geometry
geometric transformation
dilation
vector

Hello guys ,I was reading a book called Dead Famous-Albert Einstein and the Inflatable Universe written by Mike Goldsmith.
I recently came across a doubt or paradox.
The principle of relativity states that the laws of motion or nature are same for everyone undergoing a uniform motion which implies that there is no actual measurement to tell whether you are moving or not which ultimately proves that there is no such thing called Absolute Motion or Absolute time (philosphical concepts introduced by Newton) .
Reason - For example when you are travelling in a train (steady motion) you can never say that you are moving or not .(not common sense right ? :rotfl:,it sounds uncommon because the way you define common sense .  ) Consider person A and person B travelling in a train (steady) ,person A tells person B tells that they both are moving but the Person B is argumentative and not accepting the fact . The Person A says that because he sees that the trees are all moving backwards which means the train is moving across the trees . Person B says that A is wrong because it can also happen that the trees are moving (please dont argue that trees cannot move) and the train is at rest . Another example is that ,now consider the universe to consist only of the earth and sun ,A and B are standing on the earth. A watches the sun moving slowly and says that the earth is [b]Absolutely[/b] at rest . Person B (argumentative as usual  :rotfl: ) says that it can also happen that the sun is at rest and the earth is moving around it . We cannot say who is right ,can we? There is no actual test to tell whether you are moving steadily or not. You can tell that you are absolutely at rest because you are sitting and watching the computer . But you are at rest only with respect to the earth ,you are moving at some 100km per hour with respect to an observer present in the sun. So you can describe your motion only  with respect to some object . So the only thing which is present is the relative motion . You cannot say whether you are precisely moving or not without referring to anythin.
Brainy Albert had an interesting question in his mind ,he wanted to know what will happen if you travel at the speed of light .
He ultimately concluded that the [b]speed of light is same for everyone no matter how fast they travel[/b] .This means that for example when a bowler is bowling a cricket ball ,he is travelling at 10m/s(say ) ,when he is about to release the ball he imparts a velocity of 5m/s to the ball . So with respect to the ground  the net velocity of light is 15m/s.But Albert said that light is unusual ,for example when a bowler is travelling at the same 10m/s but instead of a ball he is having a torch light ,when he releases the light the velocity of light will be only be $ c$ not $ c \plus{} 10$
This was derived based on the principle of relativity (assuming that is correct ).
The book gives a proof for this .
It tells us to imaging a train travelling at the  speed of light ,and you are an observer sitting inside the train ,now it tells that the light coming from the back of the carriage of the train will not reach us because we ourselves are travelling at the speed of light ,so we wont be able to see anything from behind because we will be able to see something only if the light from it reaches our eyes.In this case that wont happen so everything will be spooky and dark :diablo:
Then now we can identify whether we are travelling or not ,that is ,we have found a foolproof test.
I mean when your naughty friend slaps you and hits you on your head making you faint and has put  you in train travelling at the speed of light .When you wake up and see behind you will observe that everything is black and assuming that you have read this post and you are a bit brainy ( :rotfl: )
you will immediately recognize that you are travelling  :wink: .THIS DISPROVES THE PRINCIPLE OF RELATIVITY. But this is a simple statement and cannot be wrong ,can it ?
Assuming that it is true ,we cannot tell whether we are travelling or not which means that light should reach our eyes . The book tells that this can happen only when light completely ignores the motion of train and travels at its usual speed .
What doubt I have is -now assume that we dont know the fact that speed of light is same for everyone . Now we are the observer inside the speedy train . If we take the frame of reference as the train (inertial) and apply newtons laws we will find that light is travelling at the speed of light that is at $ c$ so which implies that the light will reach us .
If we asuume the frame of reference to be earth- consider I am a stationary person standing on the ground (Assume that I am a layman well versed with newton laws and concept of relativity but I dont know about relativity blah blah etc)
So when I see the speed y train ,I will conclude that the passenger travels at $ c$ and the light is travelling at the speed $ c \plus{} c \equal{}2c$.
This seems to contradict what the book says
Anyone please help me out ,sorry if I am stupid .

I didn't read everything you wrote but I can tell you that $ c\plus{}c$ or $ c\plus{}10$ don't make sense.  There is an equation (which I don't remember) which takes into account people in other frames of reference.  So if you are going, say $ .6c$ and someone in another spaceship is going $ .8c$, there is a way to determine the speed at which the other person appears to be moving; it's not $ 1.4c$.  Look for a book that rigorously develops the theory (i.e., a textbook).

[quote="JRav"]I didn't read everything you wrote but I can tell you that $ c \plus{} c$ or $ c \plus{} 10$ don't make sense.  There is an equation (which I don't remember) which takes into account people in other frames of reference.  So if you are going, say $ .6c$ and someone in another spaceship is going $ .8c$, there is a way to determine the speed at which the other person appears to be moving; it's not $ 1.4c$.  Look for a book that rigorously develops the theory (i.e., a textbook).[/quote]
Yes that is what I said ,i said $ c$ and $ c\plus{}10$ doesnt make sense ,
my doubt was at the end .

The observer on the earth will observe the light traveling at a speed of c as well, because the speed of light is the same in all reference frames.

It may seem counterintuitive that two different people can observe the light to be traveling at the same rate even though they are moving with respect to each other. The reason this is possible is becasue frames of reference include time as well as spatial directions.

The observer on earth sees the light going a bigger distance, but in the earth's frame of reference, time is faster then in the trains frame of reference, so it evens out.

[quote="0714446459923"]The observer on the earth will observe the light traveling at a speed of c as well, because the speed of light is the same in all reference frames.

[/quote]
Actually the example the book gave is to prove this ,in the process of the example I had a doubt ,so you cannot assume the statement before proving it ,I mean only the proof given by the book I didnt understand so before proving it you cannot state this .Maybe you should read my post again

how come only massless objects can travel at the speed of light?? :?:

[quote="AndersoN"]I didn't read your entire post, but it seemed to me that your thought-experiment involved a train traveling at the speed of light, c. This is impossible, only massless objects can travel with the speed of light.[/quote]
Ya , I am sure that train cannot travel at the speed of light - but this is a thought experiment and it is relativity we are dealing about , common sense doesnt make that much sense at all in relativity ,so you have to imagine things at a higher horizon ,like think that you are in a very fast world (wrt to real earth) and things like $ 0.5c$ or $ 0.2c$ are very common to you

All right, time to clear this up.

As I see it, srinath is okay with all of relativity except for that one train traveling at $ c$, which gives him a headache.

His argument:
A train is moving at c. The light emitted from the back of the train will never reach observers at the front. Proof: A stationary observer notes that the light is going the same speed as the observers and that they never catch up. But then the observers in the train will not see any light and know that they are moving at c.

The problem with that:
As previously mentioned, massive objects cannot move at the speed of light.

A plausible response:
Because this is a thought experiment, it doesn't matter. In any case, we can let the train be moving at arbitrarily close to the speed of light, in which the same problem remains.

The resolution:
There is a problem of [i]time dilation[/i]. This is a result of Einstein's relativity postulate combined with the universality of the speed of light, and an explanation can be found [url=http://en.wikipedia.org/wiki/Time_dilation]here[/url].
It basically states that the faster you go, the slower your clock runs. A stationary observer watching a moving observer would see his clocks run slow, him breathe slow, gravity would act more slowly, etc. The same thing happens here. If a train is moving at a speed close to c, then the passengers will experience time more slowly, by exactly the factor needed to cancel out the time needed for light emitted from the back of the train to arrive to the front. In the limiting case where a train is moving at the speed of light, then the passengers would effectively be frozen, appearing completely stationary and unmoving (relative to the train) to stationary observers. And to him, the infinite time needed for the light to catch up to him would appear to be only a finite amount of time because of his infinitely time-dilated view on the universe. So in the end, it all works out, and the postulate of relativity is not violated.

The question I asked is ,whether the light is coming from inside the train or outside the train ?
If it is [b]outside[/b] the train ,I agree that no light can reach the speedy train . But using this we cannot prove that light travels at same speed for everyone because the book tells that since the principle of relativity is true ,light ignores the motion of train . But my doubt is light is outside the train ,obviously it doesnt have any contact with the train ,so what significance is there by telling that  [b]light ignores the motion of the train [/b] :P
So some significance exists for the statement  [b]light ignores the motion of the train [/b] only if the light is from inside the train ,now how can you tell that the passenger in the train cannot see the light .
Why I am asking this because ,if we consider the [b]frame of reference[/b] as the train ,then the light travels at $ c$ and the observer travels at $ 0$ speed ,so obviously ( :blush: ) the light will reach the passenger in the train .

Light inside and outside the train are the same. The only difference is how fast you, the observer, are moving. If you are standing on the platform, you will see that the light will never catch up to the train. If you are standing in the train, you will see that the light is traveling at speed c relative to your stationary frame of reference, the train. This means that the light is traveling at c relative to the platform and at c relative to the train.

Here is one plausible argument of contradiction:
The light is moving at c relative to the train, which is moving at c relative to the platform. Therefore light must be moving at 2c relative to the platform, due to the addition of velocities $ v_{ab} \plus{} v_{bc} \equal{} v_{ac}$, where $ v_{ab}$ represents the velocity of body a with respect to body b.

However, the problem with that argument is the formula used for vector addition. For special relativity, Einstein developed a new formula for vector addition, derived from his postulates of relativity and the universality of the speed of light. It goes along with the theory of time dilation, and gives the result $ c\equal{}c\plus{}c$.

More information on the topic can be found [url=http://en.wikipedia.org/wiki/Velocity_addition_formula]here.[/url]

[quote="bowei"]Light inside and outside the train are the same. The only difference is how fast you, the observer, are moving. If you are standing on the platform, you will see that the light will never catch up to the train. If you are standing in the train, you will see that the light is traveling at speed c relative to your stationary frame of reference, the train. . [/quote]
I accept that the observer outside will never see the light catching up with the train .
[quote="bowei"]
However, the problem with that argument is the formula used for vector addition. For special relativity, Einstein developed a new formula for vector addition, derived from his postulates of relativity and the universality of the speed of light. It goes along with the theory of time dilation, and gives the result $ c \equal{} c \plus{} c$.[/quote]
The thought experiment demonstrated is to prove this , that is speed of light is same for everyone no matter how fast they travel .Er.....I think only with this fact : The speed of light is same for everyone no matter how fast they travel ,Einstein proved that clock slows down at high speeds . I shall once again repeat whatever I have posted earlier and make my question more clearer.
[hide="once again"]It tells us to imaging a train travelling at the  speed of light ,and you are an observer sitting inside the train ,now it tells that the light coming from the back of the carriage of the train will not reach us because we ourselves are travelling at the speed of light ,so we wont be able to see anything from behind because we will be able to see something only if the light from it reaches our eyes.In this case that wont happen so everything will be spooky and dark :diablo:
Then now we can identify whether we are travelling or not ,that is ,we have found a foolproof test.
I mean when your naughty friend slaps you and hits you on your head making you faint and has put  you in train travelling at the speed of light .When you wake up and see behind you will observe that everything is black and assuming that you have read this post and you are a bit brainy ( :rotfl: )
you will immediately recognize that you are travelling  :wink: .THIS DISPROVES THE PRINCIPLE OF RELATIVITY. But this is a simple statement and cannot be wrong ,can it ?
Assuming that it is true ,we cannot tell whether we are travelling or not which means that light should reach our eyes . The book tells that this can happen only when light completely ignores the motion of train and travels at its usual speed .[/hide]
What I want to know is : If the light is outside the train ,then there is no significance by telling that light ignores the motion of train . @bowei- you said that the light inside and outside the train are same . Well ,I think that the significance contributed by light inside the train is only valid . If light is outside the train ,obviously there is no point in telling light ignores the motion of train . The book is trying to prove that  light completely ignores the motion of train and travels at its usual speed (,so for proving this we have to take into account that light travels inside the train)   so that the observer is able to see the light (if he is not able to see the light then the principle of relativity is disproved ) , but even if light takes into account of the motion of the train -that is even if the speed of light is $ c \plus{} c$ (please dont tell that this is not possible because before reading about relativity we only know that this is true and we are supposed to prove this by getting a contradiction or somethin) the observer will be able to see the light and ultimately we are not able to prove that light ignores the motion of train and still travels at $ c$ .Well ,even if anyone is able to prove this by this idea of a thought experiment ,I think my doubts will be clarified.

