A car is going to make a successful sharp turn at a high constant speed. For it to make a successful attempt which tires should rise, the inner or the outer(with respect to centre of the circular turn).

This was the actual question which anjani_kumarsingh wanted to ask in "Circular Motion"(a previous post in Physics) :D

[hide="My comments"]
What I think is that outer tyres should rise so that we get to tilt the car like our bicycle. hihi :lol: [/hide]

[quote="kubus"]I think that if the road itself is not banked, the car should lift its inner wheels so that the torque (in its inertial frame) is zero.

Not that I could imagine a car purposely rising its wheels as if it was a horse;)[/quote]
I know.
Can you tell me why should outer wheels raise?

When you're following the car around the track, you're in a noninertial frame of reference, so you can take into account the centrifugal force caused by the car's mass.
So basically, you have a force pushing the car "out" of the circle that acts on its center of mass, and friction pulling it in.  It's at rest in that frame of reference, so you can make this analogy.  when you're pushing a box(lets say it contains cereal) at its center, it doesn't move(because of friction.  But the box is like the car-- which way will it tip over?  As the direction of the "centrifugal force" acts in a direction that goes away from the center of the curve, you can say that the side that you are pushing from is the "inside" of the curve.

does that make it more clear?

So yu all didn't understand my question.

Take it this way: If the car raises its outer wheels then is it helpful to take a sharper turn at high speed(not too high)

[quote="kubus"]Well, this kind of reasoning does not properly explain the case of a bike, which is tilting inwards.

:)
[/quote]

well, in both cases, it's the torque from gravity that balances out that caused by centrifugal force--- and in both cases, the center of mass is on the "inner side" of the curve.  The bicycle leans in so that it can experience a torque from gravity to balance out centrifugal; in the case of the car, the centoer of mass is already "inwards" from the contact point(the outer wheel), so there is already a torque.

Varun,
if you had a car to better take the turns, I think it would be best to make the car's center of mass as low as possible.  It wouldn't be possible to have it both going in a circle and having the outer wheel lift up, because then both torques(grav. and centrifugal) point in the same direction, and make the car go back down.  Do you kind of get what I'm saying?[/quote]

Yes, I do and it is good thinking. :)
Ihadn't thought of the torque it will experience if its outer tyres raise.
So thanks guys.


let $a_n$ be a sequence of real numbers. Show that $\sum_1^{\infty} \epsilon_n a_n$ converges for almost every choice of the signs $\epsilon \in \{-1;+1\}$ iff $\sum_1^{\infty} a_n^2 < \infty$

What does "almost every choice" mean here?

for example one can say that the $ \epsilon_n: \Omega\to \{0,1\}$ form a sequence of independant random variables and $ \mathbb{P}(\epsilon_n \equal{} \plus{} 1) \equal{} \mathbb{P}(\epsilon_n \equal{} \minus{} 1) \equal{} \frac {1}{2}$. Then one can look at $ f(w) \equal{} \sum_1^{\infty} \epsilon_n(w)a_n$
(and sorry, this is the wrong forum .. )[color=red][EDIT : moved (from the LA forum). -mathmanman][/color]

Let $ \xi_n \equal{} x_n \varepsilon_n, n\geq 1$It's clear that they are independent. Then $ \mathbb{E}[\xi_n] \equal{} 0$
and $ \mathbb{E}[\xi_n^2] < \infty$ and from our condition we have that $ \sum \limits_{i \equal{} 1}^{\infty} \mathbb{E}[\xi_i^2] < \infty$ so we can apply One Raw Kolmogorov's theorem, we have that a.c. $ \sum \limits_{i \equal{} 1}^{\infty} \xi_i$ converges and that's exactly what we wanted to proove  :D

PS .I've omitted "iff" sufficient codition follows from the second part of Kolmogorov's theorem about one raw.


I can't figure this one out:

In the seven-sided polygon shown (look at attachment), what is the result when the sum of the outer angles CDE and AGF is subtracted from the sum of the interior angles A, B, C, E, and F of the polygon.

how did you already get the 06-07 handbook?

[hide]The sum would be 900 degrees if It was of the entire interior sum. $CDE=360-D$ $AGF=360-G$ $720-D-G$ is their sum. $900-D-G=A+B+C+E+F$
$720-D-G-900+D+G=-180$
$\boxed{-180}$[/hide]

[quote="pascal_1623"][quote="jclarkemathL314159"]Are you sure there's enough information given?[/quote]

bpms got it.[/quote]
Wow. From my answer I thought it was wrong but I didn't change it because I didn't see any error.

wow, i have a feeling this year's test will be uber hard

[hide]yes, the test will be hard this year.  That is what you should expect.

For the problem

The angles

$900-D-G=A+B+C+E+F$

Also, we subtract $360-D+360-G$ from $900-D-G$, which is 180

$180$[/hide]

I just realized that my work led to the answer [hide]-180[/hide] and my boxed answer was [hide]180[/hide]. The first one was what was intended. Am I right?

[hide]The sum of the interior angles is $900^\circ$.
$\angle CDE=360^\circ-D$ and $\angle AGF=360^\circ-G$, so that we want $A+B+C+E+F-(360-D+360-G)=900-720=\boxed{180}$.[/hide]

Wow. Talk about luck. If I could only be that lucky at the actual competition...

[quote="mathgeniuse^ln(x)"]Well, you were lucky, since you got a 33/46 at Nats, jk.

But yes, it is tricky when you see the word subtract from.[/quote]
Wait, doesn't the handbook come with solutions?

[quote="bpms"][quote="pianoforte"]I think the handbook comes with answers and solutions to some selected problems.[/quote]
But aren't all solutions in the back?[/quote]
It was to selected problems only I believe. The MathCounts handbook doesn't have a solution to this problem.


rotation
function
geometric series
calculus
calculus computations

Is there a simple closed form for $ a^{1}\plus{}a^{3}\plus{}a^{6}\plus{}\cdots$? $ |a|<1$, of course.

Well it is a geometric series so there should be. Leave $ a^1$ alone and look at $ a^0\plus{}(a^3)^1\plus{}(a^3)^2\plus{}(a^3)^3)$ etc...This forms a geometric series so just use the sum formula, then subtract $ 1$ from $ a^0$ and add $ a^1$. So the closed form would be $ (\text{geometric sum}) \minus{}1 \plus{} a$

It's not a geometric series. CatalystOfNostalgia didn't notate it very well, but his intent is clear from the title he gave this topic. The sum is

$ a^1\plus{}a^3\plus{}a^6\plus{}a^{10}\plus{}a^{15}\plus{}\cdots\equal{}\sum_{k\equal{}1}^{\infty}a^{\frac{k(k\plus{}1)}2}.$

No. If there were an elementary formula for the sum, it would be possible to extend this function analytically along almost every path on the complex plane. On the other hand, the Fabry theorem immediately implies that it cannot be extended through any arc on the unit circle.


Consider the set of digits $ C \equal{} \{1, 2, 3, 4, 5, 6\}$
a) Among all the natural numbers with four digits that it is possible to form using only elements of $ C$, find how many are multiple of $ 4$.

b) Among all the natural numbers with three different digits that it is possible to form using only elements of $ C$, find how many are multiple of $ 3$.

Can you repeat digits? Like 1111? or 2344?

[hide="Hint"]Numbers are divisible by four if their last two digits are divisible by four.[/hide]

[hide="(a)"][hide="If two of the digits can be equal"]
An integer is divisible by $ 4$ if and only if the integer in $ \mod 100$ is divisible by $ 4$. All two digit integers divisible by $ 4$ whose digits are in $ C$ are $ 12,16,24,32,36,44,52,56,64$. Thus we only need to count the number of ways to choose the first two digits of the $ 4$ digit integers, which is $ 6^2$. Then the required number is $ 6^2\times 9$.[/hide]

[hide="Otherwise"]In this case we need to consider the numbers $ 12,16,24,32,36,52,56,64$. The number of two digit integers with two distinct digits from $ C$ is $ \binom 62$. Thus our answer in this case is $ \binom 62\cdot 8$.[/hide][/hide]

[quote="123456789"]Can you repeat digits? Like 1111? or 2344?
[/quote]

In (a), yes.

(a)
[hide]
In order to be divisible by $ 4$, the last two digits must be divisible by $ 4$:
$ 12,32,52$
$ 24,44,64$
$ 16,36,56$
And the first two digits can be any two numbers, so there are in total of $ 6^2\cdot 9 \equal{} \boxed{324}$ different such four-digit numbers.[/hide]
(b)
[hide]In order to be divisible by $ 4$, the sum of digits must be divisible by $ 3$. Re-write the set in $ \mod 3$, we have
$ {0,1,2,0,1,2}$
Since each number can only chosen once, the only way to have a sum divisible by $ 3$ is: $ 0,1,2$. There are $ 2\cdot 2\cdot 2 \equal{} 8$ such combinations.
Considering the different permuations, there are indeed $ 8\cdot 3! \equal{} \boxed{48}$ such three-digit numbers[/hide]

Oh yes I see. :oops:  Thank you Joel for correcting me. :)

(b)
4 ways to get a 3-digit number divisible by 3:
1,2,3
2,3,4
3,4,5
4,5,6

3! ways to arrange each of these

So,

4*3! = 24 possible numbers

[quote="JBL"]Nayel, your "otherwise" solution is close but incorrect: it should be $ \binom{4}{2} \cdot 8$, since two digits are already taken (for the last two digits).[/quote]

Umm, for the "otherwise" solution it should be 4*3*8...

it can be ab16 or ba16 (just one example)

[quote="dvek8"](b)
4 ways to get a 3-digit number divisible by 3:
1,2,3
2,3,4
3,4,5
4,5,6
[/quote]

You're missing 4 ways.

$ 1,3,5$ for example.
[color=darkblue][b]timwu[/b][/color] get it right.


geometry
area of a triangle
geometry proposed

[quote="greatestmaths"]we have the three median of a triangle are 9, 12 and 15.
find his area. :P[/quote]
[hide="Solution"]
There is a theorem that says that the area of a triangle is $ \frac{4}{3}$ times the area of the triangle formed by the lengths of the medians.  There are many proofs for this theorem, here is a link of one of the proofs:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172424

With this in mind, we can find the area of the triangle formed by the medians to be $ 54$.  Thus, the area of the triangle is $ \frac{4}{3}(54)\equal{}\boxed{72}$.
[/hide]


So I went to insert --> picture and selected a picture that I made for a question.

I recieved an error message: ! Undefined control sequence.
1.65 \includegraphics
{<file>}

Note: this is in TeXniccenter.

Why isn't this working? It has worked in the past!

You need to have \usepackage{graphicx} in your preamble.

Also, you should post LaTeX questions under the LaTeX help forums.


A seqence satisfies $ a_{m \plus{} n} \plus{} a_{m \minus{} n} \equal{} \frac {1}{2}(a_{2m} \plus{} a_{2n})$ for all non negative $ m$ and $ n$ with $ m \ge n$.  Given $ a_1 \equal{} 1$ find $ a_{1995}$.

[hide="My question about the problem"]
This is a problem given in ACoPS.
To solve this, you could plug in values until you notice a pattern (that $ a_k \equal{} k^2$ in this case) then prove the pattern. It says that this should be done with mathematical induction, but is that necessary? Could I just show that $ (m \plus{} n)^2 \plus{} (m \minus{} n)^2 \equal{} \frac {1}{2}(4m^2 \plus{} 4n^2)$ and that $ 1^2 \equal{} 1$? The author seemed to ignore this easier method but I don't see anything wrong with it. And regardless of the method I used, would I need to show that $ a_k \equal{} k^2$ is the [i]only[/i] sequence that satisfies the condition?[/hide]

[quote="qntty"][hide="My question about the problem"]The author seemed to ignore this easier method but I don't see anything wrong with it. And regardless of the method I used, would I need to show that $ a_k \equal{} k^2$ is the [i]only[/i] sequence that satisfies the condition?[/hide][/quote]
You've answered your own question. You've shown that sequence is valid, but that doesn't find without a doubt what $ a_{1995}$ is because you don't know if other sequences are possible.

[hide="Proof of uniqueness"]Since it's said "non-negative $ m,n$", zero is a legitimate index.

