Solve this cryptarithm, where each of the letters represent a different decimal digit from 0 to 9. None of M and R can be zero.

73* (MORE)  = REACH

Why does $ M$ have to equal $ 1$, and how can it be concluded from that that $ R\equal{}7$?

$ M\equal{}1$ because if $ M\geq 2$, then $ 73M$ has more than 5 digits. It does not follow, though, that $ R$ must be 7.

From $ M\equal{}1$ then $ 7 \le R \le 9$ then the first candidate is $ R\equal{}7$

OK, now try $ R\equal{}8,9$ and show that both of them don't work.


find the minimum value of   x^4 + 1/x^2

thank you..... :)

Isn't $ x\to\Re$ in this question? AM-GM inequality only holds for all $ x\to\Re^{\plus{}}$

[quote="t0rajir0u"]But $ x^{2}\in\mathbb{R}^{\plus{}}$.  ;)[/quote]

Oh I see, you are right. :oops:


geometry
circumcircle
area of a triangle

Let $\triangle{ABC}$ be a triangle with area $1$. Let DEF be the points obtained by reflecting A, B, and C respectively in their opposite sides. Find the maximum area of $\triangle{DEF}$.

I saw this problem in the aops text book where $\triangle{ABC}$ is a right triangle. In that case we can show that the area of $\triangle{DEF}$ is a constant multiple of that of $\triangle{ABC}$. Then I wondered what would happen if $\triangle{ABC}$ is arbitrarily chosen. Experimenting with GSP, it seems that the maximum value is 5. But it's just a guess....

[quote="probability1.01"]Huh... let H be the horthocentre of ABC, then HD, HE, and HF are all at most 2R (R is circumradius of ABC). Hence the maximum area of DEF is at most the maximum area of a triangle with circumradius 2R. That maximum area occurs when said triangle is equilateral. We can find equality when ABC is equilateral, and in this case [DEF] = 4.[/quote]

Not exactly. See the attached figure below.


Who are the freshman who qualified for USAMO?  Trying to get a good idea of who I have to beat to get into MOP.  :P

Me! (of course)

Who is Jeffrey Chan? (he is listed as a 10th grader) Is that supposed to be the 8th grade Jeffrey Chan?

Ok, who are the other 45 of you? =D

yea he goes to my school.  He's an 8th grader.  They misplaced him and his sister's grades.

you, me, r.j., dennis, and mark are the only 9th graders from texas who made it.  I expected more...

eh I'm a sophomore from Texas that made usamo that should be in 9th grade agewise...ah

This post is bascially spam.  :P

[quote="chess64"][quote="Mathlete9366"][quote="eryaman"]This post is bascially spam. [/quote]

how so? I actually think this post has complete relevancy to the forum, and it actually does have content behind it... it's just I guess nobody wants to know the competition out there ...[/quote]

He was talking about his own post. :roll:[/quote]Chess64 is right, I was referring to my own post.  And this post of mine is essentially spam too.  :P

[quote="probability1.01"]Hey, guess what I'm a freshman too who made USAMO, and I'm going to OWN you all so sorry guys, no MOP for you. But seriously, don't scout out your "competition"... just try to do well on the USAMO without over-stretching. This is for the good of your own USAMO score.[/quote]

I thought you were a sophomore...

You're going against David Benjamin, who has a 300 index. But he'll probably qualify for the Black MOP anyways.

Although those who do well on the AMC and AIME aren't always the best at the USAMO. Specifically, I'm talking about Eric Larson, who was just out of Honorable Mention as an 8th grader last year (score of 23) and who qualified for MOP. And for someone who was the #1 8th grade representative at the USAMO, he finished 53rd at Mathcounts nationals.

[quote="Magixmaniak"][quote="probability1.01"]Hey, guess what I'm a freshman too who made USAMO, and I'm going to OWN you all so sorry guys, no MOP for you. But seriously, don't scout out your "competition"... just try to do well on the USAMO without over-stretching. This is for the good of your own USAMO score.[/quote]

I agree with probability1.01

you should worry about yourself more than others...

do the best you can and you might make it

dont try to beat just enough so you can make it

lol...im just trying to give you a hard time gaurav[/quote]

You still have bad karma from bruteforcing dennis  :P  and you probably won't take me down with you.  I'll already be down there. :D

there are what, about 40 freshmen in the USAMO?

there were like 38 last year without the expansion, and they only accepted 25 but thats still a majority. heck i got a 2 so I didn't make MOSP  :(

Ah well, good luck to you guys.

Yeah, obviously a guy that is so good at AIME&AMC is also going to be good at USAMO, but my point is that things can change around a bit. Eric Larson just got the floor last year, and yet he went on to do really well at USAMO. So if you're in the bottom sector of 9th grade AIME scores (top 29 9th graders got 11 or above I think), don't lose hope.

[quote="probability1.01"]I'm pretty sure the correlation between AIME and USAMO is less than 0.3. I know a few people in the past who did very well on AIME but only did mediocre on USAMO. I also know lots of people who barely made USAMO (or didn't at all) and are quite good at USAMO level problems. In fact, I think the majority of MOPpers last year were only mediocre at AIME, especially since it was an easy year.

That said, I suspect that David Benjamin as a specific individual will do pretty well on USAMO.[/quote]
I actually want to test the AIME-USAMO correlation claim. Would make a good statistics project that I have to do for AP statistics.

Well unless all 9th graders make close to the same score on USAMO there really is not much correlation as most got into USAMO, through the second selection process so most of their scores will be around 8-10, not that there aren't exceptions of course.

[quote="SamE"][quote="Phelpedo"]8th graders and below do not receive the 9th grade bonus and have to qualify like upperclassmen (I think). I know Eric Larson made it because he scored a 23, just 1 point below Honorable Mention. Although David Benjamin goes in with by far the highest index out of all the 9th graders, obviously from the example of Eric (David beat him at the AMC and AIME, but not the USAMO), the USAMO is a different ballgame.[/quote] Are you sure? I was at MOP last year and I'm pretty sure Eric got a 21 like me. That was qualifying (for Blue). Maybe we should ask him. :) But you are right, the freshman MOP is only for freshmen. And I seriously doubt David will fail to qualify for that, at the least. So some of us should see him come June...[/quote]

I may be mistaken, but I heard somewhere that he got a 23. So how does this red/black/blue MoP system work?

Black- Top 12 at USAMO
Blue- Next 18 or something that placed
Red- Top 30 or so freshamn

Cutoffs last year were 28, 21, and about 8 or 9. Yeah, and there were only 24 freshmen in red last year, not 30, but that might change.

Oh yeah, and Canadians don't count, so there were some honorable mentions in black (1 made the team - Mildorf!).

Oh wow, congratulations to everybody that qualified thanks to the new ruleset! (and of course a special congrats to you calc rulz, for putting up such a good fight =D =D)

Well, so now there's 63 of us to fight it out to get into top 30! =D

Hmmm... looks like i'll have to beat even more of you to qualify for MOsP due to the new rules set. No problem. My ninja will take care of all the competition.  :ninja:  :rotfl:

[quote="pieterminate"]Hmmmmm, is there actually 30 freshmen spots?  I thought there were just 24?[/quote]

25-30

The general approximate guideline for AMC is 30, but last year there were only 25.

Ah, cool :D
[color=white]you other freshies are going dooown![/color] :lol:

Eh, at least there won't be any weird tiebreaks with the froshes now, as the actual number is quite nebulous.


modular arithmetic
number theory
relatively prime
number theory unsolved

Already the complete problem:

Determine all pairs $(a,b)$ of integers different from $0$ for which it is possible to find a positive integer $x$ and an integer $y$ such that $x$ is relatively prime to $b$ and in the following list there is an infinity of integers:

$\rightarrow\qquad\frac{a + xy}{b}$, $\frac{a + xy^2}{b^2}$, $\frac{a + xy^3}{b^3}$, $\ldots$, $\frac{a + xy^n}{b^n}$, $\ldots$

One idea?

:arrow: [b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61319]View all the problems from XIX Mexican Mathematical Olympiad[/url][/b]

$a+xy^n\equiv 0 \pmod{b^n}$ for all $n$.
So, there is integer number $k$ that $xy^n=kb^n-a$.
$a+xy^{n+1}=a+kyb^n-ya\equiv 0 \pmod{b^{n+1}}$ .
So, there is integer number $l$ that $kyb^n-ya=lb^{n+1}-a$
From here we get $b^n | a-ya$ for all $n$.
So the first solution is $b=1,-1$.
Now, if $b$ doesn't equal to $1$ or $-1$ we get:
$a-ya=0 \to  y=1$
From here we get:
$b^n|a+x$ for all $n$.
So the final solution is:
$b=1,-1$ and all $a$.
$b\neq1,-1$ and $a<0$.

[quote="dblues"]You have only shown for the case when all the terms in the sequence $(\alpha_n)$ are integers, where $\alpha_n = \frac{a+xy^n}{b^n}$. How about the case when not all of $\alpha_n$ are integers, but still there are infinitely many $\alpha_n \in \mathbb{Z}$?[/quote]
Yes, I have forgotten about it , I will solve it. ;)

Ok, the 'patch-up' for the proof has been frustrating me, and for now, I can't think of a simpler 'patch-up' to include all cases. Here is my full solution, with as much rigour as possible:

Let $\alpha_n = \frac{a+xy^n}{b^n}$. Given $\gcd (x,b) = 1, a,b \neq 0, x >0, a,b,x,y \in \mathbb{Z}$, we claim that: the sequence $\alpha_1, \alpha_2, \alpha_3, \ldots$ contains infinitely many integers $\iff$ either $a<0, \mid b \mid >1$ or $\mid b \mid = 1$, which is the same as [i]pavel25[/i]'s solution set. (Note that both conditions are mutually exclusive.)

When $\mid b \mid =1$, obviously $\alpha_n \in \mathbb{Z} \forall n \in \mathbb{Z}^+$. When $a<0, \mid b \mid >1$, we choose $x = -a > 0$ and $y=1$. Then $\alpha_n = 0 \forall n \in \mathbb{Z}^+$. Thus the $\Leftarrow$ case is trivially true.

For the $\Rightarrow$ case, we prove an equivalent statement: If $a> 0, \mid b \mid >1$, then $\not \exists x,y \in \mathbb{Z}, x>0$ such that $\alpha_n \in \mathbb{Z}$ for infinitely many $n \in \mathbb{Z}^+$. We shall proceed to prove by contradiction.

Suppose not, then choosing a fixed pair $(a,b)$, with $a>0, \mid b \mid > 1, \exists x, y \in \mathbb{Z}, x>0$ such that $\alpha_1, \alpha_2, \alpha_3, \ldots$ contains infinitely many integers. Let $n = n_1, n_2, n_3, \ldots$ be all the integers such that $\alpha_n \in \mathbb{Z}$, where $n_i < n_j \ \forall \ i<j$. Letting $d_i = n_{i+1} - n_i$, we note that $d_1, d_2, d_3, \ldots$ is a sequence of infinite terms, all of which are positive integers.

Choose an arbitrary positive integer $d$. We note that $\alpha_{n_d} \in \mathbb{Z}$ and we have $xy^{n_d} = \alpha_{n_d}b^{n_d} - a$. We note also that $\forall \ m \geq d$, we have $\alpha_{n_d + M} \in \mathbb{Z}$, where $M = \sum_{i=d}^m d_i$. Thus $a+xy^{n_d+M} = a+y^M(\alpha_{n_d}b^{n_d} - a) = y^M\alpha_{n_d}b^{n_d} - a(y^M-1)$, which is $\equiv 0 \pmod{b^{n_d+M}} \ \forall \ m \geq d$. Taking $\mod b^{n_d}$, we have $a(y^M - 1) \equiv 0 \pmod{b^{n_d}}$. Since $\gcd (x,b) = 1 \iff \gcd(a,b) = 1$, it follows that $y^M \equiv 1 \pmod{b^{n_d}}$.

If $\alpha_{n_d} \neq 0$, then let $p_d$ be the greatest integer such that $b^{p_d} \mid \alpha_{n_d}$. Note that $p_d \geq 0$. If $p_d \geq M$, then we have $y^M \equiv 1 \pmod{b^{n_d+M}}$. If $p_d < M$, then we have $b^{n_d+p_d} \mid (y^M -1), b^{n_d+p_d + 1} \nmid (y^M -1)$.

Choosing an arbitrarily large $M$, we can have $p_d < M$, thus it follows that $b^{n_d+p_d} \mid (y^M -1), b^{n_d+p_d + 1} \nmid (y^M -1)$. We can then let $y^M = \zeta_1 b^{n_d+p_d+1} + \zeta_0 b^{n_d+p_d} + 1$ for some $\zeta_0, \zeta_1 \in \mathbb{Z}$, where $b \nmid \zeta_0$.

Letting $M_q = \sum_{i=1}^q d_{m+i}$, we also have $y^{M+M_q} \equiv 1 \pmod{b^{n_d+p_d}}, y^{M+M_q} \not\equiv 1 \pmod{b^{n_d+p_d+1}} \ \forall \ q \in \mathbb{Z}^+$. Since $y^{M+M_q} = {(\zeta_1 b^{n_d+p_d+1} + \zeta_0 b^{n_d+p_d} + 1)}^{M_q} \equiv {{M_q}\choose 1}\zeta_0 b^{n_d+p_d} + 1 \pmod{b^{n_d+p_d+1}}$, it follows that $b \nmid M_q$.

