geometry
angle bisector
geometry proposed

Given an acute-angled triangle $ABC$. Let $D$ be the midpoint of its side $BC$, and let $E$ be an arbitrary point on the segment $AD$. Let $F$ be the orthogonal projection of the point $E$ on the segment $BC$. Let $M$ be an arbitrary point on the segment $EF$. Denote by $N$ and $P$ the orthogonal projections of the point $M$ on the lines $AC$ and $AB$.

Prove that the interior angle bisectors of the angles $PMN$ and $PEN$ are parallel.

[Source: Problem 6 in http://www.oemo.at/w/index.php?namespace=Beispiele&title=Geometrie%3A+Raach+2003%2C+Teil+1 .]

Well, I must say that this problem loses a great part of its attractiveness once one sees the crucial idea of the solution, but at least it is a nice training problem (or "example") and more appealing than a lot of other olympiad problems.

  Darij

I think this problem is in some Olympiad before. I don't remember the country. And if my memory is not bad, my solution seems not to be "terrible"  :) . I 'll post it later

I think so Grobber ;)
 BTW, the language in the website is German ?

[quote="grobber"]It's from a Balkan Olympiad, I think. Both those bisectors are parallel to the angle bisector of $\angle BAC$, right?[/quote]

Yes to both of your questions ;) . So, yes for the assertion that both angle bisectors are parallel to the angle bisector of the angle BAC (in fact, this is pretty clear for the angle PMN), and yes, the language is German.

Unfortunately, you are also right about Balkan... Now I see the problem was solved on http://www.kalva.demon.co.uk/balkan/bsoln/bsol002.html , and yes, the solution is the same as mine (is there another way? probably no...).

  darij

I write here my solution:
Let N', P' be the foot of the perpendiculars drawn form E to AC and to AB, respectively, and let a parallel line to BC at E intersect AC and AB in N'' and P'' respectively. Since N'E and NM are parallel as well as MP and EP', it is enough to show that <N'EN=<P'EP which is in the same time equivalent to show that <ENM=<EPM. But from the ciclicity of the quadrilaterals NMEN'' and PMEP'' we get that <ENM=<EN''M and <EPM=<EP''M. Finally, since N''P'' is perpendicular to EF and moreover E is the midpoint of N''P'', we get that triangle MN''P'' is isosceles, that is, <EP''M=<EN''M, which is what we wanted to show.

[quote=grobber]It's from a Balkan Olympiad, I think. Both those bisectors are parallel to the angle bisector of $\angle BAC$, right?[/quote]

Can someone please explain this solution? I proved the first half of it (that the angle bisector of BAC was parallel to the angle bisector of PMN), but couldn't get that the angle bisector of BAC was also parallel to the angle bisector of PEN.

BTW, Mathematical Olympiads 2000-2001 has a neat solution to this problem that completely sidesteps this approach.


function
integration
abstract algebra
real analysis
real analysis unsolved

Assume that $u\colon\Omega\to\mathbf R$ is a harmonic function on a bounded open connected subset $\Omega$ of $\mathbf R^{n}$. Show that if $u\ge0$ then either $u=0$, or $u>0$ in $\Omega$.

If $u(x)=0$ for some $x\in\Omega$, choose $\epsilon>0$ so that $B_{\epsilon}(x)\subset\Omega$. By the mean value property, $\frac{1}{m(B_{\epsilon}(x))}\int_{B_{\epsilon}(x)}u=u(x)=0$. Since $u\geq0$ this implies that $u=0$ almost everywhere on $B_{\epsilon}(x)$. Being continuous, this is only possible if $u=0$ everywhere on this set. This shows that $B_{\epsilon}(x)\subset\mbox{ker}(u)$, so the kernel is open. The kernel is also closed. $\Omega$ is connected, so $\mbox{ker}(u)=\Omega$.


roots real of:[tex]e^x=3x[\tex]  number algebra,aren't?

Could you please explain what exactly you want¿

Do you mean: prove that the solution isn't algebraical¿
Then the answer is easy, since the theorem of Lindemann-Weierstrass gives that for every algebraical $a \neq 0$ the number $e^a$ is transcendental.

i agree with zetax
and the same question for $e+\pi$


For every integer n with $ n > 1$ prove
$ n! < (\frac{n\plus{}1}{2})^{n}$

what doesn't make sense to you?

you have to prove for every integer n>1, that its factorial is smaller than half of the next consecutive integer.

which, is, btw very untrue. take n=5, for example.

[quote="unimpossible"][quote="galois01"]For every integer n with $ n > 1$ prove
$ n! < (\frac{n\plus{}1}{2})$[/quote]
that doesn't make sense to me...

8!  is bigger than 4.5[/quote]
You are correct I made a mistake I edit immediately  :oops_sign:

[quote="HTA"][quote="13375P34K43V312"]AM-GM on 1, 2, ..., n[/quote]
yes , it's kind of weak inequality
$ 1.2.3....n\leq (\frac{1\plus{}2\plus{}...\plus{}n}{n})^{n}\equal{} (\frac{n\plus{}1}{2})^{n}$[/quote]
I solved it using AM-GM but I wonder if someone could solve it for me using induction ?

[quote="A M Y"][b]Base Case:[/b] $ 2! \equal{} 2 <\frac{9}{4}\equal{}\left(\frac{2\plus{}1}{2}\right)^{2}$

[b]Induction:[/b]
$ n! <\left(\frac{n\plus{}1}{2}\right)^{n}\iff 2^{n}n! < (n\plus{}1)^{n}\iff 2^{n}(n\plus{}1)! < (n\plus{}1)^{n\plus{}1}$

$ \left(\frac{n\plus{}2}{n\plus{}1}\right)^{n\plus{}1}\equal{}\left( 1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}\equal{}\sum_{k \equal{} 0}^{n\plus{}1}\binom{n\plus{}1}{k}\left(\frac{1}{n\plus{}1}\right)^{k}$ $ >\sum_{k \equal{} 0}^{2}\binom{n\plus{}1}{k}\left(\frac{1}{n\plus{}1}\right)^{k}\equal{} 2$

Thus, $ 2\cdot 2^{n}(n\plus{}1)! <\left(\frac{n\plus{}2}{n\plus{}1}\right)^{n\plus{}1}(n\plus{}1)^{n\plus{}1}\equal{} (n\plus{}2)^{n\plus{}1}$ $ \iff (n\plus{}1)! <\left(\frac{n\plus{}2}{2}\right)^{n\plus{}1}$[/quote]
Thank you my friend AMY for your solution  :)

[quote="HTA"][quote="13375P34K43V312"]AM-GM on 1, 2, ..., n[/quote]
yes , it's kind of weak inequality
$ 1.2.3....n\leq (\frac{1\plus{}2\plus{}...\plus{}n}{n})^{n}\equal{} (\frac{n\plus{}1}{2})^{n}$[/quote]

Well actually we want to prove that it's strictly less than, but this is easy because equality occurs iff 1=2=...=n, which is clearly not true.


Let a, b, c be positive real numbers such that $ abc\equal{}1$. Prove that
\[ \frac{a^4\plus{}1}{a\plus{}1}\plus{}\frac{b^4\plus{}1}{b\plus{}1}\plus{}\frac{c^4\plus{}1}{c\plus{}1} \ge 3
\]

[quote="zaya_yc"][quote="STARS"]Let a, b, c be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {a^4 \plus{} 1}{a \plus{} 1} \plus{} \frac {b^4 \plus{} 1}{b \plus{} 1} \plus{} \frac {c^4 \plus{} 1}{c \plus{} 1} \ge 3
\]
[/quote]

$ x^{4} \plus{} 1\geq\frac {1}{2}(x^{2} \plus{} 1)^{2} \equal{} \frac {1}{2}(x^{2} \plus{} 1)(x^{2} \plus{} 1)\geq\frac {1}{2}*2x\frac {1}{2}(x \plus{} 1)^{2} \equal{} \frac {x(x \plus{} 1)^{2}}{2}$------->$ \frac {x^{4} \plus{} 1}{x \plus{} 1}\geq\frac {\frac {x(x \plus{} 1)^{2}}{2}}{x \plus{} 1} \equal{} \frac {x^{2} \plus{} x}{2} \equal{} \frac {1}{2}(x^{2} \plus{} x)$-------> $ \frac {x^{4} \plus{} 1}{x \plus{} 1}\geq\frac {1}{2}(x^{2} \plus{} x)$------->

$ \frac {a^{4} \plus{} 1}{a \plus{} 1} \plus{} \frac {b^{4} \plus{} 1}{b \plus{} 1} \plus{} \frac {c^{4} \plus{} 1}{c \plus{} 1}\geq\frac {1}{2}(a^{2} \plus{} a) \plus{} \frac {1}{2}(b^{2} \plus{} b) \plus{} \frac {1}{2}(c^{2} \plus{} c) \equal{} \frac {1}{2}(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} a \plus{} b \plus{} c)\geq\frac {1}{2}*6\sqrt [6]{a^{3}b^{3}c^{3}} \equal{} 3$

$ a \equal{} b \equal{} c \equal{} 1$[/quote]

Nice solution, zaya. What do you think about the general problems:

Let a, b, c, x be positive real numbers. Prove that
\[ \frac {a^{x\plus{}2} \plus{} 1}{a^xbc \plus{} 1} \plus{} \frac {b^{x\plus{}2} \plus{} 1}{b^xca \plus{} 1} \plus{} \frac {c^{x\plus{}2} \plus{} 1}{c^xab \plus{} 1} \ge 3
\]


induction
geometry
real analysis
real analysis unsolved

I find this absolutely beautiful: take $ B(x_1,r_1),...,B(x_n,r_n)$ and $ B(y_1,r_1),...,B(y_n,r_n)$ two sets of balls in $ R^k$ (for some $ k$) such that $ \parallel{}x_i \minus{} x_j\parallel{}\geq \parallel{}y_i \minus{} y_j\parallel{}$ (the norm is the euclidean standard one). Then the intersection of the balls with centers $ x_1,...,x_n$ has volume smaller than or equal to the intersection of the balls with centers $ y_1,...,y_n$. It would be very nice to see a purely geometrical proof, the one I saw is quite heavy probability.

induction maybe?

 at n=2, $ \{B(x_1,r_1), B(x_2,r_2)\}$, and $ \{B(y_1,r_1), B(y_2,r_2)\}$ with $ |x_1 - x_2| \ge |y_1 - y_2|$

then  $ x_1 = x_2 + v_x$, $ y_1 = y_2 + v_y$ and $ |v_x| \ge |v_y|$

geometrically we can see that the area  $ B(a + v_x,r_1) \cap B(a,r_2)$ will be smaller than $ B(b + v_y,r_1) \cap B(b,r_2)$

so assume for n=j, and try to prove it works for n=j+1

${ \{B(x_1,r_1), B(x_2,r_2), ..., B(x_{j + 1},r_{j + 1})\ }$, and $ \{B(y_1,r_1), B(y_2,r_2), ..., B(y_{j + 1},r_{j + 1})\}$ with $ |x_i - x_l| \ge |y_i - y_l|$

then letting $ A = \cap_{i = 1}^jB(x_i,r_i)$ and $ B = \cap_{i = 1}^jB(y_i,r_i)$

so we consider $ A \cap B(x_{j + 1},r_{j + 1})$ and $ B \cap B(y_{j + 1},r_{j + 1})$ convex open sets

here $ x_{j + 1} = x_i + v_i$, $ y_{j + 1} = y_i + w_i$ where $ |v_i| \ge |w_i|$

i want to wave my hands and say the same sort of intuitive argument works here as the first case?

It is a news for me that this statement has been proved at last. A few years ago it was an open problem. The case when you can obtain $ x_j$ from $ y_j$ by a continuous motion increasing the distances all the way along is relatively simple. Around 2001, Bezdek and Connelly did the general case on the plane by a very clever trick. I haven't heard of any progress since then until this post. Are you sure that the proof is correct, harazi? If so, can you give a reference?

Well, I'm really not good at this kind of stuff, but I do love such statements. Here is a reference for this fact. I'm sorry, it's not an article, just some lectures notes.

Slepian's lemma and its proof are both classical things but the remark at the end of the note is a complete nonsense. There is no way to reduce the problem about the balls to that about the maximum of gaussians centered at $ 0$.

Oops, sorry, when I read this I thought it was clear, but actually I am completely wrong. Sorry again.  :oops:  :oops:


geometry
3D geometry
function
floor function

you are given a rectangular cube of unkown dimensions. if you place as many cubes as possible, each with volume 2, in the box, with their edges parallel to the edges of the box, you will fill exactly 40% of the box.
find the dimensions of the box.

How can you only fill 40%? Can't you just put twice as many cubes as you have at the moment to get at least 80%?

[quote="worthawholebean"]I think he means "touching the edges."[/quote]

Nope!
for example take a cube of side length 2.  then you can put only one cube w/ side length $\sqrt{2}$ in it, which contains only abot 35% of the volume.

same situation here (different numbers of course!)

[quote="1=2"]uuhhh.. The side length is $\sqrt[3]{2}$, since it's a cube and not a square...

and $\sqrt{2}^{3}=2\sqrt{2}$[/quote]

you clearly did not read my post clearly...the numbers that i used have nothing to do with the problem. i was just giving an example.  in fact $\frac{2\sqrt{2}}{8}=0.35$ (approximately) which is why i said 35%.

this problem is related to floor functions and such, it might be better if a mod moved this to pre-olympiad.

