AIME trigonometry logarithms algebra polynomial geometry

Eek… AIME in less than two weeks and right now, my main goal is not to come off like a total retard.  I haven’t taken pre-cal yet, and my friend told me that he heard there’s supposed to be a lot of it on there.  Any tips on formulas, theorems, etc to memorize?  For reference, I’ve currently got these on my list to study:

- Trig identities
- Logarithms
- Vieta’s Formula
- Stewart’s Formula
- Principle of Inclusion-Exclusion
- Fermat’s Theorem (and number theory)

I'm not sure whether Fermat's theorem is entirely useful, but who knows.
Power of a Point, and Ptolemy's Theorem, Rational Root Theorem, and all those good ol' polynomial stuff are good.

[quote="Art of Owna"]Actually, most of the problems can be solved without using pre-calc methods..except using pre-calc can make problems quicker.  There are lots of tricks for AIME..

But yea, trig identities are always helpful(including triple angles-but you can derive that..), and remember like sin15, sin75..you dont want to waste time deriving those.. and also similar triangles....and factoring tricks[/quote]

I wouldn't say that...

What about the trigonometry problems? You can't do anything without precalc... and many times, at least ~2 problems (which are 1 - 12) are geometry.

Last year on AIME 1, you could do every problem except for #12 using non pre-calc methods(if you even count trig as pre-calc) ..

And, trig is taught extensively in my algebra 2 class...(i consider precalc a basic repeat of algebra 2)

and also aime involves number theory, which isnt even taught in regular class books

So if ur a "book" person at math, most likely you wont do well at the AIME, it makes people think outside of the box..

My method is...
Brute force. :lol:
There are many problems that may be brute forced, but it significantly increases the chance of careless errors (I got four of the eight that I thought I got last year. I have a week of spring break before the AIME, so I am going to attempt (for about the 100th time) to force myself to read and solve the first two volumes of AoPS.

lol same here. they give you 3 hours, and i know i can only really do half of them. i just brute force the ones i have a chance on.

[quote="usaha"][quote]There are many problems that may be brute forced, but it significantly increases the chance of careless errors (I got four of the eight that I thought I got last year. I have a week of spring break before the AIME, so I am going to attempt (for about the 100th time) to force myself to read and solve the first two volumes of AoPS[/quote]

Really? I never brute forced any?  Can you give a couple examples of problems that are not AIME #1s?[/quote]
I brute forced #14 on last year's AIME by using Heron's formula a lot

Also, add this to the list of concepts:

How many solutions of nonnegative integers are there to $a+b+c+d=7$? In general, with any number of variables and any positive sum? Also, learn to recognize combinatorical problems (there are a lot of them!) which are really this problem in disguise

wow, i looked at that..i couldnt draw the tripod.. :blush:

It's hard to think of a specific example, but if you're just counting cases, you're likely to skip and miss one, throwing your answer off by 1 or 2. The best way to avoid missing one in my opinion is finding some sort of a pattern. Even then, you must be careful when brute forcing.

Brute force?  What's that?  Sorry, I'm kinda new to all of this :blush:

Example:

Question: How many non-negative ordered integer solutions $(a,b,c,d)$ are there to $a+b+c+d=7$?

Elegant solution: Use the balls-and-urns argument to get $\binom{10}{3}=120$.

Brute force:
If a=0, then b+c+d=7, if b=0, c+d=7, there are 8 solutions. if b=1, c+d=6, there are 7 solutions. if b=2, c+d=5, there are 6 solution In all there are 8+7+6+5+4+3+2+1=36 solutions when a=0
If a=1, then b+c+d=6 .... there are 7+6+5+4+3+2+1=28 solutions when a=1
If a=2, there are 6+5+4+3+2+1=21 solutions
If a=3, there are 5+4+3+2+1=15 solutions
if a=4, there are 10 solutions
If a=5, there are 6 solutions
If a=6, there are 3 solutions
If a=7, there is 1 solutoin
The total number of solutions is 36+28+21+15+10+6+3+1=120

What type of category usually appers the most on AIME's?

Combinatorics
Algebra
Number Theory
Geometry
Trigonometry

Eh, probably algebra, just because many a problem boils down to solving an algebra equation.
Also, if you're awesome at algebra, you can brute force many a problem.

oh, and it helps to have some exp w/ combinatorics and probability in the back of ur head...there was one year that i did which had like 4 probability questions... :maybe: ...my score dipped a lot...

Sorry for the revive guys, but there seems to be some general consensus that math competitions have been getting harder over the past few years, so would you guys say that it would be necessary to add some more concepts to this list (esp. Since it's linked to on the aops wiki for AIME studying) or no?

Honestly, in my recent days doing AIMEs, AIME is kind of like a marathon. You gotta have the stamina to bash (esp.  if you're not elegant like me) and make an intensive amount of computation while still double checking every step.

To help on AIME, I would say USE ALL of your time. I hear some kids just walk out after the first hour. DON'T. You have a ton of time to check again and again the ones you can do, and actually try the beastly-looking problems. After all, you probably can't do much after the first 8 problems, take your time and make sure you get points for problems you actually know how to do.

(I'm not sure: I'm in 6th grade and this is my first time taking AIME - tomorrow!)


Let $n,p$ be integers such that $n>1$ and $p$ is a prime. If $n\mid p-1$ and $p\mid n^3-1$, show that $4p-3$ is a perfect square.

http://www.artofproblemsolving.com/viewtopic.php?t=5863

[color=#FF0000][Mod:
Also posted here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=64244
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=297600
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=367955
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=340745

The problem has appeared in several contests, I think it originally appeared in the 2002 Czech-Polish-Slovak Olympiad. However,  maxal has pointed a stronger version appeared in a Russian journal dated 2001 (last post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=64244]here[/url]).][/color]


A shop is visited by group of 5 people. Amongst any three of these there are always two who know each other and two who do not.
Prove its possible to seat these five people at a round desk in such a way that all five have acqiuianctances sitting on both their eft and their right hand sides.

Bomb

this is rather straightforward. consider the corresponding graph of acquaintance.we claim that in this graph every vertex has degree two.if possible let a vertex $v$ have degree $0$ or $1$. then in either case among the other $4$ vertices there cannot be any edges absent else we would get three vertices with no edges induced amongst them buth that would create triangles.similarly if a degree $3$ vertex exists, then the neighbors of that vertex must all induce an edgeless graph or else a triangle is formed and again that is ruled out. hence the graph is regular of degree two and that leaves only one possibility-a cycle.


These problems are probably very easy, but I can't remember how to do them.

1. –xy + zy for x = 1, y = 2 and z = 3

2. 5x + 3 – x = 25

3. 3x – 7 + 4x = 10x + 2

4. Solve for n: 8 < n - 6

5. Solve for x: 2x – 10 > 4x + 3

6. Find the direct variation equation for the point (3,4).

Thanks :ninja:

please don't post such trivialties in an olympiad advanced section problem. they are indeed very easy.


algebra
polynomial
number theory proposed
number theory

is the set of the prime number whose form is 9k+1 infinite ?

The following problem has received elementary solutions on the forum:

Given a prime $q$ and a positive integer $n$, there are infinitely many primes of the form $kq^n+1$.

Btw, all of these results are particular cases of Dirichlet's theorem (I know, I know, it is not really elementary)...

Pierre.

but the case of $an+1$, $n$ constant, is quite easy using the n-th cyclotomic polynomial $\Omega_n(x)$ with the observation that if $\gcd(p,n)=1$, then $p|\Omega_n(x)$ for some $x$ iff $p\equiv 1\mod n$.

Well,frejer,or I can call you PNT.  frejer and PNT is ONE person , i like to tell all of you about this. Does you feel good, PNT??? :lol:  :lol:  :lol:


I'm very sad cause nobody is interested in my problem :(

i am interested in your problem, but not intelligent enough to solve it(even though i am grade 10 :rotfl:). here is my progress (which may or may not be significant):
$x+y+z=t^{2}$
$x^{2}+y^{2}-z^{2}=t^{3}$
$x^{2}+y^{2}-z^{2}=t(x+y+z)$
$\left(x-\frac{t}{2}\right)^{2}+\left(y-\frac{t}{2}\right)^{2}=\left(z+\frac{t}{2}\right)^{2}+\left(\frac{t}{2}\right)^{2}$

I observe this and i  try to solve this way, but i found another solution ( i won't tell you about it because i'm waiting to find you this).
        Sorry, but this isn't the original RMO shortlist ( i made a mistake at a sign when i translate it) but i solve it in this form.
        The original form was:
           [color=darkred][b]Find all the integers such:
$x+y+z=t^{2}$
$x^{2}+y^{2}+z^{2}=t^{3}$.[/b][/color]
        I hope you will find ( finally) a solution to my first system, but i'm interested in seeing your solution for the second one, too.
[color=red]Good luck![/color]

This problem is easy. We use inequalities to get something like $t\leq 3$.Then most of what`s next is casework.

Yes that was my idea too, and i became very surprised because no one solve it

[quote="andyciup"]This problem is [b]easy[/b]. We use inequalities to get something like $t\leq 3$.Then most of what`s next is casework.[/quote]
You are wrong - the problem isn't easy

[hide="It is very easy."]Let $x=t^{2}+u, \ \ y=t^{2}+v$ then from the 1st equation we have $z=-t^{2}-u-v$. Thus we have general solution of the 1st equation: $t, u, v$ may be arbitrary. To find the connection between them just put in the 2nd equation. After easy cancellations this gives $2uv=t^{4}-t^{3}$. For any $t$ the RHS of the last equation is even hence the factorization $2uv=t^{4}-t^{3}$ allways exists. This produses the all solutions: $t$ - arbitrary, $2uv$ - arbitrary factorization of $t^{4}-t^{3}$ and $x=t^{2}+u, \ \ y=t^{2}+v, \ \ z=-t^{2}-u-v$.[/hide]

for 8th grade isn't very easy, is medium or even easy(for NMO) :)

Yes, surely. It reqires mathematical culture.
I've started with those general solution of 1st equation: $x=t^{2}+u, \ \ y=v, z=-u-v$ considering $t$ as parameter. It requred new variable. Just to delete this extra step I've replaced the solution $(t^{2}, 0, 0)$ of the first linear equation by $(t^{2}, t^{2},-t^{2})$.

[quote="mychrom"]... and i became very surprised because no one solve it[/quote]
It is not surprising thing as many people use to see those topics in the 1st turn which are not so fresh but have min answers.  :lol:


topology
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

let $R$ be a notherian local ring having maximal ideal $m$ .let {$p_{i}$} be a countable family of prime ideals of $R$ .$a$ an ideal of $R$ ,and $x\in R$ .if $R$ is complete with respect to $m-adic$ topology ,then $x+a\in \cup_{i\geq 1}p_{i}$ implies that $x+a\in p_{j}$ for some $j\geq 1$.

I'll take $x=0$, in order to keep things simple (I don't think the general case is significantly different, but when I solved the problem, I did not take that $x$ into consideration).

First of all, just as we did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114575]here[/url], we can assume that $\{\frak p_{n},\ n\in\mathbb N^{*}\}$ is a minimal system covering $\frak a$, in the sense that no proper subset cover $\frak a$. I'll leave this as an exercise (we cannot make this assumption for any ring; it is essential that we are working in a Noetherian local ring). This means that for all $i\in\mathbb N^{*}$, we can find elements $x_{i}$ belonging to $\frak p_{i}$ but not to $\frak p_{j},\ j\ne i$.

Using the $x_{i}$ mentioned above, we can define an element $t$ of $\frak a$ which does not belong to any of the $\frak p_{i}$. This will give us the contradiction we’re looking for. We define $t$ inductively as follows:

In the first step, take $t_{1}$ to be an arbitrary element in $\frak a\setminus \frak p_{1}$. Now assume $n\ge 2$ and that $t_{n-1}\in\frak a$ has been defined such that it does not belong to any of the ideals $\frak p_{i},\ i=\overline{1,n-1}$. If $t_{n-1}$ does not belong to $\frak p_{n}$ either, then take $t_{n}=t_{n-1}+x_{1}\cdot\ldots\cdot x_{n}$. If, on the other hand, $t_{n-1}$ does belong to $\frak p_{n}$, take $t_{n}=t_{n-1}+x_{1}\cdot\ldots\cdot x_{n-1}\cdot x_{n+1}$. Either way, we get an element $t_{n}\in\frak a$ which does not belong to any of the ideals $\frak p_{i},\ i=\overline{1,n}$, and such that $t_{n}-t_{n-1}\in\frak m^{n}$. Because the ring is $\frak m$-adically complete, the Cauchy sequence $(t_{n})_{n}$ converges to an element $t$, and we can easily see from the construction of the $t_{n}$ that $t$ has the required properties: it belongs to $\frak a$ but to none of the ideals $\frak p_{i}$.

[quote="grobber"]
First of all, just as we did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114575]here[/url], we can assume that $\{\frak p_{n},\ n\in\mathbb N^{*}\}$ is a minimal system covering $\frak a$, in the sense that no proper subset cover $\frak a$. I'll leave this as an exercise.
[/quote]

My "proof" for this doesn't seem to work :(. Sorry for the inconvenience.


probability
expected value
probability and stats

The problem says:
"What is the expected value when a 1 dollar lottery ticket is bought in which the purchaser wins exactly10 million dollars if the ticket contains the six winning numbers chosen from the set {1, 2, 3, ..., 50} and the purchaser wins nothing otherwise?"

I have problem understanding the meaning of "expected value" in this particular winning money question because the value of money could be considered value as well.

I wonder if I should sum the Expected Probability x Expected Money Value or Expected Probability x Expected Number of Winning Number.

It may lead to completely two different answers.

Right now I am doing
[(6/50)^6*(1000000-1) ]+
[((44/50)^6+(44/50)^5*(6/50)^1+(44/50)^4*(6/50)^2+(44/50)^3*(6/50)^3)+(44/50)^2*(6/50)^4+(44/50)^1*(6/50)^5)*(-1)]

The probability for a winning number is (6/50)
The probability for a non-winning number is (44/50)
The first [] term has (1000000-1) in it because the expected outcome for the player will be winning ten millions minus 1 dollar spent. For the same reason, the outcome value for all the other cases is -1, since the player will lose one dollar for nothing gained.

Please let me know if I am understanding the question in the right way.

What is the expected number of the winning number?  There are six numbers.  Anyway, it's obvious that if the person wins, the possible winning sets of numbers are evenly distributed over the possible sets of numbers, so it's not a very interesting question.

