geometry
circumcircle
trigonometry
perpendicular bisector
JBMO

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$).

[i]Greece[/i]

In the above post, Iris Aliaj asked whether somebody understood the following solution (in fact, taken from Kalva):

[quote="Kalva's solution"]Solution

We show that M must lie on the line through A parallel to BC.
Let O be the center of the circumcircle of ABD and O' the center of the other circumcircle. Let X be the midpoint of AD. Take coordinates with origin the midpoint of BC, x-axis along BC and y-axis through A. It is sufficient to show that the y-coordinates of O and O' sum to that of A, or equivalently that their mean is the same as the y-coordinate of X. But it is easy to see that the x-coordinate of X less that of O equals the x-coordinate of O' less that of X. Hence the y-coordinate differences must also be equal[/quote]

Here is a simpler solution:

We will prove that the point M lies on the parallel to BC through A. Once this will be proven, it will obviously follow that the distance from M to BC equals the distance from A to BC, hence the triangles BCM and BCA have an equal altitude, and since the bases are equal, too, the areas are equal; in other words, the area of triangle BCM is equal to the area of triangle BCA, and hence independent of the position of D.

Hence it remains only to show that M lies on the parallel to BC through A.

Since BB' and CC' are diameters of corresponding circles, we have < BDB' = 90° and < CDC' = 90°. Hence, both points B' and C' lie on the perpendicular to BC at D. As a consequence, the points D, B' and C' lie on one line.

Now, in the following I will only consider the case when the point C' lies between the points D and B'. The other cases can be done similarly.

Now, since the points B, A, D and B' lie on one circle, < AB'D = < ABD. On the other hand, since the points C, A, D and C' lie on one circle, < ACD = 180° - < AC'D. Finally, since triangle ABC is isosceles, < ABC = < ACB. Hence, we have

  < AB'C' = < AB'D = < ABD = < ABC = < ACB
          = < ACD = 180° - < AC'D = < AC'B'.

Thus, the triangle AB'C' is isosceles, too, meaning that AB' = AC'. Therefore, the point A lies on the perpendicular bisector of the segment B'C'. Hence, since M is the midpoint of the segment B'C', the line AM is perpendicular to B'C'. But the line B'C' itself is perpendicular to BC (since both points B' and C' lie on the perpendicular to BC at D); hence, it follows that the line AM is parallel to BC. In other words, the point M lies on the parallel to BC through A. This is what we wanted to show.

This may look like a long proof, but the one you cited and you couldn't understand was actually longer, just the writer had left out many steps.

  Darij

Let $w_1=C(O_1,r_1),\ w_2=C(O_2,r_2)$ be the circumcircles of the triangles
$ABD,\ ACD$ respectively. Because $AD=2r_1\sin B=2r_2\sin C$ results that
the triangles $ABD,\ ACD$ have the same length $r_1=r_2=\rho$
of the circumradius. I note $x=m(\widehat {ABB'})$.

$B'D\perp BC,\ C'D\perp BC\Longrightarrow D\in B'C',\ \overline {DB'C'}\perp BC$.

$\widehat {ABB'}\equiv \widehat {ADB'}\equiv \widehat {ADC'}\equiv \widehat {ACC'}\Longrightarrow AB'=2\rho \sin x,\ AC'=2\rho\sin x\Longrightarrow$

$AB'=AC'\Longrightarrow AM\perp B'C'\Longrightarrow AM\parallel BC\Longrightarrow \sigma [BMC]=\sigma [ABC]$.

[u]Remark.[/u] $\sigma [XYZ]$ is the area of the triangle $XYZ$.

Nice problemand here is my solution :blush:

It is easy to prove B',C',D are collinear
AB=AC    BB'=CC'=AB/sin∠ADB=AC/sin∠AC'C     ∠C'CA=∠C'DA=∠ABB'   =>   ⊿ABB'≌⊿ACC'   =>   AB'=AC'    =>   AM⊥C'D
done

Since $BB'$ is a diameter, $DB'$ is perpendicular to $BD$. Since $CC'$ is a diameter, $DC'$ is perpendicular to $CD$. Thus $C'$, $B'$, and $D$ are collinear (in that order). We also have $<BAB' = 90$ and $<CAC' = 90$. Thus $<BAC = <B'AC'$. Also, $<ABC = 180 - <AB'C = <AB'C'$. Thus $ABC$ and $AB'C'$ are similar. This means that $AB'C'$ is isosceles, so $AM$ is perpendicular to $B'C'$, so $AM$ is parallel to $BC$. Thus the area of $MBC$ equals the area of $ABC$ and hence is constant.

[asy]
import olympiad;
import cse5;
pointpen = black;
pathpen = black;

size(10cm);

pair A=dir(90),
B=dir(190),
C=dir(350),
D=B/3+2C/3;

D(A--B--C--cycle);
D(circumcircle(A,B,D),green);
D(circumcircle(A,C,D),green);

pair Bp=2*circumcenter(A,B,D)-B,
Cp=2*circumcenter(A,C,D)-C,
M=(Bp+Cp)/2;

D(B--Bp,blue);
D(Cp--C,blue);
D(Bp--Cp,red);

D(Bp--A--Cp,dashed+gray);
D(A--foot(A,B,C),dotted+gray);

MP("A",D(A),A);
MP("B",D(B),B);
MP("C",D(C),C);
MP("D",D(D),dir(270));
MP("B'",D(Bp),dir(50));
MP("C'",D(Cp),Cp);
MP("O_B'",D(B/2+Bp/2),dir(300));
MP("O_C'",D(C/2+Cp/2),dir(30));
MP("M",D(M),dir(180));
MP("X",D(foot(A,B,C)),dir(270));
[/asy]
\begin{align*}
\intertext{Let $X$ be the foot of the altitude from $A$ to $BC$. Since $BB'$ and $CC'$ are diameters,}
\measuredangle BAB'=90^\circ&\text{ and }\measuredangle CAC'=90^\circ\\
\intertext{Thus,}
\measuredangle BAB'&=\measuredangle CAC'.\\
\intertext{Adding $\measuredangle B'AC=-\measuredangle CAB'$ to both sides,}
\measuredangle BAB'-\measuredangle CAB'&=\measuredangle C'AC-\measuredangle CAB'.\\
\measuredangle BAC&=\measuredangle B'AC'.\\
\intertext{Thus, there is a spiral similarity centered at $A$ with angle $90^\circ$ that maps $B\mapsto B'$, $C\mapsto C'$. Additionally, it maps $X\mapsto M$. Thus, $\measuredangle XAM=90^\circ$, and it does not depend on the choice of $D$. Hence, $AM\parallel BC$, and the height from $M$ to $BC$ is constant. Since $BC$ is fixed, we have}\\[-2em]
[MBC]&=\frac{1}{2}BC\cdot \delta(M,BC)\text{ is constant. }\\[-2.4em]
\intertext{\hfill$\Box$}
\end{align*}

Its easy to see that  $B'$, $C'$, $D$ are collinear (since angle $B'DB$ = $C'DC$ = $90^o$).

Applying Sine rule in triangle $ABC$, we get:
$Sin(BAD) / Sin(CAD) = BD / DC$

Since $BAB'D$ and $CC'AD$ are cyclic quadrilaterals, anlge $(BAD)$ = anlge $(BB'D)$
and $angle (CAD) = angle (CC'D)$

So, $Sin(BB'D) / Sin(CC'D) = BD / DC$

So $BD / Sin(BB'D) = DC / Sin(CC'D)$
Thus, $BB' = CC'$ (the circumcirlcles $\mathcal{C}_1,\mathcal{C}_2$ are congruent).

From right traingles $BB'A$ and $CC'A$, we have:
$AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}$
So $AC' = AB'$

Since $M$ is the midpoint of $B'C'$, $AM$ is perpendicular to $B'C'$ and hence $AM$ is parallel to $BC$.

So area of traiangle $MBC$ = area of traingle $ABC$ and hence is independent of position of $D$ on $BC$.


hi friends,
   any one here know where(in which website link) we can get the results of RMO-2006 :?:
   IF YOU KNOW PLEASE TELL ME :help:

  regards,
 santhosh

I don't think the results of RMO are put up on any website.
 If you have qualified , you should have been intimated about
 your selection by now.

thank u for sending the reply :)
did u get selected for rmo any time :!:
if u got tell me how u got intimated
r u from sharma institute in hyderabad :!:

The RMO results will be released on 16 Jan.

ok.
what is the website inwhich we can find the results of RMO :?:

results have been postponed.they will be released on 17 th evening

Hi all,
First you have to realise that the RMO for each region is different , by that what I mean is students in Delhi have to clear a different examination than the one that Andhra students have to clear. Not all states put up their RMO results online, but some states do put up their online. For example, the results of the RMO in Maharashtra, Mumbai are available online. Also, the results of TamilNadu RMO is also available online and his years results too have been declared.


calculus
integration
limit
logarithms
calculus computations

Let $ k\geq 1$ be a natural number. Evaluate the Riemann-integral:

$ I_{k}\equal{}\int\limits_{0}^{1}\frac{x^{k}}{(1\plus{}x)^{k}}dx$.

One simple-minded approach. Let $ u\equal{}1\plus{}x$ so that the integral becomes

$ I_k\equal{}\int_1^2\frac{(u\minus{}1)^k}{u^k}\,du.$

Multiply that out and integrate it to get

$ I_k\equal{}1\minus{}k\ln2\plus{}\sum_{j\equal{}2}^k(\minus{}1)^k\binom{k}{j}
\frac1{j\minus{}1}\left(1\minus{}\frac1{2^{j\minus{}1}}\right).$

For the first few, if I did the arithmetic right, I get:

$ I_0\equal{}1.$

$ I_1\equal{}1\minus{}\ln2.$

$ I_2\equal{}\frac32\minus{}2\ln2.$

$ I_3\equal{}\frac{17}8\minus{}3\ln2.$

$ I_4\equal{}\frac{67}{24}\minus{}4\ln2.$

So far, I'm not seeing a simplifying pattern. However, by interchanging limit and integral, we must have that $ \lim_{k\to\infty}I_k\equal{}0.$ Indeed, the handful of values given above approximate to $ 1,0.3068,0.1137,0.04556,$ and $ 0.01908.$

Another approach is to integrate by parts: $ I_k\equal{}\frac{k}{k\minus{}1}I_{k\minus{}1}\minus{}\frac{1}{(k\minus{}1)2^{k\minus{}1}}$, hence $ \frac{I_k}{k}\equal{}1\minus{}\ln 2 \minus{}\sum_{j\equal{}2}^{k}\frac{1}{j(j\minus{}1)2^{j\minus{}1}}$. But I don't know a closed form of this sum.

Another approach:The substitution $ \frac{x}{1\plus{}x}\equal{}z$.


Because of $ (1+x)^\alpha=\sum_{k=0}^\infty{\alpha \choose k}x^{k}\quad (1-z^{2})^{-\frac12}=\sum_{k=0}^\infty{-\frac12 \choose k}(-1)^{k}z^{2k}$ and this is $ \sum_{k=0}^\infty \frac{1}{2^{2k}}{2k\choose k}z^{2k}$ .

Therefore $ \arcsin z=\sum_{k=0}^\infty \frac{1}{2^{2k}}{2k\choose k}\frac{z^{2k+1}}{2k+1}$ . Divide both sides by $ z$ and set $ z=\frac12$ .


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

If $ (A,\mathfrak{m})$ is a local ring, then $ l(A/I)<\infty$ implies that $ I$ is an $ \mathfrak{m}$-primary ideal.

I don't understand this step in a complicated proof of dimension theory :)

[quote="Jacobson"]If $ (A,\mathfrak{m})$ is a local ring, then $ l(A/I) < \infty$ implies that $ I$ is an $ \mathfrak{m}$-primary ideal.[/quote]

You have to additionally assume that $ I$ is a proper ideal of $ A$ (that is, $ I\neq A$).

First, $ \sqrt {I}\subseteq\mathfrak{m}$ (because for every $ r\in\sqrt {I}$, there exists an integer $ k\geq 0$ such that $ r^k\in I$, and thus $ r^k$ is non-invertible, and thus $ r$ is non-invertible, and thus $ r\in\mathfrak{m}$ (because in the local ring $ A$, all non-invertible elements lie in the maximal ideal $ \mathfrak{m}$)).

