group theory
abstract algebra
superior algebra
superior algebra theorems

How can we prove that the set of all inner automorphisms of a finite group G is a normal subgroup of the set of all automorphisms of G? I proved 'being subgroup' part but how about the 'normality' ?

Let $G$ be a group, $\phi$ an automorphism of, and for all $g \in G$, $\theta_{g}$ the inner automorphism $x \mapsto gxg^{-1}$. Then we have that
\[\phi \theta_{g}\phi^{-1}= \theta_{\phi(g)}\]by a direct computation, which implies the inner automorphisms are normal in the automorphisms. Note, that I did not use the that $G$ is finite, as it is not needed for this result.


[quote="mr. math"]What is the fewest number of sundays that there can be in any one calender year?[/quote]

[hide]there are a total of 54 full weeks so its 54.

is this a trick question?[/hide]

[quote="mr. math"]No, i don't think its a trick question. Sorry, but you're answer isn't correct :( . i checked in the answer key.[/quote]

uugh crap
i divided 365 by 7 wrong so 52.  hope i did it rght now :rotfl:
its almost 12 here
i better get som rest :D

[hide="Sundae!"]$\frac{365}{7}=52.142857...$ This means there are 52 weeks and 1 day in a whole year. The most amount of sundays is 53 (if you started the year on a sunday) and the least is 52 (started the year on any other day).
$\boxed{52}$[/hide]


integration
calculus
superior algebra
superior algebra theorems

I know this may sound like a dumb question but I was looking through a math journal and I  saw a capitol pi symbol. it looks like a series of some sort, but I have never seen it before. where would I go to study this symbol and learn about it? I say it looks like a series because it has a capitol pi with a k on top and n=4 on the bottom of it followed by something like kx (this is an example). Thanks all

$ \Sigma$ = Sigma = S is the first letter in "Sum."

$ \Pi$ = Pi = P is the first letter in "Product."

And $ \int$ as in integral is really just another version of S.

There are few other symbols you could do this with, such as X (actually $ \times$) for "Cartesian Product" (of sets).


Sophia has 16 plants and a window. Only 5 plants can be put in the window at a time. Every plant gets the same amount of sunlight. What is the most time in the sun a plant can get in an 8 hour period?

[hide]Only five plants at any given moment so there are $5 \cdot 8 = 40$ hours total distributed among 16 plants is $\frac {40}{16} = 2.5$.[/hide]

okay thanks, because almost everyone in my school got that one wrong  ;)


Let $A$ be connected in $\mathbb{C}$ and $f: A\to\mathbb{Z}$ be a continuous function. Prove that $f$ is constant.

since f is continuous and A is connected, f(A) is connected. the only connected sets in Z are singleton sets. hence f is constant


Find all $n$ s.t. the following polynomial is irreducible over the integers:
x^n - 3 x^(n-1) + 2 x^ (n-2) + 6

people please try this problem ! it is not easy and quite nice.

The possible integer roots of $P(x)=x^n - 3 x^{n-1} + 2 x^ {n-2} + 6$ are the divisors of the constant term $\{ \pm{1}, \,\, \pm{2}, \pm{3}, \,\, \pm{6}\}$ .

We write $P(x)=x^{n-2}(x-2)(x-1)+6$. Then $P(2)=P(1)=6$ .Also $P(-2), \, P(3), \, P(6), \, P(-3), \, P(-6)$ are clearly not zero .

We only need to check $P(-1)=6[(-1)^{n-2}+1]$. If $n-2$ is odd $P(-1)=0$ but if $n-2$ is even $P(-1) \neq 0$.

So for all $n$ even the polynomial is irreducible over the integers.

 :)

sorry for the misunderstanding.
when i say irreducible i mean, it cannot be expressed as the product of two polynomials with integer coefficients.
you haven't proved it is irreducible for $n$ even, only that it has no linear factors, so your proof is wrong.

by generalised eisenstein it can not have factors of degree >1
 :D

sorry what's generalized Eisenstein's? haven't heard of it before.
does it say that if polynomials like that one are reducible, then it must have linear factors as well? because trivially for odd n, when you divide that thing by $x+1$ you get a polynomial of degree $n-1$, which can't be reduced into linear factors.

one possible generalisation of eisenstein is:
$P(x)=\sum a_i x^i$ if $p||a_0$  $p|a_i$ for $i=1,2,...,n-k-1$ then at least one of the factors of P has degree not more than k or P is irreducible...

that's very interesting, haven't heard of it before.
could you possibly provide a proof?


probability
geometry
3D geometry
calculus
quadratics
sphere
tetrahedron

Okay, I have assembled a contest (no thanks to most of you, who didn't submit problems grrr).  I will post it on this thread tomorrow probably around noon.  It's quite long and contains some very hard questions, so I will accept e-mailed solutions up through noon on Sunday, December 7.  My e-mail address is
fuzzyjoel at yahoo dot com.
Please send all of your solutions at once, in order.  Do the best you can with the geometry questions without a diagram.  Also, along with your solution send me your high-school-math-class-grade-equivalent.  Basically, if you are in or have already taken calc, say so.  If you are not yet in calc, say how many years you would take calc in your normal system assuming you do not skip any grades.  (Thus, if your are a 3rd grader taking a pre-calculus class, you would say "I will take calculus in one year."  If you are a 9th grader on track to take calculus senior year, you would say "I will take calclulus in 3 years.")
If you have any questions, I would appreciate if you would post them here as opposed to e-mailing me.  Also, try to make everything as easy to understand as possible.  Credit will be given according to both valid solution and correct answer, and I will make subjective estimates of how close you got.
Questions are weighted -- thus, it's worth (a little bit) more to get the harder questions correct.  Also, some questions have 2 parts, in which case you get points for each part.  The number of points will be written after each question.

And, of course, don't cheat.  What constitutes cheating is rather open.  However, if you go to http://www.math.unl.edu, you will be cheating.  (That is, don't go there.  It has many of the questions I'm using, since I am not a question-producing-machine.)

There are 55 points worth of questions in 16 questions, plus possible bonuses for one question.

Anyhow, if you have any questions or objections or complaints or compliments, here is a good place to put them.  If you have answers, e-mail is a good place to put them.  Yep.  I'll post them tomorrow.

Um, two questions (ya I say that even when it's online :) ). One, would we get no credit if we accidentally send partial solutions then send the complete one later? What about if we send only the first half or so of the competition in case we don't have time to do them all but sends the rest later? Do we still get credit?

Two, can we just PM 'em to you?

I would contribute problems except they are [b]way[/b] to easy for Intermediate.

It's not just intermediate -- it covers ranges from the medium end of Getting Started to the medium range of Advanced, at least in my opinion.  If I do this again, submit questions.

Reasons not for PMing -- I don't like the format nearly as much as I like my e-mail format.  And I don't know what will happen if you fill up my PM inbox.

As for e-mailing, I'm not going to be vicious on grading -- the point is to do the math after all.  However, try and have courtesy -- I'm doing this on my personal account in my spare time, and I'm not paying myself to do it :).

Oh, also, please don't send me attached files -- just write it up in the e-mail.  I don't want to have to go through the trouble that comes with attached files, which is small, but large enough to annoy me :).

Please re-read the guidelines above with respect to these questions.  The questions are in no particular order.  Have fun!

1. Given a point O.  A circle is drawn around O of radius 2 and a point A of distance 4 from O is chosen.  AB is tangent to circle O at B and C is on circle O such that AOC is a straight line.  Compute the areas of triangles BCO and ABO. [2]

2. The quadratic equation x2 + rx + s = 0 has non-zero roots r and s.  Compute the ordered pair, (r, s). [2]

3. A room contains ten people wearing badges numbered 1 to 10.  What is the probability that if three persons are chosen at random, the smallest of their badge numbers would be 5? [3]

4. A circle is inscribed in a 2-by-2 square.  Outside this circle and inside the square a second circle is drawn which is tangent to the first circle and to two sides of the square.  Find the radius of the second circle. [3]

5. [Note: re-worded due to original bad wording pointed out by TripleM]
Joel lives with his roommate.  Joel has a 70% chance of catching the flu if his roommate does not.  If Joel's roommate catches the flu, Joel's probability of catching the flu will rise to 90%.  Joel's roommate has a 70% chance of catching the flu.  What is the total probability Joel will catch the flu?  What if both Joel and his roommate get a flu shot, which will lower each of their probabilities of catching the flu to 1/4th their previous value under any circumstances? [2, 2]

6. Prove: it is possible to draw 6 congruent circles on the surface of a sphere such that each intersects 4 others and any 2 intersecting circles intersect in exactly 1 point.
It is also possible to draw m congruent circles on the surface of a sphere such that each intersects exactly n others for 4 other pairs (m, n).  Find these pairs.  (There is no need to prove this part.) [4, 2]

7. What is the probability that if 3 regular six-sided dice are rolled at least 1 shows a 6? [3]

8. Four spherical oranges are stacked with three tangent oranges forming the base and the fourth centered on top.  If each orange has radius 1, how high is the stack?  What is the volume of the smallest tetrahedron containing all 4 oranges? [4, 4]

9. If n is a positive integer and n(n + 1) is divided by 3, what are all possible values for the remainder?  What if n(n + 2) is divided by 5? [2, 2]

10. Given a circle whose area would be doubled if its radius were to be increased by 1.  What is its current radius? [2]

11. If x0.3 = 10, then x0.4 equals what? [2]

12. [Note: reworded so that the question is slightly different than it used to be.]
A chest consists of a rectangular box and a half-cylinder lid.  If the box is 3 by 5 by 2 feet (length by width by height), what are both possibilities for its total volume (counting the lid)? [2]

13. Five distinct lines are drawn in a plane.  The number of distinct points where two or more lines intersect is determined.  What are all possible values for this number? [3]

14. Given that a line segment with 2 endpoints is the 1-D analog of a square and a single point is the 0-D analog.  Define P(a shape) to be the infinite sum,
P(shape) = number of vertices + sum of edge lengths + surface area + 3-D volume + 4-D volume + ...
For example:
P(a point) = 1 + 0 + 0 + ... = 1
P(a segment of length 3) = 2 + 3 + 0 + 0 + ... = 5
P(an equilateral triangle of side s) = 3 + 3s + s2:rt3:/4 + 0 + 0 + ... = 3 + 3s + s2:rt3:/4
Find a pattern in P(an n-dimensional cube of side s) by examining the smallest 4 cases (n = 0 to 4).  Use this pattern to compute the 3-volume of a 5-dimensional cube.
How many of the individual measurements of a 4-dimensional cube that this method gives you can you justify without using the pattern? [3 + possible bonuses  don't worry too much about the second part.  It's probably not worth the work it will take you.]

15. A party of 18 people went to a restaurant.  The each chose a $21 meal but 4 of them forgot to bring their money.  In order to pay the total bill, everyone else who brought their money had to pay how much extra? [2]

16. A 6 by 6 checkerboard is covered with 18 dominoes (each covering two squares).  Prove that it is possible to cut the board either horizontally or vertically into two (not necessarily equal) pieces without cutting through any dominoes.  (Hint: first show that any horizontal or vertical slice cutting one domino has to cut two dominoes.) [4]

17. A car traveling at 50 miles per hour covers the distance between two towns in 6 hours less than a truck which travels at 35 miles per hour.  How many hours does it take the car to cover the distance? [2]

A couple of questions: I'll put them in spoiler just in case it would give anything away, it shouldn't.

[hide]

Question 5. I don't think this really makes sense.. not to me anyway. It says each have a 70% probability of catching the flu. As far as I can tell. So the total probability of Joel catching the flu is 70%. The other thing can't be taken into account.. maybe its the first sentence thats wrong, I don't know, its just confusing.

Question 12. Which orientation is the half-cylinder lid? It could be either way, both giving different results.

[/hide]

Good points, both.  Although you should have been able to figure out what I meant fornumber 5, I think :).
I didn't write number 12, so I can only imagine that there must have been some sort of picture that accompanied it.  Well, whatever.

will the individual scores be published or kept private?

Need we show work... When are these due?

yes, work, and they are due noon, dec. 7, as it says up there.

JBL - Don't we need some sort of restrictions on number 2?

All of the basic rules (what to send in, when to send it in, where to send it to) are up there on that long first message.  I understand that this is a math forum, and we don't all necessarily enjoy reading that much, but, I mean . . .

And number 2 is just fine the way it is, thanks.

Wow, those responses were both rather mean.  I'm sorry.

Actually, as regards the first question, I would like to reinforce that I would like *full solutions* for every question, including those that only ask for some kind of number.  Now, you don't necessarily have to tell me how you figured out that solution.  For example, on number 13, it is acceptable merely to list one way to achieve each of the possible values and give a brief line or two explaining why you can't reach any others.  However, you will not get much (or any) credit for just saying, "The answer is 37."  (By the way, the answer is clearly not 37.)

And with regards individual scores -- how would you like me to report them?  Each person's score to him or her via e-mail, or should I put everyone's scores up here on the forum?

In fact, how about this -- if you don't want your score to end up on the forum, please tell me in your e-mail to me that contains your answers.  Also, don't forget to write how long it will be until you take calculus.  (If that doesn't make sense, go read the first message at the top of this thread again.)

could you pick out some interesting solutions and post them, or have others post their solutions?  i did a good deal of these but dont plan to submit (based on time); i'd like to see if they were right.

"based on time"

You have, as of the time this was written, one week, 11 hours and 43 minutes.  46 minutes if I go by my watch, not my computer.  Anyhow, that should be ample time to write up solutions to 17 problems.

I certainly have no problem with opening this thread up to solutions after everyone has submitted theirs to me.  I could also produce the "best of" solutions to the 17 problems.  That would probably be more space-and-time efficient.

Do I have to prove? or can i just show it can be done on #6?

