In case you don't know, I'm banned from FTW for 10 years for [b]whispering[/b] jjx1's name to him.

But really, I don't care.  FTW is not that fun anymore as it used to be.

[b]I would be appalled if jjx1 would reply to this.[/b]

this is not the forum to complain about. complaining only extends your ban.

I don't think anybody here cares at all. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1419103#1419103]Again[/url], I might mention that it would be prudent for you to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1419113#1419113]get out of Missouri[/url].

That is not why you are banned and you know it.

This is the type of stuff I deal with all day.

/me preps to turn chat off again.

I don't see your purpose here, levans. Just give us Missourians some privacy. And as far as I'm concerned, the reason I'm banned is because I'm the person who was "being disrespectful--" and I sort of wonder why I disappeared from FTW right after saying "shut up." (Bug?) Now, I don't see another instance in which I have been blatantly disrespectful. CAPS definitely doesn't count, because there was one person who didn't want CAPS, and that person was afk like right after reporting me to jjx1, anyhow.

math, i don't think there's anything we can do.....i probably wont return to ftw.- well i can't.

anyways this is my last argument

1.  I didn't know that you couldn't do it on the game.

2.  I [b]misspelled[/b] what I was saying.

3.  Why is Missouri hated so much?  :(

math154: You should read your ban reason if you feel that is why you were banned. We aren't dealing with any thing bad right now. You should have been banned long ago. 10 mutes in the past three months are way too many.

Justin: I didn't even know you were from Missouri when I banned you. How can you not know telling people they suck isn't bad? If you aren't old enough to know that, then you aren't old enough for FTW.

math154: If you want privacy, don't use our forums.

justin. what you just said made no sense at all. theres no way you could have misspelled for the reason you were muted. It wouldn't make sense. I will pm you the reason now(though you should know it)

Oh right, levans, so why do random states have private forums? And I don't know where to find my ban reason, other than [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1418238#1418238]here[/url].

I don't know why some states have private forums. We can still access them though.

I thought the reason was displayed on the page that says you are banned. If not, I'll PM it to you.

I don't understand why you are ignoring the link? How is the person mentioned not I?

[b]Edit:[/b] Of course, thanks. :D  :D  :D

Ok then: What are some restrictions for posting on the missouri forum?  Do these 'restrictions' adhere to other forums ( even private ones ) ?

I just want to know.

Sorry, I posted that twice.

And I'm sorry jjx1, if I hurt your feelings in anyway.....

levans, I'm not saying that it isn't bad, I'm just saying that I was caused for saying that by getting muted for something so little.  However, I regret what I said to jjx1.

That's good that you regret what you said about jjx1.

You were originally muted because you used a real name in chat. A full real name at that. This is against the TOS.

AS far as restrictions on posting on the Missouri forum, follow the regular terms of service (the link is available on the FTW TOS page.) Use common sense. If you wouldn't say it in front of your parents, you probably shouldn't say it here. I would guess restrictions are the same for private forums. I really don't handle the forum issues, so I can't provide a definitive answer.

Hopefully, he randomly clicks on the Missouri Forum and reads that.  :P

what AIME15 said. yep, you're reminded everytime you enter ftw.

Nope, the random number generator decides that.

Bah, TOS should be more attractive.

I think that there should be some ideal of democracy instead of mods just muting people (cough,cough), and

proceeding to ban them (cough....oh dang now I have strep).

People who are present at the time of the mute should vote and decide if the person should be muted or not.

Of course that leaves the question of the moderator's job.  They should give one warning to the user.  The

next time an user messes with TOS, there should be a vote by other mods whispering their verdict to the

mods present at the time.

no it is pretty simple

break terms of service and mods can do whatever they want to you

they have been chosen to be mods because they are actually relatively mature and intelligent and know when people are not being mature or intelligent

such people should not be talking because all they do is annoy people

[quote="justinahmann"]Oh really?  Are all mods mature and intelligent? And are you accusing me of being stupid?[/quote]

Give an example of an immature and/or stupid mod...

[quote="pythag011"]Give an example of an immature and/or stupid mod...[/quote]

I think we all know his answer, pythag011. :wink:

Justin. we now have to be strict, because ftw is costing aops money. we can't give warnings anymore. read FAQ.

Anyway, if you didn't or don't like the way ftw works, you never or don't have to go on it. GO on AIM or something.

Topic is now about tornado's striking trailer parks.

grrr kansas sucks sucks so much, so by corollary, i am awesome.

jjx1, i never meant you!!!  That was completely twisted by the other forum writers.

And I'm just saying that mods are not perfect all the time. That's what I mean.

i beg to differ, i am perfection embodied.


geometry
rectangle
circumcircle
vector
radical axis
complex numbers
power of a point

Let $ ABDE,$ $ ACFH,$ $ BCKL$ be, three arbitrary rectangles, erected outwardly to a given triangle $ \bigtriangleup ABC,$ on $ AB,$ $ AC,$ $ BC$ respectively and we denote the points $ A'\equiv EK\cap HL,$ $ B'\equiv HL\cap DF,$ $ C'\equiv DF\cap EK.$ Prove that the lines segments $ AA',$ $ BB',$ $ CC',$ are concurrent at one point.

Kostas Vittas.

Very nice problem, dear Kostas!

For moment I will only expose some of my ideas since I've accidentally deleted my sketch and don't have the patience at this hour to do it again. So, I will be back in short time with a full solution.

Consider $ X$, $ Y$, $ Z$ the centers of the rectangles $ BCKL$, $ CAHF$ and $ ABDE$, respectively. After a straightforward angle chasing, it follows that the triangles $ XYZ$ and $ A'B'C'$ are homothetical.

Moreover, the circumcircles of $ CAHF$ and $ ABDE$ concur on $ B'C'$. Hence, the triangles $ ABC$ and $ A'B'C'$ are orthologic, having as one of the orthology centers (for moment, we talk about two of them) the radical center of the circumcircles of $ BCKL$, $ CAHF$ and $ ABDE$.

Next, as probably suggested by the above remark in the bracket, we will prove that the "second" orthology center, concides with the first one, after that, beeing well-known that two orthologic triangles having the same orthology centers are perspective.

Hence, it is enough to prove that $ KA'\perp BC$, where $ K$ is the radical center of the three rectangles...

Thank you dear [b][size=100]Pohoata[/size][/b], for your interest and successful remarks about the proof.

Both synthetic solutions I have in mind, based on the same idea you mentioned. It is enough to prove that the orthologic centers of the triangles $ \bigtriangleup ABC,$  $ \bigtriangleup A'B'C',$ are coincinded (here, with the radical center $ P,$ instead of $ K,$ of the circumcircles of the erected rectangles).

As for the result $ A'P\perp BC$ ( instead of $ A'K\perp BC$ ) you mentioned, it is enough to see the relation between the line segments $ BC,$ $ B''C'',$ where $ B'',$ is the second intersection point (the first is $ B$ ) of the circucircles of $ ABDE,$ $ BCKL,$ lies on $ A'C'$ and $ C'',$ is the second intersection point (the first is $ C$ ) of the circumcircles of $ ACFH,$ $ BCKL,$ lies on $ A'B'.$

Best regards, Kostas Vittas.

Could you post your solution, dear [b]Kostas[/b]?
I see that your problem is still true when $ ABDE$, $ ACFH$, $ BCKL$ are three similar isosceles trapeziums. I tried to prove it, but I failed  :blush: .

Thank you dear [b][size=100]April[/size][/b] for your interest about this (not well known, but nice) problem.

I am not sure if it is true the result you mentioned about the similar isodceles trapesiums and if we can have a solution by the same way. I see by the drawing that the concurrency, doesn't hold in this particular case. I will check more carefully, before a definite answer.

I will post here later, the two synthetic solutions I have in mind. Give me please a little time.

Best regards, Kostas Vittas.

[quote="vittasko"]... I am not sure if it is true the result you mentioned about the similar isodceles trapesiums and if we can have a solution by the same way. I see by the drawing that the concurrency, doesn't hold in this particular case. I will check more carefully, before a definite answer...[/quote]
Oh, I see  :blush: . You are right, dear [b]Kostas[/b]!

I'm in a hurry now, so I'll only give some hints to solve this nice and hard problem.

1. Using your strong and hard result in http://www.mathlinks.ro/viewtopic.php?p=916020#916020

2. Using [b]Jacobi[/b] theorem.

Hope that I'm not mistaken again  :D .

Best regards, April.

[size=117][color=darkblue][b]Thank you for this very nice problem, Vittasko ![/b][/color][/size]

[quote="vittasko"][color=darkred]Let $ ABDE$ , $ ACFH$ , $ BCKL$ be arbitrary rectangles erected [b]outside or inside[/b] of the triangle $ ABC$ . Denote the points $ A'\in EK\cap HL$ , $ B'\in HL\cap DF$ , $ C'\in DF\cap EK$ . Prove that the lines segments $ AA'$ , $ BB'$ , $ CC'$ are concurrently.[/color] [/quote]
[color=darkblue][b][u]Proof (vectorial method).[/u][/b] I'll choose the circumcenter $ O$ of $ \triangle\ ABC$ as the origin

of the vectorial system and denote the [b]position vector[/b] $ X\equiv\overrightarrow {OX}$ , i.e. $ \overrightarrow {XY} = Y - X$ .

Exist $ \{m,n,p\}\subset \mathcal R^*$ so that $ \{\begin{array}{c} L + C = K + B = m(B + C) \\
\ F + A = H + C = n(C + A) \\
\ E + B = D + A = p(A + B)\end{array}$ .

Thus, $ A'\in EK\cap HL$ $ \Longleftrightarrow$ exist $ \{\beta ,\gamma \}$ so that $ A' = \{\begin{array}{c} \beta \cdot E + (1 - \beta )\cdot K \\
\ \gamma\cdot H + (1 - \gamma )\cdot L\end{array}\|$ $ \implies$

$ A'=\{\begin{array}{c}
\beta\cdot [pA + (p - 1)B] + (1 - \beta )\cdot [(m - 1)B + mC]\\\\
\gamma \cdot[nA + (n - 1)C] + (1 - \gamma )\cdot[mB + (m - 1)C]\end{array}\|$ $ \implies$

$ \{\begin{array}{cc}
A\ : & p\beta=n\gamma\\\\
B\ : & (p-1)\beta +(m-1)(1-\beta )=m(1-\gamma )\\\\
C\ : & (n-1)\gamma +(m-1)(1-\gamma )=m(1-\beta )\end{array}\|$ $ \implies$  $ \frac {\beta}{n}=\frac {\gamma }{p} = \frac {1}{p(m + n) - mn}$ and

$ [p(m + n) - mn]A' = npA + m[p(m + n) - mn - p]B + m[p(m + n) - mn - n]C$ $ \implies$

$ \boxed {\ \frac {p(m + n) - mn}{m}\cdot \overrightarrow {AA'} = (n - p)A + [p(m + n) - mn - p]B + [p(m + n) - mn - n]C\ }\ .$

We can ascertain similarly the vectors $ \overrightarrow {BB'}$ , $ \overrightarrow {CC'}$ w.r.t. the position vectors $ A$ , $ B$ , $ C$ .

In conclusion,  $ A'A\cap B'B\cap C'C\ne\emptyset$ $ \Longleftrightarrow$

$ |\begin{array}{ccc} n - p & p(m + n) - mn - p & p(m + n) - mn - n \\
 \\
p - m & m(n + p) - np - m & m(n + p) - np - p \\
 \\
m - n & n(p + m) - pm - n & n(p + m) - pm - m\end{array}| = 0$ , what is truly

because the second column is sum of the first and third columns. [b]STOP[/b]

[b][u]Remark.[/u] [i] Is without importance if we'll construct the quadrilaterals outside or inside

of $ \triangle\ ABC$ . This problem is truly in the all cases : all outside ; all inside ;

one inside and another two outside ; one outside and another two inside ![/i][/b] [/color]

Thank you a lot, dear [b][size=100]Virgil[/size][/b].

I am glad when a solution of a problem I proposed, has been presented by your self and especially, when it is a solution by a method advanced to me.

Unfortunately, I can’t think by vectors and I am not happy for such that my weakness.

Best regards, Kostas Vittas.

PS. You are right, thank you again. The problem is also true as you mentioned, in all cases of the  erected rectangles on the side-segments of $ \bigtriangleup ABC.$

[quote="Virgil Nicula"][color=darkred][b][u]Remark.[/u] [i]Is without importance if we'll construct

the quadrilaterals [u]outside or inside[/u] of $ \triangle\ ABC$ .[/i][/b]

[/color][/quote]
[color=darkblue]Upper is underlined the advantage of a proof with the [b]complex numbers[/b] or with the [b]analytical geometry[/b] or with the [b]vectorial method[/b] or with the [b]proiective geometry[/b] a.s.o. Now I look for a synthetical proof of this nice and difficult problem from this view point. I like to find a same proof and in the my opinion, a synthetical proof is more nicely and more interestingly than any another, but [u]sometimes[/u] is incompletely. [b]This is the my opinion, [u]Vittasko[/u][/b] ![/color]

[quote="vittasko"]Let $ ABDE,$ $ ACFH,$ $ BCKL$ be, three arbitrary rectangles, erected outwardly to a given triangle $ \bigtriangleup ABC,$ on $ AB,$ $ AC,$ $ BC$ respectively and we denote the points $ A'\equiv EK\cap HL,$ $ B'\equiv HL\cap DF,$ $ C'\equiv DF\cap EK.$ Prove that the lines segments $ AA',$ $ BB',$ $ CC',$ are concurrent at one point.[/quote]

[b][size=100]PROOF  1.[/size][/b] – Let $ (O_{1}),\ (O_{2}),\ (O_{3})$ be, the circumcircles of the rectangles $ ABDE,\ ACFH,\ BCKL,$ respectively.

