In a $\triangle {ABC}$, there's a point $D$ in $BC$ such that $AD \perp BC$. $E$ is a point in $AD$ such that $AE=DE$. $F$ is a point in $BE$ such that $BF=EF$, and $G$ is a point in $CF$ such that $CG=FG$. If the area of $\triangle {ABC}$ is $36$, find the area of $\triangle {EFG}$.

[quote="deej21"][hide]
Let $BD = x, DC = y$.
Then $ABD = 36 \frac x {x+y} \implies BED = 18 \frac x {x+y}$
$ABC = 36 \frac y {x+y} \implies CED = 18 \frac y {x+y}$
$BEC = BED + CED = \frac{18(x+y)}{x+y} = 18 \implies FEC = 9$
$EFG = \frac 1 2 FEC = \frac 9 2$
[/hide][/quote]

Good! Another method???

[hide]
$\triangle BEC$ has half the height of $\triangle ABC$ and the same base, so its area is half that of $\triangle ABC$. Also, the area of $\triangle FEC$ is half that of $\triangle BEC$ and the area of $\triangle EFG$ is half that of $\triangle FEC$, both of which can be shown by noting that each pair of triangles have the same height and one has half the base of the other.

It follows that the area of $\triangle EFG$ is one-eighth that of $\triangle ABC$, which in this problem is $\boxed{\frac 92}$.
[/hide]

[hide]$[EBC]=1/2[ABC]=18$, since it has half the height.
$[EFG]=[BFG]=[GBC]=[EGC]$, since they have same altitudes or bases, and their sum is $[EBC]$.
$4[EFG]=[EBC]\implies[EFG]=\frac{9}{2}$.[/hide]

Thanks all!!! It's a nice problem, I was looking for nice solutions, as yours!


function
algebra
domain
real analysis
real analysis theorems

[quote="pardesi"]can someone show me the detailed proof of this fact
the solution to laplace equation is uniquely specified in a given volume it's value is given everywhere in the boundary.[/quote]

their difference is a harmonic function which is zero on its boundary.  apply the maximum principle.

[quote="sylow_theory"]I am a noob in PDE theory. I wanted to make a post to this question but I was afraid it be rated 1 star so I didnt. So instead I have a question to ask on this.

Why cant we say that since the Dirichlet problem is a Well-Posed PDE and so it has a unique solution.  :?:[/quote]

this is like saying that the solution follows from the uniqueness theorem for laplace's equation.  e.g., it is technically correct but completely useless since this combined with the poisson formula are what prove the well posedness of the dirichlet problem.

there is no a priori reason that solutions to a particular differential equation should be unique, e.g., try the differential equation $\dot{x}=x^{2}, \ x(0)=0$.

[quote="Kent Merryfield"][quote]try the differential equation $\dot{x}=x^{2}, \ x(0)=0.$[/quote]
That does have a unique solution. To make your point, try $\dot{x}=x^{1/3},\ x(0)=0$ instead.[/quote]

thanks :)

[quote="blahblahblah"]their difference is a harmonic function which is zero on its boundary.  apply the maximum principle.[/quote]
how to prove this using maximum principle

Uhm, the version blahblahblah is thinking of is that the difference function assumes both its maximum and its minimum value on the boundary.

Of course, a more detailed statement is necessary. One possibility is:
Suppose $u$ is continuous on the closure of a bounded region $\Omega$ and that it satisfies the mean value property in $\Omega$. Then $u$ takes its largest and smallest value in $\bar{\Omega}$ on $\partial \Omega$ (and if either is assumed in $\Omega$, then $u$ is constant).

Here a region is an open connected set.

Withou boundedness of the domain uniqueness is evidently not true. For example if the domain is all of $\mathbb{R}^{n}$.

one more ...is it necessary that the function be not defined outside the boundary or can the region where it is defined be broken into small spaces and still u can claim that in any region if the function attains maximum then it does so at the boundary


Prove that there are infinitely many powers of 2 in the sequence $ [n\sqrt{2}]$

we need $ 2^m\leq n\sqrt{2}<2^m\plus{}1$. that is, $ \frac{1}{\sqrt{2}}2^{m}\leq n<\frac{1}{\sqrt{2}}2^{m}\plus{}\frac{1}{\sqrt{2}}$. since $ \frac{1}{\sqrt{2}}>\frac{1}{2}$, it suffices to find infinitely many $ m$ such that $ \frac{1}{\sqrt{2}}2^m$ has fractional part at least $ \frac{1}{2}$. now write $ \frac{1}{\sqrt{2}}$ in base 2. multiplying by $ 2^m$ is the same as moving the decimal point $ m$ places to the right. since $ \frac{1}{\sqrt{2}}$ has infinitely many 1's in its base 2 representation (because it's irrational), the result follows.


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

[quote="mathematica"]The circumference of a circle is divided into $p$ equal parts by the points $A_1,A_2,...,A_p$ where $p$ is an odd prime number. How many different self-intersecting $p$-gons are there with these points as vertices if two $p$-gons are considered different only when neither of them can be obtained from the other by rotating the circle? (A self-intersecting polygon is a polygon some of whose sides intersect at other points beside the vertices.)[/quote]

And why the subject is Wilson's theorem???

Using Burnised's lemma, we just have to count the average number of fixed points of the rotations. Rotation by 0 has (p-1)!/2-1 fixed points. We start in A1 and can go through the p-1 other vertices in (p-1)! ways, each counted 2x (clockwise and counterclockwise) excluding the way A1A2A3...An. All the other rotations have (p-3)/2 fixed points, which are all the stars. So the number we're looking for is
$\frac{\frac{(p-1)!}2-1+(p-1)\frac{p-3}2}p=\frac{(p-1)!-2+p^2-4p+3}{2p}= =\frac{(p-1)!+1+p^2}{2p}-2$.
From there p|(p-1)!+1+p^2, from where one implication of Willsons theorem follows (and only for p>2 :) )

But you didn't use this: $p$ is prime. So, wilson's theorem should be true for all natural numbers!

If p was not prime, the rotations would have more fixed points (not only the stars).


number theory
least common multiple
number theory proposed

Show that $ lcm(1,2,\ldots,2n)$ is divisible by $ \frac{(2n)!}{(n!)^2}$ for any positive integer $ n$.

Let $ s_n$ be the sum of the digits of $ n$ in the base-expansion $ p$.
We know that the number of factors of a prime $ p$ in $ n!$ is:
$ \frac{n\minus{}s_n}{p\minus{}1}$
We have to show that for every prime p such that:
$ p^k\parallel{}\binom{2n}{n}$
Then $ 2n\geq p^k$

$ k\equal{}\frac{2n\minus{}s_{2n}}{p\minus{}1}\minus{}2\frac{n\minus{}s_n}{p\minus{}1}$
$ \implies k(p\minus{}1)\equal{}2s_n\minus{}s_{2n}$
First case: If $ s_n<s_{2n}$
$ \implies k(p\minus{}1)\equal{}2s_n\minus{}s_{2n}<s_{2n}$
$ \Leftrightarrow 2n\geq p^k$
as claimed.
Second case: If $ s_n\geq s_{2n}$
We have to do an analysis of $ n$, I will post my answer of it later.


Is there a function $f$ defined on the positive integers with values in the same set such that $f(n)\leq n^8$ for all $n$ but the equation $ f(m)=f(n_1)+...+f(n_k)$ does not have natural solutions? I believe the answer is yes, but I did not manage to find one. And another question: does it work for smaller values than 8?

What is $k$?  :huh:
What about a constant non-zero function?

Pierre.

$k$ is an arbitrary number at least 2. I think the conditions are that $f$ is injective and the sum of at least 2 values of the function is never a value of the function.

I still have trouble understanding the problem: may two of $n_j$ be the same? If they may, we clearly have $f(1)=1$ and therefore every value is a sum of several $f(1)$'s.  :?

No, there are no equal numbers among $n_j$.  :)


ther are many solutions to this ques. I found alredy 2 and see one more  :)
prove that there are million natural numbers that any sum of their subset isn't a square.


Tweedledum says, ``The sum of your weight and twice mine is 361 pounds.'' Tweedledee says, ``Contrariwise, the sum of your weight and twice mine is 362 pounds.''  If they are both correct, how much do Tweedledum and Tweedledee weigh together?

this problem  is equavilent  to  solve system equation :
  $ x\plus{}2y\equal{}361$ abd $ y\plus{}2x\equal{}362$
 hence  :  $ x\equal{}121$ and $ y\equal{}120$

This is a much faster way...  You are asked for x + y so need not solve for x and y separately.

x + 2y = 361.
2x + y = 362.

Adding the eqns you get
3x + 3y = 723.

Divide by 3 to get 241.

(So basically just add and divide by 3..... much simpler)

Let x be the weight of Tweedledum, and y be the weight of Tweedledee. Translating the 2 statements to equations gives us x+2y=361 and 2x+y=362. Add the two equations leaves us with 3(x+y)=723, so the sum of the weights of Tweedledum and Tweedledee is 241.

Please do not post the same solution twice, dragon96. Thank you.


Determine all $n \in \mathbb{N}$ for which [list][*] $n$ is not the square of any integer, [*] $\lfloor \sqrt{n}\rfloor ^3$ divides $n^2$. [/list]


probability
number theory
prime numbers
geometric sequence
prime factorization

Let $n$ be a natural number

$n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}\quad , p_1, p_2, ....$ are distinct prime numbers, then number of divisors is equal to:$(\alpha_1+1)(\alpha_2+1)\cdots (\alpha_k+1)$

And the sum of the divisors is ${\displaystyle \prod_{i=1}^{k} (1+p_i+p_i^2+\cdots+p_i^k)}$. You can simplify with the geometric sequence formula if you want...

[hide="Proof"]
Consider the integer $n = \prod_{i=1}^{k} p_i^{a_i}$.  Every integer $d$ such that $d | n$ must be of the form $d = \prod_{i=1}^{k} p_i^{b_i}$ where each $b_i \le a_i$.  We wish to find the number and sum of all such divisors.

Because each exponent of each $p_i$ in the prime factorization of $d$ can range from $0, 1, 2, .. a_i$ there are $a_i + 1$ possibilities for each exponent.  Simple probability tells us that the number of divisors is therefore $N = \prod_{i=1}^{k} (a_i + 1)$ (this counts $1$ and $n$).  Next, the sum:

Because we are summing over all the permutations, for each divisor $p_i$ we must sum all the divisors such that $b_i = 0$, then all divisors such that $b_i = 1$, and so on until the divisors such that $b_i = a_i$.  We repeat this logic for every divisor, and we see that this sum is given by $S = \prod_{i=1}^{k} (1 + p_i + p_i^2 + ... + p_i^{a_i})$.  Expanding this product gives the sum of all possible permutations of the divisors (up to $a_i$ for each $p_i$), precisely the sum we desire.
[/hide]

[quote="José"][quote="shadysaysurspammed"]Find a general formula for finding the nuber of divisors and sum of all divisors of a number[/quote]

K=a^p1*b^p2*c^p3*...*s^pn, for a, b, c, ..., n prime numbers.

Then, the number of divisors of K is (p1+1)*(p2+1)*(p3+1)*...*(pn+1).[/quote]

Sorry, it was for a, b, c, ..., s prime numbers, not for a, b, c, n. (It just was s instead of n).


Please vote [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=392541#392541]here[/url] too.

what do you mean 'how much'? That doesn't really make any sense . . . it's rather well-accepted that France was crucial to the outcome but that obviously doesn't mean it was the only country fighting, so do you mean like what percent of the troops it supplied or something?

[quote="stupidkid"]Why did you make two polls that are exactly the same??[/quote]I think that buffalo is trying to find the difference between what your school taught and what you believe? :?


Let be $ a,b,c,d,e$ real numbers such that :

$ |a\minus{}b|\equal{}2|b\minus{}c|\equal{}3|c\minus{}d|\equal{}4|d\minus{}e|\equal{}5|e\minus{}a|$

Prove that $ a\equal{}b\equal{}c\equal{}d\equal{}e$

[hide]Put $ x\equal{}|a\minus{}b|$. Then

$ b\equal{}a\pm x$

$ c\equal{}b\pm{x\over 2}$

$ d\equal{}c\pm{x\over 3}$

$ e\equal{}d\pm{x\over 4}$

$ a\equal{}e\pm{x\over 5}$

Summing that all up, we get

$ \pm x\pm{x\over 2}\pm{x\over 3}\pm{x\over 4}\pm{x\over 5}\equal{}0\iff x(\pm 60\pm 30\pm 20\pm 15\pm 12)\equal{}0$

However, the expression in the parentheses can never be divisible by $ 5$, hence it can't be zero. Therefore $ x\equal{}0$ and the conclusion follows.[/hide]


[quote="perfect_radio"]Little problem with the Internet connection.[/quote]

Rather with that of the AoPS servers. ;)

Moved to Support forum, as Valentin wants such things to be reported.
(ML was very lame for the last... hour?)

  darij

Look at the time of the first post (5:28pm - Romanian Time). The pages were loading pretty slowly around that time (and I was trying to post a topic, but somehow it ended up being double posted), but I just assumed the fault is "on my side", since such things happen from time to time with my Internet connection.


Has anyone heard of this show?  If so, what do you think about it?  For those who haven't heard of it, it's a game show where people can win money for truthfully answering questions such as "Did you ever cheat on your boyfriend/girlfriend?" or "Did you every steal money from your parents?"  As you can see, these questions are often personal and cause contestants to reveal some of their darkest secrets.

I don't know why someone would be willing to ruin their lives for money.  Yes, half a million dollars is nice, but is ruining your life on national television worth the money?  I think not.  Besides, I don't think anyone has ever won more than $ \$$100,000.

