analytic geometry
combinatorics unsolved
combinatorics

A merida path of order $2n$ is a lattice path in the first quadrant of $xy$- plane joining $(0,0)$ to $(2n,0)$ using three kinds of steps $U=(1,1)$, $D= (1,-1)$ and $L= (2,0)$, i.e. $U$ joins $x,y)$ to $(x+1,y+1)$ etc... An ascent in a merida path is a maximal string of consecutive steps of the form $U$. If $S(n,k)$ denotes the number of merdia paths of order $2n$ with exactly $k$ ascents, compute $S(n,1)$ and $S(n,n-1)$.

If you take $2n$ steps, then you must end up with an $x$ coordinate of $2l+d+u$, where $l,d$ and $u$ are the number of $L,D$ and $U$ moves respectively.  But $l+d+u=2n$, so $2l+d+u\ge2n$, with equality iff $l=0$, i.e. no $L$ moves.  So now we have only $U$ and $D$ moves.  Each of these takes you up one, and either to the right or to the left by one, so now we have only a two dimensional problem. It is easy to see that the $y$ coordinate after all the moves is $u-d=0$, so $u=d=n$.

Now if there are $n$ $U's$ all in one block, then there are $n+1$ possible paths (for each of the $n+1$ possible numbers of $D's$ to have before the first $U$, with all the others coming after), so $S(n,1)=n+1$.

For the $n-1$ ascent case, we number the ascents in order, $1$ to $n-1$.  Exactly one of these will be of length $2$, with the rest of length $1$, and there are $n-1$ choices for this one.  There must be at least one $D$ between each consecutive pair of $U's$, with $2$ free $D's$ to distribute.  There are $n+1\choose2$ ways to place these two (this is a classic 'multichoose' problem), so we get $S(n,n-1)=(n-1){n+1\choose2}$.

Last minute change: OK, I just saw the requirement that the path be in the first quadrant, so this means that the first step must be $U$, so the answer to the first part becomes simply $S(n,1)=1$, since the only path is $n$ $U's$ followed by $n$ $D's$.

For the second part, this means there has to be a $D$ after each of the $U's$, and the extra $D$ has to be placed after the extra $U$. There are $n\choose2$ ways to place the pair of extras with the $D$ after the $U$, so $S(n,n-1)={n\choose2}$.

Rushil, are you sure this is the correct statement of the problem?  The $L$ moves don't seem to have any role in the solution except as a red herring.

[color=#008000]2011 moderator addendum: the proof here is incorrect.  For a correct proof, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=414569[/color]


arkadaşlar aranızda bu yarışmaya girecek olanlar varsa görüşlerini bekliyorum
ayrıca çözümü anlaşılmayan geçmiş soruları da paylaşın ki ilgi biraz artsın

ewt biz Sehzade Mehmet Lisesi olarak katılacaz.bu sene sınav 25 mart ta.sınavın belli bi konusu,veya kuralı yok falan yok.ilk üç'e para odulü veriyolar.


Problem1 (by BDT_NDT): let a,b,c>0. Prove that:
$ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \plus{} \frac {1}{(c \plus{} 2a)^2}\geq\frac {1}{ab \plus{} bc \plus{} ca}$
Problem 2 (by can_hang2007): Let $ a,b,c\geq 0$ be nonnegative real numbers. Prove that
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\cdot \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a).$
When does equality occur?

[quote]Problem1 (by BDT_NDT): let a,b,c>0. Prove that:
$ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \plus{} \frac {1}{(c \plus{} 2a)^2}\geq\frac {1}{ab \plus{} bc \plus{} ca}$[/quote]
sorry, i don't think problem 2
proof of mine for Problem1
let $ b \plus{} 2c \equal{} x$,$ c \plus{} 2a \equal{} y$,$ a \plus{} 2b \equal{} z$
we have
$ ab \plus{} bc \plus{} ca \equal{} 5\sum{xy} \minus{} 2\sum{x^2}$
prove:$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^2} \ge \frac {27}{5\sum{xy} \minus{} 2\sum{x^2}}$(1)
let  :$ \sum{a} \equal{} p \equal{} 1$,$ \sum{ab} \equal{} q$,$ abc \equal{} r$

(1)<=> $ (q^2 \minus{} 2pr)(9q \minus{} 2p^2) \ge 27r^2$
<=>$ f(r) \equal{} \minus{} 27r^2 \plus{} (4 \minus{} 18q)r \plus{} 9q^3 \minus{} 2q^2 \ge 0$
$ f''(r) \le 0$=>f(r) is convex                                                                                therefore, minimal value when two of a,b,c equality or$ r \equal{} 0$,$ 4q \ge p^2$
-case $ r \equal{} 0$,$ q \ge \frac {1}{4}$
we have $ f(r) \equal{} 9q^3 \minus{} 2q^2 \ge 0$
-case $ x \equal{} y \equal{} t$,assumes $ t \plus{} c \equal{} 1$
we have$ a \equal{} \frac {4x \plus{} y \minus{} 2z}{9} \ge 0$
=> $ t \ge \frac {2}{5} z$=>$ t \ge \frac {2}{7}$
inequality become
$ \frac {2}{t^2} \plus{} \frac {1}{z^2} \ge \frac {27}{5(2tz \plus{} t^2) \minus{} 4t^2 \minus{} 2z^2} \equal{} \frac {27}{t^2 \plus{} 10zt \minus{} 2z^2}$(2)
replace $ z \equal{} 1 \minus{} t$
we have
(2)<=>$ (2t \minus{} 1)^2( \minus{} 15t^2 \plus{} 20t \minus{} 4) \ge 0$($ \frac {2}{7} \le t \le 1$)
equality when $ a\equal{}b\equal{}c$

[quote="tuantam1lan"][quote]Problem1 (by BDT_NDT): let a,b,c>0. Prove that:
$ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \plus{} \frac {1}{(c \plus{} 2a)^2}\geq\frac {1}{ab \plus{} bc \plus{} ca}$[/quote]
sorry, i don't think problem 2
proof of mine for Problem1
let $ b \plus{} 2c \equal{} x$,$ c \plus{} 2a \equal{} y$,$ a \plus{} 2b \equal{} z$
we have
$ ab \plus{} bc \plus{} ca \equal{} 5\sum{xy} \minus{} 2\sum{x^2}$
prove:$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^2} \ge \frac {27}{5\sum{xy} \minus{} 2\sum{x^2}}$(1)
let  :$ \sum{a} \equal{} p \equal{} 1$,$ \sum{ab} \equal{} q$,$ abc \equal{} r$

(1)<=> $ (q^2 \minus{} 2pr)(9q \minus{} 2p^2) \ge 27r^2$
<=>$ f(r) \equal{} \minus{} 27r^2 \plus{} (4 \minus{} 18q)r \plus{} 9q^3 \minus{} 2q^2 \ge 0$
$ f''(r) \le 0$=>f(r) is convex                                                                                therefore, minimal value when two of a,b,c equality or$ r \equal{} 0$,$ 4q \ge p^2$
-case $ r \equal{} 0$,$ q \ge \frac {1}{4}$
we have $ f(r) \equal{} 9q^3 \minus{} 2q^2 \ge 0$
-case $ x \equal{} y \equal{} t$,assumes $ t \plus{} c \equal{} 1$
we have$ a \equal{} \frac {4x \plus{} y \minus{} 2z}{9} \ge 0$
=> $ t \ge \frac {2}{5} z$=>$ t \ge \frac {2}{7}$
inequality become
$ \frac {2}{t^2} \plus{} \frac {1}{z^2} \ge \frac {27}{5(2tz \plus{} t^2) \minus{} 4t^2 \minus{} 2z^2} \equal{} \frac {27}{t^2 \plus{} 10zt \minus{} 2z^2}$(2)
replace $ z \equal{} 1 \minus{} t$
we have
(2)<=>$ (2t \minus{} 1)^2( \minus{} 15t^2 \plus{} 20t \minus{} 4) \ge 0$($ \frac {2}{7} \le t \le 1$)
equality when $ a \equal{} b \equal{} c$[/quote]
oh, it is not differen as you think
you can see solution at here [url]http://vnineqmath.hnsv.com/viewtopic.php?f=15&t=2[/url]


Can someone please explain this problem to me cause the one at Kalva is extremely poorly written and confusing:

1. Find the area of the large square divided by the area of the small square.

[img]http://www.kalva.demon.co.uk/aime/soln/diag016.jpg[/img]

[hide]Let the side of the large square be 2x, and the side of the small square be 2y. Then the radius is x :rt2:

Draw the radius from the center of the circle to one of the vertices of the small square that's on the circle.

Then draw the segment perpendicular to both squares form the center to the edge of the small square.

This forms a right triangle with hypothneuse of x :rt2: and legs of y and (x+2y)

Use Pythegorean theorem, expand out and factor to get (x+y)(x-5y)=0, so x/y=5 and x^2/y^2=ratio of areas=25[/hide]

Alright, thanks for explaining it MUCH better than Kalva! I see why the answer is 25 now.

i know that the mathematical solution is right, but couldn't you also just draw it and measure the sides or diagonals with a ruler? i guess that'd be the thing to do if you were running out of time.

and how do you know that it's exactly 5 ????????

Well, since it's an AIME question, the answer has to come out nicely. Obviously this wouldn't work outside of the AIME, but you might as well use all the advantages that you're allowed to have.

How do you know it's not [tex]\sqrt 24[/tex]

AIME answers are from 000-999 so you have a pretty good chance unless it asks for a+b+c where the answer is in the form of a+b*rt(c).  Still it s not a very good method.  If you can't find the answer and don't have much time you might as well try measuring. It wont hurt.

ratio between the diagonals is [tex]\sqrt{24}[/tex],

Also, how would you construct the picture?

[quote="JBL"]Beta has a good point in that it would be very difficult to construct the diagram in such a way that a direct measurement comparison would be possible -- I don't see how you could construct it to give the smaller square a side (or diagonal) of length 1 without knowing the answer in advance.  That makes direct measurement very difficult.[/quote]

Here's a pretty simple construction...

Yes, but how would you guarantee a nice edge length/diagonal for the small square like that?  That goes from the big square to the small square -- you need to go the other way to make it workable.

One more try. If AB is the length of the smaller square with a length of 1 inch, then CD is the length of the larger square, which can easily be measured to be 5 inches. Of course if you understand what's goin on well enough to come up with this construction, you will most likely be able to solve it without the measuring :)

We can actually get a nice, nonstandard solution from the construction in my last post.

After some (minimal) angle chasing, <FDB = <EBF, so the triangle FBD and EFB are similar, so BD/BF = FB/FE = 2, so BD = 2*sidelength. Likewise, AC = 2*sidelength, so CD is 5*sidelength and were done.

[quote="TripleM"]I presume you are allowed rulers - in which case, you should still be able to get a reasonable approximation no matter what length you use for the big square.. As long as you can do a little long division, it would probably get you the right answer.[/quote]

Rulers here do eigths of an inch -- that's a bit less than a third of a centimeter for those of you on the metric system.  Say, picking randomly, I draw the big square with a side of eight inches, just about what would fit on a piece of paper.  The side length of the small square will come down at 1 and between 1/2 and 5/8 inches, quite a bit closer to 5/8.  That could mean any number of things, unfortunately.  Even if your ruler went into 16ths, you still wouldn't be sure -- you might be able to make a guess based on "nice numbers," but it would be decidedly unconvincing (to me, at least).  The problem is that ruler gradations are very imprecise, even if they're accurate -- you lose a lot of information when you scale down, and the only way to preserve it would be to scale up, which is really hard in this problem (harder than the problem itself, even).

staple a few piece of paper together, make the length of square huge so we can make sure thr ratio is not [tex]\sqrt{24}[/tex] or [tex]\sqrt{26}[/tex]

also you need a huge compass


Let $f$ be an increasing function. Show that $\lim_{x\rightarrow 0^{+}}f(x)$ must exist.

[hide="Well..."] $f(x)$ is bounded and monotonic on the interval $[0, \epsilon]$. [/hide]

what if you have something to the effect of $f(x)=-\frac{1}{x}$ for $x> 0$, the limit does not exist

[hide="Er, excuse me..."]
$\ln x$ is increasing, yet $\lim_{x\to 0^{+}}\ln x=-\infty$.

Let's go shopping for some extra conditions.[/hide]

I think that "increasing function" meant "increasing function defined on $\mathbb{R}$."  :P

Edit:  Otherwise the problem is trivially untrue, as we could simply use functions that are not even defined around the neighborhood of $x = 0$!

OK. The function $f: R\rightarrow R$  :D

Do you think that [b]t0rajir0u[/b] didn't answered in this case?  :D


MATHCOUNTS
AMC
AMC 10
USA(J)MO
USAMO
email
inequalities

Yeah - the title says it all - post what you think the freshman MOP cutoff will be.

Much as I wish otherwise, I believe it will be 10, exactly one point above my score, given the way the score distribution is posted on the AMC website.

A lot of optimists so far, I see.  However, this reinforces my estimate, as last year's predictions were usually 7 points below the actual cutoff.

[quote="paladin8"][quote="sirorange"]I'm predicting a 9. Hopefully it's not a 15 or higher.[/quote]

15 or higher? Lol. There are barely 60 people TOTAL with that kind of score.[/quote]

Only about 80, to be exact.

for my sake, 7 would be the best.
but it seems that freshmen are doing heavenly good this year.