It is easy to show that the point $ A'',$ as the intersection point of the circles $ (O_{1}),\ (O_{2}),$ the other than $ A,$ lies on the line segment $ DF,$ from $ \angle AA''D + \angle AA''F = \angle ABD + \angle ACF = 180^{o}.$

Similarly the points $ B''\equiv (O_{1})\cap (O_{3}),\ C''\equiv (O_{2})\cap (O_{3}),$ lie on the line segments $ EK,\ HL,$ respectively.

Because of now, the line segments $ AA'',\ BB'',\ CC'',$ are the radical axis of the above circles, taken per two of them, we conclude that they are concurrent at one point so be it $ P,$ as their radical center.

So, we have the configuration of the triangles $ \bigtriangleup ABC,\ \bigtriangleup A'B'C',$ where the lines through $ A,\ B,\ C$ and perpendicular to $ B'C',\ A'C',\ A'B'$ respectively, are concurrent at one point, here the point $ P$ and then, we have that these triangles are [b][size=100]orthologic[/size][/b] with respect to each other.

Hence, we conclude that the lines through the points $ A',\ B',\ C'$ and perpendicular to $ BC,\ AC,\ AB$ respectively, are also concurrent at one point, so be it $ P'.$

As mentioned by [b][size=100]Cosmin Pohoata[/size][/b], it is enough to prove that $ P'\equiv P.$

$ \bullet$ We see that the circle $ (O_{3}),$ passes through the vertices $ B'',\ C''$ of the triangle $ \bigtriangleup PB''C''$ and intersects its sidelines $ PB'',\ PC'',$ at points $ B,\ C,$ respectively.

So, we have that the line segment $ BC,$ is the [b][size=100]antiparallel line[/size][/b] of $ B''C'',$ with respect to the angle $ \angle P$ of $ \bigtriangleup PB''C''.$

Because of now, the line segment $ AP,$ is the diameter of the circumcirle of $ \bigtriangleup PB''C'',$ from $ \angle A'B''P = \angle A'C''P = 90^{o},$ we conclude that $ AP\perp BC.$

By the same way, we can prove that $ BP\perp AC$ and $ CP\perp AB$ and then, we have that the [b][size=100]orthologic center[/size][/b] $ P'$ of $ \bigtriangleup A'B'C'$ with respect to $ \bigtriangleup ABC,$ is coincided with the point $ P.$

So, based on the below well known theorem as for the [b][size=100]orthologic triangles[/size][/b], we conclude that the triangles $ \bigtriangleup ABC,\ \bigtriangleup A'B'C',$ are also [b][size=100]perspective.[/size][/b]

Hence, based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the line segments $ AA',\ BB',\ CC',$ are concurrent at one point and the proof is completed.

[b][size=100][color=DarkBlue]THEOREM. - When the orthologic centers of two triangles, orthologic with respec to each other, are coincided, these triangles are also perspective.[/color][/size][/b]

$ \bullet$ This proof is dedicated  to [b][size=100]Nikos Dergiades.[/size][/b]

Kostas Vittas.

[quote="vittasko"]Let $ ABDE,$ $ ACFH,$ $ BCKL$ be, three arbitrary rectangles, erected outwardly to a given triangle $ \bigtriangleup ABC,$ on $ AB,$ $ AC,$ $ BC$ respectively and we denote the points $ A'\equiv EK\cap HL,$ $ B'\equiv HL\cap DF,$ $ C'\equiv DF\cap EK.$ Prove that the lines segments $ AA',$ $ BB',$ $ CC',$ are concurrent at one point.[/quote]
[b][size=100]PROOF  2.[/size][/b]  - We draw the lines through vertices $ B,\ C$ of the given triangle $ \bigtriangleup ABC$ and perpendicular to $ DL,\ FK$ respectively and we denote as $ N,$ their intersection point.

The lines through $ K,\ L$ and perpendicular to $ FK,\ DL,$ intersect each other at point so be it $ A''.$

Let $ B'',\ C''$ be, the points such that $ ANB''H,\ ANC''E$ to be parallelograms.

It is easy to show that $ NA''\perp BC$ $ ,(1)$ from $ NB\parallel A''L,\ NC\parallel A''K$ and $ BC\parallel = KL.$

Because of now, $ NB''\perp AC$ $ ,(2)$ and $ NC''\perp AB$ $ ,(3)$ we have the concurrency at point $ N,$ of the lines through vertices $ A'',\ B'',\ C'',$ of the triangle $ \bigtriangleup A''B''C''$ and perpendicular to the sidelines $ BC,\ AC,\ AB$ respectively, of $ \bigtriangleup ABC.$

So, we have that the triangles $ \bigtriangleup ABC,\ \bigtriangleup A''B''C''$ are orthologic and then, we conclude that the lines through $ A,\ B,\ C$ and perpendicular to $ B''C'',\ A''C'',\ A''B''$ respectively, are concurrent at one point and we will prove that their concurrency point, is the point $ N.$

$ \bullet$ Because of $ NA''\parallel = CK$ and $ NB''\parallel = AH\parallel = CF,$ we have that $ A''B''\parallel = FK$ $ ,(4)$ and similarly, $ A''C''\parallel = DL$ $ ,(5)$

From $ (4)$ and because of $ NC\perp FK,$ we conclude that $ NC\perp A''B''$ $ ,(6)$ and similarly, $ NB\perp A''C''$ $ ,(7)$

From $ (6),$ $ (7),$ we conclude that the point $ N$ is also the orthologic center of $ \bigtriangleup ABC,$ with respect to $ \bigtriangleup A''B''C''.$

So, because of the orthologic triangles $ \bigtriangleup ABC,\ \bigtriangleup A''B''C'',$ have their orthologic centers coincided at the same point, here the point $ N,$ we conclude that these triangles are also perspective.

Hence, based on the [b][size=100]Desarques theorem[/size][/b], we have that the line segments $ AA'',\ BB'',\ CC'',$ are concurrent at one point so be it $ P.$

$ \bullet$ Based on the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108818]An interesting collinearity[/url], we have that the point $ A'\equiv EK\cap HL,$ lies on the line segment $ AA''$ and similarly, the points $ B'\equiv HL\cap DF,\ C'\equiv DF\cap EK,$ lie on $ BB'',\ CC'',$ respectively.

Hence, we conclude that the line segments $ AA',\ BB',\ CC',$ are concurrent at one point and the proof is completed.

[b][size=100]COMMENT.[/size][/b] - After the concurrency of the line segments $ AA'',\ BB'',\ CC'',$ it is possible to complete the solution of the proposed problem, without any reference of the hard problem I mentioned before.

I will post here later a such that approach to complete the solution, based on a clever idea by a friend of mine [b][size=100]Nikos Kyriazis[/size][/b] from Thessaloniki.

Kostas Vittas.

[quote="vittasko"]  [b][size=100]COMMENT.[/size][/b] - After the concurrency of the line segments $ AA'',\ BB'',\ CC'',$ it is possible to complete the solution of the proposed problem, without any reference of the hard problem I mentioned before.

I will post here later a such that approach to complete the solution, based on a clever idea by a friend of mine [b][size=100]Nikos Kyriazis[/size][/b] from Thessaloniki.[/quote]
In the [b][size=100]proof  2[/size][/b] presented above, we used the argument of the collinearity of the points $ A,\ A',\ A'',$ which has already been posted as a separate problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108818]here[/url] and an its complicated solution has been presented by my self, in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=164548]Concurrent lines[/url], of [b][size=100]Omid Hatami.[/size][/b]

But, this result ( the collinearity of the points $ A,\ A',\ A''$ ), as [b][size=100]Nikos Kyriazis[/size][/b] pointed to me three years ago, may much more easier be proved, based on his clever idea to use the fact that the proposed problem is true in every configuration of the erected rectangles on the side-segments of the given triangle $ \bigtriangleup ABC,$ as mentioned also before by [b][size=100]Virgil Nicula[/size][/b] $ ($ all of them outwardly, or inwardly, or some of them outwardly and the other inwardly to $ \bigtriangleup ABC$ $ ).$

So, in the schema [b][size=100]t=167669(c)[/size][/b], we can say the same words as in the [b][size=100]proof  2[/size][/b], in order to conclude the concurrency of the line segments $ AA'',\ BB'',\ CC'',$ as before.

But, we see that it is an ''alternative'' configuration of the proposed problem and so, we can have a much more simpler way, to conclude the collinearity of the points $ A,\ A',\ A'',$ in the schema [b][size=100]t=167669(b)[/size][/b].

That is, the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108818]An interesting collinearity[/url] ( I spend enough nights to solve it... ), is a consequent problem of the proposed one.

Many thanks to dear [b][size=100]Nikos Kyriazis[/size][/b], for a clever idea.

Kostas Vittas.


How does this work? I'm not sure why with a certain number of slots that they must be filled before another is.

Thank you

The most common statement of this principle, also known as the Box Principle, Dirichlet's Box Principle, and so forth:

[i]When distributing $kn+1$ elements among $n$ sets, at least one sets must have at least $k+1$ elements.[/i]

The simplest way to grasp this intuitively, I think, is in terms of contradiction.  What happens if none of the sets have at least $k+1$ elements?  Then each set has at most $k$ elements.  But then there can be at most $kn$ elements in total - [b]contradiction.[/b]


How can we find the last four digits of $1996^{6}$ quickly :?:

Good point...I for some reason thought $-6 \equiv 1\mod{5}$

[quote="TRAN THAI HUNG"][quote="Gyan"][hide][quote="K81o7"]Actually, that only yields the value mod 2000.  You really just need to use

$1996^{6}=(2000-4)^{6}\equiv-6*2000*4^{5}+4^{6}\equiv-2000+4096 \equiv 2096\mod{10000}$[/quote]
Typo there, should be $-8000+4096 \equiv 6096 \bmod 10000$[/hide][/quote]
I Think $1996^{3}=7952095936\equiv 5936\mod10000$
$1996^{6}\equiv 1996{}^{3}{}^{2}\equiv 5936^{2}=35236096\equiv 6096\mod 10000$ :roll:[/quote]

um...well with that you could have just used your calculator to do $1996^{6}$ now couldn't you?

[quote="Elemennop"]Can't you just multiply 1996^6 out? It's doesn't take more than a minute[/quote]
Yes,  but  calculating  $4^{6}=4096$ could be a  little faster.
(Specially if you know  if 1K of memory = 2^10 = 4^5  is 1024 bytes)

And all  you have to do is to "modify"  the left digit  4 by adding 2 to get $6096$.

[hide]Unit digit of $4^{5}$ is $4$ (Unit digit of  fifth power of any number whose unit digit is $a$ is $a$) ( or $a^{5}\equiv a \bmod 10$) and  (-6) is same as (+4) mod 10, and you all you are doing is finding the unit digit of 4*2*4 which is 2 , so you add 2 to 4 and get $6096$.

I am, of couse, using the method  shown in the post of K18o7.
[/hide]

K1807 - Can you please edit the typo out of your post. Thanks.


With the two threads about runescape and maplestory going on, I would like to see which one you think is better.

Because nobody's ever heard of Maplestory here :P .

I voted for Runescape simply because I've never heard of Maplestory. SOme of my friends play runescape, and got really obsessed with it. I got a screen name when I went to this party but never actually played... the real reason is that I don't know how anything work... guess I'm not smart enough to play videogames/computer games....

...that's what treething mentioned...that less people have heard of Maplestory...


Taking any $a$ coprime to $n$, we find $b \equiv -a^{-1} \mod n$ such that $ab\equiv -1\mod n$, thus $a+b \equiv 0 \mod n$, giving $a^2 \equiv 1 \mod n$.
We just used $a$ coprime to $n$, thus we conclude like in http://www.problem-solving.be/pen/viewtopic.php?t=150 .

[quote="ZetaX"]giving $a^2 \equiv 1 \mod n$.[/quote]How do you do this? I see $(a+b)^2-2ab=a^2+b^2\equiv 2\pmod n$, but why do $a$ and $b$ have the same $\mod n$ remainder?

We set $b=-a^{-1}$ and get $a-a^{-1} \equiv 0 \mod n$.
Multiplication with $a$ gives $a^2-1 \equiv 0 \mod n$.

Or with your identity: $a \equiv -b \mod n$ by the condition, giving $2a^2 \equiv a^2+b^2 \equiv 2 \mod n$.

[quote="Peter"]Determine all positive integers $n$ such that $ xy+1 \equiv 0 \; \pmod{n} $ implies that $ x+y \equiv 0 \; \pmod{n}$.[/quote]

In order to fill the ??? in the source: This is part [b](b)[/b] of AMM problem S 9 [1979, 306] by M. S. Klamkin and A. Liu. Reference from JSTOR:

Solutions of Problems Dedicated to Emory P. Starke
S9
M. S. Klamkin; A. Liu; Arnold Adelberg; Jeffrey M. Cohen
The American Mathematical Monthly, Vol. 87, No. 6. (Jun. - Jul., 1980), p. 488.