Showing it can be done is the same thing as proving.  Just don't make any unverifiable claims, that's all.  If it makes sense to you, and there's nothing you say that you then think, "Why is that true?" you'll probably be on safe ground.

Don't forget -- only 62 more hours left . . .

Well, I have to admit that I was rather annoyed that only 7 of you submitted any responses, and only 5 of you submitted something resembling a full set of solutions.  Final scores:

Bugzpodder  47
Chinaboy   41
Paladin8   39
Fierytycoon   24
Bookworm271828
Topper and Confuted   6

(out of a possible 55)

gah, i forgot all about this test, will you be posting solutions, or shall we go ahead and post for random problems.

ohhh, i missed the contest. unfortunately i had no computer until last monday, so it was difficult for me to submit solutions. Joel, if you arrange a second contest, i promise to participate, and im sure a lot more people will take part in it, especially if you warn us all a lot before the solutions are due. Also, i like that problem with the n-cubes, i thought it was really cool. where is it from?

Would you like me to post solutions?  Also, if you want me to grade it, I will, if you send it rather quickly (next couple of days).

The cube problem was my own invention, actually, although I don't claim originality, just uninfluenced discovery.

Yah, please post solutions. Thanks in advance.

yeah, i have waaaaay too much work to do to achieve a decent score, Chinaboy can vouch for me, and himself. lolo. but he did the contest earlier unlike a certain syntax error i know...

so post the solutions ahoy, and thanks you too

Partial solutions:
These are solutions from the sets that various people have submitted, edited slightly.  I will suppliment them with my own solutions, or possibly solutions of other solvers, in the next few days.

Solutions:

1. Since AB is tangent to circle O, it must make a right angle. Also, since the radius is 2 and AO is 4, it makes a 30-60-90 triangle with area 2*sqrt(3). Now, since angles BOC and AOB make a straight line and AOB is 60, COB must equal 120. You can break triangle COB into two 30-60-90 triangles with hypotenuse 2. Therefore, they have area sqrt(3)/2 and triangle COB has area sqrt(3).  (Due to Paladin8)

2.  x^2 + rx+s = (x - r)(x - s)
(x - r)(x - s) = x^2 - (r + s)x + rs
rs = s
r = 1 (since s is not 0)
-r - s = r
-s = 2r = 2
s = -2
(due to Confuted)

3. We can choose the people corresponding to the numbers 5, a, b, where a,
b are in the set {6, 7, 8, 9, 10}, since each element in the set is greater than  5. Thus our probability p = C(5, 2)/C(10, 3) = 10/120 = 1/12. (Due to Fierytycoon)

5.a.   0.7 of the time your roomate gets the flu, and then you get it 0.9 of the time.  0.7*0.9=0.63.
0.3 of the time he doesn't get the flu, and then you get it 0.7 of the time.  0.3*0.7 = 0.21. The sum of 0.63 and 0.21 is 0.84 = 84%
(Due to topper)
The second part is identical, and we get a final answer of 18.375%.

6 (part 1). Imagine a cube. Now imagine all the square faces become circles. Now there are 6 circles tangent to 4 others each. Just inflate it and fill in the missing space and you get a sphere.
6 (part 2). The method mentioned above only works for regular polyhedra and there are only 5, so the other 4 are: tetrahedron (4,3) ; octahedron (8,3) ; dodecahedron (12,5) ; icosahedron (20,3). (Due to Paladin8)
I thought this problem up myself -- isn't it pretty cool?

7. The probability it doesn't show a 6 is 5^3/6^3 so he probability it will show at least one 6 is 1 - 125/216 = 91/216 (Due to Chinaboy)

10.  2*r :^2:* :pi: = (r + 1) :^2:* :pi:,  2r :^2: = (r + 1) :^2:,  r :^2: - 2r - 1 = 0, so r = :rt2: + 1 since we only want the positive root.  (Due to bookworm271828)

11) x^(3/10) = 10 <=> x = 10^(10/3). Then x^(4/10) = (10^(10/3))^(4/10) = 10^(4/3).  (Due to Fierytycoon)

13) If we move from two distinct lines into two parallel lines, we lose an intersection.  If we shift a line until two intersections coincide, (colinear) we lose another.  However, this transformation stops at 4.
0 - all parallel
1- all collinear
2- not possible
3- not possible
4- 4 parallel, 1 distinct
5- 4 collinear, 1 distinct
6- 3 parallel, 2 parallel
7- 3 parallel, 2 distinct
8- 2 parallel, 2 parallel, 1 distinct
9- 2 parallel, 4 distinct
10- no 3 collinear, no 2 parallel
(Due to Bugzpodder)
Can you find alternate methods for any of these values?

15.  There are 14 paying people (18 - 4), and a bill of 378 (18*21).  378/14 = 27.  So, each person pays $6 extra (27 - 21).  (Due to bookworm271828)

17. d = distance between two towns
d/50 + 6 = d/35
so d = 700... so it takes the car
700/50 or 14 hours to cover the distance.  (Due to Chinaboy)

did anyone get number 8? that was hard...

8a was gotten by Chinaboy and Paladin8.  Bugspodder got partial credit on 8a and b because he did something dumb on 8a and didn't bother to actually tell me the answer in 8b so I had to spend 5 minutes checking whether he was correct or not.

8.  First, let us just consider a regular tetrahedron ABCD with A the top vertex.  Dropping an altitude from A to plane BCD, we see in intersects the triangle exactly in its center.   Let us call that point G.  Then AG :^2: + GB :^2: = AB :^2: by the Pythagorean theorem.  We can find (I don't want to because I'm lazy) that if the side of the tetrahedron is s, GB = s/ :rt3:.   My wrists are starting to hurt, so I shall finish this later.  Anyone else, feel free to carry it out if you want.

14.  Let's examine the polynomials.
P(point) = 1
P(segment length s) = s + 2
P(square side s) = s :^2: + 4s + 4
P(cube side s) = s :^3: + 6s :^2: + 12s + 8

Now, am I the only one who notices that these correspond to (s + 2)^n for n = 0, 1, 2, and 3?  Well, anyhow, they do.  Thus, we can guess that for a 5D cube, we will have (s + 2) :^5: = s :^5: + 10s :^4: + 40s :^3: + 80s :^2: + 80s + 32, giving us a 3-volume of 40.  Now, for the hypercube (4D), the pattern gives us s :^4: + 8s :^3: + 24s :^2: + 32s + 16.  Let's see how much we can justify otherwise:

Well, the first coefficient comes very easily, since it's roughly the definition of 4-volume.  Now, we have 4 axes, and we need a face (a cube) for the positive and negative ends of each axis (imagine a cube or a square centered at the origin of a space or plane).  That gives us 8 cubes.  Faces I don't know how to explain, but this is what Bugzpodder said about it:  "There are 4 more ways to rotate the hypercube, which makes the number of faces 6*4=24."  I don't really know if that makes sense.  Anyhow, we can think of a hypercube as two cubes joined at their vertices (just like a cube is two squares joined at their vertices), so we easily get that the number of vertices is doubled and the number of edges is doubled, plus the extra edges necessary to connected the old cube to the new.  (We actually can get the other measurements this way, too -- for the 24 plane faces, we have 6 in the original cube and 6 in the final cube, as well as 12 formed by dragging the edges of the cube in a 4th-dimensional direction.  For the 8 cubes, we have the initial cube and final cube, as well as the cube formed by dragging each face of the original cube in a 4th dimensional direction.  This process ends up being a recursive way to find the same general formula as the pattern I asked you to find.)

[quote]
Bugspodder got partial credit on 8a and b because he did something dumb on 8a and didn't bother to actually tell me the answer in 8b so I had to spend 5 minutes checking whether he was correct or not.
[/quote]
err, sorry


geometry
cyclic quadrilateral
real analysis
real analysis unsolved

Find the area of Quadrilateral (irregular four sides shape) with sides length $ 72,23,55,59$. ?

 :)

It depends on the angles. Ask a different question.

Then I can't solve till I know angles .

I just wanna be sure only .

Thank you jmerry

The area of an arbitrary quadrilateral with sides $ a,b,c,d$ is less than or equal to the area of a cyclic quadrilateral with those sides. The area of the cyclic quadrilateral is given by Brahmagupta's formula, $ \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)},$ where $ s \equal{} \frac {a \plus{} b \plus{} c \plus{} d}2.$

Additionally you should specify the diagonal lengths

or two opposite angles of the quadrilateral.

Then you can compute the area with [url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretscheider's Formula[/url].


inequalities
FTW
Vieta
pigeonhole principle
trigonometry
induction
ratio

How many problems does the topper of KSU usually get in the second level?

Does solving 3 problems ensure u are a winner?

Befuddlers

Uh, I'm pretty sure you'll need more than that to beat the top people in the state.  Don't bother with a certain numeric goal for KSU if you're participating; just do as best as you can and keep writing pretty proofs as the goal in mind while competing in the second round. :wink:

Well, last year I got about four problems, the year before that I got all five, and before that there was no KSU tournament. :)   I don't know what the cutoff is usually like for honorable mention recognition, but it's usually around the top 30+.

By the way, I'd say getting the job done is usually more important than pretty proofs in competitions, though you should try to make all your proofs as nice as possible when you don't have severe time pressure.  I think there's usually one question on this tournament that you can get if you chug away with determination for as long as it takes.  Good style doesn't hurt, though, and bad style is often a symptom of an imperfectly done job.

Since all 2nd round tests must be postmarked by February 7th, please hold all discussion in this forum until Sunday, February 8th.

Thanks!
Tom

Last year I got 2.5, I think, but I was sufficiently high off my 144 last year so that only Miles moved above me.

The phrase "sufficiently high off my 144" sounds like 144 is some sort of illegal drug.   :wink:

Anyway, as to the whole stressing about how many to solve to get whatever-th place, with two hours and five problems, you are very likely to solve the same number of problems regardless of how much you stress about it, as long as you spend a reasonable amount of time on each problem. So, I would just relax, have fun, and remember not to show up 1 hour late like I did 2 years ago.  :ewpu: :fail:

I think that it's ok to talk about it now. Charlie emailed me asking for a link to this forum. Feel free to talk about what you liked and didn't like.

Hmm, I thought this year's proof round was easier than last year's, but I guess I shouldn't say that cuz I got number 3 wrong...

[hide="Number 1"]: Mods pwn this: 24.[/hide]

[hide="Number 2"]: Heh, I used the product of tangents equals sum of tangents identity from last year  to solve this is in a very roundabout fashion.  I'm pretty sure there was an easier way to do this: 2.
[/hide]

[hide="Number 3"]: Blargh...  Frisbeeballer (not mentioning his real name) said you could use inequalities to solve this, but I didn't solve it, so fail...[/hide]

[hide="Number 4"]: Algebra-bashing with Stewart's ftw[/hide]

[hide="Number 5"]: Vieta's and a bit of brute-forcing for the win: can't remember the triples, but there were three, I think (correct me if I'm horribly wrong)[/hide]

So yeah, I'll probably place higher than I did last year cuz I had four legit proofs instead of one and a 144 on the first round instead of $ \minus{} x$, where x is over 9000...

Then again, this year's was kinda simple, so other people might all get 5's and 150s, so nvm...

Hm so I did #3 with pidgeonhole, possibly losing some points. Otherwise I made [decently] pretty solns.

(#5: we get $ \frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}\frac{1}{r_3}\equal{}\minus{}1$, methinketh?)

(Edward was the only 150, was he not?)

[quote="Math Geek"] Heh, I used the product of tangents equals sum of tangents identity from last year  to solve this is in a very roundabout fashion.  I'm pretty sure there was an easier way to do this: 2.

 Blargh...  Frisbeeballer (not mentioning his real name) said you could use inequalities to solve this, but I didn't solve it, so fail...
[/quote]

Yes the second problem had an easier solution i think:

$ cosC*cosB$ = $ cos A$
$ cosC*cosB$ = $ cos(180 \minus{} B \minus{} C)$
$ cosC*cosB$ = $ \minus{}cos(B\plus{}C)$
$ cosC*cosB$ = $ \minus{}cosB*cosC \plus{} sinB*sinC$
$ 2cosC*cosB$ = $ sinB*sinC$
$ tanB*tanC$ = $ 2$

The third problem could be solved by pigeon hole principle which falls into the category of combinatorics rather than inequalities. :D Yeah the third problem was just Erdos and Szekeres theorem.

EDIT: Solafidefarms yes we get the expression you just stated. But by bounds you get three solutions i think

$ (10,32) ; (9,27) ; (11,36)$

Are the problems available on internet?

Also, how do you register for this competition? (I know it's probably too late but I want to do it next year.)

Yeah, that's what I did for number 5, Billy.  For number 2, I was kinda nervous at the time I was solving it, and cuz I looked at last year's solutions the night before, I was sorely tempted to prove the identity and then use it...  In the end I used the sine of the sum of two values identity though, so I guess my proof was just a roundabout version of yours.

So I'm assuming from all of these posts that pretty much everyone aced proof round...except for me...  No wonder I thought it was so easy... :)

Can anyone post their proof for number 3.  I hate these logic problems and need to get better at them...

EDIT: I almost forgot.  It turns out I did attempt number 3.  Here's my proof:

[hide]It is commonly known that a person's height will never remain constant throughout his lifetime.  Whether he be approaching the glorious peak of adulthood or the bottomless chasm of old age and senility, his height will either increase or decrease, respectively, for the whole of his life.  Ergo, if one lines up $ k^2\plus{}1$ boys shoulder to shoulder and selects a group of $ k\plus{}1$ boys to step forward, their heights will consistently increase or decrease, more likely the former because, after all, they're just boys! $ \blacksquare$[/hide]

I'll get points for the creativity of my proof, rest assured!

Grr, On #5, I accidentally overlooked the $ a \neq 1$ restriction, and put $ (1 , \minus{} t^2)$ for all integers $ t \ge 1$, as a possibility.