Now we will prove that $ \mathfrak{m}\subseteq\sqrt {I}$. In fact, let $ x\in\mathfrak{m}$ be arbitrary. Then, consider the chain of submodules $ ...\subseteq\left\langle\overline{x^3}\right\rangle \subseteq\left\langle\overline{x^2}\right\rangle \subseteq\left\langle\overline{x}\right\rangle$ of the module $ A\slash I$, where $ \overline{\lambda}$ means the canonical projection of some element $ \lambda\in A$ onto $ A\slash I$. Since the module $ A\slash I$ has finite length, some adjacent elements of this chain must be equal, i. e. there exists an integer $ n\geq 1$ such that $ \left\langle\overline{x^{n \plus{} 1}}\right\rangle \equal{} \left\langle\overline{x^n}\right\rangle$. Thus, $ \overline{x^n}\in\left\langle\overline{x^{n \plus{} 1}}\right\rangle$, so that there exists some $ y\in A$ satisfying $ \overline{x^n} \equal{} y\cdot\overline{x^{n \plus{} 1}}$. In other words, $ x^n \minus{} y\cdot x^{n \plus{} 1}\in I$. Equivalently, $ x^n\left(1 \minus{} xy\right)\in I$. Now, $ 1 \minus{} xy\not\in\mathfrak{m}$ (because if we had $ 1 \minus{} xy\in\mathfrak{m}$, then, together with $ x\in\mathfrak{m}$, this would yield $ 1\in\mathfrak{m}$, what contradicts to $ \mathfrak{m}$ being a proper ideal of $ A$), so that $ 1 \minus{} xy$ is invertible (because all elements of $ A$ not in $ \mathfrak{m}$ are invertible, since $ \left(A,\mathfrak{m}\right)$ is a local ring). Thus, $ x^n\left(1 \minus{} xy\right)\in I$ yields $ x^n\in I$ and thus $ x\in\sqrt {I}$. Since we have shown this for every $ x\in\mathfrak{m}$, we thus obtain $ \mathfrak{m}\subseteq\sqrt {I}$. Together with $ \sqrt {I}\subseteq\mathfrak{m}$, this yields $ \sqrt {I} \equal{} \mathfrak{m}$, and thus $ I$ is $ \mathfrak{m}$-primary. Qed.

By the way, instead of requiring that $ l\left(A\slash I\right) < \infty$, it was enough to require that $ A\slash I$ is an Artinian module.

What is the complicated proof where this is used?

[b]EDIT:[/b] Owk, the "known facts" you are using are pretty often restatements of the original questions. It is nice to know that there is nothing new under the sun, but I am not so sure whether this helps the original poster. Here, for instance, you use that the maximal ideal of an Artinian local ring is nilpotent.

  darij

Thanks!

Let $ A$ be a semilocal Noetherian ring and $ M$ a finite $ A$-module; then we have $ dim M \equal{} d(M) \equal{} \delta(M)$

(Matsumura -- Commutative ring theory, page 98; It is still very mysterious to me :))

Shorter proof: $ \ell(A/I) < \infty$ implies $ A/I$ artinian, hence $ \mathfrak m$ is nilpotent in $ A/I$. owk

EDIT @Darij: I just try to adapt to the prior knowledge of the OP. Someone trying to understand dimension theory knows basic facts about artinian rings (or, should know). If not, s/he should not hesitate to look them up or ask again. Of course, the problem may be easier than the result I used, but it is usually more efficient to just see that something follows trivially from known results than working out an elementary proof. owk

Thanks a lot both. A different point of views is always nice.

It's funny though that you both have a own style for mathematics. Darij his nice proofs are always long and very clear. On the other hand, owk always manages to solve something with a $ 1$-liner :D

My understanding of primary ideals was kinda blurry since I was relying on the definition

"A submodule $ N \subset M$ is $ P$-primary if $ Ass(M/N) \equal{} \{P\}$"

That brought my far away from the much more obvious (similar) definition I learned in the previous book :)


Apologize if I am making a total fool out of myself again, but could someone check question #3 on the Round 1, 4th Edition? I think there might be a little problem.  :?

Thanks!

Yeah! I thought so too...and if it's 1/2, then the problem is more plausible... :lol:

[quote="darij grinberg"]I have already written to Valentin about this, but he hasn't replied since he is probably quite busy now with correcting the 3rd edition (I can understand him very well). Anyway, the problem is wrong as stated, since the two condition $O_1O_2=\frac32 \left(r_1+r_2\right)$ implies that the circles $\Omega_1$ and $\Omega_2$ cannot intersect. I am pretty sure that this condition must be replaced by $O_1O_2=\frac12 \left(r_1+r_2\right)$; at least, if it's really $O_1O_2=\frac12 \left(r_1+r_2\right)$, then the problem is correct. And, of course, "itnersect" should be "intersects"...

  Darij[/quote]

It's really amusing because I also wrote a PM to Valentin about this, saying exactly the same thing as you, Darij, right down to the typo about "intersects" :D .

Hi everyone,

Indeed it should be $\frac 12 (r_1+r_2)$ and not $\frac 32(r_1+r_2)$ as it is now.
I will corect the file.

well, can anyone post the right question. i mean the complete correct problem.

It's the exact same problem, but only replace the constant $\frac 32$ with the constant $\frac 12$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group with $ 108$ elements. Show that $ G$ has a normal subgroup of order $ 9$ or order $ 27$. Hint: [hide] Consider the action by conjugation of $ G$ on the Sylow $ 3$-subgroups of $ G$. [/hide]

Well, if $ G$ doesn't have a normal Sylow $ 3$-subgroup, it has $ 4$ Sylow $ 3$-subgroups.  The action by conjugation is transitive and induces a map $ \phi: G\to S_4$ with $ |\phi(G)| \geq 4$.  Since $ |\phi(G)|$ divides $ (|G|,|S_4|) \equal{} 12$ we must have $ |\phi(G)|\in\{4,6,12\}$.  Looking at the kernel, the only case to worry about is $ |\phi(G)| \equal{} 6$.

Then $ G$ has a normal subgroup $ H$ of order $ 18$, with a Sylow $ 3$-subgroup $ P\lhd H\lhd G$.  Since every automorphism of $ H$ sends $ P$ to itself, we must have $ P\lhd G$.  But $ |H|\equal{}9$ and we're done.


Two mathematicians walk into a bar.
The first one ordered a full drink.
The second ordered a half of a drink.
the first then refilled to a quarter.
The second refilled to and eighth.
After this, they stopped. The first mathematician then asked, "If we had continued our patter forever, how many drinks would we have had together?"
"That is too easy," said the second. "How many drinks would we have had separately?"
Can you answer both of their questions?
 :gathering:

Together, the total number of drinks is $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}\cdots\equal{}\sum_{j\equal{}0}^{\infty}\frac{1}{2^{j}}$, which is just a standard geometric series whose sum converges to 2.

Separately, the first mathematician has $ 1\plus{}\frac{1}{4}\plus{}\frac{1}{16}\plus{}\cdots\equal{}\sum_{j\equal{}0}^{\infty}\frac{1}{4^{j}}$, which is again a standard geometric series whose sum converges to $ \frac{4}{3}$.  The second mathematician has $ 2\minus{}\frac{4}{3}\equal{}\frac{2}{3}$ of a drink.


$ abc$ is a three digit number.if $ acb$ $ \plus{}$ $ bca$ $ \plus{}$ $ bac$ $ \plus{}$ $ cab$ $ \plus{}$ $ cba$ $ \equal{}$ $ 3194$, find $ abc$

$ \sum_{sys}100a \plus{} 10b \plus{} c \equal{} 222(a \plus{} b \plus{} c)$.
and,
$ \frac {3194}{222} \equal{} 14 \plus{} \frac {86}{222}$.

Therefore, $ 100 \le 222n \minus{} 86 \le 999$  i.e.  $ n \equal{} 15, 16, 17, 18$.

$ \boxed{\text{Case} \ n \equal{} 15}$
$ 100a \plus{} 10b \plus{} c \equal{} 222 \times 15 \minus{} 86 \equal{} 136$.
$ a \plus{} b \plus{} c \equal{} 10 \neq 15$.
False.

$ \boxed{\text{Case} \ n \equal{} 16}$
$ 100a \plus{} 10b \plus{} c \equal{} 222 \times 16 \minus{} 86 \equal{} 358$.
$ a \plus{} b \plus{} c \equal{} 16 \equal{} 16$.
True.

$ \boxed{\text{Case} \ n \equal{} 17}$
$ 100a \plus{} 10b \plus{} c \equal{} 222 \times 17 \minus{} 86 \equal{} 580$.
$ a \plus{} b \plus{} c \equal{} 13 \neq 17$.
False.

$ \boxed{\text{Case} \ n \equal{} 18}$
$ 100a \plus{} 10b \plus{} c \equal{} 222 \times 18 \minus{} 86 \equal{} 802$.
$ a \plus{} b \plus{} c \equal{} 10 \neq 18$.
False.

Thus, $ abc$ is $ \boxed{358}$.


Two electric charges q1 and -q2 are at a distance r from each other.
Prove that the equipotential surface V=0 is a sphere.What's its radius?

Thanks.

PS: My physics exam is aproaching. :blush:

Because of rotational symmetry with axis through the charges, we need to look only at one plane through the axis. Assuming $ q_1, q_2 > 0.$ At a point $ P$ in this plane, with distances $ r_1, r_2$ from $ q_1, q_2,$ the potential is zero: $ V(P) \equal{} \minus{}\frac {q_1}{r_{1}} \plus{} \frac {q_2}{r_{2}} \equal{} 0$ $ \Longrightarrow$ $ \frac {r_{1}}{r_{2}} \equal{} \frac {q_1}{q_2} > 0$ (CGS units). Locus of points $ P$ with a constant ratio of distances from 2 fixed points $ Q_1, Q_2$ is a circle (see [url]http://www.mathlinks.ro/viewtopic.php?t=121062[/url]), the P-Apollonius circle of the $ \triangle Q_1Q_2P$ with sides $ r \equal{} Q_1Q_2, r_1 \equal{} PQ_1, r_2 \equal{} PQ_2,$ centered on the line $ Q_1Q_2$ and passing throught the feet $ X, Y \in Q_1Q_2$ of the internal/external bisectors of the $ \angle O_1PQ_2.$ Its diameter is  $ |XY| \equal{} \left|\frac {rr_1}{r_2 \plus{} r_1} \plus{} \frac {rr_1}{r_2 \minus{} r_1}\right| \equal{} \frac {2rr_1r_2}{|r_2^2 \minus{} r_1^2|} \equal{} \frac {2rq_1q_2}{|q_2^2 \minus{} q_1^2|}$ and its radius half of this.


1) Is $ L^{\infty}$ is complete ?
2) If $ \parallel{}g_n\parallel{}_{\infty}\leq C<\plus{}\infty$ for all $ n$ and $ g_n$ converges pointwise to some $ g$ then whether $ g\in L^{\infty}$ ?
3) If we have a $ f\in L^2$ and $ g_n$ converges to $ g$ pointwise, $ g_n$ are in $ L^s\cap L^{\infty}$ and $ \parallel{}g_n\parallel{}_{\infty}<C<\infty$ for all $ n$. Is it true that $ g_nf\to gf$ in $ L^2$ ?

Thank you

1) yes 2) yes 3) yes by [url=http://en.wikipedia.org/wiki/Dominated_convergence_theorem]dominated convergence theorem[/url]

1) thanks 2) thanks 3) thanks to [url=http://en.wikipedia.org/wiki/Dominated_convergence_theorem]dominated convergence theorem
[/url]  :lol:

You should show me the details of 2) and 3) man?

Note that "converge pointwise" should in both cases be "converge pointwise a.e." That's all that makes sense, anyway.

Take a look at #2. The statement that $ \|g_n\|\le C$ is the statement that for almost every $ x$ and every $ n$ (there's an exceptional set for each $ n,$ but the countable union of sets of measure zero is still of measure zero), $ |g_n(x)|\le C.$ So what does that tell you about the pointwise limits?


function algorithm CEMC MIT college Euler

How was CCC, everyone?  :huh:

Foundation

Lol, terrible for me. Did not make it past #2...

I thought I had a good shot of doing #3, but had a slew of technical issues ranging from I/O problems to just an incorrect algorithm. Though I basically did not prepare for this at all, I still think I could have done much better. It didn't help that someone beside me basically solved every single problem except 4 in two of the three hours given.  :rotfl:

How'd you do?

Wow... I did TERRIBLY on senior... I did S1 and S2 in the first 10 mins (literally), and for S3, by the end, I think I had a working algorithm(in pseudocode) but Turing kept giving me some random errors... so I didnt officially get that... and for S4, I basically brute forced it... and S5 I might get like 3 test cases if im really lucky...

so, overall, I got majorly pwned!  :(  :wacko:  :noo:

How did everyone else do?

Werdna, did you hard-code like all 2000 something cases for S4, or did you pseudo-hard code (a,b,c,d, switch b,a,c,d, etc)?

I would like to hear how one would approach #4. A friend of mine had a pseudo-recursive function that only works when you take one number at a time, but no more than that.

Well, I did the pseudo hard code thing, but I think I may have missed some cases... (like the ones where you have more than 1 set of parentheses...)
I didnt have enough time to finish it! But, I might get like 3  of the test cases... which would be pretty good for me... :maybe:  but otherwise.. I have absolutely no idea on how to approach it...

Man, this year was pretty screwed for me too. My first CCC was total obliteration (lol, no other words can describe it quite enough), and there people saying that this year's was so much easier than the past years. I can only sit and scream, WTF?

