[quote="MathVietNam"]chào anh em bà con đòng hương, tìm mãi mới thấy chỗ nói tiếng việt , mệt đứt cả hơi. Em nhờ các anh chị em giúp em cái khoản ánh xạ với chứ em ngu cái này lắm lắm mà hok dám hỏi thầy (ngại vì thầy mời ngoài viện toán học về dạy), thầy nói chi chi nỏ hiểu được Vậy anh chị em nào( VIP) có thể giúp em (mong mọi người nói cụ thể tí vì thầy nói rùi nhưng khó hiểu quá). Em xin cảm ơn, à  có ai quê ở Nghệ An không ạ, cụ thể hơn là học khối chuyên để anh em gặp nhau tí :P  :P  :P[/quote]
Trường bộ đó hả,muốn hiểu mấy cái ánh xạ đó chỉ có cách tìm sách mà đọc thôi,ra mấy hiệu sách đường N Vănn Cừ mà tìm ,những quyển về logic tập hợp dành cho đại học,Mình qua thời đó rồi,đọc ong cả đầu
P/s;Cho hỏi thăm Tùng bèng nhá,Vương Tùng

[quote="Allnames"]
Trường bộ đó hả,muốn hiểu mấy cái ánh xạ đó chỉ có cách tìm sách mà đọc thôi,ra mấy hiệu sách đường N Vănn Cừ mà tìm ,những quyển về logic tập hợp dành cho đại học,Mình qua thời đó rồi,đọc ong cả đầu
P/s;Cho hỏi thăm Tùng bèng nhá,Vương Tùng[/quote]
PBC hả, thank u nhiều, mình tìm đc rùi (ngồi cả buổi lun), nói chung đọc cũng tàm tạm
Quen Tùng hả


Prove that the sum of the angles of an n-sides convex polygon equals

180*(n-2).

Note: [hide]You can use the fact that a triangle has 180 degrees without having to prove that a triangle has 180 degrees.[/hide]

Here's something else to think about:

[hide]Imagine you are a bug walking around the polygon.  How many total degrees in turns do you cover?  My 6th grade teacher told me this, if he's anywhere around here .[/hide]

[hide]if you drew diagonals from one vertex to all the other vertices that this vertex wasn't connected to already (the two next to it), then you would be drawing (n-3) lines (n, the total number of sides/vertices, subtracted by 3, 2 of the vertices next to it, and the vertex itself). since you drew (n-3) lines, then you are dividing the polygon into (n-2) triangles, and we know that the sum of the angles of a triangle is 180, so 180(n-2)[/hide]

[quote="yif man12"]vertice[/quote]

Hmmm..... :D

Here s a slightly different solution:

[hide]Pick any point inside of the polygon.  Then draw lines to each vertex.  This will make n triangles.  So there are 180n degrees for all of the triangles combined.  To get the degrees of the polygon we subtract the number of degrees fromed from the angles that arent formed with a side of the polygon.  These make a circle so they measure 360 :^o: thus we get 180n-360 or 180(n-2)[/hide]

bob123 wrote:yif man12 wrote:[hide]draw diagonals from one vertice to all other vertices. you'll result with n-2 triangles. the total measures of those triangles is 180(x-2)[/hide]

i think you meant 180(n-2)

oops, my bad

JSRosen3 wrote:yif man12 wrote:vertice

Hmmm.....

lol, shut up

could you please clarify your solution, joml88?

[hide]Ok,  at first I used the point that was the center of a regular polygon.  It makes things much nicer to imagine.  If you a draw a line to vertex you will divide the polygon into n number of traingles.  Every triangle has a total of 180 degrees.  So we can see that by adding all of the degrees of all these triangles we will have an answer that is the sum of the angles of the polygon plus the sum of all the angles with the center as a vertex.  We just need to subtract the sum of all the angles with the center as a vertex to get our answer.  These angles will add to be 360.  So we have 180n-360 which factors to 180(n-2).  I just generalized this proof so that you can pick any point inside of any convex polygon.[/hide]

Please tell me if you need help understanding.  I tried to make it as understandable as possible.  It would also be helpful if you could tell me which part bothers you so I can try to explain just that part.


Find all integer solutions to the equation

1/x+1/y=1/19

$ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{19}$

$ \Rightarrow 19(x \plus{} y) \equal{} xy$

$ \Rightarrow (x \minus{} 19)(y \minus{} 19) \equal{} 19^2$

Since $ 19$ is prime the only solutions are;

$ (x,y) \equal{} (18,\minus{}342),(38,38)$ and $ (20,380)$


superior algebra
superior algebra unsolved

Let $ K$ be a division ring, and let $ a\in A$. Prove that, if $ x^2 \minus{} a \equal{} 0$ and $ x^3 \minus{} a \equal{} 0$ have solutions in $ K$, then $ x^6 \minus{} a \equal{} 0$ also has solutions in $ K$.

Let $ y^2 \equal{} a$ and $ z^3 \equal{} a$. Then $ (\frac {y}{z})^6 \equal{} \frac {y^6}{z^6} \equal{} \frac {a^3}{a^2} \equal{} a$.

EDIT: Just noticed that it's division ring, not a field, so I guess this doesn't quite work...it seems like $ y$ and $ z$ should commute with each other though, since they are both "powers" of $ a$.


Na deixete oti anamesa se 4 pragmatikous ari8mous yparxoun dyo toulaxiston (estw a,b) me thn idiothta:
$ 0\leq \frac{b\minus{}a}{a\plus{}ab} \geq \sqrt{3}$

[hide="Hint"]
Dirichlet
[/hide]

Αν μπορεις φιλε basted γραψτην σωστα γιατι ειναι ωραια ασκηση. :)

Afto akribws leei h askhsh!! Einai apo biblio. Ti pisteveis oti einai la8os? (8a pw apo poio biblio molis ly8ei h askhsh giati alliws den exei nohma).

Βασικα η διαταξη ειναι λαθος απλα φοβηθηκα μηπως και τα ακροτατα ειναι λαθος....οκ :)

Η άσκηση αυτή τους τελευταίους τρεις μήνες έχει συζητηθεί 4 φορές στο <<ελληνικά>>  :P
Ψάξτε στα ποστ μου να βρείτε λύση ... (σκεφτείτε τον τύπο εφαπτομένης)


(i) Let $ y\equal{}\sum_{n\equal{}0}^\infty a_{n}x^{n}$, where the coefficients $ a_{n}$ are independent of $ x$ and are such that this series and all others in this question converge. Show that

$ y' \equal{} \sum_{n\equal{}1}^\infty na_{n}x^{n\minus{}1},$

and write down a similar expression for $ y''$.

Write out explicitly each of the three series as far as the term containing $ a_{3}$.

(ii) It is given that $ y$ satisfies the differential equation

$ xy''\minus{}y' \plus{} 4x^{3}y\equal{}0$.

By substituting the series of part (i) into the differential equation and comparing coefficients, show that $ a_{1}\equal{}0$.

Show that, for $ n\ge 4$,

$ a_{n} \equal{} \minus{}\frac{4}{n(n\minus{}2)}a_{n\minus{}4},$

and that, if $ a_{0} \equal{} 1$ and $ a_{2} \equal{}0$, then $ y\equal{}\cos(x^{2})$.

Find the corresponding result when $ a_{0} \equal{} 0$ and $ a_{2}\equal{}1.$

I've done this, apart from the last bit: "show that for $ n \ge 4 ...$".


one more than 3 times a # is the same as 5 times a #, decreased by 15. find the#

$ 3a \plus{} 1 \equal{} 5a \minus{} 15$

$ 1 \equal{} 2a \minus{} 15$

$ 16 \equal{} 2a$

$ \boxed{8} \equal{} a$

My 2222th post! (including G & FF posts, that is)

Its 3a-1=5a-15...

ALthough you solved YOUR equationr ite... except for the last step.
it would be 15/2. not 2/15.

oh oops i always make that mistake :blush:


LaTeX
algebra
partial fractions
algebra unsolved

I don't undestand the question, and I think many of us also(like me).
LATEX form please....

$ S\equal{} \sum_{k\equal{}1}^{1004} \frac{k(k\plus{}2)}{(2k\plus{}1)(2k\plus{}3)}$

$ \frac{k^{2} \plus{} 2k}{4k^{2}\plus{}7k\plus{}3} \equal{} \frac{1}{4}[1 \plus{} \frac{k\minus{}3}{(2k\plus{}1)(2k\plus{}3)}]$.

Then partial fractions and telescoping.


[img]http://www.part.lt/img/20a1bc0c42ffcff8bd396e7801a4c8d6669.JPG[/img]

$ A,s,a,b,c$ - Area of triangle ABC, semiperimeter, lengths of BC, AC and AB respectively.

$ h_a,h_b,h_c$ - Lengths of heights of a Triangle ABC.

$ r_a,r_b,r_c$ - Length of radius of Escribed circles that are tanget to BC, AC and AB respectively.

$ R$ - Length of radius of a circumscribed circle.

$ r$ - Length of radius of an inscribed circle.

[b]PROBLEMS:[/b]

1. Prove that $ A\equal{}r_a(s\minus{}a)\equal{}r_b(s\minus{}b)\equal{}r_c(s\minus{}c)$

2. Prove that $ r_a\plus{}r_b\plus{}r_c\equal{}r\plus{}4R$

3. Prove that $ r_ar_b\plus{}r_ar_c\plus{}r_br_c\equal{}s^2$

4. Prove that $ r_ar_br_c\equal{}sA$

5. Prove that $ r\equal{}\frac{1}{3}\cdot\frac{3}{\frac{1}{r_a}\plus{}\frac{1}{r_b}\plus{}\frac{1}{r_c}}$

6. Prove that  $ \frac{1}{h_a}\plus{}\frac{1}{h_b}\plus{}\frac{1}{h_c}\equal{}\frac{1}{r_a}\plus{}\frac{1}{r_b}\plus{}\frac{1}{r_c}\equal{}\frac{1}{r}$

7. Prove that $ A^2\equal{}r\cdot{r_a}\cdot{r_b}\cdot{r_c}$

8. Prove that $ A\equal{}\frac{c\cdot{r_a}\cdot{r_b}}{r_a\plus{}r_b}$

9. Prove that $ \frac{2}{h_a}\equal{}\frac{1}{r_b}\plus{}\frac{1}{r_c}$

We can use trignometry for some nice proofs of almost all these problems!

Yes trig does it all. :P

But u can also make use of this superb idea without trignometry:
$ In a triangle ABC, if the incircle touches the sides BC, CA, AB at X,Y,Z respectively, then \frac tion {%Error. "area" is a bad command.
\triangle ABC} {r} = \frac tion{2R}{%Error. "area" is a bad command.
\triangle XYZ}$.


Fix $ b,y >0$. Prove that there exists a unique $ x$ such that $ b^x \equal{} y$.

So the key idea is to show that $ b^x$ is strictly increasing. Because this takes care of uniqueness? Also consider the set $ S \equal{} \{t: b^t < y \}$ and show $ x \equal{} \sup S$?

assuming you mean $ b>1$ (else its decreasing)

$ f(x)\equal{}b^x$ take $ \epsilon >0$
$ f(x\plus{}\epsilon)\equal{}b^{x\plus{}\epsilon}\equal{}b^x\cdot b^\epsilon > b^x$

that is, if we know (or show) $ b^\epsilon >1$ for $ \epsilon >0$

well... yeah.. we show its strictly increasing so it must be 1-to-1 on its range (when $ b>1$)

if you wanted to show that its range is $ (0,\infty)$ then we have to say

$ b\equal{}1\plus{}\delta$ for some $ \delta >0$

and since $ b^n > 1\plus{}n\delta$, we can find a large enough n to surpass any large integer $ m$,

i.e. $ \exists n_m$, $ b^{n_m}>m$

and therefore for any small number $ \frac{1}{m} > \frac{1}{b^{n_m}} \equal{} b^{\minus{}n_m}$

thus $ f_b(x)\equal{}b^x$ has $ (\frac{1}{m}, m)\subset f_b([\minus{}n_m,n_m])\subset Ran(f_b)$

so taking $ m\to \infty$, we get $ (0,\infty)\subset Ran(f_b)$

although we only proved for $ b>1$, if $ b<1$ then $ f_{\frac{1}{b}}(x)\equal{}b^{\minus{}x}$, and the same argument applies to show decreasing and onto


algorithm
number theory
Euclidean algorithm

If a team gets $ 15$ points for winning any game and $ 7$ points for drawing any game, then what is the largest mathematical point that is not attainable in an infinitely long game?

[quote="earldbest"]$ 83$

We can generalize $ (15,7)$ to any pair of integers $ (a,b)$; it is well-known that the answer is $ ab \minus{} a \minus{} b$.  :D[/quote]

Can you prove it for the general case?

[quote="earldbest"][quote="manjil"]Can you prove it for the general case?[/quote]

[b]to show $ ab \minus{} a \minus{} b$ cannot be represented:[/b]
 suppose for positive integers $ p,q$, $ ab \minus{} a \minus{} b \equal{} pa \plus{} qb$.
 taking $ ab \minus{} a \minus{} b$ modulo $ b$, we get $ q\cong \minus{} 1$(mod $ b$) .
 similarly, $ p\cong \minus{} 1$(mod $ a$).
We get $ pa \plus{} qb\geq 2ab \minus{} a \minus{} b \geq ab \minus{} a \minus{} b \Rightarrow ab \equal{} 0$.