Since the $m$ chosen was arbitrary, it follows that for a sufficiently large $m_0$, we must have $b \nmid \sum_{i=1}^{q}d_{m+i} \ \forall \ m\geq m_0, q \in \mathbb{Z}^+$.

Now consider the following $b$ numbers $\sum_{i=1}^q d_{m+i}$, for $q = 1, 2, \ldots, b$. If any of them are $\equiv 0 \pmod{b}$, we have a contradiction, thus there are $b-1$ possible residue classes for these $b$ numbers, and by Pigeon-hole Principle, $\exists$ two of them congruent to each other $\mod b$, which implies, after taking their difference, that $\exists$ a string of consecutive $d_i$'s whose sum is $\equiv 0 \pmod{b}$, therefore contradicting our original assumption.

This means that $p_d$ doesn't exist, so we must necessarily have $\alpha_{n_d} = 0$. Since the $d$ chosen was arbitrary, it follows that $\alpha_{n_i} = 0 \ \forall \ i \in \mathbb{Z}^+$, thus $a+xy^{n_i} = 0 \ \forall \ i \in \mathbb{Z}^+$. Since for each fixed pair $(a,b)$, the $x,y$ in the sequence $\alpha_1, \alpha_2, \alpha_3, \ldots$ is fixed, it follows that $y^{n_i}$ must have a constant value, so necessarily, we have $y=1$. This implies that $a+x = 0 \Rightarrow x = -a < 0$ (Contradiction)

The $\Rightarrow$ case has been proven, thus solving the whole problem.

[quote="dblues"]it follows that $y^{n_i}$ must have a constant value, so necessarily, we have $y=1$.[/quote] Oops!! Here, we can only conclude that $\mid y \mid = 1$. This means that when $a>0, \mid b \mid > 1$, we can choose $x=a > 0$ and $y = -1$, which in this case, $\alpha_n = 0 \ \forall$ odd $n$. So this problem somehow becomes trivial, with the solution set being all non-zero integers for $a,b$. This problem could be made not so trivial if we include an additional condition that $y$ is also positive, which in this case, my solution would not come to naught.

When $a=-2<0$ and $b=2$ we choose $x=-a=2$, but $gcd(x,b)=2\not=1$
 :huh:

Ok, I guessed I've overlooked this point.  :blush:  :blush:  :blush:
Let me rewrite the solution in full again. Hopefully I get it right this time..

Let $\alpha_n = \frac{a+xy^n}{b^n}$. Given $\gcd (x,b) = 1, a,b \neq 0, x >0, a,b,x,y \in \mathbb{Z}$, we want to find all possible $(a,b)$ such that the sequence $\alpha_1, \alpha_2, \alpha_3, \ldots$ contains infinitely many integers.

When $\mid b \mid =1$, obviously for any choice of $a$, $\alpha_n \in \mathbb{Z} \forall n \in \mathbb{Z}^+$. So we assume $\mid b \mid > 1$.

Now, we choose a fixed pair $(a,b)$, with $\mid b \mid > 1$ such that $\exists x, y \in \mathbb{Z}, x>0$ such that $\alpha_1, \alpha_2, \alpha_3, \ldots$ contains infinitely many integers. We shall investigate the restrictions on $a,b$.

Let $n = n_1, n_2, n_3, \ldots$ be all the integers such that $\alpha_n \in \mathbb{Z}$, where $n_i < n_j \ \forall \ i<j$. Letting $d_i = n_{i+1} - n_i$, we note that $d_1, d_2, d_3, \ldots$ is a sequence of infinite terms, all of which are positive integers.

Choose an arbitrary positive integer $d$. We note that $\alpha_{n_d} \in \mathbb{Z}$ and we have $xy^{n_d} = \alpha_{n_d}b^{n_d} - a$. We note also that $\forall \ m \geq d$, we have $\alpha_{n_d + M} \in \mathbb{Z}$, where $M = \sum_{i=d}^m d_i$. Thus $a+xy^{n_d+M} = a+y^M(\alpha_{n_d}b^{n_d} - a) = y^M\alpha_{n_d}b^{n_d} - a(y^M-1)$, which is $\equiv 0 \pmod{b^{n_d+M}} \ \forall \ m \geq d$. Taking $\mod b^{n_d}$, we have $a(y^M - 1) \equiv 0 \pmod{b^{n_d}}$. Since $\gcd (x,b) = 1 \iff \gcd(a,b) = 1$, it follows that $y^M \equiv 1 \pmod{b^{n_d}}$.

If $\alpha_{n_d} \neq 0$, then let $p_d$ be the greatest integer such that $b^{p_d} \mid \alpha_{n_d}$. Note that $p_d \geq 0$. If $p_d \geq M$, then we have $y^M \equiv 1 \pmod{b^{n_d+M}}$. If $p_d < M$, then we have $b^{n_d+p_d} \mid (y^M -1), b^{n_d+p_d + 1} \nmid (y^M -1)$.

Choosing an arbitrarily large $M$, we can have $p_d < M$, thus it follows that $b^{n_d+p_d} \mid (y^M -1), b^{n_d+p_d + 1} \nmid (y^M -1)$. We can then let $y^M = \zeta_1 b^{n_d+p_d+1} + \zeta_0 b^{n_d+p_d} + 1$ for some $\zeta_0, \zeta_1 \in \mathbb{Z}$, where $b \nmid \zeta_0$.

Letting $M_q = \sum_{i=1}^q d_{m+i}$, we also have $y^{M+M_q} \equiv 1 \pmod{b^{n_d+p_d}}, y^{M+M_q} \not\equiv 1 \pmod{b^{n_d+p_d+1}} \ \forall \ q \in \mathbb{Z}^+$. Since $y^{M+M_q} = {(\zeta_1 b^{n_d+p_d+1} + \zeta_0 b^{n_d+p_d} + 1)}^{M_q} \equiv {{M_q}\choose 1}\zeta_0 b^{n_d+p_d} + 1 \pmod{b^{n_d+p_d+1}}$, it follows that $b \nmid M_q$.

Since the $m$ chosen was arbitrary, it follows that for a sufficiently large $m_0$, we must have $b \nmid \sum_{i=1}^{q}d_{m+i} \ \forall \ m\geq m_0, q \in \mathbb{Z}^+$.

Now consider the following $b$ numbers $\sum_{i=1}^q d_{m+i}$, for $q = 1, 2, \ldots, b$. If any of them are $\equiv 0 \pmod{b}$, we have a contradiction, thus there are $b-1$ possible residue classes for these $b$ numbers, and by Pigeon-hole Principle, $\exists$ two of them congruent to each other $\mod b$, which implies, after taking their difference, that $\exists$ a string of consecutive $d_i$'s whose sum is $\equiv 0 \pmod{b}$, therefore contradicting our original assumption.

This means that $p_d$ doesn't exist, so we must necessarily have $\alpha_{n_d} = 0$. Since the $d$ chosen was arbitrary, it follows that $\alpha_{n_i} = 0 \ \forall \ i \in \mathbb{Z}^+$, thus $a+xy^{n_i} = 0 \ \forall \ i \in \mathbb{Z}^+$. Since for each fixed pair $(a,b)$, the $x,y$ in the sequence $\alpha_1, \alpha_2, \alpha_3, \ldots$ is fixed, it follows that $y^{n_i}$ must have a constant value for infinite of the terms, so necessarily, we have $\mid y \mid =1$.

Since when $\mid y^n \mid = 1$, the numerator $a+xy^n$ is bounded, while the denominator grows arbitrarily large as $n \rightarrow \infty$, it follows that if the numerator is non-zero, $\exists$ a sufficiently large $B$ such that $\mid \alpha_n \mid < 1 \forall \ b \geq B$. Thus for the sequence $(\alpha_n)$ to contain infinitely many integers, we must necessarily have either $a+x = 0$ or $a-x = 0$, which means $x = \pm a$. We finally note that $\gcd (x,b) = 1 \iff \gcd (a,b) = 1$, therefore, the only restriction we have for $a,b$ is that $\gcd (a,b) = 1$.

Indeed, when $\gcd (a,b) = 1$, then setting $x = \mid a \mid, y = -1$, we have $\alpha_n = 0$ for all odd or even terms, according to whether $a$ is positive or negative respectively.

Therefore only all $a,b$ such that $\gcd (a,b) = 1$ satisfies the given conditions. (Note that this also includes the obvious solution of all $a, b$ satisfying $\mid b \mid = 1$.)

Wonderfull!! :idea:

How was happened to you this demonstration?

Thank you very much!!

Greetings!! :D

The solutions are all $a,b$ such that $\gcd (a,b) = 1$. We will check $\gcd (a,b) = 1$ and $\gcd (a,b) \neq 1$. Let $a_n=\frac{a+xy^n}{b^n}$:

If $\gcd (a,b) = 1$, putting $x=a$ and $y=-1$ will give $a_{2k+1}=0$ for all $k$, so there are infinite integer $a_n$.

If $\gcd (a,b) \neq 1$, let p be a prime such that $p\mid \gcd (a,b)$. Then $p^n\mid b^n$. We want to find $x,y$ such that for infinite $n$, $b^n\mid a+xy^n \implies p^n\mid a+xy^n$ for infinite $n$. This is only possible if $p\mid y$. Otherwise, $v_p(p^n)=n$ will be larger than $v_p(a+xy^n)$, giving only a finite solution set. So $p\mid y$.

Since $p\mid y$, then $v_p(a+xy^n)=a$ for $n>v_p(a)$, but for sufficiently large $n$, $v_p(b^n) = n\cdot v_p(b) > v_p(a) = v_p(a+xy^n)=a \implies p^n \nmid a+xy^n$. So there is a maximal $n$ for which $a_n$ is an integer, therefore there are finite integer $a_n$ for $\gcd (a,b) \neq 1$, and we are done. :coolspeak:

@below gotta learn how to type fast  ;)

This problem could have been amazing, if limited to positive integers. But wtv, #méxico JAJAJAJA
Honestly I'm tired of how bad Number Theory is here, this is just so sad :(:(
For the original version we have that:
[hide = solution to the original Problem]
if $(a, b) = 1$, take $x = a$ and $y = -1$. This gives a numerator equal to 0 for odd $n$, and since 0 is divisible by any integer, this pairs work.
Now if $(a, b) > 1$, take a prime $p$ dividing $(a, b)$ and from here we derive that $p \mid y$. It's clear that $v_p(a + xy^n) \ge min\{v_p(a), v_p(y^n)\}$, and:
\[v_p(a + xy^n) = v_p(a)\]
for all sufficiently large $n$, and this kills it.
[/hide]

Now the actual stuff, tbh I think this problem is rather good but kinda harder than your usual p3 (in the OMM ofc), even for modern standards, I think.
[hide = solution to the modified problem]
$y$ cannot be negative, so we need to revisit the case when $(a, b) = 1$. Consider the following claim:

[b][u][color=#f00]Claim 1:[/color][/u][/b]
Fix a positive integer $y$ and a prime $p$. For all $n$ sufficiently large, the following inequality holds:
\[Ord_{p^n}(y) > n\]
[i]Proof:[/i]

We'll use LTE and the fact that $Ord_{p^n}(y) \mid \phi(p^n) = (p - 1)p^{n- 1}$. Say $O_n = Ord_{p^n}(y)$. Consider the set of all positive divisors of $p-1$, and let $k$ be the divisor that maximizes $e = v_p(y^k - 1)$. For sufficiently large $n$, $O_n$ will be equal to $k\cdot p^{n - e-1}$ by both facts I cited earlier, and we derive that from that point $O_n$ will grow exponentially, while the LHS grows linearly. This implies the claim, ofc we should be more careful when $p = 2$, however the argument is analogous.

[rule]
Now let $a_n = a + xy^n$. Regardless of the value of $b$, if the pair $a,b$ works, we can pick a prime $p$ dividing $b$ (ofc we're assuming $b>1$) and the pair $a, p$ should work too.
Let $f(n)$ be the smallest positive integer such that $p^n \mid a_{f(n)}$. Thus we have:
\[y^{f(n)} = -ax^{-1} \pmod{p^n}\]
 Now, this implies that for every $k$ in the Arithmetic Progression starting at $f(n)$ with common difference $Ord_{p^n}(y)$, it holds that $p^n \mid a_k$. Therefore we want $n$ to be in said AP infinitely many times. But this implies there's a value $i$ such that:
\[Ord_{p^n}(y) \cdot i + f(n) = n\]
\[\implies Ord_{p^n}(y) \mid n - f(n)\]
But this also implies that $Ord_{p^n}(y) \le n$, but this contradicts our claim, and therefore it dies. And the only solutions are when $b = 1$.
[/hide]
The problem is great in it's modified form, I originally misread it like that. 40 minutes later I had my argument and was like WOW, finally one nice NT problem, but then I looked at the thread, and RIP, not cool (like imagine the anti emote of what pan posts at the end of his posts hahahaha).


geometry
geometric transformation
homothety
trigonometry
circumcircle
trig identities
Law of Cosines

In $ \triangle ABC$, D is the foot of the altitude from A, E and F are the feet of the altitudes from D to AB ad AC respectively. Show the length EF is the same as if we did the problem with respect to vertices B or C rather than A. (Kalva BMO2002).
[hide="My solution"]
Let G be the orthocenter of $ \triangle ABC$, H and J the feet of the altitudes of $ \triangle ABC$ from B and C, respectively. Since $ DE \parallel CJ$ and $ DF \parallel BH$, by a homothety from A we have $ \triangle DEF \sim \triangle GJH$. Thus, $ HJ \parallel EF$. Since BJHC is cyclic, and $ \angle BJH \equal{} \angle BEF$, we also have BCFE is cyclic. Thus, $ \triangle AEF \sim \triangle ABC$, so $ \frac {EF}{BC} \equal{} \frac {AF}{AC} \equal{} \frac {AE}{AB} \equal{} \frac {AF \plus{} AE}{AB \plus{} AC}$. But since $ \triangle AFD\sim \triangle ACD$ and $ \triangle AED\sim \triangle ABD$, $ AF \equal{} \frac {AD^2}{AC}$ and $ AE \equal{} \frac {AD^2}{AB}$. It follows that $ EF \equal{} \frac {BC(\frac {AD^2}{AB} \plus{} \frac {AD^2}{AC})}{AB \plus{} AC} \equal{} \frac {BC \cdot AD^2}{AB \cdot AC}$. But $ AD^2 \equal{} \frac {4[ABC]^2}{BC^2}$, so $ EF \equal{} \frac {4[ABC]^2}{AB \cdot BC \cdot CA}$, clearly constant.[/hide]
Is my proof correct, and does someone have a better one?