What I thought was pretty clever was that one of the contestants at our school basically created a guessing program for S5, which analyzed the total amount of lines, and randomly generated, "Roland", "Patrick". The program got 2 cases correctly.  :rotfl:

Eh, well. I got some time till next year's CCC, and about two months till ECOO. Speaking of that, anyone else taking the ECOO this year?

I'm not very sure how many cases you would have gotten, as I didn't take a close look at the test cases for S5 (being that I didn't even touch S5 during the contest). However, I do know certain test cases were worth more than 3 pts. For S1, there were only 3 cases worth 5 each, but for S3/S4/S5, there were cases worth 3,4, and 5 marks.

[quote="Nawahd Werdna"]for S5 (lol) I just made my program output Roland the whole time... any idea on how many cases I might have gotten?
By the way, is each case worth 3 marks?[/quote]

I cried as soon as I read problem 5 cause I realized people were going to do exactly that and raise all the scores by a few points making it harder to make stage 2.  :|
Anyways, Massey kids did number 5 with DP. Doesn't look like anyone got number 4 perfectly though at our school. Blah.

What do you guys think the cutoff for Stage 2 will be?

DP...man, I don't even know how that works... :blush: ...have a lot to catch up since grade 8. As for cutoff, I'd wager...hmmm...57/63?

EDIT: is it just me, or did the post count just go down 28 for everyone?

Yeah.... my post count just went down!!!!

Woah... I knew 5 was DP, I just didnt know how to do it... BTW, how did you do Farnak?

EDIT : Did anyone notice that S5 = J5?!?!?!?

Wow.. I feel bad for like the one junior person who could have pwned this problem, and didnt make stage 2 because they didnt take senior, although that is an unlikely scenario...I wonder how many close to perfects there will be for junior...

[quote="Nawahd Werdna"]Yeah.... my post count just went down!!!!

EDIT : Did anyone notice that S5 = J5?!?!?!?

Wow.. I feel bad for like the one junior person who could have pwned this problem, and didnt make stage 2 because they didnt take senior, although that is an unlikely scenario...I wonder how many close to perfects there will be for junior...[/quote]

Yeah, right after the whole ordeal with the senior CCC, I took a peek at the Junior problems. Froze as soon as I noticed the J5 problem. After frantically searching for differences and loopholes, I ended up feeling relieved that I did not take the Junior test...even though I did so badly  :lol:

Lol, there was one major difference though. In J5 the string was at most 8 characters long. In S5 the string was at most 30 characters long.

Hey Nawahd! I'm just hoping I can make Stage 2 with my score, I made some careless mistakes on 3 and 4. Foundation's going to Waterloo this May for sure  :P

Ahh ..... Shalev!!!!!!! 5?!?!?!?!
You guys have more Stage 2 qualifiers than us then haha, gg gg.

[quote="shalevbd"]I think WCI kids pwned this contest, but its not marked yet for us so I'm not sure. It appears that 5(!) of us in WCI solved all problems with efficient enough algorithms, so we are making stage two unless we made stupid mistakes. That would be a record for us... but then again, this year was our CS year - everyone was training like crazy =D

I don't think we can discuss algorithms yet unfortunately, so no hinting how you did #5 guys :P

[quote]for S5 (lol) I just made my program output Roland the whole time... any idea on how many cases I might have gotten? [/quote]
Zero. The whole point of multiple tasks in one test case is that its all or nothing for that test case - this prevents people from doing what you just said.

I found the contest pretty easy compared to previous years... I expect the cutoff to be in the low 60s maybe? What do you guys think?[/quote]
Wow nice! congrats! gl @ stage 2!
I got owned, well I was never good at programming.

Not sure if we're allowed to post the problems yet, as not everyone's done marking...

[quote]I didn't take it this year, does anyone have a copy of the questions?[/quote]
The questions will be available on the CEMC website soon. We can't spread them yet :P

[quote]Wow nice! congrats! gl @ stage 2!
I got owned, well I was never good at programming.[/quote]
Thanks Alex! Too bad you didn't make stage 2 :( I guess I'll see you at MCTIC though, so its not too bad...

btw Frank how were Massey's scores? Who is making stage 2?

Peter's going for sure, and then probably me, Vick Yao, and this kid called Tom. Congrats to Waterloo too, who's going from there?
Sign up for ECOO!

It'll be nice to see you at ECOO, Shalev! It'll be a fun trip to say the least, and stage 2 preparation somewhat!

oh wow, our results are up, and we really messed up badly :S
One guy got 71, but the rest of us screwed up #4 and some of us screwed up #2 as well. I think we'll probably have 1 person in stage 2 instead of 5 lol

I think its sad how we all made mistakes... the contest is partly to blame, since there are so many places to mess up on for #4, and there was no hard enough question.

For my #4, at one point in my code I hard coded the combinations for two cards A and B as A+B, A-B, A/B, and A*B, so I forgot about B-A and B/A. This left me with a score of 62, which will probably not make stage 2 (although I'm not sure). This is the second highest score in WCI, so there is pretty much no chance of anyone else going to stage 2 :(

We did have 5 people with all correct algorithms for numbers 3,4,5... I don't know how this happened :S lol we suck :P

On a brighter note, we are trying to organize an ECOO team, so hopefully we'll see you there :)

Just wondering, how do most of you prepare for programming contests? I've never taken a class before, though I doubt that'd help. I'm interested in knowing which books/sites/contests you guys do, as I might try for senior again next year.

At Massey, we have a computer science club that our teacher actively supports. It's dedicated to helping students prepare for cs contests. If setting up a club is really impractical then there's a self-study thing, our best programmer achieved his results solely by working off the USACO training pages and contests. Other resources he used minorly were the UvA grader and some random Chinese websites that he studied code and algorithms off. The rest of us worked off USACO and old CCCs ... I guess all we did was code up question after question (without looking at solutions) and read analysis after analysis.
Programming Challenges is supposed to be a popular book. One of our alumni who went to IOI studied off this MIT book called Introduction to Algorithms.

Share with us what Waterloo does Shalev :P
Btw, One guy got 71? Nice, we now know who the top 2 in this region are.

If you are dedicated to programming, USACO is definitely the best thing to do. We also have a CS club, and this year everyone was very excited about CS and CS questions kept going around. I'm not sure where all the problems are from, but most are probably past CCCs.

Another site some guys discovered is called Project Euler, which combines CS with math. However, I don't think thats the best thing to do if your goal is just to do well in CS contests.

I have the Cornell algorithms textbook called algorithm design, which is a very good book (although I haven't read much of it).

Finally, create an account on compsci.ca and ask people there what they do :P

Thanks for the replies, I'm looking into USACO right now (never knew it existed before  :wink: ), and darn, the problems are hard. Maybe I'll work my way from the harder problems on CCC to the USACO.  Cause these look waaayyy out of my league. I'll definitely need a good book on algorithms though (I basically know nothing aside from recursion, and I'm not very hot with that either), so I'll look into Programming Challenges (any recommended bookstores? There's nothing in the TPL) and Algorithm Design.

Hopefully, by this time next year, I'll be a much more seasoned programmer. :D

Huh ... so apparently Woburn has 4 perfect scores. Dunno about the Stage 2 cutoff now ...  :maybe:

Richard Peng says that there is "NO WAY the cutoff will be over 60". I hope he's right :S he did help make the contest I think.

btw, when we say USACO we mean USACO training (at least I do), http://train.usaco.org/usacogate


If it's just a question of converting a spreadsheet to a table then try [url=http://www.tug.org/tetex/tetex-texmfdist/doc/help/Catalogue/entries/excel2latex.html]excel2latex[/url]. If by large table you mean one that goes across pages then try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=longtable]longtable package[/url].


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

Let  $G$ be a finite abelian group and x is an element of maximal order.Prove $\langle x \rangle$ is the direct summand of $G$.

ps:Please not use the decomposition of the abelian group,Because I want use this to prove it. :)

[quote="-oo-"]well, several proofs of the structure theorem use this result, so that you can look it up.[/quote]
But in my book just left it for exercise.I don't know how to prove it,can you post the proof?Thank you

Obviously we can assume that G is a p-group. A abelian p-Group is cyclic iff there is exactly one subgroup of order p.

We use induction on $|G|$.
If x has order $|G|$ were done, so we assume $ord(x)<|G|$ which means, that G is not cyclic, so there is a subgroup U of order p beside the one contained in $\langle x\rangle$.
Now $\langle x\rangle\cap U=\{0\}$ and so $x+U$ is a element of maximal order in $G/U$. Since $|G/U|<|G|$, we have a subgroup V with $U\leq V\leq G$ and $G/U = \langle x\rangle\oplus V/U$.
$\langle x\rangle\cap V$ is contained in the kernel of $G\to G/U$, so $\langle x\rangle\cap V\subseteq\langle x\rangle\cap U=\{0\}$. Hence G is the direct sum of $\langle x\rangle$ and V.

Thank you,but can I ask why can we assume G is a p-group?

for every abelian torsion group $G$, we can write $G$ as the direct sum of the $Syl_{p}(G) = \{\text{elements of order~}p^{n}\text{~for some~}n \geq 0\}$, where $p$ run's through all primes.


trigonometry
linear algebra
matrix
algebra
function
domain
vector

Let $ M(t \plus{} 1) \equal{} N\times M(t)$ and $ M^T(t) \equal{} \left[\begin{array}{ccc}1 & \sin\dfrac{\pi t}{3} & \cos\dfrac{\pi t}{3} \end{array}\right],$ where $ M^T$ represents the transposed matrix of $ M(t).$ If $ N$ is a constant matrix, calculate the value of $ \det N.$

Edit: $ M(t\plus{}1)$

[code]Solution to old version: M(t) = N.M(t)[/code]

I assume the domain of $ t$ is $ \mathbb{R}$

[hide="Solution to Old"][b]Key Observations:[/b]
1. $ N$ is a 3x3 matrix.
2. $ M(t)$ is an eigenvector of $ N$ with associated eigenvalue $ 1$ for all $ t$.

[b]Solution:[/b]
We can chose $ t_1$, $ t_2$, $ t_3$ such that: $ \text{span} \left \{ M \left( t_1\right ), M\left( t_2\right ), M\left( t_3\right ) \right \} \equal{} \mathbb{R}^3$.
By linearity, [i]any[/i] $ \mathbb{R}^3$ vector is an eigenvector of $ N$ with associated eigenvalue $ 1$.
Since the standard basis is mapped to itself, $ N \equal{} I_3 \implies \boxed{\text{det } N \equal{} 1}$[/hide]

[hide="Solution to New"]Let $ T$ be the linear map in $ \mathbb{R}^3$ that rotates a vector an angle $ \pi /3$ counterclockwise around the x-axis.

Observe that $ NM(t) \equal{} TM(t) , \; \; \forall t$

$ \therefore$ By linearity, $ Nv \equal{} Tv \; \; \forall v \in \text{span} \left \{ M(t) \; : \; t \in \mathbb{R} \right \}$

However, $ \text{span} \left \{ M(t) \; : \; t \in \mathbb{R} \right \} \supset \text{span} \left \{ M(0), M(1), M(2) \right \} \equal{} \mathbb{R}^3$

$ \therefore \; Nv \equal{} Tv \; \; \forall v \in \mathbb{R}^3$, which includes the standard basis, and so $ N$ is identically $ T$

Since $ T$ is purely rotational, $ \text{det }T \equal{} 1 \implies \boxed{\text{det }N \equal{} 1}$ [/hide]


modular arithmetic
number theory unsolved
number theory

Assume that $ m \in N$ and $ P$ be product of numbers are less than $ m$ and comprime with $ m$.Prove that $ P \equiv 1$ (mod $ m$) if $ m$=4, $ p^n,2p^n$and $ P \equiv \minus{}1$ (mod $ m$) in other case.

I can't understand your problem,anyway read the following problem maybe you'll get your answer:

[b]let $ n$ be a natural number,and let $ \{a_1,a_2,\ldots ,a_{\varphi(n)}\}$ be the set of natural numbers less than $ n$ and coprime to $ n$,prove that:

$ \left(a_1a_2\ldots a_{\varphi(n)}\right)^2\equiv 1\pmod n$[/b]

[u]proof:[/u]

[b]lemma:if $ \gcd (a,n) \equal{} 1$ then $ \gcd (a^*,n) \equal{} 1$ where $ a.a^*\equiv 1\pmod n$[/b]

[hide="proof of the lemma"]
let $ \gcd(a^*,n) \equal{} d$ then we have:

$ d\mid n$ and $ d\mid a^*\Rightarrow d\mid a.a^*$

in the other hand:

$ a.a^*\equiv 1\pmod n\Rightarrow n\mid a.a^* \minus{} 1\Rightarrow d\mid a.a^* \minus{} 1\Rightarrow d\mid 1\Rightarrow d \equal{} 1$

QED
-------------------------------[/hide]
now according to this lemma it's easy to see that:

$ \left(a_1a_2\ldots a_{\varphi(n)}\right)^2\equiv 1\times 1\times 1\times\ldots\times 1\equiv 1\pmod n$


Let a;b;c;d are positive integer numbers satisfying $ 1\leq a<b<c<d\leq 50$.Find Min :
       $ S\equal{}\frac{a}{b}\plus{}\frac{c}{d}$

hello, i have found $ S\geq\frac{53}{175}$ and the equal sign holds for $ a\equal{}1,b\equal{}7,c\equal{}8,d\equal{}50$.
Sonnhard.