Yes, it does seem like many freshmen have good scores.  Let's just hope that all non-AOPS freshmen scored 0.  :lol:

lol
its very hard for ppl who do not go to AoPS to get a high mark in junior grades.
they might not even have heard of this competition in their schools, since usually school teachers do not care about these stuff much

As a comment on nirvanatear's point, I was on the CT state mathcounts last year and I know for a fact that 2 of the kids from the team had never even heard of AMC in 8th grade, so I doubt their high schools even offered the AMC10 or AMC12.  Had they taken it, however, they probably would do well considering me and the fourth team member are both waiting on MOSP invitations...

So keep in mind there is a large pool of smart people outside the USAMO and AOPS  ;) (and mathcounts too if I think about it)

I think it will be an 8 though I scored a 10.

I think it'll be a 7 but they'll have to use the index to cut some people.

I agree with you, joml88.  The cutoff probably will be a 7.  By the way, how many people are invited to Akamai - 24 or 30?  I did a little bit of statistics and ~29 freshmen scored 7 or above based on pure, simple statistics.  Due to the fact that freshmen generally do worse than say, juniors, however, probably about 24 will score 7 or above.

What do you mean by "They'll have to use the index to cut some people."???

My index was a 236.5 so that means if they do I probably won't get cut.

One of my friends theorized that they didn't call yesterday because there was some kind of tie for red MOP. If that's true, then the cutoff is probably like 5 or something.

HUH You adjusted my prediction.  I said the tie would be at a 3.  I wonder why.   :(  :(

I'm not going to divulge my name, but with a surname initial of V and a score of 7, I'll probably be one of the very last freshmen to get the call, IF I DO.  Please post what the floor is!!!!!!!!

The most annoying point is that now you start to go over your test and say: What could I have done to get an extra point?

Since sirorange says that they don't invite extra people if some can't go, 8 may be the cutoff.  I dunno, though, since all it would take is for one freshman more than expected to score below a 7.

NO CALL YET.

Wait - just saw this in preview - Boy Soprano II - 54 total or 54 freshmen?????

I'm wondering if all of the freshman got this e-mail at the same time.  If so, my chances do not look good...  :(

I think its likely that all of the emails were sent out at once to the freshmen at least by number. For example, i bet all emails to people with $>14$ were sent out and then 13, 12, 11, and down to the floor.

Or else, hotmail might just be slow. :(

Matt276eagles:

With a sample size of 404, equal distribution is almost certainly going to be satisfied, or better.  Basically, those numbers are pretty much correct or too high, (as in <26 freshmen between 8-22), so hope for the best.

Wait - I forgot to take out 17 people from Canada - that will mean that there's even less freshmen expected to get 7-22, so my chances are better.

All the freshmen did not get the email at the same time.  Look at the fact that pieterminate posted at 3:10, and again at 3:11 saying he got it.  He must have gotten the email fairly recently, hopefully.  any info on when it was sent, pieterminate?  I'm assuming you got yours fairly recently since jrshoch got his email at least 45 minutes ago and he thinks they're doing blind copies.

The email says 54 total. And score seems like a rather arbitrary way to sort things.

hey sirorange, mind posting a copy of the email on here?

[quote="jrshoch"]sirorange I think they might make the calls by score... my dad realized yesterday that they probably go down in order of score in case some kids don't go, that way they can adjust the list accordingly.[/quote]

My son, with a score of 22, has not received a call yet, just the email. Considering 23 is an HM, if they were going by score, he would have been one of the first to be called, after the HMs.

Yes, please post the email here.

Also, what do you mean by "Score seems a rather arbitrary way of sorting things."  I can't tell what grade corresponds to each score.

Yeah, too bad for all of us freshmen who got 8's. Oh well, matt.  :(  Oh well. Next year, blue group.

I got my email at 3:14 PM, eastern time.  It says 54 people were invited, but doesn't say if they all got the email at once.  But wouldn't that make 24 freshmen...

All speculation.

sirorange how do you know - i just checked the amc site it's not there

-----------------------------

Edit: Just saw AMCDirector's post.  Oh well.

hmm, I predict 25, 19, and 9 OH MAN I HOPE I'M RIGHT

It goes like this:
[quote]
Dear MOSP-er:

Congratulations!  You are among a very small group of 54 students (from the original pool of over 226,000 AMC 10 and AMC 12 participants) invited to take part in the 2006 Mathematical Olympiad Summer Program (MOSP) scheduled to begin Sunday, June 11, and ending Saturday, July 1, 2006.  This year’s MOSP will be held on the campus of the University of Nebraska-Lincoln....[/quote]

I'm afriad the entire letter may be copyrighted though, so I'm not sure if I'm allowed to post it. :?


http://artofproblemsolving.com/Forum/search.php

Look at all the options in the top right corner.

lol, now I don't even remember what I want to search for.

look at the options in the right corner :idea: that should work


Let $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre . Suppose $ BC \equal{} AB \plus{} AI$.
Find angle $ BAC$ . This problem carries  16 marks.


Is the Algebra 3 class a continuation of the Algebra 2 class?

Because when I looked at the diagnostic tests for those classes, the material was pretty different.

But the Algebra 2 class is a continuation of the Algebra 1 class...

Thanks.

[quote="Blazer123"]Is the Algebra 3 class a continuation of the Algebra 2 class?

 [/quote]
No. Only Algebra 1 and Algebra 2 are the same book.

But it is a continuation. You learn more advanced topics in it that build up from what was taught in Algebra 2 and under. Algebra 3 is better if you really want to challenge yourself. It covers polynomials more in depth and goes into all sorts of proofs and classical inequalities.

It is a continuation in the sense that the material in Algebra 3 builds on Algebra 2.  However, it is not the case that you must (or even should) take the two sequentially.  There are several other classes you might take in-between (any or all of the three Intro classes: geometry, counting, number theory).

I found the "do you need this?" part of the algebra 2 class not very challenging, I would say it is easier than a standard algebra 2 class... Algebra 3 is a good choice.


inequalities
combinatorics proposed
combinatorics

Let m,n be positive numbers.Prove that $\frac{(m+n)!}{m!n!}\geqq{mn}+1$without induction.

Say $m\geq n$

Since you claim it's combinatorics I guess you used $C^m_{m+n} = C^n_{m+n} \geq mn + 1$?

Anyway. Fix m and let n vary. The fraction becomes: $\frac{(m+1)\cdots (m+n)}{n!} \geq mn+1$
Now if $n=1$ we got that $m+1 \geq m+1$ ok
And if $n=2$ we got that $(m+1)(m+2)\geq 4m+2$ since $m\geq n>1$
And if $n>2$ we got that $(m+3)(m+4)\cdots (m+n) \geq n!$, so that gives: \[ \frac{(m+n)!}{m!n!} \geq (m+1)(m+2) > m^2+1 \geq mn+1 \]

To Mr. Peter VDD, thank you for your reply.You are right.

Suppose m>=n
Consider the number of ways to go walking by lattice points up or to the right from
(0,0) to (m,n). Each path is determined by a list of m steps to the rigth and n steps up. so the number of ways of doing this is (m+n)!/m!n!.

Now each path pass trough only one point of the form (m',1) and after some other steps makes a right turn for the first time. for the cut with line y=1 there are m+1 possibilities, and for the first turn rigth there are n possibilities. So there are at least (m+1)n paths and the inequality follows.

Example:
(0,n)                                                          (m,n)
First turn to the right   ...........>
(0,2)                          .
(0,1) Cut with line y=1 .
(0,0)............................

[quote="kunny"]Let m,n be positive numbers.Prove that $ \frac {(m \plus{} n)!}{m!n!}\geqq{mn} \plus{} 1$without induction.[/quote]

Imagine we had 2 disjoint sets, $ A$ and $ B$,  with $ |A| \equal{} m$ and $ |B| \equal{}n$.
We can take n elements from the union in $ \binom{m\plus{}n}{n}\equal{}\tfrac{(m\plus{}n)!}{n!\cdot m!}$ ways.
Note that, to take n elements from there, we can, for example, take all elements from $ B$ and none from $ A$.
This can be done in only 1 way.

Also, we can take $ n\minus{}1$ elements from $ B$ and 1 from $ A$, this can be done in $ m\cdot n$ ways.
Thus we find : $ \binom{m\plus{}n}{n} \geq m\cdot n \plus{}1.\,\,\,\square$


ratio
geometry
rectangle
trigonometry
Asymptote
analytic geometry
trig identities

I got

[hide]$ \dfrac{15\minus{}4\sqrt{3}}{18}$.[/hide]

Explanation (very long)
[hide]We take a square, subtract four big right triangles and four small triangles to get the area of the octagon.  Since it's trisection points, the square will have sides with length 3.  Then, the right triangles each have an area of one.  To find the area of the small triangle, take two right triangles of equal size such that they overlap in one area.  The area of this overlap can be calculated by drawing a 3X1 rectangle around the whole figure and using similar triangles.  It is $ \dfrac{1}{8}$.  Subtracting this and the area of the quadrilateral in the corner yields the small triangle's area.  We can find the angles of the quad and break it into two congruent triangles.  Using law of sines and then area of triangles, we can get the area of one triangle to be $ \dfrac{\sin60^\circ\sin45^\circ}{2\sin75^\circ}$, or $ \dfrac{3\minus{}\sqrt3}{4}$.  The area of the small triangle becomes $ 1\minus{}\dfrac{1}{8}\minus{}\dfrac{3\minus{}\sqrt{3}}{2}$, or $ \dfrac{4\sqrt{3}\minus{}5}{8}$.
Now our ratio is $ \dfrac{9\minus{}4\minus{}\frac{\minus{}5\plus{}4\sqrt3}{2}}{9}$, which (I think) simplifies to the answer above.

If I made a computational mistake, feel free to tell me so I can cry about it for a whole day. :( [/hide]

sorry.  I just edited my solution.  Maybe it's better? :roll:

Also wrong. Not a computational mistake, though.

is the answer 71/81 . I can post my solution if this si correct

Assume the area of the square is 9. Then the side length of the octagon is $ \sqrt {10} \minus{} \sqrt {5}$ . Then you can easily compute the area of it. And it turns out to be $ 10\sqrt {2} \minus{} 10$. so the answer is $ \frac {10\sqrt {2} \minus{} 10}{9}$. lol, horrible solution. oh well

[quote="dnkywin"]If you want a hint, check the asymptote code for the diagram.[/quote]

Where did you get point M?

1.) all of you are wrong
2.) disregard point M. It was used to label S

How are you guys solving this without any kind of measurement?

Still no.
[hide="final hint"][asy]pair A,B,C,D,E;
A=(0,1);
B=(1,0);
C=(0,0);
D=(1/6,0);
E=(4/9,5/9);
draw(A--B--C--cycle);
fill(B--D--E--cycle,gray);[/asy]
The ratio of the shaded area to the triangle is equal to the ratio of S to ABCD[/hide]

EDIT: It is 25/54, but post your solution.

YAY!!!  (:P )

I used the fact that the medians of a triangle cut it into 3 triangles of equal area to find the area outside the octagon.

Assume the area of the square is 9, then the area of the unshaded region is
$ \frac{2}{3}\cdot{8}\minus{}\frac{1}{8}\cdot{4}\equal{}\frac{29}{6}$. Subtracting that from 9, gives that the area of the shaded region is$ \frac{25}{6}$.

I used coordinates (No thinking involved!)  :D (Mr. Green).
Let the area of the square be 9.
We find the coordinate of the lower left corner of S is (2/3,2/3), hence the area of the 4 quadrilaterals in each corner is 2/3.
Also, the coordinate of the bottom point of S is (3/2, 1,4) so the area of the small triangle right under that point is 1/8.
So the area of the 8 smallest right triangles in the figure is 1-1/8-2/3 (It is the large right triangles in the figure minus the two figures described above).
To find the shaded area, take 9 and subtract 4 of the large right triangles, and then subtract 4 of the small right triangles.
So the answer is (9-4-4(1-1/8-2/3))/9=25/54.


[color=blue]

   Let $p$ be a prime number and $a, b\in\mathbb{Z}$ such that $p=(a, b)$ and $p^{2}$ divides  $a$.
   Prove that the polynomial $P(x)=X^{n+2}+aX^{n+1}+bX^{n}+a+b$ is irreductible over $\mathbb{Z}[X]$.

  [/color]

Cezar Lupu : could you post your solution ?

Isn't this just Eisenstein? $p$ divides $a$ and $b$, but they have no other common factor and $p^2 \mid a$, so $p^2 \nmid b$. Then $p^2 \nmid a+b$. Besides, $p$ divides all coefficients of $P(x)$ except the leading one.


calculus
derivative
function
limit
trigonometry
calculus computations

Determine whether the function is differentiable at 0 if:
f(x)=x.sin1/x with x#0
     =0 with x=0
I know the answer is No but I can't prove it! I try to prove that lim(x.sin1/x) when x approaches to 0 is DNE but it does exist(=0)Therefore, I proved that this function is continuous.. On the other hand, I can't differentiate x.sin1/x, so I'm stucked at this step. Can anybody help me?