You can get rid of all the advanced techniques up until $ \triangle AEF \sim \triangle ABC$ using similar triangles. And then you can Law of Cosines for the answer.

Obviously $AEDF$ is cyclic; applying law of sines: $\frac{EF}{sin(\angle EAF)}=AD$, but $AD=\frac{2S}{BC}$, so $EF=\frac{2S\cdot sin\ \angle EAF}{BC}=\frac{S}{R}=ct$, where $S,R$ are the area and circumradius of $\triangle ABC$.

Best regards,
sunken rock


limit
integration
trigonometry
calculus
calculus computations

Sum the infinite series
\[ \frac{1}{11\cdot 13}\plus{}\frac{1}{23\cdot 25}\plus{}\frac{1}{35\cdot 37}\plus{}\frac{1}{47\cdot 49}\plus{}\frac{1}{59\cdot 61}\plus{}\dots\]

$ \sum_{k\equal{}1}^\infty \frac{1}{(12k\minus{}1)(12k\plus{}1)}\equal{}\lim_{x\to 1}\sum_{k\equal{}1}^\infty \frac{\int_0^x t^{12k\minus{}2}dt}{12k\plus{}1}$

$ \lim_{x\to 1}\sum_{k\equal{}1}^\infty  \int_0^x y\int_0^y t^{12k\minus{}2}dtdy\equal{}\lim_{x\to 1} \int_0^x y\int_0^y\left[ \sum_{k\equal{}1}^\infty t^{12k\minus{}2}\right]dtdy$

$ \equal{}\lim_{x\to 1} \int_0^x y\int_0^y\left[\frac{1}{t^2} \sum_{k\equal{}1}^\infty (t^{12})^k\right]dtdy\equal{}\lim_{x\to 1} \int_0^x y\int_0^y\frac{t^{10}}{1\minus{}t^{12}}dtdy$

which... i cant seem to integrate right now, but i remember a post about something similar?

but hey, i just handwaved the limit interchange, so maybe im going in the wrong direction altogether

[b]kenn4000[/b], try to get answer from representation $ \sin \theta \equal{} \theta\left(1\minus{}\frac{\theta^2}{\pi^2}\right)\left(1\minus{}\frac{\theta^2}{4\pi^2}\right)\left(1\minus{}\frac{\theta^2}{9\pi^2}\right)\dots\left(1\minus{}\frac{\theta^2}{n^2\pi^2}\right)\dots$


Hello:

I've gone through different sites and compiled their rankings on every player in the NFL.  I'm trying to make a master list to help me draft for my fantasy draft but I'm having a problem...

What I have is the rank for all quarterbacks for example.  On one site, Tom Brady is ranked 1, Peyton Manning 2, etc.  This site might rank the top 15.  Thus, I could give Tom Brady a "score" of 1/15th.

On another site though, they might have the same Tom Brady, Peyton Manning ranking but they might rank the top 100 players.  That way, Tom Brady now has a score of 1/15th and 1/100th.

How could I normalize the data to make one master list?  Any help would be greatly appreciated!

You could do it based on percentile.  So if you have a total of, say 250 players, you could group these players based on who is in the first percentile (first 3 or so players), second percentile (4-6), third percentile, and so on.


Source: 1996 AIME
For each permutation $a_1,a_2,a_3\dots,a_{10}$ of the integers 1,2,3,...,10, form the sum
$|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|$.
Find the average value of all of these sums.

Source:1983 AIME
For {1,2,3,....,n} and each of its nonempty subsets a unique alternating sum is defined as follows: Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately subtract and add successive numbers. (for example, the alternating sum for {1,2,4,6,9} is 9-6+4-2+1=6 and for {5} it is simply 5.) For n=7 find the sum of all alternating sums.

Source: 1995 AIME
Let $f(n)$ be the integer closest to $\sqrt[4]{n}$. Find $\displaystyle\sum_{k=1}^{1995}\frac{1}{f(k)}$.

I have no idea how to do the first and last one. The secon one i tried to find a patter by finding what it would be for n=1,2,and 3. But still am not able to solve it.
Can someone please help me.
thanks in advance.

The last one has been posted before.  The key is in integers close to $(k + \frac{1}{2})^4$ for $k$ an integer because those are the integers at which the value of $f(k)$ changes.

All these have been posted recently. Check out the [url=http://www.artofproblemsolving.com/Forum/resources.php]Resources[/url] page for the AIME problems and links to threads discussing them.


prove that there exists 16 subsets of the set$M=\{1,2,..,10^4\}$such that for every $a\in M$one can find 8 subsets whose the intersection is$\{a\}$

Put a 1 in eight subsets, put a 2 in a different set of eight subsets, etc...   Since $\binom{16}{8}>10^4$ we can continue this without repeating the same set of 8 subsets.  It is easy to check that the intersection of any 8 subsets can not contain more than one number.

Your solution isn't clear,i think either it's incorrect or you must post the complete solution.


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For the most part, I agree, but I think that there is still uncertainty and ambiguity in the wording; therefore any of the answer choices is a valid one. A single added word in the problem text would invalidate my claim, but that word was not included, and as a result, any of the given answer choices is correct.

To reiterate,
[quote="worthawholebean"]I think the intent of the problem is clear.[/quote]
You even agree with this, so what point do you have? :huh:

Acknowledged that the AMC could have provided a slightly better wording, but I don't think you have any ground on which to protest for score.

Without taking a position on your protest, I will note that a similar issue arose on the most recent AMC 8 (Question 12).  Namely, three answers were accepted because of the missing word "first".  See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249316[/url] for the problem statement and a little discussion.

If you are really interested in filing a protest, you should probably send an email message to the AMC, at the address listed on their website.

The test you should always apply in this scenario:  go with the interpretation that doesn't trivialize the problem.

I agree because I have a grasp of the practical English language.  I believe there are grounds for a protest because, if the question is held to any sort of standard beyond an "intended reading", then there is certainly an ambiguity with the wording that allows any of the choices to be correct.

Ravi: Yes, I have noted that incident and it certainly makes sense that the word "first" should be included in my opinion.

t0rajir0u: I understand your position, but at the same time, there is a clear ambiguity with a precedent already set for a similar case.

[quote="cipos2"]Being as technical as possible, "find" would be almost synonymous to discover, which would imply being first.  But if you still think the question is wrong, I encourage you to challenge it.[/quote]

This was my interpretation of the question, although you may have grounds for the challenge. Certainly if it works it would give me 6 points, but I do think the intent of the problem was clear.

[quote="dtl42"]I am posting this for my own sake and the sake of others.  It may be a foolish endeavor, but I believe it holds some validity.  Below is the text of the eleventh question of the 60th AMC 12 B (2009):

On Monday, Millie puts a quart of seeds, 25% of which are millet, into a bird feeder.  On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left.  Each day the birds eat only 25% of the millet in the feeder, but they eat all of the other seeds.  On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

My problem with the question is this, it does not say:
 “On which day, just after Mille has placed the seeds, will the birds FIRST find that more than half the seeds in the feeder are millet?”

This is very significant because it does not rule out any of the choices, and thus all of the answers are technically correct because it is not specified which week they lie in or whether they are the, for example, first Tuesday after the “Monday” referenced in the question.  I believe that any answer should be accepted for this question, as any of the given choices are, in fact correct so long as it is not specified that they are the Tuesday, Wednesday, Thursday, Friday, or Saturday immediately after the “Monday” referred to in the question.

To anyone who knows: How can I submit a formal protest?[/quote]

I think that the meaning of the question is clear. However, the current wording would make you correct. So I think it depends on AMC policy regarding issued like this.

I believe you would be justified in protesting this question. Although I agree that the problem writer probably meant "when will the birds FIRST find...", students should be expected to answer the question as it is written, not to divine the intent of the question's creator. I believe you have a good chance of succeeding in your protest, given the very similar scenario with the AMC 8 this year.

As for what someone mentioned with the word "find," that doesn't change anything; I can find a tree in my yard and still find it the next day, and the next, and so on.

[quote="ZzZzZzZzZzZz"]Look at it this way.  You are obviously arguing that the problem wasn't fair, which is reasonable since everyone is entitled to their own opinion.  If, say, you won your protest, there would obviously not be a correct answer to the question, which would help you but [b]harm[/b] other people who got it correct, which is clearly unfair to those individuals.[/quote]
How would anyone be harmed if the topic creator won his protest? The people who already answered the question according to its apparent intention wouldn't lose any points...

[quote="MrBob"]that would be like someone copying your paper on a test: your score wouldn't be harmed, but their's would be higher than it should be[/quote]
If the protest is valid (and no one has provided any legitimate reason to believe otherwise), then anyone who answered this question at all answered it correctly, and their score is not higher than it should be.

[quote="feuxfollets"]Oh really?  As I understood it, initially, 69.2% of the birds were allergic to millet seeds and 30.8% of the birds were allergic to other seeds.  In addition, because other-seeds are less nutritious than millet seeds, other-seed eaters suffered from poor vision.

When the 1/2 threshold was reached, the other-seed eaters began to have trouble differentiating between the two types of seeds, and as a result, some of them ingested the wrong seed and died.  Meanwhile, the millet-seed eaters enjoyed plenty of food and grew rapidly in population.  After some time, when enough of the other-seed eaters had died off, a large enough quantity of other-seeds had accumulated that the surviving other-seed eaters then began to enjoy a bountiful feast every day and produce more offspring, until eventually, the ratio of millet-seed eaters returned to the original ratio.  Meanwhile the millet-seed eaters now faced the faced the problem of a huge supply of other-seeds, and, unfortunately, some of them ingested the wrong seed and consequently suffered a painful death.  The same sequence of events that the other-seed eaters had just gone through then occurred, and the cycle then repeated itself.

After some calculations which this post is too small to contain, we find that this cycle has a period of 14 days.  Thus, for day within the first 7 days, the quantity of millet seeds is greater than one half of the total.  Therefore, any day from the Friday of this week to the Thursday of next week satisfies the condition.[/quote]
Your understanding clearly began with a miscalculation. The seeds started with 75% other-seeds, and 25% millet. Birds only eat 25% of the millet seeds. So if there are only birds that can only consume either millet or other-seeds, as you claim, and each bird consumes seeds at the same rate, then there should be 7.69% millet seed eaters as opposed to 92.31% other-seed eaters.

In order to properly solve the problem, there should not be miscalculations.

In addition, the problem did not imply that the other seeds were less nutritious. The fact that birds consumed the other seeds completely while only consuming 25% of the millet seeds shows that the millet seeds help the birds less. The trend ("Each day...") is said to have continued in the next few days, despite there being more and more millet seeds building up. This shows that there is a disadvantage in eating the millet seeds, not an advantage. The other-seed eaters should have no trouble differentiating between the two seeds, and since other-seeds are added at the same rate, their numbers shouldn't change.

However, like I stated earlier, once the 1/2 threshold is reached, they are repelled by the overwhelming amounts of millet seeds, and will stop eating from the feeder. Even if the millet-eaters stay like you claim, the number of millets will still drop compared to the number of other seeds to lower than one-half of all the seeds by the day after.

I also stated that they implied the days are on the same week. As a result, everything that happened after the first week is irrelevant. In fact, your description somewhat supports that Friday is the only choice in the first week.

To dtl42: I wasn't specifically targetting you (in fact, I was fully aware that you weren't doing this for the points) when I talked about those who wanted points for this question.  And if you truly didn't care, then this protest is unnecessary anyways.
I can see how much you've degraded your own topic.  It's quite ironic that right after I told feuxfollets that sarcasm does not work at all in an argument (and that is not an opinion, but a fact), you use it.  Pick whether you will continue this debate seriously or stupidly, not a mixture of both.
And I can tell that you're just siding with feuxfollets because he agrees with you.  Lightening up is the problem, is it?  Who's the one that started this whole thread?  I think barkashdanbar completely sums up how you're being: [quote]its just peevish whine, mann[/quote].
And a general word of advice for your future arguments: contradicting yourself over and over (and over) does not help your position at all.
worthawholebean, myself, and several others have tried to show you with reason just how anal and unreasonable you're being.  But obviously it has not worked.  So I'm going to stop posting on this topic until I can start to take you seriously.

ok mathandknowledge. it is interesting how you seem to be skewing and fabricating evidence to justify [i]your[/i] irrational claims.
You say that dtl42 has been contradicting himself over and over again. I'd like you to point to where.
Next, you say dtl42 doesn't even care. He obviously does. Stop making things up. Let's try to have a discussion without trying to discredit eachother, but rather about the protest of the problem.