For the sake of completeness, part [b](a)[/b] states that: Determine all positive integers $n$ such that $\gcd\left(x,n\right)=1$ implies that $x^2\equiv 1\pmod n$.

  Darij

Would this be a generalization of A.18? :wink:


algebra
polynomial
superior algebra
superior algebra unsolved

in Zn[x] how many different polynomials are there of degree <= d?

in Zn[x], how many different monic polunomials are there of degree d?

in Zn[x], how many different polynomials are there of degree d?

please advise me on steps to take... thank you

[quote="sbhs08"]please advise me on steps to take... thank you[/quote]

count...

How many possibilities are there for the constant term? For the coefficient of $x$? ...

[quote="sbhs08"]how about no sarcasm... i mean what do i need to count i am confused on the form i need.[/quote]

Excuse me! I was just trying to point out that counting is the only way to go.

solving the last one will be enough
having degree d we have d+1 coeffs
each of them can take values from {0,...,n-1}(that is n valueas)
except for the firs coeff wich cannot take value 0
so the number of polinomials of degree d in Zn[X] is
$K_d=(n-1)\cdot n^d$

now for the monic pol of degree d we have the same as above
only the first coeff is fixed 1 so all the others can take any
value in Zn so we have $n^d$ such pol

for the first problem i think one can say
$\sum_{i=1}^{d} K_i$ are the number of polinomials satisfying that
condition

i hope im right :)


trigonometry
real analysis
real analysis unsolved

We know that $\bigr(\sin(n)\bigl)_{n\in \mathbb{N}}$ is dense in $[-1,1]$.
If $k \in \mathbb{N}$, is $\bigr(n^{k}\sin(n)\bigl)_{n\in \mathbb{N}}$ dense in $\mathbb{R}$?

Hint:
The answer is no for $k=9$

For $k\ge 9$ it follows easily from Hata's theorem that $n^{k}|\sin n|$ tends to infinity as $n\to+\infty$. On the other hand, if you can tell what happens for $k=3$, say, many people including myself will be extremely interested. So, do you have a solution?

For some of the references where the related question was discussed you may look here
http://www.mathlinks.ro/Forum/viewtopic.php?t=15245&highlight=hata

fedja,
I don't have a solution unfortunately.
Thanks eugene for the reference.


group theory
abstract algebra
LaTeX
superior algebra
superior algebra unsolved

Could someone please tell me if I've correctly listed all subgroups of $ S_{4}$?

Subgroups of $ S_{4}$: $ A_{4}$, $ D_{4}$, $ S_{3}$, $ \mathbb{Z}_{2}$, $ \mathbb{Z}_{3}$, $ \mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$, $ \mathbb{Z}_{4}$, $ \mathbb{Z}_{2}\oplus\mathbb{Z}_{4}$, and the trivial group $ \left\{e\right\}$.

Also, how do you draw a subgroup lattice using LaTeX?

Not quite: all subgroups of order 8 in $ S_4$ can be shown to be conjugate, therefore isomorphic to each other. Since the dihedral group is certainly a subgroup of $ S_4$, the group $ \mathbb{Z}_2\oplus \mathbb{Z}_4$ cannot be one.

Other than that, you have all the types of proper subgroups up to isomorphism, although you might want to state how many of each type there are. (If you show that $ A_4$ has no subgroup of order 6, this can be used to show $ S_4$ has no subgroup of order 12 except $ A_4$. I mentioned the case of order 8 above. For order 6, $ S_4$ has subgroups isomorphic to $ S_3$ and no element of order 6. Finally, all groups of order 4 or less are isomorphic to a subgroup of $ S_4$.)


How can I compute my GPA when my grades are of 20?
Is there a linear relation or something else?

Different schools have different formulas for GPA; you'll have to ask your counselor.  For example, my school assigns a value of 4.0 to A, 3.0 to B, and so forth, but some schools also assign a 3.7 to an A-, a 3.3 to a B+, and so forth.

I'd multiply your grades by 5, and then assign them American letter equivalents and calculate the GPA from there. (Would make 20 a 4.3, 19 4.0, 18 3.7, 17 3.4 and so on)


number theory
prime factorization
geometric series

[quote="gokussj3"]Let H be a finite set of positive integers none of which has a prime factor
greater than 3. Show that the sum of the reciprocals of the elements of H is
smaller than 3.[/quote]

[hide="My solution"]
Okay. Well, quite obviously they're all powers of two. So, H might  be something like $ H\equal{}{2,4,6,8...1\minus{}0}$, but, another fact is that the smallest number is 2, therefore the largest reciprocal is $ 1/2$ which is obviously smaller than 3.
[/hide]

@Poincare: Umm, why can't the smallest number be $ 1$? Why must it be $ 2$? And I am not sure if you are answering the whole question. The numbers don't all have to be powers of $ 2.$ They can also be powers of $ 3,$ as the problem says that the number can't contain primes that are [i]greater[/i] than $ 3$. So $ 3$ is acceptable. Even if $ 2$ is the smallest number, how does the reciprocal of two being smaller than $ 3$ imply that the whole sum is smaller than $ 3$? You are adding a lot more terms. Why can't the sum end up being something greater than 3?

Here is I guess a more rigorous proof?

[hide]So the number must be of the form $ 2^a3^b,$ where $ a$ and $ b,$ not necessarily distinct, are non-negative integers (this represents the prime factorization of the numbers in the set). Anything else, the number would have another (greater) prime factor, which isn't allowed in this problem. Let $ a le m,$ and $ b \le n.$

So, the numbers in the set are
\[ 2^0\cdot 3^0, 2^0\cdot 3^1, \dots, 2^1\cdot 3^0, 2^2\cdot 3^0, \dots, 2^m \cdot 3^n.\]
And therefore, the sum of their reciprocals is (and from the problem, we have)
\[ \frac{1}{2^0\cdot 3^0} \plus{} \frac{1}{2^0\cdot 3^1} \plus{} \dots \plus{} \frac{1}{2^1\cdot 3^0} \plus{} \frac{1}{2^2\cdot 3^0} \plus{} \dots \plus{} \frac{1}{2^m\cdot 3^n} < 3.\]
The first term is just $ 1.$ We can group together some of the terms to give us
\[ 1 \plus{} \frac{1}{3^1} \plus{} \frac{1}{3^2} \plus{} \dots \plus{} \frac{1}{2^1} \plus{} \frac{1}{2^2} \plus{} \dots \plus{} \frac{1}{2^m\cdot 3^n} < 3,\]
so, we have the series, $ 1/3 \plus{} 1/9 \plus{} 1/27 \plus{} \dots \plus{} 1/3^n.$ We also have the series $ 1/2 \plus{} 1/4 \plus{} 1/8 \plus{} \dots \plus{} 1/2^m.$ These geometric series evaluate to $ \dfrac{1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1}$ and $ \dfrac{1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1},$ respectively.

Furthermore, we also have the series
\[ \frac{1}{2\cdot 3} \plus{} \frac{1}{2 \cdot 3^2} \plus{} \dots \plus{} \frac{1}{2^2 \cdot 3} \plus{} \frac{1}{2^2 \cdot 3^2} \plus{} \dots \plus{} \frac{1}{2^m \cdot 3^n}.\]
From all the terms that have $ 2^1$ in their denominator, we can factor out a $ 1/2$ from those fractions. Those that have $ 2^2$ in their denominator, we can factor out a $ 1/4$ from, etc. Since each of the powers of two (with positive integers, as we have already considered the case where 2^0 occurs) is multiplied by each of the positive integer powers of $ 3,$ after some factoring and grouping, we have
\[ \left(\frac{1}{2} \plus{} \frac{1}{2^2} \plus{} \dots \plus{} \frac{1}{2^m}\right)\left(\frac{1}{3} \plus{} \frac{1}{3^2}\plus{} \dots \plus{} \frac{1}{3^n}\right).\]
Each of these terms we have evaluated before. So we have
\[ \left(\dfrac{1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1}\right) \left( \frac{1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1} \right)\]
Combining everything (and subtracting the one from both sides), we now have
\[ \frac{1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1} \plus{} \frac{1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1} \plus{} \left(\frac{1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1}\right)\left(\frac{1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1}\right) < 2\]
I think we can work with that and simplify that a bit to get as desired. [/hide]

[quote="Poincare"][quote="gokussj3"]Let H be a finite set of positive integers none of which has a prime factor
greater than 3. Show that the sum of the reciprocals of the elements of H is
smaller than 3.[/quote]

[hide="My solution"]
Okay. Well, quite obviously they're all powers of two. So, H might  be something like $ H \equal{} {2,4,6,8...1 \minus{} 0}$, but, another fact is that the smallest number is 2, therefore the largest reciprocal is $ 1/2$ which is obviously smaller than 3.
[/hide][/quote]

Yeah, as limac said, there are two problems with your solution. First, you can also include powers of three, and second, even if you were assuming it was only powers of 2, you would still have to sum up $ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \cdots\plus{}\frac{1}{2^{something}}$.

To include powers of 3, limac, your solution is good, except you could probably simplify it a bit by saying that the sum is less than if the set was infinite, in which case the sum of the reciprocals is $ (1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \cdots)(1 \plus{} \frac {1}{3} \plus{} \frac {1}{9} \plus{} \cdots) \equal{} (2)(\frac {3}{2}) \equal{} 3$

[quote="andersonw"]...except you could probably simplify it a bit by saying that the sum is less than if the set was infinite, in which case the sum of the reciprocals is $ (1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \cdots)(1 \plus{} \frac {1}{3} \plus{} \frac {1}{9} \plus{} \cdots) \equal{} (2)(\frac {3}{2}) \equal{} 3$[/quote]
Wow, that is a really nice, and indeed much simpler, method!

[quote="limac"]@Poincare: Umm, why can't the smallest number be $ 1$? Why must it be $ 2$? And I am not sure if you are answering the whole question. The numbers don't all have to be powers of $ 2.$ They can also be powers of $ 3,$ as the problem says that the number can't contain primes that are [i]greater[/i] than $ 3$. So $ 3$ is acceptable. Even if $ 2$ is the smallest number, how does the reciprocal of two being smaller than $ 3$ imply that the whole sum is smaller than $ 3$? You are adding a lot more terms. Why can't the sum end up being something greater than 3?

Here is I guess a more rigorous proof?

[hide]So the number must be of the form $ 2^a3^b,$ where $ a$ and $ b,$ not necessarily distinct, are non-negative integers (this represents the prime factorization of the numbers in the set). Anything else, the number would have another (greater) prime factor, which isn't allowed in this problem. Let $ a le m,$ and $ b \le n.$

So, the numbers in the set are
\[ 2^0\cdot 3^0, 2^0\cdot 3^1, \dots, 2^1\cdot 3^0, 2^2\cdot 3^0, \dots, 2^m \cdot 3^n.\]
And therefore, the sum of their reciprocals is (and from the problem, we have)
\[ \frac {1}{2^0\cdot 3^0} \plus{} \frac {1}{2^0\cdot 3^1} \plus{} \dots \plus{} \frac {1}{2^1\cdot 3^0} \plus{} \frac {1}{2^2\cdot 3^0} \plus{} \dots \plus{} \frac {1}{2^m\cdot 3^n} < 3.\]
The first term is just $ 1.$ We can group together some of the terms to give us
\[ 1 \plus{} \frac {1}{3^1} \plus{} \frac {1}{3^2} \plus{} \dots \plus{} \frac {1}{2^1} \plus{} \frac {1}{2^2} \plus{} \dots \plus{} \frac {1}{2^m\cdot 3^n} < 3,\]
so, we have the series, $ 1/3 \plus{} 1/9 \plus{} 1/27 \plus{} \dots \plus{} 1/3^n.$ We also have the series $ 1/2 \plus{} 1/4 \plus{} 1/8 \plus{} \dots \plus{} 1/2^m.$ These geometric series evaluate to $ \dfrac{1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1}$ and $ \dfrac{1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1},$ respectively.

Furthermore, we also have the series
\[ \frac {1}{2\cdot 3} \plus{} \frac {1}{2 \cdot 3^2} \plus{} \dots \plus{} \frac {1}{2^2 \cdot 3} \plus{} \frac {1}{2^2 \cdot 3^2} \plus{} \dots \plus{} \frac {1}{2^m \cdot 3^n}.\]
From all the terms that have $ 2^1$ in their denominator, we can factor out a $ 1/2$ from those fractions. Those that have $ 2^2$ in their denominator, we can factor out a $ 1/4$ from, etc. Since each of the powers of two (with positive integers, as we have already considered the case where 2^0 occurs) is multiplied by each of the positive integer powers of $ 3,$ after some factoring and grouping, we have
\[ \left(\frac {1}{2} \plus{} \frac {1}{2^2} \plus{} \dots \plus{} \frac {1}{2^m}\right)\left(\frac {1}{3} \plus{} \frac {1}{3^2} \plus{} \dots \plus{} \frac {1}{3^n}\right).\]
Each of these terms we have evaluated before. So we have
\[ \left(\dfrac{1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1}\right) \left( \frac {1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1} \right)\]
Combining everything (and subtracting the one from both sides), we now have
\[ \frac {1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1} \plus{} \frac {1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1} \plus{} \left(\frac {1/2[(1/2)^m \minus{} 1]}{1/2 \minus{} 1}\right)\left(\frac {1/3[(1/3)^n \minus{} 1]}{1/3 \minus{} 1}\right) < 2\]
I think we can work with that and simplify that a bit to get as desired. [/hide][/quote]

whoops. I didn't read the sum part. :(


A long, cylindrical conductor of radius a has two cylindrical cavities of diameter a through
its entire length, as shown in cross-section. A current I is directed out
of the page and is uniform through a cross section of the conductor. Find the magnitude
and direction of the magnetic eld (a) at point P1 and (b) at point P2.