[quote="cincodemayo5590"]Well we can just find the volume of the box, then bound its dimensions...

It looks like the only possible way of this happening, with as many cubes packed in as possible, is having only one cube; so for the outer cube with side length $x$ we have $2=\frac25x^{3}\implies x=\sqrt[3]5$. Then, we can do whatever we want to the side lengths as long as for dimensions $l,w,h$ we have that $\sqrt[3]2 \le l,w,h < 2(\sqrt[3]2) \ \wedge \ l \cdot w \cdot h=5$.[/quote]
but l,w,h must be integers.
sorry if it was unclear from the question

someone please solve this!! :maybe:

the bounds for the sides posted above dont apply in the general case.
for example ifyou want to put two small cubes, then two of the sides must at least be $\sqrt[3]{2}$ long and one other must be at least $2\sqrt[3]{2}$ and less than $3\sqrt[3]{2}$

well,
if you are given a rectangular box of dimensions $l,w,h$ then the maximum number of cubes with volume two that you can place in it is $x\cdot y\cdot z$ where
$\lfloor x\sqrt[3]{2}+1\rfloor=l$
$\lfloor y\sqrt[3]{2}+1\rfloor=w$
$\lfloor z\sqrt[3]{2}+1\rfloor=h$

also you are given $xyz=\frac{1}{5}lwh$
someone take it form here....

ah. I see. So the dimensions must be greater than 1.

[hide="Well,"]5 2 and 3[/hide] work.

I don't have a rigorous solution except guess and check for n cubes with volume 2...

Sorry, but I'm a mere 11 year old who pwns basics, but flunks intermediate.  :wink:

this problem was also posted in Olympiad NT,
its not an intermediate level problem, its IMO 1976 # 3.
a mod should move this to pre-olympiad.


continued fraction
number theory unsolved
number theory

can someone plz help me with this:

fraction continue for a=sqrt((d^2)±1) for integer d. show both cases...

First case: $\sqrt{d^2+1}$ (WLOG $d > 0$ since $d<0$ is the same):
$d< \sqrt{d^2+1} < d+1$, thus continued fraction begins with $d$: $[d; ...]$.
Now lets look at $\sqrt{d^2+1}-d$, so we are interested in $\frac{1}{\sqrt{d^2+1}-d}=\frac{\sqrt{d^2+1}+d}{d^2+1-d^2} = \sqrt{d^2+1}+d$. Now the greatest integer smaller than that number is $2d$, and subtracting it we get again $\sqrt{d^2+1}-d$.
Thus, we will have to continue this, everytime getting $2d$, so the continued fraction is $[d;2d,2d,2d,...] = d + \frac{1}{2d+\frac{1}{2d+\frac{1}{...}}}$.

Second case: $\sqrt{d^2-1}$ (WLOG $d > 0$ again):
$d-1 < \sqrt{d^2-1} < d$, thus it begins with $d-1$ this time: $[d-1; ...]$.
Now the number we are looking at is $\sqrt{d^2-1}-(d-1)$, so we calculate $\frac{1}{\sqrt{d^2-1}-(d-1)} = \frac{\sqrt{d^2-1}+d-1}{d^2-1-d^2+2d-1} = \frac{\sqrt{d^2-1}+d-1}{2(d-1)}=1 + \frac{\sqrt{d^2-1}-(d-1)}{2(d-1)}$.
$\frac{\sqrt{d^2-1}-(d-1)}{2(d-1)}$ is clearly $>0$ and $<1$ by the previous, so we see that it continues with $1$ in the fraction and we next look at $\frac{\sqrt{d^2-1}-(d-1)}{2(d-1)}$:
$\frac{2(d-1)}{\sqrt{d^2-1}-(d-1)} = \frac{2(d-1)(\sqrt{d^2-1}+(d-1))}{d^2-1-d^2+2d-1}= \sqrt{d^2-1}+(d-1)$. By subtracting $2(d-1)$, we are where we started, at $\sqrt{d^2-1}-(d-1)$.
Thus the continued fraction is $[d-1;1,2(d-1),1,2(d-1),...]$.


One laser blast will break asteroids larger than 20 kg into three pieces,
each one third of the mass of the original. Asteroids smaller than 20
kg are shattered into dust by the laser. How many laser blasts would
be required to reduce a 2000 kg asteroid to dust?

[hide]Verify your solution here: http://www.projecteureka.org/problem/question/251[/hide]

[hide="My Solution"]
First, we find how many "rounds" of blasts there are.

$ \frac{2000}{3^x}<20$

$ 2000<20\times3^x$

$ 100<3^x$

$ x\equal{}4$

So there are 5 rounds of blasts because it goes from $ 3^0$ to $ 3^4$. To find the answer, we add these powers of 3.
$ 3^0\plus{}3^1\plus{}3^2\plus{}3^3\plus{}3^4\equal{}1\plus{}3\plus{}9\plus{}27\plus{}81\equal{}\boxed{121}$
[/hide]

Wouldn't there be 6 rounds of blasts, since you have to reduce the last ones to dust?
If that's true the answer would be 121+243=364

I guess you are right. The answer is then 364.

uh... can you please post the accuracy? I mean if you miss 99% of the time then the answer would be 3460.


Given are a circle $ w$, two fixed points $ \{A,B\}\subset w$ , the midpoint $ C$ of $ [AB]$ and the point $ D\in (AB)$ so that $ AD \equal{} 2\cdot DB$ . For a mobile point

$ M\in w$ denote the second intersections $ M$ , $ N$ of the circle $ w$ with the lines $ MC$ , $ MD$ respectively. Prove that the ratio $ \frac {AM\cdot BN}{MN}$ is constant.

[quote="Virgil Nicula"] For a mobile point

$ M\in w$ denote the second intersections $ M$ , $ N$ of the circle $ w$ with the lines $ MC$ , $ MD$ respectively. [/quote]

This section is unclear to me.

You are stating for a mobile point M belonging to the
circle, but not a mobile point N of that circle?

I do not understand "the second intersections of M, N
of the circle with the lines MC, MD respectively."

I got these out of this problem:

Points can vary in the circle to a degree, with these:

1) A & B will be fixed.
2) C is a midpoint.
3) D lies 2/3 of the way from distance from A to B.
4) M is mobile.
5) I don't know where in the circle, for sure, N is (my posts above).

Sorry, I corrected now. Thank you for your remark.

[quote="Virgil Nicula"][color=darkred]Given are a circle $ w$, two fixed points $ \{A,B\}\subset w$ , the midpoint $ C$ of $ [AB]$ and the point $ D\in (AB)$ so that $ AD \equal{} 2\cdot DB$ . For a mobile point

$ P\in w$ denote the second intersections $ M$ , $ N$ of the circle $ w$ with the lines $ PC$ , $ PD$ respectively. Prove that the ratio $ \frac {AM\cdot BN}{MN}$ is constant.[/color][/quote]

Dear Virgil,
you very nice result look like to have as a certain relation with the Haruki's result.
Please, it would be nice to explain this.
Very interesting Virgil.
Sincerely
Jean-Louis

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension[/u].[/b] Given are a circle $ w$, two fixed points $ \{A,B\}\subset w$ , two mobile points $ \{C,D\}\subset (AB)$ so that $ 2\cdot \frac {CA}{CB} \equal{} \frac {DA}{DB}$ .

For a point $ P\in w$ denote the second intersections $ M$ , $ N$ of the circle $ w$ with $ PC$ , $ PD$ respectively. Prove that $ \frac {AM\cdot BN}{MN}$ is constant.[/color][/quote]

[color=darkblue]I don't know [u][b]Haruki[/b]'s result[/u]. I"ll come back later with my proof (one row !).[/color]

Dear Virgil,
here is the Haruki's result:

let (1) be a circle, E and D two points on (1), A another point on (1), B and C two points on the arch of (1) which doesn't contain A, and I, J the second points of intersection of AB, AC with chord DE.
Prove that the ration (IE.JD)/IJ is constant.

Sincerely
Jean-Louis

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension[/u].[/b] Given are a circle $ w$, two fixed points $ \{A,B\}\subset w$ , two mobile points $ \{C,D\}\subset (AB)$ so that $ 2\cdot \frac {CA}{CB} \equal{} \frac {DA}{DB}$ .

For a point $ P\in w$ denote the second intersections $ M$ , $ N$ of the circle $ w$ with $ PC$ , $ PD$ respectively. Prove that $ \frac {AM\cdot BN}{MN}$ is constant.[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ \frac {DA}{DB}\equal{}2\cdot \frac {CA}{CB}\ \Longleftrightarrow\ \{A,C,D,B\}$ is harmonically $ \Longleftrightarrow$ $ P(A,C,D,B)$ is harmonically $ \Longleftrightarrow$

$ ABNM$ is harmonically $ \Longleftrightarrow$ $ AM\cdot BN\equal{}AB\cdot MN$ $ \Longleftrightarrow$ $ \frac {AM\cdot BN}{MN}$ is constant.[/color]

[color=darkblue]In conclusion, now we can present the "[u][b]H. Haruki & V. Nicula[/b]'s result[/u]" thus :[/color]  :o :lol:

[quote="Nicula & Haruki"][color=darkred]Given are a circle $ w$, two points $ \{A,B\}\subset w$ and two points $ \{C,D\}\subset (AB)$ so that $ C\in (A,D)$  . For a point $ P\in w$ denote

the second intersections $ M$ , $ N$ of the circle $ w$ with $ PC$ , $ PD$ respectively. Prove that $ \boxed {\ \frac {CA\cdot DB}{CD} \equal{} \frac {MA\cdot NB}{MN}\ }$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] $ \left\|\begin{array}{c} \frac {CA}{CD} \equal{} \frac {PA}{PD}\cdot\frac {MA}{MN} \\
 \\
\frac {BD}{BA} \equal{} \frac {PD}{PA}\cdot\frac {NB}{AB}\end{array}\right\|\ \bigodot\ \implies\ \frac {CA\cdot DB}{CD} \equal{} \frac {MA\cdot NB}{MN}$ .[/color]
[/quote]

Dear Virgil and mathlinkers,
nice to have presented your result in relation with Haruki...
We this new tool, we have to search new exercises... a new challenge...
Sincerely
Jean-Louis


calculus
integration
function
derivative
geometry
real analysis
calculus computations

I am having a little trouble distinguishing between the two "parts" of the fundamental theorem (as stated by wikipedia).
I am pretty sure that the first part says that if you take the definite integral of a function and obtain a new function, then the derivative of this new function is the original function.  The second part seems to say that if you take the derivative of a function and obtain a new function, then the integral of this new function is the original function.  Is this right?

One thing that I don't understand is why the fundamental theorem has two parts instead of just one, as it seems to me that the first part of the theorem directly implies the second part (and vica versa).  In other words, it almost seems redundant to me to include both parts rather than one or the other.

Finally, I am unable to figure out which part of the theorem is actually used to calculate definite integrals.  For example, in the process required to figure out the area under the graph of f(x) = 2x from x= 1 to 3, which part of the theorem is used?  Because the two parts seem essentially equivalent to me, I can't answer this question; it seems to me that you could use either part to evaluate the definite integral.

[color=green]There is no need to quote the entire post immediately preceding yours.[/color]
People have a different take on what to call which part the first part or second part so I am going to distinguish them by calling the first one(in wikipedia) the one for indefinite integrals while the other one is for definite integrals. FTC for indefinite integrals basically says that if we have $ g(x) \equal{} \int_{a}^{x} f(t) dt$ for $ g(x)$ continuous in the interval $ [a,b]$ and $ x$ is contained in that interval, then $ g'(x) \equal{} f(x)$. The one for definite integrals should be quite familiar, let's say there is a function $ f(x)$ whose antiderivative is $ F(x) \plus{} C$. Then the 2nd part or for definite integrals states that $ \int_{a}^{b} f(x) dx \equal{} F(b) \minus{} F(a)$.
It has two parts because one is strictly for definite integrals while the other is conceptually used for indefinite integrals. To answer your problem, you would use the one for definite integrals, or the 2nd one(from wikipedia) and $ \int_{1}^{3} 2x dx$ would be what you are trying to solve. Since $ x^{2}$ is the antiderivative of $ 2x$ the 2nd part states that the integral is then equivalent to $ 9 \minus{} 1 \equal{} 8$.

[quote="ckck"]To answer your problem, you would use the one for definite integrals, or the 2nd one(from wikipedia) and $ \int_{1}^{3} 2x dx$ would be what you are trying to solve. Since $ x^{2}$ is the antiderivative of $ 2x$ the 2nd part states that the integral is then equivalent to $ 9 \minus{} 1 \equal{} 8$.[/quote]

Right: without the FTC2, when you were asked to compute a definite integral you'd have to write down a Riemann sum and compute its limit.  (Maybe you did this a couple of times in your class or homework?)  This is really nasty, and it makes it effectively impossible to integrate all but the simplest functions.  However, FTC2 tells you that in order to do this, you just need to find an antiderivative and compute its values at a couple of points.  The universe of functions for which this is doable is much, much larger than the universe of functions that are realistically Riemann-summable by hand.