Yes,Sonnhard
My proof:
$ S\geq\frac{1}{b}\plus{}\frac{c}{50}$
Because $ b;c\in Z;c>b\Rightarrow c\geq b\plus{}1\Rightarrow S\geq\frac{1}{b}\plus{}\frac{b\plus{}1}{50}\geq\frac{53}{175}$


function
calculus
derivative
geometry
geometric transformation
integration
symmetry

Let $ f : R \minus{}> R$ be differentiable $ L_1$ function and let $ \tilde{f}$ be its Fourier transform. What is Fourier transfer of the function $ f(x \plus{} c)$, where
$ c \in R$ is a constant? What is Fourier transfer of the derivative of $ f$?

I prefer $ \widehat{f}$ as the notation for Fourier transform to $ \tilde{f}.$

Translation on one side (that is, $ f(x)\mapsto f(x\plus{}c)$) becomes multiplication by a complex exponential on the other side. To show that, just write out the integral for the Fourier transform and make the substitution $ u\equal{}x\plus{}c.$

Differentiation on one side (with additional hypotheses sufficient to make that make sense) becomes multiplication by a monomial on the other side. To show this, integrate by parts.

The exact notation of these results depends on your exact definition of the Fourier transform - where you put the $ 2\pi,$ where you put the minus sign.

This is quite similar to what one sees with Laplace transforms, as I worked out a bit [url=http://www.mathlinks.ro/viewtopic.php?t=284532]here[/url]. (or at least I think it is)

Laplace transforms and Fourier transforms belong to the same general family; of course their properties are quite similar. There's greater symmetry between the two sides with the Fourier transform, and the statements are a little cleaner there.

The four most familiar domains for functions on which we can define a Fourier transform are the four "Euclidean groups": $ \mathbb{R},\mathbb{T},$ (that is, periodic functions on the line), $ \mathbb{Z},$ and $ \mathbb{Z}/N\mathbb{Z}$ (that is, periodic sequences).

The Fourier transform of a function on one of these domains gives a function on the "dual" domain, where the duality works this way:
\begin{align*}\mathbb{R}&\leftrightarrow\mathbb{R}\\ \mathbb{T}&\leftrightarrow\mathbb{Z}\\ \mathbb{Z}_N&\leftrightarrow\mathbb{Z}_N\end{align*}
On all four domains, if you translate a function, you multiply its transform by an exponential, and if you multiply a function by an exponential, you translate its tranform.

On all four domains, the transform of the convolution of two functions is the pointwise product of their transforms, and the transform of the pointwise product of two functions is the convolution of their transforms. (In some cases, you need to multiply or divide by some scaling constant to make this true.)

The equivalence of convolution on one side with pointwise product on the other also works for Laplace transforms, if you define things properly.

Differentiation only makes sense on $ \mathbb{R}$ and $ \mathbb{T},$ and multiplication by a monomial (the variable) only makes sense on $ \mathbb{R}$ and $ \mathbb{Z}.$ Where these things do make sense, the transform of the derivative of a function is the same as a monomial times the transform, and [i]vice versa[/i].

Dilation only makes sense on $ \mathbb{R}$ (although you can look for analogues in other cases - especially $ \mathbb{Z}_N,$ with some number theoretic requirements observed). The transform of a dilate of a function is the dilate in the opposite sense of the transform, multiplied by a scaling factor that depends on the amount of the dilation.

I am having some troubles..

When i make the substitution $ u\equal{}x\plus{}c$, how should i handle the term $ f(u\minus{}c)$ in my fourier integral?..I don't know what goes wrong..
Also, for the differentiation..After performing integration by parts, the first part, where we should substitute the limits of integration, is something like $ \left[ f(x)e^{\minus{}2\pi ikx}\right]_{\minus{}\infty}^{\infty}$, and thus the second term tends to $ \infty$ after the substitution. Something is definitely wrong..

I know that all these questions sound stupid, but excuse me..I am trying to learn alone..
Thanks for your help..

Could you show your work regarding the first question? Kind of hard to answer at the moment.

I do not see the problem which arises from substituting the second term in that expression; note that $ i$ in the exponent! We assume $ f$ is $ \mathcal{L}^1$ and thus "$ \lim\limits_{x\to\pm \infty} f(x) \equal{} 0$", whereas that power in $ e$ is bounded between $ \minus{}1$ and $ 1$. If you have trouble with the second term, how come you don't get the same trouble with the first?

Oh, all right, I'll show georgech.

Assume $ f$ is a "nice enough" function on $ \mathbb{R}.$

If we're talking about translation, there's no integration by parts involved.

Let $ g(x) \equal{} f(x \plus{} c).$

Then $ \widehat{g}(k) \equal{} \int_{ \minus{} \infty}^{\infty}g(x)e^{ \minus{} 2\pi ikx}\,dx \equal{} \int_{ \minus{} \infty}^{\infty}f(x \plus{} c)e^{ \minus{} 2\pi ikx}\,dx$

Now we make the substitution $ u \equal{} x \plus{} c.$ We have $ x \equal{} u \minus{} c,$ which means that $ e^{ \minus{} 2\pi ikx} \equal{} e^{ \minus{} 2\pi ik(u \minus{} c)} \equal{} e^{2\pi i kc}e^{ \minus{} 2\pi iku}$

So $ \widehat{g}(k) \equal{} \int_{ \minus{} \infty}^{\infty}f(u)e^{2\pi i kc}e^{ \minus{} 2\pi iku}\,du$

$ \equal{} e^{2\pi i kc}\int_{ \minus{} \infty}^{\infty}f(u)e^{ \minus{} 2\pi iku}\,du \equal{} e^{2\pi i kc}\widehat{f}(k).$

The integration by parts arises when you try to compute the Fourier transform of $ f'.$ We're going to make some kind of "nice enough" assumptions about $ f$ that will cause the boundary terms of the integration by parts to evaluate to zero at both ends - which is what Tyl said.

[b]Exercise[/b]: Compute the Fourier transform of $ g(x)\equal{}f'(x)$ (assuming $ f$ is differentiable obviously).  This is useful in solving PDE's by means of a Fourier transform (at least it was for the heat and wave equation when I tried it).

For $ g(x)=f'(x)$, we have that ${ \hat{g}(k)=\int_{-\infty}^{\infty} f'(x)e^{-2\pi ikx} \textrm{d} x}=\left[f(x)e^{-2\pi ikx\right]_{-\infty}^{\infty} + 2\pi ik\int_{-\infty}^{\infty} f(x)e^{-2\pi ikx}\textrm{d}x}$.
My question,is about the term $ \left[f(x)e^{-2\pi ikx\right]_{-\infty}^{\infty}}$ in the last expression. When $ x$ tends to $ \infty$, this term goes to zero. But, what happens when $ x$ tends to $ -\infty$?
That's why i cannot continue.
Thanks

[quote="Tyl"]I do not see the problem which arises from substituting the second term in that expression; [b]note that $ i$ in the exponent![/b] We assume $ f$ is $ \mathcal{L}^1$ and thus "$ \lim\limits_{x\to\pm \infty} f(x) \equal{} 0$", whereas that power in $ e$ is bounded between $ \minus{} 1$ and $ 1$. If you have trouble with the second term, how come you don't get the same trouble with the first?[/quote]

georgech: Please tell me that you know that $ |e^{\minus{}2\pi ikx}|\equal{}1$ for all real $ k$ and $ x$ - that this exponential term neither grows nor shrinks in size, but just goes around in circles. In both directions, as $ x$ goes to $ \minus{}\infty$ as well as $ \plus{}\infty.$ (And that our "nice enough function" assumptions about $ f$ get us that $ f$ decays away to zero at infinity in both directions.) And please tell me that you did read Tyl's post.

Let $ F(x)=f'(x)$.  Then $ \hat{F}(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f'(x)e^{-i\omega x}\,dx$.  Let $ u=e^{-i\omega x}$, $ du=-i\omega e^{-i\omega x}\,dx$, $ dv=f'(x)\,dx$, $ v=f(x)$.  Integrating by parts gives us ${ \hat{f}(\omega}=\frac{1}{2\pi}\left(e^{-i\omega x}f(x)\bigg|_{-\infty}^{\infty}+i\omega\int_{-\infty}^{\infty}f(x)e^{-i\omega x}\,dx\right)$.  From Kent's post, the first part vanishes, and the second part is just $ \hat{f}(\omega)$ with an extra constant in front.  So ${ \hat{F}(\omega)=i\omega\hat{f}(\omega})$.  (I just used a different normalizing constant).


Hey, people! Aren't you intereseted in this problem because I called it "very easy"? Well, it has a 1-line solution, but I think it's quite nice. Maybe you should try it.

I'm very interrested in that "one line solution". :?

dzeta are u sure it's correct? it gives me $\geq 2n$  :?  :huh:

Let $P=a_1a_2 ... a_n => \sum \left ( \frac{P}{a_k} \right )^2 = 1$ and applying AM-GM we get that $P \leq 1$. Now we have: $a_1a_2 ... a_{k-1}a_{k+1} ... a_n = \frac{P}{a_k} \leq \frac {1}{a_k}$ so $a_k \geq \frac{1}{a_1a_2 ... a_{k-1}a_{k+1} ... a_n} = \frac {a_k}{P}$ for any $k$. From these results:
$a_1 + a_2 +...+a_n +a_2a_3...a_n +...+a_1a_2...a_{n-1} \geq \sum \frac {a_k}{P} + \sum \frac{P}{a_k} (*)$ and applying AM-GM we get that $(*) \geq 2n$.

hmmmm.......but this works iff $a_k >0$ for any $k$.

Am i right !?!?!? or .... am i wrong !?!?  :huh:  :?

If $P \leq 1$, then $a_{k} \leq \frac{a_{k}}{P}$.  :roll:  :rotfl:
My solution is:
By AM-GM inequality
\[ lhs=\sum_{i=1}^{n} {\left(a_{i}+ \prod_{j \neq i}{a_{j}}\right)} \geq n \sqrt[n]{\prod_{i=1}^{n} {\left(a_{i}+ \prod_{j \neq i}{a_{j}}\right)}} \eqno(1)\]
Note that
\[ \prod_{i=1}^{n} {\left(a_{i}+ \prod_{j \neq i}{a_{j}}\right)} \geq \sum_{i=1}^{n}{\left(\prod_{j \neq i} {a_{j}}\right)^{2}} = 1 \eqno(2)\]
because if you multiply all the terms of the sums in the big product you'll get a big sum which contains all the terms looking like $\left(\prod_{j \neq i} {a_{j}}\right)^{2}$ (and all the other terms are positive). The conclusion follows obviously by (1) and (2).


Could someone explain more about the trading pins? Also, if someone has the 2005 nats sprint and target, could they send it to me. I'm curious about where I stand.

OK, trading pins as explained to me by bionininquiest (sp).

A state provides each competitor a bunch of lapel pins for their state. For example, the Texas pin was a black image of Texas with the words "101% Texan" on it.

[img]http://www.lapel-pin-manufacturers.com/rimages/88/lapel-pin-emblem-badge-B.jpg[/img]
Example of lapel pins. (Just saying, these pins are random. Nothing to do with mathcounts)

Anyway, at some point in the competition dates, you are given time to trade your pins with other people.
Some pins are cool, some aren't. You probably want to have as many of the cool states' pins as possible.
Trading = going up to people from other states, maybe talking a bit, and offering one of your states' pins or a different state's pin that you have for one of his state's pins, or two of his state's pins, or two of yours for one of his, etc.

pin trading is basically a non stop activity.  Kids start trading them in the airport upon arrival, at the registration desk, during mealsand other events, in elevators and waiting for elevators.  Basically the entire competition is a big pin trading event :)

Reminds me of Odyssey of the Mind which I did in 4th grade. At World Finals, we each got 100 CT pins (2 types, one was a seal balancing a ball, one was a bear juggling) and we traded them nonstop. I traded all mine away before the trip was up, everyone else still had 10-15. Nothing with laws of economics like 8 CT pins to a VA pin or whatever, only at MathCounts I guess.

Well then, I better start now. Biffanddoc, if you give me one deleware pin at nats, then I'll give you the 04 targets.  :lol: Just kidding. I'll scan them as soon as I get time. Probably tonight or tommorow morning.

[quote="solafidefarms"]Trading pins: Don't give your pins too fast. Always extract at least 2 pins from other people, except when they are hard to get. Your pins are worth at least 10 connecticut pins. :D You can trade other states' pins too, not just your own, if you have too many.

