MATHCOUNTS
trigonometry
geometry
area of a triangle
Heron\u0027s formula

Great news for you AoPers out there.  Bobby Shen and Lilly Shen were two of four Texas students that were competing in the World Primary Mathematics Competition in Hong Kong.  As usual they won the award for the best non Asian team in the competition, but for the first time ever they tied for the World Championship,  Bobby Shen and Lilly Shen (both from First Colony MS in Sugar Land)  had perfect scores.  This was the first time in the History of the competition that a non-Asian team had won or tied for the championship!

What a great accomplishment for Bobby and Lilly who are both AoPers. :lol:  :coolspeak:  :!:

Huh? Is Lilly Shen Bobby's younger sister going to 7th grade this year?

Who is she on AOPS?

No, Lilly is in the same grade as Bobby but they are not related.  They were both on the the First Colony team that won the Mathcounts State title in Texas.  I will leave it to her to tell you her username if she chooses to.  She is one of the top ranked kids on Alcumus.

oh, i get it now...

it's great that TWO of your kids are top students in TX!!

Maybe the school u coach is really awesome for some reason... :P

[quote="isabella2296"]Wait - Bobby Shen has an account here? I didn't know that.[/quote]
Yeah, he does. All the Texas people do. (those on Mathcounts Nats)

[quote="AIME15"]I've heard of them....
If you know them, tell them congrats from AIME15
And I think it's kind of racist to call them a non-asian team...technically they're asians anyway...if theyre last name is shen[/quote]

They could be Asian-Americans, like me.

[quote="Urc"][quote="gauss1181"][quote="isabella2296"]I guess. But they are American citizens![/quote]
lol

then how do they define non asian teams?[/quote]

The team isn't based in Asia..?

I dont get how it's racist if they base it on where the team is from.

Anyway, congratulations to them both. I'm just wondering, was lilly shen in mathcounts?[/quote]
yeah but she didn't make nats

Oh, okay

nd I just noticed it said Primary 1-6, yet Bobby is in 7th grade.. :huh:

[quote="Urc"]I just noticed it said Primary 1-6, yet Bobby is in 7th grade.. :huh:[/quote]

It is not based on grade, it is based on age: to be eligible to participate in the contest, the students must be age 13 or under as of September 1.

No. USA got 3rd, Russia got 2nd. China took 1st (as expected)

Meh, the PMWC problems seem easier than MC cause they're really straight forward

[quote="Poincare"]Its not the strictness its actually the type of teaching. Here they go slow so that every single person understands but in china they think Law of Averages style. "It doesn't matter if everyone gets it or not most of them should get it" So the best students get but the students don't do that much math don't. So all you see at all the competitions is the best people not the mediocre or slow people. I speak from an indian perspective but i am sure China, Japan etc are the same. In india, our teacher used to go really fast and when we learnt addition in 1st grade i had already learnt it because i used to do everything ahead of the school but once the class was over almost no one understood it because the didn't give a flying whoop whether the class the got what the hell she was or not  :mad:  :mad:  :mad:[/quote]
It's also the competition. Since there are so many people in Asia and India there is a greater competition then in the US.
To get into the best University in Asia you have to be in the top 1% of the smartest people. I think.  :maybe:

[quote="gauss1181"][quote="RussianRocket"][quote="Poincare"]btw, india is in asia
But one thing that's bad is that the only thing that the ppl at the best univs in india give a crap about are grades.
And, in india you memorize every single formula you use. Like, my cousin is learning heron's formula before trig :!: So it doesn't matter whether you got the concept or not its like do you remember the formulas. So based on that you get a grade so if you can't memorize you done for. And the timing is so tight that you have to memorize formulas absolutely no thinking in tests at least. the only thing i can do is laugh at this situation and so i shall  :rotfl:  :rotfl:[/quote]

When I said Asia I had China in mind. Sorry.
Yah in China and India, the people actually take the time to learn the [i]why[/i] not just the what.[/quote]
So they're good at IMO and proofs :lol:[/quote]
Well suprisingly the US is good at IMO and proofs too!  :D   :D   :D

[quote="RussianRocket"][/quote][quote="gauss1181"][quote="RussianRocket"][quote="Poincare"]btw, india is in asia
But one thing that's bad is that the only thing that the ppl at the best univs in india give a crap about are grades.
And, in india you memorize every single formula you use. Like, my cousin is learning heron's formula before trig :!: So it doesn't matter whether you got the concept or not its like do you remember the formulas. So based on that you get a grade so if you can't memorize you done for. And the timing is so tight that you have to memorize formulas absolutely no thinking in tests at least. the only thing i can do is laugh at this situation and so i shall  :rotfl:  :rotfl:[/quote]

When I said Asia I had China in mind. Sorry.
Yah in China and India, the people actually take the time to learn the [i]why[/i] not just the what.[/quote]
So they're good at IMO and proofs :lol:[/quote][quote="RussianRocket"]
Well suprisingly the US is good at IMO and proofs too!  :D   :D   :D[/quote]
yay!!!!!

oh gosh how many mr .green smilies do i keep posting?

and its funny how u always reply to the same threads that i post in

[hide] :D [/hide]

[quote="gauss1181"][/quote][quote="RussianRocket"][/quote][quote="gauss1181"][quote="RussianRocket"][quote="Poincare"]btw, india is in asia
But one thing that's bad is that the only thing that the ppl at the best univs in india give a crap about are grades.
And, in india you memorize every single formula you use. Like, my cousin is learning heron's formula before trig :!: So it doesn't matter whether you got the concept or not its like do you remember the formulas. So based on that you get a grade so if you can't memorize you done for. And the timing is so tight that you have to memorize formulas absolutely no thinking in tests at least. the only thing i can do is laugh at this situation and so i shall  :rotfl:  :rotfl:[/quote]

When I said Asia I had China in mind. Sorry.
Yah in China and India, the people actually take the time to learn the [i]why[/i] not just the what.[/quote]
So they're good at IMO and proofs :lol:[/quote][quote="RussianRocket"]
Well suprisingly the US is good at IMO and proofs too!  :D   :D   :D[/quote][quote="gauss1181"]
yay!!!!!

oh gosh how many mr .green smilies do i keep posting?

and its funny how u always reply to the same threads that i post in

[hide] :D [/hide][/quote]

Its probably cause we are interested in the same things.
Plus we are one of the most active people!!!  :D

thats great russianrocket, we're both mathcounts freaks who like posting on aops :D  :D  :D

a very short List of active people

leonixln?
Yongyi
gauss1181
math154
Poincare?
isabella (i don't know the number after the name)

[quote="Poincare"]a very short List of active people

leonixln?
Yongyi
gauss1181
math154
[b]Poincare?[/b]
isabella (i don't know the number after the name)[/quote]

How do you not know whether you're active or not?!?! :o

isabella2296, by the way :)

What do you mean by active? Posting a lot in this forum?

yeah, I mean, I'm sorta active, I post a bit, but now that I'm done with mathcounts, I should probably move on...  :wink:


abstract algebra
linear algebra
linear algebra unsolved

let A,B from M_n(F) be nXn matrixes,such that A+B=E(E-unitary matrix).if rankA +rankB =n,prove that A^2=A.B^2=B.

This belongs in Superior Algebra.

We have $(A+B)x=x$ for all $x$; this means that $Ax=x$ if $Bx=0$ and $Bx=x$ if $Ax=0$.
Let $U$ be the kernel of $A$ and $V$ be the kernel of $B$. By standard theorems, $\dim U=n-\mathrm{rank}(A)=\mathrm{rank}(B)$ and $\dim V=n-\mathrm{rank}(B)=\mathrm{rank}(A).$
Since $U\subseteq\mathrm{range}(B)$, $\dim U\le\mathrm{rank}(B).$ We have equality in dimension, which means that the two spaces are equal.
Therefore, for any $x,$ $Bx\in U$ and $B^2x=Bx.$ The truth of this for all $x$ implies that $B^2=B.$ Similarly, $A^2=A$ and we are done.

Moved to Algebra.

I don't think pavle said that $A+B$ is the indentity, just that it is unitary. ($E^*E=I$).


José Arcadio Buendía flips a bird 100 times. It can either land head sides up, or tails sides up. What is the probability that at least 5 consecutive flips produce the same result?

Can also be expressed as
$ A\cup B \equal{} \{1,2,3,...,100\}$ and $ A\cap B \equal{} \emptyset$. What is the probability either $ A$ or $ B$ contains a chain of at least five consecutive integers?

Here is a program that uses the SIMD Mersenne Twister random number generator to simulate the problem:
[code]#define FLIPS 100
#define HEADS 5
#define RANDS 5000000	// uses 5000000 * 8 = 40 MB of memory, adjust if you don't have much RAM
#define TRIALS 10000000	// perform the simulation 10000000 times, adjust if you have a slow CPU
			// you probably won't have to adjust the values; it was written to run on my (very crappy) computer

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include "dSFMT.h"
#include <time.h>
#include <assert.h>

int main(void) {
  dsfmt_t dsfmt;
  double *array = memalign(16, sizeof(double) * RANDS);
  assert(array);
  unsigned long long trials = 0, successes = 0;
  int flip, side, run, pos;
  dsfmt_init_gen_rand(&dsfmt, time(NULL));
  trials = 0;
  while (trials < TRIALS) {
    dsfmt_fill_array_close_open(&dsfmt, array, RANDS);
    pos = 0;
    while (pos < RANDS - FLIPS && trials < TRIALS) {
      run = 0;
      side = 0;
      for (flip = 0; flip < FLIPS; flip++) {
	if ((array[pos++] < .5) ^ side) {
	  side = !side;
	  run = 1;
	} else {
	  run++;
	  if (run >= HEADS)
	    break;
	}
      }
      if (flip < FLIPS)
	successes++;
      trials++;
    }
    printf("%llu trials\n", trials);
  }
  printf("There were %llu trials with %llu successes, for a proportion of %llf.\n", trials, successes, (long double)successes/trials);
  return 0;
}[/code]
Running this once produced a .971746 success rate in 10,000,000 trials. Doing a statistical analysis on this, [b]I am 95% confident that the true value of the proportion of 100 flip trials that produces 5 in a row is between .97164 and .97185[/b]. I was working on a solution that does not use random simulation like this one does, but it didn't agree with this solution, so I think it's wrong. I'll keep working on the other solution.

This is a standard problem for a recursive/state diagram-type attack.  Here's a standard approach that should work in any similar problem: let $ a_1 \equal{} 2$ and for $ n \geq 2$, let $ a_n$ be the number of sequences of flips of length $ n$ containing no string of five consecutive similar flips such that the last two flips are different; let $ b_1 \equal{} 0, b_2 \equal{} 2$ and for $ n \geq 3$ let $ b_n$ be the number of sequences of flips of length $ n$ containing no string of five consecutive similar flips such that the last two flips are the same and the flip that precedes them is different; and similarly for $ c_n$, $ d_n$.  Then we can easily set up a recursive system of equations, e.g., $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n \plus{} c_n \plus{} d_n$, $ b_{n \plus{} 1} \equal{} a_n$, $ c_{n \plus{} 1} \equal{} b_n$ and $ d_{n \plus{} 1} \equal{} c_n$.  Now it takes only a little bit of effort to convert this into a recursion involving only one of the variables, and this allows for relatively efficient computation (or solving exactly, but in this case we end up looking at the roots of the cubic $ x^3 \equal{} x^2 \plus{} x \plus{} 1$, which aren't pretty).  For a value like $ n \equal{} 100$, this means you might want a computer to do it, but the computer can do it instantaneously.

The result is that the probability in question is
\[ 1 \minus{} \frac {17943803336550012914104102513}{633825300114114700748351602688} \approx 0.971689670113642....
\]

Challenge for those who are new to this method: how would you alter it if the coin landed heads with probability 2/3?  What about if it landed heads with probability $ 1/\sqrt{2}$?

[quote="Ignite168"]...flips a bird[/quote]Be careful translating problems. You ran into an unpleasant idiom here.
Just call it a coin.


I got this from a book: How many ways are there to arrange the letters of the word TOOL?

I know there's a formula for this but the book presented it this way:

The number of ways you can place the 2 O's is 4C2. The number of ways you can place the 1 T is 2C1 and 1C1 for the remaining L. The answer is 4C2x2C1 = 12.

My only question is, isn't it that when you first place the letter O, you are considering order? But when I tried to assign T a place first, then give L a place and lastly, give a place for the 2 O's, the answer is still the same: 4C1x3C1x2C2 = 12. Why is this so?

[quote="bobbym"]Hi;

This is a standard mississippi problem, I use this idea for those:

(4!) / ( 1! 1! 2! ) = 12[/quote]

Hello, thanks for the response but i already know this formula. I don't know why the alternative solution works.

By alternate solution, I am assuming you mean the book's solution?

So, we have 4 spaces for the words:
__ __ __ __

Out of these 4, you pick any combination of two to hold the O's. Note that order doesn't matter in this case. Then, after you place the O's, there are 2 spots left over. You choose any one of them to be the T, and the remaining spot goes to the L.

I am not sure what you mean by "considering order".

