I just played a game with a 10 second time limit and I got 10.14.

If that is from a "Give Up," that happens often.

Yeah, I did this in a 5 second game. (answered in 6, haha)


real analysis
vector
Functional Analysis
advanced fields
advanced fields unsolved

Let $\mathcal H$ be a complex Hilbert space with orthonormal basis $(e_{n})_{n\ge 0}$, and define the [i]unilateral shift[/i] $U$ on $\mathcal H$ by $Ue_{n}=e_{n+1},\ \forall n\ge 0$ (extended to the entire $\mathcal H$ by linearity and continuity, of course).

Is there an operator $V$ on $\mathcal H$ such that $V^{2}=U$?

Suppose $V$ exists, and let $\varphi=Ve_{0}$. Since $V$ commutes with $V^{2}=U$, have $Ve_{n}=U^{n}\varphi$ for all $n\ge 0$. This means that $V$ agrees with the convolution operator $f\mapsto f*\varphi$ on the basis vectors, and therefore on the whole space.  Hence $\varphi*\varphi=e_{1}$. Write $\varphi$ as a sequence $(\varphi_{n})_{n\ge 0}$ and observe that $\varphi_{0}^{2}=0$, yet $\varphi_{0}\varphi_{1}=1/2$. A contradiction.


Hey everyone. I was practicing some problems, and I came across one that seems pretty simple, yet I am having difficulty proving the answer. The question asks us to show that a state of pure shear causes no volume change in linear elastic deformation. I have a very good understanding on shear stresses and strains, however, this problem seems harder then it actually is. If someone could point me in the right direction I would appreciate it greatly!


algebra
polynomial
quadratics
algebra solved

Who can solved these problems from russian 2001 MO ?
P1:x3 + ax2 + bx + c has three distinct real roots, but
(x2 + x + 2001)3 + a(x2 + x + 2001)2
+ b(x2 + x + 2001) + c has no real roots. Show that 20013
+ a 20012 + b2001 + c> 1/64.
P2: The quadratics x2 + ax + b and x2 + cx + d have real coefficients and take negative values on disjoint intervals. Show that there are real numbers h, k such that h(x2 + ax + b) + k(x2 + cx + d) > 0 for all x.

The first one is trivial so I'll only post a solution
for the second one,which is more interesting.

Case I : they both have real roots

The roots of f=x^2+ax+b are :

x1=1/2(-a-sqrt(a^2-4b)) ,x2=1/2(-a+sqrt(a^2-4b))

The roots of g=x^2+cx+d are :

y1=1/2(-c-sqrt(c^2-4d)) ,y2=1/2(-c+sqrt(c^2-4d))

Since [x1,x2] and [y1,y2] are disjoint , we may
assume that WLOG :

x2 < y1 , or equivalently :

a-c > sqrt(a^2-4b) + sqrt(c^2-4d) (1)

Now , let us find real numbers u and v such that
u*f + v*g > 0 , for all x , or equivalently :

(u+v)x^2 + (u*a+v*c)x + (u*b+v*d) > 0 , for all x

But this means that the discriminant of the latter
must be negative , or :

(u*a + v*c)^2 < 4(u*b+v*d)(u+v) , which becomes:

u^2 (a^2-4b) + v^2 (c^2-4d) + 2uv(ac-2b-2d)<0 (2)

We must prove the existence of such a pair (u,v) or
if we consider (2) as a second degree equation in
u , it must have a positive discriminant :

4*v^2 (ac-2b-2d)^2 > 4*v^2(a^2-4b)(c^2-4d) <=>
(ac-2b-2d)^2 > (a^2-4b)(c^2-4d)

But now , let's square (1) and notice that it
is equivalent to :

a^2+c^2-2ac>(a^2-4b)+(c^2-4d)+2*sqrt(a^2-4b)*sqrt(c^2-4d)
<=> (2b+2d-ac) > sqrt(a^2-4b)(c^2-4d) > 0

Now squaring , we get what we want .

Case II One of them has real roots and the other doesn't.
This case should be superfluous since I understand from
the text of the problem that they do have negative values
indeed , which implies the existence of real roots!

Assume that f doesn't have real roots but g does . Let
m1=inf f(x) , when x is real , which is > 0 and let
m2=inf g(x) , when x is real , which is < 0

We take the pair (u,1) of positive numbers such that
u*m1 + m2 > 0 , and we are done .

Case III None has real roots , then there is nothing
to prove , take u=v=1.


OK First of all, I know where most of your bias will lie, as obviously this is AoPS. So really, I'm not so much looking for opinions, as reasons why I should make my decision.

It just so happens that this year's ARML will be on May 31st, which is the day of our Section V Championships in Track. Track at my school (I go to some small rural school by Rochester, NY) is this really important thing that almost everybody (Including many of my friends) does, and our team is really good at it. More so, I'm not terrible at Track myself, being about 15 seconds off the winning time in the 3000m Steeplechase, and I'm about 5-10 seconds away from the Top Returner for Section V - yes, that means I have a chance to win. So far, I have no individual sectional patches, although it's always been my goal through High School to obtain one. In addition, if I am unable to perform at Sectionals this year, I will most likely quit the team, as our team generally trains, runs, and lives for the Section V Championships, as it is our goal to win as a team as well. In other words, yes, this Sectional Championship is a really big deal to me.

On the flip side, I've been getting better and better at Math Competitions this year, and I've worked really hard especially compared to previous years. I've studied the crap out of my book of ARML and NYSML tests from 1989-1994, and I have even more materials to study on. The improvements can even be seen through the course of my test scores, culminating in an 8 at this year's NYSML. In addition, I really like the environment at NYSML, I hope to make the Upstate New York A Team, and I'm very excited to meet new people and enjoy mathematics under the roof of Penn State. Of course, I think all of us here at AoPS knows what it means.

Now, I'm only really asking people here (and at Cogito) what you think about this. Why? Well, because I feel that unlike most of my Track people who are obviously going to be very one-sided, I know that we have the intellect to objectively look at all of our options, consider all of what we have, and make a decision.

Thank you very much for your opinion!

Well things such as this have happened to me. For example, last year, I had a math competition on the same day as a soccer match, which would decide whether we made the state championship or not. It wasn't a very big competition, so I chose the game.

If I were you, since they are obviously similar in prestige, I would think to yourself about it and then choose the one that you enjoy the most. Also, another factor that you should consider your chance of winning, as if you go to one and do horribly, you will be mad at yourself for choosing it. Also try to see if it is possible to get a little of both in and see if you can maybe make the end of the track meet.

Just my two cents.

I would actually say track.

I am a track athlete as well, and I have experienced the team spirit of track, as well the atmosphere of competition. Missing such a big meet may be a bigger blow to your track record than missing ARML may be for math. Anyways, there are plenty of other math contests.

But then again, it really depends on where your passions lie. What do you do better? Would you be happier winning your event or doing well at ARML? Which would help your long term goals more? Although I said that I would choose track, that is IF your passion for track equals your passion for math. If it were me, I would go to ARML because I am more math-oriented than track-oriented.

ARML is fun. So that's what I'd choose. Take into account, though, that I've never understood the allure of competitive sports, just as most don't understand the allure of competitive math.

Hmm.
I was in the same position a few times this year with various events.

One example: I had to choose a Quiz Bowl competition over going to the Mathcounts Chapter Competition.

This bothered me the entire day that I found out that they would be on the same day.
It tore me incredibly badly. Unfortunately, most people did not understand my position--either my passion for competative mathematics or trivial pursuits (no pun intended).

What did I go with? You may have already guessed by now--but bear with me as I explain my reasoning behind it.

You see, we had just won the first quiz bowl competition of the season, and the other teams were beginning to eye us as a team that needed to...disappear. We posed a threat. We would have a good chance of winning. Our team was full of genius kids, three of the four who had photographic/near-photographic memories, and had been exposed to the right trivial stuff. (The fourth person, me, has a fairly good memory. Just not as good as the rest. :wink:)

Therefore, I wanted to go. I was good at Quiz Bowl. It greatly appealed to me. Quiz Bowl was somthing new and wonderful and adrenaline pumping. It was...fun. For lack of a better adjective.

Then, mathematics. It flows in my blood. (I'm serious. Cut me open, and the blood cells will flow in a Fibonacci sequence out of my veins. :P) But lately, with a discouraging MC coach who hated me (but couldn't damper my love for mathematics if she tried...) I wasn't quite so sure. However, this year, I was on the actual team (of four people) and chances were fairly good that I would place highly. I was getting better at mathematics all the time, and I wanted to go just as badly.

Now the dilemma. I loved both equally. However--this was the one detail that made my choice what it was. If I didn't go to the MC competition, there would be no more math competitions after that for me, and I couldn't go on to state. Then, if I didn't go to Quiz Bowl, I would still be able to attend the next one (and currently will on April 22nd).

Then I asked a "peer" of mine. We didn't really know each other, but I did need his opinion. He pointed out that he would be going to the Chapter competition for the same reason I was--to make it to the next level of competition.

What's the time frame like? Could you make both? ou might have to leave one early, but it'd be worth it, wouldn't it?

However, in your case, I believe that you need to do what...well, what everyone has said. Which would you feel help you more? Will there another math competition? Obviously, there might be another track meet, but not another one like it, judging from your description.

Mmm...I guess it really just comes down--Which do you love more? I know that'll be hard to answer, but if you really chew on it, it just might come to you.

I guess first I'll clarify a few things:

1) I'm a Senior this year. Section V Championships will be (Unless I win, which I cannot if I do not attend) the last high school sporting event *ever* for me. Similarly, ARML 2008 will be the last high school math competition I *ever* participate in.

2) I have insane passion for both. I've participated in Track since 7th grade, I've competed in math seriously starting 10th grade, but I've done small competitions since 7th as well.

To be fair, ARML is a national competition, while my Track meet is still only in our section. Also, although I've built up far longer work for Track, this year alone I've put a lot of effort into mathematics, as well as track. I guess a large part of this also has to deal with community: I've invested a lot into Track and Field over the years, and I can't stand to not attend Sectionals with my friends. Similarly, however, I find ARML fun and I would love the last chance to meet up with more mathematics folk.

Anyway, I can't thank you enough for your views.

Kevin

WELL.. It's your choice. Whatever you feel is more important to you is what you should go with. :D

If it's your last chance at ARML or the Track thing, then choose whichever one you have the last chance with.

I had a MC competition last year the same day as regionals for swimming. I had a really good chance of making the next level, but my parents would never allow me to skip MC, so I had to go to MC.
So it's your (or your parents) choice! :D

I personally [i]don't[/i] think that it should be your parents' choice. Mostly because you know yourself best and not them. Even though they DO know you well. :)

Personally, if I were you, I would go to the track.

[u]Reasons to support:[/u]
- School pride
- Bonding with school community, maybe
- You have a much better chance of doing comparatively well, judging by your track times and your most recent AMC/AIME scores
- Crossing the finish line is more exciting
- You've already been in plenty math competitions before -- AMC's are nationwide... and if you're talking about meeting people, then it's not really nationwide because there are 4 sites, I think

Of course, I'm a bit biased -- when I moved to the school I'm at now, I had to give up crew. Crew is the one thing that I really miss here, and I would give up ARML in a heartbeat to go row a race.

Wait, just asking for clarification, if you win the track, what effects will there be? One way to help you judge is to compare what benefit you get from each (including simply "i like to do it").

[quote="buzzer11"]I personally [i]don't[/i] think that it should be your parents' choice.[/quote]

I meant if his parents were leaning against one or the other. let them have at least SOME choice. :)

for arml you're just gonna get disappointed when you get steamrolled by exeter

so i mean i'd say go for track


We will be hosting an experimental Math Jam Monday, September 29, from 7:30 PM - 8:30 PM ET (4:30 PM - 5:30 PM PT).  The primary purpose of this Math Jam will be to test our new classroom.  We will be doing MATHCOUNTS Problems, and having a MATHCOUNTS Countdown Round.

The only way to access the classroom is by [url=http://www.artofproblemsolving.com/NewClassroom/AoPS_TestClassroom.php]this link[/url].  You will not be able to access this experimental Math Jam by clicking the Classroom button, as you usually do for regular Math Jams.  I look forward to seeing you Monday night, and thank you in advance for helping us debug our new classroom.

What features are included in the new classroom?
(Would it be asymptote? :P )

Yes you can use the Asymptote forum code in the new classroom.

Flash.

There will be a few new capabilities for moderators, and eventually there will be a whiteboard.  The one that I'm most looking forward to using is the ability to "sticky" a post to the top of the classroom.

