Welcome to Mafia Game 38, Assassins in the Palace!

[b]Opening Post[/b]
[i]Life at the palace was quite mundane. The khalifs, ruling the nameless, unknown nation... not doing much really. Just sitting there, attending to some foreign business... All was well. All was boring.
For the guards, it wasn't much better. After all, the life of a guard wasn't that exciting. You kinda just sat there, waiting for something to happen. But of course, you hoped nothing would happen.
Really though, life was boring as heck. One day, monotony broke as a shooting star blazed across the sky. The guards and the khalifs all wished for something exciting in their lives...
And they got what they wished for. The next day, a messenger came bursting in through the door, writhing in pain. "Assassins! Two of them attacked me... look just like guards... From... from our neighboring country... came to kill the khalifs... protect at all cost..." And as he fell to the ground dying, he handed the khalifs two guard uniforms...
That night, during the head count, there were 4 more guards than normal. And the khalifs... where had they gone? Oh, they were in the group! That was strange... but every guard knew not to say anything about it. A message was posted on the wall:
"There are 14 of you here. 10 of you, regular guards, just doing your jobs. Well, you'll definitely be doing it tonight. 2 of you are assassins, ready to kill the khalifs without regards to your own lives... Guards, if your khalifs are to survive, you've got to find out who the assassins are... and eliminate them."
One guard (or khalif... or assassin) cried out, "Eliminate? How are we gonna do that?"
Out came a response, "Well, we could vote..."
"Vote? What's that?"
"Oh, it's something I heard about before... demokercy or something like that... basically, we should kill whoever gets voted on the most!"
And so, 14 of them stood, pointing fingers, pointing spears, pointing daggers...[/i]

[u][b]Players[/b][/u]:
[b]Alive[/b]
1. mewto55555
2. crazychinesemaniac
3. Hamster1800
4. cf249
5. Peter
6. chenhsi
7. james4l
8. pacman2812
9. erabel
10. 1=2
11. zapi2007
12. jnub
13. 142857
14. BoesFX

[b]Dead[/b]
None... yet.

[b]Setup[/b]
There are 10 guards, 2 khalifs, and 2 assassins. The assassins win if the khalifs are eliminated, whether or not the assassins are alive. The khalifs and the guards win if the assassins are eliminated and at least one of the khalifs still lives. The guards know the identity of the khalifs, but not each other. The king and the assassin have no further information. Please note that this is a nightless setup.

[b]Role PMs[/b]
[quote="Guard"]You are a guard. Your job is to protect the two khalifs, who are ______ and _______. Unfortunately for you, there are two assassins in the palace making your job quite difficult. You win the game with the khalifs if the assassins are eliminated and at least one khalif is still alive.[/quote]
[quote="Khalif"]You are a khalif. Lucky you. Too bad there's two assassins out to get you! You win the game with the other khalif and the guards if the assassins are eliminated and you or the other khalif is still alive.[/quote]
[quote="Assassin"]You are the assassin. Your job is to kill the khalifs. Your partner is ______. If you are lynched, then you may make a suicide kill. If you are the last assassin alive (the second lynched), then you may make an amount of suicide kills equal to the amount of khalifs alive. You win the game if both khalifs are dead, regardless of whether or not you or your partner is still alive.
In addition, you may also suicide kill during the day by posting [b]Suicide Kill: Player[/b]. If you are the last assassin, and there are two khalifs left, you may perform 2 suicide kills. This is essentially a way to lynch yourself.[/quote]

[b][u]The Rules[/u][/b]:[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Please be attentive and [b]unvote[/b], if necessary, before casting a new vote.  This is not [i]required[/i], but I'd appreciate it.
[*]Executions will require a simple majority of votes.  Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may [i]NOT[/i] [b]vote: no lynch[/b]. (Why would you need to in a nightless game?)
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
[*]Once your death scene has been posted, you're dead.  Stop typing.
[*]Don't edit/delete previously submitted posts
[*]Don't quote any PMs from me.
[*]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, tough!  No choice will be made.
[*][b]If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread.[/b]  Treat this game as a commitment.  Be considerate - don't leave us hanging.
[*]Remember, it's a game.  It's supposed to played.  It's supposed to be [i]fun[/i].  Do your part.
[*]If I deem it necessary, I will instate a deadline. When deadline is reached, the player with the most votes is lynched. If there is a tie, whoever received the last vote first will be lynched. Please don't make me have to put the deadline.[/list]
[quote="Klebian"] Other notes. Communication outside of the thread, unless explicitly stated in your role PM, is prohibited. This is a modkillable and bannable offense. Talking after death, and discussing games that are ongoing, even if you are dead, are also prohibited, and bannable/modkillable offenses. Editing or deleting posts once they have been entered is also bannable/modkillable, even if no one has posted after you. If you need to make an edit, post again. Please play fairly, but have fun. Mafia is a good way to get better at using logic, don't ruin the game for others. Also, don't sign up if you're going to be gone for periods of time longer than about 3 days without internet access. If you can't access this site daily, it's also better to not sign up. Inactivity spoils games, and I'll make sure that those players who are inactive and need to be replaced won't be signing up for the next game immediately.[/quote]
Yeah, follow that too.

[b][u]Game Specific Rules[/u][/b]:[list][*]Communication: With the exception of assassins, no one may communicate. Assassins may daytalk by PM (or QT) for the first 120 hours of the game. After that, they cannot talk except during the phase where one of them is lynched (see below).
[*]Night: This game has no night. Seriously.
[*]When a lynch is completed, it is the [b]twilight[/b] stage. You may keep talking in the twilight stage. If the person lynched is a guard, the next day immediately begins. If the person lynched is an assassin, then the thread is pseudo-locked (please please please do not post in it) for 72 hours. During the 72 hours, the assassins may communicate (supposing both are alive) and decide to kill. Then the assassin PMs me his choice, and then the game continues. (Unless he's the last assassin, of course.)
[*]Do not make a real or fake claim about being khalif. Do not speculate in thread about who the king is (even though 10 of you know). Do so, and I will modkill you.
[*]Do not fake a day suicide kill. Do so, and I will modkill you.[/list]
I may have forgotten some rules, so use your judgment if you aren't sure.

It is now the beginning of Day 1! (and also kinda the confirmation phase, but we already did that, so it should be ok!)
Role PMs have almost been all sent out.

Go!

EDIT: Ok, all role PMs should be out now.

/confirm.

You should also mention that with 14 alive, it takes 8 votes to lynch.

Random lynch!

[b]Vote:Peter[/b] for making this thread!

confirm. Brut3Forc3: actually 12 people know who the khalifs are.

Bad idea: RANDOM LYNCH. There's a probability of more than .1 that we lynch someone very important.
Bad idea: SPAM. This is a game where we will need to look back at previous posts. Do not spam or that will be tedious.
Bad idea: INACTIVENESS. Please please PLEASE don't be replaced or this game will be made harder.
Bad idea: OMGUS VOTING. Like bad idea #1, we need some people alive to win. We can't just go lynching people.

From 1 and 4, [b]Big FoS: mewto55555[/b]

I just realized that the first and last didn't make sense. So no spamming and no inactiveness.

OMGUS=oh my gawd u suck

Oh ok. So if this setup is not good for random voting, [b]Unvote[/b].

OMGUS = oh my god you suck.

[quote="1=2"]Bad idea: RANDOM LYNCH. There's a probability of more than .1 that we lynch someone very important.[/quote]We know who the khalim are, so among the others we can only lynch guards (not 'important') and assassins (should be killed). Your words don't make sense.

Also, I believe the khalims don't know eachother, so technically only 11 people know who they are (even only 10 know both).

---
To the moderator of this game. You have a nice setup and I appreciate the new ideas, but there are some things I'd like to clear out from the start.

[quote="Brut3Forc3"]Executions will require a simple majority of votes. Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may NOT vote: no lynch. (Why would you need to in a nightless game?) [/quote]Out of curiosity: what happens when there are an even amount of players left, and they can't come to a lynch? Imagine an end game with 1 assassin and 1 khalif, then this is a draw? Or 1 assassin, 1 guard, 2 khalim and both khalim disagree on who is the last guard? Usually this is solved by a no lynch (and disbalanced by the following night kill), but here such a state would be steady.

[quote="Brut3Forc3"]Do not make a real or fake claim about being khalif. Do not speculate in thread about who the king is (even though 10 of you know). Do so, and I will modkill you.[/quote]Sorry but can you be more specific? It is vital to the town's strategy to what level we can mislead the assassins, and a fake khalim suggestion (or claim or speculation) is by no doubt the best way to do it. Please, can you formally define what you call a claim and a speculation?

@Peter: Exactly why I said that they made absolutely no sense. But I highly recommend not spamming and being reasonably active. Also in the Khalif sample PM he/she/it ( :P ) wasn't told the other khalif.

I have a sneaky suspicion that at least of one of the un-confirmed is an assassin.

1=2 and Peter: Each khalif does not know the identity of the other. So Peter is right.
[quote="Peter"][quote="Brut3Forc3"]Executions will require a simple majority of votes. Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may NOT vote: no lynch. (Why would you need to in a nightless game?) [/quote]Out of curiosity: what happens when there are an even amount of players left, and they can't come to a lynch? Imagine an end game with 1 assassin and 1 khalif, then this is a draw? Or 1 assassin, 1 guard, 2 khalim and both khalim disagree on who is the last guard? Usually this is solved by a no lynch (and disbalanced by the following night kill), but here such a state would be steady.[/quote]
Oops, in my rush to write the stuff, I forgot to put an important mechanic... :(
It's now reflected in the rules and new Role PMs. Basically, the assassin can "lynch" himself and perform suicide kill(s).

[quote="Brut3Forc3"]Do not make a real or fake claim about being khalif. Do not speculate in thread about who the king is (even though 10 of you know). Do so, and I will modkill you.[/quote]Sorry but can you be more specific? It is vital to the town's strategy to what level we can mislead the assassins, and a fake khalim suggestion (or claim or speculation) is by no doubt the best way to do it. Please, can you formally define what you call a claim and a speculation?[/quote]Okay... well, the purpose of this rule is to prevent a broken game where one townie just slips the names of the khalifs, making an assassin win inevitable. I'll make...
Claim: Saying you are a khalif. (Whether or not this is true)
Speculation: Explicitly saying something like "Oh, Player X must be a khalif". Or "Player X couldn't be a khalif"...
I will probably be pretty loose on this, since it's really something that protects the town from being too stupid... How all of you implicitly suggest who is or isn't khalif is up to you. (I will definitely let that slide.)
If you aren't certain, you can always PM me your post and I'll tell you if it's okay or not.

So, just to clarify, if the last assassin kills the last khalif with a suicide kill, the assassins win.  Correct?

Hehe yeah, should I say so?

[hide="Mod delete if not"]
zapi and ccm
[/hide]

Wow i found my role pm, i must have been wrong the entire game? It was chenhsi, guess I can't read! Who was the assassin? CCM?

I was not...peter can tell us though I think.

Zapi was pretty obvious (but not by his own fault, just the fact that people obeyed him too quickly without arguements). For the second I had still no idea (although only erabel/chensi/ccm are left at this point)

Anyway james4l was my assassin partner. You were about to lynch pacman, so even if you lynched j4l right after that, he still had a 1/3 winning chance. Not bad if you ask me.

Yeah, when I posted my "analysis", I figured zapi was pretty obvious. Then I threw in cf249 hoping he wasn't an assassin. I figured since this was one of my early games, the ploy would work and the assassin would believe cf249 was also khalif. It worked!

So you guys weren't thinking when you lynched me. I put a vote on a khalif and you all were like "OMG HE DOESN'T KNOW WHO THE KHALIM ARE LYNCH LYNCH LYNCH," and I got reflex voted. Then I thought that you (mewto) was an assassin because I didn't think that you would unwittingly give away one khalif but not both. If it was actually intentional, it was a good trick, but I couldn't believe it was intentional even after you were revealed to be a guard (and hence "knew" who the khalim were"). Most of my decisions were made without looking at who was saying them, to aviod giving away information such as what Peter picked up abuot zapi. The only time I looked up the khalim was to intentionally place a vote on one, but not enough people picked up on my strategy. I had to be vague or else the assassins would have known chenhsi to be a khalif.

Also, @mewto: my whole "think about it" thingy was completely lost on you because you didn't realize that chenhsi was a khalif. Oops.

I'm curious as to what the assassins thought of my vote on chenhsi and the subsequent discussion, since it was the last big discussion which took place, however vague it was.

[quote="alexhhmun"]Avoiding voting for a khalif is one of the surest ways to give one away.[/quote]

But by voting for one your risking giving him/her away and getting yourself killed.

[quote="Hamster1800"]I'm curious as to what the assassins thought of my vote on chenhsi and the subsequent discussion.[/quote]I didn't understand the point of the discussion, but I didn't feel like asking further info.
But it didn't change my opinions on chensi at all, I expected the khalim to get a deliberate vote somewhere.