Putting $ m = n = 0$ we get $ 2a_0 = a_0\iff a_0 = 0$

Putting $ n = 0$ we get $ 2a_m = {1\over 2}a_{2m}\iff a_{2m} = 4a_m$

Plugging that into the original equation we get

$ a_{m + n} + a_{m - n} = 2(a_m + a_n)$

Putting $ n = 1$ into the last equation we get

$ a_{m + 1} + a_{m - 1} = 2a_m + 2\iff a_{m + 1} - a_m = a_m - a_{m - 1} + 2$

Hence

\begin{eqnarray*} a_2 - a_1 & = & a_1 - a_0 + 2 \\
a_3 - a_2 & = & a_2 - a_1 + 2 \\
& \vdots \\
a_n - a_{n - 1} & = & a_{n - 1} - a_{n - 2} + 2\end{eqnarray*}

Summing that up and telescoping, we get

$ a_n - a_1 = a_{n - 1} - a_0 + 2(n - 1)\iff a_n - a_{n - 1} = 2n - 1$

Once more:

\begin{eqnarray*}a_1 - a_0 & = & 1 \\
a_2 - a_1 & = & 3 \\
& \vdots & \\
a_n - a_{n - 1} & = & 2n - 1\end{eqnarray*}

Summing up and telescoping, we get

$ a_n - a_0 = {1 + (2n - 1)\over 2}\cdot n\iff a_n = n^2$

We check that this solution works with the original equation, hence the solution is unique, and $ a_{1995} = 1995^2$[/hide]


$ 3\frac59$ gallons of water is equally divided among 16 hikers. To the nearest tenth of a gallon, how many gallons will each hiker receive?


I took a flight on a plane with 137 seats. I was the first person on the plane, and found that I lost my boarding pass, so I didn't know which seat was mine. I randomly chose a seat and sat down. As passengers boarded behind me, they took their assigned seats, unless they found it already occupied, in which case they took any available seat at random. When the last passenger boarded, what is the probability that she found her seat unoccupied?

for those of you that are preparing to start complaining that this is already posted somewhere else....i know...but no solutionw as provided and this forum is more active.

... but of course, you know the solution to this problem. POST IT!! :!:

Ans :
$ \frac {1} {2}$

its in the intro c and p book

Yes, but...

$ \text{Answer} \neq \text{Solution}$

We're still looking for a good solution... leoxnlin, I liked your "aboogas" explanation. Can you repeat it again? :D

the solution is too complex to put here but here you go(and i will have to look up sigma notation in latex):
$ \displaystyle\sum_{i\equal{}0}^n \frac {i} {137 \minus{} i}$

sorry to double post but,
i had to get my brother's calc  :P

That's OK... especially since that's not a double post!! (I intevened in between) :D

Umm...in the MOEMS forum, I have this Tourney going on, and that's one of the Round 3 questions, but not everyone has submitted their answers yet! So, um, now they know the answer...*sigh*

Next time, (^_^), please wait until AFTER everyone submits their answers. Thanks.

:jump:  :jump:  :jump: YAY! aboogas!!!

How about if the first two people have lost their boarding passes?

oh...um...sorry... :oops:

[quote="Yongyi781"]Umm... I don't think so... for one, that sum is greater than $ 1$. :!:

Here is leoxnlin's "abooga" solution.

We will term someone who has to pick a random seat an "abooga". Yongyi is therefore the first abooga. There are three places the abooga might sit: in Yongyi's (designated) seat, in the last passenger's seat, or in another random person's seat. If it is the latter, then that random person will later come in and become the next abooga (and face the same 3 choices above). If the abooga sits in Yongyi's real seat, then there will be no more aboogas, and the last person will sit in the correct seat. If the abooga sits in the last person's seat, then there will also be no more aboogas, and obviously the last person will end up sitting in the wrong seat.

Name the last abooga (there must exist one) the "ultimate abooga". Each abooga will either be the ultimate abooga (by sitting either in Yongyi's seat or in the last person's seat) or pass the task of being an abooga on to another passenger. Sitting in Yongyi's seat or in the last person's seat both have equal probabilities, therefore the answer to the question is $ \boxed{\frac {1}{2}}$.

Notice the argument holds even if Yongyi sits in the correct seat.[/quote]

um...hmmm....is this suppose to be a joke or is this a serious response?
if it is serious....i dont know whether the solution even makes sense

hwo did u do it yongyi?

:mad: The solution was written to be more rigorous than the solutions written before it!!  :mad: (In the context that my abooga solution was written in, Yongyi was the person who lost his boarding pass.) If instead of "abooga" it had said something to the effect of "victim of seat preoccupation", you would have actually read it!!! This is an example of prejudice against made up words!!!

 :D

Extension: Find a formula if the first $ n$ people lost their boarding passes

I think it's 1/(n+1) because all of the people who lose their boarding passes will start a series of "aboogas", and each of the series will have an "ultimate abooga", who must be sitting in the last person's seat or in one of the seats of the people who lost their boarding passes. There are a total of n+1 of these seats, and these are all equally likely to be the sole unoccupied seat. Therefore the probability is 1/(n+1) that the last person will be in the correct seat.


I still haven't got the hang of this. When I try to use "Split selected posts" it works fine but when I use "Split from selected posts" no matter how many I select only the first post remains and the rest are split anyways. Furthermore the original thread (with only one post) dissapeared after I split the new thread created for the second time. Any suggestions will be greatly appreciated.

(For reference the original thread was I think titles "Name it!: Carols with Dead People," the second thread created was called "911" which was later renamed "Carols with Dead People," and the third one was named "911 2" until the "911" thread was renamed, after that it was renamed to just "911.")

Um, shouldn't you only be clicking one to do split from post...  (I think split from means basically 'this is the split line - from here on goes')

Oooooh...no wonder:oops:

Will that line include itself? (Will the checked post be included?)

I believe the checked line and all that follows gets moved.


In another thread I posted a  problem with the solution that even with less than  0.00001%  popular vote, it is mathematically possible for a candidate to win the electoral college and thus become president.

Now few  othter problems: , show how it is possible to

Problem 0 :  (From above)  How it is possible for  one candidate to win electoral vote even though he wins  less than 0.00001% of popular votes polled. (Solution in the other thread)

 Problem  1:  Kerry  wins popular vote. Kerry also wins  electoral votes, (that is Kerry  wins more electoral votes, as welll as popular votes  than Bush) . Yet Bush becomes President.

Problem 2 Chenny  becomes acting president . (And Bush, Kerry and Edwards are all alive and willing)

Problem 3:   Bush - Edwards Administartion is sworn in.

Problme 4 : Edwards is sworn in as both vice president and actiing president on Jan 20, 2005.

Please post, any other such possiblities and how  that  could  happen (in spoilers)

Hint for above in spoiler:  S[hide]ee New York times of today for  problem 3 and 4.

It can also give hint for problem 1 and 2)[/hide]

Please post all solutions in spoilers... You may have heard of these questions as these are well known ...

(Anything which is not impossible, however, improbable could be considered here)

Id rather just get rid of the electoral college- thx to this and many other problems with it

[hide] 1: An independent grabs votes from Kerry. A republican controlled House votes bush into office.

2: ?

3: ?

4: Kerry gets murdered, suicidies, etc. on Inaugration Day.[/hide]

Oooh I know number 3!

Spoiler: [hide]There's an electoral tie. House of Representatives decides president, Senate decides vice president. House of Representatives is Republican-controlled, Senate is Democratic-controlled. Bush-Edwards.[/hide]

2. [hide]Bush wins in electoral college, Cheney and the cabinet declare him disabled. Cheney becomes president.[/hide]

cardwacko9999 wrote:[hide]4: Kerry gets murdered, suicidies, etc. on Inaugration Day.[/hide]

It says 2005...

frankthetank wrote:1.  [hide]Kerry has more electoral votes than Bush, but still less than 270 (Is that the number you need to win?).  Nader has some of those votes.  House of Representatives votes for Bush.[/hide]

Ahem, [hide]actually, I think you just need more electoral votes than the others to win, since we have a plurality system.  Or am I mistaken?[/hide]

It's possible for both candidates to receive 269.

ANd I actually drew a map with both candidates receiving 269 electoral votes.

MN, WI, MI, PA, FL, NH, HI, NJ go Democrat

WI, OH, IA, CO, NM,  NV, WV go Republican.

Only counting swing states.

[quote]Problem 3:  Bush - Edwards Administartion is sworn in. [/quote]

Bush is elected president. Cheney dies. Bush, for his own insane reasons, chooses Edwards to succeed as vice president (just as Nixon chose Ford when there was a vacancy though Ford was not president pro tempore), Congress approves.

Or is a new about-to-be VP chosen by some other method...?

Okay:  Soultion I had in mind for Prbblem 1:
Real life is stranger  than  these scenerios'

Till recently one of the  Ohio's  elector (Sherrod Brown)    is also a  congressman (and  thus inelligible to  be an elector according to the constitution).  Now,( NO DOUBT he reads this forum :) so  he became better informed) , recenly he  resigned from elector-college, but suppose he did not resign and with his vote: Kerry bush  tally is  270-268) .  Now it is almost certian that  Brown's vote would be disqulifed leaving the tally (269-268) in Kerry's favor. But rule says he has to win 270 votes  so  congress decides.  Congress is most likely to vote for Repupblican Bush, and here  you have it!! (not only in theory but one of those "could be")   Inspite of  winning both popular and electoral votes one candidate  still can't become president.

Now Brown did te  right thing, (he  read all this here)  and   resigned :)  so this is less likely to  happen :)

(The rule says one has to get  270 elecoral votes,  and not the majority of the votes - that is the crucial  fine print  here)

For Problem  0:  As mentioned before,  The voter tunout is the key, low voter turnout in the states where one candidate wins and take the whole state can make the winning popular count very low...

[quote]Or is a new about-to-be VP chosen by some other method...?[/quote]
In case of  no one winning 270+ electoral votes,   According to constitutions: Congress elects  President, and Senate elects VP.  So that can happen.

(Actaully Adam's and  his VP (Thomas Jefferson), if I am not mistaken, were of different party - I don't know  exactly how it happened though)

For John Adams and Thomas Jefferson, the candidate with the most votes is president and the one with the second most amount is VP.  The reason that Thomas Jefferson got the second most and not Adams' VP candidate is probably Alexander Hamilton.  I wont get into the details...you can search for the Election of 1796 if you like.

Oh, and we can't forget te case of 1876. So many votes were in dispute - tat a compromise was made; all of te disputed votes would go to Hayes and in return, Hayes would withdraw troops frm the South, thereby ending reconstruction.

Election of 1804; Aaron Burr and Thomas Jefferson tie. Congress chooses Jefferson. Then we all know what Burr did. ;)

Election of 1826; Congress chooses Jon Quincy Adams. Adams is defeated for re-election 4 years later (we can't trust Congress). ;) of course, I r teh making logical fallacy

[quote="Simfish"]

Election of 1826; Congress chooses Jon Quincy Adams. Adams is defeated for re-election 4 years later (we can't trust Congress). ;) of course, I r teh making logical fallacy[/quote]

I thought that was 1824,

Everyone SAYS they'll make peace for the world, but few intend to do it; have you not seen [i]Miss Congeniality[/i]?  It's a canned comment, and a very cliched one at that.  Kerry simultaneously claims to get our allies to participate more, and that he wants to train more US troops to go to Iraq.  Sounds to me like he's making assumptions about what the rest of the world will do.

Real realistic gameworld, real realistic.

As Robinhood says, Kerry may not have done better than Bush. It's not what you intend to do. It's what you accomplish. And Bush won the election pretty fairly.

analysis shows that voter turnout in New York and California for the dems was lower than usual because they felt their vote didn't count (source: NPR).

The election (popular vote-wise) would've been much much closer.

No dispute in the electoral college this time though.

doesnt the 26th ammendment allow 18 year olds to vote????.....i think ur talking about teh 25th where a president can choose a vice president in case of a vacancy and declare himself disabled so that the vice president will take office.

[quote="Treething"]I just had a really scary thought.

What if Kerry became president, and then he activated the 26th amendment? *shivers at the thought of Edwards being president*[/quote]

Still better than Bush or Cheney.

[quote="gameworld7"]FAIRLY? FAIRLY?????? WHAT DO YOU MEAN FAIRLY????????? THERE SHOULD'VE BEEN A RECOUNT! KERRY HAD TO WIN!!!! HE JUST HAD TO![/quote]

Yeah, Kerry just [i]had[/i] to win. I guess this explains his [i]loss[/i].

Anyways, how do we know that Kerry wouldn't be worse than Bush? Some of the stuff you guys are criticizing Bush about is kind of silly. Anyways, hindsight is 20-20.

[quote="gameworld7"]what is the 26th admenment?[/quote]

[quote="some quote that NEVER existed and was NEVER posted by athena"]doesnt the 26th ammendment allow 18 year olds to vote????....[/quote]

DUH........*stares*....*drooling*

no i don't believe you. that's why i'm asking other people.

u learn this in history. before u take the constitution test.

i didn't take it yet. my class is doing different things.

[quote="yif man12"]
Anyways, hindsight is 20-20.[/quote]

Better, I say; 20/5, at least.

well, i guess this topic doesn't really matter anymore...


