One day, on June 26th at noon, I asked the MOEMS moderator when he was going to post again. He said, "I'll post in 4000000 seconds." In what month will he post again?

THis is made up  :)

Btw, where [i]is[/i] the MOEMS moderator? He hasn't posted for a while...

[hide]4000000 sec * 1 / (60 sec/min ) * 1/  (60 min/hour) * 1 / (24 hour/day) = 46 days, which just gets into August I think[/hide]

[quote="236factorial"]
Btw, where [i]is[/i] the MOEMS moderator? He hasn't posted for a while...[/quote]
I think that the moderator has other stuff to do than AoPS, but he has only 6 posts...

Well, he does sponsor some of the Math Jams right?

YOu're right but he really isn't that active... I wish you could track when he or another user was on.


So I may have to miss either Lassiter or Furman, and if I do have to miss one I may get to choose which I miss. So here are the pros and cons of each. (Oh how I hate words beginning with con- :D)

Lassiter:
+I [i]know[/i] I'll get to see practically the whole ARML team, which is fun.
+It's a test written by Miles+Phillip+ohwhoelse?+Mrs. Poss
-It's a highschool meet, and highschoolmeets don't usually give money, nor are they usually problem-solving-tests as opposed to speed-solving-tests.

Furman:
+The team I was on last year got a trophy, because I scored something like 18 points on ciphering and on individuals we didn't do terribly.
+Atlanta/Georgia Math Club went last year, and thus a part of the ARML team will be there probably, which is probably almost as good as the greater part of the ARML team.
+Augusta Area Homeschoolers will be there, and I like Joe, Gracie, Joshua, and Matthew.
-They don't give you scrap paper (again, last year).
-The test is hard.
-The test is long.

So which should I go to? Can you give me other reasons to or not to go to one or the other? :)

(Of course, it would be best if I didn't have to miss either :), but that might not happen.)

[quote="Arvind_sn"]I would go to lassiter, probably cause I'm going... :P  But anyway, since you're varsity (right?) the lassiter test for varsity is easy. as jvish as i am, i took a practice varsity test and it was pretty easy. i guess lassiter is easier. :wink:[/quote]

I'm not required to be varsity by grade, but I'm in calc and I don't want to join Sigur's Hall of (!)Fame, so for most things I should be in V, and for grade-based jv/v distinctions I'd rather be in V.

I would go to Lassiter, for sure:

- better chance of winning.

- Furman doesn't give money FOR SURE, all you get is a plaque for team placement

Perhaps this is just me, but I find the Furman test INCREDIBLY stupid/boring:

- Junior Exam = no team placement, individual awards are basically nothing

- Senior Exam = incredibly hard test.  I wouldn't mid that, except the way it is formatted is really bad.  It's in column things, which gives you no free paper space on the contest.

About not getting paper: my friend got screwed at MathCounts that way.

I consider Furman pretty hard for the time limit.  It's one of the hardest college contests out there.  I don't really mind that, but the competition will be fierce there:  Vestavia Hills, great Georgia teams, decent NC teams (none of the killer ones from Duke...), and 3 good SC teams.  Assuming Billy's teammates are not as good as him, it would be hard to place at Furman.

I dunno, I did a practice one a while ago and it wasn't that bad. Of course, if it's the time limit that kills you I probably wouldn't have noticed, but the questions themselves aren't that hard..

@Seta: did you go without any scrap paper while you took the test? :D

@Sean: We'd be in the Small Schools division, and would do the Senior test as we did last year. We did place last year, iirc 3rd in SS, but I'm not sure that my teammates this year are as good as those last year. Then on the other hand I did badly last year - I wasn't the top score on my team - and there's a pretty good probability that I'll do better on written this year and hopefully as well on ciphering. (One ciphering question I'd have gotten 4 more points for if I had checked for reducability by 2 instead of 3, 5, and 7 :D.)

Oh, and I find the Furman senior test fun, except you don't get any scrap paper to work the problems :D.

Yes, unfortunately in high school tournaments, there are financial repercussions for a test that is too difficult :( .  But I still tried to submit some moderately nontrivial questions.  Oh well, hopefully JV people will have fun with the power round, anyhow... :D

Hmm, that's a kind of strange dilemma of yours, Billy.

what's the website for furman and when is the competition? does it [i]have [/i]a website?

i haven't been to lassiter before, so I can't offer you much on that part. However, i've been to furman for a number of years, and I like it a lot. personally, i dont think the test is too difficult, even though the time restraint is a little annoying. but keep in mind, everyone gets the same amount of time, so if you feel pressed for time, so will everyone else. also the furman campus is really nice, we feed the swans every year. again, ive never been to lassiter so i dont know how that is, but i like furman a lot

Aren't Furman and lassiter on different dates.

Lassiter is December 2nd.

Furman is March 10th.

Indeed, they are two possible SAT dates.

Here's the furman website: http://math.furman.edu/tournament/tournament.html Kinda bland :).

[quote="solafidefarms"]Indeed, they are two possible SAT dates.
[/quote]

Ouch. I just... triple... booked December 2nd. *goes to flatten my head underneath... hmm... a mountain*

I [i]needed[/i] that competition practice!

Oh, well. Maybe I should, like, stop whining and go practice AMC problems like I promised...

Or, maybe I can convince myself that my old SAT scores are acceptable... Nah, I want that 1600!

Well, I guess I have to say thanks for prompting me to realize my scheduling dillema... I don't get the choice, though...

(But aren't you losing money on various fees by waiting so late to decide?)

I [i]think[/i] my dilemma, I just won't take the SAT. I was planning on applying for dual-enrollment at Georgia Tech in the spring, which requires the new SAT score and the SATII Lit, but we decided that I wouldn't, I'd just do EPGY classes instead. So I'm not missing Lassiter or Furman for that reason :); I might for others, but not for the SAT.

Sorry Seva? :)

[quote="Karth"][quote="Elemennop"][quote="solafidefarms"]@Seta: did you go without any scrap paper while you took the test? :D[/quote]

Quit misspelling my name.[/quote]

Hmm are you Seva Tchernov? If so, congrats on UGA  :)[/quote]

No, I'm not, but I'll pass the message on to him.

I do, as the self-proclaimed Math-Related-Cool-Ness Goddess of Georgia.

Hurry up and get your 1600s, kids. Who cares about that silly writing section anyway?

[quote="jhcreinhardt"][quote="solafidefarms"]ok, who are you?[/quote]

man whats happening to america...we really need to start teaching our youth how to recognize sarcasm.

btw furman is clearly better[/quote]

Which is harder?

not lassiter.  so..that leaves..the other one

[quote="juicybooty911"][quote="cleansmyheart"]All the cool people go to Lassiter.[/quote]

who decides that?

sat is also on that date, awwww[/quote]
1) That sentence is a truth resulting from the notation we are using.  Who decides that 1+1=2, anyhow?

2) Too bad for the SAT, isn't it? :P   Lassiter had better be harder than the SAT :maybe: , so you should come to Lassiter instead of the SAT, where you can see all of your math friends, instead of being cooped up in some classroom with a bunch of random people and not talking to anyone.

[quote="akkaren"]Haha, I think I'm the only one choosing the SAT over Lassiter... I should've just actually paid attention to what I was doing when I took it two years ago.[/quote]

Not the only one as far as I know, count me in.  I have to take it since it's the last date that GT allows for its junior early admission program.

Can we protest to Lassiter to change the date? :P

[quote="akkaren"]I'd rather protest to the collegeboard... take out some pent-up animosity...[/quote]

on second thought, yay, i agree, collegeboard needs it more


$p$ is a prime.
Find the value of : $P=\sum_{k=1}^{p-1}[\sqrt[3]{kp}]$

You mean that $k$ (not $n$) is between $1$ and $p-1$. Is that true?

Alexandros


Αν $ a,b,c,d$ πραγματικοί, να αποδείξετε ότι $ (a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})^{3}\geq (2abc\plus{}2bcd\plus{}2cda\plus{}2dab)^{2}$. proposed by patilas

Η άσκηση είναι μια ωραία εφαρμογή της BCS :)
Συγκεκριμένα:
$ \displaystyle LHS\leq 4(a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2)(a^2 + b^2 + c^2 + d^2) =$
.
$ \displaystyle = 4(a^2 + c^2)(b^2 + d^2)(a^2 + b^2 + c^2 + d^2)\leq RHS$

QED :wink:

wraios!!! i dikia m lusi einai pio suntheth alla as tin grapso gia na uparxei sto arxeio...

$ (a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})^{3}\geq 4(abc\plus{}bcd\plus{}cda\plus{}dab)^{2}$
$ \Longleftrightarrow\left(\frac{1}{16}\plus{}\frac{1}{16}\plus{}\frac{1}{16}\plus{}\frac{1}{16}\right)\cdot \left(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\right)\cdot \left(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\right)^{2}\geq \left(abc\plus{}bcd\plus{}cda\plus{}dab\right)^{2}$

Apo BCS exoume $ \left(\frac{1}{16}\plus{}\frac{1}{16}\plus{}\frac{1}{16}\plus{}\frac{1}{16}\right)\cdot \left(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\right)\geq \left(\frac{\left(a\plus{}b\plus{}c\plus{}d\right)^{3}}{4^{2}}\right)^{2}$.

Opote arkei

 $ \left(\frac{\left(a\plus{}b\plus{}c\plus{}d\right)^{3}}{4^{2}}\right)^{2}\geq \left(abc\plus{}bcd\plus{}cda\plus{}dab\right)^{2}\Longleftrightarrow \left(a\plus{}b\plus{}c\plus{}d\right)^{2}\cdot \left(a\plus{}b\plus{}c\plus{}d\right)\geq 16\left(abc\plus{}bcd\plus{}cda\plus{}dab\right)$

Omws

$ \left(a\plus{}b\plus{}c\plus{}d\right)^{2}\cdot \left(a\plus{}b\plus{}c\plus{}d\right)\geq 4\left(a\plus{}b\right)\cdot \left(c\plus{}d\right)\cdot \left(a\plus{}b\plus{}c\plus{}d\right)\geq 16\left(abc\plus{}bcd\plus{}cda\plus{}dab\right)$

Opote

$ \left(a\plus{}b\right)\cdot \left(c\plus{}d\right)\cdot \left(a\plus{}b\plus{}c\plus{}d\right)\geq 4\left(abc\plus{}bcd\plus{}cda\plus{}dab\right)$

telos exoume
$ \left(a\plus{}b\plus{}c\plus{}d\right)\cdot \left(a\plus{}b\right)\cdot \left(c\plus{}d\right)\equal{}2\left(abc\plus{}bcd\plus{}cda\plus{}dab\right)\plus{}a\left(c^{2}\plus{}d^{2}\right)\plus{}b\left(c^{2}\plus{}d^{2}\right)\plus{}c\left(a^{2}\plus{}b^{2}\right)\plus{}d\left(a^{2}\plus{}b^{2}\right)\geq 4\left(abc\plus{}bcd\plus{}cda\plus{}dab\right)$ opou isxuei apo AM-GM...


geometry
rectangle
trapezoid
geometry proposed

Let $ABCD$ be a rectangle and $d$ a line that interrsects $AB, BC, CD, DA$ in $M, N, P, Q$. Prove that

                           \[ \frac{MA}{MB}\cdot\frac{NB}{NC}\cdot\frac{PC}{PD}\cdot\frac{QD}{QA}=1. \]

I mean the line intersect in the extensions of two sides.
By simple angle chasing we see that $PCN, PDQ, MBN, MAQ$ are all similar.

Hence using directed segments we get $\frac{MA}{QA} = \frac{PD}{QD} = \frac{PC}{NC} = \frac{MB}{NB}$. From which follow the thesis.
Very easy. Am I missing something?  :?

Well  - actually it holds for arbitrary convex quadrilateral - just use Menelaos twice.


geometry
incenter
trigonometry
Pythagorean Theorem
trig identities
Law of Cosines
power of a point

Given a right-angled triangle $ ABC$ such that $ AB \equal{} 3,\ BC \equal{} 4,\ CA \equal{} 5$. Let $ O$ be the incenter of the triangle and suppose that the circle touches to $ BC,\ CA,\ AB$ at $ P,\ Q,\ R$ respectively. Let $ S$ be the intersection point of the circle $ O$ and line segmemt $ AP$ with $ S\neq P$.

(1) Find the length of $ QR$ and the value of $ \sin \angle{QPR}$.

(2) Find the value of $ \tan \angle{BCS}$.

(3) Find the angle of $ \angle{PSC}$.