[quote="alexhhmun"]Avoiding voting for a khalif is one of the surest ways to give one away.[/quote]
voting for a khalif is good and a must, or else the assissins will know who is the khalif, i.e. the one that hasn't been voted.
i intentionally vote one of the khalif and get a omgus vote from the other khalif and was lynched.  :(

[quote="crazychinesemaniac"][quote="cf249"]Wow, I thought pacman was an assassin for sure...  Anyway, good game (such as it was) to all.[/quote]

Lol...pacman2812 created a really interesting situation when he voted for a khalif. I remember sitting their not knowing if I should vote or not because I didnt want to give away a khalif.[/quote]
Wrong. I didn't (Hamster was)

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


[quote="RyanLee2012"]Two circles and three straight lines lie in the same plane. If neither the circles nor the lines are coincident, what is the maximum possible number of points of intersection?

This question was from my high school's ciphering contest.
Is there a pattern or formula to get to the solution?  :D[/quote]
The number of intersection points I worked out is:
[hide]$ 17$
[/hide]

Here is how I approached it:
[hide]
Draw one circle and a line through it so the maximum number of points is $ 2$.

Draw a second line through the circle and the first line so the maximum
number of *added* points is $ 3$.

Draw a third line through the circle and intersecting the first two lines so that
the maximum number of *added* points is $ 4$.

Then draw the second circle so that it intersects the first circle for a maximum
number of $ 2$ points for those circles, *and* at the same time, make sure the
second circle intersects the three lines for a maximum number of *added*
points of $ 6$ for the second circle and the three lines.

$ 2 \plus{} 3 \plus{} 4 \plus{} 2 \plus{} 6 \equal{} 17$

$ 17$ total intersection points
[/hide]

[hide="A Similar Approach"]
A line can intersect another line can at $ <\equal{}1$ point and it can intersect a circle at $ <\equal{}2$ points.
Draw two circles, which intersect $ 2$ points, at most. Then, if we draw a line, it can intersect each circle at $ 2$ points. The next line can intersect each circle at $ 2$ points and the existing line at $ 1$ point.  The last line can again intersect each circle at $ 2$ points and each existing line at $ 1$ point. Thus, we have $ 2\plus{}2(2)\plus{}2(2)\plus{}1\plus{}2(2)\plus{}1(2)\equal{}\boxed{17}$.[/hide]

My way: Categorize based on intersection type. Note that the total PoI for each type is equal to (max number of intersections for two of them)*(number of ways to choose two of them).

Line-line= 1*3c2=3

Line-circle= 2*3*2=12

Circle-circle=1*2=2

So 17.


The ineq is equivalent to  $(b^2+1)(a^2+1)\geq (ab+1)^2$ which is true by the ineq

of Cauchy

Hmm, yes you are right. I didn't notice that it was so easy. Sorry, rubbish inequality...
Here's my solution

Working out gives \[ ab + \frac{a}{b} + \frac{b}{a} + \frac{1}{ab} \geq \frac{(ab+1)(ab+1)}{ab} \]
And \[ \displaystyle\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right) \geq \displaystyle\left(a+\frac{1}{b}\right) \left(b+\frac{1}{a}\right) \]
Simplyfies to \[ \displaystyle\frac{a}{b}+\frac{b}{a} \geq 2 \]
This is true by AM-GM.

But, you're right, it's almost literally Cauchy-Schwarz inequality, I should have seen that...


find real value of $x$ when:

a) $x^{x^{x^{x^{x^{x \cdots}}}}}= 2$
and
b) $x^{x^{x^{x^{x^{x \cdots}}}}}= 4$

Then compare the equations. How do you explain what happens? (wher do i have the mistake? 2=4?)

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

the first $x= 2^{ \frac{1}{2}}$

and the second, find $x$ is a little bit more difficult.

but there is no mistake like $2=4$

dude. we will never have a Rigth function that proof that $2=4$. if so, the function is not rigth or have some error

 c ya

no!!!

the problem is: you are using a Single variable for 2 different things

see the example

${\lim_{x\to a}\, f(x)=\lim_{x\to b}\, g(x)}$

it does not mean $a=b$ or $g=f$

and [quote]for b) x=4^{1/4}, is a solution (I think) [/quote]

this solution is wrong.

the actual solution for (b)  , I think, Is a complex number.

The search function is a wonderful thing.  Try
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122691
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79778
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39383

your problem is as follows: we have $4^{\frac{1}{4}}= \sqrt{2}$.  It can be shown (and it is not too hard) that for $x \le 2$ and $x \ge 4$, we have $2 \ge \sqrt{2}^{x}\ge x$.  Hence the sequence $\sqrt{2}^{\sqrt{2}^{\cdots}}$ can never exceed 2.

find min/max values of $C$ for which $x^{x^{\cdots}}= C$ has a solution.

[quote="Teki-Teki"]It can be shown (and it is not too hard) that for $x \le 2$ and $x \ge 4$, we have $2 \ge \sqrt{2}^{x}\ge x$.[/quote] You might want to re-think the $x \geq 4$ part of this statement.

[quote]find min/max values of $C$ for which $x^{x^{\cdots}}= C$ has a solution.[/quote] This is addressed in at least one of the threads I linked to, as well as several others on the forum.  Search for "tower" to find many of them.

do we have any techniques to address convergence if x is complex?

If $x$ is complex with non-zero imaginary part then $z^{x}$ for any $z$ has (countably) infinitely many values -- I don't think you really want to go there.

i suppose, i was just thinking from a calculator perspective...[i know the ti-89 is bad with these types of things...]...we have to adopt some convention such as taking a certain argument, etc...using what the ti-89 uses...i think it is i=e^{pi i/2}...i.e. strictly pi/2, not 5pi/2, etc

if we do x=i, etc gives 0.567*e^{.6884i} after 40 or so itterations of f(x)=i^x

so, sumarizing:(correct me if I missunderstood)

$x^{x^{x^{x^{x^{x \cdots}}}}}= 4$ has no real solution

$x^{x^{x^{x^{x^{x \cdots}}}}}$ converges in an interval, 4 isn't there

the value: $4^{1/4}=2^{1/2}$ es the unique real solution for a)

the problem was on assuming that b) has a solution wich could be obtained similary as $2^{1/2}$· for a)

Thanks all. :)


LaTeX
inequalities
function
inequalities solved

[quote="Fedor Petrov"]What is $\pi_{1,n}$? May be, geometric mean in LHS?[/quote] I think it's $\prod_1^n$.

Sorry, I don't know about LaTEX well.
It means :Product

If nobody has solution,I will post my one.

i'm not sure about it, because i dunno much about hessian test, but can this be solved by jensen's inequality on the function $f(x,y,z,w)=ln(\frac{xyzw+k}{xyzw-k})$?

I don't understand how you use Jensen's in this ineq,siuhochung!But i have a very short proof.

[quote="siuhochung"][quote="keira_khtn"]Siuhochung, i think you made a mistake!!!
Beacause here is:[tex]\frac{abcd+k}{abcd-k}[/tex] ,so we can't use Jensen'ineq[/quote]

Have you ever read about the Jensen's inequalities for multivariables?
anyway, can you post your solution? :)[/quote]
This ineq is really easy:
We always have:
[tex](abcd)^n\geq \prod a_ib_ic_id_i[/tex]
Then there exist [tex]k[/tex] such that:[tex]a_kb_kc_kd_k\leq abcd[/tex]
Therefore:[tex]L.H.S\geq \frac{a_kb_kc_kd_k+k}{a_kb_kc_kd_k-k}\geq \frac{abcd+k}{abcd-k}[/tex](Because it equivalents to [tex]a_kb_kc_kd_k\leq abcd[/tex]:it's true)

[quote="keira_khtn"]Let:[tex]a_{i},b_{i},c_{i},d_{i}\geq\ 0[/tex]
[tex]a=\frac{1}{n}\sum_{1,n}a_{i},b=\frac{1}{n}\sum_{1,n}b_{i},c=\frac{1}{n}\sum_{1,n}c_{i},d=\frac{1}{n}\sum_{1,n}d_{i}[/tex].Prove that with all [tex]k\geq\ 0[/tex]:
[tex]\prod_1^n\frac{a_{i}b_{i}c_{i}d_{i}+k}{a_{i}b_{i}c_{i}d_{i}-k}\geq\frac{abcd+k}{abcd-k}[/tex][/quote]
should the problem be $ \max \{ \frac{a_{i}b_{i}c_{i}d_{i}+k}{a_{i}b_{i}c_{i}d_{i}-k}\}\geq\frac{abcd+k}{abcd-k}$??

[quote="keira_khtn"]Because ,we always have:
[tex]\frac{a_ib_ic_id_i+k}{a_ib_ic_id_i-k}\geq 1[/tex]
with all i from 1 to n[/quote]
oh,but you still have the condition with ${a_ib_ic_id_i-k} \geq 0$


Let $A$ be a set. On its power set $\mathcal P (A)$ we define the binary operation $X \Delta Y = \left( X \setminus Y \right) \cup \left( Y \setminus X \right)$.

Prove that $\Delta$ is associative, i.e. $\left( X \Delta Y \right) \Delta Z = X \Delta \left( Y \Delta Z \right)$.

Let $\chi_{S}$ be the characteristic function of a set $S$. Then $\chi_{X\Delta Y}= \chi_{X}+\chi_{Y}\mod{2}$, so $\chi_{\left( X \Delta Y \right) \Delta Z}= \chi_{X \Delta \left( Y \Delta Z \right)}$ is equivalent to the associativety of adding modulo $2$, which is obvious.


trigonometry
geometry
circumcircle
symmetry
angle bisector

Given a triangle with $ \angle B\equal{}60,\angle C\equal{}70$. Let $ D$ be a point on side $ BC$ such that $ \angle BAD \equal{}20$. Prove that $ AB\plus{}BD\equal{}AD\plus{}DC$.

Trig solution isn't too complicated, but is there a synthetic one? That would be nice.

Well, $ AD$ contains the circumcenter of $ ABC$ :)

because $ AD$ is the isogonal conjugate of the altitude (altitude divides $ \angle DAC$ into $ 10$ and $ 20$)

I was thinking about something along the lines of superimposing lines and/or isosceles triangles.

Construct a diagram as follows:

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/e0e596a4eed72aa2c7063a717dcf30b1.png[/img]

Let $ ABE$ be an equilateral triangle.  Take points $ P$ and $ R$ such that $ AB = AP = AR = AE$ and $ \angle BAP = \angle PAR = \angle RAE = 20^\circ$.  Then triangles $ BAP$, $ PAR$, and $ RAE$ are isosceles with angles $ 20^\circ$, $ 80^\circ$, and $ 80^\circ$.

Let $ AP$ intersect $ BE$ at $ D$, and let the angle bisector of $ \angle RAE$ intersect $ BE$ at $ C$, so $ \angle RAC = 10^\circ$.  By symmetry, $ \angle CRA = \angle CEA = 60^\circ$.

Extend $ AP$ to $ Q$ so that $ \angle PBQ = 40^\circ$.  Since $ \angle BPQ = 100^\circ$, $ \angle PQB = 40^\circ$.  By symmetry, $ \angle PRQ = \angle PQR = 40^\circ$, so both triangles $ BPQ$ and $ QPR$ are isosceles.  Furthermore, $ \angle QRC = \angle PRQ + \angle ARP + \angle CRA = 40^\circ + 80^\circ + 60^\circ = 180^\circ$, so points $ Q$, $ R$, and $ C$ are collinear.

It is not hard to check that the conditions of the problem are satisfied, and they uniquely determine the diagram.  Note that triangle $ BDP$ is isosceles, since $ \angle BDP = \angle BPD = 80^\circ$, and trianlge $ QDC$ is isosceles, since $ \angle DCQ = \angle DQC = 40^\circ$.  Hence,
\[ AB + BD = AP + BP = AP + PQ = AQ,\]
and
\[ AD + DC = AD + DQ = AQ.\]

See other few solutions [hide=here] http://forum.gil.ro/viewtopic.php?f=20&t=5988&p=15561#p15561[/hide].

Best regards,
sunken rock


logarithms
limit
function
algebra proposed
algebra

Let $f:\mathbb{N^*}\rightarrow \mathbb{R}$ sastisfying the conditions:
 (i)$f(m)>f(n),\forall m>n$
(ii)$f(m\cdot n)=f(m)+ f(n),\forall m,n\in \mathbb{N^*}$
Prove that $\exists p>1:f(n)=\log_pn,\forall n>1$

Easy! (it is my favorite word today!)

Let $f(2)=k\ln 2$. Suppose $f(n)\neq k\ln n$, $n>1$. Then $f(n)=k\ln n+a$, $a\neq 0$.
We can write for all $m\in\mathbb{N}$
\[f(2^{d_m+1})\geq f(n^m)\geq f(2^{d_m}),\]
where $d_m=[\log_2 n^m]=[m\log_2 n]$.
In other words,
\[(d_m+1)\cdot k\ln 2\geq m\cdot (k\ln n+a)\geq d_m\cdot k\ln 2.\]
Divide to $m$ and take $m\to\infty$ to get contradiction, since $d_m/m\to \log_2 n=\frac{\ln n}{\ln 2}$  :)

Oh Jesus Christ! :) Don't you ever take a break, Myth? I was about to post it :D.