Hi all,
New here but in desperate need of help due to a rather unpleasant situation which has developed recently.
I discovered a note under my car windscreen wiper a couple of days ago which was in ciphertext, with each paragraph seemingly coded with a different crypt. The first three paragraphs were easy to solve (Caesar Cypher) and revealed detailed personal information about me, including my name, address, car details, medical history etc.
There are currently several people working on this inn my group of friends, and as yet three paragraphs remain uncracked.
I'd be extremely grateful if one of you people could help me out on this!

The text, as on the letter, is shown below along with the decoded sections (with the exception of the paragraph including my personal information):

[b]DBS: Tjmwfs fbsmz :l't Qspupo tbmppo / ibudicbdl (qjduvsf) Sfh. Op: L928 HZC[/b]

CAR: Silver early 90's Proton saloon / hatchback (picture) Reg. No: K817 GYB

[b]QIPOF: Npcjmf: 189:75:2933.[/b]

PHONE: Mobile: 07896491822

[b].fsvudbsg ejpiqbdt ,sfutbmq oj tlffx 72 -9 ,7113 gp sfnnvt oj utjsx uihjs flpsC :ZSPUTJI MBDJEFN[/b]

MEDICAL HISTORY: Broke their wrist in summer of 2006, 8- 16 weeks in plaster, scaphiod fracture.

[b]nsb sqj uobfzmpobt /x obtfzmpqfe zmcj zmcj tuobje ob ;sb u sftp fn ;sbtpQ tpQ uoje t ;svpD gmfsbtpQ

Qj-ef jouf uvsf. g jouvs, bfqmbt bobtbnfz fqmpvst :ZSPUTJI t.t.s, pjo. Dpg t DJEFN qipg tbnftbtft t jefqs MBDJI j-efttbt tujob g pobobnfs pg sqsftuo t zmpgs g zfbo uouo pvsqipvs bdbqipg n fzftbqipj uj-eftdut dutftbtbqs, nfbqsbs. Dpjouo ujobfmaj.

S fas, blndd oraim mero-unooler polincke narsvinge stlopas vierod deneral. Porcesv und sced as palec bly bagerine ragelori ncacen cradod rencutest llntecriz accciS ongsccakes, swalloc acars wailllef alengs suella cllef adgese wodol batem pe R quleniza-seseprge sod ci[/b]

On the back of the note, there is one line which has been typed by a normal typewriter by the looks of it:

[b]qptt. ejbhoptfe x/ x3 cjqpmbs; nbslt po bsn joejdbuf TI[/b]

poss. diagnosed w/ w2 bipolar; marks on arm indicate SH

This whole thing has been reported to the police, but since decrypting this would be a bit of a waste of money, and since you guys seem to like a challenge, I wondered if you could help out.

Thanks!

The algorithm of encrypting in the first messages is very simple, so I suppose that so is the algorithm in the other messages.

Assume that each letter represents some specific letter. That is:

We should replace [u]all[/u] 'T' by 'F', [u]all[/u] 'F' by 'S' etc

If the assumption holds, then some words would be very useful.

For example, the word [b]dutftbtbqs[/b]
[quote]Qj-ef jouf uvsf. g jouvs, bfqmbt bobtbnfz fqmpvst :ZSPUTJI t.t.s, pjo. Dpg t DJEFN qipg tbnftbtft t jefqs MBDJI j-efttbt tujob g pobobnfs pg sqsftuo t zmpgs g zfbo uouo pvsqipvs bdbqipg n fzftbqipj uj-eftdut [color=red]dutftbtbqs[/color], nfbqsbs. Dpjouo ujobfmaj. [/quote]
Notice the pattern D - U - T - F - T - B - T - B - Q - S

Or the inverse pattern: S - Q - B - T - B - T - F - T - U - D

Notice the appearence of [b]T [/b]and [b]B[/b]

[b][color=blue]Can somebody find an english word at this form?[/color][/b] (I don't know good English, so I can't find some appropriate word)

If we check the frequencies of the letters in this text,
t sould represent the letter E
b sould represent the letter T

But what is this world ?
[size=150][b]??E?ETET??[/b][/size]

g probably equals a or i as it is a 1 letter word that appears more than once Also, since it appears at the end of some words, g=a

but there are 2 other words with one letter, i'm stumped
either this is an extremely complicated code, or i'm missing something

Did anything else come with the letter?

btw did this really happen to u or did u make this up?


[quote="myc"]Whats the difference between the independent study courses and the other courses?  Do you read the book and ask the teacher if you have questions?   I'm thinking of taking either intro. to number theory or into.  to counting and probability.  How much time per day do you usually spend on these courses?  What do you recommend taking, and how did the class go if you took it?

Thanks
 :P[/quote]

Greetings,

The Intro to Counting & Probability class won't be offered again for a while.  We will probably offer Intro Number Theory this summer, so this is the course we would recommend.  Expect to spend 3-5 hours/week on the class to get the most out of it.


$ x,y,z > 0,$

(1)
$ {\frac {\left( z \plus{} x \right) \left( z \minus{} x \right) ^{2}}{x \plus{} y \plus{} z}} \plus{} {\frac {\left( x \plus{} y \right) \left( x \minus{} y \right) ^{2}}{x \plus{} y \plus{} z}} \geq \frac {1}{4}\, \left( 2\,x \minus{} y \minus{} z \right) ^{2}$

(2)

$ \frac {z \plus{} x}{x \plus{} y \plus{} z}(z \minus{} x)^{2t} \plus{} \frac {x \plus{} y}{x \plus{} y \plus{} z}(x \minus{} y)^{2t}\geq \frac {1}{4^t}(2x \minus{} y \minus{} z)^{2t} ,t \equal{} 1,2,3,....$.
BQ

[quote="xzlbq"]$ x,y,z > 0,$

(1)
$ {\frac {\left( z \plus{} x \right) \left( z \minus{} x \right) ^{2}}{x \plus{} y \plus{} z}} \plus{} {\frac {\left( x \plus{} y \right) \left( x \minus{} y \right) ^{2}}{x \plus{} y \plus{} z}} \geq \frac {1}{4}\, \left( 2\,x \minus{} y \minus{} z \right) ^{2}$

BQ[/quote]

The inequality is equivalent to

$ 4x^3\plus{}(\minus{}4z\minus{}4y)x^2\plus{}(6yz\minus{}z^2\minus{}y^2)x\plus{}3(y\plus{}z)(y\minus{}z)^2\ge 0$

Note $ 3(y\plus{}z)(y\minus{}z)^2\geq 0$

$ \Delta \equal{}\minus{}16(139y^4\plus{}48y^3z\minus{}310y^2z^2\plus{}48yz^3\plus{}139z^4)(y\minus{}z)^2\le 0$

so...............

 :lol:


Find $ a$ and $ b$ such that $ \lim_{x \to 0} \frac{\sqrt{ax\plus{}b}\minus{}2}{x} \equal{} 1$

How to do this without using L'hopitals rule?

Thanks

First of all, consider $ \lim_{x\rightarrow 0} \frac {\sqrt {ax \plus{} b} \minus{} 2}{x}\cdot x$.

Hmm how can you just multiply by x without assuming x = 1?

how do you do the Q? I am trying to manipulate $ \sqrt{ax\plus{}b}\minus{}2$ so that it cancels the x somehow but I cant do it, I've tried multiplying by the conjugate, doesn't do much either =S

[quote="Kono"]Hmm how can you just multiply by x without assuming x = 1?[/quote]

Sorry, I can't understand what you say, why do we need to assume $ x\equal{}1$?

The problem asks us very basic fact. That's why I don't like using l'Hopital's theorem at once.

[quote="Kono"]Find $ a$ and $ b$ such that $ \lim_{x \to 0} \frac {\sqrt {ax \plus{} b} \minus{} 2}{x} \equal{} 1$

How to do this without using L'hopitals rule?

Thanks[/quote]
If $ \lim_{x\to 0}(\sqrt {ax \plus{} b} \minus{} 2) \not \equal{} 0$ then our limit is of the form $ \frac {L}{0} \equal{} \plus{} \infty$ or $ \minus{} \infty$ which is not true.

So $ \lim_{x\to 0}(\sqrt {ax \plus{} b} \minus{} 2) \equal{} 0 \Longrightarrow \sqrt {b} \minus{} 2 \equal{} 0 \Longrightarrow b \equal{} 4.$

So our limit is

$ \lim_{x \to 0} \frac {\sqrt {ax \plus{} 4} \minus{} 2}{x} \equal{} \lim_{x\to 0} \frac {ax}{x(\sqrt {ax \plus{} 4} \plus{} 2)} \equal{} \lim_{x\to 0} \frac {a}{(\sqrt {ax \plus{} 4} \plus{} 2} \equal{} \frac {a}{4}.$

So $ \frac {a}{4} \equal{} 1 \Longrightarrow a \equal{} 4.$

So $ (a,b) \equal{} (4,4)$

Sorry, isn't that conclusion just necessary condition?


When I first signed up in AoPS, I didn't know how to use $ \ LaTeX$
I learned it the hard way how to use them..
And I wanted to save you guys little time by posting this.(Hope it helps!)

LaTex is commonly used to express equations or important value.
Beware not to post everything in LaTex, because it will slow the speed down tremendously(not really)for those who have low-speed connections.
Here's how to use them.
-----------------------------------------------------------------------------
How to write in LaTex:
[code]$\ {number or equation goes here}$[/code]

Special expressions:
Square Root:
[code]$\sqrt {radicand}$[/code]

Square, Cubes and more:
[code]$\ {base}^{exponent}$[/code]
Fractions:
[code]$\frac {numerator}{denominator}$[/code]

Boxed, Plus or Minus, and Right Arrow:
Boxed:
[code]\boxed {number or equation}$[/code]
Plus or Minus:
[code]$\pm {number or equation}$[/code]
Right Arrow:
[code]$\rightarrow {number or equation}$[/code]

Those are the basic stuffs..
You may entertain yourself by finding more than these.. :lol:

thanks thanks thanks thanks thanks thanks thanks thanks thanks

Haha, I can learn latex and stay lazy. Fun.

For those who want to do more, check out these sites:

[url]http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands[/url]

[url]http://www.ntg.nl/doc/biemesderfer/ltxcrib.pdf[/url]

You may want to print up DonkeyKong's post or one of these sites for reference.

Now this is only for AoPS. If you want to type latex documents using TeXnic, etc., go [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]here[/url].

Oh! I probably forgot to mention this.

[size=150][b]PLEASE DO[color=red] NOT [/color]USE LaTex ON ANY WORDS EXCEPT FOR THE VALUES(ex. x, y, z).[/b][/size]

nice
where did u learn thses stuff?
i need to learn to use them....

He learned the basics from [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url]

[quote="Temperal"]Your ideas are slightly flawed.
For example, you seem to think that putting a backslash (\) at the beginning of every LaTeX blog is required. Rather, each command, whether to modify an expression or to insert a special symbol, is preceded by it. i.e.
[code]$1+\frac{\text{Numerator}}{\text{Deominator}}$[/code]
produces
$ 1 + \frac {\text{Numerator}}{\text{Deominator}}$
while
[code]$\1+frac{text{Numerator}}{text{Deominator}}$[/code]
produces
$ \1 + frac{text{Numerator}}{text{Deominator}}$.
Also, the notation for roots in general is of the following format:
[code]$\sqrt[n]{x}$[/code]
This produces: $ \sqrt [n]{x}$.
As before, the backslash here:
[code]$\ {base}^{exponent}$[/code]
is unnecessary.
I recommend you review the symbol and command lists, and also learn the syntax at the main Wiki page.

On a side note: I wonder if asy works within LaTeX on the forum:

$ \text{\LaTeX} \begin{asy} size(100); draw((0,0) - - (100,100) - - (0,0)); \end{asy} \sqrt [n]{x}$

EDIT: Nope. Darn.

And if anyone wants to win the competition for finding more commands and all that: [[LaTeX:Symbols]] [[LaTeX:Commands]]

Hundreds right there.
[/quote]
Thanks for the additional infos. :lol:


1.........12.......25 prove that it is a whole square . There are 1997 : 1 and 1998 : 2

Junior Balkan MO - 1997 or 1998, don't know which year.

Hint
[hide]Take $a_n = 11\dots11$ ($n$ digits) and express 11...122...25 in terms of $a_n$, you should get $(3a_n + 2)^2$.[/hide]


[b][size=100][color=DarkBlue]ΠΡΟΤΑΣΗ. - Δίδονται δύο κύκλοι $ (O_{1}),$ $ (O_{2}),$ τεμνόμενοι στα σημεία $ A$ και $ B.$ Με κέντρο το σημείο $ A,$ γράφουμε τυχόντα κύκλο $ (A)$ ο οποίος τέμνει τον $ (O_{1})$ τα σημεία $ C,$ $ C'$ και τον $ (O_{2})$ στα σημεία $ D,$ $ D'.$ Οι ευθείες $ CC',$ $ DD',$ τέμνουν αντιστοίχως τους $ (O_{2}),$ $ (O_{1}),$ στα ζεύγη των σημείων $ E,$ $ E'$ και $ F,$ $ F'$ $ ($ τα $ E,$ $ F,$ κοντά στο $ A$ $ ).$ Αποδείξτε ότι οι ευθείες $ EF',$ $ E'F,$ διέρχονται δια του σημείου $ A.$[/color][/size][/b]

Κώστας Βήττας.