[quote]And if you truly didn't care, then this protest is unnecessary anyways. [/quote]

Perhaps this is where the misunderstanding begins.  My logic behind the protest is not fueled by my own selfishness, my being "anal", or any other explanation you might come up with.  It stems from my seeing that many people have been tricked and wanting to help them.  It is a simple whim that I indulge, that, at the core, is selfless and well meaning.

[quote]And I can tell that you're just siding with feuxfollets because he agrees with you. [/quote]

Nice try. I sided with him because I like his cool avatar.

[quote]And a general word of advice for your future arguments: contradicting yourself over and over (and over) does not help your position at all. [/quote]

Proof by contradiction? Ever heard of it?  But seriously, how have I contradicted myself?

[quote]But obviously it has not worked. So I'm going to stop posting on this topic until I can start to take you seriously.[/quote]

Aww, don't go away, I like playing with you.  But seriously, after I comprehensively rebut you, and prod you for more apples (examples), which you failed to deliver (not even more oranges either!), you decide to start being quiet. Interesting.

people have forgot about the original reason this thread was made:
[quote]To anyone who knows: How can I submit a formal protest?[/quote]
dtl42: I think you can make a protest on this problem and should do, by posting this whole thread on to it.
everyone else: how about we continue to argue over the validity of this, but still [size=150]tell him how to submit a protest[/size]. I mean it's the people at MAA who decide. Isn't it?
[quote](2)-Anyone who would have made USAMO "normally" will have probably gotten this question correct

These two aspects of such a compromise suggest that the prestige of USAMO will not be affected greatly, as anyone who would have "normally" made USAMO will still qualify, and almost no "extra" people will qualify solely because of the result of this protest, so the number of qualifiers will be the same. Thus the only difference is essentially that (slightly) more people qualify for AIME. However, the same number of people will qualify for USAMO, and such a compromise would not "penalize" those who answered the question correctly.[/quote]
^
Agreed. It kind of reiterates what I said in my earlier post..

There is no official procedure for submitting protests, as far as I know.

[quote="MathAndKnowledge"]I was saying that because I (and probably you as well, though you act like you know nothing) have solved enough of these contest-style problems, I can see the difference between the bad wording in the AMC 8 question and in this AMC 10/12 question.[/quote]
There is no significant difference:

AMC 8, #12: On which bounce will [the ball] not rise to a height of 0.5 meters?
AMC 12B, #11: On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

See the exact correspondence between the two questions? The AMC 12 question has exactly the same problem: it does not say that the question wants the first day that more than half the seeds are millet. Students are left to guess this on their own. The AMC 8 problem doesn't ask for the first bounce on which the ball will not reach 0.5 meters, leaving students to guess that this was what the problem was looking for. What did you have in mind when you said that the AMC 8 problem was "dramatically different" from this one? You claim that the levels of ambiguity are different; do you have any reasoning to support this?

As I pointed out earlier, the original poster of this topic has diverged from being serious to being playfully provocative towards me (and particularly sarcastic and aggressive).  Both sides of this protest are now just starting to call each other names and claim that people are being unreasonable or irrational.
So why not just submit the protest and end this discussion.  We will see who has the better (and hopefully correct) argument based on the MAA's decision (I am not saying that the MAA is always right, but this is their contest so their decision on this protest is considered "right").

And as for you, MCTP, I thank you for actually being serious and trying to discuss this issue.  But I see now that it's too difficult to explain things this complicated without missaying something or being harshly attacked.  So, again, let's see what the MAA says and then we'll see if the difference that I speak of actually exists.  Or I could just PM you to further explain my reasoning (but certainly not on this topic, because of all the immature little kids posting out here).

Personally, I don't think you can be serious. The intent of the question was completely clear.

I can [i]guarantee[/i] that the author of this topic got the question wrong. And not because he was thinking "Hmm... It doesn't specify each week! Wow, all the choices must be correct!".

When people protest random technicallities like this, it becomes blatantly obvious that the person got the question wrong, and is trying to get extra points after rereading the test.

As for the AMC 8 example, I maintained this position as well. It was clear what the problem meant.

I agree that it is not the job of the students to determine the authors intent, but in this case it was extremely clear what was intended.

I fully disagree with the ruling on the AMC 8. The intent of the questions was clear.

Please dont' attempt to grab points off small technicalities. This isn't fair to people who legitimitely solve the question. If you really were smart, you would have known what the question meant.

~BOGTRO~

MathAndKnowledge is trolling as evidenced by his little jabs like "Or I could just PM you to further explain my reasoning (but certainly not on this topic, because of all the immature little kids posting out here)." Don't let him rile you up.

BOGTRO: The topic creator has no chance for USAMO and already qualified for AIME as he himself stated. Your presumptions are wrong.

To all the people saying the intent of the problem was clear to them, on what grounds should this protest be rejected and the AMC8 protest accepted?

Stanman's Big Six

1) You all probably think the problem reads: On which day, just after Millie has placed the seeds, will more than half the seeds in the feeder be millet?

But in actuality, it reads: On which day, just after Millie has placed the seeds, will [b]the birds find that[/b] more than half the seeds in the feeder are millet?

Generally, a [i]finding[/i] is considered to be a one-time deal. The 1/2 ratio was never lost between Friday and Saturday, so how can it be found again?

2) I can only find a couple reasons why someone would want to protest such a thing.

  a) You, or a friend, had no idea how to do this problem (or did it incorrectly) and are looking to snag points on a technicality
  b) You know what the test writers were intending to communicate to you, but are just being stubborn
  c) You sat in the test room panicked for 30 minutes trying to decipher this problem and wasted your time for the rest of the test

3) I, personally, did not even notice that the word [i]first[/i] was omitted.

4) At first, I did this problem incorrectly and came up with the answer of Saturday. There was no problem with any ambiguity, I was just being stupid and somehow ended up with Saturday. I eventually checked it over and corrected this mistake, but if you are in favor of allowing Friday and Saturday to be valid answers, please consider these scenarios as well and consider that we would be rewarding people who didn't know how to do the problem.

5) I agree with BOGTRO. Although the AMC 8 decision was reversed, one could argue that the ball bounced under 2 feet or whatever on the 6th bounce or whatnot. In this problem, though, no one could argue that the birds found (initially noticed) that the feeder was more than half millet on Saturday, although MCTP or dtl might say "What if the birds have their eyes closed on Friday and don't find it until Saturday?" [quote]This would require “mind reading” [/quote]

6) Ignite: Everyone that qualifies for AIME has a chance to make USAMO. Even me!

If you're going to criticize the problem for being too "ambiguous" you might as well give up math.  The problem was based on an extremely simple assumption, and by saying its wording is incorrect, you might as well say that every single problem in math is worded incorrectly.  Math is based on certain assumptions such as "every two points can be connected by a line."  If you really, truly did not understand this problem, maybe that's a problem with your reading skills, not the problem.

dtl42:

Stop wasting your time. It is quite obvious what the problem is asking for; your protest is basically saying that the problem was trivial because of wording. However, basic logis states that MAA will not put a trivial problem on the AMC.

[quote="Ignite168"]MathAndKnowledge is trolling as evidenced by his little jabs like "Or I could just PM you to further explain my reasoning (but certainly not on this topic, because of all the immature little kids posting out here)." Don't let him rile you up.[/quote]

Take a look at page 6.  Then tell me who is riling who up again.

I'm locking this. Everything that needed to be said has been said, and people are starting to get a little angry. If you have a compelling reason to reopen, PM me.

AMCDirector can post a new thread if and when he responds.


hi,
calculate
\[ \sum_{n \equal{} 1}^{\infty}{\frac {\varphi(n)}{2^{n} \minus{} 1}}.
\]

[b]Lemma:[/b]  If $ \sum a_n x^n \equal{} \sum b_n \frac{x^n}{1 \minus{} x^n}$, then $ a_n \equal{} \sum_{d | n} b_n$.

Let $ x \equal{} \frac{1}{2}$:  then $ a_n \equal{} \sum_{d | n} \phi(d) \equal{} n$, so the sum we want to find is

$ \sum \frac{n}{2^n} \equal{} 3$.


The numbers $1$ through $n^{2}$ are randomly arranged in a square with side length $n$. Prove some adjacent (touching horizontally, vertically, or diagonally) pair of squares differs by at least $n+1$.

[quote="Phelpedo"]1 is somewhere. $n^{2}$ is somewhere. They can be connected by a path of at most $n-1$. Hey lookie...[/quote]
I don't understand...

Their difference is $n^{2}-1$, and by Phelpedo's comment and PHP, some two adjacent squares differ by at least $\frac{n^{2}-1}{n-1}=n+1$

I am so not understanding, please type slowly and use small words...
EDIT: Ok I get it but I don't think it's very rigorous..

It's completely rigorous.  $1$ must indeed be somewhere.  $n^{2}$ must indeed be somewhere.  The rest is Pigeonhole.


geometry
trapezoid
circumcircle
symmetry
geometric transformation
reflection
trigonometry

AB is the diameter of circle O. CD||AB, AD intersect the tangent line through B at E. EG is perpendicular to AC.

Prove that  AC=CG.

Let the lines $AC, BE$ intersect at a point $F$. The quadrilateral $ABCD$ is an isosceles trapezoid with circumcenter at the midpoint $O$ of the longer base $AB$. Hence, the angles $\angle ACB = \angle ADB$ spanning the semicircle $AB$ are right. By symmetry of the isosceles trapezoid, the right angle triangles $\triangle ABC \cong \triangle BDA$ are congruent. Denote $\phi = \angle ABC = \angle BAD$, The angle $\angle CAD$ is then equal to

$\angle CAD = \angle BAC - \angle BAD = 90^o - 2\phi$

From the right angle triangles $\triangle ABE, \triangle AGE$,

$\angle AEB = 90^o - \angle BAD = 90^o - \phi$

$\angle AEG = 90^o - \angle CAD = 90^o - (90^o - 2\phi) = 2\phi$

$\angle GEF = 180^o - (\angle AEB + \angle AEG) = 180^o - (90^o - \phi + 2\phi) = 90^o - \phi$

As a result, the angles $\angle GEF = \angle AEB$ are equal, which means that the line $EG$ is a reflection of the line $AE$ in the line $BE$. Therefore, the line $GE$ passes through the reflection $A'$ of the point $A$ in the line $BE$. The point $A'$ is on the normal $AB$ to the line $BE$ and $A'B = AB$, i.e., $AA' = 2AB$. The right angle triangles $\triangle AA'G \sim \triangle ABC$ have the angle $\angle A'AG \equiv \angle BAC$ common. Hence, they are centrally similar with center of similarity at their common vertex $A$ and coefficient similarity $\frac{AA'}{AB} = 2$. Consequently,

$\frac{AG}{AC} = \frac{AA'}{AB} = 2,\ \  CG = AG - AC = 2AC - AC = AC$

Can you please explain the reflecting part?? I don't quite understand it...

Sure. The angles $\angle AEB, \angle GEF$ were calculated from congruence of some right angle triangles and from the fact that the angles $\angle AEB + \angle AEG + \angle GEF = 180^o$ add up to a straight angle. These 2 angles $\angle AEB = \angle GEF$ turned to be equal. But if you reflect the line $AE$ in the line $BE$, this is exactly what happens - like a reflection of light in a mirror, the incident angle of light rays is equal to their angle of reflection. Since the line $AE$ passes through the point $A$, the reflection of this line (the line $GE$) must pass through the reflection of the point $A$ (the point $A'$, such that $A'B$ = $AB$, where $AB \perp BE$).

Regards, Yetti.

Outline of trig method:

Let $\angle BAE = \theta$, WLOG the circle has diameter 1.

We have $BD = \sin \theta$, and

$AG = \sec \theta \cos (90-2 \theta ) = \sec \theta \sin 2 \theta = 2 \sin \theta \cos \theta \sec \theta = 2 \sin \theta$

Since BD = AC by symmetry, we have $AC = \frac{AG}{2} \implies AC = CG$.

[quote="shobber"]So actually, we can prove it by some simple geometry. [/quote]
Here is another way by simple geometry:

Notice that A,B,E,G are concyclic since $\angle{ABE}=\angle{AGE}=90^\circ$, so $\angle{AGB}=\angle{AEB}$.

We also have $\angle{GAB}=\angle{ABD}=\angle{AEB}$

Therefore, the triangle ABG is isosceles.

Since BC is perpendicular to AG, C must be the midpoint of AG

QED

[quote="probability1.01"]How did you know that ABG was isosceles (pretty sure it's untrue...)?[/quote]

I showed both $\angle{GAB}$ and $\angle{AGB}$ are equal to $\angle{AEB}$.