E narode mozete li reci nesto vise o sebi i vasim dosadasnjim rezultatima.
Ko od vas ide ove godine na IMO
Je li neko bio vec prije i kakve je imao rezultate?

Da krenem od sebe - Zavrsio sam 3. razred srednje skole (Sarajevo koledz).
Ove godine sam clan BiH IMO tima i ovo mi je prva IMO na kom ucestvujem.
Prethodno sam dva puta bio ucesnik na juniorskoj balkanskoj matematickoj olimpijadi i oba puta sam osvojio bronzanu medalju.
Ove godine sam na mediteranskoj M.O. imao 23 boda i valjda bude srebro.
Skromno al opet sam prezadovoljan...

Pozdrav iz Sarajeva...

Zanima me kako su vam organizirane pripreme, ko ih drzi?

Kod nas u BiH pripreme su organizovane na PMF-u a drze ih profesori sa PMFa kao i bivsi olimpijci.
Pet dana cemo boraviti  na nekoj od planina gdje cemo imati nesto poput kampa.
Najcesce izutra radimo test a popodne imamo predavanja.

Kakva je situacija kod vas u Hrvatskoj?

Eh da i ja sebe predstavim.Imenom i prezimenom Vedran Karahodžić završio 3.razred sarajevske Druge Gimnazije.Ove godine idem na IMO u Vijetnam i to će biti moj drugi IMO.
Prošle godine u Sloveniji sam osvojio bronzanu medalju,pored IMO učestvovao sam dvaput na JBMO gdje sam osvojio jednu bronzanu i jednu srebenu medalju(bod do zlata,najbolji BiH rezultat ikad na JBMO  :D ).Što se tiče mediteranskih olimpijada imam 1 bronzanu medalju i ove godine sam tu negdje između bronze i srebra pa ćemo vidjeti... Još da dodam da sam triput zaredom na državnima za osnovne škole bio prvak što je teško ponovljiv niz  :P  i eto to je ono najvažnije...

 Djuro,što se tiče priprema otprilike je ovako što je delegat rekao samo da dodam da nam od ponedeljka počinju pripreme na Bjelašnici finalne..mislim da ove godine imamo finu ekipu

[quote]Eh da i ja sebe predstavim.Imenom i prezimenom Vedran Karahodžić završio 3.razred sarajevske Druge Gimnazije.Ove godine idem na IMO u Vijetnam i to će biti moj drugi IMO.
Prošle godine u Sloveniji sam osvojio bronzanu medalju,pored IMO učestvovao sam dvaput na JBMO gdje sam osvojio jednu bronzanu i jednu srebenu medalju(bod do zlata,najbolji BiH rezultat ikad na JBMO Mr. Green ).Što se tiče mediteranskih olimpijada imam 1 bronzanu medalju i ove godine sam tu negdje između bronze i srebra pa ćemo vidjeti... Još da dodam da sam triput zaredom na državnima za osnovne škole bio prvak što je teško ponovljiv niz Razz i eto to je ono najvažnije...

[/quote]

Eto osvojis li jos jednom bronzu i jednom srebro na IMO mislim da ces biti najbolji BiH olimpijac
Pozdrav...

Sta ovo nikog nema na podforumu :?:

da zaboravio sam pitati, sto nece mali Uljarevic biti najbolji BiH natjecatelj. vec me odero prosle godine pa nadam se da ce biti jos bolji ove.
ukratko pozelite mu srecu od Gorana ove godine.
Ovim putem bih pozelio srecu svim hrvatskim natjecateljima na IMOu(posto ne znam kad cu slijedeci put do kompjutera) a takodjer i svim natjecateljima susjednim drzava. u svakom slucaju zabavite se ;)

Hvala na lijepim zeljama, Gorane! Potrudicu se da pozdrave i lijepe zelje prenesem i Igoru...;)


linear algebra
matrix
invariant
algebra
polynomial
linear algebra unsolved

Find the Jordan form of the following matrix:

$ A=$\[ \left( \begin{array}{ccccccccc}
0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 4 & 2 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 3 & -1 & 2 & 0 & 0 & 0\\
2 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
-1 & 3 & 2 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 3 & 2 & -1\\
0 & 0 & 0 & 0 & 0 & 0 & 4 & 0 & 2 \end{array} \right)\]
[hide]I suck at canonical forms... this is the progress I got:
$ A=$\[ \left( \begin{array}{ccc}
0 & B & 0\\
C & 0 & 0 \\
0 & 0 & D \end{array} \right)\]
where $ B=$\[ \left( \begin{array}{ccc}
2 & 0 & 0\\
4 & 2 & 0 \\
3 & -1 & 2 \end{array} \right)\]
$ C=$\[ \left( \begin{array}{ccc}
2 & 4 & 0\\
0 & 2 & 0 \\
-1 & 3 & 2 \end{array} \right)\]
$ D=$\[ \left( \begin{array}{ccc}
2 & 0 & 0\\
3 & 2 & -1 \\
4 & 0 & 2 \end{array} \right)\]
$ A^{2}$=\[ \left( \begin{array}{ccc}
BC & 0 & 0\\
0 & CB & 0 \\
0 & 0 & D^{2} \end{array} \right)\]

$ |\lambda I-A^{2}|=|\sqrt{\lambda} I-A\parallel{}\sqrt{\lambda} I+A|=|\lambda I-BC\parallel{}\lambda I-CB\parallel{}\lambda I-D^{2}|=((\lambda-4)(\lambda^{2}-24\lambda+16))^{2}(\lambda-4)^{3}$
And the possible eigenvalues of $ A$ are: $ \pm2,\pm\sqrt{12\pm8\sqrt{2}}$. Now comes the part with invariant operators, eigenvectors and what not  :(  Can someone provide a detailed explanation about Jordan and rational forms?
[/hide]

The $ A$-eigenvalues are $ \minus{}2,2\plus{}2\sqrt{2},\minus{}2\minus{}2\sqrt{2},\minus{}2\plus{}2\sqrt{2},2\minus{}2\sqrt{2},2$; these eigenvalues are simple except $ 2$ which is of order $ 4$; moreover $ rank(A\minus{}2I_9)\equal{}3$.
Then $ A$-Jordan form is: $ diag(\minus{}2,2\plus{}2\sqrt{2},\minus{}2\minus{}2\sqrt{2},\minus{}2\plus{}2\sqrt{2},2\minus{}2\sqrt{2},2I_4\plus{}J_4)$ where $ J_4$ is the nilpotent Jordan matrix of dimension $ 4$. It remains to calculate the new basis $ e_1\cdots,e_9$: take  $ e_9$ s.t. $ e_9\in{ker((A\minus{}2I_9)^4)}$ and $ e_9\notin{ker((A\minus{}2I_9)^3)}$; then $ e_8\equal{}(A\minus{}2I_9)e_9,e_7\equal{}(A\minus{}2I_9)e_8,e_6\equal{}(A\minus{}2I_9)e_7$.

1) Correction: $ rank(A\minus{}2I_9)\equal{}2$ and $ rank((A\minus{}2I_9)^2)\equal{}3$; then the $ A$-Jordan form is $ diag(\minus{}2,\minus{}2\plus{}2\sqrt{2},\minus{}2\minus{}2\sqrt{2},2\plus{}2\sqrt{2},2\minus{}2\sqrt{2},2I_3\plus{}J_3,2)$; the calculation of the new basis is a little more complicate.

2) Let $ frob$ be the $ A$-Frobenius form (rational canonical form) over $ \mathbb{Q}$.
$ charpoly(A)\equal{}(x^2\plus{}4x\minus{}4)(x^2\minus{}4x\minus{}4)(x\plus{}2)(x\minus{}2)^4\equal{}p(x)$.
We know the Jordan form then we can deduce the decomposition of $ charpoly$ which gives the Frobenius form:$ p(x)\equal{}[(x^2\plus{}4x\minus{}4)(x^2\minus{}4x\minus{}4)(x\plus{}2)(x\minus{}2)^3][x\minus{}2]\equal{}q(x)r(x)$.
$ frob\equal{}diag(C(q),C(r))$ where $ C(s)$ is the companion matrix associated to the polynomial $ s$.
Let us recall that $ charpoly(r)$ must divide $ charpoly(q)$.
We find with Maple.

                [0    0    0    0    0    0    0     256    0]
                [                                            ]
                [1    0    0    0    0    0    0    -256    0]
                [                                            ]
                [0    1    0    0    0    0    0    -384    0]
                [                                            ]
                [0    0    1    0    0    0    0     448    0]
                [                                            ]
        frob := [0    0    0    1    0    0    0       0    0]
                [                                            ]
                [0    0    0    0    1    0    0    -112    0]
                [                                            ]
                [0    0    0    0    0    1    0      24    0]
                [                                            ]
                [0    0    0    0    0    0    1       4    0]
                [                                            ]
                [0    0    0    0    0    0    0       0    2]
--------------------------------------------------------------------
                  [1 ,  0 ,  4 ,  0 ,  80 ,  0 ,  1856 ,  0 , 0]
                  [                                            ]
                  [0 ,  0 ,  8 ,  0 ,  192 ,  0 ,  4480 , 0 , 0]
                  [                                            ]
                  [0 ,  0 ,  4 ,  0 ,  160 ,  0 ,  4032 , 0 , 2]
                  [                                            ]
                  [0 ,  2 ,  0 ,  40 ,  0 , 928 , 0 , 21632 , 0]
                  [                                            ]
     basischange, [0 ,  0 ,  0 ,  16 ,  0 ,  384 , 0 , 8960 , 0]
                  [                                            ]
                  [0 ,  -1 ,  0 ,  28 , 0 , 816 , 0 , 19648 , 2]
                  [                                            ]
                  [1 ,  2 ,  4 ,  8 ,  16 ,  32 ,  64 , 128 , 0]
                  [                                            ]
                  [0 , 3 , 8 , 12 , 0 , -80 , -384 , -1344 , -5]
                  [                                            ]
                  [0 , 4 , 16 , 48 , 128 , 320 , 768 , 1792 , 0]

In my first post read $ dim(ker(A\minus{}2I_9))\equal{}1$ instead of $ rank(A\minus{}2I_9)\equal{}3$.
In my second post read $ dim(ker(A\minus{}2I_9))\equal{}2$ instead of $ rank(A\minus{}2I_9)\equal{}2$ and
$ dim(ker((A\minus{}2I_9)^2))\equal{}3$ instead of $ rank((A\minus{}2I_9)^2)\equal{}3$.


Find number of ways in which one can place the numbers 1, 2, .....n^2  on the nxn chessboard, one on each such that numbers in each row and column are in

a)AP b)GP

no one replying to this??  please help me

I don't know if I'm understanding this right, but I don't see how you could ever put them so that each row and column were in GP.  For example there would be a prime in [n/2,n] that would not fit anywhere on the board.


geometry
parallelogram
trigonometry
ratio
Pythagorean Theorem

A parallelogram has consecutive sides of length $\sqrt{6}+\sqrt{2}$ and $3$. One of the angles is $15^{\circ}$. Find the area.

NO TRIG.

I get the weird area of [hide]
around 0.1363
Edit: Next time I get 0.75, still a weird area
[/hide]

Eh, find the radical value.
This problem is begging to be done with trig, I mean $\sqrt{6}+\sqrt{2}$
However, it can be done with pre-trig methods using the results from
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137296
or
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124768

How do you solve it using those, though?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137296

This is the thread to read.  I don't get the nested radical thing, but I do know that the ratio is 1 : sqrt(6) + sqrt(2)

If the hyp of the right triangle formed by drawing the altitude is sqrt(6) + sqrt(2), then the shortest side is 1.  Using the pythagorean theorem, you find that the height of the parallelogram is sqrt(7+2sqrt(2)sqrt(6))
So the area is Bh, or 3 times that.  I just don't know what to do with nested radicals.[/code]


A bag has 4 red marbles, 5 white marbles, and 6 blue marbles.  Three marbles are drawn from the bag (without replacement).  What is the probability that they are all the same color?

I got another solution that came in handy

we have $\binom{15}{3} = 455$ ways to pick 3 marbles and there is $\binom{6}{3} = 20$ ways to find how many ways you can get three blue marbles and $\binom{5}{4} = 10$ ways to find how many ways to get 3 white marbles and at last $\binom{4}{3} = 4$ ways to find how many ways possible of having 3 red marbles drawn.. this equals $\frac{\binom{6}{3} + \binom{5}{4} + \binom{4}{3}}{\binom{15}{3}} = \frac{20 + 10 + 4}{455} = \boxed{\frac{34}{455}}$

that is the same as @jbatterson 's solution.