[b]to show $ ab \minus{} a \minus{} b$ is the maximum:[/b]
 let $ c \equal{} x^{ \minus{} 1}$ (mod $ b$) and $ d \equal{} y^{ \minus{} 1}$ (mod $ a$).
 note that $ (c \minus{} 1)a \plus{} (d \minus{} 1)b \equal{} ab \minus{} a \minus{} b \plus{} 1$.

To get a representation for larger numbers, we use the extended Euclidean algorithm to find integers $ u,v$ such that $ ua \minus{} vb \equal{} 1$.

NOTE: if a coefficient becomes negative, add the appropriate multiple of this equation, reducing, if necessary, by $ ( \minus{} b)a \plus{} ab$ or  $ ba \plus{} ( \minus{} a)b$.

I had help with the second part, BTW  :blush:[/quote]

Actaully I have wondered about for more than two numbers say $ a,b,c$ then can we work out such an expression also?//

I had posted it somewhr else, maybe in PEN but I did not get any reply.

Supposedly there is a formula for the 3-variable case, but I have not been able to find it.  All I've read is that it exists and it is much more complicated.

now can someone answer my question?

[quote]sorry, but can you explain the second part of how ab - a - b is the maximum unttainable number?

thanks![/quote]


Do the problem solving classes have Problem Sets and Challenge Sets?

No.  For these courses we put many extra problems on the message board for that course - these essentially are the problem sets and challenge sets.


what is it? is it just that unit cube to the left of the URL?

I believe so.  When I open AoPS from Internet Explorer, it sometimes displays that next to the URL so I'm assuming that's it.

You are talking about the favicon?
The little cube in the address bar?
That isn't the real logo.

The favicon is a single unit cube, simplified so it looks nice.

The real logo is the one above the "Bookstore" link, and expressed in minicon's avatar.

[url]http://www.artofproblemsolving.org[/url] has a different logo...


The following problems take place on an island where all the inhabitants are either "knights" who always tell the truth or "knaves" who always lie.

1. What ONE question might be asked of an islander to find out where a road leads - to the city of knaves or country of knights?
2. What ONE question might be asked of an islander to find out whether he has a pet?
3. What ONE question might be asked of an islander to find out whether he is a knight or a knave?

In the island yes and no sound like "ha" and "hi" but we dont know which is which.

4. Now what ONE question might be asked of an islander to find out whether he is a knight or a knave?
5. What ONE question might be asked of an islander so that the answer is always "ha"?

Also I want more questions like this. Where can I find them?

[b]1)[/b] Does this road lead to your home?

[b]2)[/b] If I ask you , "Does you have a pet?" , what will your answer be?
[b]3)[/b] Same as [b]2)[/b]. If I ask you , "Is you a knight?" ,what will your answer be?

[b]4)[/b] Will your next answer be Yes?  The knaves cannot answer it.

[b]5)[/b] if I ask you "Does "ha" mean Yes?", what will your answer be?


Assuming that $ x_1,x_2\cdots x_n$ are the roots of

$ 1^x\plus{}2^x\plus{}3^x\plus{}\cdots \plus{}9^x\equal{}10^x$

Then calculate

$ S\equal{}\big[\sum_{i\equal{}1}^{n}{x_i}\big]$

Here [x] is the Guassian Function

Let $ 0<a<x$, then $ a^x$ decrease. Therefore $ f(x)\equal{}\sum_{i\equal{}1}^{9}(\frac{i}{1)})^x \equal{}1$ had only one solution ($ a\equal{}i/10$).
Because $ 1^5\plus{}2^5\plus{}...\plus{}9^5>10^5$ and $ 1^6\plus{}2^6\plus{}...\plus{}9^6<10^6$ we get $ S\equal{}5$.


The diagonals of rectangle ABCD intersect at point E. If triangle ABE is twice the size of triangle BCE and triangle CDE has area 23, what is the area of rectangle ABCD?

[hide="SameanswerasAstroPhys"]

I got the same answer, but I'll post anyways...

$ 23+23+\frac{23}{2}+\frac{23}{2}$
$ = 46+23$
$ = 69$

[b][u]Answer to Rectangle Problem:[/u][/b] $ \boxed{69}$

[/hide]


[quote="A+MATH"]Shouldn't this be in F&G?

Bonus Albatross Problem: Albatross is the owner of a store. He also employs 3 people Albatross, Himself and Bobert. The owner of the store gets 2/3 of the revenue and each of the employees evenly spilt the remaning 1/3. If the store makes $90,702 of much money does Bobert get paid.  :D[/quote]

If it was in F&G, the poll would sample the people who browse F&G. However, here, it samples the people who look at this forum.

[hide="Hint for the Albatross Problem"]There's some unneeded informantion.[/hide]

Does Himself refer to person separate from Albatross or Albatross? Cause it's capitalized like a proper noun.

[quote="A+MATH"]No himself is Albatross. What is Albatrosses wage after Bobert gets sacked for pushing Caddlebot off the Leaning Tower of Treething so Albatross has to replace him.[/quote]

[hide="Because there is only one employee..."]$90,702[/hide]

Or maybe Albatross hires Zeta, in which case the answer is different. The last sentence is kinda ambigious because Albatross is both the boss and an employee.

[quote="shinwoo"]Bloody Albatross

....

Killing his employee

That would actually be fun[/quote]

I thought he sacked Bobert?

[quote]
I thought he sacked Bobert?[/quote]

same stuff

Then A+MATH is the one who is bloody :D

I made Bobert kill Caddlebot :P . Notice how he fell off the Leaning Tower of [i]Treething[/i] :P .

[hide="Since there are two people..."]
He splits the money between himself and himself :P .
[/hide]

[quote="shinwoo"]my school don't let kids go to pre-calculus until you go up to 9th grade

that's why I'm in pre-algebra <-the highest you can go if you're in 7th grade[/quote]
that must be frustrating(being in pre-algebra, the highest you can go)

my school does the same as shinwoo's except that you take pre algebra in 7th, algebra in 8th, algebra II in 9th, pre calc/geometry in 10th, calc in 11th, and that is if ur ahead of everyone. If you do that, u have a free year of doing anything during that last 12th year. i think... :P
w00t this thing is cool  :arrow: :pilot:

[quote="-Terminator-"]im the only one in fourth grade.[/quote]
you're in 4th grade!! :o

i'm in 8th grade, and i go to Worthinton Hooker school...

I'm in 8th grade, and I am going to Vestavia Math tournament in like 2 weeks.

Bonus Albatross Problem: Albatross is the owner of a store. He also employs 3 people Albatross, Himself and Bobert. The owner of the store gets 2/3 of the revenue and each of the employees evenly spilt the remaning 1/3. If the store makes $90,702 of much money does Bobert get paid.

Bobert gets 10078. Albatross gets 2/3 of sales 60,468 +1/9 because he is an employee (albatross) which is 10078 + 1/9 because he is another employee(himself) which is also 10078 10078+10078+60468=80624 since there is only one more employee, he gets the remaining money. 90702-80624=10078

[quote="goldendomer"][quote="now a ranger"][quote="soccerstar123"]single or with a group

good luck

-soccerstar123[/quote]

I am with a group, but our team isn't very good, so I am just shooting for a top 5 finish individually.[/quote]

Good luck now a ranger! :D[/quote]
I bet you will do well now a ranger!!

hey i am not sure i am 9th but 8th but i put 9th

There really shouldn't be that many high schoolers in this forum :ninja:  :huh:

That's a warning :!:

[quote="jackdillon"][/quote][quote="goldendomer"][quote="now a ranger"][quote="soccerstar123"]single or with a group

good luck

-soccerstar123[/quote]

I am with a group, but our team isn't very good, so I am just shooting for a top 5 finish individually.[/quote]

Good luck now a ranger! :D[/quote][quote="jackdillon"]
I bet you will do well now a ranger!![/quote]

Thanks. I will need luck.  :D

High schoolers:
Please note that it says specifically Middle Schoolers ONLY
Also, try to stay on topic
(Hint, Hint, Albatross people :P )

I don't understand why it would be that big of a deal for a high schooler to be on the mathcounts forums. Although most of the problems aren't interesting for me, I still check them out because I am now the head coach for my past middle school's mathcounts team and want to be as qualified as possible. But anyways, I'm a junior who's been doing math outside of the classroom since I was a freshmen (last offline course I took was calc in 8th grade). And about that Albatross problem, Himself would be a seperate person because it was capitalized; however, I'm not sure if the Albatross being employed is the same as the owner or an employee with the same name...I don't think 'employ' is a reflexive verb...

I agree that if you are a coach or just an adult, then you should be able to view any forums and be able to post.  However, there have been many cases where a more advanced student has posted jargon in a forum such as this one that is not as advanced.  That is the reason why high schoolers are not encouraged to regularly post here.

If there is a topic where a question is being unanswered or needs clarifying, then anyone is welcome to provide an explanation.  I hope I made sense. :)


1. every sentence posted is true
2. post something that cannot be disproved by the posted sentences and prove it

Could you give an example? I'm a little confused.

well the first sentences will be of course axioms

examble:
1) The sum of angles in a triangle is 80°

then the next person cannot send the axioms of the Euclidean geometry because then the sum of angles in a triangle is 180°<>80°

[quote="modeler"][quote="Ignite168"]The sum of the angles of a quadrilateral are 270°.[/quote]

I can't understand. Why not 80° + 80° = 160°?[/quote]

No, you're going to have to work with that axiom.

[quote="modeler"][quote="Ignite168"]The sum of the angles of a quadrilateral are 270°.[/quote]

I can't understand. Why not 80° + 80° = 160°?[/quote]

[quote]examble:
1) The sum of angles in a triangle is 80° [/quote]

that was just an example.

Since the sum of angles of a quad is 270, sum of angles of a triangle is half that which is 135.

There exists a triangle ABC with two right angles and an angle of -45°

There exists a triangle with negative angles.

The product of any two integers is irratonal, and the sum of any two rationals is a Gaussian integer.

[quote="K81o7"]the sum of any two rationals is a Gaussian integer.[/quote]

[quote="randomdragoon"]42 is the additive identity.[/quote]

a.

[quote="BoesFX"]So, 42 is Gaussian integer :D[/quote]

Well hey, who said 42 is rational?  Or who said anything about additive identities.

Every number has exactly 7 distinct additive identities.

[quote="thatin"]i was thinking to post the next summer here to announce my results about this world but i couldn't resist...
the ruler and king is a function that includes all the functions...[/quote]
im the other meaning of ruler and king which is in Latin. It means overlord of everyone can make ppl do whatever I want them 2 do...

there are no people in this world
so of course you can make them do whatever you want them 2 do...

ps: keep posting... u r brilliant!!!

U r wrong because I am a person and live in this world u just contridicted me. :D  :first:

you are the only person in this world.
so you just have self-control  :D
(unlike me, i was supposed not to post here....  :blush: )

self means everybody and only means one out of the million people the rest of your sentence means the same as English.
 :D

and as everybody knows....
[size=75]1.000.000[/size]=[size=150]1[/size]

It follows that

760>759 from my hypothesis.

If I had a million dollars I'd be poor and rich.

I can fly (well technically its not flying withoout gravity...)


group theory
abstract algebra
topology
calculus
integration
induction
geometry

Let $E$ be an n-dimensional affine euclidean space and $G$ a
closed subgroup of the group of all isometries of$E$. Suppose that each element
of $G$ has a fixed point. Then the group $G$ is compact and has a fixed point.

I think I've got it (finally :) ). The proof will consist of several steps and will seem rather long, but I'm sure nobody's going to read it anyway, so there should be no problem :).

Some perliminary stuff:

I'll identify the $n$-dimensional Euclidean space with $\mathbb R^{n}$, and its group of isometries will be denoted by $E(n)$. Also, $O(n)$ will be the $n$'th orthogonal group. Notice that any element of $E(n)$ can be represented as a transformation of the form $x\mapsto Ax+a,\ \forall x\in\mathbb R^{n}$, with $A\in O(n),\ a\in\mathbb R^{n}$. Given a group $G\le E(n)$, the set $\mbox{Fix}(G)$ consists of the points in $\mathbb R^{n}$ fixed by all elements of $G$. It's an affine subspace of $\mathbb R^{n}$, so it can also be regarded as a Euclidean space of dimension at most $n$. For $g\in E(n),\ x\in\mathbb R^{n}$, I'll denote the action of $g$ on $x$ by $gx$. If a subgroup of $E(n)$ has the property that every element has a fixed point, we'll say that the group has property $(P)$.

Notice that the collection of affine subspaces of $\mathbb R^{n}$ satisfies the descending chain condition (there are no infinite properly descending chains with respect to inclusion, that is), so in order to prove that such a collection ($\mbox{Fix}(g),\ g\in G$ for some $G\le E(n)$, for instance) has non-empty intersection, it suffices to show that it has the finite intersection property (FIP; every finite subcollection has non-empty intersection).

Our group $G$ is closed in the Lie group $E(n)$ and is hence a Lie group. The topology of the groups involved will be important in the proof, so in general, for a group $G\le E(n),\ G_{1}$, will be the connected component of the identity of $G$.

One last observation before I begin the proof: the two facts we have to prove about $G$ (namely its compactness and $\mbox{Fix}(G)\ne\emptyset$) are equivalent. Indeed, if $G$ fixes a point, then $G$ can be regarded as a closed subgroup of $O(n)$, so it must be compact. Conversely, if $G$ is compact, we have integration on $G$ with respect to its normalized Haar measure $\mu$. If $x\in\mathbb R^{n}$ is any point, then its corresponding "average" $\int_{G}gx\ \mbox d\mu(g)$ is a fixed point of $G$, as can easily be checked.