[hide="Solution"]
We have $ AR \equal{} AQ \equal{} 2$, $ BR \equal{} BP \equal{} 1$, $ CP \equal{} CQ \equal{} 3$.
By the law of cosines on $ \triangle ARQ$, $ QR^2 \equal{} 2^2 \plus{} 2^2 \minus{} 2 \cdot 2 \cdot 2 \cdot \cos BAC \equal{} 8 \minus{} 8 \cdot \frac {3}{4} \equal{} \frac {16}{5}$. Thus, $ QR \equal{} \frac {4}{\sqrt {5}}$.
By the same argument, $ PQ \equal{} \frac {3 \sqrt {10}}{5}$
By the Pythagorean theorem, $ PR \equal{} \sqrt {2}$
By the law of cosines on $ \triangle PQR$, since we know $ PQ$, $ QR$, and $ PR$, we find $ \cos \angle RPQ \equal{} \frac {1}{\sqrt {5}}$. Thus, $ \sin \angle RPQ \equal{} \frac {2}{\sqrt {5}}$.
By the Pythagorean theorem, $ AP^2 \equal{} AB^2 \plus{} BP^2 \equal{} 10$, so $ AP \equal{} \sqrt {10}$
By Power of a point, $ AR^2 \equal{} 4 \equal{} AS \cdot AP \equal{} AS \sqrt {10}$, thus $ AS \equal{} \frac {4}{\sqrt {10}} \equal{} \frac {2 \sqrt {10}}{5}$.
Thus, $ SP \equal{} AP \minus{} AS \equal{} \frac {3 \sqrt {10}}{5}$.
By Stewart's theorem on $ \triangle APC$ with cevian $ CS$, $ AC^2 \cdot PS \plus{} PC^2 \cdot AS \equal{} CS^2 \cdot AP \plus{} AS \cdot SP \cdot AP$. From here we find $ CS \equal{} \frac {9}{\sqrt {5}}$.
Now we know $ PS$, $ PC$, and $ CS$, so by the law of cosines on $ \triangle PSC$, we find $ \cos \angle PCS \equal{} \frac {2\sqrt {5}}{5}$, so $ \tan \angle BCS \equal{} \frac {1}{2}$
Also by the law of cosines on $ \triangle PSC$, $ \cos \angle PSC \equal{} \frac {\sqrt {2}}{2}$, so $ \angle PSC \equal{} 45$. Also, $ \angle RSP \equal{} \angle RQP$. But we know $ RQ$, $ PQ$, and $ PR$, so by the law of cosines on $ \triangle PQR$, we find $ \cos \angle RQP \equal{} \frac {\sqrt {2}}{2}$, so $ \angle RQP \equal{} 45 \equal{} \angle RSP$. So $ \angle RSC \equal{} \angle RSP \plus{} \angle PSC \equal{} 90$.
Wait, I see kunny deleted finding $ \angle RSC$. Oh well.  [/hide]

Yeah, I just realized I left it out right before you posted, I have added it. It is just the law of cosines.

[hide="My solution"]

$ 2\triangle{OAQ}\equal{}OA\cdot \frac 12 QR\equal{}AQ\cdot OQ$, yielding $ QR\equal{}\frac{4}{\sqrt{5}}$

$ \angle{AOQ}\equal{}\frac 12 \angle{QOR}\equal{}\angle{QPR}\Longrightarrow \sin \angle{QPR}\equal{}\frac{AQ}{OA}\equal{}\frac{2}{\sqrt{5}}$[/hide]

Whoops, I made some calculation error. Do they know Menelaus in Japan? If so, we can find $ \tan \angle BCS$ this way:
Let $ CS$ intersect $ AB$ at $ X$. By Menelaus, $ \frac{CP}{CB} \cdot \frac {BX}{XA} \cdot \frac{AS}{SP}\equal{}1 \implies BX\equal{}2XA$ from my previous solution, so $ BX\equal{}2$, $ XA\equal{}1$. Thus, $ \tan \angle BCS\equal{}\frac{BX}{BC}\equal{}\frac{1}{2}$

[quote="dgreenb801"]Whoops, I made some calculation error. Do they know Menelaus in Japan? [/quote]

Yes, they learn it, so is Ceva.

[quote="dgreenb801"]
Let $ CS$ intersect $ AB$ at $ X$. By Menelaus, $ \frac {CP}{CB} \cdot \frac {BX}{XA} \cdot \frac {AS}{SP} \equal{} 1 \implies BX \equal{} 2XA$ from my previous solution, so $ BX \equal{} 2$, $ XA \equal{} 1$. Thus, $ \tan \angle BCS \equal{} \frac {BX}{BC} \equal{} \frac {1}{2}$[/quote]

Thank you for the correction. :)

By the way, Is there some name for the property [quote="kunny"]$ \angle{QPR} \equal{} \frac 12\angle{QOR}?$[/quote]

Stewart is to difficult to remember, but it is useful.


[quote="duc 95"]>0. k ,n>1
Prove that[/quote]

try  $ n \equal{} 3,k \equal{} 2,a_1 \equal{} 1,a_2\equal{}1,a_3 \equal{} 6$


The diagonal of a square is $ \sqrt{2}$ inches long. How many inches long is the side of the square?

Using the Pythagorean Theorem, the diagonal of any square with side length $ s$ is $ \sqrt{s^2\plus{}s^2}\equal{}s\sqrt2$.

$ s\sqrt2\equal{}\sqrt2$
$ s\equal{}1$

Note: You can also use this fact as a general rule for 45-45-90 triangles. The side lengths will be $ s$, $ s$, and $ s\sqrt {2}$.

EDIT: Oops, fixed. Accidentally put the $ s$ in front of the $ \sqrt{2}$.  :P

Just to clear confusion, it is not $ \sqrt{2s}$, it is $ s\sqrt{2}$. :P


When a mixture of Sodium Sulphate and Sodium Carbonate is dissolved in water and the aqueous solution is reacted with sulphuric acid only Sodium Carbonate is neutralized.
Similar thing also holds when Sodium chloride is taken in place of Sodium Sulphate.

That's because carbonate ion is a stronger base than sulphate or chloride ions, and this is because sulphuric acid and chlorydric acid are stronger acids than bicarbonate.

so on factors does neutralization depend?

In any acid-base reaction, the strongest acid will react with the strongest base in order to produce the weakest acid and the weakest base. The relative base/acid strenght can be taken from Ka tables.

but here a salt is taken .is there a way to know which will react and which won't

[quote="Pardesi"]but here a salt is taken.[/quote]

Yes, and one of the ions (or both) can have acid-base properties.

[quote="Pardesi"]is there a way to know which will react and which won't[/quote]

Look my previous post.


limit
trigonometry
real analysis
real analysis unsolved

Let $\ a\in\mathbb{R}$ $\ a\neq0$ and $\ f: \mathbb{R}\to\mathbb{R}$ such that the limit $\ \lim_{t\to0}\frac{f(xt)}{\sin at}$ exists for every $\ x\in\mathbb{R}$.
$\ a)$ Prove that $\ \lim_{x\to0}f(x)$ exists and compute this limit.
$\ b)$ Compute $\ \lim_{t\to0}\frac{f(xt)}{\sin at}$.

Proposed by Mircea Becheanu.

for point b) this is what I found $g(x)=\ \lim_{t\to0}\frac{f(xt)}{\sin at}= \lim_{t\to0}\frac{f(xt)}{\sin(axt)}\cdot \frac{\sin(axt)}{\sin(at)}=g(1)x$


In a triangle $ ABC$ , $ 2AD$=$ BC$ , where $ AD$ is the altitude on $ BC$ , If $ <ABC\equal{}75^o$, show that $ <ACB\equal{}30^o$

I believe this is a very easy geometry problem, I would like to see a purely geometric solution without trigonometry or things like that.
More precisely a proof without words just by construction is what I would like...

Thanx!

Denote E a point on AD different from A such that AD=DE. Obviously, <AEC=15. So we need to prove that if we take a point C in the extension of  the height BD of the triangle ABE to D such that BC=AE, <ACB=30. Clearly, the point C is unique. So it is enough to prove that BC=AE for <CAE=30, which is simple.

Let $CE$ be the $C-$altitude of $\triangle ABC$ and $EF$ the $E-$altitude of $\triangle BCE$. Clearly $\angle BCE=15^\circ$ and, in a right-angled triangle having an angle of $15^\circ$, its altitude is a quarter of the hypotenuse, i.e. $4EF=BC=2AD\iff 2EF=AD$; with $EF\parallel AD$ we infer $E$ is the midpoint of $AB$, or $\triangle ABC$ is isosceles, so it is $75^\circ-30^\circ-75^\circ$.

Best regards,
sunken rock


is it 17 + 8/13
or 229/13? for height, then 13 as base, so
114.5?

[hide="Hint"]Find the altitude of the right triangle.[/hide]

You could also find the other two sides and use Heron's, but that's more work than necessary.

[quote="Temperal"]This is an extension to AIME prob:

Square $ ABCD$ has side length $ 13$, and points $ E$ and $ F$ are exterior to the square such that $ BE = DF = 5$ and $ AE = CF = 12$. Find $ EF^{2}$.
[img]http://www.artofproblemsolving.com/Wiki/images/b/bb/2007_AIME_II-3.png[/img]

Here's the actual problem:

Extend the altitude of $ \triangle AEB$ to $ G$ such that $ FG \parallel DC$. Find $ [AGB]$.[/quote]
[hide="Extension Solution"]
Extend the altitude of $ \triangle AEB$ to point $ G$ so that $ FG\parallel DC$ and allow $ EG$ to intersect $ AB$ and $ CD$ at $ X$ and $ Y$ respectively.  Also, take altitude $ FZ$ of $ \triangle CFD$.  We can find that $ EX = \frac {AE(EB)}{AB} = \frac {60}{13}$.  Also, we notice that since $ EG\perp AB$ and $ AB\parallel CD\parallel FG$, we have that $ YG\perp FG$ and thus rectangle $ YGFZ$ is fomed, so $ YG = FZ = EX = \frac {60}{13}$ and notice that $ FZ = EX$ since they are altitudes of triangles $ CFD$ and $ AEB$, which are congruent.  Now, we have that $ EG = EX + XY + YG = \frac {60}{13} + AB + \frac {60}{13} = \frac {120}{13} + 13 = \frac {289}{13}$.  Now, furthermore notice that $ FG = YZ = CD - DZ - CY = 13 - DZ - BX$.  We can find that $ BX = \sqrt {BE^2 - EX^2} = \sqrt {5^2 - (\frac {60}{13})^2 = \frac {25}{13} = DZ}$.  Thus, we obtain that $ FG = 13 - \frac {25}{13} - \frac {25}{13} = \frac {169 - 50}{13} = \frac {119}{13}$.  Now, we notice that $ EF^2 = FG^2 + EG^2 = (\frac {119}{13})^2 + (\frac {289}{13})^2 = (\frac {17}{13})^2(7^2) + (\frac {17}{13})^2(17^2) = (\frac {289}{169})(338) = \boxed{578}$
[/hide]

[hide="AIME Solution"]
Well, we know all the measurements, so this is much easier.  $ GX = XY + GY = 13 + \frac {60}{13} = \frac {229}{13}$ and $ AD = 13\implies \frac {GX}{AD} = \frac {229}{169}$ notice that $ \triangle ADB$ and $ \triangle AGB$ have the same base, so the ratio of their altitudes is the ratio of their areas, so $ \frac {2[AGB]}{169} = \frac {2[AGB]}{2[ADB]}=\frac {[ABG]}{[ADB]} = \frac {GX}{AD} = \frac {229}{169}\implies 2[AGB] = 229\implies [AGB] = \boxed{114.5}$.
[/hide]

[quote="Temperal"]This is an extension to AIME prob:[quote]

Square $ ABCD$ has side length $ 13$, and points $ E$ and $ F$ are exterior to the square such that $ BE \equal{} DF \equal{} 5$ and $ AE \equal{} CF \equal{} 12$. Find $ EF^{2}$.
[img]http://www.artofproblemsolving.com/Wiki/images/b/bb/2007_AIME_II-3.png[/img][/quote]

Here's the actual problem:

Extend the altitude of $ \triangle AEB$ to $ G$ such that $ FG \parallel DC$. Find $ [AGB]$.[/quote]

[hide]Extend the altitude to G, so that $ EG // FG$.
The altitude of $ \triangle AEB$ is $ \frac {5 \cdot 12}{13} \equal{} \frac {60}{13}$ by using similar triangles. (Thanks Temperal!)
Thus $ EG \equal{} 13 \plus{} \frac {2 \cdot 60}{13} \equal{} \frac {289}{13}$.

In the AIME problem, we can immediately find $ [AGB] \equal{} \frac {1}{2}(13 \cdot \frac {229}{13}) \equal{} 114.5$.

In the extension, we have $ FG \equal{} 13 \minus{} 2(\frac {25}{13}) \equal{} \frac {119}{13}$ so $ EF^2 \equal{} (\frac {289}{13})^2 \plus{} (\frac {25}{13})^2 \equal{} \frac {84146}{169}$
...and by looking at the other posted solutions, I feel like I did something wrong.[/hide]

[hide="Maybe you made a mistake here?"]
Sorry, but where did $ (\frac {25}{13})^2$ come from?  Shouldn't it be $ FG^2 \plus{} GE^2 \equal{} EF^2\implies (\frac {289}{13})^2 \plus{} (\frac {119}{13})^2 \equal{} EF^2$ and we can simplify? [/hide]

Quatto is right.

Sorry if I wasn't clear, but the part about finding $ [AGB]$ was my extension, and the length was the original problem, not the other way around.  :wink:


When I get a problem, or go to the next one, the problem shows up, bu the input box does not.  YOu have to refresh and click get probelm a bunch of times to get it right.  Is it my comp or Alcumus itself?

[quote="levans"]What browser, browser version, and operating system are you using?[/quote]

I am using Google Chrome 0.3.154.9, and Windows Vista Home Premium SP2 32-Bit.  Is it Chrome's fault?

It's definitely a problem specific to chrome/safari and other browsers that use the webkit rendering engine. Not sure what the problem is yet.

From my trials, I could replicate the problem when Chrome first loads, but after that, everything always appeared fine (I did about 10 problems). Do you find similar or is it happening every problem? Sometimes within the same session?

Hmm...now no problems are showing up.  I get a blank screen, pretty much.  I'll attach a screenshot.

This is caused by me trying to fix the problem you are experiencing with Chrome.

The problem mainly appears when you miss the problem and click "Try again". Try that...I'm not very good so I guess I see the problem the most  :blush:

Don't worry it happens with me too sometimes :wink:

Hmm I seemed to have addressed this problem a while ago...

It still occationaly happens
I use Opera 9. something or other...
Windows XP

Is the original problem, where you can't see the input box, still occurring?

I believe it doesn't happen---both Safari and Chrome for me.  BUt I do notice one thing:  The sentence that says the question is worth $ x$ points comes after you click Get Problem again, and doesn't appear the first time.

Yeah it works for me now.

However, I switched to chrome, so I'm not sure for Opera.