Yes, essentially, you are doing the same thing as before. Except this time you are first assigning T a spot, which can be any of the four spots. Then you are assigning one of the remaining 3 spots to the L. And there are only two spots remaining, and the O's are indistinguishable, thus there is only one way to order them. This method is exactly the same as the previous method, except the things are done in a different order.


[quote="biffanddoc"]A perfect number is a number that all of its factors except itself ad up to the number. 28 and 6 are examples.
A deficient number is one that its factors minus itself add up to less than the number. 7, 21, and 8 are examples.
What is the largest 10 digit number that is exactly one too deficient to be perfect. (5 is four numbers too deficient, 20 is 2 too abundant (large))?

You can use a calculator. (try without first)[/quote]

 :?
Are we supposed to factor 10 digit numbers by hand?  Or did I interpret this wrong?

You can, but that will take a LONG time.
Theres a trick. :D

[hide="trick well, really its a big hint"]
The sum of the factors of $2$ is $1$.
The sum of the factors of $4$ is $3$.
The sum of the factors of $8$ is $7$.
[/hide]

[color=olive][size=75][You can't have apostrophes in hide tags - solafidefarms][/size][/color]

Ah. I see.

This reminded me of a trick on the sums of powers of 2.

It goes like this:

$2^0+2^1+2^2+...+2^n = 2^{n+1}-1$

You could prove it easily by induction...

[hide="Proof for above"]
$2^0 = 2-1$

Suppose
$2^0+2^1+2^2+...+2^n = 2^{n+1}-1$
Then
$2^0+2^1+2^2+...+2^n+2^{n+1} = 2^{n+1}-1+2^{n+1}$
$2^0+2^1+2^2+...+2^n+2^{n+1} = 2\cdot2^{n+1}-1$
$2^0+2^1+2^2+...+2^n+2^{n+1} = 2^{n+2}-1$
which completes the inductive step.
[/hide]

Anyways, how would you prove that it must be a power of 2?

I can't, but I assume because when you get into 10 digit numbers what are the chances?

How do you know it must be a power of 2?

[hide="A more general hint"]
If $x=p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_n^{e_n}$, where $p_i$ is prime and $e_i$ are the exponents, (basically the prime factorization) then try to find the sum of the factors in terms of $p$ and $e$.
[hide="how? a giveaway:"]
think about the product of $(p_1^0+p_1^1+p_1^2+\cdots+p_n^n)$ for all $p_i$.
[/hide]
[/hide]

[quote]Anyways, how would you prove that it must be a power of 2?[/quote]
[b]Very good question![/b]  And I don't think any one has answered it here.

Yes it is easy to see (and shown above) if it is a power of 2  than you are okay [b]but it is NOT the same thing as saying that if the number is "one too deficient" it has to be power of two.[/b]

The proof  in all likelyhood is  too advance, (or it is possible that  ot  may  not be true).. so it may bot belong in this  thread. Some one can ask the same question in one of the more advance form.

It may be fun to know here, that the question is so complex that as far as I know, NO one knows if this is  true, (The only [b]known[/b] "one  too defecient numbers to be perfect" are of the type $2^n$ and I think no one  has proved that there are no others..
Math is FUN!


Answering each question would be highly appreciated. In other words, I have placed several signs, which are substituted of other common signs, on the third and fourth questions since I cannot find any sign. Thank you.

1. A 10-ft. board is cut into 2 pieces so that one piece is 6 feet longer than 3 times the shorter piece. If the shorter piece is x feet long, fine the lengths of both pieces.

2. Four times the sum of some number plus 3 is equal to 8 times the number minus 16.

3. 7 <= (less than or equal) 3t - 2 <= 19

4. - 16 <= - 2c - 4 < - 8

5. 2(0) + 9y = 0

6. Five times the sum of a number and -105 is equal to 20. Find the number.

7. Four times a number, added to -14, is -22. Find the number.

8. -3y^0

9. (8b)^0


calculus
integration
function
trigonometry
limit
calculus computations

[quote="kunny"]Let $f(x)$ be a differentiable function such that $f'(x)+f(x)=4xe^{-x}\sin 2x,\ \ f(0)=0.$

Find $\lim_{n\to\infty}\sum_{k=1}^{n}f(k\pi).$[/quote]

We have $e^{x}f'(x)+e^{x}f(x)=4xsin2x$ or $(e^{x}f(x))'=4xsinx\Rightarrow f(x)=e^{-x}\int_{0}^{x}{4tsin2t}dt$.

We shall calculate $\int_{0}^{k\pi}{4tsin2t}dt=-\int_{0}^{k\pi}{(2t)(cos2t)'}dt=-2k\pi+2\int_{0}^{k\pi}{cos2t}dt=-2k\pi$

So $f(k\pi)=\frac{-2k\pi}{e^{k\pi}}\Rightarrow \sum_{k=1}^{n}f(k\pi)=\sum_{k=1}^{n}\frac{-2k\pi}{e^{k\pi}}$

$\sum_{i=1}^{n}{(e^{-kx})'}=\left(\sum_{i=1}^{n}{e^{-kx}\right)'=\left(e^{-x}\frac{e^{-xn}-1}{e^{-x}-1}\right)'}$

So $\sum_{k=1}^{n}f(k\pi)=2\pi\left(e^{-x}\frac{e^{-xn}-1}{e^{-x}-1}\right)'|x=\pi$ from where we got $\lim_{n\to\infty}\sum_{k=1}^{n}f(k\pi)=-2\pi\frac{e^{\pi}}{(e^{\pi}-1)^{2}}$


What is the length of the curve that is built by infinitely many semicircles with radius 1, 1/2, 1/4, etc. according to the picture?

[hide]For the first semi, 1*2pi/2=pi
2nd=1/2pi
3rd=1/4pi
...
If you do the formula for infinite gemetric series, you get:
(1/2)/(1-1/2) --> (1/2)/(1/2) --> 1
>>1<<[/hide]

Hey Dwx, shouldn't there be a  :pi: in your answer????

dwx i think you made a mistake in your calculations

[hide]the circumfrence of a semicircle would be (1/2)(pi)(d) or (r)(pi)
so the circumfrence of the curve would be (1+1/2+1/4+1/8....)(pi)
the sum of an infinite geometric series is (first term)/(1-the number your multiplying by)=1/(1-1/2)=1/(1/2)=2
The answer is 2 pi.[/hide]


Find all positive integers $ x,y,z$ such that $ 2(x^4 \minus{} y^4) \equal{} z^2$

there is a thread with such equations [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112745]here[/url]
but no solution


function
calculus
derivative
integration
complex analysis
complex analysis unsolved

f is analytin in whole complex plane and |f(z)|<= 5*sqrt(|z|)  for all |Z|>=1. Then prove that f is constant

Do you know the Cauchy estimates?

In any case, you can do this without them. Let $ f(z)\equal{}\sum a_nz^n$ (this is valid everywhere since it is analytic everywhere).

Note that the n-the derivative $ f^{n}(0)\equal{}n!a_n$.

Use Cauchy's integral formula: $ f(z)\equal{}\frac{1}{2\pi i}\int_{|\zeta\minus{}z|\equal{}R}\frac{f(\zeta)}{\zeta\minus{}z}d\zeta$.

Then $ f^{n}(z)\equal{}\frac{1}{2\pi i}\int_{|\zeta\minus{}z|\equal{}R}\frac{f(\zeta)}{(\zeta\minus{}z)^n}d\zeta$.

Thus $ |f^{n}(0)|\leq \frac{1}{2\pi}\int_{|\zeta|\equal{}R}\frac{|f(\zeta)|}{|\zeta|^n}d\zeta\leq \frac{5\sqrt{R}}{R^n2\pi}$.

But this is true for any $ R$, so let $ R\to\infty$. This shows that for any $ a_n$ with $ n\geq 1$ must be 0.

So $ f(z)\equal{}a_0$.


AMC AIME Asymptote function ARML USA(J)MO USAMO

I don't remember which problems were which, but here is SCRAMBLED ORDER:

252
25
314
80
14
17
708
47
190
32??
172
375?
40
420??

? means I'm not sure
Everything else has been checked with multiple people.

Not official answers, but here's a general list I compiled...

EDITED LIST (only number 10 changed)
1. 252
2. 025
3. 314
4. 080
5. 014
6. 017
7. 708
8. 047
9. 190
10. 032
11. 172
12. 375
13. 040
14. 432
15. 871

Some of these might be wrong.  If they are correct, I got a 9.  I should have had a 13, but I made stupid mistakes on numbers 7, 8, 9, and 14.  For number 6, I took an educated guess of 019, and for number 10, I guessed (quite randomly) 903.

I think 12 may be 374 because 375 is the asymptote, so technically we can never reach that.

[quote="krsattack"]Not official answers, but here's a general list I compiled...

1. 252
2. 025
3. 314
4. 080
5. 014
6. 017
7. 708
8. 047
9. 190
10. ??? Don't know yet ???
11. 172
12. 375
13. 040
14. 432
15. 871

Some of these might be wrong.  If they are correct, I got a 9.  I should have had a 13, but I made stupid mistakes on numbers 7, 8, 9, and 14.  For number 6, I took an educated guess of 019, and for number 10, I guessed (quite randomly) 903.[/quote]\

my school decided #6 was 016 but it doesn't matter; i didn't get it. btw #10 is 032 (please!!) but i'm not sure about the rest

I believe the following are the correct answers.

1. 252
2. 025
3. 314
4. 080
5. 014
6. 017
7. 708
8. 047
9. 190
10. 032
11. 172
12. 375
13. 040
14. 432
15. 871

I thought that was quite an easy AIME in comparision to past years. #14 was very easily trig/calc-bashable, even though you can get around the calc by making a substitution to make the function a quadratic. #15 was even easier, given that you have visualization skills (which I don't, haha).

#10 was pwnable by being like WLOG AB=BC=CD  :P , then it's obvious that E,A,D are colllinear, by 30-60-90 triangles...

#8 seemed a little weird...so many people know how to add arctangents now that there really isn't a point in asking a question like that.

A variation of #9 has been used (actually, discussed in the past couple of days) in an old AIME, but the variation was probably harder.

#12 was kinda stupid in my opinion, and the answer is indeed 375, because you can get the last car through, just not the space behind it.

[quote="Zmastr"]Pretty sure that #6 is 017...[/quote]
i might be mistaken:
sorry if i am :blush:

If those are indeed the answers, then I got an 11 :D

I still can't believe I got #12 right, because it just...didn't seem right.

I'm almost certain #12 is 375, because you can get infinitely close to 3750, but you can start and stop the camera in such a way that you get 3750 cars but not the spaces on either side of the first and last cars.  I you couldn't get 3750 then you could get 3749, which isn't divisible by 10.

I agree with the 1-12 already posted, I never even looked at 13-15.

I only got an 8 -_-... forgot to multiply by 10 on number 9, and number 10 was just. stupid. the 375 problem was too easy in my opinion.

1-6, 8, and 12. blegh.

It sounds like they test the AIME II on the 2nd, and expect the answer sheets back into the office by the 4th. From that point they grade them pretty quickly, and within about a week the AIME II it seems like they'll announce the cutoff.

[quote="ProtestanT"][quote="nr1337"][quote="100%Tuna"]Zmastr-:>> Huh...? I thought the cut-off last year was something like 185-195ish... Soo... wassup with 210 n 215 having a 50-50 chance of getting in... :?:

*CONFUSED*[/quote]

This year's AIME was easier than last year's.[/quote]
my friend thought this year's was slightly harder than last year's...[/quote]

I think for the most people, it was definitely easier, as this one involved less brute-force and less reading traps.

[quote="100%Tuna"]Actually, I juz realised... Foreigners can't make it to the USAMO anyways... Dang... so here goes everything...! but ermm... does the AIME score help ya out for univ admissions...? And, well, if it does, is a 9 considered good... :!:  :?:[/quote]

What country are you from?  Canadians can take the USAMO.  And yes, a 9 is quite good.

nr1337 :>> Actually it's kinda wierd... I'm an Indian citizen studying in Singapore... so yeah... I guess I have nothing to do wif USAMO, right? *sighs*

I got a 10!

1-9 and 15 are the ones I answered right.

I guessed on 12 and I couldn't get anywhere with the rest of them.

No offense but I'd say that unless your having difficulty with 3-D geometric problems and trig, #15 might be too easy for a typical number fifteen.

Do #10 or #14 require trig?

Number 10 was triangle inequality, and just recognizing the basic trig triangles.
Number 14 you could have used Law of Cosines.

http://www.artofproblemsolving.com/Wiki/index.php/2008_AIME_I

[quote="miller4math"]omg... i beat mathcrazed.
i got a 9
could someone explain how to do number 13?
my algebra stinks[/quote]

Nevermind I got 9 =P... I got #14 right also.  Stupid #9 -_-... 10 does not cancel out with 3's... -_-

go to math jams and click transcript

those are answers + solutions, pretty good huh?

[quote="tenniskidperson3"]Anyone else find it weird that 7 of the 15 questions had answers below 100?  Maybe because this is my first AIME or something, but it was just weird to me.[/quote]

Eh, personally I don't think that matters a whole lot.