Now for the actual proof:

[i]Step 1[/i] - This consists of proving the statement

An abelian subgroup $G$ of $E(n)$ having property $(P)$ has a fixed point.

The proof is by induction on $n$, the case $n=0$ being trivial. Now, if $G$ is trivial, there's nothing to prove. Otherwise, let $g\in G\setminus\{1\}$. $\mbox{Fix}(g)$ is fixed by $G$ as a set (not pointwise), because $G$ is abelian. Moreover, each element $h$ of $G$ has a fixed point in $\mbox{Fix}(g)$. Indeed, let $x\in\mathbb R^{n}$ be a fixed point of $h$, and let $\tilde x$ be its orthogonal projection on $\mbox{Fix}(g)$. $\tilde x$ is the unique point which realizes the minimal distance between $x$ and $\mbox{Fix}(g)$, both of which are sets fixed by the isometry $h$. This means that $\tilde x\in\mbox{Fix}(h)$. Hence, if we restrict all elements of $G$ to $\mbox{Fix}(g)$, then we can assume we're working in a Euclidean space of strictly lower dimension with the same hypotheses, and induction finishes the proof.

[i]Step 2[/i]

A [i]connected[/i] closed subgroup $G$ of $E(n)$ with property $(P)$ has a fixed point.

As noted above, each element $g\in G$ can be represented in the form $g: \ x\mapsto Ax+a,\ A\in O(n),\ a\in\mathbb R^{n}$. Notice that $g\mapsto A$ is a continuous group homomorphism of $G$ int $O(n)$. Also, notice that it is one-to-one: if $g: x\mapsto x+a$, then $g$ is a translation with a fixed point, so it must be the identity ($a=0$). We can thus regard $G$ as being a Lie subgroup of $O(n)$ (we don't yet know that it's a [i]closed subgroup[/i]; in fact, this is the conclusion we want to reach).

Being a Lie subgroup of a compact Lie group, $G$ has a Lie algebra $\mathcal L$ which admits a bi-invariant metric. It follows from this that the adjoint representation of $\mathcal L$ is semisimple (for each ideal, the orthogonal complement is again an ideal), so $\mathcal L$ must be of the form $\mathcal C\times\mathcal S$, where $\mathcal C$ is the center of $\mathcal L$ and $\mathcal S$ is semisimple with negative definite Killing form (the negative definiteness follows from the existence of an invariant metric on $\mathcal S$; it makes $\mathcal S$ act on itself through the adjoint representation by skew-symmetric operators). $G$ splits accordingly as a product $C\times S$, where $C$ and $S$ are connected Lie subgroups with Lie algebras $\mathcal C,\ \mathcal S$ respectively. $C$ is the connected component of the center of $G$, so it must be compact by Step 1. On the other hand, since the Lie algebra of $S$ (in case it is non-trivial) is semisimple with negative-definite Killing form, a general result says that every connected group having it as a Lie algebra ($S$, in particular) must be compact. We are now finished: $G=C\times S$ with both $C$ and $S$ compact, so $G$ itself must be compact.

[i]Step 3[/i] - The general case

By Step 2 we know that $G_{1}$ is compact. Since it's also normal in $G$, it follows that the non-empty set $\mbox{Fix}(G_{1})$ is fixed (not necessarily pointwise) by $G$. Let $H\le G$ be the compact subgroup consisting of elements which fix $\mbox{Fix}(G_{1})$ pointwise. Clearly, $G_{1}\le H$ and $H$ is normal in $G$. Notice now that every element of $G/G_{1}$ has finite order. Indeed, if $g\in G$ were such that $g^{t}\in G\setminus G_{1},\ \forall t\ge 1$, then the subgroup $\langle g\rangle$ of $G$ generated by $g$ would be discrete and infinite, contradicting the fact that since $g$ fixes a point, $\langle g\rangle$ is relatively compact. Because $G/G_{1}$ is torsion and $G_{1}\le H,\ G/H$ is torsion. Also, $G/H$ can be regarded as a group of isometries of $\mbox{Fix}(G_{1})$, so it can be realized as a group of matrices.

As is well-known, a torsion group of matrices over a field is locally finite. It follows from this that the group generated by $H$ and any finite collection of elements of $G$ is contained in finitely many cosets of $H$ in $G$, and is thus relatively compact. In turn, this implies that the collection of affine spaces of the form $\mbox{Fix}(g),\ g\in G$ has FIP, so it must have non-empty intersection. Q.E.D.

I hoped for a simple proof (by some clever application of the triangle inequality). Your proof is only semisimple.  :D

Just how essential is the Euclidean space structure here? Is there any reasonable locally compact metric space for which the statement fails?

[quote="mlok"]Is there any reasonable locally compact metric space for which the statement fails?[/quote]

Which statement is that? In sam-n's problem, the isometry group had a natural topology, but in general we should first topologize that group in a reasonable way, and then see whether the statement holds. We could also try to find out what happens if we drop the word [i]closed[/i] from the statement. In this form, the question makes perfect sense for every metric space:

$(*)$ Let $G$ be a group of isometries on a metric space $(X,d)$ with the property that every element fixes a point. Does it follow that the whole group fixes a point?

A simple counterexample: take $(X,d)$ to be countably infinite and discrete (so it's also locally compact), with the trivial metric $d(x,y)=1,\ \forall x\ne y$. The group of isometries is just the group $S(X)$ of permutations on $X$, and if we take $G=$ the group of permutations which move only finitely many points, we're done. I have no idea what the answer to $(*)$ for $(X,d)$ as in sam-n's problem: a Euclidean space. I think I saw this problem proposed in the Canadian Mathematical Bulletin by Djokovic (a very old issue, something like '81), but there was no solution there.

Incidentally, there's also a natural way to turn the isometry group $S(X)$ of the countably infinite discrete space $(X,d)$ used above into a well-behaved metric space (a so-called [i]Polish group[/i], i.e. a separable and complete metric group). Let $(x_{n})_{n\ge 1}$ be an enumeration of $X$, and given $\sigma,\tau\in S(X)$, set $\rho(\sigma,\tau)=\frac 1{2^{n}}$, where $n$ is the smallest positive integer for which $(\sigma(x_{n}),\sigma^{-1}(x_{n}))\ne(\tau(x_{n}),\tau^{-1}(x_{n}))$. It's easy to check that $\rho$ is a metric on $S(X)$ with the required properties. If in $(*)$ we require $G$ to be closed (so we revert to sam-n's question, but in our space $(X,d)$), then counterexamples can still be found. It's a simple combinatorial exercise to find permutations $\sigma,\tau\in S(X)$ of order two, such that (a) $\sigma,\tau$ each fix exactly one point of $X$, and these points are different, (b) $\sigma\tau$ fixes exactly two points of $X$. Now check that $G=\langle\sigma,\tau\rangle\le S(X)$ is closed in the topology induced by $\rho$, and that it's a counterexample to $(*)$.


Let $a,b,c>0$ and $a+b+c=1$, then $16abc \le b+c$.

[quote="nickolas"]$1=(a+b+c)^2\geq4a(b+c)\Rightarrow b+c\geq4a(b+c)^2\geq16abc\Rightarrow b+c\geq16abc.$[/quote]

Nice!!!

[quote="me@home"]how did you make some of those inequality changes? like from $(a+b+c)^2$ to $4a(b+c)$?[/quote]

$a+(b+c) \ge 2 \sqrt{a(b+c)}$ by AM-GM, and square both sides.


integration
Support
Functional Analysis
real analysis
real analysis unsolved

Consider $C^{k}[0,1]$, it is a Frechet space (actually a Banach space). If $I \in C^{k}[0,1]^{*}$,
Prove: $I(f)=\int f^{(k)}d\mu+\sum_{n=1}^{k-1}c_{n}f^{(n)}(0)$. with unique $\mu, c_{n}$ .

(I think the first step is to use Taylor expantion:
$f(x)=\sum_{n=1}^{k-1}\frac{f^{(n)}(0)}{n!}x^{n}+g(x)$, then let $c_{n}=I(x^{n})$.

If $g(x)$ runs trough $C[0,1]$ when $f(x)$ runs trough $C^{k}[0,1]$, then by Riesz
representation Thm, I am done. But $g(x) \in C^{k}[0,1]$ too, so $g(x)$ don't run through
$C[0,1]$!

Or maybe $g(x)$ is dense in $C[0,1]$? So we know  $g(x) \in C^{k}[0,1]$ and $g^{(n)}(0)=0$
for all $n\leq k-1$, is $g(x)$ dense in $C[0,1]$?
But then maybe $C^{k}[0,1]$ is already dense in $C[0,1]$? why do we need those
terms with single point support? So I dought this is the right direction...)

can someone give a hint?

I think the sum should start with $n=0$, not $n=1$.

Hint: $C^{k}[0,1]$ is isomorphic to $\mathbb R^{k}\oplus C[0,1]$ via the map $f\mapsto (f(0),f'(0),\dots,f^{(k-1)}(0))\oplus f^{(k)}$.

"doubt"


geometry
3D geometry
sphere
inequalities unsolved
inequalities

I'm looking at a question from the 10/2005 FAU HS Math Competition (http://www.math.fau.edu/MathematicsCompetition/Contest2005.pdf), which says:

Let a, b, and c be real numbers satisfying $a+b+c=6$ and $a^2+b^2+c^2=18$.  What are the largest and smallest possible values of $c$?

I started out squaring both sides of the first equation, landing on $ab+bc+ac=9$.  That doesn't seem to help much with the restrictions on $c$.

Then I thought about $a+b+c=6$ as a plane, $a^2+b^2+c^2=18$ as a sphere.  Of course the plane passes through $(6,0,0)$, $(0,6,0)$, and $(0,0,6)$, and the sphere is centered at the origin, so it seems like the maximum $c$ would be attained on the line $(3t,3t,6-6t)$.  Plugging into the sphere I get $18(3t-1)(t-1)=0$, which gets me $(1,1,4)$ and $(3,3,0)$.

Am I correct that the minimum and maximum values for $c$ (or $a$ or $b$) are 0 and 4 respectively?  If not, what am I doing wrong?  If so, is there a faster/nicer/more satisfying answer than what I've shown?

a + b = 6 - c;
a ^ 2 + b ^ 2= 18 - c ^ 2
Just find the discriminant when solving for a and b in terms of c, and you get the restrictions for c, I think.

put  a=6-b-c, square it and then what you get for a^2, put that in the second equation. then yuo have an equation with b and c only, and just look at the discriminant.
you'll get -c(3c-12)>=0
now it's obvious that the min for c is 0, and the max is 4.

Use Cauchy-Schwalz, we have $(1^2+1^2)(a^2+b^2)\geq (1\cdot a+1\cdot b)^2$, so  $2(18-c^2)\geq (6-c)^2 \Longleftrightarrow 0\leq c\leq 4.$


In hexagon $ AOPSML $, $ \angle O = \angle P = \angle S = \angle L $ and $ \angle A $ is supplementary to $ \angle M $.  How many degrees are in the measure of $ \angle P $?

[hide="hint"] supplementary angles add up to 180, and the total number of degrees in a regular polygon is (n-2)180 [/hide][hide=" my answer"] take
angle p=x so o,s and l are also x

720= 4x+180
540=4x
x=135

[/hide]

i hope i didnt do something stupid!


So at my high school, I had really superior peers. Because I am also in math team, the pressure to succeed is even greater. Sometimes, I felt really pessimistic. I always think that I am the laziest person in the team because I didn't achieve all those things that the people around me did. This non-stop negativity keeps me from working harder and I am afraid that if this trend continues, it would not be beneficial for me in the wrong run. I think I need to change my frame of thinking so that I could take advantage of my situation (of having strong peers) and continue reaching for my potential. Help please?

Set a goal for yourself. Take for example, at the start of the year when I started the Extension 2 Mathematics course at my high school. My half yearly exam was very bad (I think around 70 something percent) so I was a bit set back. Then I told myself, "These people started this course since last year while I only started this year."

I set a goal for myself that "By the end of the year I want to be at least at a certain level" and that motivated me to work harder, try new problems, explore new concepts (while extending knowledge I already have) and so on. If you really love Mathematics then at the end of the day, it is not about solving the most problems, it is about having a better understanding the material you are studying and being able to apply these understandings to other problems.

This way my marks improved until I topped the year by beating the one who used to be first by a margin of 21 marks. You should do the same.

First, identify you weaknesses (e.g. my weakness is combinatorics) this is crucial is improving your problem solving skills.

Secondly, try looking at how problems in these areas are solved and learn these techniques and get into a new frame of thinking.

Thirdly, practice. This is especially true for school exams. However, for things like the Olympiad you would probably seek to understand a variety of problems but always keep an open mind because there may be 50 ways to solve a problem.

Just remember, you have acknowledged the fact that they are your superiors (even if they are your peers) and you should look up to them. Look at their solutions, try to see the problem from their perspective.

Good luck :) .

When this happens, I turn everything into a competition. I try to out-do my superior peers. (This is all friendly of course, and you should get your friends to do the same, so all of you reach for the highest potential.)

[quote]To get ahead, you have to work while the other guy sleeps.[/quote]

Obviously don't study math at like 1 AM, but when there is a night where there is no homework, study harder.

Don't worry about negativity. Always face a challenge with a healthy balance of confidence. If you are expieriancing some harmful negativity you should always stop and take time to think about your problem. If you failed a test you should ask yourself-

1) "Hmm, am I failing because I do not know the material or just because I wasn't studying?"

2) "How can I prevent this from happening?"

In my AP bio class, my teacher encourages students helping others. Utilize your peers well, and if you can help them with something, that should ease your negativity.