[quote="JoeBlow"]A function $ f$ is differentiable at point $ a$ iff the limit
\[ \lim_{x\to 0}\frac {f(a \plus{} x) \minus{} f(a)}{x}
\]
exists.  So, the function $ f(x) \equal{} \begin{cases}x\sin \frac {1}{x} & x\neq 0 \\
0 & x \equal{} 0\end{cases}$ is differentiable at $ 0$ iff the limit
\[ \lim_{x\to 0}\sin \frac {1}{x}
\]
exists.  Can you prove that this limit does not exist?
[/quote]
It should be x.sin(1/x). But x.sin(1/x) does have a limit 0 when x approaches to 0. So that's why I already mentioned above that the method of proving f(x) is not continuous does not work. I already test out the vertical line but f(x) doesn't have any vertical line here. And f(x) does also have equally two-sided limits(=0). How do you prove that f(x) is not differentiable at x=0 now? I test out all of three conditions that can make the function indifferentiable.
@Joe Blow: Please read my post carefully since you seem to repeat to what I already found out and mentioned in my post above. :wink: .I need a specific solution pleased!

Wait, JoeBlow is right.

$ f'(0)\equal{}\lim_{\delta x\to 0}\frac{f(0\plus{}\delta x)\minus{}f(0)}{\delta x}\equal{}\lim_{\delta x\to 0}\frac{\delta x\sin \delta x^{\minus{}1}}{\delta x}\equal{}\lim_{\delta x\to 0}\sin \delta x^{\minus{}1}$

which doesn't exist.

Correct.  The limit you have to evaluate to determine if a function is continuous is not the same as the limit you have to evaluate to determine if a function is differentiable; JoeBlow's post required no correction.  Please think harder about the posts of people who carefully answer your questions before you decide that they haven't, in fact, answered your question.

[quote="JoeBlow"]A function $ f$ is differentiable at point $ a$ iff the limit
\[ \lim_{x\to 0}\frac {f(a \plus{} x) \minus{} f(a)}{x}
\]
exists.  So, the function $ f(x) \equal{} \begin{cases}x\sin \frac {1}{x} & x\neq 0 \\
0 & x \equal{} 0\end{cases}$ is differentiable at $ 0$ iff the limit
\[ \lim_{x\to 0}\sin \frac {1}{x}
\]
exists.  Can you prove that this limit does not exist?
[hide="Something else to think about"]Is $ f(x) \equal{} \begin{cases}x^2 \sin \frac {1}{x} & x \neq 0 \\
0 & x \equal{} 0\end{cases}$ differentiable at $ 0$?  Is it differentiable near $ 0$?  Is the derivative, as a function, continuous at $ 0$?[/hide][/quote]
I'm sorry very much JoeBlow! I didn't think of your suggestion sufficiently to see what you meant. Hope you can forgive my stupidity. :oops:


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

Let $ n \geq 4$,$ p,q\geq 2$, be some positive integers so that $ p\plus{}q\equal{}n$.Prove that $ S_n$ has a subgroup of order $ pq$ .($ S_n$ is the group of bijective functions from $ N$ to $ N$).

A little easier than your other questions, no? :)

Partition $ \{1,2,3 ..,n\}$ into two disjoint sets : $ A\equal{}\{a_1,a_2,...,a_p\}$ and $ B\equal{}\{b_1,b_2,...,b_q\}$, and consider the disjoint cycles :
$ \sigma\equal{} (a_1,a_2,..,a_p)$ and $ \tau\equal{}(b_1,b_2,..,b_q)$

We know that any two disjoint cycles commute, so $ \sigma \tau \equal{}\tau \sigma$, and moreover we know that $ ord(\sigma) \equal{}p$ and $ ord(\tau)\equal{}q$.And to finish take $ G\equal{}<\sigma, \tau>$ , and it's easy to see that $ G$ is an abelian subgroup of $ S_n$ with $ |G|\equal{}pq$.


[hide="Brute Force"]
1) Homogenize the inequality using a+b+c=1
2) Multiply out everything
3) Use AM-GM or Schur
[/hide]

[hide="Solution"]
Since $ a\plus{}b\plus{}c\equal{}1$, $ (a\plus{}b\plus{}c)^3\equal{}1\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$, so since, by AM-GM,

$ (a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)\equal{}2\ge3\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

$ \implies3(a\plus{}b)(b\plus{}c)(c\plus{}a)\le\frac89$, $ a^3\plus{}b^3\plus{}c^3\equal{}1\minus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)\ge\frac19$.

Thus, $ \frac{a^3\plus{}b^3}{c^2}\plus{}\frac{a^3\plus{}c^3}{b^2}\plus{}\frac{b^3\plus{}c^3}{a^2}\ge\frac{\frac19\minus{}a^3}{a^2}\plus{}\frac{\frac19\minus{}b^3}{b^2}\plus{}\frac{\frac19\minus{}c^3}{c^2}$

$ \equal{}\frac19\left(\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\right)\minus{}(a\plus{}b\plus{}c)\equal{}\frac19\left(\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\right)\minus{}1$.

By Cauchy-Schwarz, $ (a\plus{}b\plus{}c)\left(\frac1a\plus{}\frac1b\plus{}\frac1c\right)\equal{}\frac1a\plus{}\frac1b\plus{}\frac1c\ge(1\plus{}1\plus{}1)^2\equal{}9$, so again by Cauchy-Schwarz,

$ (1\plus{}1\plus{}1)\left(\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\right)\ge\left(\frac1a\plus{}\frac1b\plus{}\frac1c\right)^2\ge81\implies\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\ge27$.

Finally, this means that $ \frac{a^3\plus{}b^3}{c^2}\plus{}\frac{a^3\plus{}c^3}{b^2}\plus{}\frac{b^3\plus{}c^3}{a^2}\ge\frac19\left(\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\right)\minus{}1\ge3\minus{}1\equal{}2$, as desired.

Equality occurs when $ a\equal{}b\equal{}c\equal{}\frac13$.
[/hide]

[hide="Solution"]Clear the denominators: \[ \sum_{sym} a^5 b^2 \equal{} 2 a^2 b^2 c^2 \equal{} 2 (a\plus{}b\plus{}c)(a^2 b^2 c^2) \equal{} \sum_{sym} a^3 b^2 c^2 \]
which is true by Muirhead.[/hide]

Can you please give your complete proof [b]Bugi[/b]

Add $ a\plus{}b\plus{}c\equal{}1$ to both sides to get

$ (a^3\plus{}b^3\plus{}c^3)\sum\frac{1}{a^2}\ge3$

$ \Leftarrow (a^3\plus{}b^3\plus{}c^3)\frac{9}{a^2\plus{}b^2\plus{}c^2}\ge3$

$ \Leftarrow 3a^3\plus{}3b^3\plus{}3c^3\ge(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)$

Which is simple rearrangement.

It's basically TZF's solution, just I like to use AM-GM instead of Muirhead (although it doesn't matter with which method do you do).


Solve this equation in integer numbers:
\[ 1 \plus{} yx^2 \equal{} x^2 \plus{} 2xy \plus{} 2x \plus{} y
\]

We can see
$ y\equal{} \frac{x^{2} \plus{} 2x\minus{}1}{x^{2} \minus{}2x\minus{}1} \equal{} 1\plus{} \frac{4x}{x^{2}\minus{}2x\plus{}1}$
Hence $ 4x \vdots x^{2}\minus{}2x\minus{}1 \rightarrow 4 \vdots x^{2}\minus{}2x\minus{}1$
From here we can find all the solutions of the equation in $ Z$:
$ (0,1);(2,\minus{}7),(\minus{}1,\minus{}1);(3,7);(1,\minus{}1)$


Prove that for all triangle ABC, we have:
$ \sin\(\frac{\pi-A}{4}.\sin\(\frac{\pi-B}{4}.\sin\(\frac{\pi-C}{4}\ge\sin\(\frac{A}{2}.\sin\(\frac{B}{2}.\sin\(\frac{C}{2}$

The ine is equavalent to: $ \cos A\plus{}\cos B\plus{}\cos C\le \sin\frac{A}{2}\plus{}\sin\frac{C}{2}\plus{}\sin\frac{B}{2}$
We have:
$ \cos A\plus{}\cos B\equal{}2\sin\frac{C}{2}\cos\frac{A\minus{}B}{2}\le 2\sin\frac{C}{2}$
Similarly we have two other inequalitys. Add themself to drive request of problem

[quote="hien"]The ine is equavalent to: $ \cos A \plus{} \cos B \plus{} \cos C\le \sin\frac {A}{2} \plus{} \sin\frac {C}{2} \plus{} \sin\frac {B}{2}$[/quote]
Could you please explain this?


inequalities
parameterization
algebra proposed
algebra

a,b,c are real positive numbers,consider an equation:

a^2+b^2+c^2=1;

prove:
                  1         1         1
               ------ + ------+-------<4.5
                1-ab     1-bc   1-ac

The way you've written the problem is extremely confusing; should it be:

$a, b, c$ are real positive numbers such that $a^{2}+b^{2}+c^{2}= 1$; prove that

$\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ac}< 4.5$

?

Yes. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36873]That's what he meant[/url].

perfect_radio,

It is NOT what rouzbeh_k meant.  In rouzbeh_k's posted problem, he has a strict inequality for the 4.5 (or 9/2) and the a,b, and c are mentioned as being [positive reals] "real positive numbers."

However, in perfect_radio's link in his thread to where the problem was already worked on by others in a different thread, the inequality is a less than or equal to symbol for the 9/2. The a, b, and c are mentioned, in effect, that they are to be non-negative, but without mentioning what type of number system they belong to.

Well, where will we get if we have a thread for any single constellation of $<$ instead of $\leq$ etc.. Thus perfect_radio is perfectly right in linking it away.

The sooner, the better, for the two threads to be linked up no matter by whichh user, if done so correctly, is what I would have wanted.  Yes, I'm satisfied that perfect_radio linked them up.

However, it is changing the subject away from me bringing up (experienced) mathematicians here who aren't using consistent, and sometimes needed parameters to successfully solve certain equations/inequalities.


[quote="dubito"]$ {\text{Compute: }}\sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} 4\sqrt {...} } } }$[/quote]
Suppose $ n$ is a positive integer:
$ n(n\plus{}2) \equal{} n\sqrt{1\plus{}(n\plus{}1)(n\plus{}3)} \equal{} n\sqrt{1\plus{}(n\plus{}1)\sqrt{1\plus{}(n\plus{}2)(n\plus{}4)\plus{}...}}$

That's a good start, but don't you need to say more to argue that the original sequence of radicals converges to $ 3$?


If I have a $ 4\times 4$ chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn?

I recently did this question and i was wondering if their is a formula. If any one knows one please tell me.

[quote="PowerOfPi"]N1349 pointed out an error in my example (look at the last diagram). It should be:

[hide="What I got"]The chess board is in the form (I don't know asy):

 ._.._.._.._.
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

We can deal with the first row. There are 4 possible places to put the pawn. Let's say we put in in the 2nd column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

Next, we have the 2nd row. There are three choices for this. Let's say this goes in the fourth column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|_|_|_|_|
|_|_|_|_|

Next, the third row. we have 2 choices. For example, we could put it in the first column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|_|_|

Finally, there will be one spot left in the last row. In the example, it must go in the third column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|[u]X[/u]|_|

So we just multiply those and get $ 4 \cdot 3 \cdot 2 \cdot 1 \equal{} 4! \equal{} \boxed{24}$

[/hide][/quote]

Sorry for reviving, but your answer is incorrect. I looked back at the reports, and it says the answer is $ 576$.

[color=red] Moderator Says:  If you copy Alcumus's solution, please state that you have done so in thread, the following solution is copied from Alcumus. [/color]
Because our chess board is $ 4 x 4$, there must be exactly one pawn in each column and each row. Consider the ways to place one pawn in each row. In the first row, there are four potential spots for a pawn. However, no matter where we place the pawn, it takes up one column. So, in the second row, there are only three potential spots for a pawn. Similarly, there are two spots in the third row and only one in the fourth. Thus, there are $ 4\cdot 3 \cdot 2 \cdot 1 \equal{} 24$ ways in which we can place the pawns. Now, because each pawn is distinct, we have four possible pawns to place in the first slot, three in the second, two in the fourth, and one in the last. So there are 24 possible orderings of the pawns. Our final answer is thus $ 24^2 \equal{} \boxed{576}$

[quote="PowerOfPi"]N1349 pointed out an error in my example (look at the last diagram). It should be:

[hide="What I got"]The chess board is in the form (I don't know asy):

 ._.._.._.._.
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

We can deal with the first row. There are 4 possible places to put the pawn. Let's say we put in in the 2nd column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

Next, we have the 2nd row. There are three choices for this. Let's say this goes in the fourth column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|_|_|_|_|
|_|_|_|_|

Next, the third row. we have 2 choices. For example, we could put it in the first column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|_|_|

Finally, there will be one spot left in the last row. In the example, it must go in the third column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|[u]X[/u]|_|

So we just multiply those and get $ 4 \cdot 3 \cdot 2 \cdot 1 \equal{} 4! \equal{} \boxed{24}$

[/hide][/quote]

Did the same thing. The disctint part was the one I didn't understand.