An edit:
[quote=jbatterson]Another solution:

[hide]There are 15c3 possible combinations, of which 4c3 + 5c3 + 6c3 are one color.  (4c3 + 5c3 + 6c3)/(15c3) = 34/455[/hide][/quote]

Another solution:

[hide]There are $\binom{15}3$ possible combinations, of which $\binom43 + \binom53 + \binom63$ are one color.  \[\frac{\binom43 + \binom53 + \binom63}{\binom{15}3} = \frac{34}{455}.\][/hide]


Given that $ \angle ADB\equal{}\angle CBA$, $ AD\equal{}4$ and $ CD\equal{}8$, how many units are in the length of $ \overline{AB}$? Express your answer as a decimal rounded to the nearest tenth.
[asy]draw((0,0)--(3,0)--(2,1.5)--cycle);
draw((2,1.5)--(2,0));
label("$A$",(3,0),SE);
label("$B$",(2,1.5),N);
label("$C$",(0,0),SW);
label("$D$",(2,0),S);[/asy]

We have similar triangles $ ADB$ and $ ABC$.  Let $ AB\equal{}x$.  So we have the ratio $ \frac{4}{x}\equal{}\frac{x}{4\plus{}8}$.  Cross multiplying, we get $ x^2\equal{}48$.  So $ x\equal{}\sqrt{48}\equal{}4\sqrt{3}$.  $ \sqrt{3}\approx{1.732}$, so $ x\approx\boxed{6.9}$

Or, if you're cheap like me, assume that <ADB is right, and BD=4rt(2), and AB=4rt(3).


I'm having a difficult time trying to figure out the series equation for the sum of the sequence $\frac1{2n}$:

$s_1 = \frac 12$
$s_2 = \frac 12 + \frac 14 = \frac 34$
$s_3 = \frac 12 + \frac 14 + \frac 16 = \frac{11}{12}$
$s_4 = \frac 12 + \frac 14 + \frac 16 + \frac 18 = \frac{25}{24}$
 and so on...

I tried writing it out without reducing the fractions to find a pattern, but I couldn't get it.  Could anyone point me in the right direction? thanks.

There's not a nice pattern to be found there.

The next several entries are $\frac{137}{120},\frac{49}{40},\frac{363}{280},\frac{761} {560},\frac{7129}{5040}.$

Remember also that $\lim_{N\to\infty}S_N=\lim_{N\to\infty}\sum_{n=1}^N\frac1{2n}=\infty.$

There's more that can be done, including asymptotic estimates, but there's not much point to trying to find a simple formula for the partial sums.


Recently, there has been an outburst of multiple accounts in the Games and Mafia Forums. The accounts from the same IP address appear to take on this pattern:

- Both accounts join the same game (MG46 is a big example)
- One account mods and the other joins

In both cases, if both accounts are the same person, this is obviously a huge disadvantage to other people not cheating in the game. However, if they are siblings or next-door-neighbors, there is still a possible favoritism/disclosure of information, especially in the second case. (Of course, it can still happen if two IPs are not the same. For example, I know I go to the same school as some people here, but we're hung on honor code not to tell each other.)

I can think of two different solutions right now:

1. You PM a mod (james4l preferred if you only join Mafia games) telling us directly of two accounts in the same IP address. This will allow one account to join any game at a time.
2. Every game has an IP check at the start of games (not preferred).

I would much rather not post the names of multiple accounts/same IP, but I will if I have to.

What happens if we go to AoPS at school? On our break time, as they have internet for the students?

The moderators are smart enough to deduce truths. Additionally, if you only play these games during lunch, you could also potentially be "cheating" with your friends.  :wink:

Well, there are many computers in the school, so I dont see how the IP address could possibly be the same. Cause if two computers connected to the same network have the same IP address, there will be a conflict, and they would not be able to connect.

Using your logic, getting different tabs on one computer won't work either.  :wink:

Its a desktop, so if we log out, and someone else logs on... :/

Well, tell whoever in the world is using your desktop to stop.

If in doubt, just use common sense. Ask yourself, "Is what I'm doing giving me or someone else an extraneous factor that can lead to an unfair or uncalled for advantage?" If the answer corresponding to a certain action is yes, don't do it. We're obviously not going after multiple accounts themselves, we're going after those who abuse multiple accounts.


[quote="tranthanhnam"]finsd all the positive integer solutions  of the equation:
$\frac{2005}{x+y}+\frac{x}{y+2004}+\frac{y}{4009}+\frac{2004}{x+2005}=2$[/quote]
If $2005=t$ and $2004=z$ then must to solve that
$\frac{t}{x+y}+\frac{x}{y+z}+\frac{y}{z+t}+\frac{z}{t+x}=2$
But $\sum_{cyc}\frac{x}{y+z}=\sum_{cyc}\frac{x^{2}}{xy+xz}\geq\frac{(x+y+z+t)^{2}}{xy+xz+yz+yt+zt+xz+xt+yt}=$
$=\frac{(x+y+z+t)^{2}}{xy+2xz+yz+2yt+zt+xt}$ equality holds when
$x+y=y+z=z+t=t+x\Leftrightarrow x=z\wedge y=t.$
By the way $\frac{(x+y+z+t)^{2}}{xy+2xz+yz+2yt+zt+xt}\geq2\Leftrightarrow(x-z)^{2}+(y-t)^{2}\geq0.$
And here equality holds when $x-z=y-t=0.$
Hence $x=2004$ and $y=2005.$ :)


[quote="2424"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $ (2 \plus{} ab^2)^2(2 \plus{} bc^2)^3(2 \plus{} ca^2)^2 \leq\ 3456$
Equality occurs when $ (a, b, c) \equal{} (0, 1, 2)$
 :)[/quote]
It's not  hard because $ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \leq 4$ with $ a,b,c\geq0$ such that $ a \plus{} b \plus{} c \equal{} 3$ :)[/quote]
Can you post your full proof? Indeed, I also use this lenma to creat it  :)

[quote="nguoivn"][quote="2424"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a + b + c = 3$.
Prove that: $ (2 + ab^2)^2(2 + bc^2)^3(2 + ca^2)^2 \leq\ 3456$
Equality occurs when $ (a, b, c) = (0, 1, 2)$
 :)[/quote]
It's not  hard because $ ab^2 + bc^2 + ca^2 + abc \leq 4$ with $ a,b,c\geq0$ such that $ a + b + c = 3$ :)[/quote]
Can you post your full proof? Indeed, I also use this lenma to creat it  :)[/quote]
Apply lenma,we have
$ (ab^2 + 2)(bc^2 + 2)(ca^2 + 2) = 8 + 4(ab^2 + bc^2 + ca^2) + 2abc(a^2b + b^2c + c^2a) + a^2b^2c^2$
$ \leq 24 + 4abc$
$ = > P\leq16(6 + abc)^2(2 + bc^2)\leq16\frac {(14 + bc^2 + 2abc)^3}{27}$(by Am-Gm)
And $ bc^2 + 2abc = bc(c + 2a)\leq\frac {(2b + c + c + 2a)^3}{54} = 4$
$ = > P\leq3456$ :)
We are done!
[hide]Prove lenma:Assume $ min(a,b,c) \leqq\leq max(a,b,c)$
$ = > a(b - c)(b - a)\leq0$
$ = > ab^2 + ca^2 \leq a^2b + abc$
$ = > ab^2 + bc^2 + ca^2 + abc \leq b(a + c)^2 \leq\frac {4(a + b + c)^3}{27} = 4/27$(by Am-Gm)
we are done! :) [/hide]

Dear 2424,

can you please explain two step of your proof I don't understand?

$ P=(ab^2 + 2)(bc^2 + 2)(ca^2 + 2) = 8 + 4(ab^2 + bc^2 + ca^2) + 2abc(a^2b + b^2c + c^2a) + a^3b^3c^3 \leq 8+4(4-abc) +2abc(4-abc) +a^3b^3c^3= 24+4abc -2a^2b^2c^2+a^3b^3c^3=A$

Why  $ A\leq 24+4abc$ ?

And also why

$ P^2 \leq (24+4abc)^2(2+bc^2)$

Thank you very much. :)

[quote="2424"][/quote][quote="nguoivn"][quote="2424"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a + b + c = 3$.
Prove that: $ (2 + ab^2)^2(2 + bc^2)^3(2 + ca^2)^2 \leq\ 3456$
Equality occurs when $ (a, b, c) = (0, 1, 2)$
 :)[/quote]
It's not  hard because $ ab^2 + bc^2 + ca^2 + abc \leq 4$ with $ a,b,c\geq0$ such that $ a + b + c = 3$ :)[/quote]
Can you post your full proof? Indeed, I also use this lenma to creat it  :)[/quote][quote="2424"]
Apply lenma,we have
$ (ab^2 + 2)(bc^2 + 2)(ca^2 + 2) = 8 + 4(ab^2 + bc^2 + ca^2) + 2abc(a^2b + b^2c + c^2a) + a^2b^2c^2$
$ \leq 24 + 4abc$
$ = > P\leq16(6 + abc)^2(2 + bc^2)\leq16\frac {(14 + bc^2 + 2abc)^3}{27}$(by Am-Gm)
And $ bc^2 + 2abc = bc(c + 2a)\leq\frac {(2b + c + c + 2a)^3}{54} = 4$
$ = > P\leq3456$ :)
We are done!
[hide]Prove lenma:Assume $ min(a,b,c) \leqq\leq max(a,b,c)$
$ = > a(b - c)(b - a)\leq0$
$ = > ab^2 + ca^2 \leq a^2b + abc$
$ = > ab^2 + bc^2 + ca^2 + abc \leq b(a + c)^2 \leq\frac {4(a + b + c)^3}{27} = 4/27$(by Am-Gm)
we are done! :) [/hide][/quote]

[quote="manlio"]Dear 2424,

can you please explain two step of your proof I don't understand?

$ P = (ab^2 + 2)(bc^2 + 2)(ca^2 + 2) = 8 + 4(ab^2 + bc^2 + ca^2) + 2abc(a^2b + b^2c + c^2a) + a^3b^3c^3 \leq 8 + 4(4 - abc) + 2abc(4 - abc) + a^3b^3c^3 = 24 + 4abc - 2a^2b^2c^2 + a^3b^3c^3 = A$

Why  $ A\leq 24 + 4abc$ ?

And also why

$ P^2 \leq (24 + 4abc)^2(2 + bc^2)$

Thank you very much. :)

[/quote][quote="2424"][/quote][quote="nguoivn"][quote="2424"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a + b + c = 3$.
Prove that: $ (2 + ab^2)^2(2 + bc^2)^3(2 + ca^2)^2 \leq\ 3456$
Equality occurs when $ (a, b, c) = (0, 1, 2)$
 :)[/quote]
It's not  hard because $ ab^2 + bc^2 + ca^2 + abc \leq 4$ with $ a,b,c\geq0$ such that $ a + b + c = 3$ :)[/quote]
Can you post your full proof? Indeed, I also use this lenma to creat it  :)[/quote][quote="2424"]
Apply lenma,we have
$ (ab^2 + 2)(bc^2 + 2)(ca^2 + 2) = 8 + 4(ab^2 + bc^2 + ca^2) + 2abc(a^2b + b^2c + c^2a) + a^2b^2c^2$
$ \leq 24 + 4abc$
$ = > P\leq16(6 + abc)^2(2 + bc^2)\leq16\frac {(14 + bc^2 + 2abc)^3}{27}$(by Am-Gm)
And $ bc^2 + 2abc = bc(c + 2a)\leq\frac {(2b + c + c + 2a)^3}{54} = 4$
$ = > P\leq3456$ :)
We are done!
[hide]Prove lenma:Assume $ min(a,b,c) \leqq\leq max(a,b,c)$
$ = > a(b - c)(b - a)\leq0$
$ = > ab^2 + ca^2 \leq a^2b + abc$
$ = > ab^2 + bc^2 + ca^2 + abc \leq b(a + c)^2 \leq\frac {4(a + b + c)^3}{27} = 4/27$(by Am-Gm)
we are done! :) [/hide][/quote][quote="manlio"][/quote]
Dear manlio,
Because $ a,b,c\ge0$ so $ a^2b^2c^2\le2a^2b^2c^2$
then $ 2abc(a^2b + b^2c + c^2a) + a^2b^2c^2 \le 2abc(a^2b + b^2c + c^2a+abc)\le 8abc$

Thank you very much,leedt26, for your reply.

But,now,please, why $ P^{2}\leq (24\plus{}4abc)^{2}(2\plus{}bc^{2})$? :blush:

Thank you very much. :)

And here is my proof, manlio.  :)
Let $ x \equal{} 2\plus{}ab^2 , y \equal{} 2\plus{}bc^2 ,z \equal{} 2\plus{}ca^2$ then we have
\[ xy\plus{}yz\plus{}zx \equal{} 12\plus{}4(ab^2\plus{}bc^2\plus{}ca^2)\plus{}abc(a^2b\plus{}b^2c\plus{}c^2a)\equal{}P.\]
Using the well-known results $ a^2b\plus{}b^2c\plus{}c^2a \le 4$ and $ ab^2\plus{}bc^2\plus{}ca^2 \le 4\minus{}abc,$ we obtain
\[ P \le 12\plus{}4(ab^2\plus{}bc^2\plus{}ca^2\plus{}abc) \le 28.\]
We need to prove $ x^2y^3z^2 \le 3456$ with $ x, y, z > 0$ satisfy $ xy\plus{}yz\plus{}zx \le 28$. This is a simple work with Am-Gm Inequality. Indeed, applying Am-Gm Inequality for 7 numbers, we have
\[ 28 \ge 3\cdot\frac{xy}{3}\plus{}3\cdot\frac{yz}{3}\plus{}zx \ge 7\sqrt[7]{\frac{(xy)^3\cdot(yz)^3\cdot zx}{3^6}}\equal{}7\sqrt[7]{\frac{x^4y^6z^4}{3^6}},\]
Hence $ x^2y^3z^2 \le 3456$. Our proof completed. Equality occurs if and only if  $ (a, b, c) \equal{} (0, 1, 2)$.