I haven't ever seen it, but I've heard about it. It sounds pretty stupid. (Well, it is a FOX show...)
For one thing, if one refuses to answer the question or answers it with a lie, it reveals the same thing as if that person answers it. So there should be no reason not to answer every question.

I saw promos for it while watching family guy and american dad and every time I saw the promo I'd be all like MAN I WOULD BE SO GOOD AT THIS SHOW

cause seriously? I would be. Like, I don't really have any people depending on me whom I'd be afraid of letting down (like significant others or whatnot ... I don't really see how I could let down my parents in any way), and I couldn't let down myself by anything, so it seems like free money to me

It seems like a very interesting show, since for the contestants, it's basically either you get money, or you keep your relationships with your close ones.

Only one question...how do the people know exactly what questions to ask the contestants...and how do they know the history of them and their close ones (like ex's...for example)?

[quote="daboa"]You assume lie detector tests work.[/quote]

yeah, this is what my roommate (studying psychology) would respond whenever I pointed out I would be so good at the show

he'd be all like dude lie detectors work like 5% of the time

and I'd be all like dude I don't care

etc

I remember once there was a guy on and it said he lied about something and he got so pissed because he swore it was lying..... I love the family's reactions too. :)

This show sounds so stupid. If the person was on the million dollar question and the show didn't want to give the person that much money couldn't they just "say" the person was lying? Lie detectors are not reliable.

I think the show would [i]rather[/i] give the person that much money.

Meh, the questions are designed so that everyone has to answer yes to everything. Who hasn't lied, cheated on a girlfriend, stolen money, not sure if they love the person they're with and so on... Only devout religious people I guess.

Well, the clutch is if your "loved ones" know already. For example - stealing money from your parents. If your parents are present during the show, and if they know you did steal money, then it wouldn't be a big deal.

I wouldn't burn my life for money, though. There are some secrets you're supposed to keep  :wink: .

[quote="7h3.D3m0n.117"]I wouldn't burn my life for money, though. There are some secrets you're supposed to keep  :wink: .[/quote]
I agree.  There are much better ways to earn money.  Deal or No Deal, anyone?

That's because it's staged. The contestants know the questions before they are "asked" (the show even says this at the beginning)

Seriously, it's just normal TV trash. If you want to watch something worthwhile, watch The Daily Show or The Colbert Report.

I don't like "The Moment of Truth." Some people could be really calm in their brains by heart, making the Truth Detector not work and thus giving them the cash they want. And ruining your relationships for money? There was once a phrase called "Happiness is priceless."

count on fox to make something like this....

the whole premise of the show is flawed, and it's just sick.  it really shows wat some people will do for money...

what

it's about telling the truth

this doesn't seem like an inherently sick concept

[quote="MysticTerminator"]this doesn't seem like an inherently sick concept[/quote]
Well, on this show, the following equation holds:

$ \text{telling the truth}\equal{}\text{revealing dark secrets}$

Keep in mind that when the contestants reveal their darkest secrets, they simultaneously ruin their lives on national television.  You're telling me that's a good thing?

[quote="MysticTerminator"]ok? isn't that good

cleansing and all that[/quote]

Cleansing yourself while making others uncomfortable? The only way to "cleanse" yourself is to talk with someone without being forced to, and having a good discussion about it...

I saw this show once. It was hilarious. It was like:

[b]Question:[/b] "Are you a better mother to your kids than your mother was to you?"

[b]Answer:[/b] *breaks down crying*

lmao amazing. Yeah, so each person goes through a long list of questions and the lie detector determines the pre-given answers, and then they pick the people with the juciest life problems and ask them the tough questions. So great. It's just what they deserve.


If x,y,z are positive real numbers prove

[(x+y)(y+z)(z+x)]^2  >= xyz(2x+y+z)(2y+z+x)(2z+x+y)

Does anyone know the source of this problem?

I put $x=p-a$, etc., where $p$ is the semiperimeter of a triangle and $a,b,c$ are its sides. We get $(abc)^2\ge (p-a)(p-b)(p-c)(a+b)(b+c)(c+a)$. Since $abc=4RS,\ S=\sqrt{p(p-a)(p-b)(p-c)}$, we get $8(a+b+c)R^2\ge(a+b)(b+c)(c+a)$. We also know that $0\le OH^2=9R^2-(a^2+b^2+c^2)$, so it's enough to show $8(a+b+c)(a^2+b^2+c^2)\ge 9(a+b)(b+c)(c+a)$. We now use $a^2+b^2+c^2\ge \frac{(a+b+c)^2}3$ and then it's reduced to AM-GM for $a+b,b+c,c+a$.

I get the feeling I  made an unnecessary detour, but I can't pinpoint the redundant steps :).

Expanding, the inequality becomes
\[2 \sum x^3 y^3 + \sum x^2 y^4 \ge \sum x^3 y^2 z + 6 x^2 y^2 z^2,\]
where each sum is cyclic.

The inequalities
\[2 \sum x^3 y^3 \ge \sum x^3 y^2 z\]
and
\[\sum x^2 y^4 \ge 6 x^2 y^2 z^2\]
follow from Muirhead's.

This solution(not mine) avoids expanding brackets

Write  T=x+y+z

Then (x+y)(y+z) =y(T+xz/y)
        (y+z)(z+x)  =z(T+xy/z)
        (z+x)(x+y)  =x(T+zy/x)

Left side of inequality becomes   xyz(T+xy/z)(T+yz//x)(T+zx/y)

Right side of inequality becomes xyz(T+x)(T+y)(T+z)

The inequality follows after expanding brackets in these expressions and using the facts that x^2+Y^2+z^2 >=xy+yz+zx

             xy/z +yz/x +zx/y >=x+y+z
        (

Hi.
I think this ineq was proposed in Crux magazine by a greek!

 :)

the following stronger inequality is true (by popoviciu)
$[(x+y)(y+z)(x+z)]^2\geq \frac{64}{27}xyz(x+y+z)^3$

[quote="spanferkel"][quote="Albanian Eagle"]the following stronger inequality is true (by popoviciu)
$[(x+y)(y+z)(x+z)]^2\geq \frac{27}{64}xyz(x+y+z)^3$[/list][/quote]You say it's stronger. Do you mean $\frac{64}{27}(x+y+z)^3\ge (2x+y+z)(2y+z+x)(2z+x+y)$? Can you prove that?
(note: I think the fraction has to be inversed)[/quote]
 I edited
 :P
the inequality $\frac{64}{27}(x+y+z)^3\geq (2x+y+z)(2y+z+x)(2z+x+y)$ is obvious by $AM-GM$ (for three variables)  [hide] :D  :D  :D [/hide]

[quote="Albanian Eagle"]
 :P
the inequality $\frac{64}{27}(x+y+z)^3\geq (2x+y+z)(2y+z+x)(2z+x+y)$ is obvious by $AM-GM$ (for three variables)  [hide] :D  :D  :D [/hide][/quote]sorry, right away I cannot see how - maybe I'm having a bad day. [hide]EDIT: sorry, no smilies here [/hide] Please explain (you won't expand, will you :?: )

oops, when posting, I just got the message
"Too many smilies in your post: 8.
You can have at most 6 smilies in this message. Please remove some smilies before submitting this message."
Are they soo serious here??


modular arithmetic
combinatorics proposed
combinatorics

let $n$ be a positive integer,and set$A=\{n,n+1,\ldots,n+17\}$.Find all values of $n$ for which we can partiion $A$ into two subset $B$ and $C$ such that the product of the elements of $B$ equals the product of the elements of $C$.

this is a rather straightforward one. there is no such $n$.
 to start off look at the numbers modulo $19$. since there are $18$ numbers in the given set, atmost one can be divisible and for the given condition to hold, it follows that the numbers at hand are $\{1,2,3,\ldots,18\}$ (modulo $19$). since these can be partitioned into two sets with the same product, it follows that modulo $19$, this product is a square. but this product is simply $-1\pmod {19}$ and since $19\equiv 3\pmod 4$ it follows that $-1$ is not a square modulo $19$ and that is a contradiction.

Yes.This 's also my solution.And here is the generalization from Tounament of Tours problem:Show that there no such n such that:A can be partitioned into k subsets with the same property!
   K-power residue will also work!


Let $ \Delta$ ABC Prove that
8$ cos^2Acos^2Bcos^2C\plus{}cos2Acos2Bcos2C \geq0$

Let x=tanA;.....
we have x+y+z=xyz
Done<->$ (x^2\minus{}1)(y^2\minus{}1)(z^2\minus{}1)\leq8$
Note:  $ 4xy\minus{}z^2(x^2\minus{}1)(y^2\minus{}1)\equal{}(z^2\minus{}1)(x\minus{}y)^2$
we have done
 :rotfl:


$p, q$ are positive real numbers such that $\frac{1}{p}+\frac{1}{q}=1$. Prove that:

$[\sum_{k=1}^{n}x_{k}^{p}]^{\frac{1}{p}}[\sum_{k=1}^{n}y_{k}^{q}]^{\frac{1}{q}}\geq \sum_{k=1}^{n}x_{k}y_{k}$

Where $x_{k}, y_{k}$ are positive real numbers.

[quote="paladin8"][quote="Elemennop"]That's not Holder's...[/quote]

You could explain why maybe... the LHS should be a product, not a sum.[/quote]

Thanks, edited now.

How about the following proof?  By scaling, we may assume that
\[\sum_{k=1}^{n}x_{k}^{p}= 1 \]
and
\[\sum_{k=1}^{n}y_{k}^{q}= 1 \, . \]
By the weighted AM-GM inequality, we have
\[x_{k}y_{k}\le \frac{1}{p}x_{k}^{p}+\frac{1}{q}y_{k}^{q}\, . \]
Summing over all $k$ gives the desired inequality.


Yeah, it is.  By the way, the Online School forum is more appropriate for this question.

[quote="Calculus"]I was really interested in joining the independent class over the summer. Is it to late to join?[/quote]

PM rrusczyk or MCrawford about it.  They are the instructors and they will usually consider letting students join a class after the deadline.

Thanks for the help.

I am really sorry, I didn't realize there was a forum for online classes.

This is as good a place as any to post that question.  If you are very interested and willing to work hard to catch up on the work, give us a call at the office at (619) 659-1612.  It will be easier to gauge your appropriate level and explain the program quickly that way.  It would also be easier to get you started 2 1/2 weeks late with a phone conversation than any other way.


geometry
3D geometry
combinatorics proposed
combinatorics

Ok, following from the Cube-Colouring problem, I will now claim something. But please don't solve it by referring to the Cube problem.

Let S(n) be a sequence of 1 and -1 with the property
Sum(i=1 to 1000) S(i) = 0.

Hence
A = Sum(1 to 999) 1 - S(n)S(n+1) +
Sum(1 to 990) 1 - S(n)S(n+10) +
Sum(1 to 900) 1 - S(n)S(n+100).

Prove that A >= 200 for every Sequence S.

An approach could be, to give a sequence for which is 'A' = 200, and then to show that by swapping the values of S(i) and S(j) 'A' can only increase.

Well, today in the school bus I thought about it. I made a mistake.
This is NOT equal to the Cube-Coloring problem. Why did no one recognize my logical failure?  :?

But well, A >= 200 must be true - and I claim [b]A >= 222[/b]. This is to be proven.

[quote="Daniel Gutekunst"]Why did no one recognize my logical failure?[/quote]

Because you correct yourself soo fast, we hadn't even read it ;)


calculus
integration
trigonometry
function
algebra
rational function
complex analysis

I wonder how to evaluate the following integral using contour integration?

We'll do the "substitution" $z=e^{i\theta}$

Then $dz = ie^{i\theta}\,d\theta$ which we rewrite as $d\theta = \frac{dz}{iz}$

Then $\sin2\theta=2\cos\theta\sin\theta= 2\cdot\frac12\left(z+\frac1z\right)\cdot\frac1{2i}\left(z-\frac1z\right).$

Do all that; we now have the integral over the unit circle $|z|=1$ of a rational function of $z.$ Do the necessary algebraic simplifications, find the poles, determine which ones lie inside the unit circle (ignore the poles outside the unit circle), find the residues, and use the residue theorem.

After my computation, I got 2pi / sqrt(3) as the answer, is that correct? :huh:

Better: do some real calculus first. Let $\alpha = 2\theta.$ Note that the resulting function of $\alpha$ is periodic of period $2\pi,$ so replace the integral from $0$ to $4\pi$ by twice the integral from $0$ to $2\pi.$ Then do the procedure outlined in the earlier post - but the algebra is now a little simpler.

----

Yes, $\frac{2\pi}{\sqrt{3}}$ is the correct answer.


How do you insert blank lines in a \align environment?  \vspace{} doesn't seem to work there.

\intertext allows you to add a line of text between equations without affecting the numering, so \intertext{} would be a blank line.


vector
linear algebra
linear algebra unsolved

Prove that if there exists a linear map $T$ on $V$ whose null space and range are both finite-dimensional, then $V$ is finite-dimensional.

[hide="My Solution"]
I let the nullspace be denoted by $N(T)$ and the range be denoted by $V(T)$. Let $\dim N(T) = n$, $\dim V(T) = m$.

Assume that $V$ is infinite-dimensional. Let $e_1,...,e_n$ be a basis for $N(T)$. Since V is infinite-dimensional we can extend this to $e_1,\cdots,e_n,e_{n+1},\cdots,e_j$ linearly independent vectors in $V$.

I will show that $T(e_{n+1}), \cdots , T(e_j)$ are linearly independent. If
$0 = x_{n+1}T(e_{n+1}) + \cdots + x_jT(e_j) = T(x_{n+1}e_{n+1} + \cdots + x_je_j).$
then the vector $x_{n+1}e_{n+1} + \cdots + x_je_j$ belongs to $N(T)$. This is only possible if $x_{n+1} = \cdots = x_j = 0$, because a vector $x_1e_1 + \cdots + x_je_j$ belongs to $N(T)$ if and only if $x_{n+1} = \cdots = x_j = 0$.