My answers (red ones are wrong):
[color=green]1. 252
2. 025
3. 314
4. 080
5. 014[/color]
[color=red]6. 009[/color]
[color=green]7. 708
8. 047
9. 190
10. 032
11. 172
12. 375[/color]
[color=red]13. 012[/color]
[color=green]14. 432
15. 871[/color]

"Mistakes" I made:

Problem 1:
I went through two answers before finally locking in with 252.  That's because I kept messing up the algebra.  I got this down just in time.

Problem 6:
I realized a pattern, but I miscounted.

Problem 13:
I just guessed.

Problem 14:
I switched $ A$ and $ B$ by accident.  So I actually extended $ \overline{AB}$ through $ B$, not $ A$.  Nevertheless, I got the same answer.

Problem 15:
For some strange reason, I must've read the question as the square having an AREA of 100, not a SIDE LENGTH of 100.  Nevertheless, it didn't affect the answer.  After trying to build the flat pattern of the tray in Geometer's Sketchpad (I was bored in one of my classes), I realized this error.  Again, it didn't affect the answer.  I still got a height of $ \sqrt[4]{867}$.

[quote="archimedes1"]I thought the first 5 questions were a easier than usual.  Or maybe they could just be done a lot faster.

[quote="lingomaniac88"]
[color=red]6. 009[/color]

Problem 6:
I realized a pattern, but I miscounted.
[/quote]
YAaaay!  Someone else actually made the same mistake as me and got 009![/quote]
I miscounted REALLY badly..and got 257 (i thought there were 99 terms in the first row instead of 50) (I also repeated some multiples a lot more times)

On problem 15, the actual side length of the base doesn't matter as long as the side length is greater than 2sqrt(17).  The algebra in the problem doesn't involve the side length of the bases.


inequalities integration calculus geometry function

Let A = $ 1\plus{}\frac{1}{\sqrt{2}}\plus{}\frac{1}{\sqrt{3}}\plus{}...\plus{}\frac{1}{\sqrt{80}}$   find the integer n which satisfies the condition  n - 1 < A < n

What is the solution in the general case ?

[hide="Lame solution"]$ \int_1^{81}\frac{dx}{x^{1/2}}\leq\sum_{k\equal{}1}^{80}\frac 1{k^{1/2}}\leq\int_0^{80}\frac{dx}{x^{1/2}}$

$ \int\frac{dx}{x^{1/2}}\equal{}2x^{1/2}\plus{}C$

$ 2\sqrt{81}\minus{}2\equal{}18\minus{}2\equal{}16$

$ 17<2\sqrt{80}<18$.

Therefore, $ 16\leq\sum_{k\equal{}1}^{80}\frac 1{k^{1/2}}\leq 18$.

Now I guess you can notice that the integral from $ 0$ to $ 64$ would also be $ 16$ and the integral from $ 64$ to $ 81$ is just $ 2$, so it intuitively makes sense that you're accumulating a lot less area under the curve as you move into more positive $ x$. So you'd guess that the actual sum doesn't make it as far as $ 17$, which happens to be true.

(or you just pick between $ n\equal{}17$ and $ n\equal{}18$ because it has to be one of those, and hope you're right)
[/hide]

If there actually is a good solution, I'd love to see it...

[quote="grn_trtle"]
$ \int_1^{81}\frac {dx}{x^{1/2}}\leq\sum_{k \equal{} 1}^{80}\frac 1{k^{1/2}}\leq\int_0^{80}\frac {dx}{x^{1/2}}$

$ \int\frac {dx}{x^{1/2}} \equal{} 2x^{1/2} \plus{} C$

[/quote]
$ \int_1^{81}\frac {dx}{x^{1/2}}\leq\sum_{k \equal{} 1}^{80}\frac 1{k^{1/2}}\leq\int_0^{80}\frac {dx}{x^{1/2}}$

It follows to me that $ \int_1^{81}\frac {dx}{x^{1/2}} \leq \int_0^{80}\frac {dx}{x^{1/2}}$, because the first integral adds $ 2\sqrt{\frac{1}{81}},$

but subtracts away more with $ 2\sqrt{1}$, as compared to the second integral.
But how do you know/justify that the summation expression belongs in
the middle of the double-sided inequality (aside from its truth)?

[hide="My (solution for the lower bound"]
Since the $ \sqrt x$ function is concave, this implies that $ \frac 1{\sqrt x}$ is convex, so $ \sum_{x\equal{}a}^{b\minus{}1} \frac 1{\sqrt x}> (b\minus{}a)\frac 1{(\sqrt a\plus{}\sqrt {b\minus{}1})/2}>(b\minus{}a)\frac 1{(\sqrt a\plus{}\sqrt b)/2}$.

Plugging in $ a\equal{}1$ and $ b\equal{}81$, we have $ \sum_{i\equal{}1}^{80}\frac 1{\sqrt i}>80\left(\frac 2{9\plus{}1}\right)\equal{}\frac{160}{10}\equal{}16$.

Note: How would one rigorously prove the above inequality? Intuitively it makes sense, but...[/hide]

CAN YOU PROVE THIS :  $ \sum_{x\equal{}a}^{b\minus{}1}\frac {1}{\sqrt x}> (b\minus{}a)\frac {1}{(\sqrt a\plus{}\sqrt{b\minus{}1})/2}$


calculus
integration
complex analysis
algebra
partial fractions

Evaluate $ \int_\gamma \frac {2z^2 \minus{} 15z \plus{} 30}{z^3 \minus{} 10z^2 \plus{} 32z \minus{} 32} dz$
where $ \gamma$ is the circle $ |z| \equal{} 3$

Using partial fractions:
$ \int_\gamma \frac {2z^2 \minus{} 15z \plus{} 30}{z^3 \minus{} 10z^2 \plus{} 32z \minus{} 32} dz \equal{} \int_\gamma \frac {2}{z \minus{} 2} dz \plus{} \int_\gamma \frac {1}{(z \minus{} 4)^2} dz$ How would the rest of this be evaluated?

If the last equality holds (I haven't checked that) then the second integral is straightforward since $ \frac1{(z \minus{} 4)^2}$ is holomorphic on $ |z| \equal{} 3$ so it must be $ 0$. For the first one we can apply the Cauchy's formula (because $ f(z)\equiv 2$ is holomorphic on $ |z| \equal{} 3$) and we get $ 2\cdot 2i\pi \equal{} 4i\pi$. So finally
\[ \int_{\gamma}\frac {2z^2 \minus{} 15z \plus{} 30}{z^3 \minus{} 10z^2 \plus{} 32z \minus{} 32}dz \equal{} \int_{\gamma}\frac2{z \minus{} 2}dz \plus{} \int_{\gamma}\frac1{(z \minus{} 4)^2}dz \equal{} 4i\pi \plus{} 0 \equal{} 4i\pi.
\]
I think it would be more elegant to write the solution liike that:
\[ \int_{\gamma}\frac {2z^2 \minus{} 15z \plus{} 30}{z^3 \minus{} 10z^2 \plus{} 32z \minus{} 32}dz \equal{} \int_{\gamma}\frac {2z^2 \minus{} 15z \plus{} 30}{(z \minus{} 2)(z \minus{} 4)^2}dz \equal{} \int_{\gamma}\frac {\frac {2z^2 \minus{} 15z \plus{} 30}{(z \minus{} 4)^2}}{z \minus{} 2}dz \equal{}
\]
\[ \equal{} 2i\pi\left(\frac {2\cdot 4 \minus{} 15\cdot 2 \plus{} 30}{(2 \minus{} 4)^2}\right) \equal{} 4i\pi.
\]


inequalities
trigonometry
inequalities unsolved

[quote="georg"]Prove that in any triangle we have: $ \frac{ \sin^{n} A}{ \sin^{n+2} B} + \frac{ \sin^{n} B}{ \sin^{n+2} C} + \frac{ \sin^{n} C}{ \sin^{n+2} A}  \geq 4, $  $ \forall n \in N $[/quote]

$
 \left( {\frac{{\sin ^n A}}{{\sin ^{n + 2} B}} + \frac{{\sin ^n B}}{{\sin ^{n + 2} C}} + \frac{{\sin ^n C}}{{\sin ^{n + 2} A}}} \right) \\
  \ge \frac{{\left( {\frac{{\sin ^{\frac{n}{2}} A}}{{\sin ^{\frac{n}{2}} B}} + \frac{{\sin ^{\frac{n}{2}} B}}{{\sin ^{\frac{n}{2}} C}} + \frac{{\sin ^{\frac{n}{2}} C}}{{\sin ^{\frac{n}{2}} A}}} \right)^2 }}{{\sin ^2 B + \sin ^2 C + \sin ^2 A}} \\
  \ge \frac{{3^2 }}{{\frac{9}{4}}} = 4 \\
$
It remains to prove
$
\sin ^2 B + \sin ^2 C + \sin ^2 A \le \frac{9}{4}
$
which is true because
$
\sin ^2 B + \sin ^2 C + \sin ^2 A = 2 + 2\cos A\cos B\cos C \le 2 + 2\left( {\frac{{\cos A + \cos B + \cos C}}{3}} \right)^3  \le \frac{9}{4}
$


Sometimes, when I do different problems, the question asks for [u][i][b]proper factors[/b][/i][/u].  However, it seems that sometimes, the definition is a little different.

[b]A MATHCOUNTS PROBLEM[/b]
What is the sum of the proper factors of 512?

The answer to this question, according to Mathcounts, is [hide]$ 511$[/hide].  This means that $ 512$ is not counted as an addend in the sum.

According to Wolfram, [url]http://mathworld.wolfram.com/ProperFactor.html[/url], [quote]A proper factor of a positive integer n is a factor of n other than 1 or n.[/quote]

Does anyone know that Mathcounts officially wants?

I'm pretty sure generally, the proper factors of $ n$ are all factors of $ n$ except for $ n$.

Interesting enough, on MathWorld, this is a [i]proper divisor[/i] instead of a proper factor

The Mathcounts one is right, at least to my knowledge.
I wonder what's up with Wolfram.

yeah, i think the mc definition is correct, all factors of n excluding n.  if u use the other definition, then 6 wouldn't be considered a perfect number...

Which would absolutely destroy my username...


probability
function
analytic geometry
linear algebra
probability and stats

Alright...here is the question:
Use the Central Limit Therem to obtain a theoretical upper bound based on our model of a uniform probability function for the probability that an average of the squares of 900 x-coordinates is greater than or equal to 0.36.

If anyone could help, I would really appreciate it.

There isn't a probability/statistics forum, so there isn't an obvious place to post this. But it wasn't a good fit for linear algebra, so I moved it here.

There are two difficulties with this.

1. The Central Limit Theorem isn't good at giving theoretical upper bounds for probabilities. Chebyshev's Theorem can give theoretical upper bounds (often too high by multiple orders of magnitude.) The Central Limit Theorem gives best estimates for probabilities, not upper bounds.

2. You have given insufficient information to solve this problem: "an average of the squares of 900 x-coordinates" but what distribution are those coordinates chosen from? The mention of a uniform distribution is useless without knowing what set it is uniform over.


calculus
integration
trigonometry
factorial
calculus computations

let l be a positive integer,
find a general formula with proofs (involving factorials) for:
$\int_{0}^{\frac{\pi}{2}}\left( \cos x\right)^{2l+1}dx$

[quote="Kouichi Nakagawa"]Thus, $\int_{0}^{\frac{\pi}{2}}\cos^{2l+1}x \ dx = \frac{2l}{2l+1}\cdot \frac{2l-2}{2l-1}\cdot \cdots \cdot \frac{4}{5}\cdot \frac{2}{3}\cdot 1$[/quote]

So the answer is

$\frac{(2l)!!}{(2l+1)!!}$ ?


The postmark deadline is January 9, 2006.  Visit [url]http://www.usamts.org[/url] for more details.

Lisa and Bart?

[url]http://myspace-973.vo.llnwd.net/00180/37/93/180273973_l.jpg[/url]

...Again.

[quote="zanttrang"]Looks like the writers of this year's rounds really enjoy game theory problems  :lol:

Also, not that I don't enjoy them, but why has there been such a focus in math contests in general on the properties of polynomials?  Why do we care about them so much?[/quote]

Good question.. in real life we can just TI-83/6/9 any polynomial lol, but probably it's easier to make good problems out of them.

[quote="beta"]Can we submit the entry form by email?[/quote]I'm pretty sure you can. Just scan it and mail it over.  :)


How do you think I should prepare for the AIME?

I'll be trying to go through as many practice AIMEs as I can along with doing some aops v2.  The question I mostly have is how long should I spend on each question while doing practice AIMEs.  Since the actual contest is the day I get back from spring break (Tuesday, March 23rd), should I give up and read the solution after a certain amount of time or keep on truckin' to get the answer.

Please specify suggested times and any other material that would help.

I think MCrawford and some other smart people recommended some things about time spent per AIME problem somewhere in the thread entitled "Advice..." in this same forum section.

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10806]Advice thread[/url] that Fiery mentioned (and originated) has some good tips on AIME preparation. My question to you, Syntax Error, is how well do you think you need to do this year? That would dictate how you would practice. I think you are about two years older than my son, who also qualified for AIME this year. You have time, and he has time, to treat this year's AIME as practice. The strategy that I am recommending to him is to take a few "first five" AIMEs as practice.