I agree that this problem is trivial. Here's how to make it non-trivial:

Problem 1. (Erd\H{o}s) Assume that $f:\mathbb{N} \to \mathbb{R}$ is such that $f(n+1) \geq f(n)$ for all sufficiently large $n$, and $f(mn)  = f(m) + f(n)$ for all *co-prime* positive integers $m,n$, $(m,n)=1$. Show that $f(n)=c\ln n$ for some $c$.

Problem 2. (Erd\H{o}s) Replace in the above problem the monotonicity condition with the contition that $\lim_{n \to \infty} \{f(n+1) - f(n)\} = 0$. Show that the same conclusion holds.

The following is much more difficult; it was conjectured by Erd\H{o}s in the 1940's, but its proof appeared no earlier than 1970!

Problem 3. (Wirsing) Show that if we replace the monotonicity condition with the condition that $f(n+1) - f(n)$ is bounded, then there exists some $c$ such that $f(n) - c\ln n$ is bounded.

Enjoy the first two problems! (And don't try the third one unless you want to really challenge yourself.  :D )

--Vesselin

I knew the first one from Erds.
But, Vess do you have any reference for the third one?

Pierre.

[quote="pbornsztein"]I knew the first one from Erds.[/quote]
Do you mean Erdos told it to you himself?

[quote="pbornsztein"]I knew the first one from Erds.
But, Vess do you have any reference for the third one?
[/quote]

I read about it in the great memorial article "The Mathematics of Paul Erd\H{o}s" by Babai, Pomerance and Vertesi that appeared in a 1998 issue of the Notices of AMS. Here's their reference for Problem 3:

E. Wirsing,   A characterization of $\log n$ as an additive arithmetic function, Sympos. Math., Vol. IV (INDAM, Rome, 1968/69) Academic Press, London, 1970, pp. 45--57

--Vesselin

Myth : I would like... :(

Vess : Thanks!

Pierre.


a,b,c>=0
proove that:
(1/2)(n-1)(an+bn)+cn \geq 3nabc((a+b)/2)n-3
from GM nr 3/2003

Wasn't this posted before? :?
And I thought that magazine problems should be at least one year old?

I everyone breaks the rule, I will break it too and I will post a solution and a comment. First of all, the correct inequality is ((n-1)/2)(a^n+b^n)+c^n>=nabc((a+b)/2)^(n-3) (otherwise take a=b=c). Secondly,the inequality is extremely simple. Just observe that ab((a+b)/2)^(n-3)<=((a+b)/2)^(n-1)  and that ((n-1)/2)(a^n+b^n)>=(n-1)((a+b)/2)^n. Thus we have to prove that (n-1)x^n+y^n>=nx^(n-1)y, which comes down to AM-GM

well, Arne is right I posted this problem sometime ago and Mihai Fulger(amfulger) solved it ! search for it at the algebra solved problems!

cheers! :D :D

It's funny but I don't remember solving it. It's not impossible though with so many problems on my mind  :? .
Anyway I moved it from the Unsolved problems- Math for College section of this site because it does not belong there. It's IMO level (or below  :) ).


let $p$ a prime number such $p=4k+3$.
prove that if $2p+1$ is prime $\Rightarrow $  $2^{p}-1$ is not prime.


geometry
geometric transformation
reflection
analytic geometry
incenter
geometry proposed

Recently, I have proved synthetically (no computation at all)  the following fact which I find it very cool  ;). Here it is
         Let  $ABC$ be the triangle and $P$ be a point. Let $A_1B_1C_1$ be the circumcevian triangle of $P$ wrt $ABC$. Call $A', B', C'$ the reflections of $A_1, B_1, C_1$ across $BC, CA, AB$ respectively. Then $A'B'C'$ and $A_1B_1C_1$ are similar. Furthermore, the barycentric coordinates of $P$ wrt $A'B'C'$ is the same of $P$  wrt $A_1B_1C_1$.
         Now, one of the special case of this theorem is the Fuhrmann triangle. From the above, it follows that $I$ is the orthocenter of the Fuhrmann triangle, where $I$ is the incenter of $ABC$ (which is a problem in Unsolved Problem posted by jhaussmann5)

This is really nice!  Can you post your proof?

Use $\triangle AC'P\sim\triangle CA'P$ and the other like similarities.

Dear Mathlinkers,
for a synthetic proof, you can see
http://perso.orange.fr/jl.ayme  vol. 5  Le P-cercle de Hagge p. 12-14.
Sincerely
Jean-Louis


A drill team is marching in formation. At first, the marchers formed a perfect square, with an wqual number of columns and rows. Then, they changed formation and became a rectangle with 5 more columns than before. How many columns were in the rectangular formation?
a. 8
b.12
c. 14
d. 18
e. none of the above.

A six digit number of the form a2b308 is divisible by 33. How many numbers of this form exist?
a. 1
b. 2
c. 3
d. 4
e. none of the above.

For the second problem, I know for sure that (a,b) (1,1) works. However, from there, I'm not sure how to proceed. I know the divisibility rules for 3 and 11, but I haven't exactly been able to use them.

[hide="Kind of cheap solution on number 1"]
All I did was test out the numbers.

Choice a: If there are 8 columns in the rectangle, the square is 3x3, so there are 9 marchers.  9 is not divisible by 8, so this can't be the answer.

Choice b:If there are 12 columns in the rectangle, the square is 7x7, so there are 49 marchers.  49 is not divisible by 7, so this can't be the answer.

Choice c:If there are 14 columns in the rectangle, the square is 9x9, so there are 81 marchers.  81 is not divisible by 14, so this can't be the answer.

Choice d:If there are 18 columns in the rectangle, the square is 13x13, so there are 169 marchers.  169 is not divisible by 18, so this can't be the answer.

[b]Choice e[/b]: None of the above work, so this must be the answer.
[/hide]

[quote="eryaman"]After trying to solve question 1 algebraically, I tried something...but it's not working.  Can someone tell me what I'm doing wrong? :?
[hide]Let $P$ be the number of marchers.  Let $m$ be the dimensions in the original square.  We can write the problem as $P=m^{2}$ and $P=(m+5)(m-n)$, where $(m+5)$ is the number of columns in the rectangle, and $(m-n)$ is the number of rows, where $n$ is a positive integer.  Setting the equations equal to each other, we have $m^{2}=m^{2}-mn+5m-5n$.  Simplifying, we have $mn-5m+5n=0$.  Adding $-25$ to each side gives us $mn-5m+5n-25=-25$, so $m(n-5)+5(n-5)=-25$, so $(n-5)(m+5)=-25$.  Clearly $(n-5)$ and $(m+5)$ must be positive integers since $n$ and $m$ are integers.  The only possible solutions are (for $(n-5,m+5)$): $(-25,1),(1,-25),(25,-1),(-1,25),(5,-5)(-5,5)$.  But for some reason none of these fit my original equations... :( [/hide][/quote]

Looks fine to me. So the only solution in the positive integers is $(m,n) = (20,4)$.

[quote="eryaman"]Clearly $(n-5)$ and $(m+5)$ must be positive integers since $n$ and $m$ are integers. [/quote]

you can say that $m+5$ is positive since m is positive, but you CANNOt say $n-5$ is positive also... because what if $n < 5$?

Woopsies.  I just made arithmetic errors when testing out my ordered pairs.  :blush:

To reddevil, yes, I did mess up.  When writing my solution I forgot to write that m and n had to be positive integers, so I just added that whenever I saw m and n used.  So I forgot about that part.

[quote="jb05"]It appears that nobody has answered number 2:

[hide]
Using divisibility rules, we get:

5|(a+b+13)
11|(a+b-13)

It is clear that the only possibility is a+b=2 (since a+b-13 is either -11, 0, or 11 we just test each and see if it works for the first equation). Since a is non-zero, we have either a=2, b=0 or a=b=1, so answer [b]b: 2[/b] possibilities.[/hide][/quote]

Shouldn't it be 3|(a+b+13)?


inequalities
number theory unsolved
number theory

Let $e(k)$ be the number of positive even divisors of $k$, and let $o(k)$ be the number of positive odd divisors of $k$. Show that the difference between $e(1)+e(2)+...+e(n)$ and $o(1)+o(2)+...+o(n)$ does not exceed $n$.

We have that $\sum_{i=1}^{n}o(i) = \sum_{i=1}^{n}[\frac{n}{2i+1}]$ and $\sum_{i=1}^{n}e(i) = \sum_{i=1}^{n}[\frac{n}{2i}]$

From here using the obvious inequalities $[x] \leq x$, and $[x] \geq x-1$ we get what we want:

$\sum_{i=1}^{n}(o(i)-e(i)) = \sum_{i=1}^{n}(-1)^{i}[\frac{n}{i}]$

and the problem statement is equivalent to :
$\frac{-1}{2}\leq \sum_{i=1}^{n}(-1)^{i}\frac{1}{i}\leq \frac{3}{2}$
which is easy proved and well known.

DOUBT COUNT!
For some $k$ even $\leq n$ then has $[\frac{n}{k}]$ times.

:D No, I think use Fubini's principle, In fact

$\sum_{i=1}^{n}e(i)=\sum_{i=1}^{n}\sum_{k|i,2|k}1=\sum_{2|k}\sum_{k|i}1$.


Let n be the sum of the digits in a natural number A. The number A it's said to be "surtido" if every number 1,2,3,4....,n can be expressed as a sum of digits in A.

a)Prove that, if 1,2,3,4,5,6,7,8 are sums of digits in A, then A is "Surtido"
b)If 1,2,3,4,5,6,7 are sums of digits in A, does it follow that A is "Surtido"?

$ b)$ No, for example consider the number $ A\equal{}9421$ then $ S(A)\equal{}16$
and $ 1,2,3,4,5,6,7$ are sums of digits in $ A$
but $ 8$ is not a sum of digits in $ A$.

I'll post a full solution in Spanish and only a sketch in english cuz I woke up lazy af today. (I'm planing on posting a solution to a problem from the Mexican National Olympiad every day for the next two months, that is before this year's OMM which will prolly be next november!).
[hide = Solución]
Para la parte [b]a)[/b] Procederemos por inducción sobre la cantidad de dígitos de $n$. Nuestros casos base son los números surtidos con $S(n) = 8$. Ahora, Asume que $n$ es un número surtido con $S(n) \ge 8$. Si le agregamos algún dígito $d$ en cualquier posición a $n$ para así obtener $n'$, podremos aún formar todos los enteros menores o iguales a $S(n)$, y para los restantes (todos los enteros $i$ con $S(n) < i \le S(n) + d$), bastará con tomar todos los dígitos de $n'$, y restar los dígitos que sumen a cada uno de los enteros menores o iguales a 8, esto cubrirá todos los restantes pues $d \le 9$. Esto implica que $n'$ es surtido, y hemos concluido. $_{\blacksquare}$

Para la parte [b]b)[/b] basta con observar que el paso inductivo de la demostración anterior falla si $d = 9$, y por lo tanto podemos construir un ejemplo a partir de esto, como $n = 11111119$. En este caso, podemos formar todos los números menores a $16$ con excepción a $8$, y por lo tanto el enunciado no es cierto.
[/hide]

[hide = sketch in English]
Prove by induction that if $n$ is "surtido" and $S(n) \ge 8$ and we insert a new digit $d$ in any position of $n$ to create a new number $n'$, then $n'$ is "surtido".
This is equivalent to part [b]a)[/b].

For part [b]b)[/b], just see what fails if you change 8 to 7 in the above argument, and find an example accordingly, such as $n = 11111119$.
[/hide]

[hide = Comments?] This problem kinda suck but not that much? idk, I feel like I have to do more modern problems[/hide]


probability
linear algebra
matrix
probability and stats

This seems really simple, but...

Take a graph with $ k$ vertices. If you randomly color the edges with two colors independently each color occurring with equal probability what is the probability that the graph is monochromatic.

My guess was that since they are independent you could say that the probability would be $ 2\left(\frac{1}{2}\cdot\frac{1}{2}\cdots\frac{1}{2}\right) \equal{}\frac{2}{2^k}\equal{}\frac{1}{2^{k\minus{}1}}$ i.e. a $ \frac{1}{2^k}$ probability that it is all blue and a $ \frac{1}{2^k}$ probability that it is all red.

The actual probability given is $ \frac{2}{2^{\left(\begin{matrix} k \\ 2\end{matrix}\right)}}$ Any insights?

Thank you very much. That was so dumb. Even though I typed edges and wrote edges in copying the proof to my own paper, and read edges, I was still thinking we were coloring vertices.  :blush:


function
inequalities
real analysis
real analysis unsolved

Would somebody help me on this problem? I really need some help. Thanks.