I couldn't get #3, but I did fine on #1, #2, and #4, so I'm going to guess that I got a 34-38 out of 50.

Did fine on #1, #2, and #4, and made a false assumption on #5 to get 2 solutions. Failed #3.

Whoever from Walton who's likely to have a good chance all did 3 or 4 problems.

Edward got a 150, and I think he was the only one.

Here's the [url=http://math.kennesaw.edu/~ckoppelm/]KSU site[/url] from the GCTM site.

Sitan's boys in line proof FTW!!!  :rotfl:

I think I got 4 then...mebbe 4.1. xD

Okay, here is the problem 3 solution.  I think this was the hardest question on the test.  It is an early result, I think, in the study of pattern-avoiding permutations.

[hide]
We induct on $ n$.  For $ n\equal{}0$ the problem statement is trivial.

Now, suppose the problem statement holds for $ n \minus{} 1$.  Then there is a sequence of $ n$ boys whose heights are monotonically increasing or decreasing.  Call the boy at the end of this sequence $ a_1$.  Now consider the $ n^2$ boys other than $ a_1$.  By inductive hypothesis, there is another boy, $ a_2$, who is at the end of another monotonic sequence.  We may continue this process to obtain $ 2n\plus{}1$ boys who are at the end of monotonic height sequence.  Without loss of generality, let at least $ n\plus{}1$ of them be at the end of increasing height sequences; we will call these the [i]tall[/i] boys.  If any one of the tall boys is taller than a tall boy who comes before him, then we may add him to that boy's sequence to obtain a sequence of $ n\plus{}1$ boys in increasing height order.  Otherwise, the tall boys form a sequence of boys in decreasing order!  $ \blacksquare$

We note also that the problem's bound is sharp.  In particular, if we have $ n^2$ boys, who are (in height order) $ a_{11}, a_{12}, \dotsc, a_{1n}, a_{21}, a_{22} , \dotsc, a_{nn}$, then the order
\[ a_{n1}, a_{n2}, \dotsc, a_{nn}, a_{(n\minus{}1)1}, a_{(n\minus{}1)2}, \dotsc, a_{(n\minus{}1)n}, \dotsc, a_{11}, \dotsc, a_{1n} \]
contains no $ n\plus{}1$ boys in monotonic height order.  Furthermore, we may add a boy taller than any of the others between $ a_{nn}$ and $ a_{(n\minus{}1)1}$ to obtain a sequence of $ n^2\plus{}1$ boys in which there is only one sequence of $ n\plus{}1$ boys in monotonic height order.[/hide]

Also, for problem 4, I think the easiest approach is to show that the cosine of the triangle's largest angle is the cosine of twice the cosine of the triangle's smallest angle.  Therefore the triangle's largest angle is twice the triangle's smallest angle.  The problem statement then follows from AA similarity.

[quote="Math Geek"]Umm, Arvind, are you sure induction works for #3? :P[/quote]

definitely not. but hey, at least it sounds like a solution might have happened. maybe. xDDD

Easiest for number 4 in my opinion :
Triangle ABC, bisection creates point D.  Let AB=4x, BC=5x, and AC=6x.  So, by angle bisector theorem, AD=8/3x, DC=10/3x.  Then, AB/AD=4/(8/3)=6/4, and <A is shared by triangle ADB and ABC, so ABC ~ADB by SAS, and therefore ADB's sides are in a ratio of 4:5:6.

The "porblems :P " will probably come out in a few weeks on the website mentioned in one of the previous posts.

They're already out...Silly Sitan. :P

http://math.kennesaw.edu/~ckoppelm/


I have 3 problems.The problems are as follow:

1/  The color of copper (II) is blue and permanganate is purple.What about copper(II)  permanganate? Is it blue+purple? Why?

2/  Why (((sea water  ---electricity----> $H_2$ + $Cl_2$ + $NaOH$))) , but not $2H_2$ + $O_2$ + $NaCl$ ?

3/  While checking the migration of ions, why should we moisten the paper with sodium sulphate, but not water or others? Why?

1) Just because 2 ions have some colours in an aqueous solution doesn't mean the resulting compund will have the combination of colours, Colour of anything is related to the electonic transitions which result when a molecule absorb photons. So the color of the new cmpd. will be determined by the configuration of Molecular Orbitals(and other effect) which is very very difficult to determine exactly. I believe you can't simple determine the color!!!

2) $Na^ + $ has a great affinity for $ OH ^ - $ . You can explain this on the basis of $ \delta G $ or any other reason(electrostataic forces) Thus, as soon as the ions are formed, this reaction takes place.

3) I didn't understand this. Can you plz explain the context

1). my chem. teacher said that most copper compounds are blue. so blue would be my guess.

I would guess it would be a blend of copper's blue and permanganate's purple.

It has to do with the overvoltage of the oxidation of chloride, compared to that of hydroxide.

I don't quite understand the question - does this refer to a salt bridge in a voltaic cell?  Chromatography?  Something else?


induction
linear algebra
matrix
trigonometry
linear algebra unsolved

Let $ A,B\in M_{2}(R)$ so that $ (A\plus{}B)^n\equal{}A^{n}\plus{}B^{n},\forall n\geq 3$. Prove that $ AB\equal{}BA\equal{}O_{2}$.

$ (\mathbb{A}\plus{}\mathbb{B})^2 \equal{} \mathbb{A}^2 \plus{} \mathbb{B}^2$
$ \iff (\mathbb{A}\plus{}\mathbb{B})^2 \minus{} (\mathbb{A}^2 \plus{} \mathbb{B}^2) \equal{} \mathbb{O}$
$ \iff (\mathbb{A}^2 \plus{} \mathbb{AB} \plus{} \mathbb{BA} \plus{} \mathbb{B}^2) \minus{} (\mathbb{A}^2 \plus{} \mathbb{B}^2) \equal{} \mathbb{O}$
$ \iff \mathbb{AB} \plus{} \mathbb{BA} \equal{} \mathbb{O}$
$ \iff \mathbb{AB} \equal{} \minus{} \mathbb{BA}$

$ (\mathbb{A}\plus{}\mathbb{B})^3 \equal{} \mathbb{A}^3 \plus{} \mathbb{B}^3$
$ \iff (\mathbb{A}\plus{}\mathbb{B})^3 \minus{} (\mathbb{A}^3 \plus{} \mathbb{B}^3) \equal{} \mathbb{O}$
$ \iff (\mathbb{A}^3 \plus{} \mathbb{ABA} \plus{} \mathbb{BAA} \plus{} \mathbb{BBA} \plus{} \mathbb{AAB} \plus{} \mathbb{ABB} \plus{} \mathbb{BAB} \plus{} \mathbb{B}^3) \minus{} (\mathbb{A}^3 \plus{} \mathbb{B}^3) \equal{} \mathbb{O}$
$ \iff \mathbb{ABA} \plus{} \mathbb{BAA} \plus{} \mathbb{BBA} \plus{} \mathbb{AAB} \plus{} \mathbb{ABB} \plus{} \mathbb{BAB} \equal{} \mathbb{O}$
$ \iff \mathbb{ABA} \minus{} \mathbb{ABA} \plus{} \mathbb{BBA} \minus{} \mathbb{ABA} \plus{} \mathbb{ABB} \minus{} \mathbb{ABB} \equal{} \mathbb{O} \ \ \ (\because \mathbb{AB} \equal{} \minus{} \mathbb{BA})$
$ \iff \mathbb{BBA} \minus{} \mathbb{ABA} \equal{} \mathbb{O}$
$ \iff (\mathbb{B}\minus{}\mathbb{A})\mathbb{BA} \equal{} \mathbb{O}$
$ \iff \mathbb{B}\minus{}\mathbb{A} \equal{} \mathbb{O} \ \text{or} \ \mathbb{BA} \equal{} \mathbb{O}$

$ \boxed{\text{Case} \ \mathbb{B}\minus{}\mathbb{A} \equal{} \mathbb{O}}$
$ \mathbb{AB} \equal{} \minus{} \mathbb{BA}$
$ \iff \mathbb{A}^2 \equal{} \minus{} \mathbb{A}^2$
$ \iff 2\mathbb{A}^2 \equal{} \mathbb{O}$
But, $ \mathbb{A}\in M(2; \mathbb{C})$

$ \boxed{\text{Case} \ \mathbb{BA} \equal{} \mathbb{O}}$
$ \mathbb{AB} \equal{} \minus{} \mathbb{BA}$
$ \iff \mathbb{AB} \equal{} \minus{} \mathbb{O} \equal{} \mathbb{O}$
Therefore, $ \mathbb{AB} \equal{} \mathbb{BA} \equal{} \mathbb{O}$

Since calculate, the given statement is clearly true when $ n\equal{}3$.
Now we assume it is true for an arbitrary positive integer $ k>3$ : $ (\mathbb{A}\plus{}\mathbb{B})^k \equal{} \mathbb{A}^k \plus{} \mathbb{B}^k$
Then,
$ (\mathbb{A}\plus{}\mathbb{B})^{k\plus{}1} \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1})$
$ \equal{} (\mathbb{A}\plus{}\mathbb{B})^k(\mathbb{A}\plus{}\mathbb{B}) \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1})$
$ \equal{} (\mathbb{A}^k\plus{}\mathbb{B}^k)(\mathbb{A}\plus{}\mathbb{B}) \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1})$
$ \equal{} (\mathbb{A}^{k\plus{}1}\plus{}\mathbb{B}^{k\plus{}1}\plus{}\mathbb{A}^k\mathbb{B}\plus{}\mathbb{A}\mathbb{B}^k) \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1})$
$ \equal{} \mathbb{A}^k\mathbb{B}\plus{}\mathbb{A}\mathbb{B}^k$
$ \equal{} (\minus{}1)^{k\minus{}1}\mathbb{B}\mathbb{A}^k\plus{}\mathbb{A}\mathbb{B}^k$
$ \equal{} (\minus{}1)^k\mathbb{A}\mathbb{B}\mathbb{A}^{k\minus{}1}\plus{}\mathbb{A}\mathbb{B}\mathbb{B}^{k\minus{}1}$
$ \equal{} \mathbb{A}\mathbb{B}((\minus{}1)^k \mathbb{A}^{k\minus{}1}\plus{}\mathbb{B}^{k\minus{}1})$
$ \equal{} \mathbb{O}$
($ \because \ \mathbb{A}\mathbb{B} \not\equal{} \mathbb{O} \Rightarrow (\mathbb{A}\plus{}\mathbb{B})^3 \not\equal{} \mathbb{A}^3 \plus{} \mathbb{B}^3$)
Thus the formula works for $ n\equal{}k\plus{}1$.
So, by PMI, the statement is true for every integer $ n \ge 3$.

[quote]$ (\mathbb{A} \plus{} \mathbb{B})^2 \equal{} \mathbb{A}^2 \plus{} \mathbb{B}^2$[/quote]
This is not necessarily true since the property from the hypothesis is true only for $ n\geq 3$.
I also do not think that the induction part is necessary, maybe only if you are trying to prove the converse(which is easier with the binomial theorem).

The hypothesis that these are $ 2\times 2$ is essential. A $ 3\times 3$ counterexample: $ \begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}$ and $ \begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}$.

The field also matters: $ A\equal{}B\equal{}I$ is a counterexample in characteristic 2.

Any argument is guaranteed to be grubby.

Since calculate, the given statement is clearly true when $ n\equal{}3$.
Now we assume it is true for an arbitrary positive integer $ k>3$ : $ (\mathbb{A}\plus{}\mathbb{B})^k \equal{} \mathbb{A}^k \plus{} \mathbb{B}^k$
Then,
$ (\mathbb{A}\plus{}\mathbb{B})^{k\plus{}1} \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1}) \equal{} \mathbb{O}$
$ \iff (\mathbb{A}\plus{}\mathbb{B})^k(\mathbb{A}\plus{}\mathbb{B}) \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1}) \equal{} \mathbb{O}$
$ \iff (\mathbb{A}^k\plus{}\mathbb{B}^k)(\mathbb{A}\plus{}\mathbb{B}) \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1}) \equal{} \mathbb{O}$
$ \iff (\mathbb{A}^{k\plus{}1}\plus{}\mathbb{B}^{k\plus{}1}\plus{}\mathbb{A}^k\mathbb{B}\plus{}\mathbb{A}\mathbb{B}^k) \minus{} (\mathbb{A}^{k\plus{}1} \plus{} \mathbb{B}^{k\plus{}1}) \equal{} \mathbb{O}$
$ \iff \mathbb{A}^k\mathbb{B}\plus{}\mathbb{A}\mathbb{B}^k \equal{} \mathbb{O}$
$ \iff (\minus{}1)^{k\minus{}1}\mathbb{B}\mathbb{A}^k\plus{}\mathbb{A}\mathbb{B}^k \equal{} \mathbb{O}$
$ \iff (\minus{}1)^k\mathbb{A}\mathbb{B}\mathbb{A}^{k\minus{}1}\plus{}\mathbb{A}\mathbb{B}\mathbb{B}^{k\minus{}1} \equal{} \mathbb{O}$
$ \iff \mathbb{AB}((\minus{}1)^k \mathbb{A}^{k\minus{}1}\plus{}\mathbb{B}^{k\minus{}1}) \equal{} \mathbb{O}$
$ \iff \mathbb{AB} \equal{} \mathbb{O} \ \text{or} \ (\minus{}1)^k \mathbb{A}^{k\minus{}1}\plus{}\mathbb{B}^{k\minus{}1} \equal{} \mathbb{O}$