ΥΓ. Τη βρήκα σε ένα από τα παλιά τεύχη της ΕΜΕ, που αγόρασα πρόσφατα από παλαιοπωλείο.

[b][size=100][color=DarkBlue]PROBLEM. - Two circles $ (O_{1}),$ $ (O_{2})$ are given, intersect each other at points $ A$ and $ B.$ We draw an arbitrary circle $ (A)$ centered at point $ A,$ which intersects $ (O_{1})$ at points $ C,$ $ C'$ and $ (O_{2})$ at points $ D,$ $ D'.$ The line segments $ CC',$ $ DD',$ intersect ${ (O_{[}2}),$ $ (O_{1}),$ at pairs of points $ E,$ $ E'$ and $ F,$ $ F'$ respectively $ ($ the points $ E,$ $ F,$ near to $ A$ $ ).$ Prove that the line segments $ EF',$ $ E'F,$ pass through the point $ A.$[/color][/size][/b]

Kostas Vittas.

[b]ΣΧΗΜΑ(Του Κώστα)[/b]

Sorry, but on my sketch, the point $ EF'\cap E'F$ doesn't even lie on the line $ AB$... (I am talking about the English version of the problem; there may be some translation errors.)

  Darij

[quote="jonfv"]Let $ T$ be the inversion with respect to the circunference $ (A)$.
$ T$ sends the circunferences $ (O_{1})$, $ (O_{2})$ to the lines $ CC'$, $ DD'$, and vice-versa.
Then, $ T$ sends the intersections points $ E,F'$ of the circunference $ (O_{2})$ and the line $ CC'$
to the intersection points $ E',F$ of the line $ DD'$ and the circunference $ (O_{1})$.
In particular, the lines $ EF'$ and $ E'F$ pass through the point $ A$.[/quote]

[b] Very nice and simple solution !!! [/b]

Let me explain it a little simpler to the greek young stundets , who are not taught transformations:

[b]a)[/b] $ F$ belongs to both $ DF'$ and $ (O_{1})$
[b]b) [/b]The inverstion takes $ DF'$ to $ (O_{2})$ and $ (O_{1})$ to $ CC'$.
 Hence  the image of $ F$ is the intersection of $ (O_{2})$ and $ CC'$ , which is $ E'$.
 But the line defined by two inverse points , passes through the pole of the inversion. Thus lines $ EF', FE'$ pass both from the pole $ A$ of the above inversion.

[b] babis[/b]

Sorry, seems that I have let my program confuse the two points of intersection of two circles.

  darij

Η λύση που μας έστειλε ο φίλος μας [b][size=100]jonfv[/size][/b], από το μακρινό Ρίο, είναι αυτή που υπάρχει και  στο τεύχος Μαρτίου 1966, σελίδα 156, του Παραρτήματος της Ε.Μ.Ε. ( το περιοδικό ''Ευκλείδης'' της εποχής ), με υπογραφή [b][size=100]Αχιλλέας Μπουλαμάτσης[/size][/b] ( Αθήνα Β.Π.Σ. )

[b][size=100][color=DarkBlue]ΛΗΜΜΑ. – Με κέντρο τυχόν σημείο $ P$ δοσμένου κύκλου $ (O),$ γράφουμε τυχόντα κύκλο $ (P)$ και έστω $ AB,$ η κοινή χορδή των δύο κύκλων. Αποδείξτε ότι η περιφέρεια $ (O),$ είναι αντίστροφη της ευθείας $ AB,$ ως προς πόλο αντιστροφής το σημείο $ P$ και δύναμη αντιστροφής $ k^{2}\equal{} (PA)^{2}\equal{} (PB)^{2}.$[/color][/size][/b]

[b][size=100]ΑΠΟΔΕΙΞΗ.[/size][/b]

Όπως φαίνεται και στο σχήμα, αποδεικνύεται εύκολα ότι το αντίστροφο κάθε σημείου της ευθείας $ AB,$ κείται επί της περιφέρειας $ (O).$

Aπό το ορθογώνιο τρίγωνο $ \bigtriangleup APQ$ $ \Longrightarrow$ $ (PA)^{2}\equal{} (PQ)\cdot (PQ'$ $ ,(1)$

Από εγράψιμμα τετράπλευρα $ QQ'R'R$ και $ QQ'SS'$ $ \Longrightarrow$ $ (PQ)\cdot (PQ') \equal{} (PR)\cdot (PR')$ $ ,(2)$

και $ (PQ)\cdot (PQ') \equal{} (PS)\cdot (PS')$ $ .(3)$

$ \bullet$ Στο πρόβλημά μας τώρα, σύμφωνα με το παραπάνω Λήμμα, έχουμε ότι:

Το σημείο $ F'$ κείται επί της ευθείας $ DD'$ και άρα το αντίστροφό του, ως προς την αντιστροφή $ T[A, k^{2}],$
κείται επί της περιφέρειας $ (O_{2}).$

Το σημείο $ F'$ κείται επίσης επί της περιφέρειας $ (O_{1})$ και άρα το αντίστροφό του κείται επί της ευθείας $ CC'.$

Eπομένως, προκύπτει ότι το αντίστροφο του $ F'$ ως προς την ως άνω αντιστροφή, ταυτίζεται με το σημείο $ E\equiv CC'\cap (O_{2})$
και άρα η ευθεία $ EF',$ που συνδέει τα δύο αντίστροφα σημεία, διέρχεται από το σημείο $ A,$ πόλο της αντιστροφής.

Με όμοιο τρόπο, αποδεικνύεται ότι και η ευθεία $ E'F,$ διέρχεται δια του $ A$ και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.

[b][size=100]AN ABSTRACTIVE TRANSLATION[/size][/b]

Thank you dear friends [b][size=100]Darij[/size][/b] and [b][size=100]jonfv[/size][/b], for your interest.

This problem has been found in an old issue ( March of 1966 ) of the mathematical magagine of [b][size=100]HMS (Hellenic Mathematical Society)[/size][/b] and I will try to translate the solution with inversion, by [b][size=100]Achilles Boulamatcis[/size][/b] $ ($ the same as [b][size=100]jonfv’s one.[/size][/b] $ ).$

$ \bullet$ It is easy to show that the line segments $ CC',$ $ DD',$ are the images of $ (O_{1}),$ $ (O_{2})$ respectively, with respect to the inversion $ T[A, k^{2}],$ where $ k^{2}\equal{} (AC)^{2}\equal{} (AD)^{2}.$

So, as the point $ F'$ lies on $ DD',$ then its image lies on $ (O_{2}).$

Also, as the point $ F'$ lies on $ (O_{1}),$ then its image lies on $ CC'.$

Hence, the image of $ F',$ wrt $ T[A, k^{2}],$ is the point $ E\equiv (O_{2})\cap CC'.$

We conclude now, that the line segment $ EF',$ passes through the point $ A,$ as the pole of the above inversion.

Similarly, the line segment $ E'F,$ passes through the point $ A$ and the proof is completed.

Best regards, Kostas Vittas.


function
algebra
domain
calculus
calculus computations

let $f(x)=x^{x^{x^{.^{.^.}}}}$. and let the range and domain have been defined so that $f^{-1} (x)$ exist. Are $f^{-1}$ and $f$ analytic functions ?

The range should be (0,e) and then $f^{-1}(x)=\sqrt[x]x$.
For more details see
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=325561#p325561


In my neighborhood, there are six streets. There are 10 houses on each side of each street. How many houses are in my neighborhood?

[hide="Solution"]
There are 2 sides to each street with 10 houses on each side, so there are $2\times 10 = 20$ houses on each street.
There are 6 streets in the neighborhood each with 20 houses, so there are $6 \times 20 = \boxed{120}$ houses in the neighborhood.[/hide]

Wait, how is this counting? It looks to me like simple arithmetic.

edit: wowwww 1000 posts yikes

I feel like this problem should be rewritten because it is pretty ambiguous. You can get different answers by positioning the streets differently.

Yeah, that makes sense. I mean, I just did [hide=this,] $6\cdot10\cdot2$ [/hide] but I can see how it would be confusing if you took a different approach.

[quote=aryanpadarthi][quote=jbatterson]If the streets were arranged with three going N/S and three going E/W this would be an entertaining problem.  I think that (especially on AoPS) it would be reasonable to assume streets intersect.  With three going each way I think there would be just 84 houses.[/quote]

WOW!??  J BATTERSON[/quote]

:o

[quote=goldenuni678][hide=Aops Solution] There are 2 sides to each street with 10 houses on each side, so there are $2\times 10 = 20$ houses on each street. There are 6 streets in the neighborhood each with 20 houses, so there are $6 \times 20 = \boxed{120}$ houses in the neighborhood.[/hide][/quote]

Please do not post the AoPS solution, it does not contribute to the discussion.

[quote=Prod6Z][quote name="jbatterson" url="/community/p1349004"]
If the streets were arranged with three going N/S and three going E/W this would be an entertaining problem.  I think that (especially on AoPS) it would be reasonable to assume streets intersect.  With three going each way I think there would be just 84 houses.
[/quote]

:o  :wow: :omighty: :omighty: :omighty: THE CREATOR OF BA!!!

[hide=solution :)][tip=hover for solution][hide=hehe all that trouble to find that you cannot reach this :P]So since there are $6$ streets, and $10$ houses on each side of the streets, the answer should be $6(10\cdot2)$ which is $\boxed{120}.$ Pretty sure im correct, but if im not, it's ok :thumbsup:[/hide][/tip][/hide][/quote]

You are correct.


Find all ordered triples of $ (x,y,z)$ such that ${{ \frac{1}{x+\frac{1}{y+\frac{1}{z}}}}}=\frac{7}{10}$

The only integer solutions are [hide](1,2,3) and (2,-2,4). Here is the proof:
Simplification yields (yz+1)(10-7x)=7z. 10-7x can't be divis by 7, so yz+1=7n, 10-7x=z/n. From the first, z=(7n-1)/y. Subbing in gives 10-7x=(7n-1)/yn, so 1=7n+7xyn-10yn. Since n divides both sides, it must be either 1 or -1. If it equals 1, we have 6=y(10-7x). 10-7x can only equal 3 to be a factor of 6, so only y=2, x=1 works. If n=-1, we have 8=y(10-7x). 10-7x can only be -4 to be a factor of 8, so only y=-2, x=2 works. [/hide]


does below equallity have answer for $ p_i,q_i$s? except$ p_i\equal{}q_i\equal{}\frac 16$ $ (p_1x\plus{}p_2x^2\plus{}...\plus{}p_6x^6)(q_1x\plus{}q_2x^2\plus{}...\plus{}q_6x^6)\equal{}(\frac 16x\plus{}\frac 16x^2\plus{}...\plus{}\frac 16x^6)^2$

If $ p_i,p_i$ is real , R.H.S. has a factor $ (1 \plus{} x \plus{} x^2 \plus{} \dots \plus{} x^5)$ which cannot be express as product of two polynomials of lower degree containing only real coefficient except as $ (x\plus{}1)(x^2\minus{}x\plus{}1)(x^2\plus{}x\plus{}1)$.(as its roots are $ a\neq 1$ such that $ a^6\equal{}1$ we can check this)).So
[b]Case 1:[/b]$ (1 \plus{} x \plus{} x^2 \plus{} \dots \plus{} x^5)|(p_1 \plus{} p_2 x \plus{} \dots \plus{} p_6 x^5)$ , thus $ p_1 \equal{} p_2 \equal{} \dots \equal{} p_6 \equal{} k$ and $ q_1 \equal{} q_2 \equal{} \dots \equal{} q_6 \equal{} \frac {1}{36k}$ where k is real are the solutions.
[b]Case 2[/b]:$ (x\plus{}1)(x^2\minus{}x\plus{}1)^2\equal{}(1 \minus{} x \plus{} x^2 \plus{} x^3 \minus{} x^4 \plus{} x^5)|(p_1 \plus{} p_2 x \plus{} \dots \plus{} p_6 x^5)$, we get $ p_1\equal{}p_3\equal{}p_4\equal{}p_6\equal{}k, p_2\equal{}p_5\equal{}\minus{}k$  ,  $ q_1\equal{}q_6\equal{}\frac{k}{36}, q_2\equal{}q_5\equal{}\frac{k}{12}, q_3\equal{}q_5\equal{}\frac{5k}{36}$ are the solutions.
[b]Case 3[/b]:$ (x\plus{}1)(x^2\plus{}x\plus{}1)^2|(p_1 \plus{} p_2 x \plus{} \dots \plus{} p_6 x^5)$The same solution as Case 2 but $ p_i$ and $ q_i$ interchange.

