probability
function
probability and stats

NO1.Let X,Y be a random variable defined as follows:
X and Y are real valued functions on a sample space W ={(x,y): 0<=x<=1 & 0<=y<=1}, where a probability is given by width, X satisfies
X((x,y))=8x for all (x,y) in W.
and Y satisfies
Y((x,y))=7(x+y) for all (x,y) in W.
If F is a joint probability distribution of X,Y, anser  the value F3,2 , down to two places of decimals.

No.２　Let X,Y be a random variable defined as follows:
X and Y are real valued functions on a sample space W ={(x,y): 0<=x<=1 & 0<=y<=1}, where a probability is given by width, X satisfies
X((x,y))=9x for all (x,y) in W.
and Y satisfies
Y((x,y))=9(x+y) for all (x,y) in W.
If p(x,y) is a joint probability density of X,Y, anser  the value p(1,2), down to two places of decimals.

NO３
Let Y be a random variable defined as follows:
Y is a real valued function on a sample space W ={(x,y): 0<=x<=1 & 0<=y<=1}, where a probability is given by width, and Y satisfies
Y((x,y))=10(x+y) for all (x,y) in W.
anser  the variance V(Y), down to two places of decimals.

please tell me how to sove.

These all seem like pretty much the same problem.  Recall that if a joint density function is given by $ f_{X,Y}(x,y)$, then the joint probability is $ \iint_{D}f_{X,Y}(x,y)\,dx\,dy$, where $ D$ is the region you're considering.


http://www.math.upenn.edu/~wilf/DownldGF.html
i read a few pages of this. i didnt undrestand it completly. can someone please explain the example in page 4 and give some easier examples than that ?

Here's a clearer solution to the problem.

Let us take the 1 and put it into the series, with:
A(n) +1 = 2(A(n-1)+1)
Then let's define the sequence B, so that
B(n) = A(n)   +1
and notice that B(n) = 2(B(n-1))
and the sequence is 1,2,4,8,..., so we can give the solution immediately.

There is a good article about Generating Functions at
[url]http://en.wikibooks.org/wiki/High_School_Mathematics_Extensions/Counting_and_Generating_functions[/url]


discuss your individual and team rankings and scores here.

I got a 22/14=36, putting me in 20th place. Dang I was just 1 point off from cd qualification :( just b/c of one stupid mistake on the sprint round :mad:

My team took 14th with a team score of 45. Apparently we screwed up team round and got an 8.

Individual team members placed 62nd, 100th, and 102nd.

Well I think they tell you. But they tell you the team this afternoon.

The written results were all released this morning. My coach received them.

could u plz tell me the al individual and team results?

The coaches already got them this morning!?
The last 3 years I thought they were picked up a couple hours before the ceremony, after touring all day?

they don't announce it until the awards.

Right, that's the only one that's not already known (though after watching there are usually 2 clear favorites)

[quote="gauss1181"]I've somehow heard Max Schindler won?[/quote]

Can you confirm all the other results that jmerry has posted in the other thread?  Thanks.

So apparently, MathCounts was around the same difficulty as last year but without so much COUNTING AND PROBABILITY problems? My 3/8 on the target last year was deadly :rotfl:

haha i got 69th place individual with a 20/8=28 i think

our team got 8th overall and others placed 4th, 36th (i think), and 52nd

ugh. 14/12=26, putting me at 94th. I'm so pissed because on target, i KNOW i could have gotten a perfect. it was so easy. i misread one of them and needed like 30 more seconds for another. i probably could have done better at sprint too, but oh well.

haha our team got 30th (delaware), but we could have done soo much better. cause my other teammates also made stupid mistakes (like a lot of them) and on team round for the 1.499 one, i rounded it to the nearest tenth  :mad: i know i'm stupid haha.

Hmmm... got 28/14...  darn, if I had not misread #5 target I would be first written.... darn that just sucks...

Also don't watch the webcast for 2 reasons:

* pythag011 ruins it
* I failed....

 :(  :(  :(

[quote="dnkywin"]
* pythag011 ruins it
[/quote]

well i just watched it haha. maybe his parents should drop him in a public school or something, get some social skills. it's too bad - some of those things he did we just completely inappropriate. he's going to get really ragged on aops.

and btw - you didn't fail (how does that make the rest of us feel? lol)

Yeah seriously if anyone failed countdown it's I. I didn't even buzz in... :|

But the test was not legitimate in my opinion. I got really lucky and somehow made countdown with 26/14=40.

Oh well, based on written I got 3-1-3-7 throughout this year.

It can be interpreted also as 1(number of points missed on school, or the number of people who got my score)-3(-way tie for first)-3-7. :P

really you didn't fail.

you still got the IMO who knows you'll make it.

I totally failed,
I didn't make aime.

is this what serious mathy kids do? discuss how fail-y they are?
i mean i never even heard of aime before this year and in public school world i'm already considered smart.
but i must say the nats competition itself was very interesting this year.

[quote="the.bookblogger"]is this what serious mathy kids do? discuss how fail-y they are? [/quote]

yes.

[quote="the.bookblogger"]i mean i never even heard of aime before this year and in public school world i'm already considered smart.[/quote]

Same here and that's one of the reasons I epic fail, and not in a pythag011 way of fail.

[quote="the.bookblogger"]but i must say the nats competition itself was very interesting this year.[/quote]
Did you make nats this year?

[quote]Did you make nats this year?[/quote]

HA no. didnt even get countdown at states. (dont laugh) well it was also my first year so i have excuses.[/quote]

Dont think anyone failed as bad as I. Made 13 points of stupid mistakes (5 sprint, 4 target) (copying, adding, incorrect reading, not dividing by 2, wrong form). So in my practices, I got like about 32, and my goal was to get atleast 80th. I should have gotten 50th w/o 13 pts of mistakes, but that dropped me so bad. I am very disappointed my MC career had to end this way. I have gotten so much better, yet I scored lower than last year.

And yay LAST on the team. LOL

WE did okay on team, 9/10, if that is a consolation......

 :(  :(  :(

[quote="pinkmuskrat"]. Made 13 points of stupid mistakes (5 sprint, 4 target) (copying, adding, incorrect reading, not dividing by 2, wrong form). So in my practices, I got like about 32, and my goal was to get atleast 80th. I should have gotten 50th w/o 13 pts of mistakes, but that dropped me so bad. I am very disappointed my MC career had to end this way. I have gotten so much better, yet I scored lower than last year.[/quote]

So what did you get? 32?

How hard was this national test for you? Rating: 1 being super easy to 10 being insanely tough?

Eh, in my opinion it was harder than last year but not that much harder. I'll give it a 4.

Although I've noticed that my difficulty ratings are often off :(

I'd rank it like a 4ish [High expectations for the test? Low expectations for me?], but I totally blanked on the sprint round (putting stupid random answers for the whole last page, expect for the one where the answer was 6, and not checking). The sprint round made me focus for the target round though, which was easy and I did well on.

Obviously, it wasn't 2005 hard. The back page was surprisingly easy. Euh, now, it seems so.

Team round was just fun. #5 listing FTW :).

Dang... So Hawaii still never had anyone make CD.
I got (21/16?=37?), putting me in 17th place. What a fun number.
The CD cutoff was so close...

Random Question: Do the graders ever make mistakes?

ok, disregard my other thread where i said that i got a 37, i failed, and got a 35...

21 on sprint, and 14 on target... i failed.
i ended up getting 28th, and my team got 15th (michigan) because we epic failed the team round and got 7 out of 10.


Solve it
$ y^{p}\equal{}x^{2}\plus{}3$
where $ x,y,p$ are positive integers.
Any elementary solutions?

The only solutions I could find are:
1) y=2, p=2, x=1
2) p=1 and y=x^2+3
Here is a partial proof that there are no other possibilities:
[hide]First note that if x is even then y is odd and vice versa. If y is even, let y=2n and you have (2^p)(n^p)=x^2+3. If p>2, then x^2+3 must be a multiple of 8, an impossible scenario (x^2=0,1,4 mod 8). If p=2, then the only solution is y=2, p=2, x=1. After that, the gaps between two consecutive perfect squares grow monotonically larger. If p=1, then there is the solution y=x^2+3.
Now let's try y is odd and x is even. Let x=2q and you have y^p=4q+3. Since y is odd, this can be true only if p is odd (y^2=1 mod 4 if y is odd). Sorry, this is as far as I could get. If there are any other solutions, they must satisfy the conditions that p is odd, y is odd, and x is even.[/hide]

Well , I think that algebraic number theory works here but I'm not pretty sure... :maybe:
LEFTERIS


Well, there should always be opportunites where you live.  For example, at a library, nursing home, local food bank, etc.  I'm sure they would be happy to accomodate volunteers.

I myself have volunteered at the local public library, a nursing home, and a (public) school for disabled children.

if you just want a good way to spend a summer, u might also want to consider a math summer program- some are discussed in [url=http://www.artofproblemsolving.com/Forum/index.php?f=136]Other US contests and programs[/url]

[quote="banblink182"]Hi everyone,

I was wondering if any of you knew any good volunteering opportunities (either in the US or abroad) that either you've had personal experience with or have heard of from reliable sources.

Thanks[/quote]

It's usually quite gratifying giving back to the community that you live in, that way, you can continue the relations that you build during the school year.

I was a counselor at a week-long day camp over the summer. I'd definitely recommend doing something like that, it was definitely worth the time.


Violet, her brother and her parents are going to take the Metrolink from Santa Clarita to Los Angeles Union Station.  This trip covers three zones, and they need round-trip tickets.  They will save $ 30\%$ on the two adult tickets and $ 50\%$ on the two youth tickets since they are traveling during off-peak hours.  Using the fare chart, what is the total cost for Violet's family?

  \begin{tabular}{|c|c|c|}
   \hline No. of & One-Way & Round-Trip \\
   Zones & Peak & Peak \\ \hline
   $ 1$ & $ \$4.25$ & $ \$6.75$ \\ \hline
   $ 2$ & $ \$5.25$ & $ \$9.00$ \\ \hline
   $ 3$ & $ \$6.25$ & $ \$11.50$ \\\hline
  \end{tabular}

[quote="james4l"]$ \begin{tabular}{|c|c|c|}
\hline No. of & One\minus{}Way & Round\minus{}Trip \\
Zones & Peak & Peak \\ \hline
1&\$4.25 &\$6.75 \\ \hline
2&\$5.25 &\$9.00 \\ \hline
3&\$6.26 &\$11.50 \\\hline
\end{tabular}$

They pay 11.50 for a normal peak round-trip at three zones. However they pay (2*0.7+2*1)=2.4 times this amount. 2.4*11.5=$ \$27.60$[/quote]

instead of (2*0.7+2*1) it's (2*0.7+2*0.5) because its 50% off the two youth tickets, than it will add up to 2.4 times! :)


superior algebra
superior algebra unsolved

We know that([b]Z[/b][sqrt(2)],N) is an Euclidean domain. And 2 is not a atom in [b]Z[/b][sqrt(2)]. However, 2= (root 2).(root 2)= (2+ sqrt2).(2-sqrt2) would seem to contradict this. Explain this with justification?

I dont know how to explain it, can you show me or give me some hints please?

Thank you in advanced for your time on this question

One has $ 2 \pm \sqrt {2} \equal{} \sqrt {2}(\sqrt {2} \pm 1)$. What are the numbers $ \sqrt {2} \pm 1$  ?
One is the inverse of the other, so they are invertible (also called units).
Units (like $ \pm 1$ in $ \mathbb{Z}$, or them and $ \pm i$ in $ \mathbb{Z}[i]$) do not matter in factorizations.


Prove that for any subset A of the complex plane, the three sets int(A), bd(A), and int(A complement) partition the complex plane.

That is a trivial consequence of the definitions in any topological space.


Here's an original problem that I created. I just want to verify if my answer.

What is the maximum number of cylindrical logs with length 2m and a diameter of 0.8 m ca be put  in a covered rectangular container with length, width and height of 3.25 m by 4.05 m, by 3 m respectively.

Note: I put the 0.05 allowance because giving and exact number is unrealistic, the logs will not fit.

[hide="My Answer"]28[/hide]

[hide]I didn't give any allowance or anything, and I got 30.[/hide]

How did you get your answre? Would you like to show us your solution?

I get 28 as well. The logs do not go all in one direction.

I think surge assumed the logs to be rectangular prisms (0.8x0.8x2). Then he assumed the box to be 3.2x4x3. If you divide those, you get 30.

The logs should be packed hexagonally with the $2$-meter dimension of the log aligned with the $4.05$-meter dimension of the box.  I'll leave the actual calculation to someone else.

Yeah, hexagonal packing is most efficient for circles, and with that you can get 30 logs in it...

[quote="surge"]Why can't the logs go in one direction?[/quote]

At least not for the one I got 28 on...

You put two logs along the 4.05m edge, and then it's a simple matter of hexagonal packing using the circles of radius .8 and the face with dimensions 3.25m x 3m...

You should find that you can get 15 logs in that way, giving a total of 30, because it is 2 deep...


A standard, fair, 6-sided die is rolled 8 times. Given that the number 3 appears exactly three times, what is the probability that no two 3's appear on consecutive rolls?