We can choose $j$ so that we get more than $m$ linearly independent vectors $T(e_{n+1}), \cdots , T(e_j)$. A contradiction.
[/hide]

Is this correct? Is there a more constructive proof?

The dimension formula $\dim(V) = \dim(ker(T)) + \dim(Im(f))$ holds for [i]all[/i] vector spaces $V,W$ and all linear maps $V \to W$. It's just an equation of cardinal numbers, and can be proven like the popular one with the needless restriction to finite-dimensional spaces. From here, the claim is obvious.

Yes, but I don't know about cardinalities or anything. My definition of a infinite-dimensional space is that it cannot be generated by a finite amount of vectors.

I'd still like to see if my proof holds up or if I forgot something..


Given $n+1$ points in the plane, $P_1,\ldots,P_n,Q$, no three collinear, we know that for each $i\ne j$ there is a $k\ne i,j$ s.t. $Q$ lies inside $P_iP_jP_k$. Show that $n$ is odd.

We may assume that all points lies on the circle centered at $Q$. Suppose wlog that points $P_i$ form an inscribed convex polygon. Construct for each $i$ a point $P'_i$ s.t. $P_iP'_i$ is a diametr. Then we see that arc $P'_iP'_{i+1}$ neceserily contains a point from $\{P_j\}$. It follows that each such arc contains exactly one point. From here we can easily deduce the conclusion.


Prove that: $ \frac{1}{\sqrt{7\minus{}4\sqrt{3}}}\plus{}\sqrt{7\minus{}4\sqrt{3}} \cdot \sqrt{76\plus{}42\sqrt{3}}<7\plus{}\sqrt{3}.$

hello, we have $ \sqrt{7\minus{}4\sqrt{3}}\equal{}2\minus{}\sqrt{3}$, $ \sqrt{76\plus{}42\sqrt{3}}\equal{}7\plus{}3\sqrt{3}$ and so we get
$ \frac{1}{2\minus{}\sqrt{3}}\plus{}(2\minus{}\sqrt{3})(7\plus{}3\sqrt{3})\equal{}7<7\plus{}\sqrt{3}$
Sonnhard.


geometry
incenter
geometric transformation
homothety
ratio
angle bisector
projective geometry

[quote="77ant"]Pentagon $ ABCDE$ (in this order) on circle $ (O)$ with $ AC \equal{} CE$

$ BD$ meets $ AC,CE$ at $ P,Q$ respectively

circle $ (O1)$ touches $ AP,BP$,arc $ \widehat{AB}$ (not including $ C$)

circle $ (O2)$ touches $ DQ,EQ$,arc $ \widehat{DE}$ (not including $ C$)

$ (O1),(O2)$ intouch $ (O)$ at $ R,S$ respectively . Let $ RP\cap SQ \equal{} X$

Prove that $ XC$ bisects $ \angle ACE$

I know one of proofs but want to get more nice (and shorter) ones
I always appreciate your nice answers[/quote]
[hide="Solution"]
Let the incenter of $ \triangle CPQ$ be $ I$.  Notice that $ IP$ bisects $ \angle CPQ$ and $ IQ$ bisects $ \angle CQP$.  Furthermore, $ O_2Q$ bisects $ \angle EQD$ and $ O_1P$ bisects $ \angle BPA$.  Thus, $ O_1$, $ P$, and $ I$ are colinear and as are $ I$, $ Q$, and $ O_2$.  Notice that $ O$, $ O_1$, and $ R$ are collinear and as are $ O$, $ O_2$, and $ S$ because of the tangencies.  Then, since $ AC \equal{} CE$, we have that $ O$ and $ I$ both lie on the angle bisector of $ \angle ACE$.  Now, $ R$ is the center of homothety between $ (O_1)$ and $ (O)$ and $ S$ is that between $ (O_2)$ and $ (O)$, so by Monge's Theorem, we have that the center of homothety between $ (O_1)$ and $ (O_2)$ lies on $ RS$.  Yet, this is also the intersection between the common tangent to the circles, $ BD$, and the line connecting their centers, $ O_1O_2$.  It follows that $ O_1O_2$, $ RS$, and $ PQ$ concur, so $ \triangle O_1PR$ and $ \triangle O_2QS$ are perspective.  This implies that the intersection between $ O_1P$ and $ O_2Q$, which is $ I$, the intersection between $ O_1R$ and $ O_2S$, which is $ O$, and the intersection between $ RP$ and $ QS$, which is $ X$ are collinear.  Yet, $ O$ and $ I$ lie on the angle bisector of $ \angle ACE$, so $ X$ does as well.  We conclude that $ XC$ bisects $ \angle ACE$. [/hide]

EDIT: Here is a perhaps faster solution:
[hide="Alternate Solution"]
Let $ (I)$ be the incircle of $ \triangle CPQ$.  Notice that $ P$ is the center of homothety between $ (I)$ and $ (O_1)$ and $ R$ is that between $ (O_1)$ and $ (O)$.  Then, by Monge's Theorem, we have that the center of homothety between $ (O)$ and $ (I)$ lies on $ PR$.  Similarly, it lies on $ QS$, so $ X$ is this center of homothety.  Then, $ O$, $ I$, and $ X$ are collinear, as before and we finish as in the above solution. [/hide]

I am on cloud nine because of your so smart solution , The Quattomaster6000  :laugh:

[quote="The QuattoMaster 6000"]so by Monge's Theorem, we have that the center of homothety between $ (O_1)$ and $ (O_2)$ lies on $ RS$.  Yet, this is also the intersection between the common tangent to the circles, $ BD$, and the line connecting their centers, $ O_1O_2$.[/quote]
I don't understand why the 1st line is true. :maybe: (Is this theorem same to [url=http://www.cut-the-knot.org/Curriculum/Geometry/MongeTheorem.shtml]this at cut-the-knot?[/url]) I understand the second line (We know that the center of similitude is the center of homothety). Can anyone explain :?:

BTW I apologize for retrieving this old thread.

[quote="Moonmathpi496"]
I don't understand why the 1st line is true. :maybe: (Is this theorem same to [url=http://www.cut-the-knot.org/Curriculum/Geometry/MongeTheorem.shtml]this at cut-the-knot?[/url])
[/quote]
$ R$ is the center of homothety between $ (O_1)$ and $ (O)$, as it is the point of tangency between the circles, let $ Z$ be that between $ (O_1)$ and $ (O_2)$, and $ S$ is that between $ (O)$ and $ (O_2)$ since it is the point of tangency between the circles.  By Monge's Theorem, which is shown on the link you mentioned, $ R$, $ S$, and $ Z$ are collinear (if you want to prove this, apply Menelaus's Theorem on $ \triangle OO_1O_2$ with transversal $ RSZ$ by noting that $ \frac{RO_1}{RO}$, $ \frac{O_1Z}{O_2Z}$, and $ \frac{O_2S}{OS}$ are all ratios of the radii of $ (O)$, $ (O_2)$, and $ (O_1)$, so everything cancels).  Thus, $ Z$ lies on $ RS$.

I hope that was helpful, if you need anything clarified, then feel free to ask  :)


Show that for any positive integer $ a$ and $ b$,  the number $ (36a\plus{}b)(a\plus{}36b)$ cannot be a power of two.

[quote="bigman"]Show that for any positive integer $ a$ and $ b$,  the number $ (36a \plus{} b)(a \plus{} 36b)$ cannot be a power of two.[/quote]
[hide="ignore..this is wrong x_x"]Let's assume that this product can be written with powers of 2's. Set $ 36a \plus{} b \equal{} 2^m$ and $ a \plus{} 36b \equal{} 2^n$ where $ m$ and $ n$ are positive integers. Taking the first and second equations in modulo $ 2$ results in $ b\equiv 0\bmod{2}$ and $ a\equiv 0\bmod{2}$. Set $ b \equal{} 2x$ and $ a \equal{} 2y$ with both $ x$ and $ y$ being greater than $ 0$ and take the first equation to now result in $ 2^{y \plus{} 2}\cdot3^2 \plus{} 2^x \equal{} 2^m$
$ 2^{y \plus{} 2}\cdot3^2 \equal{} 2^m \minus{} 2^x$
We have reached a contradiction as the difference of powers of 2's can never be a multiple of 3 unless one of them was 1, which cannot be reached since both $ m$ and $ x$ are greater than 0. [/hide]

[quote="TachyonPulse"]Could you explain how you were able to set $ b \equal{} 2^x$ and $ a \equal{} 2^y$ just from knowing that $ a$ and $ b$ were divisible by $ 2$?[/quote]
My bad...wasn't thinking clearly when I did this...heh I meant to set $ b\equal{}2x$ and $ a\equal{}2y$

Ill try my best to prove
My wording is horrible

First, it is easy to see that both $ 36a\plus{}b$ and $ a\plus{}36b$ are powers of 2
looking at $ 36a\plus{}b\equal{}4(9a)\plus{}b$, and to make it a power of 2, this has to be a multiple of 4, and b has to be a multiple of 4 as well.

But if $ b\equal{}4k$, then
$ a\plus{}36b\equal{} a\plus{}36*4k\equal{}a\plus{}16(9k)$, and using the same reasoning, a has to be a mutiple of 16

Recursively, a b will always be greater powers of 2 without limit, and therefore, there the equation cannot be a power of 2

That seems alright but it doesn't seem complete to me. The last step in your proof needs to be tightened up. I think using infinite descent would have made your proof more rigorous.

This is the first time I'm doing such a proof so it would be nice if someone could verify my solution.

[hide="Proof"]
Suppose that $ a_1$ and $ b_1$ are the smallest integers such that $ 36a_1 \plus{} b_1 \equal{} 2^m, m \in \mathbb{N}$ and $ a \plus{} 36b_1 \equal{} 2^n, n \in \mathbb{N}$. Clearly, $ m,n \ge 6$ because $ 2^6 \equal{} 64 \ge 36a_1 \plus{} b_1, a_1 \plus{} 36b_1 \ge 37 > 2^5 \equal{} 32$. Taking $ \pmod4$ of both equations, we find that $ a_1 \equiv 0 \pmod{4}$ and $ b_1 \equiv 0 \pmod{4}$. Thus, we have $ a_1 \equal{} 4a_2$ and $ b_1 \equal{} 4b_2$ for some integers $ a_2, b_2$.

Using the first equation, we have $ 36a_1 \plus{} b_1 \equal{} 2^m\implies 4(36a_2 \plus{} b_2) \equal{} 2^m\implies 36a_2 \plus{} b_2 \equal{} 2^{m \minus{} 2}$.

In the second equation, we have $ a_1 \plus{} 36b_1 \equal{} 2^n\implies 4(a_2 \plus{} 36b_2) \equal{} 2^{n\minus{}2}\implies a_2 \plus{} 36b_2 \equal{} 2^{n \minus{} 2}$.

We have thus arrived at a contradiction because we claimed $ a_1,b_1$ were the smallest pair of integers such that $ 36a_1 \plus{} b_1$ and $ a_1 \plus{} 36b_1$ were both powers of two, but we found smaller integers $ a_2,b_2$ that satisfied the same criteria. Thus, $ (36a \plus{} b)(a \plus{} 36b)$ cannot be a power of $ 2$.
[/hide]


[quote="QuyBac"]Bạn em đánh laptex  không thấy dòng gạch đỏ phía dưới vạy mà khi em đanh lại thấy nét gạch đỏ theo sau , XIN GIẢI THÍCH CHO EM BIẾT , CẢM ƠN ANH CHỊ CÁC BẠN RẤT NHIỀU !!
 :starwars:[/quote]
Hi ...QuyBac ...Đó là do máy của bạn có sử dụng phần mềm firefox không phải là internet Explorer .
   :roll:

ZengZeng là người Việt hay sao mà nói tiếng việt thạo thế . :(  :(

[quote="NguyenDungTN"]Mình thấy ý thức của cá mem Việt Nam mình càng ngày càng đi xuống hay sao ấy. Lên mathlink chỉ đẻ spam lấy bài. Đề nghị quybac và moọt số thành viên khác ngừng ngay việc spam này lại, vì ở một forum quốc tế như thế này, các bạn ko giữ thể diện cho mình thì cũng phải giữ thể diện cho quốc gia chứ. Đừng để người nước ngoài coi thường mình là được.[/quote]
 Okê không gửi lên nữa (Trình độ ngu dốt quá không hiểu lên mới gửi bài lên thôi) .Tôi và một số người không phải là chuyên giải Toán chuyên nghiệp ,chỉ có thể nói là người hỏi chuyên nghiệp . Tôi gửi bài lên chỉ nhằm hiểu nhiều điều cùng một lúc ... Nếu việc hỏi của tôi và một số bạn làm ảnh hưởng tới thể diện cho một quốc gia tôi sẽ dừng việc hỏi ở đây ... Còn việc họ coi thường chính bản thân mình thì là việc bình thường thôi mà ... Họ coi thường mà tôi học được cái mới thì cũng được chứ suốt ngay ngu tối thì bao giờ mới học được , cách nghĩ mỗi người là một khác nhau Chứ giống nhau ai con gọi là xã hội nữa ... Goodbye ...
[quote]các bạn ko giữ thể diện cho mình thì cũng phải giữ thể diện cho quốc gia chứ. Đừng để người nước ngoài coi thường mình là được.[/quote]
 ... Ghi nhớ ... Stop here !!!  :P

Mình không có ý kiến coi thường ai cả, chỉ mong muốn là các thànhviên VN lên ML nên hạn chế các bài spam và nên post các bài có chất lượng hơn, chứ không phải chỉ post để kiếm số bài viết như các DD VN.

PS: Do dạo này hơi bức xúc với một số thành viên ở VN nên có nặng lời với bạn. Có gì quy tha lỗi nhé. Những ý kiến trên của mình chỉ mong muốn cộng đồng toán VN phát triển hơn thôi.