In parallelogram $ABDC$, $K$ is on $AB$, $L$ is on $AC$ such that $BK=CL$. $CK$ and $BL$ cut at $P$. Show that $PD$ bisects angle $BDC$.

Nice problem.[hide="A hint"] Denote M the intersection between CD and BL;N the intersection between BD and CK, and then use Menelaus for the triangle MBD. The rest is easy, considering that $BK\parallel CD$ and $CL\parallel BD$.[/hide]


Let $ z_1,z_2\in \mathbb{C}$ such that $ |z_1\plus{}z_2|\equal{}|z_1|\plus{}|z_2|$ . Prove that $ z_1,z_2,z_1\plus{}z_2$ are the affixes of an isoscele triangle and $ z_1,z_2,\minus{}z_1\minus{}z_2$ are the affixes of an equilateral triangle .

If $ |z_1\plus{}z_2|\equal{}|z_1|\plus{}|z_2|$, don't they have the same argument by the triangle inequality?


[hide="Hint"]

We can rewrite this as $ \left(\frac{1}{2}\plus{}\frac{1}{4}\plus{}\cdots \right) \plus{} \left( \frac{1}{4}\plus{}\frac{1}{8}\plus{}\cdots\right)\plus{}\left(\frac{1}{8}\plus{}\frac{1}{16}\plus{}\cdots\right)\plus{}\cdots$

[/hide]

The sum is:
$ \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \frac {4}{16} \plus{} \frac {5}{32}...$
Dividing this by two and subtracting the difference from the original we have:
$ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} \frac {1}{32}...$
This is just a geometric sequence with the sum $ 1$.
Since this was half the total sum, the answer is just $ 2(1) \equal{} \boxed 2$.

[hide="Another Way"]
Let $ S \equal{} \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \cdots$
$ 2S \equal{} 1 \plus{} \frac {2}{2} \plus{} \frac {3}{4} \plus{} \frac {4}{8} \plus{} \cdots$

$ 2S \minus{} S \equal{} 1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \cdots \equal{} 2$

Answer: 2[/hide]


How do I appeal if I don't agree with the score I was given??

Make a thread in this forum - http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174104


\[ a_n\equal{}\lfloor\frac{n\plus{}1}{\pi}\rfloor \minus{} \lfloor\frac{n}{\pi}\rfloor\]
then $ a_n\in\{0,1\}$
prove that both values are achieved infinitely many times

[hide="Solution"] Okay, so when $ [\frac {n \plus{} 1}{\pi}] \minus{} [\frac {n}{\pi}] \equal{} 1$ is when the fractional part of $ \frac {n}{\pi}$ is between $ \frac {\pi \minus{} 1}{\pi}$ and 1 (not including 1).  Similarly, when $ [\frac {n \plus{} 1}{\pi}] \minus{} [\frac {n}{\pi}] \equal{} 0$ is when the fractional part of $ \frac {n}{\pi}$ is less than $ \frac {\pi \minus{} 1}{\pi}$, but greater than 0.  So, we have that $ k \plus{} \frac {\pi \minus{} 1}{\pi}\le \frac {n}{\pi} < 1 \plus{} k$ for an integer k if $ a_n$ is 0.  Thus, we  have that $ k\pi \plus{} \pi \minus{} 1\le n < \pi \plus{} k\pi$.  It is quite clear that there is exactly one value of n for any k and there are infinitely many values of k.  So, there are infinitely many values of n such that $ a_n \equal{} 1$.  Now, we can do the same for 0.  We have that $ k_1\le \frac {n}{\pi} < k_1 \plus{} \frac {\pi \minus{} 1}{\pi}$ for $ a_n \equal{} 0$.  Thus, $ k_1\pi\le n < k_1\pi \plus{} \pi \minus{} 1$.  So, it is obvious that there are two values for n for every $ k_1$.  So, we have that there are infinitely many values of $ k_1$ and thus infinitely many values of n so that $ a_n \equal{} 0$.  So, both cases are proven.  [/hide]


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Firstly, i'm sorry if i'm writing in the wrong topic  :blush: but it has already been done..

One man came to our school and showed us a very sort solution of Fermat's great theorem using geometry.
As he said in the beginning, "It would be simplier, than Pythagorian's theorem".
Really, it was rather simple...

He showed, that the equation
x^n+y^n=z^n with n=2k+1,
has no solutions in positive integers k, and said that the case n=4 is very simple and it is already prooved.
All other cases n>2 follow from the claim.
If you tell me, how to write formulas and draw images in this forum, I'll try to tell this solution  :)

What is Fermat's Great Theorem?  Are you referring to Fermat's Last Theorem?  Fermat's Little Theorem?  If you are referring to his Last Theorem, anyone who claims they can prove the whole thing with simple geometry is a liar and should not speak about math.

could you speak simplier, i'm too weak in english  :blush:
I'm sure that the proof is not wrong. There were many people listening to that proof and they also were agree with him too

He is obviously wrong.  Fermat's Last Theorem defied proof for [i]three hundred and fifty years[/i].  And the claim that it's "simpler than the Pythagorean theorem" is absurd.

the histrory of proof is not very simple.
I think it is not surprising that noone believes - three hundred years - it is not a joke. But he spend almost all his life on that theorem. He has been trying to proove this using geometry for nearly 40 years... Nearly ten years ago he send his proof to ..как его там,не помню.. well.. and it was denied. And half-year ago he came with his own new proof.

But "simplier the Pythagorean's theorem" - it was over-statement..
If you don't believe me I'll show you the proof, but unfortenately today i can't do this as the proof is lost..  :( I'll bring it back tomorrow or the day after tomorrow and you will say if it is absurd.

I'll be grateful if you explain me how to put pictures in the forum  :)

It's completely unlikely that there is an easy proof. It's even more unlikely that some unknown man finds one. And that his proofs (how many did he send in total¿...) got denied says a lot.
Conclusion: he's probably just another crank.

If you want to add pictures:
You can add attachments, see the left side of the window when posting (there is a table called "Options" with an option "Add an Attachment").
If you only want to put formulas, it's much better (and strongly recommended!) to use $ \text{\LaTeX}$. For this, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 .

All so call geometric "proof" I have seen is just disguised inequality expressed in geometry form that can be easily proved in a few lines algebraically. The geometric construction is added only to mislead the others that it is the core of the proof, while the flaw lies elsewhere. (The most frequently used trick is ignoring the solution $ x \equal{} \minus{}a$ in $ x^2\equal{}a^2$)

Wiles' proof involved proving a case of Taniyama-Shimura conjecture stating that every elliptic curve is modular.  In this case, semistable elliptic curves.  Now I'm doubting that this guy mentioned anything about modular forms, elliptic curves, semistable curves, Galois Representations, Hecke algebra, Selmer groups, Iwasawa Theory, Flach and Kolyvagin,... so he's obviously a liar.

Are you mad? have you forgotten that Fermat Last Theorem is [i]not [/i] just for cubes? for example, what about a2342323 + b2342323 = c2342323? try to draw a geometrical solution for 2342323 dimensions!!!
note: it is just an example.

You don't need to because any attempt (even if you could draw in n-space) would end in failure.

It's very very unlikely that you can encode Fermat's Last Theorem into some relevant property of triangles.
Sorry, but this whole discussion here seems to lead nowhere.

If you have a proof, please post it without ridiculous drawings. You can explain it with words!

Else, please stop saying you have a solution.

Isn´t there a rule about make people wait for more than a week?

I don't believe what you said...Do u really believe that his correct proof(as you said) denied?....... :lol: Please if you know something write it...

[quote="Sergey"][quote]Are you mad? have you forgotten that Fermat Last Theorem is not just for cubes? for example, what about a2342323 + b2342323 = c2342323? try to draw a geometrical solution for 2342323 dimensions!!![/quote]

 :D
(By the way I hadn't found that solution yet but i'm trying not to fall in disappointment  :wink: )

The proof didnt use any "2342323"-dimension... It used only 2-nd dimension to show how the equality $ a^n \plus{} b^n \equal{} c^n$ turns into some another equality, which doesn't have a solution if $ n$ is an odd number and $ a,b,c$ are relatevely prime.. (We can suppose so, otherwise we can divide the equation on their common divisor..)

I remember quite badly but it is something connected with orthocenter of some triangle..   :huh:  Well.. If I could remember.. Have you heard of anything like that idea?[/quote]

All the special points in a triangle can be found out with coordinate geometry, and all the properties are simply equations. Introducing a geometric construction does not help.

This is clearly going nowhere.  If anyone is interested, here is the proof http://math.stanford.edu/~lekheng/flt/wiles.pdf
All 109 pages of it.  Try to find some elementary geometry.  Better, try to find anything you understand.  I think we should just put this thread to rest.

Hey! There's the proof. Can anyone find mistake in it? I really believe you that there is one but I haven't found it yet.. i'm sorry :) So, if you have decided to waste your time on reading this proof, let's begin..

$ a^n \plus{} b^n \equal{} c^n$

We can suppose that numbers a,b,c are pairly comprime. (if numbers a and b have common divisor than с has it either and gcd(a,b,c)>1)

$ \frac {a^n}{c^n} \plus{} \frac {b^n}{c^n} \equal{} 1, n \equal{} 2k \plus{} 1$
Suppose it has solutions a, b, c in positive integers, then exists $ \triangle ABC$ such that
$ AC \equal{} \frac {b^k}{c^k}$, $ BC \equal{} \frac {a^k}{c^k}$ and $ AB \equal{} 1$.

The strategy is to show that in the triangle ABC altitude CD is such that
$ AD \equal{} \frac {b^{2k \plus{} 1}}{c^{2k \plus{} 1}}$, $ BD \equal{} \frac {a^{2k \plus{} 1}}{c^{2k \plus{} 1}}$ and get new quality from law of cosines which will lead to a contradiction.
(see attachment 1.)

$ F$igure $ 1$.
let there be a line $ l$ and points $ A_1, D_1$ lying on $ l$ such that $ A_1D_1 \equal{} \frac {b^{2k \plus{} 1}}{c^{2k \plus{} 1}}$ (see attachment 2) We'll build a line $ e$ perpendicular $ l$ through point $ D_1$.
$ C_1$ is such point on $ e$ that $ A_1C_1 \equal{} \frac {b^k}{c^k}$
Let $ P_1$ be such point on $ A_1C_1$ that $ A_1P_1 \equal{} \frac {b^{k \plus{} 1}}{c^{k \plus{} 1}}$, $ B_1$ is such point that $ B_1$ lies on line $ l$ and $ B_1P_1 \bot A_1C_1$. It is easy to note that $ \triangle P_1A_1B_1 \sim \triangle D_1A_1C_1$.
Hence, $ \frac {A_1B_1}{A_1C_1} \equal{} \frac {A_1P_1}{A_1D_1}$, $ A_1B_1 \equal{} \frac {b^{k \plus{} 1} b^k c^{2k \plus{} 1}}{c^{k \plus{} 1} c^k b^{2k \plus{} 1}} \equal{} 1$

$ F$igure $ 2$.
By analogy, we'll build another configuration (see att.3) where
$ B_2C_2 \equal{} \frac {a^k}{c^k}$, $ B_2D_2 \equal{} \frac {a^{2k \plus{} 1}}{c^{2k \plus{} 1}}$, $ B_2M_2 \equal{} \frac {a^{k \plus{} 1}}{c^{k \plus{} 1}}$. From the similarity of triangles $ \triangle A_2M_2B_2 \sim \triangle C_2D_2B_2$ follows equality $ A_2B_2 \equal{} 1$.

From figures 1 and 2 it follows that this constructions coincide ($ AB \equal{} 1$)... and the claim is prooved.

From the cosines law it follows:
$ BC^2 \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AC\cdot \cos \angle CAB) \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AD)$.
Finally it follows that
$ AD \equal{} \frac {AB^2 \plus{} AC^2 \minus{} BC^2}{2AB}$
$ \frac {b^{2k \plus{} 1}}{c^{2k \plus{} 1}} \equal{} \frac {1 \plus{} \frac {b^{2k}}{c^{2k}} \minus{} \frac {a^{2k}}{c^{2k}}}{2}$
$ b^{2k \plus{} 1} \equal{} c\cdot \frac {c^{2k} \plus{} b^{2k} \minus{} a^{2k}}{2}$
$ b^{2k \plus{} 1} \equal{} c\cdot M$ , it is easy to note that M is a positive integer.

There is a contradiction since $ b$ and $ c$ are comprime positive integer numbers. So the equation doesn't have any solution if $ n$ is an odd number. The statement for even $ n$ follows from the inequality
$ a^4 \plus{} b^4 \neq c^4$ which can be prooved in an easy way.

So, that's all!

[quote="Sergey"]

$ \frac {a^n}{c^n} \plus{} \frac {b^n}{c^n} \equal{} 1, n \equal{} 2k \plus{} 1$
Suppose it has solutions a, b, c in positive integers, then exists $ \triangle ABC$ such that
$ AC \equal{} \frac {b^k}{c^k}$, $ BC \equal{} \frac {a^k}{c^k}$ and $ AB \equal{} 1$.

[/quote]

Well, firstly I'm not convinced that you can draw a triangle (at least one with any area) with side lengths like that. (I haven't finished reading the 'proof', so I'll inform you if that matters)

[quote]

From the cosines law it follows:
$ BC^2 \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AC\cdot \cos \angle CAB) \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AD)$.
Finally it follows that
$ AD \equal{} \frac {AB^2 \plus{} AC^2 \minus{} BC^2}{2AB}$

[\quote]

Where did the cosine go?