The question is, what is the expected value, [i]in dollars[/i], of the purchaser's net profit or loss.

In my opinion the question is slightly ambiguous because it does not say if the person's choice, or the choice for the winning numbers, is with or without replacement.


the sum of the two digits,in a 2-digit number is 10. when 13 is added to the number ,the sum of the digits of the resulting number becomes 5.what was the number???
please help :maybe:

Hi Earth :-)
[hide="solution"]$ 3$ such two digit numbers are possible which satisfy the above property
$ 19, 28, 37$[/hide]

[quote="earth"]did you try by forming equations???
  x and y be two digits. then,x+y=10
   now ,10x+y+13=10(x+1)+(y+3),
now as sum of digits of new number is 5,  (x+1)+(y+3)=5
 which means x+y=1 ??? leads to contradiction! :o[/quote]
Did you consider a regroup? It is $ \boxed{37, 28, or 19}$.

Firstly, the condition that the sum of the digits is $ 10$ and that the number has $ 2$ digits gives the following result:

$ N \in \{ 19,28,37,46,55,64,73,82,91\}$

$ 19\plus{}13\equal{}32 \Rightarrow s\equal{}5$
$ 28\plus{}13\equal{}41 \Rightarrow s\equal{}5$
$ 37\plus{}13\equal{}50 \Rightarrow s\equal{}5$
$ 46\plus{}13\equal{}59 \Rightarrow s\equal{}14$
$ 55\plus{}13\equal{}68 \Rightarrow s\equal{}14$
$ 64\plus{}13\equal{}77 \Rightarrow s\equal{}14$
$ 73\plus{}13\equal{}86 \Rightarrow s\equal{}14$
$ 82\plus{}13\equal{}95 \Rightarrow s\equal{}14$
$ 91\plus{}13\equal{}104 \Rightarrow s\equal{}5$

In conclusion, the solution is $ N \in \{19,28,37,91 \}$.

Remember to regroup:
Answers:
[i]19, 28, 37[/i][/i]

[quote="earth"]hey, i understand the solutions,but cannot understand why a contradiction is coming in my solution.
 i did regroup to find the new numbers.
can you explain me more clearly, what is the fault in it, that leads to a contradiction??? :([/quote]

Got that. There is actually no contradiction here. The point is you have made a minor mistake.
[b]10x+y+13=10(x+1)+(y+3)[/b].
The left hand side can be analyzed and arranged in several ways and yours is just one of those. What if I rewrite like this 10x+y+13= 10(x+2) + (y-7) . See :) .
If you wish to solve this by forming equation, though unnecessary, here is my suggested solution:
[b]Let x and y be two digits of the original number; m and n be two digits of the resulting number.
We have : x+y=10
            10x+y+13=10m+n (*)
               m+n=5 (we deduce from this that m<=5, n<5)
Thus y=10-x and n=5  -m
Replacing y and n into (*), we obtain m - x = 2. We put in m = 3,4,5 respectively to find x, then y[/b]
Hope this helps.


Given is a halfplane with points $A$ and $C$ on its edge. For every point $B$ of this halfplane, consider the squares $ABKL$ and $BCMN$ lying outside the triangle $ABC$. Prove that the lines $LM$ (as $B$ varies) pass through a fixed point.

Is this problem correct?I drawing the conditions and the line doesn't  have a fixed point

I constructed the figure using Geometer's Sketchpad, and there is a fixed point. Make sure that you're fixing A and C, and that the line you're considering connects the vertices of the squares that are furthest apart.

Unfortunately I haven't made much progress on the solution.

According to problem 2 in the topic  [url=http://www.artofproblemsolving.com/viewtopic.php?t=331337]3 geometry problems[/url], the lines AN,CK,LM concur at a point P such that <APC is right and LM is its external bisector. P moves on the semicircumference with diameter AC constructed in the given half-plane, thus LM always passes through its midpoint Q, obviously fixed.


inequalities
geometry
inradius
circumcircle
trigonometry
geometry proposed

Prove that in acute-angled traingle ABC if $r$ is inradius and $R$ is radius of circumcircle then: \[a^2+b^2+c^2\geq 4(R+r)^2\]

It is known $R + r = R(\cos A + \cos B + \cos C)$ so what we want to prove is $\sin^2 A + \sin^2 B + \sin^2 C \ge (\cos A + \cos B + \cos C)^2$ or, if $x^2 + y^2 + z^2 + 2xyz = 1, x^2 + y^2 + z^2 + xy + xz + yz \le \frac{3}{2}$. This is 19d in Old and New Inequalities.

[quote="shobber"]If $\cos^2{A}+\cos^2{B}+\cos^2{C} \leq \frac34$ is true, then the problem is true.[/quote]

But $\cos^2 A+\cos^2 B+\cos^2 C\leq\frac34$ is not true, exactly the opposite is true: $\cos^2 A+\cos^2 B+\cos^2 C\geq\frac34$.

Here is another proof of the inequality $a^2+b^2+c^2\geq 4\left(R+r\right)^2$: As MysticTerminator noted, this inequality is equivalent to $\sin^2 A+\sin^2 B+\sin^2 C\geq\left(\cos A+\cos B+\cos C\right)^2$. Since $\sin^2 x=1-\cos^2 x$ for any angle x, this becomes

$\left(1-\cos^2 A\right)+\left(1-\cos^2 B\right)+\left(1-\cos^2 C\right)\geq\left(\cos A+\cos B+\cos C\right)^2$
$\Longleftrightarrow\ \ \ 3-\left(\cos^2 A+\cos^2 B+\cos^2 C\right)\geq\left(\cos A+\cos B+\cos C\right)^2$
$\Longleftrightarrow\ \ \ \left(\cos A+\cos B+\cos C\right)^2+\left(\cos^2 A+\cos^2 B+\cos^2 C\right)\leq 3$
$\Longleftrightarrow\ \ \ \left(\cos B+\cos C\right)^2+\left(\cos C+\cos A\right)^2+\left(\cos A+\cos B\right)^2\leq 3$.

But this inequality was proved on http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 . (Another of my memorial inequalities: When it was posted in July 2004, I wasn't able to solve it; all I could do was, after Fuzzylogic gave a nice solution, to rewrite it in a different way. Later, in February 2005, the same inequality was given to me by Prof. Gronau at the IMO training camp. At that time, I had absolutely forgotten about the solutions given on ML and started solving it again. After 4 hours, I came up with a solution - it was exactly the same as the one posted by Fuzzylogic...)

  darij

As I knowed , this ineq can be written in the form : $ p^2 \geq 2R^2 +8Rr+3r^2 $ , which is found by A.W.Walker
  Because $ R \geq 2r $ and $ 2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)} \geq p^2 $ , so we just need to prove one in two ineqs :
  $ (2R+r)^2 \geq 2R^2+8Rr+3r^2 $ or $  2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)} \geq 2R^2+8Rr+3r^2 $ $ \iff R^2-2Rr-r^2 \geq 0 $ or $ R^2-2Rr-r^2 \leq 0 $ , which is true
  The equality holds for $ R=2r $ or $ R=(\sqrt2 +1)r $
 This ineq has an equivalent form : $ (HA+HB+HC)^2 \leq AB^2+AC^2+BC^2 $ , where $ H $ is the orthocenter of triangle $ ABC $  ;)

   PS : This is not my solution , it's get from my document .

Here are my solutions for this problem :
 [b]First solution :[/b]
  I'll prove that : $ (HA+HB+HC)^2 \leq a^2+b^2+c^2 $
  We have $ AH.AA' = bc.cosA = \frac{b^2+c^2-a^2}{2} $ , where $ A' $ is the feet of the altitude from $ A $ to $ BC $
  So $ AH.AA' +BH.BB' +CH.CC' = \frac{a^2+b^2+c^2}{2} $
  Apply Cauchy's ineq , we have :
   $ (HA+HB+HC)^2 = (\sum \sqrt{AH.AA'}.\sqrt{\frac{AH}{AA'}})^2 \leq \frac{a^2+b^2+c^2}{2}. \frac{AH}{AA'} = a^2+b^2+c^2 $
  [b]Second solution :[/b]
 We have : $ 2cosA.cosB = \sqrt{sin2A.cotA.sin2B.cotB} \leq \frac{1}{2}(sin2AcotB+sin2BcotA) $
 We just need to show : $ \sum{cotA(sin2B+sin2C)= -2(cos2A+cos2B+cos2C)} = 3-2(cos^2A+cos^2B+cos^2C) $ , which is true because $ cotA(sin2B+sin2C)= -2cos(C+B).cos(C-B) = -(cos2B+cos2C) $  ;)

   Now , we have over 8 solutions for this problem  :)

[quote="MysticTerminator"]$\sin^2 A + \sin^2 B + \sin^2 C \ge (\cos A + \cos B + \cos C)^2$ or, if $x^2 + y^2 + z^2 + 2xyz = 1, x^2 + y^2 + z^2 + xy + xz + yz \le \frac{3}{2}$. This is 19d in Old and New Inequalities.[/quote]

How to prove it? Can you show?

I can only prove that $xy+yz+zx\leq\frac{3}{4}$

[b][u]Remark.[/u][/b]

$4(R+r)^2\le a^2+b^2+c^2\Longleftrightarrow 2R^2+8Rr+3r^2\le s^2\ \ (1)$

$16Rr-5r^2\le 2R^2+8Rr+3r^2\Longleftrightarrow 2R^2-8Rr+8r^2\ge 0\Longleftrightarrow 2(R-2r)^2\ge 0.$

Therefore, if the triangle $ABC$ is acute, then the Blundon's inequality  $16Rr-5r^2\le s^2$ is more weak than the inequality (1).

It's not Blundon's inequality. In fact this inequality is due to Gerrstern. ;)

O.K. What is the Blundon's inequality then ?

I think $s^2\geq 16Rr-5r^2$ is Columbien -Doucet ineq(1872)

Guys, isn't it clear that some inequalities just get lots of different names? You are all talking about the same inequality, please don't worry about the name :)

What is the maximum value of $\cos{\alpha}+\cos{\beta}+\cos{\gamma}$ when $\alpha + \beta + \gamma =360 ^o$
I see that it is 3 ( $\alpha=0, \beta = 0 \gamma = 360$ )
But what if this are the angles between circumradiuses and sides of a triangle?

We have $\triangle ABC$ acute. Therefore, $a^2, b^2, c^2$ are sides of some triangle. Therefore, we can let $a^2 = y+z, \ b^2 = z+x, \ c^2 = x+y$ for some positive reals $x,y,z$. Also, let $x$ be the largest among $x,y,z$ so that $a$ is the smallest side, and therefore, $h_a$ the largest altitude.
We have, \[ \cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{x}{\sqrt{(x+y)(x+z)}}\]
and therefore, \[ \sin A = \sqrt{\frac{xy+yz+zx}{(x+y)(x+z)}} \]
Thus, we have, \[ R = \frac{a}{2 \sin A} = \sqrt{\frac{(x+y)(y+z)(z+x)}{2(xy+yz+zx)}}\]
Also, \[ 1+ \frac rR = \cos A + \cos B + \cos C = \frac{\sum x\sqrt{(y+z)}}{\sqrt{(x+y)(y+z)(z+x)}} \]
Now, let us state the problem in terms of $x,y,z$. We have to prove,
\[2(x+y+z) \ge 4(R+r)^2 = (2R)^2 (1+ \frac rR)^2 = \frac{(x+y)(y+z)(z+x)}{(xy+yz+zx)} \cdot \frac{(\sum x\sqrt{(y+z)})^2}{(x+y)(y+z)(z+x)}\]
\[ \iff 2(x+y+z)(xy+yz+zx) \ge (\sum x\sqrt{(y+z)})^2 \]
Now, we have by Cauchy Schwarz Inequality,
$(\sum x\sqrt{(y+z)})^2 \le (\sum x(y+z)) (\sum x) = 2(xy+yz+zx)(x+y+z)$
We are through. :)


Substitute each of the letters by a different decimal digit from 0 to 9 to satisfy this cryptarithmetic addition problem. None of the numbers may contain a leading zero.

SIX + NINE + ELEVEN + SIXTEEN + NINETEEN = SIXTYONE

[hide]
Lets rewrite the problem and number the columns 1-8 from left to right.
1 2 3 4  5 6 7 8
[color=white]space space[/color]S  I  X
[color=white]space spa[/color]N  I  N  E[color=white]
space [/color]E  L  E  V  E  N[color=white]
spac[/color]S  I  X  T  E  E  N
+ N  I  N  E  T  E  E  N
-----------------------
[color=white]sp[/color]S  I  X  T  Y  O  N  E

From column 1, we see that $ S=N+1$ and from column 2 that $ S=8,9$ ($ S\neq0$ and the number carried from column 3 can be 1 or 2).

$ S=8$ is easily eliminated because that forces $ N=7$ and $ X=9$ (column 8), which implies the impossible statement $ E+I+7+C_4=29$ (column 3), where $ C_n$ is the number carried from the nth column.

So we know that $ S=9$, $ N=8$, and $ X=6$. We also know that $ E+I+8+C_4=16\Leftrightarrow E+I+C_4=8$ (column 3) and $ I+8+3E+C_8=I+8+3E+3=8\Leftrightarrow I+3E\equiv7 (mod 10)$ (column 7). $ C_4=0$, $ 1$, or $ 2$ so $ E+I=8$, $ 7$, or $ 6$. $ E+I=7$ is the only that leads to integer solutions to the system of equations and $ E\neq0$, so $ E=5$ and $ I=2$.

All we have left is $ 0$, $ 1$, $ 3$, $ 4$ and $ 7$ and $ L$, $ T$, $ V$ and $ O$. From column 6, $ S+I+V+2E+C_7\equiv9+2+V+2*5+2\equiv23+V\equiv3+V\equiv O (mod 10)$ so the only possible $ (V,O)$ are $ (0,3)$, $ (1,4)$, $ (4,7)$ and $ (7,0)$.

Similarly, from column 5 we have $ N+E+2T+C_6\equiv8+5+2T+C_6\equiv13+2T+C_6\equiv3+2T+C_6\equiv Y (mod 10)$. From the cases determined above, $ C_6=2$ or $ 3$. $ C_6=3$ ($ (V,O)=(7,0)$) yields no solutions for $ (T,Y)$, so $ C_6=2$ and $ (V,O,T,Y)=(0,3,1,7)$ or $ (4,7,3,1)$. The first forces $ L=9$, an already used number, but the second forces $ L=0$.