The absolute best way to get pins is to collect everyone's email address aggressively. Make it your only goal to collect everyone's emails, and oh yeah, when you see somebody new, trade a pin with them. :)[/quote]
I heard about how you made connecticut worthless last year at nats. This was my reaction  :rotfl:  :rotfl:  :rotfl:  :rotfl: . Are you going to be here this year

How about this:
Post here if you are willing to trade pins 1-for-1 with other AoPSers.  We can "finalize" it by PMs.  That will give us an "excuse" to meet each other at Nats.  And we'll get a head start on pin collecting!  Just an idea, tell me what you think.

This year, CT pins are 1:1 non negotiable considering how little we're getting  :mad:

[quote="Klebian"]Solafide, how DID you make the CT pins worthless?[/quote] It's in one of the old threads on pin trading.

I had taken about 100 of the McDonough, my home city, pins, which were unofficial. I gave about 50 away pretty fast. Then the remaining 50 got really valuable really fast. One of the CT guys came and wanted a pin. I didn't want it. He offered me 5. I thought why not debase the CT pins? So I bid 20 CT for a GAs. Nope, he got me down to 12 or 15. So then I put all of them on a shoestring I had about my neck to put my pins on. The other CTers begged Tom not to do the 15 trade, and then all of them begged me not to put all of them on the shoestring at once. When people saw me walking around with 15 CTs and heard how little they had cost, they didn't want to part with theirs for CTs. And then there were the people who arrived late and traded 2 of their pins for a CT the first time. Until someone witnessed the trade and piped in about how worthless CTs were. So I was hearing 7 and 8 pins for a common GA pin before Nats was over. :) :) :)

(Nope, I'm 9th grade now. No more MC for me.)

Who here wants to get together and come up with a plan to COMPLETELY devalue the GA pin this year?

Younger brother of a pin-trading master (2001-2002) = thx. I managed my bro's pins in 2nd grade and I can do it again :).

[quote="solafidefarms"][quote="pgpatel"][quote="Xantos C. Guin"]Don't de-value the official GA pin.
[hide]Just the unofficial one that made it on ESPN's broadcast.  :D/ ;)  [/hide]

I traded 1:1 with the CT team. Pretty dumb/fair however you want to look at it.

The Nevada and the Defence Department pins were worth a lot last year.

BTW: Did any AoPSers from GA make nationals?[/quote]
Will the unofficial still be there this year?[/quote]

Don't think so. Why don't you ask your town about pins for Nats? Then you can repeat my feats :).[/quote] I will try o great mod.  :lol: . Bye the way, does anyone know about Fl's pins last year. Were they good?

How were ca's?

Nats sounds like SOOOOO much fun *more motivation to study for next year*

I remember that last year, I think it was Rhode Island that announced on the first day of nationals that they only bought like twenty pins total so thus were really rare. Well, everyone was excited and wanted to get one of those rare pins, and many people were trading up to three or four of their own states pins for one of theirs. Well, it turned out that Rhode Island had plenty of pins after all, but when everyone else realized that, it was too late, so the Rhode Island students had tons of duplicates and thus could trade off a lot.

I confess that I myself traded four of my Texas pins for one of theirs.  :blush:

[quote="aznness"]I'm the CA captain, i would know. Last year, we had the surfer dude pins. I think I still have around 20 lying around somewhere in the house. I used a tie a friend gave me as a display thing, and it was rather difficult to walk around with a 10 lb tie on your neck.

CA will not be trading pins until after the written and countdown rounds are done. I will be nice, and trade pins 1:1 with anybody who can come up to me and tell me one of my SN's here on AOPS. (i've used 3). If you can name 2, you get a free pin. Name all three, and you get cali pins for your team.

Last year's coach, Mrs. Shah, did a great job regulating team taboos. Miller didn't win team this year, but i feel obligated to carry on the tradition.

YAY FOR MONOPOLY!!

good luck trading pins guys.

FYI: (if you don't know me at all, you get a 1.5 to 1 ration for trading pins, at the best)

HAPPY PINNING

[hide="free pin offer see here"] I will also be giving free pins to the first 4 people i see using their ties for pincushions [/hide][/quote]

So what do you look like? :P

They gave out a yearbook last year of every student at nats.  Just look at the page for CA and find his name.  It has everyone's picture in it.  I got the yearbook last year.

AoPSers have a small advantage in pin collecting. Mostly knowing that pins exist.

Last year RI started trading with us before the 2 non-AoPSers on our team knew what pins were.

We got their pins before they raised the price.   :)

56 pins is impossible.

2 states didn't bring any: NV and DD

DD did bring some.  You may not have found or seen any.  But NV didn't though.

I asked and they said they didn't have any.
Grrrr....
That kid from the Defence Department must have been lying.
 :mad:

Or they might have only brought a few or they ran out fast.

Oh well, 55/56 is not that bad for a penguin like me.

How do you get a set number of pins though, do you have an unlimited stack...because logic tells me you must trade 55 pins to get 55 pins, 54 people don't just go and give them to you

Yeah, we usually have about 57 of our own state's pins before we get to nats.  Our coach gave them to us.

We had about 300 pins for the team. This is a good number.
Some teams bring multiple pins, so it is a good idea to bring more than needed.

Also, bring some way to keep track of what states you have.

[quote="aznness"]Look for a big tall asian guy. Shouldn't be too hard to miss.[/quote] True. Asians usually aren't tall (no offense, I'm asian too).

[quote="xxm0rg07hxx"]Arkansas is only trading fewer pins this year.  I am not going to trade because I think it's a waste of time.  And someone else on our team will probably not trade because he got 56/57 pins last year.[/quote]
You can give me your pins. :lol:

I won't even bring them to nats.  How's that.

That would be REALLY mean.  :mad: But they are your pins so whatever. I'll just aim for getting 56 states :lol:

You forgot my teammates.  They will have pins.  So try for 57.


Show that there exist two consecutive squares such that there are at least $1000$ primes between them.

It is easily be proved by the famous thoerem:
$\pi (x)\sim\frac{x}{\ln x+c}$
And I also have an elemetry proof of it(I'll post it tommorrow).
I wonder that if there's an elemetry proof of the above theorem.
And I also don't know if we can prove with elementry method the following problem:
$\frac{\pi(x)}{x^r}-->\infty,r<1$

[quote="Hawk Tiger"]And I also have an elemetry proof of it(I'll post it tommorrow).[/quote]I think, your days are very long! :P

It suffices to use Chebyshev's theorem (namely that $ \pi(x) \cdot \log(x) \geq c \cdot x$) instead of the much stronger prime number theorem.

That's still not elementary... I am curious for Hawk Tiger's elementary proof.

Sorry but you really should define your "elementary". You have now neglected a big amount of solutions for not being "elementary"...

Calm down... Maybe it is just because I am too sleepy to see the easy proof of the "Chebyshev's theorem" you seem to have.
If you have a proof of it without prerequisites, then it is of course ok. ;) (and then it is appreciated if you post the proof too, for the people not knowing the proof, like me).

As long as you use theorems I have not ever seen a proof of (like this one, of course everyone knows the fact) I will find your proof nonelementary. Yes, elementaryness is a little subjective. I'm open to more formal definitions.

Sorry for being a bit harsh...
See http://planetmath.org/?op=getobj&from=objects&id=8545 for an elementary proof. I totally agree that this may be too long or too much work et cetera.
My comment was not specially on this problem, but was meant more general. Please don't call any method that aren't part of some special courses nonelementary.
I see a very important part of any number theory book (expecially PEN) to learn new theorems and methods, not only applying old ones to solve single problems. Thus I consider it very important to teach new techniques. And after you got used to something, you are more likely to call it elementary.

Oh but I would never call a 'method' nonelementary. I'm all for learning new techniques, methods and tricks. Please point me out if I ever spoke against that, because then I probably misunderstood what it was about. :)

What I [b]am[/b] against is that sometimes, people use (often relatively deep) theorems - without proof - that nearly trivialize the problem (like you both did here). Of course this follows from Chebyshev and from the Prime Number Theorem. The reason this problem is in the book is because there presumably exists some brilliant inductive construction of the pairs without knowledge on prime distribution, not to be a standard application of some prime distribution theorem.

In the same way one can claim to have verified in Maple that 1280000401 is composite. Of course that solves the problem. But this doesn't contribute any ideas either.

I agree there is not always clear wether one is expected to find a competition-style solution (no theorems without proof and by hand) or just any solution (research style), this seems to depend from problem to problem. I will try to take care of this in the next release (which is greatly slowed down by moving to MathLinks).

I try to give as elementary proofs $ E$ often as possible, but there are cases where it is better to give an nonelementary proof $ N$:
- the result obviously is of nonelementary style.
- $ E$ is not essentially different from an "elementarisation" of $ N$; then I prefer writting down $ N$ and explaing/adding all the missing steps: otherwise the solution would pop out of nothing, and this way the reader can learn new methods and theorems.
- $ E$ relies on difficult but elementary and known theorems (like here); then one (at least I) doesn't always want to add a proof (even if it is elementary); one really good example is Zigmondy's theorem...

By the way, I think this problem can be done more directly, but for the "learning effect", e.g. how  one could do it with a bit more theorems known, the non- or half-elementary solutions are ok.

[quote=fedja]You can easily derive it from the fact that the series of inverse primes diverges and the series of inverse squares converges. Both claims are "reasonably elementary" :).[/quote]

we have two formulas for that. i am going to introduce the full solution of fedja
we know that
\[\sum \frac{1}{i^2}=\frac{\pi^2}{6}\]and also,\[ \sum \frac{1}{P_i}=\infty \]\text{where $P_i$ is the $i_{th}$ prime number}

now our idea is to use contradiction,
consider that the most prime numbers that could occur between two consecutive squares is $\leq 999$
then we list all consecutive squares:
\[\begin{array}{cccc} (1^2,2^2)\\(2^2,3^2)\\ \cdots \\ \cdots \end{array}\]

so hear we can give an upper bound for the sum of inverse of primes in interval $(t^2,(t+1)^2)$ and that is
\[\sum_{t^2 \leq P_i<(t+1)^2} \frac{1}{P_i} \leq \frac{999}{t^2}  \thinspace \thinspace \thinspace \textbf{(1)}\]
so for all primes we have :
\[ \sum_{i=1}^{\infty} \frac{1}{P_i} \leq 999\sum_{i=1}^{\infty} \frac{1}{t_i^2}  \thinspace \thinspace \thinspace  \textbf{(2)}\]
but from the formulas that i have mentioned earlier about the sum of inverses it is obviously wrong because the left hand side of $(2)$ is equal to $\infty$ but the right hand side is less than $2*999$.
hence,contradiction.. :blush:


I want to reccomend you a cool math magazine -OCTOGON edited by M Bencze from Brasov,Romania.It's in English.What's cool there are the open problems proposed and solved in  the magazine wich has two annual numbers, each of them having over 400 pages!. :)  :)

I haven't been able to find it over the Net.
I'm not sure if overseas subscription would be viable unless you can save up some lunch money for it. How do you subscribe? How much does it cost? Do you have a web site for the subscription? For overseas, I'm sure there would be extra mailing fees.

Hi,

can anybody provide more detailed information regarding these publications. When was this journal established ? And can you please provide a sample content list or a contest description.

Here what I found,

OCTOGON

Editor in chief

Mihaly BENCZE
Str.Harmanului 6,
RO-505600 Sacele
Jud. Brasov, Romania

Thanks for the information, Moubi. All I could find was his e-mail address at http://rgmia.vu.edu.au/members/Bencze.htm. He is of the members (http://rgmia.vu.edu.au/alpha_list_new.htm) of the journal "Journal of Inequalities in Pure and Applied Mathematics".


inequalities
calculus
derivative
inequalities unsolved

Find the maximal value of the expression $ x_1x_2+x_2x_3+...+x_nx_1$ where $x_1,...,x_n$ are real numbers which add up to 0 ans such that the sum of their squares is 1.

I dont have exact solution, only some thoughts.
For n = 2,3 max(2),max(3) = -1/2  it is clear I think,
for n = 4 max(4) = 0.

 (x1+x2+x3+x4)2 = (x1-x2+x3-x4)2+4(x1x2+x2x3+x3x4+x4x1)

For n = 5 max(5) = 1/6.
when x1=1/√6, x2 =1/√6, x3 = 0, x4 = - √2/√3, x5 = 0.
For n ≥ 6 max(n) = 1/2
when x1=1/2, x2 =1/2, x3 = 0, x4 = -1/2, x5 = -1/2, xj = 0 (for j ≥ 6).

I think we can the following well-known fact that for yi ≥ 0
(y1++yn)2 ≥ 4(y1y2+y2y3++yny1) for n ≥ 4.
And that our maximum is attained when we have positive numbers, than zero,
after that negative numbers, than zero.
I dont know if it is correct, as I said only thoughts.