$ \boxed{\text{Case} \ \mathbb{AB} \equal{} \mathbb{O}}$
$ \mathbb{AB} \equal{} \minus{} \mathbb{BA}$
$ \iff \mathbb{O} \equal{} \minus{} \mathbb{BA}$
$ \iff \mathbb{BA} \equal{} \mathbb{O}$
Therefore, $ \mathbb{AB} \equal{} \mathbb{BA} \equal{} \mathbb{O}$

$ \boxed{\text{Case} \ (\minus{}1)^k \mathbb{A}^{k\minus{}1}\plus{}\mathbb{B}^{k\minus{}1} \equal{} \mathbb{O}}$
$ (\mathbb{AB})^{k\minus{}1}$
$ \equal{} \sqrt{2}\sin \left(\frac{2k\minus{}1}{4}\pi\right)\mathbb{A}^{k\minus{}1}\mathbb{B}^{k\minus{}1}$
$ \equal{} (\minus{}1)^k\sqrt{2}\sin \left(\frac{2k\minus{}1}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)}$
$ \equal{} \sqrt{2}\sin \left(\frac{2k\minus{}3}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)}$

$ (\minus{}\mathbb{BA})^{k\minus{}1}$
$ \equal{} (\minus{}1)^{k\minus{}1}(\mathbb{BA})^{k\minus{}1}$
$ \equal{} (\minus{}1)^{k\minus{}1}(\minus{}1)^k\sqrt{2}\sin \left(\frac{2k\minus{}1}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)}$
$ \equal{} \sqrt{2}\sin \left(\frac{2k\minus{}1}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)}$

So,
$ (\mathbb{AB})^{k\minus{}1}\equal{}(\minus{}\mathbb{BA})^{k\minus{}1}$
$ \iff \sqrt{2}\sin \left(\frac{2k\minus{}3}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)} \equal{} \sqrt{2}\sin \left(\frac{2k\minus{}1}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)}$
$ \iff 2\sqrt{2}\sin \left(\frac{2k\minus{}3}{4}\pi\right)\mathbb{A}^{2(k\minus{}1)} \equal{} \mathbb{O}$
$ \iff \mathbb{A}^{2(k\minus{}1)} \equal{} \mathbb{O}$
But, $ \mathbb{A}\in M(2; \mathbb{C})$

Thus the formula works for $ n\equal{}k\plus{}1$.
So, by PMI, the statement is true for every integer $ n \ge 3$.


real analysis
number theory proposed
number theory

Prove that $ \frac{(2n)!}{(n)!(n\plus{}1)!}$ is an integer for all positive integers $ n$.

http://www.mathlinks.ro/viewtopic.php?p=1280283#1280283

It equals $ \binom{2n}{n} \minus{} \binom{2n}{n \plus{} 1}$.

I am tempted to give the combinatorial proof (using path counting - Catalan number definition), but I'll skip, anyone interested can look it up on wiki :)

BTW, what a coincidence, it is in my school's Real Analysis homework [ok, it got nothing to do with analysis, but this is an introductory HW so it contains some NT and combinatorical problems]  :P

Thank you very much Harun (Actually I got this problem from one of Huse's students:))


abstract algebra
superior algebra
superior algebra unsolved

Let $ B \equal{} (2,x)$ be the ideal generated by $ 2$ and $ x$ in the ring $ A \equal{} \mathbb{Z} [x]$.

Show that the element $ 2\otimes 2 \plus{} x\otimes x$ in $ B\otimes _{A} B$ is not a simple tensor.

It looks like easy problem. But I can't prove it because tensor product is quite new for me. Please help me.

The trick for problems like this is constructing bilinear maps on $ B\times B$ using the universal property of tensor products.

Say $ 2\otimes 2 \plus{} x\otimes x \equal{} a(x)\otimes b(x)$.  Since $ (f,g)\mapsto f(0)g(0)$ is a bilinear map to $ \mathbb{Z}$, there is a map $ B\otimes B\to \mathbb{Z}$ taking $ f\otimes g$ to $ f(0)g(0)$.  Applying this to our equation, we have $ a(0)b(0) \equal{} 4$.

Similarly, $ a(0)b'(0)\equal{}0$, and $ a'(0)b'(0)\equal{}1$.  But this is impossible.


let $a,b,c$ be positive constants. determine the maximum and minimum value of:$\sqrt{a^2.x^2+b^2.y^2+c^2.z^2}+\sqrt{a^2.y^2+b^2.z^2+c^2.x^2}+\sqrt{a^2.z^2+b^2.x^2+c^2.y^2}$,subject to $x^2+y^2+z^2=1$

Oh, boy, this is so nice! The extreme values are $ a+b+c $ and $\sqrt{3(a^2+b^2+c^2)} $. The maximum is trivial to find using Cauchy. For the minimum, by squaring it is enough to show that $\sum_\sqrt{a^2x^2+b^2y^+c^2z^2}\cdot\sqrt{a^2y^2+b^2z^2+c^2x^2}\geq ab+bc+ca $. This follows by adding the inequalities of the form $\sqrt{a^2x^2+b^2y^+c^2z^2}\cdot\sqrt{a^2y^2+b^2z^2+c^2x^2}\geq acx^2+aby^2+bcz^2 $ found again by Cauchy. Klamkins problems are so nice!

For the minimum, it can be done by Minkowski,

$\sqrt{a^2x^2+b^2y^2+c^2z^2}+\sqrt{a^2y^2+b^2z^2+c^2x^2}+\sqrt{a^2z^2+b^2x^2+c^2y^2}$

$\geq \sqrt { (ax+cx+bx)^2 + (by+ay+cy)^2 + (cz+bz+az)^2} = { a+b+c}$.


limit
calculus
function
inequalities
real analysis
real analysis solved

Calculus Unsolved Problems is more appropriate place for this problem ;)

Dear Myth,
Since this problem was one of questions in the exam chosing the team for VMO 2005 of my friend's school ( for high school students), I post this problem to this topic.If you feel my problem being here isn't appropriate, please move it to Calculus Unsolved Problems. Thanks!

[quote="yrch"]i'd have a technical question to this related to this problem - are the conditions given here enough to expand the function in Maclaurin series or not? if, so the solution is quite simple. if not, i'll have to think some more :) the thing is going on only around zero, so i guess we could do that, but i'm not sure, so please prove or disprove.[/quote]
Of course, you can't use Maclaurin series, since you haven't condititon, that function is differentiable.

Solution is simple enough.

Let $\epsilon > 0$ be given. Let $\eta > 0$ be such that for $|a| < \eta$ we have $ | \frac {f(3a) - f( a )} a | < \epsilon$ (from the second hypothesis).

Let $x$ such that $0 \leq x < \eta$ (the case $x < 0$ is similar).
Then for all positive integer $k$, we have :
$- \frac {\epsilon x } {3^k} < f(\frac x {3^{k-1}}) - f(\frac x {3^k}) < \frac {\epsilon x } {3^k}$. (1)

Now let's choose a positive integer $n$ sufficientely large so that, from the first hypothesis, we have $- \frac {\epsilon x} 2 < f( \frac x {3^n}) < \frac {\epsilon x} 2 $. (2)
For such $n$, add all the inequalities (1) for $k=1, \cdots ,n$.
Using (2), it leads to
$- \frac {\epsilon x} 2 - \frac {3 \epsilon x} 2 (1 - \frac 1 {3^n} ) < f(x) < \frac {\epsilon x} 2 + \frac {3 \epsilon x} 2 (1 - \frac 1 {3^n} ) $
so that
$- 2 \epsilon x < f(x) < 2 \epsilon x$
therefore $| \frac {f(x)} x | < 2 \epsilon.$ (3)

As claimed above, we may use a similar reasoning in the case where $x<0$.
Thus, from (3) the desired limit follows easily.

Pierre.

As I said above, solution is really simple.  ;)  :D

In the first line, you admit that $\lim \frac {f(x)} x$ does exist, which is not sure.
Then, you admit that $f$ is differentiable which is not given too.

Pierre.


to calculate the probablilità for a poker player to find itself served in hand: one brace of aces a poker a full

 :lol:

I'm not quite sure what you mean- I see a "pair of aces", but there are many variations on the basic game of poker with different rules, and I don't know what you want "a full" to do in that sentence.

Calculating probabilities of poker hands is a standard (if long) exercise. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=136394]Here[/url] is on version.


linear algebra
matrix
geometry
analytic geometry
projective geometry
LaTeX

As an example
A=\begin{pmatrix}0&1&0&\cdots&0&0\\
0&0&1&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&1\\
0&0&0&\cdots&0&0\end{pmatrix}

yields

$ A\equal{}\begin{pmatrix}0&1&0&\cdots&0&0\\
0&0&1&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&1\\
0&0&0&\cdots&0&0\end{pmatrix}$

Or
\left(\begin{array}{cc}-1&5\\\sqrt 2&3\end{array}\right)

yields

$ \left(\begin{array}{cc}\minus{}1&5\\\sqrt 2&3\end{array}\right)$

$ \det{A} \equal{} \left| \begin{array} {cc} \minus{}1&5\\ \sqrt 2&3 \end{array} \right|$

[code]\det{A} = \begin{vmatrix}
-1&5\\
\sqrt 2&3
\end{vmatrix}[/code] is simpler
$ \det{A} \equal{} \begin{vmatrix}
\minus{}1&5\\
\sqrt 2&3
\end{vmatrix}$

{pmatrix}

$ A\equal{}\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}$

{bmatrix}

$ A\equal{}\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$

On this forum, I usually (but not always) use bmatrix rather than pmatrix. The choice is a matter of cosmetic preference.

[quote="Kent Merryfield"]On this forum, I usually (but not always) use bmatrix rather than pmatrix. The choice is a matter of cosmetic preference.[/quote]
As an aside, on a course I used to teach for the Open University, the bmatrix style was used for matrices in projective geometry acting on homogeneous coordinates so $ \begin{bmatrix}1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9\end{bmatrix} \equal{} \begin{bmatrix}2 & 4 & 6 \\
8 & 10 & 12 \\
14 & 16 & 18\end{bmatrix}$, whereas pmatrix was used for the usual matrices and where $ \begin{pmatrix}1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9\end{pmatrix}\neq\begin{pmatrix}2 & 4 & 6 \\
8 & 10 & 12 \\
14 & 16 & 18\end{pmatrix}$. Thus great care was needed.

It did cause a lot of confusion because some of the students had used bmatrix in a previous course for ordinary matrices. I had some difficulty explaining to students how mathematics notation is not always standardised and different people like using different notation. They found this an odd concept thinking that everything in mathematics is laid down and there are no personal preferences. Giving them other examples (such as soluble vs solvable groups in British vs American usage) didn't seem to help :)


The measure of one of the smaller base angles of an isoceles trapezoid is $ 60^\circ$. The shorter base is 5 inches long and the altitude is $ 2 \sqrt{3}$ inches long. What is the number of inches in the perimeter of the trapezoid?
[asy]import olympiad; size(150); import geometry_dev; import graph; defaultpen(linewidth(0.8));
draw(origin--(9,0)--(7,2*sqrt(3))--(2,2*sqrt(3))--cycle);
draw(Label("$2\sqrt{3}$",align=E),(2,2*sqrt(3))--(2,0),dashed);
label("5",(7,2*sqrt(3))--(2,2*sqrt(3)),N);
markangle(Label("$60^{\circ}$"),(2,2*sqrt(3)),origin,(9,0),radius=10);[/asy]

The smallest length of the triangle is 2 (since its a 30-60-90 triangle), so the hypotenuse is 4. Thus, the perimeter is 5+4+2+2+5+4=22.


Having problems thinking of what the formula is for calculating the following:-

Say you you have the following:-

Gross income 2005= £8000

If this is a rise of 30% of the previous year then how would you calculate what the previous years gross was?

Same applies if it was 35% , 45% etc.

Thanks

Busby


quadratics
number theory unsolved
number theory

Let a,b be two positive integers that can be expressed as the sum of two squares of integer. Assume $ \frac{a}{b}$ is an integer, prove that $ \frac{a}{b}$ can also be expressed as the sum of two squares of integer.

[quote="greentreeroad"]Let a,b be two positive integers that can be expressed as the sum of two squares of integer. Assume $ \frac {a}{b}$ is an integer, prove that $ \frac {a}{b}$ can also be expressed as the sum of two squares of integer.[/quote]
It is good problem .
A well know lemma :
[u]Lemma [/u] A number $ n$ can represent as sum of two square if and only if $ v_p(n)\equiv 0 (\mod 2)$ for all prime  p such that $ p\equiv 3 (\mod 4)$
Now consider problem :
Because $ a,b$ is sum of two square so $ v_p(a)\equiv v_p(b)\equiv 0 (\mod 4)$ for all prime $ p\ equiv 3 (\mod 4)$
More than $ b|a$ so $ v_p(a)\geq v_p(b)$ for all prime p .
So $ v_p(\frac{a}{b})\equiv 0 (\mod 2 )$ for all prime $ p\equiv 3 (\mod 4)$
It follows that $ \frac{a}{b}$ can represent as sum of two square .

[quote="TTsphn"]
A well know lemma :
[u]Lemma [/u] A number $ n$ can represent as sum of two square if and only if $ v_p(n)\equiv 0 (\mod 2)$ for all prime  p such that $ p\equiv 3 (\mod 4)$
[/quote]

O.K, but how to prove this lemma?

It is an well know problem .
A hint :
1.$ (\frac{\minus{}1}{p})\equal{}1$ if $ p\equiv 1 (\mod 4)$
$ (\frac{\minus{}1}{p})\equal{}\minus{}1$ if $ p\equiv 3 (\mod 4)$
2. $ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2$

[quote="greentreeroad"]I can see why numbers with even exponent of 4K+3 prime can be expressed as sum of two squares, but I don't know how to prove that a number of form 4K+1 which contains odd exponent of 4K+3 prime CANNOT.(for example, 21=3*7 and 165=3*11*5)[/quote]
Yes because from $ (\frac{\minus{}1}{p})\equal{}\minus{}1$ for all prime $ p\equiv 3 (\mod 4)$ so if $ p|x^2\plus{}y^2$ then $ p|x,p|y$
It follow that $ v_p(x^2\plus{}y^2)\equal{}2.v_p (\gcd(x,y))$

Thanks a lot. At last, how do you actually get p|x, p|y just from the fact $ p|x^2\plus{}y^2$ for p a 4K+3 prime? That is, why t and p-t cannot both be a quadratic residue of p?