:P  O ke ! Nhưng chỉ có một điều là các bác viết ở diễn đàn toán học Quốc Tế nhiều quá mà không viết ở Diễn Đàn Toán Học Việt Nam ,không truyền đạt một số kinh nghiệm học tập được cho diễn đàn VN .Mình mong muốn các bạn có nhiều bài viết chất lượng cho DĐTH VN ....
  Mình tin tưởng diễn đàn toán học VN lớn mạnh hơn nhiều ...  :play_ball:


inequalities
function
inequalities proposed

i don't know if it is known but it is easy using muerhead after expanding

[quote="bodom"]...using muerhead after expanding[/quote]have you shown it's symmetric? (I wouldn't think it is, but nor do I want to expand the thing to check :blush: ) Arqady, where are you??
EDIT: The ineq is equivalent to $\sum_{cyc}(\frac{b}{a+b}+\frac14)^{2}\ge\frac{27}{16}.$ if it were symmetric, that would require $\sum_{cyc}\frac{a-b}{a+b}=0$. :(
EDIT2: OK, ugly solution:
Put $x=\frac{a}{b}$ etc. Then we have to show $\sum_{cyc}(\frac{1}{x+1}+\frac14)^{2}\ge\frac{27}{16}$ for $xyz=1$. This follows from the "Right-Convex Function Theorem" ( http://www.mathlinks.ro/Forum/viewtopic.php?p=226013#p226013 ) applied to $f(t)=(\frac{1}{e^{t}+1}+\frac14)^{2}$. Note that the original ineq has $(0,1,\infty)$ also as an equality case  :lol:

For you, spanferkel. :lol:
$2\sum_{cyc}(\frac{b}{a+b})^{2}\geq \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(3a^{4}c^{2}+a^{4}bc+a^{3}b^{3}-a^{3}b^{2}c-a^{3}c^{2}b-3a^{2}b^{2}c^{2})\geq0,$ which obviously true.

[quote="spanferkel"][quote="bodom"]...using muerhead after expanding[/quote]have you shown it's symmetric? (I wouldn't think it is, but nor do I want to expand the thing to check :blush: ) )[/quote]
sorry i was tired last night.i did just the same thing as arqady but i don't know why i wrote muerhead since i did not use it. :blush:

Try to think of a nice way to solve the ineq,besides expanding.I have a solution that uses a very known inequality with a special case[after a small trick,of course :) ].I will post it,if anyone wants to see it.

[quote="andyciup"]Let a,b,c be positive real numbers.Prove that the following inequality occurs: \[2\sum_{cyc}(\frac{b}{a+b})^{2}\geq \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\] [/quote]

Let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$. Then $(LHS)=2\sum ( \frac{b}{a+b})^{2}= 2\sum (\frac{1}{1+x})^{2}$
$= \sum (\frac{1}{(1+x)^{2}}+\frac{1}{(1+y^{2})}) \geq \sum \frac{1}{1+xy}=\sum \frac{1}{1+\frac{a}{c}}= \sum\frac{c}{a+c}$ :)

Yes,exactly!That is also my solution.Congratulations,Sung-yoon Kim!

[quote="spanferkel"]yes, beautiful solution  :first: ... er, wait a moment, $(1+x^{2})$ or $(1+x)^{2}$ ? :?:[/quote]

That's a typo. $\frac{1}{(1+y^{2})}$ should be replaced with $\frac{1}{(1+y)^{2}}$.

I know its been posted before but I can't find it after searching the forums, can someone give a nive proof of \[\sum (\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}}) \geq \sum \frac{1}{1+xy}\] Thank you

By cauchy, \[\frac{1}{(1+x)^{2}}\geq\frac{y}{(x+y)(xy+1)}\] It`s darij`s proof from here: http://www.mathlinks.ro/Forum/viewtopic.php?t=80993&highlight=dumb455ing
Also,here is a small and easy generalisation:
http://www.mathlinks.ro/Forum/viewtopic.php?t=113522


modular arithmetic
number theory proposed
number theory

Find all natural numbers such that \[ n\sigma(n)\equiv 2\pmod {\phi( n)}\]

Obviously, the only prime $ p$ dividing $ n$ and having valuation at least $ 2$ is $ 2$. Now, write $ n\equal{}2^x N$ with $ N$ a product of distinct odd primes, then $ 2^{x\minus{}1}\varphi(N)$ divides $ 2^x N (1\plus{}2\plus{}...\plus{}2^x)\sigma(N)\minus{}2$. If $ N$ has at least $ 2$ prime divisors, then by looking at the $ 2$ adic valuation we get that $ x\equal{}0$ and then a contradiction looking again at the factors of $ 2$ appearing in both sides. So in any case $ N$ is either $ 1$ or a prime. The first case is trivial, and so is the case $ N$ prime, because $ x$ is at most $ 2$.

No, the answer is 1,4,6,22 and all of the prime numbers.

You had some miscalculations in the case that n is equal to $ 2p$ in which $ p$ is prime number. You will reach to $ p\minus{}1|10$.

But Omid, I do not see something wrong in what I said: I just gave the idea of the solution, the computations are rather straightforward! I said that any such number is $ p$, $ 2p$ or $ 4p$ with $ p$ prime. The rest is below the level of intermediate students, so I guess anyone can do that.

I misunderstood:

[quote="harazi"] .. So in any case $ N$ is either $ 1$ or a prime...[/quote]

[quote="Omid Hatami"]Find all natural numbers such that
\[ n\sigma(n)\equiv 2\pmod {\phi( n)}
\]
[/quote]
if $ n\equal{}2^k\prod_{j\equal{}1}^{h}p_i^{a_i}$
$ p_i^{a_i\minus{}1}|gcd(n\sigma(n),\phi( n))$ gives $ p_i^{a_i\minus{}1}|2$ so $ a_i\equal{}1$ and by the same $ k\in\{0,1,2\}$
$ 2^{max(k\minus{}1,0)\plus{}h}|gcd(n\sigma(n),\phi( n))$ gives $ 2^{max(k\minus{}1,0)\plus{}h}|2$ then $ k\equal{}0,1 \ and\ h\le 1$ $ (n\equal{}2p,p)$ or $ k\equal{}2,h\equal{}0$ $ (n\equal{}4)$

$ n\equal{}4$ ...
$ n\equal{}1$ .....
$ n\equal{}p$ : $ p(p\plus{}1)\equal{}1.2[p\minus{}1]$ works
$ n\equal{}2p$ : $ 2p(p\plus{}1)(2\plus{}1)\equal{}2.1.2.3[p\minus{}1]\equal{}2[p\minus{}1]$ gives $ p\minus{}1|12\minus{}2$ then $ p\equal{}3,11$ so $ n\equal{}6,22$

$ n\equal{}1,4,6,22,$ and primes


geometry
3D geometry
tetrahedron
geometry unsolved

[quote="77ant"]Take P inside the tetrahedron ABCD to minimize PA + PB + PC + PD.
Show that ∠APB = ∠CPD and that the bisector of APB also bisects CPD.
No solutions,I long for synthetic proofs.
 :help: thank you.[/quote]
$ PA\plus{}PC\geq AC, PB\plus{}PD\geq BD$
Hence, $ PA\plus{}PB\plus{}PC$ attains its smallest value when P is the point where AC meets BD. :rotfl:


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Evaluate the improper integral   π/2
                                           ∫ In(sinx)dx
                                           0

Sorry, but what do you mean?

$ \int\limits_0^{\frac {\pi}{2}} \ln(\sin x) dx$?

What is this "In"? Natural logarithm?


You can angle-chase this problem easily. Another way to see it is to show that $O_{1}O_{2}O_{3}O_{4}$ has an inscired circle that touches the sides at $A, B, C, D$.

[quote="bui_haiha2000"]Let four circles $\left(O_{1}\right)$, $\left(O_{2}\right)$, $\left(O_{3}\right)$, $\left(O_{4}\right)$ be given such that:
the circle $\left(O_{1}\right)$ is tangent to the circle $\left(O_{2}\right)$ at a point $A$;
the circle $\left(O_{2}\right)$ is tangent to the circle $\left(O_{3}\right)$ at a point $B$;
the circle $\left(O_{3}\right)$ is tangent to the circle $\left(O_{4}\right)$ at a point $C$;
the circle $\left(O_{4}\right)$ is tangent to the circle $\left(O_{1}\right)$ at a point $D$.
Prove that $ABCD$ is a cyclic quadrilateral.[/quote]

I wanted to figure out that this is a consequence of http://www.mathlinks.ro/Forum/viewtopic.php?t=22480 :

We assume that all tangencies of circles described in the problem are external tangencies.

Denote by $O_{1}$, $O_{2}$, $O_{3}$, $O_{4}$ the centers of the circles $\left(O_{1}\right)$, $\left(O_{2}\right)$, $\left(O_{3}\right)$, $\left(O_{4}\right)$, respectively. Since the circles $\left(O_{1}\right)$ and $\left(O_{2}\right)$ with centers $O_{1}$ and $O_{2}$ touch each other externally at the point A, the point A must lie on the segment $O_{1}O_{2}$. Similarly, the points B, C, D lie on the segments $O_{2}O_{3}$, $O_{3}O_{4}$, $O_{4}O_{1}$, respectively. Since the points D and A lie on the circle $\left(O_{1}\right)$ with center $O_{1}$, we have $O_{1}D=O_{1}A$; similarly, $O_{2}A=O_{2}B$, $O_{3}B=O_{3}C$, $O_{4}C=O_{4}D$.

Altogether, the points A, B, C, D lie on the sides $O_{1}O_{2}$, $O_{2}O_{3}$, $O_{3}O_{4}$, $O_{4}O_{1}$ of the quadrilateral $O_{2}O_{3}O_{4}O_{1}$ and satisfy $O_{2}A=O_{2}B$, $O_{3}B=O_{3}C$, $O_{4}C=O_{4}D$, $O_{1}D=O_{1}A$. Hence, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=144687#144687]http://www.mathlinks.ro/Forum/viewtopic.php?t=22480 post #4[/url] Assertion 1, the quadrilateral $O_{2}O_{3}O_{4}O_{1}$ has an incircle, and if O is the center of this incircle, then the points A, B, C, D lie on a circle centered at O. Hence, the quadrilateral ABCD is cyclic, qed.

  Darij

This is a simple problem and really doesn’t need inversion. But, may be  good for practicing inversion is an inversive proof.

Name the tangencies ,$A,B,C,D$.

I choose $A$ as the center of inversion. ($B,C,D$, can also be the center of inversion, but here it isn’t important.)

Because two of the circles are tangent at $A$, after inversion, they turn

into two parallel lines.

And the two other circles, after inversion, will be again two circles which

are tangent to each other at $C$, and

And the two circles are tangent to the two parallel lines. One at $B$, and

the other at $D$.

Now if we prove $B,C,D$ are collinear, the problem will be solved.

And it is quite easy to prove this fact.

Now continue, and solve it.


Where can I find enourmous amounts of free/cheap problems about amc 10/12/Early AIME level with detailed solutions? Are there problem books I can buy?

kalva online, just search google for "aime problems" and it's the first site.

[quote="chess64"]you can buy the past tests from the AMC site very cheap

http://www.goodlymaths.0catch.com has an archive[/quote]
Is stuff from the ARML good practise?

I calculated the cost of [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883856409/qid=1123611159/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-5585100-3602409?v=glance&s=books&n=507846]this[/url]
at  33/(6*15+6*25)=About 13.5 cents per problem. How do the Booklets and other sources compare?(Other books?)
So far my only sources are
1)The two web sites
2)Booklets
3)The contest problem books(number 5 shown above)

Are there any more?

So at 1/25 of a dollar/problem, the booklets are cheaper.

So, what about the ARML?

ARML is harder than AMC 10. I think it's a little harder or maybe about the same as AMC 12, but don't ask me. I don't have much experience with ARML. :D

The MMPC has a near inexhaustible collection of AMC level problems, but beware of some of the erraneous early tests: http://www.delta.edu/math/mmpc/

ARML is good practice for the AMC's.  But the ARML competitions have various rounds.  The relay rounds are really good practice for the easier to mid difficulty AMC 10/12 problems.  The individual round contains problems comparable to mid AMC 10/12 to AIME difficulty.  The team round is comparable to mid/hard AMC to AIME level.  And the power round is harder to categorize.  You have to justify your work (i.e. proof).  But it isn't like USAMO (except for maybe the last part or two) difficulty wise.


One more way to type it differently: $(x\sqrt{3}-1)^2+(y+3)^2=0$  ;)  :)


function
inequalities proposed
inequalities

[quote="konlsy"]Let $r_1,r_2,...r_n\geq1$ prove
that:$\frac{1}{1+r_1}+\frac{1}{1+r_2}....\frac{1}{1+r_n}\geq\frac{n}{\sqrt[n]{r_1r_2...r_n}+1}$[/quote]
This one is wery old :) ,
juste take $r_i=e^x_i$,and use Jensen's ineq. for function $f(x)=\frac{1}{1+e^x}$.
Also this one can be solving with induction.:)

[quote="konlsy"]Tiks:
could u show more detill
 :blush:[/quote]
[b]konlsy[/b] I think the solution is done,becaus if you can show that $f$ is convex in $x\in[0,\inf)$(juste find $f\"$,and show that $f\"(x)\geq 0$,when$x\geq 0$),and the Jensen's ineq.
$f(x_1)+f(x_2)...+f(x_n)\geq nf(\frac{x_1+x_2+...+x_n}{n})$<=>
$\frac{1}{1+r_1}+\frac{1}{1+r_2}....\frac{1}{1+r_n}\geq\frac{n}{\sqrt[n]{r_1r_2...r_n}+1}$ :!:
I hope that now you can see this solution :) .

u are a kind-heart man
thank u for ur pation :P


probability quadratics algebra polynomial

1. The two hands of a clock form a 70 degree angle. What is the smallest positive number of minutes later such that the hands of the clock are guaranteed to form a 110 degree angle.