I think the previous posts have a lot of good suggestions about personal motivation, but they're not really answering the question "how can you benefit from your brilliant peers?" You should not take for granted your great fortune in having peers from whom you could learn a lot. Try to do things [i]with[/i] them, not just at the same time as them. Get your school to do team contests (like Mandelbrot team, ARML, HMMT, PUMaC, iTest, etc), and while you're working, try to observe your friends and talk to them and understand how they're so good. Practice with them. Have practice meetings where you exchange tips with them about all the little tactics everyone has to do better on contests.

Your friends are more than just a tough standard to compare yourself to and to attempt to beat. They have a wealth of experience and ideas that you can benefit from. Just ask!

Being able to compete with superior peers is always a good opportunity. You'll realize that stepping just out of your comfort zone is very helpful in the long run, since you hear and see good thinking around you. Xevarion touched upon many good ideas on how to try to benefit from your friends.

As for the low morale, kill it! First of all, you should not feel "lazy" if you are outperformed, especially if you tried a reasonable amount; you should not feel like you need to bleed your neck for your contribution to be significant. As for being outperformed, that's life.

After every competition, ask yourself: [b]What worked[/b], and what can i do to ensure this happens again? [b]What didn't work[/b], and what can I do to fix that?

If you think about these a bit, you might be able to uncover some of your own shortcomings in a more exact way (such as, "I tend to overcount" or "I tend to trust my arithmetic too much"). You then have very specific questions that you can ask your peers. In many cases, you might find that one of them has faced similar issues, and they might be able to share their strategies for overcoming their problems. Things like these will often help even more than being like, "yo, you're smart, teach me how to factor."

Also, you should practice (this always helps) and try to remember questions you have done before and even apply the results of those questions to problems you may be doing now. Take for examaple this year's IMO Question 2. Many Mathlinkers said that they have seen the converse. That gave them an advantage obviously.

Also, I sometimes admire the ways that people on this forum see subtle factorisations (which I later learn are really quite elementary  :roll: ) like $ a^{3}\plus{}b^{3}\plus{}c^{3} \minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca)$. It is those subtle things that may distinguish one person from another but I still think these things require practice.

Thank you A LOT for your perspectives. It is a bit surprising how your "what you might feel" and "what you should instead focus on" applied to me. Basically, I should not get lazy.


Find all positive integers a,b,c such that 2^a+3^b=5^c.

Bomb


[quote="paganinio"][求助]高中数学迷应该报什么大学专业?
我非常希望以数学为一生的职业。我所选择的大学只有以下几类与数学有关的专业，请大家看看是不是适合我报。

1,数学与应用数学
名字听起来很适合，但是它到底是学什么呢？喜欢高中数学的人还会不会喜欢它呢？

2,经济学类
很多人说大学的经济学“就是数学”，但是我更喜欢纯数学那一类的，而经济学掺杂了很多非数学的课程。我想知道经济学里面数学的比重到底有多大，如果不是很大我是不会报的。
另外，大学经济学所学的数学，好像不外乎就是“微积分”，“线性代数”和“概率统计”这么三样。这么狭窄的学习范围会要了我的命的！学经济学还有没有机会接触更广阔的数学知识呢？
而且我对经济学没什么好感，觉得它玷污了纯数学，只是把数学当作工具，而不是学习数学本身。

3,信息工程类
我有多年的编程经验，认为信息工程、软件工程毕竟也不是纯数学，只是它的一个应用分支，不过它的数学性总比经济学好得多。然而我发现一个专业叫做“信息与计算”，是干什么的？

恳请知情人士指点几句，谢谢。[/quote]

既然是希望以数学为一生的职业
那还是选择1吧
“数学与应用数学”一般是高考的所谓专业名称吧
一般到了大学学了三个学期基础课后还要重新分专业
基础，概率，计算，信息，金融
反正北大是这么5个专业
另外如果学了一段时间基础后不感兴趣
那也不要紧
起码读完本科还可以换专业
就出国来说，学数学的什么专业都可以申
这么多年来北大数院基础数学出国的除了学数学的还有
商学院，医学院，CS，经济的都有
想转行还是很容易
但前提是大学的课程学得好，GPA高

数学在经济学中
说白了还是工具
最重要的还是经济学思想
虽然很多经济学大师数学也很好
分析，概率，统计，拓扑，动力系统等在经济学里也有很重要的应用

至于3
信息工程类用到的fancy的数学就比2还要少了
当然一些学科比如理论计算机除外
因为那就是应用数学了

祝好运


trigonometry
inequalities
circumcircle
inequalities unsolved

If $A,B,C$ be the angles of a triangle $ABC$ and
$sin(A - \theta)sin(B - \theta)sin(C - \theta) = sin^3 \theta$
then find the maximum value of $tan \theta$

We assume that $0<\theta<min\{A,B,C\}$, so the angles $A-\theta$, etc are positive

On the circumscribed circle $(O,R)$ of $\triangle ABC$ we take the points:

$D$ on the arc $BC$
$E$ on the arc $CA$
$F$ on the arc $AB$

such that $\angle BAD = \angle CBE = \angle ACF = \theta$.

Then $\angle DAC = A-\theta, \angle EBA = B-\theta, \angle FCB = C-\theta$

[color=olive](we could choose $\angle DAC = \angle EBA = \angle FCB = \theta$, this is the second way to construct the problem geometrically)[/color]

We will see that, if $\triangle ABC$ is given, then the angle $\theta$ is unique

Using the formulas $BD = 2R\cdot\sin \theta$ etc we get

$\sin(A-\theta)\sin(B-\theta)\sin(B-\theta) = \sin^3 \theta \ \ (1) \Rightarrow$

$DC\cdot EA\cdot FB = BD\cdot CE\cdot AF \ \ (2)$

The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60520]Ceva theorem for chords[/url] says that the chords $AD,BE,CF$ concur iff $(2)$ holds. So we know that these chords are concurrent.

[hide="Here is a way to construct the angle theta if the triangle ABC is given"]

Let $P$ be the intersection point of $AD,BE$.
Since $\angle PAB = \angle PBC = \theta$ we get the conclusion that the line $CB$ is tangent to the circle $(A,P,B)$

(with $(A,P,B)$ I mean the circumscribed circle of $\triangle APB$)

Similarly, $\angle PBC = \angle PCA = \theta$, so the line $AC$ is tangent to the circle $(B,P,C)$.

The two circles intersect each other a $B,P$.
There is only one circle which passes through $A,B$ and touches $CB$ and only one circle which passes through $B,C$ and touches $AC$.
So, if the triangle $ABC$ is given then the point $P$ is unique.

Finally, the circle $(C,P,A)$ is tangent to $BA$ because $\angle PCA = \angle PAB = \theta$
So, if we try to construct a circle which passes through $A,C$ and is tangent to $BA$, then it will pass through $P$[/hide]

The angle $\theta$ depends only on the triangle $ABC$. So the problem doesn't ask to find the maximum for a specific triangle but it asks to find the maximum of $\tan\theta$ when $ABC$ belongs to the set of all triangles

$0<\theta<min\{A,B,C\} \Rightarrow \boxed{0<\theta\leq 60^o}$

So we can maximize $\theta$ instead of $\tan\theta$

We have $\triangle ABC \sim \triangle BPD$. Indeed:

$\angle BDP = \angle BCA$ (they are "looking" at the same arc $AB$)

$\angle PBD = \angle DBC + \angle CBE = \angle DBC + \angle BAD = \angle BAC$

Hence $\frac{BP}{AB} = \frac{BD}{AC} \Rightarrow \boxed{BP = BD\cdot \frac{AB}{AC}}$

Also, $\triangle ABC \sim \triangle ECP \Rightarrow \frac{PE}{AC} = \frac{EC}{AB} \Rightarrow \boxed{PE = EC\cdot \frac{AC}{AB}}$

So, the (negative) power of $P$ w.r.t. $(O,R)$ is $PE\cdot PB = BD\cdot EC$

$BD = EC = 2R\cdot\sin\theta \Rightarrow PE\cdot PB = 4R^2\sin^2\theta$

The maximum (negative) power for an interior point of a circle is $R^2$ (when the point is the center of the circle)

So $PE\cdot PB\leq R^2 \Rightarrow 4\sin^2\theta \leq 1 \Rightarrow \sin^2\theta\leq \frac{1}{4} \Rightarrow \sin\theta\leq \frac{1}{2}$

So $\theta$ can be at most 30 degrees

Now we should check if the point $P$ can be on the center $O$ of the circle $(O,R)$

Indeed, if the triagle $ABC$ is equilateral then $\angle OAB = \angle OBC = \angle OCA = \theta = 30^o$, so the condition $(1)$ holds

There is also a solution which is based on the discussion about whether we can construct the figure but it is too complicated.

the problem is not so much critical!!!
but still i appreciate ur solution
The expression for $tan \theta$can be found and on applying am-gm with jensen we get the result also.


linear algebra
matrix
algebra
polynomial
geometry
geometric transformation
superior algebra

Let $B\in\mathcal{M}_{2}(\mathbb{R})$ a fixed matrix and $M$ the set of all matrix with the folowing form: $A(x,y)=xI_{2}+yB$ for any real x,y.

(a) show that $M$ is a sabil part of  $\mathcal M_{2}(\mathbb{R})$ with "+" and "$\cdot$" and $(M,+,\cdot)$ is a comutative ring.

(b) what's the necesary contition upon B in order that $(M,+,\cdot)$ to be a field

If $B = k I_{n}$ for some $k \in \mathbb R$, then $\left( M_{B},+, \cdot \right)$ is a field isomorphic to $\left( \mathbb R,+, \cdot \right)$.

Else let $P_{B}= X^{2}-\left( \text{tr}\, B \right) X+\left( \det B \right) I_{2}$ and notice that $\left( M_{B},+, \cdot \right) \cong \mathbb R[X] / P(X)$ (use Cayley-Hamilton). That's clearly a commutative ring. If we want a field, then all polynomials of degree at most $1$ must be coprime with $P$. That happens exactly when $P$ has no real roots.

I hope it's OK.

PS: He means "parte stabila", i.e. $M$ is closed under $+$ and $\cdot$

Thanks for the translation part  :). Couldn't we take the matrix $B$ having the form $B=\left ( \begin{array}{cc}a & b \\ c &-a \end{array}\right)$ with $\det(B)=-1$? and it's pretty clear after.  :maybe:


inequalities
AMC
USA(J)MO
USAMO
induction
inequalities unsolved

assume $ {a_n}(n\in \mathbb{N^*})$ satisfying that $ a_i\plus{}a_j\leq a_{i\plus{}j}(i,j\equal{}1,2,\ldots)$,
prove that:
$ \sum_{k\equal{}1}^{n}{\frac{a_k}{k}}\leq a_n (n\in\mathbb{N}^*)$

This problem appeared three times (at least, the ones I've seen) in MO's:

USAMO 1981
China MO 1993
APMO 1998

The proof uses induction.

(During the proof, assume that $ a_0 \equal{} 0$: this does not change the proof at all)
For n=1, the inequality is obvious.

Assume that the inequality holds for 1,2,..,n-1.

By the condition and the induction hypothesis, we have $ a_1 \plus{} \frac {a_2}{2} \plus{} .. \plus{} \frac {a_k}{k} \plus{} a_{n\minus{}k} \leq a_k \plus{} a_{n \minus{} k} \leq a_n$.
Summing up from i=1,2,..,n, we get $ n( a_1 \plus{} \frac {a_2}{2} \plus{} .. \plus{} \frac {a_n}{n} ) \leq n a_n$. Q.E.D.

[quote="tobeno_1"]assume $ {a_n}(n\in \mathbb{N^*})$ satisfying that $ a_i \plus{} a_j\leq a_{i \plus{} j}(i,j \equal{} 1,2,\ldots)$,
prove that:
$ \sum_{k \equal{} 1}^{n}{\frac {a_k}{k}}\leq a_n (n\in\mathbb{N}^*)$[/quote]

a similar one I proved here before:
http://bbs.pep.com.cn/thread-500441-1-1.html


geometry
trigonometry
AMC
AIME
geometric transformation
reflection
analytic geometry

I think I will get 90 (unless they dislike my geometry=/ )... 5-8 were easier than previous years (I couldn't believe they made 8 that easy...), and the contest itself was easier than the AIME. I am hoping they accept my 9c, I was not happy with it... I did not get 10, I should have, but I messed up on 9 and had to redo it, leaving me with only 30 minutes for 10 and for double checking. I then used a really stupid approach to solving 10... Stupid... If i had just used the method I intended to, I probably could have gotten a at least.
^ Kept vauge due to the contest being finished only a few hours ago.
*doesn't want to be the first to post about the solutions in case it violates the rules*

I will post the answers I, and a couple of others got for 1-9 if it is acceptable.
*is eagerly awaiting someone to solve 10*
I can't, need to study for provs=\

I screwed up my basic trig and generally question 6 killed me. I'd be in the mid 80s had I not screwed up common sense.

Instead, I expect a score around 78-79. Grr... Lower than all my practice scores.

I better redeem myself with Hypatia tomorrow.

(smugly) I think I did all of it.   :lol:   (Thank goodness I had the sense to do some past papers beforehand!)