Because $ (\frac { \minus{} 1}{p}) \equal{} ( \minus{} 1)^{\frac {p \minus{} 1}{2}}$ (1)
If $ p\not | x$ then $ p\not |y$ so exist an integer k such that
$ x\equiv ky (\mod p)$ so
$ p|y^2(k^2 \plus{} 1)$
$ \Rightarrow p|x^2 \plus{} 1$
It gives contradiction from (1) .
So $ p|x$ ,follow that $ p|y$

We have
$ \frac{(pm\minus{}qn)^2\plus{}(pn\plus{}qm)^2}{m^2\plus{}n^2}\equal{}p^2\plus{}q^2$.
Can we prove the numbers must always take this form?


limit
logarithms
integration
calculus
calculus computations

Evaluate:

$ \lim_{n\to \plus{} \infty}\prod_{k \equal{} 1}^{n}(1 \plus{} \frac {1}{k^{3}})$

It doesn't telescope or otherwise simplify in any obvious way. It does converge - the general theory of infinite products compares its convergence to that of $ \sum_k\frac1{k^3},$ which is a convergent series.

Numerical approximation (running this through $ n\equal{}64$ and then using a Richardson extrapolation table) gives an approximate value of $ 2.428189792$

Mathematica gives $ \frac{1}{\pi} \cosh\left(\frac{\sqrt{3}}{2}\pi\right) \approx 2.42819$ and that should be OK.. It matches what Kent Merryfield suggested..

$ \lim_{n\to \plus{} \infty}\ln \prod_{k \equal{} 1}^{n}(1 \plus{} \frac {1}{k^{3}}) \equal{} \sum_{n \equal{} 1}^{\infty} \ln \left(1 \plus{} \frac {1}{n^3}\right)\leq \sum_{n \equal{} 1}^{\infty} \frac {1}{n^3} \equal{} \zeta (3)$

$ \therefore \lim_{n\to \plus{} \infty}\prod_{k \equal{} 1}^{n}(1 \plus{} \frac {1}{k^{3}})\leq e^{\zeta (3)}?$

Turns out that the product is equal to:

$ \exp\left(3\int^\infty_0 \frac {F(x)}{x \left(e^x \minus{} 1\right)}\,dx\right)$

where

$ F'''(x) \plus{} F(x) \equal{} 1 \, , \qquad F(0) \equal{} F'(0) \equal{} F''(0) \equal{} 0$


induction
number theory proposed
number theory

Foe each positive integer $ n$, let $ S(n)$ be the sum of the digits in the decimal expansion of $ n$. Prove that for all $ n$, $ S(2n) \le 2S(n) \le 10 S(2n)$ and show that there exists $ n$ such that $ S(n)\equal{}1996 S(3n)$.

Let $ 0\le a<10,n\equal{}10k\plus{}a$, then $ S(n)\equal{}a\plus{}S(k)$.
$ S(2n)\equal{}S(2a)\plus{}S(2k)$ it is true for $ a<5$ and for $ a\ge 5$.
By induktion $ S(2n)\le 2S(n)$, generally from $ S(m\plus{}n)\le S(m)\plus{}S(n)$ we get $ S(kn)\le kS(n)$.
$ S(2n)\ge S(2k)\plus{}1,a>0$. By induction $ 10S(2n)\ge 2S(n)$.
Consider $ n_k\equal{}\frac{10^k\plus{}2}{3}$. Then $ S(3n_k)\equal{}3, S(n_k)\equal{}3k\plus{}1$.
Therefore if $ n\equal{}10^{2k}n_{k\minus{}1}\plus{}10^kn_k\plus{}n_k$, then $ S(3n)\equal{}9,S(n)\equal{}3(k\minus{}1)\plus{}1\plus{}3k\plus{}1\plus{}3k\plus{}1\equal{}9k$ and $ \frac{S(n)}{S(3n)}\equal{}k$.


sorry if this is dum but how do you do from the last step? et 3/40?
from $ \frac{1}{n}-\frac{1}{n+2}$


We call a real number $x$ with $0 < x < 1$ [i]interesting[/i] if $x$ is irrational and if in its decimal writing the first four decimals are equal. Determine the least positive integer $n$ with the property that every real number $t$ with $0 < t < 1$ can be written as the sum of $n$ pairwise distinct interesting numbers.

[hide]
$n=1112$
take $t_{0}=0.1111$
interesting numbers that are smaller that $t_{0}$ begin with four zeroes so they are smaller than $0.0001$
let $t_{0}=\sum_{i=1}^{n}a_{i}$ where $a_{i}$ are interesting numbers.now we have
$t_{0}=\sum_{i=1}^{n}a_{i}<\sum_{i=1}^{n}0.0001=n*0.0001$ so $n>1111$ or $n\geq 1112$
take any $t$ such that $0<t\leq t_{0}$. take $b=\frac{t}{1112}$ we know that $b<0.0001$ and that
$t=\sum_{i=1}^{1112}b$ so all we have to do is turn the $b$'s into different irrationals. take $\phi$ such that $b-556\phi>0$ and $b+556\phi<0.0001$. if $b$ is rational pick $\phi$ irrational and vice versa. the numbers $b-556\phi,b-555\phi,...,b-\phi,b+\phi,...,b+556\phi$ satisfy.
if $t>t_{0}$ write it in this form $t=k*t_{0}+t^{'}$ where $k$ is an integer and $0<t^{'}<t_{0}$ and take one number from the representation of $t^{'}$ and add $k*t_{0}$ to it.
[/hide]

yes, you are correct. i had the same idea when i composed this (not very great) problem.

you're too modest  :wink:
the only problem is that it's really easy to see what $n$ is...


Can you program on a TI-89? If so is it different from programmind on a TI-83? How much different?

it isn't all that different.. easier, i think.. and more functions are available

LOL The hardest part for me was getting to the program editor, and then figuring out how to run programs.  There are some really cool things you can do w/it, but unfortunately I only know how to do TI-83 programming.  So I changed all my TI-83 programs into programs for my 89.  I mean you can do the same stuff that you do on 83, but also there is much more.  Its not like totally different or anything.


function
advanced fields
advanced fields unsolved

Let $ \alpha \equal{} x_1dx_1 \plus{} \ldots \plus{} x_ndx_n$ and $ \omega \equal{} dx_1 \wedge\ldots\wedge dx_n$ be differential forms on $ \mathbf R^n$. Does there exist a differential form $ \beta$ on $ \mathbf R^n$ with the property that $ \alpha\wedge\beta \equal{} \omega$?

$ \beta \equal{} \frac {1}{n}\sum_{i \equal{} 1}^n \frac {1}{x_i}dx_1\wedge \cdots \wedge d\bar{x_i}\wedge \cdots \wedge dx_n$, where the barred is omitted.

If $ n \equal{} 2$, then $ \beta \equal{} \frac {1}{2}\left(\frac {1}{x_1}dx_2 \plus{} \frac {1}{x_2}dx_1\right)$.

This way everything with a $ \frac {1}{x_i}$ in it will go to zero otherwise you get a 1 summing to n when everything is put together. Wow, weird way to explain it, you should get the point, though.

[quote="HilbertThm90"]$ \beta \equal{} \frac {1}{n}\sum_{i \equal{} 1}^n \frac {1}{x_i}dx_1\wedge \cdots \wedge d\bar{x_i}\wedge \cdots \wedge dx_n$, where the barred is omitted.[/quote]

I guess you meant $ \beta \equal{} \frac {1}{n}\sum_{i \equal{} 1}^n \frac {( \minus{} 1)^{i \plus{} 1}}{x_i}dx_1\wedge \cdots \wedge d\bar{x_i}\wedge \cdots \wedge dx_n$.

Maybe this is a stupid remark but a differential form should be smooth and this one is not (at $ 0$), or shouldn't it?

Note that $ \beta$ must be a differential form on [i]the whole of[/i] $ \mathbf R^n$, not on $ \mathbf R^n \minus{} \{0\}$ only.  :wink:

Well, that settles it: since $ \alpha$ evaluates to the zero covector at the origin, its product with anything will also be zero at the origin.

Oh right, I was thinking that, but blatantly ignored that fact that it isn't defined at the origin for some reason.

To be precise, your $ \beta$ is not defined on the union of $ n$ hyperplanes of dimension $ (n\minus{}1)$.

You may wish to solve the following more interesting generalizing one:
replace $ \alpha$ with $ f_1dx_1 \plus{} \ldots \plus{} f_ndx_n$ where $ f_1,\ldots,f_n$ are some smooth functions on $ \mathbf R^n$ such that $ f_i(x) \equal{} x_i$ on $ S^{n \minus{} 1}$. Then prove that [i]there is no[/i] differential form $ \beta$ on $ \mathbf R^n$ with the property that $ \alpha\wedge\beta \equal{} \omega$.


Let $\ a,b,c$be positive real numbers
 Prove that $\ 3(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$ $\geq$ $\ 4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})^{2}$
         Can you dvelop this inequality ?

                 Good luck  :whistling:  :D

$\sum \frac{3}{ab}\geq \frac{12}{(a+b)^{2}}\geq 4(\sum\frac{1}{a+b})^{2}$
I think it is not very hard. :wink:

oh! Yes! Thankyou!  Can you dvelop this inequality ?

Good luck  :D


Hello everyone

Sea $ n$ a positive integer greater than $ 1$. Find the minimum value of $ n$ for what is the average of the numbers $ 1^2; 2^2; 3^2; : : : ; n^2$ is a perfect square.

Greetings    :D

Itis equavalent to $ \frac{(n\plus{}1)(2n\plus{}1)}{6}\equal{}k^2$ or $ (4n\plus{}3)^2\minus{}3(4k)^2\equal{}1$ or $ 4n\plus{}3\plus{}4k\sqrt 3 \equal{}(7\plus{}4\sqrt 3)^m$. When $ m$ odd we find solution.

[quote="Rust"]Itis equavalent to $ \frac {(n \plus{} 1)(2n \plus{} 1)}{6} \equal{} k^2$ or $ (4n \plus{} 3)^2 \minus{} 3(4k)^2 \equal{} 1$ or $ 4n \plus{} 3 \plus{} 4k\sqrt 3 \equal{} (7 \plus{} 4\sqrt 3)^m$. When $ m$ odd we find solution.[/quote]

Hi Rust,

I can't understand the last passage ("$ 4n \plus{} 3 \plus{} 4k\sqrt 3 \equal{} (7 \plus{} 4\sqrt 3)^m$. When $ m$ odd we find solution.")

Can you explain please?

I did it in a completely different way, and got $ n\equal{}337$. Is it right?
If you mind, I could post all my solution :D

Cheers

[hide]
${ \frac {(n + 1)(2n + 1)}{6} = k^2}$
since $ (n + 1)(2n + 1)$ are [b]coprime[/b] we get,
$ n + 1 = 2a^{2}$ and $ 2n + 1 = 3b^{2}$
case1:
if $ 2a = k$ eliminating n we get,
$ k^2 - 3b^2 = 1$
solving, $ (2,1)(7,4)(26,15)$
since b is odd, we take $ 26$,$ 15$ to get solution $ 337$

The other case :
$ n + 1 = 6a^{2}$
$ 2n + 1 = b^{2}$
It gives $ 12a^{2} = b^{2} + 1$
This
has no solutions since there is no square leaving a remainder of $ 11$ on division with $ 12$. [/hide]

Well, I did the same thing,  but with more cases at the beginning and less calculations at the end... :D

For istance, I took into consideration also $ n \plus{} 1 \equal{} b^2$, $ 2n \plus{} 1 \equal{} 6a^2$ (impossible modulo 2) and so on...

How did you solve $ k^2 \minus{} 1 \equal{} 3b^2$ (I got some problems with it  :roll: )


in the world series, seven games are played.

the probability the the white sox win a game is 2/3

if it is known that the white sox will win at least 2 games, what is the probability that the series will end on the seventh game?

[hide="just for practice, then:"]
either the sox win 3 games or the sox win 4 games.
if the sox win 3 games, (it is implied that the astros win, so the astros have to win the last game), then the number of possibilities equals $\frac{6!}{3!3!}$
multiply that to the probability of each outcome:
$\frac{6!}{3!3!}(\frac{2}{3})^3(\frac{1}{3})^4 = \frac{160}{2187}$
if the sox win 4 games, (it is implied that the sox win, so the sox have to win the last game), then the number of possibilities equals $\frac{6!}{3!3!}$
multiply that to the probability of each outcome:
$\frac{6!}{3!3!}(\frac{2}{3})^4(\frac{1}{3})^3 = \frac{320}{2187}$
total: $\frac{480}{2187} = \frac{160}{729}$
[/hide]

[quote="4everwise"]Humph, this is in AoPS Volume II. I just did it yesterday.  :|[/quote]

really??

i seriously didnt know that

what a coincidene!!! :rotfl:


The sequence $ \{ a_n \} _{n\equal{}1}^{\infty}$ is defined by $ a_1\equal{}1, a_{n\plus{}1}\equal{}\frac{a_n}{n}\plus{}\frac{n}{a_n}, n \ge 1$. Prove that for $ n \ge 4, [a_n ^2]\equal{}n$.