The solution to this problem given to me was long. Is there a shorter way?

Hmm... this gets very complicated. I tried splitting it into cases, but then you need to split again, and again...

And I don't think you can just write out the combinations that work, either, as there would be way too many of those.

perhaps we should think just of the combinations of 3's. Maybe we should count what we don't want first, and subtract that from all possible outcomes then dividing to get the probability.

EDIT: ok, i'll keep thinking...

[quote="plokoon51"]I wouldn't want to post a wrong solution, so could someone (236factorial) tell me if my answer is right? (i doubt it is)

[hide]29/56???[/hide][/quote]

No.

[hide]How about
3-3-???? 4 spots
3--3-??? 3 spots
...
=10

-3-3-??? 3 spots
-3--3-?? 2 spots
...
=6

--3-3-?? 2 spots
--3--3-? 1 spot
=3

---3-3-? 1 spot
=1

1+3+6+10=20
divide by 8 cohoose 3 =56   to get 5/14

OR
you know you must have
3-3-3  and you have 3 -s to put where you want there are 4 spots. so partitions of three with 4 numbers
3000 4 permutations
2100 12 permutations
1110 4 permutations

4+12+4=20[/hide]

[quote="solafidefarms"][hide="Maybe, just maybe, this works"]Let's count the ones that DO have 2 or more together:

We simply count the 2 together as 1 and find how many ways you can arrange these 2 parts in the 7 places to put them. So c(2,7) ways that it doesnt' work out of the c(3,8) possible ways fail, so $\frac{c(3,8)-c(2,7)}{c(3,8)}=\frac{35}{56}$[/hide][/quote]

oops...i thought that 3 3's in a row was a separate case when all of those possibilities would be counted in c(7,2). silly me.

is that answer correct, 236factorial?

EDIT: hmmm...i guess not

[quote="solafidefarms"][hide="Maybe, just maybe, this works"]Let's count the ones that DO have 2 or more together:

We simply count the 2 together as 1 and find how many ways you can arrange these 2 parts in the 7 places to put them. So c(2,7) ways that it doesnt' work out of the c(3,8) possible ways fail, so $\frac{c(3,8)-c(2,7)}{c(3,8)}=\frac{35}{56}$[/hide][/quote]
but you could also put 3 together so you also have to subtract c(1,7) which will get 29/56, but I don't see why mine wouldn' work

the answer isn't 35/56. Should I give you the answer?


analytic geometry
vector
real analysis
real analysis unsolved

show that the following pairs of families of curves are orthogonal (a,b are arbitrary constants). That is, prove that where any member of the first family intersects a member of the second family, the curves are orthogonal.

a) r=a cos(t) ; r=b sin(t)
b) x(t) = a-ln(t), y(t)=t, t>0 ; x(s)=(s-b)/2, y(s)=root(s) s>0

I'll do b). I am still confused about a): are those equations in polar coordinates or something else?

We need to show that at the intersection point, the tangent vectors to our curves are orthogonal. The tangent vector to the first curve is $(-\frac1t,1)$ (just differentiate with respect to the parameter). The tangent vector to the second curve is $(\frac12,\frac{1}{2\sqrt s})$. Since the $y$-coordinates coincide, we have $t=\sqrt s$, so the second vector is the same as $(\frac 12,\frac1{2t})$. Now, the scalar product is $-\frac1t\cdot\frac12 +1\cdot \frac1{2t}=0$ and we are done.


probability
geometry
3D geometry
tetrahedron
algebra
function
domain

Here are some moderate to hard ciphering questions
Comments are appreciated

1. The expression (1+i)^k is a real number when k is a multiple of what number? (i is defined as the square root of –1)

2. x, y, and z are distinct real numbers that sum to 0. What is the maximum possible value of
		(xy + xz + yz)/(x^.5 + y^.5 + z^.5)

3. You are given a 4ft board. You go to the shop and receive 2 boards, each of which has a maximum possible length of 6ft. What is the probability that you can construct a triangle from the three boards?

4. xyz=1 , x+y+z=0 , and xy+yz+xz=0. Exactly one of x, y, and z is a real number. Find this value.

5. Two men are betting money on a game of tic-tac-toe. A man pays $1 dollar to randomly pick 4 spaces. If he gets tic-tac-toe, he wins$3. On average, how much can the player expect to win or lose in dollars?

6. Find sqrt(3 + sqrt(3 + sqrt(3 + ….))).

7. In triangle ABC, D lies on side BC and AB=6, BD=3, DC=7 and AC=5. The length of the segment AD can be written in the form sqrt(a/b) where a/b is a fraction in lowest terms. What is a+b?

8. The sum of the coefficients of the expansion (a+b+c+d)^9 can be written in the form 4^k where k is an integer. Find k.

9. Find the ordered 5-tuple (a,b,c,d,e) that satisfies the following system of equations:
4a + b + c + d + e = 8
a + 4b + c + d + e = 14
a + b + 4c + d + e = 24
a + b + c + 4d + e = 34
a + b + c + d + 4e = 40

10. Find the volume of a regular tetrahedron with side length 1

11. The rational numbers a,b,c, and d have the property that (ax+b)/(cx+d)=1 has no solution in x. What is the value of (a^2)/(a^2 +c^2)?

12. The number X20191817…321 is divisible by 11 if and only if X equals what positive integer? (note that X is a digit of the number; this is not multiplication)

13. x/(y-6)=y/(z-8)=z/(x-10)=3. What is the value of x+y+z?

14. Find the sum

(sqrt(1+1) – sqrt(1))/(sqrt(1^2 +1)  +  (sqrt(2+1) – sqrt(2))/(sqrt(2^2 +2) +
	(sqrt(3+1) – sqrt(3))/(sqrt(3^2 +3) + … + (sqrt(15+1) – sqrt(15))/(sqrt(15^2 +15)

15. Find the largest n such that (2004!)! is divisible by ((n!)!)!

16. Compute

(1*1! + 2*2! +3*3! + … +7*7! +1)/(6!)

17. Simplify

1(nC1) + 2(nC2) + 3(nC3) + … + n(nCn)

18. A teacher wants to divide un-graded papers into a collection of several stacks, each consisting of the same number of papers. He knows that there is somewhere between 1000 and 2000 papers. He tries dividing the stack into groups of 2, 3, 4, 5, 6,7, and 8, but repeatedly gets 1 paper left over. How many papers are there?

19. Find the sum of the series

1+ 2(1/5) + 3(1/5)^2 + 4(1/5)^3 + …

20. What is the smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 2^2^2 and 2^3^3.(note: a^b^c=(a^b)^c )

21. What is the remainder when 2^100 is divide by 15?

22. Find one value of a for which the triangle with vertices (20,30), (30,60), and (a,50) has area 20.

23. In triangle ADC, angle bisector DB is drawn. If AB=3, AD=6, and CD=8, find DB.

24. If f(x/(x-1)) for all x not equal to 0 or 1, and 0<y<(pi/2), then find f((secy)^2) in terms of y.

25. If f(x)=x^2 + x –1 for x greater than or equal to –2, and g(x)=x^2 – x for x<5, then what is the domain of g(f(x))?

26. Find a function that satisfies the functional equation
f(x+t) – f(x-t) = 4xt

27. Given f(x) such that f(1-x) + (1-x)f(x)=5, find f(5).

28. 20.41 Suppose you are given a triangle ABC with AB=5, BC=7, and AC=8. Find the length of median AD.

29. There are n people in a room. Suppose that every person shakes hands with exactly two other people. What is the minimum number of handshakes possible?

30. I am in line to get into a club. The bouncers always want the club to look crowded, so if there are n people in line, they admit rounddown(n/2) people every 5 minutes. Right after these people are admitted, another 5 people join the line. There are 99 people in line when I join the line (this is right after letting some people in, and no one else joins the line at this moment). How many minutes will I have to wait in order to get in?

31. The expressions 2*6*10*…*(4n-2) can be written in the form (s!)/(t!), where s and t are positive integers. Find s+t in terms of n.

32. There are 5 teams entered for a double elimination tournament. What is the minimum number of games that can be played in order to determine a winner?

33. Define a hypergon as a polygon drawn as follows: start at a single point, then draw a segment that is either one unit up or down. From there draw a second segment that is either two units to the left or right, then from there third segment that is either three units to the north or south, etc. Continue until the original point is reached again. What is the perimeter of the hypergon with the smallest possible perimeter? (no crossing or backtracking is allowed)

34. Find (x +1)(x^2 + 1)(x^4 +1)….., in terms of x where abs(x)<1. (abs(x) denotes the absolute value of x)

With Calculators:

35. Define the power tower function recursively by
f1(x) = x
fn(x) = (fn-1(x))^x
for n an integer greater than or equal to 2. Calculate   lim(n approaches infinity) fn(1/2) to three decimal places.

36. What are the first five leading digits of 5^200

Some of them are actually really nice; alot of them are very trivial.

Yeah I know, but keep in mind that not everyone on my school's math team is really good at math (unfortunately) and you only get about 3/2 minutes per problem. Which ones did you think were interesting?

11, 15 (An Algebra HMMT problem as ciphering? Ouch), 26,  27, 29, 33, and 34.


function
algebra
polynomial
Pascal\u0027s Triangle

How many binary sequences with length 10 containing 5 0s and 5 1s are there without no three consecutive 1s?

there are $ \frac {10!}{5!5!}$ different 10 bit sequences with 5 1s and 5 0s. Imagine the group of three 1s as one object. So there is this group and then 2 more 1s and 5 more 0s, so there are $ \frac {8!}{2!5!}$ ways of arranging these 8 objects.

Therefore there are $ 252 \minus{} 168 \equal{} 84$ different sequences with no three consecutive 1s.

EDIT: just realised that this approach doesn't work! :(

I'm looking for a more general solution. What if the problem ask a 100 or 2n digits long sequence?
      In this problem, by using case study, I found that the answer is [b]126[/b] and the computer program I coded gave this result, too.
      The number of solutions is [url=http://www.research.att.com/~njas/sequences/A005717]Sloane A005717[/url], but I want to know why.

      Thanks in advance.

Roughly: having two consecutive 0s gives you zero 1s and so corresponds to $ x^0$; having a 1 between two 0s gives you one 1 and so corresponds to $ x^1$; having two 1s between two 0s gives you two 1s and so corresponds to $ x^2$.  This works because you always have the same number of 0s.

Edit: slightly more free time means slightly more explanation.

You have $ n$ 0s.  They define $ n \plus{} 1$ spaces, each to be filled by zero, one or two 1s.  If the total number of 1s is $ k$, then the number of ways to do this is the coefficient of $ x^k$ in $ (x^0 \plus{} x^1 \plus{} x^2)^{n \plus{} 1}$.  This is reasonably standard generating function stuff -- have you seen this type of argument before, or do you need more explanation than this?

To amitie:
I count the number of sequence with 3 consecutive 1s first.
If the sequence is 111..., there are [b]7C2[/b] ways.
If the sequence is _111..., it must be 0111...cause 1111... is counted already. This give [b]6C2[/b] ways.
...
This is a bit tedious  :(. That's why I'm not satisfied.

To JBL:
Sorry but generating function is totally new to me.
If I could understand the first part, I would understand the rest. However, how can the number of 1s between 2 0s related to x^0, x^1 or x^2 ? Are you talking about the coefficients ?

Let's do a fixed example -- maybe that will be everything you need :)

For example, with $ n \equal{} 3$, we have the following 16 good sequences:
110100
110010
110001
101100
101010
101001
100110
100101
100011
011010
011001
010110
010101
010011
001101
001011

When we multiply out $ (1 \plus{} x \plus{} x^2)^4$, we get $ 1 \plus{} 4 x \plus{} 10 x^2 \plus{} 16 x^3 \plus{} 19 x^4 \plus{} 16 x^5 \plus{} 10 x^6 \plus{} 4 x^7 \plus{} x^8$, and the coefficient of $ x^3$ is indeed 16.  Now, I want to change slightly how we look both at our sequences and at our product.  First, the sequences: to each good sequence $ w$, I want to assign a new sequence $ f(w)$.  This new sequence will be of length 4 and each entry will be a 0, 1 or 2.  The first entry of $ f(w)$ is the number of 1s to the left of the first 0 in $ w$; the second entry is the number of 1s between the first and second 0 in $ w$; etc.  Thus, I can rewrite the table above, where each sequence $ w$ is followed by $ f(w)$:
110100;  2100
110010;  2010
110001;  2001
101100;  1200
101010;  1110
101001;  1101
100110;  1020
etc.
I get each sequence by choosing four numbers from the set $ \{0, 1, 2\}$ (repetitions, of course, allowed) with sum 3.

Now, consider the product $ (1 \plus{} x \plus{} x^2)^4 \equal{} (1 \plus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2)$.  How can I get an exponent of $ x^3$?  Well, every term of the product (before combining like terms) arises from choosing one term in the first factor, one term in the second factor, one term in the third factor and one term in the fourth factor.  Each term is of the form $ x^i$ where $ i \equal{} 0, 1$ or $ 2$.  If we want the exponent to be $ x^3$, we have to choose our four different terms (one from each factor) so that the sum of their exponents is 3.  So, for example, we could choose $ x^2$ from the first factor, $ x$ from the second factor, and then we'd have to choose $ 1$ and $ 1$ from the third and fourth factors.  I could record this information as 2100.  Or, I could choose $ x^2$ from the first factor, $ 1$ from the second factor, $ x$ from the third factor and $ 1$ from the fourth factor.  I could record this choice as 2010.  And so on.