Would the following example help?

There are two roads that intersect, and you are standing on one of the roads. Heading directly towards you is a plane traveling at approximately $ c$, and heading on the other road, in the path of the plane, is a very large truck, also traveling at $ c$. The two vehicles will collide if the pilot of the plane does not swerve, so the pilot indeed swerves to avoid an accident. Now, imagine the speed of light is not constant. If it were not, then you would see the plane heading towards you at $ 2c$, and the plane would swerve for no reason whatsoever, and effects would precede their causes, causing paradoxes. Therefore, the speed of light must be constant.

er....can you explain it a bit more detailed . I am not able to understand . Sorry for the trouble

Let us assume that the speed of light is not constant, and can be computed simply as a sum of two vectors; if a light source is traveling to me at $ 0.5c$, I see light coming to me at $ 1.5c$ instead of just $ c$.

Therefore, in the below example, the plane flying towards me appears to be traveling at $ 2c$. Because of this difference, when the plane decides to swerve to avoid the truck, you see this happen before the plane meets the truck, because you see the light from the plane before the light from the truck. However, this implies that the effects come before the causes. To avoid these paradoxes, the speed of light must be constant.

I think the main problem here is the thought experiment claiming to 'prove' that the speed of light is constant.

To constancy of the speed of light derives from Maxwell's Equations, not the thought experiments. The point of the thought experiments is to derive relativity based on the assumption that the speed of light is constant.

A frame of reference moving at the speed of light makes no sense. It's like dividing by 0.
You can have a frame of reference move arbitrarily close to the speed of like (just like you can divide by arbitrarily small positive numbers) but going all the way to the speed of light gives a meaningless result.

[quote="0714446459923"]

To constancy of the speed of light derives from Maxwell's Equations, not the thought experiments. The point of the thought experiments is to derive relativity based on the assumption that the speed of light is constant.[/quote]
Ok fine it is derived from maxwell equations . But there are also thought experiments to prove this like done in the book . (still no one answered what my doubt regarding the thought experiment in the book  :(  )
[quote="0714446459923"]
A frame of reference moving at the speed of light makes no sense. It's like dividing by 0.
You can have a frame of reference move arbitrarily close to the speed of like (just like you can divide by arbitrarily small positive numbers) but going all the way to the speed of light gives a meaningless result.[/quote]
Oh god  :(  :bomb:  , I know it doesnt make sense . But my or (book's) objective is to prove the speed of light is constant at whatever speed one moves through thought experiments . The experiments [b]try[/b] to prove by contradiction . There is no point in again and again telling that the laws of vector addition are not applicable for the light  (well that is exactly what we are supposed to prove) or telling that it doesnt make sense to attach a frame of reference at the speed of light .

So basically, what they are proving is that if it were possible to travel at the speed of light, weird and contradictory things would occur?
What are you confused about?

[quote="0714446459923"]So basically, what they are proving is that if it were possible to travel at the speed of light, weird and contradictory things would occur?
[/quote]
Yes that can be one way and the book proofs using the validity of principle of relativity ,that proof I am confused about and I have written that proof in the first post


The set $ T_n$ is defined as following groups of numbers:

$ T_1 \equal{} \{1\}$
$ T_2 \equal{} \{3,5\}$
$ T_3 \equal{} \{7,9,11\}$
$ T_4 \equal{} \{13,15,17,19\}$

and so on. Let $ S_n$ be the sum of the numbers within each set $ T_n$. For example:

$ S_1 \equal{} 1, S_2 \equal{} 3\plus{}5 \equal{} 8$

A) Find the sum of the numbers in the 2008th group.
B) Derive the formula for $ S_n$ and prove this formula.

[hide="Method 1"] Prove by induction that the average element of $ T_n$ is $ n^2$.  In fact, $ T_n \equal{} \{ n^2 \minus{} (n\minus{}1), n^2 \minus{} (n\minus{}2), ... n^2 \minus{} 1, n^2, n^2 \plus{} 1, ... n^2 \plus{} (n\minus{}2), n^2 \plus{} (n\minus{}1) \}$.  Then $ S_n \equal{} n^3$. [/hide]
[hide="Method 2"] Let $ R_n \equal{} \sum_{k\equal{}1}^{n} S_k$.  The largest element of $ T_n$ is $ n^2 \plus{} n \minus{} 1$ by induction (or inspection).  The sum of the first $ k$ odd numbers is $ k^2$, so $ R_n \equal{} \left( \frac{n(n\plus{}1)}{2} \right)^2 \equal{} \sum_{k\equal{}1}^{n} k^3$. [/hide]
[hide="Method 3"] $ S_n$ is obviously a polynomial in $ n$.  Interpolate it.  (This is the worst method.) [/hide]

[quote="Silverfalcon"]The set $ T_n$ is defined as following groups of numbers:

$ T_1 \equal{} \{1\}$
$ T_2 \equal{} \{3,5\}$
$ T_3 \equal{} \{7,9,11\}$
$ T_4 \equal{} \{13,15,17,19\}$

and so on. Let $ S_n$ be the sum of the numbers within each set $ T_n$. For example:

$ S_1 \equal{} 1, S_2 \equal{} 3 \plus{} 5 \equal{} 8$

A) Find the sum of the numbers in the 2008th group.
B) Derive the formula for $ S_n$ and prove this formula.[/quote]

[hide="kinda fast"]
we know the sum of the first n odd integers is $ n^2$
so we can say
$ S_1 \equal{} 1^2$
$ S_2 \equal{} 3^2 \minus{} S_1 \equal{} 3^2 \minus{} 1^2$
...
$ S_n \equal{} (\frac {n(n \plus{} 1)}{2})^2 \minus{} (\frac {n(n \minus{} 1)}{2})^2$
$ S_n \equal{} \frac {4n^3}{4} \equal{} n^3$

[/hide]

meh, this is just a different ending to t0ra's 2nd method now that i inspect it


Hi,

I am in dire need of your help. I can't seem to solve this problem:

[quote]While it is a common knowledge that Martians are green and have seven heads, few people realize that a Martian has eleven eyes. At a recent party, while I was conversing with a Martian, I innocently remarked that if one-half of the Martians at the party suddenly became Earthlings (while the other half remained Martians), there would be 295 eyes at the party. Somewhat offended, the Martian noted that if one-half of the Earthlings at the party became Martians (while the other half stayed Earthlings), there would be 410 heads at the party. How many Martians and how many Earthlings were at the party?[/quote]

Here's my trial on this:
7h+11e= martians

1/2(7h+11e) +1/2=295 eyes
1/2 + 1/2(7h+11e)= 410 heads
Then do the subtraction method from the linear equation to find one variable and go from there. The thing is I don't think my two equations are right because if I do that then all of the stuff on the left side will be 0 which is justly wrong.

CAN ANYONE HELP ME GET THOSE TWO EQUATIONS PLEASE? I know that 7h+11e= martians though. I just don't know how to go further.

Every little help counts. Please.

----------------------------------------------------------------
ANOTHER QUESTION: (that I need help on)

[quote]For years George and Martha Huber supplemented their modest income by
blackmailing their neighbours. Neither of them made very much money at this until
two years ago, when Martha found out about Susan Black’s sordid past. That year,
George and Martha together extorted $ 108,000. The following year Revenue
Canada found out about Martha, so although she took in the same amount of
money, she had to pay half of it in taxes. Meanwhile, George's activities went
unnoticed, and he took in$3000 more than he had the year before. Together they
netted $ 65,000. If Susan was paying Martha$75,000, how much Martha was
getting from other neighbours?[/quote]

My tackle on this one:
G= George's share of the money
M= Martha's share of the money

(3000+G) + (1/2M)=108,000
G+M= 65,000

I don't know if those equations are right. PLEASE HELP!

LaTeX tip:  A dollar sign must be typed as [code]$\$x$[/code]where x is the dollar amount.
Example:  $ \$34$

ANOTHER QUESTION: (that I need help on)

For years George and Martha Huber supplemented their modest income by
blackmailing their neighbours. Neither of them made very much money at this until
two years ago, when Martha found out about Susan Black’s sordid past. That year,
George and Martha together extorted 108,000. The following year Revenue Canada found out about Martha, so although she took in the same amount of money, she had to pay half of it in taxes. Meanwhile, George’s activities went unnoticed, and he took in 3000 more than he had the year before. Together they
netted 65,000. If Susan was paying Martha 75,000, how much Martha was
getting from other neighbours?

108000=75000+x+g
65000=37500+g+3000+(x/2)

We see that g=33000-x=24500-(x/2)
So x=17000 where g is george's first income and x other neighbours' income.


superior algebra
superior algebra solved

Prove that the Quaternian group is not isomorphic to the dihedral group D4

hmmm i havent thought much about this, since i have a class now but i guess you can prove it by considering how many elements have order 2 in each of those groups


For all positive real numbers $ a_1,a_2,a_3,..,a_n$ prove that,
$ n\sum^n_{k\equal{}1}a_k^2\ge \left(\sum^n_{k\equal{}1}a_k\right)^2$.

since $ n\in\,IN$ then :  $ n\equal{}1\plus{}1\plus{}.......\plus{}1$
by caushy :
$ (1\plus{}1\plus{}1...\plus{}1)\sum^n_{k\equal{}1}a_k^2\ge \left(\sum^n_{k\equal{}1}a_k\right)^2$


linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

Let $ S_3$ be a symmetric group of degree 3 and $ \mathbb{C}[S_3]$ be a group algebra over a complex field of characteristic 0.

I'd like to decompose this semisimple group algebra into irreducibles using a Maschke's theorem such that

$ \mathbb{C}[S_3] \equal{} \bigoplus_i k_iS_i$

where each $ S_i$ is a simple submodule of $ \mathbb{C}[S_3]$.

I tried to use conjugacy classes as irreducibles, but it turned out that conjugacy classes are not submodules for this algebra.

Any help will be appreciated.

[quote="t0rajir0u"]Use the fact that  $ S_3 \simeq D_3$.  That automatically gets you an irreducible representation of degree $ 2$, and you also have the sign and trivial representations.[/quote]

I've already found three representations (#of irreducible representations correspond to #of conjugacy classes in $ S_n$) and character tables for these representations.

What I am having trouble with is to decompose a semisimple algebra $ \mathbb{C}[S_3]$ into the following form

$ \mathbb{C}[S_3] \equal{} \bigoplus_i k_iS_i$

where each $ S_i$ is a simple submodule of $ \mathbb{C}[S_3]$.

Any idea?

Thanks

[quote="JackSchmidt"][quote="algtop"]Let $ S_3$ be a symmetric group of degree 3 and $ \mathbb{C}[S_3]$ be a group algebra over a complex field of characteristic 0.

I'd like to decompose this semisimple group algebra into irreducibles using a Maschke's theorem[/quote]

Decomposing it into simple two-sided ideals is well-defined.  Decomposing it into simple one-sided ideals has multiple answers, none of which is particularly canonical.

Each of the simple two-sided ideals (eχ) is easily expressed using the character table:
\[ e_\chi \equal{} \frac {\chi(1)}{|G|} \sum_{g \in G} \chi(g^{ \minus{} 1}) g\]
Let e1 = ( () + (123) + (132) + (12) + (23) + (13) )/6, here χ is the trivial character.