[quote="Sergey"]
$ \frac {a^n}{c^n} \plus{} \frac {b^n}{c^n} \equal{} 1, n \equal{} 2k \plus{} 1$
Suppose it has solutions a, b, c in positive integers, then exists $ \triangle ABC$ such that
$ AC \equal{} \frac {b^k}{c^k}$, $ BC \equal{} \frac {a^k}{c^k}$ and $ AB \equal{} 1$. [/quote]

You say there is a triangle of sides $ \frac{b^k}{c^k}$, $ \frac{a^k}{c^k}$ and 1, that is, $ a^k$, $ b^k$ and $ c^k$ are the sides of a triangle. How do you ensure that? I think you cannot take that for granted... :maybe:

:lol:  We can construct triangle whith:
$ (\frac {a}{c})^{k},(\frac {b}{c})^{k},1$
because :
if
$ (\frac {a}{c})^{2k \plus{} 1} \plus{} (\frac {b}{c})^{2k \plus{} 1} \equal{} 1$
then :
$ \frac {a}{c} < 1$
$ \frac {b}{c} < 1$
hence :
$ (\frac {a}{c})^{k} > (\frac {a}{c})^{2k \plus{} 1}$
$ (\frac {b}{c})^{k} > (\frac {b}{c})^{2k \plus{} 1}$
hence
$ (\frac {a}{c})^{k} \plus{} (\frac {b}{c})^{k} >(\frac {a}{c})^{2k \plus{} 1} \plus{} (\frac {b}{c})^{2k \plus{} 1} \equal{} 1$

hence we can construct triangle because :

$ (\frac {a}{c})^{k} \plus{} (\frac {b}{c})^{k} > 1$
and

mistake is here :
"
$ BC^2 \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AC\cdot \cos \angle CAB) \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AD)$
or you mean that : $ \angle CAB \equal{} 0$ it's false.
"

[quote="Extremal"]:lol:  We can construct triangle whith:
$ (\frac {a}{c})^{k},(\frac {b}{c})^{k},1$
because :
if
$ (\frac {a}{c})^{2k \plus{} 1} \plus{} (\frac {b}{c})^{2k \plus{} 1} \equal{} 1$
then :
$ \frac {a}{c} < 1$
$ \frac {b}{c} < 1$
hence :
$ (\frac {a}{c})^{k} > (\frac {a}{c})^{2k \plus{} 1}$
$ (\frac {b}{c})^{k} > (\frac {b}{c})^{2k \plus{} 1}$
hence
$ (\frac {a}{c})^{k} \plus{} (\frac {b}{c})^{k} > (\frac {a}{c})^{2k \plus{} 1} \plus{} (\frac {b}{c})^{2k \plus{} 1} \equal{} 1$

hence we can construct triangle because :

$ (\frac {a}{c})^{k} \plus{} (\frac {b}{c})^{k} > 1$
and

mistake is here :
"
$ BC^2 \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AC\cdot \cos \angle CAB) \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AD)$
or you mean that : $ \angle CAB \equal{} 0$ it's false.
"[/quote]

$ AC\cos \angle CAB\equal{}AD$ :P

I was thinking that
$ BC^2 \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AC\cdot \cos \angle CAB) \equal{} AB^2 \plus{} AC^2 \minus{} 2(AB\cdot AC)$
 :blush:
ok.
mistake is here
$ AD\equal{}(\frac{b}{c})^{2k\plus{}1}$ your explanation is not enough. or it's same that you suppose $ c\equal{}0$ but it's false.
really
if $ AD\equal{}(\frac{b}{c})^{2k\plus{}1}$
then
$ \cos \angle CAB\equal{}(\frac{b}{c})^{k\plus{}1}$
hence by cosinus theorem for angle  $ \angle CAB$:  we have :
$ (\frac{a}{c})^{2k}\equal{}(\frac{b}{c})^{2k}\plus{}1\minus{}2(\frac{b}{c})^{k}\cos \angle CAB$
if substitute :$ \cos \angle CAB\equal{}
(\frac{b}{c})^{2k\plus{}1}$ then after direct calculation we have :
$ (a^{2k\plus{}1}\plus{}b^{2k\plus{}1})^{\frac{1}{2k\plus{}1}}\equal{}\frac{3b^{2k\plus{}1}\plus{}a^{2k\plus{}1}}{b^{2k}\minus{}a^{2k}}$
similary for $ a$
$ (a^{2k\plus{}1}\plus{}b^{2k\plus{}1})^{\frac{1}{2k\plus{}1}}\equal{}\frac{3a^{2k\plus{}1}\plus{}b^{2k\plus{}1}}{\minus{}b^{2k}\plus{}a^{2k}}$
or it's same that $ a^{2k\plus{}1}\plus{}b^{2k\plus{}1}\equal{}0$ contradiction to the hypothesis of lemma

[quote="Sergey"][quote="Extremal"]
if substitute :$ \cos \angle CAB \equal{} (\frac {b}{c})^{2k \plus{} 1}$ then after direct calculation we have :
$ (a^{2k \plus{} 1} \plus{} b^{2k \plus{} 1})^{\frac {1}{2k \plus{} 1}} \equal{} \frac {3b^{2k \plus{} 1} \plus{} a^{2k \plus{} 1}}{b^{2k} \minus{} a^{2k}}$
similary for $ a$
$ (a^{2k \plus{} 1} \plus{} b^{2k \plus{} 1})^{\frac {1}{2k \plus{} 1}} \equal{} \frac {3a^{2k \plus{} 1} \plus{} b^{2k \plus{} 1}}{ \minus{} b^{2k} \plus{} a^{2k}}$
or it's same that $ a^{2k \plus{} 1} \plus{} b^{2k \plus{} 1} \equal{} 0$ contradiction to the hypothesis of lemma[/quote]

Oh, sorry, what do you mean under the "direct calculation"?
I see only (you have occasionally written .."if substitute $ \cos \angle CAB \equal{} (\frac {b}{c})^{2k \plus{} 1}$", but you have already noted that $ \cos \angle CAB \equal{} (\frac {b}{c})^{k \plus{} 1}$ :) maybe it is a small calculation error?  Nevertheless, I don't understand where have you taken that equality..  :maybe:[/quote]
 :blush:  I simply have not correctly written but have correctly substituted
yes you are right. I substitute: $ \cos... \equal{} (\frac {b}{c})^{k \plus{} 1}$
Direct calculation: here I also use the fact that : $ (\frac {b}{c})^{2k \plus{} 1} \plus{} (\frac {a}{c})^{2k \plus{} 1} \equal{} 1$

I see from the law of cosines and substitution if follows:

$ a^{2k}\equal{}b^{2k}\plus{}c^{2k}\minus{}2 \frac{b^{2k\plus{}1}}{c}$

Could you explain me bit more delailed?
Можете по-русски :)

The argument is indeed flawed. I'm sorry if you and your classmates were fooled, Sergey, but this guy is a total fraud (either that or he's [i]very[/i] good at deluding himself). Fermat's last theorem probably has the distinction of having "generated more incorrect proofs than any other problem in the history of mathematics." In future be very skeptical of anyone claiming to have an elementary proof of Fermat's Last Theorem. Over the past 350 years, thousands of people have claimed this. Every single one of them has been wrong (including probably Fermat).

That being said, nobody in this thread has come close to catching the flaw.

All of the specific calculations and constructions he mentions are valid. The flaw is pretty subtle.

So where is a flaw likely to be? In the part that he makes needlessly complicated of course!

It's very easy to find $ AD$ and $ BD$ directly in any arbitrary triangle. From the Pythagorean theorem we get:
$ (AC)^2 \minus{} (BC)^2 \equal{} [(CD)^2 \plus{} (AD)^2] \minus{} [(CD)^2 \plus{} (BD)^2] \equal{} (AD)^2 \minus{} (BD)^2 \equal{} (AD \plus{} BD)(AD \minus{} BD) \equal{} (AB)(AD \minus{} BD)\Rightarrow AD \minus{} BD \equal{} \frac {(AC)^2 \minus{} (BD)^2}{AB}$
And also:
$ AD \plus{} BD \equal{} AB$.
It's now easy to solve this for $ AD$ and $ BD$ (and notice that you do [b]not[/b] get the values he claimed).

So why did he do this complicated thing with constructing these figures? Because it lets him distract you with a bunch of specific details (all of which you can verify are valid), and so he can sneak the completely bogus, subtle part past you.

So, now that I've said all that stuff (that stuff was basically my though process, how you could tell in advance that there was going to be a flaw, and how you could find it), I'll finally tell you what the flaw is. It's this statement:

[quote]
From figures 1 and 2 it follows that this constructions coincide ($ AB \equal{} 1$)... and the claim is prooved.
[/quote]

You've made these two complicated diagrams, and now you want to overlay them. To do that, you need to make sure [i]all[/i] the corresponding points do indeed coincide.

So let's think about that:

We'll place the diagrams so that $ A_1$ and $ A_2$ coincide (at the point $ A$), and the [i]lines[/i] $ A_1B_1$ and $ A_2B_2$ coincide. Now from what we've proved, and the fact that
$ \frac {a^{2k \plus{} 1}}{c^{2k \plus{} 1}} \plus{} \frac {b^{2k \plus{} 1}}{c^{2k \plus{} 1}} \equal{} \frac {a^{2k \plus{} 1} \plus{} b^{2k \plus{} 1}}{c^{2k \plus{} 1}} \equal{} 1$
we easily get $ A_1D_1 \equal{} A_2D_2 \equal{} \frac {b^{2k \plus{} 1}}{c^{2k \plus{} 1}} \equal{} 1 \minus{} \frac {a^{2k \plus{} 1}}{c^{2k \plus{} 1}}$, so $ D_1$ and $ D_2$ coincide at the point $ D$. Also since $ A_1B_1 \equal{} A_2B_2 \equal{} 1$, $ B_1$ and $ B_2$ coincide (at $ B$). Now also obviously the perpendiculars to $ AB$ through $ D_1$ and $ D_2$ coincide. So far so good.

And now finally, the points $ C_1$ and $ C_2$ coincide at $ C$, because... and that's were the whole argument falls apart. Yes we know that $ C_1$ and $ C_2$ are both on the perpendicular to $ AB$ through $ D$. But that's all we know. There's no reason to think that they are actually the same point. To prove that, we would need to calculate $ C_1D_1$ and $ C_2D_2$ (and if you were to do this you would see that they are [b]not[/b] equal). So we have not actually constructed a triangle like we wanted to (see the attachment). So we can't prove anything with the law of cosines, and the entire proof is bogus.

Sorry about that, but you were tricked. Maybe now you will be a bit more skeptical next time you run into something like this. :)

[quote="Sergey"]
Можете по-русски :)[/quote]
вообще говоря противоречие в том что он в конце получил. Насчет моей последней коментарии я не уверен я быстра посчитал и написал хотя может быть и не прав я (например ошибся в расчетах (эта я часто делаю  :) )). Но это не важно. Мне не зачем искать противоречие если оно уже найдено самим автором этой задачи. Вопрос в другом: где он допустил свою ошибку когда получил противоречие. Ну из общих соображении понятно что он допустил противоречие там где объяснял тот факт что $ AD\equal{}....$ честно говоря я не вижу смысла подробно искать ошибку его (хотя я еще полностью не прочел его доказательство)
Но почему именно там он допустил ошибку ? когда доказывал что $ AD\equal{}....$ ? ну например потому что :
в данном случае мы знаем все стороны треугольника значит можем найти "все" найдем высоту  с основанием $ D$ напишем формулу герона и с другои стороны напишем формулу для вычисления площади треугольника : основание на высоту. Приравнив эти две формулы и    так найдем высоту. и дальше легко найдем  $ AD$ так как мы уже находимся в прямоугольном треугольнике. но заметь автор этой задачи так не поступает он из геометрических построении находит $ AD$  и получает противоречие, естественно он нас обманул и легко можно найти точно тот момент где! .

:rotfl: Yes, I understand! Thank you a lot!  :) It is sometimes useful to be skeptic.. But I think that his aim was not to trick anybody but he really thought he would find a short proof.. I feel sorry for him..

I'm not sure how old you are but if this guy really had a proof, do you think he would be telling it to schoolchildren?  No, he would have it published by some major journal, and give talks to mathematicians all over the world.

Wait there are no solutions to $ x\plus{}y\equal{}z$ (k=0)??!!
wow!  :|  :rotfl:

What?  $ x\plus{}y\equal{}z$?  There are infinitely many.  The theorem holds for $ x^{n}\plus{}y^{n}\equal{}z^{n}$ where $ n>2$.

but his proof for odd exponents doesn't exclude n=1

But his proof is wrong.  That's the whole point.  Of course $ x\plus{}y\equal{}z$ has solutions, infinitely many.

The placement itself is not that bad, the only question is if we want such a discussion.
As it wasn't trolling, I saw no reason to intervene. But I would have no objection if you move it (or do something else).


logarithms
function
number theory unsolved
number theory

Is it true that for each n >= 2, there are at least n primes in the first n^2 natural number?

[quote="ZetaX"]This says that there are more primes than squares, which is true.
It for example follows directly from Chebychev's Theorem.[/quote]
How???
From Chebyshev's theorem follows that there are at least n primes in  the first 2^n natural numbers.

The one I know is (copied from the Formulary):

[b]Chebychevs Theorem:[/b]
There are constants $a,b$, $0 < a < b$ such that for all big $n$ (e.g. $a=\log(2), b=\log(4)$ for $n\geq 2$) we have \[ a \cdot n \leq \pi(n) \cdot \log(n) \leq b \cdot n  \]

Using this, we get $\pi(n^2) \geq \log(2) \frac{n^2}{\log(n^2)}$, so we just need that $\frac{\log(2)}{2} \frac{n^2}{\log(n)} \geq n$, which is true (for $n\geq 8$).

OK, I understand you, you used another theorem.

$\pi(x)$ is asymptotic to $\frac{x}{\ln{x}}$, but the integer function of $\sqrt{x}$ is asymptotic to $\frac{x}{\sqrt{x}}$, and $\sqrt{x}>\ln{x}$.