How is that?

Thanks a lot. It's much clearer now. I don't know that generating function is that interesting though :blush: .

According to your presentation, the number of solutions is equal to the number of solutions (a1,a2,...,a(n+1))
a1 + a2 + ... +a(n+1) = n                (*)

I don't know if there is other solution exists, because the original problem is n=10 and multiply the polynomial manually or filling the trinomial triangle (like Pascal triangle) is actually not a good idea. Or are there any faster ways to calculate the number of solutions of (*) ?


Let $Q$ be a quadrilateral whose side lengths are $a$, $b$, $c$, $d$, in that order. Show that the area of $Q$ does not exceed $\frac{ac+bd}{2}$.

Am I being daft or is this wrong...?

Well, the area of a quadrilateral with given side-lengths is maxed out when the quadrilateral is cyclic.

If it is cyclic, then you can apply Ptolmey's theorem.

Strange, I'm getting $\frac{ab+cd}{2}$

[hide]Inscribe the quadrilateral in a circle, and draw a diagonal $e$ between two of the sides to form triangles $abe$, $cde$.  Let triangle $abe$ have inscribed angle $\theta \Rightarrow$ the inscribed angle of $cde$ will be $180-\theta$ (because they are opposite angles of a cyclic quadrilateral).

The sine area theorem is $K = \frac{1}{2}ab\sin C$, where $a$ and $b$ are sides of a triangle and $C$ is their shared angle.  Apply that to the two triangles to get

$K = \frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin (180-\theta)$

$\sin\theta = \sin (180-\theta) \Rightarrow K = \frac{\sin\theta}{2}(ab+cd)$

And $\sin\theta$ obviously maxes out at 1[/hide]

[quote="renatzu"][quote="Altheman"]i was not so clear...say we have $ABCD$, $AC\cap BD=X$

the ptolemy inequality:

$AB*CD+BC*DA\ge AC*BD= \frac{2[ABCD]}{\sin AXB}$

$\sin\angle AXB* \frac{ac+bd}{2}\ge [ABCD]$

equality occurs when the diagonals are perpendicular and when the quadrilateral is cyclic[/quote]That's interesting, I didn't know that.

But that still doesn't explain why I got $\frac{ab+cd}{2}$[/quote]

It doesn't have to explain that. The upper bounds of $\frac{ab+cd}{2}$ and $\frac{ac+bd}{2}$ are both true, but one can't be proven from the other because both are too weak; they can't be attained unless $a^{2}+b^{2}=c^{2}+d^{2}$ and $a^{2}+c^{2}=b^{2}+d^{2}$, respectively. In contrast, the sharper bound
\[[ABCD]\leq \frac{\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)}}{4}\]
(with equality for any cyclic quad) is less than or equal to both.

[quote="scorpius119"] The upper bounds of $\frac{ab+cd}{2}$ and $\frac{ac+bd}{2}$ are both true, but one can't be proven from the other because both are too weak; [/quote]
Not quite: just divide the quadrilateral by a diagonal and flip one of the resulting triangles :).

Conclusion: I was being daft. It was late, and I got exceed mixed up with a lower bound.  :blush:


Today 5th Bangladesh math olympiad has started.I am sending another problem for under 10th grade students.

A drawer in a room contains 100 black,80 blue,60 red and 40 white socks.If u take the socks randomly what is the smallest number of socks that must be taken out to guarantee that 10 pairs have been taken out?
[a pair of sock=2 socks of the same color]

try to solve
>moon

[hide]280 socks in total 20 are to be picked out....
worst case scenario
1 1 1 1
1
first pair
1111
1
1
not yet
1111
1
1
1
.
.
.
is the answer 11/70?[/hide]

[quote="Moonmathpi496"]What do u wana mean by 11/70 i dont think ur ans correct.Ok a hint try to use box principle
>moon[/quote]
[hide]i don't see how the box principle would help-
i just calculated the largest amount of socks you'd have to take(i am rather sure but this is probably where you want me to apply box principle  :( ) and put that over the amount of socks and simplified [/hide]
sorry that i am not getting this

Ok use the fact if want 1 pair of socks u have to take at least 5 socks.
And i told u to find the smallest number of socks to get 10 pair.
>moon

well what i am saying is
[hide]you already have a sock from each other than the one you have 2 of so you should add those instead for more steps....[/hide]


Just like you prove the famous $n\not |2^n-1$: let $p$ be the smallest prime divisor of $n$. We have $p\ne 2,3$, so $p|(3^{p-1}-2^{p-1},3^n-2^n)=3^{(p-1,n)}-2^{(p-1,n)}=3-2=1$, contradiction.

P.S. I assumed there is an $n$ s.t. $n|3^n-2^n$.

[quote="Myth"][quote="Collins"][quote="grobber"]Just like you prove the famous $n\not |2^n-1$.[/quote]
Can we make the same thing for $n|2^{n+1}-1$ ?[/quote]
But $n\mid 2^{n+1}-1$ for $n=3,15$, for example.[/quote]
Ok, but my question is: can we find all $n$ such that: $n|2^{n+1}-1$?

For example, all $n=2^{2^{k}}-1$ are a good choice. :)
I think, it can be all possible values for $n$  :?

I think it's not that easy,how about    n=35

i think this pro very very hard
and we have:
[b]SIMILAR-PRO[/b]: find all possitive integer $n$ such that: $2^n-1|3^n-1$
and new pro very hard :D

[quote="nguyenquockhanh"]
similar problem: find all possitive integer $n$ such that: $2^n-1|3^n-1$
and new pro very hard [/quote]
Why do you think it is similar? :? Your problem was posted by me before.Look at:
http://www.mathlinks.ro/Forum/viewtopic.php?p=134641#134641


inequalities
function
integration
inequalities proposed

Suppose $-\infty<a<b<\infty , \; x,y\in{\mathbb R},$ , $g\in C[a,b]$ and $(r,p)\in D: =(0,1)\times (0,\infty).$  Denote \[M_{r}(g;[a,b])=  \left(\frac{1}{b-a}\int_{a}^{b}|g(t)|^{r}\; dt \right)^{\frac{1}{r}}\; \; ,\; \; \; \; A_{p}[x,y]=  \left(\frac{x^{p}+y^{p}}{2}\right)^{\frac{1}{p}}\; .\] [b]If the inequalities \[M_{r}(f;[x,y])\le A_{p}\left[f(x),f(y)\right] \; \; \; ,\; \; \; \; \; \forall (x,y)\in [a,b]\times [a,b] \; ,\] are verified for all convex functions $f: [a,b]\to [0,\infty)$ ,  then prove/or disprove that the point $(r,p)$ belongs to a subset of $D$ which is not convex. [/b]
Reference:
[1] A.Lupas and L.Lupas, [i]A class of inequalities for convex functions (Romanian),[/i] R.M.T.(=Revista Matematica din Timisoara,  nr. 2 (1986) 11-23.

Hi all!

[quote]then prove/or disprove that the point (p,q)  belongs to a subset of D  which is not convex. [/quote]

What's meant by this?

I've thought a bit about the problem and came to the conclusion that whenever the inequalities hold, then we must have [tex]r<p[/tex].
Proof:
Let [tex]r\in (0,1)[/tex], then [tex]1<\frac{1}{r}[/tex]. Now let [tex]q\in {\mathbb R}, 1<q<\frac{1}{r}[/tex]. Hence, the function [tex]f:{\mathbb R}^{+}\to {\mathbb R}^{+}[/tex], [tex]f(x)=x^{q}[/tex] is convex. But since [tex]r\cdot q<r\cdot\frac{1}{r}<1[/tex] the function [tex]f^r[/tex] is concave.
Under this circumstances we get (for [tex](a,b)\in {\mathbb R}^{+}^{2}[/tex]) [tex]M_r (f,[a,b])>\left(\frac{f(a)^r+f(b)^r}{2}\right)^{\frac{1}{r}}[/tex]. But by construction, [tex]f[/tex] is a convex function, whence [tex]M_r (f,[a,b])\leq \left(\frac{f(a)^p+f(b)^p}{2}\right)^{\frac{1}{p}}=A_p([f(a),f(b)])[/tex]. Putting together these two inequalities, we get [tex]A_{r}([f(a),f(b)]) <M_r(f,[a,b]) <A_p([f(a),f(b)])[/tex], which is a contradiction if [tex]r>p[/tex].

I don't know if it's right or useful, but maybe somebody's inspired by this ;)

Hanno

[quote="Hanno"]Hi all!
[quote]then prove/or disprove that the point (p,q)  belongs to a subset of D  which is not convex. [/quote]
What's meant by this? Hanno[/quote] Thanks for remark !
Let us denote by $\; \Omega \; , \;   \Omega \subset D\; ,\; $ [b] the set of all pairs[/b] $ (r,p) $ for which the inequalities are satisfied  for all positive convex functions on $  [a,b] $ and for all $ (x,y) ,x\ne y, a\le x<y\le b  .$ It must be investigate if  $ \Omega $ is/or  not/  a convex   set.


$ A_{1}B_{1}B_{2}A_{2}$ is a convex quadrilateral, and $ A_{1}B_{1}\neq A_{2}B_{2}$. Show that there exists a point $ M$ such that
\[\frac{A_{1}B_{1}}{A_{2}B_{2}}\equal{}\frac{MA_{1}}{MA_{2}}\equal{}\frac{MB_{1}}{MB_{2}}\]

[hide="Er..."]Circles of Apollonius, which we can show have to intersect?[/hide]


induction
number theory solved
number theory

Prove that if b^n+1 divides a^n+1 for all n-positive integer, then a must be b^(2k+1) for some k positive interger (a,b positive integers, b>1).

Here is my solution. Define the series of sequences by putting $ (x_n)^{0}=\frac{b^n+1}{a^n+1}, (x_n)^{i+1}=b\cdot (x_n)^{i}-a^i\cdot (x_n)^{i} $. It is clear that the terms of these sequences are integers. Moreover, $ (x_n)^{0} $ behaves like $(\frac{b}{a})^n $, $ (x_n)^{1} $ behaves like $(\frac{b}{a^2})^n $ and so on, an easy induction shows that $ (x_n)^{k} $ behaves like $ (\frac{b}{a^{k+1}})^n $. Thus, one of these sequences becomes 0 from a certain point. Let j minimal such that $ (x_n)^{j} $ becomes 0 from a certain point. Now, we can compute the terms of the sequence $ (x_n)^{j-1} $ from that certain point and it is immediate to see that if b is not a multiple of a then $ (x_n)^{j-1} $ also becomes 0 from a certain point, contradicting the minimality of j. Thus, b is a multiple of a. Let $ b=ka $. Then $ a^n+1|k^n-1$ and we define $ (y_n)^{0}=\frac{k^n-1}{a^n+1} $ and so on (just like in the argument showing that a|b) and with the same reasoning we find that $ a|k $. Thus, $ k=ua $ and $ a^n+1| u^n+1 $. Now, you know what you have to do: a|u, we write u=sa, a|s, The conclusion follows.

Hi harazi,

could you please explain a little more your solution ? there is some points I don't understand .. thanxx  :)

Today I had a very bad day. Nothing worked. Ok, I will try to explain it more. Too bad its too long. So, define $ x_n^1=\frac{b^n+1}{a^n+1} $ and $ x_n^{i}=b\cdot x_n^{i-1}-a^i\cdot x_{n+1}^{i-1} $. By induction we can prove the following statement:
  For any $j\geq 1$ there exist constants $ c_0,c_1,, c_j $ which depend on $j$ such that $x_n^j=\frac{ c_j\cdot b^n+c_{j-1}\cdot a^{(i-1)n}++c_1\cdot a^n+c_0}{(a^n+1)(a^{n+1}+1)\cdot\cdot (a^{n+i-1}+1) } $. This is not difficult. This shows that $ x_n^j $ behaves like $(\frac{b}{a^j})^n $. Thus, for $ a^k>b $ the sequence $ x_n^k $ tends to 0, so becomes constant from a certain point. Let j minimal such that the sequence $ x_n^j $ becomes constant from a certain point $n_0$. Thus, we have $ x_n^{j-1}=(\frac{b}{a^j})^{n-n_0}\cdot x_{n_0}^{j-1} $. Suppose that $ b$ is not a multiple of $ a$. Then since the number $(\frac{b}{a^j})^{n-n_0}\cdot x_{n_0}^{j-1} $ is an integer for an infinite numbers n, we must have $ x_{n_0}=0 $ and so the sequence $ x_n^{j-1} $ becomes 0 from a certain point. Thus, j is not minimal, false. Therefore, b is a multiple of a and from now on I think it is clear.


articles inequalities AMC USA(J)MO USAMO trigonometry geometry

Just athought, maybe some aopser's that are also mopper could share their notes, especially the people that are in red.