The answer isn't 80/11.

2. Three of the vertices of a regular eleven sided polygon are randomly chosen. What is the probability that the triangle connecting the three vertices is obtuse?

3. If $ P(x)\equal{}x^4\minus{}6x^3\plus{}3x^2\plus{}18x\plus{}5\equal{}f(g(x))$, where f and g are quadratic polynomials, what are the four real roots of $ P(x)$?

3. Let $ p$, $ q$ be the solutions of $ f(x) \equal{} 0$. Then now $ g(x) \equal{} p,q$ gives all four roots.  But that means that $ p(x)$ factors as a product of two quadratics whose linear coefficients are equal. Let
\[ p(x) \equal{} (x^2 \plus{} ax \plus{} m)(x^2 \plus{} ax \plus{} n)
\]

\[ \equal{} x^4 \plus{} (2a)x^3 \plus{} (a^2 \plus{} m \plus{} n)x^2 \plus{} (am \plus{} an)x \plus{} (mn)
\]
Equating coefficients, we get
$ 2a \equal{} \minus{} 6$,
$ a^2 \plus{} m \plus{} n \equal{} 3$,
$ am \plus{} an \equal{} 18$,
$ mn \equal{} 5$.

Then $ a \equal{} \minus{} 3$, $ m \plus{} n \equal{} \minus{} 6$, $ mn \equal{} 5$, so
\[ 0 \equal{} (t \minus{} m)(t \minus{} n) \equal{} t^2 \minus{} (m \plus{} n)t \plus{} mn \equal{} t^2 \plus{} 6t \plus{} 5 \equal{} (t \plus{} 1)(t \plus{} 5)
\]
so $ (m,n) \equal{} ( \minus{} 1, \minus{} 5)$. Hence our polynomial is
\[ (x^2 \minus{} 3x \minus{} 1)(x^2 \minus{} 3x \minus{} 5)
\]
which can be solved be the quadratic formula.

#2

We fix a vertice and call it $ A$. The interior angle of one vertice of the 11-gon is $ \frac {9}{11} \cdot 180^{o} \equal{} 147 \frac {3}{11} ^{o}$. The angle "between" two adjacent vertices is $ 147 \frac {3}{11} ^{o}$ divided by 9, which is $ 16 \frac {12}{33} ^{o}$. Note that 5 times this angle is less than $ 90^{o}$.

The only way the triangle can be obtuse is if it is obtuse at $ \angle A$.

The total number of ways of choosing two other vertices is $ \binom {10}{2} \equal{} 45$. The triangle will be obtuse if the two remaining vertices chosen are 6, 7, 8, or 9 vertices apart.

Thus the answer is $ \frac {4 \plus{} 3 \plus{} 2 \plus{} 1}{45} \equal{} \boxed {\frac {2}{9}}$.

[quote="aufha"]1. The two hands of a clock form a 70 degree angle. What is the smallest positive number of minutes later such that the hands of the clock are guaranteed to form a 110 degree angle.[/quote]

So
lu [hide="tion"]Either the degrees go from 70 to 180 to 110, which is 180 degrees, or it goes from 70 to 0 to 110, which is also 180 degrees. We must figure out how many minutes it takes for the minute hand to go 180 degrees faster than the hour hand.

Let $ t$ be the number of minutes.

$ 60t\minus{}\frac{t}{2}\equal{}\frac{119t}{2}\equal{}180$

$ t\equal{}\frac{360}{119}$[/hide]

Is that right? :huh:

[quote="1=2"][quote="aufha"]1. The two hands of a clock form a 70 degree angle. What is the smallest positive number of minutes later such that the hands of the clock are guaranteed to form a 110 degree angle.[/quote]

So
lu [hide="tion"]Either the degrees go from 70 to 180 to 110, which is 180 degrees, or it goes from 70 to 0 to 110, which is also 180 degrees. We must figure out how many minutes it takes for the minute hand to go 180 degrees faster than the hour hand.

Let $ t$ be the number of minutes.

$ 60t \minus{} \frac {t}{2} \equal{} \frac {119t}{2} \equal{} 180$

$ t \equal{} \frac {360}{119}$[/hide]

Is that right? :huh:[/quote]

No, it's not right. The minute hand moves $ 6^\circ$ per min, not $ 60^\circ$ per min, as you wrote $ 60t$.


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra unsolved

If a is an irrational root of this polynomials:

f(t)= (t^4)+4(t^3)-t^2-3t-1. What's the norm and trace of a, where is considered as an element of Q(a).

One definition of norm and trace of an element $ x$ in an extension $ F$ of a field $ k$: let $ x$ act on $ F$ as a $ k$-linear map by multiplication. The norm of $ x$ is the determinant of this map, and the trace is the trace.

Now, can we find a matrix for $ a$? First, we need a basis for $ \mathbb{Q}(a)$. The natural choice is $ 1,a,\dots$, terminating when we reach the degree of the field. In this case, we can factor $ f$ as $ (t\minus{}1)(t^3\plus{}5t^2\plus{}4t\plus{}1)$, and the second factor is irreducible. The field has degree $ 3$, and we have $ a\cdot 1\equal{}a,$ $ a\cdot a\equal{}a^2,$ $ a\cdot a^2\equal{}\minus{}1\minus{}4a\minus{}5a^2$. The matrix of $ a$ with respect to the natural basis is the companion matrix $ \begin{bmatrix}0&0&\minus{}1\\1&0&\minus{}4\\0&1&\minus{}5\end{bmatrix}$. We can read off the determinant $ \minus{}1$ and the trace $ \minus{}5$ immediately.

how do we know t^3+5t^2+4t+1 is irreducible?

sry.

A polynomial of degree 3 is irreducible as long as it does not have a rational root.

Is that simply because if you try to factor out a linear factor, then that implies that the root is rational?

i.e. at+c=0 => t rational if a and c are in Q.

Yes. Now we have the rational root theorem- the list of possible linear factors is very short, and none of them work.


Is there any way to see how many people (or percent) correctly answered each question correctly on previous AIMEs/AMCs?

[quote="t0rajir0u"]They're on the AMC Website under "Item Difficulty."

http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml[/quote]
What about AIMEs?


calculus
integration
trigonometry
real analysis
real analysis unsolved

Calculate
\[ I \equal{} \int {\frac{{1 \minus{} 2{x^2}}}{{\sqrt {\left( {1 \minus{} {x^2}} \right)\left( {{x^4} \minus{} {x^2} \plus{} 1} \right)} }}} dx\], \[ x \in \left( {\minus{}1,1} \right)\]

Maple is wrong (not about definite but for indefinite). If you are able to calculate definite from 0 to  $ \sqrt(2)/2$ you might be able to calculate indefinite too

$ x\equal{}\sin \theta \Longrightarrow I\equal{}4\int_0^{\frac{\pi}{2}} \frac{\cos 2\theta}{\sqrt{4\minus{}\sin ^ 2 2\theta}}d\theta$

[quote="Dr Sonnhard Graubner"]hello, Wolfram Alpha says "no result found in terms of standard mathematical functions".
Sonnhard.[/quote]
Mhmhm wolfram is wrong too. Look better at Kunnys post. You can calcuta indefinite from it. Though i thought differently for it

hello, yes you have right,kunnys substitution works and we get zero as the result of your integral.
Sonnhard.


Point A is 12cm from line k.  How many points are 3cm from line k and 4cm from A?

They cannot be any such points.

The distance from $k$ to $P$ added to the distance from $P$ to $A$, for any point $P$, must be at least the distance from $k$ to $A$.

(This is like the "shortest distance between two points is a straight line" idea, which is more formally known as the triangle inequality)

What about the question rewritten this way:

Point A is 4cm from line k. How many points are 1cm from line k and 3cm from A?

Then there is only one point, as it lies on the line through $A$ perpendicular to $k$.

If there was another point $P$, then consider the triangle formed by $P$, $A$ and the point $M$ on $k$ such $AM$ is perpendicular to $k$: it would violate the triangle inequality.


Determine all pairs of rational numbers $(x, y)$ such that \[x^{3}+y^{3}= x^{2}+y^{2}.\]

The first we has a result :
Let $ \gcd(m,n) \equal{} 1$ then $ \gcd(m^3 \plus{} n^3,m^2 \plus{} n^2)\in \{1,2\}$
[u]Proof [/u] Let  $ d \equal{} \gcd(m^2 \plus{} n^2,m^3 \plus{} n^3)$
Imply that $ p|2x^6 \Rightarrow d|2$ so we has result.
Now consider our solution.
Let $ x \equal{} \frac {m}{p},y \equal{} \frac {n}{q}$
$ p \equal{} \frac {m^3 \plus{} n^3}{m^2 \plus{} n^2}$
Let $ d \equal{} \gcd(m,n)$ then
$ m \equal{} da,n \equal{} db$ where $ \gcd(a,b) \equal{} 1$
So $ p \equal{} d\frac {a^3 \plus{} b^3}{a^2 \plus{} c^2}$
 [u]Case 1[/u] $ \gcd(a^3 \plus{} b^3,a^2 \plus{} b^2) \equal{} 1$ so
$ a^2 \plus{} b^2|d$
Imply that $ d \equal{} k(a^2 \plus{} b^2)$
$ p \equal{} k(b^3 \plus{} a^3)$
so $ x \equal{} \frac {da}{p} \equal{} \frac {(a^2 \plus{} b^2)a}{a^3 \plus{} b^3}$
     $ y \equal{} \frac {db}{p} \equal{} \frac {(a^2 \plus{} b^2)b}{a^3 \plus{} b^3}$
[u]Case 2[/u] $ \gcd(a^3 \plus{} a^3,a^2 \plus{} b^2) \equal{} 2$ it mean that $ a\equiv b\equiv 1(\mod 2)$
Then $ a^2 \plus{} b^2|2d$ imply that
$ d \equal{} \frac {k(a^2 \plus{} b^2}{2}$
$ p \equal{} \frac {k(a^3 \plus{} b^3)}{2}$
$ x \equal{} \frac {a(a^2 \plus{} b^2)}{a^3 \plus{} b^3}$
$ y \equal{} \frac {b(b^2 \plus{} a^2)}{a^3 \plus{} b^3}$
Result :
$ x \equal{} \frac {a(a^2 \plus{} b^2)}{a^3 \plus{} b^3}$
$ y \equal{} \frac {b(a^2 \plus{} b^2)}{a^3 \plus{} b^3}$

Eh? So you say that $ a\equal{}100,b\equal{}101$ gives solution $ x\equal{}2020100/2030301$, $ y\equal{}101$? Because that really ain't no solution.

Just looking at the powers appearin in your solution says that you have a problem for big $ a,b$...

I am sorry my solution is true but I have a mistake when I write it.

You are right. My solution is wrong (and has been deleted).

I am sorry that I am asking almost a year after, but which are the solutions of this equation, because it appeared in a Bulgarian Olympiad a year ago and I don't know how to solve it.

Assume that $ x \equal{} \frac {a}{b}, y \equal{} \frac {c}{d}$ is a solution, and (a,b)=(c,d)=1. Let $ k \equal{} (b,d), t \equal{} d/k, s \equal{} b/k$.
If x=0 or y=0: it's easy to see that the possible solutions with zeros are $ (0,0),(0,1),(1,0)$. Now assume that a,b,c,d are non-zero integers.

$ x^3 \plus{} y^3 \equal{} x^2 \plus{} y^2$ is equivalent to $ (at)^3 \plus{} (cs)^3 \equal{} kts((at)^2 \plus{} (cs)^2)$, which implies that $ (at)^2 \plus{} (cs)^2 | (at)^3 \plus{} (cs)^3$.
But, as TTsphn showed, it implies that $ (at)^2 \plus{} (cs)^2 \equal{} 1$ or $ (at)^2 \plus{} (cs)^2 \equal{} 2$.

First case: $ (at)^2 \plus{} (cs)^2 \equal{} 1 \implies$, WLOG: $ |cs| \equal{} 0$, a contradiction.
Second case: $ (at)^2 \plus{} (cs)^2 \equal{} 2 \implies |a| \equal{} |t| \equal{} |c| \equal{} |s| \equal{} 1$, or: $ |a| \equal{} |c| \equal{} 1, |b| \equal{} |d|$. We can assume WLOG a,c>0, and then: a=c=1. Because $ x^2 \plus{} y^2 > 0$, we must have b=d, which gives us: $ x^3 \plus{} y^3 \equal{} \frac {2}{b^3} \equal{} x^2 \plus{} y^2 \equal{} \frac {2}{b^2}$, or: b=1.

All the solutions are: $ (0,0),(0,1),(1,0),(1,1)$

put $ \frac{x}{y}\equal{}a \in Q$ (the case $ y\equal{}0$ is trivial ) then we get :
$ (ay)^3\plus{}y^3\equal{}(ay)^2\plus{}y^2 \implies y\equal{}\frac{a^2\plus{}1}{a^3\plus{}1} \implies x\equal{}\frac{a^3\plus{}a}{a^3\plus{}1}$
so we have either
$ (x,y)\equal{}(\frac{a^3\plus{}a}{a^3\plus{}1},\frac{a^2\plus{}1}{a^3\plus{}1})$
where $ a \in Q$ and $ a \neq \minus{}1$.
and by considering the case $ a\equal{}\minus{}1$ we get the solution $ (x,y)\equal{}(0,0)$
so finally we have
$ (x,y)\equal{}(\frac{a^3\plus{}a}{a^3\plus{}1},\frac{a^2\plus{}1}{a^3\plus{}1}),(0,0),(1,0)$ where $ a \in Q$ and $ a \neq \minus{}1$ and $ (1,0)$ comes from the case $ y\equal{}0$.