Let e2 = ( () + (123) + (132) - (12) - (23) - (13) )/6, here χ is the sign character.

Let e3 = ( 2() - (123) - (132) )/3, here χ is the natural character.

Then the group ring decomposes as the sum of the cyclic two-sided ideals (e1) ⊕ (e2) ⊕ (e3).  The first two are one-dimensional over the field with bases {e1} and {e2}.  The third is four-dimensional over the field and is isomorphic to a 2×2 matrix ring.  The third has no canonical decomposition into simple submodules, but it has lots of noncanonical decompositions, each into two simple submodules, each of dimension two over the field.  One such decomposition is into the two submodules with bases { (12)+(123)-(13)-(132), ()+(13)-(12)-(123) } and { ()+(12)-(1,2,3)-(1,3), (2,3)-(1,2)+(1,2,3)-(1,3,2) }.[/quote]

Thanks a lot for your help. This is exactly what I was looking for. Anyhow, I am not able to follow this formula

\[ e_\chi \equal{} \frac {\chi(1)}{|G|} \sum_{g \in G} \chi(g^{ \minus{} 1}) g\]

I tried to find this in my group theory book, but it is not included.  I'd like to know how this formula is derived from. Any textbook or web material for this one?

Thanks again.

[quote="algtop"]I am not able to follow this formula
\[ e_\chi \equal{} \frac {\chi(1)}{|G|} \sum_{g \in G} \chi(g^{ \minus{} 1}) g\]
I tried to find this in my group theory book, but it is not included.  I'd like to know how this formula is derived from. Any textbook or web material for this one?

Thanks again.[/quote]

It is in Isaacs's Character Theory textbook, theorem 2.12 page 19.

It is in Littlewood's Theory of Group Characters, section 4.4 page 56.

It is in Huppert's Character Theory, section 4, page 55.

It is in Curtis-Reiner (1962), section 33, especially Theorem 33.8 page 236.


MATHCOUNTS is highlighted this month on LoveToKnow.com, a website committed to giving to charities. For the month of August, LoveToKnow will give a penny per hundred page views to the featured charity. So far over $ \$$950 has been raised!

Read the MATHCOUNTS interview [url=http://charity.lovetoknow.com/MATHCOUNTS]here[/url], and don't forget to visit [url=http://www.lovetoknow.com/LoveToKnow-Donations.html]Charity.LoveToKnow.com[/url] and support MATHCOUNTS!

Advertising is not what Aops is for. If you want to advertise, go to the Fun Factory.(maybe it doesn't even go there  :( )

But you know it's for a good cause and it is technically math related, so does it really matter?

Besides, I pressed ctrl+r waaaay too many times to let this go to waste :P

And maybe if I press it enough the mathcounts people will give me a ticket to nats.... hehehehehe

You can run a script to make it automatically refresh.  If you don't know how to do that,

1. Download Opera.
2. Go [url=http://www.lovetoknow.com/LoveToKnow-Donations.html]here[/url] using the Opera web browser.
3. Right-click and scroll down to "ReloadEvery".  Click "Custom", and set the timer to "1 second".  Click "OK" and let it run.

This refreshes the site every second, which gives MATHCOUNTS 6 cents every 10 minutes you use it.

Or you can download an auto-clicker by googling it (There's lots of sites where you can download it). Then keep it at 1 second and keep it on the refresh button.

Can't you just hold down F5? It refreshes the page more like 5 times a second.  :lol:

Yes, but then you need to be at the computer. You can do AIME15 or my strategy when you are eating, sleeping, etc.

Well, just get a heavy pebble and put it on the key! If you can find one that will stay and have enough pressure to push down, that is...How bout taping it down?
 :D

Wouldn't that Love.To.Know website have an alogrithm that makes sure the same IP address doesn't visit that website more than a certain number of times per hour/day, so the auto-clicker wouldn't help?

But its kinda hard to tell since every time you refresh the page, the counter goes up like 8 people  :P

So does this end in a few days, at the end of the month?

It ends at the end of the month, which is in a few days.

Hmmm last day, people. I have my back up keyboard ready incase F5 breaks.
Do it for mathcounts!

[quote="AceOfDiamonds"]...I have my back up keyboard ready incase F5 breaks...
[/quote]
My F5 already broke  :oops:

Wait do they atop giving money after at 12 million views?

If so then I think mathcounts is about to be maxed.


geometry
rectangle
calculus
integration
geometry proposed

Line t is tangent to two unit circles centered at A and B.  R1 and R2 are bounded by the circles and rectangle as shown in the following diagram.  What must AB be in order for R1=R2?
(The answer is pi/2.)

Sorry; I am having difficulties.  I hope to be able to show the diagram soon.

Here:

I have been able to prove the solution, but it uses calclulus:
I reduced the problem to one circle by splitting each region into two regions of equal area.  (See newly defined regions R1 and R2 in diagram.)

I let x be t/2.

So,
R1=$\displaystyle\int^x_0 (1-\sqrt{1-u^2})du = \left[u-\frac{u\sqrt{1-u^2}+arcsin(u)}{2}\right]^x_0 = \left(0\right)-\left(x-\frac{x\sqrt{1-x^2}+arcsin(x)}{2}\right)$

And,

R2=$\displaystyle\int^1_x (\sqrt{1-u^2})du = \left[\frac{u\sqrt{1-u^2}+arcsin(u)}{2}\right]^1_x = \left(\frac{\pi/2}{2}\right)-\left(\frac{x\sqrt{1-x^2}+arcsin(x)}{2}\right)$

Setting these equal to each other and solving for x yields $x=\frac{\pi}{4}$

So, $t=\frac{\pi}{2}$


rotation
geometry
geometric transformation

what does configuration mean in particular inversion of configuration?i am a bit confused.
let $ X$ be a optically active compound having inverse configuration as $ Y$( also optically active) does that necessarily mean nboth rotate light in opposite direction

Configuration is the particular order in which the atoms are bonded in a molecule, from a tridimensional viewpoint. It is reported by using the R, S nomenclature. An inversion of configuration occurs when an R (S) configuration is transformed in an S (R) one (the most common example is an SN2 reaction).

now i don;t think an inversion of $ R$ into $ S$ is necessary because the leaving group may have a different priority as compared to the attacking group

You're right, of course.  The R, S nomenclature system is just an arbitrary invention of man.  The important point is that the group formed by the three substituents that remain constant have been flipped, just like the spokes of an umbrella that has been blown out.  It is not necessary that, upon inversion, you rotate light in the opposite direction (indeed, there is no general correlation at all between rotation from the original and final).

Pardesi, you have asked what configuration is and I tried to provide an explanation and a common example where configuration inversion is verified. I didn't say that inversion of configuration is always observed in SN2 reactions, and I know perfectly that it doesn't.

And yes, not only configuration inversion doesn't imply rotation of polarised light in opposite direction, but also there is no correlation between R and S and the direction of rotation.

actually that's what i wanted to know corre;ation between the +- and inversion.the thing underlying this was the experiment done by cahn ingold(i suppose) :maybe:


Point P lies outside of circle O, which has a radius=x. PC is a tangent to circle O. There is some point F inside circle O. A line is drawn from point P to F. This line intersects Circle O at B and E. PB=2x+1, BF = 3, and FE = x+1. Another point G lies outside of the circle. The line GF intersects the circle at points A and D. AF = x-1 and FD = 5.

Find the legnth of PC.

EDIT: Sorry for the triple post. There was a glitch in my computer.

This is a simple question....
we know the property
$PA*PB=PC*PD$
applying here
$FB*FE=FA*FD$
we get x = 4
now applying $PA*PB=PT^{2}$
we get $(2x+1)(x+4)=PC^{2}$
solving it we get$6\sqrt{2}$


Support geometry articles search calculus videos

Basically I want to know where intelligent people stand on the subject.  I drink, smoke weed, I've had opiates, smoked salvia, and tried magic mushrooms.  I don't think any of this has hindered my ability to do well in school.  I have read that marijuana does not kill brain cells and has never killed anyone (and after seeing people smoke it thousands of times, I believe it).  But my question is, is there an intelligent way to use drugs or is it always a foolish gamble?  What do you think and have you used any drugs?

I think the only intelligent way to use drugs is to help relieve pain, or treat sickness, with medicine-drugs.

Also, I don't think anyone should use drugs as in get them into your body unless a doctor says you should, and even then you would have to be careful.

There arguably is an intelligent way to use them, but not like how you use them. Most drugs have side effects, anyway.

And yes, I have used drugs, as in simple medicine when I was sick.

[quote="7h3.D3m0n.117"]Well, I think I heard this from my Family Studies class last year but isn't marijuana the "gateway" drug into others? I dunno it mite be crack or heroine or something but I'm pretty sure it's marijuana.

It might be true that marijuana doesn't kill, but it could lead to other drugs, such as cocaine, heroine, etc, that can really be dangerous to your body.[/quote]I think that the reason marijuana is considered a gateway drug is simply because through buying it illegally you meet dealers of other illegal drugs.  If marijuana was legal, it might not lead to anything else.  Marijuana is awesome for relieving nausea and can definitely increase the quality of someone's life who is on other medication that causes nausea.  In general though, is the use of marijuana (not on medical grounds) any worse than the use of cigarettes or the use of alcohol to get a buzz?  Some of the smartest people I know smoke weed.  Should it be illegal?

[quote="7h3.D3m0n.117"]It [i]is[/i] illegal.

Marijuana has all the cons cigarettes do. For example, marijuana damages the lungs more than cigarettes do.  However, added to this, you have a higher heart rate, and if you are in poor cardiovascular health, then you can die. Also, many marijuana smokers are also cigarette smokes, so the combined habit is not good.

The addictive quality of marijuana makes the user wanting more, in this case, high doses of marijuana may face hallucinations, delusions, and depersonalization.

The danger in marijuanna is that over addictiveness can lead to cancer of the head, neck, and lungs, which ultimately kills a person.[/quote]

I know it is illegal, but should it be?  There is no addictive quality other than that it is fun.  Your body can never become dependant.  Hallucinations from marijuana don't really go beyond interpreting a sound differently or not being able to make out a distant object.  NOBODY has full blown hallucinations from smoking weed.  It does not inhibit coordination and very few car accidents are attributed to it's use (FAR less than alcohol) --> http://youtube.com/watch?v=sLmHL7YznUI.  It doesn't react badly with other drugs for the most part.  If your argument is that smoking in general is the reason marijuana is unsafe (causes cancer), then what about marijuana brownies (or pills like marinol)?

Caffeine kills more people than marijuana.

Some tidbits of my opinion, which is subject to change:  :lol:

I heard that people who smoke tend to drink more coffee.

I don't see how smoking is fun, and will never try it.

What I think is that smoking is very redolent, but not in a good way. I don't like to be around people with a malodorous odor of stale smoke, anywhere.

I demand that all cigarette companies should just give it up and stop selling them (cigarettes), no matter what the cost. All they want is money, and they are causing many deaths each year, and making people start an addiction that could be fatal, and very disturbing. Not that this is going to happen.  :P

I also read in a magazine that movies containing smoking scenes have been seen by young children, and therefore cause more people to start smoking.

Even if people don't have hallucinations from smoking, it is not good for you, and no matter how fun it may be, it's much better not to.  :ninja:

My aunt has lung cancer. It is fatal, and some people don't think that this will happen. Only 10% survive lung cancer after being diagnosed, I believe. Also, your lungs turn blackish when you smoke, and second-hand smoking also makes the air more polluted, and could cause many effects in the future.

It's illegal because conservatives believe that there's some difference between pot and alcohol/tobacco/caffeine and because alcohol and tobacco companies don't want the competition.