[color=darkblue]Slove this equation: $ 4cosx.sina \plus{} 2sinx.cosa \minus{} 3cosa \equal{} 2\sqrt {7}$.[/color]

The dots mean multiply, right?

Instead, you could have written it as:

$ 4\cos x\cdot \sin a\plus{}2\sin x\cdot\cos a\minus{}3\cos a\equal{}2\sqrt{7}$

Use \cos x and \sin x instead of cos x and sin x and use \cdot instead of .

im not really sure this is solvable...

Note that

$ 2\cos x\cdot \sin a\equal{}\sin(a\plus{}x)\plus{}\sin(a\minus{}x)$

and

$ 2\sin x \cdot \cos a\equal{}\sin(a\plus{}x)\minus{}\sin(a\minus{}x)$

$ [(4\cos x)(\sin a)\plus{}(2\sin x\minus{}3)(\cos a)]^2\le [ (4\cos x)^2\plus{}(2\sin x\minus{}3)^2]\cdot [ \sin^2 a\plus{}\cos^2 a]\equal{}[16(1\minus{}u^2)\plus{}(2u\minus{}3)^2]\le 28$

where $ u\equal{}\sin x$

and the last inequality follows from $ [16(1\minus{}u^2)\plus{}(2u\minus{}3)^2]\le 28\iff 3(1\plus{}2u)^2\ge 0$

So $ u\equal{}\frac{1}{2}$ and we use the fact that for equality in cauchy, $ [4\cos x, 2\sin x\minus{}3]$ and $ [\sin a,\cos a]$ are parallel.

[quote="Altheman"]$ [(4\cos x)(\sin a) \plus{} (2\sin x \minus{} 3)(\cos a)]^2\le [ (4\cos x)^2 \plus{} (2\sin x \minus{} 3)^2]\cdot [ \sin^2 a \plus{} \cos^2 a] \equal{} [16(1 \minus{} u^2) \plus{} (2u \minus{} 3)^2]\le 28$

where $ u \equal{} \sin x$

and the last inequality follows from $ [16(1 \minus{} u^2) \plus{} (2u \minus{} 3)^2]\le 28\iff 3(1 \plus{} 2u)^2\ge 0$

So $ u \equal{} \frac {1}{2}$ and we use the fact that for equality in cauchy, $ [4\cos x, 2\sin x \minus{} 3]$ and $ [\sin a,\cos a]$ are parallel.[/quote]
Thank Altheman!  :lol:


How do I mark a question that I've asked or answered as SOLVED?  Are we supposed to this or is it a moderator's function?

You can ask a moderator to change the subject description to SOLVED.


32. Given the equation 2x+3y=4 answer the following questions.
a. Is the slope positive or negative?
b. As x increases in value does y increase or decrease?
c. If x decreases by 2 units what is the corresponding change in y?

I rewrite the problem as y=-2/3x + 4/3
a. Negative
b. Decreases
c. y decreases by 4/3
Is that correct or have I made a big error?

All are correct except the last one. Remember x DECREASES by 2 units.

Ah man iwas not paying close attention. I kept going to 0 instead of -4. LOL.


integration
calculus
algebra
partial fractions
calculus computations

$ \int\frac{x^2}{1\plus{}x^5}dx$

Is the appropriate place to post such problems?  :maybe:

Yes, it is appropriate.

[color=green]EDIT: Ooops, I thought this was posted in Calculations and Tutorials. That would be right place for it ;)[/color]

The solution of your integral is

$ \frac {1}{20} ( - 2 \sqrt {2 (5 + \sqrt {5})} \text{arc tan}[\frac {1 + \sqrt {5} - 4 x}{\sqrt {10 - 2 \sqrt {5}}}] - 2 \sqrt {10 - 2 \sqrt {5}} \text{arc tan}[\frac { - 1 + \sqrt {5} + 4 x}{\sqrt {2 (5 + \sqrt {5})}}] +$
$ + 4 \text{log}[1 + x] - (1 + \sqrt {5} ) \text{ log}[1 + \frac {1}{2} ( - 1 + \sqrt {5}) x + x^2] + ( - 1 + \sqrt {5}) \text{ log}[1 - \frac {1}{2} (1 + \sqrt {5}) x + x^2])$

It's pretty tedious calculation (as you can see form the result), based on partial fractions (not so easy in this case). Is this a random integral, or you found it somewhere?

Answer like this???? :o  :o
I don't think the answer is correct. Can you please tell me how did you do that? :|

Actually, I've used Mathematica.

Its integrating engine is available online (see the link under, if you still don't believe  :lol: ):

http://integrals.wolfram.com/index.jsp?expr=x^2%2F(1%2Bx^5)&random=false

Copy-paste it in the adress bar, for some strange reason, it breaks when I post it ;)

OK. But have you faced such problems before?

If it comes in some Olympiad what to do?

Well, such problems are typical for the undergraduate (and high school) single variable calculus exams (although, the problems appearing in the exams are often made in such fashion that the denominator can be nicely factored, so the integral can be transformed in partial fractions fairly easily).

High school olympiads do not include calculus, so you'll never face something like this there.
And as far as college competitions are concerned, if an integral shows up, it wouldn't be something like this: it would be a problem with "a catch", with an exceptional idea to be found. (In this problem, there's no idea, just hard, hard work).

$ \int \frac {x^2}{1 \plus{} x^5}\ dx$

A possible way to solve it.
Note that  $ \int \frac {x^2 \plus{} x}{1 \plus{} x^5}\ dx \equal{} \int \frac {x}{1 \minus{} x \plus{} x^2 \minus{} x^3 \plus{} x^4}\ dx$
$ \equal{} \int \frac {4x}{(2x^2 \plus{} ( \minus{} 1 \plus{} \sqrt {5})x \plus{} 2)(2x^2 \minus{} (1 \plus{} \sqrt {5})x \plus{} 2)}\ dx$
$ \equal{} \arctan(\cdots) \plus{} \arctan(\cdots)$

and
$ \int \frac {x^2 \minus{} x}{1 \plus{} x^5}\ dx \equal{} \int \frac {a}{1 \plus{} x} \plus{} \frac {bx\plus{}c}{2x^2 \plus{} ( \minus{} 1 \plus{} \sqrt {5})x \plus{} 2} \plus{} \frac {dx\plus{}e}{2x^2 \minus{} (1 \plus{} \sqrt {5})x \plus{} 2}\ dx$
which results in a sum of logarithms.


The 2nd International (14th National) Scientific Chemistry Olympiad [b](for 3rd and 4th year university students)[/b] will be in Iran on July 15 -17, 2009.

Schedule:
Wednesday July 15 - Physical Chemistry, Inorganic Chemistry
Thursday July 16 - Organic Chemistry, Analytical Chemistry

http://olympiad.sanjesh.org/en/ch_calendar.html

Problems from the last year competition can be found here: http://olympiad.sanjesh.org/en/1203/1st-ch.PDF

[quote="OVB"]lol is the US ever gonna participate in this?[/quote]

At least in IPhO-2007 (Isfahan, Iran) they did participate: [url=http://www.ipho2007.ir/Results/results.asp]2 gold and 3 silver medals[/url]. :wink:

The problems are available online now: http://olympiad.sanjesh.org/en/1203/ch_qu.html


topology
advanced fields
advanced fields unsolved

Let M be a metric space with the discret metric. Show that M is compact $ \leftrightarrow$ M is finite.

You mean with the discrete metric?  Every subset is open, so the set of singletons is an open cover of $ M$.


I have this problem, that my teacher gave me... and im stuck. Any help would be much appreciated. Thanks (Could working please be shown as well)

find the sum of the first n terms of the series

1*3*5 + 3*5*7 + 5*7*9 + ... + (2n-1)(2n+1)(2n+3)

A brute-force solution is to expand $(2n - 1)(2n + 1)(2n + 3)$ into $8n^3 + 12n^2 -2n - 3$ and sum the individual terms for that, which isn't too hard since there are pretty well-known formulae for all the parts.  I'm not sure if there's a more "elegant" solution.

You can also telescope $(2n-3)(2n-1)(2n+1) + (2n - 1)(2n + 1)(2n + 3) = 16n^3 - 4n$, which simplifies the summation a little bit.

[hide]
You could put it in summation notation, expand, then factor it out or whatever you call that.
[/hide][hide="If you need more"]
$1*3*5+3*5*7+...+(2n-1)(2n+1)(2n+3)=$
$\displaystyle\sum_{x=1}^{n}(2x-1)(2x+1)(2x+3)=$
$\displaystyle\sum_{x=1}^{n}(8x^3+12x^2-2x-3)=$ check the expansion, i'm really tired
${8\displaystyle\sum_{x=1}^{n}x^3}+{12\displaystyle\sum_{x=1}^{n}x^2}-{2\displaystyle\sum_{x=1}^{n}x}-{\displaystyle\sum_{x=1}^{n}3}=$
${8(\frac{n(n+1)}{2})^2}+{12\frac{n(n+1)(2n+1)}{6}}-{2\frac{n(n+1)}{2}}-3n$
I'm falling asleep and don't feel like doing that. Just expand and simplify.[/hide]
Could someone please check my work.

Wow, I didn't see zanttrang's answer. So that means I spent about 20 minutes all writing that in LaTeX (I'm new so I'm not very efficient). Ugh. :mad:

[quote="sen"][quote]${8\displaystyle\sum_{x=1}^{n}x^3}+{12\displaystyle\sum_{x=1}^{n}x^2}-{2\displaystyle\sum_{x=1}^{n}x}-{\displaystyle\sum_{x=1}^{n}3}= {8(\frac{n(n+1)}{2})^2}+{12\frac{n(n+1)(2n+1)}{6}}-{2\frac{n(n+1)}{2}}-3n$[/quote]

i expanded it and got:

$2n^4+8n^3+7n^2-2n$[/quote]

That's what I got. I guess that's your answer Captain Cool.


analytic geometry
geometry
rotation
perpendicular bisector

The quadrilateral ABCD, where A is (4,5) and C is (3,-2), is a square. Find the coordinates of B and D and the area of the square.

I know B and D will be on the perpendicular bisector of AC and by inspection I can see where B and D should be B (7,1) and D (0,2) what other ways are there of finding B and D?

the midpoint of $AC$ is $(\frac{7}{2},\frac{3}{2})$. From one of my results, the equation of the line which is perpendicular on the line $ax+by+c=0$ at the point $(p,q)$ is given by
\[-bx+ay+a(p-q)+b(p+q)+c=0\] the rest is for you :wink:

Did you use [hide="this method?"]We find the midpoint of AC, call it D, then make a coordinate relationship with CD and AD, then just rotate CD and AD to see that you are indeed right with those coordinates.

To find the area of the square, we just need to find the length of AC, then divide it by the square root of 2, then square the result. Namely, we are trying to find

$\frac{AC^{2}}{2}=50/2=25$[/hide]

[quote="I"]the midpoint of $AC$ is $(\frac{7}{2},\frac{3}{2})$. From one of my results, the equation of the line which is perpendicular on the line $ax+by+c=0$ at the point $(p,q)$ is given by
\[-bx+ay+a(p-q)+b(p+q)+c=0 \]
the rest is for you :wink:[/quote]

well we need not do all that. let $O(\frac{7}{2},\frac{3}{2})$ be the midpoint of $AC$. we have $OA=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{7}{2}\right)^{2}}$. Let $D=(d_{1},d_{2})$ and $B=(b_{1},b_{2})$.
$OA=OD\Longrightarrow \left(\frac{1}{2}\right)^{2}+\left(\frac{7}{2}\right)^{2}=\left(d_{1}-\frac{7}{2}\right)^{2}+\left(d_{2}-\frac{3}{2}\right)^{2}$ after a little calculation and case check we get $d_{1}=0,d_{2}=2$. in the same way we find $b_{1}=7,b_{2}=1$ :D

sorry for the interruption. it was due to the power cut for a thunder-storm. anyways..

from the equation $\left(\frac{1}{2}\right)^{2}+\left(\frac{7}{2}\right)^{2}=\left(d_{1}-\frac{7}{2}\right)^{2}+\left(d_{2}-\frac{3}{2}\right)^{2}$ we get four cases for $d_{1}$:

[b]case 1:[/b]$d_{1}-\frac{7}{2}=\frac{1}{2}$
then $d_{1}=4$ which is not possible. because this implies $AD$ is parallel to the $y$-axis.

[b]case 2:[/b]$d_{1}-\frac{7}{2}=-\frac{1}{2}$
if so then $d_{1}=3$ another contradiction since then $CD$ is parallel to the $y$-axis.

[b]case 3:[/b]$d_{1}-\frac{7}{2}=\frac{7}{2}$
we get $d_{1}=7$ it is possible, since we can switch $B$ and $D$. it means one of them has the coordinate $7$. we can let any of them to have it.

[b]case 4:[/b]$d_{1}-\frac{7}{2}=-\frac{7}{2}$
hence $d_{1}=0$ this is another possibility. if we let $B$ have the coordinate $7$ then $D$ must have coordinate $0$.

another case check for $d_{2}$ establishes our answer.

if its not clear to you yet, let me know. i'll try my best to make it clear. always there to help :D


i have a bag of nickles and dimes. there are 3 x as many dimes as nickles. there are a total of 36.05 is there any fast ways to solve this?

[quote="cgyao15"]For every 3 dimes there's a nickel.
So 3 dimes=30 cents
1 nickel=5 cents.
30+5=35
36.05/.35
Sure you can see what to do now.  :P[/quote]
i knew how to do it i just needed a faster solution thanks(':D')


conics
hyperbola
quadratics
analytic geometry
calculus
ellipse
Asymptote

From the point $M(m,n)$, we draw two tangent line of hyperbole $(H):{x^{2}\over a^{2}}-{y^{2}\over b^{2}}=1$, meet $(H)$ at $P,Q$. Determine equation of circle $(MPQ)$.