@ Shoki :  yeah this is the General form of x,y but you should find all $a$s then...
@TTsphn : haha, so you will get that  a^3(b-1) = b^3(1-a), then what?
@bambaman : i don't know how did you use
[quote="bambaman"]
as TTsphn showed, it implies that $ (at)^2 \plus{} (cs)^2 \equal{} 1$ or $ (at)^2 \plus{} (cs)^2 \equal{} 2$.
[/quote]

cuz i don't think it's true, or from where it came from.

...

There has to be a rational parametrisation because it is a smooth projective geometrically connected curve of genus $0$ with a rational point, hence isomorphic to $\mathbf{P}^1_\mathbf{Q}$.


A possible start:
[hide]Without loss of generality, let $a$ be the least of the five.  Then $3a \geq (3a)^3$ and so $-\frac 13 \geq a$ or $\frac 13 \geq a \geq 0$.  Similarly, the largest must be in one of the intervals $[-\frac13, 0]$ or $[\frac 13, \infty)$.[/hide]

Let S = a + b + c + d + e.
The system of equations can be written as
$3a=(s-a-e)^3$,
$3b=(s-a-b)^3$,
$3c=(s-b-c)^3$,
$3d=(s-c-d)^3$,
$3e=(s-d-e)^3$.

Note that $3a=(s-a-e)^3$ if and only if $e=s-a-(3a)^{1/3}$.
Let us denote $f(t)=s-t-(3t)^{1/3}$. Then $f$ is a [i]strictly decreasing[/i] function for t.

The above system of requations can be re-written as :
$e=f(a)$,
$a=f(b)$,
$b=f(c)$,
$c=f(d)$,
$d=f(e)$.

We claim that a = b = c = d = e is the only possible case.
For otherwise, say a < b.
Then we have
$a<b \implies f(a)>f(b) \implies e>a \implies f(e) < f(a) \implies d < e \implies f(d) > f(e) \implies c > d \implies f(c) < f(d) \implies b < c \implies f(b) > f(c) \implies a > b$
[b]which is impossible[/b].

when a = b = c = d = e, we have
$3a = (3a)^3 \implies 3a = 0$ or $3a=1, -1$.
Therefore, $a= 0, \frac 1 3, -\frac 1 3$.
Answer: (a,b,c,d,e) = (0,0,0,0,0) or (1/3,1/3,1/3,1/3,1/3) or (-1/3,-1/3,-1/3,-1/3,-1/3).


inequalities
trigonometry
geometry
function
trig identities
Law of Sines
area of a triangle

If A,B,C be the angles of a triangle. Prove

1. sin A + sin B + sin C  \leq 3 \sqrt 3 /2
2. cos A + cos B + cos C  \leq 3/2
3. sin A/2 sin B/2 sin C/2  \leq 1/8
4. cot A + cot B + cot C  \geq  \sqrt 3

On second thought, ive solved 3/4, so its a 25% unsolved problem...

1. define f:[0,pi]-->R; f(x)=sin(x)
f is concave on [0,pi] so apply Jensen's inequality

2. define f: (0,pi/2)-->R, f(x)=cos(x),

and f is concave so apply Jensen's ineq!

3. just apply ln to it and then define f: (0,pi/2)-->R, f(x)=ln(sin(x))
and f is concave on (0,pi/2) so apply Jensen to it

3. another solution

just replace sin(A/2) with sqrt(yz/((x+z)(x+y)) and sin(B/2) and sin(C/2) with simmilar relations and you get :

(x+y)(x+z)(y+z)>=8xyz. which is true by AM>=GM.
cheers!:D

2. another solution

replace cos(A) with (b^2+c^2-a^2)/(2bc)<=3/2 and we have to prove that:
3abc+a^3+b^3+c^3>=ab(a+b)+bc(b+c)+ca(c+a)
and this is easy to prove .. just look:

a^3+b^3=(a^2-ab+b^2)(a+b)>=(a+b)ab
so we get using simmilar relations that:
2(sum(a^3))>=sum(ab(a+b))
and using that sum(a^2)>=sum(ab) and sum(a^3)-3abc=(sum(a))(sum(a^2)-sum(ab)) we get what we wanted!

cheers!

Liyi u r right, we could also use some identities to solve some of them, but those solutions I don't like! :D :D

I think 4 was not answered.

A solution with heavy use of trigonometry (harazi won't like it  :D).

cot A + cot B = sin(A+B)/(sinAsinB) = 2sin(A+B) / (cos (A-B) - cos (A+B)) >= 2sinC / (1+cosC)

Now 2sinC / (1+cosC) + cosC/sinC = ... (some trig.) ... = 3sinC/2(1+cosC) + (1+cosC)/2sinC >=  \sqrt 3 by AM-GM

>>2. define f: (0,pi/2)-->R, f(x)=cos(x)

It works only for triangles with acute angles,  no ?

2. yes, i thought that it is for acute angled triangles.

Observe that Cot A=(b 2 +c 2 -a 2 )/4S
So we have  \sum cot A=(a 2 +b 2 +c 2 )/4S \geq  \sqrt 3 by Weitzenbock's Inequality.

edit edit edit... don't you love this feature? :D

[quote="belenos"]I don't follow you 3X.lich ???

Being a beginner, I find Wpolly post very interesting and valuable : I didn't know cotA formula (with in fact comes directly from laws of (co)sines but is useful is this form) nor Weitzenbock's Inequality (which I've just proved using  \sum a² >= \sum ab, law of sines and convexity of 1/sinx).

Wolly solution is much better than mine  :D[/quote]

I dunno Weitzenbock's Inequality too...can anyone show me the proof of this inequality?

Proof of Weitzenbock's Inequality:
Make two isosceles triangles BCP and BCQ such that P and A on the same side of BC, let midpoint of BC be R.
By the median formula we have AP 2 +AQ 2 =PQ 2 /2+2AR 2 =3a 2/2 +2(2b 2 +2c 2 -a 2 )/4=a 2 +b 2 +c 2 .
Let the perpendicular foot from A to PQ be S.
Then we have AQ 2 -AP 2 =AS 2 +SQ 2 -AS 2 -SP 2 =SQ 2 -SP 2 =(SQ+SP)(SQ-SP)=PQ(SQ-SP)=2PQ*SR=2 \sqrt 3 a*ha=4 \sqrt 3S.
Thus the result.

(Proof from [i]Modern Geometry [/i]by R. A. Johnson)

I also have another proof using Heron's formula and  \sum a 2  \geq  \sum ab.

I believe this could be put in the Theorems & Formulas section.

Yet an other proof of Weitzenbock's Inequality :

 \sum a² >=  \sum ab = 2S \sum 1/sinA by rearangement and law of sines

f=1/sinx is convex (I like very much convex functions!) on 0.. \pi and 3f(\pi/3)=2 \sqrt 3


Find all positive integers x,y such that x^2+7=2^y.

The only solutions are (y,x)

(3,1); (4,3); (5,5); (7,11); (15,181)

I'm looking for a proof yet.

have a look here : [url=http://mathworld.wolfram.com/RamanujansSquareEquation.html]Ramanujan Square Equation[/url]

easy problem
$x^2+7=2^y $ in $Z_2$ can be written as the following
$x^2+1=0 \leftrightarrow \widehat{x^2}=\widehat{-1}$
$\leftrightarrow \widehat{x^2}=\widehat{1}$
the solutions to this equation are $x\in {-1,1}$
so from here we find $1+7=2^y$ so $y=3$
so final solutions are
$S=\{(-1,3);(1,3) \}$
correct me please if i'm wrong

I really like your avatar Blang  :D

Pierre.

[quote="spx2"]
correct me please if i'm wrong[/quote]

Yes, you're wrong. ;)
It's very insufficient to work in $\mathbb{Z}_2$ !

x 2=1 in Z2 doesn't mean that x=1 or x=-1
it means that x  :equiv: 1 [2]

What about equation $x^2+2=3^k$? It is much easier. I know it from Dusan Djukic, but am not sure he is the first. It was proposed on Kolmogorov tournament this year, without success.

so where am i wrojng and how should i go about solving this ?

It is actually hard, because we can always find $x$ such that $x^2+7=0$ mod $2^t$.


[color=indigo]
[hide]The number removed was 21[/color][/hide]

[quote]The mean of four numbers is 12. When the greatest number is removed, the mean of the remaining three numbers is 9. What number was removed?[/quote]

these kind of problems are oftne found in AMC8 or mock AMC8..or MLC, MC, etc....

very important~!!!!

<font size=15 color=red> I wish everyone wrote down their works.. </font>

[quote="MCrawford"]The mean of four numbers is 12.  When the greatest number is removed, the mean of the remaining three numbers is 9.  What number was removed?[/quote]
[hide="solution"]
The sum of the four numbers must be 4 x 12, or 48.
The sum of the three numbers must be 3 x 9, or 27.

48 - 27 = [b]21[/b]
[/hide]


geometry
trapezoid
parallelogram
rectangle
incenter
geometry proposed

[quote="April"]The bisectors of trapezoid's angles form a quadrilateral with perpendicular diagonals. Prove that this trapezoid is isosceles.[/quote]
$ \angle AHD\equal{}\angle BFC\equal{}90^o$. Let J,K be the midpoint of AD, BC
It's easy to see that $ HJ//AB, FK//AB$ but $ JK//AB$ so $ HF//AB$
Moreover $ EG\perp HF$ then $ \angle EHF\equal{}\angle EFH$
$ \Rightarrow \angle EDC\equal{}\angle ECD \Rightarrow$ QED

Sometime, your post seem quite unclearly to me .

Case1: $AD||BC$
In this case , the formed quadrilateral is a parallelogram and because  its diagonals is perp with each other hence it's a rectangore
It makes our trapezoid is also a rectangle .
Done
Case2: $AD$ is not parallel to $BC$.WLOG : $AB<CD$
Let $AD$ meets $BC$ at J .
$I$ is $\triangle JAB$'s incenter
We have $IA||ED ; IB||EC \rightarrow E \in JI$
Similarly,we get :$HF||AB$
Hence :$HF \perp EG \Leftrightarrow IJ\perp AB \Leftrightarrow ABCD$ is isosceles


function
induction
algorithm
absolute value
algebra unsolved
algebra

problem1:Find all function f: Q--> Q such that
                f( f(x) + x + y)=f(x)+x +f(y)
 problem 2:(irish 1991)
find all function f : Q+ ------->Q+ such that
i)f(x) +f(1/x)=1 for all x
ii) f(2x)=2.f(f(x))

For problem 2, after making some computations: $ f(1) \equal{} \frac 12, f(2) \equal{} \frac 23 , f(\frac 12) \equal{} \frac 1 3, ...$, I make a guess:
\[ f(\frac{m}{n}) \equal{} \frac{m}{m\plus{}n}\]
for all $ m, n>0$ such that $ \gcd(m,n) \equal{} 1$.

A proof for this might be induction on the sum $ m\plus{}n$.

I don't have solutions yet...

For problem 1, I see we can take $ g(x) \equal{} f(x) \plus{} x$, which makes the equation becomes $ g(g(x) \plus{} y) \equal{} 2g(x) \plus{} g(y)$. Now define $ h(x) \equal{} g(x) \minus{} g(0) \equal{} g(x) \minus{} c$. We have

$ h(h(x) \plus{} y) \equal{} 2h(x) \plus{} h(y)$
$ \Longleftrightarrow g(g(x) \plus{} y \minus{} c) \minus{} c \equal{} 2g(x) \plus{} g(y) \minus{} 3c$
$ \Longleftrightarrow g(g(x) \plus{} y \minus{} c) \equal{} 2g(x) \plus{} g(y) \minus{} 2c$
(here, change $ y$ with $ y \minus{} c$ in the equation we have for $ g$)
$ \Longleftrightarrow 2g(x) \plus{} g(y \minus{} c) \equal{} 2g(x) \plus{} g(y) \minus{} 2c$
$ \Longleftrightarrow g(y \minus{} c) \equal{} g(y) \minus{} 2c$
$ \Longleftrightarrow g(y) \equal{} g(y \minus{} c) \plus{} 2c$

This is obtained when we make $ x : \equal{} 0$ and $ y : \equal{} y \minus{} c$ in the equation for $ g$. So our equation is $ h(h(x) \plus{} y) \equal{} 2h(x) \plus{} h(y)$ (the same equation that we have for $ g$) and we also have $ h(0) \equal{} 0$. Now I have no idea (except the obvious conjectures $ h(x) \equal{} 2x$ and $ h(x) \equal{} 0$, that give the solutions $ f(x) \equal{} x \plus{} t$ and $ f(x) \equal{} \minus{} x$).

For problem 2, other way to state what mathangel said is $ f(x) \equal{} \frac {x}{x \plus{} 1}$... But it is obvious and maybe it doesn't help. I haven't worked in this problem.

Problem 1

1) $ f(x) \equal{} \minus{} x$

2) $ f(x) \equal{} x \plus{} c$

3) $ f(x) \equal{} r[\frac xr] \plus{} rK(\{\frac xr\}) \minus{} \{\frac xr\}$, where $ r$ is any positive rational and $ K(x): [0,r)\cap \mathbb{Q}\rightarrow \mathbb{Z}$.

Note $ [x]$ and $ \{x\}$ are integer and fraction function here.

can any one help me to solve two problems?

[quote="MellowMelon"]Full solution for 1: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=233438[/quote]

thank you so much.Can you help me to solve problem 2?