I bought all the old AIMEs still available as individual tests, and I have Contest Problem Book V, which has other old AIMEs (and a WONDERFUL bibliography of the problem-solving literature). Tomorrow A+MATH will take three hours to do a modified AIME, consisting of the first five questions from each of three different actual AIMEs. At this stage of his math development, it makes more sense for him to practice hardest on the early questions, which he should learn to master in the next few years, rather than to try taking the last questions on the AIME, which are legendarily difficult and beyond his current curriculum level. You are farther along than he in standard curriculum, I think, but you too have time to master the early questions as a strategy for this year. After all, getting a solid 6 on the AIME at your age might qualify you for USAMO, if that is your goal. Better to get 5 questions right by attempting only 5 than to get 3 right (the average AIME score) by attempting too many. I'll see after a few practice tests just how high a score is reasonable for A+MATH to expect this year--he usually does a little better on the real tests than in practice, if he gets enough practice. I will probably have him do ONE full-length AIME from beginning to end, just so that he can see what the questions from question 11 to question 15 look like.

Hope this helps! Enjoy your vacation (ours comes later) and have a good time on AIME day.

Thanks for the tips.  I would like to say that a 6 would be the AIME floor, but I'm preparing as if I need a 10.  The AIME plays much more into my favor than into mathcounts because there is a much longer timelimit and I like the questions much more.  On practice tests I really vary a whole lot.  I haven't actually taken any practice full-length tests, but I usually solve around an 8 or more just doing the contests casually.  On real tests I expect my score just because I choke on real contests.

I really just want to make USAMO because after not being close to successful at mathcounts I just want to know that there was something I succeeded in this year after working so hard.

hey syntax don't you guys have an state math contest?

I've read the stuff about advice and I think it's pretty useful. Still, i can only average about 4 or 5. Personally, I think it's way tougher than mathcounts but that's just me.

The AIME is a tough test. As far as I know, there really aren't any other contests that are comparable to it at all. Olympiads are generally harder, and ``solve it'' contests are almost always easier.

Keep in mind that when you practice for AIME (especially; this is true of other math contests as well, but I think it's a much more serious problem on the AIME), you should never solve a problem, look at the answer, see that you made a computational error, and say ``I won't do that on the actual contest.  I think the most important thing on the AIME is the carelessness/carefulness factor for people who can already solve a reasonable number of problems.

[quote="ComplexZeta"]The AIME is a tough test. . . . Keep in mind that when you practice for AIME (especially; this is true of other math contests as well, but I think it's a much more serious problem on the AIME), you should never solve a problem, look at the answer, see that you made a computational error, and say "I won't do that on the actual contest."  I think the most important thing on the AIME is the carelessness/carefulness factor for people who can already solve a reasonable number of problems.[/quote]

That's a very important point. In practicing for any kind of test, it is too easy to check an answer in practice, say, "Oh, yeah, I get it," but not practice to the level of error-free performance on that point. The AIME is a killer in that it tests, after all, on precalculus mathematics, but not typical high school precalculus mathematics. The AIME demands a THOROUGH knowledge of mathematics, and being able to calculate and error-check without an electronic calculator. It's not surprising at all that perfect papers are as [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2003-aa/03statsaime.html]rare as they are on the AIME[/url].

I usually throughly check my answers even when it's just casually; I like to practice as if it were the real thing, but I don't actually sit down for 3 hours.

There's a big difference between mathcounts and AIME, which I like.

Muahaha.

12 on a practice AIME.

Yeah, just practice by doing a lot of problems. Its fun though...I certainly enjoy spending 3+ hours on math.

Go to this website (http://www.kalva.demon.co.uk). The guy collected all AIME problems. I don't even get to take AIME because I'm horrible at math.

yes, i agree that kalva's site is very useful, but don't assume that the practice tests there will be representative of your true ability on the actual AIME. for instance, some of the problems are poorly worded to avoid copyright; obviously this won't happen as often on the AIME so keep that in mind. second of all, the answers are not all in integer form from 0-999 so that completely gets rid of one of the foundations of AIME problems; it makes it harder as well since you have no set range of possible numbers (pi, radicals, and fractions are all possible answers, unlike in the AIME.

all in all, you can probably get within 2 of your actual AIME score by taking these.

Try to get through aops 1 and 2 as much as you can this break.

[quote]I'm really, really scared. The AIMEs are in 8 days, and I couldn't do a single problem on the 1991 test.

:^(

Any advice for that?[/quote]

Take heart....the 1991 AIME was quite difficult. I got about 3 below my average on that one. If you want to boost your confidence, try the 1986 or the 2003A--the easiest so far in my opinion.

Thanks. It's about time I re-read AoPS v.1 anyway. (My state MathCounts is this Saturday.)

I'm only in 8th grade, so I'm not particularly worried about it being my last chance and I have to do well or anything like that, it's just that I have a friend who is basically level with me on these types of thing, and neither of us can let the other get too far ahead. I have to beat him, because he got 13 pts. higher than me on the AMC, beat me in Quiz Bowl, beat me in Chapter Mathcounts, and...um...that's about it. (Yes, captcha000, that's you I'm talking about.)

Hmm...well thats interesting enough...by the way fermatprime, try the 1985 test also, and both of the 2003 tests were easy in my opinion.

PS level?? about level with me?? Alright...


given are the positive real numbers a_{i} i=1..n such that sum(a_{i}) + product(a_{i}) = n+1. prove that sum(a_{i}) >= sum (sqrt(a_{i})).

If $\sum a_i <n$ then by AM-GM,
$\sum a_i+\prod a_i \leq \sum a_i +(\frac{\sum a_i}{n})^n < n+(\frac{n}{n})^n=n+1$, a contradiction. So  $\sum a_i \geq n$

Let $f(x)=x- \sqrt{x}$. Then f(x) is convex and increasing on $[1,\infty ]$. Using Jensen, $\sum a_i -\sum \sqrt{a_i}=\sum f(a_i)\geq nf(\frac{\sum a_i}{n}) \geq nf(1) =0$ and result follows.

Since  you  proved  [tex]\Sigma a_i \geq n [/tex] , the  rest  is  easy  from  old  nice  BCS : [tex](\Sigma\sqrt a_i)^2 \leq n\Sigma a_i\leq(\Sigma a_i)^2[/tex]


I just need to verify my answer...
Problem: Simplify (without using a calculator) :

        $ sec [cos^{\minus{}1} \frac{5}{13} \minus{} \frac{1}{2} tan^{\minus{}1}(\minus{}\frac{8}{15})]$

here's my solution:
[hide]
since $ \frac{1}{cos x} \equal{} sec x$, and $ cos(x\minus{} \frac{1}{2}y) \equal{} cos x cos \frac{1}{2}y \plus{} sin x sin \frac{1}{2}y$,

 let $ x\equal{}cos^{\minus{}1} \frac{5}{13}$, $ sin x \equal{} \frac{12}{13}$
 let $ y\equal{}tan^{\minus{}1}(\minus{}\frac{8}{15})]$, let $ cos y\equal{}\frac{15}{17}$ so $ cos \frac{1}{2} y\equal{} \minus{}\sqrt \frac{16}{17}$ and $ sin \frac{1}{2} y\equal{} \sqrt \frac{25}{34}$...

plugging these in the equation..and just get the reciprocal after? is my solution ok?:) thanks!

[/hide]

[quote="JRav"]It seems okay.  Does your calculator agree with you?[/quote]

err, not so much, or i think i just miscalculated it with my calculator...:))

[quote="Jakey"][...] so $ cos \frac {1}{2} y \equal{} \minus{} \sqrt \frac {16}{17}$ and $ sin \frac {1}{2} y \equal{} \sqrt \frac {25}{34}$...
[/quote]

Something went wrong here because their squares don't add up to 1.

[quote="iniquitus"][quote="Jakey"] let $ y \equal{} tan^{ \minus{} 1}( \minus{} \frac {8}{15})]$, let $ cos y \equal{} \frac {15}{17}$ so $ cos \frac {1}{2} y \equal{} \minus{} \sqrt \frac {16}{17}$ and $ sin \frac {1}{2} y \equal{} \sqrt \frac {25}{34}$...[/quote]

nah

$ cos y \equal{} \frac {15}{17}$
$ sin \frac {1}{2} y \equal{} \sqrt \frac {1}{17}$[/quote]

right! i used $ sin y$ instead of $ cos y$ when i solved for $ sin \frac{1}{2} y$. thanks!:)

\sin , \cos , \tan , etc. are nicer $ \text{\LaTeX}$.


limit
abstract algebra
probability
topology
probability and stats

Assume that random variables $ X_{n}$ converge in distribution to random variable $ X$ and $ \sup_{n}\mathbb{E}|X_{n}|^{2}<\infty$. Prove that for $ p\in(0,2)$:
\[ \lim_{n\rightarrow \infty}E|X_{n}|^{p}\equal{}E|X|^{p}\]

Is this relevant :) ?
I' ve got 3 modules in Probability theory at university, so called RP I, RP II, RP III, and RP II and RP III is optional, this is task from module RP II.

By the Skorokhod representation theorem, we can assume that the convergence is almost surely. By assumption, the family $ |X_n|^p$ is uniformly integrable, so we get the desired result.

[b]Skorokhod representation theorem[/b] If sequence $ X_n$ of random elements taking values in a separable metric space converges in distribution to $ X_0$, then there is a sequence $ Y_n$ s.t. for each $ n\ge 0$ the laws of $ X_n$ and $ Y_n$ are the same and $ Y_n\to Y_0$ almost surely.

Comment: of course, one can do it without any reference to uniform integrability or Skorokhod: split $ |X_n|^p \equal{} 1_{|X_n|^p\le C} |X_n|^p \plus{} 1_{|X_n|^p > C} |X_n|^p$. The expectation of the first converges to $ E(|X|^p1_{X\le C})$, the second converges uniformly in $ n$ to $ 0$, when $ C$ tends to infinity (use ideas of the Markov inequality).


[quote="MCrawford"]If $6\frac{4}{5} \cdot x = 1$, then what is the value of $x$?  Express your answer as a common fraction.[/quote]

[hide]
6 4/5*x=1
34/5*x=1
Divide both sides by 34/5.
x=1/(34/5)
x=5/34
[/hide]

[hide]x must be the reciprical of the number its multiplying to get 1.
6  4/5 = 34/5
reciprical of 34/5 = 5/34[/hide]

[quote="MCrawford"]If $6\frac{4}{5} \cdot x = 1$, then what is the value of $x$?  Express your answer as a common fraction.[/quote]

[hide]$6\frac{4}{5} \cdot x = 1$

$\frac{24}{5} \cdot x=1$

$x=\frac{5}{24}$[/hide]


In the grid shown, a number is to be placed in each small square so that the product of all three numbers in any row, column or diagonal is the same positive number. Find the sum of x and y.

[img]http://img291.imageshack.us/img291/6421/81345379.jpg[/img]

[hide="Solution"]Label the squares as shown:

$ A$ $ B$ $ C$
$ D$ $ E$ $ F$
$ G$ $ H$ $ I$

We know that $ 16 \cdot 8 \cdot I \equal{} 32 \cdot 8 \cdot B \equal{} 16 \cdot 32 \cdot C$

$ I \equal{} 2B \equal{} 4C$

From the diagonal top-left to bottom-right and the top row we get:

$ 32 \cdot 2B \cdot A \equal{} A \cdot B \cdot \frac {B}{2}$

$ 64AB \equal{} \frac {AB^2}{2}$

$ 64 \equal{} \frac {B}{2}$

$ B \equal{} 128$
$ I \equal{} 256$
$ C \equal{} 64$

The product of each row and column is therefore $ B \cdot 32 \cdot 8 \equal{} 32768$

$ A \cdot B \cdot C \equal{} 32768$

$ 8192A \equal{} 32768$

$ A \equal{} 4$

We can now set up equations in $ x,y$

$ 64x \equal{} 32768$
$ 16384y \equal{} 32768$

$ x \equal{} 512$
$ y \equal{} 2$

Thus the sum $ x \plus{} y \equal{} 514$[/hide]


n!= (n-1)! + (n-2)!

The below answer is incorrect becasue
i subtracted them wrong :p meh
Answer
[hide]
n!-(n-2)!=(n-1)!

so we get n(n-1)=(n-1)!

that would also mean

n=(n-2)!

now you can solve it![/hide]

[hide]
Note that $(n-1)!$ is always $\frac{n!}{n}$.
Also note that $(n-2)!$ is always equal to $n(n-1)$.

You can now re-write that equation:

$n! = \frac{n!}{n}+n(n-1)$
$n! = \frac{n!+n^{3}-n^{2}}{n}$
And it's pretty easy from there.
[/hide]

I don't think either of your solutions are valid.

8parks11: $n!-(n-2)! \neq n(n-1)$.

DiscreetFourierTransform: $(n-1)! \neq \frac{1}{n}$ and $(n-2)! \neq n(n-1)$.