I don't see any reason that marijuana is worse that the "big three" of legal drugs. Even psychedelic drugs aren't that bad (although a bit less is known about their long-term harmful effects). I've thought a fair amount about this and here are my ideas:

The basic principle upon which the rest of this is based is that the purpose of our legal system is to protect citizens from themselves. Our laws operate to prevent people from harming themselves and each other in situations where the victim didn't realize the harm being done to them. (Well, some laws also just generally keep order. But those are less relevant in this case.)

So what considerations are there for legalizing drugs? There would need to be laws keeping people who are high from harming themselves or others. But I don't think this is actually difficult:

1) Marijuana. I think if you took pretty much every alcohol-related law and replaced the word "alcohol" with "marijuana", this would work. You'd have laws against driving stoned, regulations about how old you have to be to buy, not doing it in public, etc. (Not allowing it to be sold on Sundays or at night is kinda odd, though, so I don't know if I would recommend translating those laws over.)

The only major obstacle is that it is not nearly as easy for a cop to test someone for marijuana as for alcohol. I think you need to do a blood test. People might be very resistant to the idea that a cop can pull you over and stick you with a syringe to take a blood sample if you are driving crazily but aren't drunk. But there needs to be a way to enforce the stoned-driving laws.

2) Psychedelics. These are much more dangerous because you often hallucinate wildly when you are taking these. My suggestion would be to only allow usage of legal psychedelics in licensed establishments. The establishment would be responsible for keeping the high patrons under control. You wouldn't really need lots of detailed laws, because the 'psychedelic bars' would be responsible for keeping people there until they were functioning normally again.

So you could basically have very strict blanket laws against being under the influence of psychedelic drugs outside of a licensed establishment.

Again, there's some question of how to enforce this, though. Obviously if someone is caught doing it, that's easy. But suppose someone is coming off of a shroom trip and decides to drive out for some food before he's completely back to normal. And he ends up doing something crazy and a cop pulls him over. How does the cop test him for being under the influence? Are we allowing the blood tests?

----------

Now, there's also a health issue. Is it consistent with the aims of our legal system to allow something that's potentially harmful? Well, obviously we allow people to do tons of stuff that's potentially harmful, so the answer is yes. But we would have to have some safety requirements. I don't think this is really necessary for pot -- just put Surgeon General's Warnings on the labels of everything. But for psychedelic drugs, you could require people to get a permit to do them, and in order to get the permit you would have to take a short course on the effects of psychedelic drugs.

I had a discussion about this with my father, and he commented that when a person does something that's harmful to his health, he's actually harming everyone who has health insurance, because the insurance company has to raise its premiums in order to continue profiting even though its customers keep needing medical attention for drug-related damage. He thought that allowing more drugs to be legal will increase the amount of drug-related health issues among Americans, which will worsen our already-disastrous health insurance situation.

My idea to approach this (which might also find significant opposition among ACLU-types, although I'd consider it a small concession to make in order to legalize certain drugs) is to allow insurance companies to charge different rates according to whether someone does the newly legalized drugs. That way, the users are the ones who will have to pay. * This flies in the face of what we usually do, which is forbidding insurance companies from charging people different rates based on their lifestyle and suchlike. But it would probably get insurance companies to support your legalization bill [maybe it won't affect their profits hugely, but it's generally better for them to profile people as much as possible, and this precedent would be good news for them in the long run].

Minor side-notes: legal drugs probably won't be cut with harmful substances. So there'd be less health risk in taking drugs purchased legitimately. Additionally, the degree of supervision afforded by legalization and regulation could potentially reduce overdosing.

----

One last issue: "gateway drugs". Would legalizing certain drugs increase the usage of drugs that remain illegal (and are generally the more dangerous ones)? I don't think this would happen. I think a lot of people just want to do drugs as a social activity and for the experience and feelings, and if they could do it with some drugs (like LSD and shrooms and salvia, which all have pretty powerful trips as far as I know), then they probably won't feel the need to try stuff like crack and heroin and meth. Also, if someone decides to do illegal drugs, they might have a thought like this: "Well, I've already broken tons of drug laws, why not try xxxxx too?" Whereas a legal shroom user would not be able to use that sort of argument to justify trying something like meth.

----

Well, I've already written a lot and I have quite a bit of work to do, so I'm going to stop here. But I'd be happy to answer any other objections that people have; most likely, I've already thought about it.

edit: I should answer the rest of the OP's question. I've had plenty of caffeine, tried a puff of a cigar once I think (my dad was smoking one) but didn't like it, and I've had alcohol a few times (almost always legally, i.e. from my parents). If shrooms were legal I'd try them without any hesitation.

[quote="Lazarus"]@Xevarion: I believe salvia is legal in quite a few states. Not quite sure about this, though.[/quote]
Salvia is legal. But right now drug law is moving in the opposite direction to what would be desirable. As people figure out new ways to make themselves hallucinate or otherwise have fun/feel good, the new substances get added to their watch list and then eventually they become controlled substances.
http://www.deadiversion.usdoj.gov/drugs_concern/index.html

I would prefer to have a legal system under which drugs which induce mind-altering states are controlled substances by assumption, and there was a systematic way to petition for legalization. You'd just provide a proposal with scientific studies of the effects of the drug, and some committee would decide what legal status to give it.

This sort of system would inhibit "designer drugs". People create new drugs in labs because their old ones keep getting made illegal, and the new drugs are untested and usually extremely harmful to the health and often highly addictive. But if they're different enough from existing drugs, then they're legal for a little while until the government finds out. It would be much better if those were illegal from the start and only if someone created a drug and proved that it wasn't too harmful and wasn't physically addictive would it be considered for legalization.

In order to keep drug research going, you'd have to allow certain kinds of controlled experimentation with drugs, including illegal ones. I don't know how this works right now (like with neurologists giving psilocybin to rats and stuff like this), but presumably the laws and infrastructure are already in place for this.

Commercial incentive would stimulate drug research if there was hope of new drugs being legalized (instead of new drugs being legal until discovered and then illegal; very little hope of long-term legal commercial success there. )

http://youtube.com/watch?v=sLmHL7YznUI Driving + weed isn't a huge concern.

I don't recommend doing salvia much.  Salvia is absolutely ridiculous, more intense than any illegal drug I know of (luckily the effects don't last long).  It also doesn't seem like it can be too healthy to hold smoke in for 25-30 seconds.  It seems like the legal drugs tend to be the most harmful to your health for the most part.

I think there's nothing wrong with drugs. Moralists make a big deal out of it. Moralists like to create some crap rules and enforce it on the mass. In the past moralists say sex is sin, make it a rule and enforce it on everyone. In today's world, they say drugs is illegal, make it a rule and enforce it on everyone. Can these moralists tell us why sex is sin and why drugs is illegal? No, they don't have any good reasons for their dumb rules, they just make up a dumb rule without reason, treat it as universal truth and force everyone to conform.

And the sad thing is, the mass population accept their dumb rules without questioning.

[quote="mathomorphism"]Moralists like to create some crap rules and enforce it on the mass. In the past moralists say sex is sin, make it a rule and enforce it on everyone. In today's world, they say drugs is illegal, make it a rule and enforce it on everyone. Can these moralists tell us why sex is sin and why drugs is illegal? No, they don't have any good reasons for their dumb rules, they just make up a dumb rule without reason, treat it as universal truth and force everyone to conform.[/quote]
Don't be too critical. They do have reasons. However, Kahlil Gibran wrote

"But what of those to whom life is not an ocean, and man-made laws are not sand-towers, but to whom life is a rock, and the law a chisel with which they would carve in it their own likeness? What of the cripple who hates dancers? What of the ox who loves his yoke and deems the elk and deer of the forest stray and vagrant things? What of the old serpent who cannot shed his skin, and calls all others naked and shameless? And of him who comes early to the wedding-feast, and when over-fed and tired goes his way saying that all feasts are violation and all feasters lawbreakers?

What shall I say of these save that they too stand in the sunlight, but with their backs to the sun? They see only their shadows, and their shadows are their laws. [...] But you who walk facing the sun, what images drawn on the earth can hold you? You who travel with the wind, what weather-vane shall direct your course? What man's law shall bind you if you break your yoke but upon no man's prison door? What laws shall you fear if you dance but stumble against no man's iron chains? And who is he that shall bring you to judgment if you tear off your garment yet leave it in no man's path?"

That's not to say I recommend unrestricted drug use. Be responsible. ;)

[quote="mathomorphism"]How is unrestricted drug use unresponsible? Can you enlighten me?[/quote]
Well, I mentioned some reasons in my long post above.

[b]1) Negative health effects on others.[/b] If drug use is unrestricted, then it is extremely likely that people who are high or tripping on psychedelics or something will try to drive, wander around in public places, etc. If you're in an altered state of mind like that, you may be a danger to those around you. Of course, it's very irresponsible to do psychedelics without supervision so that you might in your trip get in your car and try to drive somewhere, but arguing that people don't do irresponsible things will not get you anywhere. Some drugs make people much more aggressive (increased risk of assaulting people if they are in public), sexually aroused (risk of sexual assault), or disoriented and paranoid (risk of mistaken attempt at self-defense).

[b]2) Negative health effects on yourself.[/b] You probably want to say that what you do to your body is your own business and you can cause it long-term harm for short-term benefits if you want. But the truth is that in our current health-care system, other people will be paying for your hedonism. If you are a meth addict and you have to have expensive dental operations to fix your ruined teeth, then the insurance company is hurt. In response, they'll raise their premiums, so all of their customers have to pay more. If lots of people do meth, the expected cost of a typical customer's dental bills will go up, and dental insurance costs more even for non-meth users. I don't want to pay to repair the bodies of other people who don't feel the need to take care of themselves.

Now, if you read a few posts above, you'll see my suggestions for how to greatly increase drug-related freedoms without really introducing much of the harmful effects of (1) and (2) in this post.

[quote="mgao"]illegal things aren't always bad
the quick temporary pleasure isn't bad for your life, it's just the chemicals.[/quote]

Oh, but it is!

The "quick temporary pleasure" derives from the drugs' blocking of Dopamine reabsorption in synapses. Dopamine, when it binds with the receptor, triggers a pleasure release. The dopamine molecule is then re-released into the synapse where it can be reabsorbed by the transmitting nerve, but this process is blocked by most drugs. As a result, the Dopamine is free to continue "striking" receptors causing signals to be sent. This eventually leads to the brain compensating by removing the receptors, so that once the drugs are gone the dopamine levels are back to normal, but the receptors are still gone so you are getting less pleasure which leads to addiction!

Bottom line is this: An artificial high will be compensated by a long, drawn out "crash" which is your body "making up" for the "high" it should not have had, which ruins your life and causes addiction because now the same dose of drug will just bring you to "normal," you must take more and more for it to happen.

This only is caused by certain drugs most notably nicotine, cocaine, heroin, and other opioids. Cannabis (delta-THC) is notably absent, as is alcohol.

here is a list of things that are not really acceptable in serious discussion threads:

1)

[quote="1=2"][hide="drinking"][/hide]
[hide="drugs"][/hide][/quote]

unaccompanied youtube videos, no matter how funny/how much of a unique snowflake you think you are.

p.s. if you agree with this rule in principle but think that your link is just funny enough to merit an exception, you are wrong.

2)

[quote="theprodigy"]@1=2:is that chart real???[/quote]

stupid posts about stupid youtube posts.  you'd think that if you'd already had posts deleted in this thread that you would try to post better!