Let $ f_n$(x) = $ x$/(1+n$ x^2$). Prove that $ f_n$ converges uniformly to 0, where 0 denotes the function that is 0 everywhere.

By AM-GM inequality, $ |x| \equal{} \frac{\sqrt{nx^2}}{\sqrt{n}} \leq \frac{1 \plus{} nx^2}{2\sqrt{n}}$. This shows that $ |f_n| \leq \frac{1}{2\sqrt{n}}$, hence $ f_n$ converges uniformly to $ 0$.

[quote="sos440"]By AM-GM inequality, $ |x| \equal{} \frac {\sqrt {nx^2}}{\sqrt {n}} \leq \frac {1 \plus{} nx^2}{2\sqrt {n}}$. This shows that $ |f_n| \leq \frac {1}{2\sqrt {n}}$, hence $ f_n$ converges uniformly to $ 0$.[/quote]
Thank you very much for your help.


The diagonal of a quadrilateral are m and p units in lenght. If the
diagonals intersect at an angle B(beta), then prove the area of the
quadrilateral is A = (mp/2) sin B(beta).

[hide]Let the quadrilateral be $ABCD$, and the point of intersection of the two diagonals $E$. Let the angle of intersection of the two diagonals be $\beta$. Let $AE$ equal $m_{1}$, $BE$ equal $p_{1}$, $CE$ equal $m_{2}$, and $DE$ equal $p_{2}$, so that $m_{1}+m_{2}=m$, and likewise with $p$.
Using the sine angle formula for triangles, we may evaluate the areas of the four triangles formed by the diagonals separately. Adding them together, we get:
$[ABCD]=\frac{1}{2}(m_{1}p_{1}+m_{1}p_{2}+m_{2}p_{1}+m_{2}p_{2})\sin{\beta}$, since $\sin({180^{\circ}-\theta})=\sin{\theta}$.
This can be simplified to:
$[ABCD]=\frac{1}{2}((m_{1}+m_{2})(p_{1}+p_{2}))\sin{\beta}.$
Of course, since $m_{1}+m_{2}=m$ and $p_{1}+p_{2}=p$, we have:
$[ABCD]=\frac{1}{2}mp\sin{\beta}$ and we are done.[/hide]


limit
calculus
derivative
calculus computations

evaluate:

1. $ \lim_{h\to 0}\frac{e^{x\plus{}h}\minus{}e^x}{h}\equal{}0$

$ f '(x) \equal{} \lim_{h \to 0}\frac {f(x \plus{} h) \minus{} f(x)}{h}$.

So, $ \lim_{h \to 0}\frac {e^{x \plus{} h} \minus{} e^x}{h} \equal{} \left( e^x \right) ' \equal{} e^x$

[quote="Kouichi Nakagawa"]$ f '(x) \equal{} \lim_{h \to 0}\frac {f(x \plus{} h) \minus{} f(x)}{h}$.
[/quote]

how did you find this ?

Probably he remembered it from the first page of the section of his calculus textbook on derivatives.   :roll:

Definition :)
[url=http://mathworld.wolfram.com/Derivative.html]Derivative[/url] - MathWorld

[hide="Finishing"]
There are no real solutions to $ e^x \equal{} 0$
[/hide]


aa' is a median.
bb' divided ac 1:2.
what is ay:ya' and b'y:yb

just use coordinates

[hide]

let $A = (0,3)$, $C= (0,0)$, and $B= (2b,0)$, then $A' = (b,0)$ and $B' = (0,2)$, intersection of $AA'$ and $BB'$ is $(b/2, 3/2) = Y$ and use distance formula to get ratio $AY: YA' = 1: 1$ and $B'Y: YB = 1/3$

that should be right....
[/hide]

[hide]The ratio of two triangles with the same height is the same as the ratio of their bases.

$ba': a'c=1 \implies [aa'b]=[aa'c]$ (1), and $[ya'b]=[ya'c]$ (2).
$ab': b'c=1: 2 \implies [bb'a]=[bb'c]/2$, (3) and $[yb'a]=[yb'c]/2$ (4).

From $(3)-(4) \implies [ayb]=[cyb]/2=[a'yb]$, so $ay: ya'=1: 1$.

From $(1)-(2) \implies [ayb]=[ayc]=[ayb']/3$, so $b'y: yb=1: 3$.[/hide]

Van Obal's Theorem demolishes it:

If $AD, BE, CF$ are concurrent cevians meeting at $P$, then $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}$.


Does anyone know where I can get previous problems from the AMC 10/12 contests?

and i believe my math coach gets them from the amc people...
if you have anmore amc questions like this i suggest the amc forum

awesome i didn't know aopswiki had those tests.

i'm obviously all but blind


Let $ A,B$ are $ n$x$ n$ Hermitian matrices .
Prove inequality as :
  1) $ 2^n\det(A^2 \plus{} B^2) \geq {\det(A \plus{} B)}^2$
  2) $ Tr(A^2 \plus{} B^2)^2\geq 2(Tr(A^3B) \plus{} Tr(AB^3))$
  3) $ Tr(ABAB) \leq Tr(A^2B^2)$
[hide="Answer"]Interesting[/hide]

[quote="miwalin"]For 3)
It is known that $ tr((AB)^m)\leq tr(A^mB^m)$ for any positive integer $ m$.[/quote]
 No...General don't true...:
 you can try $ A\equal{}\begin{pmatrix}\minus{}1&0\\0&0\end{pmatrix},B\equal{}\begin{pmatrix}1&1\\1&1\end{pmatrix}$
Then $ Tr((AB)^3)\equal{}\minus{}1>Tr(A^3B^3)\equal{}\minus{}4$

Oh, sorry. If $ A$ and $ B$ are positive definite, then it is true.
But if $ m$ is even, then the requirement  "$ A$ and $ B$ are positive definite" is not necessary.
See attached.

[quote="miwalin"]If $ A$ and $ B$ are positive definite, then it is true.
But if $ m$ is even, then the requirement  "$ A$ and $ B$ are positive definite" is not necessary.
See attached.[/quote]
[b] Very Interesting ![/b] I hope that you can find many document about Inequality matrix....

For 3) Very Implies Put $ C\equal{}i(AB\minus{}BA)$ then $ C^{*}\equal{}\minus{}i(BA\minus{}AB)\equal{}i(AB\minus{}BA)\equal{}C$ ,$ C$ is Hermitian matrix
Implies $ Tr(C^2) \geq O$ or $ Tr((AB\minus{}BA)^2) \leq O$ or $ Tr((AB)^2)\plus{}Tr((BA)^2) \leq Tr(AB^2A)\plus{}Tr(BA^2B)$ we have $ Tr((AB)^2)\equal{}Tr((BA)^2)$and $ Tr(AB^2A)\equal{}Tr(BA^2B)\equal{}Tr(A^2B^2)$ the end \[ Tr(ABAB)\leq Tr(A^{2}B^{2})\]


I haven't either. I am also just wondering what is going on now.

[quote="bookaholic"]I haven't heard from Mathcamp since I registered. Will we get a list of items we need to bring and other information of that nature? If so, when?[/quote]

Yes, that information is coming in the next few days (beginning of next week at the latest), by both email and postal mail.


(A generalization of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108199]this[/url].)

A circular sector with the center $O$ and the radii $OA, OB$ is cut out of the homogenous thin cardboard and hung freely by the point $A$. Find the angle between $AO$ and the vertical if $\angle AOB=\theta\leqslant\pi$.

I could do it with calculus, i.e. determining the distance of the center of gravity from $O$, and that too with relative ease with the tool of integration...
But there has to be a simpler way...

Still no one? Here are some

[hide="keywords"]Pappus's theorem, spherical sector[/hide]

Is this really THAT difficult for Mathlinks/AoPS users?

Since obviously no one is interested in this, I have to post the

[hide="solution"]If the sector rotates around $AO$, we get the spherical sector, which has the volume $V={2\over 3}r^{2}\pi h$, with $h$ being the height of the corresponding spherical cap, in this case $h=r(1-\cos\theta)$. On the other hand, if $d$ is the distance from centroid $C$ to the line $OA$, then $V=\frac{r^{2}\theta}{2}\cdot 2d\pi$. The two expressions give $d=\frac{2r(1-\cos\theta)}{3\theta}$.

Let $C_{1}$ be the foot of the perpendicular from $C$ to $AO$. Then $OC_{1}=d\cot{\theta\over 2}$, hence

\begin{eqnarray*}\tan\varphi &=& \frac{d}{r-d\cot{\theta\over 2}}\\ &=& \frac{2r(1-\cos\theta)}{3\theta}\cdot\frac{3\theta}{3r\theta-2r(1-\cos\theta)\cot{\theta\over 2}}\\ &=& \frac{2(1-\cos\theta)}{3\theta-2(1-\cos\theta)\cot{\theta\over 2}}\\ &=& \frac{1-\cos\theta}{{3\over 2}\theta-\sin\theta}\end{eqnarray*}

because $(1-\cos\theta)\cot{\theta\over 2}=2\sin^{2}{\theta\over 2}\cot{\theta\over 2}=\sin\theta$

Therefore $\varphi=\arctan\frac{1-\cos\theta}{{3\over 2}\theta-\sin\theta}$[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

I find this problem very challenging:
  Suppose that $p$ is the least prime divisor of $|G|$ and $S$ is a Sylow $p$-subgroup of $G$, which is cyclic. Then the set of those $x\in G$ whose order is not multiple of $p$ forms a characteristic subgroup $H$ of $G$ such that $G/H$ is isomorphic to $S$. It generalizes a very classical problem concerning the groups of order $4n+2$. Does someone have an elegant solution?

Depends on what "elegant" means ;-)

1st step: Let $H\leq S$, then the normalizer equals the centralizer of H: $C_{G}(H)=N_{G}(H)$
2nd step: Every two elements of S that are conjugated in G are equal.
3rd step: Consider the transfer-homomorphism $G\to S$ and show that its restriction to S is an automorphism of S.
4th step: The kernel of this homomorphism is exactly the set of all elements with order not divisible by p. Since it is a subgroup it is in fact characteristic, because automorphisms of G preserve orders.

Hmmm, that's the solution I had, too (I could not find it myself, obviously!). I wonder if there are solutions that avoid this transfer morphism.


geometry
geometric transformation
search

Someone told me that has the book by << Calperyn and Tolpyngo(I' m not sure for the name , if it is written so!) : Olympiads from Moskaw >> in an english translation - not official edited. But he has not the figures. Probably the translation is done by AMS ( USA) .If someone has this translation or has the Russian text and wants to help us to find the figures , please write.
   USA and Canada students can ask about this their olympiad trainers.

  Thanks

 [i]Babis[/i]

There was certainly a project to translate this book into English, as of about ten years ago or more.  I don't know if it was ever completed.   For now there are hundreds of the old Russian olympiad problems in a file typed and translated into English by Vladimir Pertsel. It is in many places on the Internet.

[quote="fleeting_guest"]   For now there are hundreds of the old Russian olympiad problems in a file typed and translated into English by Vladimir Pertsel. It is in many places on the Internet.[/quote]

Could you give the internet adress where I can find the problems in english ?

thank you

[quote="Moubinool"][quote="fleeting_guest"]   For now there are hundreds of the old Russian olympiad problems in a file typed and translated into English by Vladimir Pertsel. It is in many places on the Internet.[/quote]

Could you give the internet adress where I can find the problems in english ?[/quote]

Web search produces his web page, with the following interesting anecdote:

"I had won the All-Soviet-Union school mathematical competition in 1973 and had the the second prise in 1972 and 1974. You can find the problems of those competitions in http://www.kulichki.com/puzzles/puzzles/selected/RusMath.html.
I did not participate in the international olympiad because I could not obtain an international passport, but my son did participate -- in 1996, though he was in the Israel team."

In the 1970's the Russian olympiads were difficult compared to the IMO, so he was probably denied a medal (or two or three) at the IMO.  Denial of passport refers to the travel restrictions imposed on Russian Jews as a standard anti-Semitic measure in the USSR at that time.  I think Margulis and Drinfel'd were not able to receive their Fields medals in person due to the same "passport" problem.

The translated problems are the All-Soviet Union olympiads, not the Moscow ones.
Additional years of ASU are at John Scholes' site (kalva).

Good information ! Thanks.

 In the midltime we have made some progress. Some friend found this book with Moskow Olympiads - unoficial in english - but we have not the figures. However this is not a problem , because  the solutions are very detailed and much better that those in the original text. Unfortunately I have not the text in e - form , so I can not offer it to you to enjoy it !
    [u]Babis[/u]

Thank you.
 I know and have this nice book. From the Moskow olympiads I have not only the fingures , but finally this is not a big problem. Only some compinatorics - problems are a little difficult to read , but I hope I'll find them in other sources.