Find the equation of the tangent plane at (2,3,2) to the surface below.

$ x^2 \plus{} y^2 \minus{} xyz \equal{} 1$

View the surface as the graph of a function of two variables z = g(x,y)

How can I solve this?

[quote="jmerry"][quote="EquinoX"]View the surface as the graph of a function of two variables z = g(x,y)[/quote]
Solve for $ z$. Then do calculus.[/quote]

can you elaborate on that?

$ f(x,y,z)\equal{}0$ then the equation of tangent line at $ (a,b,c)$ is
$ (x\minus{}a)f_{x}(a,b,c)\plus{}(y\minus{}b)f_{y}(a,b,c)\plus{}(z\minus{}c)f_{z}(a,b,c)\equal{}0$

[quote="1234567a"]$ f(x,y,z) \equal{} 0$ then the equation of tangent line at $ (a,b,c)$ is
$ (x \minus{} a)f_{x}(a,b,c) \plus{} (y \minus{} b)f_{y}(a,b,c) \plus{} (z \minus{} c)f_{z}(a,b,c) \equal{} 0$[/quote]

i did that way, but the question I think asks for another solution/way to do this

like jmerry said,
solve for $ z \equal{} f(x,y)$ from given equation
and the equation of the tangent line has the form
$ z \minus{} z_{0} \equal{} (x \minus{} x_{0})f_{x}(x_{0},y_{0},z_{0}) \plus{} (y \minus{} y_{0})f_{y}(x_{0},y_{0},z_{0})$
But actually, they are the same

ok what if the question asks for :

View the surface as a level surface for a function f(x,y z).

and when I put the answer to be  x-y+3z-5 = 0 it was wrong

however in part b when it asks:

View the surface as the graph of a function of two variables z = g(x,y).

and I answered z = (-x+y+5)/3, the answer seems to be correct


calculus
integration
trigonometry
complex numbers
complex analysis

I am a newby . Can you please tell me how can i compute the fresnel integral :
\[ \int_{0}^{\infty} \cos{x^2} dx
\]
using residues in complex numbers ?

At first thoughts, i take g(z) = e ^ (iz^2) but g(z) has no poles ! so it goes = 2*(pi)*i * 0 = 0 ...

but the result is different , it is sqr (2*pi) / 4...

There aren't supposed to be any poles; you just need to pick the right contour.  (Try a semiquarter circle.)


A palindrome is a number that remains unchanged when the digits are reversed. In base b, how many integers with 2n+1 digits are palindromes? Express your answer in terms of b and n.

[hide]
In base b there are $ b$ different pronumerals representing digits.
So the number palindromes with $ 2n\plus{}1$ digits is equal to the number of $ n\plus{}1$ digit numbers
So the answer is simply $ b^{n\plus{}1}$ [/hide]

[hide="Solution"]Picture it as $ n$ digits, then a central digit, and then $ n$ more digits which are the first $ n$ in reverse.
In base $ b$ our single digits are $ 0$ to $ b \minus{} 1$ (just like they're $ 0$ to $ 9$ in base $ 10$).
The first digit can be anything from $ 1$ to $ b \minus{} 1$. The next $ n$ digits can be anything from $ 0$ to $ b \minus{} 1$. The remaining $ n$ digits are fixed (they're the first $ n$ digits in reverse order).
So we have $ b \minus{} 1$ choices for the first digit, $ b$ choices for the next $ n$ digits, and the remaining digits are fixed, so we have $ \boxed{(b \minus{} 1)b^n}$ ways to form our palindrome.[/hide]

[hide="Example"]
For the sake of simplicity let's consider three digit long palindromes in base 3. We have $ 2n\plus{}1\equal{}3\implies n\equal{}1$ and $ b\equal{}3$ so the formula tells us we should have $ 2\cdot 3^1\equal{}6$ palindromes. Listing them out, we have 111, 222, 121, 212, 101, 202.[/hide]

Its basically what grn_trtle said:

[hide] The first digit can be anything from 1 to b-1 and all the rest can be from 0 to b-1. Also, since it's a palindrome, the first n digits determine the last n digits, with the middle digit being independently determined (if thats the right word). Then, you get (b-1) choices for the 1st digit, and b choices for the next n digits which results in the total number of possible integers equal to (b-1)(b^n).

P.S sup tina![/hide]


articles
LaTeX
geometry
3D geometry
prism

Whenever I put in a picture in(I already have some in but whenever i put a new one in) an error comes up saying "There was an error opening this document. This file cannot be found.". The output looks like:

[url]http://s487.photobucket.com/albums/rr233/iceflame012/?action=view&current=Capture.jpg[/url]

I think the blue is where the problem is but I am not sure.

Thanks

The log is telling you that there is a problem with that particular image. Try a different image or try opening and resaving your image in an image program or convert it to pdf, gif or png.

OK, copy here a complete [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] with a picture that doesn't work. Attach also the picture and the log files. Also tell us what method you are using to compile your document.

[u]minimal example[/u]
\documentclass{article}
\usepackage[pdftex]{graphicx}
\begin{document}
\begin{center}
	\includegraphics{C:/LaTeX/GBML4-1GuidePics/cylinder.jpg}
\end{center}
\end{document}
-----
I got the picture at [url]http://z.about.com/d/math/1/0/n/cylinder.gif[/url] at [url]http://math.about.com/library/blmeasurement.htm[/url].

I am not sure what you mean by log files.
----
I am using TeXnic Center with MiKTeX 2.7.

Thanks

[quote="Nandubetax"]I got the picture at [url]http://z.about.com/d/math/1/0/n/cylinder.gif[/quote]Is that the problem? The picture is a gif file which cannot be compiled. Convert to png first and and it should work
[quote="Nandubetax"]\includegraphics{C:/LaTeX/GBML4-1GuidePics/cylinder.jpg}[/quote]If it's a gif why did you call it a jpg? You didn't say you converted it.

Every compilation creates a log file in the same directory as your latex file. If your latex file is called [i]filename.tex[/i] then the log file is called [i]filename.log[/i], and it is also shown in the bottom window of TexncCenter.
You also didn't answer what method (profile) you used to compile as they use different exe files to do so.

Sorry about the miscommunication on my part. I am kinda new at this. I saved the picture at .jpg(i added .jpg at the end of the picture name when i went to save it). I also tried to save the pic at .png but the same error came up. (Sorry about not telling you that I converted it..if thats what i did).

The log file is GBMLGuide.log --> I could not find it on the bottom window but assuming that it has the same file name, above is just the file name with .log at the end.

I am still not sure what you means by method to compile but I think you either want to know that:
-I am using adobe reader 8
or
-I go from latex to PDF

I hope this helps and once again am sorry for any confusion on my behalf

Thanks

You can't convert a picture by just changing its name. Nor shoud you add .jpg if it isn't a jpg. You need to save it as a gif (the format of cylinder.gif) and then open it in a graphics program (such as Paint) and then save it in the new format.
You still haven't attached the log file. It is in the same directory as the latex file - you can easily find it in TexnicCenter by going to Tools, Windows Explorer.

I'm very sorry but i still cannot find the log file. The tools --> windows explorer looks like:

[url]http://s487.photobucket.com/albums/rr233/iceflame012/?action=view&current=Capture-2.jpg[/url].

Thankyou for your time ans patience

It may be the one marked Text Document. You seem to have extensions shown for some files and not others which is very strange. Check that Explorer is showing all extensions [url=http://teamtutorials.com/windows-tutorials/show-file-extension-in-windows-xp]Show File Extension in Windows XP[/url].

Ahh, that helped. Thanks. The log file is:

GBMLGuide.log

[EDIT]
I would just like to say that I converted the picture in paint and it worked!! The PDF opens. Although I am getting errors looking like:

[url]http://s487.photobucket.com/albums/rr233/iceflame012/?action=view&current=Capture-3.jpg[/url].

And how am I supposed to read that log file? There's an error in your log file but what you show is only part of it and I can't see the important part which shows the error. Please attach the [b]complete[/b] log file not just a picture of it, using [i]Add Attachment[/i] when you post*, so I can see what it says. Also attach the picture you are actually using, not just a link to it.

* AoPS doesn't allow *.log files to be attached so either change the name to GBMLGuide.txt or zip it first.

Sigh! Please read this very carefully. Attach the LOG file and the 4 PICTURES you use in the tex file you have uploaded. The log file will tell me the errors and the pictures will allow me to compile your tex file.

[b]Where is the log file?[/b]
Tri-prism-2ndpag.jpg is wrongly named as it is a png file. Rename it as Tri-prism-2ndpag.png then try again.
Also in GBMLGuide.tex missing are cylinder.jpg and volumefprism.jpg.

Looking at GBMLGuide.tex there are lots of \begin{center} and \begin{flushleft} with no \end{center} or \end{flushleft}, as the log file would tell you though you still haven't uploaded it.

When using an environment, indent your code so you get
[code]\begin{center}
    line 1
    line 2
    line 3
\end{center}[/code]
and then it's obvious if \end is missing.

I'm sorry but I thought I attached it above. If that is something else...here is the .log file in text:

P.S. I have not changed the flushleft and centers yet...i will if that if what you need

I will ask you once again. Please attach the [color=red][size=150]LOG[/size][/color] file, NOT the tex file. It is called GBMLGuide.[color=red]log[/color] NOT GBMLGuide.[color=red]tex[/color] and starts something like:
[code]This is pdfeTeX, Version 3.141592-1.30.6-2.2 (MiKTeX 2.5) (preloaded format=pdflatex 2007.1.18)  2 JAN 2009 18:29
entering extended mode[/code] and is auomatically generated when you compile the tex file ie convert to a pdf.
As AoPS does not allow log files to be uploaded change the name of GBMLGuide.[color=red]log[/color] to GBMLGuide.[color=red]txt[/color] and upload.

That is certainly the log file but there is no error in there so I presume you must have changed everything I suggested and have now got a correct output. TexnicCenter would have reported 0 Errors and 0 Warnings. So I presume the problem is sorted now.

I did make the changes and they worked. Sorry for the trouble on my part and the IMMENSE amount of aggravation you must have gone through to help me. Thanks a lot  :oops:


AMC
USA(J)MO
USAMO
geometry
induction
rectangle
geometric transformation

The week after the USAMO, people always worry about whether they'll make the MOP cutoff. If you think that you got a 28+, please post your stats. To make this organized, let's follow the collegeconfidential format.

[b]USAMO Score[/b]
Predicted Score:
Score Breakdown:

[b]Brief Explanation[/b]
Problem 1:
Problem 2:
Problem 3:
Problem 4:
Problem 5:
Problem 6:

Any furthur comments:

[b]USAMO Score[/b]
Predicted Score:28
Score Breakdown:6/6/2/7/6/2

[b]Brief Explanation[/b]
Problem 1: poorly written because I wasted too much time on #3, and I wrote up the solution at the end
Problem 2: proved the upper bound if $ n \plus{} 1$, but I'm not sure if my proof for the even case is rigorous
Problem 3: solved part a) (I think), but poorly written
Problem 4: poorly written, made the exact same mistake as day 1, spent too much time on #6
Problem 5: forgot the use directed angles
Problem 6: I thought I solved it, but there was a huge gap.

Any furthur comments: I made a horrible mistake. I should just write up the solution right after I solved a problem. I shouldn't even care about 3 and 6 until I'm absolutely sure with 1245.

USAMO Score
Predicted Score: 31
Score Breakdown:7/7/3/7/7/0

Brief Explanation
Problem 1: radical axis and orthocenter - pretty detailed solution, I thought.
Problem 2: used the set of odd numbers, then by induction proved that all good sets satisfy the conditions (first for even $ n$ then for odd $ n$
Problem 3: solved part a), pretty thorough.  Outlined an induction proof for b), not sure how much they'd give for that
Problem 4: used $ a_1\le...\le a_n$ and $ a_n \equal{} xa_1$.
Problem 5: found that $ pDRG$ is cyclic and proved one direction, then showed that $ G$ is unique on $ RS$.
Problem 6: didn't get anything

Any further comments: I felt 1 and 2 were pretty easy.  3 and 6 though were really hard.  I know the jump from 19 to 31 seems kind of weird (which does worry me).  But I don't think I'm the type to exaggerate USAMO scores - last year I underestimated my USAMO score by 1.

Edit - found out I missed the parallel case on #1, so maybe subtract 2 pts.

USAMO Score
Predicted Score:28
Score Breakdown:7/7/0/7/7/0

Brief Explanation
Problem 1: radical axes (not center!)
Problem 2: induction
Problem 3: part a for rectangles >.>
Problem 4: see thread. maybe not greatest write-up.
Problem 5: umm I switched labels for P, Q, and maybe didn't say R, S, G, X (see my solution in the thread) were collinear if GX || AB but that is obvious probably
Problem 6: proved stuff about denominators?

Any further comments: um I doubt I will win?