3)

[quote="Gza"]Physical addiction isn't forever.[/quote]

cryptic (almost) non-sequiturs.

as an aside, most of the 'reasoned' posts in this thread have expressed fairly liberal views on marijuana so complaining of persecution for your illegal drug use is pretty silly.  as for being criticized by middle schoolers with no sort of real life experience, well, you can't really do a lot about that.

the drugs that r illegal r illegal 4 a reason
if they r illegal then dont get into them, they could probably cause more problems then good

PS. this an an exerpt from the book [u]Stupid History[/u] by Leland Gregory published by Andrew McMeel Publishing, LLC in 2007

"Until 1937, consumption and sale of cannabis (as hemp for clothing and rope and in cigarette form) was made illegal without the purchase of a federal tax stamp. Congress didn't think it had the authority to outlaw a plant, so they imposed a hefty tax instead. This reduced the amount of trade, but apparently not enough. To legally make marijuana illegal, Congress simply stopped isssuing the tax stamp, and without a tax stamp, it was illegal to buy or sell cannabis."

Uncle Sam... god dam[b]n[/b]

This site [ [url]http://www.abovetheinfluence.com/facts/mythbusters.aspx[/url] ] seems to take the truth and mesh it together with extremely rare cases and use convenient statistics to scare people into believing a skewed truth.  Weed laced with cocaine, PCP, or formaldehyde?  What dealer has the money to throw away expensive drugs on unsuspecting marijuana purchasers?  Formaldehyde is never used for anything that has to do with getting high.  It's just a nickname some people use for PCP... I'm pretty sure formaldehyde would be poisonous.

[quote="JBL"][quote="unlimitd3K"]This only is caused by certain drugs most notably nicotine, cocaine, heroin, and other opioids. Cannabis (delta-THC) is notably absent, as is alcohol.[/quote]  Or in other words, your comment has nothing to do with the subject of this thread  :wink:   (Also, comments like mgao's probably aren't worth responding to, in general.)[/quote]

I just had to make the point that chemicals lead to addiction through creating "highs"

Drugs are bad, mmmk?

[quote="unlimitd3K"]I just had to [/quote]  Actually, you didn't.  (And that was my point.)  In fact, many posters could benefit from stopping, thinking, and not making low-quality posts.  For example, your assertion
[quote="unlimitd3K"]Drugs are bad, mmmk?[/quote] adds nothing at all to this discussion, so why bother making it?  (That is a rhetorical question.)

[quote="RussianRocket"]I think it is strange that marijuana is twice as strong as it was 20 years ago.[/quote]
I don't really trust that site for accurate information.  If that is true, it is probably just because marijuana use is more wide-spread now than it was and knowledge of how to grow it correctly has spread to a much wider audience via the internet.

I should probably add something cool and interesting to this thread that nerds will appreciate.  This is marijuana under something like 70x and 200x magnification respectively taken by your's truly (I hope this isn't inappropriate):
[hide="70x"][img]http://img360.imageshack.us/img360/9605/micropic70es4.jpg[/img][/hide]
[hide="200x"][img]http://img117.imageshack.us/img117/8958/micropic66bn2.jpg[/img][/hide]

Isn't it though?

I also snapped these of a different strain (I believe):
[hide="50x"][img]http://img166.imageshack.us/img166/1742/micropic59xj9.jpg[/img][/hide]
[hide="200x"][img]http://img530.imageshack.us/img530/6560/micropic39bv0.jpg[/img][/hide]

dude, what the hell. how does posting pictures of your weed at random resolutions have any relevance to our thoughts on its usage. and in general, try not to insult people by calling them nerds or whatever you may have in mind. it's kind of rude.

I know the pictures have no relevance to the topic at hand.  I just thought they were interesting pictures that you might enjoy since it's not too easy to find such pictures and it's a particularly exotic looking plant...  and I didn't mean to insult anyone by calling people here nerds...  the whole thing about math aptitude + insanity as well as using the word nerd :/ I love math myself and would probably be considered a nerd by most.  I guess I made the assumption that the word 'nerd' has a positive connotation here.  Sorry.

I think this thread is well past its expiration date, so I'm going to give Gza the last word and lock it.


geometry
circumcircle
conics
ellipse
geometry proposed

I first define the Carnot point of a line $l$ with respect to $ABC$ as follow :
       1) Let $A', B', C'$ be the projections of $A, B, C$ on $l$.
       2) Then according to  Carnot 's theorem, the perpendiculars dropped from $A', B', C'$ to $BC, CA, AB$ respectively are concurrent at a point $K$ which is called the Carnot point of the line $l$ with respect to $ABC$
   Suppose $ABC$ is a non-issoceles triangle and $P$ is any point on the plane of $ABC$. Prove that for every side of $ABC$, there exist exactly two distinct lines that pass through $P$ such that their Carnot points with respect to $ABC$ lie on that side.
   I have a strong feeling that all the Carnot points of all the lines passing through $P$ with respect to $ABC$ form an ellipse. For example, when $P$ is the circumcenter of $ABC$, these points will trace the nine-point circle of $ABC$. Does anyone know this ?  :?

It seems wrong. It isn't an ellipse as I apply the Pascal 's theorem to these 6 intercepts and it doesn't hold.  :(

I 'm sorry. I drew a wrong figure. Indeed, it seems to be an ellipse. ;)

Dear Treegoner,
I think that you're right : the locus of the Carnot's points (or orthopoles) of the lines going through P is the ellipse centered at the midpoint of HP and going through the projections of P upon the sidelines of ABC. (H = orthocenter of ABC)
The case of exception is when P lies on the circumcircle, in which case the locus is the Simson line of P.
Kind regards. JPE

Dear jpe,
    Do you have a proof of it ? I have no idea how to prove it. By the way, as you mentioned, the projections of $P$ on $BC, CA, AB$ are 3 of 6 intercepts. The 3 other intercepts are the projections of $A', B', C'$ on $BC, CA, AB$, where $A', B', C'$ are the second intersections of the circles with diameter $PA, PB, PC$ and the circumcircle.


prove for natural numbers $\sqrt[n]{n}<1+\sqrt{\frac{2}{n}}$
my strange way [hide]let's show $\sqrt[n]{n}<\left(\sqrt{\frac{2}{n}}\right)^{-\frac{2}{n}}<1+\sqrt{\frac{2}{n}}$
the left inequality is obvious, the right $1<\left(\sqrt{\frac{2}{n}}\right)^{\frac{2}{n}}+\left(\sqrt{\frac{2}{n}}\right)^{\frac{\frac{1}{2}n+2}{n}}$
we need to show $\frac{1}{2}<\left(\sqrt{\frac{2}{n}}\right)^{\frac{2}{n}}$ which is easy [/hide] show me your ways  :)

[hide]
It's clearly true for $n=1$. For $n\geq 2$, it's equivalent to
\[n<\left( 1+\sqrt{\frac{2}{n}}\right)^{n}\]
The right hand side is equal to
\[\sum_{k=0}^{n}\binom{n}{k}\left( \frac{2}{n}\right)^\frac{k}{2}\geq 1+n\sqrt{\frac{2}{n}}+\frac{n(n-1)}{2}\cdot \frac{2}{n}=n+\sqrt{2n}>n\]
[/hide]


geometry
geometric transformation
homothety
geometry unsolved

$ \mathcal{L}_b , \mathcal{L}_c$ denote the given sidelines meeting at the known vertex $ A.$ Let $ X,Y,Z$ be the unknown tangency points of $ (I)$ with $ BC,CA,AB.$  $\mathcal{L}_b , \mathcal{L}_c$ and the Gergonne ray  $ AG_e$ are double under any direct homothety with center $A.$ Thus, draw an arbitrary incircle $ (I')$ tangent to $ \mathcal{L}_b, \mathcal{L}_c,$ the ray $ AG_e$ cuts $ (I')$ at the homologous $ X'$ of $ X$ under such homothety with center $ A$ that takes $ (I) \mapsto (I').$ Tangent of $ (I')$ at $ X'$ is  then parallel to $ BC$ and cuts $ \mathcal{L}_b , \mathcal{L}_c$ at the homologous $ B',C'$  of $ B,C.$ Gergonne point $ G_e'$ of $ \triangle AB'C'$ is constructible. Since this homothety maps $ G_e$ into $ G_e',$ then draw parallels trough $ G_e$ to $ B'G_e'$ and $ C'G_e'.$ These  meet $ \mathcal{L}_b$ and  $ \mathcal{L}_c$ at  $ Y,Z$ and $ B \equiv \mathcal{L}_c \cap YG_e,$ $C \equiv \mathcal{L}_b \cap ZG_e.$


[quote]Press Ctrl-F7, or click Build, then slide the mouse over Current File, then click on Build Output. We call this process compiling the document.

[/quote]

I dont see a build button anywhere and when i press ctrl-F7 nothing happens. could someone help me?

The build button in TeXnic Center is the button that looks like a stack of paper with an arrow over it.  You can also do "build and view" by clicking on the button that looks exactly like the build button but with a magnifying glass as well.


function
limit
real analysis
geometry
geometric transformation
real analysis unsolved

Proposition 23 in chapter 3 of Royden's Real Analysis states that if $ \langle f_{n} \rangle$ is a sequence of measurable functions on a measurable set $ E$ with finite measure, and $ f$ is a real-valued function such that for each $ x \in E$, we have $ f_{n}(x) \rightarrow f(x)$, then given $ \varepsilon,\delta > 0$, there is a measurable set $ A \subset E$ with $ mA < \delta$ and an integer $ N$ such that for $ x \notin A$ and $ n \geq N$ we have $ |f_{n}(x) \minus{} f(x)| < \varepsilon$.

Show that if the integer variable $ n$ is replaced by a real variable $ t$, then the proposition fails; that is, construct a family $ \langle f_{t} \rangle$ of measurable real-valued functions on $ [0,1]$ such that for each $ x$ we have $ \lim_{t \rightarrow 0} f_{t}(x) \equal{} 0$ but there exists $ \delta > 0$ such that $ m^{*}\{ x: f_{t}(x) > 1/2 \} > \delta$.

(Everything in this post refers to Lebesgue measure)

The following is the hint given in the book:

For $ 1/2^{i\plus{}1} \leq t < 1/2^{i}$, define $ f_{t}(x)\equal{}1$ if $ x \in P_{i}$ and $ x\equal{}2^{i\plus{}1}t\minus{}1$, and $ f_{t}(x)\equal{}0$ otherwise. Here $ P_{i}\equal{}P \overset{\circ}{\plus{}} r_{i}$ where $ \langle r_{i} \rangle$ is an enumeration of the rationals in $ [0,1)$ and $ \overset{\circ}{\plus{}}$ is translation modulo 1, and $ P$ is a nonmeasurable subset of $ [0,1]$.

I still cannot figure out how this example will work.


integration
AMC
AIME
AIME II
AIME I
number theory

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

[hide="Solution"]
It is clear that there aren't any two digit numbers that satisfy this, so let's try a three digit number N=abc=100a+10B+c.  To satisfy the given condition, we must have 100a+10b+c=290b+29c, or in other words 25a-7(10b+c)=0.  Clearly 10b+c must be a multiple of 5, so we have 10a-35k=0 implies 2a=7k, from which it is clear that a=7 is the smallest value to satisfy this.  Now we just have 10b+c=25, and it is clear that b=2 and c=5, so our number is 725.
[/hide]

Ew.

[hide="Solution"]
Call the integer with the digit removed $a$.  Then $a \equiv 29a \bmod 10^k$, where $k$ is the number of digits of $29a$.  This means $28a \equiv 0 \bmod 10^k$, so $a$ must be divisible by some powers of $5$.  Check powers of $5$.

$5 \times 29 = 145$ - too big.
$25 \times 29 = \boxed{725}$ - perfect.
[/hide]

Whoa zanttrang you actually did that? I just did

[hide]
$N: =ABC_{10}=100A+10B+C$ s.t. $29(10B+C)=100A+10B+C\Longleftrightarrow 28(10B + C) = 100A$, $\therefore 7|A \implies A=7\implies N=\boxed {725}$
[/hide]

Here is my solution. Not sure if this works but yeah.