Let $ b_n\equal{}a_n^2\minus{}n\minus{}\frac 12$, then \[ b_{n\plus{}1}\equal{}\frac{(n\plus{}\frac 12\plus{}b_n)^2\plus{}\frac{n^2}{4}\minus{}n^2(n\minus{}\frac 12)b_n}{n^3\plus{}\frac{n^2}{2}\plus{}n^2b_n}.\]
We had $ a_2\equal{}2,b_2\equal{}\frac 32,a_3\equal{}2,b_3\equal{}\frac 12$, if $ | b_n|\le \frac 12, n\ge 3$, then $ |b_{n\plus{}1}|<|b_n|$.
I can prove, that $ b_n\to 0$ as $ b_n\equal{}O(\frac 1n)$.

I think this problems is wrong? Do you.....!!!
 :maybe:


group theory
abstract algebra
superior algebra
superior algebra solved

What is factor group or quotient group $G/H$ ?
what does it mean in a concrete manner.
I understand what $G/\sim$ should mean G factorized to equivalence relation $\sim$.
But it is still unclear to me what exactly would be G factorized to H with $H\leq G$

H is not related to G but in the manner that H is subgroup of G.
Now i see meaning in
$A_n=Ker(sgn)=[\sigma \in S_n | \sigma$is a even permutation meaning $sgn(\sigma)=1 ]$
So further what is the meaning of $S_n/A_n$ ?
Is it the classes of equivalence of odd and even permutations(with sgn -1 or 1) ?
And they say futher that from the first theorem of isomorphism follows
the following result
$S_n/A_n \simeq C_2$ with $C_2$  beeing i think group of complex roots of eq $z^2=1$
i think...
So i think i can grasp this...
It maybe follows from here that multiplying and odd signature permuation with an
even one gives an odd signature permutation
Also odd x odd gives even.
And also even x even gives even.

$G/H$ is defined as a group if $H$ is "normal" in $G$. $H$ is normal in $G$ if its left and right cosets coincide. That is, for every $h\in H$ and $g\in G$ there exists $h'\in H$ s.t. $gh=h'g$. Then $G/H$ is a group with all the cosets of $H$, with the operation defined as $(aH)*(bH)=(ab)H$ for any $a,b\in aH,bH$ respectively. That this operation is well-defined follows from the normality of $H$.

I suppose this also is the same as $G/\sim$ where $a\sim b$ iff $aH=bH$?

So $S_n/A_n$ is a group with two elements, $A_n$ and $\sigma A_n$ where $\sigma$ is any odd permutation. It is isomorphic to $\mathbb{Z}_2$.


Euler
search
combinatorics unsolved
combinatorics

Prove that there is an euler circuit in connected directed graph iff all vertices have the same degrees of outward and inward edges.

thank you

The proof is not appreciably different then in the undirected case.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=441425366&t=72043 for example.  (I searched for "euler* cycle" to find this thread.)


What vector representing a force $ F_4$ is needed to allow point $ O$ to be in static equilibrium?

[asy]size(200); pair O,A,B,C; O = (100,100); A = (50,100); B = (130,120); C = (90, 40); label("$O$",O,(1,0)); dot(O); draw(O--A,Arrow(A)); label("$F_1$, 10 N",A,(-1,0)); draw(O--B,Arrow(B)); label("$F_2$, 6 N",B,(1,0)); draw(O--C,Arrow(C)); label("$F_3$, 12 N",C,(1,0));[/asy]
The acute angle from $ F_2$ to the x-axis is $ 30^{\circ}$, the acute angle from $ F_3$ to the x-axis is $ 80^{\circ}$, and $ F_1$ lies on the x-axis.

Also, what's the difference between a ray and a vector?

EDIT: Oops, this was supposed to go in Intermediate topics, could someone move this?

I'll leave it to others to solve.

But to answer your question, a [b]vector [/b]is a directioned quantity. Each vector with n components corresponds to exactly one point in Euclidean n-space. This is because each point has a particular distance from the origin (the magnitude) and a particular direction with respect to the origin). Hence, a vector can be defined by [b]one Euclidean point[/b].

A [b]ray[/b], on the other hand, is a geometrical construct. It is essentially half a line. It may start anywhere and go in any direction. Rays continue to infinity and denote a direction. Hence, each ray is defined by [b]two Euclidean points in a particular order[/b].

Their representations on paper are similar, but they are totally different.


More generally, $ \forall n \in \mathbb{Z}, 3\not|\ n^2\plus{}n\plus{}2$.

[quote="Hamster1800"]More generally, $ \forall n \in \mathbb{Z}, 3\not|\ n^2 \plus{} n \plus{} 2$.[/quote]
[hide="Proof"]
If $ 3|n$, then $ n^2 \plus{} n\equiv 0\bmod 3\implies n^2 \plus{} n \plus{} 2\equiv 2\bmod 3$.  Now, if $ 3\not|\ n$, then $ n^2 \plus{} n\equiv 1 \plus{} n\equiv 2,3\bmod 3$ since $ n\equiv 1,2\bmod 3$.  Thus, $ n^2 \plus{} n \plus{} 2\equiv 1,2\bmod 3$.  So, $ n^2\plus{}n\plus{}2\equiv 1,2\bmod 3\implies 3\not|\ (n^2 \plus{} n \plus{} 2)$. [/hide]

[quote="The QuattoMaster 6000"][quote="Hamster1800"]More generally, $ \forall n \in \mathbb{Z}, 3\not|\ n^2 + n + 2$.[/quote]
[hide="Proof"]
If $ 3|n$, then $ n^2 + n\equiv 0\bmod 3\implies n^2 + n + 2\equiv 2\bmod 3$.  Now, if $ 3\not|\ n$, then $ n^2 + n\equiv 1 + n\equiv 2,3\bmod 3$ since $ n\equiv 1,2\bmod 3$.  Thus, $ n^2 + n + 2\equiv 1,2\bmod 3$.  So, $ n^2 + n + 2\equiv 1,2\bmod 3\implies 3\not|\ (n^2 + n + 2)$. [/hide][/quote]

[hide="Ugly Proof"]
\begin{align*}
3&|(n^2+n+2)\\
3&|(4n^2+4n+8) \\
3&|((2n+1)^2+7) \\
(2n+1)^2 &\equiv -1 \mod 3
\end{align*}
which is impossible.
[/hide]

[quote="Hamster1800"]More generally, $ \forall n \in \mathbb{Z}, 3\not|\ n^2 \plus{} n \plus{} 2$.[/quote]

[hide="not sure"]
suppose the contrary
consider n(n+1)(n+2), which is divisible by 3
then 2n^2 has to be divisible by 3
which implies n is divisible by 3
which means in
n^2+n+2
2 is divisble by 3
contradiction
[/hide]

can someone suggest a good elementary number theory book, which explains everything step by step. i got 2 elementary number theory books, i read but i dont get everything.

[quote="binomial_4eva"]prove or disprove that
$ \forall n\in\mathbb{Z}$, $ 3|n^2 \plus{} n \plus{} 2$[/quote]

[hide="Proof of the Converse"]If $ 3|n$, then $ n^2 \plus{} n \plus{} 2\equiv 2\pmod{3}$.

If $ 3|n \minus{} 1$, then $ n^2 \plus{} n \plus{} 2\equiv 1\pmod{3}$.

If $ 3|n \minus{} 2$, then $ n^2 \plus{} n \plus{} 2\equiv 2\pmod{3}$.

Since mod 3 runs through all integers, there are no integers that have the property that $ 3|n^2 \plus{} n \plus{} 2$.[/hide]

For the elementary number theory books:

Edit:
1. Introduction to Number Theory
2. AoPS Vol. 1 and 2. They have great chapters on Number Theory.
3. Problem Solving Strategies by Arthur Engel(Number Theory is in problem 6)

[quote="1=2"][quote="binomial_4eva"]prove or disprove that
$ \forall n\in\mathbb{Z}$, $ 3|n^2 \plus{} n \plus{} 2$[/quote]

[hide="Proof of the Converse"]If $ 3|n$, then $ n^2 \plus{} n \plus{} 2\equiv 2\pmod{3}$.

If $ 3|n \minus{} 1$, then $ n^2 \plus{} n \plus{} 2\equiv 1\pmod{3}$.

If $ 3|n \minus{} 2$, then $ n^2 \plus{} n \plus{} 2\equiv 2\pmod{3}$.

Since mod 3 runs through all integers, there are no integers that have the property that $ 3|n^2 \plus{} n \plus{} 2$.[/hide][/quote]

This is the [i]negative[/i], not the converse.  The converse of $ P \implies Q$ is $ Q \implies P$, which is still false.

This is not the negation, either.  The true statement $ \not\exists n \in {\bf Z}, 3 | n^2 \plus{} n \plus{} 2$ is not related in any simple way to the false statement given in the question.  The negation of the statement given in the question (assuming we restrict our universe to the integers) is $ \exists n \in {\bf Z}, 3 \not | n^2 \plus{} n \plus{} 2$.  This is true but much weaker than the result Hamser quoted that everyone's been proving.

Also, [i]none[/i] of the texts suggested by 1 = 2 is a book on number theory, and you shouldn't get any of them if what you're looking for is a book about number theory.


This is a problem I invented:

Determine whether or not there exists a triangle with rational side lengths that is similar to a triangle with side lengths $ \sqrt{2}, \sqrt{3}, \sqrt{6}$. If you determine there aren't, then prove so.

[hide]Well, the ratio between the largest side and the smallest side is $ \frac{\sqrt{6}}{\sqrt{2}}\equal{}\sqrt{3}$  If there exists a similar triangle with rational side lengths, then there must also exist a similar triangle with integer side lengths.  But $ \sqrt{3}$ is irrational, so it cannot be expressed as a ratio of two integers.  Thus, the longest and shortest sides cannot both be integers or rational, so no triangle similar to the one with given side lengths can have all rational side lengths.
[/hide]


function
quadratics
ceiling function
algebra
floor function
quadratic formula

Find the 'n'th term of the foll. sequences :-

1)  1, 2, 2, 2, 3, 3, 3, 3, 3, 3, ............ where k repeats k(k+1)/2 times.

2)  1, 2, 2, 2, 2, ........ where k repeats 3k-2 times.

The first one:
[hide]THe formula for adding consecutive triangluar numbers is $\frac{k(k+1)(k+2)}{6}$.  We inductively show this.
Base case of 1. $1=\frac{1(2)(3)}{6}.$
Now assume it true for k.  Prove it for k+1.
$\frac{k(k+1)(k+2)}{6}=\frac{1(2)}{2}+\frac{2(3)}{2}+...+\frac{k(k+1)}{2}$.  WE want to prove $\frac{(k+1)(k+2)(k+3)}{6}$ is equal to $\frac{1(2)}{2}+\frac{2(3)}{2}+...+\frac{(k+1)(k+2)}{2}$  Subtracting, one quickly sees both sides are equal.
Now, n equals this equation only when n is a number that at the end of a sequence of k numbers.  However, if n could still equal the same k but n could be in the middle or beginning of a sequence.  The beginning of the sequence is $\frac{(k-1)k(k+1)+6}{6}$.  THerefore, any given n value is between $\frac{(k-1)k(k+1)+6}{6}$ and  $\frac{k(k+1)(k+2)}{6}$ inclusive for some appropriately chosen k since the n value is known.  Is there a way to express f(n)=k without having to use the cubic formula?[/hide]

Or notice the patterns in Pascal's triangle.  It's the best place to look for anything involving triangular numbers.

The function $\frac{n(n+1)(n+2)}{2}$ can be combinatorically described as $\displaystyle{n+1 \choose 3}$

It's simply the formula for the nth triangular sum.

Oh, and if you want a function for something like that, I would say using the floor function would be your best bet, considering only integers are in the sequence and due to the fact that there are so many repeats.

No, because each number $n$ repeats $T_n$ times in the series...

Your formula gives that the tenth term is $330$, when we have already shown that it is $3$.

Btw, ${{n+1}\choose{3}}$ would be $\frac{n(n-1)(n+1)}{6}$ or $\frac{n(n-1)(n+1)}{(n-2)!}$

But increasing values of n should take care of that...what am I missing?

I'm not sure of all the details, but your formula will obviously increase for increasing values of $n$, and the sequence doesn't always do that.  $n_2$ through $n_4$ are all the same, for example...

THe second one:
[hide]THe formula for adding consecutive numbers in the form 3k-2 is $\frac{(3k-1)(k)}{2}$.  Using the formula for the sum of a arithmatic sequence shows that $a_0=1$, $a_{k-1}=3k-2$.  $\frac{(3k-2+1)(k)}{2}=\frac{(3k-1)(k)}{2}$.
Now, to precisely find a f(n) that equals k, set up this equation.  $n=\frac{(3k-1)(k)}{2}$.  K is only an integer in this equation $\frac{(3k-1)(k)}{2}$ when n represents the number of terms in the sequence when the last k is the last term in the sequence.  If $\frac{(3k-4)(k-1)+2}{2}\leq n<\frac{(3k-1)(k)}{2}$, the solution for k would be between k-1 and k.  However, we can see that at such n values, it still equals k, not k-1, so the round-up function is needed.  THerefore, after rearranging the equation and using the quadratic formula, $k=\lceil{\frac{1+\sqrt{1+24n}}{6}}\rceil$[/hide]

First One

[hide]Notice that $a_{k(k+1)/2}=k$ for k and integer. Let $\frac{k(k+1)}{2}=u$. Notice that if $k$ is an integer, then $u$ is always a triangle number. We use the quadratic equation to solve for $k$: $k=\frac{-1+\sqrt{1+8u}}{2}$. Thus we have that $a_u=\frac{-1+\sqrt{1+8u}}{2}$ when $u$ is a triangle number. Looking back at the sequence, we see that for values of $u$ which are not triangle numbers, the expression for $a_u$ gives decimal numbers. More precisely, when $\frac{(k-1)k}{2}<u<\frac{k(k+1)}{2}$ we have $k-1<a_u<k$ by using the above equation for $a_u$, when in reality $a_u=k$ when $\frac{(k-1)k}{2}<u<\frac{k(k+1)}{2}$. It is easy to see that we can alleviate this problem by using the round up function. A formula for our sequence is then $a_u=\displaystyle\lceil\frac{-1+\sqrt{1+8u}}{2}\displaystyle\rceil$.[/hide]

I hope you guys can understand. It was difficult to communicate the ideas.