[quote="shobber"]So actually, we can prove it by some simple geometry.
Is this the basic knowledge of Reflection??[/quote]

[color=blue]A point $A'$ is symmetrical to the point $A$ with respect to a line $l$ (the axis of symmetry), if the segment $AA'$ is perpendicular to the line $l$ and cut in half by the line $l$.[/color]

This is a definition. Symmetry with respect to a line is also called reflection in a line. Let $k$ be a line intersecting the axis of symmetry $l$ at a point $B$ and let $A$ be a point on the line $k$ different from the point $B$. Let $A'$ be the reflection of the point $A$ in the line $l$ according to the above definition. The reflection of the point $B$ in the line $l$ is the point $B$ itself. The reflection $k' \equiv A'B$ of the line $k \equiv AB$ in the line $l$ passes through the 2 reflected points. The triangle $\triangle ABA'$ is isosceles with $AB = A'B$, because its altitude (the axis of symmetry $l$) cuts its base $AA'$ in half. Let $M$ be the midpoint of the segment $AA'$ on the line $l$. The triangles $\triangle ABM \cong A'BM$ by SAS ($AM = A'M, \angle AMB = \angle A'MB = 90^o, MB$ is common). Hence, the angles $\angle ABM = \angle A'BM$ are equal. There is another definition of symmetry with respect to a line (equivalent to the above definition, which means that these 2 definitions can be deduced from each other):

[color=blue]A point $A'$ is symmetrical to the point $A$ with respect to a line $l$, if every circle centered on the line $l$ and passing through the point $A$ simultaneously passes through the point $A'$.[/color]

The advantage of the 2nd definition is that it can be easily modified for symmetry with respect to a circle. Every circle centered on the line $l$ is perpendicular to this line. Hence:

[color=blue]A point $A'$ is symmetrical to the point $A$ with respect to a circle $c$, if every circle perpendicular to the circle $c$ and passing through the point $A$ simultaneously passes through the point $A'$.[/color]

Symmetry with respect to a circle is also called reflection in a circle or inversion in a circle.

this is simple:
we have $\angle ACO=\angle CAO=\angle DBA=\angle DEB=\angle BGC$ that means
$OC \parallel BG$ or $C$ is the midpoint of $AG$

[quote="shobber"]What does" a circle perpendicular to circle" mean?[/quote]

A circle $c$ is perpendicular to a line $l$ iff the tangent $t$ to the circle $c$ at the intersection $P$ of this circle with the line $l$ is perpendicular to the line $l$. This is equivalent to the circle $c$ being centered on the line $l$.

A circle $c_1$ is perpendicular to a circle $c_2$ iff the tangents $t_1, t_2$ to the circles $c_1, c_2$ at their intersection point $P$ are perpendicular to each other. This is equivalent to the radii $O_1P, O_2P$ of the circles $c_1, c_2$ from their respective centers $O_1, O_2$ to the intersection point $P$ being perpendicular to each other.

Yetti

Since AB is a diameter, $ CB \perp AG$. Let line BE intersect the parallel at A to BC at K. Since $ AK \parallel BC \parallel GE$, we have to show $ KB \equal{} BE$. But $ \angle KAB \equal{} \angle ABC \equal{} \angle BAD$ by symmetry, so $ \triangle ABK$ is congruent to $ \triangle ABE$, as they share side AB and have two equal angles. So $ KB \equal{} BE$.


there were 500 people on the train ride to London.

- There were 200 singles on the train ride....

the train crashed after 1 hour.. into a brick wall. Every single person died, but as the police counted all the deaths they wondered how many people may have survived the crash..

-----How many people survived the trip??? and who????

-soccerstar123

excellent :-)

that's like the easiest problem ever :-)

at my camp the camp instructor told us that problem and i was the only one at my table to figure out the problem... but he didn't add in the numbers... he said:

a train crashed in a wall, every single person died, who survived.....

but i thought it would be a great problem on aops :-)

-soccerstar123


A pizza parlor offers six toppings. What is the greatest number
of four-topping pizzas that can be made such that no two pizzas
have the same topping combination?

This is simply $ \binom{6}{4}\equal{}\frac{6!}{2!4!}\equal{}\boxed{15}$.

I agree with gaussintraining, because there are 6 toppings and you need just 4 toppings. Also, it's asking for combination not permutation, so it's 6 choose 4 or 15

If you agree to his port, you don't have to post a whole new post! Just rate his post. That is just practically spam.


limit
logarithms
symmetry
geometry
geometric transformation
function
integration

hello, the limit $ \lim_{n \to \infty}\frac{1}{k^2\plus{}k\plus{}1}\equal{}\frac{1}{k^2\plus{}k\plus{}1}$ or do you meant $ \lim_{k \to \infty}\frac{1}{k^2\plus{}k\plus{}1}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the limit $ \lim_{n \to \infty}\frac {1}{k^2 \plus{} k \plus{} 1} \equal{} \frac {1}{k^2 \plus{} k \plus{} 1}$ or do you meant $ \lim_{k \to \infty}\frac {1}{k^2 \plus{} k \plus{} 1}$?
Sonnhard.[/quote]

Dr.Sonnhard,sorry that  this is my typo.Just correct it.

[quote="Kent Merryfield"]My numerical estimate is $ 0.798147281.$ I don't at first glance see that as a number with a more easily remembered name.[/quote]

Have any detailed evaluation?

Nothing much to see, just a quick calculation with a spreadsheet: partial sums out to $ n\equal{}128,$ and then a Richardson extrapolation table from that. Then I took the resulting number and divided it by $ \pi,$ mutliplied it by $ \pi,$ took its logarithm and its exponential, all without noticing anything worth commenting on.

That the series converges is not in doubt, by the usual comparison tests. It doesn't telescope, and while its partial sums are rational, they have nasty denominators and don't simplify readily.

Now, if you tried $ \sum_{k\equal{}1}^{\infty}\arctan\left(\frac1{k^2\plus{}k\plus{}1}\right)$ there might be something to see.

[quote="Kent Merryfield"]Nothing much to see, just a quick calculation with a spreadsheet: partial sums out to $ n \equal{} 128,$ and then a Richardson extrapolation table from that. Then I took the resulting number and divided it by $ \pi,$ mutliplied it by $ \pi,$ took its logarithm and its exponential, all without noticing anything worth commenting on.

That the series converges is not in doubt, by the usual comparison tests. It doesn't telescope, and while its partial sums are rational, they have nasty denominators and don't simplify readily.

Now, if you tried $ \sum_{k \equal{} 1}^{\infty}\arctan\left(\frac1{k^2 \plus{} k \plus{} 1}\right)$ there might be something to see.[/quote]
Thanks.

hello, your sum is $ \sum_{k\equal{}1}^{\infty}\frac{1}{k^2\plus{}k\plus{}1}\equal{}\minus{}1\plus{}\frac{1}{3}\sqrt{3}\pi\tanh(\frac{1}{2}\sqrt{3}\pi)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your sum is $ \sum_{k \equal{} 1}^{\infty}\frac {1}{k^2 \plus{} k \plus{} 1} \equal{} \minus{} 1 \plus{} \frac {1}{3}\sqrt {3}\pi\tanh(\frac {1}{2}\sqrt {3}\pi)$.
Sonnhard.[/quote]

Would you offer deatils,please.

$ \arctan(x\plus{}1)\minus{}\arctan(x)\equal{}\arctan\left(\frac{1}{x^{2}\plus{}x\plus{}1}\right)$
Not at that hard!!! :D

[quote="Timestopper_STG"]$ \arctan(x \plus{} 1) \minus{} \arctan(x) \equal{} \arctan\left(\frac {1}{x^{2} \plus{} x \plus{} 1}\right)$
Not at that hard!!! :D[/quote]

Thanks very much!! :P

[quote="Timestopper_STG"]$ \arctan(x \plus{} 1) \minus{} \arctan(x) \equal{} \arctan\left(\frac {1}{x^{2} \plus{} x \plus{} 1}\right)$
Not at that hard!!! :D[/quote]

 $ If \sum_{k\equal{}1}^{\infty}\arctan\left(\frac {1}{k^{2} \plus{} k\plus{} 1}\right)$can be solved , what is the relationship between it and $ \sum_{k\equal{}1}^{\infty}\frac {1}{k^{2} \plus{} k \plus{} 1}$?

Don't look for any relationship between the two series.

Going back to the series that started this problem:

The $ -1$ in the answer that Dr. Sonnhard Graubner gave suggests that it would have been more natural all along to write this as $ \sum_{k=0}^{\infty}\frac1{k^2+k+1}.$ From there, note that $ k^2+k+1=(k+\frac12)^2+\frac34$ and is symmetric about $ k=-\frac12.$ Utilizing this symmetry, we can write
\[ \sum_{k=0}^{\infty}\frac1{k^2+k+1}= \frac12\sum_{k=-\infty}^{\infty} \frac1{(k+\frac12)^2+\frac34}\]
From here, I turn to Fourier analysis and the ideas surrounding the Poisson summation formula.

Using a form of the Fourier transform on $ \mathbb{R}$ in which the $ 2\pi$ is in the exponent, I know that the Fourier transform of $ e^{-2\pi|x|}$ is $ \frac{1}{\xi^2+1}.$ Dilating that and pushing some constants around, we get that the Fourier transform of $ \frac{2\pi}{\sqrt{3}}e^{-\pi\sqrt{3}|x|}$ is $ \frac1{\xi^2+\frac34}.$

Now multiply by an exponential, causing a translation of the Fourier transform:

If $ f(x)=\frac{2\pi}{\sqrt{3}}e^{-\pi ix}e^{-\pi\sqrt{3}|x|}$ then $ \widehat{f}(\xi)=\frac{1}{(\xi+\frac12)^2 +\frac34}=\frac{1}{\xi^2+\xi+1}.$

Define a new function $ g(x)=\sum_{j=-\infty}^{\infty}f(x+j).$ $ g$ is now a function periodic of period $ 1,$ and it can be represented as a Fourier series:
\[ g(x)=\sum_{k=-\infty}^{\infty}\widehat{f}(k)e^{2\pi ix}\]
Now put $ x=0$ into this:

$ g(0)=\sum_{k=-\infty}^{\infty}\widehat{f}(k)= \sum_{k=-\infty}^{\infty}\frac1{k^2+k+1}$

$ =\sum_{j=-\infty}^{\infty}f(j)=\frac{2\pi}{\sqrt{3}} \sum_{-\infty}^{\infty} e^{-\pi ij}e^{-\sqrt{3}\pi|j|}=\frac{2\pi}{\sqrt{3}} \sum_{j=-\infty}^{\infty} (-1)^je^{-\sqrt{3}\pi|j|}$

$ =\frac{2\pi}{\sqrt{3}} (2\sum_{j=0}^{\infty}(-1)^je^{-\sqrt{3}\pi j}-1)=\frac{2\pi}{\sqrt{3}}( \frac{2}{1+e^{-\sqrt{3}\pi}}-1)$

$ = \frac{2\pi}{\sqrt{3}}\cdot \frac{1-e^{-\sqrt{3}\pi}}{1+e^{-\sqrt{3}\pi}}$

Dividing by $ 2,$ we get that
\[ \boxed{\sum_{k=0}^{\infty}\frac1{k^2+k+1}= \frac{\pi}{\sqrt{3}}\cdot \frac{1-e^{-\sqrt{3}\pi}}{1+e^{-\sqrt{3}\pi}}}\]
Although the exact form differs slightly, this is in precise agreement with the earlier statement by Sonnhard.

$ \sum_{k\equal{}0}^{\plus{} \infty} \frac{1}{k^2 \plus{} k \plus{} 1} \equal{} \frac{1}{\sqrt{3}} \int_0^{\plus{} \infty} \frac{\sin\left(\frac{\sqrt{3}}{2} x\right)}{\sinh \left(\frac{x}{2}\right)}\,dx$

[quote="Kent Merryfield"][hide]Don't look for any relationship between the two series.

Going back to the series that started this problem:

The $ - 1$ in the answer that Dr. Sonnhard Graubner gave suggests that it would have been more natural all along to write this as $ \sum_{k = 0}^{\infty}\frac1{k^2 + k + 1}.$ From there, note that $ k^2 + k + 1 = (k + \frac12)^2 + \frac34$ and is symmetric about $ k = - \frac12.$ Utilizing this symmetry, we can write
\[ \sum_{k = 0}^{\infty}\frac1{k^2 + k + 1} = \frac12\sum_{k = - \infty}^{\infty} \frac1{(k + \frac12)^2 + \frac34}
\]
From here, I turn to Fourier analysis and the ideas surrounding the Poisson summation formula.