Awesome! Congrats :D
I wish I had practiced more, I was only prepared to do simple series questions, and 10 caught me totally off gaurd=\
When do the results get released?

question 10 caught you off guard?  three years in a row, the last question has been of the sequence type!  tsk-tsk... what's funny is last year's #10 came from the book i have at home, and a few pages earlier, was this year's #10 (check out aime 1996!)... but I didn't realize it until I came home!  anyone wanna borrow my book for euclid '06? lol...

massey did ok... I got 96 unofficially... darn those light bulbs.  hope everyone else did well :)

[quote="lotus87"]question 10 caught you off guard?  three years in a row, the last question has been of the sequence type!  tsk-tsk...[/quote]
LOL. Yep. I don't know what I was thinking. I tried to solve it using modulos, but just ran out of time. I wish i had just done a simulation on a and then just proved that 2005 had to be hit on a pass. It all seems so clear now... If that is even the right solution lol.
TBH, I really wasn't expecting to get that far. In the past I have done rather poorly on the euclid (high 60s), so I wasn't expecting to do well... Still not sure how to solve it for 2005, but I am ashamed I did not get it 10)a :blush: Meh... Lets hope all my answers are complete, clear, and stylised;) and i get enough marks for a decent waterloo award... *prays*

I got everything except 9c, because I did 10 before I did 9.

I did 10 with binary numbers... 10b is easy mod 8, and 10c just intuitively makes sense once you see the patterns in binary.  Just add zeroes to the front (you can add infinitely many) and then change the 1s to 0s for values that will disappear.  It makes sense to me; I hope I explained it clearly enough that it makes sense to those marking!

So hopefully I'll get over 90, depending on how many lightbulbs I screwed up on.

On #10 my method was to use the recurrence \[f(2k) = f(2k - 1) = 2k + 1 - 2f(k)\] which is implied by the fact that after one step the lockers (1, 3, 5, ... 2k-1) which are left correspond respectively to the lockers (k, k-1, ... 2, 1) in the k-locker situation.  Then (by re-substituting) you get a very nice set of recurrences: \[f(4r - 3) = 4f(r) - 3;\] \[f(4r - 2) = 4f(r) - 3;\] \[f(4r - 1) = 4f(r) - 1;\] \[f(4r) = 4f(r) - 1.\]  From there the required properties (and the patterns you mentioned!) are fairly easy to obtain.

Hi I am a student who wrote the euclid on Tues. April. 19th 2005 aswell. I think that this is a great way to compare answers and as far as i can see there's noting in the "rules" that says that we can't post answers AFTER the date of the test. So i think we should start. Unfortunately I did not get the solution to #10 because of time restrictions. After I looked at it closer however, I noticed that the first section could have at least been done with brut force (just writing them all out and shuting the lockers yourself). I think i got 33. Anyone else get this too?

Anyone knows when will the result come out?
I did bad...
I didn't see number 9!
Or say, I left it and forgot.   :(

[quote="Elyot"]Peter Crippin told me that he marked my Euclid today and gave me either 96 or 97 (he said he couldn't remember)... and although that's unofficial, it's still pretty good.  Yay!

~ Elyot[/quote]

Oh nice.  Talked to Mike, he told me Boris got a 96 ish too, (but that was before the re-marking)

And does remarking usually help or hinder your score?  I would think it would help (ie you'd get more part marks that wouldn't be awarded to you the first time) but it could also result in having some marks taken off (for bad solutions, poor explaining, etc).

Peter Crippin told me that I write some of the best solutions he's ever seen.  I hope that gets me invited to seminar!

How high do winners on Euclid typically score?

Wow.  I've never done well on Euclids... my scores exhibit a negative trend with respect to grade.  Grade 10 - 84 (I think); Grade 11 - 82; Grade 12 ...  :blush: um, 77.  3rd in my school...  Clearly not my best contest.

At least I did well on USAMO.  Although if I'd known that I'd do so poorly on Euclid this year, I would have taken more time to rest before USAMO and maybe gotten that one more point I needed for Honourable Mention...

I got either 96 or 97... Peter Crippin said he couldn't remember. I haven't actually received my score from my teacher yet.

As for me, I had to nag my math teacher to get my score.

[quote="arm_tiger"]Hi, I'm new here.

Just have one question:

How did you guys get your results?
Through teachers or principles?

If so, is there another way (because I am finished my year and don't have any contacts with my teachers now).

Thanks a lot in advance.[/quote]

Through whoever registered you.  My principal takes care of all my math contests as opposed to a math teacher because I don't have one.

Results posted.

Arrgh ... I did worse than I thought.

I hate being an "unofficial" contestant... :S

I was being an unofficial contestant on 2004 open challenge!!! :(
that was I just came to Canada, knowing nothing--didn't sign up but showed up during the contest... :huh:

[quote="11th dimension"]What's wrong with the Euclid result booklet?  They put my age wrong as 17.  Well, guess what, at least this time they got my grade correctly and only missed my age by 1.  Look at my 2004's results, I was in grade 9 and 15 at that time, and they put me down as in grade 12 and 18 years old..... :noo:[/quote]

lol that hurts...one year they spelt my name as SRH Sun....they eventually changed it though, lol.

Holy crap I did good this year.

I totally filled in 17 as my age.  18... wtf?

everyone's age went up by 1, didn't it? everyone from my school anywayz... i was excited to hear i was 18... sadly the clubs don't agree... lol

[quote="frodo01"]Hey does anyone know when we will get our medals, certificates, money and stuff? Cause its like June already and I've gotten nothing!  :( Anyone know when????[/quote]

You have to ask your math teacher or the person who proctored the contest.

YAY! I won! Though I messed up on 3 light bulbs :lol:


What's the difference between {n \choose k} and {{n \choose k}}?

hmm...latex confusion maybe?  :?
looks the same to me

ARHGGGG...

How do you type the choose format? Disable BBText to type.

n\choose k: $n\choose k$

{n\choose k}: ${n\choose k}$

{{n\choose k}}: ${{n\choose k}}$


Bobby took a vacation to an exotic place.
He went bungee jumping off of a bridge that is 500 feet high.
When $x$ is the number of seconds elapsed, this is the equation of where he is:
$f(x)=|400-x^2|$ (There is 100 feet between him and the ground at the bungee's lowest point.)
After 50 seconds, where is he in the air? Express your answer relative to the bridge or the ground.
(The bungee keeps going up until the person reaches 3000 feet, or when they want it to stop) :D

That is an interesting formula for how far he is because he keeps getting higer and higher, and after 20 seconds, he hits the ground.
[hide]Just plug 50 into the equation to get
$f(50)=|400-50^2|$
$f(50)=|-2100|$
$f(50)=\boxed{2100}$
Therefore, he is 2100 feet away from the bridge.[/hide]

This is different from the vacation he took in my problem about Bobby.


Does anybody know if there is an international olympiad in chemistry for university students?

[quote="Carcul"]I have never heard of anyone. However, for a start you can find chalenging problems on this forum.[/quote]

I am wondering if there is a place where it would be possible to send a team to. :wink:

[quote="pardesi"]well probably a google search can help :)[/quote]

Unfortunately, not yet.


If the tens digit of a perfect square is 7, how many possible values can its units digit have?

We only need to deal with two-digit numbers, since anything above the tens place does not affect the tens place. If we let $ n \equal{} 10x \plus{} y$, with $ x$ and $ y$ being integers, $ 0 < x < 10$, and $ 0 \le y < 10$, $ n^2 \equal{} 100x^2 \plus{} 20xy \plus{} y^2$. We can ignore the $ 100x^2$. We want to find all numbers x and y such that the tens digit of $ y^2$ plus $ 2xy$ is congruent to $ 7 mod(10)$. We will go about this by plugging in values of $ y$ and seeing if we can find an $ x$ value for them. But first, we can narrow down the values of $ y$ we have to test. The tens digit of $ y^2$ cannot be even, because then the units digit of $ y^2$ plus $ 2xy$ will be even, and can never equal $ 7 mod(10)$. Therefore, we can exclude $ y \equal{} 0, 1, 2, 3, 5, 7, 8, 9$ from our list of numbers to test. The only ones left are $ y \equal{} 4$ and $ y \equal{} 6$. If $ y \equal{} 4$, then we need to find an $ x$ such that $ 1 \plus{} 8x \equiv 7 mod(10)$. This is accomplished when $ x \equal{} 2$. If $ y \equal{} 6$, then we need to find an x such that $ 3 \plus{} 12x \equiv 7 mod(10)$. This is accomplished when $ x \equal{} 2$. So, the only possible units digit of these perfect squares are the unit digits of $ 4^2$ and $ 6^2$. Since both of these units digits are 6, 6 is the only answer.

We have $ a^2 \equal{} 100b \plus{} 70 \plus{} c$
Using mod 5, we find c is 0,1,4,5,6,or 9
Using mod 4, we find c is 2,3,6, or 7
So 6 is the only possibility.


Why is $KMnO_4$ not the reagent of choice for the preparation of methyl vinyl ketone from the corresponding alcohol??

There are several interesting and unanswered topics buried in the oldest pages of this forum. This is one of them.

[hide="Answer"]Because potassium permanganate would also oxidize the double bond to the glicol (or to carbon dioxide and a carboxilic acid by oxidative cleavage if hot, alcaline KMnO4 is used).[/hide]

[hide="a better reagent"]PCC,colin's reagent[/hide]

[hide="carcul"]u starteda two year old thread  :ninja: [/hide]

[quote="pardesi"][hide="carcul"]u starteda two year old thread  :ninja: [/hide][/quote]
Well there was no response, so I think it should be allowed.

[hide="Note and interesting problem"]PCC is not very effective in oxidizing allylic alcohols. The reason of this can be seen in the oxidation of 7-methyloct-6-en-1-ol with PCC in dichlorometane. Can you find the product of this oxidation?[/hide]

[hide="Another reagent"]$ MnO_{2}$ is very effective in the oxidation of allylic alcohols to the unsaturated ketones/aldehydes.[/hide]

well ofcurse my resonse was [b]not[/b] cynical

well carcul is the compound u have given allylic alcohol :?:

No, its not an allylic alcohol, but have an hydroxil group and a double bond simultaneously, which is the important for the final outcome.

[quote="Carcul"]The reason of this can be seen in the oxidation of 7-methyloct-6-en-1-ol with PCC in dichlorometane. Can you find the product of this oxidation?[/quote]

[hide="Hint"]Don't forget that PCC is acidic. The product formed has molecular formula $ C_{9}H_{14}O$.[/hide]


Prove that sqrt2*sin2=sin({3 \pi /4})+cos({3 \pi /4}).

This problem is not hard at all, even in terms of Beginner's standards, where are they? Solve this and you will be rewarded :D .
Many people have 0 posts. You got a chance here, so come on and post one here!!!


Hi,
Not sure if I'm in the right place, but I have a problem for you math genius'

I need to find a calculator that basically works backwards.
I have a 4 digit number that is the solution and I need to know how to calculate it from a 7 digit number.
Here are the numbers.

the solution is 5395
the original number is 9421827

If I have the algorithm that I need to get 9421827 and turn it into 5395 then this will solve a lot of problems for me!!

Any help is greatly appreciated.

9421827 - 9416432 ?  :D

Why percisely do you need this?


geometry solved
geometry
circumcircle
geometric transformation
homothety
symmedian
Angle Chasing

Let $ ABC$ be an acute, scalene triangle, and let $ M$, $ N$, and $ P$ be the midpoints of $ \overline{BC}$, $ \overline{CA}$, and $ \overline{AB}$, respectively. Let the perpendicular bisectors of $ \overline{AB}$ and $ \overline{AC}$ intersect ray $ AM$ in points $ D$ and $ E$ respectively, and let lines $ BD$ and $ CE$ intersect in point $ F$, inside of triangle $ ABC$. Prove that points $ A$, $ N$, $ F$, and $ P$ all lie on one circle.

[hide="Solution sketch"]
Use the Law of Sines like 5 times to prove that $ \angle BFA\equal{}\angle AFC$. Then Use this to get another angle equal and you get $ \triangle BFA\sim\triangle AFC$. Spiral similarity sends P to N, so you get another similar triangle and you're done.[/hide]

[hide="outline"]
first do a bunch of angle chase and put angles in terms of angle BAM and CAM
then prove OEM ~ FEN and ODM ~ FDP with SAS. (used three law of sines for the ratio part, on EMN, DEF, DOE)
then you get equal angles and cyclicness easily follows from there
[/hide]

[hide]

Construct $ T$ on $ AM$ such that $ \angle BCT\equal{}\angle ACF$. Then $ \angle BCT\equal{}\angle CAM$. Then $ \triangle AMC\sim\triangle CMT$, so $ \frac{AM}{CM}\equal{}\frac{CM}{TM}$, or $ \frac{AM}{BM}\equal{}\frac{BM}{TM}$. Then $ \triangle AMB\sim\triangle BMT$, so $ \angle CBT\equal{}\angle BAM\equal{}\angle FBA$. Then we have

$ \angle CBT\equal{}\angle ABF$ and $ \angle BCT\equal{}\angle ACF$. So $ T$ and $ F$ are isogonally conjugate. Thus $ \angle BAF\equal{}\angle CAM$. Then

$ \angle AFB\equal{}180\minus{}\angle ABF\minus{}\angle BAF\equal{}180\minus{}\angle BAM\minus{}\angle CAM\equal{}180\minus{}\angle BAC$.

If $ O$ is the circumcenter of $ \triangle ABC$ then $ \angle BFC\equal{}2\angle BAC\equal{}\angle BOC$ so $ BFOC$ is cyclic. Then $ \angle BFO\equal{}180\minus{}\angle BOC\equal{}180\minus{}(90\minus{}\angle BAC)\equal{}90\plus{}\angle BAC$.

Then $ \angle AFO\equal{}360\minus{}\angle AFB\minus{}\angle BFO\equal{}360\minus{}(180\minus{}\angle BAC)\minus{}(90\plus{}\angle BAC)\equal{}90$. Then $ \triangle AFO$ is a right triangle.