I guess, with some reasons, that $x_i=kcos(\frac{2\pi i}{n}+\theta)$, where $\theta$ can be any real and $k$ is the real such that $x_1^2+...+x_n^2=1$

The idea comes from an equation which I guess is right.
$
\frac{x_n+x_2}{x_1}=\frac{x_1+x_3}{x_2}=...=\frac{x_{n-1}+x_1}{x_n}
$
I will give some more about my ideas later as soon as I can. Gotta go.

[quote="Vasc"]This is the Fan's Inequality:

If $x_1+...+x_n=0$, then $\frac{x_1x_2+...+x_nx_1}{x_1^2+...+x_n^2} \leq cos{\frac{2\pi}{n}}$[/quote]
Great. Could you give the proof please?

My idea came when considering $n=2k+2, x_n=x_k+1=0, x_1=-x_{n-1},x_2=-x_{n-2},...,$ then the equation $x_1+...+x_n=0$ is satisfied and by some ideas about differentiation I saw the best case must satisfied the equation:
$
\frac{x_n+x_2}{x_1}=\frac{x_1+x_3}{x_2}=...=\frac{x_{n-1}+x_1}{x_n}
$
then with some trying I got the result.

Darkseer, I found this inequality in the book
"Classical and New Inequalities in Analysis" by
Mitrinovic, Pecaric and Fink (2003, page 439).
I have not a proof for it. I suggest to Manlio to
create a new topic with Fun's inequality.


A dragster car can go to 0 from 100 km/h in just 1.5 seconds. When the car stops, it have a decceleration of $ \minus{}86km/s^2$. That car runs in a quarter-mile racing. It runs at the max possible speed, and stop exactly at the final line. Calculate the time elapsed between the starting and the finish.

Well first, you're in two different unit systems so you may want to clear that up.  Next, I would convert either into km/s/s, or km/h/h so you get some consistency.  Then you just use the equations of motion.

[quote="pacman2812"]A dragster car can go to 0 from 100 km/h in just 1.5 seconds. When the car stops, it have a decceleration of $ \minus{} 86km/s^2$. That car runs in a quarter-mile racing. It runs at the max possible speed, and stop exactly at the final line. Calculate the time elapsed between the starting and the finish.[/quote]

Check your data. The deceleration is nearly 9,000 g's. Always ask yourself, "Does this make sense?" If the answer is no then verify the problem statement.

I'm not going to try to answer the original poster - the problem seems to be ill-stated, and I'm not sure just how to fix the statement.

Instead, I want to talk about the actual sport of drag racing. The race track - or at least the part of the race track that counts - is indeed 1/4 of a mile long. However, the cars do not stop at the end of that quarter mile. Instead, they are trying to reach maximum speed at the end of the quarter mile; only after that do they slow down. The pavement continues for some distance, probably more than a half mile, in order to give the racers a chance to slow down. They deploy parachutes to help, so they don't have to rely entirely on brakes.

It's been a few years since I remember reading results on the sports pages, but I recall that the results of each race are expressed as two numbers: elapsed time and top speed. The winner is the one with the smallest elapsed time. In the "top fuel" category, the fastest cars tended to have elapsed times of around 5 seconds and top speeds of around 300 mph - or at least, that's what I remember.

Going from a standing start to 300 mph in 5 seconds is an average acceleration of 60 mph/sec, which is 88 feet per second squared, or about 2.75g. But that's just the average acceleration. A car undergoing uniform acceleration of 2.75g (88 ft/s/s) for 5 seconds would travel a distance of 1100 feet - but a quarter mile is 1320 feet. I experimented with some acceleration profiles that I would believe that would result in traveling 1320 feet in 5 seconds, and decided that that the acceleration is likely to be around 5g at around 1 to 2 seconds into the run, tapering off to under 2g later in the run as the speed-related aerodynamic drag increases.

That creates a problem: how can you possibly use the force of wheels on pavement to impart a horizontal force of 5 times the weight of the car? Because you need to do that in order to reach an acceleration of 5g. They use "sticky" rubber tires, but the coefficient of friction for that rubber is not going to be 5. That means that in order to produce that horizontal force, there must be a downward force on the drive wheels that is several times the weight of the car. That down force comes from a wing - a wing engineered to produce down force rather than lift, and mounted above the rear of the car. I would think that the car cannot produce maximum acceleration right at the standing start, but must get moving fast enough for the wing to take effect in order to accelerate most rapidly.

There's a second problem: the incredible torque applied to the drive axle in order to create that kind of force from the wheels would tend to lift the front end of the car off the ground. In part to combat this, top fuel dragsters are rather long vehicles, with the heavy engine mounted well forward, ahead of the driver's cockpit. There may even be a second wing up front to keep the front end down, although I'm not sure of that. Even so, accidents do sometimes happen in which the front end lifts off the ground, typically resulting in the whole vehicle becoming uncontrollably airborne.

wait, wait, I meant $ km/h^2$

I can't edit the original post, so I will write it right.

A dragster car can go to 0 from 100 km/h in just 1.5 seconds. When the car stops, it have a decceleration of $ \minus{}86km/h^2$. That car runs in a quarter-mile racing. It runs at the max possible speed, and stop exactly at the final line. Calculate the time elapsed between the starting and the finish.

That cleared up part of the problem.  You're still in two different unit systems so you should fix that just to make things simpler.  Also, what do you mean "maximum possible speed"?  There are still some problems with the way the question is being asked.

[quote="pacman2812"]wait, wait, I meant $ km/h^2$

I can't edit the original post, so I will write it right.

A dragster car can go to 0 from 100 km/h in just 1.5 seconds. When the car stops, it have a decceleration of $ - 86km/h^2. \ldots$ [/quote]

Again, are you certain?

\[{ 86\quad \text{km}/\text{h}^2 = \frac{86 \times 10^3 \,\text{m}}{(3.6 \times 10^3\,\text{s})^2} \approx 6.64 \times 10^{-3}\quad\text{m}/\text{s}^2}\]

Provided that you ever get this figured out what you want is this:
\[ v_{\text{max}}^2 = 2a_1x_1 \\ \\ 0 = v_{\text{max}}^2 + 2a_2x_2\\ \\x_2 =\frac{1}{4}\,\text{mile} - x_1\]

Then

\[ t_{\text{total}} = t_1 + t_2 \\ \\ x_1 = \frac12\left|a_1\right|t_1^2\\ \\ x_2 =\frac12\left|a_2\right|t_2^2\]


calculus
integration
trigonometry
real analysis
real analysis unsolved

Integrate 1/((1+x^2)^(n+1)) from 0 to infinity.

[hide]Considering the change of variables $ x \equal{} \tan t$, $ dx \equal{} \sec^2t$, we have

$ I \equal{} \int_0^{\plus{} \infty} \frac{1}{(1\plus{}x^2)^{n\plus{}1}}\,dx \equal{} \int_0^{\pi/2}(\cos t)^{2n}\,dt \equal{} \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n\minus{}1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)} \cdot \frac{\pi}{2}$,

where the last equality can be shown by several ways, one of them by using Beta and Gama functions.[/hide]

Thanks. I did it by integrating by parts and also got ((2n-1)!!/(2n)!!)pi/2.


limit
inequalities
real analysis
function
vector
Functional Analysis
real analysis unsolved

Let X be a normed space and $ x,x_{1},x_{2},...\in X$. Suppose that
$ f(x) \equal{}\lim_{n\rightarrow\infty}f(x_{n})$
for all $ f\in X^{*}$

Prove that $ \parallel x\parallel\leq\liminf_{n\rightarrow\infty}\parallel x_{n}\parallel$

This is what I tried:
$ |f(x)|\\ \equal{} |\lim_{n\rightarrow\infty}f(x_{n})|\\ \equal{}\lim_{n\rightarrow\infty}|f(x_{n})|\\ \leq\lim_{n\rightarrow\infty}\parallel f\parallel\parallel x_{n}\parallel\\ \equal{}\parallel f\parallel\lim_{n\rightarrow\infty}{\parallel x_{n}\parallel }\\ \equal{}\parallel f\parallel\liminf_{n\rightarrow\infty}{\parallel x_{n}\parallel }$

On the other hand
$ |f(x)|\leq\parallel f\parallel\parallel x\parallel$

So the inequality is not quite what I want.

You should think that $ x\in X$ then $ x\in\Big({X^{*}\Big)}^{*}\equal{} X^{**}$ and note that $ X^{*}$ is a complete normed space. then
$ |x(f)| \equal{} |f(x)| \equal{} |\lim_{n}f(x_{n})| \equal{}\lim_{n}|f(x_{n})|$.
So we have
$ \parallel x\parallel \equal{}\sup_{f\in X^{*}}\frac{|x(f)|}{\parallel f\parallel }\equal{}\sup_{f\in X^{*}}\lim_{n}\frac{|f(x_{n})|}{\parallel f\parallel }$
$ \leq\liminf_{n}\sup_{f\in X^{*}}\frac{|f(x_{n})|}{\parallel f\parallel }\equal{}\liminf_{n}\parallel x_{n}\parallel$

Perhaps there was a typo? was it supposed to be

$ \sup_{f \in X^*} \lim_n \frac {|f(x_n)|}{\|f\|} \leq \liminf_n \frac {|f(x_n)|}{\|f\|}$ ?

But then still it puzzles me because the definition of $ \liminf$ is the supremum of infimum and how can that be greater than or equal to supremum of limit?

am I missing something here? :huh:

Hang on,
I think it should be something like

$ \sup_{f \in X^*} \lim_n \frac {|f(x_n)|}{\|f\|} \equal{} ??? \equal{} \liminf_n \frac {|f(x_n)|}{\|f\|} \leq \liminf_n \|x_n\|$

I doubt if the following is true:

$ \sup_{f \in X^*} \lim_n \frac {|f(x_n)|}{\|f\|} \leq \liminf_n \sup_{f \in X^*} \frac {|f(x_n)|}{\|f\|}$

[quote="sit.think.solve"]thanks for the reply, there's one thing i couldn't understand:

$ \sup_{f \in X^*} \lim_n \frac {|f(x_n)|}{\|f\|} \leq \liminf_n \sup_{f \in X^*} \frac {|f(x_n)|}{\|f\|}$

Could anyone please explain this step?

sup of lim is less than or equal to liminf of sup? why?[/quote]

Here're details:

Take any $ \epsilon>0$ then there's $ f$: The LH $ \leq  \lim_n \frac {|f(x_n)|}{\|f\|}\plus{}\epsilon\leq 2\epsilon\plus{} \frac {|f(x_n)|}{\|f\|}$ for all $ n\geq n_0$.
The RH $ \equal{}\sup_n\inf_{m\geq n}\sup_{f}\frac{|f(x_m)|}{\|f\|} \geq \inf_{m\geq n}\sup_{f}\frac{|f(x_m)|}{\|f\|}\geq \sup_{f}\frac{|f(x_m)|}{\|f\|}\minus{}\epsilon$ for all $ n$ and for some $ m$ which is $ m\geq n$ and all $ f$.
So take $ n\geq n_0$ then $ m\geq n_0$ so $ RH\geq LH\minus{}3\epsilon$ for every small $ \epsilon>0$.

Hoep that's more clearer :D

wow... i would have never figured this out myself...
is there any other way to see this inequality than using these epsilons?
Not that your argument is wrong or anything...
Thanks for your help  :lol:

Just one more thing,
the last equality shouldn't it be inequality?
because, we have liminf of sup of mod f(x_n) over f, I can see that mod f(x_n) over f is less than or equal to norm of x_n so shouldn't sup of that be less than or equal to norm of x_n?
if not, why is it equality?

Also, one related question to this:
as far as i know the opposite inequality to this namely the norm of x is greater than or equal to the liminf of x_n is false.
but i'm having trouble trying to construct the counterexample.

I'm not sure what you are asking for (because of my bad English  :blush: ) but I suppose that you're familiar to some results which are  corollaries of Hahn-Banach theorem, such as:
1) $ \|x\| \equal{} \sup_{f\in X^*,f\neq0}\frac {|f(x)|}{\|f\|}$
2) the injective $ i: X\to X^{**}$ is an isometry
...

For the left part, it's enough to show for given $ x\neq0$ there exists $ f\in X^*,f\neq0$ such that $ \|x\|\equal{}\frac{|f(x)|}{\|f\|}$. Let consider the $ 1$-dimesion subspace $ E\equal{}\{t\frac{x}{\|x\|}: \;t\in {\mathbb R}\}$ and a functional linear $ \phi\Big(t\frac{x}{\|x\|}\Big)\equal{}t$. Note that $ \|\phi\|_E\equal{}1$. By Hahn-Banach theorem $ \phi$ has an extions to $ X$ with unchanged norm, so $ \|x\|\equal{}\phi(x)\equal{}\frac{|\phi(x)|}{\|\phi\|}$.

The problem is to show that the norm $ \|\cdot\|$ is weakly lower semi-continuous.

A function $ f \colon T \to \mathbf R$ on a topological space $ T$ is lower semi-continuous iff its epigraph $ \mathrm{epi}(f) \equal{} \{(x,t) \in T \times \mathbf R \mid f(x) \le t\}$ is a closed subset of $ T \times \mathbf R$.

Our function $ f(x)\equal{}\|x\|$ is a continuous convex function, hence $ \mathrm{epi}(f)$ is a closed convex subset of $ X\times\mathbf R$. But every closed convex subset of a normed vector space (or even locally convex space) is weakly closed. Thus $ \mathrm{epi}(f)$ is weakly closed, implying that $ f$ is weakly lower semi-continuous.

[quote="volodja"]The problem is to show that the norm $ \|\cdot\|$ is weakly lower semi-continuous.

A function $ f \colon T \to \mathbf R$ on a topological space $ T$ is lower semi-continuous iff its epigraph $ \mathrm{epi}(f) \equal{} \{(x,t) \in T \times \mathbf R \mid f(x) \le t\}$ is a closed subset of $ T \times \mathbf R$.

Our function $ f(x) \equal{} \|x\|$ is a continuous convex function, hence $ \mathrm{epi}(f)$ is a closed convex subset of $ X\times\mathbf R$. But every closed convex subset of a normed vector space (or even locally convex space) is weakly closed. Thus $ \mathrm{epi}(f)$ is weakly closed, implying that $ f$ is weakly lower semi-continuous.[/quote]

Oh very simply nice solution volodja


[quote="RisingStar"]Let $ ABCDEF$ be a convex hexagon, and denote by $ P, Q, R, S, T, U$ the midpoints of the sides $ AB$, $ BC$, $ CD$, $ DE$, $ EF$, $ FA$ respectively.  Suppose that the areas of the triangles $ ABR$, $ BCS$, $ CDT$, $ DEU$, $ EFP$ and $ FAQ$ are 12, 34, 56, 12, 34 and 56 respectively.

Find the area of the hexagon.

I feel like my hands are tied here  :([/quote]

[hide]
I notice the areas can be arranged in pairs
maybe that indicates the hexagon's opposite sides are congruent?
[/hide]

[hide="hints... sort of"]
I drew out the diagram accurately and it looks like $ AR \| UD$, $ FQ \| TC$, and $ AQ\|TD$. Since the areas for opposing triangles are equal, and there are parallel sides, (and the triangles look congruent), I think opposing triangles (ex. $ \triangle ABQ \cong \triangle EDT$ or $ \triangle ETD$). I can't prove anything though...

I tried shading in each triangle's area with a different color to see if there was something clever to be done with their areas. I don't think there is... Some areas are covered three times, some twice, some once, and there is a hexagon in the middle which isn't covered at all >.<

If it helps, this is a problem from the AIME class here and the topic is symmetry.
[/hide]

Bump because this is number one of a board problem set from the AIME class and I still can't solve it...


geometry
circumcircle
Gauss
conics
geometry proposed

Let $ A$, $ B$, $ C$, $ D$ be four points on a circle, and let $ M$ be a point in the same plane. Let $ E$, $ F$, $ G$, $ H$, $ K$, $ L$ be the orthogonal projections of the point $ M$ on the lines $ AB$, $ BC$, $ CD$, $ DA$, $ AC$, $ BD$, respectively. Prove that the three midpoints of the segments $ EG$, $ FH$, $ KL$ are collinear.

I don't have the actual solution, but I think I've got the main idea:

If we denote by X and Y the midpts of EG and FH respectively and by O the circumcenter of ABCD, then the lines OM and XY are antiparallelles with respect to any of the angles formed by 2 opposite sides of the quadrilateral (including the angle formed by the 2 diagonals). If we show thif for XY then it applies to YZ and XZ as well, with Z being the midpt of KL, and from here we could derive that lines XY and XZ are parallel, so they're actually one and the same because they both pass through X.

I'm trying to prove my first statement.

Any complete quadrilaterals with Gauss lines out there? Could be of help.

  dg

I tried to look for one but didn't find any.

One thing isn't hard to show, and that is that if we draw a line d through O and move M along that fixed line, then the line XY will also have a fixed direction, meaning that all the lines XY (and YZ, and XZ, of course) which are obtained for various positions of M on d are parallel to each other.

This means that it's enough to prove everything for a certain position of M, like M being on the circle, which would make some of the points E, F, G, H, L, K collinear (because of Simson's thm).

Since we can construct line M1M2 where M \in M1M2, M1 and M2 lie on circle, it is sufficient to prove statement for points on circle (due to linearity of construction ).

For point M on circle we have on figure 4 Simson's lines and statement in this case transform to following one:
Let ABC is triangle, D \in AB, E \in AC, F is point of intersection of CD and BE, then midpoints of AF, DE and BC are collinear.

This statement can be easely proved using center of mass technique.

Oh, so you did manage to find a quadrilateral (ADFE) and its Gauss line :).

Thanks to Grobber and Myth for the proof. Notably, in the special case where M is the point of intersection of the diagonals AC and BD, we get quite a nice theorem (the midpoints of EG and FH are collinear with M). The only - more or less - trivial special case is when M is the circumcenter of our quadrilateral ABCD.

  Darij

Your solution is right. The point of this problem is to prove when M is O, in (O), and based on the second case when M is in (O) to prove for every M. :D

Clearly $E,H,K$ lie on the circle with diameter $\overline{MA}$ and $E,F,L$ lie on the circle with diameter $\overline{MB}.$ Hence $\angle HEK=\angle HAK=\angle LBF=\angle LEF$ $\Longrightarrow$ $EK,EL$ are isogonals WRT $\angle HEF.$ By similar reasoning, the pairs of lines $(FK,FL),$ $(GK,GL)$ and $(HK,HL)$ are isogonals WRT $\angle EFG,$ $\angle FGH$ and $\angle GHE,$ respectively. Therefore, $K,L$ are isogonal conjugates WRT $EFGH$ $\Longrightarrow$ there is a conic $\mathcal{C}$ with foci $K,L$ inscribed in the quadrangle $EFGH.$ Thus, by Newton's theorem its center (midpoint of LK) is on the Newton line of $EFGH,$ i.e. midpoints of $LK,EG,FH$ are collinear.


function
advanced fields
advanced fields unsolved

I love doing logic problems, but these two just blew me away =|

1. No Premise
    Therefore, ((P-->Q)-->(P-->R))-->(P-->(Q-->R))

2. P-->Q . P-->R . P-->S . T-->(U-->(~V-->~S) . Q-->T . R-->(W-->U) . V-->~W . W . Therefore, ~P

Thanks in advance :)

P.S. I've attached part of my answer to Q 2

[quote="ashwinrk_jain"]fist statement is wrong
put P = true
     Q = true
     R = false
then the statement become fasle[/quote]

sorry, not quite sure wht ur saying :S

he's saying that the first formula isn't tautological, i.e. you can find truth assignments of p, q, and r that will render the formula false. however, his truth assignment does not render the formula false, so it still may be a tautology. i haven't taken the time to prove it using truth assignments or formal logic proof. here's the disproof of his disproof.

define a truth assignment phi in the following way:

phi(p) = t
phi(q) = t
phi(r) = f
for all other propositional variables, phi => f

by the definition of the truth function (-->),
(phi(a --> b) = t) <=> phi(a) = f or phi(b) = t
and
(phi(a --> b) = f) <=> phi(a) = t and phi(b) = f

so for your formula, we have
phi(q --> r) = f, so
phi(p --> (q --> r)) = f, and
phi(p --> q) = t and phi(p --> r) = f, so
phi((p --> q) --> (p --> r)) = f, thus
phi(((p --> q) --> (p --> r)) --> (p --> (q --> r))) = t
(false implies false is true)

thus, the statement is probably a tautology.

sorry, i guess i didn't make it clear before.

I was hoping someone could derive the symbolic by using the rules: Modus Ponens, Modus Tollens, Double Negation, and Repetition.

The image that i've attached is an example of the way.

Thanks for ur input

eagle


Well, i want to know about the graph theory .. as in wht are itz baiscz and stuff ...

Sir started teaching us and unfortunately i missed the first few lecturez .. :/

please would some one be kind enought to tell me the basics and itz application z ..!!  :(

Graph theory, a lot of it first 'pioneered' by Euler, is a field of discrete mathematics that, well, deals with connections, I suppose. This, for example, is a graph:

[img]http://www.ctl.ua.edu/math103/euler/Image45.gif[/img]

Common problems are things like "how many ways can you colour each node, given three colours and the fact that no two of the same colour can be connected?" (the answer depends on the graph) and looking at algorithms to solve these problems via computer, and 'route problems' like "how can I visit all the nodes in a graph in the least amount of steps, and end up where I started?". Graph theory has applications in problems that you might not think of at first glance: the "colouring nodes" problem could be applied to timetabling exams so that no two exams that people are in are run at the same time, things like that.

Euler used graph theory to solve the [url=http://en.wikipedia.org/wiki/Seven_bridges]Seven Bridges[/url] problem which wasn't so much of a problem as it was a way of allowing us to put these problems down in a mathematical way, that is, with graphs. That Seven Bridges page actually has quite a lot of stuff on basic graph theory.

If you want to learn about graph theory (and the basics will be really easy for you to catch up on): [url=http://en.wikipedia.org/wiki/Graph_theory]the Wikipedia article[/url] is not exactly a bad place to start. Unfortunately I don't think we can guide you through [i]everything[/i] to do with graph theory ourselves, and there's lots of material on the web as it is written by people a hundred times better at explaining these things than I ever will be. A quick Google reveals [url=http://www.utm.edu/departments/math/graph/]this graph theory tutorial site[/url]. That should be enough stuff to get you started.

Please stop using "z"s in place of "s"s for your plurals.  It seems like English is not your first language; this is not correct usage.


[quote="Interval"]A triangular park is formed by the intersection of three streets, Bridge
Street, Harbor Place, and College Avenue. A walkway parallel to Harbor Place goes through the park. A time
capsule has been buried in the park in a location that is equidistant from
Bridge Street and College Avenue and 5 yards from the walkway. Where is the time capsule buried?[/quote]
[hide="hint"]
AA[/hide]


ARML geometry calculus integration trigonometry

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b) : (b+c) : (c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

Holy cow I feel so godly reviving this considering it's 12 years old (OMG) and was made when I had just turned one year's old (wow) so thanks for this honor


In a round-robin tournament with no ties,what is the smallest number of teams for which it is not always possible to find two teams such that every other team has lost to at least one of them?

[quote="Treething"]Is a round robin tournament where everybody faces everybody else? (i.e. 6 people = 15 matches)[/quote]

Yes.  (This is opposed to, say, a single-elimination tournament, in which relatively few teams face any given team, or even play very many games at all.)

Maybe I don't see some clear things, but I can't invent example. :?

So, I can show 6 teams is [i]not[/i] adequate, but like Myth, I'm having trouble coming up with an example that I can show does work, of any size.  Is it possible that none exists?

For six teams, each plays 5 games.  If any team wins 5 games, we're done (take it and any other team).  If any team wins 4 games, it does not beat some team.  However, that team must have lost to someone, or else it won all of its games, so we have 2 teams that cover all others.  If no team wins 4 or 5 games, the teams with winning records all win 3 games exactly, and there must be at least 3 of them.  Call one of those teams A, and let us say A beats B, C, D and loses to E, F.  Then no team can beat E and F.  Also, each E and F can win at most 2 other games.  Thus, WLOG, we have:
E beats A, B, F
F beats A, C, D
and together they beat all 4 other teams.  (There may have been a better way of doing this -- in fact, I'm sure there is.)

I don't think this is Intermediate at all -- looks like reasonably (or maybe not so reasonably?) complicated combinatorics to me.

Actually, it's easy to do better than my last thing -- in general, if we have n teams, there cannot be a team that wins (n - 1) (obvious, since that team and any other are sufficient), (n - 2) (outlined above, for the case n = 6 but generalizes) or (n - 3), since take the 2 teams not beaten and one of them wins their head-to-head match which gives the extra team needed.  So no team can win more than n - 4 games, which means we must have at least 7 teams.  In the case of 7, all teams would need to go 3-3.  With 8 teams, any team could win as many as 4 of their 7 games, and so on.

Joel, don't tell me you forget my lectures on probabilistic methods so soon. :)  (In my handout, this problem was called unrankable tournaments.)

There is a tournament with 21 teams that satisfies the problem.  The idea is to choose a random tournament on 21 teams.  Then prove that the probability of it failing to be an example is less than 1.  I'll let you guys work out the details as a challenge problem.

I don't know whether there is a smaller example -- probably there is.

(By the way, the smallest examples would be tournaments with 0 teams or 1 team, but I assume those numbers were meant to be excluded.)

I feel there is a huge amount of details ;)
I think moderators should move it to Advanced Combinatorics Section.

Ravi B! You are absolutely right!
$n=21 $ is the first number s.t. $\frac{n(n-1)}{2}\cdot\left(\frac{3}{4}\right)^{n-2}<1$ !!!

Yes, that is the right calculation!

By the way, there are some explicit constructions known for this problem.  The constructions are easy to describe, but proving that they work requires advanced methods.

For example, consider the following "quadratic-residue" tournament.  Let p be a prime number that is 3 mod 4.  Our tournament will have p teams named 0, 1, ..., p - 1.  Say that team i beats team j if i - j is a quadratic residue mod p.  Then it is known, for sufficiently large p, that this tournament satisfies the problem.  The proof uses Weil's theorem (sometimes called "Riemann's hypothesis for finite fields").

It might be interesting to check by hand or computer whether the quadratic-residue tournament works for p = 7, 11, or 19.

Great!
$p=7$ works !!!
Thank you very much, Ravi B!

[I wrote this before I saw Myth's last post.]

Actually, the quadratic-residue tournament was easier to check than I thought.  That's because all the teams are symmetric, so we might as well assume that team 0 is part of any pair.

Unless I made a mistake, the quadratic-residue tournament with p = 7 does work.

Just to clarify, here is the tournament stated again.  The teams are numbered 0, 1, ..., 6.  Say that team i beats team j if i - j is 1, 2, or 4 mod 7.

[quote="Myth"]
$n=21 $ is the first number s.t. $\frac{n(n-1)}{2}\cdot\left(\frac{3}{4}\right)^{n-2}<1$ !!![/quote]

Oh-Oh! A real nice problem here.
For our pre-olympiad friends, just some words about Myth computations :
Following Ravi B. hint, let's consider a random tournament with $n$ teams (each tournament has equal probability to be chosen in the set of all possible tournaments).
Let $A,B$ be two distinct teams, and let $p(A,B)$ be the probability of the event $E(A,B)$ that the chosen tournament fails to be an example because of $A$ and $B$, that is every other team has lost at least one of its game against $A$ and $B$. Then, for each of the other $n-2$ teams, its result against $A$ and $B$ (in that order) is L-L or L-W or W-L (but not W-W), where L = loose and W = win. It follows that $p(A,B) = ( \frac 3 4 )^{n-2}$.

Now, the probability that the chosen tournament fails to be an example is $p = p( \cup E(A,B) )$ where the union is over all the pairs $\{A,B \}$ of teams.
Then $p \leq \sum p(A,B) = \frac {n(n-1)} 2 ( \frac 3 4 )^{n-2}$.

For $n=21$, we deduce that $p < 1$ which means that it is possible that the chosen tournament is an example. In particular, it means that such a tournament does exist.

Pierre.

Actually, I counted expected number of successful pairs (i.e. pair of team $(A,B)$ s.t. each other team lost to $A$ or $B$).

Ok.
So, let's consider that I have found another solution  :D

Pierre.

The problem in this thread was about finding a tournament that foils every pair of teams.  It can be generalized to finding a tournament that foils every triple of teams, quadruple of teams, etc., up to an arbitrary fixed limit.  That problem was raised by Schutte.  Erdos (1963) solved the problem using the probabilistic method, an early success of the method.  Graham and Spencer (1971) showed that the quadratic-residue tournament works, using Weil's theorem.

But for the original problem (pairs) in this thread, it is actually elementary to show that the quadratic-residue tournament works (for every prime p that is 3 mod 4).  It boils down to showing that every number mod p can be written as the difference of two quadratic residues.  That is pretty easy to show, and gives a nice understanding of why the tournament works.

[quote="Ravi B"]Joel, don't tell me you forget my lectures on probabilistic methods so soon. :)  (In my handout, this problem was called unrankable tournaments.)
[/quote]

Hmm...Are your lectures and the handout available?  :)
It would be nice if you could attach them.

Pierre.

I had put my handout in the Download subforum of this site a couple of months ago, but it looks like it got lost during the server crash.  So I just reposted it.  Here is the link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23507[/url].

Happy reading!


i need some help on these altough its easy. x^2+2x+1=0 . I know the answer is -1 but how do i get -1. From encarta@hotmail.com the formula for finding quadratic equation is to put ax^2+bx+c=0 . Therefore a=1 b=2 and c=1. x = ( -b ± √(b²-4ac) ) / 2a
Can somone explain the formula for solving quadratic equations?

One way to solve quadratic equations is to factor them (this only works when the answers are nice integers or rational numbers)

$ x^2 \plus{} 2x \plus{} 1 \equal{} 0$ can be factored into

$ (x\plus{}1)(x\plus{}1) \equal{} 0$. Multiply that out and you will see.

From here, it is really easy to see that $ x \equal{} \minus{}1$ will work and nothing else will.

The quadratic formula will always work. If $ 0 \equal{} ax^2 \plus{} bx \plus{} c$, then
It is $ x \equal{} \frac{\minus{}b \pm \sqrt{b^2 \minus{} 4ac}}{2a}$

Here is a quick derivation of it through completing the square.

$ ax^2 \plus{} bx \plus{} c \equal{} 0$  divide everything by $ a$

$ x^2 \plus{} \frac{b}{a} x \plus{} \frac{c}{a} \equal{} 0$ minus c/a from both sides

$ x^2 \plus{} \frac{b}{a}x \equal{} \minus{}\frac{c}{a}$ add (b/2a)^2 to both sides

$ x^2 \plus{} \frac{b}{a}x \plus{} \left(\frac{b}{2a}\right)^2 \equal{} \frac{b^2}{4a^2} \minus{} \frac{c}{a}$

$ (x \plus{} \frac{b}{2a})^2 \equal{} \frac{b^2 \minus{} 4ac}{4a^2}$

$ x \plus{} \frac{b}{2a} \equal{} \frac{\pm\sqrt{b^2\minus{}4ac}}{2a}$

$ x \equal{} \frac{\minus{}b \pm\sqrt{b^2\minus{}4ac}}{2a}$

Ya the formula  method is easy u can do any quad prob through that Its also called Shreedaracharya formula

there is another formular x=2c/(-b+or-sqrt(b^2-4ac)

That formula that 21 7 15 mentioned is another version of the quadratic formula (link: http://en.wikipedia.org/wiki/Quadratic_equation#Alternative_formula).

However, there was one problem with what you wrote.

Rather than $ \pm$, it is $ \mp$. :lol:

(See the above link, and the derivation).

Why would someone want to use the alternative formula? it looks harder to work with. What does Wikipedia mean when it says it reduces loss of precision for roots?

If $ |b|$ is close to $ \sqrt {b^2 \minus{} 4ac}$, small errors in the computation of the square root can lead to large errors in the computation of one of the roots (the one that is small in absolute value).  The alternative formula corrects this issue.  See [url=http://en.wikipedia.org/wiki/Numerical_stability]numerical stability[/url].


inequalities
induction
inequalities unsolved

For $0 < x_1 \le x_2 \le \cdots \le x_n$ and  $x_1 x_2 \cdots x_n=1$
Prove or disprove:
$x^1_1+x^2_2+\cdots+x^n_n \ge x^{-1}_1+x^{-2}_2+\cdots+x^{-n}_n$
strange :lol: ,I even unsolved when $n=3$,And I will be happy if anybody can soloved n=3 :)

[quote="lccs"]try $n=3$
For $a\ge b\ge c>0$ and $abc=1$. (1)
Prove$a^3 + b^2 + c \ge \frac{1}{{a^3 }} + \frac{1}{{b^2 }} + \frac{1}{c}$.
[/quote]

(1) implies that : $a \ge 1 \ge c$.

$b^2 + c \ge b^2 + c^2 \ge 2bc = \frac 2a \ge \frac 2{a^3}$
$a^3 + b^2 \ge a^2 + b^2 \ge 2ab = \frac 2c$

Now, 2 cases :
$\triangleright \mathbf{b \ge 1} :$
$a^3 + c \ge a^2+c^2 \ge 2ac = \frac 2b \ge \frac 2{b^2}$

$\triangleright \mathbf{b < 1} :$
${{a^3 + c = a^3(1 + \frac c{a^3}) \ge 2a^2\sqrt{\frac ca}} = 2\frac{\sqrt{c^2b}}{c^2b^2} = \frac{2\sqrt{b}}{cb^2}} \ge \frac{2\sqrt{b}}{b^3} \ge \frac 2{b^2}$

Hence, we get the desired result ;)

[b]Zhaobin[/b], I'll try to work on the general case! :)

:rotfl: ok, ok! Well, I hope you'll have other dreams like that, because this inequality is really nice! ;)

ps : at the moment, I'm also stuck on the generalization  :(

[quote="zhaobin"]thanks,mathmanman.I come up this problem when I was in dream
 :) But I haven't solved in dream,also now :)[/quote]

Respect!!! really in your dream ? great. I had readen about aolving problems in the dream and once my teacher said that this night before you go to sleep eat some amout of salt , and i said ok. Then i ate and whent to sleep. After i saw water in my dream. and then i understand the importance of solving problems in the Dream  :)  so respect for dreamers...

Well, I think that maybe we could try proving it via induction..

We know it's true for $n=1,2,3$, so let's assume it's true for some $n \in \mathbb{N^*}$.
Let $0<x_1 \le x_2 \le \dots \le x_n$ such that $x_1x_2 \dots x_n=1$.
Now, we let : $y_1=x_1, \dots, y_{n-1}=x_{n-1}x_n$.
We thus have : $0<y_1 \le \dots \le y_{n-1}$ (since $x_n \ge 1$) and $y_1 \dots y_{n-1}=1$.
So, our induction hypothesis will be : $y_1+....+y_{n-1}^{n-1} \ge \frac 1{y_1} + \dots + \frac 1{y_{n-1}^{n-1}}$
id est :
$x_1+ \dots +(x_{n-1}x_n)^{n-1} \ge \frac 1{x_1} + \dots + \frac 1{(x_{n-1}x_n)^{n-1}}$.
And hence, it remains to show :
$x_{n-1}^{n-1}+x_n^n - \frac 1{(x_{n-1}^{n-1}} - \frac 1{x_n^n} \ge (x_{n-1}x_n)^{n-1} - \frac 1{(x_{n-1}x_n)^{n-1}}$

 :roll:

Thanks,but sorry I think the last inequality is wrong take $x_{n-1}=x_n=2$,make n large enough.

Is it the induction or the proposition that's false?  :?

I think the induction in this way can't work.
And I haven't have any delepment in this inequality :(

[quote="zhaobin"]For $ 0 < x_1 \le x_2 \le \cdots \le x_n$ and  $ x_1 x_2 \cdots x_n \equal{} 1$
Prove or disprove:
$ x^1_1 \plus{} x^2_2 \plus{} \cdots \plus{} x^n_n \ge x^{ \minus{} 1}_1 \plus{} x^{ \minus{} 2}_2 \plus{} \cdots \plus{} x^{ \minus{} n}_n$
strange :lol: ,I even unsolved when $ n \equal{} 3$,And I will be happy if anybody can soloved n=3 :)[/quote]

Anybody have idea? :maybe:


ratio
geometry
circumcircle
geometry open

[quote="hamjam"]Consider a square $ ABCD$ inscribed in a circle.  Let $ P$ be an arbitrary point on the circle.  Explain why the cross-ratio of the lines $ AP, BP, CP, DP$ is independent of the choice of $ P$.  Find this cross-ratio.[/quote]
Let $ P,\ P'$ be, two arbitrary points on the minor arcs $ AD,\ AB$ respectively, of the circumcircle $ (O)$ of the given square $ ABCD.$

It is easy to show that $ (P.ABCD) \equal{} (P'.BCDA) \equal{} u$ $ ,(1)$ because of the angles formed by the homologous rays of the pencils $ P.ABCD$ and $ P'.BCDA,$ are equal.

$ ($ $ \angle APB \equal{} \angle BP'C \equal{} 45^{o},\ \angle BPC \equal{} \angle CP'D \equal{} 45^{o}$ and $ \angle CPD \equal{} \angle DP'A \equal{} 45^{o}$ $ )$

But, $ (P'.BCDA) \neq (P'.ABCD)$ $ ,(2)$

Hence, from $ (1),\ (2),$ we conclude that $ (P.ABCD) \neq (P'.ABCD)$ $ ,(3)$

That is, the [b]double ratio[/b] ( = [b]cross ratio[/b] ) of the pencil $ P.ABCD,$ is not indepented of the choice of $ P.$

If I am not mistaken, from the Double ratio theory we have $ (P'.ABCD) \equal{} \frac {u}{u \minus{} 1}$

Kostas Vittas.

[quote="hamjam"]thanks so much for the reply.  your work seems solid, however, I aksed my prof and he said it is in fact independent of the choice of $ P$.. hm..[/quote]
Sorry dear [b]hamjam[/b] and all my friends.

I see now that I was mistaken, because of we can say that $ (P'.ABCD) \equal{} (P'.A'BCD) \equal{} (P.ABCD)$ and then, indeed the [b]double ratio[/b] of the pencil $ P.ABCD,$ is independed of choice of $ P.$

Sorry again, Kostas Vittas.