One can sum the effect of the full circular cross section with that from the cavities with
a inverted current; each has symmetry that admit the use of Ampere's circuital law.

[hide="You shoud get"]
\[ H_1\; \equal{} \; I{{r^2 \minus{} 2a^2}\over{2\pi r(r^2 \minus{} a^2)}}
\]

\[ H_2\; \equal{} \; I{{r^2 \plus{} 2a^2} \over {2\pi r (r^2 \plus{} a^2)}}
\]
[/hide]


Solution in spoiler:

[hide] n is the largest value such that 3^n divides 125!.  By Lagrange, there are flr(125/3)+flr(125/3^2)+... = 41+13+4+1 = 59, so n = 59.

[/hide]

-interesting_move

[quote="DPatrick"]Find a positive integer $n$ such that $3^n \,|\, 125!$ but $3^{n+1} \not|\, 125!$  (Here $a \,|\, b$ means that $a$ divides $b$; i.e. that $b/a$ is an integer.)[/quote]

n is the amount of 3s in 125!
so the prime factorisation would have 3^n in it

there are [125/3] + [125/9] + [125/27] + [125/81] 3s in it because in every 125 numbers there are [125/3] multiples of three, [125/9] multiples of nine (which contain two threes..) etc.

interesting_more: your solution is almost correct, although I'm sorry but the formula you cited is attributed to Legendre and not Lagrange.

Umm...

[hide]

The factors of 3 in 125! are below...

There are 28 numbers that are divisible by 3^1 at most

There are 9 numbers that are divisible by 3^2 at most

There are 3 numbers that are divisible by 3^3 at most

There is 1 number that is divisible by 3^4 at most

28 + 18 + 9 + 4 = 59

n = 59

[/hide]


this equation is to be solved in which set of numbers? usually we denote by capital letters matrices, and that will also require the field / ring of the elements ... please complete the problem :)

Sorry for my erroneous notation.

A,B are real numbers.

[quote="manlio"]Find the general solution of

  A^3 + B^3 + 3AB = 1

P.S. A=-1 , B=-1 is an isolated point.[/quote]
================
I think the identity  A^3+B^3+3AB-1 = (A+B-1)(A^2+B^2+1-AB+A+B) solves this problem. The solution is A+B = 1 or A=B=-1.

Namdung


Euler calculus trigonometry function algebra polynomial Vieta

I am doing a proof in my math class that involves Euler and his Pi proof.

He proves that (Pi) squared over 6 is equal to the sum of all the squares of a harmonic series.

can anyone explain this proof to me? i just dont get it. :(

It's pretty complex, it involves taylor series, which is calculus.

lol but my teacher said there was a completely algebraic way to do it

Yes there is, but again, it involves some definitions from Calculus.
If you take my word on the rigor of these definitions, I'll outline the proof for you.

[hide="Proof"]
Basically, you see that (begin calculus assumptions here)
$\sin(x) = x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}\ldots$
If you divide by x, we get something like this:
$\frac{\sin(x)}{x}= 1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}\ldots$

Now, we know from trig that the roots of the sine function occur at integer multiples of $\pi$. From this, we know that the roots of $\frac{\sin(x)}{x}$ also occur at positive and negative integer multiples of $\pi$. This is not a difficult lemma to prove, but I'll skip it for now.

Break $\frac{\sin(x)}{x}$ into the infinite product of linear factors of it's roots, like we can do to reduce polynomials into their unique monic polynomials.

This gives $\frac{\sin(x)}{x}= (1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})(1-\frac{x}{3\pi})(1+\frac{x}{3\pi})\ldots$

Now, multiply through:
$\frac{\sin(x)}{x}= (1-\frac{x^{2}}{\pi^{2}})(1-\frac{x^{2}}{4\pi^{2}})(1-\frac{x^{2}}{9\pi^{2}})\ldots$

(This comes from binomial expansion)
From Vieta's Formulas, we can get the coefficient on $x^{2}$ in the previous formula.
Effectively, we collect the terms and get the coefficient of $x^{2}$ is $-\left(\frac{1}{\pi^{2}}+\frac{1}{\pi^{2}}+\frac{1}{4\pi^{2}}+\frac{1}{9\pi^{2}}\ldots\right)$
When we factor out a $-\frac{1}{\pi^{2}}$, we're left with this:
The coefficient on $x^{2}$ in the expansion of $\frac{\sin(x)}{x}$ is $-\frac{1}{\pi^{2}}\cdot \sum_{n=1}^{\infty}\frac{1}{n^{2}}$

Look familiar?
We're almost done.
Look at the original taylor series for $\frac{\sin(x)}{x}$.
We can IMMEDIATELY see that the coefficient on $x^{2}$ is also $-\frac{1}{3!}$, or $-\frac{1}{6}$

Now we're golden.
We have two values for the coefficient of the same formula.
Equating the sides, we get
$-\frac{1}{6}=-\frac{1}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}$
Bam.

Solve that puppy for the sum.
Multiply through by $-\pi^{2}$, and we get the closed form.
$\sum_{n=1}^{\infty}\frac{1}{n^{2}}= \frac{\pi^{2}}{6}$
Beautiful.
[/hide]

Of course, the total lack of rigour in the argument above comes in expanding $\frac{\sin x}{x}$ into a product like that. Many entire functions (functions with a power series expansion in the whole complex plane) do not admit product factorizations in the manner described here (although there is a product form). It is actually quite difficult to establish the product form.

One thing that should make the problem clear is that the initial product in DFTs post is not even absolutely convergent. This means that by changing the order of the factors I can make the product attain any value and not just $\frac{\sin x}{x}$

I concede the lack of rigor.
There is a more general solution on Wikipedia under the title
"Basel Problem"
There is a much more rigorous proof given there with the help of complex analysis, although that is WAY above normal high school level, and probably should go in the calculus forum.


When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

are you sure the question is right? I'm getting a non-integer answer.

I forgot you add the 1 in addition to the 15.  :wallbash_red:

[quote="mathgeniuse^ln(x)"][hide]Let the number of elements be n.

Let the average be x.

This means that

$\frac{nx+15}{n+1}=x+2$

So $nx+15=nx+2n+x+2$

Also, from the second condition.

$\frac{nx+16}{n+2}=x+1$.

So $nx+16=nx+n+2x+2$

So, the two equations are

$2n+x=13$ and $n+2x=14$

Solving we get that $n=4$ and $x=5$.

So the number of terms in the original list is 4

$4$[/hide][/quote]

Yes, you are correct.  :)

Thanks.

And SplashD, you should get a correct answer.

Edit:Nevermind, you realize your mistake.

Here's my method:
[hide="Algebraic method"]Let there be $n$ integers. Then their sum is $mn$, where $m$ is the mean. So, when the number $15$ is added, the sum is $mn+15$, and the number of integers is $n+1$. Therefore,

$\frac{mn+15}{n+1}= m+2$

$mn+15 = mn+m+2n+2$
$m+2n =13$

If we add 1 to this enlarged list, the sum is $mn+16$, and the number of integers is $n+2$. So,

$\frac{mn+16}{n+2}= m+1$

$mn+16 = mn+2m+n+2$
$2m+n = 14$

Solving these two equations simultaneously, subtract twice the first equation from the second to get:
$(2m+n)-(2m+4n) = 14-26$
$-3n=-12$
$n=4$

Hence, there are 4 integers. [/hide]


algebra
polynomial
function
calculus
ratio
search
algebra unsolved

These 2 problems should be a piece of cake for u guys .They probably are too [i]easy[/i]
to be here. I'll really appreciate if you could help me out.

(1) Prove that $ (X)^\( 1/3 \) $ is not a polynomial . (rigorous proof not just reasoning from defintion.)

2) Solve the functional eqn .$ (p(x))^2 = (p(x)^\(2\)) $

3) (This is a theoretical problem) As we claim in the field of Reals  that f(x) = F(inverse)(x) implies f(x)=x , can we claim that also in the field of complex nos.
Also is there any way to visualise plotting of a complex fn F(x,y,z) from the real 3-D volume to a comlex surface or anything of the like.(I'm a beginner here sp plz take it slow)

Thanks for your help.

[quote="puuhikki"]Number one is easy. $x^{1/3}$ is multiple branched function in $\mathbb{C}$ while polynomials are not.[/quote]

[b]Multiple bRANCHED ??????????? i'M A beginner[/b]

[quote="Diogene"][quote="Rushil"] [b]1) Prove that X^(1/3) is not a polynomial .(rigorous proof not just reasoning from defintion.) [/b][/quote]
Let f(x) be the function  $f(X) = X^a , a >0, a \neq natural , X\in R$ .Suppose that $f(X)$ is a polynomial
Manifestly $f(X)$ has only the root $r=0$ so  $f(X) = cX^n$ where $n=natural$
Now we have $ X^a = cX^n , \forall X\in R$, and that's impossible if $a \neq natural$

 :cool:[/quote]

how do we know it's got just one root?? how do we conclude?

OK, it seems to me that we are talking $x^{1/3}$ in the real line, so that my complex argument won't work. If you know some elementary calculus you can easily see that $\frac{d}{dx} x^\frac{1}{3}=\frac{1}{3}x^{-\frac{2}{3}}$ is not defined in $x=0$. But for an arbitrary polynomial $P(x)$ there exist real-valued $P'(0)$ contradicting the assumpion $x^{1/3}$ is a polynomial.

Or you can see that the ratio between $x^{\frac 13}$ and any non-constant monic polynomial tends to $0$ as $x\to\infty$, and this means that it should be a constant, which is clearly not true.

Neither diogene's solution nor grobber's use any calculus.

[quote="Rushil"]You guys are giving solns. based on calculus.The book wants a soln. [b]not based on calculus .[/b] Also please try to give simple solns and not those based on advanced theories.Please refer to [b]Barbeau's "Polynomials"[/b] if you like.[/quote]

i dont see anything that is unfamiliar to me, i don't see calculus there, and yet the ideas of Diogene's requires a little bit more of thinking. and you will have to be more specific because we are here only [b]to help[/b], not to force us on giving you solutions. also, not everyone has that book (i don't have it) and so you should be more specific.

the third question is the one thats really bogging me. All your solns are acceptable . I was only responding to the " mutiple branched " argumeny given.[b]So stop firing the guns !![/b]
Anyways, the third question is not from the book.(source :lol: )
Can you give me an answer for that ?!
[b]Thanks !!![/b]

[quote="Rushil"][quote="Diogene"][quote="Rushil"] 3) (This is a theoretical problem) As we claim in the field of Reals  that f(x) = F(inverse)(x) implies f(x)=x , can we claim that also in the field of complex nos.
Thanks for your help.[/quote]

[quote="Rushil"]Anyways, the third question is not from the book.(source Very Happy ) [/quote]

Indeed, it's not from the book, because the statement is false    :D  Take for exemple $f(x)= -x$ :D
 :cool:[/quote]

Are you out of your mind ? The above statement is a standard property based on the fact that the inverse of a real fn is the mirror image of f(x) in Y=X. It doesn't work when f(x) coincides with Y=X and some other cases as you have shown.What I'm asking is that is the above property true in the field of complex nos. as well?(leaving the counter-cases)[/quote]

 :|
Well...
Diogene is right, so think about your (wrong) statement...  :huh:

[quote="ZetaX"][/quote][quote="Rushil"][quote="Diogene"][quote="Rushil"] 3) (This is a theoretical problem) As we claim in the field of Reals  that f(x) = F(inverse)(x) implies f(x)=x , can we claim that also in the field of complex nos.
Thanks for your help.[/quote]

[quote="Rushil"]Anyways, the third question is not from the book.(source Very Happy ) [/quote]

Indeed, it's not from the book, because the statement is false    :D  Take for exemple $f(x)= -x$ :D
 :cool:[/quote]

Are you out of your mind ? The above statement is a standard property based on the fact that the inverse of a real fn is the mirror image of f(x) in Y=X. It doesn't work when f(x) coincides with Y=X and some other cases as you have shown.What I'm asking is that is the above property true in the field of complex nos. as well?(leaving the counter-cases)[/quote][quote="ZetaX"]

 :|
Well...
Diogene is right, so think about your (wrong) statement...  :huh:[/quote]

I hope you guys are not interpreting me wrongly . Try any fn and its inverse pair. log(x) and E^(x) ,sin(x) and it;s inverse etc.

the soln of the eqn F(x) =(f(inverse))(x) is same as f(x) = x.Is it ok now . I need some clarification

How I (and also Diogene I think) understood your statement is:
Let $f$ be any function from $\mathbb{R}$ to $\mathbb{R}$ that fulfills $f(x) = f^{-1}(x)$ for all real $x$ , where $f^{-1}$ is the inverse function to $f$, in other words the function that gives us $y$ where $f(y)=x$ for given $x$.
right¿
Then look at diogenes counterexample.
otherwise:
please explain again what you want to say.