If $ p_i,p_i$ is not real , we can express R.H.S. as $ \frac {1}{36}x^2 (x \minus{} 1^{1/6})^2 (x \minus{} 1^{2/6})^2\dots (1 \minus{} 1^{5/6})^2$ where $ 1^{1/6} \neq 1$, ... I think you should understand the rest.


blogs
geometry
rectangle
search
3D geometry
\/closed

[color=darkred][size=150][b]Signatures[/b][/size][/color]

[color=darkred][b]How do I set up a signature?[/b][/color]
1. Click the [b]Profile[/b] button near the top of the screen.
2. Under Profile, select [b]Signature Control Panel[/b]
3. Enter the signature you wish to use.
4. Click [b]Submit[/b].

[color=darkred][b]How do I attach my signature to my post?[/b][/color]
Above the submit button, check the [b]Attach signature (signatures can be changed in profile)[/b] option.

[color=darkred][b]How do I automatically attach my signature to all my posts?[/b][/color]
1. Click the [b]Profile[/b] button near the top of the screen.
2. Click [b]Preferences[/b]
3. Click [b]Posting a message[/b]
4. Select [b]Yes[/b] next to the [b]Always attach my signature[/b] option.
5. Click [b]Submit[/b].

[color=darkred][size=150][b]Posts[/b][/size][/color]

[color=darkred][b]What do the stars next to user names on posts mean?[/b][/color]
Stars are awarded for having a certain number of posts. Stars are awarded on the following basis:

<20 posts = 0 stas (New Member)
20 posts = 1/2 star (P versus NP)
50 posts = 1 star (Hodge Conjecture)
100 posts = 2 stars (Poincare Conjecture)
250 posts = 2 1/2 stars (Riemann Hypothesis)
500 posts = 3 stars (Yang-Mills Theory)
1000 posts = 4 stars (Navier-Stokes Equations)
2500 posts = 5 stars (Birch & Swinnerton Dyer)

If you know of some frequently asked questions [b][u]AND[/u][/b] you have a good response to that question, feel free to post here for incorporation into this FAQ.

[b][size=150][color=darkred]Posting a screen image[/color][/size][/b]

First, you will need to capture the screen image. The following is for Windows. I have provided information on using Microsoft Paint followed by IrfanView. Choose only one of the options. Someone else can post Mac instructions

[b]Using Microsoft Paint[/b]
1. Press the [b]Print Screen[/b] button on your keyboard. This button often also says SysReq or something similar on the keyboard. Doing this will place a copy of the screen in your clipboard. (Note: It is recommended to do Alt-Print Screen, which will only copy the currently active window instead of the entire desktop.)
2. Open Microsoft Paint by clicking the [b]Start[/b] button, clicking the [b]All Programs[/b] option, selecting [b]Accessories[/b], then selecting [b]Paint[/b]. You can also act cool and press Windows-R, then type [b]mspaint[/b].
3. Paste the Image by selecting [b]Paste[/b] from the [b]Edit[/b] menu
4. Go to the [b]File[/b] menu and select [b]Save As...[/b]
5. Set the [b]Save as type[/b] to JPEG or PNG
6. Give the image a filename, and click [b]Save[/b].

[b]Using IrfanView[/b]
The whole process is made much easier by using a program called IrfanView, available from http://www.irfanview.com.

1. Press the [b]Print Screen[/b] button on your keyboard. This button often also says SysReq or something similar on the keyboard. Doing this will place a copy of the screen in your clipboard. (Note: It is recommended to do Alt-Print Screen, which will only copy the currently active window instead of the entire desktop.)
2. Load IrfanView.
3. Select [b]Paste[/b] from the [b]Edit[/b] menu or press [b]Ctrl-V[/b]
4. Draw a rectangle around the portion of the image you want to save.
5. Crop the image by selecting [b]Crop selection[/b] from the [b]Edit[/b] menu or press [b]Ctrl-Y[/b]
6. Save the image as JPEG or PNG.

[b]Adding the image to a post[/b]
1. Create a new topic or reply to a current topic on artofproblemsolving.com
2. Near the top of the editing area, above the subject line, click [b]Add an Attachment[/b]
3. Browse for the file
4. Click [b]Add Attachment[/b]

[b][size=150][color=darkred]Posting a screen image (Mac)[/color][/size][/b]

First, you will need to capture the screen image. The following is for Mac.

[b]Capturing a Screenshot[/b]

1. Press [b]Command-Shift-3[/b] on your keyboard.  The image will be saved to the Desktop.

[b]Adding the image to a post[/b]
1. Create a new topic or reply to a current topic on artofproblemsolving.com
2. Near the top of the editing area, above the subject line, click [b]Add an Attachment[/b]
3. Browse for the file
4. Click [b]Add Attachment[/b]

[color=darkred][size=150]Visiting Someone's Website[/size][/color]

Click on the WWW button on the bottom right-hand corner of someone's post.
Or search for it on "Blogs".

[b][color=darkred][size=150]Visiting Blogs[/size][/color][/b]

  Click on the Blog button on the bottom right-hand corner of someone's post.
Or search for it on "Blogs".

I really think that this should be a sticky, with the title FAQ: READ HERE BEFORE POSTING. Constantly, there are questions repeated and repeated over and over again.

...except that it seems nowadays that a user is as likely to read a normal thread as a sticky...

It will be a sticky. I'm just wanting to get it a bit more filled out first. Feel free to add questions you've seen often and post answers.

[color=darkred][size=150][b]Board roles[/b][/size][/color]

[b][color=darkred]What are Administrators?[/color][/b]

Administrators are members assigned with the highest level of control over the entire board. These members can control all facets of board operation, including setting permissions, banning users, creating usergroups or moderators, etc., dependent upon the board founder and what permissions he or she has given the other administrators. They also have full moderator capabilities in all forums.

[b][color=darkred]What are Moderators?[/color][/b]

Moderators are individuals (or groups of individuals) who look after the forums from day to day. They have the authority to edit or delete posts and lock, unlock, move, delete and split topics in the forum they moderate. Generally, moderators are present to prevent users from going off-topic or posting abusive or offensive material.

[b][size=150][color=darkred]Friends and Foes[/color][/size][/b]

[b][color=darkred]What are my Friends and Foes lists?[/color][/b]

You can use these lists to organise other members of the board. Members added to your friends list will be listed within your Profile for quick access to see their online status and to send them private messages.

[color=darkred][b]How can I add / remove users to my Friends or Foes list?[/b][/color]

From your Profile, you can directly add users by entering their member name. You may also remove users from your list using the same page.

[color=darkred][size=150][b]How can I create the flags in my profile?[/b][/size][/color]

Go to [b]Profile[/b] -- [b]Personal[/b] and set the original and/or current.

[color=darkred][size=150][b]What does my rating mean?[/b][/size][/color]

Your rating is the average of the posts that have been rated. You can rate other people's posts by clicking on the cubes above the posts. A rating from 1-2 is bad, 3-4 is fair, and 5-6 is excelent. If your rating says "not rated yet" it means either you don't have enough posts to be rated, or you haven't been rated yet.

[color=darkred][size=150][b]Who can see rating?[/b][/size][/color]

Only you, the admins and the moderators.

[quote="Dojo"][color=darkred][size=150][b]Avatars[/b][/size][/color]

[color=darkred][b]How do I set up my Avatar?[/b][/color]
1. Click the [b]Profile[/b] button near the top of the screen.
2. Under Profile, select [b]Profile and then Edit Avatar[/b] [s][b]Avatar Control Panel[/b][/s].
3. [s]Enter either the [b]URL[/b] of the image or[/s] [b]Browse[/b] for a image on your computer.
Remember! You must follow the following:
[quote]Only one image can be displayed at a time, its width can be no greater than 80 pixels, the height no greater than 100 pixels, and the file size no more than 21[b].48[/b] KB.[/quote][/quote]
[quote="levans"]
Adding the image to a post
1. Create a new topic or reply to a current topic on artofproblemsolving.com
2. [s]Near the top of the editing area, above the subject line[/s][b]Below the Message body[/b], click [s]Add an[/s] [b]Upload[/b] Attachment
3. Browse for the file
4. Click Add [s]Attachment[/s] [b]the file[/b][/quote]
[quote="PowerOfPi"]
Click on the [b]View[/b] Blog button on the bottom[b]-right[/b] [s]right-hand corner of[/s] someone's [b]avatar[/b] [s]post[/s] .
Or search for it on "Blogs".[/quote]

cause the site was upgraded some stuff got changed you know

[quote="Dojo"][color=darkred][size=150][b]Avatars[/b][/size][/color] [color=darkred][b]How do I set up my Avatar?[/b][/color] 1. Click the [b]Profile[/b] button near the top of the screen. 2. Under Profile, select [b]Avatar Control Panel[/b]. 3. Enter either the [b]URL[/b] of the image or [b]browse[/b] for a image on your computer. Remember! You must follow the following: [quote]Only one image can be displayed at a time, its width can be no greater than 80 pixels, the height no greater than 100 pixels, and the file size no more than 21 KB.[/quote] [color=darkred][size=150][b]Blogs[/b][/size][/color] [color=darkred][b]How do I set up my Blog?[/b][/color] 1. Click the [b]Profile[/b] button near the top of the screen. 2. Under Profile, select [b]Create New Blog[/b]. 3. Fill out proprietary information. [color=darkred][b]How do I create Tags or Categories?[/b][/color] 1. Click the [b]My Blog[/b] button near the top of the screen. 2. a) For categories: Create a post, then at the bottom of the posting box, click [b]Tags and Categories[/b]. Specify your category, and a category tag should appear. After your first category, you can edit them with mod tools. b) For tags: Once again, create a post, then at the bottom of the posting box, click [b]Tags and Categories[/b]. Specify your tag. After that, you can go to [b]Blog Options[/b] => [b]Blog Layout[/b], (located under [b]Profile[/b]) and change your tag cloud settings. [color=darkred][b]Custom CSS[/b][/color] To change the CSS coding of your blog to make it look more appealing, go to Blog Customization and enter in your CSS code.[/quote]
A few things to add.
[b][size=200]How to open up shouts[/size][/b]
To open up your shouts you go to [b]profile[/b] then you go to [b]blogs.[/b]  Now you click on [b]AOPS customizations.[/b]  Scroll down to [b]Shouts to display on main blog page[/b] Now you click how many you want.
[b][size=150]I don't have enough/too much blog posts on my home page[/size][/b]
To change the amount of posts on your home page you go to [b]profile[/b] then you go to [b]blogs.[/b]  Now you click on [b]AOPS customizations.[/b]  Scroll down to [b]Entries on main blog page[/b] Now you make it how many you want(5-10 is good)
[b][size=150]How do I change the name of how many posts and to post on each entry?[/size][/b]
To  change the name of how many posts and to post on each entry you go to  [b]profile[/b] then you go to [b]blogs.[/b]  Now you click on [b]AOPS customizations.[/b]  Now you see the names for multiple posts, 1 post, etc. there.
Lets say you wanted it to say (number of comments) amount of AOPS lovers have been made.  You would make it %s amount of AOPS lovers have been made.
[b][size=150]I can't see my blog in blogs[/size][/b] First you go to profile, then you go to blogs.  Now you click on blog permissions.  Here you can choose who you want to be able to comment in your blog.  Most of the time you will have the settings off.  THen you click resync permissions.  Now wait a few minutes and check in blogs.  If you see your blog voila, if not contact an admin.

[quote="Dojo"][color=darkred][size=150][b]Avatars[/b][/size][/color]

[color=darkred][b]How do I set up my Avatar?[/b][/color]
1. Click the [b]Profile[/b] button near the top of the screen.
2. Under Profile, select [b]Avatar Control Panel[/b].
3. Enter either the [b]URL[/b] of the image or [b]browse[/b] for a image on your computer.
Remember! You must follow the following:
[quote]Only one image can be displayed at a time, its width can be no greater than 80 pixels, the height no greater than 100 pixels, and the file size no more than 21 KB.[/quote]

[color=darkred][size=150][b]Blogs[/b][/size][/color]

[color=darkred][b]How do I set up my Blog?[/b][/color]
1. Click the [b]Profile[/b] button near the top of the screen.
2. Under Profile, select [b]Create New Blog[/b].
3. Fill out proprietary information.

[color=darkred][b]How do I create Tags or Categories?[/b][/color]
1. Click the [b]My Blog[/b] button near the top of the screen.
2.
a) For categories:
Create a post, then at the bottom of the posting box, click [b]Tags and Categories[/b]. Specify your category, and a category tag should appear.