USAMO Score
Predicted Score: 29
Score Breakdown: 6/1/5/7/3/7

Explanation
Problem 1: I realized 1 minute after the exam ended that the centers could be collinear and almost wanted to kill myself.
Problem 2: found the optimal case (all odd numbers)
Problem 3: Solved part (a) and reduced part (b) to proving that any two tilings can be converted to one another by moving two dominoes at a time
Problem 4: All 2-variable AM-GM
Problem 5: Proved that the "if" part implies the "only-if" part. Also described a weird transformation that might help prove the "if" part (inversion centered at $ B$ followed by reflection in $ BG$) but had only 15 minutes left after I discovered it
Problem 6: Used Gauss's primitive polynomials lemma

Well, since this thread is pretty much dead, I will pretend that you said 27 instead of 28.

[b]USAMO Score[/b]
Predicted Score: 27
Score Breakdown: 6/1/0/7/6/7 or 6/0/1/7/6/7 or variation

[b]Brief Explanation[/b]
Problem 1: Solved, but didn't consider parallel case
Problem 2: Did not solve, maybe have one point of credit for stating a possible best set and for an attempt at solving it (considered the numbers mod $ 3n\plus{}1$, since this problem is rather classic in $ \mathbb{Z}/n\mathbb{Z}$).
Problem 3: Did not solve, wrote some bs about how an algorithm turning around distasteful squares ought to terminate, perhaps will be worth 1 point as the statement is true, although my method for proving it doesn't easily translate to a correct solution.
Problem 4: Solved, decently elegant writeup
Problem 5: Solved, but written up badly, had the inelegant solution (citing Pascal's theorem), and was rushed in general. Expecting a 6, but perhaps a low score if the writeup isn't possible to follow.
Problem 6: Solved, solution is different from everyone else's I've seen, but when typing it up to post on AoPS, I didn't notice any errors.


Hello

i am very unexperienced in latex but I am realising the advantages

suppose i would want to mail a friend something in latex
what is the best way to do so?
i use outlook express and hotmail but those are probably no options
Is there any email online thing like hotmail supporting latex

or is the best way to do it by always making a pdf
i am asking because I once got a mail from a professor full of latex language?

You have two options:

1) Send him a .tex file or a .pdf file
2) Use the PM messaging system on AoPS - MathLinks :)

[quote="fredbel6"]a texfile he has to compile himself?

Hmm, suppose my friend is actually like a professor you wish to correspond, kinda hard to get him on mathlinks just for that :P  pdf is best then

Well, that's okay, still i wonder why my professor started writing all that latex in his email :?[/quote]Well some people prefere to manually copy-paste the .tex file inside the email (not attaching it).

Others are used to ancient (more than 10 years old) systems that required them to make correspondence using some sort of language, derived from $\text{\LaTeX}$, which is sometimes called $\textrm{Notepad}-\text{\TeX}$ :)

You can also send pseudo-TeX inside the text of your message- it won't be pretty, but it will be understandable. If you want something pretty, an attached file is the way to go.

For the message itself, text is the only universal format.

Most LaTeX notation is pretty intuitive, and human-readable. You can cut a few corners, too; it doesn't have to be something that would work if you pasted it directly.


If $ (a\plus{}4b)$ is divisible by $ 7$ prove that $ 16a\plus{}43b$ is divisible by $ 7$ .

Another solution:
$ 16a\plus{}43b\equal{}14a\plus{}2a\plus{}35b\plus{}8b$
$ 7(2a\plus{}5b)\plus{}2(a\plus{}4b)$


modular arithmetic
induction
number theory unsolved
number theory

Show that any number $\geq 2$ can be expressed as a sum of numbers in form of $2^a3^b$ which none of them divide one another

Can you give your solution for $n=4$ ?  :?

:) $4=2^2\cdot 3^0$..

Since it's clear that if $t$ satisfies the assertion, then so does $3^k\cdot t,\ \forall k\ge 1$, it means that we may assume our $n$ is not divisible by $3$. We can proceed by induction. Let $m$ be the largest integer s.t. $2^m\le n,\ 2^m\equiv n\pmod 3$. We have $n-2^m=3u$. By induction, we can write $u=\sum 2^{a_i}3^{b^i}$, so $n=2^m+\sum 2^{a_i}3^{b_i+1},\ a_i,b_i\ge 0$. The only situation in which this wouldn't work would be if we had $2^m|2^{a_i}3^{b_i+1}$ for some $i$, but this means that $a_i\ge m$, so $n\ge 2^m+3\cdot 2^m=2^{m+2}$. However, this is not possible, because then $2^{m+2}\le n,\ 2^{m+2}\equiv n\pmod 3$, contradicting the maximality of $m$.

By induction:
If $n$ is odd, then $\frac{n+1}{2} < n$ and then $n+1 = 2(\frac{n+1}{2})$
If $n$ is even, we can assume $n+1$ doesn't equal $3^k$ for any $k$. Let $3^m<n<3^{m+1}$. So $\frac{n+1-3^m}{2}$ can be expressed, and so can $n+1-3^m$. Then $n+1 = 3^m +$ that sum


How is it that you were able to create the latex images and post them on the website? Same question also for the asymptote images.

I know that you put in code in [code]$$[/code], that doesn't quite answer my question. Obviously if you put that in a couple other forums, it wouldn't come out as expected here. What I am wondering is, what system was put in place in this forum so that it could process the code and turn out a image that can be interpreted well.

They just installed $ \text{\LaTeX}$ on the servers...

If you want to learn how to use latex, click it on the sidebar.

No, it is more complicated than just that. I guess only Valentin knows all the details of how exactly it is done on AoPS, but the question has been asked several times, so you may try to use the search button first.


vector
parameterization
parametric equation
calculus
calculus computations

Good day sir.

I have encountered some problems with space curves that I have found very hard to solve. One is given the curve $C$ defined by $r(t) = <1, \frac{t^{2}}{2}. \frac{t^{3}}{3}>$. Find a parametric equation for tangent line at $(1, \frac{1}{2}, \frac{1}{3})$.

What i got here for $T(t)$ is $T(t) = <{0, \frac 1{\sqrt{1+t^{2}}}, \frac{t}\sqrt{1+t^{2}}}>$.

Am i doing it right? I'm really not sure with this one. I think if i get $T(t)$ right i'm on the right track but with regards to the parametric equation for the tangent line. I dont know where to start. Hope someone could help. Thanks

A line is determined by a point and a direction vector. You have a point already. The direction vector of a tangent line is $\vec r\,'(t_{0})$, where $t_{0}$ is chosen so that $\vec r(t_{0})$ gives the point of tangency. In this case $\vec r\,'(t)=\langle 0,t,t^{2}\rangle$ and $t_{0}=1$. Using the given point $(1,1/2,1/3)$ and the direction vector $\vec r\,'(1)=\langle 0,1,1\rangle$, find the equation $\vec T(t)=\langle 1,1/2,1/3\rangle+t\langle 0,1,1\rangle$.

Thanks sir for the answer. With your explanation, I got the parametric equations for the tangent line at $(1,\frac{1}{2},\frac{1}{3})$ to be $x = 1$ ; $y = \frac{1}{2}+t$ ; $z = \frac{1}{3}+t$.

Is this the correct answer? Hope someone could help me again. Thanks.

Thanks sir for your reply.

By the way, this was actually a multi-question problem. I have answered it but i have some doubts regarding my answers. Hope that someone could check if my answers are correct. Thanks.

Using the same given (a curve $C$ with $r(t) = <1, \frac{t^{2}}{2}. \frac{t^{3}}{3}>$) as the first problem:

1) Is the moving trihedral at $t = 1$ equal to $T(1) = <0, 1/\sqrt{2}, 1/\sqrt{2}>$  $N(1) = <0, 1/\sqrt{5}, 2/\sqrt{5}>$ $B(1) = <2/\sqrt{10}, 0 , 0>$?

2) Is the equation of the normal plane at $t = 1$ equal to $\pi_{n}: y+z-5/6 = 0$?

3) Is the curvature at $t = 1$ equal to $K = \sqrt{5}/2$?

4) Is the arc length of the values of $t$ from $t = \sqrt{3}$ to $t = \sqrt{8}$ equal to $\frac{19}{3}units$?

Thanks again for all the help.

I think that $B$ should be a unit vector. Wikipedia pages on [url=http://en.wikipedia.org/wiki/TNB_frame]TNB frame[/url] and [url=http://en.wikipedia.org/wiki/Frenet-Serret_formulas]Frenet-Serret formulas[/url] have all the information you need.

Sir thanks for your reply.

With $B$ being the unit vector, should my answer be modified to $(2,0,0)$?

I just evaluated the cross product $T(t) X N(t)$ at $t =1$ to get my answer.


A box contains six cards. Three of the cards are black on both sides, one card is
black on one side and red on the other, and two of the cards are red on both sides. If
you pick a card at random from the box and see that the side facing you is red, what is
the probability that the other side is red? Express your answer as a common fraction.

If you pick a card with one side being red, then that means that you must have picked one of the three that do actually have a red side. (You can just ignore the three cards with both black sides, since it is impossible to get a red side out of those.) However, only 2 of the three cards have red on the other side.

So, the probability is $ \boxed{\frac{2}{3}}$

There are $ 5$ red sides on the cards.  Out of these, $ 4$ work so the probability is $ \frac{4}{5}$

[quote="rubik"]If you pick a card with one side being red, then that means that you must have picked one of the three that do actually have a red side. (You can just ignore the three cards with both black sides, since it is impossible to get a red side out of those.) However, only 2 of the three cards have red on the other side.

So, the probability is $ \boxed{\frac {2}{3}}$[/quote]

I thought so also but thats wrong 4/5is correct ( :(  :(  :(  :(  :( )

I believe its because you have a lower chance of picking the red-black card than the other red cards.  There's 1 way to have picked the red-black cards while there are 4 ways to have picked the red-red cards. So the probability is 4/5.  Just wondering would the answer to the problem be 2/3 if the problem already stated that the red side of the red-black side was facing towards you?

To solve this problem, all you need to know are two facts:

[hide]1. There are 2 cards with red on both sides and 1 card with one black, one red side, and

2. You are looking at an already chosen card, with a red face.

This means that you are either holding:

1. The red-black card, or

2. The red-red card(s).

Which means that you have a $\frac{2}{3}$ chance of holding a red-red card... [/hide]

Hmm...

There are $5$ red faces, and $4$ have red on the other side.  Thus, $\dfrac{4}{5}$.

Like I said, that's not right.

If you are holding a red card,  that means you're either holding the red-black card or one of the two red-red cards. Since there's 2/3 chance you're holding a red red card, that's the answer.

Incorrect.  If you pick a red side, it does not have the same likelihood of being red-black as red-red.

I never said that? It's really simple. When you have a card with a red side, you're holding one of the 3 cards that have a red side: red black, red red, and another red red. So the chance of getting a red red is 2/3 >_<

@TrueshotBarrage It is 4/5. You need to review your conditional probability...
ex. Let's say you have two coins. One always lands on heads. The other has a 1/100000000 chance of landing on heads. I randomly choose one, flip it, and get heads. By your logic, the probability that I picked the first coin is 1/2...

It's 2/3 >_< the problem doesn't consider what happened before to choose one of the three cards. It's assumed that the three cards with at least one red side has an equal chance of being picked. That means you have AN EQUAL CHANCE OF HAVING A RED RED, RED RED, OR RED BLACK. Thus it's 2/3. L2P

I do believe it is 4/5. It is not that you have 3 ways to pick a card with a red side (and 2 of those have the other side as red), it is that you have 5 ways to pick a red side, and four of those red sides have an opposite red side. It is twice as probable to choose a red/red card than the red/black card, given that you have chosen a red side rather than a red card, so you cannot say it is only 2/3.

@TrueshotBarrage Each card has an equal probability of being picked AT FIRST. If a card with 1 red side and 1 black side is picked, then there is also 1/2 chance that the black side is showing. In my example, each coin has an equal probability of being picked as well...

But the problem says you are looking at the red side. >_< so therefore it's 2/3. Read carefully!

@knittingfrenzy18 but if you already have chosen a card with the red side showing, wouldn't that mean you have 1 of the 3 cards with at least 1 red side? meaning there's a 2/3 chance of the red showing on the other side. You can't consider the side showing at all, since all we are dealing with is the opposite side. So there's 3 cards, 3 opposite sides.

Effectively, you have chosen to look at a red [i]side[/i], not a red card. Your solution is the answer to the problem "I pick a card out of the described box which has at least one red side. What is the probability the card has two red sides?", or alternatively, "Each card is labeled with sides A and B (the red/black card has red labeled as A). I choose to look at a card's A side, and I'm looking at red. What's the probability the other side is red?". However, if you are looking at a red side, four of those red sides have a red side on the other side as well. The question asks, "Each card is labeled with sides A and B. I choose to look at a card's side, any side, and I'm looking at red. What's the probability the other side is red?" You could be looking at A of red/black, A of red/red, B of red/red, A of red/red, or B of red/red.

So if I did it in real life, I would get a red side 80% of the time? O.o

Yes, following the process in the problem. I was presented with a similar problem at MathAlive! in DC once, and it was pretty cool to materialize it. Perhaps you can try it with actual cards and convince yourself the 4/5 is true.

Ok, I'll see what I can do. Thanks for actually clarifying why you think it's 4/5.