Now by homothety about $ A$ with ratio $ \frac{1}{2}$, $ B$ is taken to $ P$ and $ C$ is taken to $ N$. Thus $ O$ is taken to the circumcenter of $ \triangle APN$ and is the midpoint of $ AO$, which is also the circumcenter of $ \triangle AFO$, so $ A,P,N,F,O$ all lie on a circle.
[/hide]

I had more fun constructing the diagram than trying to get anywhere with it, which I didn't.

Plus, next time people post their solution to a geometry problem, I'd like to see at least one with a diagram.

[hide="Outline for an Inversive Solution"]

Invert by an arbitrary radius about A. We want to show that P', F', and N' are collinear. Notice that D', A, and P' lie on a circle with center B', and similarly for the other side. We also have that B', D', F', A form a cyclic quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.)[/hide]

awesomest solution ever:

F is the isogonal conjugate of the point X along AM such that MA * MX = MB^2 = MC^2.

everything else is angle chasing here

[quote="Benjamin Hu"]I tried to use analytic geometry for this problem (which was a really stupid move), but didn't get the coordinates for F in time. I'd given up on angle chasing at this point.[/quote]

Haha, I did this as well.
I actually got the coordinates of $ F$. They were really REALLY nasty.

At that point, I was running out of time so I wrote "With the coordinates of the four points of interest, it is trivial to show the points lie on one circle."

We'll see how the graders like that.

I used coordinates. I assigned the Origin to $ A$, $ (10,0)$ to $ B$ (I really should have used $ (4,0)$...), then $ (a,b)$ to $ C$, and $ a < 10$. I solved for lines and points and found the coordinates for $ F$ after a while. Then I found the equation of the circle containing $ A,$ $ N,$ and $ P$ (This was less messy and easier to do than if I tried $ A$, $ N$, $ F$ or any other way containing $ F$). At the end I said that if $ F$ was plugged into that equation, it would be true.

Outline of my solution:
Median $ AM$ of a triangle $ ABC$ implies $ \frac{\sin{BAM}}{\sin{CAM}} \equal{} \frac{\sin{B}}{\sin{C}}$.
Trig ceva for $ F$ shows that $ AF$ is a symmedian.
Then $ FP$ is a median, use the lemma again to show that $ AFP \equal{} C$, and similarly $ AFN \equal{} B$, so you're done.

[hide="my whole solution"]Sorry I don't have a picture. It's a nice diagram to draw, though, since you don't need to do any circumcircles and unless you make it almost isosceles, all the points stand out pretty well.

By the Law of Sines a whole bunch, $ \frac{\sin\angle BAM}{\sin\angle CAM}\equal{}\frac{\sin B}{\sin C}\equal{}\frac bc\equal{}\frac{b/AF}{c/AF}\equal{}\frac{\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}$. Since $ \angle ABF\equal{}\angle ABD\equal{}\angle BAD\equal{}\angle BAM$ and similarly $ \angle ACF\equal{}\angle CAM$, we cancel to get $ \sin\angle AFC\equal{}\sin\angle AFB$. Obviously, $ \angle AFB\plus{}\angle AFC>180^\circ$ so $ \angle AFC\equal{}\angle AFB$.

Then $ \angle FAB\plus{}\angle ABF\equal{}180^\circ\minus{}\angle AFB\equal{}180^\circ\minus{}\angle AFC\equal{}\angle FAC\plus{}\angle ACF$ and $ \angle ABF\plus{}\angle ACF\equal{}\angle A\equal{}\angle FAB\plus{}\angle FAC$. Subtracting these two equations, $ \angle FAB\minus{}\angle FCA\equal{}\angle FCA\minus{}\angle FAB$ so $ \angle BAF\equal{}\angle AFC$. Therefore, $ \triangle ABF\sim\triangle CAF$, so a spiral similarity centered at $ F$ takes $ B$ to $ A$ and $ A$ to $ C$. Therefore, it takes the midpoint of $ \overline{BA}$ to the midpoint of $ \overline{AC}$, or $ P$ to $ N$. So $ \angle APF\equal{}\angle CNF\equal{}180^\circ\minus{}\angle ANF$ and $ APFN$ is cyclic.[/hide]

WLOG $ AB < AC$. The intersection of $ NE$ and $ PD$ is $ O$, the circumcenter of $ \triangle ABC$.

Let $ \angle BAM \equal{} y$ and $ \angle CAM \equal{} z$. Note $ D$ lies on the perpendicular bisector of $ AB$, so $ AD \equal{} BD$. So $ \angle FBC \equal{} \angle B \minus{} \angle ABD \equal{} B \minus{} y$. Similarly, $ \angle FCE \equal{} C \minus{} z$, so $ \angle BFC \equal{} 180 \minus{} (B \plus{} C) \plus{} (y \plus{} z) \equal{} 2A$. Notice that $ \angle BOC$ intercepts the minor arc $ BC$ in the circumcircle of $ \triangle ABC$, which is double $ \angle A$. Hence $ \angle BFC \equal{} \angle BOC$, so $ FOBC$ is cyclic.

Lemma 1: $ \triangle FEO$ is directly similar to $ \triangle NEM$
\[ \angle OFE \equal{} \angle OFC \equal{} \angle OBC \equal{} \frac {1}{2}\cdot (180 \minus{} 2A) \equal{} 90 \minus{} A
\]
since $ F$, $ E$, $ C$ are collinear, $ FOBC$ is cyclic, and $ OB \equal{} OC$. Also
\[ \angle ENM \equal{} 90 \minus{} \angle MNC \equal{} 90 \minus{} A
\]
because $ NE\perp AC$, and $ MNP$ is the medial triangle of $ \triangle ABC$ so $ AB // MN$. Hence $ \angle OFE \equal{} \angle ENM$.

Notice that $ \angle AEN \equal{} 90 \minus{} z \equal{} \angle CEN$ since $ NE\perp BC$. $ \angle FED \equal{} \angle MEC \equal{} 2z$. Then
\[ \angle FEO \equal{} \angle FED \plus{} \angle AEN \equal{} \angle CEM \plus{} \angle CEN \equal{} \angle NEM
\]
Hence $ \angle FEO \equal{} \angle NEM$.

Hence $ \triangle FEO$ is similar to $ \triangle NEM$ by AA similarity. It is easy to see that they are oriented such that they are directly similar. End Lemma 1.

By the similarity in Lemma 1, $ FE: EO \equal{} NE: EM\implies FE: EN \equal{} OE: NM$. $ \angle FEN \equal{} \angle OEM$ so $ \triangle FEN\sim\triangle OEM$ by SAS similarity. Hence
\[ \angle EMO \equal{} \angle ENF \equal{} \angle ONF
\]
Using essentially the same angle chasing, we can show that $ \triangle PDM$ is directly similar to $ \triangle FMO$. It follows that $ \triangle PDF$ is directly similar to $ MDO$. So
\[ \angle EMO \equal{} \angle DMO \equal{} \angle DPF \equal{} \angle OPF
\]
Hence $ \angle OPF \equal{} \angle ONF$, so $ FONP$ is cyclic. In other words, $ F$ lies on the circumcircle of $ \triangle PON$. Note that $ \angle ONA \equal{} \angle OPA \equal{} 90$, so $ APON$ is cyclic. In other words, $ A$ lies on the circumcircle of $ \triangle PON$. $ A$, $ P$, $ N$, $ O$, and $ F$ all lie on the circumcircle of $ \triangle PON$. Hence $ A$, $ P$, $ F$, and $ N$ lie on a circle, as desired.

[quote="rem"]Let pt K is intersection of tangents to circumcircle of ABC at B and C. Draw AK. Let it intersect circumcircle of KBC at F'. This circumcircle also goes through O. Then after angle chasing get F' is the pt F (as angle BAK is equal to angle CAM). Then after a bit more angle chasing u r done.[/quote]

This is pretty much what I did. I know it's the day before the winner phone call. People are anxious about the result, so let's look at the intriguing background of symmedian from [b]perspective drawing[/b], while waiting patiently and calmly.

In my Art I class, we learned [b]perspective drawing[/b], and my teacher claimed that one vanishing point is the midpoint of two others. I immediately wondered why. After I went home and did some calculation, the problem is equivalent to the following:
[b]Problem:[/b] Let $ \triangle AA'C$ be an isosceles triangle with $ AC \equal{} A'C$(C is a vanishing point), point $ E$ on $ AC$ and $ E'$ on $ A'C$ such that $ AA'//EE'$. $ X$ denotes the intersection of $ A'E$ and $ AE'$ and the gable(an art term) pass through $ X$ is parallel to $ AA'$. Let $ D$ be a point on the gable on the different side of $ AC$ as $ A'$, and $ AD$ intersect the line through $ C$ parallel to $ AA'$ at $ B$(This is a vanishing point). $ DE$ meets $ BC$ at $ B'$(third vanishing point). Prove that $ BC \equal{} B'C$.
[b]Solution:[/b] Let the gable meet $ AC$ at $ F$, since $ \frac {DF}{BC} \equal{} \frac {AF}{AC}$ and $ \frac {DF}{B'C} \equal{} \frac {FE}{ED}$ we want to show that $ AFEC$ is a harmonic range. This follows from
\[ \frac {AF}{FE} \equal{} \frac {AA'}{EE'} \equal{} \frac {AC}{EC}
\]
Now we are going to look at a special case of the 1996 Chinese Math Olympiad Problem 1.
[b]Problem:[/b] With the same diagram as the previous problem, prove that the gable pass through the two points of tangency (namely, $ P$, $ Q$) from $ C$ to the circumcircle of $ AA'EE'$.
[b]Note:[/b] In the original problem, $ AA'EE'$ is any cyclic quadrilateral, and the problem asks to prove $ X$ and the two points of tangency are colliner. The proof is basically the same.
[b]Solution:[/b] Assume $ AC$ and $ PQ$ intersect at $ F'$, then it's well known that $ CEF'A$ is a harmonic range and by the problem above, we conclude that $ F \equal{} F'$. Since $ PQ//AA'//FX$, $ PQ$ must be the gable.
[b]Comment 1:[/b] The well know fact can be proven by
\[ \frac {AF'}{F'E} \equal{} \frac {PA^2}{PE^2} \equal{} \frac {AC}{EC}
\]
[b]Comment 2:[/b] However, we can avoid (or prove by proving the problem first) this "well known" fact and solve this problem using the following fact: if $ ABCDEF$ are six points on a circle then $ AD, BE, CF$ concur iff $ AB\cdot CD\cdot EF \equal{} BC\cdot DE\cdot FA.$ This result come from a slightly stronger version of the 1988 IMO Longlist Problem 88.

Now let's see how all these relates symmedian. On the 2007 TST problem 5, I proved the following property of symmedian using Ptolemy's Theorem, but now we can see that it's a trivial consequence of the previous problem.
[b]Problem:[/b] In the same diagram as the previous problem, prove that $ \angle PAC \equal{} \angle XAQ$ where $ AC$ is the symmedian and $ AX$ is the median.
[b]Solution:[/b] This is equivalent to showing $ PE \equal{} QE'$ which is obviously true.

Now we have proved the most import property of symmedian, and this is my [b]solution to USAMO 2008 Problem 2:[/b]
Let $ AS$ be the symmedian, and $ G$ be a point on the symmedian such that $ OG$ is perpendicular to $ AS$, then $ \angle ABG \equal{} \angle BGS \minus{} \angle BAG \equal{} \angle BAC \minus{} \angle BAG \equal{} \angle GAC \equal{} \angle BAM$, therefore, $ G$ lies on $ BF$ and similarly $ G$ lies on $ CF$, so $ G \equal{} F$, and we are done.

It's amazing to see how one art technique can relate to many Olympiad geometry problems, and that's the beauty of the math.

[quote=asdf334]don't know if this has been posted yet, but see [url=https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2018/05/two_special_points__1_.pdf]here[/url][/quote]
@asdf thank you.
Indeed I know the theorems concerning Dumpty points, but I can't see how Ru83n05 jumped to such a conclusion without stating any theorem.

it seems there's no projective solution in this thread yet, so here goes:

We claim that $AFPN$ is harmonic.
Let $F'$ be the harmonic conjugate of $A$ wrt $PN$, $D' = BF' \cap AN$ and $G = BF' \cap (APF'N)$.
Note that $(A,F';P,N) = (A,B;P, GN \cap AB) = -1$ projecting through $G$ onto  $AB$, so $GN \mid \mid AB$.
Now, $\angle ABD' = \angle F'GN = \angle F'AN = \angle BAD' \Rightarrow D'A=D'B \Rightarrow D = D'$.
This proves that $BD$ meets $(APN)$ on the harmonic conjugate of $A$ wrt $PN$. Doing this for $CE$ is completely analogous.
So $AFPN$ is cyclic, as we wanted to prove.

Let's use the power of barycentric coordinates.
Use $\bigtriangleup ABC$ as the reference triangle, with A = (1:0:0), B=(0:1:0) and C = (0:0:1). Calculate the midpoints of BC, AC and AB and we get (0:1/2:1/2), (1/2:0:1/2) and (1/2:1/2:0) the points are M, N and P. So let's calculate the perpendicular bissector of AB and AC. By the Evan's Forgotten Trick (Perpendicularity Criterion), two vectors $\vec{AB}$ with coordinates $(x_1,y_1,z_1)$ and $\vec{MN}$  with coordinates$(x_2, y_2, z_2)$ are perpendicular if and only if :

$$a^2(y_1 z_2 + y_2 z_1) + b^2(x_1 z_2 + x_2 z_1) + c^2(x_1 y_2 + x_2 y_1) = 0$$.