After your first category, you can edit them with mod tools.

b) For tags:
Once again, create a post, then at the bottom of the posting box, click [b]Tags and Categories[/b]. Specify your tag.

After that, you can go to [b]Blog Options[/b] => [b]Blog Layout[/b], (located under [b]Profile[/b]) and change your tag cloud settings.

[color=darkred][b]Custom CSS[/b][/color]
To change the CSS coding of your blog to make it look more appealing, go to Blog Customization and enter in your CSS code.[/quote]

Also, you can not set up your blog until you have a certain number of posts.

[quote="Dsquared"]
[b][size=200]How to open up shouts[/size][/b]
To open up your shouts you go to [b]profile[/b] then you go to [b]blogs.[/b]  Now you click on [b]AOPS customizations.[/b]  Scroll down to [b]Shouts to display on main blog page[/b] Now you click how many you want.
[b][size=150]I don't have enough/too much blog posts on my home page[/size][/b]
To change the amount of posts on your home page you go to [b]profile[/b] then you go to [b]blogs.[/b]  Now you click on [b]AOPS customizations.[/b]  Scroll down to [b]Entries on main blog page[/b] Now you make it how many you want(5-10 is good)
[b][size=150]How do I change the name of how many posts and to post on each entry?[/size][/b]
To  change the name of how many posts and to post on each entry you go to  [b]profile[/b] then you go to [b]blogs.[/b]  Now you click on [b]AOPS customizations.[/b]  Now you see the names for multiple posts, 1 post, etc. there.
Lets say you wanted it to say (number of comments) amount of AOPS lovers have been made.  You would make it %s amount of AOPS lovers have been made.
[b][size=150]I can't see my blog in blogs[/size][/b] First you go to profile, then you go to blogs.  Now you click on blog permissions.  Here you can choose who you want to be able to comment in your blog.  Most of the time you will have the settings off.  THen you click resync permissions.  Now wait a few minutes and check in blogs.  If you see your blog voila, if not contact an admin.[/quote]
You should post this in the appropriate forum.

Better way to take screenshot on a mac, press Command Shift 4, and you can either using the cross hair that comes up automatically, you can select an exact area you want to take photo of, and it will appear on desktop...OR you can press space bar and it will take screenshot of only the current window you're on, after you click, hence taking a photo.


In any given year, the dates (represented as month/day) 4/4,
6/6, 8/8, 10/10 and 12/12 all fall on the same day of the week.
June 3, 2020 is a Wednesday. What day of the week is
December 15, 2020?

We count from 6/3 , replacing 6/6 with 12/12.
6/4
6/5
12/12
12/13
12/14
12/15

So 6 days from Wednesday is $ \boxed{\text{Tuesday}}$.

BTW, I'm not complaining or anything, but I typoed "teusday" and still got it right... can someone fix this?

[size=75][color=blue]This isn't a glitch - it's supposed to happen, you know, be nice and whatever. :) -Isabella[/color][/size]


Find the bound : $\lim_{n\rightarrow \infty} \frac{1^p+2^p+...+n^p}{n^{p+1}}$ where $n$ and $p$ are positive integers

Let $S_n(p)$ be this sum.
Just prove that $\frac 1 {p+1} < S_n(p) < \frac {(1+ \frac 1 n )^{p+1} } {p+1}$ and the desired result will become quite obvious....

Pierre.

Alternatively, you can use Stolz-Cesaro theorem or Riemann sums.

What is Riemann sums?   :blush:   can you explain it to me [b]enescu[/b]  :D
oh thanks [b]pbornsztein[/b] from that we can find $\lim = \frac{1}{p+1}$ right ?


Can you guys help me solve this problem
For certain years from 1950 to 1990, the average annual per capita consumption C of cigaretts by Americans (18 and older) can be modeled by C=3248.89+108.64t-2.98t^2 for 0≤t≤40, where t is the year, with t=0 corresponding to 1950.
a) Use a graphing utility to graph the model.

b) Use the graph of the model to approximate the maximum average annual consumption. Beginning in 1966, all cigarette packages were required by law to carry a health warning.  Do you think the warning had any effect?  Explain.

c) In 1960, the U.S population (18 and over) was 116,530,000. Of those, about 48,500,000 were smokers.  What was the average annual cigarette smokers.  What was the average annual consumption per smoker in 1960?  What was the average daily cigarette consumption per smoker?

[hide="hint for a"]uh ok...since the question actually told you to graph it, you have to (such a waste of time).

develop a system of equations that suit your problem. make sure you have both equations set to $y=$.

graph the problem whether on paper or calculator

[/hide]

i have a hard time figuring out part c of this problem, can anyone give me any advice on how to solve for it

[quote="7h3.D3m0n.117"][hide="not sure but hint for c"]

substituting your numbers in to the original equation?

[/hide][/quote]

can u elaborate on that cause i tried subsituting numbers in and it doesnt come out to the right answer, the answer btw for c is 9703 and 27 according to the back of the book, i just have a hard time on how to come up with that answer.


Putnam
email
advanced fields
advanced fields unsolved

Hi Folks, My daughter Ashley Reiter Ahlin and I have taken on the job of editing the problems section of the Pi Mu Epsilon Journal. The journal comes out twice a year, and we need about 10 to 15 problems for each issue. The problems range from AIME/AMC12 to Putnam level, but can require more time than Putnam problems. Solutions should be less than half a page and are expected with the problem statement, although not required. We hope the posers will submit by email either in Tex or Word.
We acknowledge receipt of each problem.
I hope readers of this forum will send us your original college level problems for the Pi Mu Epsilon Journal. Email me at hbreiter@email.uncc.edu
Cheers, Harold Reiter

Hello there,

First of all welcome on the site :)
I think that this is not the best place to post this request, as people here talk about general college issues (like how to get ino college) rather than college maths.
The place you are looking for is the advanced forums / college playground.

Also, please post more details about your magazine.

Yes, a hearty welcome to the site. Your reputation as a great helper of developing young mathematicians precedes you here.

I think Valentin is correct that your very helpful announcement will gain a greater response in a forum other than the College Forum, so I will move the message to the forum he suggested, which I can do as moderator of the College Forum. Please feel free to post in whatever forum you please, as we would all like to hear more from you.


geometry
analytic geometry
trapezoid
parallelogram
geometric transformation
geometry proposed

[color=darkred]Let $ABC$ be a triangle. Denote : the middlepoint $M$ of the side $[BC]$ ; the foot $S$ of the $A$- symmedian ;

the points $\{\begin{array}{ccc}U\in AB & , & SU\parallel AC\\\ V\in AC & , & SV\parallel AB\end{array}$ ; the points $\{\begin{array}{ccc}\{S_{1},M_{1}\}\subset AB & , & SS_{1}\parallel MM_{1}\parallel CU\\\ \{S_{2},M_{2}\}\subset AC & , & SS_{2}\parallel MM_{2}\parallel BV\end{array}$.

Prove that the linnes $S_{1}M_{2}\ ,\ S_{2}M_{1}\ , BC$ are concurrently.[hide="Hint."]$1.\blacktriangleright AU\cdot AB=AV\cdot AC\ ;$
$2.\blacktriangleright$ See the my last reply from the topic
http://www.mathlinks.ro/viewtopic.php?t=140152 [/hide]
[/color]

Let us use barycentric coordinates with respect to $\triangle ABC.$  Then:

$M \equiv (0:1:1) \ , \ S \equiv (0:b^2:c^2) \ , \ V \equiv (b^2:0:c^2) \ , \ U \equiv (c^2:b^2:0)$

Parallels from $S$ to $BV \equiv c^2x-b^2z=0$ and $CU \equiv b^2x-c^2y=0$ cut $AC,AB$ at

$S_2 \equiv (b^4:0:c^4+2b^2c^2) \ , \ S_1\equiv (c^4:b^4+2b^2c^2:0)$

On the other hand, $M_1,M_2$ are the midpoints of $\overline{BU}$ and $\overline{CV}.$  Then:

$M_1 \equiv (c^2:c^2+2b^2:0) \ , \ M_2 \equiv (b^2:0:b^2+2c^2)$

$M_1S_2: (c^4+2b^2c^2)x-c^4y-b^4z=0 \ , \ M_2S_1: (b^4+2b^2c^2)x-c^4y-b^4z=0$

$M_1S_2 \cap M_2S_1 \equiv P \equiv (0:-b^4:c^4),$  which lies on $BC: x=0. \square$

[i]Remark[/i]: If we define the points $Q,R$ on $CA,AB$ as the point $P,$ then $P,Q,R$ are collinear on the trilinear polar of the third power point $X_{32} \ (a^4:b^4:c^4).$

This problem is also true for an arbitrary point $S$ on $BC.$

So, we denote the points $P\equiv BC\cap UV,\ Q\equiv BV\cap SU,\ R\equiv CU\cap SV$ and from the trapezium $BSVU$ we have that the line segment $PQ$ passes through the midpoints $M_{1},\ N_{1},$ of the segments $BU,\ SV,$ respectively.

Because of now the parallelogram $SQVS_{2},$ this line segment also passes through the point $S_{2}.$

That is, the line segment $M_{1}S_{2}$ as the problem states in general, passes through the point $P\equiv BC\cap UV.$

Similarly, the line segment $S_{1}M_{2}$ also passes through the point $P$ and the proof is completed.

Kostas Vittas.


AMC
USA(J)MO
USAMO
ARML
geometry
number theory

Just answer the question.

Competitions included.

Me: Once, it was a nightmare.  :P

more less... I dreamt I was on math training twice... something like a "neutral dream"

Huh? What happened to the poll I made?  :o

I have had like 8 math-related dreams

One where I made red MOP as a sophomore, which is of course wrong.
One where I solved 6 questions on the USAMO but messed up so that my scores would be counted unofficially.
One where I simply solved nothing on the USAMO.
One where I failed a random easy contest at my school.
Something to do with ARML
Probably more

I have had a dream predicting the next day's events at States. It was partially right.

I had a dream once about failing a math test now that I think about it (8th grade) AND IT WAS EXACTLY RIGHT. UP TO THE PERCENTAGE TOO!

I dreamed I took the USAMO. And won the prize for an elegant proof, for my stupidest yet also elegantest proof ever. I have an image of this proof. So, if that problem is on next year's USAMO, I'm going to be overconfident and botch it, or if I do write my stupid proof, I'll get the correct 0 points for it, not the elegance award.

umm... i normally forget all my dreams.

once i dreamed that i did something to my math teacher, though.
i THINK i sued her or hurt her. it relieved stress.

I often have a bit of math playing non-important roles (ie, there's a bit of math in there somewhere) just because I see a lot of math in my general every-day life. The one dream that I had in which math played an important role was one of my occasional movie-like dreams (very vivid, full storyline, dialogues, background music, etc).

To make a long story short, I was in a bus that fell off a bridge at some point in time in HS, and I got some brain damage, and then I went through therapy for a while, and got back my basic functionality. But the ending of the movie, I went into the AMC 12's, and opened the packet, and looked at the problems and I could barely get into the teens. It had sad music and stuff. And then I was sad and frustrated when I woke up.

By the way, this was not the night before the AMC's. However, when I woke up, the first thing I did was go on AoPS and make sure I was OK  :lol:

no  :D

The message is too small. Please make the message longer before submitting.

Ugh, now I sort of feel like a freak.

Sometimes I get the solutions to problems in my dreams.

ex:
At state for math team a few weeks ago, we were practicing the team (8 person, no calculators) competition and there was this one geometry problem we couldn't get. 4 of us and both of our coaches worked on it until 11, then we had to go back to our rooms to sleep. I had a dream about solving it with circles, and then when I woke up I did what I did in my dream and had it solved in 2 minutes.

Also, I often have dreams about shapes and their relationships (like triangles within pentagons and stuff), and I get alot of number theory and algebra too. It gets weird when I try to describe it to people, but when I wake up  I always forget the specifics, even when I know it's something really significant.  :(

Sciencey dreams like this (especially bio!) are common for me, as are music dreams (where there is nothing except for the music and my hands with a guitar or violin)

The fact that I can never remember what I compose in my dreams is way more frustrating compared to the math  :mad:

once i had one about flunking a test that next day... it spoke true.
there was this other one about my calculator attacking me (i woke up before he got to draw his sword.)
and plus this other one about my math teacher being an alien trying to take over planet earth... i still believe it. (anyone else who has Dr. Boyd Blackburn, aka Dr. B, would know.)
i had one recently about drowning in a sea of math tests...

yep, me and math are worst enemies.

I'll reveal my dream now that I won't feel too strange posting it...