[hide="Solution"]

Let $a$ be the leftmost digit NUMBER. Let $x$ be the rest of the number. So, our oringinal number is $a+x$. Now:

$\frac{a+x}{29} = x \rightarrow a+x = 29x$

So, $a = 28x = 4 \cdot 7 \cdot x$. Now, $a$ must have only one non-zero number. There is no way to make 7 disapper. But we can make 4 disappear by multiplying by power of 5. 5 gives 20 which doesn't work. 25 gives 100 which does work (since 7 x 100 = 700). So, $a = 28*25 = 700$. Then the original number is $700+25 = \fbox{725}$. [/hide]

[hide]
Since we all know the number is either a three or two digit number, we first look at the two digit numbers, as there are only three numbers which are multiples of 29. Trying 1, 2, and 3, we see that none yield us a number where we can drop the tens digit and maintain the same number (1 gives us 29, 2 gives us 58, and 3, 87)

So we realize the number has to be a three digit number. Call the hundreds digit, the digit being dropped $n$, where n is a [b]digit[/b], and the "leftover" two-digit number $k$. Thus the initial number can be represented as $100n + k$, and the resulting number is simply $k$

Now, all we have to do us translate the problem into a proportion.
$\frac{k}{100n+k}=\frac{1}{29}$

Simplify, to get $7k=25n$

Since the only value for which n and k are integral and n is a digit is $n=7, k=25$, giving us $\fbox{725}$
[/hide]

[quote="zanttrang"]Or you could just brute-force the problem, like I did  :lol: You know that the original number has to be a multiple of 29, so just run through all of the multiples of 29 until you hit $29 \cdot 25 = 725$, which works.[/quote]
Yup thats what i did only took like 10 minutes and i got it right  :lol:

lol...i guess i'm not the only loser that bruteforced it :P

Wow that looks nice and easy for an AIME question. I hope AIME II has some friendly ones like that, since I'm only taking the II. I took the integer to be ABC, then reduced an equation to 25A=70B+7C. Then, I needed 7C to end in a 0 or 5 and 7*5=35 was the lowest value that worked, from which 70B+35 had to be a multiple of 25. Hence B=2, A=7, and the answer is 725.

If you checked all the numbers, then to make your life easier, instead of checking each number, you should have at least realized that only time the units digit stays the same when multiplied by a 9 is 5. So, checking 5, 15 and 25 would have made more sense. I did it algebraically like one of the solutions above.

[hide="My solution"]Let's go the other way: Start with the smaller number. It has to end in 5, since if it ends in anything but 5 or 0, multiplying it by 29 doesn't preserve that last digit. If it ends in 0, you could just drop that and get a smaller number. Testing the odd multiples of 5, you get $5\cdot29=145$, $15\cdot29=435$, and $25\cdot29=\fbox{725}$.[/hide]

Wow people spent 10 minutes on this problem? It only took me <2 minutes. I'm tempted to say <1 minute, but I had to check my work too. ;) A lot of you guys are overcomplicating it. :roll:

[hide="My solution"]You know it's going to be where the $f(x)=x$ and $f(x)=29x-k$ intersect, where k is the equivalent to removing the first digit.
So..
$28x=k$
$k= 100, 200 , 300, 400, 500, 600, 700...$ (until you get an integer).
$700/28=25$

Answer is 25.[/hide]

Oh my god.  Um... I still like my solution the best, and I'm going to get rid of the guess-and-check element now too :)

[hide="New solution"]
Call the number with the digit removed $a$.  Then $a \equiv 29a \bmod 10^k \implies 28a \equiv 0 \bmod 10^k$ - now, $28a$ has to take the form of a digit multiplied by a power of $10$, and since $28 = 2^2 \times 7$ this can only occur if $28a = 700$.  Hence, $29a = \boxed{725}$.
[/hide]

Very similar to some of the things that other people have posted.

[hide="My solution"]
Assume it's a three-digit number (that's easy to figure out). Now let $x$ be the leftmost digit, and let $y$ be the rightmost two digits. Then we know that the following equation is true:
$100x + y = 29y$
Which can be rewritten as
$2^2 5^2 x = 2^2 7 y$
Clearly, a multiple of $7$ must appear on the left side, so the smallest possible value for $x$ is $7$. Likewise, a multiple of $5^2$ must appear on the right side, so the smallest possible value for $y$ is $5^2=25$. So the number is $725$.

Easy and brief.
[/hide]

Well I still think my solution is best for the following reasons:

-No guess-and-check
-Very quick (<1 minute)
-Trivializes the problem
-Makes the problem seem really easy
-Causes the problem to be easily solved

Err...yeah...

Oh and TheAmazingOne, $100x+y=29y \Longleftrightarrow 100x=28y$ and so it's basically the same solution as mine if you do that.

Our solutions are basically the same thing. The only difference is, in my solution I wrote out more words to explain things. And of course, on the AIME I wasn't actually writing out those words; I was only thinking them. So my solution likely took me just as long to do as yours.

Oh with that prime factorization thing you had going there, I thought you were doing something that overcomplicated it... sorry for misreading.

Let its leftmost digit be $x$ and the rest of the $n$ digits be $y$. Then, $y = \frac{1}{29} (10^n x +y ) \iff 2^2 \times 7y = 10^n x$. $7|x \implies x = 7 \implies n = 2  \text{ (Because the powers of two) }\implies y= 25$. The desired number is $725$.

misread the problem to mean decimal (as in, like a decimal 0.abcd... -> 0.bcd...), but it still gives an interesting problem:

[hide=Solution to misread]
The smallest number must start with $0.1\ldots$ or greater because if it started with $0.0$, the result after shifting would be larger, not smaller. Assuming that it is possible for such a number to start with $0.1$, let $x$ be the number minus $0.1$. Then, \begin{align*}
\frac{1}{29}(0.1+x) &= 10x\\
0.1 + x &= 290x\\
0.1 &= 289x\\
1 &= 2890x\\
\frac{1}{2890}&=x.\\
\end{align*}
Then the desired number is $0.1+\frac{1}{2890} = \frac{290}{2890}$.
[/hide]

[hide=Solution]
We can let our number be $100a+10b+c$ to get
\begin{align*}
100a+10b+c &= 29(10b+c) \\
100a+10b+c &= 290b+29c \\
100a &= 280b+28c \\
25a &= 70b+7c. \\
\end{align*}

We see that $25a$ must be a multiple of $7$ so $a$ must be a multiple of $7$.

We know that the only $1$ digit multiples of $7$ are $0$ and $7$.

If $a=0$, then $b=c=0$ but our number must be positive.

Therefore, $a=7$.

We substitute $a=7$ to get
\begin{align*}
25\cdot7 &= 70b+7c \\
25 &= 10b+c. \\
\end{align*}

We see that $b=2$ and $c=5$ is the only pair of $1$ digit numbers that satisfies this.

Therefore, our number is $100a+10b+c=100\cdot7+10\cdot2+1\cdot5=\boxed{725}$.
[/hide]

[hide=Quick Solution]
Notice that for all $4k+1$ multiples of 25, where $k\ge 0$ and an integer, removing the left most digit will always result in $\frac{1}{4k+1}$ of the number. So now we just have to check if $29$ makes $k$ an integer, and it does, so:

$$29\cdot 25 = \boxed{725}$$

clearly no two digit sols, so 100a + 10b + c = 29(10b + c), 100a = 28(10b + c), only 700 is a multiple of 28 so a = 7, 25 = 10b + c, we have 725

we have that $100a+10b+c=290b+29c$, hence, $280b+28c=100a$, note that $a=7,$ works, with $b=2, c=5$, hence, we have $\boxed{725},$ note that $c\equiv 0\pmod{5}$, so testing is easier.

Solved with [b]sarun[/b], [b]js104[/b], [b]crazyeyemoody[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b]

Let $N=10^cx+y$ where $x\leq9$. Then, $10^cx+y=29y$, so $10^cx=28y$. Therefore, $x=7$, $c=2$, and $y=25$, so $N=\boxed{725}$.


$ a$ , $ b$ , $ c$ are the lenghts of the  sides of   a   trinagle  , prove that :

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \le \frac{1}{a\plus{}b\minus{}c}\plus{}\frac{1}{c\plus{}a\minus{}b}\plus{}\frac{1}{b\plus{}c\minus{}a}$

[quote="y-a-s-s-i-n-e"]$ a$ , $ b$ , $ c$ are the lenghts of the  sides of   a   trinagle  , prove that :

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \le \frac {1}{a \plus{} b \minus{} c} \plus{} \frac {1}{c \plus{} a \minus{} b} \plus{} \frac {1}{b \plus{} c \minus{} a}$[/quote]
let $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} x \plus{} z$

the inequality now is equivalent to:

$ \frac {1}{2x} \plus{} \frac {1}{2y} \plus{} \frac {1}{2z} \ge \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{x \plus{} z}$

but we know that:

$ \frac {x \plus{} y}{4xy} \ge \frac {1}{x \plus{} y}$

$ \frac {y \plus{} z}{4yz} \ge \frac {1}{y \plus{} z}$

$ \frac {x \plus{} z}{4xz} \ge \frac {1}{x \plus{} z}$

and the result follows after suming the three inequalities  :)


Let $ ABCD$ be a convex quadrilateral with $ BA$ different from $ BC$ . Denote the incircles of triangles $ ABC$  and $ ADC$ by $ k_{1}$ and $ k_{2}$ respectively. Let $ k_{2}$ touch the side $ AC$ at $ E$. Suppose that there exists a circle $ k$ tangent to ray $ BA$  beyond $ A$ and to the ray $ BC$ beyond $ C$ , which is also tangent to the lines $ AD$ and $ CD$. Prove that the point $ E$ bisects the perimeter of triangle $ ABC$.

A bit easier. See the 6th paragraph of Jan's proof from [url]http://www.mathlinks.ro/viewtopic.php?p=1191735#1191735[/url] (Actually you will have to read a bit of the first part too to have a more clear view about his notations.)
[quote="Jan"]Using the degenerate form of Monge-d'Alembert on $ k_1,k,AC$, we obtain that $ B,B'',U$ are collinear, where $ B''$ is the antipode of $ B'$ in $ k_1$. Also, we can look at $ k_1,k_2,AC$, and obtain that $ U,D',B''$ are collinear. Combining these two results gives us that $ B,B'',D',U$ are collinear.[/quote]
He proves that $ BE$ (the $ E$ is from your notations) is the $ B$-Nagel cevian of $ ABC$. This proves your assertion.


LaTeX
induction
algebra unsolved
algebra

Let x(n+1)=ax(n)+bx(n-1)/2,x(0)=0,x(1)=1. Then x(n)=(q1^n-q2^n)/c,c=sqrt(a^2+2b),q1=(a+c)/2,q2=(a-c)/2. x(2k)=(c1^k-c2^k)/c,c1=q1^2=m+ac/2,c2=m-ac/2,m=(a^2+b)/2 -whole, when a and b odd (a=b=1). Therefore x(2k)=summ[C(k,2i+1)m^(k-2i-1)((ac/2)^2)^i. If y(n)=x(n)*2^[(n-1)/2], we have y(n) is whole for any n, and y(n) is odd if 4 is not divide n. If 4|n then ord2(y(n)=ord2(n)+ord2(m),m=(a^2+b)/2. Therefore x(n) is whole  for n=0,1,2,4,8, becouse x(n) is whole if only [(n-1)/2]<=ord2(n)+ord2(m). If m is odd for n>8 x(n) is not whole. If a=21,b=71 we have m=256, then x(n) is whole for n=0,1,2,4,8,12,16,20,24,32.