Soo the solution is $ S=9$, $ I=2$, $ X=6$, $ N=8$, $ E=5$, $ L=0$, $ V=4$, $ T=3$, $ Y=1$, and $ O=7$: $ 926+8285+505458+9263558+82853558=92631785$.

[/hide]


On sides $AB , BC , CA$of a triangle $ABC$ we take points $M , K , L$ with $ML // BC$ and $MK // AC$ . Segments $AK , ML$ meet at $Q$  and segments $BL , MK$ meet at  $P$ . Prove that $PQ // AB$.

[u]Babis[/u]

You must know that Moskow Olympiads is the greatest source of nice and original problems. I have a book with all these olympiads from 1935 untill 1998 ( hints, answers , solutions).
 I think that this problem is very good - Thales and its converse. I shall give the solution later , if nobody find it.
 I want to clear the topic with at least one solution.

 [u]babis[/u]

[hide]Since $AC \left|\right| MK$ then $\angle PMQ = \angle MLA$.

Since $ML \left|\right| BC$ then $\angle PMQ = \angle MKB$.

Also $\angle CAB = \angle KMB$ and so $\Delta AML \sim \Delta KMB \sim \Delta ABC$.

$\frac{AC}{AB} = \frac{PM}{MB}$ and $\frac{BK}{QM} = \frac{AB}{AM}$.

Multiply these we get $\frac{AL}{AM} = \frac{PM}{MB} \cdot \frac{BK}{QM}$.

Also $\frac{AL}{AM} = \frac{MK}{MB}$ $\Longrightarrow$ $\frac{MK}{MB} = \frac{PM}{MB} \cdot \frac{BK}{QM}$.

Hence $\frac{MK}{BK} = \frac{PM}{QM}$ and then we get $\Delta PQM \sim \Delta MBK \sim \Delta ALM$.

So $\angle QPM = \angle PMB$ and $PQ \left|\right| AB$.[/hide]

Very nice and simple solution , Andreas !

 [u]Babis[/u]


Find all continuous functions $ f$ such that $ f(x) \plus{} f(x^2) \equal{} 0$ for all real numbers $ x$.

$ f(-x)=f(x)$ so we need consider on $ (0,+\infty)$
From this we has $ f(x)=f(x^4)$
So $ f(x)=f(x^{\frac{1}{4^n})}$
Because $ \lim_{n\to\infty}x^{\frac{1}{4^n}}=1$ and $ f(x)$ is continious so $ f(x)=f(1)=f(0)=0,\forall x\in R$
[u]Result[/u] $ f(x)=0$

$ \star x \equal{} 1\longrightarrow f(1) \equal{} 0$

$ \star x \equal{} 0\longrightarrow f(0) \equal{} 0$

$ \star x\longrightarrow \minus{} x$ then $ f(x) \equal{} f( \minus{} x)$

So we should consider with $ x\ge 0$

$ \star x\ge 1$ then $ f(\sqrt [4n]{x}) \equal{} f(x),\forall x\in\mathbb{R},x\ge 1$

because $ f$ is continuous then $ f(x)\to f(1)$ when $ n\to\infty$ so $ f(x) \equal{} f(1), \forall x\in\mathbb{R},x\ge 1$

$ \star 0 < x < 1$ then $ f(x) \equal{} f(x^{4n})\to f(1)$ so $ f(x) \equal{} f(1),\forall x\in\mathbb{R},0 < x < 1$

But $ f(x)$ is continous so if we chose a sequence $ x_n\to 0$ then
$ f(x_n)\equal{}f(0)\equal{}f(1)\equal{}0$ :wink:

Because $ \lim_{n\to\infty}x^{\frac {1}{4^n}} = 1$ and $ f(x)$ is continious so $ f(x) = f(1) = f(0) = 0,\forall x\in R$

Can you prove this or explain it to me :?:

Oh yes .
For each $ x > 0$
$ \lim_{n\to\infty}x^{\frac {1}{4^n}} \equal{} 1$
So if $ x > 0$ then $ f(x) \equal{} f(1),\forall x > 0$
Now consider a sequence $ x_n\to 0$ and $ x_n > 0$ then we has $ f(1) \equal{} f(x_n) \equal{} f(0) \equal{} 0$
Implies that $ f(x) \equal{} 0$


[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge \sqrt{ 8\cdot\frac{a+b+c}{\sqrt[3]{abc}}-6 }\][/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ are positive numbers. Prove that:
\[ \sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge \sqrt{ 11\cdot\frac{a+b+c}{\sqrt[3]{abc}}-15 }\]

I believe that my transformations help.
But not without many efforts !


Given that $ x$ and $ y$ are integers and $ 2^x\cdot 3^y \equal{} 144$, what is the product of $ x$ and $ y$?


geometry
parallelogram
ratio
vector
inequalities
angle bisector
triangle inequality

Problem:

In parallelogram ABCD, the bisector of  :ang: BAC intersects BC at E. Given that BE + BC = BD, compute BD/BC

My solution:

[hide]First, lets extend AE to line DC at F. Draw FB and FC as well. FB is parallel to AB and BF parallel to AC, so ABFC is a parallelogram and thus

FC = AB = DC and FB = AC.

Lets call  :ang: BAC 2 :alpha: (And of course,  :ang: EAC =  :alpha: ). By alternate interior angles,  :ang: AFC =  :alpha: as well, and because  :ang: BAC =  :ang: BFC,  :ang: BFA =  :alpha: .

Thus,  :tri: BFE  :sim:  :tri: CAE and  :tri: CFE  :sim:  :tri: BAE, which implies that CE/BE = FE/AE = CF/AB  and that CE/BE = AC/BF = AE/FE, which leads to AE = FE and thus CE = BE.

So going back to the original problem, BD/BC = (BE+BC)/BC = 1 + 1 = 2.[/hide]

Please leave comments about my solution... criticism is well-recieved (if it has acceptable reasoning  :-x )

Please draw a diagram? If I were to draw the diagram described by your problem, I would not be able to reach the parallel statements (FB//AB, BF//AC).

Your solution is wrong- ABFC is not a parallelogram

The correct answer is [hide]BD/BC =  :sqrt:2[/hide]

Proof:[hide]

Since BE is an angle bisector in triangle ABC, CE/BE = AC/AB and BE/BC = AB/(AB+AC). We are given BD = BE + BC = BC*(AB/(AB+AC)+1) = (BC*2AB+BC*AC)/(AB+AC).

Rearranging, (2BC - BD)*AB = (BD-BC)*AC    (1)

By standard vector calculations, BD^2 + AC^2 = 2(AB^2 + BC^2)  (2)

We eliminate AC from (1) by squaring and substituting from (2):

(2BC-BD)*AB = (BC-BD)*(2AB + 2BC - BD)

AB(4BC - 4BC*BD + BD - 2BC + 4BC*BD - 2BD) = (BC-BD)*(2BC - BD)

AB(2BC - BD) = (BC-BD)*(2BC - BD)

Therefore either AB + BC = BD, AB + BD = BC, or BD = 2BC.

The first two are impossible by the triangle inequality in triangle ABD (using BC = AD), so  BD/BC =  :sqrt:2.

[/hide]

Heh... I knew it was faulty... While I was typing it up, I was telling myself that someone is going to say that that is just plain wrong. Im gonna do it later again.

My solution:

[hide]First, consider diagram 1. Take BQ on BD such that BQ = BE; Join AQ and QE. (AQ is NOT necessarily the bisector of angle BAC.) Then, we must have DQ = AD = BC. By isosceles triangle, :ang: AQD = a + x. Then :ang: BOC = x + (a + x) = a + 2x. Then :ang: OBC = 180o - a - a - 2x = 180o - 2a - 2x. Then by isosceles triangle, :ang: BQE = :ang: BEQ = a + x. Therefore :ang: BQE = :ang: AQD. This means that A, Q, and E are collinear, with Q located on BD.

Then consider diagram 2. P is the intersection of AE with BD. Clearly A, P and E are collinear, with P located on BD. Therefore P and Q are the same point. Thus we know that BP = BE, and :ang: BAE = :ang: EAC = x. Now, we know that :ang: APD = a + x; but by external angles :ang: APD = x + :ang: ABD, therefore :ang: ABD = a. Then, by parallel lines, :ang: BDC = a. Now, we have :tri: BOC :sim: :tri: BCD. Then, by ratios,

BO:BC = CB:BD. BO = half of BD, so

(1/2 * BD):BC = BC:BD,

BC2 = 1/2 * BD2,

BD:BC = square root of 2.[/hide]


1)Show that $n^4+4$ is composite number for all $n>1$.
2)Show that $n^4+n^2+1$ is composite if $n>1$.
3)Show that if $m^4+4^n$ is prime, then m is odd and n is even, except when $m = n =1$.

[quote="Rushil"]Your proofs are incomplete!!!!!![/quote]
Why? he has shown they have factors so they are composite.

What about the third one ,Is there anybody to solve it :?:
I can give you hint if you want :police:

[quote="Rushil"]he hasn't disproved that one of hose factors can't be one[/quote]
Wrong. You don't disprove. You prove all factors cannot be one.

[hide="Short proof"]1)$n^2+2n+2>1$, $n^2-2n+2=(n-1)^2+1>1$ for $n>1$
2)$n^2+n+1>1$, $n^2-n+1=(n-\frac12)^2+\frac34>1$ for $n>1$[/hide]

[quote="a_vakilian"]3)Show that if $m^4+4^n$ is prime, then m is odd and n is even, except when $m = n =1$.[/quote]

[hide]
If it's prime, it must be odd (except 2). The parity of $m^4$ is the same as the parity of $m$.
$4^n$ is even no matter what. The sum $\text {odd} + \text {even} = \text {odd}$, so that means $m$ must be odd.

Now for the second part, we must think about the "prime" part. Let's look at $4^n \bmod {10}$.

\begin{tabular}[b]{r|ccccc}\hline
n & 1 & 2 & 3 & 4 & \cdots \\\hline
4^n \bmod {10} & 4 & 6 & 4 & 6 & \cdots \\\hline
\end{tabular}

If $n$ is even, then it's 6. If $n$ is odd, then it's 4.
A prime cannot $\equiv 0, 2, 4, 5, 6, 8 \mod {10}$, so it must be $\equiv 1, 3, 7, 9$.

$m^4$, for odd $m$:
$\{ 1, 1, 5, 1, 1, ... \}$ Repeats every 5.
Let's consider each case.

$m^4 + 4^n$
The possibilities for even $n$ are $7, 1$.
The possibilites for odd $n$ are $5, 0$.

EDIT: Hmm... I also proved that the last digit of $m^4+4^n$ is 0, 1, 5, or 7, depending on the parity of $m$ and $n$! :)

Only even $n$ works. Therefore $m$ must be odd and $n$ be even and our proof, surprisingly, is complete.
[/hide]

Wow! I did it! I'm very proud of myself. :D
Where did you get this problem?

[quote="chess64"]$m^4 + 4^n$
The possibilities for even $n$ are $7, 1$.
The possibilites for odd $n$ are $5, 0$.

EDIT: Hmm... I also proved that the last digit of $m^4+4^n$ is 0, 1, 5, or 7, depending on the parity of $m$ and $n$! :)[/quote]There's a small mistake here.. the possibilities for odd $n$ are $5$ and $9$, as in $5^4+4^1=629$.

It was from An Introduction To The Theory Of Numbers (Fifth Edition)
but I have better sloution:
Abviously m is odd ,Now we consider that n is odd too, so it is"
$m^4 + 4^{2k+1}=m^4 + 4*2^{2k}$ And it seems smiliar $a^4 + 4b^4$.And it is composite,if we add and substract $4a^2b^2$ it solve:
$a^4+4b^4+4a^2b^2-(4a^2b^2) = (a^2+2b^2)^2-(2ab)^2 = (a^2+2ab+2b^2)(a^2-2ab+2b^2)$ :lol:


Let $ a_1 \equal{} a_2 \equal{} 97$ and $ a_{n \plus{} 1} \equal{} a_na_{n \minus{} 1} \plus{} \sqrt {(a^{2}_{n} \minus{} 1)(a^{2}_{n \minus{} 1} \minus{} 1)}$ $ n > 1$
Prove that:
a) $ 2 \plus{} 2a_n$ is a perfect square
b)$ 2 \plus{} \sqrt {2 \plus{} 2a_{n}}$ is a perfect square

Let $ a_n\equal{}ch(x_n)$, then $ ch(x_{n\plus{}1})\equal{}ch(x_n\plus{}x_{n\minus{}1})$ or $ x_{n\plus{}1}\equal{}x_n\plus{}x_{n\minus{}1}$. Because $ x_1\equal{}x_2$ we get $ x_n\equal{}F_nx_1$. Therefore $ 2a_n\equal{}y^{F_n}\plus{}y^{\minus{}F_n}$, were $ y\equal{}e^{x_1}$ or $ y^\minus{}194y\plus{}1$ or $ y\equal{}97\plus{}\sqrt{97^2\minus{}1}\equal{}(7\plus{}sqrt{48})^2\equal{}(2\plus{}\sqrt 3)^4.$
Therefore $ a_n\equal{}(2\plus{}\sqrt 3)^{4F_n}\plus{}(2\minus{}\sqrt 3 )^{4F_n}$. Your conditions followed from it.


calculus
integration
logarithms
trigonometry
calculus computations

Integrate $ \int ln(\frac{1\plus{}cosx}{1\minus{}cosx})dx$
I can't find out the solution, can you help me on this ???

[quote="svpam570"]Integrate $ \int \ln\frac {1 \plus{} \cos x}{1 \minus{} \cos x}\ \mathrm {dx}$ .[/quote]
 It is no possibly with the elementary functions.

Maybe definite integral, for example $ \int_0^{\pi} \ln\frac {1 \plus{} \cos x}{1 \minus{} \cos x}\ \mathrm {dx} \equal{} 0$ .

Thanks, maybe we must turn to this new problem. Could you please tell me how to compute this ?

Just substitute $ u\equal{}\pi \minus{}x$ right?

How does the lower limit of 0 work when the integrand is not defined there?