When will this be applied to the classroom? And will you be able to sticky stuff now? That's cool since now we don't have to have a new window with just the problem. :D

After this Math Jam, if there are no problems, will the other currently running classes be converted?

The sticky stuff is implemented in the new classroom now, but needs some tweaking.  I would hope to switch entirely over to the new classroom sometime in mid-October.

15 mins to check out the supercoolawesomefunnyfunfunsawesomecool1337superllamabeast classroom!

what the? it's 6:25 Central Time and it's closed.  :huh:

Why can't I get in? 7:22 EST (4:22 PST)

EDIT: [url=Link]http://www.artofproblemsolving.com/NewClassroom/ChatBlazer8.php?r=MathJam&id=1010[/url]

now it's saying

"The Virtual Classroom is loading automatically; it may take up to a few minutes to connect. If you are new to the Virtual Classroom, please review our classroom procedures.

If you reach this window but the classroom will not load, please try disabling any pop-up blockers that you may have installed. Consult your pop-up blocker's documentation for information on how to disable it. Click here for instructions on how to enable pop-ups with Internet Explorer."

and it says "done, but with errors on page" in the bottom left corner of the internet explorer page. it's taking a pretty long time...(thumbs fingers on table)  :huh:

mine isn't working either. it says on the bottom error on page. I refreshed,  it says done on bottom. Taking a really long time to load.

Can you tell us which browser and version, OS, and flash version you are using. If we can't figure it out I may ask you to help me find the problem some other time.

I had some major problems with reading the problem and going back from away to online...`

Thank you to everyone for helping us test the room!  We'll do it again in a couple weeks or so.

It looks like there's a lot of bugs:  Is there a new projected release date?  Will it be ready to be used in some of the fall classes?

For those of you who had trouble accessing the classroom, please post your operating system and browser.

Thanks again to everyone for helping us with this test.  We uncovered a lot of little issues to address before we go prime time!  We expect to get them ironed out in the next couple weeks, and we'll probably implement this new classroom by the end of October.

No, in the chat he said that it wouldn't be.


Another problem which I often meet: math conventions in an article. For example: I want to express that the image of the function f is included in the set M.
I typed:

[quote] $ Imf\subset M$[/quote]

But the Im appears in italics. How can I make it normal text? (the normal font command didn't work)?

Also, where could I find a site with math conventions? Even for that one, I'm not sure if Im should be normal text or not.
What about if I use braces (....) and inside them I have huge fractions; do I have to increase the size of the ( and ) ?

The same if I have f(....), f a function of something, and that something is a big expression; should the size of ( and ) be adjusted?

May I ask on this forum questions about such conventions too?

Thanks,
Jean

$ \text{Im }f\subset M$

\text{Im }f\subset M

--

Does that help?

However in LaTeX, the normal code for Im(z) is
[code]\Im(z)[/code]

Which gives $ \Im(z)$, but Hetidemek's way is a good example of how you would post text.

[quote="jeanvaljean"]What about if I use braces (....) and inside them I have huge fractions; do I have to increase the size of the ( and ) ?

The same if I have f(....), f a function of something, and that something is a big expression; should the size of ( and ) be adjusted? [/quote]There are 3 ways of using large versions of () and other things like [ ], { } etc
1. Use \left( and \right) to surround the mathematics. For example
(\frac{x+1}{x+2}) gives $ (\frac{x\plus{}1}{x\plus{}2})$ but \left(\frac{x+1}{x+2}\right) gives $ \left(\frac{x\plus{}1}{x\plus{}2}\right)$ The ( and ) will automatically adjust to the right size. But \left and \right must be used inside the same formula

2. Sometimes it is necessary to use a fixed larger size of ( ) where perhaps there are many formulae. Then use \big(, \Big(, \bigg(, \Bigg( which give $ \big(\Big(\bigg(\Bigg($

3.Finally, pmatrix can often help in complicated cases such as at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166865
\begin{pmatrix}
abcde \\
fghij
\end{pmatrix}
gives $ \begin{pmatrix}abcde\\ fghij\end{pmatrix}$ and the ( ) adjust themselves.

There's more on this topic in [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] which is an essential read.

Conventions vary from one country to another and from one subject to another so follow the ones in books and papers on your topics.

[quote="Hetidemek"]\text{Im }f\subset M[/quote]
Personally I would use $ \text{Im}(f)$ but again conventions differ and indeed some people do use $ \Im(f)$. If you need this a lot then best to use
\newcommand{\im}{\text{Im }} or \newcommand{\im}{\text{Im}}
which means you can use \im f or \im(f)


In a group of nine people there are no
four every two of which know each other.
Prove that the group can be partitioned
into four groups such that the people in
every group do not know each other.

if i am not wrong, it can be solved by Turan's Theorem (though with some adjustment), but then in that case it might be a bit too trivial.

See if there is any objection for this argument, I am not able to find one.I feel I can a prove a stronger statement, I am not sure : they can be partitioned into 3 groups.

If there is a r-partition ( r=1,2..,8) , we can always create a (r+1)-partition (by simply dividing a group as we are assured by pigeon hole principle that we have a partition with at least 2 people).So, we can create the required 4th partition.

If there was set of k people who knew each other, then we need at least k partitions as they have to go to different partitions.k is 4 here.

I now prove if we need at least 4 partitions, then it implies there is a set of 4 people with any pair among them knowing each other.Impossibility of accommodating the people of 4th partition into other three means to say there are people numbered say
1,2,3 in first three partitions such that they are all know an element say 4 that is currently present in 4th partition.

By same argument for first three partitions, we can say that both 1 and 2 know 3.Like this we can go on to prove , that {1,2,3,4} are all related to each other.
:?:  :wink:

Is it not a graph $K_{3,3,3}$ counterexample?

[quote=blawho12]If it is allowed to make one of the partition groups empty, then it is not a counter-example.[/quote]
Or just split one of the three groups to get a total of four.

[quote=srikanth]I feel I can a prove a stronger statement, I am not sure : they can be partitioned into 3 groups.[/quote]
In fact, this is not the case. Essentially the problem reduces to showing that a simple graph with $9$ vertices and no sub-$K_4$ has chromatic number at most $4$. If there were always a partition into $3$ groups, then the last condition could be strengthened to "a chromatic number at most $3$." However, the graph consisting of a pentagonal pyramid disjointly unified with three copies of the trivial graph has $9$ vertices, no sub-$K_4$, and a chromatic number of $4$, so said claim is false.


I heard from a friend that inequalities will no longer be on the USAMO.  Is this true???

(I am told that this information originated from Bubala.)

Bubala was probably exaggerating. It also probably won't happen, because inequalities can still be tricky and impossible to brute force. Take for example that min and max one on a USAMO in the 90's (probably 98 or so?).

But a lot of inequalities can be Jensen'd into something reasonable, then dumbassed :wink: .

I would ignore any talk of nonexistence of inequalities - it was conclusively decided at MOP last year that there would never be inequalities on the IMO EVER AGAIN and then IMO05/3, IMO06/3 were both inequalities...

plus bubala would not have any idea

The thing you might want to keep in mind is that it is perhaps more unlikely to see inequalities that you can easily dumbass (i.e. multiply out completely and use weighted AM-GM), although IMO05/3 is pretty recent...


find the minimum of
max{a+b+c,b+c+d,c+d+e,d+e+f,e+f+g}
with a,b,c,d,e,f,g >= 0,a+b+c+d+e+f+g=1


Find all integer solutions $(x,y)$ to the equation $\displaystyle \frac{x+y}{x^2-xy+y^2}=\frac{3}{7}$.

Let $y\lex$. There are not solutions with $y\le 0$. Expression had maximal means when $y=x(\sqrt 3 -1)$, it means
$\frac{7(3+2\sqrt 3)}{9x} <3/7$ if $x\ge 6$. Therefore $1\le y\le x \le 5$.It give y=4 x=5 (or x=4,y=5).

notice that $x+y>0$ or else LHS will become negative . The equation can also be written as

$7(x+y)^2=3(x^3+y^3)$

By Holder Ineq , $x^3+y^3\geq \frac{1}{4}(x+y)^3$ so $x+y\leq \frac{28}{3}$ so $0<x+y\leq 9$

Now by letting $s=x+y$ , $p=xy$ we obtain

$7s=3s^2-9p$ so $p=\frac{s(3s-7)}{9}$

We can see that $s$ must be multiple of $9$ and the only answer is when $s=9$ so $p=20$ . And this gives us

$(x,y)=(4,5),(5,4)$

For integers $x,y,$ we have $x^2-xy+y^2=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2\geq 0.$
However since $x^2-xy+y^2$ is a denominator, we have $x^2-xy+y^2\neq 0$, yielding $x^2-xy+y^2>0.$
Thus we can set $x+y=3k,\ x^2-xy+y^2=7k$ for some positive integers $k.$ Rewrite the second equation as $(x+y)^2-3xy=7k\Longleftrightarrow xy=\frac{28k-9k^2}{3}.$ Now $(x-y)^2=(x+y)^2-4xy=\frac{1}{3}k(28-9k)\geq 0,$ the natural numbers $k=1,2,3.$ Plugg these values into the  equations, the pair of integers $(x+y,xy)$ is limited to $(x+y,xy)=(9,20).$ thus the answer is $(x,y)=(4,5),\ (5,4).$

We have that $7(x+y)=3x^2+3y^2-3xy+6xy-6xy$.From there we obtain : $7(x+y)=3(x+y)^2-9xy$.Because $(x+y,xy)=1$ we have that $x+y|9$.Now we question $6$ cases and get solutions: $x=4 y=5$ and $x=5 y=4$.


geometry
circumcircle
trigonometry
geometric transformation
reflection
trapezoid
symmetry

Given that points $ D,E$ lie on the sidelines $ AB,BC$ of triangle $ ABC$, respectively, point $ P$ is in interior of triangle $ ABC$ such that $ PE \equal{} PC$ and $ \bigtriangleup DEP\sim \bigtriangleup PCA.$ Prove that $ BP$ is tangent of the circumcircle of triangle $ PAD.$

As many problems with special configuration, the analysis of construction is of paramount importance for this problem as well.
Suppose we have D given on (AB); draw DD' || BC (D' on AC) and construct the symmetrical of AC about the perpendicular bisector of DD', it will intersect BC at E. P will be on this perpendicular bisector.

From the similar triangles DEP and PCA we get DP/PA = DE/PC = PE/AC or (DP/PA)^2 =(DE/PC).(PE/AC), but PE = PC and it remains (DP/PA)^2 = DE/AC ( 1 ).

Applying the [b] sine theorem [/b] to triangles BDE and ABC with m(<BED) = 180 - m(<ACB) ( from construction ) we get BD/AB = DE/AC ( 2 ), but the tangent at P to the circle (ADP) will intersect AD at a point T with property TD/TA = (DP/AP)^2 ( 3 ). From (1), (2) and (3) we get T=B.

Best regards,
sunken rock

To avoid using the sine theorem to get BD/AB = DE/AC:

Name R the intersection of AC and DE, apply Menelaos theorem to triangle ADR with the transversal BEC, see that ER = CR.

Best regards,
sunken rock

(Sorry if the post is old, but I think I've found another solution with no computation?)

Reflect A over the perpendicular bisector of EC, call this point F; note that EP must be tangent to the circumcircle of DPF by angle condition.

Invert the diagram about P:  We get that still PA' = PF' and PC' = PE'.  The point D' is such that PF'D'E' is an isoceles trapezoid with PF' = E'D'.

It suffices to show that if B' is the point such that PB'D'A' is an isoceles trapezoid (B'D' = PA'), then C'EB'P lie on a circle, so consider this point B'.  Let O be the intersection of the perpendicular bisectors of PB' and PE'.  We have OA' = OD' = OF' so O is also on the perpendicular bisector of E'C' and so we have found O is the circumcenter of E'B'PC' and we are done.

Hello. I think that this is a nice problem. I had solved it in the way of sunken rock.

First off, one way to "complete the diagram" in some sense is to add the intersection of $ DE$ and $ AC$ (call it it $ X$).  Now you have two isosceles triangles. If we decide to start with drawing $ XCE$, $ PCE$, then it is trivial to find $ A,D$ and then construct $ B$ as the intersection of $ CE$ and $ DA$.

If we look at allen's solution, we note that his point "F" lies on $ XE$ as I've defined it. Furthermore, another way to motivate the use of inversion is that if we invert about $ P$, the similar triangles that are so difficult become congruent!  This important observation takes the two odd assumptions of the triangle similarity and the equal lengths and makes it more reasonable to manage.