[hide="one way"] We have $n! = n(n-1)(n-2)!$ and $(n-1)!+(n-2)! = (n-1+1)(n-2)! = n(n-2)!$. What do you get when you equate these?[/hide]

[quote="mr. math"]i solved this problem with guess and check, but is there any way to solve it using algebra?

n!= (n-1)! + (n-2)!

what is n?[/quote]
[hide="hint"]Factor the right side[/hide]
[hide="solution"]Divide both sides by $(n-2)!$.
This gives $(n)(n-1)=n-1+1$
$n^{2}-n=n$
$n^{2}-2n=0$. This gives the answer to be clearly n=2.[/hide]

Neat factoring icx :D

I replaced (n-2) with u  (I just liked it more this way :P)
(u+2)(u+1)(u!)=(u+1)(u!)+(u!)
(u+2)(u+1)=(u+1)+1
u^2+3u+2=u+2
u^2+2u=0
u=0,-2
n=2


induction
combinatorics proposed
combinatorics
chess

3. Plane is divided with horizontal and vertical lines into unit squares. Into each square we write a positive integer so that each positive integer appears exactly once. Determine whether it is possible to write numbers in such a way, that each written number is a divisor of a sum of its four neighbours.

[quote="Megus"]...each written number is a divisor of a sum of its four neighbours... [/quote]

What does "a sum" actually represent?

[quote="Megus"]...each written number is a divisor of a sum of its four neighbours... [/quote]

I'd rather say "each written number is a divisor of a sum of numbers written in four squares which are its neighbours"  :D

Can anyone slove this problem?I need an answer

Choose an arbitrary origin, and label the chessboard in a spiral beginning as below:

[code]13 12 11 10 25
14 3  2  9  24
15 4  1  8  23
16 5  6  7  22
17 18 19 20 21[/code]

For each $j$ (starting with $j=1$), we choose which square to fill in with $j$ using the following rule:

Let $i$ be the unfilled square with smallest label according to our labelling scheme.  If filling in square $i$ with the number $j$ will not lead to an illegal configuration (one where a square and all its neighbors are filled in, but fail to satisfy the divisor condition), we do so.  Otherwise, we place $j$ in the square in position $(j,j)$ relative to the origin ($j$ squares up and $j$ squares to the right of the square labelled 1).

Note that this construction [i]almost[/i] fills in squares by label...the only time $m<n$ and the square labelled $n$ is filled in before that labelled $m$ is if $n$ lies on the diagonal beginning $1,9,25, \dots$.

By construction each number is written in some square, and the divisor condition is satisfied.  If remains to show that each square is eventually labelled, which we do by induction on the label of the square.

The square 1 is filled in by 1.  Now assume that all squares labelled smaller than $i$ have been filled in.   Since our construction always defaults to trying to fill in square $i$, we just need to check that at some point it will be legal to do so.  There are two reasons why it would not be so:

1. All of square $i$'s neighbors have already been filled, so filling in square $i$ leads to a divisor condition which must be satisfied.  But this is impossible, since each square is adjacent to two squares with higher labels, only one of which is on that upper right diagonal.

2. Square $i$ is the last unfilled neighbor of a square whose other neighbors are filled.  It can be seen, though it's a pain to write down the proof of this, that there can only be one such filled square adjacent to $i$ (this is where it's important that our filling scheme places the out of order numbers on the upper right diagonal as opposed to some other line).  If that square is filled in with $k$, then once $j$ takes on the appropriate value modulo $k$ we can fill in square $i$ with $j$ and still have that divisor condition satisfied.

Kevinatcausa
This is an error because the neighbors of .3 are 2,4,12,14 but the sum of these is not divisible by 3. Contradiction


Who can help me with this problem : http://www.mathlinks.ro/Forum/viewtopic.php?p=155813#155813 ? It seems hard  :(

Couldn't you just post in that thread?  Why did you need to make a new one?


Determinati functiile $ f: \Re\longrightarrow\Re$ cu un numar finit de zerouri astfel incat pentru orice x,y reale sa avem
\[ f(x^4\plus{}y)\equal{}x^3f(x)\plus{}f(f(y)).\]

proof for $ f(x)\equal{}x$
plug in $ x\equal{}0$
$ f(0\plus{}y)\equal{}0f(0)\plus{}f(f(y))$
$ f(y)\equal{}f(f(y))$ Therefore $ f(y)\equal{}y$

o sa demonstrez ca $ f(x)=x$ este singura functie care verifica egalitatea din enunt. Fie $ x \longrightarrow 0$.Deci $ f(y)=f(f(y))$. Fie $ x \longrightarrow 1$, $ y \longrightarrow 0$.Atunci $ f(1)=f(1)+f(f(0))=f(1)+f(0)$ deci $ f(0)=0$ .In general daca $ f(k)=0$ atunci $ f(k+1)=f(1^4+k)=1^3 \cdot f(1)+f(f(k))=f(1)+f(k)=0$ si prin inductie rezulta ca $ f(k)=0$ oricare ar fi k natural.Dar aceasta contrazice ipoteza.Deci f(1) e diferit de 0.Presupunem ca exista un $ \alpha$ diferit de 0 si 1, cu $ f(\alpha)=0$.Atunci $ f(\alpha^4)= \alpha^3f( \alpha)=0$ si prin inductie iese ca $ f( \alpha{}^4{}^k)=0$.Cum f are un nr finit de zerouri rezulta ca $ \alpha$ poate fi 0,1,-1,ceea ce nu se poate pt ca $ \ alpha$ e diferit de -1.Deci $ \alpha=-1$ atunci $ f(1)=0$ fals.Deci f(x) dif de 0 daca x dif de 0.Fie$ P(x)=f(x^4)-x^4$.atunci rezulta ca $ P(x)=0$.Deci $ f(x^4)=x^4$, adica $ f(x)=x$ daca x pozitiv.Dar reiese si ca f(-x)=-x deci $ f(x)=x$


vector
trigonometry
geometry
parallelogram
rhombus
linear algebra
angle bisector

$ {\vec a}$ and $ {\vec b}$ are two noncollinear vectors with the same start point. Prove that the $ \left| {\vec b} \right|\vec a \plus{} \left| {\vec a} \right| \cdot \vec b$
is collinear with any vector of the bisector of the angle between $ {\vec a}$ and $ {\vec b}$ .

This being in intermediate topics, I don't know if you'll like this solution. It has a bit of linear algebra, and there's probably a simpler way.

There is always an angle-preserving mapping that can take $ \vec{a}$ and $ \vec{b}$ to vectors in a plane with one axis being the angle bisector of them. To see this, note that two noncollinear vectors (e.g. $ \vec{a}$ and $ \vec{b}$) uniquely define a plane. So simply redo the axes of the space the vectors are in so that two of the axes form that plane, and one of them is the angle bisector.

Now let's consider the plane the two vectors are on. The vectors have this form:

$ \vec{a} \equal{} |a| \langle \cos \theta, \sin \theta \rangle$
$ \vec{b} \equal{} |b| \langle \cos \theta, \minus{}\sin \theta \rangle$

Then $ |\vec{b}| \vec{a} \plus{} |\vec{a}| \vec{b} \equal{} |\vec{a}| |\vec{b}| ( \langle \cos \theta, \sin \theta \rangle \plus{} \langle \cos \theta, \minus{}\sin \theta \rangle ) \equal{} |\vec{a}| |\vec{b}| \langle 2 \cos \theta, 0 \rangle$. This vector is parallel to the axis that we defined to be the angle bisector of $ \vec{a}$ and $ \vec{b}$, completing the proof.

There's a much simpler approach:

Divide the expression by $ |\vec a|\cdot|\vec b|$ to get $ {\vec a\over|\vec a|} \plus{} {\vec b\over|\vec b|}$. This is nothing else but the vector sum of the corresponding [u]unit[/u] vectors, and by the rule of a parallelogram - which in this case turns into the rule of a rhombus - this sum bisects the appropriate angle.

(Of course, none of the vectors can be zero, otherwise the appropriate angle couldn't be defined.)

[hide="Geometric Interpretation"]Using $ \vec{v} \equal{} \| \vec{v} \| \hat{v}$, where $ \hat{v}$ is the unit vector in the direction of $ \vec{v}$, we have that:

$ \| \vec{b} \| \; \vec{a} \plus{} \vec{b} \; \| \vec{a} \| \equal{} \| \vec{b} \| \; \|\vec{a}\| \; \hat{a} \plus{} \| \vec{a} \| \; \| \vec{b} \| \; \hat{b} \equal{} \| \vec{a} \| \; \| \vec{b} \| \; (\hat{a} \plus{} \hat{b})$

Note that $ \| \vec{a} \| \; \| \vec{b} \| \; \hat{a}$ and $ \| \vec{a} \| \; \| \vec{b} \| \; \hat{b}$ span a parallelogram (as do any non-collinear vectors), and since they have equal lengths ($ \| \vec{a} \| \; \| \vec{b} \|$), this parallelogram is a rhombus. $ \| \vec{a} \| \; \| \vec{b} \| \; (\hat{a} \plus{} \hat{b})$ is the diagonal of this rhombus.

We know that the angle formed by two touching sides of a rhombus is bisected by the diagonal, hence this particular vector is an angle bisector.

We also know that positive scalar multiples of vectors have the same direction, so the result follows.[/hide]

[b]Edit:[/b] [i]See above[/i]


ratio
advanced fields
advanced fields unsolved

Alright, I am trying to find the equation to put into a 3d graphing calculator so that I can see a golden spiral in 3D space. I cannot seem to figure it out. The program I am using is graphcalc. I either end up with the usual bell curve of the golden ratio or a set that the program can't graph. Any help would be much appreciated.

Is graphing a spiral even possibile. And that's any spiral. There has to be some sort of mathematical equation that describes a spiral.

Yes it is possible. You can graph it polar I think by using the golden ratio sqrt(5)/2. So like

r = ae^[k * (1/2) * pi]

I think. I could probably derive it, but see if that works.


If six people decide to come to a basketball game, but three of them are only 2/5 sure that they will stay for the entire time (the other three are sure they'll stay the whole time), what is the probability that at the end, at least 5 people stayed the entire time?

Probability $ 5$ stay entire time:

$ \binom{3}{1} \times \left (\frac{2}{5} \right )^2 \times \frac{3}{5}\equal{}\frac{36}{125}$

Probability $ 6$ stay entire time:

$ \left (\frac{2}{5} \right )^3 \equal{} \frac{8}{125}$

Adding these yields $ \boxed{\frac{44}{125}}$.

OR

3 people stay the whole time or 4 people stay the whole time:

$\left (\frac{3}{5} \right )^3  +  \frac{2}{5} \cdot  +  \left (\frac{3}{5} \right )^2 \cdot \binom{3}{1} = \frac{81}{125}$

$1 - \frac{81}{125} = \boxed{\frac{44}{125}}$

There are two cases: 5 people and 6 people stayed.

Case 1: 5 people stayed the whole time. The probability that exactly 2 of those that are unsure stayed the entire time is $\binom{3}{2}\times \frac{2}{5}\times\frac{2}{5}\times\frac{3}{5}= 36/125$.

Case 2: 6 people stayed the whole time. The probability that all three unsure people stayed is $(2/5)^3 = 8/125$.

The sum of these probabilities is $\boxed{\frac{44}{125}}$.


[hide="hint"]we will be working in base 60.

let the initial time be $h_{1}$:$m_{1}$
and the final time be $h_{2}$:$m_{2}$

allow us to rotate the face of a clock 90 degrees ($1U_{60}$) clockwise and affix it to the complex plane.

let the hour hand and minute hand have length 1; thus, each can be represented by the equations cis $\theta_{h}$ and cis $\theta_{m}$

we have $\theta_{h}= 30(h_{2}-h_{1})+.5(m_{2}-m_{1})$
and $\theta_{m}= 360(h_{2}-h_{1})+5(m_{2}-m_{1})$

in base 60
$\theta_{h}= U_{60}(h_{2}-h_{1})+.U_{60}(m_{2}-m_{1})$
$\theta_{m}= 60_{60}(h_{2}-h_{1})+5_{60}(m_{2}-m_{1})$

now use law of cisines to solve for the difference of angles and plug into the equation $\frac{|60h-11m|}{2}$

good luck.[/hide]

[quote="slepadatu"]I need to convert a period of time (based on a start time and an end time) into a decimal number.  For example:

Start Time:  10:00
End Time:  11:30
Numeric Answer:  1.5

Does anyone know the equation for this problem?