Using a form of the Fourier transform on $ \mathbb{R}$ in which the $ 2\pi$ is in the exponent, I know that the Fourier transform of $ e^{ - 2\pi|x|}$ is $ \frac {1}{\xi^2 + 1}.$ Dilating that and pushing some constants around, we get that the Fourier transform of $ \frac {2\pi}{\sqrt {3}}e^{ - \pi\sqrt {3}|x|}$ is $ \frac1{\xi^2 + \frac34}.$

Now multiply by an exponential, causing a translation of the Fourier transform:

If $ f(x) = \frac {2\pi}{\sqrt {3}}e^{ - \pi ix}e^{ - \pi\sqrt {3}|x|}$ then $ \widehat{f}(\xi) = \frac {1}{(\xi + \frac12)^2 + \frac34} = \frac {1}{\xi^2 + \xi + 1}.$

Define a new function $ g(x) = \sum_{j = - \infty}^{\infty}f(x + j).$ $ g$ is now a function periodic of period $ 1,$ and it can be represented as a Fourier series:
\[ g(x) = \sum_{k = - \infty}^{\infty}\widehat{f}(k)e^{2\pi ix}
\]
Now put $ x = 0$ into this:

$ g(0) = \sum_{k = - \infty}^{\infty}\widehat{f}(k) = \sum_{k = - \infty}^{\infty}\frac1{k^2 + k + 1}$

$ = \sum_{j = - \infty}^{\infty}f(j) = \frac {2\pi}{\sqrt {3}} \sum_{ - \infty}^{\infty} e^{ - \pi ij}e^{ - \sqrt {3}\pi|j|} = \frac {2\pi}{\sqrt {3}} \sum_{j = - \infty}^{\infty} ( - 1)^je^{ - \sqrt {3}\pi|j|}$

$ = \frac {2\pi}{\sqrt {3}} (2\sum_{j = 0}^{\infty}( - 1)^je^{ - \sqrt {3}\pi j} - 1) = \frac {2\pi}{\sqrt {3}}( \frac {2}{1 + e^{ - \sqrt {3}\pi}} - 1)$

$ = \frac {2\pi}{\sqrt {3}}\cdot \frac {1 - e^{ - \sqrt {3}\pi}}{1 + e^{ - \sqrt {3}\pi}}$

Dividing by $ 2,$ we get that
\[ \boxed{\sum_{k = 0}^{\infty}\frac1{k^2 + k + 1} = \frac {\pi}{\sqrt {3}}\cdot \frac {1 - e^{ - \sqrt {3}\pi}}{1 + e^{ - \sqrt {3}\pi}}}
\]
Although the exact form differs slightly, this is in precise agreement with the earlier statement by Sonnhard.[/hide][/quote]

Thanks for your concreate explanations :)

[quote="Carcul"]$ \sum_{k \equal{} 0}^{ \plus{} \infty} \frac {1}{k^2 \plus{} k \plus{} 1} \equal{} \frac {1}{\sqrt {3}} \int_0^{ \plus{} \infty} \frac {\sin\left(\frac {\sqrt {3}}{2} x\right)}{\sinh \left(\frac {x}{2}\right)}\,dx$[/quote]
Could you explain more in detail please :blush:


Is "23 is not divisible by 3" or "23 are not divisible by 3"??

"Is".  The word "twenty-three" is a singular quantity.

Thanks a lot!
When I type this sentence in Word, it says that "23 are not divisible by 3" :!:

Because it doesn't understand that you're speaking mathematics: if you were discussing 23 [i]of something[/i], the plural verb would be appropriate: "Thirty of my friends are going to college.  Twenty-three are happy about their choice, and the seven others are ecstatic!"  (Note the "of them" is implicit in the second sentence.)


geometry
angle bisector
geometry proposed

Let $ C_{1},A_{1},B_{1}$ be the midpoints of sides $ AB,BC,CA$ of a scalene triangle $ ABC$, respectively. Let $ A_{2}$ be the midpoint of the polygonal line path $ BAC$. Points $ B_{2}$ and $ C_{2}$ are analogously defined. Show that the segments $ A_{1}A_{2},B_{1}B_{2},C_{1}C_{2}$ are concurrent.

let $ D$ be a point on extend of $ AB$ from $ A$ and $ M$ be midpoint of $ BD$ cause$ AB\plus{}AA_{2}\equal{}\frac{BD}{2}$ hence $ AM\equal{}AA_{2}$ hance $ MA_{2}\parallel BC\parallel A_{1}M\parallel \text{angle bisector of}\angle A$ hence $ A_{1}A_{2}$ is angle bisector of $ \angle B_{1}A_{1}C_{1}$ hence they R concurrent.


Six arbitrary points are chosen in a rectangle $3$x$4$. Prove that there are two points among them whose distance is at most $\sqrt{5}$

I'm disappointed -- 5 circles of diameter $\sqrt 5$ don't seem to quite work.

[quote="lotrgreengrapes7926"][hide="Idea"] Contradiction: Divide the rectangle into 6 rectangles, $1\times2$ each. If there are 2 points in 1 rectangle, then they are less than $\sqrt5$ apart. Therefore, the points must be in separate rectangles. To be as far from each other as possible, the 4 points in the rectangles containing the corners of the $3\times2$ rectangle must be in the corners. Drawing circles of radius $\sqrt5$ from the corners, we can see that the remaining 2 points are at most 1 unit apart.  :maybe: Does this even make sense?[/hide][/quote]
I am afraid your solution does not include the proof that placing the points in corners is the most optimal way. You can didvide the rectangle in unit squares and put every one of 6 pts in one unit square, so that the squares containing the points are in chessboard order(this can be proved using the fact that no 2 pts lie in a 1X2 rectangle). Then you draw the circles with radius of $\sqrt{5}$ and prove after that one of the squares will be overlapped with parts of the circles. This can be proved by taking 5 squares taht form an X(with one square in centre, one at each corner), then draw the circle from each of the corners of the $X$ -- the points that are furthest from the centre of the centre square. And then you are done. :wink:


calculus integration algebra function domain logarithms topology

$\oint \frac{Log z}{z-a}$ along any curve (in this case a circle) enclosing the origin.
The two cases are if $a$ is in the circle or not.

If $a$ is in the circle and I make a branch cut such that Log z is continuous and defined in the domain (e.g. 0 and the negative real axis), will this affect my integral somehow? Because a point on the circle will then be removed, and the circle will not longer be a complete circle (same end point and start point). Maybe it will considered a "complete circle" in the new domain with the branch cut? If so I could use the Cauchy integral then, right?

And using the same branch cut if $a$ is outside I assume the cauchy integral theorem could be used.

If you're enclosing zero, there's no way to make sense of the integral. Unless your curve has winding number zero around the origin, you can't define the logarithm continuously on it.

What do you mean by winding?
So I can't just create a branch cut as I wish when it comes to integreals?
Is there anything I can do with the integral?

The winding number of a curve around a point represents how many times you go around the point; it's one for a simple closed curve enclosing the point. The integral you're looking at just isn't well-defined; we need a definition of the function on an open set containing the curve to make sense of an integral. For the logarithm and a circle around the origin, there's no such definition.

If our curve doesn't enclose the origin, we can define the integral, but the value will depend on which branch we choose.

If I define it as one winding ccw and let the curve be |z-1|=2
Are you saying that this integral is not defined?
edit : I know that it isn't defined in z = 0, but I was woundering if I was able to make a branch cut and what the consequences would be for that. Like if I said:
[i] I make a branch cut at re(z) <= 0. Then the integral is defined in this new domain and I could just use (circle is enclosing $a$) the cauchy integral theorem. and get $2i\pi Log(a)$[/i]
because im not sure of this reasoning is valid and if there is some other way.

Yes. The logarithm is not defined on that circle.

I made an edit to my last message.
I have a problem for which the numbers are similar to what I'm trying to decribe, and I know that the problem can be solved with the Cauchy integral theorem. That's why I'm into these cuts and woundering if that's the way to go. Because the problem can be solved and I'm trying to figure out how.

No, it can't. The Cauchy theorem can only be used if the function is defined and analytic on an open set containing the entire disk. There are ways to deal with isolated singularities, but anything with a branch cut isn't an isolated singularity. As I've said repeatedly, there's no way to make sense of this integral.

The integral can be interpreted by means of [url=http://en.wikipedia.org/wiki/Analytic_continuation]analytic continuation[/url]. Say, we pick a point $z_{0}\in\mathbb C\setminus \{0,a\}$ and choose one of possible values of $Log(z)/(z-a)$ at that point. If $C$ is a curve that starts at $z_{0}$ and does not pass through $0$ or $a$, then the values of $Log(z)/(z-a)$ on $C$ are uniquely determined by analytic continuation, as long as $C$ does not cross itself. If $C$ crosses itself or forms a loop, we can consider it as a parametrized curve, with $Log(z)/(z-a)$ being a well-defined function of the parameter.
  This should make it possible to integrate over $C$; however, the answer is likely to depend on the choice of $Log(z_{0})/(z_{0}-a)$. Since the integrals over curves homotopic in $\mathbb C\setminus\{0,a\}$ are equal, it suffices to consider circles $\{z\colon |z|=R\}$. It's natural to take $R$ as the starting point, with $Log(R)=\log R$ (i.e. a real number). With this convention, $\int_{|z|=R}Log(z) dz=\int_{0}^{2\pi}(\log R+it)iRe^{it}dt=2\pi i R$. The one with $z-a$ is harder, but should be manageable.

The problem with that is that it depends on where you start the integral as well as the branch choice. We get a continuum of values for the circle being discussed; it's not well-defined.

Hello everyone.  I'm new here and taking Complex Analysis this semester.  This is how I'd help the original poster analyze the integral:

The complex logarithm function is multi-valued:  You know about real functions of two variables as just surfaces above the x-y plane right?  They do not "fold" or cross over themselves.  Take for example f(x,y)=Sin[xy].  For each value of (x,y), the function has a single surface over that point.  This is not so for multiple-valued functions in the complex plane.  Surfaces of these functions are called "Riemann Surfaces" and fold over themselves.  Thus for a single value of z=x+iy, the real or imaginary component of f(z) has several or infinitely many values depending on how many times the Re and/or Im surfaces fold.  Now in the case of log(z), the imaginary surface overlaps every time you go around the origin.  So if I plotted Im[log[z]]=v(x,y)=arg(z) and let z go around 5 times, I'd generate a "cork-screw" surface that overlaps 5 times.

You can see this by plotting the imaginary component using Mathematica:

p1 = ParametricPlot3D[{r Cos[t], r Sin[t],
    t}, {t, -Pi, 5 Pi}, {r, 0.01, 1}, BoxRatios -> Automatic, AspectRatio -> 1]

That's the Riemann surface for Im[log[z]] around the origin 3 times.  Now, looking at that cork-screw, can you now understand when I attempt to integrate a function containing log(z) around the origin, the value will depend where I start and end on that cork-screw surface?  Thus an integral containing log[z] or other multi-valued functions around a "closed contour" is not well-defined.  Now, note I used lower-case log(z).  By convention, we define the Principle value of log(z) as that surface between arg=-pi and arg=pi.  That's defined with upper case L as Log[z].   Can you figure out what part of the cork-screw Im[log(z)] function that corresponds to Log[z]?  I didn't plot the real component of log[z].  It's not a folded surface.  Can you?

Now do this:

Suppose I ask you to define an integral that does what you'd like your notation to represent but to do so using the principle value Log[z] function.  That is, perform the integration using Log[z] about a circular  contour of radius 1, starting at -pi and ending at pi.  Do it numerically.  For a=3, I get approx. 1.807 i.


Find all positive integers, representable uniquely as \[\frac{x^{2}+y}{xy+1},\] where $x$ and $y$ are positive integers.

Let $ n \equal{} \frac {x^2 \plus{} y}{xy \plus{} 1} > 1$, then $ x(x \minus{} ny) \equal{} n \minus{} y$ or $ y \equal{} \frac {x^2 \minus{} n}{nx \minus{} 1}$. From last we get $ x > n$.
If $ y < n\to x > ny$ we get contadition. If $ y > n$ equation $ x(ny \minus{} x) \equal{} y \minus{} n,n < x < ny$ had not integer solution (as $ y\minus{}n\equal{}x(ny\minus{}x)\ge ny\minus{}1$ and thus $ 1\minus{}n\ge (n\minus{}1)y$).
Therefore if $ n > 1$ we get $ y \equal{} n,x \equal{} n^2$ unique solution. But $ n \equal{} 1$ had many solutions $ (x,y) \equal{} (1,1),(t,t \plus{} 1)$.

[quote="Rust"]Let $ n \equal{} \frac {x^2 \plus{} y}{xy \plus{} 1} > 1$, then $ x(x \minus{} ny) \equal{} n \minus{} y$ or $ y \equal{} \frac {x^2 \minus{} n}{nx \minus{} 1}$. From last we get $ x > n$.
[/quote]
Other method:
From : $ y \equal{} \frac {x^2 \minus{} n}{nx \minus{} 1}$
The equation has solution if and only if $ nx \minus{} 1|x^2 \minus{} n$
$ \Longleftrightarrow nx \minus{} 1|n^2(x^2 \minus{} n)$
$ \Longleftrightarrow nx \minus{} 1|n^3 \minus{} 1$
[u] Case 1[/u]:$ n > 1$
$ \Longrightarrow nx \minus{} 1 \equal{} \frac {n^3 \minus{} 1}{d}$
$ \Longrightarrow nx \equal{} \frac {n^3 \minus{} 1 \plus{} d}{d}$
Imply that $ n|d \minus{} 1$
If $ d\geq n \plus{} 1$ then $ x < n$ (tradition!)
So $ d \equal{} 1$
Imply that $ x \equal{} n^2,y \equal{} n$
[u]Case 2[/u] $ n \equal{} 1$
$ (y,x) \equal{} (t \plus{} 1,t)$

we can consider the discriminant we have
$ \frac {x^{2} \plus{} y}{xy \plus{} 1} \equal{} k \implies x^2 \plus{} y \minus{} kxy \minus{} k \equal{} 0$
$ \Delta \equal{} k^2y^2 \minus{} 4(y \minus{} k) \equal{} s^2$.
the case $ k \equal{} y$ is obvious.
we have two cases:
1-$ y > k$ which implies
$ k^2y^2 > s^2 \implies (ky \minus{} 1)^2 \ge s^2 \equal{} k^2y^2 \minus{} 4y \plus{} 4k \implies \minus{} 2ky \plus{} 1 \ge \minus{} 4y \plus{} 4k \implies 4y \minus{} 4k \plus{} 1 \ge 2yk$
the rest is done with inequality.
2-$ y < k$ which implies $ k^2y^2 < s^2 \implies (ky \plus{} 1)^2 \le s^2 \equal{} k^2y^2 \minus{} 4y \plus{} 4k \implies 4yk \plus{} 1 \le \minus{} 4y \plus{} 4k$.
the rest is also trivial
we r done!


function
logarithms
real analysis
real analysis unsolved

$x+4^{x}=3$
how can we find the exact solution?