I also am curious, how did you come up for the construction of point $ F$? It is very nice in the fact that you have two isosceles trapezoids that share a diagonal the diagonal $ PD$. Also the construction of $ O$ is nice...

My motivation for F was basically what you said - since the angles DEC and ACE are equal, it's natural to extend DE and AC to meet at X, then it just makes a lot of sense to reflect A across the perpendicular bisector of EC, since you really don't have to worry about the original A that much anymore.  Then well, inversion looks too nice with the tangent circles to lines.

The construction of O was mostly from all that reflective symmetry of isosceles triangles and trapezoids.

My solution [b]:[/b]

Let $ A' $ be the reflection of $ A $ in the perpendicular bisector of $ CE $ .
Let $ R $ be the radius of $ \odot (ADA') $ and $ O $ be the circumcenter of $ \triangle ADA' $ .

From $ \angle CEA'=\angle ACE=\angle CED $ $ \Longrightarrow $ $ A', D, E $ are collinear .
From $ \triangle DEP \sim \triangle PCA $ $ \Longrightarrow $ $ EP^2=EP \cdot CP=ED \cdot CA $ .
i.e. $ EP^2=EO^2-R^2 $

From $ OP \perp BE $ $\Longrightarrow $ $ BO^2-BP^2=EO^2-EP^2=R^2 $ ,
so $ BD \cdot BA=BO^2-R^2=BP^2 $ . i.e. $ BP $ is tangent to $ \odot (PAD) $ at $ P $

Q.E.D

Let the tangent to $\odot (PAD)$ at $P$ meet $AD$ at $B'.$ Then from $\triangle B'PD \sim \triangle B'AP$, we obtain \[\frac{PD^2}{PA^2} = \frac{B'P^2}{B'A^2} = \frac{B'D \cdot B'A}{B'A^2} = \frac{B'D}{B'A}.\]

Meanwhile, note that $\angle PED = \angle PCA$ and $\angle PEC = \angle PCE.$ Consequently, $\angle BED = 180^{\circ} - \angle DEC = 180^{\circ} - \angle ACB.$ Now let $F \ne C$ lie on $BC$ with $AF = AC.$ Then it follows that $DE \parallel AF$, so from $\triangle BDE \sim \triangle BAF$, we get \[\frac{BD}{BA} = \frac{DE}{AF} = \frac{DE}{AC}.\] From $\triangle DEP \sim \triangle PCA$ we have \[\frac{DE}{AC} = \frac{DE / PE}{AC / PC} = \frac{DE^2}{PC^2} = \frac{PD^2}{PA^2}.\] Therefore, $\tfrac{BD}{BA} = \tfrac{B'D}{B'A}$, and the desired result follows because both $B$ and $B'$ lie in the exterior of $\overline{AD}.$


function
real analysis
vector
search
algebra
functional equation
real analysis unsolved

Out of curiosity, are there any non-measurable functions besides the trivial ones (by this, I mean characteristic function of a non-measurable set)?  I'm hoping for an example in $ \mathbb{R}$ but $ \mathbb{R}^{n}$ would suffice as well.  I'm not familiar with abstract measure theory so I'd like to keep the spaces relatively simple.

There are lots of trivial examples that are a trivial consequence of the existence of non-measurable sets, such as $ I_X \plus{} 2I_Y$ for any disjoint sets $ X, Y$ such that one of $ X, Y$ is non-measurable.  Indicator functions are just the simplest example; moreover, this is trivial for arbitrary measures since you can construct examples of pairs of measurable spaces such that there aren't any nontrivial measurable functions.

What I think you really wanted to ask is something along the lines of, "is there a natural way to construct a non-measurable function on $ \mathbb{R}$ with the Lebesgue measure without first constructing a non-measurable set?"

I assume indicator function and characteristic function are interchangeable.

Yes, that question seems appropriate?  Is there an answer?

Any solution in $ \mathbb{R}$ of the functional equation $ f((x\plus{}y)\equal{}f(x)\plus{}f(y)$ that is not of the form $ f(x)\equal{}ax$ is nonmeasurable, and its graph is dense in the plane. If $ \mathfrak{c}\equal{}2^{\aleph_0}$ is the cardinality of $ \mathbb{R},$ then there are $ 2^{\mathfrak{c}}$ such functions.

Of course, the "construction" of such a thing contains an appeal to some equivalent of the Axiom of Choice. You're not going to escape from that.

"By axiom of choice," take a vector-space basis for $ {\bf R}$ over $ \bf Q$.  Then define $ f$ arbitrarily on each basis vector, and extend by linearity.  This is, obviously, impossible to write down in any explicit sense, but this should be essentially true of any unmeasurable anything.

[quote="JBL"]Then define $ f$ arbitrarily on each basis vector, and extend by linearity.[/quote]
Since there are $ \mathfrak{c}$ elements in such a (Hamel) basis, that's where I got $ 2^{\mathfrak{c}}$ as the count of how many such functions there are. Of course $ \mathfrak{c}$ of those choices do result in ordinary linear functions, but then $ 2^{\mathfrak{c}} \minus{} \mathfrak{c} \equal{} 2^{\mathfrak{c}}.$

Okay, thank you.  Now to jump on something JBL wrote, is it possible to find a non-measurable (non-trivial) function that can be written down?

I'll leave the story of the full interplay of the axiom of choice, the continuum hypothesis, and the existence of non-measurable sets to those who know more logic than I do. Suffice it to say that if you could "write down" any nonmeasurable function, you would instantly have a way to "write down" a nonmeasurable set. You're not getting away with it that cheaply.

Here's another idea for a nonmeasurable function. Take the usual construction of a nonmeasurable set $ E$. Then for any $ x,$ let $ f(x)$ be the unique rational number such that $ x \minus{} f(x)\in E.$

[quote="JRav"]I am willing to accept the axiom of choice, so could you elaborate on such a construction?  What would be an example of a solution to that functional equation that is non-measurable?[/quote]

see e.g [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1844043861&t=161327]this post[/url] for a proof that measurable additive functions are linear, and hence that nonlinear additive functions are nonmeasurable.

i probably know less logic than kent does, but modulo some set-theoretic details, solovay showed that the statement 'every subset of \mathbb{R} is lebesgue measurable' is consistent with ZF without the axiom of choice.  so it is basically impossible to prove the existence of lebesgue nonmeasurable sets without using the axiom of choice.


Columbia
ratio
limit
arithmetic sequence
number theory

I don't get it...I'm basically asking for the general term,I came across this in Mathematical Induction........

The general term of what?  You need to specify your question.

Perhaps it may help, if you look at it as:

${(\frac 1 2 ) (\frac 1 1 - \frac 1 3 + \frac 1 3- \frac 1 5 + \frac 1 5- \frac 1 7 \dots ) }$

Then if you want to find the sum, only first and last terms remain  .. the limit tends to $\frac 1 2$  etc ..

(Sum does not diverge)

[quote="joml88"]That sum diverges.[/quote]

Indeed, the sum is obviously dominated by $\sum \frac{1}{n^2}$, so it can't diverge.  (it's pretty clear that he was using $.$ to denote multiplication).

[quote="joml88"]It was pretty clear to [b]me[/b] that he was asking for the sum of the reciprocals of the arithmetic sequence $\{a_n\}$ where $a_n=1.3+2.2(n-1).$  Now I understand what he was asking but I still don't understand why people use a decimal point for multiplication.  It can't be [i]that[/i] hard to type \cdot three times.

Like I said, he should have made his question more clear.[/quote]

Not to criticize you, but to make you aware.
It's not that people are lazy on the keyboard. I think it's a matter of personal preference where to place the dot, even though it is ambiguous, if it's placed like a decimal point. My number theory professor at Columbia explained that he sometimes wrote the multiplication dot like a decimal point. Luckily, he usually multiplied more than two numbers at a time, so it wasn't much of an issue.
It's not so uncommon, but I don't think it's accepted notation.

I am really sorry joml88 for all the confusion.PLease don't get angry on me,I'm just a kid and that's the way i found the question in my textbook and I have seen many questions posted using "." as multiplication over here....Sorry once again.

I guess the answer is something like: $\frac{x}{2x+1}$....is it?

$\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}$

$\sum_{n=1}^{\infty} \frac{1}{4n^2-1}$

and this converges by the ratio test...but i do not know how to evaluate this,

[quote="joml88"]It was pretty clear to [b]me[/b] that he was asking for the sum of the reciprocals of the arithmetic sequence $\{a_n\}$ where $a_n=1.3+2.2(n-1).$  Now I understand what he was asking but I still don't understand why people use a decimal point for multiplication.  It can't be [i]that[/i] hard to type \cdot three times.

Like I said, he should have made his question more clear.[/quote]

Without knowing whether it was laziness or convention, cultural standards are such that $.$ can be multiplication, and $,$ can replace the 'western' decimal point.  Of course, if he had written it out as altheman did, it would have been more clear.


The probability that Shelley will get a hit in any at-bat is 0.3.  What is the probability that she will get exactly one hit in her next two at-bats?  Express your answer as a decimal to the nearest hundredth.

[hide]We subtract the prob. of her NOT getting a hit in the next 2 and the prob. of her getting 2 in the next 2.

Case 1: $\frac{7}{10}^2$
Case 2: $\frac{3}{10}^2$
So we have $1-\frac{49+9}{10}$, so we get $\frac{42}{100}=\frac{21}{50}=.42$ probability.[/hide]

Billy

Billy, I think your right, but you have to express you answer in decimal form to the nearest hundredth.

Billy, your solutions are typically nice, but you could make them even nicer by learning TeX.  Given the amount of time you spend writing solutions on the Message Board, I think you will thank yourself after learning it.

It's not as hard to get the basics down as you might think.  The LaTeX tutorial is a nice introduction and you can always click on other posters' TeX to see how to made things look the way they did.


search
function
superior algebra
superior algebra unsolved

Please show that $ (x^2\plus{}x)^{2^n}\plus{}1$ is ireductible, $ n \in N$

search function hehe...
http://www.mathlinks.ro/viewtopic.php?search_id=1761195266&t=80920 contains many previous links :wink:


function
algebra
polynomial
search
combinatorics proposed
combinatorics

Define: $ f(n,m)$ is the number of all possible ways to give "m" candies to "n" children
but the maximum of candies to be given to each children is 2
for example $ f(3,2) \equal{} 6$  (the candies are all the same)
find the value of
$ f(2008,1) \plus{} f(2008,2) \plus{} ... \plus{} f(2008,1004)$

Sorry, but can you explain why $ f(3,2)\equal{}6$ Thanks in advance.

$ f(3,2)$  for A,B,C = children
A B C
1 1  0
1 0  1
0 1  1
0 0  2
0 2  0
2 0  0
the maximum number of the candies to be given to each of children is 2 (srry if my English is bad)

There may be an easier way, but the standard way for a problem like this would be to use generating functions:
Write down the generating function for $ f(2008,m)$, then find a polynomial that has the sum you considered as one of its coefficients, then use algebraic manipulations to find this coefficient.

vanhoadh is right, but it doesn't appear to lead to a reasonable formula for the final answer.  What seems more likely to me is that the sum was intended to go from $ f(2008,1)$ to $ f(2008, 5006)$, or in general to be $ \sum_{m \equal{} 1}^{2n} f(n, m)$.  This leads to a simple solution and a nice answer.

[hide="The answer to the question I just posed is"]$ 3^{n} \minus{} 1$.[/hide]

We only need to continue to 4016. . .

Edit: see #1 at http://www.math.ca/Competitions/CMO/reports/CMOReport2006.pdf for a similar problem which also curiously had a similar error

I'm not sure what mistake it was that caused me to conclude $ 2 \cdot 2008 \equal{} 5006$  :oops:

I knew I had seen this problem before, but the changed numbers made it impossible to search for  :mad:

thank you JBL and mathophile593  :lol: [/quote]


A group of bike riders took a 4.0-hour trip. During the first 3.0 hours, they traveled a total of 50. kilometers, but during the last hour they traveled only 10. kilometers. What was the group’s average speed for the entire trip?

(1) 15 km/hr       (3) 40. km/hr
(2) 30. km/hr      (4) 60. km/hr

[hide]Average speed=$\frac{Total distance}{Total Time}$. So the total distance is 60km. The total time is 4 hr. So its 15km/h or 1.[/hide]

[quote="Interval"]Thanks for the cool equation.

Thanks![/quote]
You can derive that equation from distance = rate * time, which is used a lot


Every digit of a 5-digit number is either 1 or a prime number. Not only that, but any number created by any 2, 3, or 4 consecutive digits of this number are prime numbers also. Find this number, and check if it is a prime number or not.