Second One

I havn't looked at dengmi's solution yet.

[hide]$a_u=\lceil\frac{-1+\sqrt{1+24k}}{6}\rceil$

very similar reasoning to first part except that you know start with $u=\displaystyle\sum_{i=1}^{k}(3i-2)$[/hide]

Some one try this after looking at my solutions (I havn't tried this yet either): Find a formula for the nth term
1,2,2,2,2,3,3,3,3,3,3,3,3,3,...
where k is repeated k^2 times

Then try this: find a formula
1,2,2,2,3,3,3,3,3,3,...
where k is repeated k(k+1)/2 times

dengmi, you left out -1 in your final expression. must have had an arithmetic error somewhere. you can check by plugging in for the first term.  ;)


abstract algebra
linear algebra
matrix
Ring Theory
superior algebra
superior algebra unsolved

1)prove that
$ R \equal{} \{\left(\begin{array}{ccc}a & b & \\
0 & d\end{array}\right) \mid a\in\mathbf{Z},b,d\in\mathbf{Q}\}$
is a right noethrian ring but it is not left noethrian.
2)Let R be a ring ,prove That $ J(M_{n}(R))\equal{}M_{n}(J(R))$.

2) I write $ Jac_R(M)$ for the jacobson radical of the left $ R$-module $ M$, i.e. the intersection of all maximal left $ R$-submodules of $ M$. in particular, $ Jac(R): \equal{}Jac_R(R)$ is the intersection of all left ideals of $ R$. I will abbreviate "left $ R$-submodule" to "submodule" and "left ideal" to "ideal".

let $ J$ be an ideal of $ M_n(R)$. this is a submodule of $ M_n(R)$ such that $ A \in J \Rightarrow E_{ij} A \in J$, where $ E_{ij}$ denotes the matrix which has $ 1$ in the entry $ (i,j)$ and zeroes elsewhere. remark that $ E_{ij} A$ consists of the $ i$th row of $ A$ in the $ j$th row, and elsewhere of zeroes. thus the set $ U$ of elements in $ R^n$, which appear as a $ i$th row in a matrix in $ J$, does not depend on the choice of $ i$, and we have $ J\equal{}(U;...;U)$, where $ (U;...;U)$ denotes the set of matrices whose rows come from $ U$ (the inclusion $ J \subseteq (U;...;U)$ is trivial, and the other one follows from $ (U;...;U)\equal{}(U;0;..)\plus{}...\plus{}(0;...;U)$). it is clear that $ U$ is a submodule of $ R^n$ and that $ (U;...;U) \subseteq (V;...;V)$ if and only if $ U \subseteq V$. given a submodule $ U$ of $ R^n$, clearly $ (U;...;U)$ is a left ideal of $ M_n(R)$.

to sum up, we have an order isomorphism

$ \{\text{left ideals of }M_n(R)\} \leftrightarrow \{\text{left submodules of } R^n\}$

therefore, the claim $ Jac(M_n(R))\equal{}M_n(Jac(R))$ is equivalent to $ Jac_R(R^n)\equal{}Jac_R(R)^n$.

let $ z \in Jac_R(R^n)$, and $ J$ be a maximal (left) ideal of $ R$. then $ R \times ... \times J \times ... \times R$, where $ J$ is the $ j$th factor, is a maximal submodule of $ R^n$, so that it contains $ z$. this shows $ z_j \in J$ and hence, varying $ J$ and $ j$, the claim $ z \in Jac_R(R)^n$. now assume $ z \in Jac_R(R)^n$, and let $ U$ be a maximal submodule of $ R^n$. we want to show $ z \in U$. if not, maximality yields $ R^n \equal{} U \plus{} R z$. we may assume $ n > 1$ and $ R \neq 0$. in particular, the $ z_i$ are not left invertible, since they lie in some maximal ideal of $ R$. find $ u \in U$ and $ r \in R$ such that $ e_1 \equal{} u \plus{} r z$. then $ 1\equal{}u \plus{} r z_1$ shows that $ u$ is left invertible (otherwise, it lies in a maximal ideal and thus also $ u \plus{} r z_1 \equal{} 1$, a contradiction), and $ 0\equal{}u  \plus{} r z_2$ shows that also $ z_2$ is left invertible, a contradiction. $ \square$


algebra
polynomial
trigonometry
calculus
calculus computations

hi,

I have a problem: find a polynomical funktion $f$ with $|f(x)-\sin(x^2)|<\frac{10^{-6}}2$  for all $x \in [-1,1].$

[Moderator edit to introduce TeX; hardly any changes needed.]

This is the particular forum where problems like this belong.

In this case, the identity $\sin^2x=\frac{1-\cos 2x}2$ can be used to simplify the calculations.

thank you very much, everything figures now...

Dear Kent may be I give a stupid question,but I can't figure out.Where we have $\sin^{2}{x}$?

Oops - I guess he did say $\sin(x^2)$ rather than $\sin^2x.$ In that case, we want to use the Taylor expansion for $\sin x$ and then substitute $x^2$ for $x$.


okay i like james idea. if you would like to be in this election(for nothing) pplease post you would like to join. after no one joins anymore. we have a vote(every person can only vote once) and see who is teh most popular AOPser. I would like to be a nominee

:rotfl: jjx as most popular aopser?  :rotfl:

:rotfl: MEWTO AS MOST POPULAR AOPSER! :huh:  :rotfl:

I know I am not. That is why I did not stoop so low as to nominate myself. However, I am infinitely more popular than you.

mewto wth. i'm not saying i am teh most popular aopser. I'm probably gonna lose. butso waht im jsut signing up for fun. geez

Hmm... tinytim as the most popular AoPSer!!
But I join anyway (and tinytim) :D

okay so teh people who are in the race so far are:

yongi
tinytim
jjx1

vote: me

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

okay

nominees:
jjx1:
yongi:
tinytim:
stevenmeow:
dragon
spammer123

and i am not counting any admins
OH AND I;M ASSUMING EVERYONE IS VOTING THEMSELVES. IF YOU'RE NOT TELL ME.OH AND ELECTION DAY ISN'T HERE

jjx1:4
tinytim:0
yongi:0
stevenmeow:0
DRAGON:0
SPAMMER:0
1=2:3
lotsofmath:0
pythag:0
ernie:0
james:0
RandomDragon:0

is he a real user?

jjx1:4
1=2:3
tinytim:0
yongi:0
stevenmeow:0
DRAGON:0
SPAMMER:0
lotsofmath:0
pythag:0
ernie:0
james:0
RandomDragon:0
bowei:0

Vote RandomDragoon.
(you misspelled his name...there's 2 o's in the last part and 1 o in the first part)

jjx1:4
1=2:3
RandomDragoon:1
tinytim:0
yongi:0
stevenmeow:0
DRAGON:0
SPAMMER:0
lotsofmath:0
pythag:0
ernie:0
james:0
bowei:0

nominate: junggi

I am "de nominator" [corny math joke, blah]

jjx1:4
1=2:3
RandomDragoon:1
tinytim:0
yongi:0
stevenmeow:0
DRAGON:0
SPAMMER:0
lotsofmath:0
pythag:0
ernie:1
james:0
bowei:0
NotA:0
MustItAlwaysBeMe:0
Junggi:0

I'm pretty sure that very few people are going to vote, but oh well.

nominate: MysticTerminator
nominate: zeb
nominate: CA Math (for doing the votecount)  :lol:

jjx1:4
1=2:3
RandomDragoon:1
tinytim:0
yongi:0
stevenmeow:0
DRAGON:0
SPAMMER:0
lotsofmath:0
pythag:0
ernie:1
james:0
bowei:0
NotA:0
MustItAlwaysBeMe:0
Junggi:0
probabilitu1.01
MysticTerminator :0
zeb:0
CA math:0

okay from now on you can't nominate anyone. if somepone wants to be a nominee they gotta post they want to join themselves.


In 1593, the Belgian mathematician Adriaan van Roomen proposed the following problem:

Find the positive roots of the equation
$x^{45} - 45x^{43} + 945x^{41} - 12300x^{39} + 111150x^{37} - 740459x^{35} + 3746565x^{33} - 14945040x^{31} + 469557800x^{29} - 117679100x^{27} + 236030652x^{25} - 378658800x^{23} + 483841800x^{21} - 488494125x^{19} + 384942375x^{17} - 232676280x^{15} + 105306075x^{13} - 34512074x^{11} + 7811375x^9 - 1138500x^7 + 95634x^5 - 3795x^3 + 45x = \sqrt{\frac{7}{4} - \sqrt{\frac{5}{16}} - \sqrt{\frac{15}{8} - \sqrt{\frac{45}{64}}}.}$

The French mathematician Viète was able to solve the equation. By hand. In just a few minutes, too, supposedly (Anecdote! One of the Bernoullis claimed to have summed the first 1000 10th powers in half of 15 minutes. My analysis professor did it in just over 8, but he was explaining it to us as he went).

Anyway, anyone here want to give it a try? Or is their 16th century intellect beyond us? :P

[quote="Xevarion"] Or is their 16th century intellect beyond us? :P[/quote]  :huh: I think its to much for us, or at least for me.

But I might give it a try. If I were to try it I would start by simplifing the radicles, right?

The only way I could see this problem being done so quickly is with trigonometric substitution...

too bad I don't know trigonometry well enough.  :D

surge: Nice insight! That is indeed the approach that Viète took. Unfortunately I still can't see exactly what he did...

i think that it has something to do with

$\Im (1+i*x)^{45}$ with $x=\tan \theta$

Viete solved this problem by repeatedly applying trigonometric double, triple, quintic angle formulas.  I read about it in Eli Maor's [u]Trigonometric Delights[/u] :)

My god! Okay, I guess those 16th century mathematicians were a lot better at brute-forcing than we are today. I'm not trying that... I do sorta wonder what the roots are, though. I bet they're not that horrible.

RHS is of course $2 \sin 12^o$
Left side is what you get if you expand $2 \sin (45 y)$    in terms of $\sin y$ and  you  put $2 \sin y = x$

so you  "easily" :)  get all the roots  (For example one such root is  $sin (12/45)^o$  and rest you can get by adding $k\cdot 8^o$ ($k = 0,1,2 \cdots$ etc.( remember 360/45=8) .

(To be honest, RHS was not that difficult to guess for any one who worked in the old days as us as  I tell my kids .. in those days   we have to do all calculations  by hand, have to remember times tables up to 100, know log tables by heart and walk  10 miles uphill both ways  in $40^o C$  (It looks even more terrible in Fahrenheit :) = $104^o F$  )  heat and 5 feet of snow ..:))

( and of course only  thing to keep in mind for LHS was to do all middle steps of calculations on slate so not to waste too much paper) :)

(Actually if you know $sin 3x = 3 sin x - 4 sin^3 x$ and $\sin 5x = 5 \sin x -20 \sin^3 x + 16 \sin^5 x$  all you have to know and  you apply first formula twice and second once.)


A neat, but somewhat tricky, problem from the 1997 Cayley:

In the diagram, ABC is equilateral, BC=2CD, AF=6, and DEF is perpendicular to AB. What is the area of the quadrilateral FBCE?

Hint:
[hide]Extend BD to G such that AG is parallel to FD.[/hide]

Actually, muy facil way to do this.

[hide="HINT"] What is CE in relation to s, the side of the equilateral triangle? Then, draw up similar triangles, and you have two triangles FBE and CBE with heights which are the same. Interesting problem. In any math competition - it's very important to know all properties of equilateral, right, and isosceles triangles.[/hide]

This problem is pretty easy.  If you related everything in terms of CD and numbers you can develop some nice equalities and then figure out everything about the traingles.  From there the area of the quadrilateral is cake.

"When you have no quote to post, post one about having no quote to post"
                           -Yours Truly

This is what I would do.

Consider $\triangle BFD$ and let $2x$ be the side length of the equilateral triangle.

Since $BFD$ is $30-60-90$, it's easy to figure out that $2BF=BD$, and we find that $x=12$.

It's easy to find now that $\triangle AEF$ has area $12\sqrt {3}$, and we just need to subtract this from the area of the equilateral triangle.


Has anybody heard of this or learned any of it?  It provides simple ways to do mental computations that  would normally be done on a caclulator.  Do you think this is something that should be taught in middle or elementary school in conjunction with, or instead of conventional teaching?

Here's a link with a few tutorials on some of the ideas behind Vedic Math
[url]http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp[/url]

I checked it out. At first, the first few tutorials were already known, but following that, I found many of them very useful. It is excellent for competition math, especially MATHCOUNTS and AMC (I heard you can't use calculators anymore?).

I have a book at home called Vedic Mathematics. Apart from simple computation, it has ways to factor quickly and solve cubic equations, along with a variety of other things. It was the same book mentioned here.
http://www.vedicmaths.org/Bookstore/05%20Vedic%20Maths/VM%20details.asp

My dad taught me one of those tricks when I was in elementary school and called it the "Indian method" for multiplication. Now I know why!

That is really great tricks for speed math! I will probably use those a lot.

I have a vedic math book.  It has lots of neat tricks, but also, more importantly, it has faster ways of multiplication and long division.  However, I haven't gotten into the habit of using them.  vedic math is based on vinculums, like "negative digits."   I should try to get used to it this summer.