[quote="grobber"]I'm pretty sure he means this: if $f(x_0)=f^{-1}(x_0)$ then $f(x_0)=x_0$. I put $x_0$ instead of $x$ to underline the fact that this is not concerned with all $x$, but a specific value of $x$. In othger words, the roots of the equation $f(x)=f^{-1}(x)$ are the saem as the roots of $f(x)=x$.[/quote]

[b]Exactly grobber!!![/b]That's what I meant .Sorry I wasn't clear

I hope now you guys can answer my problem!

that's strange, so what does this mean? that fixed point also works in complex plane?

Yes . All i am asking is that does such a "fixed point" exist for functions in the $ \mathbb{C} $ field.

[quote="grobber"]I'm pretty sure he means this: if $f(x_0)=f^{-1}(x_0)$ then $f(x_0)=x_0$. I put $x_0$ instead of $x$ to underline the fact that this is not concerned with all $x$, but a specific value of $x$. In othger words, the roots of the equation $f(x)=f^{-1}(x)$ are the saem as the roots of $f(x)=x$.[/quote]
Let $f(x)=1-x$ and pick $x_0$ randomly. You will find that $f(x_0)=f^{-1}(x_0)$, but, unless you are completely unlucky, $f(x_0)\ne x_0$.

In the case of some linear functions (as the one you have posted ) $ f^{-1} (x) \equiv f(x) $ so the eqn is true $ \forall$ x But I am asking in the general case of non-overlapping functions

whats about $f_1(x)= \frac{c}{x}$ or $f_2(x)=c-x$
the only thing u can derive out of $f(x)=f^{-1}(x)$ is that $f^2(x)=x$ , with $f^2(x)=f(f(x))$ ;)

I made a search and found this useful topic. :)

 [quote]the roots of the equation $f(x)=f^{-1}(x)$ are the same as the roots of $f(x)=x$[/quote]

A well-known [b]wrong[/b] property. :)

$f(x)=-x^3$ is a polynomial function with the properties Rushil was interested in:

- the roots of the equation $f(x)=f^{-1}(x)$  are [b]not[/b] the same as the solutions of the equation $f(x)=x$;
- the equality $f(x)=f^{-1}(x)$ is not an identity (it doesn't hold for every x in R).

(I choosed f such that its graph contains points (-1,1), (0,0), (1,-1) ).

Although it's a well-known result, since nobody mentioned it, I feel it is my duty (well, almost ;) ) to state the correct form of the property I refuted above:

If f is [b]strictly increasing[/b](that's what was missing !) and bijective, then the roots of the equation $f(x)=f^{-1}(x)$ are the same as the roots of $f(x)=x$.

Here is the proof:

Obviously $f(a)=a$ entails $f(a)=f^{-1}(a)$.

Conversely, if $f(a)=f^{-1}(a)$, then $f(f(a))=a$.
Aassume that $f(a)\neq a$.
If, say, $f(a)<a$, since f is strictly increasing it follows that $f(f(a))<f(a)$, hence $a<f(a)$, contradiction.
The same if $f(a)>a$.
It remains that $f(a)=a$.


Tran sorry for spamming in your topic but I think the ineq is false when $ c\equal{}0$ :wink:

[quote="anas"]Tran sorry for spamming in your topic but I think the ineq is false when $ c \equal{} 0$ :wink:[/quote]
Nope.I dont think so.

[quote="anas"]Tran sorry for spamming in your topic but I think the ineq is false when $ c \equal{} 0$ :wink:[/quote]
When $ c \equal{} 0$, this ineq is obviously trues, anas  :maybe:
Note that equality occurs when $ (a, b, c) \equal{} ( \minus{} 1, 1, 1)$ or $ (0,0, \sqrt {3})$...  :lol:

I'm really stupid , I switch a sign , just ignore what I wrote  :roll:

[quote="nguoivn"][quote="nguoivn"]Given $ a, b, c$ are the [b]real[/b] numbers satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:

$ 2(b^2c^2 \minus{} a^2)(c^2a^2 \minus{} b^2)(a^2b^2 \minus{} c^2) \plus{} 64a^2b^2c^2 \ge (bc \minus{} a)^2(ca \minus{} b)^2(ab \minus{} c)^2$

When does equality occur $ ?$ Hope that you'll like it, my dear friend  :)[/quote]
Can_hang and me had 2 nice and short proofs for this ineq  :lol:[/quote]

Dear nguoivn

I can use computing to solve the inequality.

substitution $ a \equal{} u x,b \equal{} u y,c \equal{} u z$($ x,y,z$are the [b]real[/b] numbers ,$ u > 0$)

from  $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ we have $ u \equal{} \sqrt {\frac {3}{x^2 \plus{} y^2 \plus{} z^2}}$

and  $ 2(b^2c^2 \minus{} a^2)(c^2a^2 \minus{} b^2)(a^2b^2 \minus{} c^2) \plus{} 64a^2b^2c^2 \geq (bc \minus{} a)^2(ca \minus{} b)^2(ab \minus{} c)^2$

$ \Longleftrightarrow f_1 \plus{} f_2\geq 0$

$ f_1 \equal{} z^4y^4x^4u^6 \minus{} z^2x^2y^2(4z^2x^2 \plus{} 3z^4 \plus{} 4y^2z^2 \plus{} 3y^4 \plus{} 4x^2y^2 \plus{} 3x^4)u^4 \plus{}$

$ (z^4x^4 \minus{} 4z^4x^2y^2 \minus{} 4z^2x^2y^4 \plus{} x^4y^4 \minus{} 4z^2x^4y^2 \plus{} z^4y^4)u^2 \plus{} 61z^2x^2y^2$

$ f_2 \equal{} 2x^3y^3z^3(x^2 \plus{} y^2 \plus{} z^2)u^5 \plus{} 2xyz(4z^2x^2y^2 \plus{} x^2z^4 \plus{} x^4z^2 \plus{} x^4y^2 \plus{}$

$ x^2y^4 \plus{} z^2y^4 \plus{} z^4y^2)u^3 \plus{} 2xyz(z^2x^2 \plus{} y^2z^2 \plus{} x^2y^2)u$

we can prove  while $ u \equal{} \sqrt {\frac {3}{x^2 \plus{} y^2 \plus{} z^2}}$

$ f_1\geq 0$

$ \Longleftrightarrow 22z^2x^2y^8 \plus{} 3z^4y^8 \plus{} 3x^4y^8 \plus{} 90x^4y^6z^2 \plus{} 6z^6y^6 \plus{} 6x^6y^6 \plus{} 90x^2y^6z^4 \plus{}$

$ 90z^6x^2y^4 \plus{} 222z^4y^4x^4 \plus{} 90z^2x^6y^4 \plus{} 3z^8y^4 \plus{} 3x^8y^4 \plus{} 90z^6x^4y^2 \plus{} 22z^2x^8y^2$

$ \plus{} 90z^4x^6y^2 \plus{} 22z^8x^2y^2 \plus{} 3x^4z^8 \plus{} 3x^8z^4 \plus{} 6x^6z^6\geq 0$

and

$ f_2 \equal{} \frac {8xyzu(6z^2x^2y^2 \plus{} x^4y^2 \plus{} x^2y^4 \plus{} z^2y^4 \plus{} z^4y^2 \plus{} x^4z^2 \plus{} x^2z^4)}{x^2 \plus{} y^2 \plus{} z^2}$

if $ xyz\geq 0$  we have $ f_1 \plus{} f_2\geq 0$ .

if  $ xyz < 0$

we have $ f1\le 0$

we only prove

$ f_1^2 \minus{} f_2^2\geq 0$

$ \Longleftrightarrow \frac {( \minus{} 3z^2x^2 \plus{} x^2y^2 \plus{} y^4 \plus{} y^2z^2)^2( \minus{} z^2x^2 \plus{} 3x^2y^2 \minus{} z^4 \minus{} y^2z^2)^2(x^4 \plus{} x^2y^2 \plus{} z^2x^2 \minus{} 3y^2z^2)^2}{(x^2 \plus{} y^2 \plus{} z^2)^6}\geq 0$

obvious true. :lol:

Wow, can you post it when you have time ?
I'm very curious. Thank you very much ^^!

[quote="nguoivn"][quote="hedeng123"]

$ (2)\Longleftrightarrow \frac {( \minus{} 3z^2x^2 \plus{} x^2y^2 \plus{} y^4 \plus{} y^2z^2)^2( \minus{} z^2x^2 \plus{} 3x^2y^2 \minus{} z^4 \minus{} y^2z^2)^2(x^4 \plus{} x^2y^2 \plus{} z^2x^2 \minus{} 3y^2z^2)^2}{(x^2 \plus{} y^2 \plus{} z^2)^6}\geq 0$

obvious true. :lol:[/quote]
Yes, that's it. This is the things I want to mention. Congratulation!  :lol:
Besides, this hint can be explain simply, my dear friend  :)[/quote]

 :lol:

Besides, we also have a similar result:
Let $ a, b, c$ be the[b] real numbers[/b] satisfying $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
\[ (ab \plus{} bc \plus{} ca)^6 \plus{} 2abc(a^2 \minus{} bc)(b^2 \minus{} ca)(c^2 \minus{} ab) \ge (a^2 \minus{} bc)^2(b^2 \minus{} ca)^2(c^2 \minus{} ab)^2\]
When does equality occurs $ ?$
 :)


geometry
probability
modular arithmetic
rectangle
MATHCOUNTS
ratio
number theory

Since people think I am basically spamming this forum, I will stop posting one at a time, and I apologize for the inconvience I have caused. Sorry.

1. What are the last two digits of $ 2^{222}$?
Doesn't tens digit repeat every 20 times? meh..
[hide="Answer"]$ 0$-tens, $ 4$-units [/hide]

2. Find the smallest (positive) integer $ n$ for which:

$ \frac{1}{1^2}\plus{}\frac{1}{2^2}\plus{} \cdots \plus{}\frac{1}{n^2}>\frac{3}{2}$
Can you do this without the calculator?
[hide="Answer"]$ n\equal{}7$[/hide]

3.One side of a triangle is divided into segments of lengths 6cm and 8cm by the point of tangency of the inscribed circle. If the radius of the circle is 4cm, what is the length, in centimeters, of the shortest side of the triangle?
[hide="Answer"]$ 13$cm[/hide]

4.Find the area of the region determined by the system:
$ y \ge|x|$

$ y \le \minus{}|x\plus{}1|\plus{}4$
Express your answer as a decimal.
[hide="Answer"]$ 7.5$[/hide]

5. The digital root of a number is found by adding the digits of the number, adding the digits of the resulting number, and so on, until the result is a single digit. If an integer $ n$ is selected randomly from 1 through 10, what is the probability that the digital root of $ n^3$ equals 1? Express your answer as a common fraction.
[hide="Answer"]$ \frac{2}{5}$[/hide]

6. In the figure, point $ A$ is the center of a 100cm-by-100cm square. Find $ x$, in centimeters, such that the shaded region has an area that is one-fifth of the area of the square.(Picture shown below)
[hide="Answer"]$ 37$cm[/hide]

for the last one, you can make the bottom-left vertice of the square be the origin, and use shoelace to find $ x$.

[hide="#1"]
Yes, you're right, the last two digits repeat after $ 2^{20}$. This is because $ 2^{21}\equal{}2097152$. So $ 2^{222} \equiv 2^{2} \pmod{100}$. Our answer is $ \fbox{04}$.[/hide]

For the last one I broke it up into one rectangle and two triangles. Hopefully you can imagine them. It is kind of hard to explain from here.
But the answer I got [hide="is"] 37 [/hide]
Edit: I can explain it. The rectangle is made from the top most right angle and the triangles are the remain pieces of the shaded region.

[hide="#5"]
This method is called "casting out nines," and it is used to check for divisibility by 9 (and also 3!). Thus we need to find all $ n^3$ in our set such that $ n^3 \equiv 1 \pmod{9}$. We find that this is satisfied when $ n$ equals 1, 4, 7, or 10. Thus our answer is $ \dfrac{2}{5}$.
[/hide]

@shentang: I figure that in our MATHCOUNTS meeting today. Thanks.

I got all except for #2,3, and 4.
Can you guys help?

[quote]2. Find the smallest (positive) integer n for which:

$ \frac{1}{1^2}\plus{}\frac{1}{2^2}\plus{} \cdots \plus{}\frac{1}{n^2}>\frac{3}{2}$
Can you do this without the calculator?[/quote]

The key to this one is the formula for the sum of a geometric sequence (yes, you do have it written down somewhere)
[hide]$ 1 \plus{} r \plus{} r^2 \plus{} r^3 ...r^n \equal{} \frac{r^{n\plus{}1} \minus{} 1}{r\minus{}1}$
Now you just have to plug in some numbers to find which ones work
[/hide]

[quote]3.One side of a triangle is divided into segments of lengths 6cm and 8cm by the point of tangency of the inscribed circle. If the radius of the circle is 4cm, what is the length, in centimeters, of the shortest side of the triangle? [/quote]

Go to [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=133[/url] and go to time stamp (16:18:52) to find help

[quote]
4.Find the area of the region determined by the system:
$ y \ge|x|$

$ y \le \minus{}|x\plus{}1|\plus{}4$
Express your answer as a decimal.[/quote]

Your answer here lies in simply graphing it.  Remember:  y cannot be negative because of statement #1.  Plotting a few points should help you find your solution.