Plugging the values we get the perpendicular bissector of AB as :

$$z(b^2-a^2) + c^2(y-x) = 0$$
and AC :
$$y(c^2-a^2)+ b^2(z-x) = 0 $$

Using the fact that the points on line $AM$ are in the form (t :1/2 : 1/2), plugging these we can calculate t and ,furthermore, calculate the coordinates o D and E :

$$ D = (\frac{b^2 + c^2-a^2}{2c^2} ; 1/2 ; 1/2) $$
$$ E = (\frac{b^2 + c^2-a^2}{2b^2} ; 1/2 ; 1/2) $$
Now, let's calculate the intersection of lines BD and CE :

First, let's use the fact that points on BD are in the form $(\frac{b^2+c^2-a^2}{2c^2} : t : 1/2)$ and points on CE are in the form : $(\frac{b^2 + c^2 -a^2}{2b^2} : 1/2 : t)$. Now, we're gonna multiply the coordinate in line BD by $2c^2$ and CE by $2b^2$, then, F gonna be (b^2+c^2-a^2 : b^2 : c^2).

Now, let's use the circle of $(ANP)$ equation that is :
$$-a^2yz -b^2xz -c^2xy + (ux + vy + wz)(x+y+z) = 0$$ and find the constants u,v,w.
A is immediate, $u=0$.
Use this on N we get that $v = \frac{c^2}{2}$ and $w = \frac{b^2}{2}$.

And this equation gonna be equal to zero if and only if F lies on this $(APN)$.

$$-a^2yz - b^2xz - c^2xy + (ux + vy + wz)(x+y+z)$$

$$\iff$$

$$-a^2(b^2)(c^2) - b^2(b^2 + c^2 -a^2)(c^2) - c^2(b^2 + c^2 -a^2)(b^2) + (b^2c^2)(2b^2 + 2c^2 - a^2)$$
$$\iff$$
$$-a^2b^2c^2 - b^4c^2 -b^2c^4 + a^2b^2c^2 -b^4c^2 - b^2c^4 + a^2b^2c^2 + 2b^4c^2 + 2b^2c^4 -a^2b^2c^2$$
which is clearly equal to zero, so, ANFP is cyclic.

[i]Solution:[/i] Let $O$ be the circumcenter of $\triangle ABC$ and let $Q := AF \cap BC$. We wish to prove that $F$ lies on    $A-$symmedian.

[asy]
    import olympiad;
    size(10cm);
    pair A = dir(120);
    pair B = dir(210);
    pair C = dir(330);
    pair M = (B+C)/2;
    pair N = (A+C)/2;
    pair P = (A+B)/2;
    pair O = circumcenter(A,B,C);
    pair D = extension(A,M,P,O);
    pair E = extension(N,O,A,M);
    pair F = extension(B,D,E,C);
    pair Q = extension(A,F,B,C);

    draw(A--B--C--A, darkmagenta);
    draw(P--N, orange);
    draw(N--M, orange);
    draw(P--M, orange);
    draw(A--M, orange);
    draw(P--O, orange);
    draw(E--N, orange);
    draw(B--D, orange);
    draw(C--F, orange);
    draw(O--F, orange);
    draw(A--Q, fuchsia);
    draw(circumcircle(A,P,N), purple + dashed);

    markscalefactor = 0.02;
    draw(anglemark(D,B,A), deepgreen);
    draw(anglemark(B,A,D), deepgreen);
    draw(anglemark(A,C,E), cyan);
    draw(anglemark(E,A,C), cyan);
    markscalefactor = 0.006;
    draw(rightanglemark(O,F,A), blue);
    markscalefactor = 0.008;
    draw(anglemark(B,F,Q));
    draw(anglemark(Q,F,C));
    markscalefactor = 0.01;
    draw(anglemark(B,F,Q));
    draw(anglemark(Q,F,C));

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$M$", M, dir(M));
    dot("$N$", N, dir(N));
    dot("$P$", P, dir(P));
    dot("$O$", O, dir(O));
    dot("$D$", D, NE);
    dot("$E$", E, dir(E));
    dot("$F$", F, NW);
    dot("$Q$", Q, dir(Q));
[/asy]

[b][color=red]Claim:[/color][/b] $FQ$ is the angle bisector of $\angle BFC$. Moreover, $\angle BFQ = \angle QFC = \angle BAC$.

[i]Proof:[/i] Start by applying Law of Sines in $\triangle FAB$ and $\triangle FAC$. This would yield
        \[\frac{\overline{FB}}{\sin(\angle BAF)} = \frac{\overline{AF}}{\sin(\angle ABF)} \qquad \text{and} \qquad \frac{\overline{AF}}{\sin(\angle ACF)} = \frac{\overline{CF}}{\sin(\angle FAC)}.\]
        Dividing both equations, eliminating $\overline{AF}$ and putting $\angle ACF = \angle MAC$ and $\angle ABF = \angle MAB$, we get
        \[\frac{\sin(\angle ACF)}{\sin(\angle ABF)} = \frac{\overline{BF}}{\overline{FC}} \cdot \frac{\sin(\angle FAC)}{\sin(\angle FAB)} = \frac{\sin(\angle MAC)}{\sin(\angle MAB)}.\]
        Applying ``Ratio Lemma'' on $\triangle ABC$ with cevian $AM$ and $AQ$ would yield us
        \[\frac{\overline{BF}}{\overline{FC}} \cdot \frac{\overline{QC}\cdot \overline{AB}}{\overline{BQ}\cdot \overline{AC}} = \frac{\overline{AB}}{\overline{AC}} \iff \frac{\overline{FB}}{\overline{FC}} = \frac{\overline{QB}}{\overline{QC}}\]
        By the converse of ``Angle Bisector Theorem'', we get that $FQ$ is the angle bisector of $\angle BFC$. For the claimed angle equality, observe that
        \begin{align*}
            \angle BFC & = \angle CFQ + \angle BFQ   \\
                       & = 2\angle BAQ + 2\angle CAQ \\
                       & = 2\angle BAC
        \end{align*}
and the claim is proven. $\square$
-----------
Clearly $O$ is the $F-$excenter of $\triangle FED$ since $EN$ and $DP$ are exterior angle bisectors by symmetry. Due to the above claim, $FD$ is exterior angle bisector of $\angle EFD$. With this, we conclude that $OF \perp AF$. Finally since
    \[\angle OFN = \angle ONA = \angle OPA = 90^\circ\]
we get that $F \in \odot(APON)$ with diameter $\overline{AO}$ as desired. $\blacksquare$

The key claim is that $\overline{AF}$ is a symmedian. (Actually, it turns out $F$ is the A-dumpty point, but we will not need this.)

Notice that $\angle BFC = 2\angle A$, so $B, F, O, C$ are concyclic, where $O$ is the circumcenter. Now, let $F'$ be the point on $(ANM)$ such that $\overline{AF'}$ is a symmedian. Then $\overline{AF'}$ passes through the midpoint of major arc $\widehat{BC}$ in $(BOC)$, and as $\angle OF'A = 90^\circ$, $F'$ lies on $(BOC)$. Finally, an angle chase reveals $\angle FBO = \angle FGO = \angle B - \angle BAD$, which is enough to characterize $F = F'$.

Barycentric coordinates finishes off in style

I claim that $F$ is the $A$-dumpty point. It is clear that the $A$-dumpty point lies on $\odot(ANP)$. Also, let $O$ denote the center of $\odot(ABC)$.

Firstly,
\begin{align*}
    \measuredangle BFC&=\measuredangle FBC+\measuredangle BCF\\
    &=(\measuredangle ABC-\measuredangle ABF)+(\measuredangle BCA-\measuredangle FCA)\\
    &=(\measuredangle ABC+\measuredangle BCA)+\measuredangle FBA+\measuredangle ACF\\
    &=\measuredangle BAC+\measuredangle BAF+\measuredangle FCA\\
    &=2\measuredangle A
.\end{align*}

So $F$ lies on $\odot(BOC)$. Now it is enough to prove that $AF$ is the $A$-symmedian.

Firstly, redefine $F$ as the $A$-dumpty point. Now note that we have $\measuredangle FCA=\measuredangle FAB=\measuredangle CAM=\measuredangle CAE=\measuredangle ECA$. This gives us that $\overline{C-E-F}$ are collinear. Similarly we also get that $\overline{B-D-F}$ are collinear and we are done.

After drawing several diagrams, we conjecture that $AF$ is the A-symmedian. We would hope that then $F$ is the midpoint of $AT$, where $T$ is the intersection of the A-symmedian with the circumcircle. Note that $T=(-a^2:2b^2:2c^2)$, as the second two coordinates are in a $b^2$ to $c^2$ ratio and it satisfies the circumcircle equation. Thus, letting $A=(-a^2+2b^2+2c^2:0:0)$ (we make them the same sum so we can just average them), we have that the midpoint of $AT$ is $$(-a^2+b^2+c^2:b^2:c^2).$$ Denote this point by $F'$, and we will try to show that $F'=F.$\\

Suppose that $BF'$ intersects $AM$ at $D'$. Then, $B=(0,1,0)$ and $F'=(-a^2+b^2+c^2:b^2:c^2)$, so the equation of line $BF'$ is $$x(c^2)=z(-a^2+b^2+c^2).$$ The equation of $AM$ is just $y=z$, so we have $$D'=(-a^2+b^2+c^2:c^2:c^2).$$

Now, we check that $D'P\perp AB.$ Note that $$P=(1:1:0)=(-a^2+b^2+3c^2:-a^2+b^2+3c^2:0)$$ and $$D'=(-2a^2+2b^2+2c^2:2c^2:2c^2).$$ This is done so that $P$ and $D'$ have the same sum now. We have $$\overrightarrow{AB}=(-1,1,0)$$ and $$\overrightarrow{D'P}=(a^2-b^2+c^2:-a^2+b^2+c^2:-2c^2)$$ (strictly speaking this is not exactly $\overrightarrow{D'P}$, but it doesn't matter as it is a real multiple that does not affect perpendicularity). This, it suffices to check that $$a^2(-2c^2)+b^2(2c^2)+c^2(2a^2-2b^2)=0,$$ which is clearly true. Hence, by EFFT, $D'P\perp AB,$ so $D'=D$. Similarly, $E'=E$. Since $F'$ lies on $BD'$ by definition (and thus $BD$), and similarly $F'$ lies on $CE$, we have $F'=F$, as desired.

As $\angle ABF = \angle BAM$ and $\angle ACF = \angle CAM$, $F$ is the $A-$ Dumpty point, which lies on $(ANP)$ since it is the midpoint of the $A-$ symmedian chord.

I have seem this point many times in the past few days  :maybe:

Apply $\sqrt{bc}$ inversion, then a homothety of scale factor $\tfrac{1}{2}$ centered at $A$. The problem becomes:
[quote=Restated problem]
In $\triangle ABC$, let $\ell$ be the $A$-symmedian and $B',C'$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. Let $X$ and $Y$ be the feet from $B$ and $C$ to $\ell$ respectively. Prove that $(AB'X)$ and $(AC'Y)$ meet on $\overline{BC}$.
[/quote]

I claim that this intersection point is in fact the midpoint $M$ of $\overline{BC}$. Consider $(AB'M)$ and suppose it intersects $\ell$ again at $X' \neq A$. Then $\measuredangle B'MX'=\measuredangle B'AX'=\measuredangle MAC=\measuredangle B'MA$, hence $B'$ is the midpoint of arc $AX'$, so $B'A=B'X'=B'B$. This implies that $\angle AX'B$ is right (since $B'$ is now its circumcenter), so $X'=X$; done. $\blacksquare$

[b]Remark:[/b] Kind of got spoiled as to the identity of the point in another thread but I think this solution is more or less motivated independently of this fact. In some sense the diagram feels kind of random (at least to me) and a $\sqrt{bc}$ inversion cleans it up significantly. The other advantage of this $\sqrt{bc}$ inversion is that it makes it much easier to guess the mystery point since midpoints are easy (not that I had to do any guessing...)

[quote=IAmTheHazard]I have seem this point many times in the past few days[/quote]

2020 iran tst G3 or something i forget

A-HM point of I_ABC :skull:

The angle conditions on $F$ uniquely determine it to be the $A$-Dumpty point, which is the midpoint of the $A$-symmedian chord, so a homothety of scale factor $\tfrac 12$ finishes. $\blacksquare$

We'll prove that $F$ is $A$-Dumpty point. Let $F'$ be $A$-Dumpty point and let $X = AF' \cap (ABC)$. Then it's well-known that $F'$ is the midpoint of $AX$. Thus $\angle ECA = \angle EAC = \angle MAC = \angle BAX = \angle BCX = \angle ACF'$, so $C, E, F'$ is collinear. Similarly $B, D, F'$ is collinear, hence $F' = F$. Then $F$ is the midpoint of $AX$, which means $F$ lies on $(APN)$, as needed. $\blacksquare$

We show that $F$ is the $A-$Dumpty Point.
Define $F_B=BF\cap AC$ and $F_C=CF\cap AB$. Then, from Menelaus' Theorem, we have
\[\frac{AD}{DM}\cdot\frac{MB}{BC}\cdot\frac{CF_B}{F_BA}=-1\]
\[\frac{AE}{EM}\cdot\frac{MC}{CB}\cdot\frac{BF_C}{F_CA}=-1\]
\[\Longrightarrow\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=\frac{DM}{AD}\cdot\frac{AE}{EM}\]
\[\frac{BF_A}{F_AC}\cdot\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=1\]
\[\Longrightarrow\frac{BF_A}{F_AC}=\frac{AD}{DM}\cdot\frac{EM}{AE}=\frac{BD}{DM}\cdot\frac{EM}{CE}=\frac{\sin(\angle FCM)}{\sin(
\angle MBF)}=\frac{BF}{FC}\]
Hence, $\angle BFF_A=\angle F_AFC$.