[quote="13375P34K43V312"]Try Mildorf's notes, they're in the articles section of AoPS, I believe. I think he was in blue.[/quote]
Mildorf was an IMO gold medalist, so he was probably in black...

I believe MOPpers are told to not distribute notes, especially not on a hugely popular open forum such as this one. However, some lecturers have posted theirs online, or even more polished versions of notes. (reid barton, kiran among others)

Mildorf hasn't posted any MOP notes (the ones under articles weren't from MOP).

[quote="krustyteklown"][b]I believe MOPpers are told to not distribute notes, especially not on a hugely popular open forum such as this one. [/b]However, some lecturers have posted theirs online, or even more polished versions of notes. (reid barton, kiran among others)[/quote]
Why :huh:

What puzzles me is why the AMC foundation doesn't make outlines or notes from MOP available. What's the point in keeping them proprietary? AoPS sells them as classes, so I understand wanting those kept secret. But I ask again, if the AMC foundation truely wants to promote nontrivial mathematics, why don't they make MOP notes available freely, or at least sell them? I can see no reason for keeping them from the public.

[quote="13375P34K43V312"][quote="krustyteklown"]for the same reason as if you take an AoPS class, you're not supposed to give the materials to anyone else

if they're not public, you shouldn't really ask individuals in order to obtain them[/quote]

I belive DPatrick said in the Classes Information forum that distributing NOTES was acceptable, as long as they are not like copied directly from the transcripts.[/quote]

As long as they are your own notes, and have just about nothing to do with material from the class. If you write down the problems and solutions from the transcripts without class conversations, I would still imagine that you can't distribute those. Basically, you shouldn't be sharing anything that was obtained from the class and might decrease someone else's incentive to take it.

@PenguinIntegral: some notes are availalbe (note Kiran Kedlaya's). I imagine that most of the notes belong to the professors, who might not want their materials flying all over the internet  :wink:

[quote="Elemennop"][quote="PenguinIntegral"]I can see no reason for keeping them from the public.[/quote]

They don't want those dirty, seedy Russians getting those documents. The Russians will do anything to win the IMO.[/quote]

....says the Russian.

[quote="PenguinIntegral"][quote="Elemennop"][quote="PenguinIntegral"]I can see no reason for keeping them from the public.[/quote]

They don't want those dirty, seedy Russians getting those documents. The Russians will do anything to win the IMO.[/quote]
I'd imagine that the Russians and other major countries already have programs of equal or greater value. I thought the IMO was for friendly competition, not information hording.

[quote="Phelpedo"]
@PenguinIntegral: some notes are availalbe (note Kiran Kedlaya's). I imagine that most of the notes belong to the professors, who might not want their materials flying all over the internet  :wink:[/quote]
Why wouldn't they want their materials flying over the internet, or being bought for a minor sum? Isn't the whole goal of the AMC foundation and it's affiliates to spread mathematical knowledge and interest?[/quote]
That's what they want you to think...

Well, despite what rumors I have read on this thread, I think the real reason that most MOPpers (not [i]mopers[/i] :wink: ) don't post their notes on the internet is that it would be a lot of tedious work to typeset more than 30 sets of notes.  Besides, there are plenty of good materials already out there (e.g., Engel, and Kiran's packet for inequalities) to give you the prerequisites for solving olympiad problems.  Indeed, much, if not most, of the lecture time at MOP is spent working [b]problems[/b], not learning material.  And this site is testimony to the fact that there are already massive quanitities of problems available for free on the internet, many of them from the AMC site itself.  If you want advice for where to start, I think the first few IMOs and (to a lesser extent) USAMOs are quite accessible.  Besides all this, there's the fact that I'm pretty sure they recycle problems (at the very least, they borrow from other sources), so it wouldn't necessarily be in the best interests of the program to have to come up with a zillion new red problems each year because everybody had already read last year's notes, now would it?

Indeed, MOP notes are not some kind of holy grail for getting better at olympiad problems. Old IMO shortlists are just as good as most MOP sheets. Anyways, if you're interested in becoming a MOPer, then you could just ignore math completely and start walking around with your head down.

[quote="Phelpedo"][quote="bpms"]
That's what they want you to think...[/quote]

Just what are you implying?[/quote]
I was joking...
Perhaps it wasn't too obvious...
Sorry if no one understood that I wasn't serious...
Next time I will use :D...

There are three reasons that the notes from MOSP are not available from the AMC.

First and foremost, the notes belong to the graders and instructors who create them.  They create them, and they own them, in the sense that any instructor owns his or her own intellectual property.  So the AMC is not free to use the material as if the AMC created or owned it.  We'd need to get permission.  Not hard, but it would take effort which would divert us form other activities.  Furthermore, some of the MOSP material comes from sources which may be copyrighted.  Using copyrighted mateiral in the classroom falls under the Fair Use category, but not if we tried to sell it or use it for more than classroom purposes.

Second, the notes are not edited.  Some may contain mistakes, impossible problems, trivial problems, old well-known problems , copyrighted material and the like.  The notes are not in a uniform format, so it would take us some effort to gather them and edit them into a uniform and useable format that would be more than the web version of a pile of papers.

Third, good versions of some of the notes do exist.  Most of the books by Titu Andreescu contain material used at MOSP in one way or another.  His books on trig and combinatorics come immediately to mind.  There is also the new book published by Springer that has every IMO short-listed problem.  Likewise for the volumes called Olympiad Problems from Around the World, published by MAA.  Also look at the books Mathematical Gems,  volums I-IV by Honsberger, published by MAA.  Many MOSP topics are covered in there, in more detail and with exposition.

Steve Dunbar
AMC Director

[quote="Boy Soprano II"]  Indeed, much, if not most, of the lecture time at MOP is spent working [b]problems[/b], not learning material.   Besides all this, there's the fact that I'm pretty sure they recycle problems (at the very least, they borrow from other sources), so it wouldn't necessarily be in the best interests of the program to have to come up with a zillion new red problems each year because everybody had already read last year's notes, now would it?[/quote] First as described by mellowmelon in http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=222065
"Le had a geometry class for us almost every day. We did start off with the basics, but later on he got into the theory of polar and projective transformations." So MOSP is rather heavy theory contrary to what you said. Also I think a lot of us that have no chance getting into MOSP or are beyond high school would really like to learn about all the cool problem solving concepts and the theories behind them. Like a lot of Titu's books don't contain that much theory and I cannot find a book that discusses pole and polar transformations. Like a lot of the problems may be on the web but without the notes they would be rather unapproachable.

please don't revive old threads, a mod should lock this

Actually, those classes were by far the heaviest on theory; that was atypical, which is one reason I noted it was my favorite lecture at MOP. Boy Soprano is right when he says most of them emphasize problem solving.


An integer sequence satisfies $a_{n+1}={a_n}^3 +1999$. Show that it contains at most one square.

$ x^3(\mod 7) \equal{} \{ \minus{} 1,1,0\}$
we prove that $ n > 2$ then
$ a_n$ can't be a square.
We has $ 7\not |a_n$
Consider the equation:
$ x^2 \equal{} y^3 \plus{} 1999$
$ x^2(\mod 7)\in \{0,1,4,2\}$
$ y^3 \plus{} 1999 (\mod 7)\in \{3,5\}$
So it has no solution when $ n > 2$
Now we prove that at most one of $ (a_1,a_2)$ can be a perfect square.
[u]Case 1[/u]$ a_1$ is a perfect square $ a_1 \equal{} m^2$
Suppose $ a_2$ is a perfect square then
$ m^6 \plus{} 1999 \equal{} x^2$ (contradiction!)
So if $ x_1$ is a perfect square then $ x_2$ is not a perfect square .
So at most one of term in this sequence can be a perfect square.

doesn't '$ m^{6} \plus{} 1999 \equal{} x^{2}$ (contradiction!)' need a little more explanation? Anyway, my solution using a different modulus:

If $ {a_{n}} \equal{} 1 (mod 4),$ then $ {a_{n \plus{} 1}} \equal{} 0 (mod 4)$
If $ {a_{n}} \equal{} 0 (mod 4),$ then $ {a_{n \plus{} 1}} \equal{} 3 (mod 4)$
If $ {a_{n}} \equal{} 3 (mod 4),$ then $ {a_{n \plus{} 1}} \equal{} 2 (mod 4)$
If $ {a_{n}} \equal{} 2 (mod 4),$ then $ {a_{n \plus{} 1}} \equal{} 3 (mod 4)$

Since squares are never 2,3 (mod 4), we must have that both $ {a_{1}}$ and $ {a_{2}}$ are squares to contradict our theorem. Assume $ {a_{1}} \equal{} x^{2}$ and $ {a_{2}} \equal{} y^{2}.$ Then: $ (y \plus{} x^{3})*(y \minus{} x^{3}) \equal{} y^{2} \minus{} (x^{2})^{3} \equal{} 1999$ . Since 1999 is prime, $ y \minus{} x^{3}$ must be 1. So:
$ 1999 \equal{} (y \plus{} x^{3})*(y \minus{} x^{3}) \equal{} 2*x^{3} \plus{} 1$
$ 999 \equal{} x^{3}$ ; impossible for an integervalued x


MIT college AMC USA(J)MO USAMO AIME calculus

I'm just an average sophomore who happens to like math.  However, these days (actually a long before) the issue of college admission starts dominating my mind and I question myself, should math be my hook.

Do you think I should find another passion of mine and start working on that?  Or build up on my current love; math.  I really hate to do something just for the sake of college resume and I don't want deceive myself about my true passion.  However math happens to be a subject where one's talent can be measured so easily through a single test paper.  And partly for this reason, it is generalised that it is the outcome (implying success in competitions) that shows how much time you've spent doing maths.  All of the winners tend to be people who've developed interest in math since earlier age.  I just don't think I'll stand a chance and compete with them.  I sound very paranoid about college and yes, I think I am.  I'm perfectly aware that not going to MIT means you've failed your life and that there are other exceptional colleges out there.  But lets not mind about college and focus on developing my passion.

I've thought about ways to show my interest in math and came up with the following ways:
1. Succeed in competitions eg: make it into USAMO (yes, very obvious)
2. Research: compete in Intel STS or Siemens.
3. Tutoring in math and help others in a form of comm. service

The road of number 1 is a possibility but I fear whether I will be able to successfully do it.  The nature of competition is that if you feel bad on the test date and screw up, tough luck.  I'm a current sophomore and got a 4 on AIME with a very low AMC score due to random guessing.  4 is a promising score and shows potential to qualify for USAMO after a year but since this years AIME was far easier than the rest, its not too encouraging.  Can anyone share their 'success stories'?  I think inspiration from the stories is what I need now to stop my habit of spending hours on CC rather than solving some problems.

I'm also focusing on number 2 and found that many of the winners for their math research aren't prodigies who went to MOSP or people who did RSI.  But I have no idea where to begin and get my initial idea; I'm just surfing thru mathworld in a glimpse of hope that I would encounter an 'Eureka' moment.  Any suggestions?

Thank you for reading my very boring, incredibly random, narrow-minded, stubborn, rude and grammatically awkward post.  Thanks.

and are there any resources I can use for the initial stage of math research?  I've read the brief descriptions of the math researches the Intel winners have done and I really don't know how in-depth you have to go.  For example, it simply says either "work on number derivatives and its special properties" or "work on abstract graphs".  They don't happen to publish their work on the internet do they?  (I've read Gabriel Carrol's poset research (came 3rd in Intel STS and he was the producer of full mark IMO paper in Washington) he posted on his website and I was baffled by all the alien language he was using.)

I've read a very interesting post about the nature of Intel at CC (just search up for "intel research" and you'll find it) and it says most of the winners work with world-class labs or professors and produce undergraduate lvl researches.  Does anyone have a full paper of any of the winners?

If you have Word Perfect, I could e-mail you my Intel paper.  (I was a finalist two years ago.)  If you only have word, I might be able to translate it, but no promises.  Send me a PM if you're interested.

Do math if you like it.  If you don't and just try to be good at it without any passion for it, its unlikely you'll get the desired results.

Both competition and research will require more problem solving skills than (I assume) you currently have, so you should try to remedy this.  Solve more problems, buy problem solving books, solve more problems, repeat.

If its research you want to do, you should also try and learn more 'advanced' math - if you're not sufficiently proficient with calculus, review it some more.  Same with linear algebra, etc, etc.

[quote="nodrog2002"]Research: compete in Intel STS or Siemens. [/quote]

What exactly is the Intel STS?  I have heard of it, but what age level is it for?  What exactly is it? Is it hard?

that is mainly for junior i believe...

i could be wrong though anyone else mind to correct me(if i am mistaken) feel free...

[url=http://www.sciserv.org/sts/]Intel STS[/url] -- basically, you submit a research project the beginning of your Senior year and compete for money.

I would guess that it would be good to get really good at some math-related skill (cryptography, computer science, physics, economics, etc.) because I don't know how you get money from being a mathematician, unless you are Euler or something.

nodrog2002, some thoughts:

Writing a research paper is a great idea but you should do it mostly for the math you'll learn, not with hopes of winning the Intel or Siemens Competition.  A mere handful of math papers win each year so your chances are very, very slim.  If you're interested in math research, you should find a mentor.  Are there any math professors or mathematicians in your community?  If not, try contacting one by e-mail.  You shouldn't expect to come up with a topic on your own.  Even graduate students don't usually choose their own dissertation topics.  Read the recent Siemens Competition Math Jam.  PoLing Loh said her RSI mentor gave her her topic.

I'm a firm believer in work.  If you decide you want to qualify for the USAMO next year, I'm sure you can.  Just set aside time every week (preferably every day) to work on problems.  If you work on problems regularly, it will be impossible for you not to improve.  If you miss the USAMO next year, there's always the year after.  In the meantime, you'll learn a lot of math, which can't hurt.

The college admissions scene is a complicated one.  I wouldn't worry too much about it right now.  The most selective colleges are so hard to get into, even for the very best students.  I wouldn't count on getting into any particular one (or four or five) but here are my thoughts on what they're looking for:

The selective colleges receive so many applications from bright students, they can afford to be very choosy.  They want students who will be the leaders of tomorrow, who will be active participants in campus life.  They do not want students who just show up for class and do their homework.  They would prefer someone who has started a software consulting business while still in high school to someone with 1600 SAT and no work experience.   Excelling at math competitions certainly wouldn't hurt your chances but you should show you have other interests too.

Hope this helps.  Relax and good luck.

if this makes you feel better, i didn't start my involvement in math till 9th grade and participation in serious competitions till 10th grade. my scores are not the greatest, but i'm very thankful for my participoation, the learning i gained, and the great people i got to meet. i have never made USAMO, though i'm very satisfied with my results overall... you dont' have to make usamo to make the college of your choice or to succeed in life!
if you do want to do an intel paper, i suggest you begin early... i left my planning to the summer after junior year and procrastinated till i never got one done. if you want to do it, you must have the initiative to begin and find someone willing to mentor you. also, there are many summer math programs that you can participate in. they look good for college because you're spending your summer doing math AND you learn many interesting things (and you can come out of them with an idea for a paper :) )
i have found tutoring to be very fun and rewarding experience. alot of understanding takes place when you're trying to explain something to others. tutoring, however, doesnt' really help you build up those skills you need for the USAMO.
all these ideas are great, hope this helps you decide what to do... best of luck in high school :)


When its digits are reversed, a particular positive two-digit
integer is increased by $ 20\%$. What is the original number?

[hide]Note that the original number must be divisible by 5. Also, since the difference of a two-digit number and its reverse is 9 times the difference of the two digits, and since the percent isn't relevant to a 9 whatsoever, the original number is also divisible by 9. Guessing $9 \times 5 = \boxed{45}$ works, since 90 obviously is too large.[/hide]


Let $ H$ a big set if  $ \exists M \subset H$ measurable such that $ \lambda(M)>0$.Is true that if $ H_1,H_2,H_3,...,H_n,...$ are big then $ \exists$ infinetly many $ (H_j)$ such that $ \bigcap_{j}H_j$ is big?

[quote="msaid"]Consider $ H_n \equal{} [n \minus{} 1,n]$ Then $ H_i$ is a Big Set for all $ i$ but $ \cap {H_i}$ is empty.[/quote]
[color=green][I think msaid intended to post that as a reply here and accidentally made a new thread instead.][/color]


Contrast Ascomycota and Basidiomycota: how are they "closely related"?

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

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

both show sexual reproduction of similar type--i'll tell u more if u want to kno

Ok. What is the similarity between their reproduction?


Hi does anyone know who proposes the problems for the RMO and the INMO?

[quote="skand"]some names

Dr.BJ Venkatachala
CR pranesachar
RB Bapat,etc
or in short,all Indian authors who write olympiads book[/quote]

Thanks!


algorithm
number theory
Euclidean algorithm
greatest common divisor

What is the greatest common divisor of 693 and 882?

We use the Euclidean algorithm here. $ 882 \minus{} 693 \equal{} 189$, and then $ 693 \minus{} 189 \equal{} 504$. $ 504 \minus{} 189 \equal{} 315$, $ 315 \minus{} 189 \equal{} 126$, $ 189 \minus{} 126 \equal{} 63$, $ 126 \minus{} 63 \equal{} 63$, and then we can easily see that the greatest common divisor is gcd(63,63) = 63.

Another method is to prime factorize each number:

$693=\boxed{2^0}\cdot \boxed{3^2}\cdot \boxed{7^1}\cdot 11^1
\\882=2^1\cdot 3^2\cdot7^2\cdot \boxed{11^0}$

Then we do the multiplication of the boxed terms:

$2^0\cdot 3^2\cdot 7^1 \cdot 11^0=1\cdot 9\cdot 7\cdot 1=\boxed{\boxed{63}}$


trigonometry
geometry
perpendicular bisector
AMC

Triangle ABC has AC = BC and ∠ACB = 106o. M is a point inside the triangle such that ∠MAC = 7o and ∠MCA = 23o. Find ∠CMB.

As I went over this problem using Kalva's solution: [color=white]Take X on the line AM so that angle XBC = 7o. Since CA = CB, X must lie on the angular bisector of ∠C. So ∠BCX = 53o. Hence ∠XCM = 53o-∠ACM = 30o. Also ∠XMC = 7o+23o = 30o. Now XB is perpendicular to MC. But XM = XC, so it must be the perpendicular bisector of MC. Hence ∠CMB = ∠MCB = 83o. [/color]I have trouble figuring out why XB is perpendicular to MC? Can somebody please explain it to me, thanks.

saturnlife wrote:Triangle ABC has AC = BC and ∠ACB = 106o. M is a point inside the triangle such that ∠MAC = 7o and ∠MCA = 23o. Find ∠CMB.

Kalva's solutions can be very confusing sometimes.  If you don't mind, might I propose an alternate one?  (A while back someone on MathLinks solved this problem, and recalling the crucial first step, it's not too bad.)

[hide]Pick a point N in triangle ABC such that :ang: NCB = 23 degrees and :ang: NBC = 7 degrees.  (N is like the equivalent of M for the side BC.)  By ASA, triangles AMC and BNC are congruent.  Therefore MC = NC, since the two are corresponding parts.  Therefore MCN is isosceles.

Now, since ACM + MCN + NCB = 106 degrees, and ACM = NCB = 23 degrees, MCN = 106 - 23 - 23 = 60 degrees.  Since MCN is isosceles, the other two angles of the triangle (CMN and CNM) are congruent and are therefore both equal to 60 degrees as well.  Thus CMN is actually equilateral, so MN = NC.  Finally, since MNC, CNB, and BNM comprise a full 360 degrres together, and MNC = 60 and CNB = 150, BNM = 360 - 60 - 150 = 150.  But CNB also equals 150, so BNM = CNB.

Now we look back at triangles MNB and CNB.  We found that BNM = CNB and MN = NC, and they share a side NB.  So by SAS, MNB is congruent to CNB.  Therefore MB = CB because they are corresponding parts.

Almost finished!  Since MB = CB, triangle MBC is isosceles.  Since angle MBC = 7 + 7 = 14, the other two angles of the triangle (BMC, the desired angle, and BCM) both have measure (180-14)/2 = 83.  Answer.

Looking back, the crucial step was constructing that congruent triangle using side BC.  I'll have a look at Kalva's solution, but I think this one is much more insightful and understandable.[/hide]

The easiest way to do this is to apply trig ceva... You do this and notice that 2sin7cos7 = sin14 and ur done.

Thanks for the solutions guys, they helped me alot!


In every turn it can be taken (c+5b,3c-5a,2b-3a) or (2a+3b,b+3c,4c+2a)  instead of (a,b,c).At the beginning if (a,b,c)=(1,2,4) , is it possible that  get (2003,2004,2005)  using  such turns?

No, it is impossible. To show this we analyze the congruence modulo 2:

terns of start               terns after first move              terns after second move

0,0,0                                    0,0,0                                       0,0,0
0,0,1                                    1,1,0                                       0,1,0
0,1,0                                    1,0,0                                       1,1,0
0,1,1                                    0,1,0                                       1,0,0
1,0,0                                    0,1,1                                       0,0,0
1,0,1                                    1,0,1                                       0,1,0
1,1,0                                    1,1,1                                       1,0,0
1,1,1                                    0,0,1                                       1,0,0

so we want obtain the tern (2003,2004,2005)=(1,0,1) and it can build only from another tern (1,0,1) but we start from a (1,0,0) so it is impossible.

Quote my last message to read the correct layout.


[quote="mdk"]On a large, flat field there are n people positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires at the person who is closest. When n is odd
prove that at least one person remains dry. Is this always true when n is even?[/quote]

[hide="Solution"]
EVEN: Fails because we can pair people together.

ODD: If $ n=3$ it is trivial. So now we proceede with induction. Say there are $ n+2$ people and we want to show at least one remains dry. Use the Extreme Principle and look at the [i]smallest distance[/i]. Then those two people must shoot at each other. Now there are $ n$ other people. If any one of those other people shoots one of those two minimized people the proof is compelte because there are more remaining people then the people shooting. Otherwise there are $ n$ people shooting amongst eachother and by induction one is dry.
[/hide]


[quote="fjwxcsl"]Let $ (x,y,z)\in{R_{ + }^3},k\in{R_{ + }},$prove that:

$ A.\sum{x^k[}$(x-y)$ +$(x-z)${ ]}\ge0;$

$ B.\sum{x^k(x^2 - yz)}\ge0.$[/quote]

or,

$ A.\sum{x^k[(x-y)+(x-z)]}\ge0 \iff \sum{(x^k-y^k)(x-y)} \ge 0$,

$ B.\sum{x^k(x^2 - yz)}\ge0 \iff \sum{[(x^k-y^k)(x - y)(x + y) + z^{k}(x - y)^2]} \ge 0$,

easy to prove that when $ k>0$ have $ (x^k-y^k)(x-y) \ge 0$, so both inequalitys holds;


modular arithmetic
number theory unsolved
number theory

Suppose that $ p$ is an odd prime, and write $ \frac {1} {1}\minus{}\frac {1} {2}\plus{}\frac {1} {3}\minus{}\ldots\minus{}\frac {1} {p\minus{}1}\equal{}\frac {a} {(p\minus{}1)!}$. Show that $ a \equiv \frac {2\minus{}2^p} {p} \pmod p$.

By $ 2 \minus{} 2^p \equal{} 2 \minus{} (\sum_{i \equal{} 0}^pC_p^i) \equal{} \minus{} (C_p^1 \plus{} \cdots \plus{} C_p^{p \minus{} 1})$ we have $ \frac {2 \minus{} 2^p}{p} \equal{} \minus{} \sum_{i \equal{} 1}^{p \minus{} 1}\dfrac{C_p^i}{p}$. Now in  $ \mathbb{F}_p$ : $ \dfrac{C_p^i}{p} \equal{} \dfrac{(p \minus{} i \plus{} 1)\cdots (p \minus{} 1)}{i!} \equal{} ( \minus{} 1)^{i \minus{} 1}\cdot\frac {1}{i}$ and we're done!.

See also http://www.mathlinks.ro/viewtopic.php?t=37352 .

  darij


[color=darkred][b]18.[/b] Let $ ABC$ have an inscribed semicircle with a diameter on $ [BC]$ which touches $ AB$ and $ AC$ at $ F$ and $ E$ respectively.
Provided that $ BF=1$ and $ CE=9$ and $ AB+AC=40$, find the area of $ \triangle ABC$.[/color]

[hide="Proof."][color=darkblue]Show easily that $ AE=AF=15$, $ AB=16$, $ AC=24$ and $ OC^{2}-OB^{2}=CE^{2}-BF^{2}=80$.

From the relation $ \frac{OB}{OC}=\frac{[OAB]}{[OAC]}$ obtain $ \frac{OB}{OC}=\frac{AB}{AC}=\frac{16}{24}=\frac{2}{3}$. Thus, $ \frac{OC^{2}}{9}=\frac{OB^{2}}{4}=\frac{OC^{2}-OB^{2}}{9-4}=\frac{80}{5}=16$, i.e.

$ OB=8$, $ OC=12$, $ BC=20$ and $ \rho=OE=OF=3\sqrt{7}$. In conclusion, $ [ABC]=\frac{1}{2}\cdot\rho\cdot (AB+AC)$ $ \implies$ $ \boxed{[ABC]=60\sqrt 7}$.

[b]Remark.[/b] The triangle $ ABC$ is similarly with the triangle $ A'B'C'$, where $ A'B'=4$, $ B'C'=5$, $ C'A'=6$. [b]Easy and nice problem, thank you ![/b][/color][/hide]

[quote="Altheman"]This is a nice solution. Here is a slightly different proposed problem that uses highlights his idea.

Given triangle $ ABC$, define $ A'\in BC$ as the point such that there is a semicircle with center $ A'$ and diameter on $ BC$ that is tangent to $ AB$ and $ AC$. Define $ B'$ and $ C'$ similarly. Prove that $ AA'$, $ BB'$, and $ CC'$ are concurrent.[/quote]

[hide]That is, $ A'$ is on the angle bisector of A?  (internal if $ B$ and $ C$ are acute, external otherwise).[/hide]

[quote="Altheman"]I never said that problem was difficult. I guess that I should have specified that it is acute.

$ \frac{BA'}{A'C}=\frac{[BAA']}{[CAA']}=\frac{AB}{AC}$ which implies that $ AA'$ bisects angle $ A$.[/quote]  It's just based on the fact that $ A'$ is equidistant from $ AB$ and $ AC$.  I thought I'd give that way of seeing it: from a locus standpoint.

Then again, it would still use Virgil Nicula's method on the original problem.


superior algebra
superior algebra solved

Find, with proof, the number of automorphisms of the group $Z_{n}$ with itself under modular addition.

That's easy: Every endomorphism of $\mathbb{Z}_{n}$ is uniquely determined by the image of 1. Hence there are exactly n endomorphims which can be seen as multiplication in the ring $\mathbb{Z}_{n}$. The multiplication with a is an automorphism iff a is invertible mod n iff gcd(a,n)=1.
Hence there are $\phi(n)$ automorphisms of $\mathbb{Z}_{n}$.


In a book written by J. I. Diaz, namely, Nonlinear partial differential equations and free boundaries, Lemma 4.10, there are 2 inequalities as follows

\[ \begin{gathered} \left| {a - b} \right|^p \leqslant C\left( {\left| a \right|^{p - 2} a - \left| b \right|^{p - 2} b} \right)\left( {a - b} \right) {\rm \quad for \quad } p \geqslant 2, \hfill \\ \left| {a - b} \right|^2 \leqslant C\left( {\left| a \right|^{p - 2} a - \left| b \right|^{p - 2} b} \right)\left( {a - b} \right)\left( {\left| a \right| + \left| b \right|} \right)^{2 - p} {\rm \quad for \quad }1 < p < 2, \end{gathered}  \]
where $a$ and $b$ are in $\mathbb{R}^n$.

Because I have not this book, so, I want to ask all of us for which conditions on C, these above inequalities hold. Thanks for each comment(s)...


[quote]Could not query weblogs information

DEBUG MODE

SQL Error : 1030 Got error -1 from storage engine

SELECT w.*, u.*, e.* FROM phpbb_users u, phpbb_weblogs w LEFT JOIN phpbb_weblog_entries e ON e.weblog_id = w.weblog_id WHERE u.user_weblog = w.weblog_id AND w.deleted = 0 AND e.entry_deleted = 0 AND e.entry_time < 1110337280 ORDER BY e.entry_time DESC

Line : 736
File : functions_weblog.php [/quote]

*blinks* Well, it works now...Funny thing is i was nowhere near the weblogs...

Bet it didn't keep working for long.  We had a little database problem tonight, but we've fixed it.  Please let me know if you see any odd database behavior on the site.

Yea. It crashed again before I hit the next screen. Everything looks fine now. :-)


algebra
polynomial
linear algebra
linear algebra unsolved

The same idea as the one I used to solve the problem posted by you for $m=2$ (as problem from training agregation) works here as well. Of course, we have to place ourselves in the algebra generated by $A_i$, which is commutative.


Let k be a positive integer.  $\sum_{i=1}^{n}x_i=k$.  What are the possible values for $\sum{x_i^2}$

from what i remember, the original problem didn't require $k$ to be an integer

For the lower limit we apply the power mean - arithmetic mean inequality:

$\sqrt{\frac{\sum x_i^2}{n}} \geq \frac{\sum x_i}{n}$

$\Rightarrow \sum x_i^2 \geq \frac{k^2}{n}$

with equality iff $x_1 = x_2 = ... = x_n$.  Note that this implies a lower limit of 0 as $n \rightarrow \infty$, with an arbitrarily small non-zero sum able to be constructed by setting equality of all variables and adjusting $n$ accordingly.

For the upper limit we note that

$\sum x_i^2 \leq \left(\sum x_i \right)^2 = k^2$

with equality iff $n = 1$ (we note that for $n > 1$, all $x_i$ are greater than zero, so the inequality is strict).

We therefore see that

$0 < \sum x_i^2 \leq k^2$

where we are able to tend toward the lower limit as $n$ increases, and toward the upper limit as $n$ decreases (with equality iff $n = 1$).

$k$ can be any positive real. Sorry about that.

Can you also show how to get the value $k$ by construction?

[quote="paladin8"]If the problem doesn't specify $x_i$ to be positive, the expression can go to infinity...[/quote]
I think the problem did specify that (looking at Kalva's version).

The problem here is stated incorrectly (we're dealing with strictly infinite sums).  The best way of approaching it is probably to deal with geometric series.

Yes. The original problem (paraphrased):
Let $x_1,x_2,\dots$ be a sequence of positive real numbers with $\sum_{n=1}^\infty x_n=A$. What are all possible values for $\sum_{n=1}^\infty x_n^2$?

It was a fairly easy problem, but it's easy to miss part of it if you're not careful.


As you probably know, a lot of terms in math are not pronounced how they would seem.  In this thread, please post any terms you have heard pronounced wrong or once did pronounce wrong, and how to pronounce it correctly.

I'll start with a few:

Euler - OI ler [like an "oiler"]
Ptolemy - tal e mi [p is silent]
Poisson - PWA so
Poussin - Pu-(nasal sen)

And I have a question on "cyclic":
Cyclic - sick-lick? sigh-clik? are both acceptable?

[url]http://www.merriam-webster.com/cgi-bin/dictionary?book=Dictionary&va=cyclic[/url]

Both are acceptable. I use both strangely enough.

I've only ever heard sick-lick.

How do you pronounce Godel?

I believe it is approximately "guur-del."  Anyhow, most of these aren't unusual pronunciations, they are just foreign pronunciations.

I've always pronounced it like the English word "girdle", which I think is close to what Joel said.

[url=http://en.wikipedia.org/wiki/Close-mid_front_rounded_vowel]Wikipedia[/url] has a quicktime sound file for the first vowel in his name -- it's close to what Ravi and I said, but with perhaps a little bit less "r" sound.  It is, technically, the "close-mid front rounded vowel."  Go figure.

The one I haven't been able to figure out is Sylow.

Yes, I would also like to be definite on how to pronounce Sylow.  I've heard it both like "silo" and like "see-low", and am under the impression it's the latter, but am not sure.

I remember my professor, Prof. Wolmer Vasconcelos, pronounce Sylow as "see-low". Perhaps that helps.

[quote="ComplexZeta"]I have heard two pronunciations: Zee-lof and See-low. Do we have any Norwegian speakers who can clarify the matter for us?[/quote]There are a couple of Norwegians on the site, however I'm not sure that they are active in this part  :?

[quote="ComplexZeta"]I have heard two pronunciations: Zee-lof and See-low. Do we have any Norwegian speakers who can clarify the matter for us?[/quote]

The first pronunciation would be much more plausible for a Norwegian pronunciation of "Sylow." Alas, Sylow is not in the edition of Webster's New Biographical Dictionary I have at hand, which would provide a definitive answer for speakers of American English. Maybe you could send an email to the math department of St. Olaf College, which would surely know the answer to this question.    ;)

Personally, I always thought "Cauchy" was a fairly tricky name, but it is easy to look up, and it has lots of oral tradition transmitting the preferred pronunciation to most math students.

Cauchy is pronounced co-she, right?

How about Heron as in Heron's formula?
Or phi?

phi can be pronounced either fee or fie, but not fo or fum.

For the pronunciation of Cauchy, is it

"KO she"

or

"ko SHE"

?

Never could figure this one out...


calculus
integration
analytic geometry
trigonometry
calculus computations

I just started learning how to evaluate evaluating double integrals by converting to polar coordinates. my question is how do you set the limits on each integrand?

for example i know that to evaluate $\int_{0}^{\infty}e^{x^{2}}$ you need to use $I^{2}=\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^{2}+y^{2})}dxdy$
and then when u convert to polar you have
$I^{2}=\int\int re^{-r^{2}}drd\theta$ but how do i find the limits?

It's $e^{-x^{2}}$ in the first integral. $I^{2}$ is taken over the first quadrant, so you need a description of this quadrant in terms of $r$ and $\theta$. With a very simple sketch you can convince yourself that $0\le\theta\le\pi/2$ while $r$ is unrestricted ($0\le r<\infty$).

thanks. but in general how do you go about determining the limits?

(1) draw the region of integration;
(2) determine the range of $\theta$ within the region;
(3) find the smallest and largest value of $r$ along the half-line that makes angle $\theta$ with the positive $x$-axis.

Two examples from my midterm exams:
[quote]Evaluate the integral $\int_{-1}^{1}\int_{0}^{\sqrt{1-x^{2}}}(x^{2}+y^{2})^{5/2}dy\,dx$ using polar coordinates.[/quote]
(1) the region of integration is the upper half of the disk of radius $1$.
(2) $0\le\theta\le\pi$
(3) for each $\theta$ as in (2), the polar distance $r$ is between $0$ and $1$.
Obtain $\int_{0}^{\pi}\int_{0}^{1}r^{6}\,dr\,d\theta$, which is $\pi/7$.

[quote]Convert the integral $\int_{-1}^{1}\int_{|x|}^{1}xy \,dy\,dx$ into polar coordinates.[/quote]
(1) the region of integration is the triangle with vertices $(0,0)$, $(1,1)$, and $(-1,1)$.
(2) $\frac{\pi}{4}\le\theta\le\frac{3\pi}{4}$
(3) for each $\theta$ as in (2), the smallest value of $r$ is $0$, while the largest occurs on the line $y=1$ (hypotenuse of the triangle). In polar coordinates this line has equation $r\sin\theta=1$, hence $r=\csc\theta$.
Answer: $\int_{\pi/4}^{3\pi/4}\int_{0}^{\csc\theta}r^{3}\cos\theta\sin\theta\,dr\,d\theta$.


function
Euler
calculus
integration
number theory

this person i know wants to find out what the Riemann Zeta Function is, and its uses.  He does have an username on aops, but seems to be hibernating. so, i'm asking for him.

if this is too easy or hard then please can a mod move it?

Dzeta function is defined for Re(s)>1 as Dzeta(s)=1/1^s+1/2^s+1/3^s+...; By unique factorization theorem a so called Euler formula holds: Dzeta(s)=Product(1/(1-1/p_n^s)), where the product is taken over p_n prime; Then it can be proved that it has meromorphic continuation to the whole plane (if I'm not mistaken it's done by the use of formula Dzeta(s)=1/Gamma(s)*integral from 0 to infinity of x^(s-1)/(e^x-1) dx; ) It can be then shown that it is regular everywhere in the plane, except a simple pole at s=1;

This function (because of existance of Euler formula) has connection to distribution of primes, and therefore is used in number theory; But it is not elementary; The famous Riemann conjecture asserts that all nontrivial zeros of Dzeta lie on the line Im(s)=1/2; (trivial zeroes are zeroes at -2,-4,-6...)

Another facts can be found in [url=http://mathworld.wolfram.com/RiemannZetaFunction.html]Mathworld[/url] site :lol:

Where are the zeros of zeta of s?
to the tune of "Sweet Betsy from Pike"; words by Tom Apostol
(Without any comments -  Saw it some where...)

Where are the zeros of zeta of s?
G.F.B. Riemann has made a good guess,
They're all on the critical line, sai he,
And their density's one over 2pi log t.

This statement of Riemann's has been like trigger
And many good men, with vim and with vigor,
Have attempte to find, with mathematical rigor,
What happens to zeta as mod t gets bigger.

The efforts of Landau and Bohr and Cramer,
And Littlewood, Hardy and Titchmarsh are there,
In spite of their efforts and skill and finesse,
(In) locating the zeros there's been no success.

In 1914 G.H. Hardy did find,
An infinite number that lay on the line,
His theorem however won't rule out the case,
There might be a zero at some other place.

Let P be the function pi minus li,
The order of P is not known for x high,
If square root of x times log x we could show,
Then Riemann's conjecture would surely be so.

Related to this is another enigma,
Concerning the Lindelof function mu(sigma)
Which measures the growth in the critical strip,
On the number of zeros it gives us a grip.

But nobody knows how this function behaves,
Convexity tells us it can have no waves,
Lindelof said that the shape of its graph,
Is constant when sigma is more than one-half.

Oh, where are the zeros of zeta of s?
We must know exactly, we cannot just guess,
In orer to strengthen the prime number theorem,
The integral's contour must not get too near 'em.


hi,
im stuck with this problem please help,

Anne borrows 3200. Loan is paid 7 months later by a lump sum that includes 168 in interest. What was the simple rate of interest on the loan?

Im not sure which formulate to use? its not the simple interest formula right?i dont think it is

thanks

ans. (168/7)multiplied by 12 gives you amt. of int. that accrues for a year

rate of int={(168/7)*12}/3200 best of luck.........

Believe it or not, it IS the simple interest thm. The percentage of the interest over the borrowed amount is 5.25 percent over the whole time period. If you want to calculate the monthly interest, you divide that by 7, to get 0.75 percent interest. If you want the yearly interest, you will multiply the monthly interest by 12 to get 9 percent simple interest. Hopefully this helps =D


\noindent The numbers 1, 2 and 3 are written in these nine unit squares.
\begin{itemize}
\item Each of the numbers appears three times, and
    there is only one number placed in each of the
    nine unit squares.
\item Each number is in a unit square horizontally
    or vertically adjacent to a unit square with the same number.
\item The sum of the numbers in the leftmost column and the sum
    of the numbers in the top row are each 7.
\end{itemize}
What is the sum of the numbers in the four shaded squares?

