[color=blue]"It was news – big news! A professor at Carthsinge College was found murdered in his bed – with a dagger struck in his heart! Everybody in Carthsinge knew the good professor – he had invented some thingermajigit for the poor starving children in Africa or something. Anyways, he was quite rich before his death, and now all of the money he had stored in his house was gone! The media was quite excited by this unexpected murder. Especially since the main police force was out in Africa looking for the Masks, instead of being here to look for the murderers. The media shouted:
“So who’s going to look into this case?”
Interesting enough, a band of twelve individuals have volunteered to go to the residence of the late Professor Grant and investigate this murder. Including the Mayor! How exciting! However, these twelve are not just any random bunch of people. They have doctors, investigators, policemen, and even people with a criminal record of murder! My, my, how interesting!" [/color]
---------------------------

Welcome to List Mafia! Role PM's are being sent out now. Please confrim that you have recieved the PM by responding.

Here are the rules. Please note that you must post once every two days, else I'm going to replace you. If you cannot, please note so in thread:

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

Normal Rules
(Edited by Me, stolen from LFR, stolen from Mr.Stoofer)

[01] Votes must be in bold. If you do not bold your vote, it will not be counted.
[02] Please be attentive and unvote, if necessary, before casting a new vote. This is not required, but I’d appreciate it.
[03] Lynching will require a simple majority of votes. Once a player has reached the necessary majority, their pleas are useless and any attempts to unvote will be unheeded. Feel free to post in twilight, though.
[04] You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.
[05] If a deadline is reached with no majority, the day will end without a lynch.
[06] The game is not to be discussed outside the thread unless your role specifically states that you may do so – and then only at night.
[07] When I set a day deadline, the deadline lasts until I actually lock the thread.
[08] Once your death scene has been posted, you’re dead. Stop typing. Not even a “Bah!” post, please. Save it for after the game is over.
[09] No small or invisible text. Don’t edit/delete previously submitted posts either.
[10] Don’t quote any PMs from me. If you do, I will destroy you.
[11] If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, I’ll make it for you and you won’t like it.
[12] If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread. Treat this game as a commitment. Be considerate – don’t leave us hanging.
[13] Unlike other mods I am fallible. Please point out any mistakes I make gently. In particular, if I post a vote count that does not accurately reflect your vote, it is your responsibility to correct me. If you don’t your vote may be treated as it appears in the vote count.
[14] If you have any questions at all, please PM me or post the question in the thread.
[15] Remember, it’s a game and it’s supposed to be fun. Be nice to your fellow players and your mod. I have zero tolerance for abusive posts.
[16] Roleplaying is strongly encouraged, but not required.
[17] The key to this game is to pay attention. So do it!
[Quote=”Town Win Condition”]You win when all of the murderers of Professor Grant are brought to justice.[/Quote]

[b][u]The Still Living:[/u][/b]

[b][color=red] Math_Explorer (Perfect628), Sibel Ivelin, Mafia Partner[/color][/b]
[b][color=red] Peter, Fenno Arxin, Mafia Partner[/color][/b]
[b][color=red] Zapi2007, Marik Durth, Mafia Partner[/color][/b]

[u][b]The Dead:[/b][/u]

[b][color=blue]Distracted523, Alyss Board, Town Partner, shot on accident, MODKILLED Day 1.[/color][/b]
[color=blue][b]Chenhsi, Rushford Greens, Town Partner, suicided, lynched Day 1[/b][/color]
[color=blue][b] Dattebayo, Mayor McMagasin, Town Partner, stabbed with a poison dagger, killed Night 1[/b][/color]
[color=green][b] #H34N1, Filth Camare, Neutral Partner, shot with a shotgun, killed Night 1[/b][/color]
[color=blue][b]142857, Luict Oogle, Town Partner, shot by the butler, lynched Day 2 [/b][/color]
[color=blue][b]Pacman2812, Gimbley Mench, Town Partner, shot with a shotgun, killed Night 2[/b][/color]
[color=blue][b]mz94, Griffith Stone, Town Partner, shot in his room, killed Night 2[/b][/color]
[color=blue][b]Brut3Forc3, Brick Oswarl, Town Partner, strangled by his partner, ENDGAMED[/b][/color]
[color=blue][b] BOGTRO (Shaggy75), Carkas Finz, Town Partner, shot and bound, ENDGAMED[/b][/color]

In case I didn't make it clear:

[size=150][b]Please confirm that you recieved your role by PM![/b][/size]

Can't you see in your Sentbox who received the PMs?

Anyway, I'm here.

@Peter: Yeah, but I want for people to have a chance to ask me about stuff they don't understand.

Plus, you have to confirm on most other Mafia sites, and I wanted to keep it conventional.

[size=150][b]Please don't post until everyone confirms. Everybody has confirmed except for Shaggy75.[/b][/size]

Shaggy 75 has still not confirmed. I'll give him 24 more hours before I start looking for a replacement.

[quote="Peter"]If he fails to reply in 24 hours, why would he even be active?
[/quote]

He could've had something come up. It could just be a one time thing, and I'll give him the benefit of the doubt as he has recieved his PM.

@Shaggy75: This is a warning. You need to be active, or notify me if something comes up.

-----------------------------------------------------------
[size=150][i][b]C[/b][/i][/size]arkas Finz walked slowly into the grand hallway. Beside him were eleven other individuals, all talking quietly to each other. He stepped over to the left side of the hallway, next to a tall bronze statue of a man. The man was standing there with his knee held high, in perpetual dignity. How heroic looking. I wish I looked that cool…

Without warning, Carkas jumped. Someone had touched his shoulder. He turned around and saw the short Gimbley Mench staring back at him:

“Oh, it’s just you.”

The short man said softly:

“Come on, let’s go.”

Carkas followed the man’s advice and continued onward, along with the rest of the party. The party was quite a group. By the looks of it you’d say that it was more of a procession than a serious group. The Mayor McMagasin was here, followed by that mindless dog of his. And some other tough looking men were there too. And Alyss, in all her fair beauty! What a group!

When the group reached the living room, a young lady in the front gasped: A man covered entirely with gold paint was standing before them. Before any of them could move, they’re mysterious host spoke:

“Don’t be alarmed – I woke this morning and found that I was the victim of a cruel practical joke. Strange – I thought I was the only one here in the Grant Mansions. But nevertheless, don’t allow me to ruin your entrance to the Grant Mansion. Ever since my master died, it has been awfully lonesome around here and I do wish that you guys can sort out this nasty murder business. Anyway, my name is James, so if you need any help, you can call.”

As a strong young man moved to talk to the butler, Carkas slinked away. They wouldn’t want him there anyhow. He got no further than the stairs when the room was broken with a cold laughter:

[b]“Ah ha ha ha! Seems like I scared you guys! Well, have no fear; I’m here to help. In a way, at least. I know that three of you here are the murderers. Yes, the murderers are amongst you! And not only that, the storm outside has conveniently isolated you from the rest of the world! So it is up to you to figure out among yourselves who are the murderers. To help with this task, each of you must go to the rooms I tell you. Mr. McMagasin to Room 1. Mr. Stone to Room 2. Mr. Finz to Room 3, Mr. Mench to Room 4, Mr. Camare to Room 5, Mr. Arxin to Room 6, Mr. Durth to Room 7, Mr. Oswarl to Room 8, Miss Board to Room 9, Mr. Greens to Room 10, Mrs. Oogle to Room 11, and Mrs. Ivelin to Room 12. Now go! Listen to me on pain of death! Ha haa ha!”[/b]

Shocked, the investigators obediently went to their respective rooms. Once safely inside, nine of them were each shocked to see a small slip of paper.

[b]“Ah ha! Each of you have seen the List! Yes, each of you gets a list. Of the four names on the list, one and only one is a name of a murderer. However, the murderers, you guys have found that you do not get Lists! You must rely on your own cunning to make up one! And beware, for one of the lists is purposely left incorrect!
And now, I leave you guys alone. Alone to find the living death that is among you. Remember for the real investigators to leave alive, they must kill all of the investigators. And for the murderers to escape, they must outnumber the murderers! Good luck – you’ll need it!”[/b]

[b]It is now Night 0. It will be a no-kill night. Night will end at Thursday June 19 at 8:00 AM EST.[/b]

Thanks, that's what I meant.

@Pacman2812 and everybody else: Just a friendly reminder that PM quoting is not allowed, ever.

i might be gone from time to time, like some hours at a time, just due to taking three courses over the summer (death!)  :(

Some hours at a time is fine. If it's more than 48 continious hours, you need to notify me so I know why you're not posting.

Don't forget to get you Night Actions in before the deadline! You can still use all nonkill Night Actions on Night 0.

[quote="iLord"]Remember for the real investigators to leave alive, they must kill all of the investigators.[/quote]

Err....

[quote="iLord"]@Pacman2812 and everybody else: Just a friendly reminder that PM quoting is not allowed, ever.[/quote]
:dry:
The message contains only quote and code tags. Please add some text of your own.

[quote="Distracted523"]Err....
[/quote]

Err...

I mean murderers.

My bad again - that's like the seventh typo I've found.

[quote="mz94"][quote="pacman2812"]Wait, if we (doctors) protect mz94, and distracted isn't modkilled, it would be 4-3, and we would be at LYLO right now and I would anyways be dead.

Peter, when you killed me, it was
a) random?
b) you knew I was doctor?
c) you knew I had any power role?
d) you thought  I had a power role?

And, if BOGTRO and me would pay attention to the "math_reveal", then we might have protected mz94 (with my scumlist).[/quote]

mehh...and just when i had all 3 scum...

by deduction on distracted's list, and my list, and 142857's claims, now that he was town, i deduced the 3 mafia were zapi, m_e, and Peter/pacman. so i inspected Peter night 2. we couldve prolonged it....but the doctor (BOGTRO) wouldve had to be very lucky to protect the right person, adn if they went for him, well, we lose

now that its over...

EDIT: and kudos to Peter, who played an excellent game. i seriously never suspected him. he played very well. so, good job[/quote]

I was the other doctor. And I screwed up all by suspecting on you. If you played better, I would have protected you, be killed, and we have the scum information safe in your hands after my death (I couldn't protect myself), and without distracted's modkill, we would be at 4-3 (lylo)

EDIT: And Peter vigged me, so the only opportunity to us was Brut3Forc3 forgetting to send, and we protecting mz94. That would also be 4-3, and with me and BOGTRO, and with your help and convincing Brut3 to bandwagon zapi (most suspicious for me).

[quote="pacman2812"][quote="mz94"][quote="pacman2812"]Wait, if we (doctors) protect mz94, and distracted isn't modkilled, it would be 4-3, and we would be at LYLO right now and I would anyways be dead.

Peter, when you killed me, it was
a) random?
b) you knew I was doctor?
c) you knew I had any power role?
d) you thought  I had a power role?

And, if BOGTRO and me would pay attention to the "math_reveal", then we might have protected mz94 (with my scumlist).[/quote]

mehh...and just when i had all 3 scum...

by deduction on distracted's list, and my list, and 142857's claims, now that he was town, i deduced the 3 mafia were zapi, m_e, and Peter/pacman. so i inspected Peter night 2. we couldve prolonged it....but the doctor (BOGTRO) wouldve had to be very lucky to protect the right person, adn if they went for him, well, we lose

now that its over...

EDIT: and kudos to Peter, who played an excellent game. i seriously never suspected him. he played very well. so, good job[/quote]

I was the other doctor. And I screwed up all by suspecting on you. If you played better, I would have protected you, be killed, and we have the scum information safe in your hands after my death (I couldn't protect myself), and without distracted's modkill, we would be at 4-3 (lylo)[/quote]

meh. this was my first real mafia game in 3 months, so yeah a little rusty on my part.

[quote="iLord"]Either the mafia have 4 or 2 people (Either too strong or too weak) or their were scum masons.[/quote]Actually, 2 mafia seems not too weak, seeing we were much more that just regular mafia. Basicly we made 2 kills every night, and I was immune to inspection on top of that.

[quote="iLord"]But the town had a lot of power on their side too. It would take the scum two shots to kill any power role.[/quote]Yeah but they know nothing about how powerful the mason pairs are. From the town's side, it's not an unlikely assumption.

uhh... town had only one doc pair and one cop pair (one of the cops was a miller) and we had three scum-town masons... and 1-2 scum nightkills... that is NOT balanced.

[quote="zapi2007"]Maybe two mafia, or two scum mason together+ one powerless (other than kill) mafia would have been a little more balanced[/quote]That would have been impossible to win from mafia side.

well, if one of them was a godfather though...

That one can still be killed by the vigilante.

@Pacman:

The town had a doc pair, [b]3[/b] cop pairs, a pair of double voters, and a vig pair.

Once the town figured out that they were divided into pairs, and that the scum were among them, they would gain the added advantage of being able to see their partner's reactions to their suggestions of doing certain things.

The scum had almost no control over the targets of their partners, except for what they could pursuade. If they purposely choose a different target, then their partner would be instantly suspicious.

The only unbalanced part would be if Brick died, and Fenno had sole control over the vig kill.

Luckily, that did not come up, and I have noted the change in my records.

[quote="iLord"]The only unbalanced part would be if Brick died, and Fenno had sole control over the vig kill.

Luckily, that did not come up, and I have noted the change in my records.[/quote]And the only reason for that was the wording in your role pm.

I won't even call it unclear, for me it was clear in the opposite way: if you say we have to send in both to succeed, that implies that the power is gone as soon as one of us dies. Otherwise I'd have asked the mafia to kill Brick immediately on night 1.

(Until the post above, I was convinced that a mason's night actions were lost if the partner died.)

Maybe we just made it look so lopsided  :P

[quote="Peter"]And the only reason for that was the wording in your role pm.