I look forward to seeing your solutions for these problems!

you said that number 2. was a geometric series, if so, what is the common ratio?

i'm pretty sure that it is not a geometric series.
if i were solving that problem or a problem similar to that, the first thing i would do is check small numbers. the definition of small depends on the problem.

since the answer to this problem is a small number, i would like to propose a similar problem

it's the same as number two but the series has to be grater than 5. find n.
hint:or show that no n exists.

Sorry, I misread #2

I thought it said
$ \frac {1}{2^0}, \frac {1}{2^1},\frac {1}{2^2}...?$

6. In the figure, point  is the center of a 100cm-by-100cm square. Find , in centimeters, such that the shaded region has an area that is one-fifth of the area of the square.(Picture shown below)

[hide]One-fifth of the area would be 2000 cm.
Divide the quadrilateral into two triangles by connecting the center to the corner. The part with base of 17 has an area of 425. So, the second part has to have area 1575 and the base of it must be 63. x then = 100-63 = 37.[/hide]

[quote="mathking123"]
it's the same as number two but the series has to be grater than 5. find n.
hint:or show that no n exists.[/quote]

lol, the value of $ \frac{1}{1^2}\plus{}\frac{1}{2^2}\plus{}\frac{1}{3^2}\plus{}...$ is $ \frac{\pi^2}{6}$, which is about 1.6 something, so no $ n$ exists.

[quote="abacadaea"][quote="alanchou"][quote="mathking123"]
it's the same as number two but the series has to be grater than 5. find n.
hint:or show that no n exists.[/quote]

lol, the value of $ \frac {1}{1^2} \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ...$ is $ \frac {\pi^2}{6}$, which is about 1.6 something, so no $ n$ exists.[/quote]

erm, thats like saying the sum of terms in a series is 7. it is impossible to pick the first n terms and have a sum of 6.[/quote]

how so?


geometry calculus trigonometry Stanford college MATHCOUNTS Duke

Does anyone know of a school where either of the "Art of Problem Solving" books is used as the primary textbook for a math class?  From what I can tell, these books seem far superior to standard textbooks, at least for good students.

Well, the Stuyvesant math team is run like a class, and the Art of Problem Solving is used as one of several resources by the captains (who teach the classes) in preparing lessons.  I know I used it a lot when teaching the freshman level (algebra 1/geometry as their real math class) math team.  I've never heard of a part of the math curriculum anywhere that was competition-based, however.

Thanks for the info.  Even if a school was not particularly interested in competitions, I would still say that the Art of Problem Solving makes a good textbook.

Hey Joel, Stuy has the integrated math stuff but like what's the equivalent math they teach at which grades like at 9th, 10th or whatever.

Um . . . I'm not entirely sure what your question is, but I'll take a stab:
almost everyone comes into Stuy in the equivalent of geometry in freshman year, which is technically a 10th-grade class.  A few (20-60) come in taking the equivalent of algebra 1.  However, a rather large number of students end up taking both precalculus and calculus in their junior year and then taking multivariable calculus and differential equations senior year.  (There are about 150 or so students who are taking multi and diffeq this year, I think -- not all seniors.)

[quote="JBL"]The traditional high school curriculum, to the best of my understanding, is Algebra, Geometry, Algebra, and then I think Precalculus.  No?[/quote]

Here in the United States, that has been a common pattern. In the school district I grew up in in Minnesota in the 1960s and 1970s, the sequence was one year of algebra, one year of geometry, another year of algebra ("algebra 2"), and one year of precalculus (called "analysis" in my school). I lived in two school districts growing up, in suburban Minneapolis and suburban Milwaukee. The Minnesota school district did exactly the sequence I described, starting "enriched" (that is, tracked) students in algebra in eighth grade, and starting pretty nearly all students in algebra by ninth grade. The Wisconsin district started the top students in algebra in ninth grade, and most all the students started algebra in tenth grade. (The next course after algebra in my Wisconsin school district was NOT geometry, however, but something more like an "algebra 2" course. That is the next most common traditional sequence in America: two years of algebra, followed by a year of geometry, followed by a year of precalculus--which is usually called "trigonometry" in such a sequence.)

One consequence of my move from Minnesota to Wisconsin just after I started ninth grade is that I had the same algebra course TWICE, taught by a rather better teacher the second time, but with the identical textbook. I'm glad that today there are correspondence courses in mathematics for kids who want to move ahead and don't live somewhere with a school that allows math acceleration. The Stanford EPGY course sequence is algebra 1, algebra 2, geometry, precalculus for the span of secondary mathematics in question here. The local University of Minnesota Talented Youth Mathematics Program, founded just after I graduated from high school [sigh], has the sequence algebra 1 & 2 (first year), geometry & analysis (2nd year), followed by two years of calculus and then a year that is half advanced calculus and half advanced topics. Some kids here complete that whole sequence by the time they are tenth-grade age, but most are a year or two older, and not a lot of kids complete the whole sequence in any event.

I would like to know, JBL, at your convenience, a LOT more about the school where you got your secondary education. I have been to that fine city many times on business trips in subsequent years, but I don't know much about the school system there. Your alma mater is where most of the very first IMO team came from, right?

[quote="Leeyuhang"]geometry 10th grade? wow that's something new i learned today.[/quote]

I take it you are referring to the fact that in most of east Asia geometry is a junior-high subject, right? I have textbooks at home from Taiwan and from China (I can read Chinese), and have seen the translated seventh-grade textbook from Japan, and they all have more geometry in junior high years than one would see in the United States until recently.

Answering the original question about the book as a textbook - many schools use it as a textbook for a 'math team' or 'problem solving' course.  It is also used as a text at some summer programs.  Some places (other than Stuy) that we have sent the book to in this capacity in the last couple years include AAST, ICAE (weekend program in Michigan that put all 4 students on the state MATHCOUNTS team last year), the Duke TIP program, numerous middle schools in Fairfax County, numerous middle schools in GA & FL, most of the top math teams in AL, and at least a couple dozen other schools scattered throughout the country.

Also, there are a few schools that have allowed their top students to do independent study by working through our book instead of staying with the main curriculum.

Thanks for the response, and the great books.


calculus
integration
limit
calculus computations

find
$ \lim_{a \to 0,b \to 0} \int_{0}^{2(a\plus{}b)} \frac{dx}{\sqrt{2E\minus{}(x\minus{}a)^{2}\minus{}2c(x\minus{}a)^{n}}}$ where $ E\equal{}\frac{a^{2}}{2}\plus{}ca^{n}\equal{}\frac{b^{2}}{2}\plus{}cb^{n}$

The problem is not well defined.
E.g. for $ n \equal{} 2$, $ c \equal{} \minus{} \frac12$
$ \lim_{a \to 0,b \to 0} \int_{0}^{2(a \plus{} b)} \frac {dx}{\sqrt {2E \minus{} (x \minus{} a)^{2} \minus{} 2c(x \minus{} a)^{n}}} \equal{} \lim_{a \to 0,b \to 0} \int_{0}^{2(a \plus{} b)} \frac {1}{0}\, dx \equal{}$?

Moreover, I do not understand the meaning of $ \lim_{a \to 0,b \to 0}$ when $ a$ and $ b$ are dependent.

sorry i had forgotten to add n is odd and that a+b tends to zero and that a and b are independent.

$ E\equal{}\frac{a^{2}}{2}\plus{}ca^{n}\equal{}\frac{b^{2}}{2}\plus{}cb^{n}$
$ \to (b\minus{}a)(a\plus{}b)\equal{}\minus{}2c(b^n\minus{}a^n) \to b\equal{}a$ or $ \frac{a\plus{}b}{\minus{}2\sum_{k\equal{}1}^{n\minus{}1}a^kb^{n\minus{}k}}\equal{}c$

so c isnt a constant i take it?

For $ n\equal{}1$:
$ \lim_{b \to 0}\lim_{a \to 0} \int_{0}^{2(a\plus{}b)} \frac{dx}{\sqrt{2E\minus{}(x\minus{}a)^{2}\minus{}2c(x\minus{}a)^{n}}} \equal{}
\lim_{b \to 0}\int_{0}^{2b}\frac{dx}{\sqrt{\minus{}x^{2}\plus{}bx}}$,
but for $ x>b$ the right integrand is not defined.

For $ n>1$, $ n$ odd:
$ \lim_{b \to 0}\lim_{a \to 0} \int_{0}^{2(a\plus{}b)} \frac{dx}{\sqrt{2E\minus{}(x\minus{}a)^{2}\minus{}2c(x\minus{}a)^{n}}} \equal{}
\lim_{b \to 0}\int_{0}^{2b}\frac{dx}{\sqrt{\minus{}x^{2}\plus{}\frac{x^n}{b^{n\minus{}2}}}}$,
but for $ 0<x<b$ the right integrand is not defined.


[quote="Valentin Vornicu "]Ce părere aveţi de olimpiada judeţeană din acest an?[/quote]

Reluand topicul de anul trecut al lui Valentin.

Frumoase, dar cam usurele subiectele la toate clasele, mai putin la clasa a XII-a, unde au fost ca de obicei destul de grele.

Nu vreau sa par nu stiu cum, dar in comparatie cu anul trecut subiectele la clasa a-12-a au fost mult mai usoare.
 Cel putin analiza si problema 2 de algebra erau facubile dupa umila mea parere.  :)

Mie nu mi s-au parut foarte grele la a 11a numai ca am cam dat-o in bara la ultima problema (1p  :oops: ) dar in final a iesit bine , am luat 20,5 din cauza punctajelor de genul 6,5 :mad:

eu: clasa a XI-a: probl 1: 6.5
                         probl 2: 6.5
                         probl 3: 6
                         probl 4: 4

 :oops:

void -suceava age 17 a.k.a stefan popovici

[quote="Cezar Lupu"]Spix, tu esti din Suceava? Void tu de unde esti? De fapt ar fi o idee sa divulgati numele vostru real. ;)[/quote]

Da sunt. Parintii imi zic Marius Gherman.  :)

S-au centralizat cumva rezultatele pe tara?

eu clasa a X-a : pb1 7
                        pb2 7
                        pb3 7
                        pb4 6
27 p  :) oricum au fost foarte simple, de obicei  iau 21- 22 la judeteana (Satu Mare)

Ma bucur ca toate lumea este de parere ca subiectele au fost abordabile si ca s-au facut punctaje mari. Din pacate, nu este cazul si in Bacau, unde s-au obtinut rezultate destul de proaste la clasele mari. La a X-a si la a XII-a cele mai mari punctaje au fost in jur de 8 si nu va merge nimeni. La celelalte clase sunt punctaje peste 20. Astept si eu curios sa vad rezultatele pe tara.

Tu cat ai luat Adriane la astea de a 11-a? 28 presupun ... :)

Eu am luat 23 pcte. la a 12-a. Timisoara:) Nu stiti rezultate generale de prin tara? Vad ca in Buc s-a luat si cu 28. In Arad am auzit  ca la a 12-a s-a luat cu 16 pcte, Prahova 27,5, Mehedinti.... nu are rost sa zic, Constanta, 23 mi se pare. Sper sa nu ma insel.

la buzau primul  la a 12-a a luat 26 pct! frozen din pacate nu sunt eu! ( Octavian Rinciog)

au fost usoare in general, desi la noi(suceava) la clasele 9-10 s-au luat maxim 14-15...iar la a XI-a doar 2 oameni peste 20( io cu spix)

app
stie careva cand e exact onm?
stie cineva programu exact(cand se da proba)???
Mkr dati un link

Eu sunt din clasa 12a, Timisoara ; am luat 20 . Se gasesc pana la urma pe net rezultatele centralizate din toata tara?

eu am luat un 14 la a X-a cu toate ca subiectele au fost foarte abordabile.   :(

...sa vedem nationala

Hmm, nu stiu ce sa zic. Oricum problema mea si cea de-a doua care a fost de I.V. Maftei din cate am inteles puteau fi facute fara sa te gandesti prea mult la ele, iar problema 3 nu prea avea legatura cu clasa a-10-a, dar era facubila. ;)

am fost si eu la mate anu asta la jud , am mai fost si anu trecut dar eram cu chimia pe vremea aia ... anu asta m-am descurcat la judeteana , si am trecut mai departe .... n-am avut pcj f mare ... la a doua problema de a XI-a am luat doar 1 pt ... pt ca am folosit chestii de genu ca daca o A inversabila atunci rang(B)=rang(AB) , oricare ar fi matricea B ... ceea ce stiu sigur ca este adevarat . dar nu conteaza . voi p-aci sunteti veterani presupun cu nationalele m ? :)

oricum scuzati offtopicu ... dar voiam sa zic ca baietii cu matematica sunt cei mai inteligenti oameni .. am vazut care-i treaba cu chimia, cu fizica ... nu se compara cu matematica zic eu . respectu meu pt Valentin Vornicu !! si pt Zahariuc Adrian sau cum il cheama, baieti destepti , ce sa mai !!  :idea:

da tu cine esti ? si despre ce subiecte penibile vorbesti ? aha ... edit : hai k m-am uitat, am vazut ca ai fost la Yakutsk ... probabil il stii pe Andrei Stefan de la chimie ;) si pe Deneanu . oricum , ai fost cam snob la faza asta. sunt totusi subiecte de judeteana ... adik oricat de simple ar fi , au totusi o dificultate in ele . zic eu , the human

[quote="octavian_158"]ba da voi astia de prin tara de va ganditi k lati luat pe unu sau pe altu cu 2-3 pct chiar credeti k e obligatoriu sai luati si la nationala; stati ma calmi k la subiectele astea penbiile chiar nu conteaza ceai facut  ;)[/quote]

1.cine a zis asa ceva?
2.ce vrei sa zici cu 'astia de prin tara'?

el a fost la Yakutsia si acolo au fost mai multe tari :P China, Rusia , si inca vreo 4 . si deja s-a obisnuit cu aerul international si se refera la ceilalti olimpici muritori de rand ca fiind 'din tara' B)

Cred ca Octavian se referea la olimpicii care nu sunt din Bucuresti si probabil postul era mai mult in gluma (vezi emoticonul de final).

acuma daca eu scriu 1 pagina numa cu injuraturi si la sfarsit pun un  ;)  inseamna ca totu a fost o gluma? nu creeeeeeeeeed :P

[quote="pureblack2003"]acuma daca eu scriu 1 pagina numa cu injuraturi si la sfarsit pun un  ;)[/quote]Nu exagera. Haideti inapoi ontopic, sau se inchide taverna aici, ok? ;)

Noi, lotu' de Bucuresti am plecat  :D

Bafta tuturor participantilor si sa aveti parte de perfect score-uri din plin ;)

Hai sa fim seriosi. Acum, daca zici de subiectele de la judeteana ca sunt [b] penibile [/b] asta inseamna, in primul rand ca ai jignit comisia care le-a ales. Ma simt si eu jignit avand in vedere ca mi-au fost date 2 probleme. Totusi vreau sa cred ca Octavian a vrut sa zica ca adevarata valoare se va vedea la nationala.  :)


Here's the new IIT Format. The syllabus, by the way, is absolutely unchanged. Give your views!!!