There are $5$ different faces that contain a red face. Of them, $4$ of them are double sided red, so the probability that the card has a red face on the back side is $4/5$.

I ran into the same problem as most others did.

I am stupid and put $\frac{2}{3}$ first lol.

i thought it is 2/3 but is 4/5 because u have 5 red faces and only 1 of those red faces have a black face behind so 1- 1/5 is 4/5


I need to know how many cubic yards are in 2005 square feet? How would i come up with that answer???  please help

i dont understand how can cubic be in square??? 3d in 2d?

You can't convert cubic to square. They have to be in the same dimension to be convertable.

[quote="lotrgreengrapes7926"]A more reasonable question: How many square yards are in 1728 square feet?[/quote]

[hide]
[size=200][color=blue]192 Square Yards[/color][/size][/hide]

[quote="lotrgreengrapes7926"]A more reasonable question: How many square yards are in 1728 square feet?[/quote]
[hide]$\frac{1728}{9}=192$.[/hide]

rvictoryia, if you meant square yards, the answer is 2005/9. if you meant cubic feet, the answer's 2005/27

[hide]can cubic convert into square?[/hide]

No, mathsfun_Tony, it can't. On the other hand, like everyone else said, you can convert in the same dimension.  :)

[quote="vishalarul"]Yeah. All of them are right. It's like trying to convert something in volume into something into area. You can only convert things if they're in the same dimension, like volume to volume, area to area, length to length, etc...

P.S. If you find an answer, tell me. You've successfully bent the laws of mathematics (and nature)! :stretcher:[/quote]
Well using calculus if you integrate a 2d plane you get three space... therefore it seems reasonable to say that $10^{-100}m^{3}=\infty m^{2}$

OK. That's enough.

You two can have a PM war about this topic. :)

Hi  :) ,
         I think U can refer an online resource named educationtrak.com  for the problem u specified get solved.
Bye :)

Ref    http://www.educationtrak.com

I would kindly like to thank all those who responded to my information through mail.


calculus geometry inequalities AMC USA(J)MO USAMO integration \/closed

I just stumled, as I got a message from one of the beginners in math olympiad telling me that a beginners section is not required, and that it would be best for me NOT to do it ...

what do you think about it?

PS the effect of that message was exactly the opposite :D:D

i think that u shouldn't make a beginners section.
i like it very much now and it helps me very much :D
Kuba

a begginners section would not in any way affect the site. it will be separate from the current sections, and it is designed to accomodate the users that are not as good as some people on the forum.

many users just look around and do not post because they feel that they might say something stupid and will be taxed for that.

I want to change that.

Also a broad band 1Gb/s connection will soon be available and there will be no speed problems also.

I cannot see where the negative argument will fall in place. You don't have to check out the beginner's section, but some of us will, because I remember that I was sometime ago a beginner and needed help.

PS And please do not post as guest anymore. :D

I think it would be a good idea (especially for me  :))

I think it would be a great idea!  Personally, I feel a bit overwhelmed by the problems on this site . . . and therefore, I rarely try them.  I don't try them because I don't know the difficulty of the problems and how involved they really are.  As of right now, I am just behind Calculus, so my math level is pretty low for this site.

It would be nice to add a "beginner's area" to encourage people like me to try things harder later.

My 2 cents  :)

Thanks Valentin

I agree with integral0.  I'm generally overwhelmed by the problems here, so a beginners corner would be great.  Good idea, Valentin.

- mathfanatic

I think a beginner's section would be a good idea, it seems as though the majority of problems on this site are posted by and solved by only a few people.  Adding a beginner's section may entice more people into participating and help them to build up the skills required for more advanced problems.

I still think that the beginner's section should be strictly olympiad type problems, just ones from easier national and local olympiads.

Why shouldn't there be a beginners section? And why do you ask us? It's your site and as long as you don't change our section there's no need for a vote.

my idea is that not all of us are very good in maths. 4 example me. so, i think we need that beginner's corner. i would mention that it shouldn't be a sepparate forum, because an 'advanced' allways can help a newbie  :)
and we must give a chance to the beginners, because one day, they will be like us one day :)

Definitely. Didn't all of you remember how was that when you started?

[quote="zscool"]...it seems as though the majority of problems on this site are posted by and solved by only a few people... [/quote]

Correct

384 members on mathlinks
Only 40 members has more than 30 post (problems & solutions)
242 members has 0 post

http://www.mathlinks.ro/phpBB/memberlist.php?mode=posts&order=DESC&start=0

Thats normal for a math forum with 250+ members  :(

I think this is a good idea ! A lot of us are not so good, and a beginner's corner'll not destroy the forum,... it'll make it better !

due to the crushing majority of votes, posts, and private messages I have received I declare this motion passed. In other words a new beginner's corner will be created.

I would like you to continue this little discussion as to what the corner should contain: I was thinking about dividing it into the usual
Algebra
Inequalities
Geometry
Number Theory
Combinatorics
Other stuff - Questions of any kind

the first five are obvious. I will make the forums self pruning - that is topics older than say a month or so will be automatically deleted. Of course you can re-ask some of the questions.

the latter will be for different advices given by the advance people to others.

what do you say?

WELL:
We  were  all  beginers  once  upon  the  time....dont  foget  it. We  all  neaded  help  on our  first  math  steps. So  a  baginers  forum  will  help  some  people  and  this is  not bad! For  the  rest  of  you , who  dont  wish  a  beginers  forum , i  have  to  say '' just  dont  click  the  posts''.
Beginers  need  help  and  later  they  will  join  other  forums...i dont  see  the  reason  why  they  dont  deserve  a forum...

[quote="Valentin Vornicu"]... I will make the forums self pruning - that is topics older than say a month or so will be automatically deleted. Of course you can re-ask some of the questions. the latter will be for different advices given by the advance people to others. what do you say?[/quote]

I completely disagree that old topics should be deleted. What do you mean when by your pre-last statement, Valentin ?

I am the guest who posted the comment before.

One possibility is that we might introduced a new section for problems which were solved at least one month ago. Maybe only the first post of the topic should be allowed to view. I think it is stupid idea to delete all the old solved topics as many questions get reasked again and again. This is quite annoying. If talented students are really interested in learning problem-solving will resolve the problems again. So it is a good idea only to provide the first post of the topic as an instantly available solution won't spoil the problem. Read also the intriguing account due to [quote="Richard Rusczyk"] I was invited to the Math Olympiad Summer Program (MOP) in the 10th grade. I went to MOP certain that I must really be good at math In my five weeks at MOP, I encountered over sixty problems on various tests. I didnt solve a single one. Thats right  I was 0-for-60+. I came away no longer confident that I was good at math. I assumed that most of the other kids did better at MOP because they knew more tricks than I did. My formula sheets were pretty thorough, but perhaps they were missing something. By the end of MOP, I had learned a somewhat unsettling truth. The others knew fewer tricks than I did, not more. They didnt even have formula sheets!

At another contest later that summer, a younger student, Alex, from another school asked me for my formula sheets. In my local and state circles, students formula sheets were the source of knowledge, the source of power that fueled the top students and the top schools. They were studied, memorized, revered. But most of all, they were not shared. But when Alex asked for my formula sheets I remembered my experience at MOP and I realized that formula sheets are not really math. Memorizing formulas is no more mathematics than memorizing dates is history or memorizing spelling words is literature. I gave him the formula sheets. (Alex must later have learned also that the formula sheets were fools gold  he became a Rhodes scholar.)

The difference between MOP and many of these state and local contests I participated in was the difference between problem solving and what many people call mathematics. For these people, math is a series of tricks to use on a series of specific problems. Trick A is for Problem A, Trick B for Problem B, and so on. In this vein, school can become a routine of learn tricks for a week  use tricks on a test  forget most tricks quickly. The tricks get forgotten quickly primarily because there are so many of them, and also because the students dont see how these tricks are just extensions of a few basic principles.

I had painfully learned at MOP that true mathematics is not a process of memorizing formulas and applying them to problems tailor-made for those formulas. Instead, the successful mathematician possesses fewer tools, but knows how to apply them to a much broader range of problems. We use the term problem solving to distinguish this approach to mathematics from the memorize-use-forget approach.

After MOP I relearned math throughout high school. I was unaware that I was learning much more. When I got to Princeton I enrolled in organic chemistry. There were over 200 students in the course, and we quickly separated into two groups. One group understood that all we would be taught could largely be derived from a very small number of basic principles. We loved the class  it was a year long exploration of where these fundamental concepts could take us. The other, much larger, group saw each new destination not as the result of a path from the building blocks, but as yet another place whose coordinates had to be memorized if ever they were to visit again. Almost to a student, the difference between those in the happy group and those in the struggling group was how they learned mathematics. The class seemingly involved no math at all, but those who took a memorization approach to math were doomed to do it again in chemistry. The skills the problem solvers developed in math transferred, and these students flourished.

We use math to teach problem solving because it is the most fundamental logical discipline. Not only is it the foundation upon which sciences are built, it is the clearest way to learn and understand how to develop a rigorous logical argument. There are no loopholes, there are no half-truths. The language of mathematics is precise, as is right and wrong (or proven and unproven). Success and failure are immediate and indisputable; there isnt room for subjectivity. This is not to say that those who cannot do math cannot solve problems. There are many paths to strong problem solving skills. Mathematics is the shortest.

Problem solving is crucial in mathematics education because it transcends mathematics. By developing problem solving skills, we learn not only how to tackle math problems, but also how to logically work our way through any problems we may face. The memorizer can only solve problems he has encountered already, but the problem solver can solve problems shes never seen before. The problem solver is flexible; she can diversify. Above all, she can create. [/quote]

I agree with Orl: Beginner's Corner forums shouldn't be self-pruned.  If they are, then any concepts that get explained in these forums will have to be explained over and over again.  Furthermore, good questions can get asked there, too, and if pruning exists, then those questions will be deleted.

Another possibility is if we have a Beginners Archive forum that all of the particularly good ("useful") posts in Beginners Corner forum get moved to.  This Beginners Archive would not be pruned, so we would have a way of saving old posts from otherwise certain oblivion.   :)

I was thinking about the questions of the type: how do you prove Cauchy? kinna stuff which will be asked regularly. the archive is a better idea indeed :)

http://www.physicsforums.com does a good job at archiving past posts . . . please follow up on how they do it (make yours better  :D )

http://www.physicsnetwork.com

I am sorry to tell you integral0 but not the site itself does the job, but Vbulletin 2.x which is quite expensive. I prefere to tweak this open source phpBB. :D

no problem at all. just for the record I have spent hundreds of hours in managing / developing for this site :) and I can tell you it's not an easy job ...

[quote="Syntax Error"]Yay! Finally I can now understand the goings on of this forum.  Preferably USAMO level (usa math olympiad).  I bet not as many people here are 13 years  old either[/quote]
you'd be amazed to know how many :D:D anyway - it's done ...

[quote="PiDeltaPhi"]phpBB is the best free forum software available,so its obvious choice.Happy tweaking  :) ![/quote]

phpbb freak as well? I keep getting pleasantly surprised here :lol:

well I think the beginner's section is incredibly good!!
It's pretty cool to see a bit increasing in your skills

when I came here solve% (advance - begin) was like 15% - 60%
Now it's like 35% - 95%

Plus the beginners' is a perfect training stage for national olympics (hope I'm finally going to be selected for IMO :lol: )


2 circles Y1 and Y2 intersect at A and B. A point C lies on Y1 and points D and E lie on Y2 such that CAD and CBE are straight lines. Points P and Q are constructed on AB (produced or otherwise) such that $\angle QDA=\angle PEB=\angle ACB$. QD and PE are extended to cut Y2 again at S and R respectively. Prove that SR = AB iff DE = AR = SB = d, where d is the diameter of Y2.

We can see that both $S$ and $R$ are fixed point, because $\hat{ADS}= \hat{PER}= \hat{ACB}=const$, and with this we also have $AB=SR$ iff $(Y_{1})$ and $(Y_{2})$ are perpendicular.
When $(Y_{1})$ and $(Y_{2})$ are perpendicular :
+ $CB \perp BD \Rightarrow EB \perp BD \Rightarrow DE=d$
+ $ABRS$ is a rectangle so $AR=BS=d$
[hide="Image"][img]http://i108.photobucket.com/albums/n9/PDatK40SP/4534535.jpg[/img][/hide]


Functional Analysis
advanced fields
advanced fields unsolved

$E$ a Banach space and $T$ an isometry of $E$. Prove that if the spectrum of $T$ is not included in the unit circle of $\mathbb{C}$, then it is the closed unit ball of $\mathbb{C}$.

Clearly, the spectrum is contained in the closed unit disk of the complex plane.

If $T$ is invertible, then the reciprocals of the elements of the spectrum form the spectrum of the isometry of $T^{-1}$, so they're also contained in the unit disk, and it follows that the spectrum of $T$ is a subset of the unit circle.