What most people forget if that the 4 pawns are distinguishable, and that each order in different. You should be able to realize that you can put the pawns (in general) in 4! places (if you put one pawn on one square, you can put the next pawn anywhere BUT the square adjacent to the one you put your first pawn in, and so on). Additionally, there are multiple ways to put the pawns in each formation. Using the same method, the first pawn could be one of 4 pawns, the second can be one of 3, the third can be 1 of 2, and finally, the last pawn can only be 1 pawn, since there is only one left. Therefore, the answr is $(4!)^{2}=576$

Because our chess board is $4$ x $4$, there must be exactly one pawn in each column and each row. Consider the ways to place one pawn in each row. In the first row, there are four potential spots for a pawn. However, no matter where we place the pawn, it takes up one column. So, in the second row, there are only three potential spots for a pawn. Similarly, there are two spots in the third row and only one in the fourth. Thus, there are $4\cdot 3 \cdot 2 \cdot 1 = 24$ ways in which we can place the pawns. Now, because each pawn is distinct, we have four possible pawns to place in the first slot, three in the second, two in the fourth, and one in the last. So there are $24$ possible orderings of the pawns. Our final answer is thus $24^2 = \boxed{576}$.

[quote=PowerOfPi][hide="What I got"]The chess board is in the form (I don't know asy):

 ._.._.._.._.
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

We can deal with the first row. There are 4 possible places to put the pawn. Let's say we put in in the 2nd column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

Next, we have the 2nd row. There are three choices for this. Let's say this goes in the fourth column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|_|_|_|_|
|_|_|_|_|

Next, the third row. we have 2 choices. For example, we could put it in the first column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|_|_|

Finally, there will be one spot left in the last row. In the example, it must go in the third column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|_|[u]X[/u]|

So we just multiply those and get $ 4 \cdot 3 \cdot 2 \cdot 1\equal{}4!\equal{}\boxed{24}$

[/hide][/quote]

this is wrong, i just happened to come across it playing alcumus

dude this is dumb

the "distinct" part is so confusing arghhhhhh I got it wrong cuz of that :(  :mad:

[quote=N1349]Did anyone realize that if you solve this problem by figuring out how many places there are, and then subtracting to find the amount of places the next pawn can be placed, the answer is 16*9*4*1?[/quote]

Complementary counting for the problem is much more difficult. [color=#f00]If somebody finds a sol using complementary counting please post in the thread.[/color] The alcumus problem is from constructive counting so I used a constructive counting approach This problem is actually really similar to $n$-rooks problems, which you can read about in the first below link. An extended and much more difficult version of the $n$-rooks problem is the $n$-queens problems, which you can read about in the second below link. The difference between the alcumus problems and the $n$-rooks problem is that the $n$-rooks problems assumes the rooks are distinct; like many users in this thread, the "distinct" part of the problem made me find a wrong sol. Fortunately, I noticed the problem and studied the problem correct for the second sol.

$n$ rooks problem:[url]https://en.wikipedia.org/wiki/Rook_polynomial[/url]

$n$ queens problem:[url]https://en.wikipedia.org/wiki/Eight_queens_puzzle[/url](the $n$ queens problem is often bashed using computers :P )

[quote=PowerOfPi][hide="What I got"]The chess board is in the form (I don't know asy):

 ._.._.._.._.
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

We can deal with the first row. There are 4 possible places to put the pawn. Let's say we put in in the 2nd column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|_|
|_|_|_|_|
|_|_|_|_|

Next, we have the 2nd row. There are three choices for this. Let's say this goes in the fourth column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|_|_|_|_|
|_|_|_|_|

Next, the third row. we have 2 choices. For example, we could put it in the first column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|_|_|

Finally, there will be one spot left in the last row. In the example, it must go in the third column.

 ._.._.._.._.
|_|[u]X[/u]|_|_|
|_|_|_|[u]X[/u]|
|[u]X[/u]|_|_|_|
|_|_|_|[u]X[/u]|

So we just multiply those and get $ 4 \cdot 3 \cdot 2 \cdot 1\equal{}4!\equal{}\boxed{24}$

[/hide][/quote]

Nice chessboard!! XD

[quote=jasperE3][quote=ycz9000]IMO, "distinguishable" is a better word to use here than "distinct"; as worded, it's a bit confusing.[/quote]

What's IMO? International Math Olympiad?[/quote]

yes

I put it in as a probability :(

Anyway, this is how you do it.

[hide=Solution] For placing the first pawn, there are $16$ squares to put it. For the second one, there are$9$, as every row and column has $4$ squares, one being the one where the first pawn was placed, so there are $9$ left for the second one. Continuing this pattern, there are $4$ and $1$ options left for the last two pawns.

In total, this is $16\cdot9\cdot4\cdot1=576$. (Note that it doesn't matter which pawn we put down first.) [/hide]

[quote=jasperE3][quote=ycz9000]IMO, "distinguishable" is a better word to use here than "distinct"; as worded, it's a bit confusing.[/quote]

What's IMO? International Math Olympiad?[/quote]

IMO=In my opinion

Well, this is easy...

[hide = Solution]

For the first pawn, we have a total of $4$ columns and $4$ rows. Therefore, we have $\boxed{16}$ total ways to arrange the first pawn.

Similarly, since a row and column is taken, we have $3$ columns and $3$ rows. Therefore for the second pawn we have $\boxed{9}$ choices.

...and so on.

Continuing in this manner, we have a total of $16\cdot 9 \cdot 4 \cdot 1 = \boxed{\boxed{\boxed{576}}}$

[/hide]

[quote=N1349]Did anyone realize that if you solve this problem by figuring out how many places there are, and then subtracting to find the amount of places the next pawn can be placed, the answer is 16*9*4*1?[/quote]

Yeah i did this lol

alcumus should add a second solution


Find all $x \in \mathbb{R}$ dependently on $n$ satisfying $\sum \limits^n_{k=1} (cos(k \cdot x))^2 = 1$.

We will show that $S=\sum\limits_{k=1}^{4}(\cos(kx))^2>1$. Actually, $2S-4=\sum\limits_{k=1}^4\cos(2kx)=2(4t^4+2t^3-3t^2-t)$, where $t=\cos(2x)$. So we must prove $4t^4+2t^3-3t^2-t+1>0$. It is quite easy task.
For $n$=1, 2 and 3 we obtain simple linear, quadratic and cubic equations for $t$.

This solution shows that no matter how ugly a problem could appear,, there is always a super-smart ( congratulations, Myth, you always enjoy me with your solutions!) solution which makes the problem beautiful!

I'm sad you consider this sneer. But it's your opinion. Anyway, those were my sincere feelings and I really am impressed by your solutions. But if you don't like to believe me, it's your choice.


AMC USA(J)MO USAMO geometry circumcircle function topology

I can't remember where I saw this, but here it goes:

We have a map on the table. We take a picture of it. Suppose we can put the picture on the map such that they overlap perfectly. Show that there is at least one point on the map which is directly under its image on the picture.

P.S. I think it's necessary to put the picture on the map facing upwards.

No ideas AT ALL? Any help is welcome  :D

hey,not a solution, but i've heard of some Brouwer's fixed point thorem saying something like this. and then ther's USAMO 1978
Two square maps cover exactly the same area of terrain on different scales. The smaller map is placed on top of the larger map and inside its borders. Show that there is a unique point on the top map which lies exactly above the corresponding point on the lower map. How can this point be constructed?
cheers

Let be ABCD the big map and A'B'C'D' the small one.
Since the two maps are similar, a spiral similirality transforms the first one in the second one. So a fixed point exists. I heard that proving that this point is inside the little map is quite difficult.
Suppose that this point, O,  is inside the small map.
Let \gamma  be the angle of the similirality.
((AB),(A'B'))=  \gamma  . So O belongs to the circle which goes by A and A' and by the intersection of (AB) and (A'B'), X.
O belongs to the circumcircle of XAA'.
The same thing done with (CD) and (C'D') gives O (the intersection of the two circumcircles)

(My english is pretty bad...)

[quote="grobber"]We have a map on the table. We take a picture of it. Suppose we can put the picture on the map such that they overlap perfectly.
[/quote]

How does picture and the map "perfectly overlap"? It means that you take a picture of some part of the map with the shape of it and zoom in to the size of the whole map?

If so:

Let's have a function
picture->map, point x on the picture->corresponding point x' on the map.
Because picture lies exactly on the map, it is a function:
f: picture->picture, point x on the picture->point on the picture y which lies over the corresponding point x' on the map

Because the picture was enlarged, this is a contraction. The picture is closed subset of R^2, so it is a complete metric space. So Banach's theorem says that there is exactly one point x, so that f(x)=x, and this the point which we were looking for.

Indeed, the first question of this USAMO problem is a (direct) corollary of the two-dimensional version of Brouwer's Fixed Point Theorem. This proposition can be stated, in the language of Algebraic Topology, as follows:

Denote by E^n the standard n-cell of IR^n (the "n-ball"). Then any continuous map (ie function) f: E^n --> E^n has at least one fixed point.

How is that applied? Of course E^2 is homeomorphic to the (square) map in question (and to its image under the enlargement). Hence, as long as this enlargement is applied using the same scale everywhere, it is continuous and hence must have some fixed point (topologically at least, the rotation that follows the enlargement, by placing the small map arbitrarily on the big map, has no effect on what we are discussing). Hence, there exists a fixed point, as desired. Now we've shown existence, the explicit construction of the point in question is a task for classical geometers ;-D.

Cheers,
Thanos

This is USAMO 1978 #2.  Here are two simple arguments (without large theorems).

[b]Existence:[/b]  Fix one side of the original square map as the "top" and the opposite side as the "bottom."  Consider the side as a line segment $ L$ and consider its image $ L'$ in the smaller map.  Clearly $ L'$ is below $ L$.

Now suppose we continuously move $ L$ downward until it is at the "bottom" of the map.  Now clearly $ L'$ is above $ L$.

By the mean value theorem, $ L$ intersects $ L'$ somewhere in between.  In fact, first they intersect at one end of $ L'$ and then at the other.  Again by the mean value theorem, somewhere in between they will intersect at the (unique) fixed point.

[b]Construction:[/b]  This is something of a cop-out.  Center the original map at the origin of the complex plane.  We can then represent the spiral similarity as $ f(x) \equal{} cx \plus{} d$ where $ c, d \in \mathbb{C}$ and $ |c| < 1$.  Its (unique) fixed point is therefore

$ x_0 \equal{} \frac {d}{1 \minus{} c}$.

Perhaps this can be expressed in terms of inversions or similar classically, but I haven't the patience.   :P

[quote="t0rajir0u"]In fact, first they intersect at one end of $ L'$ and then at the other.  [/quote]
One more thing may happen - if $ L$ and $ L'$ are paralell, $ L$ will intersect with the ends of $ L'$ simultaneously, but at that fixed moment we may make 2 points to go from the ends of $ L$ and $ L'$ continiously to the other ends, and again by the MVT, they will meet at the fixed point.


In a far away galaxy in a special little solar system on the ninth planet from the sun live humanoids that also use a base ten numeral system.  On this planet, the inhabitants have a very lucky number which is the least positive multiple of nine such that each digit is even and each digit is a unique (different) digit.  Find this very lucky number.

 :)  :|

Is that even possible? The sum of the digits in a number divisible by 9 have to add up to 9, and no number of even digits can add up to an odd number. Unless I'm missing something.

Corrected:

[quote="Waffle"]Is that even possible? The sum of the digits in a number divisible by 9 [b]have to add up to[/b] $ 9b, b\in \mathbb{N}$[/quote]

i.e. is $ 9|99$?  :wink:

[quote]On this planet, the inhabitants have a very lucky number which is the least positive multiple of nine such that each digit is even and each digit is a unique (different) digit. Find this very lucky number.
[/quote]
each digits are even and different
and it's the least positive multiple of nine that fits the above
like 180, i think
but i'm not sure if that's the answer  :|

A bunch of adding in my calculator reveals the answer as $ \boxed{468}$.  How I did:

can't  be 1 or 2 digit
cant be 100s
cant be 300s
checked all the 200s, none
checked 400s, found this

An easier way, take 2 _ _ and fill in blanks with different even digits and see if divisible by 9.

good job
i think that's correct

[quote]like 180, i think [/quote]
whoa
1 is not even. i'm so stupid
fail
 :wallbash:

Wow  :oops:

I completely forgot about that. I feel stupid now.


Let x_1 , x_2 , ... , x_2008 be real numbers in the interval [-1 , 1 ] such that x_1^3 + x_2^3 + ... +x_2008^3 = 0 . Find the maximum value of the x_1 + x_2 + ... +x_2008


How do I make the 13-gon in question five in LaTeX?

Please do not discuss this any further while the round is in progress.