This is as far as I've gotten. [hide] Let $y=sx+n-ms$ be the tangent lines, with $s$ as the slope. Substituting into the equation of the hyperbola, we get the quadratic $\left(\frac{1}{a^{2}}-\frac{s^{2}}{b^{2}}\right)x^{2}+\left(\frac{2ms^{2}-2ns}{b^{2}}\right)x+\left(\frac{2mns-n^{2}-m^{2}s^{2}-b^{2}}{b^{2}}\right)=0$. We need the determinate equal to 0, to be tangent. The x-coordinate of one of the points of tangency is $\frac{a^{2}ns-a^{2}ms^{2}}{b^{2}-a^{2}s^{2}}$.

From $y=sx+n-ms$, we have $x=\frac{y-n+ms}{s}$. Substituting, we get the quadratic $\left(\frac{1}{a^{2}s^{2}}-\frac{1}{b^{2}}\right)y^{2}+\left(\frac{2ms-2n}{a^{2}s^{2}}\right)y+\left(\frac{n^{2}+m^{2}s^{2}-2mns-a^{2}s^{2}}{a^{2}s^{2}}\right)=0$. Again, the determinate must be 0. And the y-coordinate is $\frac{b^{2}n-b^{2}ms}{b^{2}-a^{2}s^{2}}$.

I'm still not sure how to get two lines of tangency.  :maybe: [/hide]

Lines of tangency with curves is calculus (I am not going to pretend that I know how to do it)...

If I am not mistaken, I remember you said you will be taking it next year.

You can use some alternate definitions and/or common properties of an ellipse to find lines of tangency without actually using calculus, if I am not mistaken.  :)

I don't think there are any tangent lines when the point $(m,n)$ is between the asymptotes.

I'm confused about what a tangent line for a hyperbola is.

If we draw a line parallel to an asymptote, it intersects the hyperbola in one point.  Is that line then tangent to the hyperbola?

I think when a line is tangent, it touches the curve but does not cross it.

If crossing the line only one qualified as being tangent, then x=0 would be tangent to y=x^2 at (0,0) and I think y=0 is the line that is tangent to y=x^2 at (0,0)

But my signature tells the truth

[quote="ehehheehee"]I think when a line is tangent, it touches the curve but does not cross it.[/quote]

This is true for conic sections, but not true more generally.  A curve may have a tangent which crosses it at some entirely different point, like many tangents to a sine curve, but also may have a tangent which crosses it at the point of tangency.  Consider the curve $f(x) = x^{3}$ -- it has the tangent line $y = 0$ at the point $(0, 0)$, and this line also crosses the curve there.

[quote="matt276eagles"]If we draw a line parallel to an asymptote, it intersects the hyperbola in one point.  Is that line then tangent to the hyperbola?[/quote]No.


i'm having problems solving this i don't remember how to do it but i'd like someoen to help me figure out how to solve it... it's probably really simple but... yeah lol 14 is less than 5-3f which is greater than or equal to 53... i feel really stupid

edit: nvm i understnad now but thanks to thsoe for looking and i guess possibly tryin to help

[quote="Eros88"]:?: I have no idea what are you trying to say  :([/quote]

Yes, can you be more clear?

The post was never edited...

Is this just trolling?


"Nice" Angles and Polygons:
Definition 1: We call an angle A "nice" if both sin(A) and cos(A) are rational.
Definition 2: We call a convex polygon "nice" if all of its interior angles are "nice."

III. A convex quadrilateral has the following properties:
1. Its sides are integers whose product is a square.
2. It can be inscribed in a circle, and can be circumscribed about (another) circle.

Prove that this quadrilateral is "nice."

Such a quadrilateral is called a [url=http://mathworld.wolfram.com/BicentricQuadrilateral.html]bicentric quadrilateral[/url] and has the property that $ K \equal{} \sqrt {abcd}$.


Let $ a<b<c$ be real numbers such that $ a\plus{}b\plus{}c\equal{}6$ and $ ab\plus{}bc\plus{}ca\equal{}9$. Prove that $ 0<a<1<b<3<c<4$.

Let $ abc \equal{} k$, $ a, b\ c$ are the roots of the cubic equation $ x^3 \minus{} 6x^2 \plus{} 9x \minus{} k \equal{} 0\Longleftrightarrow x(x \minus{} 3)^2 \equal{} k$.
Let $ f(x) \equal{} x(x \minus{} 3)^2$, Let move the line $ y \equal{} k$ in parallel to the $ x$ axis, in such a way that $ y \equal{} f(x)$ has 3 intersection points with $ y \equal{} k$,
the condition is $ 0<k<4$. Thus the attached graph gives the proof.

wouldn't it be better to find an expression for $ abc$?

the best i have so far is
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} 2abc(a \plus{} b \plus{} c) \equal{} 81$

from $ (ab \plus{} bc \plus{} ca)^2$

It's obvious that $a<2<c$. From the conditions, we have:
$$(b-3)^2=ac \Rightarrow c>b>a>0$$
and
$$\begin{cases} 3(a-1)(a-3)=(a-b)(a-c)>0 \Rightarrow a<1\\3(b-1)(b-3)=(b-a)(b-c)<0\Rightarrow 1<b<3\\3(c-1)(c-3)=(c-a)(c-b)>0\Rightarrow 3<c\end{cases}$$
Moreover, $(c-1)(c-4)=(a-1)(b-1)<0 \Rightarrow c<4$. Therefore, $0<a<1<b<3<c<4$.


inequalities
trigonometry
inequalities unsolved

Given $ \gamma \plus{} \beta \ge x \plus{} \alpha \ge \delta \plus{} y$,
[color=red][b]Prove or Disprove:[/b][/color]

$ \sin{\frac {x \plus{} y}{2}}\sin{\frac {\alpha \plus{} \beta}{2}}\sin{\frac {\gamma \plus{} \delta}{2}} \ge \sin{\frac {\gamma \plus{} \beta}{2}}\sin{\frac {x \plus{} \alpha}{2}}\sin{\frac {\delta \plus{} y}{2}}$.

[$ \alpha, \beta,\gamma,\delta,x,y \in (0, \frac {\pi}{2})$]

Try alpha = beta = pi/4 and x = y = gamma = delta = 0,01.

It is useful to know how to refute conjectures. But maybe you haven't tried hard enough because you would like your conjecture to be true. In this case, you should [url=http://terrytao.wordpress.com/career-advice/be-sceptical-of-your-own-work/]be more sceptical of your own work[/url].


trigonometry
function
algebra unsolved
algebra

You already have a recurrent formula for the sequence!

[quote="caothujjj"]Find the recurrence formula of sequence {a\_{n}}. a(0)=2; a(n+1) = 2. a(n)^2 - 1[/quote]

I think you are looking for a "closed expression" not a recurrence formula.

It's easy to see that $ a_n>1$ $ \forall n$
Let then $ a_n\equal{}\cosh(u_n)$.

We have $ u_{n\plus{}1}\equal{}2u_n$ and so $ a_n\equal{}\cosh(\alpha2^n)$

With $ a_0\equal{}2$, we have $ \cosh(\alpha)\equal{}2$ and so $ \alpha\equal{}\ln(2\plus{}\sqrt{3})$

And we have $ a_n\equal{}\cosh(\ln(2\plus{}\sqrt{3})2^n)$ $ \equal{}\frac{1}{2}((2\plus{}\sqrt{3})^{2^n}\plus{}(2\minus{}\sqrt{3})^{2^n})$

[quote="pco"][quote="caothujjj"]Find the recurrence formula of sequence {a\_{n}}. a(0)=2; a(n+1) = 2. a(n)^2 - 1[/quote]

I think you are looking for a "closed expression" not a recurrence formula.

It's easy to see that $ a_n > 1$ $ \forall n$
Let then $ a_n \equal{} \cosh(u_n)$.

We have $ u_{n \plus{} 1} \equal{} 2u_n$ and so $ a_n \equal{} \cosh(\alpha2^n)$

With $ a_0 \equal{} 2$, we have $ \cosh(\alpha) \equal{} 2$ and so $ \alpha \equal{} \ln(2 \plus{} \sqrt {3})$

And we have $ a_n \equal{} \cosh(\ln(2 \plus{} \sqrt {3})2^n)$ $ \equal{} \frac {1}{2}((2 \plus{} \sqrt {3})^{2^n} \plus{} (2 \minus{} \sqrt {3})^{2^n})$[/quote]

           what is "cosh"

[quote="caothujjj"]  what is "cosh"[/quote]

$ \cosh(x)\equal{}\frac{e^x\plus{}e^{\minus{}x}}{2}$, hyperbolic cosine (see [url]http://en.wikipedia.org/wiki/Hyperbolic_function[/url])

You have $ \cosh(2x)\equal{}2\cosh(x)^2\minus{}1$

If you don't know $ \cosh(x)$, you may solve the problem using $ a_n\equal{}\frac{1}{2}(v_n\plus{}\frac{1}{v_n})$ and you obtain the same result.

[quote="pco"][quote="caothujjj"]  what is "cosh"[/quote]

$ \cosh(x) \equal{} \frac {e^x \plus{} e^{ \minus{} x}}{2}$, hyperbolic cosine (see [url]http://en.wikipedia.org/wiki/Hyperbolic_function[/url])

You have $ \cosh(2x) \equal{} 2\cosh(x)^2 \minus{} 1$

If you don't know $ \cosh(x)$, you may solve the problem using $ a_n \equal{} \frac {1}{2}(v_n \plus{} \frac {1}{v_n})$ and you obtain the same result.[/quote]

  using $ a_n \equal{} \frac {1}{2}(v_n \plus{} \frac {1}{v_n})$ . What do you do then . Can you tell me to understand

[quote="caothujjj"]  using $ a_n \equal{} \frac {1}{2}(v_n \plus{} \frac {1}{v_n})$ . What do you do then . Can you tell me to understand[/quote]

First, you notice that $ a_n>1$ $ \forall n$. So you can define a unique sequence $ v_n>1$ such that $ a_n\equal{}\frac{1}{2}(v_n\plus{}\frac{1}{v_n})$

Then $ 2a_n^2\minus{}1\equal{}\frac{1}{2}(v_n\plus{}\frac{1}{v_n})^2\minus{}1$ $ \equal{}\frac{1}{2}(v_n^2\plus{}\frac{1}{v_n^2}\plus{}2)\minus{}1$ $ \equal{}\frac{1}{2}(v_n^2\plus{}\frac{1}{v_n^2})$

And so $ \frac{1}{2}(v_{n\plus{}1}\plus{}\frac{1}{v_{n\plus{}1}})\equal{}\frac{1}{2}(v_n^2\plus{}\frac{1}{v_n^2})$ and so $ v_{n\plus{}1}\equal{}v_n^2$ (since $ v_n>1$ $ \forall n$)

And so $ v_n\equal{}v_0^{2^n}$.
We have also $ a_0\equal{}2\equal{}\frac{1}{2}(v_0\plus{}\frac{1}{v_0})$ and so $ v_0\equal{}2\plus{}\sqrt{3}$ (since $ v_n>1$)

And the result $ a_n\equal{}\frac{1}{2}((2\plus{}\sqrt{3})^{2^n}\plus{}(2\minus{}\sqrt{3})^{2^n})$

It is easy to check that this solution works.

And in case you want to see some [url=http://www.mathlinks.ro/viewtopic.php?p=109517#109517]more[/url] property of this sequence ... :P


Is it possible to construct a triangle with length $\sqrt{5}, \sqrt{10}, \sqrt{15}$ with only ruler and a pencil?

Length must be EXACT.

[hide="Hint 1"]
$(\sqrt{15})^2 = (\sqrt{5})^2 + (\sqrt{10})^2$ [/hide]

[hide="Hint 2"]
$\sqrt{5} = \sqrt{1^2 + 2^2}$

$\sqrt{10} = \sqrt{1^2 + 3^2}$ [/hide]

[hide="Solution"]
Hence, if you construct two right triangles with length 1,2, $\sqrt{5}$ and 1,3, $\sqrt{10}$ (This is possible since if you draw two lengths accurate with ruler and make sure they are perpendicular), and connect two hypotenuse to form another right triangle, we're done. [/hide]

Then by same reason, is it possible to have a triangle with one length $\sqrt[3] {2}$?

[quote="Silverfalcon"]Is it possible to construct a triangle with length $\sqrt{5}, \sqrt{10}, \sqrt{15}$ with only ruler and a pencil?[/quote]

if the ruler had irrational lengths on it o.0  :P
but yea $\sqrt{5}+\sqrt{10}>\sqrt{15}$

edit: o.0 your solution wasn't there before..  :?  sorry, misunderstood the question o.0

[quote="ch1n353ch3s54a1l"][quote="Silverfalcon"]Is it possible to construct a triangle with length $\sqrt{5}, \sqrt{10}, \sqrt{15}$ with only ruler and a pencil?[/quote]

if the ruler had irrational lengths on it o.0  :P
but yea $\sqrt{5}+\sqrt{10}>\sqrt{15}$[/quote]

Okay, it is possible but show the way to do it without looking at hints. And the ruler is regular ruler you buy from Walmart.

Are we following [i]de rigeur[/i] rules of construction? If not, then a compass must be required if this is going to be a strict construction.

You can't construct cube roots. So no to the second part of the problem.

Although, I am not sure how to prove that you can't construct cube roots.


geometry
3D geometry
tetrahedron
AMC
AIME
trigonometry
USA(J)MO

I noticed that problems involving tetrahedron often come up on the AIME. What formulas besides V = (1/3)bh and, for regular tetrahedron, V = Sqrt(2)/12 * s should I know? Now I know someone is going to reply and say that memorizing formulas is useless, etc..., but I am just asking since I have failed to solve tetrahedron problems on the AIME before simply because I didn't know the right formula. Thanks.