[hide="Hint"]
Obviously, there's not gonna be any 2s or 5s, except possibly in the first digit.
[/hide]

[quote="worthawholebean"][hide="Answer"]31373[/hide][/quote]''

[hide]To answer the second part of the question, this number isn't prime, it's 137*229...[/hide]

In my perspective, the problem just asks to tell [i]if[/i] the number is prime, not check to see if it is actually prime.


Prove that in any triangle:
$ 6(sin{\frac{A}{2}}\plus{}sin{\frac{B}{2}}\plus{}sin{\frac{B}{2}}) \le cosA\plus{}cosB\plus{}cosC\plus{}\frac{15}{2}$

[quote="NguyenDungTN"]Prove that in any triangle:
$ 6(sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {B}{2}}) \le cosA \plus{} cosB \plus{} cosC \plus{} \frac {15}{2}$[/quote]

$ \cos A\plus{}\cos B\plus{}\cos C\equal{}1\plus{}4\sin \frac A2\sin \frac B2\sin \frac C2\equal{}3\minus{}2(\sin^2 \frac A2\plus{}\sin^2 \frac B2\plus{}\sin^2 \frac C2)$

$ 6(sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {B}{2}}) \le cosA \plus{} cosB \plus{} cosC \plus{} \frac {15}{2}$
$ \Leftrightarrow 6(sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {B}{2}}) \plus{} 2(\sin^2 \frac A2\plus{}\sin^2 \frac B2\plus{}\sin^2 \frac C2) \le \frac {21}{2}$

I could'nt seem to get any further beyond this..... :(

But very obviously, Jensen inequality gives us the required answer. :)

[quote="hell_ever"][quote="NguyenDungTN"]Prove that in any triangle:
$ 6(sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {B}{2}}) \le cosA \plus{} cosB \plus{} cosC \plus{} \frac {15}{2}$[/quote]

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} 4\sin \frac A2\sin \frac B2\sin \frac C2 \equal{} 3 \minus{} 2(\sin^2 \frac A2 \plus{} \sin^2 \frac B2 \plus{} \sin^2 \frac C2)$

$ 6(sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {B}{2}}) \le cosA \plus{} cosB \plus{} cosC \plus{} \frac {15}{2}$
$ \Leftrightarrow 6(sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {B}{2}}) \plus{} 2(\sin^2 \frac A2 \plus{} \sin^2 \frac B2 \plus{} \sin^2 \frac C2) \le \frac {21}{2}$

I could'nt seem to get any further beyond this..... :([/quote]
I think you can't use Jensen inequality. My solution is very long, about 1 page; I will post my solution if you want.

since $ \sin \frac {A}{2}$ is concave we have
$ \sum \sin \frac {A}{2} \leq \frac {3}{2}$ by jensen's
so continuing from hell_ever left
we have to prove $ 21 \minus{} 12\sum \sin \frac {A}{2} \minus{} 4\sum \sin^{2} \frac {A}{2} \geq 0$
observe that
$ 0 \leq 5\sum (1 \minus{} 2 \sin \frac {A}{2}) \leq \sum (7 \plus{} 2 \sin \frac {A}{2})(1 \minus{} 2 \sin \frac {A}{2}) \equal{}21 \minus{} 12\sum \sin \frac {A}{2}\minus{} 4\sum \sin^{2} \frac {A}{2}$

[quote="pardesi"]since $ \sin \frac {A}{2}$ is concave we have
$ \sum \sin \frac {A}{2} \leq \frac {3}{2}$ by jensen's
so continuing from hell_ever left
we have to prove $ 21 \minus{} 12\sum \sin \frac {A}{2} \minus{} 4\sum \sin^{2} \frac {A}{2} \geq 0$
observe that
$ 0 \leq 5\sum (1 \minus{} 2 \sin \frac {A}{2}) \leq \sum (7 \plus{} 2 \sin \frac {A}{2})(1 \minus{} 2 \sin \frac {A}{2}) \equal{} 21 \minus{} 12\sum \sin \frac {A}{2} \minus{} 4\sum \sin^{2} \frac {A}{2}$[/quote]

Beautiful pardesi


Find a closed form for the expression:

$ \displaystyle\sum_{k \ge 0} \binom{n}{3k}$

Hint: [hide]Generating functions and roots of unity.[/hide]

I assigned this as a practice problem [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490#comments]here[/url] as part of a general framework of discussing Newton transforms.  See also the matrix approach [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url].  In the second post I raise the question whether this summation can be put in closed form by a purely combinatorial argument; any thoughts?

Wow, that's fantastic. I've never thought about this problem in those contexts.


inequalities
search
inequalities theorems

[quote="mps"]I 'm just only a newbie. A month ago, I knew Schur's Theorem, many people say that
Schur's is very useful . I can't apply that very much [b]so please give me the inqualities which can use Schur' to done it[/b]. And [b]applying of Schur's theorem[/b].

                                                                                Thank you very much ,mps.[/quote]

just search for it......
 :)


A thin, uniform rod has mass $ m$ and length $ L$.  Let the acceleration due to gravity be $ g$. Let the rotational inertia of the rod about its center be  $ md^2$.

31.   Find the ratio L / d.  (3 pts)

The rod is suspended from a point a distance $ kd$ from the center, and undergoes small oscillations with an angular frequency $ \beta\sqrt\frac gd$.

32. Find an expression for  β  in terms of k.

33. Find the maximum value of  β .

Ok, so #31 is pretty easy, but I'm stuck on #32. I've followed the answer solution up the the point where it started talking about differential equations. Is there any way to solve this with minimal calculus?

(If you want the answer choices, ask)

Do you know how the position equation for a spring obeying Hooke's Law is derived? If so, you should be able to make a direct comparison between the that derivation and this problem.

Umm... what position equation is this?

On a side note, do you think the 2010 problems will be of this difficulty, since it only has 25 versus 38 problems?

For spring motion, we have $ \minus{}kx \equal{} ma \equal{} m \frac{d^2 x}{dt^2}$. Rearranging, this becomes
\[ \minus{}\frac{k}{m} x \equal{} \frac{d^2 x}{dt^2}.\]
When we solve this differential equation, we find that $ \omega^2 \equal{} \frac{k}{m}$. So, in cases where you have something similar to that differential equation above, just follow the same pattern.

You may need something called the parallel axis theorem (perhaps you know that from AP Physics), to find the rotational inertia of the rod about that axis.

$ I$(around some arbitrary axis) $ \equal{} I$(around some axis for which you know the formula) $ \plus{} md^2$ (where $ d$ is the distance from the known axis to the desired axis)

As this is angular S.H.M you'd have the equation as $ \frac{d^2\theta}{dt^2}\equal{}\minus{}{\omega}^2\theta$ $ \implies$ $ \alpha\equal{}\minus{}{\omega}^2\theta$ $ \implies$ $ \omega\equal{}\sqrt{\mid \frac{\alpha}{\theta} \mid}$
SO the problems comes down to finding relation between $ \alpha$ and $ \theta$

Did anyone get $ \beta \equal{} \sqrt{\frac{1}{1\plus{}k^2}}$?

Assuming this is right, $ \beta$ achieves the maximum value as the absolute value of $ k$ decreases. Since that is $ k \equal{} 0$, shouldn't $ \beta_{max} \equal{} 1$? Somehow that's not the correct answer.

I'm still completely lost for #32.

[quote="123456789"]For spring motion, we have $ \minus{} kx \equal{} ma \equal{} m \frac {d^2 x}{dt^2}$. Rearranging, this becomes
\[ \minus{} \frac {k}{m} x \equal{} \frac {d^2 x}{dt^2}.\]
When we solve this differential equation, we find that $ \omega^2 \equal{} \frac {k}{m}$. So, in cases where you have something similar to that differential equation above, just follow the same pattern.[/quote]

Ok, so how did you get $ \omega^{2}\equal{}\frac{k}{m}$?

Looking at $ \frac{d^2x}{dt^2} \equal{} \minus{}\frac{k}{m}x$, we see that two solutions are possible:

1) $ x \equal{} Ae^{Bt}$

2) $ x \equal{} Asin(Bt)$

Since it's undergoing harmonic motion, we can rule out 1, so the general solution is 2.

Going back to that differential equation: we know that differentiating $ x \equal{} Asin(Bt)$ twice should give $ a \equal{} constant \times x$. That $ constant$ will be your $ w \equal{} \sqrt{\frac{k}{m}}$.

[quote="Quarky"]Did anyone get $ \beta \equal{} \sqrt {\frac {1}{1 \plus{} k^2}}$?

Assuming this is right, $ \beta$ achieves the maximum value as the absolute value of $ k$ decreases. Since that is $ k \equal{} 0$, shouldn't $ \beta_{max} \equal{} 1$? Somehow that's not the correct answer.[/quote]

I got $ \beta \equal{} \sqrt{\frac{k}{1\plus{}k^2}}$, which has a maximum of $ \frac{1}{\sqrt{2}}$.


trigonometry
induction
algebra
polynomial
floor function
geometry
circumcircle

For each of the following, determine (with proof) an equivalent, closed form expression that contains no trigonometric functions.

[tex]\displaystyle{1.\quad\prod_{i=1}^{n-1} \cos\left(\frac{i\pi}{n}\right)}[/tex]

[tex]\displaystyle{2.\quad\prod_{i=1}^{n-1} \sin\left(\frac{i\pi}{n}\right)}[/tex]

[tex]\displaystyle{3.\quad\sum_{i=1}^{n-1}
\frac{1}{\sin^{2}\left(\frac{i\pi}{n}\right)}}[/tex]

[tex]\displaystyle{4.\quad\tan\left(\frac{3\pi}{11}\right) + 4\sin\left(\frac{2\pi}{11}\right)}[/tex]

Note: I have arranged them in what I believe to be increasing order of difficulty.

Mildorf wrote:[tex]\displaystyle{1.\quad\prod_{i=1}^{n-1} \cos\left(\frac{i\pi}{n}\right)}[/tex]

I start with the easiest one

[hide]Induction doesn't seem to work nicely here ?? I used Viete's formulas instead.

If n is even the product contains cos (n/2*pi/n) so is null.

If n is odd, the cos (i*pi/n) are the roots of the real part of polynomial X^n-1=0, with X=a+i*b and a+b=1

The product of the roots is -an/a0.

Developping and collecting we can see that here a0 = -1 and that an =  :Sigma: C(n,k) for k even = 2^(n-1).

P(n) = ((-1)^n - 1)/2^n for all n[/hide]

The first one, differently:

[hide]Re-write the kth cosine term as an exponential, (e^(2ki*pi/n) + 1)/e^(k*pi*i/n).  When you get the product of these things, the denominator can easily be found to be 2^(n - 1)*e^(i*pi*(n - 1)) = 2^(n - 1)*(-1)^(n - 1).  For the numerator, we can express it as [tex]\displaystyle \sum_{S\subset\{1, 2, ..., n - 1\}}e^{\sum_{x \in S}x}[/tex], or the sum, over all subsets S of {1, 2, ..., n - 1} of terms which have the form e^{\sum_{x \in S}x} (sorry I can't make it look any nicer).  Now, by simply examining here what happens when we take S and SC, it should be clear that this will be 0 when n is even and 1 when n is odd.  So for even n, it is 0 and for odd n it is 2^(1 - n) (since the negative bit goes away).[/hide]

I believe we will be able to do something basically the same for the second one, although the minus signs could theoretically cause problems.

Good work.  Both of you have found the same formula, but somewhere in the process lost [tex]\displaystyle{(-1)^{\left\lfloor \frac{n}{2} \rfloor}}[/tex], the factor that keeps track of the sign, so that:

[tex]\displaystyle{\prod_{i=1}^{n-1} \cos\left(\frac{i\pi}{n}\right) = \frac{1 - (-1)^{n}}{2^{n}} \cdot (-1)^{\left\lfloor \frac{n}{2} \rfloor}}[/tex]

My proof uses a trigonometric argument like that illustrated by the example below:

[tex]\displaystyle{\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)

\cos\left(\frac{4\pi}{7}\right)}[/tex] [tex]= \displaystyle{\frac{\sin\left(\frac{\pi}{7}\right)}{\sin\left(\frac{\pi}{7}\right)}\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)

\cos\left(\frac{4\pi}{7}\right)}[/tex]

[tex]= \displaystyle{\frac{1}{2\sin\left(\frac{\pi}{7}\right)}\sin\left(\frac{2\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)

\cos\left(\frac{4\pi}{7}\right)}[/tex]

[tex]= \displaystyle{\frac{1}{4\sin\left(\frac{\pi}{7}\right)}\sin\left(\frac{4\pi}{7}\right)

\cos\left(\frac{4\pi}{7}\right)}[/tex] [tex]= \displaystyle{\frac{1}{8\sin\left(\frac{\pi}{7}\right)}\sin\left(\frac{8\pi}{7}\right)

} = -\frac{1}{8}[/tex]

We have the product of [tex]\cos\left(\frac{i\pi}{n}\right)[/tex] for the non-zero residues [tex]i[/tex] on modulo [tex]n[/tex].  Clearly, for even [tex]n[/tex] the product is 0.  Otherwise, [tex]n[/tex] is odd and relatively prime to 2.  If we ignore signs, the double-angle formula forms the loops of multiplication by 2 in mod [tex]n[/tex], dividing by 2 for each residue.  This is sufficient to establish the magnitude for all [tex]n[/tex], and a simple counting argument gives the correct sign.