Thanks.[/quote]
[hide]$h_{2}-h_{1}=a$

$\frac{m_{2}-m_{1}}{60}=b$

The difference in decimal form is $a.b.$[/hide]

couldnt you like...subtract? usin base 60


logarithms
trigonometry
function
algebra proposed
algebra

$ a \equal{} \sin{x}$
$ b \equal{} \cos {y}$ where $ a,b > 0$
Easy to check $ a \equal{} b$ now consider the equation:
$ \log_{2} {\sqrt {a \plus{} 3}} \equal{} \log_{3}(3a)$
$ \Leftrightarrow \log_{2}(a \plus{} 3) \equal{} 2(\log_{3}a \plus{} 1)$
Consider function:
$ f(a) \equal{} \log_{2}(a \plus{} 3) \minus{} 2\log_{3}(a) \minus{} 2$
$ f'(a) \equal{} \frac {1}{\ln{2}(a \plus{} 3)} \minus{} 2\frac {1}{\ln{3}a} < 0$
$ f(1) \equal{} 0$ imply that it has only solution $ a \equal{} b \equal{} 1$

[quote="zaizai-hoang"]Solve the system equations
\[ \left\{\begin{array}{l} \log _2 \sqrt {\sin x \plus{} 3} \equal{} \log _3 (3\cos y) \\
\log _2 \sqrt {\cos y \plus{} 3} \equal{} \log _3 (3\sin x) \\
\end{array} \right.
\]
[/quote]
 work only $ sinx\equal{}cosy\equal{}1.$

yeah, but it is not complete  :D We have $ a\equal{}b$ and then we will solve the following system equations:
\[ \left\{ \begin{array}{l}
 a \equal{} b \\
 \log _2 (a \plus{} 3) \equal{} 2\log _3 a \plus{} 2 \\
 \end{array} \right.
\]


function
linear algebra
matrix
algebra
polynomial
logarithms

The first one is $ y\equal{}1.2t^{2}$, the second is $ y\equal{}5t^{3}$, and the third is (approximately) $ y\equal{}2.80\cdot 1.7^{t}$.

How do I know?  Well determining the third one is easy, because if tells you that $ h(0)\neq 0$ so based on the choice of functions, that could only mean an exponential function.  Then the next two you can just determine by checking the first two $ t$ values.

Think you meant: $ y \equal{} 2.08\cdot 1.7^{t}$

Yes, typo.  I think you get the idea, though.

In practice, there are a couple of other useful techniques to distinguish polynomial from exponential growth:

- Finite differences.  This only works well if you have reason to believe that your sequence is actually generated by a polynomial, i.e. if you're counting something, or by a simple recurrence (for example, doing this on the Fibonacci sequence gets you the Fibonacci sequence back).

- Plotting $ \log y$ vs. x or $ \log y$ vs. $ \log x$.  The former reveals exponentials and the latter reveals power functions.  This is most useful if you're dealing with actual data, i.e. you need to find a good model and you want to apply linear regression.  This is, for example, the standard method to determine the order of a chemical reaction.

[quote="t0rajir0u"]This is, for example, the standard method to determine the order of a chemical reaction.[/quote]
In this case, you typically look at a plot of $ t$, the reaction time, vs. $ \ln{t}$ or some other function and you look to see which graph gives a linear relationship.  Once you know this, it's easy to pick out the order of the reaction.


Let $ f(x) \equal{} \sqrt {x}$. Find a number $ b$ such that $ \frac {\Delta f}{\Delta x}$ on the interval $ [1,b]$ is $ \frac {1}{7}$.

$ \frac {\Delta f}{\Delta x}$ means just slope

so $ \frac {\sqrt {b} \minus{} 1}{b \minus{} 1} \equal{} \frac {1}{7}$

$ \sqrt {b} \equal{} \frac {b \plus{} 6}{7} \implies b \equal{} \frac {b^2 \plus{} 12b \plus{} 36}{49}$

$ 49b \equal{} b^2 \plus{} 12b \plus{} 36 \implies b^2 \minus{} 37b \plus{} 36 \equal{} 0 \implies (b \minus{} 1)(b \minus{} 36) \equal{} 0$

but if when $ b \equal{} 1$, we have $ \frac {0}{0} \equal{} \frac {1}{7}$

so the only value of b is 36

answer : 36

A space (on a number line).

Square brackets $ [0,1]$ mean "all the numbers between 0 and 1, including 0 and 1 (inclusive).

Parenthethes $ (0,1)$ mean "all the numbers between 0 and 1, excluding 0 and 1 (exclusive).

[url=http://en.wikipedia.org/wiki/Interval_(mathematics)]Intervals[/url]


Let $ \textbf{V}$ and $ \textbf{W}$ be vector spaces, and let $ T$ and $ U$ be non-zero linear transformations from $ \textbf{V}$ into $ \textbf{W}$.  If $ R(T)\cap R(U)\equal{}{0}$, prove that $ {T,U}$ is a linearly independent subset of $ \mathcal{L}(\textbf{V},\textbf{W})$.

My main question is, how do I show that linear transformations are linearly independent?

For a two-element set, linear independence is just saying that neither is a constant multiple of the other. If they were constant multiples, either one would be zero or their images would be the same.

Yes, that makes sense.  That was my main problem, trying to show that they are not multiples of each other.


function
limit
topology
real analysis
real analysis solved

Qualifying Examination
problem 5/Day 2/October 2003

Give an example or show that none exist

a/ $f:R\rightarrow R$ whose set of discontinuities is $\mathbb Q$

b/  $f:R\rightarrow R$ whose set of discontinuities is $\mathbb R- \mathbb Q$

There's a classical example for a): $f(x)=0$ if $x\in\mathbb R\setminus \mathbb Q$, and $f(\frac pq)=\frac 1q$, where $q\in\mathbb N^*, p\in\mathbb Z,\ (p,q)=1$. It really is not that hard to show that it's continuous in all irrationals, and it's clearly discontinuous in all rationals.

I don't know what the answer is for b), but I have a feeling there's no such function. I'll give it more thought.

I might be mistaken, but this looks awfully familiar: it might have been discussed before.

i guess there is such a function; why shouldn't it exist? from a topological point of view, $\mathbb{R}-\mathbb{Q}$ and $\mathbb{Q}$ have just the same properties and continuity is something defined just by topology. i couldn't yet construct a counterexample; maybe something like
$\sum_{k=0}^{\infty} \frac{\sin(2k\pi x)}{k}$ for all irrational $x$; else some other constant value, 1 or something.
i'm not quite sure whether it works; i don't even know whether it converges for irrational $x$.

I cannot argue with you by saying that a function does not exist, for I don't know if this is true or not, but there is one crucial difference between the two sets: one of them is uncountable, while the other one is countable. There can be big differences (topologically) between such sets.

is the difference topologically that crucial? there is exactly none topological property except the size(which isn't really a topological property) which is different between the two sets. thus i don't see why there should be a proof that the one function doesn't exist but the other one does.
however, that's philosophy, it's better to argue about mathematically rigor proofs of either case.

something of that fashion(well, not exactly what we need): http://mathworld.wolfram.com/WeierstrassFunction.html
a more nice-looking one:
http://mathworld.wolfram.com/BlancmangeFunction.html

Hmmm...this is strange because I'm quite sure having already answered to this question before, but I cannot find the post...

There is no real function such that its set of discontinuity is $\mathbb R - \mathbb Q$, because it is a classical result that the set of discontinuity of any real function is a countable union of closed sets (so that the set of continuity of any real function is a countable intersection of open sets).

Assuming that such a function does exist, it would follows that both $\mathbb R - \mathbb Q$ and $\mathbb Q$ would be countable union of closed sets, with empty interior.
Therefore, $\mathbb R$ would be a countable union of closed sets too. But, from Baire's theorem, it would imply that $\mathbb R$ has empty interior, which is clearly absurd.

Note that the converse of the above classical result is also true. That is :
Let $A$ be any subset of $\mathbb R$ which is a countable union of closed sets. Then, there esits a bounded real function $f$ such that $A$ is exactly the set of discontinuity of $f$..

Pierre.

hi Pierre,

could you post a proof of this wondeful result ? I find it very interesting.
thanx

I also remember you saying something very similar to this, now that you've mentioned it..

What do you exactly want Alekk?

Pierre.

the proof of
" it is a classical result that the set of discontinuity of any real function is a countable union of closed sets (so that the set of continuity of any real function is a countable intersection of open sets). "

thanx Pierre.

[quote="pbornsztein"]There is no real function such that its set of discontinuity is $\mathbb R - \mathbb Q$[/quote]
Well, this is a classical problem in measure theory... It happened to be a homework exercise to me...

[quote="alekk"]
it is a classical result that the set of discontinuity of any real function is a countable union of closed sets (so that the set of continuity of any real function is a countable intersection of open sets). "
[/quote]

Define
$\omega_f(x,\delta) = \sup_{[x-\delta,x+\delta]} f(x) - \inf_{[x-\delta,x+\delta]} f(x)$
and the oscillation function
$\omega_f(x) = \inf_{\delta>0} \omega_f(x,\delta)$

$f$ is continuous at $x$ if and only if $\omega_f(x) = 0$.

Let $E(\epsilon) = \{x | \omega_f(x) \geq \epsilon\}$. We are able to prove that $E(\epsilon)$ is closed.

Suppose $x$ is limit point of $E(\epsilon)$. If $I$ is an open interval containing $x$ then $I$ must contain some point $y$ of $E(\epsilon)$ since $x$ is a limit point of $E(\epsilon)$. $\omega_f(x) \geq \omega_f(I) \geq \omega_f(y) \geq \epsilon$. thus $x\in E(\epsilon)$ and hence it is closed.
(In fact, \omega_f(x) is upper semi-continuous.)

$\{x | f \hbox{ is discontinuous at } x\}  = \{x | f(x) > 0\}  = \bigcup_{i=1}^\infty \{x | f(x) \geq \frac{1}{n}\}$

Therefore the discontinuity set is $F_\sigma$ (union of countable closed sets).

====================================================

It seems the conclusion holds when $f$ is defined on an open set. What about when it is defined on a closed set?

[quote="pbornsztein"]

Let $A$ be any subset of $\mathbb R$ which is a countable union of closed sets. Then, there esits a bounded real function $f$ such that $A$ is exactly the set of discontinuity of $f$..

[/quote]

A proof of this nice result
thx

Thanks to Liyi  I will not have to write it  :D
No thanks to Moubi : I will have to write a longer one... :(

Let $\dsp A = \cup_{n>0} A_n$ be a subset of $\mathbb R$ where $A_n$ is a closed set for all $n$.
Let $A_n = I_n \cup F_n$ where $I_n$ is the interior of $A_n$ and $F_n$ is the boundary of $A_n$ (and then a closed set with empty interior).
Wlog, we may assume that $A_n \subset A_{n+1}$ for all $n$ (otherwise consider the family $\dsp A'_n = \cup_{k \leq n} A_k$.

We define $f$ as follows :
- For $x \notin A$, let $f(x) = 0$
- For $x \in A$, let $n = \min \{ k | x \in A_k \}$.
Then $f(x) = - \frac 1 n $ for $x \notin \mathbb Q$ and $x \in I_n$,
$f(x) = \frac 1 n$ for $x \in \mathbb Q \cap I_n$ or $x \in F_n$.

1) If $x \in A$ and let $n$ defined as above.
- First case : If $x \in I_n$.
Then $x \in I_n-A_{n-1}$ (where $A_0 = \emptyset$).
But $I_n-A_{n-1}$ is an open set, so that there exists an open interval $I$ which contains $x$ and is contained in $I_n-A_{n-1}$.

If $x \in \mathbb Q$ then there exists a sequence $(x_p)$ of irrational numbers in $I$ which converges to $x$. Then $f(x_p) = - \frac 1 n$ so that $ \lim x_p = - \frac 1 n  \neq \frac 1 n = f(x)$. Thus $f$ is not continuous at $x$.

If $x \notin \mathbb Q$ we use a similar reasoning to prove the same conclusion.

Second case : If $x \in F_n$.
Then $f(x) = \frac 1 n$.
But $F_n$ is a closed set with empty interior thus its complementary set is a dense open set. Then : for all $p>0$ there exists $x_p \notin F_n$ such that $|x-x_p| < \frac 1 {2p}$, and there xists $\epsilon >  0$ such that $]x_p - \epsilon ; x_p + \epsilon [ \cap F_n = \emptyset.$
Then, for all $p$ there exists $y_p \notin \mathbb Q$ and $y_p \notin F_n$ such that $|x - y_p | \< \frac 1 p$.
The sequence $(y_p)$ has limit $x$ but $f(y_p) \leq 0$ or $f(y_p) = \frac 1 k \neq \frac 1 n$. In all cases, $f(y_p)$ does not have limit $f(x)$.
Thus, $f$ is not continuous at $x$.

Then $A$ is contained in the set of discontinuity of $f$.

2) if $x \notin A$ :
Then $f(x) = 0$.
Let $\epsilon > 0$ and $n_0 > 0$ such that $\frac 1 {n_0} < \epsilon$.
The set $A_{n_0}$ is closed then its complementary is an open set. It follows that there exists an open interval $O$ such that $x \in O$ and $O _cap A_{n _0} = \emptyset.$.
It follows that for all $y \in O$ we have $y \notin A_{n_0}$ and $|f(y)| < \frac 1 {n_0} < \epsilon.$
That means :
For all $\epsilon > 0 $ there exists an open interval $O$ containing $x$ such that for all $y \in O$ we have $|f(x) - f(y)| < \epsilon.$
Thus $f$ is continuous at $x$, and we are done....

Pierre.

[quote="pbornsztein"]
No thanks to Moubi : I will have to write a longer one... :(

Pierre.[/quote]

Yes, but you write perfectly long proof Pierre  ;)

thx again

I've a good training in very long proofs and I'm still trying to improve my style  :D

Pierre.


superior algebra
superior algebra unsolved

For a p-group of order $ p^4$, assume the center of $ G$ has order $ p^2$. Determine the number of conjugacy classes of $ G$.