It's an improper integral, and it grows slowly enough that we have convergence in the usual sense.


inequalities
vector
inequalities unsolved

Let $\bar{x}=(x_1,\cdots,x_{n+1})$ and $\bar{y}=(y_1,\cdots,y_{n+1})$ be real vectors and $f(\bar{x},\bar{y})=\sum_{i=1}^n x_iy_i-x_{n+1}y_{n+1}$. Suppose further that $f(\bar{x},\bar{x})=f(\bar{y},\bar{y})=-1,x_{n+1}>0,y_{n+1}>0$. Prove that $f(\bar{x},\bar{y})\le -1$ with equality if and only if $\bar{x}=\bar{y}$.

easy. Put $a: =x_{n+1},b: =y_{n+1}$. To show: $\sum_{i=1}^n x_iy_i\le ab-1$.
By Cauchy, we have $(\sum_{i=1}^n x_iy_i)^2\le\sum_{i=1}^n x_i^2\sum_{i=1}^ny_i^2=(a^2-1)(b^2-1)=(ab-1)^2-(a-b)^2\leq(ab-1)^2$.
For the eqaulity case, the CS ineq yields $x_i=y_i (i=1,...,n)$ and the rightmost one $a=b$. Done.  :roll:


Let $ a_1 , a_2 , ... , a_n$ be positive numbers such that $ \sum_{i \equal{} 1}^{n}a_{i} \equal{} 1$.
Prove that
\[ \sum_{i \equal{} 1}^{n}\frac {a_{1}}{\sqrt {1 \minus{} a_{1}}} \ge \frac {1}{\sqrt {n \minus{} 1}}(\sqrt {a_{1}} \plus{} ... \plus{} \sqrt {a_{n}})
\]

The problem is from 4th CMO.(Chinese Mathematical Olympic)
We can prove $ f(x) \equal{} x/(sqrt(1 \minus{} x))$,thenf''(x)<0
Then We use Jesen……

Oh,I just have another method.$ sumxi/sqrt(1 \minus{} xi) > \equal{} n*sumxi/sumsqrt(1 \minus{} xi)$(chbyshev)
Cauchy,
$ (sumsqrt(1 \minus{} xi))^2 < \equal{} n*sum(1 \minus{} xi) \equal{} n^2 \minus{} n . (sumsqrt(xi))^2 < \equal{} n*sumxi < \equal{} n$
proof.


calculus
integration
geometry
3D geometry
sphere
vector
calculus computations

Evaluate $ \int \int (curl F) \times dS$ where $ S$ is the portion of the surface of the sphere $ x^2 \plus{}y^2 \plus{} z^2 \equal{} 1$ above the plane $ x \plus{} y \plus{} z \equal{} 1$ and $ F \equal{} (x \mathbf{i} \plus{} y\mathbf{j} \plus{} z\mathbf{k}) \times (\mathbf{i\plus{}j\plus{}k}).$

Of course, the moment you see a problem that looks like this, the very first thought you have is, "How can I use either Stokes' Theorem or the Divergence Theorem to my advantage?" Don't assume you have to integrate on the curved surface.

Isn't it just 0?
Because $ \vec{\nabla}\cdot(\vec{\nabla} \times \vec{F}) \equal{} 0$

This isn't a closed surface, so it's not that easy.

We might as well actually compute $ \mathbf{F}$ and $ \nabla\times\mathbf{F}.$ We get

$ \mathbf{F}\equal{}(y\minus{}z)\mathbf{i}\plus{}(z\minus{}x)\mathbf{j}\plus{}(x\minus{}y)\mathbf{k}$

$ \nabla\times\mathbf{F}\equal{}\minus{}2(\mathbf{i}\plus{}\mathbf{j}\plus{}\mathbf{k}).$

Now, we can argue, either by Stokes' Theorem or the divergence theorem, that the surface integral of $ \nabla\times\mathbf{F}$ over the curved surface in question is the same as the surface integral over the flat surface that closes it off - in other words, the portion of the plane $ x\plus{}y\plus{}z\equal{}1$ that lies inside the sphere. This is a circular disk, but we must determine its radius, hence area. The points $ (1,0,0),(0,1,0),$ and $ (0,0,1)$ lie on that circle and form an equilateral triangle with side length $ \sqrt{2}.$ A geometric computation shows that a circle that circumscribes such a triangle has radius $ \sqrt{\frac23}$ and hence area $ \frac{2\pi}{3}.$

The oriented surface element is $ d\mathbf{S}\equal{}\mathbf{n}\,dS.$ In the case of the plane $ x\plus{}y\plus{}z\equal{}1,$ the unit normal vector is $ \mathbf{n}\equal{}\frac1{\sqrt{3}}(\mathbf{i}\plus{}\mathbf{j}\plus{}\mathbf{k}).$

Hence, on that plane, $ \nabla\times\mathbf{F}\cdot\mathbf{n}\,dS\equal{} \minus{}\frac2{\sqrt{3}}(\mathbf{i}\plus{}\mathbf{j}\plus{}\mathbf{k})\cdot (\mathbf{i}\plus{}\mathbf{j}\plus{}\mathbf{k})\,dS$

$ \equal{}\minus{}2\sqrt{3}\,dS.$

To integrate that constant, we need only multiply by the area of the circle. Our final answer is $ \boxed{\frac{\minus{}4\pi}{\sqrt{3}}}$


superior algebra
superior algebra unsolved

[quote="manuel"]Let $ R$ be a ring, $ I$ an ideal, and suppose $ R/I$ is flat. Prove $ I^2 \equal{} I$.[/quote]

Assuming R is a commutative (unital, associative) ring, tensor the exact sequence 0→II→I→I/II→0 with R/I.  The result is exact since R/I is flat, but II⊗R/I is zero since generators of II are divisible by I and elements of R/I are I-torsion.  Also I/II⊗R/I is zero since it is isomorphic to II/II.  Hence I⊗R/I=0 if R/I is flat.

Now just tensor 0→I→R→R/I→0 with I to get I⊗I≅II=I (two special properties of I, not just one).


Actually, I solved. Never mind..
$ \partial \partial A \equal{} \partial (cl A \cap cl A^c) \equal{} \partial (A \cap \partial A^c) \equal{} \partial A$

[quote="car202"]If A is closed, show that $ \partial A \equal{} \partial(\partial A)$[/quote]

Strange.  If $ A$ is a closed interval in the line, $ \partial A$ is two points and $ \partial\partial A$ is empty.  It didn't work!  What's wrong?

Boundary points = points of which every neighborhood intersects the complement, no?  They need not be limit points of the set in question.  (In the example you gave, I claim $ \partial\partial A$ is still two points.)


trigonometry
real analysis
real analysis solved

Study the convergence of the sequence:

x_(n+1)=(-x_n+sqrt(3(1-(x_n)^2)))/2; n is natural.

cheers! :D :D

This seems to be a problem in trigonometry. After a few iterations, we have $0\leq x_n\leq \sqrt{3}/2$. Setting $x_n=\sin y_{n}$, the given recurrence is equivalent to $y_{n+1}=\pi/3-y_{n}$, from which everything follows.


geometry rectangle rotation integration conics parabola trigonometry

find the shaded area .length and breadth of rectangle are 8 by 4

hello
[hide="solution"]
just rotate this u will get the figure below .
The corresponding equations of curves are ::
$ y^2\equal{}4(x\plus{}2)$
$ y^2\equal{}\minus{}4(x\minus{}2)$
area of shaded part =$ 2\int^{\minus{}2}_{0}2\sqrt{x\plus{}2}dx$
                           =$ \frac{16\sqrt2}{3}$
[/hide]
thank u

Consider the parabola with the vertex $ (0,\ 2)$ that passes through the point $ (4, \minus{} 2)$, we have the equation of the parabola $ y \equal{} \minus{} \frac {1}{4}x^2 \plus{} 2$.

Thus the desired area $ S$ equals to the area bounded the parabola and the $ x$ axis, we get $ S \equal{} 2\cdot \frac {1}{3}\{2\sqrt {2} \minus{} ( \minus{} 2\sqrt {2})\}\cdot 2 \equal{} \boxed{\frac {16\sqrt {2}}{3}}$ whiich is kabi's answer.

good solution.
anyway figure doesn't say its a parabola or circle, if circle ans=$ 4 (\minus{}2 \sqrt {3} \plus{} (8 \Pi)/3)$

[quote="JRav"]What is "x grade question"?[/quote]
x class question

[quote="ayush_2008"][quote="ith_power"]good solution.
anyway figure doesn't say its a parabola or circle, if circle ans=$ 4 ( \minus{} 2 \sqrt {3} \plus{} (8 \Pi)/3)$[/quote]\
kabi its x grade question
pl try to do in that way[/quote]
sorry buddy .If u r x std this is not the right place.
BTW i saw the solution in goiit,it was right.
thanks

[quote="ith_power"]good solution.
anyway figure doesn't say its a parabola or circle, if circle ans=$ 4 ( \minus{} 2 \sqrt {3} \plus{} (8 \Pi)/3)$[/quote]

Hum, why do we consider the garph other than parabola? If we should take the graph as you think, there are more graphs to be considered, say, something like $ y \equal{} |\sin x|, x \equal{} a(\theta \minus{} \sin \theta), y \equal{} a(1 \minus{} \cos \theta)$, or cycloid , $ y\equal{}\frac{e^x\plus{}e^{\minus{}x}}{2}$, or Catenary e.t.c.

[quote="kunny"][quote="ith_power"]good solution.
anyway figure doesn't say its a parabola or circle, if circle ans=$ 4 ( \minus{} 2 \sqrt {3} \plus{} (8 \Pi)/3)$[/quote]

Hum, why do we consider the garph other than parabola? If we should take the graph as you think, there are more graphs to be considered, say, something like $ y \equal{} |\sin x|, x \equal{} a(\theta \minus{} \sin \theta), y \equal{} a(1 \minus{} \cos \theta)$, or cycloid , $ y \equal{} \frac {e^x \plus{} e^{ \minus{} x}}{2}$, or Catenary e.t.c.[/quote]
edit:Its a X(10th((Prehigh school))) class question, :D you expect it to be catenary!!! :P

What's X class? How difficult is X class? :wink:

About 15 years of age and also it is a school question means that a not-too-exceptional 15 year's question.
And in India (I believe Ayush is an Indian name) the only curved figure included in most tenth syllabus is circle.
@Ayush: Wrong forum.
And to do the question (if it is circle), some hints:

[hide="1"]What we have to do is calculate the area of the portion of the circle with center $ Z$ enclosed by the angle $ \angle XZW$. And similarly for the circle with center $ W$. Then add the 2 areas and then substract the area of the rhombus $ XZYW$ Then we will have the area of the shaded region.[/hide]

[hide="2"]To calculate the area of these figures, we need to know the angle $ \angle YZC$[/hide]

[hide="3"]Drop perpendicular $ YP$. Its length wud be half the width of the rectangle, which is $ 2$ also $ ZY \equal{} 4$. So considering the right triangle $ ZYP$, which has its hypotenuse twice another side, $ \angle YZC \equal{} 30^0$[/hide]

[hide="4"]Now $ \angle XZY \equal{} 120^0$ we can now calculate the area of the portions of the circles. And $ WYZ \equal{} 60^0$. We have that $ YWZ$ is a equilateral triangle with side length $ 4$. So we can now calculate the area of that rhombus.[/hide]


Please show us some work so we can see if your answer makes sense or not. :)

[quote="nat mc"][quote="Mo²"][quote="G-UNIT"]Whats the smallest 3-digit from which you cannot create a prime number by changing one of its digits?[/quote]

I think 102.[/quote]
you can change the 2 to a 1, to get 101, which is prime.
You are looking for the set of 10 numbers where the hundreds and tens digits are the same, and the units digit can change but make no prime.  My answer is
[hide]200, because there are no primes that start with 20-.[/hide][/quote]

I thought you can only change their places :s

Oh, I see where you're coming from.  Lets have G-UNIT make the clarification


logarithms
function
inequalities unsolved
inequalities

Well after you take the natural logaritm you should prove $f(m)=\frac{\ln{(2^{m}+1)}-ln 2}{m}$ is strictly increasing on $\mathbf{N}$.Now take the derivative.

I think is something wrong with your solution...at least with the typing

[quote="mateivld"]Well after you take the natural logaritm you should prove $f(m)=\frac{\ln{2^{m}+1}-ln 2}{m}$ is strictly increasing on ${\mathbf{N}\{0}$.Now take the derivative.[/quote]
The function is not differentiable. It is not even continuous.


algebra
polynomial
calculus
search
algorithm
integration
algebra unsolved

The leading coefficient is actually $ \frac {1} {n\plus{}1}$, and this must have been discussed and proved a thousand times ...

[quote="mavropnevma"]The leading coefficient is actually $ \frac {1} {n \plus{} 1}$, and this must have been discussed and proved a thousand times ...[/quote]Yes, it is. My mistake.

Would you care to show me a proof, preferably one that does not use calculus?

[quote="mavropnevma"]Check [url]http://mathworld.wolfram.com/PowerSum.html[/url]
or [url]http://www.mathlinks.ro/viewtopic.php?search_id=21828740&t=294843[/url][/quote]Thank you. I'll study these, though they look rather complicated.

I thought there'd be a high-school level solution, because I discovered the algorithm for finding those polynomials when I was thirteen. I know integrals render this problem trivial, but still...

[quote="Elixir10"]Prove that for each $ n,x \in \mathbb{N}$,
\[ {\sum^{x}_{i \equal{} 1}i^n}\]
can be expressed as a polynomial of degree $ n \plus{} 1$ with leading coefficient $ \frac {1}{(n \plus{} 1)}$.[/quote]

Bernulli says that $ \sum^{x}_{i \equal{} 1}i^n \equal{} \frac {x^{n \plus{} 1}}{n \plus{} 1} \plus{} \frac {x^n}{2} \plus{} \sum^{n}_{i \equal{} 2}\binom{n \plus{} 1}{i}B_{i}\frac {x^{n \plus{} 1 \minus{} i}}{n \plus{} 1}$ where $ B_{k} \equal{} 1\cdot 2\cdot 3\cdot \cdot \cdot k \plus{} 2\cdot 3\cdot 4\cdot \cdot \cdot (k \plus{} 1) \plus{} ... \plus{} n(n \plus{} 1)(n \plus{} 2)\cdot \cdot \cdot (n \plus{} k \plus{} \minus{} 1)$. So the problem is solved


[quote="mr. math"]how many odd perfect square factors does $2^{4}\cdot 3^{6}\cdot 5^{10}\cdot 7^{9}$ have?

how would u do this problem?[/quote]

Do we have to use casework or is there a better way? :maybe:

[quote="mr. math"]how many odd perfect square factors does 2^4 x 3^6 x 5^10 x 7^9 have?

how would u do this problem?[/quote]

This is actually not that hard. The number must have 0 2s; 0, 2, 4, or 6 3s; 0, 2, 4, 6, 8, or 10 5s; and 0, 2, 4, 6, or 8 7s. Therefore, there are $1 \cdot 4 \cdot 6 \cdot 5 = 120$ odd perfect square factors of $2^{4}3^{6}5^{10}7^{9}$.


real analysis
complex numbers
real analysis unsolved

Here is a nice problem that was discussed today and to which Fedja gave a nice proof using diophantine approximations:
    For any complex numbers $z_1,...,z_k$ the superiour limit of the sequence $\sqrt[n]{|z_1^n+...+z_k^n|}=max_{i=1,k}{|z_i}$.
    Prove this without any use of diophantine approximation.