Now try the sign product.  : )

If you are stuck:[hide] see if you can make an argument based on a regular n-gon.[/hide]

#2. Let [tex]x=1[/tex] in the following identity and take absolute value on both sides.

[tex]\displaystyle{\prod_{k=1}^{n-1}(x-e^{{2k\pi \over n}i})={x^n-1\over x-1} = x^{n-1}+\cdots+x+1}[/tex].

Since [tex]\displaystyle{|1-e^{{2k\pi \over n}i}|=2\sin{k\pi\over n}[/tex], it follows that [tex]\displaystyle{\prod_{k=1}^{n-1}\sin{k\pi\over n}={n\over 2^{n-1}}[/tex].

[quote="ComplexZeta"]Solutions are appreciated.[/quote]

Solution for #3.

Comparing the imaginary part of the following identity:
$\displaystyle \cos{n\alpha}+i\sin{n\alpha} = (\cos{\alpha}+i\sin{\alpha})^n = \sum_{k=0}^n\binom{n}{k}\cos^{n-k}\alpha(i\sin\alpha)^k,$

we have
$\displaystyle \sin{n\alpha}=\binom{n}{1}\cos^{n-1}{\alpha}\sin{\alpha} - \binom{n}{3}\cos^{n-3}{\alpha}\sin^3{\alpha} + \binom{n}{5}\cos^{n-5}{\alpha}\sin^5{\alpha} - \cdots $

$\displaystyle =\sin^n\alpha\bigg(\binom{n}{1}\cot^{n-1}{\alpha} - \binom{n}{3}\cot^{n-3}{\alpha} + \binom{n}{5}\cot^{n-5}{\alpha} - \cdots\bigg).$

Let $\alpha={k\pi\over n}$ and $r_k=\cot{k\pi\over n}$ for $k=1,2,\ldots,n-1$, then $r_k$ ($k=1,2,\ldots,n-1$) are the roots of the following polynomial:

$\displaystyle \binom{n}{1}x^{n-1}  - \binom{n}{3}x^{n-3}  + \binom{n}{5}x^{n-5} - \cdots.$

Hence $\displaystyle \sum_{k=1}^{n-1} r_k = 0$, and $\displaystyle \sum_{1\leq i < j \leq n-1} r_ir_j = {- \binom{n}{3} / \binom{n}{1}}$.
Therefore,
$\displaystyle \sum_{k=1}^{n-1}\frac{1}{\sin^{2}\left(\frac{k\pi}{n}\right)}
= \sum_{k=1}^{n-1} \left(1 + \cot^2\left(\frac{k\pi}{n}\right)\right) = (n-1) + \sum_{k=1}^{n-1} r_k^2$
$\displaystyle = (n-1) + \left(\sum_{k=1}^{n-1} r_k\right)^2 - 2\left(\sum_{1\leq i < j \leq n-1} r_ir_j \right) = {(n^2-1)\over 3}$.

Excellent!  Those are the only ways I know about to establish the identities.  Of course, there is the related product

[tex]

\displaystyle{\prod_{i=1}^{n-1} \tan\left(\frac{i\pi}{n}\right) = \frac{2n}{1 - (-1)^{n}} \cdot (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}}

[/tex]

obtained from combining 1 and 2.  From the sine product, we can establish that regular [tex]n[/tex]-gon [tex]A_{1}A_{2}\dots A_{n}[/tex] of circumradius [tex]R[/tex] obeys the identity

[tex]\displaystyle{\prod_{i=2}^{n} A_{1}A_{i} = nR^{n-1}}[/tex]

by applying the law of sines.  I stumbled across the 4th expression as part of an identity at MathWorld and spent a long time trying prove it.

If you are stuck: [hide]try using Gaussian sums to prove that the given expression is exactly equal to sqrt(11).[/hide]

[quote="Mildorf"] From the sine product, we can establish that regular [tex]n[/tex]-gon [tex]A_{1}A_{2}\dots A_{n}[/tex] of circumradius [tex]R[/tex] obeys the identity

[tex]\displaystyle{\prod_{i=2}^{n} A_{1}A_{i} = nR^{n-1}}[/tex]

by applying the law of sines.[/quote]

I did exactly the reverse. I solved the polygon problem first, then found out that this is related to #2. See my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12714]A polygon problem[/url] and a solution provided by [b]belenos[/b].

The only complete solution that I know of (originally explained to me by Mr. Vandervelde of the Mandelbrot contest) for the last problem is a bit high tech, so I will boil it down to the trigonometry.  Let [tex]\chi[/tex] be a primitive Dirichlet character with conductor [tex]f[/tex], and let [tex]W(\chi)[/tex] denote its Gauss sum, which is given by

[tex]\displaystyle{W(\chi ) = \sum_{k=0}^{f-1} \chi (k) \cdot z^k}[/tex]

where [tex]z= e^{\frac{2\pi i}{f}}[/tex] is a primitive [tex]f[/tex]-th root of unity.  Let [tex]\chi *[/tex] denote the conjugate character; i.e. [tex]\chi*[/tex][tex](k)[/tex] is the complex conjugate of [tex]\chi (k)[/tex] for all [tex]k \in \mathbb{Z}[/tex].  Then a property about Gauss sums is that

[tex]\displaystyle{W(\chi ) \cdot W(\chi *) = \chi (-1) \cdot f}[/tex].

Use the character [tex]\chi (k) = 0, 1, -1, 1, 1, 1, -1, -1, -1, 1, -1[/tex]  for [tex]k = 0, 1, 2, \dots, 10[/tex] which has conductor [tex]f = 11[/tex].  Since its values are all real,  [tex]\chi *[/tex] is the same as [tex]\chi[/tex].  Also in our case, [tex]\chi (-1) = \chi (10) = -1[/tex].  Hence [tex]W(\chi )^2 = -11[/tex], and a simple sign check shows that [tex]W(\chi ) = i*\sqrt{11}[/tex].  Inspecting the complex part of the Gauss sum, we have a trig equation for [tex]\sqrt{11}[/tex].

Suffice it to show that

[tex]\displaystyle{ \sin\left(\frac{2\pi}{11}\right) - \sin\left(\frac{4\pi}{11}\right) + \sin\left(\frac{6\pi}{11}\right) + \sin\left(\frac{8\pi}{11}\right)}[/tex] [tex]\displaystyle{ + \sin\left(\frac{10\pi}{11}\right) - \sin\left(\frac{12\pi}{11}\right) - \sin\left(\frac{14\pi}{11}\right) - \sin\left(\frac{16\pi}{11}\right)}[/tex]

[tex]\displaystyle{+ \sin\left(\frac{18\pi}{11}\right) - \sin\left(\frac{20\pi}{11}\right) = \tan\left(\frac{3\pi}{11}\right) + 4\sin\left(\frac{2\pi}{11}\right)}[/tex]

A different trig problem, evaluate exactly:

[tex]
\displaystyle{5. \quad \sin\left(\frac{2\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right) - \sin\left(\frac{6\pi}{7}\right)}[/tex]

[quote="Mildorf"]The only complete solution that I know of (originally explained to me by Mr. Vandervelde of the Mandelbrot contest) for the last problem is a bit high tech[/quote]

I found the following on a Brazilian math olympiad mailing list.

[quote][tex]\displaystyle(\tan(3s) + 4\sin(2s))^2 = 11 - \frac{(\tan(8s) + \tan(3s))\cos(8s)}{\sin(s)\cos(3s)}[/tex][/quote]

Since [tex]\displaystyle\tan(8s)+\tan(3s)[/tex] vanishes when [tex]\displaystyle s=\frac{\pi}{11}[/tex], your problem 4 is a special case of this form.

The mailing list gives several additional identities, and states:  "This problem was given on a 1895 Tripos Exam, and it is in the book by Bromwich on infinite series (p. 270)."  I have not attempted to confirm the truth of this identity.  And I do not have access to the references.

Wow!, This is pretty Hi-tech.

[quote="Mildorf"]
A different trig problem, evaluate exactly:

[tex]
\displaystyle{5. \quad \sin\left(\frac{2\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right) - \sin\left(\frac{6\pi}{7}\right)}[/tex][/quote]

Is this still in the increasing order of difficulty? :-)

This can be transformed into a product: [tex]\displaystyle{4\sin\frac{\pi}{7} \sin\frac{2\pi}{7} \sin\frac{3\pi}{7}}[/tex]. Let [tex]n=7[/tex] in your trig identity #2 would give us the result. The answer is [tex]\displaystyle\frac{\sqrt{7}}{2}[/tex].

[quote="rcv"]
[quote][tex]\displaystyle(\tan(3s) + 4\sin(2s))^2 = 11 - \frac{(\tan(8s) + \tan(3s))\cos(8s)}{\sin(s)\cos(3s)}[/tex][/quote]

The mailing list gives several additional identities, and states:  "This problem was given on a 1895 Tripos Exam, and it is in the book by Bromwich on infinite series (p. 270)."  I have not attempted to confirm the truth of this identity.  And I do not have access to the references.[/quote]

By multiplying [tex]\sin(s)\cos^2(3s)[/tex] on both sides, the above identity is equivalent to:

[tex]\sin(s)(\sin(3s) + 4\sin(2s)\cos(3s))^2 = 11\sin(s)\cos^2(3s) - \sin(11s)[/tex]

Is it possible to use Taylor Series Expansions to verify this?

Here is another possible way to verify (I don't own any math software, I am pretty sure Mathematica or Maple can do this):

Express [tex]\sin(ns), \cos(ns)[/tex] in terms of [tex]\sin(s), \cos(s)[/tex] using [tex]\cos{ns}+i\sin{ns} = (\cos{s}+i\sin{s})^n = \sum_{k=0}^n\binom{n}{k}\cos^{n-k}{s}(i\sin{s})^k[/tex].

Express [tex]\sin(s), \cos(s)[/tex] in terms of [tex]x = \tan{s\over2}[/tex] using the identities [tex]\sin(s) = \frac{2x}{1+x^2}[/tex] and [tex]\cos(s) = \frac{1-x^2}{1+x^2}[/tex].

It becomes a verification of an identity of rational functions of [tex]x[/tex].

[b]Note: A proof of the identity has been found, see attached PDF file[/b]

Is it possible to verify with Taylor expansions? My guess is that it probably is.  Pair up like powers of [tex]s[/tex] in the power series expansion of the left:

[tex]\displaystyle{\left(\sum_{k=0}^{\infty}\frac{s^{2k+1}\cdot(-1)^{k}}{(2k+1)!}\right)}[/tex] [tex]\displaystyle{\Bigg(\sum_{k=0}^{\infty}\frac{(3s)^{2k+1}\cdot(-1)^{k}}{(2k+1)!} + 4\left(\sum_{k=0}^{\infty}\frac{(2s)^{2k+1}\cdot(-1)^{k}}{(2k+1)!}\right)}[/tex][tex]\displaystyle{ \left(\sum_{k=0}^{\infty}\frac{(3s)^{2k}\cdot(-1)^{k}}{(2k)!}\right)\Bigg)^{2}}[/tex]

and that of the right:

[tex]\displaystyle{11 \Bigg(\sum_{k=0}^{\infty}\frac{s^{2k+1}\cdot(-1)^{k}}{(2k+1)!}\Bigg) \Bigg(\sum_{k=0}^{\infty}\frac{(3s)^{2k}\cdot(-1)^{k}}{(2k)!}\Bigg)^{2}}[/tex] [tex]\displaystyle{ - \sum_{k=0}^{\infty}\frac{(11s)^{2k+1}\cdot(-1)^{k}}{(2k+1)!}}[/tex]

Though it will probably take some serious combinatorics and a lot of patience.  It does appear to be true, checking exactly for nice values of [tex]s[/tex].  When I when graphed the equation for [tex]11[/tex] in terms of trig, it appeared to be a straight line. :)

* Yes, problems 5 and 6 are out of order. ;)

[tex]\displaystyle{6. \quad \sum_{k=1}^{\infty} \cot^{-1}\left(k^{2} + k + 1\right) \quad = \quad ?}[/tex]

[quote="Mildorf"]* Yes, problems 5 and 6 are out of order. ;)

[tex]\displaystyle{6. \quad \sum_{k=1}^{\infty} \cot^{-1}\left(k^{2} + k + 1\right) \quad = \quad ?}[/tex][/quote]
Telescoping the sum [tex]\displaystyle{\sum_{k=1}^{n} \cot^{-1}\left(k^{2} + k + 1\right) = \sum_{k=1}^{n} \left(\cot^{-1}\frac{1}{k+1}-\cot^{-1}\frac{1}{k}\right)[/tex] [tex]\displaystyle= \cot^{-1}\frac{1}{n+1} - \cot^{-1}1[/tex] and letting [tex]n\rightarrow\infty[/tex].