[asy]draw((0,0)--(30,0)--(30,30)--(0,30)--cycle);
draw((10,0)--(10,30));
draw((20,0)--(20,30));
draw((0,10)--(30,10));
draw((0,20)--(30,20));
fill((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7));
fill((0,20)--(10,20)--(10,30)--(0,30)--cycle,gray(0.7));
fill((20,0)--(30,0)--(30,10)--(20,10)--cycle,gray(0.7));
fill((20,20)--(20,30)--(30,30)--(30,20)--cycle,gray(0.7));[/asy]

One way to fill this box is:

$ 3$ $ 2$ $ 2$
$ 3$ $ 3$ $ 2$
$ 1$ $ 1$ $ 1$ (i believe this is legal.)

If this is so, then the corners add up to $ 7$.

Also, since you know the sum of the numbers in the left most column and the sum of the numbers in the top column are equal to 7, you can tell that the corners are equal to 7. I think. I'm not quite sure about this.


group theory
abstract algebra
superior algebra
superior algebra unsolved

How to show there is no simple group of order $ 210$ , without using the odd-test of simple group .

I always stuck in the techniques I know.

1) Caley's Generalized theorem.

2) the method of counting elements

3) playing by center of the group , and normalizer .

I really don't know .

the odd test say :

there is no simple group of order $ 2n$ , where $ n$ is odd.

 :)

Done, Simon  :wink:

I just thinking about to get a contradiction from Sylow $ 7$-subgroup and Sylow $ 5$-Subgroup .

Since If we assume $ n_7\ne 1 \Rightarrow n_7\equal{}15$($ H_1,\cdots, H_{15}$), and I assume there more than one Sylow $ 5$-subgroup say $ K$

Could we show there is no other than identity element common in $ H_iK, \forall i\equal{}1,2\cdots , 15$.?

 :)


Determine function $ f : R^{\plus{}} \minus{}\minus{}>R^{\plus{}}$ such that $ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}4 (f(x)f(y))^{1/2}$ for every $ x>y>0$.

There is no such function.

1) $ f$ is bounded
Let $ a\ne1$ be positive, we know
$ f(1\plus{}a)<4\sqrt{f(1)f(a)}$,
$ P(1\plus{}a,1): f(a)<4\sqrt{f(1\plus{}a)f(1)}$.
Hence,
$ f(a)^2<16f(1\plus{}a)f(1)<64f(1)^{3/2}f(a)^{1/2}\Rightarrow f(a)<16f(1)$.
Since $ a$ can be arbitrary, $ f$ is bounded.

Now let $ A\equal{}\sup_{x>0}f(x)<\infty$.
Case 1)
If $ A$ is not attainable, we can find $ x>y$ such that $ f(x),f(y)$ are arbitrarily close to $ A$.
Then $ 4\sqrt{A^2}\le 2A$. Contradiction!
Case 2)
If $ f(b)\equal{}A$. Then
$ 4\sqrt{f(b)f(3b)}\le 2A$ and $ 4\sqrt{f(b)f(2b)}\le 2A$ imply
$ 4\sqrt{f(3b)f(2b)}\le A$.
But
$ 4\sqrt{f(3b)f(2b)}\equal{}f(5b)\plus{}f(b)>A$. Contradiction!

maybe you made mistke :maybe:

$ f(x)\equal{}kx^2$ can be a solution

Maybe you made mistake.

$ LHS \equal{} 2k(x^2 \plus{} y^2)$
$ RHS \equal{} 4kxy$.

If the original problem gives
$ f(x\plus{}y)\minus{}f(x\minus{}y)\equal{}4\sqrt{f(x)f(y)}$, not
$ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}4\sqrt{f(x)f(y)}$,
then it should be easy to derive the only solution $ f(x)\equal{}kx^2$.


algebra
polynomial
induction
binomial theorem
algebra unsolved

A polynomial $ f(x)\equal{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx$ with $ a,b,c,d>0$ is such that $ f(x)$ is an integer for $ x \in \{ \minus{}2,\minus{}1,0,1,2 \}$ and $ f(1)\equal{}1$ and $ f(5)\equal{}70$.
$ (a)$ Show that $ a\equal{}\frac{1}{24}, b\equal{}\frac{1}{4},c\equal{}\frac{11}{24},d\equal{}\frac{1}{4}$.
$ (b)$ Prove that $ f(x)$ is an integer for all $ x \in \mathbb{Z}$.

b) is a standard result in finite differences; any five consecutive integer values determines the finite difference table of the polynomial, from which all the other values of $ f(x), x \in \mathbb{Z}$ can be easily computed.

Interesting problem, I think. (Sorry for the length; I wanted to make my work more clear, and included a lot of space...)

[hide="Solution"](a).From the givens, we have the following equations in $ a$, $ b$, $ c$, $ d$:
[list]
[*]$ 16a\minus{}8b\plus{}4c\minus{}2d\in \mathbb{Z}$ (1)
[*]$ a\minus{}b\plus{}c\minus{}d\in \mathbb{Z}$ (2)
[*]$ a\plus{}b\plus{}c\plus{}d\equal{}1\in \mathbb{Z}$ (3)
[*]$ 16a\plus{}8b\plus{}4c\plus{}2d\in \mathbb{Z}$ (4)
[*]$ 625a\plus{}125b\plus{}25c\plus{}5d\equal{}70$ (5)[/list]
Note that sums and differences of integer multiples of integers remain integral. Thus, we can add and subtract (2) and (3) and do the same for (1) and (4):
[list]
[*]$ 2a\plus{}2c\in \mathbb{Z}$ (6)
[*]$ 2b\plus{}2d\in \mathbb{Z}$ (7)
[*]$ 32a\plus{}8c\in \mathbb{Z}$ (8)
[*]$ 16b\plus{}4d\in \mathbb{Z}$ (9)[/list]
Subtracting multiples of (6) from (8) and doing likewise for (7) and (9) yields the following:
[list]
[*]$ 24a\in \mathbb{Z} \Leftrightarrow 24c\in \mathbb{Z}$ (10)
[*]$ 12b\in \mathbb{Z} \Leftrightarrow 12d\in \mathbb{Z}$ (11)[/list]
Note that if $ 2a\plus{}2c>2$, then $ a\plus{}b\plus{}c\plus{}d>1$, contradiction. Also if $ 2a\plus{}2c<0$, then at least one of $ a$ or $ c$ is negative, contradiction. Similarly, if $ 2a\plus{}2c\equal{}0$, then we cannot have $ a,c>0$, contradiction. In addition, if $ 2a\plus{}2c\equal{}2$, then $ b\plus{}d\equal{}0$ via (3). We conclude that because $ 2a\plus{}2c\in \mathbb{Z}$, we have $ 2a\plus{}2c\equal{}1\Leftrightarrow 2b\plus{}2d\equal{}1$ (via (3)). We can substitute $ c\equal{}1/2\minus{}a$ and $ d\equal{}1/2\minus{}b$ into (5) to obtain (simplifying) $ 120a\plus{}24b\equal{}11$. If $ a\geq 3/24$, $ b<0$, which is a contradiction. If $ a\equal{}1/12$, $ b\equal{}1/24$, contradicting (11). Therefore, as $ a>0$ and $ 24a\in \mathbb{Z}$, we must have $ a\equal{}1/24\Leftrightarrow c\equal{}11/24$ and $ b\equal{}1/4\Leftrightarrow d\equal{}1/4$, as desired. We can check again that the properties are satisfied by this definition of $ f(x)$.

(Note: It is not necessary to have $ a,b,c,d>0$; at least $ a,b,c,d\geq 0$ or perhaps something even sharper suffices with small modifications to this proof.)

(b). t0rajir0u is correct. However, we can also prove this by writing $ f(x)$ as $ \frac{(x)(x\plus{}1)(x\plus{}2)(x\plus{}3)}{24}$ and noting that among the four consecutive numbers $ x$, $ x\plus{}1$, $ x\plus{}2$, and $ x\plus{}3$, there must be at least $ 1$ multiple of $ 3$, exactly $ 2$ multiples of $ 2$, and exactly $ 1$ multiple of $ 4$ (which is one of the multiples of $ 2$). Therefore, the numerator of $ f(x)$ is divisible by $ 3\cdot 2\cdot 4\equal{}24$, as desired. We can also notice that $ f(x)$ equals the binomial coefficient $ \binom{x\plus{}3}{4}$ which must be an integer (we might have to make special considerations if we want to avoid $ a$ being negative in the definition of $ \binom{a}{b}$.[/hide]

Here's an elementary proof of the fact that given a polynomial $ \deg P(x)\equal{}n$ so that $ P(t)\in\mathbb{Z}$ for a set of $ n\plus{}1$ consecutive integers, $ P(t)\in\mathbb{Z}$ for all integers $ t$.

We proceed by induction on $ n$; consider $ n\equal{}0\Rightarrow P(x)\equal{}c$. If we have that $ c\equal{}t$ for an integer $ t$, then clearly $ P$ is an integer everywhere, so our base case is established.

Now assume we have this result for $ 1,2,\ldots,k\minus{}1$; we show that it's true for $ n$. To establish this, let the integers so that $ P(x)$ is an integer be $ a_i\equal{}a\plus{}i,0\le i \le n,a\in \mathbb{Z}$. Observing that $ j(b\plus{}1)^{k}\minus{}jb^{k}$ is a $ k\minus{}1$-degree polynomial as a trivial result of the binomial theorem, we define the polynomial $ Q(x)\equal{}P(x)\minus{}P(x\minus{}1)$; from our result above $ \deg Q(x)\equal{}k\minus{}1$. Setting $ x\equal{}a_1,\ldots,a_n$, we have that $ Q(x)$ is an integer (as the difference of two other integers) for $ k$ consecutive integers, so our inductive step is complete and the result is true.


[quote="cezar lupu"]Find the last three digits of the number 2005^2005.[/quote]

(2000 + 5)^2005 == x mod (1000);
5^2005 == x mod (1000);
cycle of 125 and 625 repeat for odd and even; hence last 3 digits are 125.


Let $ x,y,z$ be real numbers, such that $ x\plus{}y\plus{}z\equal{}1$.
Prove that the following inequality holds

$ \left.\frac{x^2}{3x^2\minus{}2x\plus{}3}\plus{}\frac{y^2}{3y^2\minus{}2y\plus{}3}\plus{}\frac{z^2}{3z^2\minus{}2z\plus{}3}\geq \frac{1}{8}\right.$

[quote="rachid"]
 $ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \geq \frac {x^2}{2(x^2 \plus{} y^2 \plus{} z^2 \plus{} 1)}$
[/quote]
Without this remark, written for $ x,y,z$, it's really difficult to prove the inequality! In fact, I don't think that it's possible. I invite you to find new solutions.
Thanks, [b]rachid[/b]! Your solution is also my solution!

Could you be more specific, Mr. xzlbq ? Is it mentained the initial condition $ x\plus{}y\plus{}z\equal{}1$? What do you mean? :mad:

I think the new condition should be $ |x| \plus{} |y| \plus{} |z| \equal{} 1$ so that the inequality of xzlbq is true.  :)

Let's try to analyze this a bit further.