It had something intimately to do with powers of three and the end of the world. It was really scary, and there was opera music in the background. This is basically how it goes (sorry for the vagueness):

There is a whole sea of little circular blobs. They represent people. The winner has to win out of the whole sea in the end. Two of them are part of a little group of 3. They switch places. Then two little groups of 3, which are part of a group of 3 groups of 3, switch. Then two groups of 9, which are both part of a group of 3 groups of 9, switch. And so on. The larger the switch, the more (economically) disastrous. Somehow it was really very, very, frightening. There was this scary opera music in the background to the tempo of my pulse, which I could feel in my head against the pillow. Soon the whole world ends. My hand can feel all that is going on.

 :blush:
I must be a very strange person to be having these kinds of dreams. Everyone else's dreams seem very normal.[/hide]

No, it's ok, I have very psychedelic dreams sometimes. Like one time, I was sitting on top of a snow-capped mountain in my swimming shorts and sunglasses, meditating. Except I was on fire. And I was also underwater. And at the same time, I was dreaming about this weird place in space with all these wooden boards and turning gears, much like some odd Mario stage. Dual-core, dude...


How many positive integers less than $ 100$ can be expressed as a power of $ 3$?

There's $ 3^0\equal{}1$, $ 3^1\equal{}3$, $ 3^2\equal{}9$, $ 3^3\equal{}27$, and $ 3^4\equal{}81$, for a total of $ \fbox{5}$ such positive integers less than $ 100$.

Could it be a sum of powers? If so, all numbers less than 100 can be expressed as a sum of powers of 3.

It's not the sum of powers...the problem didn't say that. You just have to find the powers of 3.

How come 0 counts? I thought 0 was not a positive integer?

Nevermind, i just realized


The numbers $1,2,...,12$ are written around a circle. We can swap any two adjacent numbers if their difference is greater than $1$. Show that we can arrange the numbers in the natural order $1,2,...,12$.

Simple Solution, just start by moving 1 clockwises around the circle untilit hits 2, then move the 1,2 string arnd, swapping 2 first, then 1, till the 2 hits a three, continue in this way till all 12 numbers are in order.

[quote="me@home"]I think theres a typo... it says "if their difference is greater than one" but isn't this always the case?[/quote]

Not if the difference is equal to one...

Oh... too many inequalities... I just saw greater than and thought greater than or equal...


I finished making my Mock AMC that I started a while back  :lol: .   There's not much time before the real AMC so I'm sure that many of you are really starting to practice...you can also see the increased AMC mood by the increased number of posts in this forum.  I am setting the official date and time to [b]Sunday January 9, 2005 at 6 PM EST[/b].  Sorry, but I am not ready to let people take it yet but soon I will allow people to take it early if they need/want to.

Are you going to have it ready before january 3rd? i'd like to take it before school starts again.

I like to take it before the school starts which is next Monday. Is it possible?

And IF you have time, can you tell how hard this might be? Like easy, medium, hard, super hard, or hard that you don't want to dream about it :D

I prefer on the Saturday if possible.

Happy Holidays~

yeah: before schol starts... for me Tuesday the 4th. b

Nooooooooooooooooooooooooooooooooooooooooooooooo!

Not another Mock AMC! They're taking over the world!

Just kidding. :lol: Expect me to do it. I need to get my 118/105 up to 130/115.

Count me in

best before school start again (JAN 4th for me)

Ok, if you need to take the test early you can now.  Just tell me if you want it early and I'll send it to you.  As always, you can take the test anytime before the next Sunday when I administer this test to everyone.  Just make sure to follow the rules.  This is on the honor system so please don't cheat.

captcha000,  shake9991, and Zratonx I will send you a copy early so you don't have to contact me about getting it early.

As for difficulty....hmm...I dunno...I guess we'll find out  :D I think its at about the difficulty level of most of the other mock AMC's, so we'll see.

O, I should also note that my cable (and thus also my internet connection) has been bad today.  So if you need the contest early I'll try my best to get you it.   My cable usually isn't like this and it should be steady by tomorrow if not tonight.

Thanks!

I think I posted too.

I need to take it early. Send me one please.

Thanks.

[quote="joml88"]
captcha000,  shake9991, and Zratonx I will send you a copy early so you don't have to contact me about getting it early.[/quote]

thanks

I hope I can do it, but I'll need to take it early.

Is there anyone who [b]isn't[/b] going to take it early?

And I need it early too  :D .

i'd take it on the 9th except i'll be realllllly busy once school starts.

O, yeah.  I would also like to inform everyone of my idea for a solution packet on this one.  I wont have time to write up all the solutions by myself so I'll probably do something were you can send in a solution to anyone you want after the test is over.  I will look over your solutions and pick the best ones.  This isn't like a contest or anything it's just so that if someone wants to look at old Mock tests later on they can refer to a solution packet instead of reviving old threads.  After next Sunday I will post a list of the problems and mark out which ones have been answered so that you don't do one that someone else did.

I'll might get a little bored so I'll start writing solutions to some of them before next Sunday.

Can I join the ever-growing group of people who want to take it early?

This is just a reminder that the Mock AMC H will be posted Sunday (look on the first page of this thread for exact time).  I would also like to remind those who got the exam early to please try to send in answers.  This exam appears to harder than the usual AMC so do not be discouraged if you do not do as well as you usually do or hope you will do.

Yah, I believe I have finally figured out a reliable way to get the pdf working.  So if you are still planning on taking the test early and want to use the pdf version please PM me about it (that is if I have already sent you the test).  There is also one error that I know about on the version I sent in the PM's to people so please contact me before you take it if you plan to use that version (the error is on #6).

Of course, if you plan on taking it tomorrow at the usual time, you shouldn't have to worry about anything.

That really was brutal. :(  Joml88, how much would my real amc 12 scores be compared to the 74 i got?


[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ ab \plus{} ac \plus{} bc \equal{} 1.$ Prove that:
\[ \frac {1}{{ab}} \plus{} \frac {1}{{bc}} \plus{} \frac {1}{{ca}} \plus{} 3 \geq2\left(\sqrt {\frac {1}{{a^2 }} \plus{} 1} \plus{} \sqrt {\frac {1}{{b^2 }} \plus{} 1} \plus{} \sqrt {\frac {1}{{c^2 }} \plus{} 1}\right)
\]
[/quote]
 $ RHS \equal{} 2 \left( \frac {\sqrt {(a \plus{} b)(a \plus{} c)}}{a} \plus{} \frac {\sqrt {(a \plus{} b)(b \plus{} c)}}{b} \plus{} \frac {\sqrt {(c \plus{} b)(a \plus{} c)}}{c} \right) \leq \frac {2a \plus{} b \plus{} c}{a} \plus{} \frac {2b \plus{} a \plus{} c}{b} \plus{} \frac {2c \plus{} a \plus{} c}{c} \equal{} 3 \plus{} \frac {a \plus{} b \plus{} c}{abc} \equal{} LHS$
 Occur  equality  if and only if:  $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt{3}}$


a program finds the answer
-.9932892
, which is not an integer

I haven't worked on this problem yet, but Lord Venom, it can't be negative because you can cancel out a bunch of terms and be left with a sum of positive numbers. Did the program perhaps run in radians?

Ah, didn't realise the problem was supposed to be in degrees. That makes it considerably easier (though still difficult).

You can cancel out the terms sin 110 through sin 250, so you are left with
sin 5 + sin 10 + sin 15 + ... + sin 105. Is this right?

hmm, I might have made a mistake typing the problem. The question says that its value is m/n, find m+n. But when I tell mathematica to evaluate it, it seems that m+n is ALOT greater than 999. So, someone please check the problem?

And Im also wondering, do you use the formula SinA + SinB=2Sin[(A+B)/2]Cos[(A-B)/2]?

Consider as the imaginary portion of cis 5 + cis 10 + ... + cis 250 = e^(i * pi/36) + e^(i * pi/36 * 2) + ... + e^(i *pi/36 * 50), a geometric series. Sum the series, follow the will of Heaven and you can't go wrong.

Actually it would probably make more sense to start from cis 0 so that the series sums more easily, and it doesn't alter the answer. It will be very messy though. There should be a nicer way to do it.

The sum of sin 5 + sin 10 + ... + sin 65 + sin 70 and sin 185 + sin 190 + ... + sin 245 + sin 250 comes out to 0...and sin 180 is already 0, so you actually just have to evaluate sin 75 + sin 80 + ... + sin 170 + sin 175. I think.

And the answer's definitely not an integer. Still, that doesn't help a lot.

yup. answer's not.

its in m/n. AIME tells u to find m+n

Whoa there guys, wrong problem. The problem is to sum sin 5 + sin 10 + ... + sin 175 and express the answer as tan(m/n).

#11 on here:

http://www.kalva.demon.co.uk/aime/aime99.html

Oh yeah, that's why the geometric series works out... as the sum (using a = 1 as Simon Rubinstein-Salzedo has suggested), you get e^(pi * i) - 1 = -2, over e^(pi/36 * i) - 1, which is -2/(cos 5 - 1 + i*sin 5), the imaginary portion of which is 2sin 5/(cos^2 5 - 2cos 5 + 1 + sin^2 5) = sin 5/(1 - cos 5). Anybody who's anybody knows that this is the tangent of (90 - 5/2) = 175/2. So m + n = 177.

[quote]sin 5/(1 - cos 5). Anybody who's anybody knows that this is the tangent of (90 - 5/2) = 175/2. So m + n = 177.[/quote]

Wha? Yes..well...I am in the small class of nobodys, so how did you get tan(90 - 5/2) from sin 5/(1 - cos 5)? The tan subtraction formula doesn't work here since tan90 is undefined, so I dunno.

Use the half-angle identity for cotangent. Don't worry. I have to rederive it every time.

[quote]Use the half-angle identity for cotangent.[/quote]

Cotangent? You mean the tangent formula tan(x/2) = sin(x)/(1+cos(x)).

By the way, if you have to derive it everytime, how would you have recognized sin 5/(1 - cos 5) as something random as tan(a/2)?  8-)

Well, usually, when hooyounh (teratomato) says something is obvious, it means it took him 30min to figure out...

lol

Seriously though, you should remember those formulas. They make your life a lot easier than wading through square roots of cosines and whatnot.

Oh, and by the way, that is the half-angle formula for cotangent (see how it's flipped?). However, since cot theta = tan (90- theta) it really doesn't make much difference.


[quote="ssupermath"]If $x^x=81$ find x[/quote]

neither solution for x is very nice, you have to plug this one into a calc to solve it....
$f(x)=x^x$ does not generally have nice solutions unless the problem is very contrived

looking at a graph one solution is around 3.5 ....

I plugged the equation into Mathematica, and technically there is a "solution," but it's a complicated expression in terms of the exponential, the product log function, and a lot of complex numbers, which is just another way of saying that there's no easily expressible solution.


I didn't really see the joke, bad or not, ... but try solving $x^x = 0.2078795763$ instead.


function
limit
calculus
integration
real analysis
real analysis unsolved

If $ X=B.V.[0,1]$ (Bounded variation functions on $ I=[0,1]$ ) and if we define :

$ \parallel{}f\parallel{}_I=Total\ Variation \ of \ (f) =V\limits_{0}^1 (f)$

Is that a normed space ?

 :)

No, you need to normalize the BV functions, (for example any constant function has total variation zero). I.e. you need to quotient out the $ f$ such that $ \parallel{}f\parallel{} \equal{} 0$.

The quotient process is as follows: First we may look only at representatives such that $ f(0) \equal{} 0$. Then let $ g \sim g'$ mean that $ g$ and $ g'$ only differ at points where they're both discontinuous. Then $ \parallel{} \left[g\right] \parallel{} \equal{} inf \parallel{}f\parallel{}_I$ where the infimum is taken over $ f \in [g]$ defines a norm on this quotient space.

Then for each class we may choose a unique representative by taking $ g$ and letting $ g_0(t) \equal{} \lim_{x \to 0^\plus{}} g(t\plus{}x) \minus{} \lim_{x \to 0^\plus{}} g(x)$. This representative is the unique element of $ [g]$ that is right continuous in $ [0,1)$ and left continuous at 1, and the norm of $ [g]$ is the norm of $ g_0$

After factorization, it's also a dual Banach space. The duality is given by the Stieltjes integral $ \langle f,g\rangle \equal{} \int_0^1 f\,dg$, where $ f\in C[0,1]$ and $ g\in BV$.


absolute value
complex numbers
imaginary numbers

I'm not sure I learned some of the operations on radicals correctly...I'd really appreciate it if you guys could help me out on these.  :)

1. 4th root of (a+b)^5.  My guess is that |a+b|^5/4.

2. Why is (sqr root x)^2 |x|?  Is it because if x is negative, let's say -25, then the square root of x will be 5i, 5i^2 is -25.  In these situations, principal answers (positive) are looked for, so if x is negative, then the opposite of x must be found...eh I'm not sure.