This is the idea similar to idea of Myth, solving another problem.
By multiplying all $ z_{1}, \ldots , z_{k} $ by a complex number,
and rearranging $ z_{1}, \ldots , z_{k} $
we can assume that $ max_{i=1,\ldots,k} |z_{i}| = 1 $, and that
$ z_{1} = 1 $. Define $ a_{n} : = z_{1}^{n} + \ldots + z_{k}^{n} $.
For every $ z $, $ z+z^{2}+ \ldots + z^{n} = \frac{z^{n+1} - z}{z-1} $ if
$ z \neq 1 $  and $ n $ if $ z = 1 $. Now , if $ z \neq 1 $ and $ |z| \leqslant 1 $,
then $ \frac{z^{n+1} - z}{z-1} $ is bounded , independently of $ n $.
Therefore if we define $ b_{n} := \frac{a_{1}+ \ldots + a_{n}}{n} $ then
$ b_{n} \rightarrow m $, $ n \rightarrow \infty $, where $ m \geqslant 1 $
is the number of $ 1 $ - s between $ z_{1}, \ldots , z_{k} $ .
Therefore $ lim_{n \rightarrow \infty } b_{n} = m \geqslant 1 $,
so $ limsup_{n \rightarrow \infty} |a_{n}| \geqslant m \geqslant 1 $. Therefore,
of course, $ limsup_{n \rightarrow \infty} \sqrt[n]{|a_{n}|} \geqslant 1 $.
On the other hand, $ |a_{n}| \leqslant k $ for every $ n $,
so $ limsup_{n \rightarrow \infty} \sqrt[n]{|a_{n}|} \leqslant 1 $.
Therefore $ limsup_{n \rightarrow \infty} \sqrt[n]{|a_{n}|} = 1 $.

Nice! One can also consider $f(z)=\frac{1}{1-z_1z}+...+\frac{1}{1-z_nz}$ and observe that if the superiour limit is smaller than 1, then we can make $z$ very close to $\frac{1}{z_1}$. But on the other hand, $f$ is bounded.

Here is another question related to this:
We have that for every $ z_{1}, \ldots , z_{k} \in \mathbb{C} $,
we have $ limsup_{n \rightarrow \infty}
\sqrt[n]{|z_{1}^{n} + z_{2}^{n} + \ldots + z_{k}^{n}|} = max_{i=1,\ldots,k} |z_{i}| $.
Now suppose that $ |z_{1}| = |z_{2}| = \ldots = |z_{k}| = 1 $. Then we have
$ limsup_{n \rightarrow \infty}
\sqrt[n]{|z_{1}^{n} + z_{2}^{n} + \ldots + z_{k}^{n}|} = 1 $.
Take now a torus $ \mathbb{T}^{k} = S^{1} \times S^{1} \times \ldots \times S^{1}
\subset \mathbb{C}^{k} $ and take $ z = (z_{1},z_{2}, \ldots , z_{k}) \in \mathbb{T}^{k} $.
Call $ z \in \mathbb{T}^{k} $ good if we have
$ lim_{n \rightarrow \infty}
\sqrt[n]{|z_{1}^{n} + z_{2}^{n} + \ldots + z_{k}^{n}|} = 1 $.
Call $ z \in \mathbb{T}^{k} $ bad, if it is not good.
Prove that the set of bad points in $ \mathbb{T}^{k} $ is of measure $ 0 $.


Three 3-digit numbers are formed using the digits 1 through 9 exactly once each.  The hundreds digit of the first number is 1.  The tens digit of the second number is 8.  The  units digit of the third number is 5.  The ratio of the first number to the second number to the third number is 1:3:5, respectively.  What is the sum of the three numbers?

[hide="ANS"]1161[/hide]

***Is there a faster way to do this besides guess and check?

Consider the number in the tens digit of the first number, it must be either 2, 3, 4, 6, 7, or 9. But note that for the tens digit of the second number to be 8, it must be either 2, 6, or 9. If it is 9, the hundreds digit of the second number must be 5, contradiction. If it is 6, then the hundreds digit of the second number must be 4, and 16A*3=48B, A and B are digits. We need (A,B) to be (1,3), (2,6) or (3,9), and the first two give repeated numbers so the first two numbers must be 163, 489, and the third number must be 163*5=815, but then 8 and 1 are used twice and we lose.

So the tens digit of the first number is 2, meaning the hundreds digit of the second and third numbers are 3 and 6, respectively. We now have to use 4, 7, 9. We need 12A*3=38B, and if A and B are 4, 7, or 9, we need A=9, B=7, so the three numbers are 129, 387, 645, which we verify to work, and they sum to 1161.


function
complex analysis
calculus
integration
algebra
domain
geometry

If $ f$ is holomorphic on the strip $ S\equal{}\{z\in\mathbb{C} : \minus{}1<Im(z)<1\}$ and is bounded $ |f(z)|\leq \frac{1}{1\plus{}(Re(z))^2}$ for all $ z\in S$ with $ |Re(z)|>100$, then there are functions $ f_\plus{}$ holomorphic on the half-plane $ S_\plus{}\equal{}\{z : Im(z)>\minus{}1\}$ and $ f_\minus{}$ holomorphic on $ S_\minus{}\equal{}\{z : Im(z)<1\}$ such that on $ S$ we have $ f\equal{}f_\plus{}\minus{}f_\minus{}$.

Actually, I don't even need a solution. I'm pretty sure I could work this out if I had just a general idea what theorem or technique I'd end up needing. So if anyone is fairly certain they see what will work, but doesn't want to work out the specifics that would be greatly appreciated.

A typical approach when you see that you want two holomorphic functions whose sum is your original function is to use Cauchy's Integral Formula, but I don't see what curves could be useful here.

A typical approach when you see "bounded" is Maximum Modulus Principle. This also seems to be a dead-end. That is why I'm stumped.

This can be done for any doubly connected domain, just like in proving the Laurent series expansion. Now this domain is not doubly connected, but that's what the extra condition is for. Note that $ f$ is not only bounded, but is continuous with value 0 at infinity. So it should be possible to use Cauchy's theorem with a curve consisting of two lines parallell the real axis, connecting at infinity.

Fairly sure the functions in the decomposition will be given by integration over a straight line in the strip. Don't guarantee that this is correct though, because I've never understood integration at infinity.

edit: The intuition here is not correct, because the curve above is not a curve; I thought of infinity as two different points. So it's an area bounded by two curves, connecting at a point. I don't quite understand this area, but nevertheless I think just integrating over a big rectangular block, using cauchy's formula here, and just passing a limit will do the trick.

I feel like there is subtlety I'm missing, that I'll explain at the end. Fix $ 0 < \epsilon < 1$.

Let $ \gamma_\epsilon^ \plus{} \equal{} \{z: Im(z) \equal{} 1 \minus{} \epsilon\}$ and $ \gamma_\epsilon^ \minus{} \equal{} \{z: Im(z) \equal{} \minus{} 1 \plus{} \epsilon\}$. So they are the two lines just inside the strip.

Goal: $ f(z) \equal{} \frac {1}{2\pi i}\int_{\gamma_\epsilon^ \plus{} \minus{} \gamma_\epsilon^ \minus{} }\frac {f(\zeta)}{\zeta \minus{} z}d\zeta$.

Let $ \gamma_n$ be the curve that runs along $ \gamma_\epsilon^ \plus{}$ from n to -n, then is the straight line down along $ Re(z) \equal{} \minus{} n$ to $ \gamma_\epsilon^ \minus{}$ then runs along this line from -n to n, then is the straight line up to where we started.

Assume $ n > 100$. Then

$ \Big|\frac {1}{2\pi i}\int_{\gamma_n}\frac {f(\zeta)}{\zeta \minus{} z}d\zeta \minus{} \frac {1}{2\pi i}\int_{\gamma_\epsilon^ \plus{} \minus{} \gamma_\epsilon^ \minus{} }\frac {f(\zeta)}{\zeta \minus{} z}d\zeta\Big| < \frac {1}{n^2}$ after a bunch of exactly what you would think estimates.

Now since $ f(z) \equal{} \frac {1}{2\pi i}\int_{\gamma_n}\frac {f(\zeta)}{\zeta \minus{} z}d\zeta$ for each $ n$ and $ z$ inside the curve $ \gamma_n$ we have that the sequence converges to $ \frac {1}{2\pi i}\int_{\gamma_\epsilon^ \plus{} \minus{} \gamma_\epsilon^ \minus{} }\frac {f(\zeta)}{\zeta \minus{} z}d\zeta$, moreover by Weierstrass we have that the limit is analytic and so each of the curves separately is analytic.

Since this was independent of $ \epsilon$ we can take $ \epsilon$ to 0.

Define: $ f_ \plus{} \equal{} \frac {1}{2\pi i}\int_{Im(z) \equal{} \minus{} 1}\frac {f(\zeta)}{\zeta \minus{} z}d\zeta$ which by Cauchy is analytic off the curve and hence in $ S_ \plus{}$ and $ f_ \minus{} \equal{} \frac {1}{2\pi i}\int_{Im(z) \equal{} 1}\frac {f(\zeta)}{\zeta \minus{} z}d\zeta$ and we are done.

Problem. It seems as if taking $ \epsilon$ to 0 is fishy. Why do things stay analytic? Why is it still equal to f in the strip? These things would be resolved if I did the $ n\to\infty$ and $ \epsilon\to 0$ simultaneously somehow.

Why do you need to pass $ \epsilon$ to zero? Assigning boundary values to $ f$ is rather difficult, and is most likely even impossible because we have no information on what happens when it approaches the boundary in $ |\Re z| < 100$. So defining $ f_\plus{}$ as an integral over the boundary makes little sense.  Rather we want to define $ f_ \plus{} \equal{} \frac {1}{2\pi i}\int_{\gamma^ \plus{} _\epsilon}$ The argument shows (if you allow yourself to only vary the $ \epsilon$ in $ \gamma_\epsilon^ \plus{}$ and fix the  "epsilon" for the negative curve) that this uniquely defines a function in $ \Im z > \minus{} 1$. That is if you want to evaluate $ f_ \plus{}$ in $ z$ you just choose a small enough epsilon and calculate the line integral.

I guess I don't really understand why the $ f_\plus{}$ is a legitimate function if the $ \epsilon$ is varying. Also, if the "negative" one is fixed and inside the strip, what is the value of $ f(z_0)$ if $ Im(z_0)\equal{}\minus{}1\plus{}\epsilon$? This doesn't seem to give a function in all of  $ Im(z)>\minus{}1$. That was why I wanted it to tend to the boundary.

I think I'm missing something, since you seem to have it working with the lines inside the strip...  :blush:

For each $ \epsilon$ we get a function which is defined and analytic in $ \Im z > \minus{} 1 \plus{} \epsilon$, let's call them $ f^ \plus{} _\epsilon$. The point is this: let $ \epsilon_1 < \epsilon_2$. Then $ f^ \plus{} _{\epsilon_1}$ is an analytic extension of $ f^ \plus{} _{\epsilon_2}$. So by analytic extension $ f^ \plus{}$ is a well-defined function on $ \Im z > \minus{} 1$.

They are analytic extensions of each other because by what you wrote $ f(z) \equal{} f^ \plus{} _{\epsilon_1}(z) \minus{} f^ \minus{} _\eta(z)$ and $ f(z) \equal{} f^ \plus{} _{\epsilon_2}(z) \minus{} f^ \minus{} _\eta(z)$, for $ \minus{} 1 \plus{} \epsilon_2 < \Im z < 1 \minus{} \eta$. So clearly they have the same values on that strip, and hence in all of $ \minus{} 1 \plus{} \epsilon_2 < \Im z$.

So the same argument shows that $ f^ \minus{}$ is well-defined, and $ f \equal{} f^ \plus{} \minus{} f^ \minus{}$ at least in a smaller strip, but hence also everywhere.

This is a very standard type of idea, when you can't integrate on the boundary. Typically we would be integrating over a rectifiable curve and then the equality of $ f^ \plus{} _{\epsilon_1}$ and $ f^ \plus{} _{\epsilon_2}$ would usually follow by a homotopy argument.

Ah. Sorry about that. That makes sense. In fact, I have seen it before. This is the same idea we used to prove Laurent series in an annulus. Thanks!


[color=darkblue]Prove that in any $\triangle ABC$ there is the following inequality :

$\boxed{\ 3\cdot\left(\frac{a}{h_{a}}+\frac{b}{h_{b}}+\frac{c}{h_{c}}\right)\ge 4\cdot\left(\frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}\right)+\frac{2(2R-r)(R^{2}-4r^{2})}{R^{2}p}\ }\ge$ $4\cdot\left(\frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}\right)+\frac{2\sqrt 3}{3}\cdot\left[1-\left(\frac{2r}{R}\right)^{2}\right]\ge$ $4\cdot\left(\frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}\right)$.

See[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=151123

[quote="Virgil Nicula"][color=darkred]Prove that in any $\triangle ABC$ there is the following inequality :

$\boxed{\ 3\cdot\left(\frac{a}{h_{a}}+\frac{b}{h_{b}}+\frac{c}{h_{c}}\right)\ge 4\cdot\left(\frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}\right)+\frac{2(2R-r)(R^{2}-4r^{2})}{R^{2}p}\ }\ge$

$4\cdot\left(\frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}\right)+\frac{2\sqrt 3}{3}\cdot\left[1-\left(\frac{2r}{R}\right)^{2}\right]\ge$ $4\cdot\left(\frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}\right)$.