    [u]Babis[/u]

[quote="stergiu"][quote="thecool"][quote="RiemannZetaFunction"]I dont know if this is what your looking for but dover publishes a book called: [u] The USSR Olympiad Problem Book[/u] the authors are D.O. Shklarsky, N.N. Chentzov, and I.M. Yaglom its got like 320 olympiad problems.[/quote]

Could you provide me with a scanned copy. I need this book very much  :([/quote]

You must write to DOVER Publications to ask if the book is still in print.Otherwise you must find a photocopy.

 [u]babis[/u][/quote]

Thanks for your quick reply. I don't have a credit card so I am searching for a scanned copy or some other electronic form. If you could help, it would be fantastic.

Well ,
I had not opened the topic.
Sure I send you a photocopy. Where do you live. Send a PM.
[u] Babis[/u]

[quote="thecool"][/quote][quote="stergiu"][quote="thecool"][quote="RiemannZetaFunction"]I dont know if this is what your looking for but dover publishes a book called: [u] The USSR Olympiad Problem Book[/u] the authors are D.O. Shklarsky, N.N. Chentzov, and I.M. Yaglom its got like 320 olympiad problems.[/quote]

Could you provide me with a scanned copy. I need this book very much  :([/quote]

You must write to DOVER Publications to ask if the book is still in print.Otherwise you must find a photocopy.

 [u]babis[/u][/quote][quote="thecool"]

Thanks for your quick reply. I don't have a credit card so I am searching for a scanned copy or some other electronic form. If you could help, it would be fantastic.[/quote]

There are at least 3 versions of this book, and they are available in many libraries all over the world.  One the original Russian, another the English version translated in the USSR, and finally the Dover reprint of the USA version translated by Americans.  Also many translations into other languages.  There are also many free on-line sources of problems that cover similar material (check e.g. Pertsel's site that I posted above, or the resources on this site).


algebra
polynomial
induction
real analysis
real analysis solved

hey evry1
help me solve this problem regarding limits
let x_o = a>1 be a real number.define the sequence
x_(n+1) = 1 + log[ (x_n)^3 + 3x_n] - log[3(x_n)^2 +1)
prove that this sequence has a limit and find it
thanx.
(hey ,does ne1  here know of a website where i can find the solutions to some of the romanian selection tests,1999 and 2000)

here are the problems.i dont have the yearsbut they are after 1997
Let n > 3 and A1;A2; : : : ;An points on a circle. Find the greatest number of acute triangles that can be considered with the vertices in these points.
2)Consider in the plane a finite set of segments such that the sum of their lengths is less than  \sqrt 2. Prove that there exists an infinite unit square grid covering all the plane such that the lines defining the grid do not intersect any of the segments.
3)3. Show that for any n belonging N the polynomial
 f(x) = (x^2+x)^(2^n) +1 is irreducible over Z[X].

Well, concerning your problem, you can show that the sequence is greater than 1 (by induction), and you can show that it's decreasing, using the fact that ex-1>=x, for any real x. That'll mean that the sequence is bounded and monotone, so it's convergent, and it's only limit can be 1 (again by using ex-1>=x)

[quote="vineet"]
help me solve this problem regarding limits
let x_o = a>1 be a real number.define the sequence
x_(n+1) = 1 + log[ (x_n)^3 + 3x_n] - log[3(x_n)^2 +1)
prove that this sequence has a limit and find it
thanx[/quote]
The variations of f(x) = 1+ln(x^3+3x)-ln(3x^2+1) on ]1;+oo[ show f is
increasing f(1)=1 so f(x) is positive, all x_n are positive.
The variation of g(x) = f(x)-x  is negative on ]1;+oo[ so
x_(n+1)-x_n = g(x_n)  \leq 0
x_n is decraesing, positive lim x_n =L exist

L= f(L) => L=1


inequalities
algebra
polynomial
algebra proposed

Determine the lowest degree of a polynomial with integer coefficients which have a root:cosx x=p/q*pi  pi=3.1415926......
and p,q positive integers (p,q)=1

I believe that question is answered [url=http://icm.mcs.kent.edu/reports/1998/ICM-199802-0001.pdf]here[/url].  Note also that this is the degree of the real cyclotomic extension $ \mathbb{Q} \left[ \zeta \plus{} \zeta^{\minus{}1} \right]$ where $ \zeta \equal{} e^{ \frac{2 \pi i}{q} }$.

Thank you for the excellent article :lol:
My answer is $ \frac {\varphi(2q)}{2}$
That was involved in the PDF!


Note that $ n^{\frac {1}{2}} \equal{} \sqrt {n}$ and in general, $ n^{\frac{1}{x}} \equal{} \sqrt[x]{n}$.

[quote="isabella2296"]Note that $ n^{\frac {1}{2}} \equal{} \sqrt {n}$ and in general, $ n^{\frac {1}{x}} \equal{} \sqrt [x]{n}$.[/quote]

Hm i think that is a bit [b]too[/b] general. Does it apply to imaginary numbers? What about negative bases? Or 0? Izzy, please be more specific in these "helpful" hints.  :D

Good point.

Correction: For all real numbers excluding 0, blah, blah, blah.


trigonometry
email
AoPSwiki
summer program
Mathcamp
logarithms
percent

[u][b]Teams[/b][/u]

Each team will consist of 4-5 players. It is suggested only people who already know each other make teams of 5 people as it can be harder to communicate with more people. Players can agree between themselves and create their own teams. Others who haven't made their own teams will be randomized into teams.

During the tournament, a team cannot request to have another player added to their team unless one of their players cannot participate anymore. Players cannot switch teams, so choose wisely if you pick your own teams.

[u][b]Rules[/b][/u]

A team must post a move within 2 days after the other team's move. Each time a team fails to do so (unless there is a good reason), a warning will be issued. 5 warnings will result in disqualification. Moves should be posted in the proper Wiki article.

The use of computers, chess programs, and other devices that give your team an advantage are not allowed. This rule cannot be strictly enforced, but hopefully people will play fairly as this tournament is for fun only.

[u][b]Tournament Format[/b][/u]

Depending on the amount of players and teams, there will be several rounds of Swiss, followed by single elimination for the final rounds. For the final game, 3 days will be allowed per move.

[u][b]Signing Up[/b][/u]

Before signing up, please make sure you will be able to participate. I will accept sign ups via post, PM or email.

[url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info]Click here[/url] for the AoPS Wiki project page.

[hide="List of Players (35)"][list]i_like_pie
7h3.D3m0n.117
cincodemayo5590
BoesFX
Temperal
shaggy75
Tenoreoz
abacadaea
nayel
not_trig
bubka
AnonAmoz
davidyko
toadoncart
Ars
bos1234
Hamster1800
TheDropOutBoy
mz94
rem
ProtestanT
Farrell
D3m0n Shad0w
azjps
Lawrence Wu
miyomiyo
ZeroFive1
chessplayingballer
CircleSquared
Rafaelloaa
nitish
DonkeyKong
moogra
snee
dbkarp[/list][/hide]
[hide="List of Teams"][list][b]Team Alpha ($ \alpha$)[/b]
[list]i_like_pie
7h3.D3m0n.117
cincodemayo5590
D3m0n Shad0w
TheDropOutBoy[/list]
[b]Team Beta ($ \beta$)[/b]
[list]Temperal
CircleSquared
abacadaea
miyomiyo
davidyko[/list]
[b]Team Gamma ($ \gamma$)[/b]
[list]nayel
mz94
BoesFX
shaggy75[/list]
[b]Team Delta ($ \delta$)[/b]
[list]nitish
AnonAmoz
Rafaelloaa
Lawrence Wu[/list]
[b]Team Epsilon ($ \epsilon$)[/b]
[list]not_trig
Hamster1800[/list][/list][/hide]

i will obviously participate... or do we PM you to sign up?

You can sign up by post, PM, or email, which ever is more convenient for you.

By the way, I'm glad you dropped the captains idea.  :)

I figured the entire idea of team chess was to work together, and having captains would give them a little edge of control over the team, which defeated the purpose.

On another note: I'll try to update the player list at least once a day, but don't worry if I don't add you right away.

And one more thing: I'm going to San Diego this weekend, so this won't be updated for about 4 days starting Friday.

[quote="Temperal"]The number of teams will have to be a power of two if it's knockout. Will you decide the teams?[/quote]
Yes, it will have to be a power of 2. This can be easily arranged no matter how many people we have, however, because so far, there have been no team requests made (except my team), so I will be able to make teams of 2 or 3 people.

As for choosing teams, I will do it completely randomly; most likely with a random sorting program I have.

[quote="#H34N1"]Maybe we should put our ratings so you can pair like a good/bad user up or something?[/quote]

The idea was that [i]you[/i] decide the teams, not the TDs.

[quote="not_trig"][quote="1=2"]I can be on the substitute list! :D My dad is really encouraging this, so I will definitely be very active![/quote]

I wish [i]my[/i] parents encouraged [i]me[/i] to get on AoPS... I apparently waste too much time reading things and not doing problems...[/quote]
Sigh...my parents think I'm wasting my time by going on here :(

The substitute list will be compiled privately so teams may not view them and plan to sabatoge others. If you wish to be added onto a substitute list (you must not be a member of the current teams) then send a private message to me. You will recieve a pm if you are selected to replace an existing member.

New rules:

If you have not been following the game in G&FF a couple of months ago, please use [url]http://www.apronus.com/chess/wbeditor.php[/url] as a chess board generator.

Update for the Holidays: Please allow the other team some slack if their moves are late. Some people might be on vacation.

Happy Holidays!

Now that the holidays are over, deadlines are as normal again. Each team as until January 7th to post their next move.

Can I get randomed into a team?

I dont know anyone who goes on aops.

It's been at least 10 days since the last move post. If no one's going to move soon then the tournament's going to be canceled...

[quote="thanhnam2902"]Hi. Who can help me? I want to join![/quote]

Team Eta desperately needs some replacements...(I'm from team Beta)

Well I think it looks like its been cancelled......far more than 10 days passed.
JOIN MY TOURNAMENT! :D

Hmm it looks like the tournament has been canceled due to inactivity.

[b]Please finish your remaining games.[/b]

I want to join, if I still can do it... :lol:

I like chess...

something about the fact that the last post was two months ago should have tipped you off.


Am I too late to participate in USACO? If not, how do I join?

Thanks


AMC USA(J)MO USAMO email algebra polynomial induction

So, what is your value of $\frac{\text{Your Expected Score}-\text{Your Score}}{\text{Your Score}}$? Mile is 2/3 :blush:...

By a huge factor, $\frac{-1}{3}$ here.
The USAMO graders are super nice it seems.
I mean I got 1 point just for pointing out that $R \geq 2r$

Well, pre-Red MOP i estimated like 8 and I got an 11. So, -3/11

What happens if we scored a 0? (I didn't, just wondering)

I underestimated by $-\frac{5}{12}$.

I got 11.  My estimate was 10.5, so $\frac{-1}{22}$.  He, I was pretty close.  :D

Predicted - 15? That's what I remember saying.
Actual - 18. One away from HM :(

overestimated by $\frac{-1}{6}$

[quote="undefined117"]diophantient, how do you know you got a point for that? do they send you an email?

hehe solafidefarms should've included an option for undefined.. division by 0 XD

although I also pointed out $R \geq 2r$.. i wonder why I didn't get a point like diophantient did.. :huh:[/quote]
Hmm.
I did some other stuff by dividing both sides by 8 and using AM-GM, but I thought that was useless.
Anyways, I completely solved number 1, and then I think I got 1 point on number 5. Unless they're giving 2 points for getting to the polynomial having to be complicate.

well, mine is $\frac{2}{7}$, same as last year. 9-7 and 18-14.

I had predicted somewhere near a 10, and ended up getting 19. $-\frac{9}{19}\approx-\frac{1}{2}$

-.21, which is pretty cool. (11-14)
I'm still not too happy though.

[quote="MysticTerminator"]predicted: death
actual: 25

so uh (death - 25) / 25 ?[/quote]
Errr ok so you predicted that you would just keel over and die? Anyway predicted--35, actual--28, x thingy--0.25, except eh I think a few turkeys were involved...

really? I'd have predicted 36 given the monumentally large turkey size of #5

and dude yes the predicted score was indeed death

*sigh* I predicted 10. Got 16. And got depressed.

Predicted: 21, actual: 17.

Anybody know how much the graders took off for not proving that putting three circles tangent to each other in problem two creates a "minimal" space in which to fit your circle of radius >= 1/root2? I'm not sure if I lost most of my points for that omission or for my generally crappy proof writing. (I answered 1, 2, 4 completely, and got the n + 3 factors that everybody got on 5).

Predicted: About 24-25
Actual: 23

Hm...I thought what I said on #2 would have warranted at least two points...if this was the case, this means what I wrote on #5 probably warranted a point (it was essentially what most other people wrote) and what i wrote on #6 nothing...unless I lost some points on my complete solutions for 1, 3, 4...

Predicted before the test: 14
Predicted after the test: 14
Actual score: 14

Man I am good at this.

My predicted score for next year will be 21.

Predicted: 15
Actual: 18
Thingy: -1/6 (assuming hamster can do arithmetic - I got the same predicted and actual)

So I guess I got 3/4 points on #2 (which sort of seems unlikely given that these are really rare scores) and 1/0 points on #5 (I basically did the same thing as everyone else).

My score is 5/3 of what I predicted, making my answer to the poll question -2/5. Wow. Not expected. I'm happy now.

[quote="gighiuhui"]I've had my fair share of experience with overstimating scores.  Two years ago, I predicted I'd score around a 10.  Last year, I tried to be conservative and estimated 6 points.  Both times I got 3.

If nothing else, the lesson I learned was what is and what is not a proof.  Overestimation of USAMO scores stems from people thinking they've proved something when really they've only intuitively grasped why it [i]should [/i] be true.

I was advised to learn how to write a clean proof, but more fundamental than that is learning how to determine whether or not one even has a proof.  Most times beginners struggle with writing proofs not because of faulty organization, but because of faulty logic.

This year, I estimated my score right on the button.[/quote]

Thanks for this post, gighiuhui.  I think it pretty well explains why I only got 2 points on #4, but eh, 9 is not bad for an 8th grader.

So what do you vote for if you predict 0 and get 0?  :roll:

Actually I was predicting 10-14 and got 14.

Division is too complicated for me, so I'll go by last year's system and say +2.

$-\frac1{10}$. I got 1 more than i predicted i would

I got $\frac{3}{11}$. It seems like the graders only gave 1-2ish points each on #2 and #4 (and in both cases I nearly had a proof  :( ). I
 got
#1 totally right
#2 up to the point with the circle, and didn't realize that a circle with radius $\frac{1}{\sqrt{2}}$ had to contain a lattice point
#4 got the answer, set up induction to prove 8026 wouldn't work, and nearly got the answer when time was called
#5 did the induction, didn't have the factorization
#6 proved that $R \ge 2r$ and a couple other things

Predicted - 4
Actual - 10

Overestimated by $-\frac{3}{5}$.


search
combinatorics proposed
combinatorics

[quote="delegat"]Ten persons were buying books. It is known that:
(a) Every person bought four different books
(b) Every two persons bougt at least one book in common
Consider the book bought by largest number of these people.Find the smallest possible value of that number.[/quote]

The answer is 4.
[hide="Solution"]
Determine that book as "favorite book"(if it is few such books get randomly one of them). Assume that first person bought book 1,2,3,4. Then 9 other persons bought one of this 4 books(becouse they have common one book with first person). Therefore at least one of this books was bought at least 4 times. So number of people who had bought the favorite book is at least 4. So it's left to show example of buying books by 10 person such that favorite book will be bought exactly by 4 people.
Let's determine 1,2,3,4,5,6,7,8,9,10 -is the serial number of 10 books that will be bought, $ P_1,P_2,P_3,P_4,P_5,P_6,P_7,P_8,P_9,P_{10}$ is the set of books that was bought by this 10 person.
So
$ P_1${1,2,3,10}
$ P_2${1,4,5,6}
$ P_3${1,7,8,9}
$ P_4${1,4,5,6}
$ P_5${2,4,8,9}
$ P_6${2,4,7,10}
$ P_7${2,5,7,9}
$ P_8${3,5,8,10}
$ P_9${3,6,7,8}
$ P_{10}${3,6,9,10}
this example satisfy condition (a) and (b), so favorite book was bought 4 times, and therefore smallest possible such number is 4.
[/hide]

The problem represents this so much:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1586306467&t=67465


probability
geometry
3D geometry
sphere
tetrahedron
Putnam
algebra solved

Any idea about this one, anybody?

We have a sphere and we randomly take 4 points on it. What's the probability that the center of the sphere lies inside the tetrahedron formed by the 4 points?

hey u can look up john scholes' site.the basic idea being to consider three points as a spherical triangle and getting the condition that for the centre to be inside the tetrahedron the fourth point must be antipodal to sum point in the spherical triangle.then out of a set of 4 points there are triangles possible.and their probabilities are the same as u can find automorphisms between any sets of three points.so reqd prob 1/8

Hey, the probability is really 1/8? I remember I solved this once(also geometrically) but I didn't know anything about spherical triangles then, so it was insome other way) but I can't remember how. I'm trying to really hard. I remember I also got 1/8 though..

link is:

http://www.kalva.demon.co.uk/putnam/psoln/psol926.html


in ABC triangle D is a point on BC such that $ \frac{sinBAD}{sinCAD}\equal{}k$
$ ,\forall k\in R$ can you draw AD?

$ BD/CD \equal{} (AB/AC).(sin < BAD/sin < CAD) \equal{} k(AB/AC)$

yes it Is constructible as long as k is a constructible number wit the unit segment


A $3\times 3 \times 3$ cube is formed by gluing together 27 standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the $3\times 3 \times 3$ cube is \[ \text {(A) } 60\qquad \text {(B) } 72\qquad \text {(C) } 84\qquad \text {(D) } 90\qquad \text {(E) } 96 \]

[hide]The dies can show 0 sides, 1 side, 2 sides, or 3 sides. There is 1 die that appear on the surface of the cube. There are 6 dies that show 1 side, one on each face of the cube. These will show the number 1. There are 8 cubes (the vertices) that show 3 sides, and these will show 1, 2, and 3. The remaining 12 cubes will show 2 sides, 1 and 2.

The sum is $6\cdot1+12\cdot3+8\cdot6=90\implies\boxed D$.[/hide]

What a waste of 27 perfectly good dice...
[hide]
Position the dice so that 1, 2, and 3 are shown at the 8 corners, 1 and 2 are shown at the 12 edges, and 1 is shown on each of the 6 faces. Thus, we get $8 \cdot 6 + 12 \cdot 3 + 6 \cdot 1$ or $90$, so the answer is D.
[/hide]

[hide="Answer"]We obviously want all of the centers to show $1$, all of the edges to show $1$ and $2$, and all of the corners to show $1$, $2$, and $3$. We have eight corners, 12 edges, and six centers, so our answer is $8(6)+12(3)+6(1)=90\Rightarrow \boxed{D}$.[/hide]

[hide]Let the 6 dice that have only one face showing show a 1. Let the 12 dice that show two faces show 1 and 2. Let the 8 dice that 3 faces visible show 1, 2, and 3. THe total is 6(1)+12(3)+8(6)=90. [/hide]

[hide]2 of the faces of the cube will have all ones, the 4 other sides will be a repeat of the form:
$213$
$212$
$322$
$-----$  adding we get
$7+4+7 = 18$
Multiply for 4 sides...
$18*4 = 72$
Add the extra 18 for the 1's on both sides...
$72+ 18 = 90$[/hide]


A horizontal turntable in the form of a uniform disc of mass 150 kg is mounted on a light, frictionless vertical axis at its centre. Two men of mass 75 kg stand at opposite ends of a diameter, and they and the turntable are at rest. The men then move round the table in the same direction and at the same constant speed. Calculate the angle they have turned through in space when they have made one complete circuit of the table?

Use conservation of angular momentum. Let the disc be your reference frame.

Like:
$ I\omega \equal{} 2mvr \\ \\
2\pi \equal{} \left(\frac{v}{r} \plus{} \omega\right)t$


how many $(x,y,z)$ exist that $xyz=2+x+y+z,  xy+yz+xz=12$. we know that $(x,y,z)\in\mathbb{R}^+$
A)$0$  B)$1$   C)$3$     D)$6$    E)$12$

The answer is $B)$.

W.L.O.G $1\leqq x\leqq y\leqq z$, $x+y+z+2\geqq 3x+2,\ xyz\leqq x^3$, by the given condition
we have $3x+2\leqq x^3\Longleftrightarrow (x-2)(x+1)^2\geqq 0$.Since $x>0$, we obtain $x\geqq 2$.
Similarly since $y>0,z>0,$we have $3x^2\leqq xy+yz+zx\leqq 3z^2\Longleftrightarrow x\leqq 2\leqq z$.
Therefore we obtain $x=2$.
Plugg this into the given two equation, we get the system of equation $2yz=y+z+4$ and $2(y+z)+yz=12$.
Solving for $y+z,yz$, we have $(y+z,yz)=(4,4)\Longleftrightarrow (y,z)=(2,2)$,
Consequently We obtain $(x,y,z)=(2,2,2)$

Wait... if $1 \le x \le y \le z$, doesn't that mean $xyz \ge x^3$ and not the other way around?

The answer is $E)$.

Let $xyz=k$, $x,y,z$ are roots of the cubic equation for $t$, $t^3-(k-2)t^2+12t-k=0$.

Solving for $k$, we have $k=\frac{t^3-t^2+12t}{t^2+1}=t-1+\frac{11t+1}{t^2+1}\cdots [E]$.

Therefore $k$ is a positive number $\Longrightarrow t^2+1\leqq 11t+1$,yielding $1\leqq t\leqq 11$.
Then plugging $t=1,2,\cdots 11$ into $[E]$, we obtain the two pair of positive integers $(t,k)=(1,6),(11,11)$.
Consequently, we get $xyz=6$ or $xyz=11\Longleftrightarrow {x,y,z}\in\{1,1,6\},\{1,2,3\},\{1,1,11\}$.
Taking into consideration the permutation of these pairs, we have $3+6+3=12$pairs. ;)

no,  the answer is incorrect :(
[hide="hint"]
AM-GM[/hide]

[quote="paladin8"][hide]
$xyz = 2+x+y+z \ge 2+3\sqrt[3]{xyz}$

Let $a = \sqrt[3]{xyz}$

Then $a^3 \ge 2+3a \Rightarrow a^3-3a-2 = (a+1)^2(a-2) \ge 0 \Rightarrow a \ge 2$.

So $\sqrt[3]{xyz} \ge 2$.

We also have

$(xyz)^{\frac{2}{3}} \le \frac{xy+yz+zx}{3} = 4 \Rightarrow \sqrt[3]{xyz} \le 2$.

Thus $\sqrt[3]{xyz} = 2$ and since equality on AM-GM holds iff $x=y=z$, we have $x=y=z=2$. And there is only one solution. So [b]B[/b].[/hide][/quote]
 :)


analytic geometry
inequalities
pigeonhole principle
triangle inequality
complex numbers
geometry proposed
geometry

Each point of the plane with two integer coordinates is the center of a disk with radius $ \frac {1} {1000}$.
Prove that there exists an equilateral triangle whose vertices belong to distinct disks.
Prove that such a triangle has side-length greater than 96.

again a romanian problem, taken from the isl 2003. boy, we used a french problem last year in our tst-s, and now you guys are paying us back by using two romanian problems in this year's tst-s? :P:P so nice ..  :lol:

Note that, at least one of the contestants ('Raf' on mathlinks) has solved it with the improved bound of 124 instead of 96 (as noticed by Darij) and a very nice and easy solution.
I'll give it if he doesn't.

Pierre.

I belive that raf should have valsed through this imo tst, since I already solved 4 questions in less than 20 minutes. :D

He seems to have a made a minor error in the pb 3 but he will probably score around 40-42/42.
He said he spent many times to find his error because he cannot believe to have solve the problem so easily with 124 instead of 96. Maybe, it was the most difficult part for him :D , everyone has its own problems....

Pierre.

Valentin, Raphal went to the IMO last year, do you remember him by any chance?

well I remember the whole french team, cuz I kept seeing them on the airports on my way on and back (bucharest -> charles de gaule -> tokio and back) :P but I couldn't pin point him by name  :blush:

So, my solution for the second part of the problem (first part is very easy)

Let ABC be an equilateral triangle with A,B end C in differents disks.
Let O1 be the center of the disk wich contain A.
O2 th center of the disk wich contain B
and O3 the center of the disk wich contain C.

Lemma: There isn't an equilateral triangle PQR with P,Q and R at integers points.

Proof: We can assume that P(0,0)
Let be Q(u,v) and R(u',v') the coordinates of Q and R.
If 2 divides u,v,u' and v' points Q'(u/2,v/2) and R'(u'/2,v'/2) are such that PQ'R' is equilateral too. So we can assume that u is odd for example.
Then you look at the two cases of v modulo 2, you get a contradiction with:
u^2+v^2=u'^2+v'^2=(u-u')^2+(v-v')^2

So the lemma is true.

Now by triangle inequality:

O1O2-O1A-O2B=<AB=<O1O2+O1A+O2B
Let a=Ab=AC=BC
Moreover we have O1O2-O1A-O2B>=O1O2-2/1000 and O1O2+O1A+O2B=<O1O2+2/1000
So we get a-2/1000=<O1O2=<a+2/1000
We get with the sames arguments:
a-2/1000=<O1O2,O2O3,O1O3=<a+2/1000
So O1O2^2,O2O3^2 and O1O3^2 are all in [(a-2/1000)^2,(a+2/1000)^2]
But at least two of O1O2^2,O2O3^2 and O1O3^2 are different because of the lemma and there are all integers.
But the size of the interval [(a-2/1000)^2,(a+2/1000)^2] is 8a/1000 and contain at least two integers differents so we have 8a/1000>=1
so a>=125.

i remember a long hair boy of the romanian team and I think there is a boy of your team at Louis le Grand now in France, no?

yes Manolescu is following classe preparatoire and you probably remember our dear Kappa :) which you will probably meet again in Greece :D

This problem was used in one of the Swedish correspondence tests. For the first part there were two different official solutions: one using Kroenecker's Approximation Theorem with complex numbers, and the other using Pigeonhole together with some other stuff.

Wheras the second part, the prove was by contradiction.

It is quite strange how the same question can appear on almost every Team TST in Europe  :)

... and I have used the [url=http://mathlinks.ro/viewtopic.php?t=5537]"well-known theorem of Dedekind"[/url] (yeah, it is the Kronecker density theorem, but not for complex numbers)...

  Darij


I need ebook about solve locus problem by vector.Can somebody help me?

http://www.ebook-search-queen.com/ebook/locu/locus+problem.all.html - here's what I managed to find. I don't know much about these issues, thus check it yourself for more specific understanding   :!: And let me know if this link was of any help to you.


probability
combinatorics unsolved
combinatorics

Each citizen in a town knows at least $ 30$% of the remaining citizens. A citizen votes in elections if he/she knows at least one candidate. Prove that it is possible to schedule elections with two candidates for the mayor of the city so that at least $ 50$% of the citizen can vote.

Choose the candidates uniformly at random from all pairs of distinct candidates.

Lets say a given voter fails to know $ q$ of the $ n$ citizens.  The probability he fails to know both candidates is $ (\frac {q}{n}) (\frac {q \minus{} 1}{n \minus{} 1}) \leq (\frac {q}{n})^2 \leq 0.49$ (I'm assuming he knows and can vote for himself if he happens to be chosen as a candidate).

Therefore on average 51% of the citizens can vote, so there must be a particular pair of candidates for which 51% of the citizens can vote.


MATHCOUNTS
geometry
3D geometry
sphere
rectangle
inequalities
modular arithmetic

Please keep the level MATHCOUNTS - AMC 12ish, proofs are okay as long as they're not incredibly, incredibly hard. The rules are simple: solve the problem with a solution and then post a new problem.

Let's start off easy...

[size=167]Problem 1[/size]

A soccer ball with radius three is inscribed in a cubical box. Compute the box's surface area.

Soccer balls aren't perfect spheres. They are actually polyhedra.  :P

Assuming that they were spheres, the answer is $ (3\cdot{2})^2\times{6}\equal{}\boxed{216}$.

Bah. When I saw there were new posts in the New Jersey forum, I got all excited, thinking the NJ people finally got NJ-spirit.  :P

Yeah, it was supposed to be a sphere. Please post the next problem.

Fine, I'll post the next one.

[size=167]Problem 2[/size]

Two perpendicular lines are dropped in a hexagon so that two triangles and a rectangle are created, as shown below. If a diagonal of the rectangle has length 13 centimeters, find the area of the hexagon.

 [geogebra]bc81eb1f31354125bf766c586969a6819e861667[/geogebra]

The hexagon can be divided into 6 equilateral triangles with side length $ \frac{13}{2}$. Thus the area of the hexagon is ${ 6(\frac{\sqrt{3}}{4}})(\frac{13}{2})^2=6(\frac{169\sqrt{3}}{16})=\boxed{\frac{507\sqrt{3}}{8}}$.

I'll post the next one now.

[size=150]Problem 3[/size]

What is the sum of the last two digits of $ 1337^{42}$?

we only need to focus on the last $ 2$ digits. hopefully we will see it repeat somewhere.
the last $ 2$ digits of $ {1337}^1$ is (obviously) $ 37$. then we find that the last $ 2$ digits of $ {1337}^2$ is $ 69$ (because we just square $ 37$). similarly, we can generate a list of last $ 2$ digits by multiplying that last $ 2$ digits of the previous number by $ 37$
$ 37, 69, 53, 61, 57, 09, 33, 21, 77, 49, 13, 81, 97, 89, 93, 41, 81, 61$ PHEW! it finally repeats. since there are $ 11$ terms before the repeat, we need to find the $ 31st$ term in that series starting from $ 81$. it repeats after $ 5$ terms so we do $ \frac{31}{5} \equal{} 6 R 1$. so the last $ 2$ digits of $ {1337}^{42}$ are $ 81$ and the sum of $ 8 \plus{} 1 \equal{} \boxed{9}$

UGH I keep forgetting to post new problem  :|

NEW PROBLEM:

Prove the Power Mean Inequality Chain:

$ \sqrt {\frac {a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2}{n}} \ge \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \ge \sqrt[n]{a_1a_2...a_n} \ge \frac {n}{\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n}}$

When does equality hold?

P.S. Is this too hard?

Bleh when all the $ a_i$ are equal.

[hide="solution to problem 3"] $ \phi({100}) \equal{} 40$ so $ 1337^{42}\equiv7^{42}\equiv7^2\equiv49\pmod{100}$, so our answer is 4+9=13.[/hide]

I refuse to post a problem.

EDIT: Yeah uh I only read the equality part  :wink: .
The rest is too hard.

eh fine, then I'll post one.

What is the largest integer that is a divisor of

$ (n \plus{} 1)(n \plus{} 3)(n \plus{} 5)(n \plus{} 7)(n \plus{} 9)$

for all positive [b]even[/b] integers $ n$?

haha since I'm bored I will post another one.

Suppose $ x < 0$. What is the value of $ \sqrt{\frac{x}{1 - \frac{x - 1}{x}}}$ in terms of $ x$?

............ and another one.............

How many positive integers $ (a, b)$ are there such that $ GCF(a, b) = 1$ and $ \frac{a}{b} + \frac{14b}{9a}$ is an integer?

sigh
This is a pitiful attempt to revive the marathon, which consisted of $ < 10$ problems.
[hide="solution to first problem"]Among $ 5$ consecutive odd numbers, at least one is divisible by $ 3$ and exactly one that is divisible by $ 5$. This means the product is always divisible by $ \boxed{15}$. This is the largest[/hide]
[hide="solution to second problem"]$ \sqrt {\frac {x}{1 \minus{} \frac {x \minus{} 1}{x}}} \equal{} \sqrt {\frac {x}{\frac {x}{x} \minus{} \frac {x \minus{} 1}{x}}} \equal{} \sqrt {\frac {x}{\frac {1}{x}}} \equal{} \sqrt {x^2} \equal{} |x|$. Since $ x < 0,$ this is equal to $ \boxed{ \minus{} x}$[/hide]
[hide="solution to third problem"]Rewrite
\[ \frac {a}{b} \plus{} \frac {14b}{9a} \equal{} m
\]
in two ways.
\[ \frac {14b^2}{9a} \equal{} bm \minus{} a
\]
Because $ a, b, m$ are integers, and $ GCD(a, b) \equal{} 1$, that means $ a|14$. Similarly,
\[ \frac {9a^2}{b} \equal{} 9am \minus{} 14b
\]
, so $ 9|b$.
The possible values for $ a$ are $ 1, 2, 7, 14$, and for $ b$ they are $ 1, 3, 9$, which gives a total of $ 12$ solutions. Checking each one gives the answers $ (1, 3), (2,3), (7, 3), (14,3)$. The answer is thus $ \boxed{4}$[/hide]

[hide="correction for problem 3"]xpmath basically had the answer. Since you want the last two digits, you have to leave two digits. That would mean you can't just leave the $ 7^2$, you have to leave the $ 37^2\equal{}1369$ so the sum of the digits is $ \boxed{15}$. (eh you could $ n$spire bash this)[/hide]

[b]new problem[/b]
A regular hexagon has perimeter $ 12 \plus{} 84\sqrt {3}$. What is its area?

That means each side has length $ 2 \plus{} 14\sqrt{3}$. Since we can split the hexagon into 6 congruent, equilateral triangles, the total area is $ 6\left(\frac{(2\plus{}14\sqrt{3})^2\sqrt{3}}{4}\right) \approx 1790$ square units.

[b]Problem 9[/b]:

Let $ p_1 \equal{} 2$, $ p_2 \equal{} 3$, $ p_3 \equal{} 5$, and so on, so that it's the sequence of prime numbers. Determine the arithmetic mean of the three smallest values of $ n$ such that $ p_1 \plus{} p_2 \plus{} \dots \plus{} p_n$ is not prime.

[hide="in heah"]brute force gives n is either 3, 5, or 7. their mean is 5.[/hide]

[b]Problem 10:[/b]

um... How many integer solutions are there to $ \frac{1}{x^2}\plus{}\frac{1}{y^3}\equal{}\frac{1}{24}$?

[hide="This is Probably Wrong..."]The highest degree of $y$ is is $3$, therefore, there are $3$ values for $y$.  The highest degree of $x$ is $2$, therefore, there are $2$ values for $x$.

For every $2$ values of $x$ there are $3$ values for $y$.  Therefore, there are $2(3)=\boxed{6}$ integer solutions?

This is wrong...I can tell.  Can someone please post a correct solution?  :oops:

Thank you.[/hide]

[size=200][b]Problem 11:[/b][/size]

A cable is tied around the Earth at the equator. A longer second cable is placed so that it is always exactly 8 feet directly above the Earth's equator. What is the positive difference in the number of feet in the lengths of the two cables?

[hide="Solution"]
It's
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n(n+1)}$
$\frac{1}{n(n+1)}$ simplifies to $\frac{1}{n}-\frac{1}{n+1}$.
The series becomes $\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...$.
All of the alternating terms cancel each other out, leaving with $1+\frac{1}{n+1}$ as the sum for $n$ terms of the sequence.
As $n$ reaches infinity, $\frac{1}{n+1}$ approaches $0$.  So the answer is just $1+0=\boxed{1}$.[/hide]

[size=200][b][u]Problem #17:[/u][/b][/size]

For how many positive integer values of $n$ is $\sqrt{144^2-n}$ an integer?

Solution:

[hide]
The value 144^2-n must be a perfect square, anywhere from 143^2 to 0^2 (n cannot equal zero and the value must be non-negative), so there are 144 solutions.
[/hide]

Someone else post a problem for me because I don't want whirly to verbally attack one of mine again.

Hard problems are not made from a bunch of random stuff mixed together so that you end up with a long and boring problem... I know I used to do that earlier (it may not have been your intention). Time should not be equated to difficulty (just because it takes a long time does not mean it's necessarily harder...). For the problem to be truly "hard", well, then it has to be mentally stimulating... (the solutions may turn out to be beautiful, but that's besides the point). Bashing is not stimulating.

A good example of a problem that is short and is kind of [i]beautiful [/i] is 2010 USAMO #4/USAJMO #6. Well, this one wasn't too hard, but it was somewhat hard enough that it could be put onto USAMO. Needless to say, as I believe someone earlier on the AMC forums say, "it's a one sentence solution."

Anyways, I think this may not have been colinhy's intention. xD. If not, then I was just randomly ranting.

Let's see...
I posted this on my blog, but I could post it over here too. I know the people who've posted answers/problems here so far (in recent time) are 8th graders, but I don't know how experienced you are with proof writing, so try this one (it's a basic proof):

[b][u][size=200]Problem #18[/size][/u][/b]

Find all integer triples $(x, y, z)$ that satisfy $x^3+y^3+z^3=2011$.

[hide="Problem #18 (Badly Worded) Solution"]
Modulo 9 is good for cubes.  The only residues modulo 9 for cubes are $-1, 0, 1$ (which I should prove but I'm too lazy).  We see that $2011\equiv 4\pmod{9}$ therefore we must add three of any of the mentioned residues to create a number congruent to $4$ modulo $9$.  This is impossible seeing that there are only three residues that can be added and 3 is the largest residue you can go.  If we add the $-1$'s together the best you can do is 6.  Thus, you cannot reach $4$ modulo $9$ with the sum of cubes.  Therefore, there are $\boxed{0}$ integer triples.
[/hide]

Sorry for the terrible wording, and I can't think of a new problem at the moment so can someone post one for me?

[hide="Solution"]
In the denominators, cancel out $\sqrt{x + 1}$ to get $\sqrt{x + 6} = \frac{\sqrt{7x + 15}}{\sqrt{x + 1}}$.  Then this becomes $x^2 + 7x + 6 = 7x + 15$, so $x^2 = 9$ and $x = \pm 3$.  $-3$ doesn't work because then $\sqrt{-3 + 1}$ is not real.  So $\boxed{3}$.
[/hide]

New problems need some more emphasis, so:

[b][size=200][color=blue][u]PROBLEM #20:[/u][/color][/size][/b]

Let $ABCD$ be a quadrilateral. The diagonals $AC$ and $BD$ meet at $O$. Suppose $S_{1}, S_{2}, S_{3},$ and $S_{4}$ are the areas of triangles $AOB$, $BOC$, $COD$, and $AOD$ respectively. If $S_{1}=2S_{2}$ and $S_{2}+2=S_{3}+5=S_{4}$, compute the area of $ABCD$ is

I expect this to be done very soon.  It should less than 3 minutes because that's how long it took me kthxbai.

[hide="Solution"]Note that $\frac{s_1}{s_2}=\frac{s_4}{s_3}=2$. Thus, $s_3=5, s_4=10, s_2=8, s_1=16$ and the answer is $\boxed{39}$.[/hide]
[b]Problem 21[/b]
Find the largest possible radius of a sphere inscribed in a cone with diameter and height $24$.

[hide="Problem 21 Solution"]
$ \text {3-D} \implies \text {2-D} $

We make [b]similar triangles[/b] here.

We get: \[ \cfrac {24-r}{r}=\cfrac{24\sqrt{5}}{12} \] We solve for $r$ to get \[ r = \boxed {6 \sqrt {5} - 6} \]
[/hide]

[b][u]Problem 22[/u][/b]

Adam draws a triangle ABC. His sister, Julie, then draw a point D on AB and a point E on AC, such that DE || BC. If AD = 4 and DE = 3 and BC = 7, then what is BD?

[quote="ahaanomegas"]

We get: \[ \cfrac {24-r}{r}=\cfrac{24\sqrt{5}}{12} \] We solve for $r$ to get \[ r = \boxed {6 \sqrt {5} - 6} \]
[/quote]
Shouldn't the numerator of the RHS be $12\sqrt5$ not $24\sqrt{5}$ since it's the leg of the isosceles triangle which is $\sqrt{12^2+24^2}=\sqrt{12^2\times5}=12\sqrt5$.
[quote="applepi2000"]Note that $\frac{s_1}{s_2}=\frac{s_4}{s_3}=2$.[/quote]
How do we get that the ratio of the areas separated by the diagonals have constant ratios?

It's been a long time.  I've missed the activity.  ;)
[hide="Solution"]
After drawing the diagram, we see similar triangles.
$\frac{3}{4}=\frac{7}{BD+4}$

$28=3\cdot BD+12$

$16=3\cdot BD$

$\boxed{BD=\frac{16}{3}}$[/hide]

[b][u][size=175]Problem 23![/size][/u][/b]
If $a-b=7$ and $ab=5$, what is the value of $a^3-b^3$?

[hide="Problem 24 Solution"]
\[ \begin {eqnarray*} \sqrt {3 - \sqrt {2 + x}} &=& x \\ 3 - \sqrt {2 + x} &=& x^2 \\ \sqrt {2 + x} &=& 3 - x^2 \\ 2 + x &=& 9 - 2x^2 + x^4 \] We solve the equation and get [url=http://www.akiti.ca/Quad4Deg.html]these solutions[/url]
[/hide]

Can someone show us how to do that problem without a calculator?

Also, new problem:

[b]Problem 25[/b]

Let $S$ be a set of $ \{ 1, 2, \cdots, 1989 \} $ such that no two elements differ by $4$ or $7$. Find the greatest possible number of elements in $S$.

Hmm, well there [i]is[/i] the quartic formula. :P


Find the greatest real $ p > 1$,such that for all $ \dfrac{1}{p}\leq a,b,c \leq p$,the following inequality holds:
$ 9(ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2)\geq(a \plus{} b \plus{} c)^4$.

I've posted this inequality a long time ago,but I got no reply.I asked if the answer is $ p \equal{} \sqrt {2}$ as the inequality holds if $ a,b,c$
are sides of triangle (the fact can be easily proved by making the triangle substitution $ a \equal{} x \plus{} y$,etc.).I hope now I'll see a solution.
Please help! :D

Have you forgotten a condition ?(such as $ ab\plus{}bc\plus{}ca\equal{}3$ or $ a\plus{}b\plus{}c\equal{}3$)

I've forgotten that $ \dfrac{1}{p}\leq a,b,c \leq p$. The rest is OK. :D

Mixing varibles ? .
;) .We only need to prove this ineq with :$ a \equal{} b$;).
The ineq is equivalent to:
<=>$ (a \minus{} c)^2( 2\frac {a^2}{c^2} \plus{} 8\frac {a}{c} \minus{} 1) \ge 0$
Setting :$ t \equal{} \frac {a}{c}\rightarrow t \in [\frac {1}{p^2},p^2]$
The equation :$ 2t^2 \plus{} 8t \minus{} 1 \equal{} 0$ has two roots .$ \{\frac { \minus{} 4 \plus{} 3\sqrt {2}}{2},\frac { \minus{} 4 \minus{} 3\sqrt{2}}{2}\}$
=>$ \frac {1}{p^2} \ge \frac { \minus{} 4 \plus{} 3\sqrt {2}}{2}$
Done ;)
[b]P/s[/b]:Sorry for my gammar and my solution(if i have some mistake )

I don't understand what is your answer.Does it give you a number greater than $ \sqrt {2}$?

Yeah ;) p must safity that :$ \minus{}(a\minus{}b)^2\plus{}2(2a\plus{}2b\minus{}c)c \ge 0$ ;)
[b]P/s[/b] I don't think my solution is true ;) .Because, ...(see my signature )

Sorry, but $ p>1$, or I don't understand your solution.Can someone help or post a complete solution?
P.S.: I am a more stupid boy. :(

Actually,the question is why if you the case $ a\equal{}b$, then you you check all cases ($ a$,$ b$,$ c$ are distinct)? :?:


geometry
trigonometry
vector
rotation
analytic geometry

I really don't think we're given enough info to solve this problem:
http://gauss.math.oakland.edu/mmpc/graphics/exams/2003.II.problems.pdf , #5.

Any clever solutions?

What more info could you ask for?  There's plenty in the problem.

[hide="Solution"] Let $P$ be the point of tangency between the two circles, which is a very important intermediate to take into consideration.  After rolling an angle $\theta$, $P$ is at $(\cos \theta, \sin \theta)$.

Where is the center of the large circle?  Let that point be $C$.  One look at the diagram should convince you that it is opposite the point $P$ with respect to the smaller circle; in other words, at $(-\cos \theta,-\sin \theta)$.

Finally, where is our mysterious black point $X$?  Consider the arc $PX$ on the large circle.  What angle does this arc subtend?

Since it rolled on the small circle, the length of the corresponding arc on the smaller circle is the same as this length, which is $\theta$.  Since the radius of the large circle is $2$, arc $PX$ subtends an angle $\frac{\theta}{2}$.

The vector $\vec{v}$ that gives the line segment from $C$ to $P$ is given by

$\vec{v}= \left< 2 \cos \theta, 2 \sin \theta \right>$

We need to rotate this to the right by an angle equal to the angle subtended by arc $PX$.  It shouldn't take long to convince you (again, from the diagram) that this is

$\vec{u}= \left< 2 \cos \frac{\theta}{2}, 2 \sin \frac{\theta}{2}\right>$

This gives the vector from $C$ to $X$.  We are almost done!  The vector from the origin to $X$, which gives the coordinates of $X$, is

$\vec{x}= \left< 2 \cos \frac{\theta}{2}-\cos \theta, 2 \sin \frac{\theta}{2}-\sin \theta \right>$

... I think. [/hide]


Three hockey pucks , $A,BandC$ ,lie on a playing field.A hockey player hits one of them in such a way that it passes between the other two.He does this 25 times.Can he return the three pucks to their starting points?

Each move changes the orientation of the triangle $ABC$ in the plane, so in order to bring the pucks back to their initial positions, a necessary condition is that we make an even number of moves. Since $25$ is odd, the answer is negative.


Is there some way to efficiently compute the minimum weight tree that spans at least a given $V' \subseteq V$ using edges from the weighted graph $G = (V,\, E)$? (e.g. you get the usual MST problem if $V' = V$)

And if that's too abstract, suppose further that the graph is just a (rectangular) grid of finitely many white and black cells (black = $V'$, white = $V-V'$) with unit edge weights.

Well, if we can guarantee $ V'$ will be large (say, $ |V \setminus V'| = O(\log |V|)$) then you could just enumerate over each of the induced subgraphs of $ G$ whose vertex set contains $ V'$ and compare the results.  (Sorry that this is probably not very helpful.)

What about running some all-shortest-paths algo on the graph and then making a new graph of shortest paths between vertices. Then you could just run MST on $ V\prime$.

--Artur

That's an interesting thought.  Just to clarify, you want to find the shortest path between each pair of vertices in $ V'$ (I don't know if that can be done efficiently, actually), then construct a complete weighted graph $ G$ on $ V'$ in which the distance between a pair of vertices is the shortest path between them in the original graph, then construct a spanning tree of $ G$ and take the union of the associated paths?  It's a nice-sounding idea, but I don't think it necessarily gives you the desired structure because every time you include an edge, that might affect the shortest path between two other vertices (since you now get that edge for free).  In fact, I don't think it will even necessarily give you a tree.

Yes, that's the idea.
BTW, computing shortest paths between all pairs of vertices can be done in $ O(n^{3})$ time either by using Floyd-Warshall Algorithm or in $ O(n^{2}\log n)$ with the repeated Dijkstra's algorithm.

Still, that was just a rough idea. There are few little details that come to mind(and make it seem reasonable):

1. Vertices from $ V\prime$ can only appear at the beginning or end of every shortest path (no $ v\subseteq V'$ can be in between).
2. While executing the MST algorithm we have to keep track of the used vertices ( some shortest paths might go through the same vertex).

That should assure that we're creating the tree. But I'm not entirely convinced that it's going to be the minimum.

Anyways, the problem sounds similar to k-minimum spanning tree which is NP-Complete.

--Artur


I've been wondering, is there a math magazine you guys recommend subscription to?  I would really like to have something fresh to read about math every month.   :)

Well there's
http://www.pims.math.ca/resources/publications/pi-sky
http://reflections.awesomemath.org/
http://www.math.ust.hk/excalibur/
http://www.komal.hu/info/bemutatkozas.e.shtml
http://www.maa.org/subpage_2.html

Here is some Mathematical issues of an University in Belgium :

   http://www.math.leidenuniv.nl/~naw/home/


calculus
integration
function
inequalities
geometry
trapezoid
real analysis

Let M_n be the nth Midpoint Rule approximation for a function f.
Let T_n be the nth Trapezoidal Rule approximation.
Show that if f''(x) >= 0 (i.e. f is convex on [a,b]), then for any natural numbers m,n, we have
M_n <= integral f(x)dx <= T_m.
If f''(x) <= 0 on [a,b], this inequality is reversed.

I've been playing around with the respective error bounds forever. How do you make this work algebraically? I'm having trouble simplifying the expression and working out the absolute values.


Alcumus videos probability geometry Support

How far can you go? I solve more than a 100 and get 99% right, I still get no videos now, I have like 75 videos, but what is the name of the last vide yu can reach?

There is no "last video" as everyone progresses differently. There are around 75 videos in total.

Another video question--I have gotten Probability with Combinations parts 1, 2, and 4. No number 3. Why is this? And should this be happening? It's somewhat annoying.

Depending on how you are doing, you get 1, 2, and 4 first or 3 first, and then get the other. I got 3 first.

Hi,

i have a question: How many Problem do I need to solve before I can get the geometry videos?

See you

mathe760

You have to finish the diagnostic phase, which I think is either 30 or 60 questions, you will start getting videos.

EDIT: Whoops. Didn't see geometry.

Alcumus is currently focused on Introduction to Counting and Probability. There are currently no geometry videos.

ok I didn`t know that, but will there come some geometry videos later on? Do you propably know some other sides with geometry videos?

See you

mathe760

Geometry might be next on the list after AoPS finishes implementing the Algebra part.

There are some "using geometry for probability" videos.

But I don`t meant this art of geometry. I meant this geometry with circles, tangents, triangles and so on, where you have to prove something...

See you

mathe760 :)

I got a geometry problem once; I looked at the "Related Readings" text after solving it and it referenced "Introduction to Geometry".  I couldn't find the problem on my Report under "Probability - Using Geometry".  Was that an Introduction to Geometry problem?


I need to learn C++ by the end of this summer. Any good online tutorials you might suggest?

In my experience, I have always learned better from books than online tutorials. So my recommendation would be to check out some books at your local library (maybe Idiot's Guide to C++, or similar beginning books), especially if you do not know any other programming languages.

1. Yes. Books(paper) are usually better-written. I like to read books [i]first[/i], and go to online tutorials/articles for [i]further[/i] information. For example I wasn't able to understand STL just by reading my book so I went online, and I found a tutorial that met my needs.

2. I think most people learn, and [i]should[/i] learn C before they learn C++. But there are exceptions. I learned a lot of Pascal, and a little of C, before I learned a looooooooot of C++. It worked for me. I should warn you that many C++ textbooks [i]require[/i] knowledge of C.

3. Books give you more freedom -- you don't have to sit by a computer all the time. However, you do need to spend a lot of time using computer(a C++ compiler software) during the learning process.

Sams' Teach Yourself C++ in 24 Hours is pretty good.

The Deitel & Deitel books are a bit too complex for beginners, but you can look at htose too. I don't have any experience in online tutorials.