3. sqr root of x^4 doesn't need the abs value and can simply be written as x^2 because any number squared is always a positive number.

4. A value such as:
2|a|^3
all over
a
can be canceled to 2a^2, absolute value doesn't matter because the expression in the numerator is always positive.

Are any of these statements wrong?  What are some common errors when doing operations relating to abs. value/radicals/exponents, etc that I should remember for future use.

Thanks!

[quote="frt"][quote="brady00n"]2. Why is (sqr root x)^2 |x|?  Is it because if x is negative, let's say -25, then the square root of x will be 5i, 5i^2 is -25.  In these situations, principal answers (positive) are looked for, so if x is negative, then the opposite of x must be found...eh I'm not sure.[/quote]
You're on the right track, but it is not so much about imaginary numbers.

Say $x=25$. There are two numbers for $x$ such that $x^2=25$. One is $x=5$ and the other is $x=-5$.
Now, for $x=5$, $\sqrt{5^2}=\sqrt{25}=5$, so the sign doesn't change. However, for $x=-5$, $\sqrt{(-5)^2}=\sqrt{25}=5$, so the sign does change. It is because $\sqrt{}$ cannot be negative. That is why $\sqrt{x^2}$ has to be $|x|$, not just $x$. ($\because$ $|x|$ is non-negative)

I hope you understand my explanation...[/quote]

Oops, I meant when the square root of x is found before the quantity is squared.  The thought I was going through was that when the negative number is square rooted, say -25, it would become 5i.  5i^2 becomes -25, but negative numbers aren't wanted, so the abs value is needed...was wondering if square rooting first would be different, and if my line of logic was on the right track.

I'm sorry, I better start learning Latex... :blush:

Oops, I meant when the square root of x is found before the quantity is squared.  The thought I was going through was that when the negative number is square rooted, say -25, it would become 5i.  5i^2 becomes -25, but negative numbers aren't wanted, so the abs value is needed...was wondering if square rooting first would be different, and if my line of logic was on the right track.

I'm sorry, I better start learning Latex... :blush:[/quote]

if you begin with the square root of a negative, then use the imaginary number, then your "allowed" to result in a negative number in your answer..since then it goes thorugh the process of

(taking it apart)

root of -25=i rad 25=5i.  (this is where you apply the abs value thingy, resulting in 5i again).

step 2 of your equation

squaring it, 5^2*i^2=25(-1)=-25.

hope that makes sense?


If $ x\plus{}y\equal{}12$ and $ x\minus{}y\equal{}8$, what is the value of $ 2x\minus{}xy$?

x+y=12
x-y=8
Elimination yields x=10 and y=2
subbing in,
2(10)-10(2)
20-20
$ 0$


[hide="Yay for multiple choice!"]Divide each of the answer choices by 1.04. Only 13 and 26 give terminating decimals when divided by 1.04. 13 is smaller, so [b]B[/b].

I'm sure there's an algebraic way, but this was easier... :D [/hide]

[quote="4everwise"]A store prices an item in dollars and cents so that when 4% sales tax is added, no rounding is necessary because the result is exactly $n$ dollars where $n$ is a positive integer. The smallest value of $n$ is

$\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100$[/quote]

[hide]in order for something to be multiplied by 1.04 and get a whole number, the result needs to be able to be divided by 1.04 and get a number that doesnt have to be rounded centwise
by guess and check, $\frac{13}{1.04}$ is the smallest i think (B)[/hide]

[quote="4everwise"]Anyone have an algebraic method?  :D

[hide="hint"]1.04x=n[/hide][/quote]

[hide="Maybe..."]$\frac{26}{25}x=n$
$26x=25n$
$x=\frac{25}{26}n$

n must be a factor of 26, which are 1,2,13 and 26. For x to be a terminating decimal, n can be 13 or 26. 13 is smaller.[/hide]

That works.

What I did got me exactly 13 though.  :)


The second one doesn`t hold...We apply x1^2 / a1 + x2^2 / a2 ...+ xn ^ 2 / an >= (x1+x2 +..xn) ^2 / a1+a2+..+an...for positivez number and it doesn`t hold

I don`t think that it doesn`t hold..It`s just that Cauchy in  Engel isn`t strong is this case..If Cauchy in Engel is weaker than the ineq,that doesn`t mean that the ineq is wrong..

[quote="snake_pissken"]Let $a,b,c$ be positive real numbers. Prove that
$2)$ $\frac{a^{2}}{ a^{2}+ab+b^{2}}+\frac{b^{2}}{ b^{2}+bc+c^{2}}+\frac{c^{2}}{ c^{2}+ca+a^{2}}\geq 1$[/quote]
Good joke!
Let $x={a \over b}, \cdots$ then $xyz=1$. Expand all, it remains to: \[x^{2}+y^{2}+z^{2}\ge xy+yz+zx.\] PC.

for 2) you can see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110854]here[/url]


[b]Nordic university-level mathematics team-competition, NMC[/b]

[quote]The fourth Nordic mathematics competition for university students will be arranged in February 2010. The participants in this team-based competition should be Bachelor's or Master's level students in Nordic universities. Faculty in these universities are asked to encourage and support the participation of able students. There is no fee for participation, but teams are required to register, see contact persons. Details can be found in the competition rules.

Participating teams will solve a set of mathematical problems at their own institutions during one week's time and submit the solutions by e-mail. The level of the problems will be chosen so as to provide even the most able of students with sufficient challenge. For an appreciation of the kind of problems possible we recommend you to have a look at the problems from the International Mathematics Competition for University Students and previous year's competitions, which can be accessed below.

[b]Schedule[/b]
February 2 Team registration closes
February 4 Competition begins at 12 noon CET
February 8 Competition ends at 6 p.m. CET
~ February 15 Results announced

[b]Rules of the 2010 competition[/b]
1 §. Participation
The competition is open for teams of 3 to 5 students. Team members should be bachelor or master students enrolled in universities (or polytechniques, etc.). A team may consist of students form different universities. A person may not participate in more than one team.
2 §. Start and finish
Before noon CET, April 14th, each team should send an e-mail with the list of the team members to the organizers stating their intention to participate. The problems will be available for downloading at the competition web-site on Thursday, February 2th, at 12 noon CET. Each team must submit their solutions by e-mail on Monday, February 8th, at 6 p.m. CET.
3 §. Evaluation
The jury will grade the problems. The results will be made public as soon as possible after the closing of the competition period.
4 §. The problems
The jury will select 8-10 problems for the competition. Persons not participating in the competition may propose problems to the jury. The problems may refer to the material of basic university-level courses, corresponding to the first two years of a five year curriculum: Algebra, Calculus, Combinatorics, Linear algebra, Metric Topology, Number theory, and Probability.
The problems may also be based on more advanced mathematics, provided that the base for these problems may be presented in a brief form (a few pages).
If a problem has multiple parts, then the score of each part will be indicated.
5 §. Problem solving
The team is free to choose any division of labor in solving the problems. One may refer to sources in the solution, as long as the source is commonly available. The jury should be consulted about any unclear statements in the formulation of the problems. Other than this, no-one outside the team may be consulted.
6 §. Submitting solutions
The solutions, preferably prepared on computer, should be sent by e-mail to the co-ordinators by the closing time of the competition. Solutions should be submitted in English.

[b]Contact persons[/b]
[img]http://www.ljplus.ru/img4/d/m/dmitin/emails.png[/img] [/quote]

http://cc.oulu.fi/~phasto/competition/


binomial coefficients
combinatorics proposed
combinatorics

if someone can prove that

                   -n[size=150]C[/size]r  =  (-1)^r  (n+r-1)[size=150]C[/size]r

say what? when r is even, LHS = negative < positive = RHS. I can't seem to understand problems correctly these days... :?

You want a proof of the "upper negation" identity

$ \binom{ \minus{} n}{r} \equal{} \left( \minus{} 1\right)^r\binom{n \plus{} r \minus{} 1}{r}$.

Indeed, look at the definition of binomial coefficients:

$ \binom{ \minus{} n}{r} \equal{} \frac {\left( \minus{} n\right)\cdot\left( \minus{} n \minus{} 1\right)\cdot ...\cdot\left( \minus{} n \minus{} r \plus{} 1\right)}{r!}$
$ \equal{} \left( \minus{} 1\right)^r\frac {n\cdot\left(n \plus{} 1\right)\cdot ...\cdot\left(n \plus{} r \minus{} 1\right)}{r!}$
$ \equal{} \left( \minus{} 1\right)^r\frac {\left(n \plus{} r \minus{} 1\right)\cdot ...\cdot\left(n \plus{} 1\right)\cdot n}{r!} \equal{} \left( \minus{} 1\right)^r\binom{n \plus{} r \minus{} 1}{r}$.

@ al. M. V.: Binomial coefficients are defined by the formula $ \binom{x}{u} \equal{} \frac {x\cdot\left(x \minus{} 1\right)\cdot ...\cdot\left(x \minus{} u \plus{} 1\right)}{u!}$; hereby, u must be a positive integer, but x can be any arbitrary real number, in particularly it can be negative.

  Darij

oh, it's (-n) C r, i thought it was - (n C r). Thanks for the clarification, Darij, nonetheless.


What is the correct way to do the following problem?

To take quizzes, 30 students in a class are divided into groups of three (means three students in one group, total 10 groups). If the process is done randomly, what is the probability that Margo and her best friend, Irma will be in the same group? Express your answer as a common fraction.

You use guess and check. Either that or list all of the possibilites :) .

[hide="how I did it"]
Well, her friend can be one of the two people in Margo's group, or she can be one of the 27 people not in Margo's group.

2/27
[/hide]

The best way would probably be
[hide]Making a list of ways this could happen[/hide]

[quote]To take quizzes, 30 students in a class are divided into groups of three
(10 groups of 3 students).  If the process is done randomly, what is the probability
that Margo and her best friend, Irma will be in the same group?
Express your answer as a common fraction.[/quote][hide]
Margo is definitely in some group of three students.
What is the probability that Irma is one of the other two members of her group?

The answer is $\frac{2}{29}.$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[b]Proof[/b]

There are $C(29,2) = \frac{29!}{2!27!} =406$ ways to select the other two students.

There is $1$ way to select Irma; there are $28$ choices for the third student.
Hence, there are $1 \times 28 = 28$ ways to select Irma and a third student.

The probability is: $\frac{28}{406} = \frac{2}{29}.$[/hide]

[hide]we know that Margot will be in one of the groups... we don't really care which one.
so assume that taht we now are picking the other members of this group from the 29 remaining people
so there are 29C2 ways this can be done
there are 28 ways Irma can be in the group (getting paired with twenty eight other possible people) so our answer is:
28/(29*28/2) = 2/29
[/hide]

here is another type of problem:
A graduating class is taking their class picture. In the front row there are 10 chairs for 10 people to sit in. What is the probability that best friends George and Richard sit next to each other in the front row?
Another one:
A group of Broadway dancers are lining up for their final number. There are twelve total dancers and each one is wearing either a green or blue costume. There are 6 dancers in green and there are six in blue. How many ways are there so that no green dancer is standing next to another one.

For the George and Richard problem.  There are 10! ways to seat the students in the front row.  Consider George and Richard to be a package.  Thus, there are 9 ways the package, "GeorgeRichard", and 9 ways the package "RichardGeorge" can occur.  So the probability they are seated together is 18/10!.

[hide]If you start by finding the probability of the 1st group being picked in the order margo, irma, someone else, you'd have
1/30 * 1/29 * 28/28 = 1/870

If you them multiply by 6, for the different orders those three people could get picked, you get 1/145

If you then multiply by 10 for the ten different groups, you get 10/145 = 2/29

Several other people have had simpler ways at getting 2/29, so I'd go with 2/29[/hide]

Treething- how did you use guess and check on this problem?

[hide="my answer to the George and Richard problem. "]
For the George and Richard problem. There are 10! ways to seat the students in the front row. Consider to tie George and Richard together. Thus, there are (1+8)! ways to seat all. Then we untie George and Richard to get 2! ways. So the answer is 9!*2/10! = 1/5.  The chance is pretty big.[/hide]

[hide="my answer to the counting problem"]

There are C(30,3) total ways to group any three students. There are C(2,2)*C(28,1) ways to put Margo and Irma into any one  group.  There are ten groups. So the solution is

C(2,2)*C(28,1)*10/C(30,3) = 2/29[/hide]

Thanks for taking time to respond. I am glad to see so many different and good solutions.

I created this problem based on Mr. Crawford's Mathcounts class weekly problems. That problem stated:

To take quizzes, each of 30 students in a class is paired with another student. If the pairing is done randomly, what is the probability that Margo is paired with her best friend, Irma? Express your answer as a common fraction.

Mr. Crawford, could you please tell us if our answers are correct? Thank you.