See[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=151123 [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $\left\{\begin{array}{ccc}\sum bc=p^{2}+r^{2}+4Rr & \implies & \sum a^{2}=2p^{2}-2r(4R+r)\\\\ p^{2}\le 4R(R+r)+3r^{2}& \implies & 9R^{2}\ge \sum a^{2}+\left(R^{2}-4r^{2}\right)\\\\ (x+y+z)^{2}\ge 3(xy+yz+zx) & \implies & \left(\sum a^{2}\right)^{2}\ge 3\cdot\sum b^{2}\cdot c^{2}\\\\ p^{2}+5r^{2}\ge 16Rr & \implies & \sum a^{2}\ge 12r(2R-r)\end{array}\right\|$ $\implies$

$9R^{2}\cdot\sum a^{2}\ge$ $\left(\sum a^{2}\right)^{2}+\left(R^{2}-4r^{2}\right)\cdot\sum a^{2}\ge$ $3\cdot\sum b^{2}c^{2}+12r(2R-r)\left(R^{2}-4r^{2}\right)$. Divide by $3R^{2}$ $\implies$

$3\cdot \sum a^{2}\ge \sum\frac{b^{2}c^{2}}{R^{2}}+\frac{4r(2R-r)(R^{2}-4r^{2})}{R^{2}}$. Divide by $abc=4Rpr$ $\implies$

$3\cdot \sum \frac{a}{bc}\ge \sum \frac{bc}{aR^{2}}+\frac{(2R-r)(R^{2}-4r^{2})}{R^{3}p}$. Replace $bc=2Rh_{a}$ $\implies$

$3\cdot\sum \frac{a}{2Rh_{a}}\ge \sum \frac{2Rh_{a}}{aR^{2}}+\frac{(2R-r)(R^{2}-4r^{2})}{R^{3}p}$. Multiply by $2R$ $\implies$

$\boxed{\ 3\cdot\sum\frac{a}{h_{a}}\ge 4\cdot\sum\frac{h_{a}}{a}+\frac{2(2R-r)(R^{2}-4r^{2})}{R^{2}p}\ }$.

[b][u]Remark.[/u][/b] The relations $3\sqrt 3\cdot r\le p\le \frac{3\sqrt3}{2}\cdot R$ are well-known and $\left\{\begin{array}{c}R\ge \frac{2p}{3\sqrt 3}\\\\-r\ge-\frac{p}{3\sqrt 3}\end{array}\right\|$ $\implies$ $2R-r\ge \frac{3p}{3\sqrt 3}$ $\implies$ $\frac{2R-r}{p}\ge \frac{\sqrt 3}{3}$.[/color]


probability
function
algebra
polynomial
expected value

An exam is given to students that consists of 5 questions worth 20 points each.
Suppose for an average student, the probability for the number of points earned on each question is:

20 points - 40%
15 points - 30%
10 points - 20%
0 points - 10%

What is the most probable total score for an average student?

[hide="My attempt"]
The expected value for each point is $ E \equal{} (.4)(20)\plus{}(.3)(15)\plus{}(.2)(10)\plus{}(.1)(0) \equal{} 14.5$. Thus the total for 4 points is $ (14.5)(5)\equal{}72.5$, however since we cannot attain 72.5 from a sum of 10, 15, and 20, we consider the closest: $ \boxed{75}$ and $ \boxed{70}$.
[/hide]

And why are you more likely to get a question right, rather than get a question wrong?

EDIT: Thanks b-flat. Fixed both

[hide]The expected value of an average student's score on each question is $ \left(\frac{4}{10}\right)20\plus{}\left(\frac{3}{10}\right)15\plus{}\left(\frac{2}{10}\right)10\plus{}\left(\frac{1}{10}\right)0\equal{}\frac{29}{2}$.

Thus, the expected value of the total score on the exam is $ \frac{145}{2}$.  The two closest obtainable scores are 70 and 75.[/hide]

@Anaxerzia:  58 is cannot be the most probable score, because 58 cannot be obtained.  Also, there are 5 questions.

Exactly- so its the closest attainable score - which are 70 and 75 (both are equally close to 145/2)

[quote="Jean-Louis"]I was given the following hint :

for the 5-question exam where the total marks = 5*20 = 100.
The total points in the 5 points can be any one of the following -- 0, 10,
15, 20, 25, 30, 35, 40, 45, 50, .... , 95 and 100. Then, find out what's the
probability for each of these marks and then proceed to calculate the
weighted average.
It is not correct to say that the expected value is = 14.5 * 5 = 72.5.

Let p(n) be the probability to get n marks out of 100.

p(0) = (.1)*(.1)*(.1)*(.1)*(.1) = 0.00001 = 10^(-5)
p(10) = (.2)*(.1)*(.1)*(.1)*(.1) + (.1)*(.2)*(.1)*(.1)*(.1) +
(.1)*(.1)*(.2)*(.1)*(.1) + (.1)*(.1)*(.1)*(.2)*(.1) +
(.1)*(.1)*(.1)*(.1)*(.2) = C(5,1) * 0.00002 = 5*2*{10^(-5)} = 10^(-4).

Then proceed to calculate p(15), p(20), p(25), p(30), ....[/quote]

That would take quite a while...  However, you are correct in that the expected value won't produce the most probable event every time...

[quote="Anaxerzia"][quote="MellowMelon"]You can't conclude right away that $ 70$ and $ 75$ are equally probable from the expected value unless the distribution is symmetric. In fact, they may not even be the most likely score (though one of them is in this case).

If you brute force it, you get a score of $ 70$ having probability $ .1329$ and $ 75$ having probability $ .14163$.[/quote]

Then what is the ideal solution to the problem?[/quote]
The most elegant solution I can find to this problem is to compute $ (0.4x^{20}\plus{}0.3x^{15}\plus{}0.2x^{10}\plus{}0.1)^{5}$ and see which term has the greatest coefficient. Without combining, that expansion has 1024 terms. In short, I don't think this problem is nice.

I see, but out of curiosity why would you use that technique? (I mean how does it work?)

You might want to read up on generating functions (perhaps the AOPS wiki?). The most intuitive explanation I can come up with is that polynomial when expanded represents all possible ways to combine the scores on each question. Probably not best for me to have brought this idea up in HSB though.

Just came up with a few surprises though:
- A score of $ 80$ has a chance of $ .1566$. That's higher than the chance to get a $ 75$.
- From $ 85$ on, there are very few combinations of question totals that work, so the chance starts dropping. A score of $ 85$ has a chance of $ .1200$, and $ 90$ has $ .0832$.

I'm guessing $ 80$ is correct. Terribly inelegant way to do it, though.

Oh thanks :-) I think I see why the normal expected values approach fails- one must also consider the number of many ways one even can get such a combination of numbers from the given scores to make a score, no?

My math teacher told me that what i did is wrong.

The most probable is the mode of the distribution, not necessarily the same as the mean.  For a symmetrical distribution they will be the same but not for a distribution with any degree of skewness.


function
real analysis
real analysis unsolved

prove that : Don't exist  funtion $ f(x) \neq 0$ such that :
 $ f(x)f(y) \equal{} f(x\minus{}y)$

Something must be wrong, since $ f\equal{}1$ verifies the statement of your problem.

Edit:  My mistake.  See scorpius' solution.

[quote="Hong Quy"][quote="Hong Quy"]prove that : Don't exist  funtion $ f(x) \neq 0$ such that :
 $ f(x)f(y) \equal{} f(x \minus{} y)$[/quote]
 sory , i think that : $ f(x) \neq Const$[/quote]

[hide]
Suppose such a function exists. First, plug in $ x\equal{}y\equal{}0$ to get $ f(0)^2\equal{}f(0)$. So either $ f(0)\equal{}0$ or $ f(0)\equal{}1$.

Also, plugging in $ x\equal{}y$ gives $ f(x)^2\equal{}f(0)$ for all $ x$. If $ f(0)\equal{}0$, then $ f(x)\equal{}0$ for all $ x$, making $ f$ constant.

If $ f(0)\equal{}1$, then for each $ x$, $ f(x)\equal{}1$ or $ f(x)\equal{}\minus{}1$. For each $ a$, we show $ f(a)\equal{}1$, which will make $ f$ constant.

Plug in $ x\equal{}0,y\equal{}a/2$ to get $ f(a/2)\equal{}f(\minus{}a/2)$. In particular, each is 1 or -1, so their product is 1. But $ a\equal{}(a/2)\minus{}(\minus{}a/2)$ implies $ f(a)\equal{}f(a/2)f(\minus{}a/2)\equal{}1$. So $ f(a)\equal{}1$ for all $ a$.
[/hide]


ok so all of the '07 people know about the Frozen Burrito incident, as well as some '08ers. There were (as far as I know) three incidents in '08. The first was the Billiard Ball incident, where some people were in the clubhouse playing with billiards, and one person shot a billiard a little to hard.
Another incident I know of was the Big Rock incident. korammath decided to launch a huge rock off the third story balcony. Luckily nobody was hurt but that earned him a strike.
The third incident was the Jack of Hearts and the Rose incident. A whole suite of AMSP students decided to play a little prank. They emptied their room of everything but a jack of hearts and a rose. Then the counselors(all but mine) were shown the room(by mine), and the six students hid in my(with alexhhmun and emannes) room. Then there was this big chase, with Brian T. Basham screaming, and it ended with the six students jumping out and hugging Brian.

does anyone have any more incidents?

What's the frozen burrito one?  I wasn't at amsp...

haha i know few more, but those are just unsuitable for the public to know haha

alll the basham pranks were pretty funny. although i feel reallly bad for brian for having to go through so much yike-ness

ROFLLL.
darnnn. i wanna go back and relive those memories XP
omg the basham joker was was hilarious. i loveee watching the vid on fb :D
oh yea...
remember when tanya took basham's shoe???
shoot. there are a crapload more. i just cant remmeberrrrr. ><

Pranking Basham wasn't an instance, it was just hilarious.

That game of mafia with everyone as mafia except Brian...

omg.
like.
mafia in general was amazing. esp the 2nd night! :D

That prank was SOOOOO FUNNNN to do....=)

It took us about 2 days to clean up the whole dorm though...The videotape was hilarious too...except their was a lot of um...stuff said by Basham

We did pranks like almost everyday that whole week anyways (trip wires...falling boxes...balloon popping...lights turning off randomly, etc.)

lol the "stuff" said by basham made it even funnier lol xP  :rotfl:

Craig, our counselor, was supposedly outside as well.

And our dorm completely missed it, I think. We woke up to a sleepy counselor and we were all just like, "Huh?"

And probably not nearly as interesting was when Juan signed yearbooks while duct taped to the outside wall of the clubhouse (on the last day).

lol that was pretty amazing too. and then brian dunked juan in the pool and failed XP

[quote="BTBasham"]You are all dead.[/quote]

It wasn't MY fault that my counselor helped them! (if somebody didn't steal my yearbook i would remember his name...)

NEways I think i still have pics of Basham and Juan in the pool(I don't think Juan was pwning Basham at the time i was taking pics). I'm glad i had my cell phone or else Juan would have thrown me in too. :)

1=2,
u should get a fb.
add the photos.
and then share the funness XP


sum d series 1+3x+5x2+7x3+9x4+... up to n terms

This is an [url=http://www.mathlinks.ro/Wiki/index.php/Arithmetico-geometric_series]Arithmetico-geometric series[/url].

I'll show how to derive the closed form later, when I have time.

[hide="Solution"]
Simple summation:

$ 1\cdot3\plus{}2\cdot5\plus{}3\cdot7\plus{}4\cdot9\plus{}\cdots\plus{}n\cdot(2n\plus{}1)\equal{}\displaystyle\sum_{k\equal{}1}^{n}k(2k\plus{}1)\equal{}2\displaystyle\sum_{k\equal{}1}^{n}k^2\plus{}\displaystyle\sum_{k\equal{}1}^{n}k$

$ \equal{}2\cdot\frac{n(n\plus{}1)(2n\plus{}1)}{6}\plus{}\frac{n(n\plus{}1)}{2}$

Hence, the expression is equal to $ \frac{n(n\plus{}1)(4n\plus{}5)}{6}$.[/hide]

That's an $ x$, not a times.  See the combinatorial proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url] of the general result.


floor function
Sequence
algebra
Fibonacci
Fibonacci sequence
Inequality
IMO Shortlist

Let $ \lfloor x \rfloor$ denote the greatest integer less than or equal to $ x.$ Pick any $ x_1$ in $ [0, 1)$ and define the sequence $ x_1, x_2, x_3, \ldots$ by $ x_{n\plus{}1} \equal{} 0$ if $ x_n \equal{} 0$ and $ x_{n\plus{}1} \equal{} \frac{1}{x_n} \minus{} \left \lfloor \frac{1}{x_n} \right \rfloor$ otherwise. Prove that

\[ x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n < \frac{F_1}{F_2} \plus{} \frac{F_2}{F_3} \plus{} \ldots \plus{} \frac{F_n}{F_{n\plus{}1}},\]

where $ F_1 \equal{} F_2 \equal{} 1$ and $ F_{n\plus{}2} \equal{} F_{n\plus{}1} \plus{} F_n$ for $ n \geq 1.$

the sequence $ {x_i}$ satisfies 'off hand' criteria for "chaotic", in that it acts by stretching and folding. so $ {x_i}$ is just a collection of maybe not so random points in $ [0,1)$.

on the RHS you may remember that $ \frac{F_{i\plus{}1}}{F_i} \approx \varphi \equal{} \frac{1\plus{}\sqrt{5}}{2}\approx 1.618$.

with gross abuse of approximations we get

$ x_1 \plus{} x_1 \plus{} \ldots \plus{} x_n \approx  \frac{n}{2} < 0.618n \approx \frac{n}{\varphi} \approx \frac{F_1}{F_2} \plus{} \frac{F_2}{F_3} \plus{} \ldots \plus{} \frac{F_n}{F_{n\plus{}1}}$

certainly not a proof, but motivation(can be somewhat verified with programming). RHS looks usable, but LHS needs work.

i will prove it by induction
when $ n\equal{}1,2$, it is obvious.
if $ x_1<\equal{}\frac{F_n} {F_{n\plus{}1}}$, then by induction, it is done.
if $ x_1>\frac{F_n} {F_{n\plus{}1}}$, then since
$ x_1\equal{}\frac{1} {a\plus{}x_2}<\frac{1} {1\plus{}x_2}$, hence
$ x_2<\frac{F_{n\minus{}1}} {F_n}$.
so $ x_1\plus{}x_2<\frac{1} {1\plus{}x_2}\plus{}x_2<\frac{F_n} {F_{n\plus{}1}}\plus{}\frac{F_{n\minus{}1}} {F_n}$.
then we can use induction.