[quote="Fierytycoon"]V = Sqrt(2)/12 * s[/quote]

Don't you mean :rt2:/12 * s^3?

But no, memorizing formula's is NOT useless, as long as you understand them to a certain extent.

[quote]Don't you mean /12 * s^3? [/quote]

Yeah, my bad.

So...any formulas I should know?

Formulas eh?

I mathworlded it, here's the link

http://mathworld.wolfram.com/Tetrahedron.html

It has all the formulas you're ever going to need and more.

Enjoy!
IW.OAV

Ok, thanks. I was sorta wondering which ones I should actually bother to memorize though, since I don't think many test-makers would expect students to know the law of cosines for tetrahedron.

Oh, sorry. Well, when it comes to memorizing formulas (though not totally math-tastic but still helpful on competitions) I would suggest knowing the traditional ones. That would be area, volume, height, etc. things like that. Use sound judgement. Don't go memorizing the formula for the "midcircle" or whatever that is but try to use something that you think would help you.

Honestly, the only way to do that is to look at past tests and solve the problems. Normally you can find the trend in the formulas you need (law cosines can help you a lot on AHSME and so can cyclic quads etc.) So look for a trend.

Maybe someone else can share their expertise?

Anyways hope that helps...
IW.OAV

WEll, i don't think memorizing formulas is that important, well obviously you need to memorize the basics, but for more abstract formulas, such as the formula for the volume of a regular tetrahedron in terms of side length, or even Ceva's Theorem, it's important to undersatnd where those theorems came from, and ultimately, if it comes to it, derive the formula yourself on the test. Surely, it takes me less than 2min to derive the tetrahedron one.

You do have a point. However, when it comes to the AMC 12 2 minutes can make a lot of difference (at least for me).

If you ask me, it all depends on who you are. If you work fast then you don't have to worry about memorizing too many formulas. However, if you need the time like me then memorizing formulas can help somewhat.

As you can see, gauging whether or not you need formulas depends on how you perform test-wise on the AMC. For those who are new the world of AMC I would suggest taking an old AMC yourself and finding out exactly where you stand.

Just some thoughts...
IW.OAV

Yeah, I would have to agree with you there.

Hey, do you know where I can find old AMC tests?

I think you have to mostly buy them. I bought one of them for a dollar.
Old IMOs and USAMOs, along with other problems are in the [url=http://www.unl.edu/amc/problems.html]AMC homepage[/url].

uh...Tare...that isn't the AMC homepage. It is at http://www.unl.edu/amc . You linked to the TI-83 site.  :lol:

:?: Somebody must have changed my post... (or maybe i was smokin' some weed!)
Anyways, i fixed it.

no one should worry about really abstract formulas being necessary, at least on the aime. generally every geometry problem can be solved by the things you learned in geometry class. if you look at a "hard" problem like i think it was aime 01 12, there are two methods, one if you know ptolemy's thorem and one if you dont. i dont think memorizing formulas is useful. if you understand how to derive it, you can quickly derive it for every test, especially the more advanced tests where there is more time.

also equally important but perhaps MORE important than knowing formulas:

knowing how to apply them.

It's one thing knowing what a formula is, its another thing being able to recognize when and how to use the formula. After all, test makers want to see not only if you know what formulas to use, but whether you know HOW to use them. So don't get so caught up in just straight up memorizing formuals becasue if you don't understand them and don't know how to use them, i guarantee that you will still do poorly on math contests.

Yes guys, I know that memorizing formulas without understanding them is useless--that is why I made that little disclaimer at the beginning. But teratomato, I think you are going a bit far. I mean, if someone went into a contest not knowing the formula for sin(x+y), heron's formula, and other such basic things, I can also assure you that he/she will not do well.

[quote="teratomato"]Formulas are useless. In order to do well on math contests, you don't need to memorize any formulas, never mind understand them. Doing well on math contests is about going with the flow and not acting counter to the will of nature. This doesn't mean do nothing, it simply means do not fight the natural way of things. Formulas are another pathetic and ultimately futile device of so-called science by which ignorant men attempt to further analyze, classify and document that which we cannot even contemplate, let alone understand.[/quote]

not to pick any fights, but how do you get along without knowing fundamental laws of math? basically, i dont understand your point...

I think he is refering to the zen forces of nature. You see, he relies upon "mind power" to solve all.  :lol:

NO.

I have a feeling that tetratomato is really Hooyoung Chung. Anyone heard of him?

Spoiler:

[hide]He's a genius a math and computers, but as you can see, he's got problems...[/hide]

I kind of get what he's saying. Also I have to say, math is more materialistic than spiritual when it is at it's basis. The "flow" comes into play when you study one subject to its very end. There, only your instincts can lead you to the solution. Of course, I'm talking "there," as in "a point where you're the first one to solve the question." In something such as tests where the solution is already found, the spiritual stuff doesn't matter.
The only time you have to have "mind power" during a test is when you need to stop yourself from panicing and keeping the cool.

and darkquantum, just because one is a philosopher, doesn't mean they're mental. It just mean they show curiosity to life's ultimate questions and want to know the "truth."

Whoa, I think this is thread is kinda going off some where, where it ought not to go. Remember guys play it cool here we're all fans of math.

My opinion... if you can handle memorizing formulas then do it... I mean what harm is there in memorizing formulas? If its a pain for you then don't, and just derive it... Just do what works for you.

IW.OAV

[quote="darkquantum"]hmm, true, but generally, the AMC tests the mroe well known formulas, ones you have memorized well.

IT is really on the AIME and USAMO where obscure formulas pop up, but on those tests you do have the time to derive them (or you can be like me and "accidentally" derive Ceva's formula even though i knew the formula in the beginning and could have saved myslef 3 minutes!!  :evil: ).

Yea, ok...[/quote]

Actually, it seems like the really obscure formula are [b]not[/b] on the AIME and USAMO, and that is what makes the solutions to those contests really cool. Whenever I read a solution, I think WOW! That's so simple! The hard part is figuring out how to use one's elementary knowledge.
P.S. Ceva and Menelaus (sp?) are not that obscure.

Hmm, well, yea.

I was just saying that cuz i only heard about them over this summer...


Given hexagon ABCDEF ,M,N,,P,Q,R,S are midpoint AB,BC,CD,DE,EF,FA respectively.suppose every angle of hexagon
MNPQRS is 2$\frac\pi{3}$.Prove that $\frac{P(ABCDEF)}{P(MNPQRS)}\geq$$\frac{2}{\sqrt{3}}$

See http://www.mathlinks.ro/Forum/viewtopic.php?p=18340 .

  darij


Hi, I was wondering is there ANY way you can factor $\log(x+y)$ into the product of two relations $f(x)$ and $g(y)$ s.t $\log(x+y)=f(x)*g(y)$? If not, then it is just because of our inability or is it proved to be impossible to do this?

what is the realtion that corresponds to $f(x)$ and $g(x)$ in this case?

EUCLID 3.14, what you should've concluded from the given example is that such factoring is impossible, NOT because of our inability but because that's the way it is.

Really? How?

P.S. I kinda see now what Ramanujan was talking about, but I am a little confused because I don't know what do the f's and g's stand for in the above post.

You introduced $f,g$, remember¿
And how: read the thread again... or tell what's not clear.

I don't see how can Ramanujan conclude that $\log(2)\cdot\log(4) = f(1)f(2)g(1)g(2)$? I mean, I don't even know what f and g are (I mean they could be anything like $f(x) = 2x^{2}$ and $g(x)={5 \over x^{2}}$ or something).

You have to [u]assume[/u] that such $f,g$ exists, then find a contradiction from this.

OK. So you assume that $\log(2) \cdot \log(4) = f(1)f(2)g(1)g(2)$ but then why is it OK to assume that $\log(3) \cdot \log(3) = f(1)f(2)g(1)g(2)$ ?

You assume that such $f,g$ as you said in the first post exist, not more and not less. Then you set different values into $x,y$...

Oh, I see!
$\log(2) \cdot \log(4) =(f(1)g(1)) \cdot (f(2)g(2)) = f(1)f(2)g(1)g(2)$
and
$\log(3) \cdot \log(3) = (f(1)g(2)) \cdot (f(2)g(1)) = f(1)f(2)g(1)g(2)$

but then you reach a contradiction because the two expression are suppose to be equal but they aren't.


I know this question has come up many times... but the [url=http://en.wikipedia.org/wiki/Wikipedia:Wikiportal/Mathematics]Wikipedia Mathematics Portal[/url] seems very comprehensive. Are we trying to form articles that differ from these "basics" of math and are more in-depth of the strategies of problem solving?

Ideally, we'd like to make our wiki better than that one :D  Hopefully the articles in our wiki will be accessible to less experienced students.  They should be easier to learn from and should hopefully contain links to forum posts (GOOD forum posts) and example problems.

Well, that is just one thing that AoPSWiki will hopefully provide.  Potentially it can offer much more.


Two infinite plane parallel plates with distance h between them are connected with conductor.There is a charge $q$ in the middle point between the plates.If we displace the charge $q$ to the distance $x$  perpendicular to the plates,what charge$Q$ will path through the conductor?

Well, I thought that since the induced change on the plates won't change, I can 'smear out' the point charge into a plane of total charge $q$, and then it's straightforward: you calculate the potentials of each plate, demand that the potential difference between the connected plates be zero, and find the induced charges on each plate... I did a lot of guessing, I admit, and I have no idea whether this is correct, so don't try to analyze the above statement unless the answer is correct. Is it?


calculus
integration
inequalities
function
real analysis
real analysis theorems

While doing a problem, I came across the following equality:

$\int_0^{2\pi} \sqrt{a^2 sin^2 \theta + b^2 cos^2 \theta} \; d\theta = \int_0^{2\pi} \sqrt{a^2 sin^2 \theta + (a^2+c^2) cos^2 \theta} \; d\theta$

Could someone explain to me why $b^2 = a^2 + c^2$ must always be true?

Thank you.

If they are sides of a right triangle, it's ok!
But  what exactly are  $a,b,\ and\ c$ ???

Sorry, I forgot to say that a, b, c are constants.

My problem arose while I was doing B2 here:

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1995.pdf

If you integrate an inequality, you get an inequality; obviously the side with the larger coefficient is larger, since the function being integrated is pointwise greater.

[quote="blahblahblah"]Because the inequality $f(x)\geq g(x)\geq 0$, $f,g\in C([a,b])$ implies the inequality

$\int^b_a f(x)dx\geq \int^b_a g(x)dx\geq 0$.  Moreover, the inequality is strict iff the original inequality is strict.[/quote]

Hmm...well, in this case, we do not know if $f(x)\geq g(x)$. Are we allowed to make the argument that if $b^2 \geq a^2 + c^2$, then $f(x)\geq g(x)$ and $\int^b_a f(x)dx\geq \int^b_a g(x)dx$, but if instead $b^2 \leq a^2 + c^2$, then $f(x)\leq g(x)$ and $\int^b_a f(x)dx\leq \int^b_a g(x)dx$, but in either case equality holds only when $b^2= a^2 + c^2$, $f(x)=g(x)$, and $\int^b_a f(x)dx= \int^b_a g(x)dx$?

By the way, f(x) and g(x) do not have to be greater than or equal to 0, right?


The numbers 1 through 25 are written on 25 cards with one number on each card.  Sara picks one of the 25 cards at random.  What is the probability that the number on her card will be a multiple of 2 or 5?

[hide]there are 12 multiples between 1 and 25 and 5 multiples of 5
so 12+5=17
$\boxed{\frac{3}{5}}$is the answer
EDIT: i see what i did wrong.  Since 10 and 20 are both multiples of 2 and 5 they don't count[/hide]

[hide]

multiples of 2 from 1-25
2,4,6,8,10,12,14,16,18,20,22,24 total=12

multiples of 5
5,10,15,20,25 total=5

10 and 20 do not count because they're multiples of 2  [b]and[/b]5

total=12+5-2=15

$\frac{3}{5}$
[/hide]

[b]multiples of 2 from 1-25, we can get so easily.[/b]
          Multiples of 2 = floor(25/2)=12(integer)

[b]And multiples of 5 from 1-25 is :[/b]
         Multiples of 5 = floor(25/5)=5(integer)
Then calculate :
         P(2 or 5) = p(2)+p(5)-p(2 and 5)
                       = 12+5-2=15
Here [b]p(2 or 5)[/b] means all number betn them.
And [b]p(2 and 5)[/b] means just common betn them.

So the probablity will be 15/25=3/5
Thats all.


Determine the minimum value of

    abc(a 2 +b 2 +c 2)

where a+b+c=1 and a,b,c \geq 0.

Maybe it's about the maximum value, since the minimum value is of course 0. The problem about the maximum is indeed very nice. We will show that it is 1/81. So we must prove that abc(1-2 sum ab)<=1/81 when a+b+c=1, or 1/(81abc)+ab+bc+ca+ab+bc+ca>=1. But the RHS is at least  3((ab+bc+ca)^2/81abc)^1/3, which is at least 1, because (ab+bc+ca)^2>=3abc(a+b+c)=3abc.
   here is another problem, which I think is significantly difficult:
   Find the maximum value of x_1*x_2*...*x_n(x^2_1+...+x^2_n) when sum x_i=1

[quote="harazi"]Here is another problem, which I think is significantly difficult:
   Find the maximum value of x_1*x_2*...*x_n(x^2_1+...+x^2_n) when sum x_i=1[/quote]

========
This problem can be solved by following lemma:
xy(A+x^2+y^2)<= (1/4)(x+y)^2(A+(x+y)^2/2).

Namdung


function
algebra
polynomial
geometry
geometric transformation
dilation
complex analysis

Let $ f: \mathbb {D}\to\mathbb{C}$ be an analytic univalent function from the open unit disc which continous on $ \overline{\mathbb{D}}$.Is it true that there exists the sequence of univalent in $ \mathbb{D}$ polynomials that uniformly converges to $ f$ on $ \mathbb {D}$.

