday and Irodov :yinyang:

Solving Irodov doesn't help much. I'm a live example of this  :P

For problems try Krotov.
For reading, Feynman's Lectures (but most of it goes over the head, not because of physics but maths. It is full of concepts like divergence, curl abt which we have zero knowledge)

Irodov is very good . I ahvent used krotov but heard its got bizzare problems.

I hope bizzare is good. I will try krotov,griffith..Thanks.
I already use resnick halliday, irodov (monster maths)

Do Feynaman's lectures come assembled in a book, Please provide any links regarding the same.

Are there any books that discuss theory of relativity starting from basics. I am really interested but don't know where to begin.

[quote="_MM_"]
Do Feynaman's lectures come assembled in a book, Please provide any links regarding the same.
[/quote]

They come in a three volume set. The name is 'Feynman's Lectures in Physics'
First deals with mechanics/optics
Second mostly on electromagnetism
Third on quantum theory.

I have books one and two. The third one is mostly impossible to understand because of new concepts and weird maths.
Although don't keep much expectations from the first two. It involves a lot of maths which we don't know.

@manaswi...seeing ur interest,i'll suggest u only 1 book.."At the speed of light" by G.Venkataraman

even the others of series "Vignettes in Physics" r good..

but i just cant stop crying.. :(

Arnav Tripathy(US IMO team 2006-2007) owed much thanks to Resnick Halliday as it helped him a lot while preparing for physics Olympiads.
HC Verma may provide a good introductory reading. :D

try out 'UNIVERSITY PHYSICS' by HUGH.D.YOUNG,ROGER.A.FREEDMAN,.....
ITS A VERY NICE BOOK FOR BASICS...

Actually,Feynman's lectures are not meant for the 11th and the 12th. :D

I will try to get my hand on feynman's lectures.

@scary: I want to read out of interest and as I have successfully passed out of 12th this year I am allowed to read it right?

Hey _MM_,

I just said what my Dad told me about Feynman.I have never seen the book myself till today! :rotfl: Please forgive me if I have offended u. :maybe: Besides,let me tell u that I am not very remarkable in Physics,I only know as much physics is required for the ICSE! :D

:rotfl:
I am not at all offended, infact its impossible to offend me.

So does anyone else have any other good books. I also wanted to ask if the feynman series is available online as an ebook to download.

@_MM_: thank you for the advice. Where are you studying now?

[quote="_MM_"][quote="Carolstar9"]@anish7: I don't think the answer of such a question would be much useful after 2 and a half years. :wink:  :P[/quote]

Yup, not really useful.. I have gone way out of physics in the past two and half years. And i suggest Berkley course.. Was a wonderful read and really covers basics of all the topics..especially if one can get a hand on all the 5 volumes.[/quote]

you see it depends .....

Umm, maybe, but in a case like this, it does not depend usually.

to study physics, it is more to analyse nature and derive the formulas yourself without knowing it before,
than reading more books.
consult not more than some classical books like of fyneman and irodov etc..

as for stydying physics in any collage,
i think[b] iisc bangaloru[/b] is better than other (including iit kanpur ,delhi, bombay)
as it is non commercial and truely study and research based.

actually irodov wont help much. i tried it. the problems are mechanical , they require less physics and more maths ( vectors , calculas etc.) for good theory , one should go for feynman lectures. resnick and haliday is toooo easy :D

I simply cant believe it that no one here has suggested Goldstein Classical Mechanics !!!! That is the best book with vector calculus treatment of mechanics and was an awesome read. Ofcourse for electrodynamics one could use Griffiths :D

Resnick and walter lewin + ramamurthy (yale ) lectues .
Hc verma
Later Go for Feynman . you can solve Irodov but I dont see any point in it once you have done problem solving through HC as there isnt much left to learn in these topics after that . And irodov basically involves a great deal of mathematics which is anyways kind of handicapped in physics taking dx as a small length to integrate :)

[quote="shubh123"]i think that one must be a good mathematician before being a student of physics.
maths denotes one's thinking ability(IQ).'
einstine,stephen haking, newton etc..were all a brilliant mathematician during collage days.[/quote]
einstein was pretty average in math in his college days.

try introduction to classical mechanics by david morin.it is far far harder than irodov and requires too much maths and too much physics.

too much maths & too much physics?? What did you really mean?

you need a lot creativity in maths and phy to solve some questions in them.