Let $[a_0; a_1, a_2, \cdots, a_n, \cdots]$ be the continued fraction representation of $x_1$, in other words

$$x_1 = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \cdots}}},$$

where $a_0 = 0$ and $a_1, a_2, \cdots, a_n > 1.$

Then, in terms of continued fraction representations, we wish to prove that $[0; a_1, a_2, \cdots] + [0; a_2, a_3, \cdots] + [0; a_3, a_4, \cdots] + \cdots + [0, a_n, \cdots] < \frac{F_1}{F_2} + \frac{F_2}{F_3} + \cdots + \frac{F_n}{F_{n+1}}.$

Since we clearly have $[0; a_1, a_2, \cdots] < [0; a_1, a_2, \cdots, a_n], [0; a_2, a_3, \cdots] < [0; a_2, a_3, \cdots, a_n], \cdots$, we can just show that

$$[0;a_1, a_2, \cdots a_n] + [0; a_2, a_3, \cdots, a_n] + \cdots + [0; a_n] \le \frac{F_1}{F_2} + \frac{F_2}{F_3} + \cdots + \frac{F_n}{F_{n+1}}.$$

Observe that the RHS is the same as the LHS when all $a_i$'s are one, and so let us just show that the LHS is maximized when all $a_i$'s are one. We will prove this via strong induction on $i$, with our hypothesis being that $a_1, a_2, \cdots, a_i$ must all be one in any "maximal" choice of $a_i$'s.

For $i=1$, it is immediate, since $a_1$ has no influence on any of the terms but $[0; a_1, a_2, \cdots, a_n]$ and it's clear that $a_1 = 1$ maximizes $[0; a_1, \cdots, a_n].$

Let's now suppose that it holds for $i \le t.$ In this way, we have that $a_1 = a_2 = \cdots = a_t = 1.$ Let us show that $a_{t+1} = 1$ in any maximal configuration. Clearly, $a_{t+1}$ is independent of $[0; a_{t+2}, \cdots, a_n], [0, a_{t+3}, \cdots, a_n], \cdots$. Hence, we can let $q = [0; a_{t+2}, \cdots, a_n]$ and pick the optimal $a_{t+1}$ in these conditions. Viewing this as a function in terms of $x = a_{t+1}+q$, the relevant quantity (not including later stuff not influenced by $a_{t+1}$) is precisely:

$$\frac{1}{x} + \sum_{j=1}^{t} \frac{F_jx + F_{j-1}}{F_{j+1}x + F_j}.$$

We now make the spectacular observation that:

$$\frac{F_jx + F_{j-1}}{F_{j+1}x + F_j} = \frac{F_j}{F_{j+1}} + \frac{F_{j+1}F_{j-1} - F_j^2}{F_{j+1} (F_{j+1}x + F_j)},$$

and so hence if we plug this into the expression we actually just need to maximize:

$$\frac1x + \sum_{j=1}^{t} \frac{(-1)^j}{F_{j+1} (F_{j+1}x + F_j)} = \frac{1}{x(x+1)} + \sum_{j=2}^{t} \frac{(-1)^j}{F_{j+1} (F_{j+1}x + F_j)}. \qquad (1)$$

Keeping in mind that $x$ is positive, it's obvious that any fraction of the form $\frac{\alpha}{\beta x^2 + \gamma x + \delta}$ is maximized when $x$ is minimized, and similarly for fractions of the form $\frac{\alpha}{\beta x + \gamma}$. Now, it just remains to partition the sum in $(1)$ into functions of these form. Indeed, $\frac{1}{x(x+1)}$ is of this form, and so is $\frac{1}{F_{j+1}(F_{j+1} x + F_j)} - \frac{1}{F_{j+2} (F_{j+2} x+ F_{j+1})}$ for any $j$ even. Finally, if $t$ is even, then we will need a final piece which is $\frac{1}{F_{t+1} (F_{t+1}x + F_t)},$ which is also maximized as $x$ is minimized.

In conclusion, it's clear that $x$ should be minimized to maximize the sum in $(1)$, so since $x = a_{t+1} + q$, we have that $a_{t+1} = 1$ in any maximal choice of $a_i$'s. The induction is complete.

$\square$


function
algebra
domain
real analysis
polynomial
algebra unsolved

An arithmetic function is a real-valued function whose domain is the set of positive integers. Define the convolution product of two arithmetic functions $ f$ and $ g$ to be the arithmetic function $ f * g$, where \[ (f * g)(n) \equal{} \sum_{ij\equal{}n} f(i) \cdot g(j),\] and $ f^{*k} \equal{} f * f * \ldots * f$  ($ k$ times) We say that two arithmetic functions $ f$ and $ g$ are dependent if there exists a nontrivial polynomial of two variables $ P(x, y) \equal{} \sum_{i,j} a_{ij} x^i y^j$ with real coefficients such that

\[ P(f,g) \equal{} \sum_{i,j} a_{ij} f^{*i} * g^{*j} \equal{} 0,\]

and say that they are independent if they are not dependent. Let $ p$ and $ q$ be two distinct primes and set

\[ f_1(n) \equal{} \begin{cases} 1 & \text{ if } n \equal{} p, \\
0 & \text{ otherwise}. \end{cases}\]

\[ f_2(n) \equal{} \begin{cases} 1 & \text{ if } n \equal{} q, \\
0 & \text{ otherwise}. \end{cases}\]

Prove that $ f_1$ and $ f_2$ are independent.

I wonder how this problem managed to get into ILL 1989. No more is needed to solve it than directly applying the definition.

It can be easily checked that:

$ f_{1}^{*i}*f_{2}^{*j}(n) = \begin{cases}1 \,\,  \text{if} \,\, n=p^{i}q^{j} \\ 0 \,\,  \text{otherwise} \end{cases} $.

This implies that $f_1$ and $f_2$ are independent.

Problems get into ILL simply because there were submitted that year; no selection by the Problem Selection Committee is done. So the only responsibility (or lack of) is that of the submitting country.


For all positive real number $ x,y,z$. Prove that:
\[ \left(a^2\plus{}b^2\plus{}c^2\right)^2 \ge 3\left(a^3b\plus{}b^3c\plus{}c^3a\right)\]

[quote="Martin N."]It is not your friend's own (assuming you don't mean Vasc), it is a very well-known inequality of Vasc...see for example [url=http://www.mathlinks.ro/viewtopic.php?t=6026]here[/url]...[/quote]

thats right. This is one of the famous prof. Cirtoaje's inequalities.
see also here for another solution:
http://gbas2010.wordpress.com/2010/01/08/problem-19vasile-cirtoaje/


Determine all the values of p such that (p-1)!+1 is a power of p.

I would really like to know where the problem comes from. It came up in the German TST 2004 and I solved it by rewriting p as p-1+1 and writing (p-1+1)^k in the form with binomial coefficients and divided the equation by p-1. then you get (p-1)|k for p>5 which gives p<k+2 which leads to a contradiction. So the only solutions are 2,3,5. Do you have an other solution?

It comes from a number theory book, but judging the level of the problems there, which  generally is not so high it sound astonishing that this problem has been in a TST.
There is a solution in the back of the book but it gives only the result which is the same as yours but it doesn't show how one comes to this solution(which generally implies that the problem isn't that hard(though i dont have a soln for it) )

This problem has a strange story... In March 2004, a Belgian - Antoine Verroken - sent me the problem and asked whether I could solve it. I couldn't and therefore, on March 12, 2004, I posted the problem into the [url=http://mathlinks.ro/Forum/viewtopic.php?p=14710]number theory section of MathLinks[/url]. Some hours after posting that problem, I left for the second IMO training seminar at Bad Homburg, where I had no internet access; hence, I did not see the solution which Harazi posted (still in the same day). Two days after that, we wrote our 3rd TST, where this problem occured. I knew the problem, but since I hadn't seen Harazi's reply, I didn't know the solution, and I also didn't succeed to solve the problem on myself during the exam. The next day, when I returned home, I saw the solution posted by Harazi. Actually the problem is not new; Harazi says he has already seen it on the forum, and another friend of mine wrote that "this problem is really quite well-known, I saw it in some British correspondence course from about 4 years ago...". Also note that one of our leaders claimed that the three problems from the 3rd TST were from the Shortlist 2003. Actually all of them turned out to be NOT from the Shortlist. Then, where were they from, from the Longlist?

  Darij


geometry
circumcircle
Euler
analytic geometry
perpendicular bisector
complex numbers
geometry unsolved

Given a nondegenerate triangle $ABC$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|OH| < 3R$.

Given a triangle ABC we construct a new triangle DEF such that DE parallel to AB and contains C, EF is parallel to BC and contains A and similarly, FD is parallel to AC and sontains B.

Clearly the triangle ABC is similar to DEF with the enlargment soefficient equal to 2. Moreover each of the triangles AFB, CBD and EAC is congruent to ABC. Therefore FB=Bd, DC=CE and EA=AF.

Since H is the orthocentre of ABC, BH is perpendicular to AC.
Hence BH is perpendicular to FD as FD is parallel to AC.
Thus BH is the perpendicular bisector to FD.
Similarly HA and HC are the perpendicular bisectors to EF and ED respectively.
Therefore H is the circumcentre os the perpendicular bisectors  to EF and ED respectively.
Therefore H is the circumcentre of the triangle DEF. Hence FH is the circumradius of DEF. Therefore FH=2R
In the triangle BOH, OH<BH+BO
Since FBH is a right triangle BH<FH. Hence BH<2R
Besides BO=R
Therefore OH < BH + BO < 2R + R = 3R

Davron

Easy to see $G$ is in the interior of triangle $ABC$,so it's in the interior of circle $O$,
$OG<R$,and $OH=3OG<3R$

let A(a),B(b),C(c) and |a|=|b|=|c|=R then OH=|a+b+c|
we must prove that OH<3r <=> |a+b+c|<|a|+|b|+|c| and that is wellknown ineq

The problem is obvious if we use well-known formula: $|OH|^2 = 9R^2 - a^2 - b^2 - c^2$.

I can do stronger
$OH \leq HA+HB+HC-3R < 3R$
thence $HA+HB+HC \geq 3R \geq GA+GB+GC$

This one is the basic formula, every senior student who studies in the Math Major class in my country can solve this easily :lol: .
The main idea of proving this formula is using another well- known formula- The leibniz formula for the centroid point of triangle ABC:
OA^2+OB^2+OC^2=GA^2+GB^2+GC^2+3OG^2
From the above, we easily get this result:OG=1/3sqrt( 9R^2-(a^2+b^2+c^2).
But OH=3OG (From Euler line of triangle ABC).
Then we get the result :lol:

[quote="shobber"]Given a nondegenerate triangle $ ABC$, with circumcentre $ O$, orthocentre $ H$, and circumradius $ R$, prove that $ |OH| < 3R$.[/quote]
In complex coordinates let $ a,b,c,o,h$ be $ A,B,C,O,H$. Then $ h \equal{} a \plus{} b \plus{} c \minus{} 2o$, and thus $ |OH| \equal{} |h \minus{} o| \equal{} |(a \minus{} o) \plus{} (b \minus{} o) \plus{} (c \minus{} o)|$. But $ 3R \equal{} |a \minus{} o| \plus{} |b \minus{} o| \plus{} |c \minus{} o|$, so it is left to prove: $ |(a \minus{} o) \plus{} (b \minus{} o) \plus{} (c \minus{} o)| < |a \minus{} o| \plus{} |b \minus{} o| \plus{} |c \minus{} o|$, which is just the triangle inequaltiy :) (There can't be equality since $ a \minus{} o,b \minus{} o,c \minus{} o$, never could be in line in a non-degenerated triangle)

[quote="dgreenb801"][quote="TomciO"]The problem is obvious if we use well-known formula: $ |OH|^2 \equal{} 9R^2 \minus{} a^2 \minus{} b^2 \minus{} c^2$.[/quote]

Can you prove this formula?[/quote]

You can Stewart's bash using triangle $ HAO$ and cevian $ AG$. It's not too hard to express the sides of $ \triangle HAO$ and $ AG$ in terms of $ OH$, $ R$, $ a$, $ b$, and $ c$.

$ \overrightarrow{AB} \equal{} \overrightarrow{OB} \minus{} \overrightarrow{OA}\Rightarrow c^2 \equal{} |AB|^2 \equal{} \left(\overrightarrow{OB} \minus{} \overrightarrow{OA}\right)^2 \equal{} 2R^2 \minus{} 2.\overrightarrow{OA}.\overrightarrow{OB}\Rightarrow \\
\Rightarrow 2.\overrightarrow{OA}.\overrightarrow{OB} \equal{} 2R^2 \minus{} c^2\mbox{ and analogue: } \\
2.\overrightarrow{OA}.\overrightarrow{OC} \equal{} 2R^2 \minus{} b^2; \\
2.\overrightarrow{OB}.\overrightarrow{OC} \equal{} 2R^2 \minus{} a^2\Rightarrow \\
\Rightarrow \overrightarrow{OH} \equal{} 3.\overrightarrow{OG} \equal{} \overrightarrow{OA} \plus{} \overrightarrow{OB} \plus{} \overrightarrow{OC}\Rightarrow \\
\Rightarrow |OH|^2 \equal{} 9.|OG|^2 \equal{} 3R^2 \plus{} 2.(\overrightarrow{OA}.\overrightarrow{OB} \plus{} \overrightarrow{OA}.\overrightarrow{OC} \plus{} \overrightarrow{OB}.\overrightarrow{OC}) \equal{} \\
\equal{} 3R^2\plus{}2\left[(2R^2 \minus{} c^2) \plus{} (2.R^2 \minus{} b^2) \plus{} (2.R^2 \minus{} a^2)\right] \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2).$

Even easier with complex numbers :P
Assume wlog that $ \triangle ABC$ has circumcenter in $ o$.
Then use the real product: $ OH^2 \equal{} |a \plus{} b \plus{} c|^2 \equal{} (a \plus{} b \plus{} c) \cdot (a \plus{} b \plus{} c) \equal{} 9R^2 \minus{} |a \minus{} b|^2 \minus{} |b \minus{} c|^2 \minus{} |c \minus{} a|^2$ $ \equal{} 9R^2 \minus{} \alpha^2 \minus{} \beta^2 \minus{} \gamma^2$

One liner (?)

[hide]
$OH < AH+AO = R+2R \cos  A < R+2R < 3R$

Toss the problem on the complex plane with the circumcircle of $\triangle ABC$ as the unit circle. Then it suffices to show $$|h| < |a|+|b|+|c| \rightarrow |a+b+c| < |a|+|b|+|c|$$ which is clearly true. $\blacksquare$


Coul anybody show me the proof for all the equilaterals morley's triangles or tell me where can I find it (Also some problems which can be solved whit morley theorem)

please!