Suppose now that $T$ isn't invertible, but that the spectrum $\sigma(T)$ of $T$ doesn't contain the entire unit disk. It means then that $\sigma(T)$ has at least one element $|\lambda|<1$ on its border. Let $\lambda_n\to\lambda$ be a sequence of elements which do not belong to $\sigma(T)$. We have $\left\|(T-\lambda_n)^{-1}\right\|\to\infty$ (this is easy to prove), so there is a sequence $\|x_n\|=1$ such that $(T-\lambda_n)x_n\to 0$. It follows from this and from $\lambda_n\to\lambda$ that $(T-\lambda)x_n\to 0$. This, however, is impossible, because $|\lambda|<1$ and $\|(T-\lambda)x\|\ge|\|x\|-\|\lambda x\||=(1-|\lambda|)>0$ whenever $\|x\|=1$.


linear algebra
matrix
linear algebra unsolved

It seems to me it was posted

Any matrix A in M(3,C) is trigonalisable call a,b,c the eigenvalues

$det(A^2+I)=(a^2+1)(b^2+1)(c^2+1)=(det A)^2 + \sum a^2b^2 + \sum a^2 +1$ (1)

with $trA^2=0$ we have $\sum a^2 = 0$
with trA= 0 taking the square and simplify to get $\sum ab = 0$ taking again the square

$0=\sum a^2b^2 +2abc(\sum a)$ so $\sum a^2b^2 =0$

with (1) $det(A^2+I)=(det A)^2 +1$

Yes, Moubi! That's my solution too. :)  ;)


Assume $ x_{i}$ are real numbers that satisfy
1) $ \sum\limits_{i\equal{}1}^{n} x_{i}\equal{}0$
2) $ x_{i} \equal{}1$ with some $ i \in (1,2,..,n)$
Prove that
                 $ Max(|x_{1}\minus{} \frac{x_{n}\plus{}x_{2}}{2}|;|x_{1}\minus{} \frac{x_{n}\plus{}x_{2}}{2}|;..;|x_{1}\minus{} \frac{x_{n}\plus{}x_{2}}{2}|) \geq \frac{8}{n^{2}}$

Can anyone solve this problem? It is really hard to me!

It is not true. For example n=2 $ x_1\equal{}\minus{}1,x_2\equal{}1,n\equal{}2$.

[quote="duytungct"]
Prove that
                 $ Max(|x_{1} \minus{} \frac {x_{n} \plus{} x_{2}}{2}|;|x_{1} \minus{} \frac {x_{n} \plus{} x_{2}}{2}|;..;|x_{1} \minus{} \frac {x_{n} \plus{} x_{2}}{2}|) \geq \frac {8}{n^{2}}$[/quote]

Sorry!I'm very sorry! I can't understand why I didn't realize that big mistake earlier :(
This is the true problem

[quote]Assume $ x_{i}$ are real numbers that satisfy
1) $ \sum\limits_{i \equal{} 1}^{n} x_{i} \equal{} 0$
2) $ x_{i} \equal{} 1$ with some $ i \in (1,2,..,n)$
Prove that
                 $ Max(|x_{1} \minus{} \frac {x_{n} \plus{} x_{2}}{2}|;|x_{2} \minus{} \frac {x_{1} \plus{} x_{3}}{2}|;..;|x_{n} \minus{} \frac {x_{1} \plus{} x_{n \minus{} 1}}{2}|) \geq \frac {8}{n^{2}}$[/quote]

To Rust: with $ n\equal{}2,x_{1}\equal{}\minus{}1,x_{2}\equal{}1$ then $ |x_{1} \minus{} \frac {x_{n} \plus{} x_{2}}{2}|\equal{}|x_{1} \minus{} \frac {x_{2} \plus{} x_{2}}{2}|\equal{}2$. It is true

Sorry everyone again!  :blush:


There are 10 prizes. Five A's, three B's, and two C's are each placed in identical envelopes for 10 mathematics winners to pick from. They are to pick randomely. What is the probability that when the eighth contestant goes to pick their prize, there is 1 A, 1 B, and 1 C to choose from?

sorry , but this is just the road.it´s actually quite intuitive...sorry for the sloppiness and  thye lack of details.
Here you have to count the possiblities. it will have been picked 4 $ A$ , 2$ B$ and 1$ C$.now all we have to do is count all possible orders and divide by all possibilities for the $ 7$ first picked cards.This wiil be the probability we´re searching for.
(you´ve gotta check if you´re not overcounting or "less counting", but then are just little modifications...)
i´m not doing the computations.Sorry...

At first I tried taking $ \frac{7!}{4!2!1!}$ and dividing by the total number of ways to pick the first 7, but I didn't know how to count the total number of ways to pick the first 7. So I looked for the total number of possibilities of all 10 letters that included our requirements and divided by the total number of ways to pick 10 letters. The desired ways is $ (\frac{7!}{4!2!1!})(3!)\equal{}630$ because there are $ 3!$ ways to arrange the last A, B, and C. And the total number of 10 letter possibilities is $ \frac{10!}{5!3!2!}\equal{}2520$. So $ 630/2520\equal{}1/4$


function
induction
strong induction
algebra unsolved
algebra

If N* includes 0 it's pretty easy.  :lol:
If it doesn't, is it possible to somehow extend the function so that it does and then take 0 of again?
(I really don't think so but...)

$f(n+2) = \frac{ f(f(n+1)) f(n+1) - 1 }{ f(f(n)) }$,

We can show by strong induction that $f(2) = 2 \implies f(n) = n$.

Now I think the general solution depends on $f(2)$.

Yeah.. I can't think of any way to calculate f(2)

it is a problem in my school ,I think it be hard ,I hope anyone will solve it in this week.maybe it is number theory problem

Okay, so casework on $f(2)$!

Case:  $f(2) = 1$.

$f(3) + 1 = 1$, so $f(3) \not \in \mathbb{N}^{*}$.

Case:  $f(2) = 3$.

$f(3) + 1 = 3 f(3)$, so again $f(3) \not \in \mathbb{N}^{*}$.

Case:  $f(2) = k > 3$.

$f(3) + 1 = k f(k)$

Now if $f(3)$ exists then $f(3) \ge 3$.

I think we can derive contradiction from here; specifically, $f(3) \ge 3 \implies f(k) \ge k \implies f(3) \ge 8 \implies ...$

But I am not sure.  Anyone want to fill in the details?   :|

I think [i]t0rajir0u[/i] is wrong .why if $f(3)$ exist then $f(3)>3$

http://www.mathlinks.ro/Forum/viewtopic.php?t=85638

[quote="dieuhuynh"]I think [i]t0rajir0u[/i] is wrong .why if $f(3)$ exist then $f(3)>3$[/quote]

Oops, that should be $f(3) \ge 3$.  :|


I guess now (> 23,00, at least on my clock) we should be allowed to discuss the 1st Round problems of the Contest.

Well, I solved the 2nd one quite simply, and I guess the other should have a similar solution. I have been thinking for a long time about the 1st one, but I still don't have any idea. Similarly for the 3rd one.

Where do you have such difficult problems from, Valentin?

  Friendly,
  Darij

Unfortunately  you  made a mistake in problem 2's solution :blush: I had to give you a 0p for it :(.  The problem was created by me.

The last problem is a proposal of Harazi, and the first one I picked it up from one of my old notes, so it must have come from an old olympiad or something.

The first part of the 2nd problem was easy (AM-GM), but the second part...  :?

The third I brought down to this:

Let a,b,c relativley prime natural numbers. Furthermore we can assume: c is a prime and a > b and a > c.

Now prove that there exists a n_0 such that
a^n - b^n != 0  (mod c^n)
is true for all n >= n_0.

the 2nd part of the inequality in problem 2 can be done with brute force,use good old Muirhead's inequality.i gave some thought to question 3 and made little progress.
as fro my not entering into the contest,i dont actually see the point in doing it becoz i'm no more eligible for the math olympiads :(  :(  and then goddamnit exams i'm having these days

Curses - now you all see how ignorant I am in this field, I have applied AM-GM with reversed sign!!

Sorry Valentin, I could have not sent the solution to you as well, this would save your time... But was the format readable?

@ galois: Is there a good means to memorize Muirhead's inequality?

Thanks.
  Sincerely,
  Darij Grinberg

I checked the pdf file. It was pretty obvious that you applied AM-GM reversely :D, so don't worry about me spending too much time correcting it.

I will have soon some soln online, don't worry.

Problem 2 was nice. Let's give some tricky solutions without brute force (god, I hate that one):
   The first one:
   Put a=x^2,... and so on. Now use the fact that (u+v)(v+w)(u+w) \geq 8/9(u+v+w)(uv+vw+wu). So, it is enough to prove that 8/9 \sum x^2* \sum (xy) 2  \geq 8/27(xy+yz+zx)^3. This follows from  \sum x2  \geq  \sum xy and  \sum (xy) 2  \geq ( \sum xy) 2 /3. And we are done!

   The second one:

   We have xy/(x^2+y^2)+ y^2/(y^2+z^2)+x^2(z^2+x^2) \geq 3xy/ PRODUCT^1/3. Write also y^2/(x^2+y^2)+yz/(y^2+z^2)+z^2/(z^2+x^2) and a similar one and add them up, using also that xy/(x^2+y^2)<=1/2. You will find excatly the proposed inequality.

  Third one:

   It follows from Carlson inequality: ((a+b)(b+c)(c+a))^1/3 \geq  \sqrt (ab+bc+ca)/3. This one can be proved exactly using the first approach.

 Fourth approach:

   It follows directly from an inequality already discussed:
  prod (a^2+ab+b^2) \geq (ab+bc+ca)^3 using that a^2+ab+b^2<=3/2(a^2+b^2).

   A beautiful problem, with so many solutions. And I bet there are more. Very nice!

aha,plenty of ways to prove this inequality.nice inequality ,valentin
actually i thought about using carlson but somehow went off the track proving carlson's inequality is a task in itself ,though my proof is not brute force,it's much more beautiful :D

yeah Valentin.. nice problem... the first part was pretty easy... but the second one... i tried a LOT to solve that... damn.... i tried so many things and nothing. i also reduced it in an inequality with 2 variables, or with triangle sides... ot one time i was obsess to solve it... in the end I renounced because I had to prepare myself for the district olympiad.

I didn't participate, bacause I didn'trealize the contest had begun until 2 days before the end, but I did think a bit about the problems. The answer for the 1'st problem is "yes", isn't it? I'm asking because I might be remembering it all wrong because I can't look at the problems right now (I'm in an ICafe and I can't download the problems, since they don't give customers access to the hard drive of the PC).


real analysis
invariant
geometry
geometric transformation
vector
rotation
linear algebra

I'm working through Rudin's Principles of Mathematical Analysis. This question might be better answered by someone familiar with the text, but in case you aren't or don't have a copy handy, here's the exercise in question (#16 from Chapter 1):
Let x,y be elements of Euclidean k-space (k>=3) such that |x-y|=d>0. Show:
a) if 2r>d, there exist an infinite number of z such that |z-x|=|z-y|=r.
b) if 2r=d, there exists one such z.
c) if 2r<d, there exists no such z.
Rudin also asks about the relevance of the condition that k>=3.

Intuitively, these statements seem clear. I can picture (in 3-space) the solution set to part a); it would form a circle equidistant from x and y with radii perpendicular to the line between x and y. I'm somewhat at a loss when it comes to explicitly proving that such a solution set would be infinite (in general k-space, no less). Any help, especially hints as to proof techniques, would be much appreciated.

I flipped past this problem last week, actually.  The technique here is to show two things:

* If $ v_z \equal{} z \minus{} \frac{x\plus{}y}{2}$, then any $ z$ such that $ v_z\cdot(x\minus{}y)\equal{}0$ and $ |v_z|^2\equal{} r^2\minus{}\frac{d^2}{4}$ satisfies the condition.

* There are infinitely many $ z$ with this property.

I think that you'll find both statements significantly easier to prove than the original problem.

What I've done is to convert the geometric picture into something purely algebraic.  Try it again, and let me know if you need more details.

Well, I was definitely thinking along those lines, but I'm having difficulty showing the connection between the collection of points you name and the condition required by Rudin. Maybe I just need to fool around with it more, but every time I try to show their equivalence I either hit a dead end or the algebra gets so messy I assume I'm taking the wrong approach.

Never mind, as soon as I posted that I found a workable approach: |z-x|^2=|v+(y-x)/2|^2=(v+(y-x)/2)*(v+(y-x)/2)=|v|^2+v*(y-x)+(|y-x|^2)/4
=r^2-(d^2)/4+0+(d^2)/4=r^2.

I'm still working on showing that there are infinitely many z with those properties. There are no similar proofs in the text, and I really have no idea where to start.

It's not hard to show that the problem is invariant under translations, so we can assume that $ x\plus{}y\equal{}0$, then we're looking for infinitely many vectors $ v$ of a fixed length that are perpendicular to a fixed vector $ w$.

Lots of ways to do this.  I can think of at least 2:

* Show that we can rotate $ w$ to some multiple of $ e_1$, preserving distance and inner products.  This is potentially more annoying than it sounds without using the theory of matrices.  Perhaps explicitly demonstrate a matrix that zeros out the last coordinate, then one that zeros out the second to last, and so forth...  Anyway, after you're able to assume $ w\equal{}e_1$, the problem is easy.

* Show that the set $ V\equal{}\{v|v\cdot w\equal{}0\}$ is a vector space, and that any element of $ \mathbb{R}^k$ can be written as a linear combination of $ w$ and an element of $ V$.  It follows that the dimension of $ V$ is $ k\minus{}1 \geq 2$, so that it contains a circle of the desired radius.