I won't even call it unclear, for me it was clear in the opposite way: if you say we have to send in both to succeed, that implies that the power is gone as soon as one of us dies. Otherwise I'd have asked the mafia to kill Brick immediately on night 1.
[/quote]

That was another error that I fixed.

lastpost (this is used to preserve order of locked topic)


geometry
area
linear algebra
Triangle
IMO Shortlist

Let $ABC$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ (or extensions thereof) in $D$, $E$, $F$, respectively. Suppose further that the areas of triangles $PBD$, $PCE$, $PAF$ are all equal. Prove that each of these areas is equal to the area of triangle $ABC$ itself.

[hide]Because affine transformations preserve ratios of areas, it is sufficient to prove this when $A = (0,0)$, $B = (1,0)$, and $C = (0,1)$, and we may suppose without loss of generality that $P$ and $A$ do not lie on the same side as $BC$. Hence, we may suppose that $P = (p,-q)$, where $q$ is a positive real number.

Consider the attached diagram with colors. $P$ cannot lie in the left pink region, since then $\triangle PBD$ lies inside $\triangle PAF$; $P$ cannot lie in the right pink region, since then $\triangle PCE$ lies inside $\triangle PBD$; $P$ cannot lie in the orange region, since then $\triangle PBD$ lies on $\triangle PAF$; and $P$ cannot lie in the yellow region, since then $\triangle PBD$ lies inside $\triangle PAF$, so $P$ must lie in the green region. Therefore, $p \geq 1$ and $p - q \leq 1$.

We compute the coordinates of $D$, $E$, and $F$. Line $AP$ has equation $y = \frac{1+q}{p}x + 1$, which intersects the line $y=0$ at $D = \left(\frac{p}{q+1}, 0\right)$; line $CP$ has equation $y = \frac{q}{1-p}x + \frac{q}{1-p}$, and it intersects the line $x=0$ at $F = \left(0, \frac{q}{1-p}\right)$; and line $BP$ has equation $y = \frac{-q}{p}x$, and intersects the line $x+y=1$ at $E = \left(\frac{p}{p-q}, \frac{-q}{p-q}\right)$. It is easy to see that the area of $\triangle PBD$ is $\frac{pq}{2(q+1)}$ and that the area of $\triangle PFA$ is $(1/2)p\left(\frac{q}{p-1} - 1\right) = \frac{p(q-p+1)}{2}$. The distance from $P$ to the line $x+y=1$ is $\frac{1+q-p}{\sqrt{2}}$, and the length of $CE$ is $\frac{\sqrt{2} q}{p-q}$, so the area of $\triangle PCE$ is $\frac{(1+q-p)q}{2(p-q)}$. Equating these areas, we find that
\begin{align*} \frac{pq}{q+1} = \frac{p(1+q-p)}{p-1} = \frac{(1+q-p)q}{p-q} = 2K, \tag{1} \end{align*}
where $K$ is the area of $\triangle PBD$, $\triangle PCE$, and $\triangle PAF$.

From the first and second equations, we get $\frac{q}{q+1} = \frac{q}{p-1} - 1$, so $\frac{q}{p-1} = \frac{2q+1}{q+1}$, so $p-1 = \frac{q^2 + q}{2q+1}$, so
\begin{align*} p = \frac{q^2 + 3q + 1}{2q+1}. \tag{2} \end{align*}
From the second and third equations, we get $\frac{p}{p-1} = \frac{q}{p-q}$, or $p^2 - 2pq + q = 0$.
Substituting in the equation from (2) and multiplying both sides by $(2q+1)^2$, we find that
\begin{align*}
0
&= (2q+1)^2 (p^2 - 2pq + q) \\
&= (p(2q+1))^2 - 2q(2q+1)(p(2q+1)) + q(2q+1)^2 \\
&= (q^2 + 3q + 1)^2 - (4q^2 + 2q)(q^2 + 3q + 1) + q(4q^2 + 4q + 1) \\
&= (q^4 + 6q^3 + 11q^2 + 6q + 1) - (4q^4 + 14q^3 + 10q^2 + 2q) + 4q^3 + 4q^2 + q \\
&= -3q^4 - 4q^3 + 5q^2 + 5q + 1 \\
&= -(3q+1)(q^3+q^2-2q-1).
\end{align*}
Because $q > 0$, $q \neq \frac{-1}{3}$, so $q^3 + q^2 - 2q - 1 = 0$. Hence, $q^3 + 3q^2 + q = 2q^2 + 3q + 1 = (q+1)(2q+1)$, so $\frac{q+1}{q} = \frac{q^2 + 3q + 1}{2q+1} = p$, by (2), so $\frac{pq}{q+1} = 1$. But from (1), we get that $K = \frac{1}{2}$, which is the area of $\triangle ABC$, so we are done. [/hide]

Since coordinates are so nasty and ugly, we use

[hide="Solution"]BARYCENTRIC COORDINATES!

So $A$ is $(1,0,0)$, $B$ is $(0,1,0)$, $C$ is $(0,0,1)$, and $P$ is $(p,q,r)$, with $p+q+r=1$.

Now, the equation of line $AP$ is just the line $qz=ry$, $BP$ is just $pz=rx$, and $CP$ is $qx=py$.

Also, $AB$ is just $z=0$, $BC$ is $x=0$, and $AC$ is $y=0$.

Thus, the coordinates of $D$ is $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$. Similarly, $E$ is at $\left(\frac{p}{p+r},0,\frac{r}{p+r}\right)$ and $F$ is at $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$

Now, the ratio $[PBD]$ to $[ABC]$ is just

$\begin{vmatrix}
p & 0 & 0 \\
q & 1 & \frac{q}{q+r}\\
r & 0 & \frac{r}{q+r}
\end{vmatrix}= \frac{pr}{q+r}$

The other ratios are similarly $\frac{pq}{p+r}$ and $\frac{qr}{p+q}$

Since $p+q+r=1$, we have $\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K$ and we want to show that $|K|=1$.

Thus, we have $\frac{pqr}{p(1-r)}=\frac{pqr}{r(1-q)}=\frac{pqr}{q(1-p)}$.

Since none of $p,q,r=0$ (else $P$ would be on one of the sides of $ABC$):

$p(1-r)=r(1-q)=q(1-p)$.

We know $r=1-p-q$. Substuting:

$p^2+pq=1-p-2q+pq+q^2=q-pq$.

From the first and third, we get that $q(1-2p)=p^2 \implies q=\frac{p^2}{1-2p}$

Now consider first and second;

$p^2+p-1=q^2-2q$

Subbing back in $q$:

$(p^2+p-1)(1-2p)^2=p^4-2p^2(1-2p)$

which rearranges to

$0=3p^4-4p^3-5p^2+5p-1=(3p-1)(p^3-p^2-2p+1)=0$

If $p=\frac{1}{3}$, then $q=r=\frac{1}{3}$, so $P$ is in the triangle (as all of $p,q,r>0$) contradiction.

Thus, we have $p^3-p^2-2p+1=0$

So, $1-2p=p^2(1-p) \implies q=\frac{p^2}{1-2p}=\frac{1}{1-p}$

Thus, $K=\frac{pq}{1-q}=\frac{\frac{p}{1-p}}{1-\frac{1}{1-p}}=\frac{p}{1-p-1}=-1$

Therefore, if $[PBD]=[PCE]=[PAF]$, necessarily $[PBD]=[PCE]=[PAF]=[ABC]$.[/hide]

[quote="Mewto55555"]
Since $p+q+r=1$, we have $\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K$ [/quote]
I'm confused. Because, it is possible that some of these ratios are positive and some of them are negative. For example: it is possible: $\frac{qr}{1-r}=\frac{pq}{1-q}=1$ and $\frac{pr}{1-p}=-1$.

[quote="Dragonboy"]I'm confused. Because, it is possible that some of these ratios are positive and some of them are negative. For example: it is possible: $\frac{qr}{1-r}=\frac{pq}{1-q}=1$ and $\frac{pr}{1-p}=-1$.[/quote]

Just notice the directions, $[PBD], [PCE],[PAF]$, which are opposite to direction of $[ABC]$. So, we can safely assume that ratios of their areas will have the same sign (in this case, negative, as the direction is opposite).

Is there a solution by complex coordinates?
I think we can easily calculate areas by matrices

WLOG assume $P$ lies inside $\angle ABC$.
We first deal with a few configuration issues.
[color=#00f]CLAIM 1. [/color]$P$ and $B$ lies in different sides of $AC$.
[color=#00f]Proof. [/color]
Otherwise $\triangle PBD$ lies strictly inside $\triangle PAF$, hence $[PBD]<[PAF]$, contradiction. $\blacksquare$
[color=#00f]CLAIM 2.[/color] $A$ lies between $F$ and $B$.
[color=#00f]Proof. [/color]
Otherwise $\triangle PCE$ lies strictly inside $\triangle PAF$, hence $[PBD]<[PAF]$, contradiction. $\blacksquare$

Now let $A=(1:0:0)$, $B=(0:1:0)$ and $C=(0:0:1)$. Let $P=(x:y:z)$, where $x+y+z=1$, we have
$$D=(0:y:z),\hspace{5pt}E=(x:0:z)\text{ and } F=(x:y:0)$$
Then from the above assumptions, $x>1>z>0>y$
Therefore,
\begin{align*}
\frac{[PCE]}{[ABC]}&=\begin{vmatrix}x&0&\frac{x}{x+z}\\y&0&0\\z&1&\frac{z}{x+z}\end{vmatrix}\\
&=\frac{xy}{x+z}
\end{align*}
Notice that $\frac{xy}{x+z}<0$, hence $|[PCE]|=-[ABC]\cdot\frac{xy}{x+z}$, similarly we have
$|[PBD]|=-[ABC]\cdot\frac{xz}{y+z}$ and $|[PAF]|=-[ABC]\cdot\frac{yz}{x+y}$
Therefore we have
$$\frac{xy}{x+z}=\frac{xz}{y+z}=\frac{yz}{x+y}\hspace{5pt}(1)$$
This becomes $x(x+y)=y(y+z)=z(x+z)$. Hence
$$z=\frac{x^2+xy-y^2}{y}=x^2+x-y \hspace{5pt}(2)$$
Therefore from $x+y+z=1$ we have $2x+\frac{x^2}{y}=1$ hence
$$y=\frac{x^2}{1-2x}\hspace{5pt}(3)$$
Substituting $(2)$ into $(1)$ we have
$$z=\frac{1-3x+x^2}{1-2x}\hspace{5pt}(4)$$
Now we have $x(x+y)=z(z+x)$, hence
$$x\left(x+\frac{x^2}{1-2x}\right)=\frac{1-3x+x^2}{1-2x}\cdot\left(\frac{1-3x+x^2}{1-2x}+x\right)$$
Clearing denominators, we have
\begin{align*}
3x^4-4x^3-5x^2+5x-1&=0\\
(3x-1)(x^3-x^2-2x+1)&=0\\
x^3-x^2-2x+1&=0
\end{align*}
Therefore,
$$\frac{xz}{y+z}=\frac{x(1-3x+x^2)}{(1-x)(1-2x)}=-1$$
This completes the proof.

A more symmetric way to solve the equations after using bary:

[hide]
If $P=(x,y,z)$ with $x+y+z=1$, then examining the possible regions $P$ could be in (to decide the orientations of the triangles), we get the following equations:
\[\frac{xz}{1-x}=\frac{yx}{1-y}=\frac{zy}{1-z}=t\]
for some constant $t$, and we want $|t|=1$. This implies $y(1-x)=z(1-y)=x(1-z)=\frac{xyz}{t}$. SO $y-z=y(x-z)$, and multiplying this cyclicly and using $x,y,z$ distinct, we get $xyz=-1$. Now note
\[t^3=\prod_{cyc} \frac{xz}{1-x}=\frac{(xyz)^2}{\prod_{cyc}(1-x)}=\frac{1}{\prod_{cyc}(1-x)}\]
or
\[1-x-y-z+xy+xz+yz-xyz=1-1+xy+xz+yz+1=1+xy+xz+yz=t^{-3}\]
so $xy+xz+yz=t^{-3}-1$. Now, write
\[t^{-1}=\sum_{cyc}zt^{-1}=\sum_{cyc}\frac{1-x}{x}=-3+\frac{\sum_{cyc}xy}{xyz}=-3-(t^{-3}-1)=-2-t^{-3}\]
so $(t+1)(2t^2-t+1)=0$, or $t=-1$.
[/hide]