For each strictly smaller disc, all taylor polynomials of $ f$ with large enough degree are univalent there. Just use the argument principle. So dilate the taylor polynomials a bit :)

Of course, you are right about Taylors polynomials.Using Bieberbch theorem we can get a nice universal value for the radius of disc (when $ f(0)\equal{}1$),where Taylor's ppolynomials are univalent functions. We can prove dirrectly the rezult for some nice functions($ e^z$...). But how to modify these Taylor's polynomials in general case? :maybe:

Oh, I did not notice you want uniform convergence on the entire disc. Hmm, maybe I am overlooking something, but is this question not the same question as if whether or not there at all is a polynomial sequence converging uniformly to $ f$ on the unit disc? And the fact that such a sequence exists is the content of Mergelyan's theorem.

If $ f_n$ is a sequence of polynomials converging uniformly to $ f$ on the disc, then...

Step 1. By counting the number of times $ f_n$ achieves some particular value using the argument principle, it follows that for each $ r < 1$, there is an $ N$ s.t. $ f_n$ is injective on $ |z| < r$ whenever $ n \geq N$.

Step 2. Choose $ n_1 \leq n_2 \leq n_3 \leq \ldots$ such that $ f_{n_k}$ is univalent on $ |z| < 1 \minus{} \frac {1}{k}$, or equivalently such that $ g_k(z) \equal{} f_{n_k}(( 1 \minus{} \frac {1}{k})z )$ is univalent on the unit disc. Now $ g_k$ converges uniformly to $ f$ on the unit disc (because $ f_n$ does, and $ f$ is uniformly continuous).

:blush:
Of course,you are right, I thought that this problem is much more difficult and tried to use ,,nuclear weapon"(Bieberbach theorem) :) Thanks a lot


calculus
integration
function
real analysis
real analysis unsolved

[quote="cckek"]Find all functions $f: R^{2}\to R,if \int_{0}^{\infty}f(t,y)dt=y^{2}-1$[/quote]

$f(t,y)=e^{-t}(y^{2}-1)$ is a solution, and $f(t,y)=g(t)(y^{2}-1)$ is solution too if $\int_{0}^{\infty}g(t,y)dt=1$


Hi

I've got a two-column report, and i need to add a footnote that spans across the full width (both columns) - any idea how i can achieve this???

Thanks!


没有北京的么??可惜呀......
这6个人中有没有mathlinks的成员啊?
是就快坦白! :D

[quote="zhaoli"]真是人才济济呀，今年中国数学奥林匹克四个满分的人中只有罗晔被选中。
六位选手中有五位在今年的中国数学奥林匹克中获得一等奖，一位获得二等奖。[/quote]
Wpolly是其中的一位么??

Good luck to all those people, sorry my Chinese is too bad for me to post in Chinese.

China is the best!!! :D I hope we win.

难道大部分学校都把竞赛放在正课里讲么?!

中国队今年肯定是 第一名, 只是要看领先美国或俄国多少分了...

[quote="8th Army"]中国队今年肯定是 第一名, 只是要看领先美国或俄国多少分了...[/quote]

我觉着至少20分。兄弟们， 你们怎么认为？

Control Panel -> Regional Settings.
Add the language of Simplified / Tranditional Chinese,
or Asian languages.

任庆春  42分  金牌
刁晗生  42分  金牌
罗烨    42分  金牌
邵烜程  42分  金牌
康嘉引  35分  金牌
赵彤远  32分  银牌

不错! 但是第一题丢分有点不应该.

感谢zhaoli!
P.S. zhaoli你是从那里知道的这个呀?

非常简单！

(1) 数学竞赛之窗网站上有每个人的得奖情况

(2) imo2005官方网站上有中国队六个队员的得分，但没有名字

(3) 剩下的就是从大到小的排序

http://www.sxjszc.com/

or

http://www.aoshoo.com/index.asp

[quote="Arwen"][quote="wpolly"]I got the 8th place............ :([/quote]

已经很强了～

陕西省真的是不行，我们都说铁一中在陕西省的数学竞赛里最强，可是最近几年也是没有什么建树。我现在在西工大附中，发现高中部把竞赛弄到星期六和星期天的辅导课去了，教材用浙大的那套书…………我想应该是这里的教育问题，我们的老师很难自己解出IMO的题（我没怪他什么）～附中侧重高考，别的学校就更不用说了。全省六个理科试验班除了师大附中和长安一中的两个化学试验班还出成绩外，其余四个数理的都不行了……

我周围除了天赋特别高的同学，其余的都目的性很强要不然专弄竞赛（人越来越少，好像都是知难而退），要不然一心高考。也不知道再过两个月我将何去何从。[/quote]

说实话，在来到西安之前，我对西工大附中几乎没有了解，因为竞赛并不突出。铁一中还可以，2002年出过一个IMO满分。来了之后，据西安的同学讲，西工大附中高考特别强，每年清华北大录取占去了陕西省一半的名额，铁一中只是那一个班强。竞赛方面，去年联赛陕西出了很大的问题，捅到了老胡那里，被中国科协批评，你们今年要小心了。

[quote="8th Army"]美国对和加拿大对中大部分都是中国移民,[/quote]

美国队今年只有一个中国人,而且还是当地出生的.

[quote="nirvanatear"][quote="jmtlf"][quote="8th Army"]美国对和加拿大对中大部分都是中国移民,[/quote]
美国队领队冯祖鸣、加拿大领队刘江枫都是中国人。
还有，美国数学奥林匹克还用中国人出的题，比如2002年的那个函数方程就是吴伟朝（广州大学）出的。[/quote]
加拿大国家队还有非华人吗?
呵呵.[/quote]

Not really. ;)
Coz you should've said亚洲人instead of 中国人
LOL
Last year, there was Elyot Grant as an non-asian guy, this year, eh... dunno, could be Alex Wice.

[quote="Kalimdor"][/quote][quote="nirvanatear"][quote="jmtlf"][quote="8th Army"]美国对和加拿大对中大部分都是中国移民,[/quote]
美国队领队冯祖鸣、加拿大领队刘江枫都是中国人。
还有，美国数学奥林匹克还用中国人出的题，比如2002年的那个函数方程就是吴伟朝（广州大学）出的。[/quote]
加拿大国家队还有非华人吗?
呵呵.[/quote][quote="Kalimdor"]

Not really. ;)
Coz you should've said亚洲人instead of 中国人
LOL
Last year, there was Elyot Grant as an non-asian guy, this year, eh... dunno, could be Alex Wice.[/quote]
JUST KIDDING.
知道还有其他人...
但看看MAJORITY
现在获奖名单从大到小都已经变成DAVID ZHAO, ALICIA LI, JASON ZHOU之类的了...

[quote="wenxin"]2005中国国家队名单

任庆春
 天津耀华中学

刁晗生
 上海华东师大二附中

罗  晔
 江西师范大学附中

康嘉引
 广东深圳中学

邵烜程
 上海复旦大学附属中学

赵彤远
 河北省石家庄市二中[/quote]
国内那几所高中最棒?
有谁能将历年来各校进入国家队的人数统计一下. :oops: :)

(我们上海有两个. :) 我今年初三,试问上海哪所学校最好?)

康嘉引
广东深圳中学

这个是我们母校的牛B生.
我们学校也不算很好,在广东也只能算老二.


Hi everyone,
I'm new to this forum. I have a few questions to those who are more aquainted with the subject. I'm currently at an American university studying Mathematics. My area of expertise is continuous math (I'm goodat complicated computation). Growing up I was taught math ass-backwards (excuse my language). I was taught how to solve problems to get the answer, instead of learning the truth of what I was doing. Because of this fact, I have trouble with discrete mathematics.

I have this book:

http://www.amazon.com/gp/product/0130144126/sr=8-1/qid=1149257749/ref=pd_bbs_1/104-6071302-5107930?%5Fencoding=UTF8

but it didn't seem to do the trick. When I look at a problem (like those on the Putnam Competition and such) I have trouble seeing a way to solve the problem. I enjoy problem-solving and have been good at brain teasers all my life; but my low quality K-12 education hinders my potential (I think).

Does anyone have any advice for me? (books I should read?) Does one develop this ability through problem sets or through a firm understanding of fundamental mathematics? I'm confident I can be good at these types of problems, I just need a push in the right direction.

I would recommend [url=http://www.amazon.com/exec/obidos/ASIN/0471135712/artofproblems-20]The Art and Craft of Problem Solving[/url] by Paul Zeitz.

Specifically for the Putnam, we offer three [url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php]collections of old Putnam problems[/url] with annotated solutions.  The most recent of these -- the 1985-2000 volume by Kedlaya, Poonen and Vakil -- has especially good solutions and problem-solving commentary.

Problem solving is definitely an ability that is developed through practice.

[quote="DPatrick"]Problem solving is definitely an ability that is developed through practice.[/quote]

I hope you're right. :)

Hi Mr.Patrick, hi know_not

Yea, problem solving definitely comes through practice.

Right here, this forum site helped me tremendously with my problem solving skills. At first, I felt that 'yea classroom math is too easy, and now what?' I felt as if there was no challenge. After I came across this site when my cousin showed me, I was so amazed to find that math can be so challenging, I couldn't even solve more than 80% of the problems on this forum!!! Now, I have developed my skills (and still improving!) and believe me, practice does make a difference. I may not be the Reid Barton or some top-notch IMO participant but this site definitely helped immensely.

Although I am nowhere near as authority to recommend, but why not start from the basics? Having a strong foundation helps you to build a solid performance in problem solving. The books recommended by Mr.Patrick are excellent. Meanwhile, come along and try the different MO ( math olympiads) problems in the resource section.

Lastly, :welcomeani:


now does this book exixt or not??? i can seem to find even a mention of it anywhere across the net .
I am interested in this book. someone plz help!!

I do not think you will find that book somewhere in the internet , I think there is no electronic form of the book .
I am also intersted to buy it, so If you find any way how to buy it please tell me .

$ \mbox{GIL Publishing House: }$
http://www.gil.ro/oferta.pdf
http://www.gil.ro/ONI2.php
$ \mbox{or private mesage to: Mircea Lascu: }$ gil1993@zalau.astral.ro

Mihai, he was looking for other book, even the authors aren't the same

I'm looking for [b]Secrets in Inequalities ,second volume [/b] by Pham Kim Hung ,GIL publishing house.

I sent a private message in mathlinks to mr.Mircea Lascu but he did not reply me  :(

same here but the fact is i dint expect there to be an e-book version BUT THR SEEMS TO BE ABSOLUTELY NO MENTION IN THE NET!!!!!
for buying anyone knows any alternative? like the book can bouhgt in so and so place or something..

[quote="geniusbliss"]Why havent anyone replied
i wud be thankful to anyone who gives a lead on this matter :maybe:[/quote]

You must buy it from GIL publishing house . That's the best way .


inequalities
function
inequalities proposed

Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac{2(a^2\plus{}b^2\plus{}c^2)}{(ab\plus{}bc\plus{}ca)^2} \geq\ \frac{a^2\plus{}b^2}{(a^2\plus{}ab\plus{}b^2)^2} \plus{} \frac{b^2\plus{}c^2}{(b^2\plus{}bc\plus{}c^2)^2} \plus{} \frac{c^2\plus{}a^2}{(c^2\plus{}ca\plus{}a^2)^2}$
Have fun, my brother  :)

Since a,b,c are symmetric in this inequality, let's assume that a>=b>=c>=0, then try to translate the right side into the same denominator using ">=", then u can easily find out the solution. Then apply the same way on b>=c>=a>=0 and c>=a>=b>=0, now it's complete.

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {a^2 \plus{} b^2}{(a^2 \plus{} ab \plus{} b^2)^2} \plus{} \frac {b^2 \plus{} c^2}{(b^2 \plus{} bc \plus{} c^2)^2} \plus{} \frac {c^2 \plus{} a^2}{(c^2 \plus{} ca \plus{} a^2)^2}$
Have fun, my brother  :)[/quote]
We can prove this inequality easily if we note that the function
\[ f(x) \equal{} \frac{x(x\plus{}1)(x^2\plus{}1)}{(x^2\plus{}x\plus{}1)^2}\]
is increasing for all $ x>0.$ ;) :)

It can be solved by classic ineq. I think you want to try  :)

I haven't tried (the inequality is easy) but I am curious to see your AM-GM solution. ;)

Let a+b+c=1 Rewrite :
$ \sum (a^2\plus{}b^2) (\frac{1}{(ab\plus{}bc\plus{}ca)^2}\minus{} \frac{1}{(a^2\plus{}ab\plus{}b^2)^2})$
 $ \equal{} \sum (a^2\plus{}b^2)(a^2\minus{}bc\plus{}b^2\minus{}ac)(a\plus{}b) \frac{1}{(ab\plus{}bc\plus{}ca)^2(a^2\plus{}b^2\plus{}ab)^2}$
$ \equal{} \sum (a^2\minus{}bc)(b\plus{}c) \frac{(a^2\plus{}b^2)(a\plus{}b)\plus{}(c^2\plus{}a^2)(c\plus{}a)}{(b\plus{}c)(a^2\plus{}ab\plus{}b^2)^2(ab\plus{}bc\plus{}ca)^2}$
We have $ \sum (a^2\minus{}bc)(b\plus{}c) \equal{}0$
By Chebyshev,We have solution.

Indeed, I don't think it's easy to prove:
$ \frac {(a^2 \plus{} b^2)(a \plus{} b) \plus{} (c^2 \plus{} a^2)(c \plus{} a)}{(b \plus{} c)(a^2 \plus{} ab \plus{} b^2)^2} \geq\ ...$   :wink:
And ok, here is my proof which I said  :)