J.F. Jolly (1809-1884), a Munich teacher of physics, measured with his very sensitive self-made balance that at a depth of 5.3 m the weight of a body is greater by one and a half millionth than that on the surface. How can that be reconciled with Newton's gravitational law?

What do you mean by "reconciled"?  By Newton's Law of Gravitation, $ F\equal{}G\frac{Mm}{r^{2}}$.  On the surface, we can say the distance is $ r$.  At a depth of 5.3m, we can say it is $ r\minus{}5.3$.  When this is plugged into the equation, you'll find that $ F$ is very slightly greater.  This is because the distance between the two bodies (earth and person) is smaller, so the force between them must be greater.

What you said about the force Jrav is only true when the particle is outside of the earth, then the gravitational force on it is the same as if we put a "point" mass with the mass of the earth at the center. But now we put the particle inside the earth, hence the gravitational field will be differant then in the outside, so the force on the particle will not be what you said.


I asked already very many people and anyone can find mistake in my solution of task 5 from LVII Polish Mathematical Olympiad. Good solution of this task is in attachment om57_2r.pdf. I proved in complex.pdf that thesis doesn't work in infinite number of cases. Can someone help me?

At the first glance of your note (I don't have much time right now), it seems to be correct except of one mistake: The perpendicularity of the lines $z_1z_2$ and $0z$ is not equivalent to the statement that $z_1-z_2$ is equal to $iz$ or to $-iz$, but it is equivalent to the statement that $z_1-z_2$ is equal to $rz$ for some real $r$. Your proof can be easily corrected.

By the way, it would be useful if you post the problem statement in English, as well:

[color=blue][b]Problem.[/b] Let C be the midpoint of a segment AB. Some circle $o_1$ passes through the points A and C, and some circle $o_2$ passes through the points B and C. The circles $o_1$ and $o_2$ intersect at some point D (apart from C). Let P be the midpoint of the arc AD on the circle $o_1$ which doesn't contain the point C, and let Q be the midpoint of the arc BD on the circle $o_2$ which doesn't contain the point C. Prove that $PQ\perp CD$.[/color]

  Darij

Sorry, I'm an idiot :blush:. But I already found this mistake and a few days ago I solved this problem. See complex.pdf.


For an object, with the mass $ m\equal{}2kg$, bound at one extremity by an horizontal spring fixed on the other extremity, apply an initial horizontal speed in the sense of the stretch of the spring. The coefficient of the rub between the object and the horizontal surface is $ 0,1$. The elastic constant of the spring is $ k\equal{}100 \frac {N} {m}$. Determine the first three amplitudes of the spring.

$ R: A_1\equal{}0,5m,A_2\equal{}\frac {7} {15}m,A_3\equal{}\frac {13} {30}m$

I will solve the exercise, but with Romanian notations:

[u][b]first amplitude[/b][/u]

$ E_f\minus{}E_i\equal{}L \Rightarrow \frac {kx^2} {2}\minus{}\frac {mv_0^2} {2}\equal{}\minus{}F_fx \Rightarrow \frac {kx^2} {2}\minus{}\frac {mv_0^2} {2}\equal{}\minus{}\mu mgx \Rightarrow A_1\equal{}x\equal{}0,5$

so $ L\equal{}\minus{}F_fx\equal{}\minus{}2 \cdot \frac {1} {2}\equal{}\minus{}1$

[u][b]second amplitude[/b][/u]

$ E_i\equal{}\frac {mv_0^2} {2}\plus{}2L\equal{}16\minus{}2\equal{}14 \Rightarrow \frac {ky^2} {2}\minus{}14\equal{}\minus{}F_f \cdot y \Rightarrow A_2\equal{}y\equal{}\frac {7} {15} \Rightarrow L'\equal{}\minus{}F_f \cdot y\equal{}\minus{}\frac {14} {15}$

[u][b]third amplitude[/b][/u]

$ E_i\equal{}\frac {mv_0^2} {2}\plus{}2L\plus{}2L'\equal{}\frac {182} {15} \Rightarrow \frac {kz^2} {2}\minus{}\frac {182} {15}\equal{}\minus{}F_f \cdot z \Rightarrow A_3\equal{}z\equal{}\frac {13} {30}$


Let $a,b,c,d \ge 0$ and $a+b+c+d=1$ . Prove that
$abc+bcd+cda+abd \ge \frac{1}{27} +\frac{176}{27}abcd$.

Maybe not that hard.
First, make the inequality homogenous.
\[(a+b+c+d)(abc+bcd+cda+abd) \ge \frac{1}{27}(a+b+c+d)^4 +\frac{176}{27}abcd\]
It follows that,
\[27(a+b+c+d)(abc+bcd+cda+abd) \ge (a+b+c+d)^4 +176abcd\]
This seems to be done if you use Muirhead.

Thanks [b]siuhochung[/b], I need it.

[quote="lightrhee"][quote="blahblahblah"]This is incredibly obviously false. $a=1,b=c=d=0$[/quote]
He is right. Maybe it is the other way?[/quote]
$a=b=c=d=\frac{1}{4}$


is there an easier way to memorize the formula for permutations and combinations? I would be happy to know some suggestions!

:)

Well, it always helps to understand a formula before you begin to memorize it so here are a few explanations:
[hide="Permutations"] Say that there are n people and p chairs (assume that the children have names, so one child will be different from another child).  We want to find out how many ways there are to seat these children.  n People can be seated in the first chair.  For the second chair, only n-1 people can be seated (since one boy is done).  Then, n-2, and n-3, and so on.  There will be n terms, thus the number of combinations is
\[P(n,p)=(n)(n-1)(n-2)...(n-p+1)=\frac{n!}{p!}\]
[/hide]
[hide="Combinations"] Say there are n boys and m girls and you wish to seat these children.  Also assume the boys are not different from one another and the girls are not different from one another.  Let's seperate this into boys and girls.  There is one way to organize n boys (since they are all the same), but if the boys are different, then there are n! ways.  There is one way to seat the girls, but if the gilrs were different, then there would be m! ways.  Thus, for every one [b]combiniation[/b] of boys, there are n! [b]permutations[/b] of boys.  Similarly, for every one [b]combination[/b] for boys, there are m! [b]permutations[/b] of girls.  Thus, for every one combination of a way boys and girls can be seated, there are m!n! permutations.  Now, there are (m+n)! permutations in which the boys and girls can be seated (since there are m+n people total), thus,
\[C(m+n,m)=C(m+n,n)=\frac{(n+m)!}{n!m!}\]
[/hide] Well, I hope this helped, but I am really bad at explaining.

[quote="Quevvy"][quote="davidlizeng"]im sure that these two formulas are worth memorizing because they appear so much. spend 5 minutes to memorize them

Combinations k objects out of n objects: $\frac{n!}{k!(n-k)!}$
Permutations k objects out of n objects: $\frac{n!}{(n-k)!}$

of course understand them too....[/quote]
In the first one, it that basically "n choose k"?[/quote]

yeah remember in your future years that

$\binom{9}{3}$ means to 9 choose 3, and you use the formula  $\frac{n!}{k!(n-k)!}$ to figure out

combinations is also

(this is pretty easu)

# of permutation
______________
(#u are picking)!

-jorian

choose is important but memorizing the permutation formula is pointless

just use logic

then again you really dont need to memorize choose but it pops up much more often

memorizing kills your brain

did you know that your brain slows down because your brain is too full of information as you get older? Don't stick useless stuff that you don't need to memorize exactly in your brain.  :P


Sea ABC un triangulo acutangulo, y sea O su circuentro. El circulo a traves de C, O y B se llama S. Las rectas AB y AC cortan al circulo S de nuevo en P y Q respectivamente. Prueba que las rectas AO y PQ son perpendiculares.

Issam: es usted de Marruecos? sie s asi porque habla español? interesante...

Si, hablo español desde que era un crio  :P

sea m<BAO=x, m<OAC=y y m<OBC=z. como m<BOC=180-2z entonces m(arcoBC)=4z, asi m<BQC=2z  y se sigue que m<CBQ=z+y-2z=y-z. Luego m<BPQ=y+z y como PAO=x se sigue que AO y Pq son perpendiculares.


integration
calculus
inequalities unsolved
inequalities

Let $a \leq b \leq c$. If f is continuous on $[a,c]$ and $f'$ increases on $(a,c)$ then: $(a-b)f(c)+(b-c)f(a)+(c-a)f(b) \leq 0$.

[tex]\frac{f(b)-f(a)}{b-a}=\frac1{b-a}\int_a^b f'(x)dx\leq\frac1{b-a}\int_a^b f'(b)dx=\frac1{c-b}\int_b^c f'(b)dx\leq\frac1{c-b}\int_b^c f'(x)dx=\frac{f(c)-f(b)}{c-b}[/tex]
This gives the inequality.

Hmmm... and a not integral solution?  (if that exists) :D

thank you very much.


limit
logarithms
algebra unsolved
algebra

$ a_n \equal{} \frac {1.3.5.\ldots.(2n \minus{} 1)}{2.4.6.\ldots.(2n)}$ for all $ n\in \mathbb{Z^ \plus{} }$

Find $ \lim_{n\to\infty}a_n$.

Let us have $ b_{n}\equal{}\frac{2}{3}*\frac{4}{5}*...*\frac{2n}{2n\plus{}1}$
Of course $ a_{n}<b_{n}$ so $ 0<a_{n}^{2}<a_{n}b_{n}\equal{}\frac{1}{2n\plus{}1}$

thus $ \lim_{n\to \infty}0<\lim_{n\to \infty}a_{n}<\lim_{n\to \infty}\frac{1}{\sqrt{2n\plus{}1}}$
hence $ \lim_{n\to \infty}a_{n}\equal{}0$ :wink:

Another solution...
[hide]
Taking $ ln$ for $ a_n$, we will have:
$ \ln a_n \equal{} \ln \frac {1}{2} \plus{} \ln \frac {3}{4} \plus{} ... \plus{} \ln \frac {2n \minus{} 1}{2n}$;
$ \minus{} \ln a_n \equal{} \sum_{k \equal{} 1}^{n}\ln \frac {2k}{2k \minus{} 1} \equal{} \sum_{k \equal{} 1}^{n}\ln (1 \plus{} \frac {1}{2k \minus{} 1})$.

But serie $ \sum_{n \equal{} 1}^{\infty}\ln (1 \plus{} \frac {1}{2n \minus{} 1}) \equal{} \infty$, because $ \ln (1 \plus{} \frac {1}{2n \minus{} 1}) \equal{} O(\frac {1}{n})$, so $ a_n\rightarrow 0$.
[/hide]


geometry
inequalities
circumcircle
inequalities unsolved

Let $ G$ the centroid of triangle $ ABC$ and $ D,E,F$ the points on sides $ BC,CA,AB$ such that $ DG,EG,FG$ are perpendicular with $ BC,CA,AB$ respectively
Prove that $ \frac{4}{27}<\frac{\left[ DEF\right] }{\left[ ABC\right] }\leq \frac{1}{4}$

Since $ \Delta DEF$ is a pedal triangle, we have $ FE\equal{}a(\frac {AG}{2R})$
$ DE\equal{}c(\frac {CG}{2R}),FD\equal{}b(\frac {BG}{2R})$

$ [DEF]\equal{}\frac {abc}{4r}\frac {AG.CG.BG}{2R.2R.2R}$,where $ r$ is the circumradius of $ \Delta DEF$

$ [ABC]\equal{}\frac {abc}{4R}$

$ \frac {4}{27}\leq\frac {[DEF]}{[ABC]}\leq\frac 14\Leftrightarrow\frac {4}{27}\leq\frac {AG.CG.BG}{4r.2R^2}\leq\frac 14\Leftrightarrow\frac {4}{27}\leq\frac {8}{27}\frac {m_am_bm_c}{4r.2R^2}\leq\frac 14$
I tried a co-ordinate approach but my calculations were turning out to be too tedious..... :(

Let AA', BB', CC' the altitudes of the triangle ABC

AE=AB'+B'E=AB'+(AC/2-AB')·2/3=(AC+AB')/3     and analogous  AF=(AB+AC')/3

[DEF]=[ABC]-[AFE]-[ECD]-[BFD]=[ABC]·∑(AE·AF/AC·AB)

        =[ABC]-[ABC]1/9·∑(cos²A+1+AB'/AC+AC'/AB)

        =[ABC]-[ABC]1/9(cos²A+cos²B+cos²C+6)

        =[ABC](sen²A+sen²B+sen²C)/9

Then [DEF]/[ABC]=(sen²A+sen²B+sen²C)/9

Then the steatement is equivalent to

4/3<sen²A+sen²B+sen²C≤9/4

The RHS is well-know but the LHS is false. Indeed sen²0+sen²0+sen²180=0


Let's talk about the free lunch Rickards is giving out.
What will it be?

Eli Ross.
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oh yes.  of course you are. i forgot about that time when i cloned myself  :wink:

but really, who are you?

[quote="gatorlover124"]XxCharizardxX, do you even live in FL?????anyway, the Rickards Invit. is a competition for MAO that the 'whole' state is going to...i think...and why do you care about free lunch???

BUCHHOLZ!!!!!!!!!:)[/quote]

steve, xxcharizardxx GOES to rickards, lmao. he's just jk. :]

[quote="gatorlover124"]o rly??wow...i feel rly dumb...buchholz is betta than rickards anyway...o and bhs cant go to rickards invite thanks to frazer jess, so even if u do visit...[/quote]

wait, you guys arent going to rickards invite?! WHY NOT?


The only solution is $ x=2$.
Notice that the equation is equivalent with
$ \left(\frac{10}{13}\right)^{x}+\left(\frac{11}{13}\right)^{x}+\left(\frac{12}{13}\right)^{x}=1+\left(\frac{14}{13\right)}^{x}$
with lhs is decreasing,and rhs is increasing in x.Thus,it has exactly 1 solution. :D