Note that \[\angle BFC=\angle BAC+\angle FBA+\angle ACF=\angle BAF+\angle FAC+\angle FBA+\angle ACF=2\angle BAC.\] Hence, $B,F,O,C$ are concyclic. Moreover, \[\angle BAF+\angle FBA=\angle FAC+\angle ACF=\angle BAC=\angle BAF+\angle FAC\Longrightarrow \angle FBA=\angle FAC,\angle BAF=\angle ACF\]
This means $F$ is the $A-$Dumpty by definition. So, $F\in (ANPO)$, as required. $\blacksquare$


inequalities
geometry
3D geometry
tetrahedron

1. If a2 + b2 = 1 and c2 + d2 = 1, prove that ac + bd :le: 1.

2. Prove that for all real numbers x, y, and z, x2 + y2 + z2 :ge: xy + yz + zx.

[hide]Cauchy "straightforwardly" proves both[/hide]

Why not actually provide a proof probability1.01?

I use the trivial inequality [tex]a^2 \geq 0[/tex]

1. [hide] First, [tex]a^2 + b^2 \geq 2ab[/tex] This is true since [tex]a^2 -2ab + b^2 = (a-b)^2 \geq 0[/tex]. So we have [tex]ab \leq (a^2+b^2)/2 = 1/2, cd \leq (c^2 + d^2)/2 = 1/2.[/tex] Then indeed [tex] ab + cd \leq 1 .[/tex][/hide]

2. [hide] We show [tex]2x^2 + 2y^2 + 2z^2 \geq 2xy + 2xz + 2yz[/tex]. This is the same as [tex] (x-y)^2 + (z-x)^2 + (y-z)^2 \geq 0[/tex][/hide]

JBL wrote:tetrahedron -- you didn't quite prove what was asked for.

his/her proof seems to answer the right question to me...

Edit: oh yeah #1 is not quite, but using the same principles, one can find that [hide]a:^2:+b:^2:+c:^2:+d:^2:=2 and 2ac :le: a:^2:+c:^2: and 2bd :le: b:^2:+d:^2:[/hide]

Sorry, didn't specify.  Look at number 1, what you proved versus what the question asked you to prove.


calculus
integration
real analysis
calculus computations

Prove that if f is integrable on R, then $f^{2}$ is also integrable on R.

This only makes sense as a Riemann integral question, and then $R$ must refer to some bounded set over which we are calculating integrals of some bounded function.

If you admit either Lebesgue or improper integrals, then we have such examples as $\frac{1}{\sqrt{x}}$ is integrable over $(0,1]$ but its square is not integrable. (Or multi-dimensional versions of essentially the same example.)


abstract algebra
category theory
calculus
derivative
topology
algebra
function

Hi,

I am very new in module theory; :oops:  if my question is too advanced please move it to the Advanced section

1)We are studying left derivation of functors.
Those functors are from the category of R modules to the category Ab (abelian groups) and they are additive.

It requires theory of complexes over a module M (if you do not know what that is , please ask), and especially of projective resolutions

When one applies an additive covariant functor F on a pojective resolution C over M, one gets a complex over FM, and that new complex is denoted by FC

now a n-th left derivative of a functor F maps each object (thus each R module) onto the n-th homology module of
the complex FM

My question : how does this work?  I mean, I used a projective resolution over my module M in order to construct my functor.  What is my category?  My professor said that'maybe' it's not really the category of R modules but the category of all R modules, together with a projective resolution for each module?

2)And then a second question : my syllabus also says :
if you have two projective resolutions C and C' over a module M, there is a unique isomorphism between the n th homology module in FC, and the nth homology module in FC'.
Now uniqueness, I'am almost certain that is not correct.  That means my modules on both end have trivial automorphism group.  My guess: it must either be omitted , or the uniqueness is correct when one gives more restrictions

Truly, ALL ideas or references are welcome :oops:

Hi,

1) The whole theory can be developped in arbitrary abelian categories (with enough injective/projective objects). The n-th left derivative of a right-exact functor F : C -> D is just also a functor C -> D.

2) I guess you're right: The uniqueness doesn't hold in general. Usually, the proof of existence is done by homotopy equivalence, which induces an isomorphism in homology. But the uniqueness of such an isomorphism is not required. It does not even have to come from homotopy.

[quote="-oo-"]Hi,

1) The whole theory can be developped in arbitrary abelian categories (with enough injective/projective objects). The n-th left derivative of a right-exact functor F : C -> D is just also a functor C -> D.
[/quote]

Thanks, but can you elaborate a bit?  So in may case just a functor from R mod to Ab?

How to define the functor in a non proj resolution dependent way??

Yes, in your case it's a functor R-mod -> Ab.

The functor is well-defined because all projective resolutions are homotopy equivalent and therfore yield the same homology, up to isomorphism. So ... strictly speaking, it's a functor R-Mod -> Ab / ~, where ~ is the equivalence relation "isomorphic to". But in applications, you just say R-Mod -> Ab.

I was fearing the actual category was gonna be Ab modulo some equivalence relation...but a functor also maps morphisms onto morphism

Take a morphism $A\buildrel{f}\over{\rightarrow}B$

what will $F(f)$ be now?
its domain and codomain will be isomorphic regardless of what you choose as resolutions..but it's not because a and b are map that are up to isomorphism the same, that their composition will also be like that

Which functor do you denote now by $F$? :-)

Of course, the derivative of a functor also operates only "up to isomorphism" on the morphisms.

hmm yeah that was confusing I actually meant the derived 'functor'

suppose I give you a morphism ' up to isomorphism', and then another 'up to isomorphism'

I ask for the composition, you will probably say, can't do that.
I'll say: but I only care for it up to isomorphism.

You still won't be able to do that in my opinion : consider in $\mathbb{R}^2$ the morphisms of projection onto X and Y axis, respectively a and b
a and b are the same up to isomorphism.

But a after a is not the same, up to isomorphism, with a after b

You understand my problem with the category you proposed?

Hi,

a friend told me the solution of this problem ;-)

Let $F : C \to D$ a right-exact functor between abelian categories (in your case, $C = R-Mod, D = Ab$). For each object $M$ of $C$, you have to [i]choose[/i] a [i]fixed[/i] projective resolution $P(M)$ of $M$ (which is possible by the axiom of choice). Then, you define the n.th derivative of $F$ by the n.th homology of the chain complex $F(P(M))$:
\[ (L_n F)(M) = H_n(F(P(M)))  \]
If $f : M \to N$ is a morphism in $C$, the fundamental theorem of homological algebra states the existence of a chain morphism $P(M) \to P(N)$ over $f$, and it's uniqueness up to homotopy. So, it induces a unique morphism in homology, i.e. a morphism $(L_n F)(f) : (L_n F)(M) \to (L_n F)(N)$. That makes $L_n F$ a functor $C \to D$. You don't have to worry about the arbitrariness of the choice of $P$, because other choices yield isomorphic $L_n F$. The fixed choice just makes it possible to (well)define our functor. But in applications this is rather unimportant, because ony the isomorphism-type of $(L_n F)(M)$ is of interest.

Hmm, thanks, that inspired me to propose this unifying statement  :ninja:  :

when $F$ is a functor (additive and covariant) from $R-mod$ to $S-mod$

consider now the category $D=R-modwpr$ , the category of R modules M along with a projective resolution

the morphisms in this latter category are chain morphisms along with a morphism from the first $M_1$ to the second $M_2$

it is worth noting that in a morphism between two objects in $D$, the chain morphism is determined up to homotopy by the morphism between $M_1$ and $M_2$

now our left derived functor $L_n F$ is a functor from $D$ to $R-mod$

What do you think up until now?

Well, the statement in my syllabus, stating that if you define $L_n F M= H_n (FC)$ , not only it is not dependent up to isomorphism of the chosen complex , but that there is a unique isomorphism between $H_n(FC)$ and $H_n(F C')$ simply cannot be correct as I said earlier (very few modules have only one automorphism)

I am sure there is a very precise way to correct this though, in light of what I tried earlier here.

What do you think?

I think that both formalisations work ...


calculus
integration
trigonometry
analytic geometry
real analysis
real analysis unsolved

Reverse the integration order:

$ \int_0^{\pi/2}\!\!\int_0^{\cos\theta}\cos\theta\,dr\,d\theta.$

Thanks.

Does it matter? This is a meaningful integral whether we interpret $ r$ and $ \theta$ as polar coordinates for the plane or as Cartesian coordinates for another plane.
In particular, simply interchanging the order of $ r$ and $ \theta$ doesn't depend on the interpretation.

Ahhh, then how should it be $ d\theta\,dr$?

The interesting part is transforming the limits. Draw the region we're integrating over in the plane, and then solve for $ \theta$ in those curves to find the new inner limits.

The boundaries of the region are the curve $ r\equal{}\cos\theta$ and a pair of straight lines; one of those lines is $ r\equal{}0$.


Ring Theory
superior algebra
superior algebra unsolved

Let $ k$ be an algebraically closed field, let $ R$ and $ S$ be finitely generated $ k$-algebras and let $ \varphi: R \to S$ be a $ k$-algebra homomorphism. Show that the inverse image under $ \varphi$ of any maximal ideal in $ S$ is a maximal ideal in $ R$.

Or from the geometrical point of view: closed points map to closed points. And we don't need that $ k$ is algebraically closed.
Let $ \mathfrak n$ be maximal in $ S$ and let $ \mathfrak m \equal{} \varphi^{\minus{}1}(\mathfrak n)$. Now $ R / \mathfrak m$ injects to $ S/\mathfrak n$, and the latter is a a finite algebraic field extension of $ k$ (by a version of the Nullstellensatz). But now every element of $ R / \mathfrak m$ is algebraic over $ k$, thus it itself is a field, so $ \mathfrak m$ is maximal.

Don't you need the fact that $ k$ is algebraically closed to use the Nullstellensatz?

No, that's a weaker version that applies to all fields. A proof is e.g. in Liu's book (about algebraic geometry).
The more often stated corollary is that $ S / \mathfrak n$ is already $ k$ if $ k$ is alg. closed; this obviously follows from the version above.


What's the contents(or knowledge) needed in PhOs of your nations?

Mechanics and E&M, basically... although i saw some advanced stuff there as well... the past USAPhO test i mean... What about in your country, Alexander Lenin?

[quote="shobber"]that's true, i think most of the countries have the similar knowledge, just like China.[/quote]

But as I know, many nations have more contents involved in PhOs. I can't ensure whether that's true.

[quote="Alexander Lenin"]Here is the contents of CPhO: http://www.shenzhong.net/ljwuli/Article_Show.asp?ArticleID=263

I'm sorry I don't have time to translate them into English right now, and I don't have a dictionary at hand now, maybe I will try to translate them a few days later.[/quote]

you dont have to translate them. :lol:  but it would be nice to have some sort of guideline/criteria of the test... basic things needed to know to take the test.. etc. basically, your original question.

it is said that the most difficult things in PhOs in China are the machenics problems. Is that true?


modular arithmetic
greatest common divisor
number theory
relatively prime
number theory proposed

Given an integer $ n>6$. Consider those integers $ k\in (n(n\minus{}1),n^{2})$ which are coprime with $ n$. Prove that the greatest common divisor of the considered numbers is $ 1$.

I have solve problem and I find a result:
There exist (a,b,c) such that they pair with relative prime.
A solution quite long .I will post when have a person post solution for the first.

Am I misinterpreting the problem? It seems like only $ n\geq 3$ is necessary.
[hide="if not"]
The integers $ n^{2}\minus{}n\plus{}1$ and $ n^{2}\minus{}1$ lie in $ (n(n\minus{}1),n^{2})$ and are relatively prime to $ n$. Any common divisor between these two divides their difference, $ n\minus{}2$, and then must also divide $ n^{2}\minus{}1\minus{}(n\plus{}2)(n\minus{}2)\equal{}3$. So if $ n\not\equiv 2\pmod{3}$, $ n^{2}\minus{}n\plus{}1$ and $ n^{2}\minus{}1$ are relatively prime.

In the case that $ n\equiv 2\pmod{3}$, $ n(n\minus{}1)$ is not divisible by 3, so $ n^{2}\minus{}n\plus{}3$ is relatively prime to $ n$, and it too lies in the interval $ (n(n\minus{}1),n^{2})$. But then $ n^{2}\minus{}n\plus{}1$ and $ n^{2}\minus{}n\plus{}3$ are relatively prime, since any common divisor between them must divide their difference, 2, but $ n^{2}\minus{}n\plus{}1$ is always odd.
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suppose that prime p divides all of these numbers.
let q be the smallest prime such that q doesnt divide n.
if n>2 ,q<n because any prime divisor of n-1 doesnt divide n.
so p divides n^2-1 and n^2-q because they are coprime with n and they are belonging to the interval (n^2,n^2-n).
so p divides q-1 hence p is smaller then q and doesnt divide n.
Contradiction.


geometry
combinatorics unsolved
combinatorics

Is there an infinite subset S of Euclidean 3-space such that any plane P intersects a finite subset of S?

Thanks ZeTaX. Sorry I don't follow Geometry and Combinatorics sections much.