As a reminder, you should not be discussing [b]any[/b] aspects of the problems while the round is in progress.


floor function
combinatorics unsolved
combinatorics

In an island, there exist 7 towns and a railway system which connected some of the towns. Every railway segment connects 2 towns, and in every town there exists at least 3 railway segments that connects the town to another towns. Prove that there exists a route that visits 4 different towns once and go back to the original town. (Example: $ A\minus{}B\minus{}C\minus{}D\minus{}A$)

OK so we assume it is possible not to have a 4-cycle. Name the towns $ T_{1},...,T_{7}$. Assume one town is connected to at least $ 4$ other towns (w.l.g., let this be $ T_{1}$ connected to at least $ T_{2},...,T_{5}$). If one of $ T_{2},...,T_{5}$ is connected to two of $ T_{2},...T_{5}$, then we can find the cycle. Therefore, since each town is connected to at least $ 3$ other towns, each of $ T_{2},...,T_{5}$ is connected to at least one of $ T_{6}$,$ T_{7}$ (assuming none of $ T_{6}$, $ T_{7}$ is connected to $ T_{1}$. If one is connected the case is trivial). Then, by Pigeon hole, there are at least two of $ T_{2},...,T_{5}$ connected to one of $ T_{6}$, $ T_{7}$ (assume $ T_{2}$ and $ T_{3}$ are connected to $ T_{6}$). Then $ T_{1},T_{2},T_{6},T_{3},T_{1}$ forms the cycle, which is a contradiction.

Then every town is connected to exactly $ 3$ other towns. We get that twice the number of vertices in the graph is $ 7(3)\equal{}21$, which is a contradiction since $ 21$ is an odd number. Then it is not possible for every town to be connected to exactly $ 3$ other towns, and we get that a 4-cycle always exists.

Some more info about the problems:
The proposer intended to submit the general case, which there is $ n$ towns and at least $ 2n\plus{}1$ railway segment from each town, but this following special case made it to the INAMO, and not the general case :(
Any of you wanted to prove the general case?

You can solve this problem by applying the following lemma :
Let G be a graph wich $ n>4$ vertices and m edges . Prove that if $ m>\frac{n.(\sqrt{4n\minus{}3}\plus{}1)}{4}$ then this graph has 4-cycle .

[quote="wangsacl"]there is $ n$ towns and at least $ 2n \plus{} 1$ railway segment from each town[/quote]I think it's the other way around.  :wink:

I'll think about this problem.

edit: After a little thought, actually the general problem is almost exactly the same as the special case. :D

The theorem

[color=blue]If a graph on $ n$ vertices contains no $ 4$-cycle, then the number of its edges is not larger than $ \left \lfloor \frac {n} {4} (1 \plus{} \sqrt{4n \minus{} 3}) \right \rfloor$[/color]

TTsphn referred to, is by Istvan Reiman, and a beautiful proof is in "Proofs from the Book" (Aigner and Ziegler).

Then, having $ 2n\plus{}1$ vertices of degree at least $ n$ each is an overkill ...


Please check...seems too easy.

Prove that $ f(z) \equal{} |z|$ is not holomorphic.

We have $ f(z) \equal{} \sqrt {x^2 \plus{} y^2} \plus{} i0$, so $ \frac {\partial{u}}{\partial{x}} \equal{} x(x^2 \plus{} y^2)^{ \minus{} \frac {1}{2}},\frac {\partial{u}}{\partial{y}} \equal{} y(x^2 \plus{} y^2)^{ \minus{} \frac {1}{2}}, \frac {\partial{v}}{\partial{x}} \equal{} \frac {\partial{v}}{\partial{y}} \equal{} 0$

One sees that if either x or y, but not both, were zero, either condition would be satisfied and the other not. So there are no such points, for which the Riemann conditions would hold.

Since holomorphic functions are complex-differentiable, and f is not differentiable, it follows that f is not holomorphic.

It strikes me that the equivalent condition $ \frac {\partial{f}}{\partial{\bar{z}}} \equal{} 0$ is not trustworthy, because using only this one could say that f is differentiable only in the origin, which is false if we work with Riemann. Why is it so?

[quote]using only this one could say that f is differentiable only in the origin, which is false if we work with Riemann.[/quote]
I'm confused by this statement. $ f$ is real-differentiable everywhere [b]except[/b] for the origin. In particular, $ \frac {\partial f}{\partial \bar z}$ exists everywhere except for the origin. But $ \frac {\partial f}{\partial \bar z} \equal{} \frac { \minus{} 1/2}{|z|\bar z}$ is never equal to zero. Therefore, $ f$ is not complex-differentiable anywhere.

[quote="mlok"] But $ \frac {\partial f}{\partial \bar z} \equal{} \frac { \minus{} 1/2}{|z|\bar z}$ is never equal to zero. Therefore, $ f$ is not complex-differentiable anywhere.[/quote]

When I use the definition:

$ \frac{\partial f}{\partial \overline{z}}\equal{}1/2\left[(u_x\minus{}v_y)\plus{}i(v_x\plus{}u_y)\right]$

I get:

$ \frac{\partial f}{\partial \overline{z}}\equal{}1/2\left(\frac{x\plus{}iy}{\sqrt{x^2\plus{}y^2}}\right)\equal{}1/2\frac{z}{|z|}$

which doesn't seem to be equal to $ \minus{}1/2\frac{1}{|z|\overline{z}}$.  Can someone explain where I went wrong?

Your expression is correct, shawtoos. It still isn't defined at zero, so it's not zero anywhere.

I differentiated $ \frac {1}{|z|}$ instead.  :blush:


The variable x varies directly as the cube of y, and y varies directly as the square root of . If x equals 1 when z equals 4, what is the value of z when x equals 27?

when you say varies directly do you mean directly proportional?

Proportionally. Yeah. :D

[hide]
So $ x \equal{} k_1y^3$ and $ y \equal{} k_2\sqrt z$ for arbitrary constants $ k_n$. Then that means that $ x \equal{} k_1k_2z^{\frac 32} \equal{} kz^{\frac 32}$ for some arbitrary constant $ k \equal{} k_1k_2$. This means that $ x$ is directly proportional to $ z^{\frac 32}$.

Now we need to find $ k$. Using the fact that $ x \equal{} 1$ when $ z \equal{} 4$, that means that $ z^{\frac 32} \equal{} 8$ and then $ k \equal{} \frac 18$.

When $ x \equal{} 27$, we need to solve $ 27 \equal{} \frac 18z^{\frac 32}$. This means that $ z^{\frac 32} \equal{} 216$, $ \sqrt z \equal{} 6$, and $ z \equal{} \boxed {36}$.
[/hide]

[hide]we know that $ x \equal{} (a)y^3$ and that $ y \equal{} (b)\sqrt {z}$ a and b representing the proportions.
From the two equations we can see that $ x \equal{} (a)(b^3)(z^\frac {3}{2})$.
By plugging in $ x \equal{} 1$ and $ z \equal{} 4$ we see that

$ (a)(b^3) \equal{} \dfrac{1}{8}$.  From this we set up the equation...

                                                     $ x \equal{} \dfrac{1}{8}(z^\frac {3}{2})$

Replacing $ x$ with 27, we can get $ z$.

                                                     $ 27 \equal{} \dfrac{1}{8}(z^\frac {3}{2})$

                                                     $ z \equal{} 216^\frac {2}{3}$

                                                     $ z \equal{} 36$[/hide]

[quote="Poincare"]I don't really like that problem too much classroom type math but the solution was really clear Yongyi!  :winner_first:  :thumbup:[/quote]Well, it's a bit beyond typical classroom math, but make sure you rate posts when they are good just as you do when they are bad.


MATHCOUNTS AMC AIME USA(J)MO USAMO search

There's no such thing. There are quite a few different kinds of problems, much less different problems, all the way from MathCounts level in middle school to AIME/USAMO/IMO level in high school - although, AIME and USAMO have been taken by middle schoolers as well. For MathCounts problems, check out the [url=http://www.artofproblemsolving.com/Forum/index.php?f=132]MathCounts Forum[/url]; the [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]AMC Forum[/url] for AMC/AIME/USAMO problems; and this forum and the Intermediate Forum for things in-between, although easier AIME problems and a number of AMC problems have been asked on the Intermediate Forum, as the difficulty level overlaps.

1 other thing.  There are forms of problems that they substitute numbers into every time that they do the contest, but that's mainly in MathCounts (I think)

:D  :D If you want lots of good practice problems, read chinese workbooks.  :D  :D

Singapore word problems are great if you're at that level. ;)

Foreign problems get complex as they are in another language. :)
(But I understand Chinese, so that's a plus! :) )

[quote="4everwise"]I'm going to do some Chinese workbooks soon. I have so many that I haven't even looked at yet.  :D[/quote]
What are they?

[quote="chess64"]Where is everyone getting these Chinese and Singapore problems?? I want some... :([/quote]
Search for it. Or you can come to China to buy some books. You can easily find a lot of Olympiad books in Chinese book store that are more that 1000 pages.

Chinese books are really good.

They even have a grade school practicing book for the IMO!  :)

A lot of the books are almost as good as the Art of Problem Solving books!

The only bad thing is that you need to understand Chinese in order to read the books.

Good thing for me, I'm Chinese!

 :P


Βρειτε τους πρωτους $ p$ ώστε το σημείο $ A(\sqrt{p},\sqrt{p\plus{}1})$ να ισαπεχει απο σημεια του καρτεσιανού επιπέδου με ακέραιες συντεταγμένες.
Την σκάρωσα για επίπεδο Θαλή- Ευκλείδη.....περιμένω εκτιμήσεις

Από πόσα σημεία να ισαπέχει ? Δύο (μέσο τμήματος), τρία (κέντρο περιγεγραμμένου κύκλου τριγώνου), τέσσερα (κέντρο περιγεγρεμμένου κύκλου τετραπλεύρου) κτλ. ?  Κάτι δεν πιάνω....Μάλλον δν είναι αυτός ο τρόπος προσέγγισης...

Δεν νομιζω οτι θέλει διευκρίνηση αλλα αν βοηθάει να το διατυπωσω και διαφορετικά:
Να βρείτε.......ωστε να υπάρχουν σημεία με ακέραιες συντεταγμενες που να ισαπέχουν απ΄το $ A$.
Billy οταν ανεβεις σε περιμένω....

Κατάλαβα thanx....είναι γνωστός τύπος ασκησης...
ναι εννοειται για το δεύτερο....!

Mου σταλθηκε μια λυση με fax απο εναν ''μικρο'' αλλα καταπληκτικο στην Θ.Α.
καλα θα ειναι ''μικρε'' να ξεκινησεις να γραφεις στο forum

Εστω οτι το Α ισαπεχει απο τα Β,C που εχουν ακεραιες συντεταγμενες και οριζουν την ευθεια ε.Τοτε το μεσο Μ του BC εχει συντεταγμενες ρητους και η ε εχει συντ.διευθυνσεως ρητο.Τοτε το Α κειται στην μεσοκαθετο του BC,εστω την d με τυπο y=ax+b,οπου φανερα τα a,b ειναι ρητοι.Τοτε το Α επαληθευει τον τυπο της d.Αρα ισχυει $ \sqrt{p\plus{}1}\equal{}a\sqrt{p}\plus{}b$.Αρκει να υψωσουμε τη σχεση στο τετραγωνο και ευκολα προκυπτει οτι $ \sqrt{p}$ ειναι ρητος που ειναι ατοπο για καθε πρωτο p.

[quote="Sakis"]Εστω οτι το Α ισαπεχει απο τα Β,C που εχουν ακεραιες συντεταγμενες και οριζουν την ευθεια ε.Τοτε το μεσο Μ του BC εχει συντεταγμενες ρητους και η ε εχει συντ.διευθυνσεως ρητο.Τοτε το Α κειται στην μεσοκαθετο του BC,εστω την d με τυπο y=ax+b,οπου φανερα τα a,b ειναι ρητοι.Τοτε το Α επαληθευει τον τυπο της d.Αρα ισχυει $ \sqrt {p \plus{} 1} \equal{} a\sqrt {p} \plus{} b$.Αρκει να υψωσουμε τη σχεση στο τετραγωνο και ευκολα προκυπτει οτι $ \sqrt {p}$ ειναι ρητος που ειναι ατοπο για καθε πρωτο p.[/quote]

Νομίζω πως πρέπει να είσαι πιο προσεκτικός ...  :wink:

Έστω σημεια $ B(c,d)$ και $ C(e,f)$ τέτοια ώστε: $ AB \equal{} AC$ δλδ $ AB^2 \equal{} AC^2$ δλδ $ (\sqrt {p} \minus{} c)^2 \plus{} (\sqrt {p \plus{} 1} \minus{} d)^2 \equal{} (\sqrt {p} \minus{} e)^2 \plus{} (\sqrt {p \plus{} 1} \minus{} f)^2$ δλδ $ (c \minus{} e)(c \plus{} e \minus{} 2\sqrt {p}) \equal{} (d \minus{} f)(2\sqrt {p \plus{} 1} \minus{} d \minus{} f)$ Ονομάζω: $ c \plus{} e \minus{} 2\sqrt {p} \equal{} a$ and $ \sqrt {p \plus{} 1} \minus{} d \minus{} f \equal{} b$ . Τώρα προφανώς ο $ \sqrt {p}$ είναι άρρητος και  $ \sqrt {p \plus{} 1}$ είναι ή αρρητός ή ακέραιος θετικός. Άρα:

Στους $ a,b$  ο ένας πρέπει να είναι ρητό πολλαπλάσιο του άλλου. Έτσι, έστω $ R \equal{} \frac {(c \minus{} e)}{(d \minus{} f)}$Τότε: $ 2\sqrt {p \plus{} 1} \plus{} 2R\sqrt {p} \equal{} Q$ (πραγματικός) αρα ας  πούμε θα ισχύει και για εναν $ Q_1$ πραγματικό ότι: $ \sqrt {p \plus{} 1} \plus{} \sqrt {p} \equal{} Q_1 (1)$ και ο συζυγής του
$ \sqrt {p \plus{} 1} \minus{} \sqrt {p} \equal{} \frac {1}{Q_1} (2)$  Άρα από  $ (1) \plus{} (2)$ είναι  $ \sqrt {p \plus{} 1} \equal{} Q_1 \plus{} \frac {1}{Q_1} \equal{} k$ and  από $ (1) \minus{} (2)$ είναι $ \sqrt {p} \equal{} Q_1 \minus{} \frac {1}{Q_1} \equal{} l$ όπου $ k,l$ πραγματικοί… Για την περίπτωση που ανέφερα πιο πάνω δλδ για $ \sqrt {p \plus{} 1}$ να είναι θετικός ακέραιος είναι $ p \plus{} 1 \equal{} k^2$ που σημαίνει $ p \equal{} 3$....

 EDIT: συγνώμη για το τυπογραφικό και για την κακή εκφραση...ελπίζω να είναι λίγο πιο ξεκάθαρο τώρα....

Billy mαλλον ενοείς $ \sqrt{p\plus{}1}$ και επισης ο $ \sqrt{p\plus{}1}$ δεν ειναι παντα αρρητος π.χ. για ρ=3.....

Νομιζω πως καταλαβα τι πας να αποδειξεις και πραγματικα το $ p$ ειναι 3 ,
αλλα πολλές απ τις σχεσεις σου δεν εντοπισα πως βγηκαν ,επισης δεν διερευνεις τον παρανομαστη στο $ R$ και γενικα η λυση κουραζει αυτον που προσπαθει να καταλαβει τι κάνεις π.χ. γραφεις ''αρα ας πουμε θα ισχυει εναν $ Q1$ πραγματικο" τι θελει να πει ο ποιητης.......
Επισης για τον  Sakis αν δεν καταλαβες τι λεει ο δημητρης σου υπενθυμιζω οτι υπαρχουν και ευθειες που δεν οριζεται ο συντ. διευθυνσης..

Θα βάλλω μια λύση ... Έστω ότι το σημείο $ A$ ισαπέχει από τα σημεία $ B(x_1,y_1)$ , $ C(x_2,y_2)$
Τότε $ (\sqrt {p} \minus{} x_1)^2 \plus{} (\sqrt {p \plus{} 1} \minus{} y_1)^2 \equal{} (\sqrt {p} \minus{} x_2)^2 \plus{} (\sqrt {p \plus{} 1} \minus{} y_2)^2$

και μετά τις απλοποιήσεις πρέπει
$ \minus{} 2\sqrt {p}x_1 \plus{} x_1^2 \plus{} y_1^2 \minus{} 2\sqrt {p \plus{} 1}y_1 \equal{} \minus{} 2\sqrt {p}x_2 \plus{} x_2^2 \plus{} y_2^2 \minus{} 2\sqrt {p \plus{} 1}y_2$

Τώρα αφού το $ \sqrt {p}$ είναι άρρητος (prove it !) θα έχουμε ότι πρέπει $ x_1 \equal{} x_2$ (γιατί ?)
Επομένως  $ y_1^2 \minus{} 2\sqrt {p \plus{} 1}y_1 \equal{} y_2^2 \minus{} 2\sqrt {p \plus{} 1}y_2$ και παραγοντοποιώντας έχουμε

$ (y_1 \minus{} y_2)(y_1 \plus{} y_2 \minus{} 2\sqrt {p \plus{} 1})\equal{}0$ . Τώρα αφού τα σημεία $ B,C$ είναι διαφορετικά θα ισχύει

$ y_1 \plus{} y_2 \equal{} 2\sqrt {p \plus{} 1}$  Επομένως πρέπει $ p \plus{} 1 \equal{} q^2$ για κάποιον ακέραιο $ q$ . Αρα $ p \equal{} (q \minus{} 1)(q \plus{} 1)$
Επομένως θα πρέπει $ q \minus{} 1 \equal{} 1$ δηλαδή $ q \equal{} 2$ και $ p \equal{} 3$
QED :)

[quote="silouan"]
Τώρα αφού το $ \sqrt {p}$ είναι άρρητος (prove it !) θα έχουμε ότι πρέπει $ x_1 \equal{} x_2$ (γιατί ?)
[/quote]

Βαζω αποδειξεις στις 2 προτασεις που χρησημοποιει ο Σιλ
(Ειμαι σε ιντερνετ καφε και δεν εχω πολυ χρονο ουτε χαρτι και μολυβι οποτε συγχωρεστε με αν κανω κανενα λαθος :P)

Για το $ \sqrt{p} \notin Q^*_ \plus{}$ εχουμε:

Εστω $ \sqrt {p} \in Q^*_ \plus{} \Leftrightarrow \sqrt {p} \equal{} \frac {a}{b}, a,b \in Z^*_ \plus{} , gcd(a,b) \equal{} 1$
Τοτε $ b^2p \equal{} a^2$, και επηδη $ p$: πρωτος και διαιρει τον $ a^2$ θα διαιρει και τον $ a$ και o $ p^2$ θα διαιρει επισεις τον $ a^2$, αρα προφανος ο $ p$ θα διαιρει και τον $ b^2$ και επηδη ειναι πρωτος διαιρει και τον $ b$, οποτε καταληγουμε στο οτι $ p|gcd(a,b) \equal{} 1 \Leftrightarrow p \equal{} 1$ ατοπο.

Για τη 2η προταση, αφηνουμε τις ριζες στο 1ο μελος και υψωνουμε στο τετραγωνο, οποτε αν $ x_1 \neq x_2$ και $ y_1 \neq y_2$, παιρνουμε οτι ο $ \sqrt {p}\sqrt {p \plus{} 1} \in Q^*_ \plus{}$, δηλαδη $ \sqrt {p}\sqrt {p \plus{} 1} \equal{} \frac {a}{b}, a,b \in Z^*_ \plus{} , gcd(a,b) \equal{} 1$, με υψωση στο τετραγωνο παιρνουμε $ pb^2 \equal{} (pb \minus{} a)(pb \plus{} a)$, αρα ο $ p$ διαιρει τον $ a$ και $ a \equal{} kp$ οποτε η σχεση γινεται $ p^2(k^2 \minus{} b^2) \equal{} pb^2$ απο οπου (με βαση την προηγουμενη προταση) εχουμε οτι ο $ p$ διαιρει και τον $ b$, αρα τελικα $ p|gcd(a,b) \equal{} 1 \Leftrightarrow p \equal{} 1$ ατοπο κλπ.


If ABCD is rhomb and BM=p and DN=q,then AB=?
[hide="ANSWER"] $ \sqrt{pq}$ [/hide]

[hide="Solution"]Let $ AD \equal{} x$. Therefore, by similar triangles, $ \frac{ND}{DC} \equal{} \frac{NA}{AM}$. This is equivalent to $ \frac{p}{x} \equal{} \frac{p\plus{}x}{q\plus{}x}$, so $ pq \plus{} px \equal{} px \plus{} x^2$, so $ x^2 \equal{} pq$, so $ \boxed{x\equal{}\sqrt{pq}}$.[/hide]


Could anyone explain how to use cyclic sum notation? I see it used occasionally, and I have no idea what it means. I tried Googling it, but all the pages that came up scared me...

Have a look at one of these:
http://artofproblemsolving.com/Forum/viewtopic.php?t=1659
http://artofproblemsolving.com/Forum/viewtopic.php?t=106


ratio
geometry
3D geometry
tetrahedron
geometric sequence
geometric series

Scrambled wrote:This is an interesting problem similar to the one on the side.

1/x + 1/x^2 + 1/x^3 + 1/x^4+...=1/y

What is y in terms of x?
And prove it.

Trivially, y = 1/(1/x + 1/x^2 + 1/x^3 + 1/x^4+...). That was easy.

[hide]

1/y = 1/x*(1+1/x+...)

x/y - 1 = 1/x + 1/x^2 + ...

x/y - 1 - 1/y = 0

(x-1)/y = 1

y = (x-1)  (note x can not be 1 because 1/0 is undefined)

[/hide]

But that's not quite right. It doesn't work for all x.

Mathfanitic got it.  The answer if its one half is 1.

Actually, it is.  it is 0.5 +0.25 +0.125 + 0.0625...

So
It will eventually be adding 0.000000...5.  So the area you covered is 0.999999999.....8
 _
.9 is equal to 1.

What happens if [size=150][b](x = 1/2)[/b][/size]? Not 1/x.

The question is flawed in that if x < 1, the series goes to infinity. (Which is meaningless.) So I suppose one needs to put the restriction x > 1 in the question. Then one can simply plug in the formula for infinite geometric series and get the answer in approximately five seconds. o_o

(GOD THAT WAS STUPID OF ME...)

Er, for those of you who don't know the infinite geometric series formula, here it is:

For an infinite geometric sequence with the first term a and the common ratio r satisfying 0 < r < 1,

a / (1 - r)

is the sum of the sequence.

Proof: The sum of the first n terms in a geometric sequence with first term a and common ratio r is given by

Sn = a (1 - rn) / (1 - r).

When 0 < r < 1 and n approaches infinity, rn approaches 0, so

limn -> :inf: a (1 - rn) / (1 - r)
= a (1 - 0) / (1 - r)
= a / (1 - r).

Osiris -- "Getting Started," right?  Like, for people who haven't done this kind of thing a million times.

Also, you left out the negative case and had a typo. ;)  Go answer some tetrahedron questions in the Advanced forum.

JBL wrote:Osiris -- "Getting Started," right?  Like, for people who haven't done this kind of thing a million times.

I fully agree.

By the way, everybody who answered (except tare) ive seen browses the advanced/intermediate forum.

[hide]LET SOME BEGINNER TRY IT OUT WITHOUT BEING OVERWHELMED BY THE TEMPTATION TO READ SPOILERS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/hide]

Yeah, I initially answered it thinking I had clicked on the intermediate forum... then I left it. Sorry.

The answer was already given, so I figured I'd provide some additional insight... meh.

[quote="Osiris"]Er, for those of you who don't know the infinite geometric series formula, here it is:

For an infinite geometric sequence with the first term a and the common ratio r satisfying 0 < r < 1,

a / (1 - r)

is the sum of the sequence.

Proof: The sum of the first n terms in a geometric sequence with first term a and common ratio r is given by

Sn = a (1 - rn) / (1 - r).

When 0 < r < 1 and n approaches infinity, rn approaches 0, so

limn -> :inf: a (1 - rn) / (1 - r)
= a (1 - 0) / (1 - r)
= a / (1 - r).[/quote]

Can r ever be negative?

Interesting, I never thought about it, but apparently it does work for |r| < 1. It does make sense though.

That gets me thinking... what happens when r < -1?

Are you asking us or are you asking a rhetorical question?

And while you are thinking about Osiris's question, you can also think about what happens when r is a complex number.

No, actually if r < -1 the partial sums will oscillate between positive and negative. Take r = -2 for example:

1 - 2 + 4 - 8 + 16 - 32 + 64 - 128 +...

mathfanatic wrote:Quote:This is an interesting problem similar to the one on the side.

1/x + 1/x^2 + 1/x^3 + 1/x^4+...=1/y

What is y in terms of x?
And prove it.

[hide]1/x + 1/x^2 + 1/x^3 + ... is an infinite geometric sequence with common ratio 1/x and first term 1/x.  (I'm going to ignore the fact that |1/x| doesn't have to be < 1.  So slap me.)

Using the formula for infinite series, the sum is (1/x)/(1-1/x) = (1/x)/[(x-1)/x] = 1/(x-1), so we get

1/(x-1) = 1/y

Note that x =/= 1, since this would make 1/(x-1) undefined.

Now we simplify a little.

x - 1 = y

Yay.[/hide]

formula for infinite series? wat is the formula   ?

For a geometric sequence with first term a and common ratio r, the sum to infinity is a/(1-r)..


abstract algebra
vector
advanced fields
advanced fields unsolved

Hi all

I am taking a graduate algebra course as we are going over the basics of category theory. Most people including myself have never done this stuff before.

Anyways, in class our professor said he had a "joke" for people and he wrote the following problem on the board

Determine if **: k-modules to k-modules (k a field)  is essentially surjective. Where ** is the usual double dual functor.

I don't think anyone got it because no one laughed haha but I was wondering if someone could explain this to me. Obviously for finite vector spaces the objects and their image are isomorphic to each other but for the infinite case I don't really know why this should be "obvious". It is not for marks or anything but I would like to try to understand this.

Thanks for the help, hopefully you will find this funnier then my class did haha.

The Joke is that Category Theory is a joke ;)

In infinite dimensions, the double dual is much larger than the original space.

the question is: can every vector space be represented (up to isomorphism) as a bidual space? the answer is no:

if $ V$ is a vector space with an infinite basis $ B$, we have $ \dim(V^*) \equal{} \dim(K^B) \equal{} |K^B| \geq 2^{\aleph_0}$ by a theorem of erdös-kaplansky. in particular, the dimension of duals are finite or uncountable.

I see, so any vector space that is countably infinite cannot be naturally equivalent to a dual since all infinite duals are necessarily uncountable? Very interesting, thanks for the help.