Indeed, induction proves the claim. Suppose we know that $ f(\frac {a}{b}) \equal{} \frac {a}{a \plus{} b}$ if $ a \plus{} b < n$ and suppose that $ x \plus{} y \equal{} n$. Then

$ x < y \Rightarrow f(\frac {x}{y}) \equal{} f(f(\frac {x}{y \minus{} x}))$ (induction hypothesis)
$ f(f(\frac {x}{y \minus{} x}) \equal{} \frac {1}{2}f(\frac {2x}{y \minus{} x})$ (ii)

Thus $ f(\frac {x}{y}) \equal{} \frac {1}{2}f(\frac {2x}{y \minus{} x}) (*)$.

Now take a rational number $ r$ consider the following algorithm:

* If we have $ f(r) \equal{} u \plus{} tf(\frac {p}{q})$ and $ p < q$ use $ (*)$ to get a new equation.

* If we have $ f(r) \equal{} u \plus{} tf(\frac {p}{q})$ and $ p > q$ rewrite the equation as $ f(r) \equal{} u \plus{} 1 \minus{} tf(\frac {q}{p})$

* If $ p \equal{} q$ stop (in this case we have found $ f(r) \equal{} u \plus{} tf(1) \equal{} u \plus{} 0.5t$, so we can ignore it).

There is a finite number of solutions for $ x \plus{} y \equal{} n$, then the transformations $ (x, y) \rightarrow (2x, y \minus{} x)$ and $ (x, y) \rightarrow (y, x)$ will make a cycle. Take a rational $ r'$ in this cycle. Then we have constructed an equation $ f(r') \equal{} u \plus{} tf(r')$. The absolute value of $ t$ can be, each time, cut by half or mantained. It is only mantained until the end if we don't use $ (*)$, i.e., if the algorithm just replaced $ r'$ by $ \frac{1}{r'}$, case in which $ r' \equal{} 1$. If it is not mantained, we have $ t \not \equal{} 1$, so we can compute $ f(r')$. Reverting the process, we compute $ f(r)$.

Conclusion: $ f$ is unique. Therefore $ f(x) \equal{} \frac {x}{x \plus{} 1}$.

Is it clear, math man?


For $ a,b,c$ positive numbers such that $ abc\equal{}1$ prove that

$ \sum \frac{b\plus{}c}{a^4\plus{}1} \geq 3$

[quote="manlio"]For $ a,b,c$ positive numbers such that $ abc \equal{} 1$ prove that

$ \sum \frac {b \plus{} c}{a^4 \plus{} 1} \geq 3$[/quote]
If $ f(a,b,c)\equal{}\sum_{cyc} \frac {b \plus{} c}{a^4 \plus{} 1}\minus{}3$ and $ a\equal{}\min\{a,b,c\}$ then, $ f(a,b,c)\minus{}f(a,\sqrt{bc},\sqrt{bc})\geq0.$ :wink:

It seems that the following inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
\[ \frac{b\plus{}c}{a^7\plus{}1}\plus{}\frac{a\plus{}c}{b^7\plus{}1}\plus{}\frac{a\plus{}b}{c^7\plus{}1}\geq3\]


When all possible diagonals are drawn in a [b]Regular[/b] 50-sided convex polygon, how may triangles are formed?

That depends a great deal on how the polygon is arranged.  Two possible questions: 1) the polynomial is arranged so that no three diagonals intersect in a point.  (This problem has been discussed before on the forum.)  2) The polynomial is regular.  (I think this problem is fairly hard but solved in general, although I could be wrong.  It may also have been discussed on the forum.)

I was thinking of the condition to make it regular, but did not think it would be necessary.
As far as positioning the vertices, that could be a problem.
I guess to fix the position, I'll add the condition of regularity.
Can you guys tell me whether or not that is too restrictive?

To see why it might depend on which polygon you choose, just consider a hexagon: a regular hexagon has three long diagonals that are concurrent at the center.  If you nudge a single one of the vertices just slightly, you create a new triangular region in the center.

Unfortunately, the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url] seems to be down at the moment, so I can't go find your sequence for you.  In addition, there are still several possible interpretations of your question.  To clarify: if we replaced "50" with "4," what would the answer be?  What if we replaced "50" with "5?"

I see your point.
In a $ 4$ sided polygon it is $ 4$ triangles.
In a $ 5$ sided polygon it is either $ 6$ or $ 7$.

So the position of the vertices affects the answer.

1) Is a regular polygon invariant under zoom (with respect to number of triangles)?
2) If so, when we set the polygon's vertices to that of a regular one, would the number of triangles be maximum?

3) Also I seem to think there is a pattern emerging from these numbers but can't seem to generalize. Any hints?

I don't see how one can count either 6 or 7 triangles in a 5-gon -- could you elaborate?

Sorry I meant $ 6$ or $ 7$ unique triangles.
Using symmetry I multiply by $ 5$ to get $ 35$ triangles in a pentagon.
then I used two parallel lines, put two points on one and three on the other, drawing all the diagonals I got $ 6\times 5 \equal{} 30$ triangles.

I made a mistake with the number in the square. Their are $ 2$ unique triangles, and $ 2\times4\equal{}8$ triangles.

I was thinking that in a regular polygon the number of unique triangles is maximized, and that there was a pattern for that number. I will think about this problem some more till I can 'see' more of it.

Here is a link:

http://www.cs.uwaterloo.ca/journals/JIS/sommars/newtriangle.html


call the altitudes of a triangle: $a_1, a_2, a_3$ and the radius of its inscribed circle $r$.

Find the minimum value of $(a_1 + a_2 + a_3)/r$  (over all triangles).

Which triangle achieves this minimum value?

[quote="xxxyyyy"][quote="matt276eagles"][hide]Let the sides be $s_{1},s_{2},s_{3}$ and let $K$ be the area.   Since $K=\frac{a_{i}s_{i}}{2}$, we have $a_{i}=\frac{2K}{s_{i}}$.  Also, since $K=\frac{r(s_{1}+s_{2}+s_{3})}{2}$, we have $r=\frac{2K}{s_{1}+s_{2}+s_{3}}$.  Therefore,

$\frac{a_{1}+a_{2}+a_{3}}{r}=\frac{\frac{2K}{s_{1}}+\frac{2K}{s_{2}}+\frac{2K}{s_{3}}}{\frac{2K}{s_{1}+s_{2}+s_{3}}}=(s_{1}+s_{2}+s_{3})\left(\frac{1}{s_{1}}+\frac{1}{s_{2}}+\frac{1}{s_{3}}\right)$
$=3+\frac{s_{1}}{s_{2}}+\frac{s_{2}}{s_{1}}+\frac{s_{1}}{s_{3}}+\frac{s_{3}}{s_{1}}+\frac{s_{2}}{s_{3}}+\frac{s_{3}}{s_{2}}$

By AM-GM, $\frac{s_{1}}{s_{2}}+\frac{s_{2}}{s_{1}}\ge 2$, and the same for the other pairs of fractions.  The minimum is $9$, and the triangle is equilateral.[/hide][/quote]

Follow up question to the above one...

What would be the minimum and maximum value of $(a1 + a2 + a3)/R$ where $R=circum-radius$.

And for which triangle?[/quote]

anyone?


[quote="Philmac"]I've already wasted a good hour and a half getting absolutely nowhere on this problem, so I know I'm getting nowhere too soon.

Given the vectors a = (w+1, 3w+1), and b = (2, w+2).

a) If {a,b} is [i]not[/i] a basis for v2 find the value(s) for the scalar w.

b) If {a,b} is a basis for v2 find the value(s) for the scalar w.[/quote]
Your question is posted in wrong forum, it should be in [b]Linear Algebra[/b] forum, even though this is a basic question.

[hide="Hints"]For both parts all you need to know is the definition of a basis. For part a two conditions are checked. If one or the other or both conditions fail, what happens?
For part b, since it is a basis, what condition holds to set up an equation and allows you to find the value(s) of scalar w?[/hide]

For other users in this forum, please don't post answers. This is a basic question and should not be answered right away. If you post the solution whether to show off or to help (you think), you are hurting him instead of helping. Best way to help him is to lead him to [b]think[/b] and get his own answers, that is the best reward a learner can get.

Hrrm this is a pretty basic problem that follows from the definition of bases.

Use the fact that a basis is linearly dependent...you don't need the fact that it's a spanning set.

This is all assuming that you have son understanding of linear algebra.


$1.$ In a rectangular examination hall of arbitrary capacity no seat is vacant and no examinee is answering the same question paper as any of his/her eight neighbours. Find the minimum number of question papers being used in the hall.

$2.$ A friend said, "I travelled 20 km due south, then 20 km due east and then 20 km due north. I found myself in the same place I started from." Can he be speaking the truth?

$3.$ A boat made of wood $1.5$ times heavier than water has an internal volume $2/3$ of its external volume. When filled with oil to the brim, it barely floats in water. Find the density of the oil.

$4.$ A rope hangs over a pulley. On one side a weight is tied. The weight is resting on ground when a monkey of equal weight starts climbing up the other side of the rope. Will the monkey reach the top first or the weight? Assume the rope and pulley to be lossless and the rope weightless

$5.$We have a triangle with each side $n$ cm long. On each
side we mark $n-1$ points to divide it into $n$ equal parts of $1$ cm each.
From each of these points we draw two lines parallel to the other  sides to meet the other sides. This creates a mesh of triangles.
Total how many triangles are there, in this diagram?

[quote="boostage"][quote="xxxyyyy"]$1.$ In a rectangular examination hall of arbitrary capacity no seat is vacant and no examinee is answering the same question paper as any of his/her eight neighbours. Find the minimum number of question papers being used in the hall.
[/quote]
[hide="Answer"] 4 I think, because 1, 2 and 3 do not work, and I can find several configurations with 4.[/hide][/quote]

If an examinee and all of their EIGHT neighbors have different papers, wouldn't a minimal answer be nine?

[quote]If an examinee and all of their EIGHT neighbors have different papers, wouldn't a minimal answer be nine?[/quote]
Not necessarily. Your eight neighbors must have different papers from your paper, but some of them can have the same papers as each other. For example, the person in front of you can have the same paper as the one behind you, as long as it is different from yours.

Oh, good point. Haven't thought of that  :maybe:

[quote="xxxyyyy"]$3.$ A boat made of wood $1.5$ times heavier than water has an internal volume $2/3$ of its external volume. When filled with oil to the brim, it barely floats in water. Find the density of the oil.

$5.$We have a triangle with each side $n$ cm long. On each
side we mark $n-1$ points to divide it into $n$ equal parts of $1$ cm each.
From each of these points we draw two lines parallel to the other  sides to meet the other sides. This creates a mesh of triangles.
Total how many triangles are there, in this diagram?[/quote]
[hide]
$3$) $\frac{1}{3}$ of the boat is wood. Give the wood a volume of $1$, and the oil a volume of $2$. With the oil's density as $x$,
$1(1.5)+2(x)=3 \iff \boxed{x=\frac{3}{4}}$.

$5$) Look at the rows formed, starting at a vertex. The first row (a 1x1 triangle) has $1$ triangle drawn. The second row has $3$, then $5$, etc. So the total number is $\sum_{k=1}^{n}{2k-1}= \boxed{n^{2}}$.

And i have no idea for #4...  it seems like they'd get there at the same time... [/hide]

[quote="xxxyyyy"]$1.$ In a rectangular examination hall of arbitrary capacity no seat is vacant and no examinee is answering the same question paper as any of his/her eight neighbours. Find the minimum number of question papers being used in the hall.
[/quote]
[hide="Answer"]
I think the answer is 9 because for every group of 9 people the same law could apply where the people across horizantally or vertically had the same paper and the diagonals all had the same paper.(diagram below)

1  2  1
3  4  3
1  2  1

or

1  2  1
3  4  3
2  1  2

the numbers denote different test numbers.[/hide]

[quote="lotrgreengrapes7926"][quote="xxxyyyy"]$2.$ A friend said, "I travelled 20 km due south, then 20 km due east and then 20 km due north. I found myself in the same place I started from." Can he be speaking the truth?[/quote][hide] Yes, at the North Pole and $20+\frac{10}{\pi}$ miles from the South Pole. :D [/hide][/quote]

How do you prove for south pole case?

[quote="xxxyyyy"][quote="lotrgreengrapes7926"][quote="xxxyyyy"]$2.$ A friend said, "I travelled 20 km due south, then 20 km due east and then 20 km due north. I found myself in the same place I started from." Can he be speaking the truth?[/quote][hide] Yes, at the North Pole and $20+\frac{10}{\pi}$ km from the South Pole. :D [/hide][/quote]

How do you prove for south pole case?[/quote]
How did I find it? Or how do I know it works?

The first question: I saw it in a book before.  :blush:

The second question: After walking 20 km south, the person must walk 20 km east and end in the same place. Thus, we need a circle with circumference $20$ with center at the South Pole, and therefore, it's radius is $\frac{10}{\pi}$.

If a person started at the north pole, walked 20km south, then 20km east, he would still be 20 km south of the north pole. So, walking 20 km north would get him back to where he started. We don't need a circle with a radius of $\frac{10}{\pi}$ km.

[quote="NUMB3RS"][quote]4. A rope hangs over a pulley. On one side a weight is tied. The weight is resting on ground when a monkey of equal weight starts climbing up the other side of the rope. Will the monkey reach the top first or the weight? Assume the rope and pulley to be lossless and the rope weightless [/quote]

[hide]The monkey reaches the top first. I haven't taken physics so I don't know if this is definitely true, but here's my reasoning:

Take a rope and loop it around a tree (making the tree a "pulley") and take an end of the rope in each hand. Then pull each end of the rope with an equal amount of force. What happens? Nothing. Because the two forces cancel each other out.

Similarly, since the monkey weighs the same as the weight, neither object would move if the monkey was just hanging on to the rope, but not climbing. However, since the monkey is climbing the rope, it will reach the top first. The weight will not even move because at any given point in time, there will be an equal amount of weight on each side = an equal amount of force being exerted in each direction = the forces cancel out.