We use barycentric coordinates. Let $P$ be $(\tfrac{x}{x+y+z},\tfrac{y}{x+y+z},\tfrac{z}{x+y+z})$. We note that $\triangle BDP$, $\triangle AFP$, and $\triangle CEP$ are all oppositely oriented to $\triangle ABC$, so where $[s]$ denotes signed area, we'll show that
\[\frac{[AFP]}{[ABC]}=\frac{[BDP]}{[ABC]}=\frac{[CEP]}{[ABC]}=t\]
implies $t=-1$. Note that $PC$ intersects $AB$ at $F=(\tfrac{x}{x+y},\tfrac{y}{x+y},0)$, and $PA$ intersects $BC$ at $D=(0,\tfrac{y}{y+z},\tfrac{z}{y+z})$, and $PB$ intersects $CA$ at $E=(\frac{x}{z+x},0,\frac{z}{z+x})$. We have
\[\frac{[AFP]}{[ABC]}
=
\begin{vmatrix}
1 & 0 & 0\\
\frac{x}{x+y} & \frac{y}{x+y} & 0\\
\frac{x}{x+y+z} & \frac{y}{x+y+z} & \frac{z}{x+y+z}
\end{vmatrix}
=
\frac{yz}{(x+y+z)(x+y)} \]
\[
\frac{[BDP]}{[ABC]}
=
\begin{vmatrix}
0 & 1 & 0\\
0 & \frac{y}{y+z} & \frac{z}{y+z}\\
\frac{x}{x+y+z} & \frac{y}{x+y+z} & \frac{z}{x+y+z}
\end{vmatrix}
= \frac{zx}{(x+y+z)(y+z)} \]
\[
\frac{[CEP]}{[ABC]}
=
\begin{vmatrix}
0 & 0 & 1\\
\frac{x}{z+x} & 0 & \frac{z}{z+x}\\
\frac{x}{x+y+z} & \frac{y}{x+y+z} & \frac{z}{x+y+z}
\end{vmatrix}
= \frac{xy}{(x+y+z)(z+x)}\]
Since everything is homogeneous, we can assume $x+y+z=1$. Now, we have
\[\frac{xyz}{x(x+y)}=\frac{xyz}{y(y+z)}=\frac{xyz}{z(z+x)}\]
so $x(x+y)=y(1-x)=(1-x-y)(1-y)$. Expanding, we have $x^2+xy=y-yx=1-x-2y+xy+y^2$. We have $x^2=y(1-2x)\implies y=\frac{x^2}{1-2x}$ because obviously $x\neq \tfrac12$ and we have $x^2+x-1=y^2-2y=\frac{x^2(x^2+4x-2)}{(1-2x)^2}$. This implies that $(3x-1)(x^3-x^2-2x+1)=3x^4-4x^3-5x^2+5x-1=0$. $x=\tfrac13$ implies $y=\tfrac13$ and $z=\tfrac13$ but at least one of $x$, $y$, $z$ must be negative. Thus,
\[x^2\left(x-1+\frac{1}{y}\right)=x^3-x^2-2x+1=0\] Clearly $x$ cannot be zero, so $1-x=\tfrac1y$. Thus, $y(y+z)=y(1-x)=1$ so we have
\[\frac{zx}{(x+y+z)(y+z)}=xyz\]
so it remains to show that $xyz=-1$. The cyclic equivalents of $1-x=\tfrac1y$ are $1-y=\tfrac1z$ and $1-z=\tfrac1x$. If $x=y$ then we get $x^2-x-1=0$ and it is easy to see that this is not a solution to $x^3-x^2-2x+1=0$. Thus, $x$, $y$, $z$ are mutually distinct. Thus, being the three roots of $f(x)=x^3-x^2-2x+1$, by Vieta's formulas we must have $xyz=-1$. We're done.


Which pieces you know have the most technical and musical difficulties?

All the Rachmaninoff Piano concertos require flexible arm technique, fast fingers, and musical senses.
The Prokofiev toccata is very quick and has a lot of chromatic material.
Ravel's Jeux d'eau is a very flowing piece, and requires delicacy in touch and smoothness in finger excecution.

The list goes on and on.  What do you guys think?

-Ligeti etudes, especially Escalier de Diable, and Die Zauberlehring. Touches Bloquees is also quite tough (pardon my spelling)

-Liszt Transcendental Etudes

-Hammerklavier sonata

Less Technical difficulties, more musical:

_Anything_ by Mozart (almost no one can get him right)

[quote="rookworm"]_Anything_ by Mozart (almost no one can get him right)[/quote]

lol, I fall asleep everytime I listen to his pieces, except his symphony no.40 :P

Technical Difficulties:

Chopin  Preludes, Dmaj. Ebmaj ,Dmin
Liszt  Les Jeux d'eau a la Villa d'Este
Beethoven  Sonata 28 op101-last movement

Musical Difficulties:

Chopin  Prelude in Fsharp,Berceuse,Nocturnes op48.op15(Gmin) Polonaise in Cmin(op40)
Beethoven Sonata29(Hammerklavier)-3rd movement,Sonate Pathetique(Grave opening).Piano part in Violin Sonata No10
Schubert Sonata in Gmaj(1st movt) Sonata in Amaj(D664)-2nd movt.
Schuman Phantasiestucke Op12 especially Warum .Albumblatt op100(the piece Brahms wrote variations on )

This is just a short selection but in my opinion the difficulty of mastering  just these few could occupy a lifetime.


Let $ f: R\plus{}... R\plus{}$
      $ f(3x)\geq f(f(2x)) \plus{}x$ for all $ x \in R$
Find the max value of k to $ f(x)\geq kx$ for $ all x\in R$

I proved that k = 1/2...
what do you think about it? :)

Obviosly $ f(x)\ge \frac x3 \ \forall x\ge 0$.
Let proved $ f(x)\ge ax \ \forall x\ge 0$, then $ f(x)\ge f(f(\frac{2x}{3}))\plus{}\frac x3 \ge a f(\frac{2x}{3})\plus{}\frac x3 \ge (2a^2\plus{}1)\frac x3$.
It give if $ k\equal{}a$ work, then $ k\equal{}\frac{2a^2\plus{}1}{3}$ work too. From $ k\equal{}\frac 13 \to k\equal{}\frac{11}{27} \to k\equal{}\frac{971}{2187}\to ...$
Limit is $ k\equal{}\frac 12$. For example may be $ f(x)\equal{}\frac x2$.

Nice,thanks,my proof is similar to :lol:

Maybe I missed something,but what if $ k \equal{} 1$,I mean if we suppose that $ f(x) \equal{} x$,then
$ f(3x) \equal{} f(f(2x)) \plus{} x$.

[quote]Maybe I missed something,but what if ,I mean if we suppose that ,then...
.[/quote]
you should find k for all functions not just for one


linear algebra
matrix
Ring Theory
superior algebra
superior algebra unsolved

Let $ p$ be a prime number. Prove that a unitary ring with $ p^{2}$ elements is commutative and that there exists a non-unitary ring of $ p^{2}$ elements that is not commutative.

Also prove that there exists unitary rings with $ p^{3}$ elements that are not commutative.

$ A: \equal{}\left(\begin{matrix}1&1\\0&0\end{matrix}\right)$ and $ B: \equal{}\left(\begin{matrix}0&0\\1&1\end{matrix}\right)$ satisfy $ A^2\equal{}AB\equal{}A$ and $ B^2\equal{}BA\equal{}B$.
So $ \mathbb{F}_p A\plus{}\mathbb{F}_p B\equal{}\{\left(\begin{matrix}a&a\\b&b\end{matrix}\right): a,b\in\mathbb{F}_p\}$ is a not commutative subrng of $ \mathcal{M}_2(\mathbb{F}_p)$ with $ p^2$ elements and
$ \mathbb{F}_p I_2\plus{}\mathbb{F}_p A\plus{}\mathbb{F}_p B\equal{}\{\left(\begin{matrix}a\plus{}c&a\\b&b\plus{}c\end{matrix}\right): a,b,c\in\mathbb{F}_p\}$ is a not commutative unitary subring of $ \mathcal{M}_2(\mathbb{F}_p)$ with $ p^3$ elements.


geometry
articles
blogs
search
number theory

I was sitting in a lecture by Carlos Kenig on harmonic analysis, who is well-known in PDE, and I was surprised that he often cited Terry Tao. I have thought of Tao as someone who works in arithmetic combinatorics, analytic number theory, algebraic combinatorics, but yes, now that I look: PDE and harmonic stuff too. My question is, how come he works in these seemingly different areas? Are there prominent linkages between harmonic analysis and combinatorics, or harmonic analysis and number theory? I suppose when you are someone as gifted as he, you can afford to work in such diverse areas, as normally one's work would almost surely be watered down, rather than strengthened, by this diversity.

Yours,

Los from Finnland

People learn math from multiple sources, and odd clusters can pop up that way. Combinatorics and harmonic analysis? I learned a big chunk of harmonic analysis from my father (same thesis adviser as Carlos Kenig) and then went toward combinatorics in school.
Terry Tao is special in that he can take all of these to a research level, but this sort of thing comes up all the time.

Math is interconnected enough that it's usually possible to find some way of linking even seemingly unrelated subjects; see my posts on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268054]this thread[/url], mixing the discrete Fourier transform with q-binomial coefficients.

You should read [url=http://arxiv.org/abs/math.HO/0702396]this article of his[/url].  Gowers expresses similar sentiments:  even if two branches of mathematics aren't closely related, working in one can hone habits of thought that help in another.  In this case, if you read a lot of Tao's blog posts you'll realize he uses surprisingly similar techniques in both additive number theory and harmonic analysis.  A lot of his more technical results in both fields are the result of various delicate bounds on various norms associated with the problem.

I also don't think this phenomenon is unique to Terence Tao, by any means.  Have you looked at the major fields that other prominent mathematicians work in?

One of Tao's interests is Harmonic Analysis. More specially he began with Harmonic Analysis. His PhD thesis was in Harmonic Analysis (''Three Regularity Results in Harmonic Analysis-Terence Tao, 1996'')  under the supervision of Elias Stein....

see here:[url]http://www.math.ucla.edu/~tao/[/url]... :)

[quote="Los"]I have thought of Tao as someone who works in arithmetic combinatorics, analytic number theory, algebraic combinatorics, but yes, now that I look: PDE and harmonic stuff too. [/quote]

he's originally a PDE/harmonic analysis guy.  you can see [url=http://www.math.ucla.edu/~tao/preprints/]here[/url] that he has several papers in that area from the late 90s and early 00s.

[quote="Los"]
My question is, how come he works in these seemingly different areas? Are there prominent linkages between harmonic analysis and combinatorics, or harmonic analysis and number theory?[/quote]

certainly.   you may want to read about 'arithmetic combinatorics' or 'additive combinatorics' to find out more.  [url=http://en.wikipedia.org/wiki/Arithmetic_combinatorics]here[/url] is a wikipedia page, [url=http://books.google.com/books?id=xpimQMtn5-IC&dq=additive+combinatorics&printsec=frontcover&source=bn&hl=en&ei=AYjvSenfIZu6sgPbgpniAQ&sa=X&oi=book_result&ct=result&resnum=4]here[/url] a relatively recent book on the subject by tao and vu.  a quick search through the book shows that you can read a bit about the link between the fourier transform and combinatorics in chapter 4.

Thank you jmerry, blahblahblah, and t0rajir0u: these were just the sort of answers that I was hoping to find!

Los

Number theory is a bit strange as a branch of mathematics, in that it is defined more by a family of related problems than a family of related methods.  Terry Tao is definitely an analyst.  (Just look at his blog.)  It just so happens that his work in number theory is his most famous.  But that work is still contiguous with the rest of his body of work (which is pretty darn broad).


geometry
angle bisector
geometry proposed

(A.Zaslavsky, 9--10) Quadrilateral $ ABCD$ is circumscribed arounda circle with center $ I$. Prove that the projections of points $ B$ and $ D$ to the lines $ IA$ and $ IC$ lie on a single circle.

Denote by $ B_{1}$, $ B_{2}$ the orthogonal projections of $ B$ on the lines $ IA$, and $ IC$, respectively, and similarly, let $ D_{1}$, $ D_{2}$ be the projections of $ D$ on the same lines $ IA$, and $ IC$, respectively. We consider the following preliminary result:

[b]Lemma[/b]. Let $ ABC$ be a triangle, and denote by $ M$ the midpoint of segment $ BC$. If $ X$, $ Y$ are the orthogonal projections of the vertices $ B$, $ C$ on the internal angle bisector of angle $ BAC$, then $ MX \equal{} MY \equal{} |b \minus{} c|/2$.

[i]Proof[/i]. We shall resume to proving that $ MX \equal{} |b \minus{} c|/2$. For this, let $ B^{\prime}$ be the intersection of the line $ BX$ with the sideline $ CA$. Since the triangle $ ABB^{\prime}$ is isosceles, the length of segment $ CB^{\prime}$ is $ |b \minus{} c|$. On other hand, $ X$, $ M$ are the midpoints of segments $ BB^{\prime}$, and $ BC$, respectively. Thus $ XM \equal{} CB^{\prime}/2 \equal{} |b \minus{} c|/2$. This proves our Lemma.

Returning to the problem, let $ T$ be the midpoint of the diagonal $ BD$. According to the Lemma, applied for the triangle $ BAD$, we have that $ TB_{1} \equal{} TD_{1} \equal{} |AB \minus{} AD|/2$. Similarly, according to the same Lemma, this time applied in triangle $ BCD$, we have that $ TB_{2} \equal{} TD_{2} \equal{} |BC \minus{} CD|/2$. On the other hand, the quadrilateral $ ABCD$ beeing circumscribed, $ AB \plus{} CD \equal{} AD \plus{} BC$ (Pithot's theorem), and hereby, we conclude that $ TB_{1} \equal{} TB_{2} \equal{} TD_{1} \equal{} TD_{2}$, i.e. the projections of $ B$, and $ D$ on the lines $ IA$, and $ IC$ lie on a single circle (with center $ T$, the midpoint of diagonal $ BD$).

I think it is easy to calculate angles of triangle IB1B2 an ID1D2 to get them similar which gives result


Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1$, what is the least possible total for the number of bananas?

[hide="Answer"]We label the monkeys $m_1$, $m_2$, and $m_3$. From the given information, we have:

$m_1=\frac{3x}{4}+\frac{3y}{8}+\frac{11z}{24}$
$m_2=\frac{x}{8}+\frac{y}{4}+\frac{11z}{24}$
$m_3=\frac{x}{8}+\frac{3y}{8}+\frac{z}{12}$

where $x+y+z=M$, the minimum possible number of bananas. Also, we have the equations in ratio $3: 2: 1$.

First, we set the first and second equations equal by multiplying the first by $2$ and the second by $3$, and we have:

$\frac{3x}{2}+\frac{3y}{4}+\frac{11z}{12}=\frac{3x}{8}+\frac{3y}{4}+\frac{11z}{8}\Rightarrow 36x+22z=9x+33z\Rightarrow x=\frac{11z}{27}$

Next, we set the second and third equations equal by multiplying the third by $2$, so we have:

$\frac{x}{8}+\frac{y}{4}+\frac{11z}{24}=\frac{x}{4}+\frac{3y}{4}+\frac{z}{6}\Rightarrow 3x+6y+11z=6x+18y+4z\Rightarrow$
$3x+12y-7z=0$

Substituting in our previous equation, we have $3\cdot \frac{11z}{27}+12y-7z=0\Rightarrow 27y=13z\Rightarrow y=\frac{13z}{27}$. From these two equivalencies, we have that $z$ must be divisible by $27$, since all three are integer values, and from $m_1$ and $m_2$, we have that $z$ must be divisible by $24$. $LCM(24,27)=216$, so we set $z=216$. From there, we substitute back into the other two and we have $(x,y,z)=(88,104,216)$, and $x+y+z=88+104+216=\boxed{408}$.