José Carlos

Don't you know that this is the OPEN QUESTIONS section!? Do you think that Morley's Theorem is an Unsolved problem!? Please, read before you post.

Morley's Theorem:
http://mathworld.wolfram.com/MorleysTheorem.html
http://www.cut-the-knot.org/triangle/Morley/newman.shtml

What did I have to do  :|  I thought that opne question was to order ask about theorems you don't know the solution

Open Questions: Post problems that you can't solve and you don't know if there is a solution.

A theorem certainly has a solution.

The fact that the interior Morley triangle is equilateral is the well-known Morley theorem and was proved in multiple ways at http://www.cut-the-knot.org/triangle/Morley/index.shtml . By the way, you could have done a search on MathLinks for the keyword "Morley" and found this link ;) .

As you are speaking of "[i]all[/i] the equilateral Morley triangles", you probably mean the "extraversions" of the Morley triangle, which are obtained by replacing some of the internal trisectors by external ones (note that there are two different possibilities for an external trisector) - but not arbitrarily; in fact, the external trisectors should satisfy some conditions in order to have the new triangles equilateral. Anyway, if I am not mistaken, in all cases when the triangles obtained are equilateral, you can prove this completely in the same way as for the interior Morley triangle. A systematic account of all cases, revealing which of them are "analogous" and which are substantially different, comes down to a study of extraversion; there are some posts by John H. Conway and Steve Sigur in the Hyacinthos archive related to this. but I cannot locate them right now.

[b]EDIT:[/b] Here is a short note by Steve Sigur: http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/TB%20pictures/morley.html . And here is some Hyacinthos material: [url=http://groups.yahoo.com/group/Hyacinthos/messages/799?xm=1&m=e&viscount=11]Hyacinthos messages #799, #800, #801, #802, #803, #804, #805, #806, #808, #809[/url].

[b]PS.[/b] Problems like this should definitely go into "Unsolved Problems". The "Open Problems" section is for [b]conjectures[/b] which you don't even know whether they are true or false (so you have no proof, and you don't know whether anybody else has a proof). Thus, most of your questions should go to "Unsolved Problems", except of a few ones which belong into "Theorems and Formulas" since they are questions of very general nature (like "can you tell me all about ..." or "what are the most important properties of ..." - however, few of us have the time to answer these questions).

See http://www.mathlinks.ro/Forum/viewtopic.php?p=330787#p330787 for a more detailed explanation of the intention of the subforums (I agree, the "official" definition is rather ambiguous).

  Darij

I have one book which says that we can form 27 triangles in this way and 18 of them are always equilateral.

I have read an article wich said all morley's triangles have parallel sides :s


[quote="hitek"]Show that
\[{ \sqrt {1 - a} + \sqrt {1 - b} + \sqrt {1 - c} + \sqrt {1 - d}\geq \sqrt a + \sqrt b + \sqrt c + \sqrt d},
\]
whenever $ a,b,c,d$ are $ \geq0$ and $ a^2 + b^2 + c^2 + d^2 = 1$.[/quote]
${ \sqrt {1 - a} + \sqrt {1 - b} + \sqrt {1 - c} + \sqrt {1 - d}\geq \sqrt a + \sqrt b + \sqrt c + \sqrt d}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(\sqrt{1- a}-\sqrt{a}+\sqrt2(a^2-\frac{1}{4}))\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a-1)(\frac{2a+1}{2\sqrt2}-\frac{1}{\sqrt{a}+\sqrt{1-a}})\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(2a-1)((2a+1)^2(\sqrt a+\sqrt{1-a})^2-8)}{(\sqrt a+\sqrt{1-a})((2a+1)(\sqrt a+\sqrt{1-a})+2\sqrt2)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(2a-1)((2a+1)^2(1+2\sqrt{a(1-a)})-8)}{(\sqrt a+\sqrt{1-a})((2a+1)(\sqrt a+\sqrt{1-a})+2\sqrt2)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(2a-1)((2a+1)^2(-1+2\sqrt{a(1-a)})+2(2a+1)^2-8)}{(\sqrt a+\sqrt{1-a})((2a+1)(\sqrt a+\sqrt{1-a})+2\sqrt2)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(2a-1)^2(4a+6-\frac{(2a+1)^2(2a-1)}{(\sqrt a+\sqrt{1-a})^2})}{(\sqrt a+\sqrt{1-a})((2a+1)(\sqrt a+\sqrt{1-a})+2\sqrt2)}\geq0.$
If $ a\leq\frac{1}{2}$ then our inequality is true.
But if $ \frac{1}{2}\leq a\leq1$ we obtain:
$ 4a+6-\frac{(2a+1)^2(2a-1)}{(\sqrt a+\sqrt{1-a})^2}\geq4a+6-(2a+1)^2(2a-1)\geq0.$
The last inequality easy to check out.

how did you come up with $ \sqrt2 (a^2\minus{}\frac{1}{4})$ arqady?

[quote="behemont"]how did you come up with $ \sqrt2 (a^2 - \frac {1}{4})$ arqady?[/quote]
We know that the equality occurs when $ a=b=c=d=\frac{1}{2}.$
${ \sqrt {1 - a} + \sqrt {1 - b} + \sqrt {1 - c} + \sqrt {1 - d}\geq \sqrt a + \sqrt b + \sqrt c + \sqrt d}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}f(a)\geq0,$ where $ f(a)=\sqrt{1-a}-\sqrt{a}+\lambda\left(a^2-\frac{1}{4}\right)$ for all $ \lambda\in\mathbb R.$
If we wish to get $ f(a)=(2a-1)^2g(a)$ we need $ f'\left(\frac{1}{2}\right)=0.$ It gives the value of $ \lambda.$

and why exactly $ (2a\minus{}1)^2$? :blush:

[quote="behemont"]and why exactly $ (2a \minus{} 1)^2$? [/quote]
We can  wish about $ (2a\minus{}1)^3$ or $ (2a\minus{}1)^4$ but if $ f(a)\equal{}(2a\minus{}1)^2g(a)$ and $ g(a)\geq0$ it proves our inequality already.
This is very not strong method, but it helps sometimes.

[quote="zaya_yc"][quote="hitek"]Show that
\[{ \sqrt {1 - a} + \sqrt {1 - b} + \sqrt {1 - c} + \sqrt {1 - d}\geq \sqrt a + \sqrt b + \sqrt c + \sqrt d},
\]
whenever $ a,b,c,d$ are $ \geq0$ and $ a^2 + b^2 + c^2 + d^2 = 1$.[/quote]

$ \frac {1}{4}\sqrt {1 - a} + \frac {1}{4}\sqrt {1 - b} + \frac {1}{4}\sqrt {1 - c} + \frac {1}{4}\sqrt {1 - d}\geq \sqrt {1 - (\frac {1}{4}a + \frac {1}{4}b + \frac {1}{4}c + \frac {1}{4}d)}$ ...[/quote]

i think it False, $ x ->\sqrt{1-x}$ it's concave not convex .

Vasile, I saw the same problem proposed by you in Maths Reflections. Is this coincidence? It is my teacher's inequality.

[i]There is a good solution

   Just using these demo :

   $ 2(1\minus{}a)(1\minus{}b) \equal{} 1 \minus{} 2a \minus{} 2a \plus{} 2ab \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$

$ \equal{} ( a \plus{} b \minus{}1 )^2 \plus{} c^2 \plus{} d^2 \geq c^2 \plus{} d^2 \geq 2cd$

 $ \to (1\minus{}a)(1\minus{}b) \geq cd$

   Square two side of the ineq and we easily finish :lol: [/i]

${ (\sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c}+\sqrt{1-d})^2\geq(\sqrt a+\sqrt b+\sqrt c+\sqrt d})^2$
and $ \sum_{cyc}\sqrt{(1-a)(1-b)}\geq\sum_{cyc}\sqrt{cd}$
and $ \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\frac{a+b+c+d}{4}\Leftrightarrow$
$ 1-a+1-b+1-c+1-d\geq\ a+b+c+d$ so it ineq easily finish.

Sqrt(1-x) is not convex. We can only apply jensen to the rear part which yields
Sum( sqrt(a)) <= 4 sqrt( (a+b+c+d)/4) = 2sqrt(a+b+c+d). Also observe 2>=a+b+c+d>=1. Then I'm stuck.

[quote=hitek]Show that
\[{ \sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c}+\sqrt{1-d}\geq \sqrt a+\sqrt b+\sqrt c+\sqrt d},\]
whenever $ a,b,c,d$ are $ \geq0$ and $ a^2+b^2+c^2+d^2=1$.[/quote]

[img]http://s12.sinaimg.cn/middle/006ptkjAzy75GAPV2eD7b&690[/img]

[quote=hitek]Show that
\[{ \sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c}+\sqrt{1-d}\geq \sqrt a+\sqrt b+\sqrt c+\sqrt d},\]
whenever $ a,b,c,d$ are $ \geq0$ and $ a^2+b^2+c^2+d^2=1$.[/quote]
$$\sqrt{1-a}-\sqrt{a} \geq \frac{1-4a^2}{2\sqrt{2}} $$
[url]https://artofproblemsolving.com/community/c6h1570701p9643700[/url]


geometry
geometric transformation
rotation

let a massless rod of be clamped at the centre with two small ball each of mass m be attached at the ends of the rod.let a small ball of mass m be dropped from a height h

what should be the minimum value of h so that the rod will make a full rotation

the axis is horizontal so that it can rotate freely about it
intially the rod with the both particles is at rest
the ball comes from a height h and sticks to one of the balls

First of all, the inertia of the rod system + the new ball is $3mr^{2}$.

Next, the speed of the ball right before it hits the rod is $\sqrt(2gh)$.

Therefore, $mr \sqrt(2gh) = 3mr^{2}\omega$.
Thus, $r \omega = \frac{\sqrt(2gh)}{3}$.

The kinetic energy is then $I \omega^{2}= 3m (r \omega)^{2}= \frac{2mgh}{3}$.

This must be at least the difference in potential energy from this point to the highest energy position.

The energy difference is $2mgr-mgr = mgr$.

Thus, $\frac{2mgh}{3}\geq  mgr$,
so $h \geq  \frac{3}{2}r$.

The minimum value is $\frac{3}{2}r$.


[quote="kolahalb"]I am to show that the differential equataion D^2f+[A+V(x)]f=0
{A is a constant and V(x+m)=V(x)}
has the solutions of the form f=exp[ikx]U(x)where U(x+m)=U(X)

I tried to differentiate the given solution and put it in the equation...but the method is not working.Can you please help?[/quote]
Try the calculus computations forums...
and latex:
[quote="kolahalb"]If $A$ is a constant and $V(x+m)=V(x)$, I am to show that the differential equation $D^{2}f+\left[A+V(x)\right]f=0$
has solutions of the form $f=e^{ikx}U(x)$ where $U(x+m)=U(X)$

I tried to differentiate the given solution and put it in the equation...but the method is not working.Can you please help?[/quote]
Did I interpret it correctly?


We're in the process of setting the schedule for this fall, and I'm wondering what day of the week students prefer to have classes on.

In this thread, please post your preferences for days of the week of classes, noting both days that you strongly prefer, and days that you very much don't want to have class.  Assume all classes are 7:30 - 9 PM ET (4:30-6 PM PT).

Friday would be best for me, or maby Wednesday... I go with Wednesday :)

It would be cool if they could have a WOOT session on Fridays. I'm considering enrolling.

I would like Fridays, because my parents don't like me staying late to review the class notes.

Also for Fridays, I have homework on other days and procrastinating homework for the weekends and doing an AoPS class wouldbe cool for Fridays.

I'd prefer WOOT being on Fridays the most as I'm probably gonna take that :D

No, not Friday, we always have our piano lesson on Friday. Saturday is best, any of the other week days except Friday.

I would also like Friday or Satuday.

I will be very busy on every other day.

I think the original reason no Fridays was because of "social activity" but I think they forget us AoPsers have no social activity.  :D  :(

Thursdays OR TUESDAYS
(THE CAPITAL LETTERED ONE MORE FAVORED)

That would force them to work on weekends...

I like Fridays, though really any day of the week should be fine.

Mondays and Tuesdays, after that I'm too busy catching up on what I procrastinated earlier in the week :D

As a future note - Saturdays are never good for me. I'm at a competition almost every Saturday since January - May

science olympiad? just a guess. its the same case for me

fridays are bad for the same reason #H34N1 stated

mondays-wensdays are good. there should be a poll....

Tuesdays and Thursdays, and maybe Monday?

[quote="piggyinc"]Not Friday. We need at least one day of the week where we can sit back, relax, and let our brains take a break.  :lol:[/quote]

Can't agree more...

Not Monday, Wednesday, or Thursday. Anything else is all right.

Mondays and Thursdays are pretty bad for me because I have afterschool Math team stuff. WOOT would be pretty sweet if it was on Friday! :)

Monday and Friday are the best days for me.

Friday = end of the weekdays
Monday = start of the weekdays

$ \rightarrow$Friday$ \leftarrow$ or Wednesday.
arrows indicate preferred.

Since the class is supposed to be offered twice per week, I would think Friday and some other week day would be best, so that students can plan ahead and decide which day to pick so it doesn't interfere with their school work or other activities. As abacadaea already said, we should get a poll going, to get a better idea of what days are generally good and bad.

And if you consider doing math on a Friday night to be a bother, well then, I have no idea what's wrong with you.  :lol:

Well, not Mondays 'cuz that's the beginning of the week and i feel kinda lazy on mondays....Tuesday is good, Wednesday....i guess, Thursday, no, not really, and Friday is good. Saturday is fine with me too.

So any day excdpt monday and thursday are good, i gues...

Hmm... I was under the impression that class would be held at 9-11 ET on one of the two days (so it says on the WOOT schedule page.) I would actually prefer this, since I live on the west coast and 4:30-6:30 isn't very convenient on weekdays. But if the time-frame is fixed, I guess Wednesdays or Thursdays would be fine.


let x belong to a group. If x^2 ≠ e and x^6 = e, prove that x^4 ≠ e and x^5 ≠ e. What can we say about the order of x?
need some help on this one!
thanks
greg

The order of $x$ divides $6$ and is $\neq 2$ ...

i still don't see it. then what about X^5... i don't see where you are headed

Let be $o$ the order of $x$ and so also of $x^{-1}$. Then as said before, $o|6$, but also $o\nmid 2$, so $o \in \{3;6\}$.
Now $x^5=x^{-1}, x^4=x^{-2}$, which can't be $e$ by the previous said.