This is my logic. There's a good chance that I'm wrong. If someone has taken physics, they might be able to help with this problem. Or maybe I'm using the wrong approach to solving this problem...[/hide][/quote]

I'm not sure if the monkey reaches the top first.

If the monkey pulls itself up, then it will have to pull down on the rope (Newton's Laws... one of them:  If the monkey is pulled up, the rope is pulled down. Equal and opposite forces.). This will transfer over to the rock via the pulley, and the rock will be lifted up too. Since their masses are equal, it doesnt make any sense that the rock wouldnt move, as when the monkey isnt doing anything (no forces other than gravity) the rock and monkey will be at the same height (they will level on the pulley). So, if the same force pulls on the same mass, then the same acceleration will occur. Soo...  I think they get there at the same time.

I think.

Or the rope breaks and they both fall.

(Crazy monkey...  why doesnt he just use the branches?)

[quote="cincodemayo5590"]
I'm not sure if the monkey reaches the top first.

If the monkey pulls itself up, then it will have to pull down on the rope (Newton's Laws... one of them:  If the monkey is pulled up, the rope is pulled down. Equal and opposite forces.). This will transfer over to the rock via the pulley, and the rock will be lifted up too. Since their masses are equal, it doesnt make any sense that the rock wouldnt move, as when the monkey isnt doing anything (no forces other than gravity) the rock and monkey will be at the same height (they will level on the pulley). So, if the same force pulls on the same mass, then the same acceleration will occur. Soo...  I think they get there at the same time.

I think.

Or the rope breaks and they both fall.

(Crazy monkey...  why doesnt he just use the branches?)[/quote]

the trick in this problem is that the F_t (on the rock) is the monkey's weight + him pulling himself up

[hide="why"]the system has more GPE at the end, therefore, the monkey has to output energy, this means that force also was used, rope to monkey friction pulls him up, and as sam said using newton's third law, F_t increases on the rock[/hide]

basically, they meet at the same time


function
integration
parameterization
geometry
calculus
logarithms
real analysis

The function $f(x)$, which gives the width of the Eiffel Tower as a function
of the distance from top, is a solution of the following equation:

(E)  $af(x)\int_{0}^{x}f^2(t)dt=x\int_{0}^{x}f(t)dt-\int_{0}^{x}tf(t)dt$

a>0 fixed parameter

Find $f$ continuous on $R$

To know how to arrive to (E), have a look at

[url= https://www.chouard-eiffel-equation.com/] see here [/url]

https://www.chouard-eiffel-equation.com/

[url= https://www.chouard-eiffel-equation.com/reponse-eiffel.htm] see also [/url]

https://www.chouard-eiffel-equation.com/reponse-eiffel.htm

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

Picture  Eiffel Tower
[url= http://christophe.chouard.free.fr/eiffel/] pictures Eiffel Tower [/url]
http://christophe.chouard.free.fr/eiffel/

$f(x)=C\sqrt x$ (in this case each term is some constant times $x^{5/2}$). But if my eyes do not deceive me, this solution has nothing to do with the actual shape of the tower!

if i didn't make any silly mistake, it turned me out something like \[ f = Cx^{1/4}\] or something like this. I'll try to recover all of my papers (blackboards, to be more precise :P) and write it down...

It's possible to semplify the equation by noting:
\[
\rho gA=-\frac {d} {dy} (\sigma A)
\]
where $ \rho $ is the density of the material; A = the area of the square cross section =$ 2x^2 $; g = acceleration of gravity; y = the vertical axe; x = the orizontal axe and $\sigma$ = the compressive stress which is constant. Infact the tower bends outward as the elevation, y, decreases.
Integration of equation gives:
\[
y=-\frac {2\sigma} {\rho g}\ln \frac{x} {x_0}
\]
$  x_0 $ is half of the lenght of the base.
Thus a logarithimic equation describes the profile of the Eiffel Tower.
Then
\[
x= x_0 exp \left ({\displaystyle -\frac {y\rho g} {2 \sigma}\right)
}\]
gives the width as a function of height.

All of you have found something different.

I am going to watch carefully which function represent Eiffel Tower soon.

Well, it is easy to see that ma_go's solution is wrong (just plug it into the equation and integrate 3 power functions). Nene's answer bears the closest resemblance to the actual shape of the tower but she solves a completely different equation coming from a different engineering problem of keeping the stress constant throughout the tower that has nothing to do with the original Moubinool question. Perhaps, Nene is right and the actual shape is logatithmic but that means that Chouard is wrong in his assumptions about what problem was being solved by Eiffel when choosing the shape.

yes, i've noticed my solution is wrong...
i've found an idiot error in my papers, and i've fixed it...
but it leads to a differential equation i cannot solve, even if it's much easier than the first one :)
i'll try again :)

[quote="ma_go"]
but it leads to a differential equation i cannot solve
[/quote]
That's strange because, as I said, $C\sqrt x$ with appropriately chosen $C=C(a)$ is a solution (just plug it in!). The problem is to show that there is no other continuous solution that is strictly positive for $x>0$. So, one doesn't really need to solve the equation, just to prove the uniqueness theorem for it...

Many of you think the equation in the first post of this topic is not correct.

So propose your equation for Eiffel Tower ?


limit
function
probability
calculus
calculus computations

The task is from another topic but we didn't manage to find any proof so here is the problem which is to be solved:
Prove that $\lim_{x\to\infty}[1-F_{X}(x)]g(x) = 0$ where $g$ is continuous function, $X$ is a random variable with probability disrtibution function $F_{X}(x)$ and $X\geq0$ Moreover we know that $E[|g(X)|]$ exists (it's finite). I was trying to do it in some ways but I didn't get the aim.
Thanks for help

You mean that $F$ is the cumulative distribution function- this makes no sense otherwise.

This isn't true without some additional conditions on $g$. For example, let $g(x)$ be zero between $2n-\frac12$ and $2n+\frac12$, with a peak of height $2n+1$ at $2n+1$. We can choose $X$ so that $g(X)=0$, but $1-F(x)$ goes to zero very slowly at $\infty$.

If you restrict $g$ to be increasing, this is true. (the unbounded case is the only nontrivial one)


modular arithmetic
geometry
3D geometry
number theory unsolved
number theory

the number of solutions of the equations $4x^{3}$-$7y^{3}$=2003 is

[quote="ThAzN1"]No solutions? Take both sides mod 7 so $4x^3 \equiv 1 \pmod 7$ but cubes are $0, \pm 1 \pmod 7$ so contradiction...[/quote]

yes i think you are correct.it is given in one of the options.but can you explain me the last sentence i.e

but cubes are $0, \pm 1 \pmod 7$ so contradiction...[/quote]

You can verify (all of this is mod 7)

$0^3 \equiv 0$
$1^3 \equiv 1$
$2^3 \equiv 8 \equiv 1$
$3^3 \equiv 27 \equiv -1$
and so on

then $4x^3 \equiv 0, \pm 4$ which cannot be 1 mod 7...


function
combinatorics solved
combinatorics

A system $X=\{x_1,x_2,...,x_m\}$ of real numbers is called monotonic, iff  for all  $i\in \{1,2,...,m-1\}$ we have
\[ \Delta x_i: = x_{i+1}-x_i>0 \; \; ,\; \; \Delta x_i\ge 0 \; ,\; \Delta x_i=0 \; ,\; \; \Delta x_i \le 0\; \; \; \; \mbox{\rm or}\; \; \Delta x_i <0\; .  \]
If  $1\le i_1<i_2<\ldots \ldots i_k\le m$ we say that $\{x_{i_1},x_{i_2},\ldots,x_{i_k}\}$ is a subsystem of $X$.
Prove or disprove : [b]  Any system of at least $M=100$ real numbers has a monotonic subsystem of $5$ numbers.[/b]

[color=green] [b]When M= 17 the statement it's true ?[/b] [/color]

Note: I have'nt a complete solution. Perhaps the question is well-nown.

I think you only need 26 numbers.

A very famous theorem of Erdos states that any sequence of $mn+1$ real numbers has a non-increasing subsequence of length at least $m+1$ or a non-decreasing subsequence of length at least $n+1$ (in fact it deals with increasing and decreasing subsequences, for pairwise distinct numbers).

Let $U_1, \cdots , U_{mn}$ be a sequence of $mn+1$ numbers.
For each $i$, let $L_i^+$ (resp. $L_i^-$) be the length of the longest non-decreasing (resp. non-increasing) subsequence starting with $U_i$.

Suppose for a contradiction that for all $i$, we have $L_i^+ \leq n$ and $L_i^- \leq m$.
Let $f(i) = (L_i^+ , L_i^-)$. The function $f$ is from $\{1, \cdots , mn+1 \}$ onto $\{1, \cdots , m \} \times \{1, \cdots , n \}$.
Moreover it is an injective function since if $i < j$ we have that :
if $U_i \leq U_j$ then $L_i^+ > L_i^+$ thus $f(i) \neq f(j)$
if $U_i \geq U_j$ then $L_i^-  > L_i^-$ thus $f(i) \neq f(j)$
Since $mn+1 > m \times n$ we have the desired contradiction.

Pierre.

Hi  Pierre  ,
thank you for your reference regardin Erdos result. Can you give the exact
reference  in literature , title, journal, year?
Thanking in advance, Alex

If I'm not wrong, it is the famous paper from Erds and Szekeres, which contains their rediscovery of Ramsey's theorem and application to set of points in convex position.

P.Erds, G.Szekeres, 'A combinatorial problem in geometry', Composito Math;, 2 (1935), 464-470.

Pierre.

[quote="pbornsztein"]I think you only need 26 numbers.[/quote]

Hmm... you need to find a monotonic row of 5 numbers... so according to erdos's theorem you only need 17, not 26 right? Or did I misunderstand something? :?

You are right. I confused between $m$ and $m+1$. :D

Pierre.

Are you sure that the result belongs to Erdos? I always thought
that the result you mentioned (or rather its generalized form pertaining
to ordered sets) is called Dilworth theorem. As to the paper by Erdos and Szekeres, it deals with points in the plain forming convex polygon...

The theorem about monotonic subsequences is due to Erdos. I think Dilworth's theorem states that in a poset with $mn+1$ elements we either have an $n+1$-chain or an $m+1$-antichain. This generalizes Erdos-Szekeres's theorem.


Let $ X$ the set of all sequences $ \{a_1, a_2,\ldots , a_{2000}\}$, such that each of the first 1000 terms is 0, 1 or 2, and each of the remaining terms is 0 or 1. The [i]distance[/i] between two members $ a$ and $ b$ of $ X$ is defined as the number of $ i$ for which $ a_i$ and $ b_i$ are different.

Find the number of functions $ f : X \to X$ which preserve the distance.

[quote="carlosbr"]Let $X$ the set of all sequences $\{a_{1}, a_{2},\ldots , a_{2000}$, such that each of the first 1000 terms is 0, 1 or 2, and each of the remaining terms is 0 or 1. The [i]distance[/i] between two members $a$ and $b$ of $X$ is defined as the number of $i$ for which $a_{i}$ and $b_{i}$ are different.

Find the number of functions $f : X \to X$ which preserve the distance.[/quote]
We notice that the number of same elements is also preserved(that's quite obvious), therefore every two same elements of two sequences $a$ and $b$ get mapped on the other same two elements(so that we can form a bijection between the elements). Also, two different elements get mapped on other two different elements. So, in this function, we can only have to things: the place of a given element changes, and/or its value changes(or neither). Because the range of values for first 1000 terms is different from the range of values for least 1000 terms, then the transformtions(change of value and place) will occur in two disjoint subsets: the first 1000 terms and the last 1000 terms.
All we have to do now is calculate the number of different sets of transformations we can make. Wolog we will first change the values of the elements. For the first 1000 terns, this can be done in 6 ways: $\{0,1,2\}$ can be mapped on one of $\{0,1,2\},\{1,0,2\},\{2,0,1\},\{0,2,1\},\{2,1,0\},\{1,2,0\}$ because every two same elements are mapped on same values. while every two different valued elements are mapped on different values. For the last two terms, the change of values can be made in 2 ways: $\{0,1\}$ can be mapped on $\{0,1\}$ or $\{1,0\}$
Now the number of changes of places is equal to te number of permutations of the terms; because we do not know how many terms are the same, the number of permutations for first 1000 terms is $1000!$ and last 1000 terms, it is $1000!$.
So the total number of functions is equal to the number of transformations on first 1000 terms times the number of transformations on last 1000 terms, which is:
$1000!*6*1000!*2=12*(1000!)^{2}$

[quote="rem"]

We notice that the number of same elements is also preserved(that's quite obvious), therefore every two same elements of two sequences $a$ and $b$ [b]get mapped [/b]on the other same two elements(so that we can form a bijection between the elements). Also, two different elements [b]get mapped[/b] on other two different elements. [/quote]
This "get maped" part is not obvious .
[hide]
[tex]0000...0 \Rightarrow 1100...0[/tex]
[tex]1100...0 \Rightarrow 1100...0[/tex]
[tex]2200...0 \Rightarrow 1001...0[/tex][/hide]
I hope to see a more detailed proof for this problem .


Al, Betty, and Clare split $ \$1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of $ \$1500$. Betty and Clare have both doubled their money, whereas Al has managed to lose $ \$100$. What was Al’s original portion?

$ \textbf{(A)}\ \$ 250 \qquad
\textbf{(B)}\ \$ 350 \qquad
\textbf{(C)}\ \$ 400 \qquad
\textbf{(D)}\ \$ 450 \qquad
\textbf{(E)}\ \$ 500$

[hide]
$ a\plus{}b\plus{}c \equal{}1000$
$ 2a\plus{}2b\plus{}c\minus{}100\equal{}1500$

$ \Rightarrow a\plus{}b\equal{}600 \Longrightarrow c\equal{} 400 \longrightarrow \boxed{C}$[/hide]


I really think you should collect all your problems in one topic...