That third monkey is pretty generous. :lol:[/hide]

[hide="Solution"]
Let there be $d$ bananans in the pile, so $d=a+b+c$, where $a, b$ and $c$ are the number of bananans the first, second, and third monkeys take (respectively).
We get that the first monkey has a total of $\frac{3}{4}a+\frac{3}{8}b+\frac{11}{24}c$, the second monkey has a total of $\frac{1}{8}a+\frac{1}{4}b+\frac{11}{24}c$, and the third monkey has a total of $\frac{1}{8}a+\frac{3}{8}b+\frac{1}{12}c$.

We get $\frac{\frac34 a+\frac38 b+\frac{11}{24} c}{\frac18 a+\frac 14b+\frac{11}{24} c}=\frac{3}{2}$.  This is the same thing as $\frac{18a+9b+11c}{3a+6b+11c}=\frac{3}{2}\implies 9a+18b+33c=36a+18b+22c\implies 27a=11c$.

Also, we know that $\frac{\frac34 a+\frac38 b+\frac{11}{24} c}{\frac18 a+\frac38 b+\frac{1}{12} c}=3$.  This is the same as $\frac{18a+9b+11c}{3a+9b+2c}=3$.  This implies that $18a+9b+11c=9a+27b+6c\implies 9a+5c=18b$.

Lastly,we know that $\frac{\frac{1}{8}a+\frac{1}{4}b+\frac{11}{24}c}{\frac{1}{8}a+\frac{3}{8}b+\frac{1}{12}c}=2$.  This is the same as $\frac{3a+6b+11c}{3a+9b+2c}=2$.  This implies that $3a+6b+11c=6a+18b+4c\implies 3a+12b=7c$.

From our first equation, we have $c\equiv 0\pmod{27}$ or $c=27m$.  Substituting this into the equation $3a+12b=7c$ gives us $12b=3(63m-a)$, which implies that $4b=63m-a$.  Also, susbtitute this into our second equation gives us $a+15m=2b$. This implies that we need $a+m\equiv 0\pmod2$ and $m-a\equiv 0\pmod 2$.  However, from our first class $a+m\equiv 0\pmod 2\implies m-a\equiv 0\pmod 2$, because $a+m\equiv 0\pmod 2\iff a\equiv m\pmod{2}$.  Hence, we need to have $a+m\equiv 0\pmod 2$.  Note that we need $a\equiv 0\pmod{8}, b\equiv 0\pmod{8}$ and $c\equiv 0\pmod{24}$.  Since $c=27m$ and $a=11m$ (our first equation), we get $a+m\equiv 0\pmod{2}$.  Now, all we need is $a\equiv 0\pmod{8}, b\equiv 0\pmod{8}, c\equiv 0\pmod{8}$, which is minimized when $a=88, c=216$, and substituting this into the second equation $9(88)+5(216)=18b\implies b=44+60=104$.  Therefore, the minimial value is $d=a+b+c=88+216+104=\boxed{408}$.

Also, note that the last equation is irrelevant (i.e we don't have to check it), as the first two equations being true is enough for the third equation to be true.  [/hide]

p16: The first monkey takes $x$ bananas, so he now has $\frac{3}{4}x$ bananas and the other two have $\frac{1}{8}x$ bananas. The pile now has $p-x$ bananas. The second monkey takes $y$ bananas, so the first monkey now has $\frac{3}{4}x+\frac{3}{8}y$ bananas, the second monkey now has $\frac{1}{8}x+\frac{1}{4}y$ bananas, and the third monkey now has $\frac{1}{8}x+\frac{3}{8}y$ bananas. Lastly, the third monkey takes $z$ bananas, so the first monkey now has $\frac{3}{4}x+\frac{3}{8}y + \frac{11z}{24} = \frac{18x+9y+11z}{24}$, the second monkey now has $\frac{1}{8}x+\frac{1}{4}y+\frac{11z}{24} = \frac{3x+6y+11z}{24}$, and the third monkey now has $\frac{1}{8}x+\frac{3}{8}y+\frac{z}{12} = \frac{3x+9y+2z}{24}$. We can set up the equations $\frac{18x+9y+11z}{3x+9y+2z} = 3, \frac{3x+6y+11z}{3x+9y+2z} = 2, \frac{3x+6y+11z}{3x+9y+2z}=\frac{3}{2}$. The integer solution is $x = 11p, y = 13p, z = 27p$. But $x,y$ must be multiples of $8$ and $z$ must be a multiple of $8$, so the minimum is $8(11+13+27) = \boxed{408}$.

[b]Solution:[/b] Let the first monke take $x$ bananas initially, the second take $y$ initially, and the third take $z$. Then we have that the first monke has $S_1 = \frac{3}{4}x + \frac{3}{8}y + \frac{11}{24}z, S_2 = \frac{1}{8}x + \frac{1}{4}y + \frac{11}{24}z, S_3 = \frac{1}{8}x + \frac{3}{8}y + \frac{1}{12}z$. Then noting the fact that $S_1 = \frac{3}{2}S_2$ gives us that $27x = 11z$. Similarly, $S_2 = 2S_3$ gives us $13x = 11y$. This makes the smallest possible values $x=11, y=13, z=27$. However, remember that each monke must receive a whole number of bananas whenever the bananas are divided. So the minimum possible values are actually $8$ times all of those. So our answer is $8(11 + 13 + 27) = \boxed{408}$.


linear algebra
matrix
linear algebra unsolved

Let $M_n(\mathbb C)$ denote the set of all $n\times n$ complex matrices. Define the centralizer of a fixed $A\in M_n(\mathbb C)$ as \[ C(A)= \{X\in M_n(\mathbb C): \: AX=XA\},  \] which is a subspace of $M_n(\mathbb C)$.

I saw the following proposition on AoPS for many times:

[quote]$\dim C(A)$ is the sum of squares of some positive integers whose sum is $n$, in which $n$ is the size of $A$.[/quote] To prove this, consider the Jordan canonical form $J=T^{-1}AT$, and let $Y=T^{-1}XT$. Apparently, $AX=XA$ iff $JY=YJ$.

To describe the result I've already got, I have to explain something about the expressions at first. For an $r\times r$ matrix $P$ and an $s\times s$ matrix $Q$, we denote by $P\oplus Q$ the $(r+s)\times (r+s)$ matrix \[ \begin{bmatrix} P & 0 \\ 0 & Q \end{bmatrix}.  \]
For $\lambda\in\mathbb C$ we denote by $J(\lambda,r)$ the $r\times r$ Jordan block \[ \begin{bmatrix} \lambda & 1 \\ & \ddots & \ddots \\ & & \ddots & 1 \\ & & & \lambda \end{bmatrix}.  \]
Accordingly, the Jordan canonical form of $A$ can be written as \[ J = \bigoplus_{i=1}^s \bigoplus_{j=1}^{r_i} J(\lambda_i,n_{ij})\,,  \] where $\lambda_i\ne \lambda_k$ if $i\ne k$, and $n_{i1}\ge n_{i2}\ge \cdots\ge n_{ir_i}$ for each $i$.

The main result I've got is the following one:
\[ \dim C(J) = \sum_{i=1}^s \sum_{j=1}^{r_i} n_{ij} (2j-1)\,.  \] (Do It Yourself, please!) And, of course, the size of $J$ is
\[ n = \sum_{i=1}^s \sum_{j=1}^{r_i} n_{ij}\,.  \]
Therefore, the proposition, $\dim C(J)$ is the sum of squares of some positive integers whose sum is $n$, seems to be a problem in Number Theory. I am not good at that, so I need some HELP!

[quote="alekk"]but the statement was for an arbitrary field. So to solve the pb in generality, Jordan form can't work ?[/quote]
Let us just consider the problem in Complex Field at first, OK? Please help me.


Fie ($x_{n } $) si ($y_{n } $) doua siruri de nr reale cu proprietatile: \[
\mathop {\lim }\limits_{n \to \infty } \frac{{x_n }}{{n^3 }} = 4,\mathop {\lim }\limits_{n \to \infty } \frac{{y_n }}{{n^4 }} = 3.
\]Calculati:\[
\mathop {\lim }\limits_{n \to \infty } \frac{{(x_1 ^2  + x_2 ^2  + ... + x_n ^2 )(y_1 ^2  + y_2 ^2  + ... + y_n ^2 )}}{{(x_1 y_1  + x_2 y_2  + ... + x_n y_n )^2 }}=?
\]Indicatie:fie a>0 si ($x_{n } $)  si :\[
\mathop {\lim }\limits_{n \to \infty } \frac{{x_n }}{{n^a }} = A \in R;
\]
Calculati:\[
\mathop {\lim }\limits_{n \to \infty } \frac{{x_1  + x_2  + ... + x_n }}{{n^{a + 1} }} = ?;
\]

Can you solve the probleme using the indication?
I should be more interesting and easy.
 :lol:

$\mbox{limita ceruta esteA/(a+1) folosind Cesaro si binomul lui Newton}$
$\mbox{limita se scrie astfel}$
$\lim\frac{\frac{x_1^2+x_2^2+.....+x_n^2}{n^7}*\frac{y_1^2+....+y_n^2}{n^9}}({\frac{x_1*y_1+...x_n*y_n}{n^8})^2}=\frac{64}{63}$

nu prea are rost sa pui \mbox pe acolo :)
si nici displaystyle-ul decat in chestii etajate pt ca il are implicit (de ex $\frac{a+b}{c+d}$ =\frac{a+b}{c+d} va fi afisata la fel si cu \displaystyle in fata :) )

ai $\frac{x_n^2}{n^6}\to 16$ deci cu criteriul lui Cesaro Stolz obtii $\lim_{n\to\infty}\frac{x_1^2+\cdots+x_n^2}{n^7}=\lim_{n\to\infty}\frac{x_{n+1}^2}{(n+1)^7-n^7}=\frac{16}7$.
apoi la fel obtii ca $\lim_{n\to\infty}\frac{y_1^2+\cdots y_n^2}{n^9}=1$
iar $\lim_{n\to\infty}\frac{x_1y_1+\cdots+x_ny_n}{n^8}=\frac 32$ in final obtii $\frac {64}{63}$


In $\triangle ABC$, the excircle opposite $C$ touches $AB,BC$ at $P,Q$, and the excircle opposite $B$ touches $AC,BC$ at $R,S$. $PQ$ meets $RS$ at $T$. Show that $AT$ is perpendicular to $BC$.

Nice problem...
We well know that : $QB=BP=CS=SR=u-a$.İntersections of $AH=h$ (altitude) with $QP,SR$ are $M,M'$, respectively.We need to prove $M=M'$.
$m(PQH)=m(MQH)=\frac{B}{2}$ and $m(RSH)=m(M'SH)=\frac{C}{2}$so, $MH=QH.tan\frac{B}{2}$ and $M'H=SH.tan\frac{C}{2}$
$MH=M'H$
$\Longleftrightarrow QH.tan\frac{B}{2}=SH.tan\frac{C}{2}$
$\Longleftrightarrow (QB+BH).tan\frac{B}{2}=(SC+CH).tan\frac{C}{2}$
$\Longleftrightarrow (u-a). tan\frac{B}{2}+h.cotB.tan\frac{B}{2}=(u-a). tan\frac{C}{2}+h.cotC.tan\frac{C}{2}$
$\Longleftrightarrow (u-a).tan\frac{B}{2}+\frac{h}{2}.(1-tan^{2}\frac{B}{2})=(u-a). tan\frac{C}{2}+\frac{h}{2}.(1-tan^{2}\frac{C}{2})$
$\Longleftrightarrow (u-a).(tan\frac{B}{2}-tan\frac{C}{2})=\frac{h}{2}. (1-tan^{2}\frac{C}{2}-1+tan^{2}\frac{B}{2})$
$\Longleftrightarrow (u-a).(tan\frac{B}{2}-tan\frac{C}{2})=\frac{h}{2}.(tan\frac{B}{2}-tan\frac{C}{2}).(tan\frac{B}{2}+tan\frac{C}{2})$
$\Longleftrightarrow (u-a)=\frac{h}{2}.(tan\frac{B}{2}+tan\frac{C}{2})$
$\Longleftrightarrow (u-a)=\frac{h}{2}.(\frac{r}{u-b}+\frac{r}{u-c})$
$\Longleftrightarrow (u-a)=\frac{h}{2}.(\frac{(u-c+u-b).r}{(u-b)(u-c)})$
$\Longleftrightarrow u(u-a)(u-b)(u-c)=\frac{(ur)(ah)}{2}$
$\Longleftrightarrow Area(ABC)^{2}=Area(ABC)^{2}$
Clearly, this is true. :)

Lokman GÖKÇE

Denote that $AT$ and $BC$ meet at $X$,and the centers of the two excircles are $C'$ and $B '$ respectively.

On one hand, by applying Menelaus thoerem,

$\frac{AP}{PB}.\frac{BQ}{QX}.\frac{XT}{TA}=1$

and

$\frac{AR}{RC}.\frac{CS}{SX}.\frac{XT}{TA}=1$

For we have already known that $BP=BQ$and$CR=CS$,we obtain that

$\frac{AP}{AR}=\frac{QX}{SX}$.

On the other hand,$\triangle APC'$∽$\triangle ARB'$ $\Rightarrow$ $\frac{AP}{AR}=\frac{QX}{SX}=\frac{C'A}{B'A}=\frac{C'Q}{B'S}$.