For any $ x\in G\minus{}Z(G)$, $ C_G(x)$ must have index $ p$. (it is properly contained in $ G$ and properly contains $ Z(G)$)

It follows that the conjugacy class of $ x$ has order $ p$.

That would make the number of conjugacy classes equal to $ p^3 \plus{} p^2 \minus{} p$, no?

[quote="Zubr"]I wasn't sure in my answer but he is same so now think that it is correct. Thank you.[/quote]

Of course, you should only think your answer is right if you think that the logic is correct.  We could both be wrong. :)

I think C_G(x)  does not include Z(G)
Otherwise
 gx = ag = ga implies x=a for some a in Z(G) and g in G

$ C_G(x)$ include $ Z(G)$, because for all $ z\in Z(G)$ and for all $ g \in G$ $ g z g^\minus{}1\equal{}z$, so there is conjugacy class, which include only z.


Our design team drew all $48$ line segments that connect $6$ [i]equally spaced[/i] points on one side of rectangle $R$ to $8$ [i]equally spaced[/i] points on the opposite side. In at most how many different points inside $R$ can these $48$ line segments intersect?

sorry i don't have an answer to this :blush: .. any thoughts?
and also if you would like, generalize to $m$, $n$ points and $mn$ line segments

[quote="GlenthemanN"]the answer was
$\binom{8}{2}\cdot \binom{6}{2}= 420$
which I dont fully comprehend :oops:

you added in that theyd be equally spaced.. I dont think this would make a difference.[/quote][quote="goldengoal"]Here's a hint.  Hopefully this will shed some light on the problem.[hide="hint"] Consider picking any 2 points out of the 8 on one line and similarly picking any 2 points out of 6 on the other line. Draw in the 4 lines between these points and note that this gives one intersection point in the rectangle.  See if this makes sense and try to finish from there. [/hide][/quote][quote="Kamil Witek"]I did this on the meet. The trick is to notice htat the number that we are looking for is (1+2+3+4+5)(1+2+3+4+5+6+7). Lets first draw all the segments from the first point on the top to all of the points on the bottom. then draw the same thing for the second point on the top. then the segment we draw to the left most bottom point could intersect all the previously drawn segments at at most 7 points. for the segment we draw connecting the second points of the top and the bottom it is 6. and the pattern keeps on going. just try this approach yourself and you're bound to get it.

By the way what did you get? I got 5 (screwed up on the trig question)[/quote][quote="boxedexe"]My idea during the contest (good for missing class  :D): the maximum number of intersections is the maximum number of quadrilaterals (each quadrilateral has one intersection).  Clearly, the maximum number of quadrilaterals in $\binom{6}{2}\binom{8}{2}=420$.[/quote][quote="Zellex"]Between any two points on the side with $8$ points, if you connect those two points with the $6$ points on the opposite side, there are a total of $15$ intersection points.

Thus, you choose any two sets of points from $8$ points, and multiply that by $15$.[/quote]yoyoyo hold up for just one min'
what if the points are equally spaced? wouldn't some line segments have concurrency? :huh:


function
calculus
integration
algebra
domain
calculus computations

Anyone can demonstrate it???

\[ \Gamma (n + 1) = n\Gamma (n)  \]

thanks

From the integral definition: integrate by parts. There are few enough choices that it's hard not to find the correct proof that way.

\[ \Gamma(n+1) = \int_0^\infty t^ne^{-t} \ dt = \left[-e^{-t}t^n\right]_0^\infty +\int_0^\infty nt^{n-1}e^{-t} \ dt = n\Gamma(n) \]

Jimmy, you've forgotten to mention the domain for Gamma function defined using the integral formula --- $n \geq 1$

Actually, $n$ doesn't have to be an integer, and the proof above works for any $n>0.$ In fact, that could be any complex $n$ such that $\Re(n)>0.$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63000]this thread[/url] for more.


quadratics function algebra polynomial calculus derivative

This problem is derived from a famous book written by Sung-Dae Hong. He is a Korean mathematician who has come up with a lot of good mathematics questions in beginners and intermediate level (of Olympiad, of course).

"10th grade - 2nd term

Example 16-6: Find the minimum and maximum value of the following.

         y = a cos(x) - {sin(x)}^2

"

HInt for those who are not advanced: [hide]y = a cos(x) - {sin(x)}^2 = a cos(x) - [1 - {cos(x)}^2] = {cos(x)}^2 + a cos(x) - 1 => quadratic equation!![/hide]

Mod edit--I put the hint in spoiler. In general, hints and solutions in the intermediate forum should be put in spoiler. Thanks! -JSRosen3

This is better suited as an easy intermediate level problem. Can a mod please move this?

Sorry! I am a newcomer, so I do not exactly know how good "beginners" are. I guess it depends whether people here are "beginners" of this subject or of Olympiad. By the way, hi! Nice to meet you. I am Seung Hyun (or Sam) from South Korea. I study in the states and am looking forward to participating in IMO sometime in my high school career.

Is there some way I can move this problem to other sections? You told me it may be too hard for beginners.

Hi! I'm Jeff, from Connecticut. Recently there has been a push to get the problems in the Getting Started forum to an easier level (a set of challenge problems covering the approximate realm of the Getting Started forum can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13611]here[/url]. If a Getting Started moderator thinks that this problem is too hard for Getting Started, they will move this thread to Intermediate.

Good luck in your quest for the IMO  ;)

This is an excellent problem, and I especially appreciate seeing examples of problems from non-Western cultures. I will move this to Intermediate Topics, because the meaning of "Getting Started" is "getting started at problem solving in general," and thus includes pupils who have yet to learn algebra. A problem involving tenth-grade subjects (from most countries) usually fits comfortably into Intermediate Topics.

Thanks for sharing the problem.

[quote="liyi"]eh....To Chinese students, this is just a simple and basic problem in the school math work.[/quote]

Hmm, so maybe it should be in the advanced forum ;).

Yeah, well, you get my point. Intermediate looks right.

This question may be simple for Asian students who study for the standardaized tests, but that does not mean that Western students are not as good. In North East Asia (Korea, Japan, China), standardized tests are extremely difficult because they are basically the only measure of students' level. So what happens in all these countries is many students usually "memorizes" the traditional patterns of questions. This is one of sad abuses resulted from absurdly difficult standardized tests.

   You see, that is why these countries (except for China that has enormous population) do not do as good on IMO as the United States or Russia where school mathematics is much easier but focused on understanding of concepts. Not only that, this is why there are uncomparably more of able mathematicians in the Western society.

   To my Chinese friend, I guess we shouldn't be proud of this fact.

What is "spoiler" that you are talking about? Hmm... too many regulations. (T.T;;)

[quote="Seung-Hyun Yang"]
"10th grade - 2nd term
Example 16-6: Find the minimum and maximum value of the following
         y = a cos(x) - {sin(x)}^2[/quote]
if so,
since sinx=-sin(x)^2+a sin(90-x)=y. Because sin x will always be less than 1, we define x as 0 to get the highest amount which is 1??  and  the mininum we set x to be ninety to get -1?
(am i seriously misunderstanding the question here)?

Sorry again for my mistake here. I completely forgot to add this crucial sentense: "answer in terms of a." The value of the equation cannot be defined as a certain number, and I think that confused you. For example, (and this is not true answer, so don't worry) the answer could be "when a>10, the maximum value is 9.2a - a^2; when a=10, ......"

I have encountered some Olympiad questions involving maximum and minimum value, and that made me forget the necessity of details that might confuse other people. Again, sorry about that. I think I shouldn't translate questions anymore...

If you still have further question, please quote! It is my mistake to arouse confusion due to my lack of proficiency in translating, and it is my responsibility to calm any turmoil created because of me.

I think that the maximum is |a| and the minimum is -|a|.
Provided a>0 (which you can assume, since if a<0 it is the reverse across the y axis of the function with |a|)
You can rewrite it as y=cos^2(x)+acos(x)-1 or as y=(cos(x)-1)(cos(x)+1)+acos(x).  Then my intuition tells me that the max will be getting cos^2(x) to be 1, which will make acos(x)=a.  And the minimum will be when acos(x)=a and cos^2(x)=1.

Actually, I'm not sure of this result anymore, since things get weird when 0<a<1.  When a is in that range, the result changes.

So the maximum is |a| and the minimum is either -|a| when |a|>1 or something else close to -1 when 0<a<1.

Big hint:

[hide]As the spoiler notes, it can be turned into the polynomial f(x)=cos^2(x)+a*cos(x)-1.  Setting cos x=m, we see that the range of this polynomial is the same of g(m)=m^2+a*m-1 when -1<=m<=1.  Therefore, we need only to examine the maximum and minimum values of this function over that domain.[/hide]

solutions on the interval 0<=x<=2*pi:

[hide]

when a=0: max at x=0, pi ; min at x=pi/2, 3*pi/2

when -2<a<0: max at x=pi ; min at cos(x)=-a/2

when 0<a<2: max at x=0 ; min at cos(x)=-a/2

when a<-2: max at x=pi ; min at x=0

when a>2: max at x=0 ; min at x=pi

[/hide]

explanation:

[hide]

taking the derivative,

y'=-2cos(x)sin(x)-a*sin(x)

   =(sin(x))*(-2cos(x)-a)

therefore the critical points occur when sin(x)=0 or cos(x)=-a/2:

the former solution gives x=0, pi

the latter solution gives x=arccos(-a/2) and is restricted to -2<=a<=2.

testing each of the 3 critical points for values of a gives the solutions

[/hide]


symmetry
probability
function
real analysis
expected value
geometric series
probability and stats

Why does the Cantor distribution have an expected value of 1/2? Wikipedia says that it's by "symmetry", what do they mean by that?

If $ X$ and $ 1\minus{}X$ have the same distribution, then $ 2E(X)\equal{}E(X)\plus{}E(1\minus{}X)\equal{}E(1)\equal{}1$

Thanks mlok, I completely understand your input, but what's the explanation of the expectation given the probability mass function is $ P\{X_{n} \equal{} 0\} \equal{} P\{X_{n} \equal{} 2 \} \equal{} \frac{1}{2}$ (i.e. why just $ X_n \equal{} 0, 2$?) and that the random variable $ X \equal{} \sum_{n \equal{}1}^{\infty} \frac{X_{n} }{3^{n}}$?

Because I've tried $ E[X] \equal{} E[\sum_{n \equal{}1}^{\infty} \frac{X_{n} }{3^{n}}] \equal{} \sum_{n \equal{}1}^{\infty} \frac{E[X_{n}]}{3^{n}}$ and then I didn't really know where to go from there to get $ E[X] \equal{} \frac{1}{2}$.

[quote="mstrip"] then I didn't really know where to go from there to get $ E[X] \equal{} \frac {1}{2}$.[/quote]
If you just substitute $ E[X_{n}]\equal{}1$ in the above what you get is a geometric series whose value is $ 1/2$.Isn't it?

[quote="mstrip"]Thanks mlok, I completely understand your input, but what's the explanation of the expectation given the probability mass function is $ P\{X_{n} \equal{} 0\} \equal{} P\{X_{n} \equal{} 2 \} \equal{} \frac {1}{2}$ (i.e. why just $ X_n \equal{} 0, 2$?)[/quote]

The condition that $ X_n$ takes values in $ \{0,2\}$ a.s. ensures that the random variable $ X$ is supported on the Cantor set (as one possible definition of the Cantor set is the set of numbers whose ternary expansion doesn't contain the number $ 1$).

@limsup: Oh, right, the sums for a geometric series are different if $ n$ starts at 0 vs. starting at 1. Thanks!


[hide="Hint"] Let $a = \frac{1+\sqrt{5}}{2}, b = \frac{1-\sqrt{5}}{2}$.  Then $ab =-1, a+b = 1$.

[hide="Furthermore..."] We have the identity

$a^{n+1}\pm b^{n+1}= (a+b)(a^{n}\pm b^{n})-ab(a^{n-1}\pm b^{n-1})$ [/hide] [/hide]

can you explain to me how this came

$a^{n+1}\pm b^{n+1}= (a+b)(a^{n}\pm b^{n})-ab(a^{n-1}\pm b^{n-1})$

and thank you

egheghehg is this really highschool basics?

This is the famous Fibonacci sequence (but given by Binet's formula) so let's write $f_{n}$ instead of $s_{n}$.

Let $\phi=\frac{1+\sqrt{5}}{2},\ \psi=\frac{1-\sqrt{5}}{2}$. Then $\phi,\psi$ are the solutions of $x^{2}=x+1$ what makes 1) obvious (by induction).

2. It could be written in equivalent form
\[f_{k+l+1}\ =\ f_{k}f_{l}+f_{k+1}f_{l+1}\]
Prove this inductively (induction by $k$, with $l$ - fixed natural number)

3. Put $l=k=n-1$ in 2.

4. This also can be easily obtained from 2.

@Mr.Mathematics:  Only the most basic proofs go in High School Basics; this would probably do better in Intermediate.


what do you guys consider to be a good college math resume? achievements(usamo etc.), internships, sat's, etc?
Here is mine so far:
going into sophmore year $ \implies$ taking BC calc and then 2 years of math at brown, 2007-2008 top freshman in mathleauge, 2007-2008 member of Math A-Team, 2007-2008 AMC 10A 108 (top in school)Summer: Finance course at brown and then internship in finance with national bank of Australia. I would appreciate input- dont be afraid to be harsh!
Note: I have been studying alot of competition math and should do about 7 on the AIME and 140+ on the 10

Define "good."  Who are you trying to please?  (Ideally, the answer should be "no one.")

im trying to pleaz myself but also get into a good university

If you're trying to please yourself, why ask our opinion?