[hide="Hint"]For $a,b,c,d > 0,$ we have
$\frac{ab}{a+b}+\frac{cd}{c+d}\leq \frac{(a+c)(b+d)}{a+b+c+d}.$[/hide]


express $ \boxed{\sqrt[{\;(\textbf g+\imath\cdot\textbf h)}]{\;\;\log_{(\textbf a+\imath\cdot\textbf b)}\;\;\{\;(\textbf c+\imath\cdot\textbf d)^{(\textbf e+\imath\cdot\textbf f)}\;\}\;\;}}$ as $ \boxed{\textbf x+\imath\cdot\textbf y}$

[ where $ \textbf{\;\{a,b,c,d,e,f,g,h,x,y\}\;}\;\in\;\mathbb R$ ] :D

I don't see the trick Misan.  Looks messy to me.  Little log or big Log?  I'd work it by extracting the principle value of everything (big Log).  That is, solve:

$ \sqrt[w]{Log_{s}(z^{r})}=f+ig;\quad Log(z)\equiv ln|z|+iArg[z];\quad-\pi<Arg(z)\leq\pi$

with:

$ \{w,s,z,r\}\in \mathbb{C}$ and $ \{f,g\}: \mathbb{R}^{8}\to\mathbb{R}$ and

$ Log_{s}(w)\equiv \frac{Log[w]}{Log[s]}$ again, the principle values.

so . . . start from the right and work towards the left.  Also, I'd want some empirical verification I have it straight:  I'd input an expression into Mathematica such as:

In[15]:=
w = 1 + I;
s = -3 + 4*I;
z = 2 + 2*I;
r = 4 - 6*I;
N[Log[s, z^r]^(1/w)]

Out[19]=
0.303023-0.929694i

then work it piece by piece myself to see if I get the same answer as Mathematica.

Maybe though that's not what you're looking for.

Actually Mathematica can calculate the entire result explicitly.  If you have it, type in the following:

Simplify[ComplexExpand[Log[a+bI,(c+dI)^(e+fI)]^(1/(g+hI))]]

As anyone might guess, the result is fairly complicated and not especially instructive.


algebra
polynomial
vector
topology
real analysis
real analysis unsolved

Let $ \mathcal{P}_{n}$ denote the set of polynomials of degree at most $ n$, considered as a subset of $ C[a,b]$.  Obviously, $ \mathcal{P}_{n}$ is a subspace of $ n\plus{}1$.  How can I prove that $ \mathcal{P}_{n}$ is closed in $ C[a,b]$?  By this, I mean that a sequence of polynomials in $ \mathcal{P}_{n}$ necessarily has a limit in $ \mathcal{P}_{n}$.

Because a sequence of polynomials is exactly like a sequence of n+1-dimensional vectors in $ %Error. "mathBB" is a bad command.
{R}^{K+1}$, and $ %Error. "mathBB" is a bad command.
{R}^{k+1}$ is a complete metric space, so the result directly follows.

True. :)   What would the appropriate term for "exactly like a sequence" be?  Obviously the map would be $ \sigma: \mathcal{P}_{n}\to\mathbb{R}^{n\plus{}1}$ defined by $ \sigma(p)\equal{}(p_{0},p_{1},\cdots,p_{n})$, but what type of map is this?

Actually, I guess they're equivalent as vector spaces and all finite-dimensional vector spaces are complete since $ \mathbb{R}^{n}$ is.


find all fuction $f: R-\to R$ such that $f(x-f(y))=f(x+y^{2002}) +f(f(y)+y^{2002}) +1$
 For  all $x,y$ is real.

[quote="dieuhuynh"]$f: R-\to R$[/quote]

do you mean $f: \mathbb{R}^{-}\to \mathbb{R}$?

noone has solution.I remember[i] tranthanhnam[/i] be posted


I was solving a problem and it said the numbers 1 to 49 inclusive. What does this mean? I googled it and it simply talked about calender inclusive counting which I did not understand. Does this mean I don't include 1 and 49? What does it mean? Thanks

It just means you include 1 and 49. When you say, for example, the numbers between 1 to 49, it means you want the numbers 2,3,...,48 because the word [b]between[/b] was used. The inclusive is added to signify that you also want the upper and lower extremes included in the list.


trigonometry
inequalities proposed
inequalities

[quote="pvthuan"]Prove or disprove that for $a,b,c\geq0$ and $a^2+b^2+c^2=4$,

\[a^3b+b^3c+c^3a\leq 1+\tfrac32(ab+bc+ca)\][/quote]

Either I misunderstand something, or $a=\frac{4}{\sqrt5}$, $b=\frac{2}{\sqrt5}$, c = 0 is a counterexample.

  darij

If $x,y,z$ be nonnegative real numbers such that $x^{2}+y^{2}+z^{2}=4$, and $t\geq\frac13+\frac{2\sqrt{7}}{3}\cos\left(\frac{\pi}{3}-\frac13\arccos\frac{1}{2\sqrt{7}}\right)$,then
\[\frac{t^{4}-1}{4t^{2}}(x^{3}y+y^{3}z+z^{3}x)\leq\frac{8t}{t^{2}+1}+(t^{2}-3)(yz+zx+xy), \]
with equality if $x: y: z=t: 1: 0$.
Where the constant $\frac13+\frac{2\sqrt{7}}{3}\cos\left(\frac{\pi}{3}-\frac13\arccos\frac{1}{2\sqrt{7}}\right)=1.8019377358048382524722\cdots$ is the best possible.
Specially, let $t=1+\sqrt2$, we obtain
\[x^{3}y+y^{3}z+z^{3}x\leq 2+2(yz+zx+xy). \]

[quote="darij grinberg"][quote="pvthuan"]Prove or disprove that for $a,b,c\geq0$ and $a^{2}+b^{2}+c^{2}=4$,
\[a^{3}b+b^{3}c+c^{3}a\leq 1+\tfrac32(ab+bc+ca) \]
[/quote]

Either I misunderstand something, or $a=\frac{4}{\sqrt5}$, $b=\frac{2}{\sqrt5}$, c = 0 is a counterexample.

  darij[/quote]

Where did the root 5 come from? Why not a simple value?

[quote="me@home"]Where did the root 5 come from? [/quote]
If $x,y,z\geq0$ and $x^{2}+y^{2}+z^{2}=4$, then

$x^{3}y+y^{3}z+z^{3}x\leq \frac{256}{75}+\frac{16}{15}(yz+zx+xy)$,

with equality if $x=\frac{4}{\sqrt5},y=\frac{2}{\sqrt5},z=0$.


Here's a new problem in the Getting Started Series, for beginning problem-solvers. You can learn more about the series in my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10857]introduction to the Getting Started Series[/url]. The basis idea is to post some [b]easy[/b] problems that allow beginners practice in problem-solving.

The problem below, in unmodified form, has been described as "a complex algebraic and geometry equation," but that is baloney. You can solve it with prealgebra methods. Here are the rules: read the problem statement below, and [b]using no pencil or paper calculations (or calculator)[/b] come up with an answer in no more than five minutes. So only use mental math, and you may look at the problem statement while thinking about the problem, but solve it in five minutes. That supposedly is very hard, but I think not. When you have an answer, write it in hidden text in a reply to this message. Show how you got your answer, in everyday language.

Scroll down for problem statement, which you may look at for five minutes only.

Albatross visited his friend JBL every year on the same holiday at Harvard, and planted an ivy plant there. A year later, after one year, Albatross noticed the ivy plant had grown up on a building wall to a height of twelve feet. After two years, it had grown up to eighteen feet high. After three years, it had grown to twenty-seven feet high.

How high was it after four years?

Please show your answer in hidden text. Every beginner is welcome to try out this problem.

[hide]39 Feet Because the increase from the first year of 12-18 is 6 feet then the second year was 18-27 which was 9 feet then 3rd to the fourth would be 27-39 witha growth of 3 feet a year.[/hide]

[hide="Another Method and Another Answer"]We could say that it grows increases by 50% every year. 18-12=12/2, 27-18=18/2. By that reasoning the answer would be 40.5=27+27/2.[/hide]

[hide]The plant is 150% of what it was last time so its 150 * 27 = 40.5.[/hide]


Charlie's mom just had a baby. He notices that they use an average of nine diapers per day. At this rate, how many total diapers will they use during June and July of this year?

There are $ 30$ days in June and $ 31$ days in July. Combined, it is a toal of $ 30\plus{}31\equal{}61$. Therefore, the number of diapers used will be $ 61(9)\equal{}\boxed{549}$.

Please stop posting the same thing that the second post portrayed (you literally do this in almost every FTW topic).

Would you also like me to define almost, Einstein? Use that thing you have in your head and figure it out.


There is a dartboard with only two regions worth 9 points and 4 points. If you could throw any number of darts, what is the largest positive score you cannot get?

[hide]By the famous Chicken McNugget Theorem, which states that the greatest total that cannot be acheived with two numbers m and n is $mn-m-n$, this is $9*4-9-4 = 23$.[/hide]

don't forget that m and n have to be relatively prime.  So if the dart board was 12 and 16 think in terms of 3 and 4 then multiply by 4.

does anyone have proof for the chicken mcnugget theorem  :P

[hide]
call the maximum number we can get n.  then we can somehow get n+1,n+2,n+3,n+4 (from which point we could keep adding 4 to generate the rest of the natural numbers). thus, as soon as I find a sequence of 4 numbers that I CAN get, i'm done.  playing around gives the first sequence is 24,25,26,27, so 23 is the largest number we cant get.

[/hide]

[quote="Barnacle"]does anyone have proof for the chicken mcnugget theorem  :P
[/quote]

Look here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Chicken+McNugget&t=30040[/url]

[hide]Using Chicken McNuggets you get 9*4-9-4=36-13=[b]23[/b][/hide]

[quote="myspacesucks"]technically this is an infinity problem with no solution because the answer changes with how many darts you throw, so the answer is really infinity.[/quote]
No it's not.
Can [i]you[/i] make 23? No.


vector
parameterization
calculus
calculus computations

Consider two points located at $ \bold{r}_1$ and $ \bold{r}_2$ separated by a distance $ r \equal{} |\bold{r}_1 \minus{} \bold{r}_2|$. Find a vector $ \bold{A}$ from the origin to a point on the line between $ \bold{r}_1$ and $ \bold{r}_2$ at distance $ xr$ from the point at $ \bold{r}_1$, where $ x$ is some number.

So would it be $ \bold{r}_1 \plus{} \bold{r}_2$?

(a) $ r_1\plus{}r_2$ doesn't depend on $ x$; your answer must depend on that parameter.
(b) $ r_1\plus{}r_2$ is not on the line.

Could it be $ \frac{x}{\sqrt{2}} \bold{r}_1 \plus{} \frac{x}{\sqrt{2}} \bold{r}_2$?

No. That's not on the line either, except for one particular value of $ x$. At that value, it's the wrong distance.

You're looking at this completely wrong. Start here: what is the line?

The line is $ (1\minus{}t)\bold{r}_1 \plus{} t \bold{r}_{2}$

Yes, that's a parametrization of the line. Now, how are $ t$ and $ x$ related?

No, $ t$ does not equal $ x.$ We do have that $ t$ is proportional to $ x.$ (At least, that's true for $ t>0.$) Ask yourself this: if $ t \equal{} 1,$ what would $ x$ be?

If $ t\equal{} 1$ then it would be $ \bold{r}_2$.

Actually, that $ xr$ distance fixes the proportionality constant. We do have $ x\equal{}|t|$.

So then its $ (1\minus{}|x|)\bold{r}_1 \plus{} |x|\bold{r}_2$?


Any help would be appreciated, Thank you

This is puzzles and brainteasers.....this should be moved, and the previous one.


Let $X=x^{\frac{1}{2}}+x^{-\frac{1}{2}}$, We can rewrite the given equation as $\frac{X^3-3X+2}{X^4-4X^2+5}=\frac{2}{5}$, yielding $X=\boxed{3}.\ (\because X\geq 2).$

[quote="shobber"]How did you solve the equation to get $X=3$, kunny?[/quote]

Sorry I made a typo.

kunny

[quote="shobber"]Well, I still don't know how you solve the equation. Can you give me some hints?[/quote]

Isnt that just by simplyfying and get

$2X^4-5X^3-8X^2+15X=0$ $\implies X(X-3)(2X^2+X-5)=0$


In some computer languages (such as APL), when there are no parentheses in an algebraic expression, the operations are grouped from left to right.   Thus, $ a \times b \minus{} c$ in such languages means the same as $ a(b\minus{}c)$ in ordinary algebraic notation.  If $ a \div b \minus{} c \plus{} d$ is evaluated in such a language, the result in ordinary algebraic notation would be

$ \textbf{(A)}\ \frac{a}{b} \minus{} c \plus{} d \qquad \textbf{(B)}\ \frac{a}{b} \minus{} c \minus{} d \qquad \textbf{(C)}\ \frac{d \plus{} c \minus{} b}{a} \qquad \textbf{(D)}\ \frac{a}{b \minus{} c \plus{} d} \qquad \textbf{(E)}\ \frac{a}{b\minus{}c\minus{}d}$

$ a\div{b}\minus{}c\plus{}d$ would equal $ a\div{(b\minus{}(c\plus{}d))}\equal{}\frac{a}{b\minus{}c\minus{}d}\implies{(E)}$

Sorry for bringing up a 5 year old thread, but shouldn't the question say grouped from "right to left" instead of "left to right"?


Perhaps this problem is too hard for Per-Olympiad level?   :o

Let $ a\equal{}10$ and $ b\equal{}7$.
If we switch $ a$ and $ b$ the variables cycle the other way so we may assume that $ z\le y\le x$.

$ x^{a\plus{}b}\plus{}y^{a\plus{}b}\plus{}z^{a\plus{}b}\minus{}x^{a}y^b\minus{}y^a z^b\minus{}z^ax^b\equal{}z^b(z^a\minus{}y^a)\plus{}y^b(y^a\minus{}x^a)\plus{}x^{a\plus{}b}\minus{}z^ax^b\le 0\plus{}0\plus{}1\plus{}0\equal{}1$ with equality

$ z^b(z^a\minus{}y^a)\equal{}0$, $ y^b(y^a\minus{}x^a)\equal{}0$, $ x^{a\plus{}b}\equal{}1$, $ z^ax^b\equal{}0$. So $ x\equal{}1$, $ y\equal{}z\equal{}0$. All solutions are $ (x,y,z)\equal{}(1,0,0),(0,1,0),(0,0,1)$.

Something went wrong, since it looks like $ (1,1,0)$ works. Are you sure you can get that ordering?

[quote="Altheman"]Let $ a \equal{} 10$ and $ b \equal{} 7$.
If we switch $ a$ and $ b$ the variables cycle the other way so we may assume that $ z\le y\le x$.

$ x^{a \plus{} b} \plus{} y^{a \plus{} b} \plus{} z^{a \plus{} b} \minus{} x^{a}y^b \minus{} y^a z^b \minus{} z^ax^b \equal{} z^b(z^a \minus{} y^a) \plus{} y^b(y^a \minus{} x^a) \plus{} x^{a \plus{} b} \minus{} z^ax^b\le 0 \plus{} 0 \plus{} 1 \plus{} 0 \equal{} 1$ with equality

$ z^b(z^a \minus{} y^a) \equal{} 0$, $ y^b(y^a \minus{} x^a) \equal{} 0$, $ x^{a \plus{} b} \equal{} 1$, $ z^ax^b \equal{} 0$. So $ x \equal{} 1$, $ y \equal{} z \equal{} 0$. All solutions are $ (x,y,z) \equal{} (1,0,0),(0,1,0),(0,0,1)$.[/quote]
$ x\equal{}1,z\equal{}0$ is correct. But note that if $ y(y^a\minus{}x^a)\equal{}0$, $ y\equal{}0$ or $ y\equal{}x\equal{}1$.


A board 90cm long just fits inside the bottom of a box 80cm long and 60cm wide. How wide is the board?

[quote="i_like_pie"][hide][img]http://img240.imageshack.us/img240/2835/boxis9.png[/img]

In the diagram, $c$ is the width of the board, $a$ is the distance from where the board touches the 80cm side to the corner, and $b$ is the distance from where the board touches the 60cm side to the corner.

Right away, we can say that $a^{2}+b^{2}=c^{2},$ and $(80-a)^{2}+(60-b)^{2}=90^{2}.$

The length of the diagonal is 100cm. This produces a third equation: $\left(\frac{c}{2}\right)^{2}+5^{2}=a^{2}.$

Now, the system of equations is solved...

First, we isolate $b$ in the second equation:

$(80-a)^{2}+(60-b)^{2}=90^{2}$

$a^{2}-160a+6400+(60-b)^{2}=8100$

$(60-b)^{2}=-a^{2}+160a+1700$

$60-b=\sqrt{(a+10)(170-a)}$

$b=60-\sqrt{(a+10)(170-a)}$

Then, we isolate $c^{2}$ in the third equation:

$\left(\frac{c}{2}\right)^{2}+5^{2}=a^{2}$

$\frac{c^{2}}{4}+25=a^{2}$

$c^{2}+100=4a^{2}$

$c^{2}=4a^{2}-100$

Now, we can plug $b$ and $c^{2}$ back into the first equation to find $a.$

$a^{2}+\left(60-\sqrt{(a+10)(170-a)}\right)^{2}=4a^{2}-100$

$\left(60-\sqrt{(a+10)(170-a)}\right)^{2}=3a^{2}-100$

Does anybody want to continue? :ninja:[/hide][/quote]

Would it help to draw [hide]the diagonal of the board? You would get the equation $60^{2}+(80-2a)^{2}= (90^{2}+c^{2})$, which when solved simultaneously with $\left(\frac{c}{2}\right)^{2}+5^{2}=a^{2}$ gives $\boxed{ c= \frac{15}{2}}$.[/hide]

Sorry but I'm not seeing where those two equations came from... :|

Pie, how do you know its 5 on each side of the board if $a \ne b$. How do you know it hits at exactly $\frac{c}{2}$, i'd see that were true if a = b.

[quote="i_like_pie"]
If you look at the diagram, you notice that the diffenece between the length of the board and the length of the diagonal is 10. This means it is 5 on each side of the board.

The diagonal of the box is the $\perp$ bisector of $c,$ so the side length of the traingle formed by $a,$ the diagonal of the box, and $c$ is $\frac{c}{2}.$ Therefore, we have the equation $\left(\frac{c}{2}\right)^{2}+5^{2}=a^{2}.$
[/quote]

Couldn't one, with same kind of reasoning, say: $\left(\frac{c}{2}\right)^{2}+5^{2}=b^{2}$
That would (together with $\left(\frac{c}{2}\right)^{2}+5^{2}=a^{2}$) imply $a=b$ (assuming $a > 0$ and $b > 0$)

[quote="i_like_pie"][quote="hipsishopsis"]I'm not either seeing how you get $\big(\frac{c}{2}\big)^{2}+5^{2}=a^{2}$

Why would the diagonal halve $c$ if $a \neq b$?[/quote]
If you look at the diagram, you notice that the diffenece between the length of the board and the length of the diagonal is 10. This means it is 5 on each side of the board.

The diagonal of the box is the $\perp$ bisector of $c,$ so the side length of the traingle formed by $a,$ the diagonal of the box, and $c$ is $\frac{c}{2}.$ Therefore, we have the equation $\left(\frac{c}{2}\right)^{2}+5^{2}=a^{2}.$

I hope that clears up my reasoning, but I still don't understand cincodemayo's solution. :huh:[/quote]

My second equation comes from
[hide] drawing the board's diagonal and dropping down an altitude, from the board's upper left corner, to the 80-ft side of the larger box. You get a triangle with $60$ as one side (vertical), $80$ minus two times the $a$ value in your diagram as another side (horizontal), and the board's diagonal as the hypotenuse ($\sqrt{90^{2}+c^{2}}$).[/hide]

Though i'm questioning the validity of your first equation... anyway, if you're correct on the first eqn, then my $c$ value should be correct.

[quote="i_like_pie"][quote="cincodemayo5590"]My second equation comes from
[hide] drawing the board's diagonal and dropping down an altitude, from the board's upper left corner, to the 80-ft side of the larger box. You get a triangle with $60$ as one side (vertical), $80$ minus two times the $a$ value in your diagram as another side (horizontal), and the board's diagonal as the hypotenuse ($\sqrt{90^{2}+c^{2}}$).[/hide]

Though i'm questioning the validity of your first equation... anyway, if you're correct on the first eqn, then my $c$ value should be correct.[/quote]
The equation of the right triangle formed by the diagonal and two sides is $80^{2}+60^{2}=90^{2}.$ I don't see where $2a$ and $c$ come in. Or maybe I'm just stupid...[/quote]

This triangle:


This is a stupid question, but is a point that's ON the edge of a polygon considered an interior point? Or does it have to be strictly inside the polygon?

Thanks in advance.

Generally speaking, a simple (non-intersecting) closed curve (of which a polygon is an example) divides the plane into 3 regions: the interior, the exterior and the curve itself.  Thus, the boundary of a region is not part of the interior of the region and so the edges and vertices of a polygon are not part of the interior.

The only definition of interior point I know is this: can you fit an 'open' ball (e.g, a circle not containing its boundary) of some non-zero radius completely inside the figure you're talking about?  If yes, then the point is an interior one, otherwise, the point is on the boundary or the exterior.

Using my (the usual point-set topology) definition, the answer is no.

Using the usual metric in $\mathbb{R}^2$, if we center an open ball on an edge of the polygon, part of the open ball will lie strictly outside the polygon independent of radius.

[quote="jen7"]But would a point on the boundary be considered within or strictly within the figure?[/quote]

Within--I'm not sure
Strictly within--Can not be on boundary (think of it as the < or > sign).


Prove that the following inequalities hold for all  a,b,c - sides of a triangle.-
$ 1.    \frac {a}{3a \minus{} b \plus{} c} \plus{} \frac {b}{3b \minus{} c \plus{} a} \plus{} \frac {c}{3c \minus{} a \plus{} b} \ge 1$
$ 2.    \frac {c}{3a \minus{} b \plus{} c} \plus{} \frac {a}{3b \minus{} c \plus{} a} \plus{} \frac {b}{3c \minus{} a \plus{} b} \ge 1$
the first one is by Samin Riasat (nayel?) and the second is something that i got so i dont know the second could be wrong but then give counter-example.

hello, to 1) simplifying and factoring the first inequality is equivalent to
$ \frac{\minus{}2ab^2\minus{}2ca^2\minus{}2bc^2\plus{}3abc\plus{}a^3\plus{}b^3\plus{}c^3}{(3a\minus{}b\minus{}c)(3b\minus{}c\plus{}a)(3c\plus{}b\minus{}a)}\geq0$
setting $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ in the numerator we get
$ xy^2\plus{}zx^2\plus{}yz^2\geq3xyz$
which is true by AM-GM.
Sonnhard.

The second inequality is true and nice.

By using Cauchy Schwartz, we have

$ \sum_{cyc} \frac{c}{3a\minus{}b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{(a\plus{}b\plus{}c)^2}\equal{}1$

We end our proof here.

i think this is what u have used - $ \sum_{cyclic} (3a\minus{}b\plus{}c)c \equal{} (a\plus{}b\plus{}c)^2$  am i right?(this is also how i proved it! jinx!)

[quote="geniusbliss"]i think there are easier proofs for the first(anyway i am just saying cuz i cant factorize things like that) and what about the second? is it true?[/quote]

Well, it's not that bad to bash...substituting $ a \equal{} x \plus{} y$, etc. for nonnegative $ x,y,z$, we get

$ \sum\frac {x \plus{} y}{4x \plus{} 2y}\ge1\Longleftrightarrow\sum\frac {x \plus{} y}{2x \plus{} y}\ge2\Longleftrightarrow\sum\frac x{2x \plus{} y}\le1$

$ \Longleftrightarrow\sum x(2y \plus{} z)(2z \plus{} x) \equal{} 12xyz \plus{} 4(x^2y \plus{} y^2z \plus{} z^2x) \plus{} x^2z \plus{} y^2x \plus{} z^2y$

$ \le9xyz \plus{} 4(x^2y \plus{} y^2z \plus{} z^2x) \plus{} 2(x^2z \plus{} y^2x \plus{} z^2y)$

$ \Longleftrightarrow x^2z \plus{} y^2x \plus{} z^2y\ge3xyz$.

There should be a better way but I can't find it...

ok i am getting something weird
for the first one let $ a\geb\gec$ then $ 2c \le a+b$ or we have $ \frac{c}{c - (a+b-2c)}$ i.e the denominator is less than c the numerator so one term is definitely greater than one so the whole thing is too?? :huh:  (all denominators are positive)
this looks too easy something has to be wrong

what are you doing with the first one, genius bliss?

i dont know what you mean - you can see my previous post.
anyway with the assumption i am trying to show that one of the terms is greater than or equal to one while the others are positive  so making the LHS \ge 1.


calculus
integration
trigonometry
real analysis
real analysis solved

Evaluate

\[\sum_{n=1}^\infty \frac{1}{n+1}\int^{\pi}_{0} x(\sin nx+n\pi\cos nx) dx \ (n=1,2,\cdots)\]

$\displaystyle \int_0^\pi x(\sin
nx+n\pi\cos nx)dx=\int_0^\pi x\sin nxdx+n\pi\int_0^\pi x\cos nx
dx=-\frac1n\int_0^\pi x(\cos nx)'dx+\pi\int_0^\pi x(\sin
nx)'dx=-\frac1nx\cos nx|_0^\pi+\frac1n\int_0^\pi\cos nx dx+\pi x\sin
nx|_0^\pi-\pi\int_0^\pi \sin nxdx=(-1)^{n+1}\frac{\pi}n+\frac\pi
n\cos nx|_0^\pi=(-1)^{n+1}\frac\pi n+(-1)^n\frac\pi n-\frac\pi
n=-\frac\pi n$

So our sum is $\displaystyle\sum_{n\ge1}\frac1{n+1}\int_0^\pi x(\sin
nx+n\pi\cos nx)dx=-\pi$.

I edited my post.. hope it's ok now.

Thank you for your reply, Kuba.

Regrettably,you are wrong in calculation.
Please try again!

kunny

Sorry for my delaying reply, Kuba.

You are right.

kunny