Finally, because of $C'Q//B'S$,we can conclude $AX//C'Q//B'S$,so$AT$⊥$BC$.


calculus
integration
calculus computations

A cylindrical water tank 12 feet high with a radius of 8 feet is located on a tower so that the bottom of the tank is 20 feet above ground level.  How much work is done in filling the tank half full of water through a hole in the bottom of the tank, using water from ground level.

The ans ,I'm getting is 32*58*6*pi*d*g

Where d = density of water and

g = acceleration due to gravity.

Could you by any chance post your solution?

Let the level of water in the tank increase by dy with addition of every infinitesimal volume dV of water lifted from the ground level.

Then dV = pi(8)^2dy

The mass of this volume of water = dM = PdV,where P = density of water.

The force required to lift this mass of water against gravity to a height y above the ground is given by

 dF = gdM ,where g = acceleration due to gravity

 The corresponding work done by this force is therefore

 dW = dF . y = ydF=y(gdM) = yg(PdV) = ygP (pi(8)^2dy)

=>  dW = 64pi.gP(ydy)

Total work done = W = definite integral (64pi.gP(ydy)) between y = 20 + 12/2  and y = 20 + 12

= 64pi.gP (32^2 - 26^2)/2

= 32pi.gP * 58 * 6

I didn't see any errors in your work but then again, I just skimmed it.  Nevertheless, it appears that your method is correct.


Find all integers $ x$ and $ y$ such that
$ x^{6}\plus{}x^{3}y\equal{}y^{3}\plus{}2y^{2}$

[p.s. If posted before please give a link]

[hide]

\[ x^{6}\plus{}x^{3}y\equal{}y^{3}\plus{}2y^{2}\]

Now this implies that both $ x,y$ are even since $ 2|(x^{6}\plus{}x^{3}y\minus{}y^{3})$

$ x\equal{}y\equal{}0$ is an obvious Solution.

Making $ x\equal{}0$ gives another solution:

$ y^{3}\plus{}2y^{2}\equal{}0\implies y\equal{}\minus{}2$

Now we have two Solutions $ (x,y) \equal{} (0,0) ; (0,\minus{}2)$

Let $ x^{3}\equal{}t$

Then,

$ t^{2}\plus{}yt\minus{}(y^{3}\minus{}2y^{2})\equal{}0\implies t\equal{}\frac{\minus{}y\pm\sqrt{y^{2}(4y\plus{}9)}}{2}$

Since $ y^{2}\geq 0$, it follows that $ 4y\plus{}9\geq 0$ and therefore $ y\geq\minus{}2$ (Considering only integers).

[/hide]


Let $ m,n$ be natural number such that$ m> n$; $ 2|m \plus{} n$ (it means $ m,n$ are both evens or odds) .Prove that
if
\[ m^2 \minus{} n^2 \plus{} 1|n^2 \minus{} 1
\]
Then $ A$ is a perfect square,with
\[ A \equal{} \frac {n^2 \minus{} 1}{m^2 \minus{} n^2 \plus{} 1}
\]

I have just edited(missed an important condition) :lol:

I believe the right statement is $ A\plus{}1\equal{}\frac{m^2}{m^2\minus{}n^2\plus{}1}$ is perfect square.
It should be easy to verify by letting $ (m,n)\equal{}(30,26)$.
The proof is as follows:

Let $ (m_k,n_k)$ solve equation $ A(m^2\minus{}n^2\plus{}1)\equal{}n^2\minus{}1$ with $ A>0$, $ m_k\ge n_k>0$ and $ 2|m_k\plus{}n_k$.
We can construct a new solution $ (m_{k\plus{}1},n_{k\plus{}1})$ with
$ m_{k\plus{}1}\equal{}|(2A\plus{}1)m_k\minus{}(2A\plus{}2)n_k|$ and
$ n_{k\plus{}1}\equal{}(2A\plus{}1)n_k\minus{}2Am_k$.

We can verify $ m_{k\plus{}1}\le m_k$ and $ n_{k\plus{}1}\le n_k$ and $ 2|m_{k\plus{}1}\plus{}n_{k\plus{}1}$.
$ (m_{k\plus{}1},n_{k\plus{}1})\equal{}(m_k,n_k)$ if and only if $ m_k\equal{}n_k$.

Since we can only get finitely many distinct solution via this operation, $ m_k\equal{}n_k$ for some $ k$.
Then $ A\plus{}1\equal{}m_k^2$ is a perfect square.


inequalities
function
inequalities solved

[quote="vodanh"]Let $a,b,c,k$ be positive reals such that $abc = 1$.
Find the maximum and minimum of the expression
\[\frac{1}{\sqrt{1 + ka}} + \frac{1}{\sqrt{1 + kb}} + \frac{1}{\sqrt{1 + kc}}.\]

[color=red]Moderator edit: Is that the inequality you meant?[/color][/quote]
When $k\rightarrow 0, \frac{1}{\sqrt{1 + ka}} + \frac{1}{\sqrt{1 + kb}} + \frac{1}{\sqrt{1 + kc}}\rightarrow 3$. This is maximum.
When $k\rightarrow \infty, \frac{1}{\sqrt{1 + ka}} + \frac{1}{\sqrt{1 + kb}} + \frac{1}{\sqrt{1 + kc}}\rightarrow 0$. This is minimum.
But I think this is not the correct expression... :huh:

I don't know, I edited it and it was pretty much what he wrote, he just didn't use LaTeX. But I agree, it looks weird. Maybe you are somehow supposed to find the minimum and maximum as a function of $k$?

I suppose he wants us to write it in terms of k?

Let a,b,c be positive reals such that abc=1 and k is constant (k>0)
Find the maximum and minimum of the expression follow k

1/[(1+ka)]^1/2+1/[(1+kb)]^1/2+1/[(1+kc)]^1/2


If $ p \neq 5$ is an odd prime, prove that either $ p^{2} \minus{} 1$ or $ p^{2}\plus{} 1$ is divisible by $ 10$.

hello, since $ p$ is an odd prime with $ p\ne5$ so we get $ p\equiv 1,3,7,9 \mod 10$ and from here we get
$ p^2\equiv1,9 \mod 10$ therefore $ p^2-1\equiv 0\mod 10$ or $ p^2+1\equiv 0 \mod 10$.
Sonnhard.


superior algebra
superior algebra unsolved

On $M$ we define an asociative law "$\cdot$" with the following properties:

a) $x^{3}=x$ for any $x\in M$
a) $xy^{2}x=yx^{2}y$ for any $x,y\in M$

Show that "$\cdot$" is comutative. (this kind of problems are  wierd :furious: )

Generalize. (for any integer $k\geq2$ and  $x^{k+1}=x, xy^{k}x=yx^{k}y$ the law is comutative)

$x(xy)^{2}x=(xy)x^{2}(xy)=xyx^{3}y=xyxy$
Then
$x(xy)=x(xy)^{3}=(xy)(xy)y$
Then
$xxyy=(xy)xy^{3}=xyxy$
and
$xy=x^{-1}xxyyy^{-1}=x^{-1}xyxyy^{-1}=yx$.

The same idea also works for the general case.

ups sorry  :blush: . See the comment below

Nice problem, Slizzel.

It is possible to reduce the generalized case to the case k=2, by observing $x^{k}= x^{2}$. (Play with the equation $x^{m}(x^{n})^{k}x^{m}= x^{n}(x^{m})^{k}x^{n}$.) Then we get the laws $x^{3}= x$ and $x y^{2}x = y x^{2}y$. With only 8 applications of these two laws (in a long equation chain) one can show $xy = yx$.

Barra, we don't have inverses here, just associativity and the two laws.

LiMa

LiMa can you expand a little bit, please?

Lets give the laws numbers.
1: $a^{3}= a$ for all a.
2: $abba = baab$ for all a, b.

You start at $xy$ and appy the laws  1 ($a=xy$) and 1 ($a=x$) to get to $yxxxyxyx$, then 2 ($a=x, b=xy$) and 1 to reach $yxxyxxxyxyxy$, then 2 and 2 to get $xyxyxxyxxyxy$, and finally 1 and 1 to get $yx$. Of course I could tell you the exact equations, but that would spoil the fun of finding it for yourself. :)

For the generalisation you take the equation from my last post with $m=k, n=1$, and reduce both sides with $x^{k+1}= x$.

LiMa


real analysis
limit
function
real analysis unsolved

let $ A\subset[0,1]$ be a measurable set of positive measure and let $ A_n \equal{} A \plus{} \frac {1}{n}$ for nonzero $ n\in\mathbb{Z}$.

Prove or disprove:
\[ \lim_{n\to\infty}\lambda(A\Delta A_n) \equal{} \lim_{n\to \minus{} \infty}\lambda(A\Delta A_n) \equal{} 0
\]

It's sometimes easier to think of functions instead of sets, replacing a set $ A$ with its characteristic function $ \chi_A$. In the context of functions, we have $ f\in L^1[0,1]$ and its translates $ \tau_{s}f(x): \equal{}f(x\minus{}s)$. The question is whether or not $ \|\tau_{s}f\minus{}f\|_{L^1}\to 0$ as $ s\to 0$.

The answer is clearly yes if $ f$ is continuous: indeed, $ f$ is uniformly continuous by compactness of $ [0,1]$, which means that $ \tau_{s}f\minus{}f\to 0$ uniformly.

Next we invoke the fact that continuous functions are dense in $ L^1$: for any $ f\in L^1$ and any $ \epsilon>0$ there is $ g\in C[0,1]$ such that $ \|f\minus{}g\|_{L^1}<\epsilon$. This implies $ \limsup_{s\to 0}\|\tau_s f\minus{}f\|\le 2\epsilon$, and since $ \epsilon$ is arbitrary, we have $ \lim_{s\to 0}\|\tau_s f\minus{}f\|\equal{}0$.

The above may seem overly complicated, but I just can't think of a straightforward measure-theoretic solution right now.  :maybe:


Generally, how does one go about starting a project in research in college? If anybody's heard about UROP, this is what I mean. I'm not a college student yet (I'm still a senior, but am awaing results), but would still like to know so I can see what sort of classes I'm interested in, etc.

I'm not sure about what colleges you may go to, but I go to an engineering/science-focused college.  It's relatively small (~ 3500 undergraduates AND graduates combined), so there's a lot of faculty-student interaction.

I'm still a freshman, but I started asking professors if I could get involved in their labs or research, etc. after I got adjusted to college life (~ 3 months into the academic year).  Two of them said yes (I had both of them for courses); all I did was approach them and ask "Can I work in your lab?" and their response was "Sure."  One of them got busy, so I'm working in the other's lab.  That simple!  Just go around to the professors and ask; some can't provide for you working in their lab, but others surely will.

My college also actively advertises internship, fellowship, and other research opportunities; I'm probably going to apply to one that's over this summer.

However, I'm not sure what UROP is... I do know that some colleges do offer high school programs over the summers, but I would think you would need to look that up yourself depending on what is convenient for you...

Thank you for your response  :) I'll definitely go to MIT this year should I be accepted. If not, I'll go to Georgia Tech. Any ideas based on these colleges?

Thanks Zakary
Possibly silly question: who do you approach if you are interested in research in Math etc.?

Karth:

I'm not familiar with Georgia Tech, but from what I know about MIT from my friends, you are basically swimming in research opportunities there (in addition to other activities... I *believe* MIT has THE largest intramural sports program in the US).  Although, the courses at MIT give you a lot of work to do, so you'd want to make sure you can balance out your life even if you don't into research.  (I personally chose not to go to MIT because being a perfectionist + lots of work + having to pull all-nighters frequently = not a good combo for me.)  I believe MIT is very capable of presenting research opportunities and making people aware of possible internships; its prestige practically guarantees its graduating students a job anywhere.  I'm not 100% familiar with MIT, so someone from MIT might have a better idea of all this.

shreyas_patankar:

Well, if you're interested in doing research in straight-up math, I would suggest talking to professors within a math department.  However, professors in other departments surely use math; thus, if you would like to apply math to something like mathematical models, you *may* be able to look around in other departments, particularly Physics -- Physics typically involves a lot of math because a lot of its concepts and relationships are are expressed, well, mathematically.  If you can't find a professor who you can do research with and there is a math club/society around, they probably know of opportunities in research.  At my college, there is a Career Development Center that helps people apply for internships and the like; there probably is an equivalent at most other colleges.


function
complex analysis
complex analysis unsolved

Let $ g_{n}$ be a sequence of entire functions which have real zeros only. If $ g_{n}$ converges uniformly on compact sets to a function $ g$
then  prove that $ g$ as well has real zeros only.

Not true since $ g_n$ can converge to $ 0$. If it doesn't converge to $ 0$, then use the argument principle to show that $ g$ cannot have zeros in the upper or lower half-plane.

Yes I forgot to  mention that $ g$ is nonzero :blush:
I used Hurwitz's convergence theorem to prove the result. Argument principle works as well quite in the same way.
When posting the problem I had the following generlization in my mind:
Let $ g_{n},g$ be as in the initial problem(g does not vanish identically).Then the set of zeros,say $ Z$ of all the functions $ g_{n}$ coincides with the set of limit points of $ Z$


function
trigonometry
algebra unsolved
algebra

find all continuous function $f: R \to R$  such that:
$f(x)f(y)-f(x+y) = sin x siny$ $\forall x,y \in R$

[b]Cuopbientcd[/b], it seems you didnt use the assumption that $f$ is continious, that is weird. And why you write $sin(x)=-sin(x)f(\pi)$? How it follows from above?