Same question:  define "good university."

Hey good job. I wish I knew that many pencil spinning tricks.

haha thanks  :lol: Ive been spinning for a few months-- its pretty fun, but distracting in class lol

This site isn't the best place to look for praise for your accomplishments. From my view, your accomplishments are great. If you continue as you do, you'll get into a fine university. But just realize that your resume doesn't look like the one of a top student.


Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be $ 6$. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts $ 1$. If it comes up tails, he takes half of the previous term and subtracts $ 1$. What is the probability that the fourth term in Jacob's sequence is an integer?

$ \textbf{(A)}\ \frac{1}{6} \qquad
\textbf{(B)}\ \frac{1}{3} \qquad
\textbf{(C)}\ \frac{1}{2} \qquad
\textbf{(D)}\ \frac{5}{8} \qquad
\textbf{(E)}\ \frac{3}{4}$

This is pretty straightfoward. Set up a diagram with all $ 8$ outcomes, and count how many are integers. There are $ 5$ of them, so the probability is $ \frac{5}{8}\Rightarrow \boxed{D}$.


A baseball team has won 50 games out of 75 played. The team has 45 games remaining in the season. What is the smallest number of remaining games that the team must win in order to have won at least $ 60\%$ of its games this season?

It will play a total of 120 games. 60% of that is 72. Thus they need to win 22 more.


algebra polynomial function LaTeX conics parabola

[quote="Interval"]For the polynomail function below, identify all zeros and their multiplicities.

$f(x) = 5(x-2)(x+3)^{2}(x-(\frac{1}{2}))^{4}$

Also, how do I graph the graph above?

I need to learn [size=150]HOW [/size]to graph such complex functions WITHOUT a calculator.

[edit by mod]: learn latex already...it is very handy and you should know because youve posted over 450.  You can learn how by clicking latex on the left side of the screen[/quote]

x=2, -3, 1/2.
multiplicity is how many times it appears in the solution
so x=2 happens 1 time, so the multiplicity is 1, and x=-3 happens 2 times, so the multiplicity is 2. With the total number of solutions you can find the "end-behavior" which is what the main shape looks like ($x^{some even number}$ looks like a parabola kidna of shape and $x^{odd}$ looks like a $x^{3}$ kind of shape. Look for the x-intercept and y-intercept

[quote="Interval"]
$f(x) = 5(x-2)(x+3)^{2}(x-\left(\frac{1}{2}\right))^{4}$

Also, how do I graph the graph above?
[/quote]

ok im not exactly sure of what multiplicities are but i do know how to find the zeros...

$0=5$

$0=x-2$

$0=(x+3)^{2}$

$0=\left(x-\frac{1}{2}\right)^{4}$

solve for x and those are your zeros

if you are wondering how i got that check your current math textbook for the Zero Product Property....

the coefficient of the highest power is positive so the right end gets higher.  the power of the equation is 7 so it looks like a cubic graph.  It starts low, then passes through the x-axis at x=2,-3 and .5 then goes to infinity

[quote="Interval"]For the polynomail function below, identify all zeros and their multiplicities.

$f(x) = 5(x-2)(x+3)^{2}(x-(\frac{1}{2}))^{4}$

Also, how do I graph the graph above?

I need to learn [size=150]HOW [/size]to graph such complex functions WITHOUT a calculator.

[edit by mod]: learn latex already...it is very handy and you should know because youve posted over 450.  You can learn how by clicking latex on the left side of the screen[/quote]
[hide]
multiplicity is the power of the root.
$x-2$ has a multiplicity of 1
$x+3$ has a multiplicity of 2
etc.
graph polynomial equations by first putting all the x-intercepts on the graph. then mark which side of the x axis it goes to each time it crosses it. you can do this by 2 ways.
one way is to know the side (positive or negative) when you plug in $-\infty$ for $x$. (or something hugely negative, lower than the lowest root) then you change signs whenever you encounter a intercept with an odd multiplicity, or keep the sign whenever you encounter an intercept with an even multiplicity.
another way is to test a point in between each two intercepts, and mark the sign.
lastly, graph 1-2 points in between each intercept, and connect all the points with a [b]SMOOTH[/b] curve.
make arrows pointing upward or downward at each end, depending on which sign it is at each end.
[/hide]

Thank you all for your answers and tips.


number theory
greatest common divisor
relatively prime

Suppose $ a$ and $ b$ are integers that divide the integer $ c$. If $ a$ and $ b$ are relatively prime, show that $ ab$ divides $ c$. Show that if $ a$ and $ b$ are not coprime, then $ ab$ does not necessarily have to divide $ c$.

So $ c \equal{} ak$ and $ c \equal{} bj$ for $ k,j \in \mathbb{Z}$. Now $ akx\plus{}bjy \equal{} 1$ for $ x,y \in \mathbb{Z}$. Also $ ab \equal{} \frac{c^2}{kj}$. Thus $ c^2 \equal{} abkj$. Now what? Any counter-examples for the second case?

[quote="lordmahone123"]Now $ akx \plus{} bjy \equal{} 1$ for $ x,y \in \mathbb{Z}$.[/quote]  No.  (The equation you've written suggests that you believe $ c$ is relatively prime to itself  :wink: )

[quote="hsiljak"]$ 2|12$ and $ 4|12$, but $ 2\times4$ doesn't divide $ 12$.[/quote]
Or even, 2 divides 2 and 2 divides 2, but 4 does not divide 2.  (By the way, sorry for taking so long to respond to your e-mail!  Hopefully later this afternoon....)

[hide="Part 1"]
I'm making this up as I go, so it's probably not the most elegant proof in the world.

We assume $ a,b\ne 1$ (if either one is $ 1$, the problem is pretty trivial).

$ c\equal{}ak\equal{}bj\iff bj\equiv 0\bmod a$. Clearly $ j\equiv 0\bmod a$, so $ j\equal{}am$.

$ c\equal{}ak\equal{}bam$ (haha, bam), so $ \frac{c}{a}\equal{}k\equal{}bm$. $ k\equiv 0\bmod b\implies k\equal{}bn$.

$ \frac{c}{a}\equal{}bn\equal{}bm\iff\frac{c}{ab}\equal{}n\equal{}m$. Clearly $ n,m\in\mathbb{Z}$, so we're done.[/hide]

[hide="Part 2"]
Counterexample: $ a\equal{}5, b\equal{}10, c\equal{}30$. $ \text{gcd}(a,b)\equal{}5$. $ 5|30$, $ 10|30$, $ 50\nmid 30$.
[/hide]


geometry
3D geometry
pyramid
trigonometry
Pythagorean Theorem

Consider a pyramid $P-ABCD$ whose base $ABCD$ is a square and whose vertex $P$ is equidistant from $A$, $B$, $C$, and $D$. If $AB=1$ and $\angle APD=2\theta$ then the volume of the pyramid is

$\text{(A)} \ \frac{\sin \theta}{6} \qquad \text{(B)} \ \frac{\cot \theta}{6} \qquad \text{(C)} \ \frac1{6\sin \theta} \qquad \text{(D)} \ \frac{1-\sin 2\theta}{6} \qquad \text{(E)} \ \frac{\sqrt{\cos 2\theta}}{6\sin \theta}$

[hide]Draw the altitude from $P$ to square $ABCD$ and let the point of intersection be $Q$. We need to find $PQ$ so that the volume will be $\frac13 PQ$.

In triangle $ABP$, draw the altitude from $P$ to $AB$ and let that be called $R$. Notice that $\sin \theta=\frac{AR}{AP}\implies AP=\frac{1}{2\sin \theta}$. Because $ABCD$ is a square, we know that $AQ=\frac{1}{\sqrt 2}$.

Now we can apply the Pythagorean Theorem to find $PQ$. Simplifying, $PQ=\frac{\sqrt{\cos 2\theta}}{2\sin theta}$, and the volume is $PQ=\frac{\sqrt{\cos 2\theta}}{6\sin theta}$[/hide]

[hide="Question"]
How does 4everwise get $ \sin\theta=\frac{AR}{AP}\implies AP=\frac{1}{2\sin\theta}?$ Also what are we setting equal here?
Sorry, not very experienced with problem solving trig techniques yet! :( [/hide]

I am not going to tell you the answer but I will lead you up to it. Basically, 4everwise first wants to find an expression for the height of one of the side faces. Then that height can be used as a hypotenuse.

[hide="Another Question"]
[url=http://grab.by/oQYQ]AMC Solution[/url]
How does the AMC Solution get that $PM=\frac12 \cot{\theta}?$[/hide]

how did you simplify to get the final value of PQ


trigonometry
symmetry
geometry unsolved
geometry

A,B,C are coaxal circles,P is on circle C, PG,PF are tangents to circles A,B.Prove:when P moves on circle C,|GH|:|FI| is a centain value.

Does coaxal means that the centres of the three circle lie on the same line?

Use the sine law for $GH$ in the circle $\mathcal A$ and for $FI$ in the circle $\mathcal B$ to conclude that what we must prove is equivalent to $\frac{\sin\angle PGH}{\sin\angle PFI}$ being constant, which, in turn, is equivalent to $\frac{PG}{PF}$ being constant. Call this result $(\#)$.

Let's look at it backwarrds now: we are given the circles $\mathcal A,\mathcal B$, and we want to prove that the locus of the points $X$ for which $\mathcal P_{\mathcal A}(X)=k\mathcal P_{\mathcal B}(X)\ (*)$ for some fixed $k$ is a circle coaxal with $\mathcal A,\mathcal B$, where $\mathcal P_{\mathcal O}(T)$ is, in general, the power of $T$ wrt the circle $\mathcal O$. $(\#)$ will follow from here.

Let $r_a,r_b$ be the radii of $\mathcal A,\mathcal B$ respectively, let $A_1,A_2$ be the intersections of $\mathcal A$ with $AB$, and define $B_1,B_2$ similarly.

We can write $(*)$ as $PA^2-r_a^2=k(PB^2-r_b^2)$, and it's well-known that the locus of the points satisfying this condition is a circle (you can apply Stewart's Theorem, for example, to deduce this). Because of the symmetry of the figure wrt $AB$, thic circle has its center on $AB$, so if we prove that the two points where it cuts $AB$ (call them $C_1,C_2$) are maped to each other by the involution which maps $A_1$ to $A_2$ and $B_1$ to $B_2$, we will have shown that the locus we're looking for is, indeed, coaxal with $\mathcal A,\mathcal B$.

The above follows quickly (we have to perform some computations, but it's not hard to do that) from $\frac{\overline{C_1A_1}}{\overline{C_1B_1}}:\frac{\overline{C_1B_2}}{\overline{C_1A_2}}=\frac{\overline{C_2A_1}}{\overline{C_2B_1}}:\frac{\overline{C_2B_2}}{\overline{C_2A_2}}$, which is equivalent to $(*)$ when we replace $P$ with $C_1,C_2$ and eliminate $k$.


Well, I tried the first USACO training problem, and when I compiled it using my own machine, it went fine. But when I submitted it to the USACO grading, I got this:

[quote]
  > Run 1: Execution error: Your program exited with exit status `1'.

      ------ Data for Run 1 ------
      COMETQ
      HVNGAT
      ----------------------------

Your program printed data to stderr.  Here is the data:
-------------------
Exception_in_thread_"main"_java.lang.NoClassDefFoundError:_ride
-------------------
[/quote]

And I have no idea what that means  :maybe:

Here's the code that I submitted:

[quote]
*/
import java.io.*;
import java.util.*;

class YourRideIsHere {
    public static void main (String [] args) throws IOException {
        // Use BufferedReader rather than RandomAccessFile; it's much faster
        BufferedReader f = new BufferedReader(new FileReader("ride.in"));
        // input file name goes above
        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("ride.out")));

        String s = f.readLine();   // gets entire line

        char[] c = s.toCharArray();
        int[] arr = new int[c.length];
        for(int i=0; i<c.length; i++)
            {
            arr[i]=c[i]-'A'+1;
            }

        int cometNum = 1;
        int groupNum = 1;

        for(int i=0; i<6; i++)
            {
            cometNum = cometNum * arr[i];
            }

    for(int i=6; i<c.length; i++)
        {
        groupNum = groupNum * arr[i];
        }

        if (cometNum%47 == groupNum%47)
            {
        out.write("GO");                    // output result
            }
        else
            {
                out.write("STAY");
            }
        out.close();                                  // close the output file
        System.exit(0);                               // don't omit this!
    }
}
[/quote]

Could someone explain to me what I did wrong?

You need to name your class "ride" not "YourRideIsHere"

In general, name your class what is listed as Task name

(the problem statement says Task name: ride)

Oh, oops. Thanks for pointing that out, I got it to work (after changing some parts of the code I included above, I thought readLine() got everything).