Last two trig problems:

[tex]7.\quad[/tex] Regular heptagon [tex]AB[/tex][tex]CDE[/tex][tex]FG [/tex] is inscribed in unit circle [tex]\omega[/tex], which has center [tex]O[/tex].  [tex]M[/tex] is the midpoint of line segment [tex]BC[/tex], and [tex]P[/tex] is the midpoint of minor arc [tex]AB[/tex] of [tex]\omega[/tex].  If [tex]S[/tex] is the midpoint of line segment [tex]OP[/tex], then, using only trig (even though complex numbers are tempting here!), determine the exact value of [tex]MS[/tex].

[tex]8.\quad[/tex] Determine an equivalent closed form expression in terms of [tex]\lambda[/tex]:

[tex]
\displaystyle{ \sum_{k=1}^{\infty} \tan^{-1}\left( \frac{\lambda}{k^{2}} \right) } [/tex]

* I have only been able to solve 8 for specific [tex]\lambda[/tex]; I leave the general form as an open challenge.

[quote="Mildorf"]
[tex]8.\quad[/tex] Determine an equivalent closed form expression in terms of [tex]\lambda[/tex]:

[tex]
\displaystyle{ \sum_{k=1}^{\infty} \tan^{-1}\left( \frac{\lambda}{k^{2}} \right) } [/tex]

* I have only been able to solve 8 for specific [tex]\lambda[/tex]; I leave the general form as an open challenge.[/quote]

Let [tex]\displaystyle{f(\lambda)=
 \sum_{k=1}^{\infty} \tan^{-1}\left( \frac{\lambda}{k^{2}} \right) } [/tex] be a function of [tex]\lambda[/tex]. Taking the derivative and [b]assume we can switch the differentiation and the summation[/b] (not sure about this?), then
[tex]\displaystyle f^{\prime}(\lambda)=  \sum_{k=1}^{\infty}\frac{k^2}{k^4+\lambda^2}[/tex]. Does anybody know how to find a closed form for this sum?


For how long can you edit something? Sometimes, I try to edit something and it says You cannot edit this post. Please ask a moderator to do it for you if it is really nessecary or something like that?

Also, could an admin consider lengthening the time?

Forty-eight hours. And no - be more confident in your posts next time.

Well sometimes, like when you have a sign-up sheet, people take more than 48 hours... do you just double post then?

You can generally ask a mod to change the original post. That means, include what you want the post to look like so all we have to do is copy and paste.


quadratics
number theory proposed
number theory

From the proof of the famous Dirichlet's theorem it follows that if $p$ is a prime number of the form $4k+3$ the sum of the quadratic residues is strictly smaller than the sum of the non quadratic residues. However, the only proof I know uses some nice analytic arguments. I tried for hours to prove it elementarily, but with no success. Is there an elementary proof?

Let $I_{k}$ is the sum of the k degree residues divide p(in your case k=2). If d=(p-1,k), then $I_{d}=I_{k}$, therefore we can consider only case $k|(p-1)$.
It is easy to show, that $I_{k}$ is integer if $(p-1)\not |k$ (In this case 1/p).
Let $m=\frac{p-1}{k}$. Let a is residue, suth that for all x exist r: $x=a^{r}(mod \ p)$. If m is even, then $a^{rk}(mod \ p)+a^{(r+m/(2k))k}(mod \ p)=p.$ Therefore $I_{k}=\frac{m}{2}$. When m is odd, can be showed, that $|I_{k}-\frac{m}{2}|<k\sqrt{m}(\ln(m+1))$.
If $k>2$ exist p, suth that : $I_{k}(p)>\frac{m}{2}$ and exist prime q, suth that $I_{k}<\frac{m}{2}$. Only for k=2 $I_{k}=\frac{m-h}{2}$, were h is class number of field $Q(\sqrt{-p})$, when m is odd (m-odd mean p=3(mod 4)). There are not elementary proof, that $I_{2}<m/2$.


I'm getting nervous, because all the violinists I know are starting to...  APPRECIATE THE VIOLA!!!

Fortunately my stand partner, little miss Concert Mistress (she's very nice actually), and I continue on the tradition of cracking jokes at the violas.

If you hate the viola, tell me, so I don't lose faith in the system (which as you should know goes:

Conductor
Concert-Master
Assistant Concert-Master
The rest of the 1st Violins
2nd Violins
Cellos
String Basses
The chairs everyone sits on
The floor they're on
The dust on the floor
The bacteria on the dust
The individual organelles of the bacteria
Violists) <--- Wow that was a long parenthesis!

Oh and post some good viola jokes!

Aren't violas and cellos just different types of instrumrnts of the same genre as violins?

I believe cellos go above 2nd violins, but otherwise, your system's good.  :wink:

[quote="futuremontrealian"]Ok I can believe that a cello is better than a 2nd violin.

And a viola is no where near the league of a cello![/quote]
I agree  :wink:
violins and cellos are MUCH nicer than violas

I think this topic is best summed up in this picture

[img]http://mahopa.de/bilder/lustige-forenbilder/room.jpg[/img]

See, this is exactly the kind of behavior I'm talking about.  It's as though people have forgotten one of the basic rules of nature and are now forsaking an ancient prejudice (ok sure, the violin family wasn't invented til the 1500's, but they're hatred for each other goes back a long way)!

I love violas because right now I hear a violin.....sweet low sound.........*Drops down dead* My ears are extremely sensitive to high noises. Also I think the orchestra needs less string and more woodwinds(well not stringed orchestras.)

viola sounds more badass than violin though

violins sound much better, because violas are low, but not low enough to make them sound good.

I like woodwinds. :D

Go mess with your string instruments. WOODWINDS PWN

Bach and Haydn preferred the viola to the violin.

And as for viola jokes, how about this one
Which would you rather be: a good violist, a bad violist, or a good violinist?
Answer:  a good violist -- the other two don't exist

Both are good in their own way...it depends entirely on the type of piece.  The darker, richer sound of the viola can be better at times...and no, I'm not saying a violin doesn't produce a rich sound.
I'm a violinist, in case you're wondering...

Lol the good viola joke is almost the same as a "bad viola" joke I've heard.

Here's one:

The Chicago Symphony's concert hall was condemned due to structural faults, and the symphony could not afford to hire builders to build their new hall, so the orchestra took it upon themselves to do the building.  Over time, the class structure of the orchestra began to reveal itself in the jobs assigned to different sections.  At the top, the violinists planned and drew the designs for the building and surveyed the work, and at the bottom, the violists dug pits for the foundation.

Two violists, shoveling next to each other, began a conversation:

"Say, why do we have to do all the work while the trumpets up there just get to watch us and boss us around?" said the first violist.

"I dunno.  Why don't you go ask one?" said the second.

So the violist climbed out of the ditch and found a trumpeter who was idle.

"Hey, why do you guys get to just sit around and boss us around?" inquired the violist.

"Well, because we're smarter than you." answered the trumpet.

"I don't understand... " said the violist.

"Well, let me explain to you," said the trumpeter, backing up to a nearby tree and holding his hand in front of it.  "Okay, now hit my hand," he said.

The violist swung at it and the trumpeter took away his hand so the violist's hit the tree at full speed.

"Ouch!  Okay, I see what you mean," said the violist.

The violist ambled back down the pit and began shoveling again next to his partner.

"So what did he say?" asked his partner.
"Well, apparently they're smarter than us," replied the violist.

"I don't understand..." said the partner.

"Here, let me explain," said the first violist, taking his hand and, for lack of a tree, held it in front of his face.  "Okay, now hit my hand with your shovel..."

Good eh?  Yes it's cruel, but it's all in good fun.  I can take a good violin joke too.

no it's stupid it's like wtf can you not handle that violas just sound cooler

Who tells viola jokes? Violists. We don't take ourselves too seriously.

My favorite string joke:

What do you call the members of a string quartet?
[hide="Answer"]A good violinist, a bad violinist, a former violinist, and someone who hates violinists.[/hide]

[quote="K81o7"]Just a remark...some of the best violinists I know play viola as well.[/quote]

I agree.  Even I play the viola sometimes when I can't stand listening to the upper register any longer.  It's particularly fun to record yourself playing a song on both and then putting them together to make an ensemble :).

But that doesn't mean a joking prejudice isn't acceptable.  My violist friend and I exchange jokes all the time!  neither is really better, although the violin does have better parts usually (except of course in a lot of bach songs and "Santa Plays the viola" lol.)


inequalities
algebra
polynomial
triangle inequality
superior algebra
superior algebra solved

If all the conjugates $x^{(i)}$ of an algebraic number $x$ have $|x^{(i)}| = 1$, then $x$ is a root of unity.

To be an algebraic number is not enough, it has to be an algebraic integer.

Assume $x$ is an algebraic integer of degree $n$ with conjugates $x_1,\ldots, x_n$, all of modulus 1.
Let $m$ be an arbitrary integer. We have $x^m\in\mathbb Z[x]\subset\mathbb Q(x)$. We then have $[\mathbb Q(x^m): \mathbb Q]\leq[\mathbb Q(x): \mathbb Q]=n$. So $x^m$ is an algebraic integer of degree at most $n$.
Its conjugates are a subset of $\{x_1^m,\ldots,x_n^m\}$, so they are all of modulus $1$. Using Viete's relations and the triangle inequality, it is easy to prove that all the coefficients of the minimal polynomial $P_m$ of $x^m$ are bounded by $2^n$. They are also integers and it is easy to prove that the set $\{P_m|m\in\mathbb Z\}$ is finite. It is also easy to prove that there exist $s\neq r\in\mathbb Z$ and $i\in \{1,\ldots,n\}$ such that $x_i^r=x_i^s$, from which $x^{r-s}=1$ follows.

$\frac{3+4\imath}5$ is an algebraic number, not an algebraic integer (because $\mathbb Z[\imath]$ is integrally closed) such that all its conjugates ($\frac{3\pm4\imath}5$) have modulus 1. I think it can be shown that it is not a root of unity.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=42893 for the last one.


Every round, throw your egg at somebody. One person has an egg bomb, and the person that the egg bomb is thrown at dies. Last one standing wins.

12 players to start.

1. Chompy
2. kianex
3. EggyLv.999
4. cf249
5. King!
6. jjfun1
7. spaceguy524
8. ernie
9.
10.
11.
12.

in[hide=" "]No I dont watch spongebob[/hide]

1. Chompy
2. kianex
3. EggyLv.999
4. cf249
5. King!
6. jjfun1
7. spaceguy524
8. ernie
9. cgyao15
10. polkadotman1
11. westiepaw
12.

One spot left!

I shall be cool and throw the egg at myself.

I am like disqualified or something?

Srry that I was REALLY inactive these past few days..

too much HW!

Oh, boo.

Eenie meenie miney mo...

[b]Throw at ernie[/b]

"One, two, three, THROW!" lotsofmath shouts.

Eggs (and a bomb) go flying:

cf249 throws a [color=olive]Rotten[/color] egg at ernie
jjfun1 throws a [color=white]White[/color] egg at cf249
ernie throws a [color=yellow]Yellow[/color] egg at cf249

All of the eggs not thrown by cf249 were thrown at cf249. The yellow one didn't break. "Uh oh," cf249 thinks as the [b][size=200]BOOM[/size][/b] is heard.

"Two competitors left! Please walk into the second chamber," lotsofmath says.

jjfun1 and ernie walk into the chamber, wondering what in the world is going on. When they get there, they see a carton of eggs.