You can't, at least with elementary functions. It is of course easy to find numerical approximations.

[quote="gunes"]It's not important that the method is elementary or not, is there a method to find the exact x ?[/quote]
Completely missing the point. jmerry didn't say "elementary method," he said "elementary function."

We already have an [i]exact[/i] name for this number: "the unique real number $x$ such that $x+4^{x}=3.$" (The existence of this number is guaranteed by the Intermediate Value Theorem; the uniqueness by some theorem in the Rolle/Mean Value family depending on your tastes in the proof.) Exact, but not very useful. (But useful enough in that it justifies attempting to approximate it.)

Clearly you want a better name for this number than that. So what would be "better"? There's where the notion of "elementary function" comes in: take a certain conventionally agreed upon list of functions: powers and roots, trigonometric functions and inverse trigonometric functions, logarithms and exponentials. An elementary function is any function that can be built from these pieces using a finite number of additions, subtractions, multiplications, divisions, and compositions.

I'm not sure I've ever heard the expression "elementary number," but there should be some notion for of the numbers that occur as values of elementary functions evaluated at rational numbers. For instance, $\pi$ would certainly count as an "elementary number." It's $4\arctan 1,$ among other things. And $e, \ln{28},$ and $\sqrt{2+\sqrt[3]{4+\sqrt[5]{6}}}$ would also be "elementary numbers."

But there are more real numbers than there are elementary numbers - or as I like to phrase it, there are more real numbers then there are numbers you know the names of.

The great wonder of the completeness of the real numbers is that you can discuss those numbers whose names you do not know, because it is possible to assert the convergence of some limit without [i]a priori[/i] knowing the limit. I quoted the Intermediate Value Theorem above - that's one of the more direct consequences of the completeness of the real numbers.

About 6 or 7 steps of Newton's method on a spreadsheet gives me $x\approx 0.624205136472531.$ As both jmerry and Extremal said, it's easy enough to do that.

$x$ can be expressed in terms of the Lambert $W$ function. The Lambert $W$ function is defined by the equation
\[W(x)\cdot e^{W(x)}= x. \]
(see  http://mathworld.wolfram.com/LambertW-Function.html )
It is easy to check that the solution of the equation
\[y+a e^{y}= b \]
is $y =-W(a e^{b})+b$. Your equation can be reduced to this one by setting $y = \ln 4 \, x$.

The expression for $x$ in terms of $W$ is [i]exact[/i] but $W$ is not an [i]elementary[/i] function so $x$ is not an "elementary number" (by the definition of Kent Merryfield).


calculus
inequalities
inequalities unsolved

I've shown (through calculus) that this inequality is equivalent to

For three arbitrary positive real numbers $ x,y,z$ such that $ x\plus{}y\plus{}z\plus{}4xyz\equal{}1$, prove that

$ xy\plus{}yz\plus{}zx \leq 1$.

Can anyone prove this?

[quote="qwerty414"]I've shown (through calculus) that this inequality is equivalent to

For three arbitrary positive real numbers $ x,y,z$ such that $ x \plus{} y \plus{} z \plus{} 4xyz \equal{} 1$, prove that

$ xy \plus{} yz \plus{} zx \leq 1$.

Can anyone prove this?[/quote]
$ xy\plus{}xz\plus{}yz\leq\frac{1}{3}(x\plus{}y\plus{}z)^2\leq\frac{1}{3}<1.$ :D

A more beautiful one:

Prove that $ \forall_{x,y,z \in R_{ \plus{} }}$ satisfying the condition
$ 3(a \plus{} b \plus{} c \plus{} d)^3 \plus{} 16(abc \plus{} bcd \plus{} cda \plus{} dab) \equal{} 64(a \plus{} b \plus{} c \plus{} d)$
the following inequality holds:
$ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \leq 6$ :wink:

denote that $ f(a,b,c,d) \equal{} ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd$
since if we replace a by b and b by a , the inequality still remain the same so without lossing our generation we can
assume that $ a \equal{} max{a,b,c,d}$ and $ b \equal{} max{b,c,d}$
we will prove that
$ f(a,b,c,d) \minus{} f(a,\frac {8}{3} \minus{} a,0,0) \leq 0$
it is left to prove
$ ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd \minus{} a(\frac {8}{3} \minus{} a) \leq 0$
or $ a^2 \plus{} ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd \minus{} a(a \plus{} b \plus{} c \plus{} d \plus{} \frac {4}{3}(abc \plus{} bcd \plus{} cda \plus{} dab)) \leq 0$
$ bc \plus{} cd \plus{} bd\leq \frac {4}{3}(a^2bc \plus{} abcd \plus{} a^2cd \plus{} a^2bd)$
or $ \frac {4}{3}(a^2 \minus{} 1)(bc \plus{} cd \plus{} bd) \plus{} \frac {4}{3}abcd \geq 0$
it suffices to prove that $ a \geq 1$ with $ a \equal{} max{a,b,c,d}$
assume that a < 1 then $ 8 \equal{} 3(a \plus{} b \plus{} c \plus{} d) \plus{} 4(abc \plus{} bcd \plus{} cda \plus{} dab) < 3 \plus{} 4 \equal{} 7$ which is not true , which means $ a \geq 1$ , so we are done

It looks good. But any more nice solutions?

[quote="swensen"]Let $ a,b,c,d$ be positive real numbers such that

$ 3(a \plus{} b \plus{} c \plus{} d) \plus{} 4(abc \plus{} bcd \plus{} cda \plus{} dab) \equal{} 8$.

Prove that $ ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd\leq 2$.[/quote]
By Rolle's theorem, there exist $ x,y,z \ge 0$ such that
\[ 3(a \plus{} b \plus{} c \plus{} d) \equal{} 4(x \plus{} y \plus{} z)
\]

\[ abc \plus{} bcd \plus{} cda \plus{} dab \equal{} 4xyz
\]

\[ ab \plus{} a \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 2(xy \plus{} yz \plus{} zx)
\]
Then we get
\[ x \plus{} y \plus{} z \plus{} 4xyz \equal{} 2
\]
and we need to prove
\[ xy \plus{} yz \plus{} zx \le 1
\]
We can prove it by $ pqr$ technique like this $ p \equal{} x \plus{} y \plus{} z,q \equal{} xy \plus{} yz \plus{} zx,r \equal{} xyz$ then from Schur's Inequality for third degree, we get
\[ r \ge \frac {p(4q \minus{} p^2)}{9}
\]

\[ \Rightarrow p \plus{} \frac {4p(4q \minus{} p^2)}{9} \le 2
\]

\[ \Leftrightarrow 4q \minus{} p^2 \le \frac {9(2 \minus{} p)}{4p}
\]

\[ \Leftrightarrow 4q \le p^2 \plus{} \frac {9(2 \minus{} p)}{4p}
\]
Notice that from $ p \plus{} 4r \equal{} 2$, we get $ 2 \ge p \ge \frac {3}{2}$,We will show that
\[ p^2 \plus{} \frac {9(2 \minus{} p)}{4p} \le 4
\]

\[ \Leftrightarrow 9(2 \minus{} p) \le 4p(4 \minus{} p^2)
\]

\[ \Leftrightarrow 9 \le 4p(2 \plus{} p)
\]
which is trivial since $ p \ge \frac {3}{2}$, thus
\[ 4q \le p^2 \plus{} \frac {9(2 \minus{} p)}{4p} \le 4
\]

\[ \Leftrightarrow q \le 1
\]
We have done. :)

[quote="can_hang2007"]
By Rolle's theorem, [/quote]

This is quite advanced. Any easier?

Thanks anyway  :)

That problem (2nd) was also on polish board (matematyka.pl), and there was also an answer:

[quote="King James"]$ 64(a \plus{} b \plus{} c \plus{} d)(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd) \leqslant$
$ 18(a \plus{} b \plus{} c \plus{} d)^3 \plus{} 96(abc \plus{} bcd \plus{} cda \plus{} dab)$
$ \iff$

$ 32(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2d \plus{} dc^2 \plus{} d^2a \plus{} da^2 \plus{} a^2c \plus{} ac^2 \plus{} b^2d \plus{} bd^2$
$ \plus{} 3abc \plus{} 3bcd \plus{} 3cda \plus{} 3dab) \leqslant 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 102(abc \plus{} bcd \plus{} cda \plus{} dab)$
$ \plus{} 27(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2d \plus{} dc^2 \plus{} d^2a \plus{} da^2 \plus{} a^2c \plus{} ac^2 \plus{} b^2d \plus{} bd^2)$
$ \iff$

$ 5(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2d \plus{} dc^2 \plus{} d^2a \plus{} da^2 \plus{} a^2c \plus{} ac^2 \plus{} b^2d \plus{} bd^2) \leqslant 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 6(abc \plus{} bcd \plus{} cda \plus{} dab)$

For x,y,z>=0 we have:

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz \geqslant xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x) \ \ (*)$

Proof: WLOG: $ \min(x,y,z) \equal{} x$ and:

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz \minus{} xy(x \plus{} y) \minus{} yz(y \plus{} z) \minus{} zx(z \plus{} x) \equal{}$
$ (y \minus{} z)^2(y \plus{} z \minus{} x) \plus{} x(z \minus{} x)(y \minus{} x)\geqslant 0$

Taking inequality (*) for sequences (a,b,c), (b,c,d), (c,d,a), (d,a,b) and sum them up:

$ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} cda \plus{} dab) \geqslant$
$ 2(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2d \plus{} dc^2 \plus{} d^2a \plus{} da^2 \plus{} a^2c \plus{} ac^2 \plus{} b^2d \plus{} bd^2)$

So we need to prove:

$ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3)\geqslant a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2d \plus{} dc^2 \plus{} d^2a \plus{} da^2$
$ \plus{} a^2c \plus{} ac^2 \plus{} b^2d \plus{} bd^2$
$ \iff$
$ (a \plus{} b)(a \minus{} b)^2 \plus{} (b \plus{} c)(b \minus{} c)^2 \plus{} (c \plus{} d)(c \minus{} d)^2 \plus{} (d \plus{} a)(d \minus{} a)^2$
$ \plus{} (a \plus{} c)(a \minus{} c)^2 \plus{} (b \plus{} d)(b \minus{} d)^2 \geqslant 0$

which is true[/quote]

See here for a generalization:
http://www.mathlinks.ro/Forum/viewtopic.php?t=3463


The nitrate of an unknown metal is dissolved in water and electrolysed. The unknown metal is deposited on the cathode, which increases in mass by $ 1.79g$ while the copper anode loses $ 6.34g$. What is the metal?

$ A) Sr$
$ B) Ge$
$ C) Al$
$ D) Na$
$ E) Cu$

[hide="Anwer"]The quantity of copper that reacted is $ n(Cu) = \frac{6.34 g}{63.546 g/mol}= 0.0998 mol$.

From the equation for the copper oxidation, $ Cu(s) \longrightarrow Cu^{2+}(aq)+2e^{-}$, we now that each mole of copper supplies 2 moles of electrons. Then, in our case a total of $ 0.0998 \times 2 = 0.1996 mol$ of electrons were supplied.
The metal cannot be copper, since then the mass increase of the cathode would be the same as the mass decrease of copper anode (and because in solution Cu(I) is unstable with respect to disproportionation). The reduction of Sr requires two moles of electrons:

$ Sr^{2+}(aq)+2e^{-}\longrightarrow Sr(s)$,

and so for 0.1996 moles of electrons would produce $ \frac{0.1996}{2}= 0.0998$ moles of Sr, which is equivalent to 8.74 g of Sr. So A) can be eliminated. Doing similar calculations, we may conclude that C) and D) can also be eliminated. So the answer seems to be Germanium.

[/hide]


algorithm
number theory
Euclidean algorithm

Say $S_n(j)=\displaystyle\sum^{n-1}_{i=0}\left(2^i\right)^j$. For examply, $S_2(j)=\displaystyle\sum^{1}_{i=0}\left(2^i\right)^j=2^j + 1$ and $S_3(j)=\displaystyle\sum^{2}_{i=0}\left(2^i\right)^j=4^j + 2^j + 1$, and from there, I think it makes sense. Anyways, prove, for all $n\in N$ and $n\geq 2$, that $S_n(j)$ can't be prime unless $j$ is a power of $n$. (Disclaimer: You DON'T need to prove that $S_n(j)$ IS prime if $j$ IS a power of $n$, just that it isn't prime if $j$ isn't a power of $n$.(Disclaimer on the Disclaimer: That may be obvious to most, but not to all.:lol:))

I don't understand siuhochung's solution. could someone please clarify it? I think it's missing some details that would make it more complicated...

I have a different solution, which is quite simple, but involves cyclotomic polynomials.