If I make $ x \rightarrow \infty$ , $ y \rightarrow \minus{}\infty$ and $ z \rightarrow 1$ then the expression will come to a value that exceeds $ \frac{1}{4}$ . Hence, the inequality happens (not holds) when  $ |x| \plus{} |y| \plus{} |z| \equal{} 1$

[quote="marin.bancos"]Let $ x,y,z$ be real numbers, such that $ x \plus{} y \plus{} z \equal{} 1$.
Prove that the following inequality holds

$ \left.\frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \plus{} \frac {y^2}{3y^2 \minus{} 2y \plus{} 3} \plus{} \frac {z^2}{3z^2 \minus{} 2z \plus{} 3}\geq \frac {1}{8}\right.$[/quote]

$ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \ge \frac{6x\minus{}1}{24}$

$ \iff \frac{(3x\minus{}1)^2(3\minus{}2x)}{24(3x^2\minus{}2x\plus{}3)} \ge 0,$

so

$ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \plus{} \frac {y^2}{3y^2 \minus{} 2y \plus{} 3} \plus{} \frac {z^2}{3z^2 \minus{} 2z \plus{} 3}$

$ \geq \frac{6x\minus{}1}{24}\plus{}\frac{6y\minus{}1}{24}\plus{}\frac{6z\minus{}1}{24}$

$ \equal{}\frac18.$

[quote="great math"]Kuing, the inequality states that[b][u][i] $ x,y,z$ are reals [/i][/u][/b][/quote]

oh...sorry for my mistake... :(

thank you for reminding me... :blush:

[size=150][quote="marin.bancos"]Could you be more specific, Mr. xzlbq ? Is it mentained the initial condition $ x \plus{} y \plus{} z \equal{} 1$? What do you mean? :mad:[/quote]
no positive,Thanks.
BQ[/size]

[quote="kuing"]

$ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \ge \frac {6x \minus{} 1}{24}$

$ \iff \frac {(3x \minus{} 1)^2(3 \minus{} 2x)}{24(3x^2 \minus{} 2x \plus{} 3)} \ge 0,$

so

$ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \plus{} \frac {y^2}{3y^2 \minus{} 2y \plus{} 3} \plus{} \frac {z^2}{3z^2 \minus{} 2z \plus{} 3}$

$ \geq \frac {6x \minus{} 1}{24} \plus{} \frac {6y \minus{} 1}{24} \plus{} \frac {6z \minus{} 1}{24}$

$ \equal{} \frac18.$[/quote]
This way still usefull, kuing. We only need more sentences.
If $ x, y, z$ are in $ [0, \frac{3}{2}]$, the way above effect.
If in $ x, y, z$ at least have a number $ \ge \frac{3}{2}$, example assume $ 3\minus{}2x \le 0$
Now, easily to see that  $ \frac{x^2}{3x^2\minus{}2x\plus{}3} \ge \frac{x^2}{3x^2}\equal{}\frac{1}{3}$
So, $ \left.\frac{x^{2}}{3x^{2}\minus{}2x\plus{}3}\plus{}\frac{y^{2}}{3y^{2}\minus{}2y\plus{}3}\plus{}\frac{z^{2}}{3z^{2}\minus{}2z\plus{}3}\ge  \frac{1}{3}\plus{}0\plus{}0 \ge \frac{1}{8}\right.$
The proof completed  :)

[quote="nguoivn"][quote="kuing"]

$ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \ge \frac {6x \minus{} 1}{24}$

$ \iff \frac {(3x \minus{} 1)^2(3 \minus{} 2x)}{24(3x^2 \minus{} 2x \plus{} 3)} \ge 0,$

so

$ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \plus{} \frac {y^2}{3y^2 \minus{} 2y \plus{} 3} \plus{} \frac {z^2}{3z^2 \minus{} 2z \plus{} 3}$

$ \geq \frac {6x \minus{} 1}{24} \plus{} \frac {6y \minus{} 1}{24} \plus{} \frac {6z \minus{} 1}{24}$

$ \equal{} \frac18.$[/quote]
This way still usefull, kuing. We only need more sentences.
If $ x, y, z$ are in $ [0, \frac {3}{2}]$, the way above effect.
If in $ x, y, z$ at least have a number $ \ge \frac {3}{2}$, example assume $ 3 \minus{} 2x \le 0$
Now, easily to see that  $ \frac {x^2}{3x^2 \minus{} 2x \plus{} 3} \ge \frac {x^2}{3x^2} \equal{} \frac {1}{3}$
So, $ \left.\frac {x^{2}}{3x^{2} \minus{} 2x \plus{} 3} \plus{} \frac {y^{2}}{3y^{2} \minus{} 2y \plus{} 3} \plus{} \frac {z^{2}}{3z^{2} \minus{} 2z \plus{} 3}\ge \frac {1}{3} \plus{} 0 \plus{} 0 \ge \frac {1}{8}\right.$
The proof completed  :)[/quote]

thank you... :)

Very interesting your remarks, dear friends! Thanks for them. I'd like also to thank for the alternative solution to this inequality.

[quote="enndb0x"]Another Solution :

We can rewrite given inequality as:
\[ 1 \minus{} \frac {1}{8} \geq \sum_{cyc}{\frac {1 \minus{} \frac {2x}{3}}{3x^2 \minus{} 2x \plus{} 3}}\]
Since $ 3x^2 \minus{} 2x \plus{} 3 \geq \frac {8}{3}$  It is obvious that :
\[ 1 \minus{} \frac {1}{8} \equal{} \sum_{cyc}{\frac {1 \minus{} \frac {2x}{3}}{\frac {8}{3}}}\geq \sum_{cyc}{\frac {1 \minus{} \frac {2x}{3}}{3x^2 \minus{} 2x \plus{} 3}}\]
Done ![/quote]

Your solution is similar with kuing's and you still need nguoivn's remark.

Using the substitutions $ x\equal{}\frac{a\plus{}1}{3}$, $ y\equal{}\frac{b\plus{}1}{3}$, $ z\equal{}\frac{c\plus{}1}{3}$, where $ a\plus{}b\plus{}c\equal{}0$, we can write the inequality as
$ \sum \frac {(a\plus{} 1)^2}{a^2 \plus{} 8} \ge \frac 3{8}$.
Without loss of generality, assume that $ bc\ge 0$. By the Cauchy-Schwarz inequality, we have
$ \frac {(b\plus{} 1)^2}{b^2 \plus{} 8} \plus{} \frac {(c\plus{} 1)^2}{c^2 \plus{} 8}\ge \frac {(b\plus{}c\plus{}2)^2}{b^2 \plus{} c^2 \plus{} 16} \equal{} \frac {(a\minus{}2)^2}{a^2 \plus{} 16 \minus{} 2bc}\ge \frac {(a\minus{}2)^2}{a^2 \plus{} 16}$.
Thus, it suffices to show that
$ \frac {(a\plus{} 1)^2}{a^2 \plus{} 8} \plus{} \frac {(a\minus{}2)^2}{a^2 \plus{} 16}\ge \frac 3{8}$,
that is
$ a^2(13a^2\minus{}16a\plus{}160)\ge 0$.


AMC
USA(J)MO
USAMO
AwesomeMath
summer program

Hello,
I am a girl who is interested in the China Girls Math Olympiad. I have been unable to find any information regarding how the team(s?) will be selected or if the US is even participating in it this year. I have already sent an e-mail to rjensen@maa.org more than 3 weeks ago but have not received any reply (although I realize that I may simply need to wait longer). It seems that last year's participants were already informed before the 2007 USAMO. Is this also the case this year, or is the 2008 USAMO factored in for team selection? Does anyone on this forum have information regarding the CGMO 2008?

Thank you.

CGMO is being run differently this year. I think USAMO 2008 scores will be used... so no one has been notified yet.

Thank you for your information!
Is there any place I could find more information on it? Any additional details that you might know would also be great since I can find nothing whatsoever online regarding CGMO 2008.

Tell me how this goes!

I've taken the Chinese Math Olympiad before and I have a stash of girl's olympiad problems, so PM me if you'd like some practice!

My understanding -- and this is unofficial, since I don't know for sure -- is that they will use a combination of 2007 and 2008 USAMO results to select the team.  Last year the team trained at AwesomeMath before going to China, and I assume it's the same this year (but again, I don't know for sure).


Prove that for a prime $p$ there is another prime $q$ for which $n^p-p$ is never divisible by $q$ for any integer value of $n$.

It's problem $6$ from IMO 2003. It's been discussed before, and there's even a generalization around here somewhere. I think the generalization states that there are infinitely many such primes $q$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=98

 And there is also something about the density of the primes satisfiyng that condition.. now i don't remember exactly

Oh! thanks every one! i really did n't know that it is IMO problem :blush: !

Anyhow thanks a lot!


[quote="ch1n353ch3s54a1l"][quote="chien than"]Solve of the system of equations
$y+xy^{2}=6 x^{2}$
$1+x^{2}y^{2}=5x^{2}$[/quote]
by the second original equation, $y=5-\frac{1}{x^{2}}$[/quote]

Erm, how did you get that?

[quote="Zellex"]Erm, how did you get that?[/quote]

Pah. The answer works though >_>

[hide]$xy=\sqrt{5x^{2}-1}$, $6x^{3}=(xy)^{2}+xy$, $5x^{2}-1+\sqrt{5x^{2}-1}=6x^{3}$, which...well, conveniently has the solution $x=\frac{1}{2}$...I guess you can just test the reciprocals of the factors of 18...[/hide]

$a=xy$

$a+a^{2}=6x^{3}$

$1+a^{2}=5x^{2}$

$[(a+a^{2})/6]^{2}=[(1+a^{2})/5]^{3}$

now we have an awesome 6th degree polynomial to solve...

at which point...i bust out the calculator, $a=2,1/2, 18a^{4}+45a^{3}+86a^{2}+45a+18=0$

that factor happens to have nonreal solutions...which you can solve for...make the substiution $u=a+1/a$, etc

the problem is now solved conceptually...too much computation to finish


geometry
3D geometry
trigonometry
integration
calculus

We have a cone shaped glass dish. We put the open end of it in a another dish full of mercury. The distance between the mercury in the big dish and the tip of the cone is $H$. There is some mercury in the cone too. In the closed part above the mercury there is nytrogen of mass $m$. The outside pressure is equal to the hidrostatic pressure of a mercury pillar of height $H$. how much heat does the nytrogen pick up if the temperature rises slowly by $\Delta T$ ? (The change of the mercury level in the big dish is negligible, and the level of mercury in the cone doenst go below the level of the big dish)

Is the coefficient of thermal expansion of mercury given?

$\Delta T$ is small , so heat expansion is negligible.

So then, does that mean that the nitrogen's temperature rises by $\Delta T$, and there are no other effects such as the thermal expansion of mercury?

Yes. all heat expansion of mercury is negligibe.

If nytrogen molecule is considered with 5 degrees of freedom, the heat should be

  $Q\, ={13 \over 4}{m \over M}R\Delta T$ where R stands for Clapeyron's constant and M for nytrogen's mol.
    I enjoyed it: quite interesting. Could you tell us where did you find it?

Yes Immanuel bonfils's answer is correct. you can post the solution if you want. :D

Can someone please post the solution to this problem?

Thanks.

Here is a solution:
let the height of the gase cone be $h$.
So the volume of the gas:
$v=\frac{\pi h^{3}\tan^{2}\theta}{3}$
Where $\theta$ is the half opening angle of the cone.
So the height of the gas cone:
$h=V^{\frac13}\sqrt[3]{\frac{3}{\pi \tan^{2}\theta}}$
now lets find the pressure of the gas.
The outside pressure:
$p_{0}=\rho g H$ (\rho is the mass density of mercury)
In the tub the pressure is equal throughout the surface:
$p_{0}=p+\rho g (H-h)$
So the pressure of the gas:
$p=\rho gh$
Using the formula we got for $h$:
$p=\rho g V^{\frac13}\sqrt[3]{\frac{3}{\pi \tan^{2}\theta}}$
now lets make the following substitution:
$\psi =\sqrt[3]{\frac{3}{\pi \tan^{2}\theta}}$
so the pressure:
$p=\psi V^{\frac13}$
The work of the gas:
$W=\int_{V_{1}}^{V_{2}}pdV =\psi \int_{V_{1}}^{V_{2}}V^{\frac13}dV$
After evaluating the integral:
$W=\frac34 (\psi V_{2}^{\frac13}V_{2}-\psi V_{1}^{\frac13}V_{1})=\frac34 (p_{2}V_{2}-p_{1}V_{1})=\frac34 nR\Delta T$
According to the first law of thermodynamics, the heat taken up by the gas (nytrogen)
$Q=\Delta E+W = \frac{5}{2}nR\Delta T+\frac34 nR \Delta T =\frac{13}{4}nR \Delta T$


vector
geometry
function
complex numbers
real analysis
real analysis unsolved

This is, actually, a geometry question that came up during an attempt to understand something about random analytic functions but, since there is no geometry subforum on the College Playground and since it can be asked without mentioning complex numbers at all, I'm placing it here.

Let $n$ be a large number and $\delta$ be a very small number (you may think that $\delta=e^{-n}$). Suppose that $v_1,\dots,v_n$ and $w_1,\dots,w_n$ are unit vectors in the Euclidean space such that all the scalar products $(v_i,w_j)$ are less than $\delta$ in absolute value. Is it true that then there exist unit vectors $v'_1,\dots,v'_n$ and $w'_1,\dots,w'_n$ such that $|v_i-v'_i|$ and $|w_j-w'_j|$ are all very small (say, do not exceed $n^{100}\sqrt[10]\delta$ or something similar) and each vector $v'_i$ is orthogonal to each vector $w'_j$?

[quote="fedja"]and each vector $ v_i $ is orthogonal to each vector $ w_j $?[/quote]

You mean $ v_i' $ is orthogonal to $ w_j' $?

Yes. I've edited. Sorry for the misprint.  :blush:

Gm. Consider a compact set $K=(S_{k})^{2n}$ of $2n$-uples of unit $n$-dimensional vectors, lying in a subspace generated by $v_i$ and $w_i$ ($k$ is a dimension of the span of these vectors). Define a function $f$ on $K$ as $\sum_{i,j=1}^n |(x_i,y_j)|$, where $(x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n)$ is our $2n$-uple. Note that this function is continious, so the preimage of 0 is closed, denote it by $K_0$ and its open $\epsilon$-neighborhood by $K_{\varepsilon}$. Then the complement of $K_{\epsilon}$ is closed, so $f$ is bounded by below on it. It means that if $v_1,\dots,v_n,w_1,\dots,w_n$ are far from any system of vectors with $(v_i',w_j')=0$, then the scalar product of some two $(v_i,w_j)$ is not too small.

You need some concrete estimates, yes?

Yes, I need the distances $|v_j-v'_j|$ and $|w_j-w'_j|$ to be less than $e^{-n/10}$ when $\delta=e^{-n}$.
($10$ can be replaced by any other absolute constant, but the estimate should be exponential in $n$).