"NEW RULES! There are 12 eggs in that carton. Each person picks one egg (with numbers) and throws it at the other person. jjfun1, getting a double kill, picks first. One of the eggs is a bomb if thrown by jjfun1, and one is a bomb if thrown by ernie. The first person to throw their bomb wins. If both players throw the other person's bomb, they both go on to the next round." lotsofmath says.

Remaining players:
6. jjfun1
8. ernie

jjfun1 must pick an egg by Saturday, October 24, at 8 AM EST. Ernie must wait until jjfun1 picks his egg, then he can pick.

OK. Haha, gl ernie! :P
I pick and throw[b]egg 3[/b].

jjfun1 threw egg 3 at ernie. It broke. It was an ordinary egg.

Ernie must pick an egg by Saturday, October 24, at 6 PM EST.

Wow. this game was epic i suggest we start a new one after this is over.

IDC if I win or lose, CT PWNS.  [b]Throw: Egg 8[/b]

haha thaz right!!
LAST TWO PEOPLE BOTH FROM CT!! W00T!!
heh.
/me crosses her fingers and prays now :P

ernie throws egg 8 at jjfun1. It...did...not...break. An "OH NO" was heard, right before the [size=200][b]BOOM[/b][/size].

Wow, that was fast.

"Last man standing!" lotsofmath says. "Please proceed to the final chamber."

ernie walks to the last chamber, wondering what could possibly be next.

There was another carton of eggs on the table, and a hooded figure next to it.

"This is my personal dummy. It will throw random eggs from that carton. The same rules apply, except this time, you both have the same bomb. That bomb has the number of jjfun1's bomb in the previous round. Ernie gets to go first."

(Find jjfun1's bomb's number before it does)

Gah, the dummy better be from CT... or TX.

[b]Throw: Egg 6[/b]

OH...MY...GOSH...ernie gets the right egg on the first try twice in a row! The egg hits the dummy and emits a big [b][size=200]BOOM[/size][/b]. Strangely enough, the dummy didn't disappear, and that's because it wasn't...a...dummy.

"You win! You may now leave through that door," lotsofmath says.
"It was YOU?" ernie says
"Yes, the "dummy" was actually me." lotsofmath says.

ernie left through the door, elated that he actually did "beat the mod". There were two paths. He first took the left one, and found all his friends!

"Did you win?" jjfun1 asked.
"Yes, but why are you all alive?" ernie says.
"Oh...about that. The bombs just opened up vortices, which just teleported us to here. We didn't die."
"Well, let's go!"

Ernie takes all the rest of the AoPSians down the right path, and they all go back to doing tons of math.

[b]GAME OVER! ERNIE WINS![/b]


probability
Putnam
probability and stats

for $ \alpha \in (0,1)$ with a fair coin. we want to find a way for people name A and B that the possibity of winning of A is $ \alpha$. Is it possible to find this way?

they can play an infinite game by looking at the result of $ S \equal{} \sum_{1}^{\infty} \frac{X_n}{2^n}$ and say that $ A$ wins if $ S < A$  :D  not very efficient isn't it ..

Actually this is extremely efficient if you finish the game when it becomes clear that one person will win. ;)


How many distinguishable ways are there to write 9 as the sum of 1s, 2s and 4s, where the order of the addends matters? For example, $ 4 \plus{} 4 \plus{} 1$ and $ 1 \plus{} 4 \plus{} 4$ are two different ways.

[quote=lcsmart]Is there a faster way to do this problem? Without listing them out?[/quote]

[quote=paperplates]^^ same, is there any way?
Haha, you posted that the day before pi day.[/quote]

I think you can use generating functions to partition $9$. However, I'm not too familiar with that stuff, so you might want to look into it more.

[quote=MrMXS][quote=lcsmart]Is there a faster way to do this problem? Without listing them out?[/quote]

[quote=paperplates]^^ same, is there any way?
Haha, you posted that the day before pi day.[/quote]

[quote=giacomorizzo]No, you just have to do Constructive Counting. List each one out, and find how many ways you can rearrange each sum.[/quote]

There is another way.

Since the numbers in this problem are pretty small, this method may not be much faster. However, as the numbers get a bit larger, this method will probably be much faster. Additionally, there is no searching involved.

[hide=Recursion]Let $S_{9}$ be the number of ways to write $9$ as the sum of $1\text{'s},$ $2\text{'s},$ and $4\text{'s},$ where the order of the addends matters. We know that each sum starts with either a $1$, $2$, or $4$.

[b]If the sum starts with 1:[/b] The rest of the sum must add up to $8$. Thus, the number of sums that start with a $1$ is $S_{8}$.

[b]If the sum starts with 2:[/b] The rest of the sum must add up to $7$. Thus, the number of sums that start with a $2$ is $S_{7}$.

[b]If the sum starts with 4:[/b] The rest of the sum must add up to $5$. Thus, the number of sums that start with a $4$ is $S_{5}$.

We now have an equation: $S_{9}=S_{8}+S_{7}+S_{5}$. More generally, we have $S_{n}=S_{n-1}+S_{n-2}+S_{n-4}$.

Clearly, $S_{1}=1$, $S_{2}=2$ and $S_{3}=3$. The sums that add up to $4$ start with either a $1$, a $2$, or we can just have a single $4$. Thus, $S_{4}=S_{3}+S_{2}+1$, where the $1$ represents the case where the sum is just $4$. This gives $S_{4}=3+2+1=6$. Solving for all $S_{n}$s until $S_{8}$ gives us
\begin{align*}
S_{1}&=1,\\
S_{2}&=2,\\
S_{3}&=3,\\
S_{4}&=6,\\
S_{5}&=10,\\
S_{6}&=18,\\
S_{7}&=31,\\
&\text{and}\\
S_{8}&=55.\\
\end{align*}

Therefore, $S_{9}=S_{8}+S_{7}+S_{5}=55+31+10=\boxed{96}$.

I realize that this solution [i]looks[/i] slower. Personally, though, this was much faster then writing out all $9$ sets of $1\text{'s},$ $2\text{'s},$ and $4\text{'s},$ that add up to $9$ where the order of the addends doesn't matter. :)[/hide][/quote]

Brilliant!

wait:

[quote=the solution]There are $3!/2!=3$ distinguishable orders for the first sum, $4!/2!=12$ for the second sum, $5!/3!=20$ for the third sum, $6!/5!=6$ for the fourth sum, $5!/4!=5$ for the fifth sum, $6!/3!3!=20$ for the sixth sum, $7!/5!2!=21$ for the seventh sum, $8!/7!=8$ for the eighth sum, and $1$ for the last sum. In total, there are $\boxed{96}$ distinguishable ways are there to write $9$ as the sum of $1\text{'s},$ $2\text{'s}$ and $4\text{'s}.$[/quote]

Aren't the arrangements supposed to be distinguishable? So no dividing?

@above
I don't quite understand what you mean.

I think the reason why the solution divided is because for the first sum, we have the numbers $1,4,4$ being added. However, as stated in the problem, $1+4+4$ and $4+4+1$ are different sums. Therefore, we want the number of ways to rearrange $1,4,4$, and that will give us the first sum, which there are $\cfrac{3!}{2!}$ ways of doing. We then do the same for $4,2,2,1$, which there are $\cfrac{4!}{2!}$ ways of doing, and so on for all the other sums.

Hopefully this clears it up. :)

Yeah but we divide by 2 cause order doesn't matter, right?  :huh:

The order of the addends matters. Also, for certain sums like $2+2+1+1+1+1+1$, we must divide by $2!5!$ because there are $2!$ ways of rearranging the $2$s and $5!$ ways of rearranging the $1$s. However, since the $2$s and $1$s are identical, we don't want to count these rearrangements.

Oh, ok! Like when you inter-switch the 2's and 1's...

The thing about this problem is that you have to just use "strategic" trial and error to find the cases.

[hide = organized casework]i used case work for my trial and error............
case no 4's:
         sub case 0 2's
         sub case 1 2
         sub case 2 2's
         $2 +2 +1 +1 +1 +1 +1$ = $\frac{7!}{5!2!} = \boxed{21}$ ways
         sub case 3 2's
         sub case 4 2's
case 1 $4$:
         sub case 0 2's
         sub case 1 2
         sub case 2 2's
case 2 $4$'s:

for each of these sub cases the rest are filled out with ones, then you use fixing overcounting with division to find the amount of ways you can arrange each sum.
I have given one example
THis seems like it takes long but it doesn't
this helps you stay organized...
adding all the cases you should get [size=200]$\boxed{96}$[/size]
[/hide]

i saw this and immediately thought generating functions, but idrk how to do that. can some1 post a sol for it

ref: intermediate c+p, chap 14

There is also an alternative, but I haven't finished the solution.

We can treat this like a "Leo the Rabbit" problem (from YouCubed), and use combinatorics to solve for each individual combination of numbers  (2 4s and 1 1, or one 4 and two 2s and one 1, etc). Simply use combinatorics to find the number of possible positions of each one.

I find this solution very accurate, but can be pretty lengthy and annoying to do with more than 2 variables.


I wasn't sure where to put this... but anyway
    My private messages usually get sent right away when I hit submit... but when I send them to a certain person they get stuck in my outbox. I have tried to send several messages to them in the past month and they're all still sitting in my outbox. What's wrong?

:arrow: Move to the Questions, Suggestions, and Announcements forum?

This belongs in the Support forum.

Maybe the people who you sent it to haven't been on for a while, so they didn't read it yet? Can you receive messages?

if they're in ur outbox it just means the person u sent it to hasn't read it yet. :D it should go to sentbox when the person has read it. i think...

Exactly. Many members don't log in very often, so messages sent to them will stay in the outbox. My outbox contains a message from 2007.


help.... i just saw mods and i need help plz explain wat mods are thnx

I'm probably not the one to exlain it, but I'll take a crack at it
mod is just a fancy way of sying what the remainder is when something is divided by something else
for example:
43=3 (mod 8)
43=1 (mod 6)
and so on.

now nobody ever explained this I just got it from the contexts of problems, so if I'm wrong somebody come explain it.

Here is a topic on this: http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28395 .

mods are useful in number theory and MATHCOUNTS a lot so check out that thread


linear algebra
matrix
calculus
calculus computations

Whats the easiast way to solve the following system:

[img]http://i25.tinypic.com/1z4lonp.jpg[/img]

Thanks.

These can be done with the matrix exponential ([url]http://en.wikipedia.org/wiki/Matrix_exponential[/url]) very easily if the matrix is diagonalizable. If it's an initial value problem you can also adapt the Laplace transform for systems, and instead of solving one algebraic equation to find the solution you have to solve three.

Note: this matrix is not diagonalizable. It's eigenvalues are $ 0$ and $ \minus{} 1,$ but the eigenvalue $ 0$ has algebraic multiplicity 2 and geometric multiplicity 1.

The answer will be of the form
\[ \mathbf{y}(x) \equal{} \mathbf{u} \plus{} x\mathbf{v} \plus{} e^{ \minus{} x}\mathbf{w}.
\]
Someone else can work out the details of what $ \mathbf{u},\mathbf{v},$ and $ \mathbf{w}$ are.


Hi,I am a Chinese,I wish to see the 47th IMO preelection Problems,who can help me?
My English is poor,can you understand?


quadratics
number theory unsolved
number theory

In a quadratic table with $ 2007$ rows and $ 2007$ columns is an odd number written in each field.
For $ 1\leq i\leq2007$ is $ Z_i$ the sum of the numbers in the $ i$-th row and for $ 1\leq j\leq2007$ is $ S_j$ the sum of the numbers in the $ j$-th column.
$ A$ is the product of all $ Z_i$ and $ B$ the product of all $ S_j$.
Show that $ A\plus{}B\neq0$

[quote="valerie"]In a quadratic table with $ 2007$ rows and $ 2007$ columns is an odd number written in each field.
For $ 1\leq i\leq2007$ is $ Z_i$ the sum of the numbers in the $ i$-th row and for $ 1\leq j\leq2007$ is $ S_j$ the sum of the numbers in the $ j$-th column.
$ A$ is the product of all $ Z_i$ and $ B$ the product of all $ S_j$.
Show that $ A \plus{} B\neq0$[/quote]
It it easy to show that $ Z_j$ and $ S_j$ are odd numbers
Let $ x,y$ be the numbers of $ Z_j$ which have form $ 4k \plus{} 1$ and $ 4k \plus{} 3$ .
Let $ z,t$ be the number of $ S_j$ which have form $ 4k \plus{} 1$ and $ 4k \plus{} 3$
Then the some of all numbers on this board congruence $ x \plus{} 3y$ and $ z \plus{} 3t$ to  $ (\mod 4)$ ,thus $ x \plus{} 3y\equiv z \plus{} 3t (\mod 4)$ .Because $ x \plus{} y \equal{} z \plus{} t \equal{} 2007$ ,we obtain $ y\equiv t (\mod 2)$ .
It's obvious to show $ A\equiv ( \minus{} 1)^y (\mod 4)$ and $ B\equiv ( \minus{} 1)^t (\mod 4)$ ,therefore $ A \plus{} B\equiv 2 (\mod 4)$ , it means that this number can't be equal 0 .Apply this method we can prove that this problem is also true for all $ (2n \plus{} 1).(2n \plus{} 1)$ square .


If in any triangle $ ABC$, we have $ \frac{AB}{AC}\equal{}\frac{BE}{CF}$ where $ BE$ and $ CF$ are the two cevians then $ BE$ and $ CF$ are the two internal angle bisectors.

An Addition:

The above is clearly tru for an isoceles traingle and an equatorial triangle, I havnt checked out other cases! :lol:

This is the same problem you had post today in the Hyacinthus group...

You must be joking. If in a given isosceles triangle $ ABC$ with base $ BC$, you take a point $ P$ lying on the line bisector of $ BC$, then the $ B$ and $ C$ - cevians through $ P$ are congruent, and so they satisfy your $ AB / AC \equal{} BE / CF$ without being the internal angle bisectors.

Thats why I put a ? mark, and yes Cosmin it was given in your paper with Nicula! Sorry, for this mistake!


geometry
geometric transformation
rotation
3D geometry
counting
distinguishability
tetrahedron

Find the number of distinct rigid re-orientations of a cube.

In other words, in how many distinct ways can a cube be rotated so that its vertices will lie on top of its original set of vertices?

Could you specify a little more? What kind of cube is this? Is each face distinguishable?

Well, if faces weren't distinguishable, this question would have a pretty easy answer ...

Yeah, I know.  I guess I'm just being stupid  :fool:

My description of the problem probably wasn't very good. Assume the faces of the cube are distinct though.

[hide="uuh"]Look at the cube from eye level. There can be a total of six 'top' faces. For each face there are four possible rotations that change the face you're looking at. Once those two faces are determined, the cube is fixed. So, 4*6=24?[/hide]

I think youre right. Another way to do it is to mark one face with a nonsymmetrical figure, then that figure can be in one of 6 different positions and 4 different rotations, 24.  :)

Or my way, since each corner is distinct (the three faces touching it are distinct), we have 8 positions for a corner, and 3 rotations for it, equals 24!

You're all correct!

Now you can use the same methods to find the number of symmetries (called isometries) of the tetrahedron, the octahedron, the dodecahedron, and the icosahedron! What do you notice about these numbers? Why are some of them the same?

A tetrahedron is a 4-faced shape with each of its faces a regular triangle, so it would have 4 distinct faces to be set down on and 3 different positions/rotations for each of those, or 12 "distinct rigid re-orientations" right? And an octahedron is an 8-faced shape.. which I cannot quite picture, but I'm assuming all the faces would be regular and the same as each other; then there are 8 different faces to set in down on and... how many sides would each face have? Anyone know?

Now I'm curious, is there such thing as a regular pentahedron? a five-faced shape with sides all the same..?

There are only 5 [url=http://mathworld.wolfram.com/PlatonicSolid.html]Platonic solids[/url].  None of them have 5 faces.

The octahedron is basically two square-based pyramids glued base-to-base, so it also has equilateral triangles for faces. The dodecahedron is 12-faced with pentagons. The icosahedron is 20-faced with (I think) equilateral triangles. So for each of these the isometries would be (number of faces)*(edges per face) or (number of vertices)*(faces per vertex) (unless not every rotation of the dodecahedron or icosahedron works? I don't see how that can be but I can't make certain since I can't immediately draw a dodecahedron or icosahedron.)

These 'platonic solids' are very enlightening. I didn't think a 5-faced shape like that made sense, but then I used to think you could make a triangle with any three side lengths. (1, 2, and 4, sure why not?.. lost my train of thought. oh well I have both a major History and English project to work on anyway.) :)


geometry
incenter
geometric transformation
homothety
ratio
geometry theorems

[quote][color=darkred][b]Problem.[/b] Let $ I$ and $ G$ be the incenter and the centroid of the given triangle $ ABC.$ Let $ M,\,N,\,P$ be the midpoint of $ BC,\,CA,\,AB,$ respectively and let $ J$ be the incenter of triangle $ MNP.$
Then we have: $ I,\,G,\,J$ are collinear and $ GI=2\cdot GJ$[/color] [/quote]
[color=indigo][b]Proof.[/b]
We have $ \triangle ABC\sim\triangle MNP\implies \triangle IAB\sim\triangle JMN\implies \begin{cases}IA\parallel JM\\ IA=2\cdot JM\end{cases}$
Denote $ G'=AM\cap IJ\implies AG'=2\cdot G'M\implies G\equiv G'$
Hence $ I,\,G,\,J$ are collinear and $ GI=2\cdot GJ$[/color]

[size=150]Who can proposed some applications of this result?[/size]

Lemma 1: The incenter of a triangle is the Nagel point (N) of the medial triangle
Lemma 2: I, G and N are on the same line and $ NG \equal{}"IG$

so in this problem in the medial triangle J is the incenter, I the nagel point and G the centroid  :wink:

Dear Mathlinkers,
for a proof see
Honsberger R., Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, New Mathematical Library, 1995 p. 12
Sincerely
Jean-Louis

EG=DK
EG/DL=EK/KL=DH/DL=CE/CL   =>  CK is the bisector of angle C
similar  K is on the bisector of angle A, B
K is incenter of ABC
done!

[quote="ishfaq420haque"]can anyone kindly give the proof of lemma 1 by ¬[ƒ(Gabriel)³²¹º]¼[/quote]
Lemma 1: The incenter of a triangle is the Nagel point (N) of the medial triangle

We know that $ I,G,N$ are collinear with $ GN\equal{}2IG$. So a homothety with ratio $ \minus{}\frac 12$ and center $ G$ maps $ N \to I$ and the Lemma is proved. :)


A deck of seven cards numbered $ 0$ to $ 6$ are shuffled randomly and the following game is played. A card is turned face-up. If the card is a zero, the player loses. If the card is $ k>0$, then the next $ k$ cards in the deck are discarded facedown (or all of the cards, if fewer than $ k$ remain). The process is repeated until either a zero is turned faceup (a loss) or all of the cards are either turned up or discarded (a win). Compute the probability that the player wins.

[hide]
Let's count the complement (it's easier).

If $ 0$ is the first card, we lose with probability $ 1$.

If $ 0$ is the second card, we lose with probability $ 0$, as we are guaranteed to discard it.

If $ 0$ is the third card, we lose iff $ 1$ is the first card, which happens with probability $ \frac{1}{6}$.

If $ 0$ is the fourth card, we lose  iff $ 2$ is the first card, which happens with probability $ \frac{1}{6}$.

If $ 0$ is the fifth card, we lose iff $ 3$ is the first card, which happens with probability $ \frac{1}{6}$.

If $ 0$ is the sixth card, we lose if $ 4$ is the first card, if $ 1$ is first and $ 2$ is third, or if $ 2$ is first and $ 1$ is fourth.  One of these happens with probability $ \frac{1}{6}\plus{}2\left(\frac{1}{6}*\frac{1}{5}\right)\equal{}\frac{7}{30}$.

If $ 0$ is the seventh card, we lose if $ 5$ is the first card, if $ 1$ is first and $ 3$ is third, or if $ 3$ is first and $ 1$ is fifth.  One of these happens with probability $ \frac{1}{6}\plus{}2\left(\frac{1}{6}*\frac{1}{5}\right)\equal{}\frac{7}{30}$.

So the probability that we lose is $ \frac{1}{7}\left(1\plus{}0\plus{}\frac{1}{6}\plus{}\frac{1}{6}\plus{}\frac{1}{6}\plus{}\frac{7}{30}\plus{}\frac{7}{30}\right)\equal{}\frac{59}{210}$.

Thus the probability that we win is $ 1\minus{}\frac{59}{210}\equal{}\frac{151}{210}$.
[/hide]

[hide]Case 1 card: Either a 5 or 6 is chosen. 2/7 possibility.

Case 2 cards:
   Card is 1: then next card is 3-6
   Card is 2: then next card is 2-6
   Card is 3: then next card is 1-6
   Card is 4: then next card is 1-6

For total of $ (1/7)(4/6 \plus{} 4/6 \plus{} 5/6 \plus{} 5/6) \equal{} 3/7$.

Case 3 cards: the order is either 1,2,3 or 1,3,2, so $ 2(1/7)(1/6)(1/5) \equal{} 1/210$.

$ 2/7\plus{}3/7\plus{}1/210 \equal{} 151/210$.

[/hide]

Edit: Gah...messed up

E^(pi*i)=-1's is correct.
I also did complementary counting, but instead considered the possibilities depending on which card is first.


abstract algebra
limit
superior algebra
superior algebra unsolved

Let $ (N_i)_{i\in I}$ be a direct system of $ A$-modules and $ M$ an $ A$-module. Show that $ \lim_\to (N_i\otimes_AM)$ verifies the universal property of the tensor product $ \lim_\to (N_i)\otimes_AM$ and that there exists a canonical isomorphism $ \lim_\to (N_i\otimes_AM)\cong (\lim_\to N_i)\otimes_AM$.

[hide="Hint"]Play around with the representation of the corresponding universal properties or objects in terms of Hom's and plug them together.[/hide]


we have :
$1/x+1/y \geq 4/(x+y)$ .
  hence , we must pove that :
$2(sina+sinb )-2cosc \leq 3$
$2sina+2sinb-2cosc \leq 4cos\frac{c}{2}-4cos^{2}\frac{c}{2}+2 =-(cosc/2+1/2)^{2}+5/4 \leq 3$.
  oh , i think that it... that right

I think that you must to prove that:
$2(sina+sinb )-2cosc \leq 3$.

OK! Key of solutions is:
$(sina+sinb-1)^{2}+(cosa-cosb)^{2}\geq 0$

 :wink:

[quote="kostas"]What do you mean Macedonian olymbiad??? .you mean F.Y.R.O.M  olymbiad :!:[/quote]

I'm albanian, but I live in Macedonia. In this forum, i dont like te speak for politics!
 :wink:

See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=481962
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2819019

[quote=Fitim]If $a,b,c$ are angel from one triangle, prove that:
$\frac{1}{sina}+\frac{1}{sinb}\geq \frac{8}{3+2cosc}$[/quote]
In a triangle $ABC$, prove that
\[\frac{1}{sinA}+\frac{1}{sinB}\ge\frac{2}{cos\frac{C}{2}}\ge\frac{8}{3+2cosC}.\]
[url=https://artofproblemsolving.com/community/c6h497950p2819019]Triangle inequality (13)[/url]
[url=https://artofproblemsolving.com/community/c6h461515p2588679]Trigonometric inequality (4)
[/url]

[quote=sqing]In a triangle $ABC$, prove that
\[\frac{1}{sinA}+\frac{1}{sinB}\ge\frac{2}{cos\frac{C}{2}}.\]
[/quote]
$$\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC}\ge\frac{1}{cos\frac{A}{2}}+\frac{1}{cos\frac{B}{2}}+\frac{1}{cos\frac{C}{2}}$$
单壿:

[quote=sqing]In a triangle $ABC$, prove that
\[\frac{1}{sinA}+\frac{1}{sinB}\ge\frac{2}{cos\frac{C}{2}}.\]
$$\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC}\ge\frac{1}{cos\frac{A}{2}}+\frac{1}{cos\frac{B}{2}}+\frac{1}{cos\frac{C}{2}}$$
[/quote]
In a triangle $ABC$, prove that
$$\frac{1}{sin^2A}+\frac{1}{sin^2B}\ge\frac{2}{cos^2\frac{C}{2}}.$$
$$\frac{1}{sin^2A}+\frac{1}{sin^2B}+\frac{1}{sin^2C}\ge\frac{1}{cos^2\frac{A}{2}}+\frac{1}{cos^2\frac{B}{2}}+\frac{1}{cos^2\frac{C}{2}}$$

This problem was used in oral examination Ecole Polytechnique, France.

[quote=Fitim]If $a,b,c$ are angel from one triangle, prove that:

$$\frac{1}{sina}+\frac{1}{sinb}\geq \frac{8}{3+2cosc}$$[/quote]
[url=https://artofproblemsolving.com/community/c1068820h2006000p14040387]Macedonian 1996[/url]