[quote="cuopbientcd"]I know my post be have small mistake but you can solve it from my solution!I belive you do it :lol:[i] Yustas[/i][/quote]
Your solution is absolutely correct. Let me shorten it a little. :wink:

(*) $f(x)f(y)-f(x+y)=\sin x\sin y$

$y=0$ gives $f(x)(f(0)-1)=0$ for all $x\in\mathbb{R}$. And $f\not \equiv 0$ gives $f(0)=1$.
Then $x=-{\pi\over 2},y={\pi\over 2}$ gives $f(-{\pi\over 2})f({\pi\over 2})=0$.
$x=-{\pi\over 2},y=\pi$ gives $f(-{\pi\over 2})f(\pi)=f({\pi\over 2})$.
Together this gives $f({\pi\over 2})=0$.
And now $y={\pi\over 2}$ gives $-f(x+{\pi\over 2})=\sin x$ and $f(x)=-\sin (x-{\pi\over 2})=\cos x$ for all $x\in\mathbb{R}$


AMC
AIME
calculus
geometry
analytic geometry
linear algebra

i would like to take a CC courses.
at the same time, i want to take ones that could help me on AIME/ USAMO. (AIME especially)

Calculus With Analytic Geometry II,Finite Mathematics II, Applied Calculus I, Linear Algebra, Discrete MAthematics, Differential Equations  <--- is there any useful classes among them??

thx

AIME/USAMO don't use any math beyond Precalculus.  The most important thing is using that math well.  None of those classes are going to help.


algebra
polynomial
real analysis
real analysis unsolved

Show that the polynomial P(x)=4x^3-3x-1/2 has no rational roots.

This is a double post, and it is in the wrong forum. Your previous post has been moved to a more appropriate forum. Please do not double post.

To find your previous post, click on your own user name to get to your profile, then click on "view all posts by ... "


What is the value of $x$ if
\begin{eqnarray*} x &=& y + 5 \\ y &=& z + 10 \\ z &=& w + 20 \\ w &=& 80 \end{eqnarray*}

[hide]If $w=80$ then $z$ must equal 100. If $z=100$ then $y=110$.  If $y=110$ then x must equal $\boxed{115}$[/hide]


inequalities
function
induction
inequalities proposed

For any positive real $a, b, c$ and positive integer $n$, show that
\[\frac{2}{(n+1)(n+2)} \sum_{i+j+k=n} a^i b^j c^k \geq \left(\frac{a+b+c}{3}\right)^n\]
where the sum is taken over all non-negative integers $i, j, k$ with $i+j+k=n$.

I think, we can write similar inequality in the general case for $k$ variables, can't we?  ;)

Very possibly, but I can't see exactly how. My algebra got quite messy when generalising to 4 variables. If you have a proof of the general case, please post it.

By the way, I looked up Hadamard's inequality on Mathworld and got one involving matrices (which I didn't really absorb). So my subject line may well be inaccurate. (Initially I just looked at Harazi's comment).

Stop.
Are you sure your inequality is correct?

I have a proof, which I hope is correct. I can't see any obvious counter-examples either. If you have a counter-example, feel free to post it (and make me check my results more carefully in future). In any case, I will re-check my argument in response to your query.

I asked you, because it seems my argumentation, which I used for initial inequality, doesn't work, though I am not sure... :?

Even if no-one is interested in this (as they appear not to be), I will post my solution rather than waiting for Myth to dig it up in 18 months' time...

Firstly, we split the sum up into
\[(b^n+b^{n-1}c+...+c^n)+a(b^{n-1}+...+c^{n-1})+...+a^n\]
\[\geq (n+1)(\frac{b+c}{2})^n+...+a^n\]
Now letting $z=\frac{b+c}{2}$, we need
\[(n+1)z^n+nz^{n-1}a+...+2za^{n-1}+a^n \geq {n+2 \choose 2}(\frac{a+2z}{3})^n\]
Let the LHS be $f(z, a)$. Then the RHS is $f(\frac{a+2z}{3}, \frac{a+2z}{3})$. Rearranging, we have
$f(z, a)=(z^n+z^{n-1}a+...+a^n)+...+z^n \geq (n+1)(\frac{z+a}{2})^n+...+z^n = f(\frac{z+a}{2}, z)$.

Applying this repeatedly, we can see that we can approach arbitrarily close to the required result. By the continuity of the functions involved this implies the result.

[quote="Fiachra"]Even if no-one is interested in this (as they appear not to be), I will post my solution rather than waiting for Myth to dig it up in 18 months' time...[/quote]
Hmm... It is so roughly ;)  :P

Anyway, my congratulations! I don't know why it didn't work when I tried similar approach.

What about general case?

[b]General Case:[/b]

If $k$ and $n$ are positive integers and $x_1, x_2,...,x_k$ are positive real numbers, then
\[\sum (\prod_{i=1}^k x_i^{a_i}) \geq {n+k-1 \choose k-1}(\frac{\sum_{i=1}^k x_i}{k})^n\]
where the sum is taken over all distinct sequences of non-negative integers $(a_1,a_2,...,a_k)$ with $\sum_{i=1}^k a_i = n$.

Lemma: For all positive integers $n$ and $k$ and positive reals $a, b$,
\[\sum_{i=0}^n {i+k-2 \choose k-2}a^ib^{n-i} \geq {n+k-1 \choose k-1}(\frac{(k-1)a+b}{k})^n\]
We proceed by induction on $k$. We have already shown that this holds for $k=3$. Suppose it holds for $r-1$. Let $f(a, b)=\sum_{i=0}^n {i+r-2 \choose r-2}a^ib^{n-i}$ and note that $f(\frac{(r-1)a+b}{r},\frac{(r-1)a+b}{r})={n+r-1 \choose r-1}(\frac{(r-1)a+b}{r})^n$. Now $f(a, b)$ can be rearranged into
\[\sum_{j=0}^n(\sum_{i=j}^n {i+r-3 \choose r-3}a^ib^{n-i})\] which is $\geq \sum_{j=0}^n (\frac {(r-2)a+b}{r-1})^ja^{n-j}$ by the induction hypothesis. So $f(a, b) \geq f(\frac{(r-2)a+b}{r-1}, a)$. Applying this repeatedly we get arbitrarily close to $f(\frac{(r-1)a+b}{r},\frac{(r-1)a+b}{r})$. This proves the lemma.

Now suppose the result is true for $r-1$. We can break the LHS up according to the index on $x_1$ and apply the induction hypothesis. Letting $x_1=a$ and $\frac{\sum_{i=0}^{r-1} x_i}{r-1}=b$ we have that the LHS is at least $f(a, b)$. By the lemma we have that the RHS is at most $f(a, b)$. Result follows.

Dear Fiachra,

before studying your nice proof of the general case I would like to know how can you deduce from

$\displaystyle f(z, a) \geq f(\frac{z+a}{2}, z) $

that

$\displaystyle f(z,a) \geq f(\frac{a+2z}{3}, \frac{a+2z}{3}) $

I think it is not obvious as $f(z,a)$ is a non-symmetric function and I would like to see a rigorous proof of this fact that it is the central and very difficult point of your proof.

Thank you very much :)

[quote="manlio"]how can you deduce from
$\displaystyle f(z, a) \geq f(\frac{z+a}{2}, z) $
that
$\displaystyle f(z,a) \geq f(\frac{a+2z}{3}, \frac{a+2z}{3}) $[/quote]
It is just iteration of $f(z,a)\geq f\left(\frac{z+a}{2},z\right)$:
\[f(z,a)\geq f\left(\frac{z+a}{2},z\right)\geq f\left(\frac{\frac{z+a}{2}+z}{2},\frac{z+a}{2}\right)=f\left(\frac{3z}{4}+\frac{a}{4},\frac{z+a}{2}\right)\geq ...\to f\left(\frac{2z+a}{3},\frac{2z+a}{3}\right).\]


Pythagorean Theorem
geometry
similar triangles

Let M be the midpoint of side AB of equilateral triangle ABC, and let points N, S, and K divide side BC into four equal segments. Given that P is the midpoint of CM, prove that $\angle MNB = \angle KPN = 90^\circ$

[hide="for MNB"]
Since it's an equilateral triangle, and the bottom is cut into 4 equal sections, we can conclude that the top of the triangle is directly above the midpoint of the base, so if a line was drawn to the midpoint of the base, then it would split the triangle into two right triangles. N is between K (midpoint of base) and B, so it would be right below the midpoint of AB, and that's M, so we can see that if a line was drawn to n from M, it would form the right triangle MNB.[/hide]

As for KPN, my lightbulb isn't on at full power yet  :)

[hide]For $KPN$, we can find the lengths of the sides of triangle $KPN$.  If we call the length of segment $MB$    $x$, then $AB$ must be length $2x$ and it is easy to see that $KN$ must be of length $x$.  To make things simpler, we now send perpendiculars to segment $AM$ from $K$ and $N$, which will intersect segment $AM$ at $K'$ and $N'$, respectively. ($S$ is already connected at $P$ (midline theorem).)  To find the length of $KP$, we have right triangle $KK'P$.  Because $AKK'$ is a similar triangle to $ABM$, we know that the length of $KK'$ is $\frac{1}{4}x$, and we also know that the length of $K'P$ is $\frac{x\sqrt{3}}{4}$.  By the Pythagorean theorem, $KP=\frac{1}{2}x$.  Now we turn our attention to segment $NP$, for which we will use right triangle $NN'P$.  The length of $N'P$ is $\frac{x\sqrt{3}}{4}$, and the length of $NN'$ is $\frac{3}{4}x$, all because of the similar triangles and porportional segments, which leaves us with $NP$ of length $\frac{x\sqrt{3}}{2}$.  Now, a simple application of the Pythagorean theorem, or the fact that the triangles $KNP$ and $BMN$ are congruent (SSS), we can show that angle $KPN$ is a right angle, and therefore is congruent to angle $BNM$.
N.B. :  All this will become much more clear if you draw a large diagram and follow along.[/hide]


[quote="math man"]Find P(x,y,z) is a polynomial such that: exist Q(x,y,z) so that
(x+y+z).P(x,y,z)= $ Q(x^2,y^2,z^2)$[/quote]

$ (x\plus{}y\plus{}z)(x\minus{}y\minus{}z)\equal{}x^2\minus{}(y\plus{}z)^2\equal{}x^2\minus{}y^2\minus{}z^2\minus{}2yz$

$ (x^2\minus{}y^2\minus{}z^2\minus{}2yz)(x^2\minus{}y^2\minus{}z^2\plus{}2yz)\equal{}(x^2\minus{}y^2\minus{}z^2)^2\minus{}4y^2z^2$

And so you can take $ P(x,y,z)\equal{}(x\minus{}y\minus{}z)(x^2\minus{}y^2\minus{}z^2\plus{}2yz)$ and $ Q(x,y,z)\equal{}(x\minus{}y\minus{}z)^2\minus{}4yz$

[quote="math man"]thank you for your example,but i want a solution.[/quote]

I dont understand. Your question is "find a polynomial which ..." and I found one. What do you want else ?  What does mean "a solution"? Do you want all polynomails such ??? ? If so, just say it.

Clearly $ x \plus{} y \plus{} z$ divides $ Q(x^2,y^2,z^2)$.  Since we may exchange $ x$ by $ \minus{} x$ and the RHS is not changed, clearly $ \minus{} x \plus{} y \plus{} z$ also divides $ Q(x^2,y^2,z^2)$, or $ Q(x^2,y^2,z^2)$ is divisible by $ (x \plus{} y \plus{} z)(x \plus{} y \minus{} z)(x \minus{} y \plus{} z)( \minus{} x \plus{} y \plus{} z) \equal{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2 \minus{} x^4 \minus{} y^4 \minus{} z^4$ (necessary).  This is clearly a possible form of $ Q(x^2,y^2,z^2)$, as is the result of multiplying this polynomial by any polynomial of $ x^2,y^2,z^2$ (sufficient).  Hence any and all solutions are $ P(x,y,z) \equal{} (x \plus{} y \minus{} z)(x \minus{} y \plus{} z)( \minus{} x \plus{} y \plus{} z)Q'(x^2,y^2,z^2)$, where $ Q'(x^2,y^2,z^2)$ is any polynomial of $ x^2,y^2,z^2$ (including a constant).

[quote="hqthao"]pca so funny. solve the equation, we must find all the roots, right? why solve the polymial problem, you just find only a polymial  :)[/quote]

I'm sorry. Surely my english is quite poor.

I thought that "find a polynomial such ... " meant "find a polynomial such ..." and not "find all polynomials such ."

 :blush:

Can you help me to solve this problem:
Find n belong to N such that exist Polynomials P ,Q - n variables so that:
$ (x_1\plus{}x_2\plus{}....\plus{}x_n).P(x_1,x_2,x_3....,x_n)\equal{}Q(x_1^2,x_2^2,.......,x_n^2)$

[quote="hqthao"]pca so funny. solve the equation, we must find all the roots, right? why solve the polymial problem, you just find only a polymial  :)[/quote]
math man's problem, as stated, does not have a quantifier on $ P$.  It's slightly ambiguous because there is no "a" before $ P(x, y, z)$, but it's reasonable to parse the problem as "find a polynomial $ P$ such that..." as this is just a slight shuffling of what was actually written.  So it only asks to find an example of one such polynomial.

[quote="math man"]Can you help me to solve this problem:
Find n belong to N such that exist Polynomials P ,Q - n variables so that:
$ (x_1 \plus{} x_2 \plus{} .... \plus{} x_n).P(x_1,x_2,x_3....,x_n) \equal{} Q(x_1^2,x_2^2,.......,x_n^2)$[/quote]

can somebody help me with this ?


One of my homework questions....

[color=red]Part 1:[/color] In the following song, how many presents are received in total over the twelve days of christmas?

On the first day of Christmas,
my true love sent to me
A partridge in a pear tree.

On the second day of Christmas,
my true love sent to me
Two turtle doves,
And a partridge in a pear tree.

On the third day of Christmas,
my true love sent to me
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the fourth day of Christmas,
my true love sent to me
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the fifth day of Christmas,
my true love sent to me
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the sixth day of Christmas,
my true love sent to me
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the seventh day of Christmas,
my true love sent to me
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the eighth day of Christmas,
my true love sent to me
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the ninth day of Christmas,
my true love sent to me
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the tenth day of Christmas,
my true love sent to me
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the eleventh day of Christmas,
my true love sent to me
Eleven pipers piping,
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the twelfth day of Christmas,
my true love sent to me
Twelve drummers drumming,
Eleven pipers piping,
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree!

[color=blue]Part 2:[/color] How many presents would be given if the poem was about "The Twenty days of christmas"?

[color=green]Part 3:[/color] See if you can explain how would you work out how many presents would be recieved over any other number (n) of days.

Please help! :|

Do "two turtle doves" count as 1 or 2 presents?

The $ n$th verse has $ 1\plus{}2\plus{}\cdots\plus{}n$ gifts, which is the $ n$th triangular number. A method for the sum of triangular numbers can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=306743]here[/url].

See if you can go from there.

2 days..
mgmmm
We wish you a merry christmas, we wish you a merry christmas, we wish you a merry christmas, and a happy new year :P

[quote="darkspartan931"]I get part one and two but i need help on part 3 please[/quote]

Just read Power of Pi's blog entry that he posted carefully.

You can also derive the formula for the sum of the first $ n$ triangular numbers using Pascal's Triangle.  It's obvious that the second diagonal of Pascal's triangle is just the counting numbers: 1, 2, 3, 4, 5, ...
Using the hockey stick  identity, we can show that the third diagonal of Pascal's Triangle is the triangular numbers: 1, 3, 6, 10, 15,...
Using the hockey stick identity again, we can see that the forth diagonal gives the sum of the first $ n$ triangular numbers: 1, 4, 10, 20, 35,...
Note that the forth diagonal doesn't start until row 3 (The first row of Pascal's Triangle is row zero).  It is not hard to derive a formula for the nth number of the forth digaonal; it is 3rd element of row n+2 (The kth element of a row is counted from zero too).  We have a formula for this: it is C(n+2,3) which simplifies to
n(n+1)(n+2)/6.

And if all you were concerned with was proving the formula, you could also just use induction.

Could someone help me with part 3 in detail......please. I still dont understand. You guys have wonderful answers...but i still need help with part 3.

[hide="here are the steps"]\begin{align}S & =\frac{n(n+1)}{2}+\frac{(n-1)((n-1)+1)}{2}+\cdots+\frac{1(1+1)}{2}\\
\nonumber \\
& =\frac{n(n+1)+(n-1)((n-1)+1)+\cdots+1(1+1)}{2}\\
\nonumber \\
& =\frac{(n^{2}+n)+((n-1)^{2}+(n-1))+\cdots+(1^{2}+1)}{2}\\
\nonumber \\
& =\frac{(n^{2}+(n-1)^{2}+\cdots+1^{2})+(n+(n-1)+\cdots+1)}{2}\\
\nonumber \\
& =\frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{2}\\
\nonumber \\
& =\frac{\frac{n(n+1)(2n+1)}{3}+\frac{3n(n+1)}{3}}{4}\\
\nonumber \\
& =\frac{\frac{n(n+1)(2n+1+3)}{3}}{4}\\
\nonumber \\
& =\frac{ n(n+1)(2n+4)}{12}\\
\nonumber \\
& =\boxed{\frac{ n(n+1)(n+2)}{6}}\end{align}

1. We just add the first $ n$ triangular numbers.
2. Combine denominators.
3. Expand a little.
4. Regroup.
5. Use formulas for the sum of the first $ n$ squares and the sum of the first $ n$ positive integers.
6. Make common denominators.
7. Combine fractions.
8. Simplify fraction.
9. Divide both top and bottom by 2.[/hide]

If you don't understand any of the steps, please say the number.

Thank you power of pi. I dont understand steps 3, 4, 5, 6, 7, 8, and 9.

Oh and one more thing......this is the answer to part 3 right? :huh:

Thanks  :D

[quote="AdithyaGanesh"]Isn't the sum of the first $ n$ triangular numbers simply $ \frac {n(n \plus{} 1)}{2}$?

Aren't triangular numbers just natural numbers?  According to wikipedia, they are.[/quote]

I don't understand this post at all. The sum of the first n triangular number cannot be $ \frac{n(n\plus{}1)}{2}$, because that is the formula for the sum of the first n numbers.

Also, triangular numbers are not natural numbers. I don't know where on wikipedia you saw this.

http://en.wikipedia.org/wiki/Triangular_number

should give you some information.

:huh:  :huh:  :huh:

what is the correct answer ????????

[quote="BOGTRO"][quote="AdithyaGanesh"]Isn't the sum of the first $ n$ triangular numbers simply $ \frac {n(n \plus{} 1)}{2}$?

Aren't triangular numbers just natural numbers?  According to wikipedia, they are.[/quote]

I don't understand this post at all. The sum of the first n triangular number cannot be $ \frac {n(n \plus{} 1)}{2}$, because that is the formula for the sum of the first n numbers.

Also, triangular numbers are not natural numbers. I don't know where on wikipedia you saw this.

http://en.wikipedia.org/wiki/Triangular_number

should give you some information.[/quote]

This is straight from that very wikipedia page.

[quote]A triangle number is, equivalently, the sum of the n natural numbers from 1 to n.

[img]http://upload.wikimedia.org/math/2/4/9/249ba13e3f16a0de187bdf9da33cc49c.png[/img][/quote]

Unless I am misunderstanding it?

EDIT: Never mind, I get it.  That would mean the sum of the first triangular numbers is:

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

No, they are talking about the SUM of the first consecutive triangular numbers.

that is, $ 1 \plus{} 3 \plus{} 6 \plus{} 10 \plus{} 15 \cdots$

yes, you are right, a triangular number can be expressed as $ \binom{n \plus{} 1}{2}$

Thus, if we wanted the sum of the first n triangular numbers, we would get:

$ \binom{2}{2} \plus{} \binom{3}{2} \plus{} \binom{4}{2} \plus{} \binom{5}{2} \cdots \binom{n \plus{} 1}{2}$

which simply equals (by the hockey-stick identity)

$ \binom{n\plus{}2}{3} \equal{} \dfrac{(n\plus{}2)(n\plus{}1)(n)}{6}$

I believe the sum of the first $ n$ triangular numbers is also the $ n$th [url=http://en.wikipedia.org/wiki/Tetrahedral_number]tetrahedral number[/url].

Hey guys, you guys helped a lot, but i found my answer on this website:
[url]
http://www.squarecirclez.com/blog/the-twelve-days-of-christmas-how-many-presents/1686#comment-56275[/url]

:roll: thanx for telling the answer i was really confused........

that was a very confusing problem.
ans is all these added up
12x1
11x2
10x3
9x4
8x5
7x6
6x7
...
1x12
which is 12, 22, 30, 36, 40, 42 added together then x2
which is 364 in the first twelve days.
i wish i had that many presents =D


trigonometry
integration
calculus
real analysis
real analysis unsolved

For what $ \alpha \in \mathbb{N}$ does $ \sum\limits_{n \equal{} 1}^{\infty}(\sin(\pi e n!))^\alpha$ converge?  When does it converge absolutely?

I proved that it did not converge absolutely by expressing $ \sin(\pi e n!) \equal{} \int_0^\pi n! e \cos (e n! x)dx$.  If we assume it converges absolutely, we can use Fubini to flip the sum and integral, getting a contradiction.  This also proves it does not converge for even $ \alpha$.

To prove it converges for $ \alpha \equal{} 1$, it's pretty straightforward to show that the series satisfies the hypotheses of the alternating series test.  By comparison, then, it converges for odd $ \alpha$.

Is there a simpler way to prove this?

Of course, the trick is that $ e\equal{}\sum_{k\ge 0}\frac 1{k!}$. Fubini has nothing to do with the question. The condition $ \sum_k\Bigl|\int f_k\Bigr|<\plus{}\infty$ is very far from sufficient for swapping the sum and the integral. Besides, I really wonder how you computed the integrals and how you managed to "prove" that it does not converge for even $ \alpha$ using that integral representation.:? It is true, that if you show that a series $ \sum_k a_k$ is alternating and $ |a_k|$ decrease to $ 0$, then the series $ \sum_k a_k^\alpha$ has the same property and, therefore, also converges but to say that the second series converges "by comparison" is sloppy language, at best (it makes one think that you believe that if $ \sum_ka_k$ converges and $ b_k$ have the same signs as $ a_k$ and satisfy $ |b_k|<|a_k|$, then $ \sum_k b_k$ also converges).

So, think of the problem a bit more. So far your answer is wrong and your techniques are questionable :(

Thank you for pointing out my error with applying Fubini.

"By comparison" just meant that it satisfied the hypotheses of the alternating series test for the same reasons.

For big $ n$: $ a_n\equal{}\sin(\pi e n!)\equal{}\sin \left(\pi\plus{}\frac{\pi}{n\plus{}1}\plus{}O(n^{\minus{}2}) \right)\equal{}\minus{}\frac{\pi}{n\plus{}1}\plus{}O(n^{\minus{}2})$

$ a_n\approx \minus{}\frac{\pi}{n}$

Now, if $ \alpha$ is even then $ a_{n}^{\alpha}\approx \frac{\pi^{\alpha}}{n^{\alpha}}$ and converges if $ \alpha > 1$.

If $ n$ is odd then $ a_{n}^{\alpha}\approx \minus{}\frac{\pi^{\alpha}}{n^{\alpha}}$ and series alsow converges if $ \alpha > 1$.

So, it converges for all naturals $ \alpha\geq 2$.


limit
real analysis
real analysis unsolved

Let k be a positive integer, 0<q<1. Find the limit:
$ \lim_{n\rightarrow \plus{} \infty}(k \plus{} (k \plus{} 1)(^k_1)q \plus{} (k \plus{} 2)(_2^{k \plus{} 1})q^2 \plus{} ... \plus{} (k \plus{} n)(_n^{k \plus{} n \minus{} 1})q^n)$

Uh- all of those binomial coefficients are zero, at least for $ k>1$. Did you perhaps mean $ \binom{j\plus{}k\minus{}1}{j}$?

[quote="jmerry"]Uh- all of those binomial coefficients are zero, at least for $ k > 1$. Did you perhaps mean $ \binom{j \plus{} k \minus{} 1}{j}$?[/quote]
oh, I've just edited it  :D

\[ \sum_{j \equal{} k}^\infty j\binom{j \minus{} 1}{j \minus{} k}q^{j \minus{} k} \equal{} k q^{ \minus{} k}\sum_{j \equal{} k}^\infty \binom{j}{k}q^{j}\]
But
\[ \sum_{k \equal{} 0}^\infty x^k \sum_{j \equal{} k}^\infty \binom{j}{k}q^{j} \equal{} \sum_{j \equal{} 0}^\infty q^j \sum_{k \equal{} 0}^j \binom{j}{k} x^k \equal{} \sum_{j \equal{} 0}^\infty q^j (1 \plus{} x)^j \equal{} \frac1{1 \minus{}(q \plus{} qx)} \equal{} \sum_{k \equal{} 0}^\infty x^k \frac {q^k}{(1 \minus{} q)^{k \plus{} 1}},\]
whence
\[ \sum_{j \equal{} k}^\infty \binom{j}{k}q^{j} \equal{} \frac {q^k}{(1 \minus{} q)^{k \plus{} 1}}.\]
Finally,
\[ \sum_{j \equal{} k}^\infty j\binom{j \minus{} 1}{j \minus{} k}q^{j \minus{} k} \equal{} \frac {k}{(1\minus{} q)^{k \plus{} 1}}.\]
(There is also a shorter solution through Taylor formula.)

zhoraster, Thank you so much!
But I haven't found the solution using Taylor formular. Can you show me it? :lol:

By the way, there was some mess with signs -- check the edited post.
\[ \sum_{j \equal{} 0}^\infty\binom{j \plus{} k}{k}q^{j} \equal{} \sum_{j \equal{} 0}^\infty \frac {(k \plus{} j)(k \plus{} j \minus{} 1)\dots(k \plus{} 1)}{j!} q^j \equal{} \sum_{j \equal{} 0}^\infty \frac {f^{(j)}(0)}{j!}  q^j \equal{} f(q),\]
where $ f(q) \equal{} (1 \minus{} q)^{k \plus{} 1}$.


I know there was discussion about Mu Alpha Theta long time ago (I remember reading it) but this question is little bit different.

Question: Among the many questions available at MAO at site, which questions are good questions that I can try on?

I found this page:

http://www.mualphatheta.org/National_Convention/MAO_Past_Tests.htm

It has bunch of tests...

But i don't know which one i should do...

The reason I'm planning to do MAO problems is that because I found they're way harder than old AHSME's and I'm getting okay and easy with AHSME part (at least on algebraic) and I know current AMC's are easier than AHSME so I really don't worry much.

But I'm just thinking if I can handle MAO, I will SURE won't have much problem on AMC.

Suggestions?

Thanks.

Just glancing over the MAT questions, I found them to be very simple and regurgitative in nature.  I could do all of those fairly easily; however, it's unlikely that I would score 150 on an AMC (though possible on a good day).

AMC, though easier than the AHSME, is still significantly harder than that.

I've written lots of Mu Alpha Theta (MAO) problems (many state conventions, and three national conventions), and will give you my opinion of their difficulty.

First, there are three divisions for MAO competitors: Mu, Alpha, and Theta.  These have meant different things at different times, but currently (and for the past few years) have been stable.  Mu is for students who are taking (or have taken) a calculus class.  Alpha is for students who are taking (or have taken) a class _beyond_ Algebra II (sometimes called Algebra 3/4).  Theta is for all other students (in lower classes).  From a difficulty standpoint, Mu is harder than Alpha, which is harder than Theta.

Second, there are many different test formats at MAO national conventions.  Every student takes an "Individual Test", so this has both easy questions (so that everyone can get a modest score) and hard questions (so that the most experienced competitors can be distinguished from one another for awards).  The "Individual Test" is usually X problems in 2X minutes.  Every student also takes a "Ciphering Test".  This usually consists of 10-15 questions administered individually, with students receiving different numbers of points depending on the speed with which they solved the problem.  Time limits are usually 1, 2, and 3 minutes.  In many years, this has been a very difficult test, while in others it has included both easy and hard problems (it is desirable to avoid having many zero scores :-).

There are also many "Topic Tests" at the convention.  These are in a format similar to the individual test, but all the problems relate to a specific topic (e.g. Geometry, Number Theory, or Advanced Calculus).  These are easy or hard depending on how familiar you are with the topic, and all of them include problems to challenge the most experienced competitors.  "Gemini Tests" are for two people, and frequently are more difficult than "Individual Tests".  The "School Bowls" are tests for the top four students from each school, while the "State Bowls" are tests for the top four students from each state, and typically are quite difficult.  The formats of these tests change somewhat from year to year.  There is usually a test simultaneous with the "School Bowl" that is for the other students (not the top four in their school), so it tends to be easier.  There is also an "Interschool Test" which typically takes place over several days and allows your entire team to collaborate.  This test usually includes very difficult and creative problems.

Mu Alpha Theta has asked my company to administer the 2004-2005 Mu Alpha Theta "Log1" Contest, a free contest consisting of three rounds modeled on the "Individual Test", "Topic Tests", and "Ciphering Test" at national conventions.  Our Ciphering Time Trials is in the same format as the "Ciphering Round" of the "Log1" Contest, too.  Finally, our Collaborative Problem-Solving Contest gives teams one week to work together to solve fifteen complex and difficult problems, and has the flavor of the "Interschool Tests" at their national conventions.  Samples of all of these contests are available on our website.

So, what test on Mu Alpha Theta should I try?

I've noticed that the questions on AoPS book that is from MAO are very good questions to look.

Silverfalcon

As a serious MAO competitor, the difference between National Convention tests (the site you found) and regional tests is pretty large. For example, students who are practicing for the regional competitions (once every month starting in Jan.) don't use National Convention tests. Likewise, students studying for National Convention don't practice with regional tests.

I have found, and note that this is my opinion, that MAO tests are very regurgitative in nature. The ways to solve most questions usually involve rote processes and the top placers usually are the students who take the most practice tests or remember the most rote methods. I myself do it. I know how it works. It IS the path to success in MAO competitions.

MAO math does offer good things though. I know many of the rote methods, but I also understand how most formulas and processes work. This helps understand non-MAO math. It is the same abstract thinking. So as long as you follow the right side of the MAO math, it will end up to your benefit.

Also, don't assume that every MAO test is regurgitative. It always depends on the test writers. When a test that doesn't involve these rote methods but rather advanced problem solving skills, the scores tend to be lower. There are many MAO tests out there that are not MAO-like.

on my website, http://www.jptmath.com , under [i]Links[/i]. There is a section that has links to downloadable practice tests (meaning past MAO tests). You can try those.

I agree with Chion that the difference between National Convention tests (the site you found) and regional tests is pretty large.  In fact, the difference between any national test and another is pretty large.  :-)  That said, the students I know who do well regionally and nationally tend to use any test they can get their hands on, rather than focus just on the one competition that is approaching.  I think the important thing is to get exposure to lots of math, rather than to get exposure just to the math that is most likely to show up at the next contest.  For regional tests, there are some topics that are more heavily emphasized (especially in Florida), but for the national ones it depends a lot on the test-writers.  Because I'm in charge of the MAO National Convention tests this summer, they will probably be more like the 2001 Denver tests than others, for example.

Chion said that [quote]MAO tests are very regurgitative in nature.[/quote]  This is certainly true for the initial problems on each test, and is the intention of most test-writers.  There are several purposes for these tests, one of which is to encourage students to enjoy mathematics, and exceptionally low scores for less-experienced competitors are therefore considered undesirable.  The problems on most of these tests cover a range from very easy to very difficult, with the last five problems typically requiring very insightful approaches.  Chion also said that  [quote]the top placers usually are the students who take the most practice tests or remember the most rote methods.[/quote]  I must confess that I believe this is a fantastic way to succeed at anything:  practice.  I did this when I learned to ride a bike, I did this when I learned to drive, and I would encourage people to do this when learning mathematics.  The more math you're exposed to, the more you'll learn and the better your scores will be.  This includes "rote methods", which are often helpful in solving non-rote problems.

Chion also said that [quote]I also understand how most formulas and processes work.[/quote]  I think this is the key to taking a rote problem and extending it to non-rote problems.  All those formulae that beginning competitors memorize are based on fundamental relationships.  If you use a formula repeatedly with an eye towards trying to learn what it is based upon, this will often give you deeper insights into the fundamental relationships for problems of this type, and will allow you to solve non-rote problems that extend upon typical rote problems.


modular arithmetic
number theory unsolved
number theory

Determine all triplets off positive integers $ (a,b,c)$ for which $ \mid2^a\minus{}b^c\mid\equal{}1$

[quote="Matematika"]Determine all triplets off positive integers $ (a,b,c)$ for which $ \mid2^a \minus{} b^c\mid \equal{} 1$[/quote]

If $ c\equal{}1$, we have the trivial solutions $ (a,b,c)\equal{}(a,2^a\pm 1,1)$
Consider now $ c>1$

1) $ b^c\equal{}2^a\plus{}1$
$ b$ is odd $ >1$ (else $ 2^a\equal{}0$) and so $ b\equal{}2^ku\plus{}1$ with $ u$ odd and $ k>0$
$ (2^ku\plus{}1)^c\equal{}2^a\plus{}1$ $ \implies$ $ 2^kcu\equal{}2^a\pmod{2^{k\plus{}1}}$ and, since $ k<a$ (since $ c>1$) $ c\equal{}2d$ even.
The equation becomes $ (b^d\minus{}1)(b^d\plus{}1)\equal{}2^a$ and so $ b^d\minus{}1\equal{}2$ and so $ b\equal{}3$ and $ d\equal{}1$ and the unique solution $ (a,b,c)\equal{}(3,3,2)$

2) $ b^c\equal{}2^a\minus{}1$
If $ b\equal{}1$, we have the solution $ (a,b,c)\equal{}(1,1,c)$
Else, $ a>1$ and $ b$ is odd $ >1$ and so $ b\equal{}2^ku\plus{}1$ with $ u$ odd and $ k>0$
$ (2^ku\plus{}1)^c\equal{}2^a\minus{}1$ $ \implies$ $ 2^a\equal{}2\pmod{2^{k}}$ and so (since $ u>0$ and $ c>1$ imply $ k<a$) $ k\equal{}1$
$ (2u\plus{}1)^c\equal{}2^a\minus{}1$ $ \implies$ $ 2cu\equal{}2\pmod 3$ and so $ c\equal{}2d\plus{}1$ odd.
The equation becomes $ b^{2d\plus{}1}\plus{}1\equal{}2^a$ and so $ b\plus{}1\equal{}2^u$
$ (2^u\minus{}1)^{2d\plus{}1}\equal{}2^a\minus{}1$ $ \implies$ $ (2d\plus{}1)2^u\equal{}2^a\pmod{2^{u\plus{}1}}$ and so $ a\equal{}u$ and $ c\equal{}1$, impossible (since we considered $ c>1$)

3) Synthesis of solutions :
$ (a,b,c)\equal{}(a,2^a\pm 1,1)$
$ (a,b,c)\equal{}(3,3,2)$
$ (a,b,c)\equal{}(1,1,c)$

Hi pco,

part (1) of your solution is correct. But unfortunately, part (2) is wrong.

2) $ b^c\equal{}2^a\minus{}1$
If $ b\equal{}1$, we have the solution $ (a,b,c)\equal{}(1,1,c)$  SO FAR EVERYTHING IS FINE.

Else, $ a>1$ and $ b$ is odd >1 and "so $ b\equal{}2^ku\plus{}1$ with $ u$ odd and $ k>0$"

The claim within quotes is wrong. How does one know that $ b$ when divided by $ 2^k$ leaves a remainder of 1.

I am including herewith my solution to the second part.

We have $ a \geq 2$. Hence, $ 2^a \minus{} 1 \equiv 3 (\;\text{mod} \;4)$. If $ c$ is even, then $ b^c \equiv 1 (\;\text{mod} \;4)$.
Therefore, $ c$ is odd and we have
$ 2^a \equal{} b^c \plus{}1 \Rightarrow 2^a \equal{} (b\plus{}1)(b^{c\minus{}1} \minus{}b^{c\minus{}2} \plus{}...\plus{}1) \Rightarrow (b\plus{}1) \equal{} 2^{l}$ where $ l \leq a$.
$ b \equal{} 2^l \minus{}1 \Rightarrow b^c \equal{} (2^l \minus{} 1)^c \equal{} 2^{lc} \minus{} c 2^{l(c\minus{}1)} \plus{}...\plus{}c 2^{l} \minus{} 1 \equal{} 2^a \minus{}1$
$ \Rightarrow 2^{lc} \minus{} c 2^{l(c\minus{}1)} \plus{}...\plus{}c 2^{l} \equal{} 2^a$
$ \Rightarrow 2^{l(c\minus{}1)} \minus{} c 2^{l(c\minus{}2)} \plus{}...\minus{}\frac{c(c\minus{}1)}{2}2^{l} \plus{} c  \equal{} 2^{a\minus{}l}$
If $ l < a$, then the LHS is an odd integer and the RHS is an even integer. Hence, $ l\equal{}a$. This implies $ b \equal{} 2^{a} \minus{} 1 \Rightarrow c \equal{} 1$. Hence, the solutions for $ a \geq 2$ are $ (a,b,c) \equal{} (a,2^{a}\minus{}1,1)$.

[quote="nicetry007"]unfortunately, part (2) is wrong.

2) $ b^c \equal{} 2^a \minus{} 1$
If $ b \equal{} 1$, we have the solution $ (a,b,c) \equal{} (1,1,c)$  SO FAR EVERYTHING IS FINE.

Else, $ a > 1$ and $ b$ is odd >1 and "so $ b \equal{} 2^ku \plus{} 1$ with $ u$ odd and $ k > 0$"

The claim within quotes is wrong. How does one know that $ b$ when divided by $ 2^k$ leaves a remainder of 1.
[/quote]

If $ b$ is odd$ >1$, $ b\equal{}v\plus{}1$ with $ v$ even$ >0$. Let then $ k$ be the greatest power of $ 2$ which divides $ v$. We can then write $ v\equal{}2^ku$ with $ k>0$ and $ u$ odd.

So $ b\equal{}2^ku\plus{}1$ for some integer $ k>0$ and some odd positive integer $ u$.

And, fortunately, this part of my demo, at least, is right.  :wink:

Hi Pco,

  I'm sorry about that. But I still cannot understand how you were able to conclude
"$ (2u\plus{}1)^c\equal{}2^a\minus{}1 \implies 2cu\equal{}2\pmod 3$ and so $ c\equal{}2d\plus{}1 odd$". Could you please explain it.

[quote="nicetry007"]Hi Pco,

  I'm sorry about that. But I still cannot understand how you were able to conclude
"$ (2u \plus{} 1)^c \equal{} 2^a \minus{} 1 \implies 2cu \equal{} 2\pmod 3$ and so $ c \equal{} 2d \plus{} 1 odd$". Could you please explain it.[/quote]

No prob. Dont hesitate to ask. I'm always glad when I see that someone reads my demos.

Here, there is a typo error : it's mod 4, not mod 3 :

$ (2u \plus{} 1)^c \equal{} 2^a \minus{} 1$  $ \implies$ $ 2cu \equal{} 2\pmod 4$  :
$ (2u\plus{}1)^c\equal{}2cu\plus{}1\pmod 4$ since all other terms of $ (2u\plus{}1)^c$ are $ \equal{}0\pmod 4$ and $ 2^a\equal{}0\pmod 4$ since $ a>1$ so $ 2cu\plus{}1\equal{}\minus{}1\pmod 4$ and $ 2cu\equal{}\minus{}2\equal{}2\pmod 4$

 and so $ c \equal{} 2d \plus{} 1$ odd (else $ 2cu\equal{}0\pmod 4$)

The solutions are $(a,2^a\pm1,1),(1,1,c),(3,3,2)$. If $c=1$, then $|2^a-b|=1$, so $b=2^a\pm 1$, this yields the solutions $(a,2^a\pm1,1)$. Assume $c>1$. If $a=1$, then $b=1$. This gives the solutions $(1,1,c)$. Finally, if $a>1$, then we have two consecutive perfect powers. Mihaelescu's theorem then asserts they must be $8$ and $9$, so $a=3,b=3,c=2$. $\blacksquare$
To avoid appealing to Mihaelescu, we can do the following:
a) If $b^c-1=2^a$, we have $(b-1)(b^{c-1}+\ldots+b+1)=2^a$. Evidently, $b$ is odd. Also, $b-1$ and $b^{c-1}+\ldots+b+1$ are powers of $2$. Finally, $\gcd(b-1,b^{c-1}+\ldots+b+1)=\gcd(b-1,c)$. Since $b-1,b^{c-1}+\ldots+b+1>1$, it follows $c$ is even. Hence, $(b^{c/2}-1)(b^{c/2}+1)=2^a$, so $b^{c/2}=3$, giving $b=3,c=2$, and $a=3$.
b) If $b^c+1=2^a$, we know by Zsigmondy that $b^c+1$ has a primitive prime factor or $b=2,c=3,2^a=9$. However, since the latter case is impossible and $b^c+1$ is a power of $2$, this must mean $b+1$ is odd. However, $b$ is clearly odd, so this case yields no solutions.

Take $a,c>1$ and use catalans conjecture


Prove that in 3-space, a unit sphere cannot be simultaneously tangent to 13 other nonoverlapping unit spheres.

EDIT> Thanks for the link!