[quote="siuhochung"]Write $S_n=\frac{2^{nj}-1}{2^j-1}$.
Write $j=n^k y$ so that $(y,n) < n$, then the numerator has a factor $2^{n^{k+1}}-1$, which forces $n|y$, contradiction.

To detail the solution better, one needs the lemma $(2^a -1, 2^b -1) = 2^{(a,b)}-1$. Let $a=qb+r$, then $(2^a-1, 2^b-1) =(2^a-1, 2^a-2^b) = (2^a-1,2^{a-b}-1)= \cdots =(2^a-1, 2^r-1)$ Then repeat the process like euclidean algorithm does and this finishes the proof.[/quote]

When we write $j=n^k y$ so that y is not a multiple of n, the numerator equals $2^{n^{k+1} y}-1$ and it obviously has a factor $2^{n^{k+1}}-1$. If $(2^j-1, 2^{n^{k+1}}-1) < 2^{n^{k+1}}-1$, then there are at least two divisors in the numerator after simplification, so it is not a prime anymore.

Thus we must have ${(2^j-1, 2^{n^{k+1}}-1) = 2^{n^{k+1}}-1}$. However, it is well known that $(2^a -1, 2^b -1) = 2^{(a,b)}-1$. So we have $(j, n^{k+1})=n^{k+1}$, which means $j$ is a multiple of $n^{k+1}$. However, $j=n^k y $ and y is not a multiple of n so this is impossible. Contradiction.

Is it clear now?


ahhhh its coming up
predictions/comments/interesting insights anyone?

Um...random, but three people on my team (including me) are staying at the same different hotel. (oxymoronic lol)

uhhh not really oxymoronic because one adjective applies to the other
oh well
i predict top ten will be stevenmeow, budi713, ra5249, distracted523,  dingzhou, hamt30, Lawrence Wu, Holymushrooms, a_k_ilic, michael jin (don't know his username), and random tenth person (policecap?)

budi713 isn't in mathcounts anymore. He's in 9th grade. He was in 8th last year.

uhhh wtf
budi713 is like my best buddy
he is in 8th grade
he was 7th grader last year

Are you posting in your predicted ranking? I'd be really scared if I made fourth written. I assume budi=Kevin Li?

what i implied was that budi sounds like buddy and isn't ra5249, kevin li?

budi is kevin li
ra (me) is kevin tian
and that isn't the predicted ranking, just the predicted top ten
sorry for the misunderstanding

really? cause it seems like it..... :maybe:


Find the volume of the parellelpiped defined by the vectors:

$ \minus{}5i\plus{}0j\minus{}5k$, $ i\minus{}4j\minus{}4k$, and $ i\minus{}5j\plus{}3k$

[hide="HINT"]
Volume of a parallelepiped is
$ | \vec{x} \left ( \vec{y} \times \vec{z} \right) |$
 or
$ \left| det \left( \begin{array}{ccc} x_{i} & x_{j} & x_{z} \\ y_{i} & y_{j} & y_{z} \\ z_{i} & z_{j} & z_{z} \end{array} \right) \right|$.
[/hide]

that's what i tried but for some reason i got $ 35$ but that is not correct


inequalities
trigonometry
function
real analysis
real analysis unsolved

I get a conjecture on Sin-ineq. which will be stronger than the Virgil' left inequality if ture:
[See  http://www.mathlinks.ro/Forum/viewtopic.php?t=70753 ]

(Sin-inequality)Conjecture:

Let $0<x<\frac{\pi}{2}$,then
\[ \frac{\frac{sinx}{x}-\frac{2}{\pi}}{cosx}>1-\frac{2}{\pi}+(\frac{1}{3}-\frac{1}{\pi})\cdot{x^2}  \]

Who can prove or disprove?[/quote]

NO one can answer me, Virgil, can you  try(to answer)?   :rotfl:


Let $ \Gamma$ be a circle, $ A$, $ B$, $ C$, $ D$ and $ E$ five points on this circle. Let $ G_1$ be the centroïd of $ ABC$ and $ d_1$ the straight line perpendicular to
(DE) passing through  $ G_1$. Let $ G_2$ be the centroïd of $ ABD$ and $ d_2$ the straight line perpendicular to (CE) passing through  $ G_2$, etc...

I want to show that the ten lines $ d_{i}$ are concurrent.

Thanks in advance for some hint.

Denote $ G'$ the centroid of $ \Delta CDE, G$ the centroid of $ ABCDE, L,K$ the midpoints of $ EC, AB$, respectively, $ H$ the intersection of $ OG$ and the line which passes through $ G'$ and  perpendicular to $ AB, J$ the centroid of $ ABCD$.
Obviously, $ J$ is the midpoint of $ KL$. We will show that $ D,G,J$ are collinear.
We have $ \vec{DJ}\equal{}\frac{1}{4}(\vec{DA}\plus{}\vec{DB}\plus{}\vec{DC}\plus{}\vec{DE})$
$ \equal{}\frac{1}{4}(4\vec{DG}\plus{}\vec{GA}\plus{}\vec{GB}\plus{}\vec{GC}\plus{}\vec{GE})\equal{}\vec{DG}\plus{}\frac{1}{4}\vec{DG}\equal{}\frac{5}{4}\vec{DG}$
Therefore $ D,J,G$ are collinear.
Applying Menelaus's theorem for triangle $ KLG'$ with line $ (J,G,D)$ we get $ \frac{KG}{GG'}\equal{}\frac{3}{2}$. But $ OK//G'H$ then $ \frac{OG}{GH}\equal{}\frac{KG}{GG'}\equal{}\frac{3}{2}.$
Similarly, other lines $ d_i$ pass through $ H$. We are done.

Let the center of the circle be the origin, and take $ X$ so that $ \vec{X} = (\vec{A} + \vec{B} + \vec{C} + \vec{D} + \vec{E})/3$.  Then $ \overrightarrow{G_1 X} = \vec{X} - \vec{G}_1 = (\vec{D} + \vec{E})/3$, and
\[ \overrightarrow{G_1 X} \cdot \overrightarrow{DE} = \frac{\vec{D} + \vec{E}}{3} \cdot (\ved{E} - \vec{D}) = |\vec{E}|^2 - |\vec{D}|^2 = 0.\]
Hence, $ G_1 X$ is perpendicular to $ DE$, i.e. $ X$ lies on $ d_1$.  By symmetry, $ X$ lies on all ten lines $ d_i$.


geometry
3D geometry
tetrahedron
sphere
IMO
IMO 1966

Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.

[hide]$ A(-a,-a,-a),B(-a,a,a),C(a,-a,a),D(a,a,-a),O(0,0,0),X(x,y,z)$

$ AO+BO+CO+DO=4a\sqrt3$

AM-QM:\begin{align*}AX+BX+CX+DX&=2\sqrt{AX^2+BX^2+CX^2+DX^2}\\&=2\sqrt{4(x^2+y^2+z^2)+12a^2}\\&\ge4a\sqrt3\end{align*}

$ \text{Equality holds}\Leftrightarrow x^2+y^2+z^2=0\Leftrightarrow x=y=z=0\Leftrightarrow X=O$[/hide]

[quote="matematikros"][hide]$ A(\minus{}a,\minus{}a,\minus{}a),B(\minus{}a,a,a),C(a,\minus{}a,a),D(a,a,\minus{}a),O(0,0,0),X(x,y,z)$

..... [/quote]
[img]http://img37.picoodle.com/i55i/duytuan119/id_53a_u9n46.jpg[/img]
This explanation does not seem right. because, subject to all of this tetrahedron $ABCD$ is [color=#FF0000]regular tetrahedron[/color].
 With your solution, just a [i][u]special case [/u][/i]of the tetrahedron $ABCD$, [u][b]not true for the general[/b][/u]:$MA+MB+MC+MD \ge 4R$.

[quote=matematikros][hide]AM-QM:\begin{align*}AX+BX+CX+DX&=2\sqrt{AX^2+BX^2+CX^2+DX^2}\end{align*}[/hide][/quote]

Why does this hold??

because quadratic mean greater than or equal to algebric mean

But I wonder why he make equivalent of AM with QM

I'm pretty sure the above solution is incorrect, since AM<=QM, not the other way around.

Let $ABCD$ be the tetrahedron and $P$ be any point in space. We want to find the optimal point $P$. We let $P$ vary on the plane passing through it parallel to $ABC$. Then obviously, the distance $PD$ is minimal when $P$ lies on line $DO$, where $O$ is the centroid of the tetrahedron. We claim that $PA+PB+PC$ is then minimal as well, which will finish the problem (since then $P$ analogously lies on $BO$, so $P=O$).

To show this, we now fix $PC$, so that $P$ varies on a fixed circle, the center of which is equidistant from $A,B$. Now consider the set of points $M$ for which $AM+BM$ is fixed. This is an ellipsoid and thus, a convex set. We now consider growing ellipsoids and the minimum $AP+BP$ for $P$ on the circle is obviously achieved for the point where the ellipsoid is first tangent to the circle. Assume that this point does not lie symmetrically with the plane $\mathcal P_{CDO}$. Then we could reflect the minimum across this plane. By symmetry, this point has the same sum of distances. By convexity, their midpoint lies inside the ellipsoid and thus, it has a smaller sum of distances. However, the point where $\mathcal P_{CDO}$ intersects the circle (the intersection point closer to $A,B$), is obviously closer to $A,B$ than this midpoint. Hence, we conclude $O$ is indeed the desired point.


Man, that gets crazy trying to keep everything striaght and understandable, this is why I love math, and hate (verbal) language. It's fairly simple once you sort through all the words, ect. Anyway, Ann's 21 and Roy's 17, right?

bookworm271828 wrote:3 years ago Ann was 1/2 again as old as Roy was when Ann was a year younger than Roy is now.  When Roy is 5 years older than Ann will be when Roy is as old as Ann is now Ann will be twice as old as Roy is now.

How old is Roy? Ann?

Spoiler (highlight the blank spot and all shall be reveal'd to thee!)

[hide]Let R equal Roy's age and R+X equal Ann's age.

From the second statement,

When Roy is as old as Ann is now, X years will have passed.

So Ann will be R+2X at that point.

In order for Roy to be 5 years older than Ann X years from now, Roy will be R+2X+5; that is to say, 2X+5 years will have passed.

At this point, Ann will be (R+X)+2X+5 = R+3X+5

Since this age is twice Roy's age right now,

R+3X+5 = 2R

R = 3X+5.

So right now, Ann is R+X = (3X+5)+X = 4X+5 and Roy is 3X+5.

Now, from the first statement, 3 years ago,

Ann was 4X+5-3, which equalled 1.5*[Roy X+1 years ago]

So 4X+5-3 = 1.5[3X+5 - (X+1)]

4X+2 = 1.5(2X+4)

4X+2 = 3X+6

X = 4

So right now, Ann is 4X+5 = 21 and Roy is 3X+5 = 17.[/hide]

I concur, just keep everything organized on paper (or meditate like me :) ) and you should be all set.


[color=darkblue][b]Construct (with a compasses and a line) the position of the point[/b] $X$ [b]which belongs to a given segment[/b] $AB$ [b]so that[/b] $AX^{2}=AB\cdot XB$.[/color]

[hide="i found this in our text book"]construct $BD\perp AB$. take the point $C$ on $BD$ such that $BC=\frac{AB}{2}$. take $E$ on $AC$ such that $CE=BC$. take $X$ on $AB$ such that $AX=AE$. then $X$ is our desired point.[/hide]

[quote="nayel"][hide="i found this in our text book"]construct $BD\perp AB$. take the point $C\in BD$ such that $BC=\frac{AB}{2}$. take $E\in AC$ such that $\boxed{AE=BC}$. take $X\in AB$ such that $AX=AE$. then $X$ is our desired point.[/hide][/quote]
$\left\{\begin{array}{c}BC=\frac{AB}{2}\\\ AE=BC\\\ AX=AE\end{array}\right\|$ $\implies$ $AX=\frac{AB}{2}$, i.e. the point $X$ is the middlepoint of the segment $AB$ ?!

I think $E\in AC$ so that $\boxed{CE=BC}$. Indeed, for $AB=2$ we have $BC=EC=1$, $AC=\sqrt 5$, $AX=AE=AC-EC=\sqrt 5-1$,

$XB=AB-AX=2-\left(\sqrt 5-1\right)=3-\sqrt 5$ and $AX^{2}=AB\cdot XB$ $\Longleftrightarrow$ $\left(\sqrt 5-1\right)^{2}=2\cdot \left(3-\sqrt 5\right)$, what is truly.

sorry it was just a typo :blush: . i've corrected it.


There is a cylindrical shell with radius R and Length H.Also there is a magnetic field of strength B in the direction of cylinder axis.N number of Electrons move in the radial direction(perpendicular to the plane of cylinder axis and B) with velocity $v$ .What will be the total pressure produced by electrons on the cylindrical shell.$\frac{mv}{Be}<<R$

Do you mean that we have a distribution $N(v)$, or we're just concerned with $N$ electrons with velocity $v$?

Just concerned N electrons with velocity v.


Use the laws of exponents to solve:
$(a^{b})^{c}=a^{bc}$
$(ab)^{c}=a^{c}*b^{c}$
$\frac{a^{x}}{a^{y}}=a^{x-y}$

If you are unfamiliar with $\LaTeX$, use this site's Latex guide,
([url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]Here[/url]; you don't have to download any software to use Latex on AoPS)
 or hold your mouse over other users' Latex to see their code.