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=297[/img]

If you checked " The changes are not up to the mark; I suggest : " above, please suggest your changes as well!!!

if i am not wrong there 'll be more competition now cos there 's just 1 exam, even though there are still the same number of candidates?

i maybe wrong

Well,since there is -ve marking,so competition almost remains same as the mugging ones(Like ketan) can't be so good at guessing.

yup thats right

but now all people have to concentrate on only 1 exam rather than originally the 2 types of exams(mains and screening) .Now they can sort of ( not exactly) but ..... mug the screening type of questions and get in ?

I dont get the point of making IIT like this,
1. it is gonna get harder ALOT harder!!!
2. I dont no HOw on earth they are going to check the papers, cos there are like 16,000 people on 1 similar mark and there will be even more if there is 1 exam
3. The IIT NOW will contain only MCQ's that will be easier than the mains

[quote="shadysaysurspammed"]Maybe now they will give theory type questions in the screening,i mean whaat's the harm in doing so?Ultimately you have to get to the answer...

By the way ,why was this rule impolemented?To make IIT easier or something?[/quote]

i think it was to reduce the gowing use of IIT tutions like bansal and Kota-mania and stuff,you know. And people were beginning to forget school and concentrate more on these tutions , which in their eyes was wrong. Also to reduce the competition, i guess ;)

Actually if they see it like that then I think it is quite right, cos people of india will become SPECIALISED TOO much, they will not get knowledge in any other field except their own. So it is good in that way....

It is not at all necessary that IIT becomes so easy that there is no Kota stuff...I don't think anyone is going to start preparing less for IIT,its still going to be the same presure....ALso i think India has not produced an extraordinary mathematician or physicsist(Except for eminem and riddler)....no one studies only to get into IIT but to be the best of them all.I don't feel any difference even though the thing has changed,maybe some muggers do....eminem,do you feel any change in your studies for IIT? :D

shady , i think you maybe right about the mugging part , no can just mug and get through i guess , but still,  the preparation , i think we do have to change it , the MCQ 's DO require logic and quickness( as tution sir said). MCQ's are different from theory questions , so i guess  you have to prepare keeping in mind the types of questions that can come as $MCQ's$. Mind you i dont mean you $mug$ MCQ's , but subconsciusly prepare yourself for them ...............

IIT JEE had never been easy nor will be; because of its standard.
In the past few yrs it has created a lot of panic in the mind of students and their family members.
We should be happy that in some or the other way it is an indication that we are developing.

If you still believe it is easy. Then you would not given FIITJEE's
PART TEST I maths paper.

IIT its great. that's what i can say.
Recently I visited IIT Roorkee and saw the exposure students had there. No one looked like a book worm or just studying. Everyone was playing but also were very sincere.
Now i realize its importance and so next year I will try surely to be in IIT(now i'm in 12th).

IIT will always b great.

the fact that they've changed d pattern is riddiculus.

there'll b same pressure n compitition among the students. but there'll b no fun in solving mcqs. it was grt fun to solve subjective type physics questions

you are correct.
but now when i have solved some very good objective questions i got to know various things in physics which i couldn't have known without doing the objective questions.

in doing subjective we work hard to reach at the answer. but objective is the real test. so friend try to solve some excellent objective and you will feel the real challenge. ( i can suggest you the multiple choice more than one option of H.C.Verma book.) they are excellent.

ya. i've solved many of them. they're really good.

If anyone know site like AoPS for physics please post it here. i'm searching for these sites.


Thirteen points have been placed in a plane so that no three points are collinear. What is the number of different lines determined by these points?

There are 13C2= $ 78$ lines, since we can choose any pair of them.


linear algebra
matrix
vector
linear algebra unsolved

Suppose $A$ is a symmetric $n$x$n$ matrix, and let $V$ be a subspace of $R^{n}$ with the property that $Ax$ is in $V$ for all $x$ in $V$. Prove that $Ax$ is in $V$ perpendicular for all $x$ in $V$ perpendicular.

I'm trying to be as detailed as possible...

What is known:
- $A \in \mathcal M_{n}\left( \mathbb R \right)$;
- $A = A^{T}$;
- $Ax \in V$ for all in $x \in V$, where $V$ is some subspace of $\mathbb R^{n}$.

What is to be proven:
- If $x \in V^\perp$ then $Ax \in V^\perp$.

[i]Solution:[/i]
Take $x \in V^\perp$. That means that $\left< x, v \right> = 0$ for all $v \in V$. In matrix notation that translates to $v^{T}x = 0$ for all $v \in V$.
Let's equivalently transform the conclusion:
$Ax \in V^\perp$
$\Longleftrightarrow \left< Ax, v \right> = 0$ for all $v \in V$
$\Longleftrightarrow v^{T}(Ax) = 0$ for all $v \in V$
$\Longleftrightarrow \left( v^{T}A x \right)^{T}= 0$ for all $v \in V$
$\Longleftrightarrow x^{T}A^{T}v = 0$ for all $v \in V$
$\Longleftrightarrow x^{T}A v = 0$ for all $v \in V$
$\Longleftrightarrow \left< Av, x \right> = 0$ for all $v \in V$.
We know that $x$ is orthogonal to any vector from $V$. But the hypothesis tells us that $Av \in V$, from where it follows that $\left< Av, x \right> = 0$ for all $v \in V$.


show that $ n$ does not divide $ 2^n\minus{}1$.

I used trial and error and looked for a pattern.  There might be a better way.

I assume the restriction $ n>1$.  When $ n\equal{}1$, $ 2^1\minus{}1\equal{}1$, which is divisible by $ 1$.

[hide="my attempt at a solution"]
For all even $ n$, we know that $ 2^n\minus{}1$ is odd, and thus $ n$ does not divide $ 2^n\minus{}1$

For $ n\equal{}3$, $ 2^n\minus{}1\equal{}7\equiv1 \pmod{3}$

For all odd $ n$, such that $ n\ge5$, if $ n\equiv 3 \pmod{6}$, $ (2^n\minus{}1)/n \equiv 7 \pmod{n}$.  If $ n\not\equiv 3 \pmod{6}$, $ (2^n\minus{}1)/n \equiv 1 \pmod{n}$.

[/hide]

[quote="e^pi"]show that $ n$ does not divide $ 2^n \minus{} 1$.[/quote]
I think it's wrong!See this link :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232846

EDIT: Mod, Pleased delete this post for me!Sorry for my careless mistake!

ghjk, that's a totally different problem.  This result is in fact true.

Discrete_Math:  even if those statements were true (which I doubt; test more cases), you still need to prove them since they're not at all obvious.

[quote="e^pi"]show that $ n$ does not divide $ 2^n \minus{} 1$.[/quote]
The result also holds for $ 3^n \plus{} 1$ as long as $ n$ is odd.  Anyways, here's the solution for the given problem:
[hide="Solution"]
Assume that $ 2^n \minus{} 1\equiv 0\bmod n\iff 2^n\equiv 1\bmod n$.  Let $ p$ be the smallest prime divisor of $ n$, so we have that $ 2^n\equiv 1\bmod p$.  Let $ k$ be the order of $ 2\bmod p$, so $ k|n$.  Yet, $ k|\phi (p) \equal{} p \minus{} 1$.  Clearly, the prime factors of $ p \minus{} 1$ are all less than $ p$, so the prime factors of $ k$ are all less than $ p$.  However, $ p$ is the smallest prime factor of $ n$, so the prime factors of $ k$ are all greater than or equal to $ p$ since $ k|n$, which is a contradiction.  Thus, $ 2^n \minus{} 1$ cannot be a multiple of $ n$ for natural $ n$ above $ 1$.  [/hide]

[hide]An alternative approach:
Assume that there exists a positive integer n>1 such that 2^n-1 is divisible by n.We easily infer that n must be an odd number and n is a composite number(according to Fermat Little Theorem). Now let p be the smallest prime divisor of n, so p must be odd and (2,p)=1. Since 2^n-1 is divisible by n, it's also divisible by p.So, we have: 2^n-1 is divisible by p, which contradicts to the Fermat Little Theorem(2^n is equal to 2 module p).Therefore, our assumption is wrong, which leads to the conclusion(Q.E.D)[/hide]

[quote="ghjk"]2^n-1 is divisible by p, which contradicts to the Fermat Little Theorem(2^n is equal to 2 module p).[/quote]
This is not true.  All you know is that if $ n \equal{} pq$ then $ 2^n \equiv 2^q \bmod p$, and this quantity is not necessarily congruent to $ 2 \bmod p$.


induction
combinatorics unsolved
combinatorics

Prove that a 3-connected and planar graph has a spanning tree with vertex degree at most 3.

what do you mean by ' vertex at most 3? do you mean with max degree $\leq 3$?

i think the following thing should work.
If $G$ is $3$-connected, then we can prove that for some edge $e,G\cdot e$, (followed by deletion of multiple edges) the contraction of $G$ by $e$ is also $3$-connected. this graph is a 3-connected graph with fewer vertices, so by induction on $V(G)$ ,we can get a spanning tree of $G\cdot e$ with all vertex degrees being $\leq 3$. Now in $G$ it is easy to see that this extends to a spanning tree with the degree condition also satisfied.

yes true..
i knew this result because i read it when studying the proof of kuratowski's theorem..
a good result for applying induction!  ;)

actually, i think that i have to rephrase saying that we can contract $\textit{some}$ edge and still retain 3-connectivity-for instance if we take the 'wheel graph'.i.e. a cycle with another central vertex adjacent to all the vertices of the cycle, then we can contract any of the edges of the cycle but not any of the edges adjacent to the central vertex-that will kill 3-connectivity.
 actually there is also a theorem of Tutte saying that any 3-connected graph can be built from a wheel by eitehr adding new edges or 'splitting a vertex',i.e. doing the inverse of contraction in some sense.

yes yes, i've seen that one too..
and yes it is : there exists an edge $e$ such that....


Program a reverse polish calculator in C without using any functions other than ones [b]you yourself wrote.[/b]

Do you mean you have to write your own prinf and scanf function and so on? I think the easiest way to do the program is use some function others have written and then step by step replace those function by your own functions.

They probably meant you had no right to put in ready-to-use parsing and evaluation functions from any library.

There are many algorithms described here and there on the internet to achieve your goal.

First, you must establish a good structure to represent your expressions, and then define evaluation functions over such objects, and finally write the parsing stuffs.


Hello Soarer. The following inequality is created last night for you...

(i). Prove that for all non-negative real numbers $a,b,c$ for which $ab+bc+ca=1$ then
\[ \frac{a^2\sqrt{b+c}}{\sqrt{1+a^2}}+\frac{b^2\sqrt{c+a}}{\sqrt{1+b^2}}+\frac{c^2\sqrt{a+b}}{\sqrt{1+c^2}} \ge \sqrt{2}.  \]

(ii). Prove that for any non-negative real numbers $a,b,c,d$ we always have
\[ a^4+b^4+c^4+d^4-4abcd \ge 2(a-b)(b-c)(c-d)(d-a)  \]
And if $(a-b)(c-d) \le 0$ then we have a stronger one
\[ a^4+b^4+c^4+d^4-4abcd \ge 17(a-b)(b-c)(c-d)(d-a)  \]

(iii) Prove that for all non-negative real numbers $a,b,c$ such that $a^2+b^2+c^2=3$ then
\[ \frac{a^2\sqrt{b+c}}{b^2+c^2}+\frac{b^2\sqrt{c+a}}{c^2+a^2}+\frac{c^2\sqrt{a+b}}{a^2+b^2} \ge \frac{3}{\sqrt{2}} \]
I hope you will enjoy it.  :lol:

PS. Sorry, the first one was false. Try to solve the second one. Now it's correct. I also add problem $(iii)$.

Thanks  :P
It is equivalent to $ab+bc+ca = 1 \Rightarrow 2 \ge (a+b)(b+c)(c+a)$, which is not true :( Take $a = 2, b = \frac{1}{2}, c = 0$

I think we should add a constraint that $a,b,c\leq 1$
let $a=tan{\frac{A}{2}}$
b,c is similar.
then if A,B,C are the angles of a cute triangle,then  $a,b,c\leq 1$

Sorry Soarer very much. I will recheck it carefully...

Hi Soarer and Albanian Eagle and other

Try to prove $(ii)$.

[quote="Hawk Tiger"][quote="Albanian Eagle"]actually $LHS\geq \sqrt{\frac{3\sqrt{3}}{2}}$ ;)[/quote]
do you mean the first one?
I think there is no low bound (or low bound is 0)for the first ineq.[/quote]
I was refering to the inequality before being edited $\sum \frac{a\sqrt{b+c}}{\sqrt{1+a^2}}...$ ;)

[quote="arqady"][quote="hungkhtn"]
(ii). Prove that for any non-negative real numbers $a,b,c,d$ we always have
\[ a^4+b^4+c^4+d^4-4abcd \ge 17(a-b)(b-c)(c-d)(d-a)  \]
[/quote]
I think this is wrong: $a=c=0,$ $b=d=1.$ ;)[/quote]

HIx... Nothing to say,  :blush:

I will check this one again anh thank for your counter example. I think the problem will be true if we replace $17$ by $2$. I will answer tomorrow.

[quote="hungkhtn"]
(ii). Prove that for any non-negative real numbers $a,b,c,d$ we always have
\[ a^4+b^4+c^4+d^4-4abcd \ge 2(a-b)(b-c)(c-d)(d-a)  \]
[/quote]
Yes this is true.

OK, Arqady. At first my solution is false because I only check a specific case. If we replace $17$ by $2$, it's not difficult. Now try to prove problem $(ii)$ with additional condition and problem $(iii)$.


function
floor function
number theory unsolved
number theory

1. Prove that for $ n \in N$

$ \left \lfloor \sqrt{n} \plus{} \sqrt{n\plus{}1} \right \rfloor \equal{} \left \lfloor \sqrt{4n\plus{}2} \right \rfloor$

I've just brainstormed the following...

Consider first the perfect squares: $ \sqrt{1}, \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}... \equal{} 1, 2, 3, 4, 5...$

It can be seen that the largest difference between the perfect squares is 1.

Thus for $ \sqrt{n}$ and $ \sqrt{n\plus{}1}$ the difference between them will be smaller than 1.

This implies that $ \left \lfloor \sqrt{n} \right \rfloor \equal{} \left \lfloor \sqrt{n\plus{}1} \right \rfloor$ IFF $ n\plus{}1$ is NOT a perfect square.

And also....

$ RHS \equal{} \left \lfloor \sqrt{4(n\plus{}\frac{1}{2})} \right \rfloor \equal{} \left \lfloor 2\sqrt{(n\plus{}\frac{1}{2})} \right \rfloor \equal{} \left \lfloor \sqrt{(n\plus{}\frac{1}{2})} \plus{} \sqrt{(n\plus{}\frac{1}{2})} \right \rfloor$

I'm stuck now, can anyone help me prove this?

Thank you!

Assume that there is an integer, say $ m$, between $ \sqrt{n}\plus{}\sqrt{n\plus{}1}$ and $ \sqrt{4n\plus{}2}$
Then $ \sqrt{n}\plus{}\sqrt{n\plus{}1}<m<\sqrt{4n\plus{}2}$
which implies $ 4n(n\plus{}1)<(m\minus{}2n\minus{}1)^2<4n(n\plus{}1)\plus{}1$ which is impossible
This, with the observation that $ 4n\plus{}2$ is never a perfect square gives the result

sorry. it has to be $ 4n(n\plus{}1)<(m^2\minus{}2n\minus{}1)^{2}<4n(n\plus{}1)\plus{}1$ in the third line above


analytic geometry
geometry
ratio
geometry theorems

I never use barycentric coordinates itself, I use center-of-mass method. And there are a lot of problems which can be solved with this method.

Lots of such stuff can be found on [url=http://www.math.fau.edu/yiu/Geometry.html]Paul Yiu's website[/url], especially the "Introduction to the Geometry of the Triangle". It contains many applications of barycentric coordinates. Or also look at [url=http://www.math.fau.edu/yiu/barycentric.pdf]his barycentric coordinates paper[/url].

  Darij

Myth, can you tell me more about that method?
Darij, thanks for the links. I appreciate your help :)

What can I tell you? I haven't appropriate links to papers.
However, I am sure Darij knows many sources :)

My only adviсe is to try to solve geometry problems using center of mass.

are there any ways to find out distance of two points by means of barycentric coordinates?

[quote="prowler"]Look : Geometry > Geometry Teorems and Formulas >
a question about baricentric coordinates.
 Darij Grinberg wrote there a formula of the distance
between two points with normalized coordinates.[/quote]

Thanks very much!

[quote="fleeting_guest"][quote="siuhochung"]Can anybody show me some examples of using barycentric coordinates or trilinear coordinates to solve geometry problems? Thanks![/quote]

Here is a problem which has a very nice statement and proof (one line!).

Given 3 points $P_i$ with barycentric coordinates $(x_i,y_i,z_i)$ relative to a triangle $ABC$.
Find the ratio of area $[P_1P_2 P_3]$ to area $[ABC]$.[/quote]
The answer is
x_1 y_1 z_1
x_2 y_2 z_2
x_3 y_3 z_3

(Sorry that i don't know how to type determinant)
But what's the one-line proof? I'm curious.

The one-line proof is to work in three dimensions, and compute the ratio of volumes of tetrahedra $OP_1 P_2 P_3$ and $OABC$.   Barycentric coordinates are just the ordinary 3-D Euclidean coordinates in the plane $x+y+z=1$.  Volumes of the tetrahedra are (proportional to) 3x3 determinants of their coordinates.  That's all.


I am seeing that congruence is used a lot for proofs.
what's the general topic that includes congruence?
(I want to get a book to study.)

And what other topics do I need to know?

If you are talking about number theory, the Niven and Zuckermann book "An Introduction into the Theory of Numbers" is excellent. This book goes pretty deep into some number theory that you wouldn't even need for math comps, but it is interesting nevertheless. Also, after you read this book, I doubt you would have trouble understanding when general number theory can and should be used.

The Engel book "Problem Solving Strategies" also has a chapter on number theory and problem solving techniques.


topology
geometry
geometric transformation
rotation
real analysis
real analysis unsolved

Let $ X,Y$ be real normed linear spaces and $ f: X \longrightarrow Y$ be a [b]map[/b] satifying $ \parallel{}f(x)\parallel{} \equal{} \parallel{}x\parallel{}, \forall x \in X$ . Prove that $ f(x) \equal{} Ax \plus{} c,\forall x \in X$. ($ c\in Y$ and $ A$ is a homeomorphism satifying $ \parallel{}Ax\parallel{} \equal{} \parallel{}x\parallel{}, \forall x \in X$)

---
Sorry, f is only a map, not homeomorphism!!!

[quote="mlok"]Is $ A$ supposed to be a linear map (an isomorphism)?[/quote]

$ A$ is called a homeomorphism iff it is an isomorphism linear map such that $ A$ and $ A^{\minus{}1}$ are continuous

Then I do not see what is there to prove: take $ A\equal{}f$, $ c\equal{}0$ :P
Seriously, pay attention to how you pose the problems, please :) We are not here to exercise in mind reading :roll:

[quote="fedja"]Then I do not see what is there to prove: take $ A \equal{} f$, $ c \equal{} 0$ :P
Seriously, pay attention to how you pose the problems, please :) We are not here to exercise in mind reading :roll:[/quote]

I've just edited. $ f$ is a map satifying...

And now the problem is just false :( Take $ X\equal{}Y\equal{}\mathbb R^2$ and take the map $ f$ to be the counterclockwise rotation by angle $ t$ on the circle of radius $ t$, say.


calculus
integration
derivative
conics
ellipse
analytic geometry
real analysis

Hi,
I want to find the midpoint of an elliptical arc.
I have the following input
1.Start point
2.End point
3.Center point
4.Angle between center and start point
5.Angle between center and end point
6.Distance between center and start point
7.Distance between center and end point

Please refer the attached file for clear knowledge.
Can you sort out a procedure to find the midpoint of an elliptical arc from above input?

Thank you
M Rajee

There is no simple solution, as elliptic arc length is given by an elliptic integral.

Hi,
   Can you please provide a example or method to find midpoint of an elliptical arc using integral.

Thanks,
Rajee

Parametrize the ellipse. Angles from the center are not recommended.
Write down the arclength integral $ s(T)\equal{}\int_{T_0}^{T}\frac{ds}{dt}\,dt$.
Solve $ s(T)\equal{}\frac12s(T_1)$ numerically. Since we can find the derivative $ \frac{ds}{dt}$ of $ s$ easily, Newton's method together with approximate integration works.

[quote="jmerry"]Parametrize the ellipse. Angles from the center are not recommended.
Write down the arclength integral $ s(T) \equal{} \int_{T_0}^{T}\frac {ds}{dt}\,dt$.
Solve $ s(T) \equal{} \frac12s(T_1)$ numerically. Since we can find the derivative $ \frac {ds}{dt}$ of $ s$ easily, Newton's method together with approximate integration works.[/quote]

Can you please work out the solution for following example:

(X- 230)^2 / 15.02 ^2   + (Y -181) ^2 / 8.75 ^2 = 1       // Ellipse Equation

Where
Center Point =  (230 , 181)
Minor radius in Y coordinate: b = 8.75
Major radius in X coordinate: a = 15.052

Mid Point =  (MidX , MidY)=?

Thank You


a,b,c,x,y,z>0 Prove that:
ax+by+cz+2 \sqrt (ab+bc+ca)(xy+yz+zx)  \leq (a+b+c)(x+y+z)

Hi Maverick,

it is a one-line proof!!!

Use the Cauchy-Schwarz ineq with the two 5x1-vectors
(a, b, c, sqrt(bc+ca+ab), sqrt(bc+ca+ab)) and
(x, y, z, sqrt(yz+zx+xy), sqrt(yz+zx+xy))
and you will get the desired ineq immediately.

-Eckard

i don't think that there is another solution ... easier than this one ... topic closed :)


linear algebra
matrix
algebra
polynomial
system of equations

I'm trying to solve this system of equations for all values of a,b,c,d without resorting to any brute-force techniques. I'd appreciate any hints to get me further.

$ a \equal{} a^2 \plus{} bc$
$ b \equal{} ab \plus{} bd$
$ c \equal{} ac \plus{} cd$
$ d \equal{} bc \plus{} d^2$

so far I've got that
$ b\plus{}c\equal{}(a\plus{}d)(c\plus{}d)$
$ a\minus{}d\equal{}(a\plus{}d)(a\minus{}d)$

so $ a\plus{}d\equal{}1$
and $ b\plus{}c\equal{}(1)(c\plus{}d) \implies b\equal{}c$

What exactly do you mean by "brute-force" techniques?  The solution I see is straightforward with a little bit of organization.

The eigenvalues satisfy $ \lambda^2 \equal{} \lambda$, so they're all either 0 or 1.  If both eigenvalues are 1, your matrix is similar to $ \begin{pmatrix}1 & x \\
0 & 1\end{pmatrix}$ for some $ x$ and so the square of your matrix is similar to $ \begin{pmatrix}1 & 2x \\
0 & 1\end{pmatrix}$ (with the same matrix of similarity) and thus $ x \equal{} 0$ and so we've got the identity matrix.  If both eigenvalues are 0, your matrix is the all-zeroes matrix.  Otherwise, your matrix is similar to $ \begin{pmatrix}1 & 0 \\
0 & 0\end{pmatrix}$, and there's a bit more work here.

If none of that makes any sense to you, it's okay and I just guessed wrong about how you stumbled onto this particular set of equations :).

By the way, from $ a \minus{} d \equal{} (a \plus{} d)(a \minus{} d)$ you can't conclude immediately that $ a \plus{} d \equal{} 1$ -- there are two cases there.

[quote="JBL"]If none of that makes any sense to you, it's okay and I just guessed wrong about how you stumbled onto this particular set of equations :). [/quote]
Tricky.  JBL thinks that you are trying to solve

$ \left[ \begin{array}{cc} a & b \\
c & d \end{array} \right]^2 \equal{} \left[ \begin{array}{cc} a & b \\
c & d \end{array} \right]$

which seems like a reasonable enough hypothesis.  How much linear algebra do you know?

It's usually not a good idea to work componentwise with matrices.  Here the reason is that "matrix polynomials" have a very large class of symmetries:  if a matrix satisfies $ A^2 \equal{} A$, then so does every matrix similar to it.  In other words,

- If there is a non-identity solution, there are infinitely many solutions.
- You only need to think about the solutions "up to similarity."

Then the obvious tool here is an analysis using Jordan normal form.  Another reason to focus on the matrices and not the components is to think about [url=http://en.wikipedia.org/wiki/Projection_(linear_algebra)]projections[/url], which give a good geometric picture as to what all such $ A$ should look like.

(Yet another reason:  if you try this for any matrix size larger than $ 2 \times 2$ the equations quickly get extremely cumbersome, but the analysis using Jordan normal form has only a few additional cases.)

Did you proved that $ a\minus{}d \neq 0$ ?