Proved
x(n) whole (n>2) <---> (4|n and n/2-1<=ord2(n*m), where m=(a^2+b/2),
for any sequence x(n+1)=ax(n)+bx(n-1)/2, a and b is odd numbers.

Rust,I see you are a very eager mathematician and you are very active at $\text{AOPS}$.Why dont you learn $\text{LATEX}$ and contribute to $\text{AOPS}$
more effectively? :?
Udaci tebe

[quote="nthd"]The answer is $n\in$ {$0,1,2,4,8$}[/quote][b]What?[/b]


If AB and BC are both tangent to circle O so that E lies on AB and D lies on BC and ED is tangent to circle O
what is the perimeter of triangle EBD given AB=20?

Is it safe to assume that A and C are the points of tangency and that B lies outside the circle?

Suppose that $ ED$ meets $ (O)$ at $ F$
The perimeter of $ \triangle{EBD}$ is:
$ BE\plus{}ED\plus{}BD\equal{}BE\plus{}EF\plus{}FD\plus{}BD\equal{}(BE\plus{}EF)\plus{}(BD\plus{}DF)\equal{}(BE\plus{}EA)\plus{}(BD\plus{}DC)\equal{}AB\plus{}BC\equal{}2AB\equal{}40$
[hide]
It's too basic that it should be moved to the basic topic :maybe: [/hide]

Then maybe you can do this problem for me, it's a lot more difficult.

Two quarter circles ABD and CBD are drawn into square ABCD. A new circle is inscribed between the two arcs of the quarter circles such that the circle's center is concurrent with the center of square ABCD. If AB=5, find the radius of this inscribed circle.

I did this question by using the quarter circle ABD's radius of 5 the subtracted it by half of AC's diagonal to get the inscribed circle's radius so $ 5\minus{}\frac{5\sqrt{2}}{2}$ or $ \frac{10\minus{}5\sqrt{2}}{2}$. Is there any other way of doing it?


Anyone have it? How is it, i'm probably getting one in a few weeks.

i played madden 06 on it and the graphics were amazing...its really worth gettin it

Why would I want an Xbox 360, when (at least for now) all of the good games (at least, the ones I want to play) are coming out on current systems?  I don't care how good games look, I'm more interested in how fun they are to play.

I heard that Microsoft loses money on the xbox.

400 dollars for a system whose games cost 60 is too much, no matter how good the graphics are.

I think it's too much too but i'm gettin mine for my birthday. So...yah I'm cool with that.

I'd much rather have the PS3. Halo 3 is the only thing I'm looking forward to on the Xbox 360.

PS3 looks great cept for the part where it's 500 dollars and 60 dollars per game  :what?:

PS3 looks nice, except for the controller, it looks very uncomfortable.

[quote="maniacman384"]with the xbox 360 controller u can turn the 360 on, open the disk drive and turn it off, so more being lazy. but the PS3 controller handles are mad long so that might be annoying. especially for little kids who can't hold the controller fully.[/quote]

Heh, then there's the [url=http://www.gamespot.com/news/6133335.html?q=Revolution%20controller]Revolution's controller[/url].

That'll either turn out to be very good or very, very bad.

I got a 360, and I love it. The games are great, and the graphics are awesome.

No. I only play Call of Duty 2 and some Full Auto online. I don't know why I'm not getting Halo 2.


This is more of an accounting (taxation) knowledge question. What I am going to ask, is going to probably sound like I am an idiot, but here it goes,

The individual income tax is currently now in 4 or 5 brackets, I cant remember which, and this is the federal income tax as I understand. The lowest bracket is 10%, but the percentage taken from every paycheck is only a couple of % for the federal tax, what is going on?

Like I said, I am not understanding something here, but I just had to ask.

Your first few thousand dollars of income are not federally taxed at all.  This means your effective rate of taxation is lower than the brackets suggest.

Additionally, somebody in, say, bracket 2, pays no income tax on their first few thousand, and only the bracket 1 rate on their next few thousand.  (At least, I'm pretty sure that's what I've been told.)


AMC
USA(J)MO
USAMO
AMC 8
email
geometry
AIME

You don't have to reveal your name (though it will eventually posted in their book and online) but does anyone know if any AoPS's earned a perfect score this year, and if so, what state(s) are you/they from?

I think Temperal got a 25...
Or so he says...
 :wink:
I accidentally skipped a problem, and I got another one wrong, so I think I got a 23...
Not sure though.

[quote="karatemagic7"]Hey, do you know the Average score on the AMC 8 is I think between a 9 and 10? (you can check AMC website for exact stats) - so you guys who are getting scores in the 20's are FAR from "stupid"!!!!![/quote]

Yes and they're probably making those who scored lower than them feel "stupid."

Only 4 ppl in MA for perfect scores: 2 male, 2 female. When will the perfect score section be released on the AMC web site anyway?

It is usually posted around January. If you got a perfect score and they haven't asked for your picture, don't expect them to put up the names for a while. Obviously, they won't be able to put up your picture if they don't have it.  :wink:

[quote="diophantient"][quote="xpmath"][quote="1=2"]I'm pretty sure that xpmath and hunter34 also got perfect.[/quote]
No. I messed up dividing, and was being stupid on another.
Oh well, I think hunter did get perfect.[/quote]
A 23 is a good score on the AMC 8, no worries. The last 8th grader I knew who got a 23 on the AMC 8 got a 9 on the USAMO the same year.[/quote]
that makes me feel allllooooot better ;-)

Um, the only three people who got perfect in my state were all in my class (one of them being me :lol: ).  Sorry, I'm really bad at modesty. :whistling:

I got a 24 because I messed up on #18 :(

argh! i got 22.   :mad:
i don't think i'll make aime because ive been doing amc 10s from previous years and the highest i've gotten is like a 105 or something.

do you know what the qualifying score is for aime? is it 120?

It's either the top 1% or above 120 (which ever is lower).

Last year, with the 2.5 to 1.5 change, the cutoffs were 117 and 115.5 for 10A and 10B,respectively.
I wonder if they'll change because of the calculator policy? I don't expect them to too much.

Ahhhh!!! Last year, I made one stupid mistake...I was thinking about spheres and cubes in the second dimension, so I messed up a calculation...ended up with 112.5...if I had gotten it right, I would've made the cutoff...

Anyways, this year I've learned a lot more about problem solving, and I've gotten a lot better, so I hope I do better.

[quote="xpmath"]
I wonder if they'll change because of the calculator policy? [/quote]

i HOPE so! :o

I say. This is getting quite the popular topic.

But really it looks like it didn't change.

wow, 2 people in my MC chapter got a perfect score. they are brothers. 3 people in PA got perfect. i am scared(should i be?)

No. The AMC 8 is trivial; even more so than Mathcounts.

I got 24.

If only I had filled in that other bubble...

I'm from HI and 2 other people in my algebra 1 class also got 24, one of which uses these forums as well.

The AMC8 was definitely not trivial for me--it was the first time I discovered that I had talent in math competitions/math itself!

I remember being sooo pleasantly surprised by that 24 and then 25! (Back then they didn't have pictures of the perfect scorers on the website though! You lucky young'uns...)

Remember, guys, the AMC8 is a stepping stone, not an end in itself. =]

I'm surprised I didn't post in this yet o.o. Maybe I did? -_-  I think I got a perfect... That's what the picture thing says... Hmmmm...


Wait, doesn't the first one diverge?  :maybe:

There's a sum for the second one but the last one diverges.

[quote="computer_nuke"][quote="Kalle"]How does this topic fit here? I highly doubt the second sum can be calculated here, since calculus is not generally known and rearrangement tricks don't bite on conditionally convergent series since you can't rearrange them.[/quote]

The second one isn't too tricky though,

You have $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}=\sum_{n=1}^{\infty}(\frac{1}{2n-1}-\frac{1}{2n})=\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)}$[/quote]

which equals.....


abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove that if E and F are fields, and E is an extension of F, then char(E) = char(F).

Let a and b be natural numbers.

I can easily prove it in cases where char(F) = a and char(E) = 0, or char(F) = a, char(E) = b, and a < b.

However, it is for the cases char(F) = 0 and char(E) = a, or char(F) = a, char(E) = b, and b < a that I am having a bit of trouble with.

I figure I'm just overlooking some basic fact about fields.  Any suggestions?

By definition, you have an injection $\iota: F \to E$ with $\iota(1)=1$. Note that the characteristic is defined as the (nonnegative) generator of the kernel of the homomorphism $\mathrm{char}_E: \mathbb{Z} \to E$, respectively $\mathrm{char}_F: \mathbb{Z} \to F$, given by $1 \mapsto 1$.
So you have to show that the kernel of $\mathrm{char}_E: \mathbb{Z} \to E$ and $\mathrm{char}_F = \iota \circ \mathrm{char}_E : \mathbb{Z} \to F$ are the same, which follows from injectivity.

If $H$ is a subgroup of a group $G$ and $h \in H$, then the order of $h$ in $H$ is equal to the order of $h \in G$. Apply this to the additive groups of $F,E$ and $h=1$, and you're done.

-oo-, are you saying something like this:

Let the additive order of 1 in F be n.  Then it must be that the addtive order of 1 in E is n.  Thus, 1*n = 0 in both E and F.

But I don't see why there can't be an element k in E such that k < n and 1*k = 0.

Zetax, I don't understand the last part of your post:

$\mathrm{char}_F = \iota \circ \mathrm{char}_E : \mathbb{Z} \to F$

What does the $\circ$ mean?  I've never seen that before.

$\mathrm{char}_F (n) = \iota( \mathrm{char}_E(n))$, just the concatenation of two functions.

@Ricky: Remember  the definition of the order of an element ...

Oy, two bad mistakes... I'm on a role.

I'm not sure why I didn't recongize that as composition.
And thanks -oo-.  I always seem to think to advanced and miss the most basic facts.


[quote="G-UNIT"]What is the sum of the first $N$ triangular numbers?

1, 3, 6, 10, 15, 21, 28......$N$

  A traingular number is expressed as $\frac{n(n+1)}{2}$[/quote]

I messed up somewhere here :x ...care to find the error?

[hide="not a solution"]
If you use the formula $\frac{n(n+1)}{2}$ (n = 1) and add up to the $t$th term, you get:

$\frac{n(n+1)+(n+1)(1+2)+(n+2)(n+3)+...(n+t)(n+(t+1))}{2}$

which can be put in the form

$\frac{(n^2+a_1n+b_1)+(n^2+a_2n+b_2)...(n^2+a_tn+b_t)}{2}$

Because in $(n+t)(n+(t+1)) = n^2+an+t$, a = $2t+1$ and b = $(t)(t+1)$, we need to find

$\sum_{i=1}^{t}=2t+1$ and $\sum_{i=1}^{t}=(t)(t+1)$

Since the sum of the first y integers is $\frac{y(y+1)}{2}$, $\sum_{i=1}^{t}=2t+1=t(t+1)+t=t(t+2)$

The second one can be expanded to get $t^2+t$.  The sum of the first y perfect squares is $\frac{y(y+1)(2y+1)}{6}$, so $\sum_{i=1}^{t}=(t)(t+1) = \frac{t(t+1)(2t+1)}{6}+\frac{t(t+1)}{2}$

So the sum is $\frac{t(t+1)n^2}{2}+t(t+2)n+(\frac{t(t+1)(2t+1)}{6}+\frac{t(t+1)}{2})$
Subsituting n=1 and simplyfing (sort of), $\frac{(6t)(t+1)+6t(t+2)+t(t+1)(2t+1)}{6}$
[/hide]

[hide]I can derive it out one way, but i'll do it the easy way  :D

The first few sums are 1, 4, 10, 20, 35. They look like n choose 3.