Putnam factorial algebra polynomial logarithms function probability

Alice and Bob play a game in which they take turns removing stones from a heap that initially has $n$ stones. The number of stones removed at each turn must be one less than a prime number. The winner is the player who takes the last stone. Alice plays first. Prove that there are infinitely many such $n$ such that Bob has a winning strategy. (For example, if $n=17,$ then Alice might take $6$ leaving $11;$ then Bob might take $1$ leaving $10;$ then Alice can take the remaining stones to win.)

My argument was based on [hide]density, considering the sets $\{n_{i}+p-1\}$ where $n_{i}$ were the (finite, by supposition) collection of numbers from which Bob could win and $p$ ran over the primes.  The primes have zero density, and a finite union of zero-density sets has zero density, so all numbers of this form have zero density, but by hypothesis they include all of $\mathbb Z_{\geq 0}$, a contradiction.[/hide]

A friend suggested that there is a more elementary proof which just requires you to find [hide]a very long string of composite numbers -- longer than the largest difference between any two of the $n_{i}$.[/hide]

I tried to argue this: Take all numbers of the form 2^x such that the number isn't itself a prime minus one, and that the number is not the sum of a prime minus one with a smaller number that's already in this set.  All of these numbers should have guaranteed wins for Bob if he plays his cards right.  For almost all of the possibilites Alice can choose from, Bob is left with a prime minus one so he can take them all and win.  For the remainder, within a move or two Bob can corner Alice into a situation where she has three stones to choose from (such a situation is a guaranteed win for Bob because whether Alice takes one or two, Bob can take the rest).  I ran out of time on this problem (I wasted most of my time on the darned problem with the sequence of 2006 numbers :roll: ).  I didn't have time to think up a proof of why my set is infinite, or indeed if it even works past the examples that I came up with.  Does anyone have any ideas?  Thanks :)

[quote="JBL"]Your set is infinite because $2^{2k+1}+1$ is divisible by 3 for all positive integers $k$; however, I'm intensely skeptical of the claim that all of those numbers really are wins for Bob.

In fact, I think I have a counter-example: from 64 stones (65 not being prime), Alive removes 53 to leave 11 and wins.[/quote]

But 54 is not prime, so Alice cannot remove 53 (if I understand the question correctly).  I'm pretty sure that 3, 8, 32, and 64 were wins for Bob, but 16 (which is 17-1) and 128 (which is 96+32) are loses for him.  That's why I added the caveat that n can't be a prime minus 1, and n can't be the sum of a prime minus 1 and a smaller guaranteed win for Bob (because then Alice could take away that prime minus 1 and leave Bob cornered)  I didn't have time to prove if this works beyond the examples I did, though.  I think it would be really cool if it did work :)

Sorry, you are indeed correct that my counterexample was completely wrong.  (54 is of course not prime; I got a bit confused about what I was doing  :oops: )

I also didn't read carefully:  you may be correct that your set, as described, is infinite.  However, it is not going to be easy to show.  In particular, the powers of 2 have no very nice relation to the primes, so there's nothing really convenient for you to expoit.

[quote="Psi"]... such that the number isn't itself a prime minus one, and that the number is not the sum of a prime minus one with a smaller number that's already in this set.[/quote]
This is exactly the description of the set of winning numbers for Bob, if we leave off the part about powers of 2. Trying to find more than a few of these winning numbers explicitly is a fool's errand- the additive structure of primes is really hard to understand.

my idea was exactly similar to the one mentioned by JBL.

This is a neat problem. I solved it (unofficially, of course) using the factorial giving arbitrarily long sequences of consecutive composite numbers. An interesting question that occurred to me while doing this was:

Is there an asymptotic approximation to $|B_{n}|$, where $B_{n}$ is the set of all numbers $\le n$ that Bob has a winning strategy on?

[quote="Kent Merryfield"][quote="paladin8"]Is there an asymptotic approximation to $|B_{n}|$, where $B_{n}$ is the set of all numbers $\le n$ that Bob has a winning strategy on?[/quote]
It's a good thing they [i]didn't[/i] ask that question. We only have three hours for six questions, after all![/quote]

Yes, I figured that it would be a lot more difficult than the given problem, but I was just curious about it :P.

I proved that for any number N, there exists two consecutive primes p_k, p_{k+1} with p_{k+1}-p_k>N.

The answer to the question I believe follows pretty directly from this lemma.

I ended up finding the third solution in the official solutions--multiplying everything together and subtracting one to get a new Bob win.  Once I realized that primes were critical, I knew there wasn't a nice formula for Bob's wins, so I figured it must be a proof by contradiction.  From there, I latched onto the idea of making a proof that looked like the proof that there are infinitely many primes, and that led me to the solution.

dear Mr Kent:

   I dont understand your solution to putnam 2006 a2:  [quote="Kent Merryfield"]How about this: let $T$ be the set of starting numbers for which Bob has a winning strategy. I find it useful to include $0$ as a member of $T.$  We can show that the members of $T$ that are smaller than $10$ are $0,3,$ and $8.$ (The next two are $11$ and $32,$ but we won't need them for this proof.)

After that, $n\in T$ if the difference between $n+1$ and any member of $T$ is a composite number.

So assume $T$ is finite. Let $t=\max(T).$ Consider $t!-1.$ Since we have that $t\ge 8,$ we then have that $t!-x$ is a composite number for any $x\le t.$[/quote]

so did u just show t!-1 is in T?

Does t!-x equals a composite number guarantees that it is in T?

Please teach me.

thank you very much

I didn't actually show that $t!-1$ is a member of $T.$  What I showed was this: under the assumption that $T$ was finite, if you start with $t!-1,$ it is impossible to bring it in one legal move to a previously known member of $T.$

For instance: suppose that the only members of $T$ that we know for sure are $0,3,8,11.$

We consider $n=11!-1.$ Then we cannot bring it in one legal step to any of the numbers $0,3,8$ or $11.$

We can't subtract $11!-1$ to get zero because $11!$ is composite.
We can't subtract $11!-4$ to get $3$ because $3\mid(11!-3),$ making $11!-3$ composite.
We can't subtract $11!-9$ to get $8$ because $8\mid(11!-8),$ making $11!-8$ composite.
We can't subtract $11!-12$ to get $11$ because $11\mid(11!-11),$ making $11!-11$ composite.

But that does not mean that $11!-1\in T.$ One could write a recursive computer program to check for membership in $T.$ Unless you're cleverer than I can see, it might take a fairly long time to execute this program for a moderately large value of $n.$ I think the odds are that $11!-1$ is not itself a member of $T,$ but that it can be brought in one legal step to a member of $T$ that isn't on our original short list. This argument falsifies the assumption that $T$ is finite but does not directly produce a member of $T.$

This is perfectly analogous to Euclid's classic proof of the infinitude of primes. Suppose there were finitely many primes. Multiply them together and add 1. This produces a number not divisible by any of our allegedly complete list of primes. This most emphatically does not prove that our newly constructed number is itself prime (although it might, accidentally, be so), but rather that it has prime factors not on our previous list.

So it is impossible to bring it in one legal move to a previously known member of T.  But what is being contradicted?  It is a proof by contradiction and I cannot see what is being contradicted.   I am really sorry for being so dumb.

What is being contradicted is the assumption that $T$ is finite (and that we've listed it all.)

It might be better to phrase it this way: the first $m$ elements of $T$ cannot be all of $T,$ no matter how big $m$ is. We prove this by taking those first $m$ elements of $T$ and constructing a number that implies there must be at least one element of $T$ outside our original list.

I just tried a semi-hand computation. Errors are highly likely, but if I somehow avoided errors, the list of members of $T$ starts out:

$0,3,8,11,32,35,56,59,64,67,118,\dots$

I just posted this as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153460]a computer programming challenge.[/url]

nice, ....,  what i am still not clear is that if "it is impossible to bring it in one legal move to a previously known member of T", then that implies T is finite?  It seems like that if T is finite, there is one more assumption in your proof that you did not state explicitly, namely, that every integer can be written as the sum of a member of T and p-1.  Can you like write out every single step in your proof?  I am still confused.

p.s.  I looked over the offiical solution and there seemed to be a typo on line 5 of the solution, instead of set B, they wrote set S, and the next statement:"Then for every nonnegative interger not in S, Alice must have some move on a heap of n stones leading to a position in which the second player wins." this statement makes no sense to me.

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2006s.pdf

Thank you so much for taking the time to answer my questions already, mr.  Kent.

Here's the other part of the argument:

Consider a game between two players $A$ and $B$ with the following conditions:
(1) We start with some number of objects in one or more piles.
(2) Each turn, a player removes objects from one or more piles. A move is legal or not depending only on the current piles and their contents, and there is always at least one legal move unless all piles are empty.
(3) The player to remove the last object wins.
(4) A goes first.

Under these conditions, we can classify all positions as "winning" or "losing", based on whether the player playing now can win with perfect strategy. The winning and losing positions can be constructed inductively as follows:
"Empty" is a losing position.
If a legal move is possible to a losing position, the current position is a winning position- just make that move.
If all legal moves would lead to winning positions, the current position is a losing position.

These two possibilities exhaust everything, and we can induct on size- if we have already determined which positions with less than $n$ objects win and lose, we can look at all moves from a particular position with $n$ objects, and determine whether it wins or loses.

In this language, $T$ is the set of losing positions. By the description above, any $n$ that we can't go to $T$ from in one move is a member of $T$. That's where the contradiction comes from; we have assumed that there aren't any more members of $T$.

Thanks to TripleM on the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153460]programming topic[/url], here are some conjectures. Can you prove them?

1. $T$ can be completely partitioned into pairs of the form $(n,n+3).$ That is, we can put parentheses into the list so it goes $(0,3),(8,11),(32,35),(56,59),(64,67),(118,121),\dots$

2. The set $T$ is "sparse" in the sense that $\sum_{n\in T\setminus\{0\}}\frac1n$ converges.

3. Asymptotically, the cardinality of $T\cap[0,N]$ is between $O(N^{-1/3})$ and $O(N^{-1/4}).$

finally i think i am starting to understand it.

I don't really believe that asymptotic guess- I think it's smaller than $n^{b}$ for any $b>0$. My guess would be that it's polynomial in $\log n$, or something close to that. $\log^{2}n$ doesn't look too bad based on the empirical data, although I think it's a hair bigger.

The count for (exactly) $k$-digit numbers:
$\begin{array}{cc}k&\#\\1&3\\2&7\\3&18\\4&20\\5&44\\6&48\\7&66\\8&82\end{array}$

A probabilistic argument for the asymptotics:
Let $T(n)$ be the number of elements of $T$ less than or equal to $n$, and suppose $T$ can be approximated by a nice function $f$.
The probability that a random integer close to $x$ is prime is approximately $\frac1{\log x}$, and the probability that a random integer close to $x$ is in $T$ is approximately $f'(x)$.
Therefore, $f'(n)\approx\prod_{0\le k<n}\left(1-f'(k)\cdot\frac1{\log(n+1-k)}\right)\approx \exp\left(-\int_{0}^{n-1}f'(x)\cdot\frac1{\log(n+1-x)}\,dx\right)$
Taking logarithms, $-\log f'(n)\approx\int_{0}^{n-1}\frac{f'(x)}{\log(n+1-x)}\,dx<\frac1{\log n}\int_{0}^{n}f'(x)\,dx=\frac{f(n)}{\log n}$

If we test $f(n)=cn^{s}$, we get that the left hand side is $O(\log n)$ and the right hand side is $O(\frac{n^{s}}{\log n})$, which is much larger when $s>0$. The guess $f(n)=c(\log n)^{s}$ makes the right hand side $O(\log n)$ and the left hand side $O(\log n)^{s-1}$, so $s=2$ should be close. I think there's still a $\log\log n$ factor or something like it, but this method's too crude to find that out.

There's some bad notation in this post (I basically use $t$ for everything), but the idea should be clear.

Claim: for all $t \in T$, we have that $t$ even implies $t+3 \in T$ and $t$ odd implies $t-3 \in T$.

Proof: By induction.  By calculation, it's true for $t \leq 12$.  Suppose that it's true for all $t < 2n$, and that $2n \in T$.  Then $2n+1, 2n+2 \not\in T$ (use $2n$ to get 2, 3).  Now, we want to consider $(2n+3)+1-t$ for each $t \in T_{<2n+3}$.  When $t = 2n$, we get 4, which is composite.  Otherwise, if $t$ is even then $2n+3+1-t$ is an even number larger than 4 and therefore composite.  If $t$ is odd then $t-3 \in T$, by hypothesis, so $2n+3+1-t = 2n+1-(t-3)$ is composite by the assumption $2n \in T$.  Thus $2n+3 \in T$ as well.

Now, suppose the assumption holds for $t < 2n+3$ but that $2n \not\in T$.  Then there is some $t \in T_{<2n}$ such that $2n+1-t$ is prime.  If $t$ is even, then by assumption $t+3$ is in $T$ and so $(2n+3)+1-(t+3)$ is prime and $2n+3 \not\in T$.  If $t$ is odd then we must have $t = 2n-1$.  But then $2n+3+1-(2n-1) = 5$ is prime, so $2n+3 \not \in T$.

By induction, we're done.

Assume there are finite losing positions $n = x_1 \le x_2 \le ,..., \le x_n$. We can find another losing position in the following way: suppose $a$ is the losing position we wish to find, then take $a$ to be much larger than $x_n$. We can then simultaneously satisfy $a \not\equiv x_1 -1 (mod p_i), a\not\equiv x_2 -1 (mod p_i)... a\not\equiv x_n -1 (mod p_i)$ for  large primes $p_i$. This $a$ is not a winning position since it does not lead to a losing one, so it must be a losing position.

Call a integer $n$ a [i]Loosing position[/i] when some player finds a heap of $n$ stones,and no matter what he does,he can't win the game.Call rest of the numbers [i]Winning position[/i].Observe that $x$ is a [i]Loosing position[/i] if and only if $0\ne x-(p-1)$ is a [i]Winning position[/i] for all $p<x$.

FTSOC,assume $\{a_1,a_2,\dots,a_r\}$ be the only [i]Loosing positions[/i] in increasing order.Also assume $0$ is a [i] Loosing position[/i].(So $a_1=0$).Hence for any integer $x>a_r$ there is a prime $p<x$ such that $x-(p-1)$ is either 0 or a [i]Loosing position[/i]. Let $$A_i=\{n|n=a_i+p-1\text{where p is a prime}\}$$

Since any number $x$ greater than $a_r$ is a [i]Winning position[/i] it must belong to some of $A_i$ where $i\in \{1,2,\dots,r\}$.Hence $$\mathbb N-\{1,2,\dots a_r\}\subset \cup_{i=1}^r A_i$$

Now pick a large natural number $N$.Number of elements of $A_i$ which are less than $N$ is $\pi(N-a_i+1)$ where $\pi(x)$ counts the number of primes less than $x$.Since $\pi(x)\sim \frac{x}{\ln x}$ hence for sufficiently large $N$
$$ \sum_{i=1}^r \pi(N-a_i+1)\le c \sum_{i=1}^r \frac{N-a_i+1}{\ln (N-a_i+1)} <N-a_r-r$$
which contradicts the assumption that all $x>a_r$ belongs to some of $A_i$ $\blacksquare$


geometry
angle bisector
congruent triangles
similar triangles

[quote="Interval"]Describe the locus of points that are equidistant from sides AB and AC of Triangle ABC.[/quote]

[hide="Not entirely sure"]Angle bisector of angle A?

If you drop perpendiculars from the angle bisector to side AB and AC, you get congruent triangles because you have similar triangles by AA and they share one side, so the altitudes are equivalent so the points on the line are equidistant from sides AB and AC.[/hide]

[quote="ch1n353ch3s54a1l"][quote="ragnarok23"][quote="Interval"]Book's answer:

Intersection of the perpendicular bisector of AB and AC.

Who is right?[/quote]
That only describes one point of the locus though...[/quote]

The locus doesn't have to be more than one point...[/quote]
Really? I didn't know that...

[quote="ragnarok23"][/quote][quote="ch1n353ch3s54a1l"][quote="ragnarok23"][quote="Interval"]Book's answer:

Intersection of the perpendicular bisector of AB and AC.

Who is right?[/quote]
That only describes one point of the locus though...[/quote]

The locus doesn't have to be more than one point...[/quote][quote="ragnarok23"]
Really? I didn't know that...[/quote]
Sure, look at this question:
What is the locus of points one to the right of (8,9)?

At risk of belaboring the obvious, the locus should be the sum of the points such that the length of the perpendicular to AB equals the length of the perpendicular to BC.  I don't think that the distance is defined in terms of the bisector of the segments.  Therefore, the locus should be a line segment rather than a single point.

I don't know how to prove this, but it seems to me that the most likely candidate is the bisector of angle BAC.

It is possible to prove that it is the angle bisector.  The proof involves setting up a right cyclic quadrilateral.  Also, it will not necessarily extend exactly to the third side; it depends on the various angles of the triangle.

Note that, if you are looking for the locus of points equidistant from all [i]three[/i] sides of the triangle, then the locus will be a single point.  That, however, will be the intersection of all three angle bisectors.

actually, it is a pretty simple proof.
no cyclic quadrilaterals needed :D
[hide]draw the angle bisector (extending infinitely).  if , from an arbitrary point, we draw an altitude from this line to each side, we have HL congruence (altitudes --> right triangle).  thus. it is the angle bisector.[/hide]

The right cyclic quadrilateral is a way to prove this, and also can be used to prove HL congruence.


[hide="hint"]Draw an altitude, set it equal to $\frac{x}{\sqrt{2}}$ and set $AD=x$ solve for that[/hide]

I've done that... (I might just be making a dumb mistake, its prettly late...)

[quote="kohjhsd"][quote="happyme"]I have no clue why I'm getting some strange answer on this, but:

Trapezoid ABCD is such that AB=49 cm is parallel to CD=187 cm. The angle ADC is equal 45 degrees. If the area of the trapezoid is 10620 sq cm, what is the number of centimeters in the length of line segment BC ?
Enter your answer in this box[/quote]

[hide]You can find the height by calculating

$\frac{10620}{\frac12 \cdot (49+187)}=90$.

This is also the length of $AH$ if $AH$ is the altitude drawn from $A$ to $DC$.

Then we know that $BK$ is $90$ as well if $BK$ is the altitude drawn from $B$ to $DC$. Then $HK=49$ and $KC=187-90-49=48$.

Then in triangle $BKC$, you find $BK=90$ and $KC=48$, so the length of hypotenuse $BC=\boxed{102}$.[/hide][/quote]

[hide]I tried this using my own method and I got BC is almost equal to 102.47438704378767836924283930106[/hide]

[quote="kohjhsd"][quote="happyme"]I have no clue why I'm getting some strange answer on this, but:

Trapezoid ABCD is such that AB=49 cm is parallel to CD=187 cm. The angle ADC is equal 45 degrees. If the area of the trapezoid is 10620 sq cm, what is the number of centimeters in the length of line segment BC ?
Enter your answer in this box[/quote]

[hide]You can find the height by calculating

$\frac{10620}{\frac12 \cdot (49+187)}=90$.

This is also the length of $AH$ if $AH$ is the altitude drawn from $A$ to $DC$.

Then we know that $BK$ is $90$ as well if $BK$ is the altitude drawn from $B$ to $DC$. Then $HK=49$ and $KC=187-90-49=48$.

Then in triangle $BKC$, you find $BK=90$ and $KC=48$, so the length of hypotenuse $BC=\boxed{102}$.[/hide][/quote]

I got that answer.


Two distinct numbers are chosen at random from $ \{1,2,3,4,5,6\}$. What is the probability that the quotient of the smaller number divided by the larger number is a terminating decimal? Expressing your answer as a common fraction.

Here are the possibilities:

1/2
1/4
1/5
2/4
2/5
3/4
3/5
3/6
4/5

So there are 9 ways, out of 15 total possibilities, thus the answer is $ \boxed{\frac{3}{5}}$.

right.  So choosing the numerator, then the denominator, we have 6*5=30 choices of fractions.  Those that aren't terminating are those that. when simplified, have a denominator of either 3 or 6.

Those fractions would be: 1/3, 2/3, 4/3, 5/3
                    as well as:  1/6, 2/6, 4/6, and 5/6.

There are a total of 8 repeating fractions; there must be 22 fractions with terminating decimals.  The probablitily is then

22/30=[b]11/15[/b]

to izzy: 3/6=.5 :lol:


Diophantine equation
number theory proposed
number theory

[quote="josef ondrej"]Prove that there are infinitely many positive integers n, such that all prime divisors of n^2 + 1 are smaller than n.[/quote]

Take any solution of the Pell equation $ n^2\equal{}2m^2\minus{}1$ with $ n>2$

$ 2<n$ and any prime $ p\neq 2$ which divides $ n^2\plus{}1$ divides $ m$ and so $ p\leq m\equal{}\sqrt{\frac{n^2\plus{}1}{2}}<n$

(and there are infinitely many solutions to this Pell equation which can be obtained from $ (m,n)\equal{}(1,1)$ with the transformation $ (m,n)\to(2n\plus{}3m,3n\plus{}4m)$)


trigonometry
inequalities
inequalities unsolved

Given a triangle ABC.None of its angles is obtuse.Prove the ineq:
$ \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}$≥$ 5R$
Determine when the equality holds.

[quote="SUPERMAN2"][quote="tuandokim"]try a=b=c :)[/quote]
Suppose a=b=c=1=>LHS=3
R=a/2sinA=1/(2sin60)=0.577350269
=>RHS=2.886751346
So the ineq is true.[/quote]
sorry for my stupidity
 :blush: the equality holds when a=b,c=$ \sqrt{2}a$

[quote="tuandokim"][quote="SUPERMAN2"][quote="tuandokim"]try a=b=c :)[/quote]
Suppose a=b=c=1=>LHS=3
R=a/2sinA=1/(2sin60)=0.577350269
=>RHS=2.886751346
So the ineq is true.[/quote]
sorry for my stupidity
 :blush: the equality holds when a=b,c=$ \sqrt {2}a$[/quote]
That's right.I think this ineq is nice.Try it.

I like your ineq very much .I try to solve this ineq so many times but I still can't solve it.
Can you tell me your solution or something ....?

[quote="SUPERMAN2"]Given a triangle ABC.None of its angles is obtuse.Prove the ineq:
$ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b}$≥$ 5R$
Determine when the equality holds.[/quote]

$ a^2 \plus{} b^2 \equal{} c^2 \plus{} 2ab\cos\gamma$ and  $ ab \equal{} \frac {2S}{\sin\gamma}$ , $ R \equal{} \frac {c}{2\sin\gamma}$

So we can write : $ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \equal{} \frac {ab}{c} \plus{} \frac {c}{ab}(c^2 \plus{} 2ab\cos\gamma) \equal{} \frac {2S}{c\sin\gamma} \plus{} \frac {c^3\sin\gamma}{2S} \plus{} 2c\cos\gamma \ge \frac {5c}{2\sin\gamma}$

$ \frac {2S}{c^2\sin\gamma} \plus{} \frac {c^2\sin\gamma}{2S}\ge \frac {5}{2\sin\gamma} \minus{} 2\cos\gamma$

Let $ t \equal{} \frac {2S}{c^2\sin\gamma}$  and we must prove : $ t \plus{} \frac {1}{t} \ge \frac {5}{2\sin\gamma} \minus{} 2\cos\gamma$

We can assume :$ \boxed{\frac {\pi}{2} \ge \gamma \ge \frac {\pi}{3}}$

$ S \le \frac {c^2\sin\gamma}{8\sin^2\gamma} \equal{} > t \le \frac {1}{4\sin^2(\frac {\gamma}{2})} \le 1$

$ f(t) \equal{} t \plus{} \frac {1}{t} \ge f(\frac {1}{4\sin^2(\frac {\gamma}{2}}) \equal{} \frac {1}{4\sin^2(\frac {\gamma}{2})} \plus{} 4\sin^2(\frac {\gamma}{2}) \ge \frac {5}{2\sin\gamma} \minus{} 2\cos\gamma < \equal{} >$  $ \frac {1}{4\sin^2(\frac {\gamma}{2})} \ge \frac {5}{2\sin\gamma} \minus{} 2$

Let $ f \equal{} ctg (\frac {\gamma}{2}) \ge 1$

Then :$ < \equal{} > f^3 \minus{} 5f^2 \plus{} 9f \minus{} 5 \equal{} (f \minus{} 1)(f^2 \minus{} 4f \plus{} 5)\ge 0$

Equality will be for $ a \equal{} b$ and $ \gamma \equal{} \frac {\pi}{2}$ :)

[quote="SUPERMAN2"]Given a triangle ABC.None of its angles is obtuse.Prove the ineq:
$ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b}$≥$ 5R$
Determine when the equality holds.[/quote]
This inequlity is equivalent to $ \frac{sinAsinB}{sinC}\plus{}\frac{sinBsinC}{sinA}\plus{}\frac{sinCsinA}{sinB}\geq\frac{5}{2}$
what exactly is$ \frac{1}{cotA\plus{}cotB}\plus{}\frac{1}{cotB\plus{}cotC}\plus{}\frac{1}{cotC\plus{}cotA}\geq\frac{5}{2}$
Let $ cotA\equal{}x,cotB\equal{}y,cotC\equal{}z$,In a triangle,we have $ xy\plus{}yz\plus{}zx\equal{}1$
then we have to prove that $ \frac{1}{x\plus{}y}\plus{}\frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x}\geq\frac{5}{2}$ with the condition $ xy\plus{}yz\plus{}zx\equal{}1$
That's a very old problem . And I'm sure it have been posted in the forum already :)

How to prove the last inequality
$ \frac{1}{x\plus{}y}\plus{}\frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x}\geq\frac{5}{2}$ when $ xy\plus{}yz\plus{}zx\equal{}1$

[quote="Jianxing113725"][quote="SUPERMAN2"]Given a triangle ABC.None of its angles is obtuse.Prove the ineq:
$ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b}$≥$ 5R$
Determine when the equality holds.[/quote]
This inequlity is equivalent to $ \frac {sinAsinB}{sinC} \plus{} \frac {sinBsinC}{sinA} \plus{} \frac {sinCsinA}{sinB}\geq\frac {5}{2}$
what exactly is$ \frac {1}{cotA \plus{} cotB} \plus{} \frac {1}{cotB \plus{} cotC} \plus{} \frac {1}{cotC \plus{} cotA}\geq\frac {5}{2}$
Let $ cotA \equal{} x,cotB \equal{} y,cotC \equal{} z$,In a triangle,we have $ xy \plus{} yz \plus{} zx \equal{} 1$
then we have to prove that $ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x}\geq\frac {5}{2}$ with the condition $ xy \plus{} yz \plus{} zx \equal{} 1$
That's a very old problem . And I'm sure it have been posted in the forum already :)[/quote]
Very nice solution!! The last is pretty easy to prove :)


I just bought the book "200 functional equations" by Mihai Onucu Drimbe (it's in Romanian; I've translated the title) and found some cute ones (I mean problems). They're not half as difficult as the one Lagrangia posted, but they're still nice:

1. Show that there are no functions f:N->N s.t. f(f(n)-n)=f(n)+1

2. Find all f:N\{0}->N\{0}, strictly increasing, s.t. f(nf(m))=m^2*f(mn).
Maybe I'll post some more a bit later

1) was solved before by me,search for it in the solved problems corner!!! :D
2) as for the 2nd one it is from the Latin American Contest also given at a local contest in a district in romania! the answer is f(n)=n^2, it's not very hard to prove, I don't have time now to post it, I'll do it later! :D

P.S. I don't have that book! :( :(
cheers! :D :D

i'm back!!!! :D

if n=1 then f(f(m))=m^2f(m)
if n=f(n) then f(f(n)f(m))=m^2f(f(n)m)
but f(f(n)m)=n^2f(mn) and we get that f(f(n)f(m))=m^2n^2f(mn) and with
f(f(m))=m^2f(m) we get that f(f(m)f(n))=f(f(mn)) iff f(m)f(n)=f(mn) (from the injectivity of f)

we prove easily that f(n)=n^2. considering that f(n)<n^2 or f(n)>n^2 we get to a contradiction so the only function is f(n)=n^2.

cheers!! :D


Consider the sequence $ a_{1}\equal{}1, a_{2}\equal{}2, a_{3}\equal{}5/2, ...$ satisfying $ a_{n\plus{}1}\equal{}a_{n}\plus{}(a_{n})^{\minus{}1}$ for $ n>1$. Show that $ a_{10000}>141$

Agh 141 was the name of my mechanics course! Agh!
Must........ gouge out......... own eyes........


Geometry, number theory, and algebra are nice in that once you figure out what seems to be a solution, checking your work is easy.  Counting problems, even at an elementary level, are very annoying to me because there is no way to check!

I divide problems into cases, and solve it.  It seems completely logical.  However, I try to check my answer by using a different method, and the answer turns out to be different.  Both methods seem logically correct.  When I am taking a test, I can't decide as to which one is correct.  What I really seem to need is a genius problem solver that can immediately say why this method was wrong.

Any tips for counting problems in general that are relevant to my problems?

Try your counting methods for smaller cases (ones that you can brute-force) and see if it still holds up. If it doesn't, see if you can figure out where the method over/under counts.

I think that with counting problems, once you find a method of counting for a specific problem, the main troubles arise from over- and under-counting, i.e.:

[b](1) [/b]Are you absolutely sure that your counting exhausts all the cases?

[b](2) [/b] Have you accounted for any overlap between cases?

[b](3) [/b] Are you counting something in order, when the order of something is irrelevant?

When there is a discrepancy between the answers that two counting methods give, it's usually because of accidentally over- or under-counting.  :)

Cool, I see.  Does overcounting/undercounting in this context mean something specific, or merely getting a bigger or smaller number?  Is it usually a problem of missing it by a multiple as opposed to a small number?


function
Asymptote
analytic geometry
graphing lines
slope
calculus
calculus computations

Find the relative extrema of: f(x)=9-x(x=<3)
                                            =x^2-3(x>3)(so, f(x) is a piece-wise function)
I'm not really sure about the answer. I get that f(x) has the relative minimum at x=3 and it has no relative maximum. Can anybody help me solve this problem? I know it's a bit easy for most of MLKer's. By the way, how do we prove that the function does not have absolute minimum or absolute maximum? My teacher told me that I have to compare the absolute extrema with the horizontal asymptote, but I don't know whether I should find both limit of f(x) when x approaches to negative/positive infinity or just one. Since in some problems, it does exist one absolute maximum or minimum, but not both(like x+1/x with x belongs to (0,positive infinity).

There is no maximum.  If you plot the function, it's pretty easy to see.  Up to $ x\leq 3$, you have a negatively-sloped line so it has no extremes.  Then for $ x>3$, you have an upward-turning parbola which is already on the upturn.

Remember, that if you are working on an interval, which in this case you are, you can test the endpoints to see if they're extremem values.  $ x\equal{}3$ is a minimum.