The formula is, actually, the sum of the first n triangular numbers is ${n+2\choose3}$[/hide]


Find the number of sequences [b]a1,a2,...,a(k+1)[/b]

of non-negative integers with the properties

[b]a1=0[/b] and [b]|a(i+1) - ai|=1[/b] for [b]i=1,2,...,k[/b]

hey hey, check out Dyke paths and Catalan numbers

i think your problem is wrong
the sequence is not unique!!!!

I suggest a possible approach, I dont have a solution.
Let $ f(n)$ be the number of sequences with n terms and $ n_k$ be the number of sequences with n terms and ending with $ k$.
\[ f(n)\equal{}n_0\plus{}n_1\plus{}\dots\plus{}n_{n\minus{}1}\]
We can construct a recursion as follows,
$ (n\plus{}1)_0\equal{}n_1$
$ (n\plus{}1)_1\equal{}n_0\plus{}n_2$
$ \dots$
$ (n\plus{}1)_k\equal{}(n\plus{}1)_{k\minus{}1}\plus{}(n\plus{}1)_{k\plus{}1}$
$ \dots$
$ (n\plus{}1)_{n\minus{}1}\equal{}n_{n\minus{}2}$
$ (n\plus{}1)_n\equal{}n_{n\minus{}1}$

We have  $ f(n\plus{}1)\equal{}(n\plus{}1)_0\plus{}\dots\plus{}(n\plus{}1)_n$
$ \Rightarrow f(n\plus{}1)\equal{}n_0\plus{}2(n_1\plus{}\dots\plus{}n_{n\minus{}1})$

How can we proceed from here ?

$ n_k$ is a pretty troublesome designation since you're also using $ n$.

In any case, the answer is given by [url=http://mathworld.wolfram.com/CatalanNumber.html]Catalan numbers[/url] as pleurestique noted.  :)

It's actually [i]not[/i] given by the Catalan numbers: if we required the sequence to have even length and the last term to be $ 0$, then it would have a really obvious bijection with Dyck paths.  (As a side note, a substantial fraction of the other questions hastings has posted [i]are[/i] just typical Catalan problems.)  If you'll calculate the first few values, you may find a sequence you recognize -- it's even a bit simpler than the Catalan numbers, in some sense.


Well a quite different one :
$\vartheta0=0,
\vartheta1=\frac{1}{2},
\vartheta2=\frac{\sqrt3}{2},
\vartheta3=1,
\vartheta4=\frac{\sqrt3}{2},
\vartheta5=\frac{1}{2},
\vartheta6=0$
and for a good help i add that :
$\vartheta7=-\frac{1}{2}$
Now Guess the
$\vartheta12=?$

Well ,basicly I am sorry for saying that statement ,it seems that you are a bit insulted by it :) .
I know this form is for middle school .i post those problem cause i by myself have seen them when i was in middle  school.
In my country,we have this kind of problems for our  middle school students in our books(I by myself was not to successful  in solving them  :) ),so i just thought it would be intersting for you guys too.

Hmm,about this new one (post33),
I have seen it when i was going to highschool(grade 8 to 9) so it shouldnt be called basic but it is
too interesting.
I will give some guidences to make in more sensable for here:
Well it has something to do with the $sin$.
To be exact $\vartheta N=sin(\mathcal l{k}n)$ so you guys just try too find the $\mathcal l{k}$
A better guide line is this that :
$sin(30)=\frac{1}{2},sin{60}=\frac{\sqrt3}{2},sin{90}=1$
Forgive me if this is not suitable for here :) .

PS.For using a adj about making the life a bit sad ,i think we should use
$bitter$ beside of $sour$,But i am not sure , we should ask a native speaker anyway.

OK I'm not argue with you about life bitter or sour. Look, you must read other problems and find level of forum. I did too and I post correct level. Maybe Iran have different system, in Macau they too have different system. But you must find out level of forum. $\sin$ is trigonometry, even as easy as $\sin 0=0$, trigonometer cannot be posted here, OK? I hope you understand me. I don't want to make this a big problem but if you must, I will.

Trig is not allowed in any of the middle school forums, not even mathcounts. Yeah, I learned the trig functions in 4th grade, but the majority of the students across the nation don't learn it until high school.

Trig belongs in the high schools forums and above.

The level of this sequence is well above the appropriate level for the Middle School Basics Forum.

I refer you to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=148301#p148301]very first post[/url] ever made to the Middle School Basics Forum.  [b]The MS Basics forum should only be used by middle school students and teachers.[/b]

I am somewhat confused by the problem. However, I would say that trig is well inside the capabilities of middle-schoolers who like math. It may not be Basics level, and it might not be absolutely necessary in Mathcounts, but it can help.

What's $\vartheta$? I've seen $\theta$, but never in cursive :D

I'm not absolutely sure, I think both are theta, but $\vartheta$ was how it was written first.

To write that in latex, you use \vartheta, I don't know what the "var" part is.

Sorry guys ,I didnt see your answers up to now (why?????)).
You asked what is $\vartheta$
$\vartheta$ is the Simbol of function .
Usually ,letter $f,g$ are used for functions but i wanted it to be a little Cool.
So $\vartheta\theta=\vartheta(\theta)=Sin(n\theta)$

Some thing cool:
When you want to use a letter ,for example $F$ ,put a """"\mathcal"""befor it and it
will turn to $\mathcal F$.

Some Example:
$F \longrightarrow \mathcal F$
$R \longrightarrow  \mathcal R$
$W \longrightarrow \mathcal W$ :D  :D  :D  :D

Actually, I believe that there were previous intructions to keep all trigonometry problems Intermediate and above. I'm not sure if that policy has changed...

The answer is: Anything. Because we can construct a sequence consists of:
\[0, \frac12, \frac{\sqrt{3}}{2}, 1, \frac{\sqrt{3}}{2}, \frac12, 0, -\frac12, 100, 100, 100, 100, 100 \cdots\]

He never said it was an arithmetic or geometric sequence. It could just be a random one. :P

[quote="236factorial"]Uh,, what's the rule in that?[/quote]

The rule is:

$a_n = sin\text{ }30n \text{ if } n < 8$

and

$a_n = 100 \text{ if } n \geq 8$


How would I determine the number of valence electrons from the configuration $ 1s^22s^22p^63s^23p^4$?

Thanks in advance!

Well, how many electrons are in the outer most shell? The question should not be hard unless you don't know what a valence electron is, or you don't know how to read an electron configuration.

Valence electrons are the electrons that occupy the level with the highest principal quantum number. In this case, that is the level n = 3 in which there are 6 electrons. These electrons are so called because they are responsible for most of the chemical properties of an element.

just see the last principle quantum no. in urs case it is n=3.
then go to the orbitals i.e 3s and 3p and just the electrons present in them.

the number of valence electrons are the electrons in the outermost energy level which is 3 in this case count.  count the number electrons in both the 3s and 3p to get the number of valence elctrons


quadratics
algebra
polynomial
Vieta
number theory unsolved
number theory

For which positive integers $ n$ does the equation
\[ w\plus{}x\plus{}y\plus{}z\equal{}n\sqrt{wxyz}\]
have a nonempty set of positive integer solutions (w,x,y,z).

For any natural n with a solution, take the solution whose sum is minimal: $ (x_{1},y_{1},z_{1},w_{1}) (x_{1}\plus{}y_{1}\plus{}z_{1}\plus{}w_{1}$ is minimal), WLOG $ x_{1} \ge y_{1} \ge z_{1} \ge w_{1}$.
We can look at the equation as a quadratic equation in x: $ x^{2}\plus{}x(2y\plus{}2w\plus{}2z\minus{}n^{2}ywz)\plus{}(y\plus{}w\plus{}z)^{2}\equal{}0$. If we put $ y\equal{}y_{1}, z\equal{}z_{1}, w\equal{}w_{1}$, we have the solution $ x\equal{}x_{1}$. By vieta, we get another natural solution: $ x_{1}\plus{}x_{2}\equal{}n^{2}y_{1}z_{1}w_{1}\minus{}2(y_{1}\plus{}z_{1}\plus{}w_{1}) \implies x_{2} \in Z$, $ x_{1}x_{2}\equal{}(y_{1}\plus{}z_{1}\plus{}w_{1})^{2} \implies x_{2}>0 \implies x_{2} \in N$.
because of the minimality of $ x_{1}$: $ x_{2} \ge x_{1} \ge y_{1}$, therefore, if we plug $ y_{1}$ in the quadratic equation, we get a non-negative value: $ 0 \le y_{1}^{2}\plus{}y_{1}(2y_{1}\plus{}2w_{1}\plus{}2z_{1}\minus{}n^{2}y_{1}w_{1}z_{1})\plus{}(y_{1}\plus{}w_{1}\plus{}z_{1})^{2} \le y_{1}^{2}\plus{}(6y_{1}\minus{}n^{2}y_{1}z_{1}w_{1})y_{1}\plus{}(3y_{1})^{2}\equal{}16y_{1}^{2}\minus{}n^{2}(y_{1}^{2}w_{1}z_{1}) \implies n^{2}w_{1}z_{1} \le 16 \implies n \le 4$. It's easy to find solutions for n=1,2,3,4: (1,1,1,1), (2,2,2,2), (4,4,4,4), (1,1,2,2)

Awesome. Thanks. I like the trick of putting the $ y_1<x_1,x_2$ and it therefore being positive in the quadratic. :)


What is a neat way to prove the following:

If $ b_n > 0$ is such that $ \sum b_n < \infty$. Then there are $ a_n > 0$ such that $ a_n \to \infty$ and $ \sum a_n b_n < \infty$

For each $ j \equal{} 1, 2, 3, \cdots$, there is an integer $ N_j$ such that $ \sum_{k\equal{}N_j}^{\infty} b_k < 3^{\minus{}N_j}$. We may choose $ N_j$ to be strictly increasing.

Define $ N_0 \equal{} 1$ and $ a_n \equal{} 2^{j}$ for $ N_{j} \leq n < N_{j\plus{}1}$ where $ j \equal{} 0, 1, 2, \cdots$. Then $ \sum_{n\equal{}1}^{\infty}a_n b_n \equal{} \sum_{j\equal{}0}^{\infty} \sum_{n\equal{}N_j}^{N_{j\plus{}1}\minus{}1} 2^j b_n \leq \sum_{n\equal{}1}^{\infty} b_n \plus{} \sum_{j\equal{}1}^{\infty} (2/3)^j < \infty.$

[quote="sos440"]For each $ j \equal{} 1, 2, 3, \cdots$, there is an integer $ N_j$ such that $ \sum_{k \equal{} N_j}^{\infty} b_k < 3^{ \minus{} N_j}$....[/quote]
I dont' understand this moment well. Suppose $ b_{k} \equal{} 3^{ \minus{} k}$ then $ \sum b_{k} < \infty$ and $ b_{k} > 0$
but then there is no such $ N_{j} \in \mathbb{N}$ when we have
$ \sum_{k\geq N_{j}} b_{k} \leq 3^{ \minus{} N_{j}}$ because $ \sum_{k\geq N_{j}} b_{k} \minus{} 3^{ \minus{} N_{j}} > 0$ fow all $ N_{j}$
 :maybe:

Oh, thanks for pointing out my my mistake. It have to be replaced by $ 3^{ \minus{} j}$.


calculus
integration
trigonometry
complex analysis
complex analysis unsolved

(2) The poles are $\pm \frac{\pi}2, \pm \frac{3 \pi}2$, all with residues $-1$, so the integral is $-8 \pi i$. At least I hope so :lol:

(4,5) I think that we should first make the substitution $y = \frac1{z}$ to get rid of the plethora of poles. But I don't know complex analysis (namely the change of variable theorems), so I'm not really sure.

Maybe if you write in detail, we can see better. :wink: